![8. For any a there is a number x in R such that x þ a ¼ 0.
x is called the inverse of a with respect to addition and is denoted by a.
9. For any a 6¼ 0 there is a number x in R such that ax ¼ 1.
x is called the inverse of a with respect to multiplication and is denoted by a1
or 1=a.
Convention: For convenience, operations called subtraction and division are defined by
a b ¼ a þ ðbÞ and a
b ¼ ab1
, respectively.
These enable us to operate according to the usual rules of algebra. In general any set, such as R,
whose members satisfy the above is called a field.
INEQUALITIES
If a b is a nonnegative number, we say that a is greater than or equal to b or b is less than or equal to
a, and write, respectively, a A b or b % a. If there is no possibility that a ¼ b, we write a b or b a.
Geometrically, a b if the point on the real axis corresponding to a lies to the right of the point
corresponding to b.
EXAMPLES. 3 5 or 5 3; 2 1 or 1 2; x @ 3 means that x is a real number which may be 3 or less
than 3.
If a, b; and c are any given real numbers, then:
1. Either a b, a ¼ b or a b Law of trichotomy
2. If a b and b c, then a c Law of transitivity
3. If a b, then a þ c b þ c
4. If a b and c 0, then ac bc
5. If a b and c 0, then ac bc
ABSOLUTE VALUE OF REAL NUMBERS
The absolute value of a real number a, denoted by jaj, is defined as a if a 0, a if a 0, and 0 if
a ¼ 0.
EXAMPLES. j 5j ¼ 5, j þ 2j ¼ 2, j 3
4 j ¼ 3
4, j
ffiffiffi
2
p
j ¼
ffiffiffi
2
p
, j0j ¼ 0.
1. jabj ¼ jajjbj or jabc . . . mj ¼ jajjbjjcj . . . jmj
2. ja þ bj @ jaj þ jbj or ja þ b þ c þ þ mj @ jaj þ jbj þ jcj þ jmj
3. ja bj A jaj jbj
The distance between any two points (real numbers) a and b on the real axis is ja bj ¼ jb aj.
EXPONENTS AND ROOTS
The product a a . . . a of a real number a by itself p times is denoted by ap
, where p is called the
exponent and a is called the base. The following rules hold:
1. ap
aq
¼ apþq
3. ðap
Þr
¼ apr
2.
ap
aq ¼ apq
4.
a
b
p
¼
ap
bp
CHAP. 1] NUMBERS 3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-12-320.jpg)
![POINT SETS, INTERVALS
A set of points (real numbers) located on the real axis is called a one-dimensional point set.
The set of points x such that a @ x @ b is called a closed interval and is denoted by ½a; b. The set
a x b is called an open interval, denoted by ða; bÞ. The sets a x @ b and a @ x b, denoted by
ða; b and ½a; bÞ, respectively, are called half open or half closed intervals.
The symbol x, which can represent any number or point of a set, is called a variable. The given
numbers a or b are called constants.
Letters were introduced to construct algebraic formulas around 1600. Not long thereafter, the
philosopher-mathematician Rene Descartes suggested that the letters at the end of the alphabet be used
to represent variables and those at the beginning to represent constants. This was such a good idea that
it remains the custom.
EXAMPLE. The set of all x such that jxj 4, i.e., 4 x 4, is represented by ð4; 4Þ, an open interval.
The set x a can also be represented by a x 1. Such a set is called an infinite or unbounded
interval. Similarly, 1 x 1 represents all real numbers x.
COUNTABILITY
A set is called countable or denumerable if its elements can be placed in 1-1 correspondence with the
natural numbers.
EXAMPLE. The even natural numbers 2; 4; 6; 8; . . . is a countable set because of the 1-1 correspondence shown.
Given set
Natural numbers
2 4 6 8 . . .
l l l l
1 2 3 4 . . .
A set is infinite if it can be placed in 1-1 correspondence with a subset of itself. An infinite set which
is countable is called countable infinite.
The set of rational numbers is countable infinite, while the set of irrational numbers or all real
numbers is non-countably infinite (see Problems 1.17 through 1.20).
The number of elements in a set is called its cardinal number. A set which is countably infinite is
assigned the cardinal number Fo (the Hebrew letter aleph-null). The set of real numbers (or any sets
which can be placed into 1-1 correspondence with this set) is given the cardinal number C, called the
cardinality of the continuuum.
NEIGHBORHOODS
The set of all points x such that jx aj where 0, is called a neighborhood of the point a.
The set of all points x such that 0 jx aj in which x ¼ a is excluded, is called a deleted
neighborhood of a or an open ball of radius about a.
LIMIT POINTS
A limit point, point of accumulation, or cluster point of a set of numbers is a number l such that
every deleted neighborhood of l contains members of the set; that is, no matter how small the radius of
a ball about l there are points of the set within it. In other words for any 0, however small, we can
always find a member x of the set which is not equal to l but which is such that jx lj . By
considering smaller and smaller values of we see that there must be infinitely many such values of x.
A finite set cannot have a limit point. An infinite set may or may not have a limit point. Thus the
natural numbers have no limit point while the set of rational numbers has infinitely many limit points.
CHAP. 1] NUMBERS 5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-14-320.jpg)
![From the point of view of an axiomatic foundation of complex numbers, it is desirable to treat a
complex number as an ordered pair ða; bÞ of real numbers a and b subject to certain operational rules
which turn out to be equivalent to those above. For example, we define ða; bÞ þ ðc; dÞ ¼ ða þ c; b þ dÞ,
ða; bÞðc; dÞ ¼ ðac bd; ad þ bcÞ, mða; bÞ ¼ ðma; mbÞ, and so on. We then find that ða; bÞ ¼ að1; 0Þ þ
bð0; 1Þ and we associate this with a þ bi, where i is the symbol for ð0; 1Þ.
POLAR FORM OF COMPLEX NUMBERS
If real scales are chosen on two mutually perpendicular axes X 0
OX and Y 0
OY (the x and y axes) as
in Fig. 1-2 below, we can locate any point in the plane determined by these lines by the ordered pair of
numbers ðx; yÞ called rectangular coordinates of the point. Examples of the location of such points are
indicated by P, Q, R, S, and T in Fig. 1-2.
Since a complex number x þ iy can be considered as an ordered pair ðx; yÞ, we can represent such
numbers by points in an xy plane called the complex plane or Argand diagram. Referring to Fig. 1-3
above we see that x ¼ cos , y ¼ sin where ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
p
¼ jx þ iyj and , called the amplitude or
argument, is the angle which line OP makes with the positive x axis OX. It follows that
z ¼ x þ iy ¼ ðcos þ i sin Þ ð2Þ
called the polar form of the complex number, where and are called polar coordintes. It is sometimes
convenient to write cis instead of cos þ i sin .
If z1 ¼ x1 þ iyi ¼ 1ðcos 1 þ i sin 1Þ and z2 ¼ x2 þ iy2 ¼ 2ðcos 2 þ i sin 2Þ and by using the
addition formulas for sine and cosine, we can show that
z1z2 ¼ 12fcosð1 þ 2Þ þ i sinð1 þ 2Þg ð3Þ
z1
z2
¼
1
2
fcosð1 2Þ þ i sinð1 2Þg ð4Þ
zn
¼ fðcos þ i sin Þgn
¼ n
ðcos n þ i sin nÞ ð5Þ
where n is any real number. Equation (5) is sometimes called De Moivre’s theorem. We can use this to
determine roots of complex numbers. For example, if n is a positive integer,
z1=n
¼ fðcos þ i sin Þg1=n
ð6Þ
¼ 1=n
cos
þ 2k
n
þ i sin
þ 2k
n
k ¼ 0; 1; 2; 3; . . . ; n 1
CHAP. 1] NUMBERS 7
_4
X¢ X
_3 _2 _1 1 2 3 4
4
Y
Y¢
3
2
1
_1
_2
_3
O
Q(_3, 3)
S(2, _2)
P(3, 4)
T(2.5, 0)
R(_2.5, _1.5)
Fig. 1-2
X′ X
O
Y
Y′
ρ
φ
y
x
P(x, y)
Fig. 1-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-16-320.jpg)
![from which it follows that there are in general n different values of z1=n
. Later (Chap. 11) we will show
that ei
¼ cos þ i sin where e ¼ 2:71828 . . . . This is called Euler’s formula.
MATHEMATICAL INDUCTION
The principle of mathematical induction is an important property of the positive integers. It is
especially useful in proving statements involving all positive integers when it is known for example that
the statements are valid for n ¼ 1; 2; 3 but it is suspected or conjectured that they hold for all positive
integers. The method of proof consists of the following steps:
1. Prove the statement for n ¼ 1 (or some other positive integer).
2. Assume the statement true for n ¼ k; where k is any positive integer.
3. From the assumption in 2 prove that the statement must be true for n ¼ k þ 1. This is part of
the proof establishing the induction and may be difficult or impossible.
4. Since the statement is true for n ¼ 1 [from step 1] it must [from step 3] be true for n ¼ 1 þ 1 ¼ 2
and from this for n ¼ 2 þ 1 ¼ 3, and so on, and so must be true for all positive integers. (This
assumption, which provides the link for the truth of a statement for a finite number of cases to
the truth of that statement for the infinite set, is called ‘‘The Axiom of Mathematical Induc-
tion.’’)
Solved Problems
OPERATIONS WITH NUMBERS
1.1. If x ¼ 4, y ¼ 15, z ¼ 3, p ¼ 2
3, q ¼ 1
6, and r ¼ 3
4, evaluate (a) x þ ðy þ zÞ, (b) ðx þ yÞ þ z,
(c) pðqrÞ, (d) ðpqÞr, (e) xðp þ qÞ
(a) x þ ðy þ zÞ ¼ 4 þ ½15 þ ð3Þ ¼ 4 þ 12 ¼ 16
(b) ðx þ yÞ þ z ¼ ð4 þ 15Þ þ ð3Þ ¼ 19 3 ¼ 16
The fact that (a) and (b) are equal illustrates the associative law of addition.
(c) pðqrÞ ¼ 2
3 fð 1
6Þð3
4Þg ¼ ð2
3Þð 3
24Þ ¼ ð2
3Þð 1
8Þ ¼ 2
24 ¼ 1
12
(d) ðpqÞr ¼ fð2
3Þð 1
6Þgð3
4Þ ¼ ð 2
18Þð3
4Þ ¼ ð 1
9Þð3
4Þ ¼ 3
36 ¼ 1
12
The fact that (c) and (d) are equal illustrates the associative law of multiplication.
(e) xðp þ qÞ ¼ 4ð2
3 1
6Þ ¼ 4ð4
6 1
6Þ ¼ 4ð3
6Þ ¼ 12
6 ¼ 2
Another method: xðp þ qÞ ¼ xp þ xq ¼ ð4Þð2
3Þ þ ð4Þð 1
6Þ ¼ 8
3 4
6 ¼ 8
3 2
3 ¼ 6
3 ¼ 2 using the distributive
law.
1.2. Explain why we do not consider (a) 0
0 (b) 1
0 as numbers.
(a) If we define a=b as that number (if it exists) such that bx ¼ a, then 0=0 is that number x such that
0x ¼ 0. However, this is true for all numbers. Since there is no unique number which 0/0 can
represent, we consider it undefined.
(b) As in (a), if we define 1/0 as that number x (if it exists) such that 0x ¼ 1, we conclude that there is no
such number.
Because of these facts we must look upon division by zero as meaningless.
8 NUMBERS [CHAP. 1](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-17-320.jpg)
![1.3. Simplify
x2
5x þ 6
x2 2x 3
.
x2
5x þ 6
x2
2x 3
¼
ðx 3Þðx 2Þ
ðx 3Þðx þ 1Þ
¼
x 2
x þ 1
provided that the cancelled factor ðx 3Þ is not zero, i.e., x 6¼ 3.
For x ¼ 3 the given fraction is undefined.
RATIONAL AND IRRATIONAL NUMBERS
1.4. Prove that the square of any odd integer is odd.
Any odd integer has the form 2m þ 1. Since ð2m þ 1Þ2
¼ 4m2
þ 4m þ 1 is 1 more than the even integer
4m2
þ 4m ¼ 2ð2m2
þ 2mÞ, the result follows.
1.5. Prove that there is no rational number whose square is 2.
Let p=q be a rational number whose square is 2, where we assume that p=q is in lowest terms, i.e., p and q
have no common integer factors except 1 (we sometimes call such integers relatively prime).
Then ðp=qÞ2
¼ 2, p2
¼ 2q2
and p2
is even. From Problem 1.4, p is even since if p were odd, p2
would be
odd. Thus p ¼ 2m:
Substituting p ¼ 2m in p2
¼ 2q2
yields q2
¼ 2m2
, so that q2
is even and q is even.
Thus p and q have the common factor 2, contradicting the original assumption that they had no
common factors other than 1. By virtue of this contradiction there can be no rational number whose
square is 2.
1.6. Show how to find rational numbers whose squares can be arbitrarily close to 2.
We restrict ourselves to positive rational numbers. Since ð1Þ2
¼ 1 and ð2Þ2
¼ 4, we are led to choose
rational numbers between 1 and 2, e.g., 1:1; 1:2; 1:3; . . . ; 1:9.
Since ð1:4Þ2
¼ 1:96 and ð1:5Þ2
¼ 2:25, we consider rational numbers between 1.4 and 1.5, e.g.,
1:41; 1:42; . . . ; 1:49:
Continuing in this manner we can obtain closer and closer rational approximations, e.g. ð1:414213562Þ2
is less than 2 while ð1:414213563Þ2
is greater than 2.
1.7. Given the equation a0xn
þ a1xn1
þ þ an ¼ 0, where a0; a1; . . . ; an are integers and a0 and
an 6¼ 0. Show that if the equation is to have a rational root p=q, then p must divide an and q
must divide a0 exactly.
Since p=q is a root we have, on substituting in the given equation and multiplying by qn
, the result
a0pn
þ a1pn1
q þ a2pn2
q2
þ þ an1pqn1
þ anqn
¼ 0 ð1Þ
or dividing by p,
a0pn1
þ a1pn2
q þ þ an1qn1
¼
anqn
p
ð2Þ
Since the left side of (2) is an integer, the right side must also be an integer. Then since p and q are relatively
prime, p does not divide qn
exactly and so must divide an.
In a similar manner, by transposing the first term of (1) and dividing by q, we can show that q must
divide a0.
1.8. Prove that
ffiffiffi
2
p
þ
ffiffiffi
3
p
cannot be a rational number.
If x ¼
ffiffiffi
2
p
þ
ffiffiffi
3
p
, then x2
¼ 5 þ 2
ffiffiffi
6
p
, x2
5 ¼ 2
ffiffiffi
6
p
and squaring, x4
10x2
þ 1 ¼ 0. The only possible
rational roots of this equation are 1 by Problem 1.7, and these do not satisfy the equation. It follows that
ffiffiffi
2
p
þ
ffiffiffi
3
p
, which satisfies the equation, cannot be a rational number.
CHAP. 1] NUMBERS 9](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-18-320.jpg)
![Sn ¼
1
2
þ
1
4
þ
1
8
þ þ
1
2n1
Let
1
2
Sn ¼
1
4
þ
1
8
þ þ
1
2n1
þ
1
2n
Then
1
2
Sn ¼
1
2
1
2n : Thus Sn ¼ 1
1
2n1
1 for all n:
Subtracting,
EXPONENTS, ROOTS, AND LOGARITHMS
1.15. Evaluate each of the following:
ðaÞ
34
38
314
¼
34þ8
314
¼ 34þ814
¼ 32
¼
1
32
¼
1
9
ðbÞ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð5 106
Þð4 102
Þ
8 105
s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5 4
8
106
102
105
s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:5 109
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
25 1010
p
¼ 5 105
or 0:00005
ðcÞ log2=3
27
8
¼ x: Then 2
3
x
¼ 27
8 ¼ 3
2
3
¼ 2
3
3
or x ¼ 3
ðdÞ ðloga bÞðlogb aÞ ¼ u: Then loga b ¼ x; logb a ¼ y assuming a; b 0 and a; b 6¼ 1:
Then ax
¼ b, by
¼ a and u ¼ xy.
Since ðax
Þy
¼ axy
¼ by
¼ a we have axy
¼ a1
or xy ¼ 1 the required value.
1.16. If M 0, N 0; and a 0 but a 6¼ 1, prove that loga
M
N
¼ loga M loga N.
Let loga M ¼ x, loga N ¼ y. Then ax
¼ M, ay
¼ N and so
M
N
¼
ax
ay ¼ axy
or loga
M
N
¼ x y ¼ loga M loga N
COUNTABILITY
1.17. Prove that the set of all rational numbers between 0 and 1 inclusive is countable.
Write all fractions with denominator 2, then 3; . . . considering equivalent fractions such as 1
2 ; 2
4 ; 3
6 ; . . . no
more than once. Then the 1-1 correspondence with the natural numbers can be accomplished as follows:
Rational numbers
Natural numbers
0 1 1
2
1
3
2
3
1
4
3
4
1
5
2
5 . . .
l l l l l l l l l
1 2 3 4 5 6 7 8 9 . . .
Thus the set of all rational numbers between 0 and 1 inclusive is countable and has cardinal number Fo
(see Page 5).
1.18. If A and B are two countable sets, prove that the set consisting of all elements from A or B (or
both) is also countable.
Since A is countable, there is a 1-1 correspondence between elements of A and the natural numbers so
that we can denote these elements by a1; a2; a3; . . . .
Similarly, we can denote the elements of B by b1; b2; b3; . . . .
Case 1: Suppose elements of A are all distinct from elements of B. Then the set consisting of elements from
A or B is countable, since we can establish the following 1-1 correspondence.
CHAP. 1] NUMBERS 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-20-320.jpg)
![(b) Since no member of the set is greater than 1 and since there is at least one member of the set (namely 1)
which exceeds 1 for every positive number , we see that 1 is the l.u.b. of the set.
Since no member of the set is less than 0 and since there is at least one member of the set which is
less than 0 þ for every positive (we can always choose for this purpose the number 1=n where n is a
positive integer greater than 1=), we see that 0 is the g.l.b. of the set.
(c) Let x be any member of the set. Since we can always find a number x such that 0 jxj for any
positive number (e.g. we can always pick x to be the number 1=n where n is a positive integer greater
than 1=), we see that 0 is a limit point of the set. To put this another way, we see that any deleted
neighborhood of 0 always includes members of the set, no matter how small we take 0.
(d) The set is not a closed set since the limit point 0 does not belong to the given set.
(e) Since the set is bounded and infinite it must, by the Bolzano–Weierstrass theorem, have at least one
limit point. We have found this to be the case, so that the theorem is illustrated.
ALGEBRAIC AND TRANSCENDENTAL NUMBERS
1.22. Prove that
ffiffiffi
2
3
p
þ
ffiffiffi
3
p
is an algebraic number.
Let x ¼
ffiffiffi
2
3
p
þ
ffiffiffi
3
p
. Then x
ffiffiffi
3
p
¼
ffiffiffi
2
3
p
. Cubing both sides and simplifying, we find x3
þ 9x 2 ¼
3
ffiffiffi
3
p
ðx2
þ 1Þ. Then squaring both sides and simplifying we find x6
9x4
4x3
þ 27x2
þ 36x 23 ¼ 0.
Since this is a polynomial equation with integral coefficients it follows that
ffiffiffi
2
3
p
þ
ffiffiffi
3
p
, which is a
solution, is an algebraic number.
1.23. Prove that the set of all algebraic numbers is a countable set.
Algebraic numbers are solutions to polynomial equations of the form a0xn
þ a1xn1
þ þ an ¼ 0
where a0; a1; . . . ; an are integers.
Let P ¼ ja0j þ ja1j þ þ janj þ n. For any given value of P there are only a finite number of possible
polynomial equations and thus only a finite number of possible algebraic numbers.
Write all algebraic numbers corresponding to P ¼ 1; 2; 3; 4; . . . avoiding repetitions. Thus, all algebraic
numbers can be placed into 1-1 correspondence with the natural numbers and so are countable.
COMPLEX NUMBERS
1.24. Perform the indicated operations.
(a) ð4 2iÞ þ ð6 þ 5iÞ ¼ 4 2i 6 þ 5i ¼ 4 6 þ ð2 þ 5Þi ¼ 2 þ 3i
(b) ð7 þ 3iÞ ð2 4iÞ ¼ 7 þ 3i 2 þ 4i ¼ 9 þ 7i
(c) ð3 2iÞð1 þ 3iÞ ¼ 3ð1 þ 3iÞ 2ið1 þ 3iÞ ¼ 3 þ 9i 2i 6i2
¼ 3 þ 9i 2i þ 6 ¼ 9 þ 7i
ðdÞ
5 þ 5i
4 3i
¼
5 þ 5i
4 3i
4 þ 3i
4 þ 3i
¼
ð5 þ 5iÞð4 þ 3iÞ
16 9i2
¼
20 15i þ 20i þ 15i2
16 þ 9
¼
35 þ 5i
25
¼
5ð7 þ iÞ
25
¼
7
5
þ
1
5
i
ðeÞ
i þ i2
þ i3
þ i4
þ i5
1 þ i
¼
i 1 þ ði2
ÞðiÞ þ ði2
Þ2
þ ði2
Þ2
i
1 þ i
¼
i 1 i þ 1 þ i
1 þ i
¼
i
1 þ i
1 i
1 i
¼
i i2
1 i2
¼
i þ 1
2
¼
1
2
þ
1
2
i
ð f Þ j3 4ijj4 þ 3ij ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð3Þ2
þ ð4Þ2
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð4Þ2
þ ð3Þ2
q
¼ ð5Þð5Þ ¼ 25
CHAP. 1] NUMBERS 13](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-22-320.jpg)
![1.28. Evaluate (a) ð1 þ
ffiffiffi
3
p
iÞ10
, (b) ð1 þ iÞ1=3
.
(a) By Problem 1.27(b) and De Moivre’s theorem,
ð1 þ
ffiffiffi
3
p
iÞ10
¼ ½2ðcos 2=3 þ i sin 2=3Þ10
¼ 210
ðcos 20=3 þ i sin 20=3Þ
¼ 1024½cosð2=3 þ 6Þ þ i sinð2=3 þ 6Þ ¼ 1024ðcos 2=3 þ i sin 2=3Þ
¼ 1024 1
2 þ 1
2
ffiffiffi
3
p
i
¼ 512 þ 512
ffiffiffi
3
p
i
(b) 1 þ i ¼
ffiffiffi
2
p
ðcos 1358 þ i sin 1358Þ ¼
ffiffiffi
2
p
½cosð1358 þ k 3608Þ þ i sinð1358 þ k 3608Þ. Then
ð1 þ iÞ1=3
¼ ð
ffiffiffi
2
p
Þ1=3
cos
1358 þ k 3608
3
þ i sin
1358 þ k 3608
3
The results for k ¼ 0; 1; 2 are
ffiffiffi
2
6
p
ðcos 458 þ i sin 458Þ;
ffiffiffi
2
6
p
ðcos 1658 þ i sin 1658Þ;
ffiffiffi
2
6
p
ðcos 2858 þ i sin 2858Þ
The results for k ¼ 3; 4; 5; 6; 7; . . . give repetitions of these. These
complex roots are represented geometrically in the complex plane
by points P1; P2; P3 on the circle of Fig. 1-5.
MATHEMATICAL INDUCTION
1.29. Prove that 12
þ 22
þ 33
þ 42
þ þ n2
¼ 1
6 nðn þ 1Þð2n þ 1Þ.
The statement is true for n ¼ 1 since 12
¼ 1
6 ð1Þð1 þ 1Þð2 1 þ 1Þ ¼ 1.
Assume the statement true for n ¼ k. Then
12
þ 22
þ 32
þ þ k2
¼ 1
6 kðk þ 1Þð2k þ 1Þ
Adding ðk þ 1Þ2
to both sides,
12
þ 22
þ 32
þ þ k2
þ ðk þ 1Þ2
¼ 1
6 kðk þ 1Þð2k þ 1Þ þ ðk þ 1Þ2
¼ ðk þ 1Þ½1
6 kð2k þ 1Þ þ k þ 1
¼ 1
6 ðk þ 1Þð2k2
þ 7k þ 6Þ ¼ 1
6 ðk þ 1Þðk þ 2Þð2k þ 3Þ
which shows that the statement is true for n ¼ k þ 1 if it is true for n ¼ k. But since it is true for n ¼ 1, it
follows that it is true for n ¼ 1 þ 1 ¼ 2 and for n ¼ 2 þ 1 ¼ 3; . . . ; i.e., it is true for all positive integers n.
1.30. Prove that xn
yn
has x y as a factor for all positive integers n.
The statement is true for n ¼ 1 since x1
y1
¼ x y.
Assume the statement true for n ¼ k, i.e., assume that xk
yk
has x y as a factor. Consider
xkþ1
ykþ1
¼ xkþ1
xk
y þ xk
y ykþ1
¼ xk
ðx yÞ þ yðxk
yk
Þ
The first term on the right has x y as a factor, and the second term on the right also has x y as a factor
because of the above assumption.
Thus xkþ1
ykþ1
has x y as a factor if xk
yk
does.
Then since x1
y1
has x y as factor, it follows that x2
y2
has x y as a factor, x3
y3
has x y as a
factor, etc.
CHAP. 1] NUMBERS 15
P2
P3
P1
165°
285°
45°
√2
6
Fig. 1-5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-24-320.jpg)
![(d) It is possible for L to have no largest number and for R to have no smallest number. What
type of number does the cut define in this case?
(a) Let a be the largest rational number in L, and b the smallest rational number in R. Then either a ¼ b or
a b.
We cannot have a ¼ b since by definition of the cut every number in L is less than every number
in R.
We cannot have a b since by Problem 1.9, 1
2 ða þ bÞ is a rational number which would be greater
than a (and so would have to be in R) but less than b (and so would have to be in L), and by definition a
rational number cannot belong to both L and R.
(b) As an indication of the possibility, let L contain the number 2
3 and all rational numbers less than 2
3, while
R contains all rational numbers greater than 2
3. In this case the cut defines the rational number 2
3. A
similar argument replacing 2
3 by any other rational number shows that in such case the cut defines a
rational number.
(c) As an indication of the possibility, let L contain all rational numbers less than 2
3, while R contains all
rational numbers greaters than 2
3. This cut also defines the rational number 2
3. A similar argument
shows that this cut always defines a rational number.
(d) As an indication of the possibility let L consist of all negative rational numbers and all positive rational
numbers whose squares are less than 2, while R consists of all positive numbers whose squares are
greater than 2. We can show that if a is any number of the L class, there is always a larger number of
the L class, while if b is any number of the R class, there is always a smaller number of the R class (see
Problem 1.106). A cut of this type defines an irrational number.
From (b), (c), (d) it follows that every cut in the rational number system, called a Dedekind cut,
defines either a rational or an irrational number. By use of Dedekind cuts we can define operations
(such as addition, multiplication, etc.) with irrational numbers.
Supplementary Problems
OPERATIONS WITH NUMBERS
1.35. Given x ¼ 3, y ¼ 2, z ¼ 5, a ¼ 3
2, and b ¼ 1
4, evaluate:
ðaÞ ð2x yÞð3y þ zÞð5x 2zÞ; ðbÞ
xy 2z2
2ab 1
; ðcÞ
3a2
b þ ab2
2a22b2 þ 1
; ðdÞ
ðax þ byÞ2
þ ðay bxÞ2
ðay þ bxÞ2
þ ðax byÞ2
:
Ans. (a) 2200, (b) 32, (c) 51=41, (d) 1
1.36. Find the set of values of x for which the following equations are true. Justify all steps in each case.
ðaÞ 4fðx 2Þ þ 3ð2x 1Þg þ 2ð2x þ 1Þ ¼ 12ðx þ 2Þ 2 ðcÞ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ 8x þ 7
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x þ 2
p
¼ x þ 1
ðbÞ
1
8 x
1
x 2
¼
1
4
ðdÞ
1 x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 2x þ 5
p ¼
3
5
Ans. (a) 2, (b) 6; 4 (c) 1; 1 (d) 1
2
1.37. Prove that
x
ðz xÞðx yÞ
þ
y
ðx yÞðy zÞ
þ
z
ðy zÞðz xÞ
¼ 0 giving restrictions if any.
RATIONAL AND IRRATIONAL NUMBERS
1.38. Find decimal expansions for (a) 3
7, (b)
ffiffiffi
5
p
.
Ans. (a) 0:_
4
4_
2
2_
8
8_
5
5_
7
7_
1
1, (b) 2.2360679 . . .
CHAP. 1] NUMBERS 17](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-26-320.jpg)
![CHAP. 1] NUMBERS 19
COUNTABILITY
1.58. (a) Prove that there is a one to one correspondence between the points of the interval 0 @ x @ 1 and
5 @ x @ 3. (b) What is the cardinal number of the sets in (a)?
Ans. (b) C, the cardinal number of the continuum.
1.59. (a) Prove that the set of all rational numbers is countable. (b) What is the cardinal number of the set in (a)?
Ans. (b) Fo
1.60. Prove that the set of (a) all real numbers, (b) all irrational numbers is non-countable.
1.61. The intersection of two sets A and B, denoted by A B or AB, is the set consisting of all elements belonging
to both A and B. Prove that if A and B are countable, so is their intersection.
1.62. Prove that a countable set of countable sets is countable.
1.63. Prove that the cardinal number of the set of points inside a square is equal to the cardinal number of the sets
of points on (a) one side, (b) all four sides. (c) What is the cardinal number in this case? (d) Does a
corresponding result hold for a cube?
Ans. (c) C
LIMIT POINTS, BOUNDS, BOLZANO–WEIERSTRASS THEOREM
1.64. Given the set of numbers 1; 1:1; :9; 1:01; :99; 1:001; :999; . . . . (a) Is the set bounded? (b) Does the set have
a l.u.b. and g.l.b.? If so, determine them. (c) Does the set have any limit points? If so, determine them.
(d) Is the set a closed set?
Ans. (a) Yes (b) l:u:b: ¼ 1:1; g:l:b: ¼ :9 (c) 1 (d) Yes
1.65. Give the set :9; :9; :99; :99; :999; :999 answer the questions of Problem 64.
Ans. (a) Yes (b) l:u:b: ¼ 1; g:l:b: ¼ 1 (c) 1; 1 (d) No
1.66. Give an example of a set which has (a) 3 limit points, (b) no limit points.
1.67. (a) Prove that every point of the interval 0 x 1 is a limit point.
(b) Are there are limit points which do not belong to the set in (a)? Justify your answer.
1.68. Let S be the set of all rational numbers in ð0; 1Þ having denominator 2n
, n ¼ 1; 2; 3; . . . . (a) Does S have
any limit points? (b) Is S closed?
1.69. (a) Give an example of a set which has limit points but which is not bounded. (b) Does this contradict the
Bolzano–Weierstrass theorem? Explain.
ALGEBRAIC AND TRANSCENDENTAL NUMBERS
1.70. Prove that (a)
ffiffiffi
3
p
ffiffiffi
2
p
ffiffiffi
3
p
þ
ffiffiffi
2
p , (b)
ffiffiffi
2
p
þ
ffiffiffi
3
p
þ
ffiffiffi
5
p
are algebraic numbers.
1.71. Prove that the set of transcendental numbers in ð0; 1Þ is not countable.
1.72. Prove that every rational number is algebraic but every irrational number is not necessarily algebraic.
COMPLEX NUMBERS, POLAR FORM
1.73. Perform each of the indicated operations: (a) 2ð5 3iÞ 3ð2 þ iÞ þ 5ði 3Þ, (b) ð3 2iÞ3
ðcÞ
5
3 4i
þ
10
4 þ 3i
; ðdÞ
1 i
1 þ i
10
; ðeÞ
2 4i
5 þ 7i
2
; ð f Þ
ð1 þ iÞð2 þ 3iÞð4 2iÞ
ð1 þ 2iÞ2
ð1 iÞ
:
Ans. (a) 1 4i, (b) 9 46i, (c) 11
5 2
5 i, (d) 1, (e) 10
37, ( f ) 16
5 2
5 i.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-28-320.jpg)
![1.92. ðcos þ i sin Þn
¼ cos n þ i sin n. Can this be proved if n is a rational number?
1.93. 1
2 þ cos x þ cos 2x þ þ cos nx ¼
sinðn þ 1
2Þx
2 sin 1
2 x
, x 6¼ 0; 2; 4; . . .
1.94. sin x þ sin 2x þ þ sin nx ¼
cos 1
2 x cosðn þ 1
2Þx
2 sin 1
2 x
; x 6¼ 0; 2; 4; . . .
1.95. ða þ bÞn
¼ an
þ nC1an1
b þ nC2an2
b2
þ þ nCn1abn1
þ bn
where nCr ¼
nðn 1Þðn 2Þ . . . ðn r þ 1Þ
r!
¼
n!
r!ðn rÞ!
¼ n Cnr. Here p! ¼ pðp 1Þ . . . 1 and 0! is defined as
1. This is called the binomial theorem. The coefficients nC0 ¼ 1, nC1 ¼ n, nC2 ¼
nðn 1Þ
2!
; . . . ; nCn ¼ 1 are
called the binomial coefficients. nCr is also written
n
r
.
MISCELLANEOUS PROBLEMS
1.96. Express each of the following integers (scale of 10) in the scale of notation indicated: (a) 87 (two), (b) 64
(three), (c) 1736 (nine). Check each answer.
Ans. (a) 1010111, (b) 2101, (c) 2338
1.97. If a number is 144 in the scale of 5, what is the number in the scale of (a) 2, (b) 8?
1.98. Prove that every rational number p=q between 0 and 1 can be expressed in the form
p
q
¼
a1
2
þ
a2
22
þ þ
an
2n þ
where the a’s can be determined uniquely as 0’s or 1’s and where the process may or may not terminate. The
representation 0:a1a2 . . . an . . . is then called the binary form of the rational number. [Hint: Multiply both
sides successively by 2 and consider remainders.}
1.99. Express 2
3 in the scale of (a) 2, (b) 3, (c) 8, (d) 10.
Ans. (a) 0:1010101 . . . ; (b) 0.2 or 0:2000 . . . ; (c) 0:5252 . . . ; (d) 0:6666 . . .
1.100. A number in the scale of 2 is 11.01001. What is the number in the scale of 10.
Ans. 3.28125
1.101. In what scale of notation is 3 þ 4 ¼ 12?
Ans. 5
1.102. In the scale of 12, two additional symbols t and e must be used to designate the ‘‘digits’’ 10 and 11,
respectively. Using these symbols, represent the integer 5110 (scale of 10) in the scale of 12.
Ans. 2e5t
1.103. Find a rational number whose decimal expansion is 1:636363 . . . .
Ans. 18/11
1.104. A number in the scale of 10 consists of six digits. If the last digit is removed and placed before the first digit,
the new number is one-third as large. Find the original number.
Ans. 428571
1.105. Show that the rational numbers form a field.
1.106. Using as axioms the relations 1–9 on Pages 2 and 3, prove that
(a) ð3Þð0Þ ¼ 0, (b) ð2Þðþ3Þ ¼ 6, (c) ð2Þð3Þ ¼ 6.
CHAP. 1] NUMBERS 21](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-30-320.jpg)
![LIMIT SUPERIOR, LIMIT INFERIOR
A number
l
l is called the limit superior, greatest limit or upper limit (lim sup or lim) of the sequence
fung if infinitely many terms of the sequence are greater than
l
l while only a finite number of terms are
greater than
l
l þ , where is any positive number.
A number l is called the limit inferior, least limit or lower limit (lim inf or lim) of the sequence fung if
infintely many terms of the sequence are less than l þ while only a finite number of terms are less than
l , where is any positive number.
These correspond to least and greatest limiting points of general sets of numbers.
If infintely many terms of fung exceed any positive number M, we define lim sup fung ¼ 1. If
infinitely many terms are less than M, where M is any positive number, we define lim inf fung ¼ 1.
If lim
n!1
un ¼ 1, we define lim sup fung ¼ lim inf fung ¼ 1.
If lim
n!1
un ¼ 1, we define lim sup fung ¼ lim inf fung ¼ 1.
Although every bounded sequence is not necessarily convergent, it always has a finite lim sup and
lim inf.
A sequence fung converges if and only if lim sup un ¼ lim inf un is finite.
NESTED INTERVALS
Consider a set of intervals ½an; bn, n ¼ 1; 2; 3; . . . ; where each interval is contained in the preceding
one and lim
n!1
ðan bnÞ ¼ 0. Such intervals are called nested intervals.
We can prove that to every set of nested intervals there corresponds one and only one real number.
This can be used to establish the Bolzano–Weierstrass theorem of Chapter 1. (See Problems 2.22 and
2.23.)
CAUCHY’S CONVERGENCE CRITERION
Cauchy’s convergence criterion states that a sequence fung converges if and only if for each 0 we
can find a number N such that jup uqj for all p; q N. This criterion has the advantage that one
need not know the limit l in order to demonstrate convergence.
INFINITE SERIES
Let u1; u2; u3; . . . be a given sequence. Form a new sequence S1; S2; S3; . . . where
S1 ¼ u1; S2 ¼ u1 þ u2; S3 ¼ u1 þ u2 þ u3; . . . ; Sn ¼ u1 þ u2 þ u3 þ þ un; . . .
where Sn, called the nth partial sum, is the sum of the first n terms of the sequence fung.
The sequence S1; S2; S3; . . . is symbolized by
u1 þ u2 þ u3 þ ¼
X
1
n¼1
un
which is called an infinite series. If lim
n!1
Sn ¼ S exists, the series is called convergent and S is its sum,
otherwise the series is called divergent.
Further discussion of infinite series and other topics related to sequences is given in Chapter 11.
CHAP. 2] SEQUENCES 25](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-34-320.jpg)
![LIMIT OF A SEQUENCE
2.3. A sequence has its nth term given by un ¼
3n 1
4n þ 5
. (a) Write the 1st, 5th, 10th, 100th, 1000th,
10,000th and 100,000th terms of the sequence in decimal form. Make a guess as to the limit of
this sequence as n ! 1. (b) Using the definition of limit verify that the guess in (a) is actually
correct.
ðaÞ
n ¼ 1 n ¼ 5 n ¼ 10 n ¼ 100 n ¼ 1000 n ¼ 10,000 n ¼ 100,000
:22222 . . . :56000 . . . :64444 . . . :73827 . . . :74881 . . . :74988 . . . :74998 . . .
A good guess is that the limit is :75000 . . . ¼ 3
4. Note that it is only for large enough values of n that
a possible limit may become apparent.
(b) We must show that for any given 0 (no matter how small) there is a number N (depending on )
such that jun 3
4 j for all n N.
Now
3n 1
4n þ 5
3
4
¼
19
4ð4n þ 5Þ
when
19
4ð4n þ 5Þ
or
4ð4n þ 5Þ
19
1
; 4n þ 5
19
4
; n
1
4
19
4
5
Choosing N ¼ 1
4 ð19=4 5Þ, we see that jun 3
4 j for all n N, so that lim
n!1
¼ 3
4 and the proof is
complete.
Note that if ¼ :001 (for example), N ¼ 1
4 ð19000=4 5Þ ¼ 1186 1
4. This means that all terms of the
sequence beyond the 1186th term differ from 3
4 in absolute value by less than .001.
2.4. Prove that lim
n!1
c
np ¼ 0 where c 6¼ 0 and p 0 are constants (independent of n).
We must show that for any 0 there is a number N such that jc=np
0j for all n N.
Now
c
np when
jcj
np , i.e., np
jcj
or n
jcj
1=p
. Choosing N ¼
jcj
1=p
(depending on ), we
see that jc=np
j for all n N, proving that lim
n!1
ðc=np
Þ ¼ 0.
2.5. Prove that lim
n!1
1 þ 2 10n
5 þ 3 10n ¼
2
3
.
We must show that for any 0 there is a number N such that
1 þ 2 10n
5 þ 3 10n
2
3
for all n N.
Now
1 þ 2 10n
5 þ 3 10n
2
3
¼
7
3ð5 þ 3 10nÞ
when
7
3ð5 þ 3 10nÞ
, i.e. when 3
7 ð5 þ 3 10n
Þ 1=,
3 10n
7=3 5, 10n
1
8 ð7=3 5Þ or n log10f1
3 ð7=3 5Þg ¼ N, proving the existence of N and thus
establishing the required result.
Note that the above value of N is real only if 7=3 5 0, i.e., 0 7=15. If A 7=15, we see that
1 þ 2 10n
5 þ 3 10n
2
3
for all n 0.
2.6. Explain exactly what is meant by the statements (a) lim
n!1
32n1
¼ 1, (b) lim
n!1
ð1 2nÞ ¼ 1.
(a) If for each positive number M we can find a positive number N (depending on M) such that an M for
all n N, then we write lim
n!1
an ¼ 1.
In this case, 32n1
M when ð2n 1Þ log 3 log M; i.e., n
1
2
log M
log 3
þ 1
¼ N.
(b) If for each positive number M we can find a positive number N (depending on M) such that an M
for all n N, then we write lim
n!1
¼ 1.
In this case, 1 2n M when 2n 1 M or n 1
2 ðM þ 1Þ ¼ N.
CHAP. 2] SEQUENCES 27](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-36-320.jpg)
![2.11. If lim
n!1
bn ¼ B 6¼ 0, prove there exists a number N such that jbnj 1
2 jBj for all n N.
Since B ¼ B bn þ bn, we have: (1) jBj @ jB bnj þ jbnj.
Now we can choose N so that jB bnj ¼ jbn Bj 1
2 jBj for all n N, since lim
n!1
bn ¼ B by hypothesis.
Hence, from (1), jBj 1
2 jBj þ jbnj or jbnj 1
2 jBj for all n N.
2.12. If lim
n!1
an ¼ A and lim
n!1
bn ¼ B, prove that lim
n!1
anbn ¼ AB.
We have, using Problem 2.10,
janbn ABj ¼ janðbn BÞ þ Bðan AÞj @ janjjbn Bj þ jBjjan Aj ð1Þ
@ Pjbn Bj þ ðjBj þ 1Þjan Aj
But since lim
n!1
an ¼ A and lim
n!1
bn ¼ B, given any 0 we can find N1 and N2 such that
jbn Bj
2P
for all n N1 jan Aj
2ðjBj þ 1Þ
for all n N2
Hence, from (1), janbn ABj 1
2 þ 1
2 ¼ for all n N, where N is the larger of N1 and N2. Thus, the
result is proved.
2.13. If lim
n!1
an ¼ A and lim
n!1
bn ¼ B 6¼ 0, prove (a) lim
n!1
1
bn
¼
1
B
, (b) lim
n!1
an
bn
¼
A
B
.
(a) We must show that for any given 0, we can find N such that
1
bn
1
B
¼
jB bnj
jBjjbnj
for all n N ð1Þ
By hypothesis, given any 0, we can find N1, such that jbn Bj 1
2 B2
for all n N1.
Also, since lim
n!1
bn ¼ B 6¼ 0, we can find N2 such that jbnj 1
2 jBj for all n N2 (see Problem 11).
Then if N is the larger of N1 and N2, we can write (1) as
1
bn
1
B
¼
jbn Bj
jBjjbnj
1
2 B2
jBj 1
2 jBj
¼ for all n N
and the proof is complete.
(b) From part (a) and Problem 2.12, we have
lim
n!1
an
bn
¼ lim
n!1
an
1
bn
¼ lim
n!1
an lim
n!1
1
bn
¼ A
1
B
¼
A
B
This can also be proved directly (see Problem 41).
2.14. Evaluate each of the following, using theorems on limits.
ðaÞ lim
n!1
3n2
5n
5n2 þ 2n 6
¼ lim
n!1
3 5=n
5 þ 2=n 6=n2
¼
3 þ 0
5 þ 0 þ 0
¼
3
5
ðbÞ lim
n!1
nðn þ 2Þ
n þ 1
n3
n2 þ 1
( )
¼ lim
n!1
n3
þ n2
þ 2n
ðn þ 1Þðn2 þ 1Þ
( )
¼ lim
n!1
1 þ 1=n þ 2=n2
ð1 þ 1=nÞð1 þ 1=n2Þ
( )
¼
1 þ 0 þ 0
ð1 þ 0Þ ð1 þ 0Þ
¼ 1
ðcÞ lim
n!1
ð
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
ffiffiffi
n
p
Þ ¼ lim
n!1
ð
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
ffiffiffi
n
p
Þ
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
þ
ffiffiffi
n
p
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
þ
ffiffiffi
n
p ¼ lim
n!1
1
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
þ
ffiffiffi
n
p ¼ 0
CHAP. 2] SEQUENCES 29](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-38-320.jpg)
![CHAP. 2] SEQUENCES 31
(b) Let x ¼ required limit. Since lim
n!1
unþ1 ¼ lim
n!1
ffiffiffiffiffiffiffi
3un
p
, we have x ¼
ffiffiffiffiffiffi
3x
p
and x ¼ 3. (The other
possibility, x ¼ 0, is excluded since un A 1:Þ
Another method: lim
n!1
31=2þ1=4þþ1=2n1
¼ lim
n!1
311=2n
¼ 3
limð11=2nÞ
n!1 ¼ 31
¼ 3
2.17. Verify the validity of the entries in the following table.
Sequence Bounded
Monotonic
Increasing
Monotonic
Decreasing
Limit
Exists
2; 1:9; 1:8; 1:7; . . . ; 2 ðn 1Þ=10 . . . No No Yes No
1; 1; 1; 1; . . . ; ð1Þn1
; . . . Yes No No No
1
2 ; 1
3 ; 1
4 ; 1
5 ; . . . ; ð1Þn1
=ðn þ 1Þ; . . . Yes No No Yes (0)
:6; :66; :666; . . . ; 2
3 ð1 1=10n
Þ; . . . Yes Yes No Yes (2
3)
1; þ2; 3; þ4; 5; . . . ; ð1Þn
n; . . . No No No No
2.18. Prove that the sequence with the nth term un ¼ 1 þ
1
n
n
is monotonic, increasing, and bounded,
and thus a limit exists. The limit is denoted by the symbol e.
Note: lim
n!1
1 þ
1
n
n
¼ e, where e ffi 2:71828 . . . was introduced in the eighteenth century by
Leonhart Euler as the base for a system of logarithms in order to simplify certain differentiation
and integration formulas.
By the binomial theorem, if n is a positive integer (see Problem 1.95, Chapter 1),
ð1 þ xÞn
¼ 1 þ nx þ
nðn 1Þ
2!
x2
þ
nðn 1Þðn 2Þ
3!
x3
þ þ
nðn 1Þ ðn n þ 1Þ
n!
xn
Letting x ¼ 1=n,
un ¼ 1 þ
1
n
n
¼ 1 þ n
1
n
þ
nðn 1Þ
2!
1
n2
þ þ
nðn 1Þ ðn n þ 1Þ
n!
1
nn
¼ 1 þ 1 þ
1
2!
1
1
n
þ
1
3!
1
1
n
1
2
n
þ þ
1
n!
1
1
n
1
2
n
1
n 1
n
Since each term beyond the first two terms in the last expression is an increasing function of n, it follows that
the sequence un is a monotonic increasing sequence.
It is also clear that
1 þ
1
n
n
1 þ 1 þ
1
2!
þ
1
3!
þ þ
1
n!
1 þ 1 þ
1
2
þ
1
22
þ þ
1
2n1
3
by Problem 1.14, Chapter 1.
Thus, un is bounded and monotonic increasing, and so has a limit which we denote by e. The value of
e ¼ 2:71828 . . . .
2.19. Prove that lim
x!1
1 þ
1
x
x
¼ e, where x ! 1 in any manner whatsoever (i.e., not necessarily along
the positive integers, as in Problem 2.18).
If n ¼ largest integer @ x, then n @ x @ n þ 1 and 1 þ
1
n þ 1
n
@ 1 þ
1
x
x
@ 1 þ
1
n
nþ1
.
Since lim
n!1
1 þ
1
n þ 1
n
¼ lim
n!1
1 þ
1
n þ 1
nþ1
,
1 þ
1
n þ 1
¼ e](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-40-320.jpg)
![2.23. Prove the Bolzano–Weierstrass theorem (see Page 6).
Suppose the given bounded infinite set is contained in the finite interval ½a; b. Divide this interval into
two equal intervals. Then at least one of these, denoted by ½a1; b1, contains infinitely many points.
Dividing ½a1; b1 into two equal intervals, we obtain another interval, say, ½a2; b2, containing infinitely
many points. Continuing this process, we obtain a set of intervals ½an; bn, n ¼ 1; 2; 3; . . . ; each interval
contained in the preceding one and such that
b1 a1 ¼ ðb aÞ=2; b2 a2 ¼ ðb1 a1Þ=2 ¼ ðb aÞ=22
; . . . ; bn an ¼ ðb aÞ=2n
from which we see that lim
n!1
ðbn anÞ ¼ 0.
This set of nested intervals, by Problem 2.22, corresponds to a real number which represents a limit
point and so proves the theorem.
CAUCHY’S CONVERGENCE CRITERION
2.24. Prove Cauchy’s convergence criterion as stated on Page 25.
Necessity. Suppose the sequence fung converges to l. Then given any 0, we can find N such that
jup lj =2 for all p N and juq lj =2 for all q N
Then for both p N and q N, we have
jup uqj ¼ jðup lÞ þ ðl uqÞj @ jup lj þ jl uqj =2 þ =2 ¼
Sufficiency. Suppose jup uqj for all p; q N and any 0. Then all the numbers uN; uNþ1; . . .
lie in a finite interval, i.e., the set is bounded and infinite. Hence, by the Bolzano–Weierstrass theorem there
is at least one limit point, say a.
If a is the only limit point, we have the desired proof and lim
n!1
un ¼ a.
Suppose there are two distinct limit points, say a and b, and suppose b a (see Fig. 2-1). By definition
of limit points, we have
jup aj ðb aÞ=3 for infinnitely many values of p ð1Þ
juq bj ðb aÞ=3 for infinitely many values of q ð2Þ
Then since b a ¼ ðb uqÞ þ ðuq upÞ þ ðup aÞ, we have
jb aj ¼ b a @ jb uqj þ jup uqj þ jup aj ð3Þ
Using (1) and (2) in (3), we see that jup uqj ðb aÞ=3 for infinitely many values of p and q, thus
contradicting the hypothesis that jup uqj for p; q N and any 0. Hence, there is only one limit
point and the theorem is proved.
INFINITE SERIES
2.25. Prove that the infinite series (sometimes called the geometric series)
a þ ar þ ar2
þ ¼
X
1
n¼1
arn1
(a) converges to a=ð1 rÞ if jrj 1, (b) diverges if jrj A 1.
Sn ¼ a þ ar þ ar2
þ þ arn1
Let
rSn ¼ ar þ ar2
þ þ arn1
þ arn
Then
ð1 rÞSn ¼ a arn
Subtract,
CHAP. 2] SEQUENCES 33
a b
b _ a
3
b _ a
3
Fig. 2-1](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-42-320.jpg)
![1 þ n þ n2
¼ ð1 þ unÞn
¼ 1 þ nun þ
nðn 1Þ
2!
u2
n þ
nðn 1Þðn 2Þ
3!
u3
n þ þ un
n
Then 1 þ n þ n2
1 þ
nðn 1Þðn 2Þ
3!
u3
n or 0 u3
n
6ðn2
þ nÞ
nðn 1Þðn 2Þ
:
Hence, lim
n!1
u3
n ¼ 0 and lim
n!1
un ¼ 0: Thus lim
n!1
ð1 þ n þ n2
Þ1=n
¼ lim
n!1
ð1 þ unÞ ¼ 1:
2.30. Prove that lim
n!1
an
n!
¼ 0 for all constants a.
The result follows if we can prove that lim
n!1
jajn
n!
¼ 0 (see Problem 2.38). We can assume a 6¼ 0.
Let un ¼
jajn
n!
. Then
un
un1
¼
jaj
n
. If n is large enough, say, n 2jaj, and if we call N ¼ ½2jaj þ 1, i.e., the
greatest integer @ 2jaj þ 1, then
uNþ1
uN
1
2
;
uNþ2
uNþ1
1
2
; . . . ;
un
un1
1
2
Multiplying these inequalities yields
un
uN
1
2
nN
or un 1
2
nN
uN:
Since lim
n!1
1
2
nN
¼ 0 (using Problem 2.7), it follows that lim
n!1
un ¼ 0.
Supplementary Problems
SEQUENCES
2.31. Write the first four terms of each of the following sequences:
ðaÞ
ffiffiffi
n
p
n þ 1
; ðbÞ
ð1Þnþ1
n!
( )
; ðcÞ
ð2xÞn1
ð2n 1Þ5
( )
; ðdÞ
ð1Þn
x2n1
1 3 5 ð2n 1Þ
( )
; ðeÞ
cos nx
x2
þ n2
:
Ans: ðaÞ
ffiffiffi
1
p
2
;
ffiffiffi
2
p
3
;
ffiffiffi
3
p
4
;
ffiffiffi
4
p
5
ðcÞ
1
15
;
2x
35
;
4x2
55
;
8x3
75
ðeÞ
cos x
x2 þ 12
;
cos 2x
x2 þ 22
;
cos 3x
x2 þ 32
;
cos 4x
x2 þ 42
ðbÞ
1
1!
;
1
2!
;
1
3!
;
1
4!
ðdÞ
x
1
;
x3
1 3
;
x5
1 3 5
;
x7
1 3 5 7
2.32. Find a possible nth term for the sequences whose first 5 terms are indicated and find the 6th term:
ðaÞ
1
5
;
3
8
;
5
11
;
7
14
;
9
17
; . . . ðbÞ 1; 0; 1; 0; 1; . . . ðcÞ 2
3 ; 0; 3
4 ; 0; 4
5 ; . . .
Ans: ðaÞ
ð1Þn
ð2n 1Þ
ð3n þ 2Þ
ðbÞ
1 ð1Þn
2
ðcÞ
ðn þ 3Þ
ðn þ 5Þ
1 ð1Þn
2
2.33. The Fibonacci sequence is the sequence fung where unþ2 ¼ unþ1 þ un and u1 ¼ 1, u2 ¼ 1. (a) Find the first 6
terms of the sequence. (b) Show that the nth term is given by un ¼ ðan
bn
Þ=
ffiffiffi
5
p
, where a ¼ 1
2 ð1 þ
ffiffiffi
5
p
Þ,
b ¼ 1
2 ð1
ffiffiffi
5
p
Þ.
Ans. (a) 1; 1; 2; 3; 5; 8
LIMITS OF SEQUENCES
2.34. Using the definition of limit, prove that:
CHAP. 2] SEQUENCES 35](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-44-320.jpg)
![Ans. (a) 1
3 ; 1; 0; 0 ðbÞ 1; 1; 1; 1 ðcÞ none, none, þ1, 1 (d) none, 1; þ1; 1
2.51. Prove that a bounded sequence fung is convergent if and only if lim un ¼ lim un.
INFINITE SERIES
2.52. Find the sum of the series
X
1
n¼1
2
3
n
. Ans. 2
2.53. Evaluate
X
1
n¼1
ð1Þn1
=5n
. Ans. 1
6
2.54. Prove that
1
1 2
þ
1
2 3
þ
1
3 4
þ
1
4 5
þ ¼
X
1
n¼1
1
nðn þ 1Þ
¼ 1. Hint:
1
nðn þ 1Þ
¼
1
n
1
n þ 1
2.55. Prove that multiplication of each term of an infinite series by a constant (not zero) does not affect the
convergence or divergence.
2.56. Prove that the series 1 þ
1
2
þ
1
3
þ þ
1
n
þ diverges. Hint: Let Sn ¼ 1 þ
1
2
þ
1
3
þ þ
1
n
. Then prove
that jS2n Snj 1
2, giving a contradiction with Cauchy’s convergence criterion.
MISCELLANEOUS PROBLEMS
2.57. If an @ un @ bn for all n N, and lim
n!1
an ¼ lim
n!1
bn ¼ l, prove that lim
n!1
un ¼ l.
2.58. If lim
n!1
an ¼ lim
n!1
bn ¼ 0, and is independent of n, prove that lim
n!1
ðan cos n þ bn sin nÞ ¼ 0. Is the result
true when depends on n?
2.59. Let un ¼ 1
2 f1 þ ð1Þn
g, n ¼ 1; 2; 3; . . . . If Sn ¼ u1 þ u2 þ þ un, prove that lim
n!1
Sn=n ¼ 1
2.
2.60. Prove that (a) lim
n!1
n1=n
¼ 1, (b) lim
n!1
ða þ nÞp=n
¼ 1 where a and p are constants.
2.61. If lim
n!1
junþ1=unj ¼ jaj 1, prove that lim
n!1
un ¼ 0.
2.62. If jaj 1, prove that lim
n!1
np
an
¼ 0 where the constant p 0.
2.63. Prove that lim
2n
n!
nn ¼ 0.
2.64. Prove that lim
n!1
n sin 1=n ¼ 1. Hint: Let the central angle, , of a circle be measured in radians. Geome-
trically illustrate that sin tan , 0 .
Let ¼ 1=n. Observe that since n is restricted to positive integers, the angle is restricted to the first
quadrant.
2.65. If fung is the Fibonacci sequence (Problem 2.33), prove that lim
n!1
unþ1=un ¼ 1
2 ð1 þ
ffiffiffi
5
p
Þ.
2.66. Prove that the sequence un ¼ ð1 þ 1=nÞnþ1
, n ¼ 1; 2; 3; . . . is a monotonic decreasing sequence whose limit
is e. [Hint: Show that un=un1 @ 1:
2.67. If an A bn for all n N and lim
n!1
an ¼ A, lim
n!1
bn ¼ B, prove that A A B.
2.68. If junj @ jvnj and lim
n!1
vn ¼ 0, prove that lim
n!1
un ¼ 0.
2.69. Prove that lim
n!1
1
n
1 þ
1
2
þ
1
3
þ þ
1
n
¼ 0.
CHAP. 2] SEQUENCES 37](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-46-320.jpg)
![39
Functions, Limits, and
Continuity
FUNCTIONS
A function is composed of a domain set, a range set, and a rule of correspondence that assigns
exactly one element of the range to each element of the domain.
This definition of a function places no restrictions on the nature of the elements of the two sets.
However, in our early exploration of the calculus, these elements will be real numbers. The rule of
correspondence can take various forms, but in advanced calculus it most often is an equation or a set of
equations.
If the elements of the domain and range are represented by x and y, respectively, and f symbolizes
the function, then the rule of correspondence takes the form y ¼ f ðxÞ.
The distinction between f and f ðxÞ should be kept in mind. f denotes the function as defined in the
first paragraph. y and f ðxÞ are different symbols for the range (or image) values corresponding to
domain values x. However a ‘‘common practice’’ that provides an expediency in presentation is to read
f ðxÞ as, ‘‘the image of x with respect to the function f ’’ and then use it when referring to the function.
(For example, it is simpler to write sin x than ‘‘the sine function, the image value of which is sin x.’’)
This deviation from precise notation will appear in the text because of its value in exhibiting the ideas.
The domain variable x is called the independent variable. The variable y representing the corre-
sponding set of values in the range, is the dependent variable.
Note: There is nothing exclusive about the use of x, y, and f to represent domain, range, and
function. Many other letters will be employed.
There are many ways to relate the elements of two sets. [Not all of them correspond a unique range
value to a given domain value.] For example, given the equation y2
¼ x, there are two choices of y for
each positive value of x. As another example, the pairs ða; bÞ, ða; cÞ, ða; dÞ, and ða; eÞ can be formed and
again the correspondence to a domain value is not unique. Because of such possibilities, some texts,
especially older ones, distinguish between multiple-valued and single-valued functions. This viewpoint
is not consistent with our definition or modern presentations. In order that there be no ambiguity, the
calculus and its applications require a single image associated with each domain value. A multiple-
valued rule of correspondence gives rise to a collection of functions (i.e., single-valued). Thus, the rule
y2
¼ x is replaced by the pair of rules y ¼ x1=2
and y ¼ x1=2
and the functions they generate through the
establishment of domains. (See the following section on graphs for pictorial illustrations.)
Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-48-320.jpg)
![If m @ f ðxÞ @ M in an interval, we call f ðxÞ bounded. Frequencly, when we wish to indicate that a
function is bounded, we shall write j f ðxÞj P.
EXAMPLES. 1. f ðxÞ ¼ 3 þ x is bounded in 1 @ x @ 1. An upper bound is 4 (or any number greater than 4).
A lower bound is 2 (or any number less than 2).
2. f ðxÞ ¼ 1=x is not bounded in 0 x 4 since by choosing x sufficiently close to zero, f ðxÞ can be
made as large as we wish, so that there is no upper bound. However, a lower bound is given by
1
4 (or any number less than 1
4).
If f ðxÞ has an upper bound it has a least upper bound (l.u.b.); if it has a lower bound it has a greatest
lower bound (g.l.b.). (See Chapter 1 for these definitions.)
MONOTONIC FUNCTIONS
A function is called monotonic increasing in an interval if for any two points x1 and x2 in the interval
such that x1 x2, f ðx1Þ @ f ðx2Þ. If f ðx1Þ f ðx2Þ the function is called strictly increasing.
Similarly if f ðx1Þ A f ðx2Þ whenever x1 x2, then f ðxÞ is monotonic decreasing; while if f ðx1Þ f ðx2Þ,
it is strictly decreasing.
INVERSE FUNCTIONS. PRINCIPAL VALUES
Suppose y is the range variable of a function f with domain variable x. Furthermore, let the
correspondence between the domain and range values be one-to-one. Then a new function f 1
, called
the inverse function of f , can be created by interchanging the domain and range of f . This information is
contained in the form x ¼ f 1
ðyÞ.
As you work with the inverse function, it often is convenient to rename the domain variable as x and
use y to symbolize the images, then the notation is y ¼ f 1
ðxÞ. In particular, this allows graphical
expression of the inverse function with its domain on the horizontal axis.
Note: f 1
does not mean f to the negative one power. When used with functions the notation f 1
always designates the inverse function to f .
If the domain and range elements of f are not in one-to-one correspondence (this would mean that
distinct domain elements have the same image), then a collection of one-to-one functions may be created.
Each of them is called a branch. It is often convenient to choose one of these branches, called the
principal branch, and denote it as the inverse function, f 1
. The range values of f that compose the
principal branch, and hence the domain of f 1
, are called the principal values. (As will be seen in the
section of elementary functions, it is common practice to specify these principal values for that class of
functions.)
EXAMPLE. Suppose f is generated by y ¼ sin x and the domain is 1 @ x @ 1. Then there are an infinite
number of domain values that have the same image. (A finite portion of the graph is illustrated below in Fig. 3-2(a.)
In Fig. 3-2(b) the graph is rotated about a line at 458 so that the x-axis rotates into the y-axis. Then the variables are
interchanged so that the x-axis is once again the horizontal one. We see that the image of an x value is not unique.
Therefore, a set of principal values must be chosen to establish an inverse function. A choice of a branch is
accomplished by restricting the domain of the starting function, sin x. For example, choose
2
@ x @
2
.
Then there is a one-to-one correspondence between the elements of this domain and the images in 1 @ x @ 1.
Thus, f 1
may be defined with this interval as its domain. This idea is illustrated in Fig. 3-2(c) and Fig. 3-2(d).
With the domain of f 1
represented on the horizontal axis and by the variable x, we write y ¼ sin1
x, 1 @ x @ 1.
If x ¼ 1
2, then the corresponding range value is y ¼
6
.
Note: In algebra, b1
means
1
b
and the fact that bb1
produces the identity element 1 is simply a rule of algebra
generalized from arithmetic. Use of a similar exponential notation for inverse functions is justified in that corre-
sponding algebraic characteristics are displayed by f 1
½ f ðxÞ ¼ x and f ½ f 1
ðxÞ ¼ x.
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 41](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-50-320.jpg)
![Absolute extrema are not necessarily unique. For example, if the graph of a function is a horizontal
line, then every point is an absolute maximum and an absolute minimum.
Note: A point of inflection also is represented in Fig. 3-3. There is an overlap with relative extrema in
representation of such points through derivatives that will be addressed in the problem set of Chapter 4.
TYPES OF FUNCTIONS
It is worth realizing that there is a fundamental pool of functions at the foundation of calculus and
advanced calculus. These are called elementary functions. Either they are generated from a real variable
x by the fundamental operations of algebra, including powers and roots, or they have relatively simple
geometric interpretations. As the title ‘‘elementary functions’’ suggests, there is a more general category
of functions (which, in fact, are dependent on the elementary ones). Some of these will be explored later
in the book. The elementary functions are described below.
1. Polynomial functions have the form
f ðxÞ ¼ a0xn
þ a1xn1
þ þ an1x þ an ð1Þ
where a0; . . . ; an are constants and n is a positive integer called the degree of the polynomial if
a0 6¼ 0.
The fundamental theorem of algebra states that in the field of complex numbers every
polynomial equation has at least one root. As a consequence of this theorem, it can be proved
that every nth degree polynomial has n roots in the complex field. When complex numbers are
admitted, the polynomial theoretically may be expressed as the product of n linear factors; with
our restriction to real numbers, it is possible that 2k of the roots may be complex. In this case,
the k factors generating them will be quadratic. (The corresponding roots are in complex
conjugate pairs.) The polynomial x3
5x2
þ 11x 15 ¼ ðx 3Þðx2
2x þ 5Þ illustrates this
thought.
2. Algebraic functions are functions y ¼ f ðxÞ satisfying an equation of the form
p0ðxÞyn
þ p1ðxÞyn1
þ þ pn1ðxÞy þ pnðxÞ ¼ 0 ð2Þ
where p0ðxÞ; . . . ; pnðxÞ are polynomials in x.
If the function can be expressed as the quotient of two polynomials, i.e., PðxÞ=QðxÞ where
PðxÞ and QðxÞ are polynomials, it is called a rational algebraic function; otherwise it is an
irrational algebraic function.
3. Transcendental functions are functions which are not algebraic, i.e., they do not satisfy equations
of the form (2).
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 43
Fig. 3-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-52-320.jpg)
![6. Inverse hyperbolic functions. If x ¼ sinh y then y ¼ sinh1
x is the inverse hyperbolic sine of x.
The following list gives the principal values of the inverse hyperbolic functions in terms of
natural logarithms and the domains for which they are real.
ðaÞ sinh1
x ¼ lnðx þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1
p
Þ; all x ðdÞ csch1
x ¼ ln
1
x
þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ 1
p
jxj
!
; x 6¼ 0
ðbÞ cosh1
x ¼ lnðx þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
1
p
Þ; x A 1 ðeÞ sech1
x ¼ ln
1 þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p
x
!
; 0 x @ 1
ðcÞ tanh1
x ¼
1
2
ln
1 þ x
1 x
; jxj 1 ð f Þ coth1
x ¼
1
2
ln
x þ 1
x 1
; jxj 1
LIMITS OF FUNCTIONS
Let f ðxÞ be defined and single-valued for all values of x near x ¼ x0 with the possible exception of
x ¼ x0 itslef (i.e., in a deleted neighborhood of x0). We say that the number l is the limit of f ðxÞ as x
approaches x0 and write lim
x!x0
f ðxÞ ¼ l if for any positive number (however small) we can find some
positive number (usually depending on ) such that j f ðxÞ lj whenever 0 jx x0j . In such
case we also say that f ðxÞ approaches l as x approaches x0 and write f ðxÞ ! l as x ! x0.
In words, this means that we can make f ðxÞ arbitrarily close to l by choosing x sufficiently close to
x0.
EXAMPLE. Let f ðxÞ ¼
x2
if x 6¼ 2
0 if x ¼ 2
. Then as x gets closer to 2 (i.e., x approaches 2), f ðxÞ gets closer to 4. We
thus suspect that lim
x!2
f ðxÞ ¼ 4. To prove this we must see whether the above definition of limit (with l ¼ 4) is
satisfied. For this proof see Problem 3.10.
Note that lim
x!2
f ðxÞ 6¼ f ð2Þ, i.e., the limit of f ðxÞ as x ! 2 is not the same as the value of f ðxÞ at x ¼ 2 since
f ð2Þ ¼ 0 by definition. The limit would in fact be 4 even if f ðxÞ were not defined at x ¼ 2.
When the limit of a function exists it is unique, i.e., it is the only one (see Problem 3.17).
RIGHT- AND LEFT-HAND LIMITS
In the definition of limit no restriction was made as to how x should approach x0. It is sometimes
found convenient to restrict this approach. Considering x and x0 as points on the real axis where x0 is
fixed and x is moving, then x can approach x0 from the right or from the left. We indicate these
respective approaches by writing x ! x0þ and x ! x0.
If lim
x!x0þ
f ðxÞ ¼ l1 and lim
x!x0
f ðxÞ ¼ l2, we call l1 and l2, respectively, the right- and left-hand limits of
f at x0 and denote them by f ðx0þÞ or f ðx0 þ 0Þ and f ðx0Þ or f ðx0 0Þ. The ; definitions of limit of
f ðxÞ as x ! x0þ or x ! x0 are the same as those for x ! x0 except for the fact that values of x are
restricted to x x0 or x x0, respectively.
We have lim
x!x0
f ðxÞ ¼ l if and only if lim
x!x0þ
f ðxÞ ¼ lim
x!x0
f ðxÞ ¼ l.
THEOREMS ON LIMITS
If lim
x!x0
f ðxÞ ¼ A and lim
x!x0
gðxÞ ¼ B, then
1: lim
x!x0
ð f ðxÞ þ gðxÞÞ ¼ lim
x!x0
f ðxÞ þ lim
x!x0
gðxÞ ¼ A þ B
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 45](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-54-320.jpg)
![EXAMPLES. 1. If f ðxÞ ¼
x2
; x 6¼ 2
0; x ¼ 2
then from the example on Page 45 lim
x!2
f ðxÞ ¼ 4. But f ð2Þ ¼ 0. Hence
lim
x!2
f ðxÞ 6¼ f ð2Þ and the function is not continuous at x ¼ 2.
2. If f ðxÞ ¼ x2
for all x, then lim
x!2
f ðxÞ ¼ f ð2Þ ¼ 4 and f ðxÞ is continuous at x ¼ 2.
Points where f fails to be continuous are called discontinuities of f and f is said to be discontinuous at
these points.
In constructing a graph of a continuous function the pencil need never leave the paper, while for a
discontinuous function this is not true since there is generally a jump taking place. This is of course
merely a characteristic property and not a definition of continuity or discontinuity.
Alternative to the above definition of continuity, we can define f as continuous at x ¼ x0 if for any
0 we can find 0 such that j f ðxÞ f ðx0Þj whenever jx x0j . Note that this is simply the
definition of limit with l ¼ f ðx0Þ and removal of the restriction that x 6¼ x0.
RIGHT- AND LEFT-HAND CONTINUITY
If f is defined only for x A x0, the above definition does not apply. In such case we call f continuous
(on the right) at x ¼ x0 if lim
x!x0þ
f ðxÞ ¼ f ðx0Þ, i.e., if f ðx0þÞ ¼ f ðx0Þ. Similarly, f is continuous (on the left)
at x ¼ x0 if lim
x!x0
f ðxÞ ¼ f ðx0Þ, i.e., f ðx0Þ ¼ f ðx0Þ. Definitions in terms of and can be given.
CONTINUITY IN AN INTERVAL
A function f is said to be continuous in an interval if it is continuous at all points of the interval. In
particular, if f is defined in the closed interval a @ x @ b or ½a; b, then f is continuous in the interval if
and only if lim
x!x0
f ðxÞ ¼ f ðx0Þ for a x0 b, lim
x!aþ
f ðxÞ ¼ f ðaÞ and lim
x!b
f ðxÞ ¼ f ðbÞ.
THEOREMS ON CONTINUITY
Theorem 1. If f and g are continuous at x ¼ x0, so also are the functions whose image values satisfy the
relations f ðxÞ þ gðxÞ, f ðxÞ gðxÞ, f ðxÞgðxÞ and
f ðxÞ
gðxÞ
, the last only if gðx0Þ 6¼ 0. Similar results hold for
continuity in an interval.
Theorem 2. Functions described as follows are continuous in every finite interval: (a) all polynomials;
(b) sin x and cos x; (c) ax
; a 0
Theorem 3. Let the function f be continuous at the domain value x ¼ x0. Also suppose that a function
g, represented by z ¼ gðyÞ, is continuous at y0, where y ¼ f ðxÞ (i.e., the range value of f corresponding to
x0 is a domain value of g). Then a new function, called a composite function, f ðgÞ, represented by
z ¼ g½ f ðxÞ, may be created which is continuous at its domain point x ¼ x0. [One says that a continuous
function of a continuous function is continuous.]
Theorem 4. If f ðxÞ is continuous in a closed interval, it is bounded in the interval.
Theorem 5. If f ðxÞ is continuous at x ¼ x0 and f ðx0Þ 0 [or f ðx0Þ 0], there exists an interval about
x ¼ x0 in which f ðxÞ 0 [or f ðxÞ 0].
Theorem 6. If a function f ðxÞ is continuous in an interval and either strictly increasing or strictly
decreasing, the inverse function f 1
ðxÞ is single-valued, continuous, and either strictly increasing or
strictly decreasing.
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 47](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-56-320.jpg)
![Solved Problems
FUNCTIONS
3.1. Let f ðxÞ ¼ ðx 2Þð8 xÞ for 2 @ x @ 8. (a) Find f ð6Þ and f ð1Þ. (b) What is the domain of
definition of f ðxÞ? (c) Find f ð1 2tÞ and give the domain of definition. (d) Find f ½ f ð3Þ,
f ½ f ð5Þ. (e) Graph f ðxÞ.
(a) f ð6Þ ¼ ð6 2Þð8 6Þ ¼ 4 2 ¼ 8
f ð1Þ is not defined since f ðxÞ is defined only for 2 @ x @ 8.
(b) The set of all x such that 2 @ x @ 8.
(c) f ð1 2tÞ ¼ fð1 2tÞ 2gf8 ð1 2tÞg ¼ ð1 þ 2tÞð7 þ 2tÞ where t is such that 2 @ 1 2t @ 8, i.e.,
7=2 @ t @ 1=2.
(d) f ð3Þ ¼ ð3 2Þð8 3Þ ¼ 5,
f ½ f ð3Þ ¼ f ð5Þ ¼ ð5 2Þð8 5Þ ¼ 9.
f ð5Þ ¼ 9 so that f ½ f ð5Þ ¼ f ð9Þ is not defined.
(e) The following table shows f ðxÞ for various values of x.
Plot points ð2; 0Þ; ð3; 5Þ; ð4; 8Þ; ð5; 9Þ; ð6; 8Þ; ð7; 5Þ; ð8; 0Þ;
ð2:5; 2:75Þ; ð7:5; 2:75Þ.
These points are only a few of the infinitely many points
on the required graph shown in the adjoining Fig. 3-5. This
set of points defines a curve which is part of a parabola.
3.2. Let gðxÞ ¼ ðx 2Þð8 xÞ for 2 x 8. (a) Discuss the difference between the graph of gðxÞ and
that of f ðxÞ in Problem 3.1. (b) What is the l.u.b. and g.l.b. of gðxÞ? (c) Does gðxÞ attain its
l.u.b. and g.l.b. for any value of x in the domain of definition? (d) Answer parts (b) and (c) for
the function f ðxÞ of Problem 3.1.
(a) The graph of gðxÞ is the same as that in Problem 3.1 except that the two points ð2; 0Þ and ð8; 0Þ are
missing, since gðxÞ is not defined at x ¼ 2 and x ¼ 8.
(b) The l.u.b. of gðxÞ is 9. The g.l.b. of gðxÞ is 0.
(c) The l.u.b. of gðxÞ is attained for the value of x ¼ 5. The g.l.b. of gðxÞ is not attained, since there is no
value of x in the domain of definition such that gðxÞ ¼ 0.
(d) As in (b), the l.u.b. of f ðxÞ is 9 and the g.l.b. of f ðxÞ is 0. The l.u.b. of f ðxÞ is attained for the value
x ¼ 5 and the g.l.b. of f ðxÞ is attained at x ¼ 2 and x ¼ 8.
Note that a function, such as f ðxÞ, which is continuous in a closed interval attains its l.u.b. and g.l.b.
at some point of the interval. However, a function, such as gðxÞ, which is not continuous in a closed
interval need not attain its l.u.b. and g.l.b. See Problem 3.34.
3.3. Let f ðxÞ ¼
1; if x is a rational number
0; if x is an irrational number
. (a) Find f ð2
3Þ, f ð5Þ, f ð1:41423Þ, f ð
ffiffiffi
2
p
Þ,
(b) Construct a graph of f ðxÞ and explain why it is misleading by itself.
(a) f ð2
3Þ ¼ 1 since 2
3 is a rational number
f ð5Þ ¼ 1 since 5 is a rational number
f ð1:41423Þ ¼ 1 since 1.41423 is a rational number
f ð
ffiffiffi
2
p
Þ ¼ 0 since
ffiffiffi
2
p
is an irrational number
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 49
2 4 6 8
x
2
4
6
8
f (x)
Fig. 3-5
x 2 3 4 5 6 7 8 2.5 7.5
f ðxÞ 0 5 8 9 8 5 0 2.75 2.75](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-58-320.jpg)
![to y is multivalued. (d) Indicate possible principal values for tan1
x. (e) Using your choice,
evaluate tan1
ð1Þ.
(a) The graph of f ðxÞ ¼ tan x appears in Fig. 3-10 below.
(b) The required graph is obtained by interchanging the x and y axes in the graph of (a). The result, with
axes oriented as usual, appears in Fig. 3-11 above.
(c) In Fig. 3-11 of (b), any vertical line meets the graph in infinitely many points. Thus, the relation of y to
x is multivalued and infinitely many branches are available for the purpose of defining tan1
x.
(d) To define tan1
x as a single-valued function, it is clear from the graph that we can only do so by
restricting its value to any of the following: =2 tan1
x =2; =2 tan1
x 3=2, etc. We
shall agree to take the first as defining the principal value.
Note that no matter which branch is used to define tan1
x, the resulting function is strictly
increasing.
(e) tan1
ð1Þ ¼ =4 is the only value lying between =2 and =2, i.e., it is the principal value according
to our choice in ðdÞ.
3.8. Show that f ðxÞ ¼
ffiffiffi
x
p
þ 1
x þ 1
, x 6¼ 1, describes an irrational algebraic function.
If y ¼
ffiffiffi
x
p
þ 1
x þ 1
then ðx þ 1Þy 1 ¼
ffiffiffi
x
p
or squaring, ðx þ 1Þ2
y2
2ðx þ 1Þy þ 1 x ¼ 0, a polynomial
equation in y whose coefficients are polynomials in x. Thus f ðxÞ is an algebraic function. However, it is not
the quotient of two polynomials, so that it is an irrational algebraic function.
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 51
_p/2
_p p/2 p 3p/2 2p
x
y = f (x) = tan x
Fig. 3-10
_p
p
_p/2
p/2
3p/2
x
f
_1(x) = tan
_1x
Fig. 3-11
f (x)
x
y =
x
y = _
x
1/2p 1/p
Fig. 3-8
_3 _2 _1 1 2 3 4 5
x
f (x)
Fig. 3-9](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-60-320.jpg)
![Then the graph, shown in the adjoining Fig. 3-12,
consists of the lines y ¼ 1, x 3; y ¼ 1, x 3 and
the point ð3; 0Þ.
(b) As x ! 3 from the right, f ðxÞ ! 1, i.e., lim
x!3þ
f ðxÞ ¼ 1,
as seems clear from the graph. To prove this we must
show that given any 0, we can find 0 such that
j f ðxÞ 1j whenever 0 x 1 .
Now since x 1, f ðxÞ ¼ 1 and so the proof con-
sists in the triviality that j1 1j whenever
0 x 1 .
(c) As x ! 3 from the left, f ðxÞ ! 1, i.e.,
lim
x!3
f ðxÞ ¼ 1. A proof can be formulated as in (b).
(d) Since lim
x!3þ
f ðxÞ 6¼ lim
x!3
f ðxÞ, lim
x!3
f ðxÞ does not exist.
3.13. Prove that lim
x!0
x sin 1=x ¼ 0.
We must show that given any 0, we can find 0 such that jx sin 1=x 0j when
0 jx 0j .
If 0 jxj , then jx sin 1=xj ¼ jxjj sin 1=xj @ jxj since j sin 1=xj @ 1 for all x 6¼ 0.
Making the choice ¼ , we see that jx sin 1=xj when 0 jxj , completing the proof.
3.14. Evaluate lim
x!0þ
2
1 þ e1=x
.
As x ! 0þ we suspect that 1=x increases indefinitely, e1=x
increases indefinitely, e1=x
approaches 0,
1 þ e1=x
approaches 1; thus the required limit is 2.
To prove this conjecture we must show that, given 0, we can find 0 such that
2
1 þ e1=x
2 when 0 x
2
1 þ e1=x
2 ¼
2 2 2e1=x
1 þ e1=x
¼
2
e1=x þ 1
Now
Since the function on the right is smaller than 1 for all x 0, any 0 will work when e 1. If
0 1, then
2
e1=x þ 1
when
e1=x
þ 1
2
1
, e1=x
2
1,
1
x
ln
2
1
; or 0 x
1
lnð2= 1Þ
¼ .
3.15. Explain exactly what is meant by the statement lim
x!1
1
ðx 1Þ4
¼ 1 and prove the validity of this
statement.
The statement means that for each positive number M, we can find a positive number (depending on
M in general) such that
1
ðx 1Þ4
4 when 0 jx 1j
To prove this note that
1
ðx 1Þ4
M when 0 ðx 1Þ4
1
M
or 0 jx 1j
1
ffiffiffiffiffi
M
4
p .
Choosing ¼ 1=
ffiffiffiffiffi
M
4
p
, the required results follows.
3.16. Present a geometric proof that lim
!0
sin
¼ 1.
Construct a circle with center at O and radius OA ¼ OD ¼ 1, as in Fig. 3-13 below. Choose point B on
OA extended and point C on OD so that lines BD and AC are perpendicular to OD.
It is geometrically evident that
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 53
f (x)
x
(3, 0)
1
1
Fig. 3-12](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-62-320.jpg)
![(b) We have
j f ðxÞgðxÞ ABj ¼ j f ðxÞ½gðxÞ B þ B½ f ðxÞ Aj ð4Þ
@ j f ðxÞjjgðxÞ Bj þ jBjj f ðxÞ Aj
@ j f ðxÞjjgðxÞ Bj þ ðjBj þ 1Þj f ðxÞ Aj
Since lim
x!x0
f ðxÞ ¼ A, we can find 1 such j f ðxÞ Aj 1 for 0 jx x0j 1, i.e.,
A 1 f ðxÞ A þ 1, so that f ðxÞ is bounded, i.e., j f ðxÞj P where P is a positive constant.
Since lim
x!x0
gðxÞ ¼ B, given 0 we can find 2 0 such that jgðxÞ Bj =2P for
0 jx x0j 2.
Since lim
x!x0
f ðxÞ ¼ A, given 0 we can find 3 0 such that j f ðxÞ Aj
2ðjBj þ 1Þ
for
0 jx x0j 2.
Using these in (4), we have
j f ðxÞgðxÞ ABj P
2P
þ ðjBj þ 1Þ
2ðjBj þ 1Þ
¼
for 0 jx x0j where is the smaller of 1; 2; 3 and the proof is complete.
(c) We must show that for any 0 we can find 0 such that
1
gðxÞ
1
B
¼
jgðxÞ Bj
jBjjgðxÞj
when 0 jx x0j ð5Þ
By hypothesis, given 0 we can find 1 0 such that
jgðxÞ Bj 1
2 B2
when 0 jx x0j 1
By Problem 3.18, since lim
x!x0
gðxÞ ¼ B 6¼ 0, we can find 2 0 such that
jgðxÞj 1
2 jBj when 0 jx x0j 2
Then if is the smaller of 1 and 2, we can write
1
gðxÞ
1
B
¼
jgðxÞ Bj
jBjjgðxÞj
1
2 B2
jBj 1
2 jBj
¼ whenever 0 jx x0j
and the required result is proved.
(d) From parts (b) and (c),
lim
x!x0
f ðxÞ
gðxÞ
¼ lim
x!x0
f ðxÞ
1
gðxÞ
¼ lim
x!x0
f ðxÞ lim
x!x0
1
gðxÞ
¼ A
1
B
¼
A
B
This can also be proved directly (see Problem 3.69).
The above results can also be proved in the cases x ! x0þ, x ! x0, x ! 1, x ! 1.
Note: In the proof of (a) we have used the results j f ðxÞ Aj =2 and jgðxÞ Bj =2, so that the final
result would come out to be j f ðxÞ þ gðxÞ ðA þ BÞj . Of course the proof would be just as valid if we
had used 2 (or any other positive multiple of ) in place of . A similar remark holds for the proofs of ðbÞ,
(c), and (d).
3.20. Evaluate each of the following, using theorems on limits.
ðaÞ lim
x!2
ðx2
6x þ 4Þ ¼ lim
x!2
x2
þ lim
x!2
ð6xÞ þ lim
x!2
4
¼ ðlim
x!2
xÞðlim
x!2
xÞ þ ðlim
x!2
6Þðlim
x!2
xÞ þ lim
x!2
4
¼ ð2Þð2Þ þ ð6Þð2Þ þ 4 ¼ 4
In practice the intermediate steps are omitted.
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 55](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-64-320.jpg)
![These results follow at once from the proofs given in Problem 3.19 by taking A ¼ f ðx0Þ and B ¼ gðx0Þ
and rewriting 0 jx x0j as jx x0j , i.e., including x ¼ x0.
3.25. Prove that f ðxÞ ¼ x is continuous at any point x ¼ x0.
We must show that, given any 0, we can find 0 such that j f ðxÞ f ðx0Þj ¼ jx x0j when
jx x0j . By choosing ¼ , the result follows at once.
3.26. Prove that f ðxÞ ¼ 2x3
þ x is continuous at any point x ¼ x0.
Since x is continuous at any point x ¼ x0 (Problem 3.25) so also is x x ¼ x2
, x2
x ¼ x3
, 2x3
, and
finally 2x3
þ x, using the theorem (Problem 3.24) that sums and products of continuous functions are
continuous.
3.27. Prove that if f ðxÞ ¼
ffiffiffiffiffiffiffiffiffiffiffi
x 5
p
for 5 @ x @ 9, then f ðxÞ is continuous in this interval.
If x0 is any point such that 5 x0 9, then lim
x!x0
f ðxÞ ¼ lim
x!x0
ffiffiffiffiffiffiffiffiffiffiffi
x 5
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x0 5
p
¼ f ðx0Þ. Also,
lim
x!5þ
ffiffiffiffiffiffiffiffiffiffiffi
x 5
p
¼ 0 ¼ f ð5Þ and lim
x!9
ffiffiffiffiffiffiffiffiffiffiffi
x 5
p
¼ 2 ¼ f ð9Þ. Thus the result follows.
Here we have used the result that lim
x!x0
ffiffiffiffiffiffiffiffiffi
f ðxÞ
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lim
x!x0
f ðxÞ
q
¼
ffiffiffiffiffiffiffiffiffiffiffi
f ðx0Þ
p
if f ðxÞ is continuous at x0. An ,
proof, directly from the definition, can also be employed.
3.28. For what values of x in the domain of definition is each of the following functions continuous?
(a) f ðxÞ ¼
x
x2
1
Ans. all x except x ¼ 1 (where the denominator is zero)
(b) f ðxÞ ¼
1 þ cos x
3 þ sin x
Ans. all x
(c) f ðxÞ ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
10 þ 4
4
p Ans. All x 10
(d) f ðxÞ ¼ 101=ðx3Þ2
Ans. all x 6¼ 3 (see Problem 3.55)
(e) f ðxÞ ¼ 101=ðx3Þ2
; x 6¼ 3
0; x ¼ 3
Ans. all x, since lim
x!3
f ðxÞ ¼ f ð3Þ
( f ) f ðxÞ ¼
x jxj
x
If x 0, f ðxÞ ¼
x x
x
¼ 0. If x 0, f ðxÞ ¼
x þ x
x
¼ 2. At x ¼ 0, f ðxÞ is undefined. Then f ðxÞ is
continuous for all x except x ¼ 0.
ðgÞ f ðxÞ ¼
x jxj
x
; x 0
2; x ¼ 0
8
:
As in ð f Þ, f ðxÞ is continuous for x 0. Then since
lim
x!0
x jxj
x
¼ lim
x!0
x þ x
x
¼ lim
x!0
2 ¼ 2 ¼ f ð0Þ
if follows that f ðxÞ is continuous (from the left) at x ¼ 0.
Thus, f ðxÞ is continuous for all x @ 0, i.e., everywhere in its domain of definition.
ðhÞ f ðxÞ ¼ x csc x ¼
x
sin x
: Ans: all x except 0; ; 2; 3; . . . :
(i) f ðxÞ ¼ x csc x, f ð0Þ ¼ 1. Since lim
x!0
x csc x ¼ lim
x!0
x
sin x
¼ 1 ¼ f ð0Þ, we see that f ðxÞ is continuous for all x
except ; 2; 3; . . . [compare (h)].
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 57](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-66-320.jpg)
![Suppose that f ðaÞ 0 and f ðbÞ 0. Since f ðxÞ is continuous there must be an interval ða; a þ hÞ, h 0,
for which f ðxÞ 0. The set of points ða; a þ hÞ has an upper bound and so has a least upper bound which
we call c. Then f ðcÞ @ 0. Now we cannot have f ðcÞ 0, because if f ðcÞ were negative we would be able to
find an interval about c (including values greater than c) for which f ðxÞ 0; but since c is the least upper
bound, this is impossible, and so we must have f ðcÞ ¼ 0 as required.
If f ðaÞ 0 and f ðbÞ 0, a similar argument can be used.
3.33. (a) Given f ðxÞ ¼ 2x3
3x2
þ 7x 10, evaluate f ð1Þ and f ð2Þ. (b) Prove that f ðxÞ ¼ 0 for some
real number x such that 1 x 2. (c) Show how to calculate the value of x in (b).
(a) f ð1Þ ¼ 2ð1Þ3
3ð1Þ2
þ 7ð1Þ 10 ¼ 4, f ð2Þ ¼ 2ð2Þ3
3ð2Þ2
þ 7ð2Þ 10 ¼ 8.
(b) If f ðxÞ is continuous in a @ x @ b and if f ðaÞ and f ðbÞ have opposite signs, then there is a value of x
between a and b such that f ðxÞ ¼ 0 (Problem 3.32).
To apply this theorem we need only realize that the given polynomial is continuous in 1 @ x @ 2,
since we have already shown in (a) that f ð1Þ 0 and f ð2Þ 0. Thus there exists a number c between 1
and 2 such that f ðcÞ ¼ 0.
(c) f ð1:5Þ ¼ 2ð1:5Þ3
3ð1:5Þ2
þ 7ð1:5Þ 10 ¼ 0:5. Then applying the theorem of (b) again, we see that the
required root lies between 1 and 1.5 and is ‘‘most likely’’ closer to 1.5 than to 1, since f ð1:5Þ ¼ 0:5 has a
value closer to 0 than f ð1Þ ¼ 4 (this is not always a valid conclusion but is worth pursuing in practice).
Thus we consider x ¼ 1:4. Since f ð1:4Þ ¼ 2ð1:4Þ3
3ð1:4Þ2
þ 7ð1:4Þ 10 ¼ 0:592, we conclude
that there is a root between 1.4 and 1.5 which is most likely closer to 1.5 than to 1.4.
Continuing in this manner, we find that the root is 1.46 to 2 decimal places.
3.34. Prove Theorem 10, Page 48.
Given any 0, we can find x such that M f ðxÞ by definition of the l.u.b. M.
Then
1
M f ðxÞ
1
, so that
1
M f ðxÞ
is not bounded and hence cannot be continuous in view of
Theorem 4, Page 47. However, if we suppose that f ðxÞ 6¼ M, then since M f ðxÞ is continuous, by
hypothesis, we must have
1
M f ðxÞ
also continuous. In view of this contradiction, we must have
f ðxÞ ¼ M for at least one value of x in the interval.
Similarly, we can show that there exists an x in the interval such that f ðxÞ ¼ m (Problem 3.93).
Supplementary Problems
FUNCTIONS
3.35. Give the largest domain of definition for which each of the following rules of correspondence support the
construction of a function.
(a)
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð3 xÞð2x þ 4Þ
p
, (b) ðx 2Þ=ðx2
4Þ, (c)
ffiffiffiffiffiffiffiffiffiffiffiffi
sin 3x
p
, (d) log10ðx3
3x2
4x þ 12Þ.
Ans. (a) 2 @ x @ 3, (b) all x 6¼ 2, (c) 2m=3 @ x @ ð2m þ 1Þ=3, m ¼ 0; 1; 2; . . . ;
(d) x 3, 2 x 2.
3.36. If f ðxÞ ¼
3x þ 1
x 2
, x 6¼ 2, find: (a)
5f ð1Þ 2f ð0Þ þ 3f ð5Þ
6
; (b) f f ð 1
2Þg2
; (c) f ð2x 3Þ;
(d) f ðxÞ þ f ð4=xÞ, x 6¼ 0; (e)
f ðhÞ f ð0Þ
h
, h 6¼ 0; ( f ) f ðf f ðxÞg.
Ans. (a) 61
18 (b) 1
25 (c)
6x 8
2x 5
, x 6¼ 0, 5
2, 2 (d) 5
2, x 6¼ 0; 2 (e)
7
2h 4
, h 6¼ 0; 2
( f )
10x þ 1
x þ 5
, x 6¼ 5; 2
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 59](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-68-320.jpg)
![3.46. (a) Prove that tan1
x þ cot1
x ¼ =2 if the conventional principal values on Page 44 are taken. (b) Is
tan1
x þ tan1
ð1=xÞ ¼ =2 also? Explain.
3.47. If f ðxÞ ¼ tan1
x, prove that f ðxÞ þ f ðyÞ ¼ f
x þ y
1 xy
, discussing the case xy ¼ 1.
3.48. Prove that tan1
a tan1
b ¼ cot1
b cot1
a.
3.49. Prove the identities:
(a) 1 tanh2
x ¼ sech2
x, (b) sin 3x ¼ 3 sin x 4 sin3
x, (c) cos 3x ¼ 4 cos3
x 3 cos x, (d) tanh 1
2 x ¼
ðsinh xÞ=ð1 þ cosh xÞ, (e) ln jcsc x cot xj ¼ ln j tan 1
2 xj.
3.50. Find the relative and absolute maxima and minima of: (a) f ðxÞ ¼ ðsin xÞ=x, f ð0Þ ¼ 1; (b) f ðxÞ ¼ ðsin2
xÞ=
x2
, f ð0Þ ¼ 1. Discuss the cases when f ð0Þ is undefined or f ð0Þ is defined but 6¼ 1.
LIMITS
3.51. Evaluate the following limits, first by using the definition and then using theorems on limits.
ðaÞ lim
x!3
ðx2
3x þ 2Þ; ðbÞ lim
x!1
1
2x 5
; ðcÞ lim
x!2
x2
4
x 2
; ðdÞ lim
x!4
ffiffiffi
x
p
2
4 x
; ðeÞ lim
h!0
ð2 þ hÞ4
16
h
;
ð f Þ lim
x!1
ffiffiffi
x
p
x þ 1
:
Ans. ðaÞ 2; ðbÞ 1
7 ; ðcÞ 4; ðdÞ 1
4 ; ðeÞ 32; ð f Þ 1
2
3.52. Let f ðxÞ ¼
3x 1; x 0
0; x ¼ 0
2x þ 5; x 0
8
:
: ðaÞ Construct a graph of f ðxÞ.
Evaluate (b) lim
x!2
f ðxÞ; ðcÞ lim
x!3
f ðxÞ; ðdÞ lim
x!0þ
f ðxÞ; ðeÞ lim
x!0
f ðxÞ; ð f Þ lim
x!0
f ðxÞ, justifying your
answer in each case.
Ans. (b) 9, (c) 10, (d) 5, (e) 1, ( f ) does not exist
3.53. Evaluate (a) lim
h!0þ
f ðhÞ f ð0þÞ
h
and (b) lim
h!0
f ðhÞ f ð0Þ
h
, where f ðxÞ is the function of Prob. 3.52.
Ans. (a) 2, (b) 3
3.54. (a) If f ðxÞ ¼ x2
cos 1=x, evaluate lim
x!0
f ðxÞ, justifying your answer. (b) Does your answer to (a) still remain
the same if we consider f ðxÞ ¼ x2
cos 1=x, x 6¼ 0, f ð0Þ ¼ 2? Explain.
3.55. Prove that lim
x!3
101=ðx3Þ2
¼ 0 using the definition.
3.56. Let f ðxÞ ¼
1 þ 101=x
2 101=x
, x 6¼ 0, f ð0Þ ¼ 1
2. Evaluate (a) lim
x!0þ
f ðxÞ, (b) lim
x!0
f ðxÞ, (c) lim
x!0
f ðxÞ, justifying
answers in all cases.
Ans. (a) 1
2, (b) 1; ðcÞ does not exist.
3.57. Find (a) lim
x!0þ
jxj
x
; ðbÞ lim
x!0
jxj
x
. Illustrate your answers graphically.
Ans. (a) 1, (b) 1
3.58. If f ðxÞ is the function defined in Problem 3.56, does lim
x!0
f ðjxjÞ exist? Explain.
3.59. Explain exactly what is meant when one writes:
ðaÞ lim
x!3
2 x
ðx 3Þ2
¼ 1; ðbÞ lim
x!0þ
ð1 e1=x
Þ ¼ 1; ðcÞ lim
x!1
2x þ 5
3x 2
¼
2
3
:
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 61](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-70-320.jpg)
![3.72. Prove that lim
x!x0
f ðxÞ ¼ l if and only if lim
x!x0þ
f ðxÞ ¼ lim
x!x0
f ðxÞ ¼ l.
CONTINUITY
In the following problems assume the largest possible domain unless otherwise stated.
3.73. Prove that f ðxÞ ¼ x2
3x þ 2 is continuous at x ¼ 4.
3.74. Prove that f ðxÞ ¼ 1=x is continuous (a) at x ¼ 2, (b) in 1 @ x @ 3.
3.75. Investigate the continuity of each of the following functions at the indicated points:
ðaÞ f ðxÞ ¼
sin x
x
; x 6¼ 0; f ð0Þ ¼ 0; x ¼ 0 ðcÞ f ðxÞ ¼
x3
8
x2 4
; x 6¼ 2; f ð2Þ ¼ 3; x ¼ 2
ðbÞ f ðxÞ ¼ x jxj; x ¼ 0 ðdÞ f ðxÞ ¼
sin x; 0 x 1
ln x 1 x 2
; x ¼ 1:
Ans. (a) discontinuous, (b) continuous, (c) continuous, (d) discontinuous
3.76. If ½x ¼ greatest integer @ x, investigate the continuity of f ðxÞ ¼ x ½x in the interval (a) 1 x 2,
(b) 1 @ x @ 2.
3.77. Prove that f ðxÞ ¼ x3
is continuous in every finite interval.
3.78. If f ðxÞ=gðxÞ and gðxÞ are continuous at x ¼ x0, prove that f ðxÞ must be continuous at x ¼ x0.
3.79. Prove that f ðxÞ ¼ ðtan1
xÞ=x, f ð0Þ ¼ 1 is continuous at x ¼ 0.
3.80. Prove that a polynomial is continuous in every finite interval.
3.81. If f ðxÞ and gðxÞ are polynomials, prove that f ðxÞ=gðxÞ is continuous at each point x ¼ x0 for which gðx0Þ 6¼ 0.
3.82. Give the points of discontinuity of each of the following functions.
ðaÞ f ðxÞ ¼
x
ðx 2Þðx 4Þ
ðcÞ f ðxÞ ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 3Þð6 xÞ
p
; 3 @ x @ 6
ðbÞ f ðxÞ ¼ x2
sin 1=x; x 6¼ 0; f ð0Þ ¼ 0 ðdÞ f ðxÞ ¼
1
1 þ 2 sin x
:
Ans. (a) x ¼ 2; 4, (b) none, (c) none, (d) x ¼ 7=6 2m; 11=6 2m; m ¼ 0; 1; 2; . . .
UNIFORM CONTINUITY
3.83. Prove that f ðxÞ ¼ x3
is uniformly continuous in (a) 0 x 2, (b) 0 @ x @ 2, (c) any finite interval.
3.84. Prove that f ðxÞ ¼ x2
is not uniformly continuous in 0 x 1.
3.85. If a is a constant, prove that f ðxÞ ¼ 1=x2
is (a) continuous in a x 1 if a A 0, (b) uniformly
continuous in a x 1 if a 0, (c) not uniformly continuous in 0 x 1.
3.86. If f ðxÞ and gðxÞ are uniformly continuous in the same interval, prove that (a) f ðxÞ gðxÞ and (b) f ðxÞgðxÞ
are uniformly continuous in the interval. State and prove an analogous theorem for f ðxÞ=gðxÞ.
MISCELLANEOUS PROBLEMS
3.87. Give an ‘‘; ’’ proof of the theorem of Problem 3.31.
3.88. (a) Prove that the equation tan x ¼ x has a real positive root in each of the intervals =2 x 3=2,
3=2 x 5=2, 5=2 x 7=2; . . . .
CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 63](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-72-320.jpg)
![(b) Illustrate the result in (a) graphically by constructing the graphs of y ¼ tan x and y ¼ x and locating
their points of intersection.
(c) Determine the value of the smallest positive root of tan x ¼ x.
Ans. ðcÞ 4.49 approximately
3.89. Prove that the only real solution of sin x ¼ x is x ¼ 0.
3.90. (a) Prove that cos x cosh x þ 1 ¼ 0 has infinitely many real roots.
(b) Prove that for large values of x the roots approximate those of cos x ¼ 0.
3.91. Prove that lim
x!0
x2
sinð1=xÞ
sin x
¼ 0.
3.92. Suppose f ðxÞ is continuous at x ¼ x0 and assume f ðx0Þ 0. Prove that there exists an interval
ðx0 h; x0 þ hÞ, where h 0, in which f ðxÞ 0. (See Theorem 5, page 47.) [Hint: Show that we can
make j f ðxÞ f ðx0Þj 1
2 f ðx0Þ. Then show that f ðxÞ A f ðx0Þ j f ðxÞ f ðx0Þj 1
2 f ðx0Þ 0.]
3.93. (a) Prove Theorem 10, Page 48, for the greatest lower bound m (see Problem 3.34). (b) Prove Theorem 9,
Page 48, and explain its relationship to Theorem 10.
64 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-73-320.jpg)
![RIGHT- AND LEFT-HAND DERIVATIVES
The status of the derivative at end points of the domain of f , and in other special circumstances, is
clarified by the following definitions.
The right-hand derivative of f ðxÞ at x ¼ x0 is defined as
f 0
þðx0Þ ¼ lim
h!0þ
f ðx0 þ hÞ f ðx0Þ
h
ð3Þ
if this limit exists. Note that in this case hð¼ xÞ is restricted only to positive values as it approaches
zero.
Similarly, the left-hand derivative of f ðxÞ at x ¼ x0 is defined as
f 0
ðx0Þ ¼ lim
h!0
f ðx0 þ hÞ f ðx0Þ
h
ð4Þ
if this limit exists. In this case h is restricted to negative values as it approaches zero.
A function f has a derivative at x ¼ x0 if and only if f 0
þðx0Þ ¼ f 0
ðx0Þ.
DIFFERENTIABILITY IN AN INTERVAL
If a function has a derivative at all points of an interval, it is said to be differentiable in the interval.
In particular if f is defined in the closed interval a @ x @ b, i.e. ½a; b, then f is differentiable in the
interval if and only if f 0
ðx0Þ exists for each x0 such that a x0 b and if f 0
þðaÞ and f 0
ðbÞ both exist.
If a function has a continuous derivative, it is sometimes called continuously differentiable.
PIECEWISE DIFFERENTIABILITY
A function is called piecewise differentiable or piecewise smooth in an interval a @ x @ b if f 0
ðxÞ is
piecewise continuous. An example of a piecewise continuous function is shown graphically on Page 48.
An equation for the tangent line to the curve y ¼ f ðxÞ at the point where x ¼ x0 is given by
y f ðx0Þ ¼ f 0
ðx0Þðx x0Þ ð7Þ
The fact that a function can be continuous at a point and yet not be differentiable there is shown
graphically in Fig. 4-3. In this case there are two tangent lines at P represented by PM and PN. The
slopes of these tangent lines are f 0
ðx0Þ and f 0
þðx0Þ respectively.
DIFFERENTIALS
Let x ¼ dx be an increment given to x. Then
y ¼ f ðx þ xÞ f ðxÞ ð8Þ
is called the increment in y ¼ f ðxÞ. If f ðxÞ is continuous and has a continuous first derivative in an
interval, then
y ¼ f 0
ðxÞx þ x ¼ f 0
ðxÞdx þ dx ð9Þ
where ! 0 as x ! 0. The expression
dy ¼ f 0
ðxÞdx ð10Þ
is called the differential of y or f(x) or the principal part of y. Note that y 6¼ dy in general. However
if x ¼ dx is small, then dy is a close approximation of y (see Problem 11). The quantity dx, called the
differential of x, and dy need not be small.
CHAP. 4] DERIVATIVES 67](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-76-320.jpg)
![On the other hand, dy and y are related. In particular, lim
x!0
y
x
¼ f 0
ðxÞ means that for any 0
there exists 0 such that
y
x
dy
dx
whenever jxj . Now dx is an independent variable
and the axes of x and dx are parallel; therefore, dx may be chosen equal to x. With this choice
x y dy x
or
dy x y dy þ x
From this relation we see that dy is an approximation to y in small neighborhoods of x. dy is called
the principal part of y.
The representation of f 0
by
dy
dx
has an algebraic suggestiveness that is very appealing and will appear
in much of what follows. In fact, this notation was introduced by Leibniz (without the justification
provided by knowledge of the limit idea) and was the primary reason his approach to the calculus, rather
than Newton’s was followed.
THE DIFFERENTIATION OF COMPOSITE FUNCTIONS
Many functions are a composition of simpler ones. For example, if f and g have the rules of
correspondence u ¼ x3
and y ¼ sin u, respectively, then y ¼ sin x3
is the rule for a composite function
F ¼ gð f Þ. The domain of F is that subset of the domain of F whose corresponding range values are in
the domain of g. The rule of composite function differentiation is called the chain rule and is represented
by
dy
dx
¼
dy
du
du
dx
½F 0
ðxÞ ¼ g0
ðuÞf 0
ðxÞ.
In the example
dy
dx
dðsin x3
Þ
dx
¼ cos x3
ð3x2
dxÞ
The importance of the chain rule cannot be too greatly stressed. Its proper application is essential
in the differentiation of functions, and it plays a fundamental role in changing the variable of integration,
as well as in changing variables in mathematical models involving differential equations.
IMPLICIT DIFFERENTIATION
The rule of correspondence for a function may not be explicit. For example, the rule y ¼ f ðxÞ is
implicit to the equation x2
þ 4xy5
þ 7xy þ 8 ¼ 0. Furthermore, there is no reason to believe that this
equation can be solved for y in terms of x. However, assuming a common domain (described by the
independent variable x) the left-hand member of the equation can be construed as a composition of
functions and differentiated accordingly. (The rules of differentiation are listed below for your review.)
In this example, differentiation with respect to x yields
2x þ 4 y5
þ 5xy4 dy
dx
þ 7 y þ x
dy
dx
¼ 0
Observe that this equation can be solved for
dy
dx
as a function of x and y (but not of x alone).
CHAP. 4] DERIVATIVES 69](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-78-320.jpg)
![DERIVATIVES OF ELEMENTARY FUNCTIONS
In the following we assume that u is a differentiable function of x; if u ¼ x, du=dx ¼ 1. The inverse
functions are defined according to the principal values given in Chapter 3.
1.
d
dx
ðCÞ ¼ 0 16.
d
dx
cot1
u ¼
1
1 þ u2
du
dx
2.
d
dx
un
¼ nun1 du
dx
17.
d
dx
sec1
u ¼
1
u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 1
p
du
dx
þ if u 1
if u 1
3.
d
dx
sin u ¼ cos u
du
dx
18.
d
dx
csc1
u ¼
1
u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
1
p
du
dx
if u 1
þ if u 1
4.
d
dx
cos u ¼ sin u
du
dx
19.
d
dx
sinh u ¼ cosh u
du
dx
5.
d
dx
tan u ¼ sec2
u
du
dx
20.
d
dx
cosh u ¼ sinh u
du
dx
6.
d
dx
cot u ¼ csc2
u
du
dx
21.
d
dx
tanh u ¼ sech2
u
du
dx
7.
d
dx
sec u ¼ sec u tan u
du
dx
22.
d
dx
coth u ¼ csch2
u
du
dx
8.
d
dx
csc u ¼ csc u cot u
du
dx
23.
d
dx
sech u ¼ sech u tanh u
du
dx
9.
d
dx
loga u ¼
loga e
u
du
dx
a 0; a 6¼ 1 24.
d
dx
csch u ¼ csch u coth u
du
dx
10.
d
dx
loge u ¼
d
dx
ln u ¼
1
u
du
dx
25.
d
dx
sinh1
u ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2
p
du
dx
11.
d
dx
au
¼ au
ln a
du
dx
26.
d
dx
cosh1
u ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 1
p
du
dx
12.
d
dx
eu
¼ eu du
dx
27.
d
dx
tanh1
u ¼
1
1 u2
du
dx
; juj 1
13.
d
dx
sin1
u ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 u2
p
du
dx
28.
d
dx
coth1
u ¼
1
1 u2
du
dx
; juj 1
14.
d
dx
cos1
u ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 u2
p
du
dx
29.
d
dx
sech1
u ¼
1
u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 u2
p
du
dx
15.
d
dx
tan1
u ¼
1
1 þ u2
du
dx
30.
d
dx
csch1
u ¼
1
u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
þ 1
p
du
dx
HIGHER ORDER DERIVATIVES
If f ðxÞ is differentiable in an interval, its derivative is given by f 0
ðxÞ, y0
or dy=dx, where y ¼ f ðxÞ. If
f 0
ðxÞ is also differentiable in the interval, its derivative is denoted by f 00
ðxÞ, y00
or
d
dx
dy
dx
¼
d2
y
dx2
.
Similarly, the nth derivative of f ðxÞ, if it exists, is denoted by f ðnÞ
ðxÞ, yðnÞ
or
dn
y
dxn, where n is called the
order of the derivative. Thus derivatives of the first, second, third, . . . orders are given by f 0
ðxÞ, f 00
ðxÞ,
f 000
ðxÞ; . . . .
Computation of higher order derivatives follows by repeated application of the differentiation rules
given above.
CHAP. 4] DERIVATIVES 71](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-80-320.jpg)
![lim
x!x0
f ðxÞ
gðxÞ
¼ lim
x!x0
f 0
ðxÞ
g0
ðxÞ
ð19Þ
whenever the limit on the right can be found. In case f 0
ðxÞ and g0
ðxÞ satisfy the same conditions
as f ðxÞ and gðxÞ given above, the process can be repeated.
2. If lim
x!x0
f ðxÞ ¼ 1 and lim
x!x0
gðxÞ ¼ 1, the result (19) is also valid.
These can be extended to cases where x ! 1 or 1, and to cases where x0 ¼ a or x0 ¼ b in which
only one sided limits, such as x ! aþ or x ! b, are involved.
Limits represented by the so-called indeterminate forms 0 1, 10
, 00
, 11
; and 1 1 can be
evaluated on replacing them by equivalent limits for which the above rules are applicable (see Problem
4.29).
APPLICATIONS
1. Relative Extrema and Points of Inflection
See Chapter 3 where relative extrema and points of inflection were described and a diagram is
presented. In this chapter such points are characterized by the variation of the tangent line, and
then by the derivative, which represents the slope of that line.
Assume that f has a derivative at each point of an open interval and that P1 is a point of the graph of
f associated with this interval. Let a varying tangent line to the graph move from left to right through
P1. If the point is a relative minimum, then the tangent line rotates counterclockwise. The slope is
negative to the left of P1 and positive to the right. At P1 the slope is zero. At a relative maximum a
similar analysis can be made except that the rotation is clockwise and the slope varies from positive to
negative. Because f 00
designates the change of f 0
, we can state the following theorem. (See Fig. 4-6.)
Theorem. Assume that x1 is a number in an open set of the domain of f at which f 0
is continuous and
f 00
is defined. If f 0
ðx1Þ ¼ 0 and f 00
ðx1Þ 6¼ 0, then f ðx1Þ is a relative extreme of f . Specifically:
(a) If f 00
ðx1Þ 0, then f ðx1Þ is a relative minimum,
(b) If f 00
ðx1Þ 0; then f ðx1Þ is a relative maximum.
(The domain value x1 is called a critical value.)
This theorem may be generalized in the following way. Assume existence and continuity of
derivatives as needed and suppose that f 0
ðx1Þ ¼ f 00
ðx1Þ ¼ f ð2p1Þ
ðx1Þ ¼ 0 and f ð2pÞ
ðx1Þ 6¼ 0 ( p a posi-
tive integer). Then:
(a) f has a relative minimum at x1 if f ð2pÞ
ðx1Þ 0,
(b) f has a relative maximum at x1 if f ð2pÞ
ðx1Þ 0.
CHAP. 4] DERIVATIVES 73
Fig. 4-6](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-82-320.jpg)
![where the two representations of the slope of the tangent line have been equated. The solution of this
relation for x1 is
x1 ¼ x0
f ðx0Þ
f 0
ðx0Þ
Starting with the tangent line to the graph at P1½x1; f ðx1Þ and repeating the process, we get
x2 ¼ x1
f ðx1Þ
f 0ðx1Þ
¼ x0
f ðx0Þ
f 0ðx0Þ
f ðx1Þ
f 0ðx1Þ
and in general
xn ¼ x0
X
n
k¼0
f ðxkÞ
f 0ðxkÞ
Under appropriate circumstances, the approximation xn to the root r can be made as good as
desired.
Note: Success with Newton’s method depends on the shape of the function’s graph in the neighbor-
hood of the root. There are various cases which have not been explored here.
Solved Problems
DERIVATIVES
4.1. (a) Let f ðxÞ ¼
3 þ x
3 x
, x 6¼ 3. Evaluate f 0
ð2Þ from the definition.
f 0
ð2Þ ¼ lim
h!0
f ð2 þ hÞ f ð2Þ
h
¼ lim
h!0
1
h
5 þ h
1 h
5
¼ lim
h!0
1
h
6h
1 h
¼ lim
h!0
6
1 h
¼ 6
Note: By using rules of differentiation we find
f 0
ðxÞ ¼
ð3 xÞ
d
dx
ð3 þ xÞ ð3 þ xÞ
d
dx
ð3 xÞ
ð3 xÞ2
¼
ð3 xÞð1Þ ð3 þ xÞð1Þ
ð3 xÞ2
¼
6
ð3 xÞ2
at all points x where the derivative exists. Putting x ¼ 2, we find f 0
ð2Þ ¼ 6. Although such rules are
often useful, one must be careful not to apply them indiscriminately (see Problem 4.5).
(b) Let f ðxÞ ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x 1
p
. Evaluate f 0
ð5Þ from the definition.
f 0
ð5Þ ¼ lim
h!0
f ð5 þ hÞ f ð5Þ
h
¼ lim
h!0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h
p
3
h
¼ lim
h!0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h
p
3
h
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h
p
þ 3
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h
p
þ 3
¼ lim
h!0
9 þ 2h 9
hð
ffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
9 þ 2h
p
þ 3Þ
¼ lim
h!0
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h
p
þ 3
¼
1
3
By using rules of differentiation we find f 0
ðxÞ ¼
d
dx
ð2x 1Þ1=2
¼ 1
2 ð2x 1Þ1=2 d
dx
ð2x 1Þ ¼
ð2x 1Þ1=2
. Then f 0
ð5Þ ¼ 91=2
¼ 1
3.
4.2. (a) Show directly from definition that the derivative of f ðxÞ ¼ x3
is 3x2
.
(b) Show from definition that
d
dx
ffiffiffi
x
p
Þ ¼
1
2
ffiffiffi
x
p .
CHAP. 4] DERIVATIVES 75](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-84-320.jpg)
![Since lim
x!0
f 0
ðxÞ ¼ lim
x!0
cos
1
x
þ 2x sin
1
x
does not exist (because lim
x!0
cos 1=x does not exist), f 0
ðxÞ
cannot be continuous at x ¼ 0 in spite of the fact that f 0
ð0Þ exists.
This shows that we cannot calculate f 0
ð0Þ in this case by simply calculating f 0
ðxÞ and putting x ¼ 0,
as is frequently supposed in elementary calculus. It is only when the derivative of a function is
continuous at a point that this procedure gives the right answer. This happens to be true for most
functions arising in elementary calculus.
4.6. Present an ‘‘; ’’ definition of the derivative of f ðxÞ at x ¼ x0.
f ðxÞ has a derivative f 0
ðx0Þ at x ¼ x0 if, given any 0, we can find 0 such that
f ðx0 þ hÞ f ðx0Þ
h
f 0
ðx0Þ when 0 jhj
RIGHT- AND LEFT-HAND DERIVATIVES
4.7. Let f ðxÞ ¼ jxj. (a) Calculate the right-hand derivatives of f ðxÞ at x ¼ 0. (b) Calculate the left-
hand derivative of f ðxÞ at x ¼ 0. (c) Does f ðxÞ have a derivative at x ¼ 0? (d) Illustrate the
conclusions in (a), (b), and (c) from a graph.
ðaÞ f 0
þð0Þ ¼ lim
h!0þ
f ðhÞ f ð0Þ
h
¼ lim
h!0þ
jhj 0
h
¼ lim
h!0þ
h
h
¼ 1
since jhj ¼ h for h 0.
ðbÞ f 0
ð0Þ ¼ lim
h!0
f ðhÞ f ð0Þ
h
¼ lim
h!0
jhj 0
h
¼ lim
h!0
h
h
¼ 1
since jhj ¼ h for h 0.
(c) No. The derivative at 0 does not exist if the right and
left hand derivatives are unequal.
(d) The required graph is shown in the adjoining Fig. 4-8.
Note that the slopes of the lines y ¼ x and y ¼ x are 1 and 1 respectively, representing the right and
left hand derivatives at x ¼ 0. However, the derivative at x ¼ 0 does not exist.
4.8. Prove that f ðxÞ ¼ x2
is differentiable in 0 @ x @ 1.
Let x0 be any value such that 0 x0 1. Then
f 0
ðx0Þ ¼ lim
h!0
f ðx0 þ hÞ f ðx0Þ
h
¼ lim
h!0
ðx0 þ hÞ2
x2
0
h
¼ lim
h!0
ð2x0 þ hÞ ¼ 2x0
At the end point x ¼ 0,
f 0
þð0Þ ¼ lim
h!0þ
f ð0 þ hÞ f ð0Þ
h
¼ lim
h!0þ
h2
0
h
¼ lim
h!0þ
h ¼ 0
At the end point x ¼ 1,
f 0
ð1Þ ¼ lim
h!0
f ð1 þ hÞ f ð1Þ
h
¼ lim
h!0
ð1 þ hÞ2
1
h
¼ lim
h!0
ð2 þ hÞ ¼ 2
Then f ðxÞ is differentiable in 0 @ x @ 1. We may write f 0
ðxÞ ¼ 2x for any x in this interval. It is
customary to write f 0
þð0Þ ¼ f 0
ð0Þ and f 0
ð1Þ ¼ f 0
ð1Þ in this case.
CHAP. 4] DERIVATIVES 77
y
x
y
=
x
y
=
_
x
Fig. 4-8](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-86-320.jpg)
![d
dx
uv ¼ lim
x!0
ðu þ uÞðv þ vÞ uv
x
¼ lim
x!0
uv þ vu þ uv
x
¼ lim
x!0
u
v
x
þ v
u
x
þ
u
x
v
¼ u
dv
dx
þ v
du
dx
where it is noted that v ! 0 as x ! 0, since v is supposed differentiable and thus continuous.
4.13. If y ¼ f ðuÞ where u ¼ gðxÞ, prove that
dy
dx
¼
dy
du
du
dx
assuming that f and g are differentiable.
Let x be given an increment x 6¼ 0. Then as a consequence u and y take on increments u and y
respectively, where
y ¼ f ðu þ uÞ f ðuÞ; u ¼ gðx þ xÞ gðxÞ ð1Þ
Note that as x ! 0, y ! 0 and u ! 0.
If u 6¼ 0, let us write ¼
y
u
dy
du
so that ! 0 as u ! 0 and
y ¼
dy
du
u þ u ð2Þ
If u ¼ 0 for values of x, then (1) shows that y ¼ 0 for these values of x. For such cases, we
define ¼ 0.
It follows that in both cases, u 6¼ 0 or u ¼ 0, (2) holds. Dividing (2) by x 6¼ 0 and taking the limit
as x ! 0, we have
dy
dx
¼ lim
x!0
y
x
¼ lim
x!0
dy
du
u
x
þ
u
x
¼
dy
du
lim
x!0
u
x
þ lim
x!0
lim
x!0
u
x
¼
dy
du
du
dx
þ 0
du
dx
¼
dy
du
du
dx
ð3Þ
4.14. Given
d
dx
ðsin xÞ ¼ cos x and
d
dx
ðcos xÞ ¼ sin x, derive the formulas
ðaÞ
d
dx
ðtan xÞ ¼ sec2
x; ðbÞ
d
dx
ðsin1
xÞ ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p
ðaÞ
d
dx
ðtan xÞ ¼
d
dx
sin x
cos x
¼
cos x
d
dx
ðsin xÞ sin x
d
dx
ðcos xÞ
cos2 x
¼
ðcos xÞðcos xÞ ðsin xÞð sin xÞ
cos2
x
¼
1
cos2
x
¼2
x
(b) If y ¼ sin1
x, then x ¼ sin y. Taking the derivative with respect to x,
1 ¼ cos y
dy
dx
or
dy
dx
¼
1
cos y
¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sin2
y
q ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p
We have supposed here that the principal value =2 @ sin1
x @ =2, is chosen so that cos y is
positive, thus accounting for our writing cos y ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sin2
y
q
rather than cos y ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sin2
y
q
.
4.15. Derive the formula
d
dx
ðloga uÞ ¼
loga e
u
du
dx
ða 0; a 6¼ 1Þ, where u is a differentiable function of x.
Consider y ¼ f ðuÞ ¼ loga u. By definition,
dy
du
¼ lim
u!0
f ðu þ uÞ f ðuÞ
u
¼ lim
u!0
logaðu þ uÞ loga u
u
¼ lim
u!0
1
u
loga
u þ u
u
¼ lim
u!0
1
u
loga 1 þ
u
u
u=u
CHAP. 4] DERIVATIVES 79](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-88-320.jpg)
![If h 0, then
f ð þ hÞ f ðÞ
h
@ 0 and
lim
h!0þ
f ð þ hÞ f ðÞ
h
@ 0 ð1Þ
If h 0, then
f ð þ hÞ f ðÞ
h
A 0 and
lim
h!0
f ð þ hÞ f ðÞ
h
A 0 ð2Þ
But by hypothesis f ðxÞ has a derivative at all points
in ða; bÞ. Then the right-hand derivative (1) must be
equal to the left-hand derivative (2). This can happen only if they are both equal to zero, in which case
f 0
ðÞ ¼ 0 as required.
A similar argument can be used in case M ¼ 0 and m 6¼ 0.
4.20. Prove the mean value theorem.
Define FðxÞ ¼ f ðxÞ f ðaÞ ðx aÞ
f ðbÞ f ðaÞ
b a
.
Then FðaÞ ¼ 0 and FðbÞ ¼ 0.
Also, if f ðxÞ satisfies the conditions on continuity and differentiability specified in Rolle’s theorem, then
FðxÞ satisfies them also.
Then applying Rolle’s theorem to the function FðxÞ, we obtain
F 0
ðÞ ¼ f 0
ðÞ
f ðbÞ f ðaÞ
b a
¼ 0; a b or f 0
ðÞ ¼
f ðbÞ f ðaÞ
b a
; a b
4.21. Verify the mean value theorem for f ðxÞ ¼ 2x2
7x þ 10, a ¼ 2, b ¼ 5.
f ð2Þ ¼ 4, f ð5Þ ¼ 25, f 0
ðÞ ¼ 4 7. Then the mean value theorem states that 4 7 ¼ ð25 4Þ=ð5 2Þ
or ¼ 3:5. Since 2 5, the theorem is verified.
4.22. If f 0
ðxÞ ¼ 0 at all points of the interval ða; bÞ, prove that f ðxÞ must be a constant in the interval.
Let x1 x2 be any two different points in ða; bÞ. By the mean value theorem for x1 x2,
f ðx2Þ f ðx1Þ
x2 x1
¼ f 0
ðÞ ¼ 0
Thus, f ðx1Þ ¼ f ðx2Þ ¼ constant. From this it follows that if two functions have the same derivative at all
points of ða; bÞ, the functions can only differ by a constant.
4.23. If f 0
ðxÞ 0 at all points of the interval ða; bÞ, prove that f ðxÞ is strictly increasing.
Let x1 x2 be any two different points in ða; bÞ. By the mean value theorem for x1 x2,
f ðx2Þ f ðx1Þ
x2 x1
¼ f 0
ðÞ 0
Then f ðx2Þ f ðx1Þ for x2 x1, and so f ðxÞ is strictly increasing.
4.24. (a) Prove that
b a
1 þ b2
tan1
b tan1
a
b a
1 þ a2
if a b.
(b) Show that
4
þ
3
25
tan1 4
3
4
þ
1
6
.
(a) Let f ðxÞ ¼ tan1
x. Since f 0
ðxÞ ¼ 1=ð1 þ x2
Þ and f 0
ðÞ ¼ 1=ð1 þ 2
Þ, we have by the mean value
theorem
CHAP. 4] DERIVATIVES 81
f (x)
a ξ b
x
M
Fig. 4-9](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-90-320.jpg)
![from which we see that
f ðxÞ
gðxÞ
¼
f 0
ðÞ
g0ðÞ
1 gðx1Þ=gðxÞ
1 f ðx1Þ=f ðxÞ
ð1Þ
Let us now suppose that lim
x!x0þ
f 0
ðxÞ
g0ðxÞ
¼ L and write (1) as
f ðxÞ
gðxÞ
¼
f 0
ðÞ
g0ðÞ
L
1 gðx1Þ=gðxÞ
1 f ðx1Þ=f ðxÞ
þ L
1 gðx1Þ=gðxÞ
1 f ðx1Þ=f ðxÞ
ð2Þ
We can choose x1 so close to x0 that j f 0
ðÞ=g0
ðÞ Lj . Keeping x1 fixed, we see that
lim
x!x0þ
1 gðx1Þ=gðxÞ
1 f ðx1Þ=f ðxÞ
¼ 1 since lim
x!x0þ
f ðxÞ ¼ 1 and lim
x!x0
gðxÞ ¼ 1
Then taking the limit as x ! x0þ on both sides of (2), we see that, as required,
lim
x!x0þ
f ðxÞ
gðxÞ
¼ L ¼ lim
x!x0þ
f 0
ðxÞ
g0ðxÞ
Appropriate modifications of the above procedure establish the result if x ! x0, x ! x0,
x ! 1, x ! 1.
4.27. Evaluate (a) lim
x!0
e2x
1
x
ðbÞ lim
x!1
1 þ cos x
x2
2x þ 1
All of these have the ‘‘indeterminate form’’ 0/0.
ðaÞ lim
x!0
e2x
1
x
¼ lim
x!0
2e2x
1
¼ 2
ðbÞ lim
x!1
1 þ cos x
x2 2x þ 1
¼ lim
x!1
sin x
2x 2
¼ lim
x!1
2
cos x
2
¼
2
2
Note: Here L’Hospital’s rule is applied twice, since the first application again yields the ‘‘indeter-
minate form’’ 0/0 and the conditions for L’Hospital’s rule are satisfied once more.
4.28. Evaluate (a) lim
x!1
3x2
x þ 5
5x2 þ 6x 3
ðbÞ lim
x!1
x2
ex
All of these have or can be arranged to have the ‘‘indeterminate form’’ 1=1.
ðaÞ lim
x!1
3x2
x þ 5
5x2
þ 6x 3
¼ lim
x!1
6x 1
10x þ 6
¼ lim
x!1
6
10
¼
3
5
ðbÞ lim
x!1
x2
ex
¼ lim
x!1
x2
ex ¼ lim
x!1
2x
ex ¼ lim
x!1
2
ex ¼ 0
4.29. Evaluate lim
x!0þ
x2
ln x.
lim
x!0þ
x2
ln x ¼ lim
x!0þ
ln x
1=x2
¼ lim
x!0þ
1=x
2=x3
¼ lim
x!0þ
x2
2
¼ 0
The given limit has the ‘‘indeterminate form’’ 0 1. In the second step the form is altered so as to give
the indeterminate form 1=1 and L’Hospital’s rule is then applied.
4.30. Find lim
x!0
ðcos xÞ1=x2
.
Since lim
x!0
cos x ¼ 1 and lim
x!0
1=x2
¼ 1, the limit takes the ‘‘indeterminate form’’ 11
.
CHAP. 4] DERIVATIVES 83](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-92-320.jpg)
![MISCELLANEOUS PROBLEMS
4.34. If x ¼ gðtÞ and y ¼ f ðtÞ are twice differentiable, find (a) dy=dx; ðbÞ d2
y=dx2
.
(a) Letting primes denote derivatives with respect to t, we have
dy
dx
¼
dy=dt
dx=dt
¼
f 0
ðtÞ
g0ðtÞ
if g0
ðtÞ 6¼ 0
ðbÞ
d2
y
dx2
¼
d
dx
dy
dx
¼
d
dx
f 0
ðtÞ
g0ðtÞ
¼
d
dt
f 0
ðtÞ
g0
ðtÞ
dx=dt
¼
d
dt
f 0
ðtÞ
g0
ðtÞ
g0ðtÞ
¼
1
g0ðtÞ
g0
ðtÞf 00
ðtÞ f 0
ðtÞg00
ðtÞ
½g0ðtÞ2
¼
g0
ðtÞf 00
ðtÞ f 0
ðtÞg00
ðtÞ
½g0ðtÞ3
if g0
ðtÞ 6¼ 0
4.35. Let f ðxÞ ¼ e1=x2
; x 6¼ 0
0; x ¼ 0
. Prove that (a) f 0
ð0Þ ¼ 0; ðbÞ f 00
ð0Þ ¼ 0.
ðaÞ f 0
þð0Þ ¼ lim
h!0þ
f ðhÞ f ð0Þ
h
¼ lim
h!0þ
e1=h2
0
h
¼ lim
h!0þ
e1=h2
h
If h ¼ 1=u, using L’Hospital’s rule this limit equals
lim
u!1
ueu2
¼ lim
u!1
u=eu2
¼ lim
u!1
1=2ueu2
¼ 0
Similarly, replacing h ! 0þ by h ! 0 and u ! 1 by u ! 1, we find f 0
ð0Þ ¼ 0. Thus
f 0
þð0Þ ¼ f 0
ð0Þ ¼ 0, and so f 0
ð0Þ ¼ 0.
ðbÞ f 00
þ ð0Þ ¼ lim
h!0þ
f 0
ðhÞ f 0
ð0Þ
h
¼ lim
h!0þ
e1=h2
2h3
0
h
¼ lim
h!0þ
2e1=h2
h4
¼ lim
u!1
2u4
eu2 ¼ 0
by successive applications of L’Hospital’s rule.
Similarly, f 00
ð0Þ ¼ 0 and so f 00
ð0Þ ¼ 0.
In general, f ðnÞ
ð0Þ ¼ 0 for n ¼ 1; 2; 3; . . .
4.36. Find the length of the longest ladder which can be carried around the corner of a corridor, whose
dimensions are indicated in the figure below, if it is assumed that the ladder is carried parallel to
the floor.
The length of the longest ladder is the same as the shortest
straight line segment AB [Fig. 4-10], which touches both outer
walls and the corner formed by the inner walls.
As seen from Fig. 4-10, the length of the ladder AB is
L ¼ a sec þ b csc
L is a minimum when
dL=d ¼ a sec tan b csc cot ¼ 0
a sin3
¼ b cos3
or tan ¼
ffiffiffiffiffiffiffiffi
b=a
3
p
i.e.;
sec ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2=3 þ b2=3
p
a1=3
; csc ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2=3 þ b2=3
p
b1=3
Then
L ¼ a sec þ b csc ¼ ða2=3
þ b2=3
Þ3=2
so that
Although it is geometrically evident that this gives the minimum length, we can prove this analytically
by showing that d2
L=d2
for ¼ tan1
ffiffiffiffiffiffiffiffi
b=a
3
p
is positive (see Problem 4.78).
CHAP. 4] DERIVATIVES 85
a
b
B
O
A
a
s
e
c
G
b
c
s
c
G
G
G
Fig. 4-10](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-94-320.jpg)
![4.51. If f ðxÞ ¼ x2
þ 3x, find (a) y; ðbÞ dy; ðcÞ y=x; ðdÞ dy=dx; and (e) ðy dyÞ=x, if x ¼ 1 and
x ¼ :01.
Ans. (a) .0501, (b) .05, (c) 5.01, (d) 5, (e) .01
4.52. Using differentials, compute approximate values for each of the following: (a) sin 318; ðbÞ lnð1:12Þ,
(c)
ffiffiffiffiffi
36
5
p
.
Ans. (a) 0.515, (b) 0.12, (c) 2.0125
4.53. If y ¼ sin x, evaluate (a) y; ðbÞ dy. (c) Prove that ðy dyÞ=x ! 0 as x ! 0.
DIFFERENTIATION RULES AND ELEMENTARY FUNCTIONS
4.54. Prove: (a)
d
dx
f f ðxÞ þ gðxÞg ¼
d
dx
f ðxÞ þ
d
dx
gðxÞ; ðbÞ
d
dx
f f ðxÞ gðxÞg ¼
d
dx
f ðxÞ
d
dx
gðxÞ,
ðcÞ
d
dx
f ðxÞ
gðxÞ
¼
gðxÞ f 0
ðxÞ f ðxÞg0
ðxÞ
½gðxÞ2
; gðxÞ 6¼ 0:
4.55. Evaluate (a)
d
dx
fx3
lnðx2
2x þ 5Þg at x ¼ 1; ðbÞ
d
dx
fsin2
ð3x þ =6Þg at x ¼ 0.
Ans. (a) 3 ln 4; ðbÞ 3
2
ffiffiffi
3
p
4.56. Derive the formulas: (a)
d
dx
au
¼ au
ln a
du
dx
; a 0; a 6¼ 1; ðbÞ
d
dx
csc u ¼ csc u cot u
du
dx
;
ðcÞ
d
dx
tanh u ¼ sech2
u
du
dx
where u is a differentiable function of x:
4.57. Compute (a)
d
dx
tan1
x; ðbÞ
d
dx
csc1
x; ðcÞ
d
dx
sinh1
x; ðdÞ
d
dx
coth1
x, paying attention to the
use of principal values.
4.58. If y ¼ xx
, computer dy=dx. [Hint: Take logarithms before differentiating.]
Ans. xx
ð1 þ ln xÞ
4.59. If y ¼ flnð3x þ 2Þgsin1
ð2xþ:5Þ
, find dy=dx at x ¼ 0:
Ans:
4 ln 2
þ
2 ln ln 2
ffiffiffi
3
p
ðln 2Þ=6
4.60. If y ¼ f ðuÞ, where u ¼ gðvÞ and v ¼ hðxÞ, prove that
dy
dx
¼
dy
du
du
dv
dv
dx
assuming f , g; and h are differentiable.
4.61. Calculate (a) dy=dx and (b) d2
y=dx2
if xy ln y ¼ 1.
Ans. (a) y2
=ð1 xyÞ; ðbÞ ð3y3
2xy4
Þ=ð1 xyÞ3
provided xy 6¼ 1
4.62. If y ¼ tan x, prove that y000
¼ 2ð1 þ y2
Þð1 þ 3y2
Þ.
4.63. If x ¼ sec t and y ¼ tan t, evaluate (a) dy=dx; ðbÞ d2
y=dx2
; ðcÞ d3
y=dx3
, at t ¼ =4.
Ans. (a)
ffiffiffi
2
p
; ðbÞ 1; ðcÞ 3
ffiffiffi
2
p
4.64. Prove that
d2
y
dx2
¼
d2
x
dy2
dx
dy
3
, stating precise conditions under which it holds.
4.65. Establish formulas (a) 7, (b) 18, and (c) 27, on Page 71.
MEAN VALUE THEOREMS
4.66. Let f ðxÞ ¼ 1 ðx 1Þ2=3
, 0 @ x @ 2. (a) Construct the graph of f ðxÞ. (b) Explain why Rolle’s theorem is
not applicable to this function, i.e., there is no value for which f 0
ðÞ ¼ 0, 0 2.
CHAP. 4] DERIVATIVES 87](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-96-320.jpg)
![4.82. A particle travels with constant velocities v1 and v2 in mediums I and II,
respectively (see adjoining Fig. 4-11). Show that in order to go from point
P to point Q in the least time, it must follow path PAQ where A is such
that
ðsin 1Þ=ðsin 2Þ ¼ v1=v2
Note: This is Snell’s Law; a fundamental law of optics first discovered
experimentally and then derived mathematically.
4.83. A variable is called an infinitesimal if it has zero as a limit. Given two
infinitesimals and , we say that is an infinitesimal of higher order (or the same order) if lim = ¼ 0 (or
lim = ¼ l 6¼ 0). Prove that as x ! 0, (a) sin2
2x and ð1 cos 3xÞ are infinitesimals of the same order,
(b) ðx3
sin3
xÞ is an infinitesimal of higher order than fx lnð1 þ xÞ 1 þ cos xg.
4.84. Why can we not use L’Hospital’s rule to prove that lim
x!0
x2
sin 1=x
sin x
¼ 0 (see Problem 3.91, Chap. 3)?
4.85. Can we use L’Hospital’s rule to evaluate the limit of the sequence un ¼ n3
en2
, n ¼ 1; 2; 3; . . . ? Explain.
4.86 (1) Determine decimal approximations with at least three places of accuracy for each of the following
irrational numbers. (a)
ffiffiffi
2
p
; ðbÞ
ffiffiffi
5
p
; ðcÞ 71=3
(2) The cubic equation x3
3x2
þ x 4 ¼ 0 has a root between 3 and 4. Use Newton’s Method to
determine it to at least three places of accuracy.
4.87. Using successive applications of Newton’s method obtain the positive root of (a) x3
2x2
2x 7 ¼ 0,
(b) 5 sin x ¼ 4x to 3 decimal places.
Ans. (a) 3.268, (b) 1.131
4.88. If D denotes the operator d=dx so that Dy dy=dx while Dk
y dk
y=dxk
, prove Leibnitz’s formula
Dn
ðuvÞ ¼ ðDn
uÞv þ nC1ðDn1
uÞðDvÞ þ nC2ðDn2
uÞðD2
vÞ þ þ nCrðDnr
uÞðDr
vÞ þ þ uDn
v
where nCr ¼ ðn
rÞ are the binomial coefficients (see Problem 1.95, Chapter 1).
4.89. Prove that
dn
dxn ðx2
sin xÞ ¼ fx2
nðn 1Þg sinðx þ n=2Þ 2nx cosðx þ n=2Þ.
4.90. If f 0
ðx0Þ ¼ f 00
ðx0Þ ¼ ¼ f ð2nÞ
ðx0Þ ¼ 0 but f ð2nþ1Þ
ðx0Þ 6¼ 0, discuss the behavior of f ðxÞ in the neighborhood
of x ¼ x0. The point x0 in such case is often called a point of inflection. This is a generalization of the
previously discussed case corresponding to n ¼ 1.
4.91. Let f ðxÞ be twice differentiable in ða; bÞ and suppose that f 0
ðaÞ ¼ f 0
ðbÞ ¼ 0. Prove that there exists at least
one point in ða; bÞ such that j f 00
ðÞj A
4
ðb aÞ2
f f ðbÞ f ðaÞg. Give a physical interpretation involving
velocity and acceration of a particle.
CHAP. 4] DERIVATIVES 89
P
Q
A
G1
G2
Medium I
velocity L1
Medium II
velocity L2
Fig. 4-11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-98-320.jpg)
![Subdivide the interval a @ x @ b into n sub-intervals by means of the points x1; x2; . . . ; xn1 chosen
arbitrarily. In each of the new intervals ða; x1Þ; ðx1; x2Þ; . . . ; ðxn1; bÞ choose points 1; 2; . . . ; n
arbitrarily. Form the sum
f ð1Þðx1 aÞ þ f ð2Þðx2 x1Þ þ f ð3Þðx3 x2Þ þ þ f ðnÞðb xn1Þ ð1Þ
By writing x0 ¼ a, xn ¼ b; and xk xk1 ¼ xk, this can be written
X
n
k¼1
f ðkÞðxk xk1Þ ¼
X
n
k¼1
f ðkÞxk ð2Þ
Geometrically, this sum represents the total area of all rectangles in the above figure.
We now let the number of subdivisions n increase in such a way that each xk ! 0. If as a result
the sum (1) or (2) approaches a limit which does not depend on the mode of subdivision, we denote this
limit by
ðb
a
f ðxÞ dx ¼ lim
n!1
X
n
k¼1
f ðkÞxk ð3Þ
This is called the definite integral of f ðxÞ between a and b. In this symbol f ðxÞ dx is called the integrand,
and ½a; b is called the range of integration. We call a and b the limits of integration, a being the lower
limit of integration and b the upper limit.
The limit (3) exists whenever f ðxÞ is continuous (or piecewise continuous) in a @ x @ b (see Problem
5.31). When this limit exists we say that f is Riemann integrable or simply integrable in ½a; b.
The definition of the definite integral as the limit of a sum was established by Cauchy around 1825.
It was named for Riemann because he made extensive use of it in this 1850 exposition of integration.
Geometrically the value of this definite integral represents the area bounded by the curve y ¼ f ðxÞ,
the x-axis and the ordinates at x ¼ a and x ¼ b only if f ðxÞ A 0. If f ðxÞ is sometimes positive and
sometimes negative, the definite integral represents the algebraic sum of the areas above and below the x-
axis, treating areas above the x-axis as positive and areas below the x-axis as negative.
MEASURE ZERO
A set of points on the x-axis is said to have measure zero if the sum of the lengths of intervals
enclosing all the points can be made arbitrary small (less than any given positive number ). We can
show (see Problem 5.6) that any countable set of points on the real axis has measure zero. In particular,
the set of rational numbers which is countable (see Problems 1.17 and 1.59, Chapter 1), has measure
zero.
An important theorem in the theory of Riemann integration is the following:
Theorem. If f ðxÞ is bounded in ½a; b, then a necessary and sufficient condition for the existence of
Ðb
a f ðxÞ dx is that the set of discontinuities of f ðxÞ have measure zero.
PROPERTIES OF DEFINITE INTEGRALS
If f ðxÞ and gðxÞ are integrable in ½a; b then
1.
ðb
a
f f ðxÞ gðxÞg dx ¼
ðb
a
f ðxÞ dx
ðb
a
gðxÞ dx
2.
ðb
a
A f ðxÞ dx ¼ A
ðb
a
f ðxÞ dx where A is any constant
CHAP. 5] INTEGRALS 91](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-100-320.jpg)
![Of course, the question of for what value x ¼ the average is attained is not answered; and, in fact,
in general, only existence not the value can be demonstrated. To see that there is a point x ¼ such that
f ðÞ represents the average value of f on ½a; b, recall that a continuous function on a closed interval has
maximum and minimum values, M and m, respectively. Thus (think of the integral as representing the
area under the curve). (See Fig. 5-2.)
mðb aÞ @
ðb
a
f ðxÞ dx @ Mðb aÞ
or
m @
1
b a
ðb
a
f ðxÞ dx @ M
Since f is a continuous function on a closed interval, there exists a point x ¼ in ða; bÞ intermediate
to m and M such that
f ðÞ ¼
1
b a
ðb
a
f ðxÞ dx
While this example is not a rigorous proof of the first mean value theorem, it motivates it and
provides an interpretation. (See Chapter 3, Theorem 10.)
1. First mean value theorem. If f ðxÞ is continuous in ½a; b, there is a point in ða; bÞ such that
ðb
a
f ðxÞ dx ¼ ðb aÞ f ðÞ ð4Þ
2. Generalized first mean value theorem. If f ðxÞ and gðxÞ are continuous in ½a; b, and gðxÞ does not
change sign in the interval, then there is a point in ða; bÞ such that
ðb
a
f ðxÞgðxÞ dx ¼ f ðÞ
ðb
a
gðxÞ dx ð5Þ
This reduces to (4) if gðxÞ ¼ 1.
CHAP. 5] INTEGRALS 93
y
b
a x
b _ a
A
D
F E
M
m
B
C
y = f (x)
Fig. 5-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-102-320.jpg)
![If the open interval on which f is continuous includes a and b, then we may write
ðb
a
f ðxÞ dx ¼ FðbÞ FðaÞ: (See Problem 5.11)
This is the usual form in which the theorem is used.
EXAMPLE. To evaluate
ð2
1
x2
dx we observe that F 0
ðxÞ ¼ x2
, FðxÞ ¼
x3
3
þ c and
ð2
1
x2
dx ¼ 23
3 þ c
13
3 þ c
¼ 7
3. Since c subtracts out of this evaluation it is convenient to exclude it and simply write
23
3
13
3
.
GENERALIZATION OF THE LIMITS OF INTEGRATION
The upper and lower limits of integration may be variables. For example:
ðcos x
sin x
t dt ¼
t2
2
#cos x
sin x
¼ ðcos2
x sin2
xÞ=2
In general, if F 0
ðxÞ ¼ f ðxÞ then
ðvðxÞ
uðxÞ
f ðtÞ dt ¼ F½vðxÞ ¼ F½uðxÞ
CHANGE OF VARIABLE OF INTEGRATION
If a determination of
Ð
f ðxÞ dx is not immediately obvious in terms of elementary functions, useful
results may be obtained by changing the variable from x to t according to the transformation x ¼ gðtÞ.
(This change of integrand that follows is suggested by the differential relation dx ¼ g0
ðtÞ dt.) The funda-
mental theorem enabling us to do this is summarized in the statement
ð
f ðxÞ dx ¼
ð
f fgðtÞgg0
ðtÞ dt ð6Þ
where after obtaining the indefinite integral on the right we replace t by its value in terms of x, i.e.,
t ¼ g1
ðxÞ. This result is analogous to the chain rule for differentiation (see Page 69).
The corresponding theorem for definite integrals is
ðb
a
f ðxÞ dx ¼
ð
f fgðtÞgg0
ðtÞ dt ð7Þ
where gð Þ ¼ a and gð Þ ¼ b, i.e., ¼ g1
ðaÞ, ¼ g1
ðbÞ. This result is certainly valid if f ðxÞ is con-
tinuous in ½a; b and if gðtÞ is continuous and has a continuous derivative in @ t @ .
INTEGRALS OF ELEMENTARY FUNCTIONS
The following results can be demonstrated by differentiating both sides to produce an identity. In
each case an arbitrary constant c (which has been omitted here) should be added.
CHAP. 5] INTEGRALS 95](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-104-320.jpg)
![SPECIAL METHODS OF INTEGRATION
1. Integration by parts.
Let u and v be differentiable functions. According to the product rule for differentials
dðuvÞ ¼ u dv þ v du
Upon taking the antiderivative of both sides of the equation, we obtain
uv ¼
ð
u dv þ
ð
v du
This is the formula for integration by parts when written in the form
ð
u dv ¼ uv
ð
v du or
ð
f ðxÞg0
ðxÞ dx ¼ f ðxÞgðxÞ
ð
f 0
ðxÞgðxÞ dx
where u ¼ f ðxÞ and v ¼ gðxÞ. The corresponding result for definite integrals over the interval
½a; b is certainly valid if f ðxÞ and gðxÞ are continuous and have continuous derivatives in ½a; b.
See Problems 5.17 to 5.19.
2. Partial fractions. Any rational function
PðxÞ
QðxÞ
where PðxÞ and QðxÞ are polynomials, with the
degree of PðxÞ less than that of QðxÞ, can be written as the sum of rational functions having the
form
A
ðax þ bÞr,
Ax þ B
ðax2
þ bx þ cÞr where r ¼ 1; 2; 3; . . . which can always be integrated in terms of
elementary functions.
EXAMPLE 1.
3x 2
ð4x 3Þð2x þ 5Þ3
¼
A
4x 3
þ
B
ð2x þ 5Þ3
þ
C
ð2x þ 5Þ2
þ
D
2x þ 5
EXAMPLE 2.
5x2
x þ 2
ðx2
þ 2x þ 4Þ2
ðx 1Þ
¼
Ax þ B
ðx2
þ 2x þ 4Þ2
þ
Cx þ D
x2
þ 2x þ 4
þ
E
x 1
The constants, A, B, C, etc., can be found by clearing of fractions and equating coefficients of like powers of x
on both sides of the equation or by using special methods (see Problem 5.20).
3. Rational functions of sin x and cos x can always be integrated in terms of elementary functions by
the substitution tan x=2 ¼ u (see Problem 5.21).
4. Special devices depending on the particular form of the integrand are often employed (see
Problems 5.22 and 5.23).
IMPROPER INTEGRALS
If the range of integration ½a; b is not finite or if f ðxÞ is not defined or not bounded at one or more
points of ½a; b, then the integral of f ðxÞ over this range is called an improper integral. By use of
appropriate limiting operations, we may define the integrals in such cases.
EXAMPLE 1.
ð1
0
dx
1 þ x2
¼ lim
M!1
ðM
0
dx
1 þ x2
¼ lim
M!1
tan1
x
M
0
¼ lim
M!1
tan1
M ¼ =2
EXAMPLE 2.
ð1
0
dx
ffiffiffi
x
p ¼ lim
!0þ
ð1
dx
ffiffiffi
x
p ¼ lim
!0þ
2
ffiffiffi
x
p 1
¼ lim
!0þ
ð2 2
ffiffi
ffi
p
Þ ¼ 2
EXAMPLE 3.
ð1
0
dx
x
¼ lim
!0þ
ð1
dx
x
¼ lim
!0þ
ln x
1
¼ lim
!0þ
ð ln Þ
Since this limit does not exist we say that the integral diverges (i.e., does not converge).
CHAP. 5] INTEGRALS 97](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-106-320.jpg)
![ARC LENGTH
As you walk a twisting mountain trail, it is possible to determine the distance covered by using a
pedometer. To create a geometric model of this event, it is necessary to describe the trail and a method
of measuring distance along it. The trail might be referred to as a path, but in more exacting geometric
terminology the word, curve is appropriate. That segment to be measured is an arc of the curve. The
arc is subject to the following restrictions:
1. It does not intersect itself (i.e., it is a simple arc).
2. There is a tangent line at each point.
3. The tangent line varies continuously over the arc.
These conditions are satisfied with a parametric representation x ¼ f ðtÞ; y ¼ gðtÞ; z ¼ hðtÞ; a @ t @ b,
where the functions f , g, and h have continuous derivatives that do not simultaneously vanish at any
point. This arc is in Euclidean three space and will be discussed in Chapter 10. In this introduction to
curves and their arc length, we let z ¼ 0, thereby restricting the discussion to the plane.
A careful examination of your walk would reveal movement on a sequence of straight segments,
each changed in direction from the previous one. This suggests that the length of the arc of a curve is
obtained as the limit of a sequence of lengths of polygonal approximations. (The polygonal approx-
imations are characterized by the number of divisions n ! 1 and no subdivision is bound from zero.
(See Fig. 5-4.)
Geometrically, the measurement of the kth segment of the arc, 0 @ t @ s, is accomplished by
employing the Pythagorean theorem, and thus, the measure is defined by
CHAP. 5] INTEGRALS 99
Fig. 5-4
Fig. 5-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-108-320.jpg)
![This method of generating a volume is called the disk method because the cross sections of revolution
are circular disks. (See Fig. 5-5(a).)
EXAMPLE. A solid cone is generated by revolving the graph of y ¼ kx, k 0 and 0 @ x @ b, about the x-axis.
Its volume is
V ¼
ðb
0
k2
x2
dx ¼
k3
x3
3
b
0
¼
k3
b3
3
Shell Method
Suppose f is a continuous function on ½a; b, a A 0, satisfying the condition f ðxÞ A 0. Let R be a
plane region bound by f ðxÞ, x ¼ a, x ¼ b, and the x-axis. The volume obtained by orbiting R about the
y-axis is
V ¼
ðb
a
2x f ðxÞ dx
This method of generating a volume is called the shell method because of the cylindrical nature of the
vertical lines of revolution. (See Fig. 5-5(b).)
EXAMPLE. If the region bounded by y ¼ kx, 0 @ x @ b and x ¼ b (with the same conditions as in the previous
example) is orbited about the y-axis the volume obtained is
V ¼ 2
ðb
0
xðkxÞ dx ¼ 2k
x3
3
b
0
¼ 2k
b3
3
By comparing this example with that in the section on the disk method, it is clear that for the same
plane region the disk method and the shell method produce different solids and hence different volumes.
Moment of Inertia
Moment of inertia is an important physical concept that can be studied through its idealized geo-
metric form. This form is abstracted in the following way from the physical notions of kinetic energy,
K ¼ 1
2 mv2
, and angular velocity, v ¼ !r. (m represents mass and v signifies linear velocity). Upon
substituting for v
K ¼ 1
2 m!2
r2
¼ 1
2 ðmr2
Þ!2
When this form is compared to the original representation of kinetic energy, it is reasonable to
identify mr2
as rotational mass. It is this quantity, l ¼ mr2
that we call the moment of inertia.
Then in a purely geometric sense, we denote a plane region R described through continuous func-
tions f and g on ½a; b, where a 0 and f ðxÞ and gðxÞ intersect at a and b only. For simplicity, assume
gðxÞ A f ðxÞ 0. Then
CHAP. 5] INTEGRALS 101
Fig. 5-5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-110-320.jpg)
![5.4. Evaluate lim
n!1
1
n þ 1
þ
1
n þ 2
þ þ
1
n þ n
.
The required limit can be written
lim
n!1
1
n
1
1 þ 1=n
þ
1
1 þ 2=n
þ þ
1
1 þ n=n
¼ lim
n!1
1
n
X
n
k¼1
1
1 þ k=n
¼
ð1
0
dx
1 þ x
¼ lnð1 þ xÞj1
0 ¼ ln 2
using Problem 5.2 and the fundamental theorem of the calculus.
5.5. Prove that lim
n!1
1
n
sin
t
n
þ sin
2t
n
þ þ sin
ðn 1Þt
n
¼
1 cos t
t
.
Let a ¼ 0; b ¼ t; f ðxÞ ¼ sin x in Problem 1. Then
lim
n!1
t
n
X
n
k¼1
sin
kt
n
¼
ðt
0
sin x dx ¼ 1 cos t
and so
lim
n!1
1
n
X
n1
k¼1
sin
kt
n
¼
1 cos t
t
using the fact that lim
n!1
sin t
n
¼ 0.
MEASURE ZERO
5.6. Prove that a countable point set has measure zero.
Let the point set be denoted by x1; x2; x3; x4; . . . and suppose that intervals of lengths less than
=2; =4; =8; =16; . . . respectively enclose the points, where is any positive number. Then the sum of
the lengths of the intervals is less than =2 þ =4 þ =8 þ ¼ (let a ¼ =2 and r ¼ 1
2 in Problem 2.25(a) of
Chapter 2), showing that the set has measure zero.
PROPERTIES OF DEFINITE INTEGRALS
5.7. Prove that
ðb
a
f ðxÞ dx @
ðb
a
j f ðxÞj dx if a b.
By absolute value property 2, Page 3,
X
n
k¼1
f ðkÞxk @
X
n
k¼1
j f ðkÞxkj ¼
X
n
k¼1
j f ðkÞjxk
Taking the limit as n ! 1 and each xk ! 0, we have the required result.
5.8. Prove that lim
n!1
ð2
0
sin nx
x2 þ n2
dx ¼ 0.
ð2
0
sin nx
x2
þ n2
dx @
ð2
0
sin nx
x2
þ n2
dx @
ð2
0
dx
n2
¼
2
n2
Then lim
n!1
ð2
0
sin nx
x2 þ n2
dx ¼ 0, and so the required result follows.
CHAP. 5] INTEGRALS 103](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-112-320.jpg)
![Another method:
By Problem 5.10 and Problem 4.3, Chapter 4, it follows that F 0
ðxÞ exists and so FðxÞ must be con-
tinuous.
CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION
5.13. Prove the result (7), Page 95, for changing the variable of integration.
Let FðxÞ ¼
ðx
a
f ðxÞ dx and GðtÞ ¼
ðt
f fgðtÞg g0
ðtÞ dt, where x ¼ gðtÞ.
Then dF ¼ f ðxÞ dx, dG ¼ f fgðtÞg g0
ðtÞ dt.
Since dx ¼ g0
ðtÞ dt, it follows that f ðxÞ dx ¼ f fgðtÞg g0
ðtÞ dt so that dFðxÞ ¼ dGðtÞ, from which
FðxÞ ¼ GðtÞ þ c.
Now when x ¼ a, t ¼ or FðaÞ ¼ Gð Þ þ c. But FðaÞ ¼ Gð Þ ¼ 0, so that c ¼ 0. Hence FðxÞ ¼ GðtÞ.
Since x ¼ b when t ¼ , we have
ðb
a
f ðxÞ dx ¼
ð
f fgðtÞg g0
ðtÞ dt
as required.
5.14. Evaluate:
ðaÞ
ð
ðx þ 2Þ sinðx2
þ 4x 6Þ dx ðcÞ
ð1
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 2Þð3 xÞ
p ðeÞ
ð1=
ffiffi
2
p
0
x sin1
x2
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x4
p dx
ðbÞ
ð
cotðln xÞ
x
dx ðdÞ
ð
2x
tanh 21x
dx ð f Þ
ð
x dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ x þ 1
p
(a) Method 1: Let x2
þ 4x 6 ¼ u. Then ð2x þ 4Þ dx ¼ du, ðx þ 2Þ dx ¼ 1
2 du and the integral becomes
1
2
ð
sin u du ¼
1
2
cos u þ c ¼
1
2
cosðx2
þ 4x 6Þ þ c
Method 2:
ð
ðx þ 2Þ sinðx2
þ 4x 6Þ dx ¼
1
2
ð
sinðx2
þ 4x 6Þdðx2
þ 4x 6Þ ¼
1
2
cosðx2
þ 4x 6Þ þ c
(b) Let ln x ¼ u. Then ðdxÞ=x ¼ du and the integral becomes
ð
cot u du ¼ ln j sin uj þ c ¼ ln j sinðln xÞj þ c
ðcÞ Method 1:
ð
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 2Þð3 xÞ
p ¼
ð
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6 þ x x2
p ¼
ð
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6 ðx2
xÞ
p ¼
ð
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
25=4 ðx 1
2Þ2
q
Letting x 1
2 ¼ u, this becomes
ð
du
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
25=4 u2
p ¼ sin1 u
5=2
þ c ¼ sin1 2x 1
5
þ c
Then
ð1
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 2Þð3 xÞ
p ¼ sin1 2x 1
5
1
1
¼ sin1 1
5
sin1
3
5
¼ sin1
:2 þ sin1
:6
CHAP. 5] INTEGRALS 105](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-114-320.jpg)
![(a) Use integration by parts, letting u ¼ ln x, dv ¼ xn
dx, so that du ¼ ðdxÞ=x, v ¼ xnþ1
=ðn þ 1Þ. Then
ð
xn
ln x dx ¼
ð
u dv ¼ uv
ð
v du ¼
xnþ1
n þ 1
ln x
ð
xnþ1
n þ 1
dx
x
¼
xnþ1
n þ 1
ln x
xnþ1
ðn þ 1Þ2
þ c
ðbÞ
ð
x1
ln x dx ¼
ð
ln x dðln xÞ ¼
1
2
ðln xÞ2
þ c:
5.18. Find
ð
3
ffiffiffiffiffiffiffiffi
2xþ1
p
dx.
Let
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x þ 1
p
¼ y, 2x þ 1 ¼ y2
. Then dx ¼ y dy and the integral becomes
ð
3y
y dy.
Integrate by parts, letting u ¼ y, dv ¼ 3y
dy; then du ¼ dy, v ¼ 3y
=ðln 3Þ, and we have
ð
3y
y dy ¼
ð
u dv ¼ uv
ð
v du ¼
y 3y
ln 3
ð
3y
ln 3
dy ¼
y 3y
ln 3
3y
ðln 3Þ2
þ c
5.19. Find
ð1
0
x lnðx þ 3Þ dx.
Let u ¼ lnðx þ 3Þ, dv ¼ x dx. Then du ¼
dx
x þ 3
, v ¼
x2
2
. Hence on integrating by parts,
ð
x lnðx þ 3Þ dx ¼
x2
2
lnðx þ 3Þ
1
2
ð
x2
dx
x þ 3
¼
x2
2
lnðx þ 3Þ
1
2
ð
x 3 þ
9
x þ 3
dx
¼
x2
2
lnðx þ 3Þ
1
2
x2
2
3x þ 9 lnðx þ 3Þ
( )
þ c
ð1
0
x lnðx þ 3Þ dx ¼
5
4
4 ln 4 þ
9
2
ln 3
Then
5.20. Determine
ð
6 x
ðx 3Þð2x þ 5Þ
dx.
Use the method of partial fractions. Let
6 x
ðx 3Þð2x þ 5Þ
¼
A
x 3
þ
B
2x þ 5
.
Method 1: To determine the constants A and B, multiply both sides by ðx 3Þð2x þ 5Þ to obtain
6 x ¼ Að2x þ 5Þ þ Bðx 3Þ or 6 x ¼ 5A 3B þ ð2A þ BÞx ð1Þ
Since this is an identity, 5A 3B ¼ 6, 2A þ B ¼ 1 and A ¼ 3=11, B ¼ 17=11. Then
ð
6 x
ðx 3Þð2x þ 5Þ
dx ¼
ð
3=11
x 3
dx þ
ð
17=11
2x þ 5
dx ¼
3
11
ln jx 3j
17
22
ln j2x þ 5j þ c
Method 2: Substitute suitable values for x in the identity (1). For example, letting x ¼ 3 and x ¼ 5=2 in
(1), we find at once A ¼ 3=11, B ¼ 17=11.
5.21. Evaluate
ð
dx
5 þ 3 cos x
by using the substitution tan x=2 ¼ u.
From Fig. 5-7 we see that
sin x=2 ¼
u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2
p ; cos x=2 ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2
p
CHAP. 5] INTEGRALS 107
x/2
u
√1 + u
2
1
Fig. 5-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-116-320.jpg)
![x
2
fy0 þ 2y1 þ 2y2 þ 2y3 þ y4g ¼
0:25
2
f1:0000 þ 2ð0:9412Þ þ 2ð0:8000Þ þ 2ð0:6400Þ þ 0:500g
¼ 0:7828:
(b) Simpson’s rule gives
x
3
fy0 þ 4y1 þ 2y2 þ 4y3 þ y4g ¼
0:25
3
f1:0000 þ 4ð0:9412Þ þ 2ð0:8000Þ þ 4ð0:6400Þ þ 0:5000g
¼ 0:7854:
The true value is =4 0:7854:
APPLICATIONS (AREA, ARC LENGTH, VOLUME, MOMENT OF INTERTIA)
5.25. Find the (a) area and (b) moment of inertia about the y-axis of the region in the xy plane
bounded by y ¼ 4 x2
and the x-axis.
(a) Subdivide the region into rectangles as in the figure on
Page 90. A typical rectangle is shown in the adjoining
Fig. 5-8. Then
Required area ¼ lim
n!1
X
n
k¼1
f ðkÞ xk
¼ lim
n!1
X
n
k¼1
ð4 2
kÞ xk
¼
ð2
2
ð4 x2
Þ dx ¼
32
3
(b) Assuming unit density, the moment of inertia about the y-
axis of the typical rectangle shown above is 2
k f ðkÞ xk.
Then
Required moment of inertia ¼ lim
n!1
X
n
k¼1
2
k f ðkÞ xk ¼ lim
n!1
X
n
k¼1
2
kð4 2
kÞ xk
¼
ð2
2
x2
ð4 x2
Þ dx ¼
128
15
5.26. Find the length of arc of the parabola y ¼ x2
from x ¼ 0 to x ¼ 1.
Required arc length ¼
ð1
0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ ðdy=dxÞ2
q
dx ¼
ð1
0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ ð2xÞ2
q
dx
¼
ð1
0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ 4x2
p
dx ¼
1
2
ð2
0
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2
p
du
¼ 1
2 f1
2 u
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2
p
þ 1
2 lnðu þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u2Þ
p
gj2
0 ¼ 1
2
ffiffiffi
5
p
þ 1
4 lnð2 þ
ffiffiffi
5
p
Þ
5.27. (a) (Disk Method) Find the volume generated by revolving the region of Problem 5.25 about the
x-axis.
Required volume ¼ lim
n!1
X
n
k¼1
y2
kxk ¼
ð2
2
ð4 x2
Þ2
dx ¼ 512=15:
(b) (Disk Method) Find the volume of the frustrum of a paraboloid obtained by revolving f ðxÞ ¼
ffiffiffiffiffiffi
kx
p
,
0 a @ x @ b about the x-axis.
CHAP. 5] INTEGRALS 109
Fig. 5-8](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-118-320.jpg)
![5.30. Evaluate lim
x!0
Ðx
0 sin t3
dt
x4
.
The conditions of L’Hospital’s rule are satisfied, so that the required limit is
lim
x!0
d
dx
ðx
0
sin t3
dt
d
dx
ðx4
Þ
¼ lim
x!0
sin x3
4x3
¼ lim
x!0
d
dx
ðsin x3
Þ
d
dx
ð4x3
Þ
¼ lim
x!0
3x2
cos x3
12x2
¼
1
4
5.31. Prove that if f ðxÞ is continuous in ½a; b then
ðb
a
f ðxÞ dx exists.
Let ¼
X
n
k¼1
f ðkÞ xk, using the notation of Page 91. Since f ðxÞ is continuous we can find numbers Mk
and mk representing the l.u.b. and g.l.b. of f ðxÞ in the interval ½xk1; xk, i.e., such that mk @ f ðxÞ @ Mk.
We then have
mðb aÞ @ s ¼
X
n
k¼1
mkxk @ @
X
n
k¼1
Mkxk ¼ S @ Mðb aÞ ð1Þ
where m and M are the g.l.b. and l.u.b. of f ðxÞ in ½a; b. The sums s and S are sometimes called the lower and
upper sums, respectively.
Now choose a second mode of subdivision of ½a; b and consider the corresponding lower and upper
sums denoted by s0
and S0
respectively. We have must
s0
@ S and S0
A s ð2Þ
To prove this we choose a third mode of subdivision obtained by using the division points of both the first
and second modes of subdivision and consider the corresponding lower and upper sums, denoted by t and T,
respectively. By Problem 5.84, we have
s @ t @ T @ S0
and s0
@ t @ T @ S ð3Þ
which proves (2).
From (2) it is also clear that as the number of subdivisions is increased, the upper sums are monotonic
decreasing and the lower sums are monotonic increasing. Since according to (1) these sums are also
bounded, it follows that they have limiting values which we shall call
s
s and S respectively. By Problem
5.85,
s
s @ S. In order to prove that the integral exists, we must show that
s
s ¼ S.
Since f ðxÞ is continuous in the closed interval ½a; b, it is uniformly continuous. Then given any 0,
we can take each xk so small that Mk mk =ðb aÞ. It follows that
S s ¼
X
n
k¼1
ðMk mkÞxk
b a
X
n
k¼1
xk ¼ ð4Þ
Now S s ¼ ðS SÞ þ ðS
s
sÞ þ ð
s
s sÞ and it follows that each term in parentheses is positive and so is less
than by (4). In particular, since S
s
s is a definite number it must be zero, i.e., S ¼
s
s. Thus, the limits of
the upper and lower sums are equal and the proof is complete.
Supplementary Problems
DEFINITION OF A DEFINITE INTEGRAL
5.32. (a) Express
ð1
0
x3
dx as a limit of a sum. (b) Use the result of (a) to evaluate the given definite integral.
(c) Interpret the result geometrically.
Ans. (b) 1
4
5.33. Using the definition, evaluate (a)
ð2
0
ð3x þ 1Þ dx; ðbÞ
ð6
3
ðx2
4xÞ dx.
Ans. (a) 8, (b) 9
CHAP. 5] INTEGRALS 111](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-120-320.jpg)
![CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION
5.49. Evaluate: (a)
ð
x2
esin x3
cos x3
dx; ðbÞ
ð1
0
tan1
t
1 þ t2
dt; ðcÞ
ð3
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4x x2
p ; ðdÞ
ð
csch2 ffiffiffi
u
p
ffiffiffi
u
p du,
(e)
ð2
2
dx
16 x2
.
Ans. (a) 1
3 esin x3
þ c; ðbÞ 2
=32; ðcÞ =3; ðdÞ 2 coth
ffiffiffi
u
p
þ c; ðeÞ 1
4 ln 3.
5.50. Show that (a)
ð1
0
dx
ð3 þ 2x x2Þ3=2
¼
ffiffiffi
3
p
12
; ðbÞ
ð
dx
x2
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
1
p ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 1
p
x
þ c.
5.51. Prove that (a)
ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
a2
p
du ¼ 1
2 u
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
a2
p
1
2 a2
ln ju þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
a2
p
j
(b)
ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 u2
p
du ¼ 1
2 u
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 u2
p
þ 1
2 a2
sin1
u=a þ c; a 0.
5.52. Find
ð
x dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 2x þ 5
p : Ans.
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 2x þ 5
p
ln jx þ 1 þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 2x þ 5
p
j þ c.
5.53. Establish the validity of the method of integration by parts.
5.54. Evaluate (a)
ð
0
x cos 3x dx; ðbÞ
ð
x3
e2x
dx: Ans. (a) 2=9; ðbÞ 1
3 e2x
ð4x3
þ 6x2
þ 6x þ 3Þ þ c
5.55. Show that (a)
ð1
0
x2
tan1
x dx ¼
1
12
1
6
þ
1
6
ln 2
ðbÞ
ð2
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ x þ 1
p
dx ¼
5
ffiffiffi
7
p
4
þ
3
ffiffiffi
3
p
4
þ
3
8
ln
5 þ 2
ffiffiffi
7
p
2
ffiffiffi
3
p
3
.
5.56. (a) If u ¼ f ðxÞ and v ¼ gðxÞ have continuous nth derivatives, prove that
ð
uvðnÞ
dx ¼ uvðn1Þ
u0
vðn2Þ
þ u00
vðn3Þ
ð1Þn
ð
uðnÞ
v dx
called generalized integration by parts. (b) What simplifications occur if uðnÞ
¼ 0? Discuss. (c) Use (a) to
evaluate
ð
0
x4
sin x dx. Ans. (c) 4
122
þ 48
5.57. Show that
ð1
0
x dx
ðx þ 1Þ2
ðx2 þ 1Þ
¼
2
8
.
[Hint: Use partial fractions, i.e., assume
x
ðx þ 1Þ2
ðx2
þ 1Þ
¼
A
ðx þ 1Þ2
þ
B
x þ 1
þ
Cx þ D
x2 þ 1
and find A; B; C; D.]
5.58. Prove that
ð
0
dx
cos x
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1
p ; 1.
NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS
5.59. Evaluate
ð1
0
dx
1 þ x
approximately, using (a) the trapezoidal rule, (b) Simpson’s rule, taking n ¼ 4.
Compare with the exact value, ln 2 ¼ 0:6931.
5.60. Using (a) the trapezoidal rule, (b) Simpson’s rule evaluate
ð=2
0
sin2
x dx by obtaining the values of sin2
x
at x ¼ 08; 108; . . . ; 908 and compare with the exact value =4.
5.61. Prove the (a) rectangular rule, (b) trapezoidal rule, i.e., (16) and (17) of Page 98.
5.62. Prove Simpson’s rule.
CHAP. 5] INTEGRALS 113](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-122-320.jpg)
![114 INTEGRALS [CHAP. 5
5.63. Evaluate to 3 decimal places using numerical integration: (a)
ð2
1
dx
1 þ x2
; ðbÞ
ð1
0
cosh x2
dx.
Ans. (a) 0.322, (b) 1.105.
APPLICATIONS
5.64. Find the (a) area and (b) moment of inertia about the y-axis of the region in the xy plane bounded by
y ¼ sin x, 0 @ x @ and the x-axis, assuming unit density.
Ans. (a) 2, (b) 2
4
5.65. Find the moment of inertia about the x-axis of the region bounded by y ¼ x2
and y ¼ x, if the density is
proportional to the distance from the x-axis.
Ans. 1
8 M, where M ¼ mass of the region.
5.66. (a) Show that the arc length of the catenary y ¼ cosh x from x ¼ 0 to x ¼ ln 2 is 3
4. (b) Show that the length
of arc of y ¼ x3=2
, 2 @ x @ 5 is 343
27 2
ffiffiffi
2
p
113=2
.
5.67. Show that the length of one arc of the cycloid x ¼ að sin Þ, y ¼ að1 cos Þ, ð0 @ @ 2Þ is 8a.
5.68. Prove that the area bounded by the ellipse x2
=a2
þ y2
=b2
¼ 1 is ab.
5.69. (a) (Disk Method) Find the volume of the region obtained by revolving the curve y ¼ sin x, 0 @ x @ ,
about the x-axis. Ans: ðaÞ 2
=2
(b) (Disk Method) Show that the volume of the frustrum of a paraboloid obtained by revolving
f ðxÞ ¼
ffiffiffiffiffiffi
kx
p
, 0 a @ x @ b, about the x-axis is
ðb
a
kx dx ¼
k
2
ðb2
a2
Þ. (c) Determine the volume
obtained by rotating the region bound by f ðxÞ ¼ 3, gðxÞ ¼ 5 x2
on
ffiffiffi
2
p
@ x @
ffiffiffi
2
p
. (d) (Shell Method)
A spherical bead of radius a has a circular cylindrical hole of radius b, b a, through the center. Find the
volume of the remaining solid by the shell method. (e) (Shell Method) Find the volume of a solid whose
outer boundary is a torus (i.e., the solid is generated by orbiting a circle ðx aÞ2
þ y2
¼ b2
about the y-axis
(a b).
5.70. Prove that the centroid of the region bounded by y ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x2
p
, a @ x @ a and the x-axis is located at
ð0; 4a=3Þ.
5.71. (a) If ¼ f ðÞ is the equation of a curve in polar coordinates, show that the area bounded by this curve and
the lines ¼ 1 and ¼ 2 is
1
2
ð2
1
2
d. (b) Find the area bounded by one loop of the lemniscate
2
¼ a2
cos 2.
Ans. (b) a2
5.72. (a) Prove that the arc length of the curve in Problem 5.71(a) is
ð2
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
þ ðd=dÞ2
q
d. (b) Find the length
of arc of the cardioid ¼ að1 cos Þ.
Ans. (b) 8a
MISCELLANEOUS PROBLEMS
5.73. Establish the mean value theorem for derivatives from the first mean value theorem for integrals. [Hint: Let
f ðxÞ ¼ F 0
ðxÞ in (4), Page 93.]
5.74. Prove that (a) lim
!0þ
ð4
0
dx
ffiffiffiffiffiffiffiffiffiffiffi
4 x
p ¼ 4; ðbÞ lim
!0þ
ð3
dx
ffiffiffi
x
3
p ¼ 6; ðcÞ lim
!0þ
ð1
0
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p ¼
2
and give a geo-
metric interpretation of the results.
[These limits, denoted usually by
ð4
0
dx
ffiffiffiffiffiffiffiffiffiffiffi
4 x
p ;
ð3
0
dx
ffiffiffi
x
3
p and
ð1
0
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p respectively, are called impro-
per integrals of the second kind (see Problem 5.29) since the integrands are not bounded in the range of
integration. For further discussion of improper integrals, see Chapter 12.]
5.75. Prove that (a) lim
M!1
ðM
0
x5
ex
dx ¼ 4! ¼ 24; ðbÞ lim
!0þ
ð2
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xð2 xÞ
p ¼
2
.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-123-320.jpg)
![CHAP. 5] INTEGRALS 115
5.76. Evaluate (a)
ð1
0
dx
1 þ x3
; ðbÞ
ð=2
0
sin 2x
ðsin xÞ4=3
dx; ðcÞ
ð1
0
dx
x þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1
p .
Ans. (a)
2
3
ffiffiffi
3
p ðbÞ 3 ðcÞ does not exist
5.77. Evaluate lim
x!=2
ex2
= e=4 þ
Ð=2
x esin t
dt
1 þ cos 2x
. Ans. e=2
5.78. Prove: (a)
d
dx
ðx3
x2
ðt2
þ t þ 1Þ dt ¼ 3x3
þ x5
2x3
þ 3x2
2x; ðb
d
dx
ðx2
x
cos t2
dt ¼ 2x cos x4
cos x2
.
5.79. Prove that (a)
ð
0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ sin x
p
dx ¼ 4; ðbÞ
ð=2
0
dx
sin x þ cos x
¼
ffiffiffi
2
p
lnð
ffiffiffi
2
p
þ 1Þ.
5.80. Explain the fallacy: I ¼
ð1
1
dx
1 þ x2
¼
ð1
1
dy
1 þ y2
¼ I, using the transformation x ¼ 1=y. Hence I ¼ 0.
But I ¼ tan1
ð1Þ tan1
ð1Þ ¼ =4 ð=4Þ ¼ =2. Thus =2 ¼ 0.
5.81. Prove that
ð1=2
0
cos x
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ x2
p dx @
1
4
tan1 1
2
.
5.82. Evaluate lim
n!1
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p
þ
ffiffiffiffiffiffiffiffiffiffiffi
n þ 2
p
þ þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
2n 1
p
n3=2
( )
. Ans. 2
3 ð2
ffiffiffi
2
p
1Þ
5.83. Prove that f ðxÞ ¼
1 if x is irrational
0 if x is rational
is not Riemann integrable in ½0; 1.
[Hint: In (2), Page 91, let k, k ¼ 1; 2; 3; . . . ; n be first rational and then irrational points of subdivision and
examine the lower and upper sums of Problem 5.31.]
5.84. Prove the result (3) of Problem 5.31. [Hint: First consider the effect of only one additional point of
subdivision.]
5.85. In Problem 5.31, prove that
s
s @ S. [Hint: Assume the contrary and obtain a contradiction.]
5.86. If f ðxÞ is sectionally continuous in ½a; b, prove that
ðb
a
f ðxÞ dx exists. [Hint: Enclose each point of disconti-
nuity in an interval, noting that the sum of the lengths of such intervals can be made arbitrarily small. Then
consider the difference between the upper and lower sums.
5.87. If f ðxÞ ¼
2x 0 x 1
3 x ¼ 1
6x 1 1 x 2
8
:
, find
ð2
0
f ðxÞ dx. Interpret the result graphically. Ans. 9
5.88. Evaluate
ð3
0
fx ½x þ 1
2g dx where ½x denotes the greatest integer less than or equal to x. Interpret the result
graphically. Ans. 3
5.89. (a) Prove that
ð=2
0
sinm
x
sinm
x þ cosm x
dx ¼
4
for all real values of m.
(b) Prove that
ð2
0
dx
1 þ tan4 x
¼ .
5.90. Prove that
ð=2
0
sin x
x
dx exists.
5.91. Show that
ð0:5
0
tan1
x
x
dx ¼ 0:4872 approximately.
5.92. Show that
ð
0
x dx
1 þ cos2 x
¼
2
2
ffiffiffi
2
p :](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-124-320.jpg)
![116
Partial Derivatives
FUNCTIONS OF TWO OR MORE VARIABLES
The definition of a function was given in Chapter 3 (page 39). For us the distinction for functions of
two or more variables is that the domain is a set of n-tuples of numbers. The range remains one
dimensional and is referred to an interval of numbers. If n ¼ 2, the domain is pictured as a two-
dimensional region. The region is referred to a rectangular Cartesian coordinate system described
through number pairs ðx; yÞ, and the range variable is usually denoted by z. The domain variables are
independent while the range variable is dependent.
We use the notation f ðx; yÞ, Fðx; yÞ, etc., to denote the value of the function at ðx; yÞ and write
z ¼ f ðx; yÞ, z ¼ Fðx; yÞ, etc. We shall also sometimes use the notation z ¼ zðx; yÞ although it should be
understood that in this case z is used in two senses, namely as a function and as a variable.
EXAMPLE. If f ðx; yÞ ¼ x2
þ 2y3
, then f ð3; 1Þ ¼ ð3Þ2
þ 2ð1Þ3
¼ 7:
The concept is easily extended. Thus w ¼ Fðx; y; zÞ denotes the value of a function at ðx; y; zÞ [a
point in three-dimensional space], etc.
EXAMPLE. If z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ðx2 þ y2Þ
p
, the domain for which z is real consists of the set of points ðx; yÞ such that
x2
þ y2
@ 1, i.e., the set of points inside and on a circle in the xy plane having center at ð0; 0Þ and radius 1.
THREE-DIMENSIONAL RECTANGULAR COORDINATE SYSTEMS
A three-dimensional rectangular coordinate system, as referred to in the previous paragraph,
obtained by constructing three mutually perpendicular axes (the x-, y-, and z-axes) intersecting in
point O (the origin). It forms a natural extension of the usual xy plane for representing functions of
two variables graphically. A point in three dimensions is represented by the triplet ðx; y; zÞ called
coordinates of the point. In this coordinate system z ¼ f ðx; yÞ [or Fðx; y; zÞ ¼ 0] represents a surface,
in general.
EXAMPLE. The set of points ðx; y; zÞ such that z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ðx2
þ y2
Þ
p
comprises the surface of a hemisphere of radius
1 and center at ð0; 0; 0Þ.
For functions of more than two variables such geometric interpretation fails, although the termi-
nology is still employed. For example, ðx; y; z; wÞ is a point in four-dimensional space, and w ¼ f ðx; y; zÞ
[or Fðx; y; z; wÞ ¼ 0] represents a hypersurface in four dimensions; thus x2
þ y2
þ z2
þ w2
¼ a2
represents
a hypersphere in four dimensions with radius a 0 and center at ð0; 0; 0; 0Þ. w ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 ðx2 þ y2 þ z2Þ
p
,
x2
þ y2
þ z2
@ a2
describes a function generated from the hypersphere.
Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-125-320.jpg)
![NEIGHBORHOODS
The set of all points ðx; yÞ such that jx x0j , j y y0j where 0, is called a rectangular
neighborhood of ðx0; y0Þ; the set 0 jx x0j , 0 j y y0j which excludes ðx0; y0Þ is called a
rectangular deleted neighborhood of ðx0; y0Þ. Similar remarks can be made for other neighborhoods,
e.g., ðx x0Þ2
þ ð y y0Þ2
2
is a circular neighborhood of ðx0; y0Þ. The term ‘‘open ball’’ is used to
designate this circular neighborhood. This terminology is appropriate for generalization to more
dimensions. Whether neighborhoods are viewed as circular or square is immaterial, since the descrip-
tions are interchangeable. Simply notice that given an open ball (circular neighborhood) of radius
there is a centered square whose side is of length less than
ffiffiffi
2
p
that is interior to the open ball, and
conversely for a square of side there is an interior centered of radius of radius less than =2. (See Fig.
6-1.)
A point ðx0; y0Þ is called a limit point, accumulation point, or cluster point of a point set S if every
deleted neighborhood of ðx0; y0Þ contains points of S. As in the case of one-dimensional point sets,
every bounded infinite set has at least one limit point (the Bolzano–Weierstrass theorem, see Pages 6 and
12). A set containing all its limit points is called a closed set.
REGIONS
A point P belonging to a point set S is called an interior point of S if there exists a deleted
neighborhood of P all of whose points belong to S. A point P not belonging to S is called an exterior
point of S if there exists a deleted neighborhood of P all of whose points do not belong to S. A point P
is called a boundary point of S if every deleted neighborhood of P contains points belonging to S and
also points not belonging to S.
If any two points of a set S can be joined by a path consisting of a finite number of broken line
segments all of whose points belong to S, then S is called a connected set. A region is a connected set
which consists of interior points or interior and boundary points. A closed region is a region containing
all its boundary points. An open region consists only of interior points. The complement of a set, S, in
the xy plane is the set of all points in the plane not belonging to S. (See Fig. 6-2.)
Examples of some regions are shown graphically in Figs 6-3(a), (b), and (c) below. The rectangular
region of Fig. 6-1(a), including the boundary, represents the sets of points a @ x @ b, c @ y @ d which
is a natural extension of the closed interval a @ x @ b for one dimension. The set a x b, c y d
corresponds to the boundary being excluded.
In the regions of Figs 6-3(a) and 6-3(b), any simple closed curve (one which does not intersect itself
anywhere) lying inside the region can be shrunk to a point which also lies in the region. Such regions are
called simply-connected regions. In Fig. 6-3(c) however, a simple closed curve ABCD surrounding one of
the ‘‘holes’’ in the region cannot be shrunk to a point without leaving the region. Such regions are called
multiply-connected regions.
CHAP. 6] PARTIAL DERIVATIVES 117
Fig. 6-1 Fig. 6-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-126-320.jpg)
![LIMITS
Let f ðx; yÞ be defined in a deleted neighborhood of ðx0; y0Þ [i.e.; f ðx; yÞ may be undefined at
ðx0; y0Þ]. We say that l is the limit of f ðx; yÞ as x approaches x0 and y approaches y0 [or ðx; yÞ
approaches ðx0; y0Þ] and write lim
x!x0
y!y0
f ðx; yÞ ¼ l [or lim
ðx;yÞ!ðx0;y0Þ
f ðx; yÞ ¼ l] if for any positive number we
can find some positive number [depending on and ðx0; y0Þ, in general] such that j f ðx; yÞ lj
whenever 0 jx x0j and 0 j y y0j .
If desired we can use the deleted circular neighborhood open ball 0 ðx x0Þ2
þ ð y y0Þ2
2
instead of the deleted rectangular neighborhood.
EXAMPLE. Let f ðx; yÞ ¼
3xy if ðx; yÞ 6¼ ð1; 2Þ
0 if ðx; yÞ ¼ ð1; 2Þ
. As x ! 1 and y ! 2 [or ðx; yÞ ! ð1; 2Þ], f ðx; yÞ gets closer to
3ð1Þð2Þ ¼ 6 and we suspect that lim
x!1
y!2
f ðx; yÞ ¼ 6. To prove this we must show that the above definition of limit with
l ¼ 6 is satisfied. Such a proof can be supplied by a method similar to that of Problem 6.4.
Note that lim
x!1
y!2
f ðx; yÞ 6¼ f ð1; 2Þ since f ð1; 2Þ ¼ 0. The limit would in fact be 6 even if f ðx; yÞ were not defined at
ð1; 2Þ. Thus the existence of the limit of f ðx; yÞ as ðx; yÞ ! ðx0; y0Þ is in no way dependent on the existence of a value
of f ðx; yÞ at ðx0; y0Þ.
Note that in order for lim
ðx;yÞ!ðx0;y0Þ
f ðx; yÞ to exist, it must have the same value regardless of the
approach of ðx; yÞ to ðx0; y0Þ. It follows that if two different approaches give different values, the
limit cannot exist (see Problem 6.7). This implies, as in the case of functions of one variable, that if a
limit exists it is unique.
The concept of one-sided limits for functions of one variable is easily extended to functions of more
than one variable.
EXAMPLE 1. lim
x!0þ
y!1
tan1
ð y=xÞ ¼ =2, lim
x!0
y!1
tan1
ð y=xÞ ¼ =2.
EXAMPLE 2. lim
x!0
y!1
tan1
ð y=xÞ does not exist, as is clear from the fact that the two different approaches of Example
1 give different results.
In general the theorems on limits, concepts of infinity, etc., for functions of one variable (see Page
21) apply as well, with appropriate modifications, to functions of two or more variables.
118 PARTIAL DERIVATIVES [CHAP. 6
Fig. 6-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-127-320.jpg)
![ITERATED LIMITS
The iterated limits lim
x!x0
lim
y!y0
f ðx; yÞ
and lim
y!y0
lim
x!x0
f ðx; yÞ
, [also denoted by lim
x!x0
lim
y!y0
f ðx; yÞ and
lim
y!y0
lim
x!x0
f ðx; yÞ respectively] are not necessarily equal. Although they must be equal if lim
x!x0
y!y0
f ðx; yÞ is to
exist, their equality does not guarantee the existence of this last limit.
EXAMPLE. If f ðx; yÞ ¼
x y
x þ y
, then lim
x!0
lim
y!0
x y
x þ y
¼ lim
x!0
ð1Þ ¼ 1 and lim
y!0
lim
x!0
x y
x þ y
¼ lim
y!0
ð1Þ ¼ 1. Thus
the iterated limits are not equal and so lim
x!0
y!0
f ðx; yÞ cannot exist.
CONTINUITY
Let f ðx; yÞ be defined in a neighborhood of ðx0; y0Þ [i.e.; f ðx; yÞ must be defined at ðx0; y0Þ as well as
near it]. We say that f ðx; yÞ is continuous at ðx0; y0Þ if for any positive number we can find some
positive number [depending on and ðx0; y0Þ in general] such that j f ðx; yÞ f ðx0; y0Þj whenever
jx x0j and jy y0j , or alternatively ðx x0Þ2
þ ð y y0Þ2
2
.
Note that three conditions must be satisfied in order that f ðx; yÞ be continuous at ðx0; y0Þ.
1. lim
ðx;yÞ!ðx0;y0Þ
f ðx; yÞ ¼ l, i.e., the limit exists as ðx; yÞ ! ðx0; y0Þ
2. f ðx0; y0Þ must exist, i.e., f ðx; yÞ is defined at ðx0; y0Þ
3. l ¼ f ðx0; y0Þ
If desired we can write this in the suggestive form lim
x!x0
y!y0
f ðx; yÞ ¼ f ð lim
x!x0
x; lim
y!y0
yÞ.
EXAMPLE. If f ðx; yÞ ¼
3xy ðx; yÞ 6¼ ð1; 2Þ
0 ðx; yÞ ¼ ð1; 2Þ
, then lim
ðx;yÞ!ð1;2Þ
f ðx; yÞ ¼ 6 6¼ f ð1; 2Þ. Hence, f ðx; yÞ is not contin-
uous at ð1; 2Þ. If we redefine the function so that f ðx; yÞ ¼ 6 for ðx; yÞ ¼ ð1; 2Þ, then the function is continuous at
ð1; 2Þ.
If a function is not continuous at a point ðx0; y0Þ, it is said to be discontinuous at ðx0; y0Þ which is then
called a point of discontinuity. If, as in the above example, it is possible to redefine the value of a
function at a point of discontinuity so that the new function is continuous, we say that the point is a
removable discontinuity of the old function. A function is said to be continuous in a region r of the xy
plane if it is continuous at every point of r.
Many of the theorems on continuity for functions of a single variable can, with suitable modifica-
tion, be extended to functions of two more variables.
UNIFORM CONTINUITY
In the definition of continuity of f ðx; yÞ at ðx0; y0Þ, depends on and also ðx0; y0Þ in general. If in a
region r we can find a which depends only on but not on any particular point ðx0; y0Þ in r [i.e., the
same will work for all points in r], then f ðx; yÞ is said to be uniformly continuous in r. As in the case
of functions of one variable, it can be proved that a function which is continuous in a closed and
bounded region is uniformly continuous in the region.
PARTIAL DERIVATIVES
The ordinary derivative of a function of several variables with respect to one of the independent
variables, keeping all other independent variables constant, is called the partial derivative of the function
with respect to the variable. Partial derivatives of f ðx; yÞ with respect to x and y are denoted by
CHAP. 6] PARTIAL DERIVATIVES 119](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-128-320.jpg)
![The form dz ¼ fxðx0; y0Þdx þ fyðx0; y0Þdy signifies a linear function with the independent variables dx
and dy and the dependent range variable dz. In the one variable case, the corresponding linear function
represents the tangent line to the underlying curve. In this case, the underlying entity is a surface and
the linear function generates the tangent plane at P0. In a small enough neighborhood, this tangent
plane is an approximation of the surface (i.e., the linear representation of the surface at P0). If y is held
constant, then one obtains the curve of intersection of the surface and the coordinate plane y ¼ y0. The
differential form reduces to dz ¼ fxðx0; y0Þdx (i.e., the one variable case). A similar statement follows
when x is held constant. See Fig. 6-4.
If f is such that f (or zÞ can be expressed in the form (4) where 1 and 2 approach zero as x and
y approach zero, we call f differentiable at ðx; yÞ. The mere existence of fx and fy does not in itself
guarantee differentiability; however, continuity of fx and fy does (although this condition happens to be
slightly stronger than necessary). In case fx and fy are continuous in a region r, we shall say that f is
continuously differentiable in r.
THEOREMS ON DIFFERENTIALS
In the following we shall assume that all functions have continuous first partial derivatives in a
region r, i.e., the functions are continuously differentiable in r.
1. If z ¼ f ðx1; x2; . . . ; xnÞ, then
df ¼
@f
@x1
dx1 þ
@f
@x2
dx2 þ þ
@f
@xn
dxn ð6Þ
regardless of whether the variables x1; x2; . . . ; xn are independent or dependent on other vari-
ables (see Problem 6.20). This is a generalization of the result (5). In (6) we often use z in place
of f .
2. If f ðx1; x2; . . . ; xnÞ ¼ c, a constant, then df ¼ 0. Note that in this case x1; x2; . . . ; xn cannot all
be independent variables.
CHAP. 6] PARTIAL DERIVATIVES 121
Fig. 6-4](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-130-320.jpg)
![IMPLICIT FUNCTIONS
In general, an equation such as Fðx; y; zÞ ¼ 0 defines one variable, say z, as a function of the other
two variables x and y. Then z is sometimes called an implicit function of x and y, as distinguished from a
so-called explicit function f, where z ¼ f ðx; yÞ, which is such that F½x; y; f ðx; yÞ 0.
Differentiation of implicit functions requires considerable discipline in interpreting the independent
and dependent character of the variables and in distinguishing the intent of one’s notation. For
example, suppose that in the implicit equation F½x; y; f ðx; zÞ ¼ 0, the independent variables are x and
y and that z ¼ f ðx; yÞ. In order to find
@f
@x
and
@f
@y
, we initially write (observe that Fðx; t; zÞ is zero for all
domain pairs ðx; yÞ, in other words it is a constant):
0 ¼ dF ¼ Fx dx þ Fy dy þ Fz dz
and then compute the partial derivatives Fx; Fy; Fz as though y; y; z constituted an independent set of
variables. At this stage we invoke the dependence of z on x and y to obtain the differential form
dz ¼
@f
@x
dx þ
@f
@y
dy. Upon substitution and some algebra (see Problem 6.30) the following results are
obtained:
@f
@x
¼
Fx
Fz
;
@f
@y
¼
Fy
Fz
EXAMPLE. If 0 ¼ Fðx; y; zÞ ¼ x2
z þ yz2
þ 2xy2
z3
and z ¼ f ðx; yÞ then Fx ¼ 2xz þ 2y2
, Fy ¼ z2
þ 4xy.
Fz ¼ x2
þ 2yz 3z2
. Then
@f
@x
¼
ð2xz þ 2y2
Þ
x2 þ 2yz 3z2
;
@f
@y
¼
ðz2
þ 4xyÞ
x2 þ 2yz 3x2
Observe that f need not be known to obtain these results. If that information is available then (at
least theoretically) the partial derivatives may be expressed through the independent variables x and y.
JACOBIANS
If Fðu; vÞ and Gðu; vÞ are differentiable in a region, the Jacobian determinant, or briefly the Jacobian,
of F and G with respect to u and v is the second order functional determinant defined by
@ðF; GÞ
@ðu; vÞ
¼
@F
@u
@F
@v
@G
@u
@G
@v
¼
Fu Fv
Gu Gv
ð7Þ
Similarly, the third order determinant
@ðF; G; HÞ
@ðu; v; wÞ
¼
Fu Fv Fw
Gu Gv Gw
Hu Hv Hw
is called the Jacobian of F, G, and H with respect to u, v, and w. Extensions are easily made.
PARTIAL DERIVATIVES USING JACOBIANS
Jacobians often prove useful in obtaining partial derivatives of implicit functions. Thus, for
example, given the simultaneous equations
Fðx; y; u; vÞ ¼ 0; Gðx; y; u; vÞ ¼ 0
CHAP. 6] PARTIAL DERIVATIVES 123](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-132-320.jpg)
![Under the transformation (10) a closed region r of the xy plane is, in general, mapped into a closed
region r0
of the uv plane. Then if Axy and Auv denote respectively the areas of these regions, we can
show that
lim
Axy
Auv
¼
@ðx; yÞ
@ðu; vÞ
ð11Þ
where lim denotes the limit as Axy (or Auv) approaches zero. The Jacobian on the right of (11) is
often called the Jacobian of the transformation (10).
If we solve (10) for u and v in terms of x and y, we obtain the transformation u ¼ f ðx; yÞ, v ¼ gðx; yÞ
often called the inverse transformation corresponding to (10). The Jacobians
@ðu; vÞ
@ðx; yÞ
and
@ðx; yÞ
@ðu; vÞ
of these
transformations are reciprocals of each other (see Problem 6.43). Hence, if one Jacobian is different
from zero in a region, so also is the other.
The above ideas can be extended to transformations in three or higher dimensions. We shall deal
further with these topics in Chapter 7, where use is made of the simplicity of vector notation and
interpretation.
CURVILINEAR COORDINATES
If ðx; yÞ are the rectangular coordinates of a point in the xy plane, we can think of ðu; vÞ as also
specifying coordinates of the same point, since by knowing ðu; vÞ we can determine ðx; yÞ from (10). The
coordinates ðu; vÞ are called curvilinear coordinates of the point.
EXAMPLE. The polar coordinates ð; Þ of a point correspond to the case u ¼ , v ¼ . In this case the
transformation equations (10) are x ¼ cos , y ¼ sin .
For curvilinear coordinates in higher dimensional spaces, see Chapter 7.
MEAN VALUE THEOREM
If f ðx; yÞ is continuous in a closed region and if the first partial derivatives exist in the open region
(i.e., excluding boundary points), then
f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ 0 1 ð12Þ
This is sometimes written in a form in which h ¼ x ¼ x x0 and k ¼ y ¼ y y0.
Solved Problems
FUNCTIONS AND GRAPHS
6.1. If f ðx; yÞ ¼ x3
2xy þ 3y2
, find: (a) f ð2; 3Þ; ðbÞ f
1
x
;
2
y
; ðcÞ
f ðx; y þ kÞ f ðx; yÞ
k
;
k 6¼ 0.
ðaÞ f ð2; 3Þ ¼ ð2Þ3
2ð2Þð3Þ þ 3ð3Þ2
¼ 8 þ 12 þ 27 ¼ 31
ðbÞ f
1
x
;
2
y
¼
1
x
3
2
1
x
2
y
þ 3
2
y
2
¼
1
x3
4
xy
þ
12
y2
CHAP. 6] PARTIAL DERIVATIVES 125](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-134-320.jpg)
![Trace on xz plane ðy ¼ 0Þ is the hyperbola
x2
a2
z2
c2
¼ 1, y ¼ 0.
Trace on any plane z ¼ p parallel to the xy plane is the ellipse
x2
a2ð1 þ p2=c2Þ
þ
y2
b2ð1 þ p2=c2Þ
¼ 1
As j pj increases from zero, the elliptic cross section increases in size.
The surface is a hyperboloid of one sheet (see Fig. 6-8).
LIMITS AND CONTINUITY
6.4. Prove that lim
x!1
y!2
ðx2
þ 2yÞ ¼ 5.
Method 1, using definition of limit.
We must show that given any 0, we can find 0 such that jx2
þ 2y 5j when 0 jx 1j ,
0 j y 2j .
If 0 jx 1j and 0 j y 2j , then 1 x 1 þ and 2 y 2 þ , excluding
x ¼ 1; y ¼ 2.
Thus, 1 2 þ 2
x2
1 þ 2 þ 2
and 4 2 2y 4 þ 2. Adding,
5 4 þ 2
x2
þ 2y 5 þ 4 þ 2
or 4 þ 2
x2
þ 2y 5 4 þ 2
Now if @ 1, it certainly follows that 5 x2
þ 2y 5 5, i.e., jx2
þ 2y 5j 5 whenever
0 jx 1j , 0 j y 2j . Then choosing 5 ¼ , i.e., ¼ =5 (or ¼ 1, whichever is smaller), it
follows that jx2
þ 2y 5j when 0 jx 1j , 0 j y 2j , i.e., lim
x!1
y!2
ðx2
þ 2yÞ ¼ 5.
Method 2, using theorems on limits.
lim
x!1
y!2
ðx2
þ 2yÞ ¼ lim
x!1
y!2
x2
þ lim
x!1
y!2
2y ¼ 1 þ 4 ¼ 5
6.5. Prove that f ðx; yÞ ¼ x2
þ 2y is continuous at ð1; 2Þ.
By Problem 6.4, lim
x!1
y!2
f ðx; yÞ ¼ 5. Also, f ð1; 2Þ ¼ 12
þ 2ð2Þ ¼ 5.
Then lim
x!1
y!2
f ðx; yÞ ¼ f ð1; 2Þ and the function is continuous at ð1; 2Þ.
Alternatively, we can show, in much the same manner as in the first method of Problem 6.4, that given
any 0 we can find 0 such that j f ðx; yÞ f ð1; 2Þj when jx 1j ; j y 2j .
6.6. Determine whether f ðx; yÞ ¼
x2
þ 2y; ðx; yÞ 6¼ ð1; 2Þ
0; ðx; yÞ ¼ ð1; 2Þ
.
(a) has a limit as x ! 1 and y ! 2, (b) is continuous at ð1; 2Þ.
(a) By Problem 6.4, it follows that lim
x!1
y!2
f ðx; yÞ ¼ 5, since the limit has nothing to do with the value at ð1; 2Þ.
(b) Since lim
x!1
y!2
f ðx; yÞ ¼ 5 and f ð1; 2Þ ¼ 0, it follows that lim
x!1
y!2
f ðx; yÞ 6¼ f ð1; 2Þ. Hence, the function is
discontinuous at ð1; 2Þ:
6.7. Investigate the continuity of f ðx; yÞ ¼
x2
y2
x2 þ y2
ðx; yÞ 6¼ ð0; 0Þ
0 ðx; yÞ ¼ ð0; 0Þ
8
:
at ð0; 0Þ.
Let x ! 0 and y ! 0 in such a way that y ¼ mx (a line in the xy plane). Then along this line,
lim
x!0
y!0
x2
y2
x2
þ y2
¼ lim
x!0
x2
m2
x2
x2
þ m2
x2
¼ lim
x!0
x2
ð1 m2
Þ
x2
ð1 þ m2
Þ
¼
1 m2
1 þ m2
CHAP. 6] PARTIAL DERIVATIVES 127
Fig. 6-8](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-136-320.jpg)
![ðaÞ x ¼
@
@x
¼
@
@x
ðx3
y þ exy2
Þ ¼ 3x2
y þ exy2
y2
¼ 3x2
y þ y2
exy2
ðbÞ y ¼
@
@y
¼
@
@y
ðx3
y þ exy2
Þ ¼ x3
þ exy2
2xy ¼ x3
þ 2xy exy2
ðcÞ xx ¼
@2
@x2
¼
@
@x
@
@x
¼
@
@x
ð3x2
y þ y2
exy2
Þ ¼ 6xy þ y2
ðexy2
y2
Þ ¼ 6xy þ y4
exy2
ðdÞ yy ¼
@2
@y2
¼
@
@y
ðx3
þ 2xy exy2
Þ ¼ 0 þ 2xy
@
@y
ðexy2
Þ þ exy2 @
@y
ð2xyÞ
¼ 2xy exy2
2xy þ exy2
2x ¼ 4x2
y2
exy2
þ 2x exy2
ðeÞ xy ¼
@2
@y @x
¼
@
@y
@
@x
¼
@
@y
ð3x2
y þ y2
exy2
Þ ¼ 3x2
þ y2
exy2
2xy þ exy2
2y
¼ 3x2
þ 2xy3
exy2
þ 2y exy2
ð f Þ yx ¼
@2
@x @y
¼
@
@x
@
@y
¼
@
@x
ðx3
þ 2xy exy2
Þ ¼ 3x2
þ 2xy exy2
y2
þ exy2
2y
¼ 3x2
þ 2xy3
exy2
þ 2y exy2
Note that xy ¼ yx in this case. This is because the second partial derivatives exist and are
continuous for all ðx; yÞ in a region r. When this is not true we may have xy 6¼ yx (see Problem 6.41,
for example).
6.11. Show that Uðx; y; zÞ ¼ ðx2
þ y2
þ z2
Þ1=2
satisfies Laplace’s partial differential equation
@2
U
@x2
þ
@2
U
@y2
þ
@2
U
@z2
¼ 0.
We assume here that ðx; y; zÞ 6¼ ð0; 0; 0Þ. Then
@U
@x
¼ 1
2 ðx2
þ y2
þ z2
Þ3=2
2x ¼ xðx2
þ y2
þ z2
Þ3=2
@2
U
@x2
¼
@
@x
½xðx2
þ y2
þ z2
Þ3=2
¼ ðxÞ½ 3
2 ðx2
þ y2
þ z2
Þ5=2
2x þ ðx2
þ y2
þ z2
Þ3=2
ð1Þ
¼
3x2
ðx2 þ y2 þ z2Þ5=2
ðx2
þ y2
þ z2
Þ
ðx2 þ y2 þ z2Þ5=2
¼
2x2
y2
z2
ðx2 þ y2 þ z2Þ5=2
@2
U
@y2
¼
2y2
x2
z2
ðx2 þ y2 þ z2Þ5=2
;
@2
U
@x2
¼
2z2
x2
y2
ðx2 þ y2 þ z2Þ5=2
:
Similarly
@2
U
@x2
þ
@2
U
@y2
þ
@2
U
@z2
¼ 0:
Adding,
6.12. If z ¼ x2
tan1 y
x
, find
@2
z
@x @y
at ð1; 1Þ.
@z
@y
¼ x2
1
1 þ ð y=xÞ2
@
@y
y
x
¼ x2
x2
x2
þ y2
1
x
¼
x3
x2
þ y2
CHAP. 6] PARTIAL DERIVATIVES 129](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-138-320.jpg)
![Solution:
ðaÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ
¼ fðx þ xÞ2
ð y þ yÞ 3ð y þ yÞg fx2
y 3yg
¼ 2xy x þ ðx2
3Þy
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
ðAÞ
þ ðxÞ2
y þ 2x x y þ ðxÞ2
y
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
ðBÞ
The sum (A) is the principal part of z and is the differential of z, i.e., dz. Thus,
ðbÞ dz ¼ 2xy x þ ðx2
3Þy ¼ 2xy dx þ ðx2
3Þ dy
Another method: dz ¼
@z
@x
dx þ
@z
@y
dy ¼ 2xy dx þ ðx2
3Þ dy
ðcÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ ¼ f ð4 0:01; 3 þ 0:02Þ f ð4; 3Þ
¼ fð3:99Þ2
ð3:02Þ 3ð3:02Þg fð4Þ2
ð3Þ 3ð3Þg ¼ 0:018702
dz ¼ 2xy dx þ ðx2
3Þ dy ¼ 2ð4Þð3Þð0:01Þ þ ð43
3Þð0:02Þ ¼ 0:02
Note that in this case z and dz are approximately equal, because x ¼ dx and y ¼ dy are
sufficiently small.
(d) We must find f ðx þ x; y þ yÞ when x þ x ¼ 5:12 and y ¼ y ¼ 6:85. We can accomplish this by
choosing x ¼ 5, x ¼ 0:12, y ¼ 7, y ¼ 0:15. Since x and y are small, we use the fact that
f ðx þ x; y þ yÞ ¼ f ðx; yÞ þ z is approximately equal to f ðx; yÞ þ dz, i.e., z þ dz.
Now z ¼ f ðx; yÞ ¼ f ð5; 7Þ ¼ ð5Þ2
ð7Þ 3ð7Þ ¼ 154
dz ¼ 2xy dx þ ðx2
3Þ dy ¼ 2ð5Þð7Þð0:12Þ þ ð52
3Þð0:15Þ ¼ 5:1:
Then the required value is 154 þ 5:1 ¼ 159:1 approximately. The value obtained by direct com-
putation is 159.01864.
6.16. (a) Let U ¼ x2
ey=x
. Find dU. (b) Show that ð3x2
y 2y2
Þ dx þ ðx3
4xy þ 6y2
Þ dy can be
written as an exact differential of a function ðx; yÞ and find this function.
(a) Method 1:
@U
@x
¼ x2
ey=x
y
x2
þ 2xey=x
;
@U
@y
¼ x2
ey=x 1
x
dU ¼
@U
@x
dx þ
@U
@y
dy ¼ ð2xey=x
yey=x
Þ dx þ xey=x
dy
Then
Method 2:
dU ¼ x2
dðey=x
Þ þ ey=x
dðx2
Þ ¼ x2
ey=x
dðy=xÞ þ 2xey=x
dx
¼ x2
ey=x x dy y dx
x2
þ 2xey=x
dx ¼ ð2xey=x
yey=x
Þ dx þ xey=x
dy
(b) Method 1:
Suppose that ð3x2
y 2y2
Þ dx þ ðx3
4xy þ 6y2
Þ dy ¼ d ¼
@
@x
dx þ
@
@y
dy:
Then (1)
@
@x
¼ 3x2
y 2y2
; (2)
@
@y
¼ x3
4xy þ 6y2
From (1), integrating with respect to x keeping y constant, we have
¼ x3
y ¼ 2xy2
þ FðyÞ
CHAP. 6] PARTIAL DERIVATIVES 131](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-140-320.jpg)
![6.20. Prove that dz ¼
@z
@x
dx þ
@z
@y
dy even if x and y are dependent variables.
Suppose x and y depend on three variables u; v; w, for example. Then
ð1Þ dx ¼ xu du þ xv dv þ xw dw ð2Þ dy ¼ yu du þ yv dv þ yw dw
zx dx þ zy dy ¼ ðzxxu þ zyyuÞ du þ ðzxxv þ zyyvÞ dv þ ðzxxw þ zyywÞ dw
Thus,
¼ zu du þ zv dv þ zw dw ¼ dz
using obvious generalizations of Problem 6.19.
6.21. If T ¼ x3
xy þ y3
, x ¼ cos , y ¼ sin , find (a) @T=@, (b) @T=@.
@T
@
¼
@T
@x
@x
@
þ
@T
@y
@y
@
¼ ð3x2
yÞðcos Þ þ ð3y2
xÞðsin Þ
@T
@
¼
@T
@x
@x
@
þ
@T
@y
@y
@
¼ ð3x2
yÞð sin Þ þ ð3y2
xÞð cos Þ
This may also be worked by direct substitution of x and y in T.
6.22. If U ¼ z sin y=x where x ¼ 3r2
þ 2s, y ¼ 4r 2s3
, z ¼ 2r2
3s2
, find (a) @U=@r; ðbÞ @U=@s.
ðaÞ
@U
@r
¼
@U
@x
@x
@r
þ
@U
@y
@y
@r
þ
@U
@z
@z
@r
¼ z cos
y
x
y
x2
ð6rÞ þ z cos
y
x
1
x
ð4Þ þ sin
y
x
ð4rÞ
¼
6ryz
x2
cos
y
x
þ
4z
x
cos
y
x
þ 4r sin
y
x
ðbÞ
@U
@s
¼
@U
@x
@x
@s
þ
@U
@y
@y
@s
þ
@U
@z
@z
@s
¼ z cos
y
x
y
x2
ð2Þ þ z cos
y
x
1
x
ð6s2
Þ þ sin
y
x
ð6sÞ
¼
2yz
x2
cos
y
x
6s2
z
x
cos
y
x
6s sin
y
x
6.23. If x ¼ cos , y ¼ sin , show that
@V
@x
2
þ
@V
@y
2
¼
@V
@
2
þ
1
2
@V
@
2
.
Using the subscript notation for partial derivatives, we have
V ¼ Vxx þ Vyy ¼ Vx cos þ Vy sin ð1Þ
V ¼ Vxx þ Vyy ¼ Vxð sin Þ þ Vyð cos Þ ð2Þ
Dividing both sides of (2) by , we have
1
V ¼ Vx sin þ Vy cos ð3Þ
Then from (1) and (3), we have
V2
þ
1
2
V2
¼ ðVx cos þ Vy sin Þ2
þ ðVx sin þ Vy cos Þ2
¼ V2
x þ V2
y
6.24. Show that z ¼ f ðx2
yÞ, where f is differentiable, satisfies xð@z=@xÞ ¼ 2yð@z=@yÞ.
Let x2
y ¼ u. Then z ¼ f ðuÞ. Thus
CHAP. 6] PARTIAL DERIVATIVES 133](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-142-320.jpg)
![6.28. If x ¼ 2r s and y ¼ r þ 2s, find
@2
U
@y @x
in terms of derivatives with respect to r and s.
Solving x ¼ 2r s, y ¼ r þ 2s for r and s: r ¼ ð2x þ yÞ=5, s ¼ ð2y xÞ=5.
Then @r=@x ¼ 2=5, @s=@x ¼ 1=5, @r=@y ¼ 1=5, @s=@y ¼ 2=5. Hence we have
@U
@x
¼
@U
@r
@r
@x
þ
@U
@s
@s
@x
¼
2
5
@U
@r
1
5
@U
@s
@2
U
@y @x
¼
@
@y
@U
@x
¼
@
@r
2
5
@U
@r
1
5
@U
@s
@r
@y
þ
@
@s
2
5
@U
@r
1
5
@U
@s
@s
@y
¼
2
5
@2
U
@r2
1
5
@2
U
@r @s
!
1
5
þ
2
5
@2
U
@s @r
1
5
@2
U
@s2
!
2
5
¼
1
25
2
@2
U
@r2
þ 3
@2
U
@r @s
2
@2
U
@s2
!
assuming U has continuous second partial derivatives.
IMPLICIT FUNCTIONS AND JACOBIANS
6.29. If U ¼ x3
y, find dU=dt if (1) x5
þ y ¼ t; ð2) x2
þ y3
¼ t2
.
Equations (1) and (2) define x and y as (implicit) functions of t. Then differentiating with respect to t,
we have
ð3Þ 5x4
ðdx=dtÞ þ dy=t ¼ 1 ð4Þ 2xðdx=dtÞ þ 3y2
ðdy=dtÞ ¼ 2t
Solving (3) and (4) simultaneously for dx=dt and dy=dt,
dx
dt
¼
1 1
2t 3y2
5x4
1
2x 3y2
¼
3y2
2t
15x4y2 2x
;
dy
dt
¼
5x4
1
2x 2t
5x4
1
2x 3y2
¼
10x4
t 2x
15x4y2 2x
Then
dU
dt
¼
@U
@x
dx
dt
þ
@U
@y
dy
dt
¼ ð3x2
yÞ
3y2
2t
15x4
y2
2x
!
þ ðx3
Þ
10x4
t 2x
15x4
y2
2x
!
:
6.30. If Fðx; y; zÞ ¼ 0 defines z as an implicit function of x and y in a region r of the xy plane, prove
that (a) @z=@x ¼ Fx=Fz and (b) @z=@y ¼ Fy=Fz, where Fz 6¼ 0.
Since z is a function of x and y, dz ¼
@z
@x
dx þ
@z
@y
dy.
Then dF ¼
@F
@x
dx þ
@F
@y
dy þ
@F
@z
dz ¼
@F
@x
þ
@F
@z
@z
@x
dx þ
@F
@y
þ
@F
@z
@z
@y
dy ¼ 0.
Since x and y are independent, we have
ð1Þ
@F
@x
þ
@F
@z
@z
@x
¼ 0 ð2Þ
@F
@y
þ
@F
@z
@z
@y
¼ 0
from which the required results are obtained. If desired, equations (1) and (2) can be written directly.
6.31. If Fðx; y; u; vÞ ¼ 0 and Gðx; y; u; vÞ ¼ 0, find (a) @u=@x; ðbÞ @u=@y; ðcÞ @v=@x; ðdÞ @v=@y.
The two equations in general define the dependent variables u and v as (implicit) functions of the
independent variables x and y. Using the subscript notation, we have
CHAP. 6] PARTIAL DERIVATIVES 135](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-144-320.jpg)
![@u
@x
¼
@ðF; GÞ
@ðx; vÞ
@ðF; GÞ
@ðu; vÞ
¼
Fx Fv
Gx Gv
Fu Fv
Gu Gv
¼
3 1
1 4v
2u 1
1 4v
¼
1 12v
1 8uv
provided 1 8uv 6¼ 0. Similarly, the other partial derivatives are obtained.
6.33. If Fðu; v; w; x; yÞ ¼ 0, Gðu; v; w; x; yÞ ¼ 0, Hðu; v; w; x; yÞ ¼ 0, find
ðaÞ
@v
@y x
; ðbÞ
@x
@v w
; ðcÞ
@w
@u y
:
From 3 equations in 5 variables, we can (theoretically at least) determine 3 variables in terms of the
remaining 2. Thus, 3 variables are dependent and 2 are independent. If we were asked to determine @v=@y,
we would know that v is a dependent variable and y is an independent variable, but would not know the
remaining independent variable. However, the particular notation
@v
@y x
serves to indicate that we are to
obtain @v=@y keeping x constant, i.e., x is the other independent variable.
(a) Differentiating the given equations with respect to y, keeping x constant, gives
ð1Þ Fu uy þ Fv vy þ Fw wy þ Fy ¼ 0 ð2Þ Gu uy þ Gv vy þ Gw wy þ Gy ¼ 0
ð3Þ Hu uy þ Hv vy þ Hw wy þ Hy ¼ 0
Solving simultaneously for vy, we have
vy ¼
@v
@y x
¼
Fu Fy Fw
Gu Gy Gw
Hu Hy Hw
Fu Fv Fw
Gu Gv Gw
Hu Hv Hw
¼
@ðF; G; HÞ
@ðu; y; wÞ
@ðF; G; HÞ
@ðu; v; wÞ
Equations (1), (2), and (3) can also be obtained by using differentials as in Problem 6.31.
The Jacobian method is very suggestive for writing results immediately, as seen in this problem and
Problem 6.31. Thus, observe that in calculating
@v
@y x
the result is the negative of the quotient of two
Jacobians, the numerator containing the independent variable y, the denominator containing the
dependent variable v in the same relative positions. Using this scheme, we have
ðbÞ
@x
@v w
¼
@ðF; G; HÞ
@ðv; y; uÞ
@ðF; G; HÞ
@ðx; y; uÞ
ðcÞ
@w
@u y
¼
@ðF; G; HÞ
@ðu; x; vÞ
@ðF; G; HÞ
@ðw; x; vÞ
6.34. If z3
xz y ¼ 0, prove that
@2
z
@x @y
¼
3z2
þ x
ð3z2
xÞ3
.
Differentiating with respect to x, keeping y constant and remembering that z is the dependent variable
depending on the independent variables x and y, we find
3z2 @z
@x
x
@z
@x
z ¼ 0 and ð1Þ
@z
@x
¼
z
3z2
x
Differentiating with respect to y, keeping x constant, we find
3z2 @z
@y
x
@z
@y
1 ¼ 0 and ð2Þ
@z
@y
¼
1
3z2 x
CHAP. 6] PARTIAL DERIVATIVES 137](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-146-320.jpg)
![ðaÞ
@ðu; vÞ
@ðx; yÞ
¼
ux uy
vx vy
¼
1 þ y2
ð1 xyÞ2
1 þ x2
ð1 xyÞ2
1
1 þ x2
1
1 þ y2
¼ 0 if xy 6¼ 1:
(b) By Problem 6.35, since the Jacobian is identically zero in a region, there must be a functional relation-
ship between u and v. This is seen to be tan v ¼ u, i.e., ðu; vÞ ¼ u tan v ¼ 0. We can show this
directly by solving for x (say) in one of the equations and then substituting in the other. Thus, for
example, from v ¼ tan1
x þ tan1
y we find tan1
x ¼ v tan1
y and so
x ¼ tanðv tan1
yÞ ¼
tan v tanðtan1
yÞ
1 þ tan v tanðtan1
yÞ
¼
tan v y
1 þ y tan v
Then substituting this in u ¼ ðx þ yÞ=ð1 xyÞ and simplifying, we find u ¼ tan v.
6.37. (a) If x ¼ u v þ w, y ¼ u2
v2
w2
and z ¼ u3
þ v, evaluate the Jacobian
@ðx; y; zÞ
@ðu; v; wÞ
and
(b) explain the significance of the non-vanishing of this Jacobian.
ðaÞ
@ðx; y; zÞ
@ðu; v; wÞ
¼
xu xv xw
yu yv yw
zu zv zw
¼
1 1 1
2u 2v 2w
3u2
1 0
¼ 6wu2
þ 2u þ 6u2
v þ 2w
(b) The given equations can be solved simultaneously for u; v; w in terms of x; y; z in a region r if the
Jacobian is not zero in r.
TRANSFORMATIONS, CURVILINEAR COORDINATES
6.38. A region r in the xy plane is bounded by x þ y ¼ 6, x y ¼ 2; and y ¼ 0. (a) Determine the
region r0
in the uv plane into which r is mapped under the transformation x ¼ u þ v, y ¼ u v.
(b) Compute
@ðx; yÞ
@ðu; vÞ
. (c) Compare the result of (b) with the ratio of the areas of r and r0
.
(a) The region r shown shaded in Fig. 6-9(a) below is a triangle bounded by the lines x þ y ¼ 6, x y ¼ 2,
and y ¼ 0 which for distinguishing purposes are shown dotted, dashed, and heavy respectively.
Under the given transformation the line x þ y ¼ 6 is transformed into ðu þ vÞ þ ðu vÞ ¼ 6, i.e.,
2u ¼ 6 or u ¼ 3, which is a line (shown dotted) in the uv plane of Fig. 6-9(b) above.
CHAP. 6] PARTIAL DERIVATIVES 139
Fig. 6-9](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-148-320.jpg)
![Similarly, x y ¼ 2 becomes ðu þ vÞ ðu vÞ ¼ 2 or v ¼ 1, which is a line (shown dashed) in the
uv plane. In like manner, y ¼ 0 becomes u v ¼ 0 or u ¼ v, which is a line shown heavy in the uv
plane. Then the required region is bounded by u ¼ 3, v ¼ 1 and u ¼ v, and is shown shaded in Fig. 6-
9(b).
ðbÞ
@ðx; yÞ
@ðu; vÞ
¼
@x
@u
@x
@v
@y
@u
@y
@v
¼
@
@u
ðu þ vÞ
@
@v
ðu þ vÞ
@
@u
ðu vÞ
@
@v
ðu vÞ
¼
1 1
1 1
¼ 2
(c) The area of triangular region r is 4, whereas the area of triangular region r0
is 2. Hence, the ratio is
4=2 ¼ 2, agreeing with the value of the Jacobian in (b). Since the Jacobian is constant in this case, the
areas of any regions r in the xy plane are twice the areas of corresponding mapped regions r0
in the uv
plane.
6.39. A region r in the xy plane is bounded by x2
þ y2
¼ a2
, x2
þ y2
¼ b2
, x ¼ 0 and y ¼ 0, where
0 a b. (a) Determine the region r0
into which r is mapped under the transformation
x ¼ cos , y ¼ sin , where 0, 0 @ 2. (b) Discuss what happens when a ¼ 0.
(c) Compute
@ðx; yÞ
@ð; Þ
. (d) Compute
@ð; Þ
@ðx; yÞ
.
(a) The region r [shaded in Fig. 6-10(a) above] is bounded by x ¼ 0 (dotted), y ¼ 0 (dotted and dashed),
x2
þ y2
¼ a2
(dashed), x2
þ y2
¼ b2
(heavy).
Under the given transformation, x2
þ y2
¼ a2
and x2
þ y2
¼ b2
become 2
¼ a2
and 2
¼ b2
or
¼ a and ¼ b respectively. Also, x ¼ 0, a @ y @ b becomes ¼ =2, a @ @ b; y ¼ 0, a @ x @ b
becomes ¼ 0, a @ @ b.
The required region r0
is shown shaded in Fig. 6-10(b) above.
Another method: Using the fact that is the distance from the origin O of the xy plane and is the
angle measured from the positive x-axis, it is clear that the required region is given by a @ @ b,
0 @ @ =2 as indicated in Fig. 6-10(b).
(b) If a ¼ 0, the region r becomes one-fourth of a circular region of radius b (bounded by 3 sides) while r0
remains a rectangle. The reason for this is that the point x ¼ 0, y ¼ 0 is mapped into ¼ 0, ¼ an
indeterminate and the transformation is not one to one at this point which is sometimes called a singular
point.
140 PARTIAL DERIVATIVES [CHAP. 6
Fig. 6-10](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-149-320.jpg)
![ðcÞ
@ðx; yÞ
@ð; Þ
¼
@
@
ð cos Þ
@
@
ð cos Þ
@
@
ð sin Þ
@
@
ð sin Þ
¼
cos sin
sin cos
¼ ðcos2
þ sin2
Þ ¼
(d) From Problem 6.43(b) we have, letting u ¼ , v ¼ ,
@ðx; yÞ
@ð; Þ
@ð; Þ
@ðx; yÞ
¼ 1 so that, using ðcÞ;
@ð; Þ
@ðx; yÞ
¼
1
This can also be obtained by direct differentiation.
Note that from the Jacobians of these transformations it is clear why ¼ 0 (i.e., x ¼ 0, y ¼ 0) is a
singular point.
MEAN VALUE THEOREMS
6.40. Prove the mean value theorem for functions of two variables.
Let f ðtÞ ¼ f ðx0 þ ht; y0 þ ktÞ. By the mean value theorem for functions of one variable,
Fð1Þ ¼ Fð0Þ ¼ F 0
ðÞ 0 1 ð1Þ
If x ¼ x0 þ ht, y ¼ y0 þ kt, then FðtÞ ¼ f ðx; yÞ, so that by Problem 6.17,
F 0
ðtÞ ¼ fxðdx=dtÞ þ fyðdy=dtÞ ¼ hfx þ kfy and F 0
ðÞ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ
where 0 1. Thus, (1) becomes
f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ ð2Þ
where 0 1 as required.
Note that (2), which is analogous to (1) of Problem 6.14 where h ¼ x, has the advantage of being
more symmetric (and also more useful), since only a single number is involved.
MISCELLANEOUS PROBLEMS
6.41. Let f ðx; yÞ ¼
xy
x2
y2
x2 þ y2
!
ðx; yÞ 6¼ ð0; 0Þ
0 ðx; yÞ ¼ ð0; 0Þ
8
:
.
Compute (a) fxð0; 0Þ; ðbÞ fyð0; 0Þ; ðcÞ fxxð0; 0Þ; ðdÞ fyyð0; 0Þ; ðeÞ fxyð0; 0Þ; ð f Þ fyxð0; 0Þ:
ðaÞ fxð0; 0Þ ¼ lim
h!0
f ðh; 0Þ f ð0; 0Þ
h
¼ lim
h!0
0
h
¼ 0
ðbÞ fyð0; 0Þ ¼ lim
h!0
f ð0; kÞ f ð0; 0Þ
k
¼ lim
k!0
0
k
¼ 0
If ðx; yÞ 6¼ ð0; 0Þ,
fxðx; yÞ ¼
@
@x
xy
x2
y2
x2 þ y2
!
( )
¼ xy
4xy2
ðx2 þ y2Þ2
!
þ y
x2
y2
x2 þ y2
!
fyðx; yÞ ¼
@
@y
xy
x2
y2
x2
þ y2
!
( )
¼ xy
4xy2
ðx2 þ y2Þ2
!
þ x
x2
y2
x2
þ y2
!
CHAP. 6] PARTIAL DERIVATIVES 141](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-150-320.jpg)
![which simplifies to
@2
V
@x2
¼ cos2
@2
V
@2
þ
2 sin cos
2
@V
@
2 sin cos
@2
V
@ @
þ
sin2
@V
@
þ
sin2
2
@2
V
@2
ð7Þ
Similarly,
@2
V
@y2
¼ sin2
@2
V
@2
2 sin cos
2
@V
@
þ
2 sin cos
@2
V
@ @
þ
cos2
@V
@
þ
cos2
2
@2
V
@2
ð8Þ
Adding ð7Þ and ð8Þ we find, as required,
@2
V
@x2
þ
@2
V
@y2
¼
@2
V
@2
þ
1
@V
@
þ
1
2
@2
V
@2
¼ 0:
6.43. (a) If x ¼ f ðu; vÞ and y ¼ gðu; vÞ, where u ¼ ðr; sÞ and v ¼ ðr; sÞ, prove that
@ðx; yÞ
@ðr; sÞ
¼
@ðx; yÞ
@ðu; vÞ
@ðu; vÞ
@ðr; sÞ
.
(b) Prove that
@ðx; yÞ
@ðu; vÞ
@ðu; vÞ
@ðx; yÞ
¼ 1 provided
@ðx; yÞ
@ðu; vÞ
6¼ 0, and interpret geometrically.
ðaÞ
@ðx; yÞ
@ðr; sÞ
¼
xr xs
yr ys
¼
xuur þ xvvr xuus þ xvvs
yuur þ yvvr yuus þ yvvs
¼
xu xv
yu yv
ur us
vr vs
¼
@ðx; yÞ
@ðu; vÞ
@ðu; vÞ
@ðr; sÞ
using a theorem on multiplication of determinants (see Problem 6.108). We have assumed here, of
course, the existence of the partial derivatives involved.
ðbÞ Place r ¼ x; s ¼ y in the result of ðaÞ: Then
@ðx; yÞ
@ðu; vÞ
@ðu; vÞ
@ðx; yÞ
¼
@ðx; yÞ
@ðx; yÞ
¼ 1:
The equations x ¼ f ðu; vÞ, y ¼ gðu; vÞ defines a transformation between points ðx; yÞ in the xy plane
and points ðu; vÞ in the uv plane. The inverse transformation is given by u ¼ ðx; yÞ, v ¼ ðx; yÞ. The
result obtained states that the Jacobians of these transformations are reciprocals of each other.
6.44. Show that Fðxy; z 2xÞ ¼ 0 satisfies under suitable conditions the equation
xð@z=@xÞ yð@z=@yÞ ¼ 2x. What are these conditions?
Let u ¼ xy, v ¼ z 2x. Then Fðu; vÞ ¼ 0 and
dF ¼ Fu du þ Fv dv ¼ Fuðx dy þ y dxÞ þ Fvðdz 2 dxÞ ¼ 0
ð1Þ
Taking z as dependent variable and x and y as independent variables, we have dz ¼ zx dx þ zy dy. Then
substituting in (1), we find
ð yFu þ Fvzx 2Þ dx þ ðxFu þ FvzyÞ dy ¼ 0
Hence, we have, since x and y are independent,
ð2Þ yFu þ Fvzx 2 ¼ 0 ð3Þ xFu þ Fvzy ¼ 0
Solve for Fu in (3) and substitute in (2). Then we obtain the required result xzx yzy ¼ 2x upon dividing by
Fv (supposed not equal to zero).
The result will certainly be valid if we assume that Fðu; vÞ is continuously differentiable and that Fv 6¼ 0.
CHAP. 6] PARTIAL DERIVATIVES 143](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-152-320.jpg)
![6.56. Evaluate each of the following limits where they exist.
(a) lim
x!1
y!2
3 x þ y
4 þ x 2y
ðcÞ lim
x!4
y!
x2
sin
y
x
ðeÞ lim
x!0
y!1
e1=x2
ðy1Þ2
ðgÞ lim
x!0þ
y!1
x þ y 1
ffiffiffi
x
p
ffiffiffiffiffiffiffiffiffiffiffi
1 y
p
ðbÞ lim
x!0
y!0
3x 2y
2x 3y
ðdÞ lim
x!0
y!0
x sinðx2
þ y2
Þ
x2 þ y2
ð f Þ lim
x!0
y!0
2x y
x2 þ y2
ðhÞ lim
x!2
y!1
sin1
ðxy 2Þ
tan1ð3xy 6Þ
Ans. (a) 4, (b) does not exist, (c) 8
ffiffiffi
2
p
; ðdÞ 0; ðeÞ 0; ð f Þ does not exist, (g) 0, (h) 1/3
6.57. Formulate a definition of limit for functions of (a) 3, (b) n variables.
6.58. Does lim
4x þ y 3z
2x 5y þ 2z
as ðx; y; zÞ ! ð0; 0; 0Þ exist? Justify your answer.
6.59. Investigate the continuity of each of the following functions at the indicated points:
ðaÞ x2
þ y2
; ðx0; y0Þ: ðbÞ
x
3x þ 5y
; ð0; 0Þ: ðcÞ ðx2
þ y2
Þ sin
1
x2 þ y2
if ðx; yÞ 6¼ ð0; 0Þ, 0 if ðx; yÞ ¼ ð0; 0Þ;
ð0; 0Þ:
Ans. (a) continuous, (b) discontinuous, (c) continuous
6.60. Using the definition, prove that f ðx; yÞ ¼ xy þ 6x is continuous at (a) ð1; 2Þ; ðbÞ ðx0; y0Þ.
6.61. Prove that the function of Problem 6.60 is uniformly continuous in the square region defined by 0 @ x @ 1,
0 @ y @ 1.
PARTIAL DERIVATIVES
6.62. If f ðx; yÞ ¼
x y
x þ y
, find (a) @f =@x and (b) @f =@y at ð2; 1Þ from the definition and verify your answer by
differentiation rules. Ans. (a) 2; ðbÞ 4
6.63. If f ðx; yÞ ¼
ðx2
xyÞ=ðx þ yÞ for ðx; yÞ 6¼ ð0; 0Þ
0 for ðx; yÞ ¼ ð0; 0Þ
, find (a) fxð0; 0Þ; ðbÞ fyð0; 0Þ.
Ans. (a) 1, (b) 0
6.64. Investigate lim
ðx;yÞ!ð0;0Þ
fxðx; yÞ for the function in the preceding problem and explain why this limit (if it exists)
is or is not equal to fxð0; 0Þ:
6.65. If f ðx; yÞ ¼ ðx yÞ sinð3x þ 2yÞ, compute (a) fx; ðbÞ fy; ðcÞ fxx; ðdÞ fyy; ð f Þ fyx at ð0; =3Þ.
Ans. (a) 1
2 ð þ
ffiffiffi
3
p
Þ; ðbÞ 1
6 ð2 3
ffiffiffi
3
p
Þ; ðcÞ 3
2 ð
ffiffiffi
3
p
2Þ; ðdÞ 2
3 ði
ffiffiffi
3
p
þ 3Þ; ðeÞ 1
2 ð2
ffiffiffi
3
p
þ 1Þ,
( f ) 1
2 ð2
ffiffiffi
3
p
þ 1Þ
6.66. (a) Prove by direct differentiation that z ¼ xy tanðy=xÞ satisfies the equation xð@z=@xÞ þ yð@z=@yÞ ¼ 2z if
ðx; yÞ 6¼ ð0; 0Þ. (b) Discuss part (a) for all other points ðx; yÞ assuming z ¼ 0 at ð0; 0Þ.
6.67. Verify that fxy ¼ fyx for the functions (a) ð2x yÞ=ðx þ yÞ, (b) x tan xy; and (c) coshðy þ cos xÞ, indicating
possible exceptional points and investigate these points.
6.68. Show that z ¼ lnfðx aÞ2
þ ðy bÞ2
g satisfies @2
z=@x2
þ @2
z=@y2
¼ 0 except at ða; bÞ.
6.69. Show that z ¼ x cosðy=xÞ þ tanðy=xÞ satisfies x2
zxx þ 2xyzxy þ y2
zyy ¼ 0 except at points for which x ¼ 0.
6.70. Show that if w ¼
x y þ z
x þ y z
n
, then:
CHAP. 6] PARTIAL DERIVATIVES 145](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-154-320.jpg)
![6.83. Find (a) dy=dx and (b) d2
y=dx2
if x3
þ y3
3xy ¼ 0.
Ans. (a) ðy x2
Þ=ðy2
xÞ; ðbÞ 2xy=ðy2
xÞ3
6.84. If xu2
þ v ¼ y3
, 2yu xv3
¼ 4x, find (a)
@u
@x
; ðbÞ
@v
@y
. Ans: ðaÞ
v3
3xu2
v2
þ 4
6x2
uv2
þ 2y
; ðbÞ
2xu2
þ 3y3
3x2
uv2
þ y
6.85. If u ¼ f ðx; yÞ, v ¼ gðx; yÞ are differentiable, prove that
@u
@x
@x
@u
þ
@v
@x
@x
@v
¼ 1. Explain clearly which variables
are considered independent in each partial derivative.
6.86. If f ðx; y; r; sÞ ¼ 0, gðx; y; r; sÞ ¼ 0, prove that
@y
@r
@r
@x
þ
@y
@s
@s
@x
¼ 0, explaining which variables are independent.
What notation could you use to indicate the independent variables considered?
6.87. If Fðx; yÞ ¼ 0, show that
d2
y
dx2
¼
FxxF2
y 2FxyFxFy þ FyyF2
x
F3
y
6.88. Evaluate
@ðF; GÞ
@ðu; vÞ
if Fðu; vÞ ¼ 3u2
uv, Gðu; vÞ ¼ 2uv2
þ v3
. Ans. 24u2
v þ 16uv2
3v3
6.89. If F ¼ x þ 3y2
z3
, G ¼ 2x2
yz, and H ¼ 2z2
xy, evaluate
@ðF; G; HÞ
@ðx; y; zÞ
at ð1; 1; 0Þ. Ans. 10
6.90. If u ¼ sin1
x þ sin1
y and v ¼ x
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 y2
p
þ y
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p
, determine whether there is a functional relationship
between u and v, and if so find it.
6.91. If F ¼ xy þ yz þ zx, G ¼ x2
þ y2
þ z2
, and H ¼ x þ y þ z, determine whether there is a functional relation-
ship connecting F, G, and H, and if so find it. Ans. H2
G 2F ¼ 0.
6.92. (a) If x ¼ f ðu; v; wÞ, y ¼ gðu; v; wÞ, and z ¼ hðu; v; wÞ, prove that
@ðx; y; zÞ
@ðu; v; wÞ
@ðu; v; wÞ
@ðx; y; wÞ
¼ 1 provided
@ðx; y; zÞ
@ðu; v; wÞ
6¼ 0. (b) Give an interpretation of the result of (a) in terms of transformations.
6.93. If f ðx; y; zÞ ¼ 0 and gðx; y; zÞ ¼ 0, show that
dx
@ð f ; gÞ
@ðy; zÞ
¼
dy
@ð f ; gÞ
@ðz; xÞ
¼
dz
@ð f ; gÞ
@ðx; yÞ
giving conditions under which the result is valid.
6.94. If x þ y2
¼ u, y þ z2
¼ v, z þ x2
¼ w, find (a)
@x
@u
; ðbÞ
@2
x
@u2
; ðcÞ
@2
x
@u @v
assuming that the equations
define x; y; and z as twice differentiable functions of u, v; and w.
Ans: ðaÞ
1
1 þ 8xyz
; ðbÞ
16x2
y 8yz 32x2
z2
ð1 þ 8xyzÞ3
; ðcÞ
16y2
z 8xz 32x2
y2
ð1 þ 8xyzÞ3
6.95. State and prove a theorem similar to that in Problem 6.35, for the case where u ¼ f ðx; y; zÞ, v ¼ gðx; y; zÞ,
w ¼ hðx; y; zÞ.
TRANSFORMATIONS, CURVILINEAR COORDINATES
6.96. Given the transformation x ¼ 2u þ v, y ¼ u 3v. (a) Sketch the region r0
of the uv plane into which the
region r of the xy plane bounded by x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 1 is mapped under the transformation.
(b) Compute
@ðx; yÞ
@ðu; vÞ
. (c) Compare the result of (b) with the ratios of the areas of r and r0
.
Ans. (b) 7
CHAP. 6] PARTIAL DERIVATIVES 147](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-156-320.jpg)
![@ðx; y; zÞ
@ðr; s; tÞ
¼
@ðx; y; zÞ
@ðu; v; wÞ
@ðu; v; wÞ
@ðr; s; tÞ
6.110. Given the equations Fjðx1; . . . ; xm; y1; . . . ; ynÞ ¼ 0 where j ¼ 1; 2; . . . ; n. Prove that under suitable condi-
tions on Fj,
@yr
@xs
¼
@ðF1; F2; . . . ; Fr; . . . ; FnÞ
@ð y1; y2; . . . ; xs; . . . ; ynÞ
,
@ðF1; F2; . . . ; FnÞ
@ð y1; y2; . . . ; ynÞ
6.111. (a) If Fðx; yÞ is homogeneous of degree 2, prove that x2 @2
F
@x2
þ 2xy
@2
F
@x @y
þ y2 @2
F
@y2
¼ 2F:
(b) Illustrate by using the special case Fðx; yÞ ¼ x2
lnðy=xÞ:
Note that the result can be written in operator form, using Dx @=@x and Dy @=@y, as
ðx Dx þ y DyÞ2
F ¼ 2F. [Hint: Differentiate both sides of equation (1), Problem 6.25, twice with respect
to .]
6.112. Generalize the result of Problem 6.11 as follows. If Fðx1; x2; . . . ; xnÞ is homogeneous of degree p, then for
any positive integer r, if Dxj @=@xj,
ðx1Dx1
þ x2Dx2
þ þ xnDxn
Þr
F ¼ pðp 1Þ . . . ðp r þ 1ÞF
6.113. (a) Let x and y be determined from u and v according to x þ iy ¼ ðu þ ivÞ3
. Prove that under this
transformation the equation
@2
@x2
þ
@2
@y2
¼ 0 is transformed into
@2
@u2
þ
@2
@v2
¼ 0
(b) Is the result in ða) true if x þ iy ¼ Fðu þ ivÞ? Prove your statements.
CHAP. 6] PARTIAL DERIVATIVES 149](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-158-320.jpg)
![150
Vectors
VECTORS
The foundational ideas for vector analysis were formed indepen-
dently in the nineteenth century by William Rowen Hamilton and
Herman Grassmann. We are indebted to the physicist John Willard
Gibbs, who formulated the classical presentation of the Hamilton
viewpoint in his Yale lectures, and his student E. B. Wilson, who
considered the mathematical material presented in class worthy of
organizing as a book (published in 1901). Hamilton was searching for
a mathematical language appropriate to a comprehensive exposition
of the physical knowledge of the day. His geometric presentation
emphasizing magnitude and direction, and compact notation for the
entities of the calculus, was refined in the following years to the benefit of expressing Newtonian
mechanics, electromagnetic theory, and so on. Grassmann developed an algebraic and more philo-
sophic mathematical structure which was not appreciated until it was needed for Riemanian (non-
Euclidean) geometry and the special and general theories of relativity.
Our introduction to vectors is geometric. We conceive of a vector as a directed line segment PQ
!
from one point P called the initial point to another point Q called the terminal point. We denote vectors
by boldfaced letters or letters with an arrow over them. Thus PQ
!
is denoted by A or ~
A
A as in Fig. 7-1.
The magnitude or length of the vector is then denoted by jPQ
!
j, PQ, jAj or j ~
A
Aj.
Vectors are defined to satisfy the following geometric properties.
GEOMETRIC PROPERTIES
1. Two vectors A and B are equal if they have the same magnitude and direction regardless of their
initial points. Thus A ¼ B in Fig. 7-1 above.
In other words, a vector is geometrically represented by any one of a class of commonly
directed line segments of equal magnitude. Since any one of the class of line segments may be
chosen to represent it, the vector is said to be free. In certain circumstances (tangent vectors,
forces bound to a point), the initial point is fixed, then the vector is bound. Unless specifically
stated, the vectors in this discussion are free vectors.
2. A vector having direction opposite to that of vector A but with the same magnitude is denoted
by A [see Fig. 7-2].
Q
P
A or A
B
Fig. 7-1
Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-159-320.jpg)
![3. The sum or resultant of vectors A and B of Fig. 7-3(a) below is
a vector C formed by placing the initial point of B on the
terminal point of A and joining the initial point of A to the
terminal point of B [see Fig. 7-3(b) below]. The sum C is
written C ¼ A þ B. The definition here is equivalent to the
parallelogram law for vector addition as indicated in Fig.7-3(c)
below.
Extensions to sums of more than two vectors are
immediate. For example, Fig. 7-4 below shows how to obtain
the sum or resultant E of the vectors A, B, C, and D.
4. The difference of vectors A and B, represented by A B, is that vector C which added to B gives
A. Equivalently, A B may be defined as A þ ðBÞ. If A ¼ B, then A B is defined as the null
or zero vector and is represented by the symbol 0. This has a magnitude of zero but its direction
is not defined.
The expression of vector equations and related concepts is facilitated by the use of real
numbers and functions. In this context, these are called scalars. This special designation arises
from application where the scalars represent object that do not have direction, such as mass,
length, and temperature.
5. Multiplication of a vector A by a scalar m produces a vector mA with magnitude jmj times the
magnitude of A and direction the same as or opposite to that of A according as m is positive or
negative. If m ¼ 0, mA ¼ 0, the null vector.
ALGEBRAIC PROPERTIES OF VECTORS
The following algebraic properties are consequences of the geometric definition of a vector. (See
Problems 7.1 and 7.2.)
CHAP. 7] VECTORS 151
A
_ A
Fig. 7-2
B
A
A A
B
B
C = A + B
C = A + B
(a) (b) (c)
Fig. 7-3
A
A
C
C
B
B
D D
E = A + B + C + D
Fig. 7-4](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-160-320.jpg)
![If A, B and C are vectors, and m and n are scalars, then
1. A þ B ¼ B þ A Commutative Law for Addition
2. A þ ðB þ CÞ ¼ ðA þ BÞ þ C Associative Law for Addition
3. mðnAÞ ¼ ðmnÞA ¼ nðmAÞ Associative Law for Multiplication
4. ðm þ nÞA ¼ mA þ nA Distributive Law
5. mðA þ BÞ ¼ mA þ mB Distributive Law
Note that in these laws only multiplication of a vector by one or more scalars is defined. On Pages
153 and 154 we define products of vectors.
LINEAR INDEPENDENCE AND LINEAR DEPENDENCE OF A SET OF VECTORS
A set of vectors, A1; A2; . . . ; Ap, is linearly independent means that a1A1 þ a2A2 þ þ apAp þ þ
apAp ¼ 0 if and only if a1 ¼ a2 ¼ ¼ ap ¼ 0 (i.e., the algebraic sum is zero if and only if all the
coefficients are zero). The set of vectors is linearly dependent when it is not linearly independent.
UNIT VECTORS
Unit vectors are vectors having unit length. If A is any vector with length A 0, then A=A is a unit
vector, denoted by a, having the same direction as A. Then A ¼ Aa.
RECTANGULAR (ORTHOGONAL) UNIT VECTORS
The rectangular unit vectors i, j, and k are unit vectors having the direction of the positive x, y, and z
axes of a rectangular coordinate system [see Fig. 7-5]. We use right-handed rectangular coordinate
systems unless otherwise specified. Such systems derive their name from the fact that a right-threaded
screw rotated through 908 from Ox to Oy will advance in the positive z direction. In general, three
vectors A, B, and C which have coincident initial points and are not coplanar are said to form a right-
handed system or dextral system if a right-threaded screw rotated through an angle less than 1808 from A
to B will advance in the direction C [see Fig. 7-6 below].
152 VECTORS [CHAP. 7
Fig. 7-5 Fig. 7-6](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-161-320.jpg)
![COMPONENTS OF A VECTOR
Any vector A in 3 dimensions can be represented with
initial point at the origin O of a rectangular coordinate
system [see Fig. 7-7]. Let ðA1; A2; A3Þ be the rectangular
coordinates of the terminal point of vector A with initial
point at O. The vectors A1i; A2j; and A3k are called the
rectangular component vectors, or simply component vec-
tors, of A in the x, y; and z directions respectively. A1; A2;
and A3 are called the rectangular components, or simply
components, of A in the x, y; and z directions respectively.
The vectors of the set fi; j; kg are perpendicular to one
another, and they are unit vectors. The words orthogonal
and normal, respectively, are used to describe these charac-
teristics; hence, the set is what is called an orthonormal basis.
It is easily shown to be linearly independent. In an n-dimensional space, any set of n linearly indepen-
dent vectors is a basis. The further characteristic of a basis is that any vector of the space can be
expressed through it. It is the basis representation that provides the link between the geometric and
algebraic expressions of vectors and vector concepts.
The sum or resultant of A1i; A2j; and A3k is the vector A, so that we can write
A ¼ A1i þ A2j þ A3k ð1Þ
The magnitude of A is
A ¼ jAj ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2
1 þ A2
2 þ A2
3
q
ð2Þ
In particular, the position vector or radius vector r from O to the point ðx; y; zÞ is written
r ¼ xi þ yj þ zk ð3Þ
and has magnitude r ¼ jrj ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2 þ z2
p
:
DOT OR SCALAR PRODUCT
The dot or scalar product of two vectors A and B, denoted by A B (read A dot B) is defined as the
product of the magnitudes of A and B and the cosine of the angle between them. In symbols,
A B ¼ AB cos ; 0 @ @ ð4Þ
Assuming that neither A nor B is the zero vector, an immediate consequence of the definition is that
A B ¼ 0 if and only if A and B are perpendicular. Note that A B is a scalar and not a vector.
The following laws are valid:
1. A B ¼ B A Commutative Law for Dot Products
2. A ðB þ CÞ ¼ A B þ A C Distributive Law
3. mðA BÞ ¼ ðmAÞ B ¼ A ðmBÞ ¼ ðA BÞm, where m is a scalar.
4. i i ¼ j j ¼ k k ¼ 1; i j ¼ j k ¼ k i ¼ 0
5. If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, then
A B ¼ A1B1 þ A2B2
þ A3B3
The equivalence of this component form the dot product with the geometric definition 4 follows
from the law of cosines. (See Fig. 7-8.)
CHAP. 7] VECTORS 153
Fig. 7-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-162-320.jpg)
![6. jA Bj ¼ the area of a parallelogram with sides A and B.
7. If A B ¼ 0 and neither A nor B is a null vector, then A and B are parallel.
TRIPLE PRODUCTS
Dot and cross multiplication of three vectors, A, B, and C may produce meaningful products of the
form ðA BÞC; A ðB CÞ; and A ðB CÞ. The following laws are valid:
1. ðA BÞC 6¼ AðB CÞ in general
2. A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ ¼ volume of a parallelepiped having A, B, and C as
edges, or the negative of this volume according as A, B, and C do or do not form a right-
handed system. If A ¼ A1i þ A2j þ A3k, B ¼ B1i þ B2j þ B3k and C ¼ C1i þ C2j þ C3k, then
A ðB CÞ ¼
A1 A2 A3
B1 B2 B3
C1 C2 C3
ð6Þ
3. A ðB CÞ 6¼ ðA BÞ C (Associative Law for Cross Products Fails)
4. A ðB CÞ ¼ ðA CÞB ðA BÞC
ðA BÞ C ¼ ðA CÞB ðB CÞA
The product A ðB CÞ is called the scalar triple product or box product and may be denoted by
½ABC. The product A ðB CÞ is called the vector triple product.
In A ðB CÞ parentheses are sometimes omitted and we write A B C. However, parentheses
must be used in A ðB CÞ (see Problem 7.29). Note that A ðB CÞ ¼ ðA BÞ C. This is often
expressed by stating that in a scalar triple product the dot and the cross can be interchanged without
affecting the result (see Problem 7.26).
AXIOMATIC APPROACH TO VECTOR ANALYSIS
From the above remarks it is seen that a vector r ¼ xi þ yj þ zk is determined when its 3 components
ðx; y; zÞ relative to some coordinate system are known. In adopting an axiomatic approach, it is thus
quite natural for us to make the following
Definition. A three-dimensional vector is an ordered triplet of real numbers with the following
properties. If A ¼ ðA1; A2; A3Þ and B ¼ ðB1; B2; B3Þ then
1. A ¼ B if and only if A1 ¼ B1; A2 ¼ B2; A3 ¼ B3
2. A þ B ¼ ðA1 þ B1; A2 þ B2; A3 þ B3Þ
3. A B ¼ ðA1 B1; A2 B2; A3 B3Þ
4. 0 ¼ ð0; 0; 0Þ
5. mA ¼ mðA1; A2; A3Þ ¼ ðmA1; mA2; mA3Þ
In addition, two forms of multiplication are established.
6. A B ¼ A1B1 þ A2B2 þ A3B3
7. Length or magnitude of A ¼ jAj ¼
ffiffiffiffiffiffiffiffiffiffiffi
A A
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2
1 þ A2
2 þ A2
3
q
8. A B ¼ ðA2B3 A3B2; A3B1 A1B3; A1B2 A2B1Þ
Unit vectors are defined to be ð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þ and then designated by i; j; k, respectively,
thereby identifying the components axiomatically introduced with the geometric orthonormal basis
elements.
If one wishes, this axiomatic formulation (which provides a component representation for vectors)
can be used to reestablish the fundamental laws previously introduced geometrically; however, the
CHAP. 7] VECTORS 155](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-164-320.jpg)
![dA
du
¼
dA1
du
i þ
dA2
du
j þ
dA3
du
k
Higher derivatives such as d2
A=du2
, etc., can be similarly defined.
3. If Aðx; y; zÞ ¼ A1ðx; y; zÞi þ A2ðx; y; zÞj þ A3ðx; y; zÞk, then
dA ¼
@A
@x
dx þ
@A
@y
dy þ
@A
@z
dz
is the differential of A.
4. Derivatives of products obey rules similar to those for scalar functions. However, when cross
products are involved the order may be important. Some examples are
ðaÞ
d
du
ðAÞ ¼
dA
du
þ
d
du
A;
ðbÞ
@
@y
ðA BÞ ¼ A
@B
@y
þ
@A
@y
B;
ðcÞ
@
@z
ðA BÞ ¼ A
@B
@z
þ
@A
@z
B (Maintain the order of A and BÞ
GEOMETRIC INTERPRETATION OF A VECTOR DERIVATIVE
If r is the vector joining the origin O of a coordinate system and the point ðx; y; zÞ, then specification
of the vector function rðuÞ defines x, y; and z as functions of u (r is called a position vector). As u changes,
the terminal point of r describes a space curve (see Fig. 7-11) having parametric equations
x ¼ xðuÞ; y ¼ yðuÞ; z ¼ zðuÞ. If the parameter u is the arc length s measured from some fixed point
on the curve, then recall from the discussion of arc length that ds2
¼ dr dr. Thus
dr
ds
¼ T ð7Þ
is a unit vector in the direction of the tangent to the curve and is called the unit tangent vector. If u is the
time t, then
dr
dt
¼ v ð8Þ
is the velocity with which the terminal point of r describes the curve. We have
v ¼
dr
dt
¼
dr
ds
ds
dt
¼
ds
dt
T ¼ vT ð9Þ
CHAP. 7] VECTORS 157
Fig. 7-11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-166-320.jpg)
![from which we see that the magnitude of v is v ¼ ds=dt. Similarly,
d2
r
dt2
¼ a ð10Þ
is the acceleration with which the terminal point of r describes the curve. These concepts have important
applications in mechanics and differential geometry.
A primary objective of vector calculus is to express concepts in an intuitive and compact form.
Success is nowhere more apparent than in applications involving the partial differentiation of scalar and
vector fields. [Illustrations of such fields include implicit surface representation, fx; y; zðx; yÞ ¼ 0, the
electromagnetic potential function ðx; y; zÞ, and the electromagnetic vector field Fðx; y; zÞ.] To give
mathematics the capability of addressing theories involving such functions, William Rowen Hamilton
and others of the nineteenth century introduced derivative concepts called gradient, divergence, and curl,
and then developed an analytic structure around them.
An intuitive understanding of these entities begins with examination of the differential of a scalar
field, i.e.,
d ¼
@
@x
dx þ
@
@y
dy þ
@
@z
dz
Now suppose the function is constant on a surface S and that C; x ¼ f1ðtÞ; y ¼ f2
ðtÞ; z ¼ f3ðtÞ is a
curve on S. At any point of this curve
dr
dt
¼
dx
dt
i þ
dy
dt
j þ
dz
dt
k lies in the tangent plane to the surface.
Since this statement is true for every surface curve through a given point, the differential dr spans the
tangent plane. Thus, the triple
@
@x
,
@
@y
,
@
@z
represents a vector perpendicualr to S. With this special
geometric characteristic in mind we define
r ¼
@
@x
i þ
@
@y
j þ
@
@z
k
to be the gradient of the scalar field .
Furthermore, we give the symbol r a special significance by naming it del.
EXAMPLE 1. If ðx; y; zÞ ¼ 0 is an implicity defined surface, then, because the function always has the value zero
for points on it, the condition of constancy is satisfied and r is normal to the surface at any of its points. This
allows us to form an equation for the tangent plane to the surface at any one of its points. See Problem 7.36.
EXAMPLE 2. For certain purposes, surfaces on which is constant are called level surfaces. In meteorology,
surfaces of equal temperature or of equal atmospheric pressure fall into this category. From the previous devel-
opment, we see that r is perpendicular to the level surface at any one of its points and hence has the direction of
maximum change at that point.
The introduction of the vector operator r and the interaction of it with the multiplicative properties
of dot and cross come to mind. Indeed, this line of thought does lead to new concepts called divergence
and curl. A summary follows.
GRADIENT, DIVERGENCE, AND CURL
Consider the vector operator r (del) defined by
r i
@
@x
þ j
@
@y
þ k
@
@z
ð11Þ
Then if ðx; y; zÞ and Aðx; y; zÞ have continuous first partial derivatives in a region (a condition which is
in many cases stronger than necessary), we can define the following.
158 VECTORS [CHAP. 7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-167-320.jpg)
![1. Gradient. The gradient of is defined by
grad ¼ r ¼ i
@
@x
þ j
@
@y
þ k
@
@z
¼ i
@
@x
þ j
@
@y
þ k
@
@z
ð12Þ
¼
@
@x
i þ
@
@y
j þ
@
@z
k
2. Divergence. The divergence of A is defined by
div A ¼ r A ¼ i
@
@x
þ j
@
@y
þ k
@
@z
ðA1i þ A2j þ A3kÞ ð13Þ
¼
@A1
@x
þ
@A2
@y
þ
@A3
@z
3. Curl. The curl of A is defined by
curl A ¼ r A ¼ i
@
@x
þ j
@
@y
þ k
@
@z
ðA1i þ A2j þ A3kÞ ð14Þ
¼
i j k
@
@x
@
@y
@
@z
A1 A2 A3
¼ i
@
@y
@
@z
A2 A3
j
@
@x
@
@z
A1 A2
þ k
@
@x
@
@y
A1 A2
¼
@A3
@y
@A2
@z
i þ
@A1
@z
@A3
@x
j þ
@A2
@x
@A1
@y
k
Note that in the expansion of the determinant, the operators @=@x; @=@y; @=@z must precede
A1; A2; A3. In other words, r is a vector operator, not a vector. When employing it the laws of
vector algebra either do not apply or at the very least must be validated. In particular, r A is a new
vector obtained by the specified partial differentiation on A, while A r is an operator waiting to act
upon a vector or a scalar.
FORMULAS INVOLVING r
If the partial derivatives of A, B, U, and V are assumed to exist, then
1. rðU þ VÞ ¼ rU þ rV or grad ðU þ VÞ ¼ grad u þ grad V
2. r ðA þ BÞ ¼ r A þ r B or div ðA þ BÞ þ div A þ div B
3. r ðA þ BÞ ¼ r A þ r B or curl ðA þ BÞ ¼ curl A þ curl B
4. r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ
5. r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ
6. r ðA BÞ ¼ B ðr AÞ A ðr BÞ
7. r ðA BÞ ¼ ðB rÞA Bðr AÞ ðA rÞB þ Aðr BÞ
8. rðA BÞ ¼ ðB rÞA þ ðA rÞB þ B ðr AÞ þ A ðr BÞ
9: r ðrUÞ r2
U
@2
U
@x2
þ
@2
U
@y2
þ
@2
U
@z2
is called the Laplacian of U
and r2 @2
@x2
þ
@2
@y2
þ
@2
@z2
is called the Laplacian operator:
10. r ðrUÞ ¼ 0. The curl of the gradient of U is zero.
CHAP. 7] VECTORS 159](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-168-320.jpg)
![11. r ðr AÞ ¼ 0. The divergence of the curl of A is zero.
12. r ðr AÞ ¼ rðr AÞ r2
A
VECTOR INTERPRETATION OF JACOBIANS,
ORTHOGONAL CURVILINEAR COORDINATES
The transformation equations
x ¼ f ðu1; u2; u3Þ; y ¼ gðu1; u2; u3Þ; z ¼ hðu1; u2; u3Þ ð15Þ
[where we assume that f ; g; h are continuous, have continuous partial derivatives, and have a single-
valued inverse] establish a one-to-one correspondence between points in an xyz and u1u2u3 rectangular
coordinate system. In vector notation the transformation (17) can be written
r ¼ xi þ yj þ zk ¼ f ðu1; u2; u3Þi þ gðu1; u2; u3Þj þ hðu1; u2; u3Þk ð16Þ
A point P in Fig. 7-12 can then be defined not only by rectangular coordinates ðx; y; zÞ but by coordinates
ðu1; u2; u3Þ as well. We call ðu1; u2; u3Þ the curvilinear coordinates of the point.
If u2 and u3 are constant, then as u1 varies, r describes a curve which we call the u1 coordinate curve.
Similarly, we define the u2 and u3 coordinate curves through P.
From (16), we have
dr ¼
@r
@u1
du1 þ
@r
@u2
du2 þ
@r
@u3
du3 ð17Þ
The collection of vectors
@r
@x
;
@r
@y
;
@r
@z
is a basis for the vector structure associated with the curvilinear
system. If the curvilinear system is orthogonal, then so is this set; however, in general, the vectors are
not unit vectors. he differential form for arc length may be written
ds2
¼ g11ðdu1Þ2
þ g22ðdu2Þ2
þ g33ðdu3
Þ2
where
g11 ¼
@r
@x
@r
@x
; g22 ¼
@r
@y
@r
@y
; g33 ¼
@r
@z
@r
@z
The vector @r=@u1 is tangent to the u1 coordinate curve at P. If e1 is a unit vector at P in this
direction, we can write @r=@u1 ¼ h1e1 where h1 ¼ j@r=@u1j. Similarly we can write @r=@u2 ¼ h2e2 and
@r=@u3 ¼ h3e3, where h2 ¼ j@r=@u2j and h3 ¼ j@r=@u3j respectively. Then (17) can be written
160 VECTORS [CHAP. 7
Fig. 7-12](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-169-320.jpg)
![dr ¼ h1 du1e1 þ h2 du2e2 þ h3 du3e3 ð18Þ
The quantities h1; h2; h3 are sometimes caleld scale factors.
If e1; e2; e3 are mutually perpendicular at any point P, the curvilinear coordinates are called
orthogonal. Since the basis elements are unit vectors as well as orthogonal this is an orthonormal
basis. In such case the element of arc length ds is given by
ds2
¼ dr dr ¼ h2
1 du2
1 þ h2
2 du2
2 þ h2
3 du2
3 ð19Þ
and corresponds to the square of the length of the diagonal in the above parallelepiped.
Also, in the case of othogonal coordinates, referred to the orthonormal basis e1; e2; e3, the volume of
the parallelepiped is given by
dV ¼ jgjkjdu1du2du3 ¼ jðh1 du1e1Þ ðh2 du2e2Þ ðh3 du3e3Þj ¼ h1h2h3 du1du2du3 ð20Þ
which can be written as
dV ¼
@r
@u1
@r
@u2
@r
@u3
du1du2du3 ¼
@ðx; y; zÞ
@ðu1; u2; u3Þ
du1du2du3 ð21Þ
where @ðx; y; zÞ=@ðu1; u2; u3Þ is the Jacobian of the transformation.
It is clear that when the Jacobian vanishes there is no parallelepiped and explains geometrically the
significance of the vanishing of a Jacobian as treated in Chapter 6.
Note: The further significance of the Jacobian vanishing is that the transformation degenerates at the
point.
GRADIENT DIVERGENCE, CURL, AND LAPLACIAN IN ORTHOGONAL
CURVILINEAR COORDINATES
If is a scalar function and A ¼ A1e1 þ A2e2 þ A3e3 a vector function of orthogonal curvilinear
coordinates u1; u2; u3, we have the following results.
1: r ¼ grad ¼
1
h1
@
@u1
e1 þ
1
h2
@
@u2
e2 þ
1
h3
@
@u3
e3
2: r A ¼ div A ¼
1
h1h2h3
@
@u1
ðh2; h3A1Þ þ
@
@u2
ðh3h1A2Þ þ
@
@u3
ðh1h2A3Þ
3: r A ¼ curl A ¼
1
h1h2h3
h1e1 h2e2 h3e3
@
@u1
@
@u2
@
@u3
h1A1 h2A2 h3A3
4: r2
¼ Laplacian of ¼
1
h1h2h3
@
@u1
h2h3
h1
@
@u1
þ
@
@u2
h3h1
h2
@
@u2
þ
@
@u3
h1h2
h3
@
@u3
These reduce to the usual expressions in rectangular coordinates if we replace ðu1; u2; u3Þ by ðx; y; zÞ,
in which case e1; e2; and e3 are replaced by i, j, and k and h1 ¼ h2 ¼ h3 ¼ 1.
SPECIAL CURVILINEAR COORDINATES
1. Cylindrical Coordinates (; ; z). See Fig. 7-13.
Transformation equations:
x ¼ cos ; y ¼ sin ; z ¼ z
CHAP. 7] VECTORS 161](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-170-320.jpg)
![Solved Problems
VECTOR ALGEBRA
7.1. Show that addition of vectors is commutative, i.e., A þ B ¼ B þ A. See Fig. 7-15 below.
OP þ PQ ¼ OQ or A þ B ¼ C;
OR þ RQ ¼ OQ or B þ A ¼ C:
and
Then A þ B ¼ B þ A.
7.2. Show that the addition of vectors is associative, i.e., A þ ðB þ CÞ ¼ ðA þ BÞ þ C. See Fig. 7-16
above.
OP þ PQ ¼ OQ ¼ ðA þ BÞ and PQ þ QR ¼ PR ¼ ðB þ CÞ
OP þ PR ¼ OR ¼ D; i:e:; A þ ðB þ CÞ ¼ D
Since
OQ þ QR ¼ OR ¼ D; i:e:; ðA þ BÞ þ C ¼ D
we have A þ ðB þ CÞ ¼ ðA þ BÞ þ C.
Extensions of the results of Problems 7.1 and 7.2 show that the order of addition of any number of
vectors is immaterial.
7.3. An automobile travels 3 miles due north, then 5 miles
northeast as shown in Fig. 7-17. Represent these displace-
ments graphically and determine the resultant displacement
(a) graphically, (b) analytically.
Vector OP or A represents displacement of 3 mi due north.
Vector PQ or B represents displacement of 5 mi northeast.
Vector OQ or C represents the resultant displacement or sum
of vectors A and B, i.e., C ¼ A þ B. This is the triangle law of
vector addition.
The resultant vector OQ can also be obtained by constructing
the diagonal of the parallelogram OPQR having vectors OP ¼ A
and OR (equal to vector PQ or B) as sides. This is the parallelo-
gram law of vector addition.
(a) Graphical Determination of Resultant. Lay off the 1 mile unit
on vector OQ to find the magnitude 7.4 mi (approximately).
CHAP. 7] VECTORS 163
A
A
B
B
P
R
O
Q
C
=
A
+
B
C
=
B
+
A
Fig. 7-15
(A
+ B)
(
B
+
C
)
A
B
C
D
O
P Q
R
Fig. 7-16
N
S
O
P
R
Q
A
A
B
B
W E
Unit = 1 mile
45°
135°
C
=
A
+
B
Fig. 7-17](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-172-320.jpg)
![By the Pythagorean theorem,
ðOPÞ2
¼ ðOQÞ2
þ ðQPÞ2
where OP denotes the magnitude of vector OP, etc. Similarly, ðOQÞ2
¼ ðORÞ2
þ ðRQÞ2
.
Then ðOPÞ2
¼ ðORÞ2
þ ðRQÞ2
þ ðQPÞ2
or A2
¼ A2
1 þ A2
2 þ A2
3, i.e., A ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2
1 þ A2
2 þ A2
3
q
.
7.9. Determine the vector having initial point Pðx1; y1; z1Þ
and terminal point Qðx2; y2; z2Þ and find its magnitude.
See Fig. 7-21.
The position vector of P is r1 ¼ x1i þ y1j þ z1k.
The position vector of Q is r2 ¼ x2i þ y2j þ z2k.
r1 ¼ PQ ¼ r2 or
PQ ¼ r2 r1 ¼ ðx2i þ y2j þ z2kÞ ðx1i þ y1j þ z1kÞ
¼ ðx2 x1Þi þ ðy2 y1Þj þ ðz2 z1Þk
Magnitude of PQ ¼ PQ
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx2 x1Þ2
þ ðy2 y1Þ2
þ ðz2 z1Þ2
q
:
Note that this is the distance between points P and Q.
THE DOT OR SCALAR PRODUCT
7.10. Prove that the projection of A on B is equal to A b, where b is a
unit vector in the direction of B.
Through the initial and terminal points of A pass planes perpen-
dicular to B at G and H respectively, as in the adjacent Fig. 7-22: then
Projection of A on B ¼ GH ¼ EF ¼ A cos ¼ A b
7.11. Prove A ðB þ CÞ ¼ A B þ A C. See Fig. 7-23.
Let a be a unit vector in the direction of A; then
Projection of ðB þ CÞ on A ¼ projection of B on A þ projection
of C on A
ðB þ CÞ a ¼ B a þ C a
CHAP. 7] VECTORS 165
C
A B
D d E
a
c
b
b
1
2
a
1
2
Fig. 7-19 Fig. 7-20
A
B
H
G
E F
G
Fig. 7-22
Fig. 7-21
E F G A
B C
(B + C)
Fig. 7-23](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-174-320.jpg)
![Multiplying by A,
ðB þ CÞ Aa ¼ B Aa þ C Aa
ðB þ CÞ A ¼ B A þ C A
and
Then by the commutative law for dot products,
A ðB þ CÞ ¼ A B þ A C
and the distributive law is valid.
7.12. Prove that ðA þ BÞ ðC þ DÞ ¼ A C þ A D þ B C þ B D.
By Problem 7.11, ðA þ BÞ ðC þ DÞ ¼ A ðC þ DÞ þ B ðC þ DÞ ¼ A C þ A D þ B C þ B D.
The ordinary laws of algebra are valid for dot products where the operations are defined.
7.13. Evaluate each of the following.
ðaÞ i i ¼ jijjij cos 08 ¼ ð1Þð1Þð1Þ ¼ 1
ðbÞ i k ¼ jijjkj cos 908 ¼ ð1Þð1Þð0Þ ¼ 0
ðcÞ k j ¼ jkjjjj cos 908 ¼ ð1Þð1Þð0Þ ¼ 0
ðdÞ j ð2i 3j þ kÞ ¼ 2j i 3j j þ j k ¼ 0 3 þ 0 ¼ 3
ðeÞ ð2i jÞ ð3i þ kÞ ¼ 2i ð3i þ kÞ j ð3i þ kÞ ¼ 6i i þ 2i k 3j i j k ¼ 6 þ 0 0 0 ¼ 6
7.14. If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, prove that A B ¼ A1B1 þ A2B2 þ A3B3.
A B ¼ ðA1i þ A2j þ A3kÞ ðB1i þ B2j þ B3kÞ
¼ A1i ðB1i þ B2j þ B3kÞ þ A2j ðB1i þ B2j þ B3kÞ þ A3k ðB1i þ B2j þ B3kÞ
¼ A1B1i i þ A1B2i j þ A1B3i k þ A2B1j i þ A2B2j j þ A2B3j k
þ A3B1k i þ A3B2k j þ A3B3k k
¼ A1B1 þ A2B2 þ A3B3
since i j ¼ k k ¼ 1 and all other dot products are zero.
7.15. If A ¼ A1i þ A2j þ A3k, show that A ¼
ffiffiffiffiffiffiffiffiffiffiffi
A A
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2
1 þ A2
2 þ A2
3
q
.
A A ¼ ðAÞðAÞ cos 08 ¼ A2
. Then A ¼
ffiffiffiffiffiffiffiffiffiffiffi
A A
p
.
Also, A A ¼ ðA1i þ A2j þ A3kÞ ðA1i þ A2j þ A3kÞ
¼ ðA1ÞðA1Þ þ ðA2ÞðA2Þ þ ðA3ÞðA3Þ ¼ A2
1 þ A2
2 þ A2
3
By Problem 7.14, taking B ¼ A.
Then A ¼
ffiffiffiffiffiffiffiffiffiffiffi
A A
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2
1 þ A2
2 þ A2
3
q
is the magnitude of A. Sometimes A A is written A2
.
THE CROSS OR VECTOR PRODUCT
7.16. Prove A B ¼ B A.
A B ¼ C has magnitude AB sin and direction such that A, B, and C form a right-handed system [Fig.
7-24(a)].
B A ¼ D has magnitude BA sin and direction such that B, A, and D form a right-handed system
[Fig. 7-24(b)].
Then D has the same magnitude as C but is opposite in direction, i.e., C ¼ D or A B ¼ B A.
The commutative law for cross products is not valid.
7.17. Prove that A ðB þ CÞ ¼ A B þ A C for the case where A is perpendicular to B and also
to C.
166 VECTORS [CHAP. 7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-175-320.jpg)
![Since A is perpendicular to B, A B is a vector perpendicular to the plane of A and B and having magnitude
AB sin 908 ¼ AB or magnitude of AB. This is equivalent to multiplying vector B by A and rotating the
resultant vector through 908 to the position shown in Fig. 7-25.
Similarly, A C is the vector obtained by multiplying C by A and rotating the resultant vector through
908 to the position shown.
In like manner, A ðB þ CÞ is the vector obtained by multiplying B þ C by A and rotating the resultant
vector through 908 to the position shown.
Since A ðB þ CÞ is the diagonal of the parallelogram with A B and A C as sides, we have
A ðB þ CÞ ¼ A B þ A C.
7.18. Prove that A ðB þ CÞ ¼ A B þ A C in the general case where A, B, and C are non-
coplanar. See Fig. 7-26.
Resolve B into two component vectors, one perpendicular to A and the other parallel to A, and denote
them by B? and Bk respectively. Then B ¼ B? þ Bk.
If is the angle between A and B, then B? ¼ B sin . Thus the magnitude of A B? is AB sin , the
same as the magnitude of A B. Also, the direction of A B? is the same as the direction of A B.
Hence A B? ¼ A B.
Similarly, if C is resolved into two component vectors Ck and C?, parallel and perpendicular respec-
tively to A, then A C? ¼ A C.
Also, since B þ C ¼ B? þ Bk þ C? þ Ck ¼ ðB? þ C?Þ þ ðBk þ CkÞ it follows that
A ðB? þ C?Þ ¼ A ðB þ CÞ
Now B? and C? are vectors perpendicular to A and so by Problem 7.17,
A ðB? þ C?Þ ¼ A B? þ A C?
A ðB þ CÞ ¼ A B þ A C
Then
CHAP. 7] VECTORS 167
Fig. 7-24
Fig. 7-25 Fig. 7-26](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-176-320.jpg)
![TRIPLE PRODUCTS
7.23. Show that A ðB CÞ is in absolute value equal to the volume
of a parallelepiped with sides A, B, and C. See Fig. 7-28.
Let n be a unit normal to parallelogram I, having the direction of
B C, and let h be the height of the terminal point of A above the
parallelogram I.
Volume of a parallelepiped ¼ ðheight hÞðarea of parallelogram IÞ
¼ ðA nÞðjB CjÞ
¼ A fjB Cjng ¼ A ðB CÞ
If A, B and C do not form a right-handed system, A n 0 and the volume =jA ðB CÞj.
7.24. If A ¼ A1i þ A2j þ A3k, B ¼ B1i þ B2j þ B3k, C ¼ C1i þ C2j þ C3k show that
A ðB CÞ ¼
A1 A2 A3
B1 B2 B3
C1 C2 C3
A ðB CÞ ¼ A
i j k
B1 B2 B3
C1 C2 C3
¼ ðA1i þ A2j þ A3kÞ ½ðB2C3 B3C2Þi þ ðB3C1 B1C3Þj þ ðB1C2 B2C1Þk
¼ A1ðB2C3 B3C2Þ þ A2ðB3C1 B1C3Þ þ A3ðB1C2 B2C1Þ ¼
A1 A2 A3
B1 B2 B3
C1 C2 C3
:
7.25. Find the volume of a parallelepiped with sides A ¼ 3i j; B ¼ j þ 2k; C ¼ i þ 5j þ 4k.
By Problems 7.23 and 7.24, volume of parallelepiped ¼ jA ðB CÞj ¼ j
3 1 0
0 1 2
1 5 4
j
¼ j 20j ¼ 20:
7.26. Prove that A ðB CÞ ¼ ðA BÞ C, i.e., the dot and cross can be interchanged.
By Problem 7.24: A ðB CÞ ¼
A1 A2 A3
B1 B2 B3
C1 C2 C3
; ðA BÞ C ¼ C ðA BÞ ¼
C1 C2 C3
A1 A2 A3
B1 B2 B3
Since the two determinants are equal, the required result follows.
7.27. Let r1 ¼ x1i þ y1j þ z1k, r2 ¼ x2i þ y2j þ z2k and r3 ¼ x3i þ y3j þ z3k be the position vectors of
points P1ðx1; y1; z1Þ, P2ðx2; yx; z2Þ and P3ðx3; y3; z3Þ. Find an equation for the plane passing
through P1, P2; and P3. See Fig. 7-29.
We assume that P1, P2, and P3 do not lie in the same straight line; hence, they determine a plane.
Let r ¼ xi þ yj þ zk denote the position vectors of any point Pðx; y; zÞ in the plane. Consider vectors
P1P2 ¼ r2 r1, P1P3 ¼ r3 r1 and P1P ¼ r r1 which all lie in the plane. Then
P1P P1P2 P1P3 ¼ 0
CHAP. 7] VECTORS 169
Fig. 7-28](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-178-320.jpg)
![7.29. If A ¼ i þ j, B ¼ 2i 3j þ k, C ¼ 4j 3k, find (a) ðA BÞ C, (b) A ðB CÞ.
ðaÞ A B ¼
i j k
1 1 0
2 3 1
¼ i j 5k. Then ðA BÞ C ¼
i j k
1 1 5
0 4 3
¼ 23i þ 3j þ 4k:
ðbÞ B C ¼
i j k
2 3 1
0 4 3
¼ 5i þ 6j þ 8k. Then A ðB CÞ ¼
i j k
1 1 0
5 6 8
¼ 8i 8j þ k:
It can be proved that, in general, ðA BÞ C 6¼ A ðB CÞ.
DERIVATIVES
7.30. If r ¼ ðt3
þ 2tÞi 3e2t
j þ 2 sin 5tk, find (a)
dr
dt
; ðbÞ
dr
dt
; ðcÞ
d2
r
dt2
; ðdÞ
d2
r
dt2
at t ¼ 0 and
give a possible physical significance.
ðaÞ
dr
dt
¼
d
dt
ðt3
þ 2tÞi þ
d
dt
ð3e2t
Þj þ
d
dt
ð2 sin 5tÞk ¼ ð3t2
þ 2Þi þ 6e2t
j þ 10 cos 5tk
At t ¼ 0, dr=dt ¼ 2i þ 6j þ 10k
ðbÞ From ðaÞ; jdr=dtj ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð2Þ2
þ ð6Þ2
þ ð10Þ2
q
¼
ffiffiffiffiffiffiffiffi
140
p
¼ 2
ffiffiffiffiffi
35
p
at t ¼ 0:
ðcÞ
d2
r
dt2
¼
d
dt
dr
dt
¼
d
dt
fð3t2
þ 2Þi þ 6e2t
j þ 10 cos 5tkg ¼ 6ti 12e2t
j 50 sin 5tk
At t ¼ 0, d2
r=dt2
¼ 12j.
ðdÞ From ðcÞ; jd2
r=dt2
j ¼ 12 at t ¼ 0:
If t represents time, these represent respectively the velocity, magnitude of the velocity, acceleration,
and magnitude of the acceleration at t ¼ 0 of a particle moving along the space curve x ¼ t3
þ 2t,
y ¼ 3e2t
, z ¼ 2 sin 5t.
7.31. Prove that
d
du
ðA BÞ ¼ A
dB
du
þ
dA
du
B, where A and B are differentiable functions of u.
Method 1:
d
du
ðA BÞ ¼ lim
u!0
ðA þ AÞ ðB þ BÞ A B
u
¼ lim
u!0
A B þ A B þ A B
u
¼ lim
u!0
A
B
u
þ
A
u
B þ
A
u
B
¼ A
dB
du
þ
dA
du
B
Method 2: Let A ¼ A1i þ A2j þ A3k, B þ B1i þ B2j þ B3k. Then
d
du
ðA BÞ ¼
d
du
ðA1B1 þ A2B2 þ A3B3Þ
¼ A1
dB1
du
þ A2
dB2
du
þ A3
dB3
du
þ
dA1
du
B1 þ
dA2
du
B2 þ
dA3
du
B3
¼ A
dB
du
þ
dA
du
B
CHAP. 7] VECTORS 171](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-180-320.jpg)
![ðdÞ div ðAÞ ¼ r ðAÞ ¼ r ðx3
yz4
i x2
y3
z3
j þ 2x4
y2
z3
kÞ
¼
@
@x
ðx3
yz4
Þ þ
@
@y
ðx2
y3
z3
Þ þ
@
@z
ð2x4
y2
z3
Þ
¼ 3x2
yz4
3x2
y2
z3
þ 6x4
y2
z2
ðeÞ curl ðAÞ ¼ r ðAÞ ¼ r ðx3
yz4
i x2
y3
z3
j þ 2x4
y2
z3
kÞ
¼
i j k
@=@x @=@y @=@z
x3
yz4
x2
y3
z3
2x4
y2
z3
¼ ð4x4
yz3
3x2
y3
z2
Þi þ ð4x3
yz3
8x3
y2
z3
Þj ð2xy3
z3
þ x3
z4
Þk
7.35. Prove r ðAÞ ¼ ðrÞ A þ ðr AÞ.
r ðAÞ ¼ r ðA1i þ A2j þ A3kÞ
¼
@
@x
ðA1Þ þ
@
@y
ðA2Þ þ
@
@z
ðA3Þ
¼
@
@x
A1 þ
@
@y
A2 þ
@
@z
A3 þ
@A1
@x
þ
@A2
@y
þ
@A3
@z
¼
@
@x
i þ
@
@y
j þ
@
@z
k
ðA1i þ A2j þ A3kÞ
þ
@
@x
i þ
@
@y
j þ
@
@z
k
ðA1i þ A2j þ A3kÞ
¼ ðrÞ A þ ðr AÞ
7.36. Express a formula for the tangent plane to the surface ðx; y; zÞ ¼ 0 at one of its points
P0ðx0; y0; z0Þ.
Ans: ðrÞ0 ðr r0Þ ¼ 0
7.37. Find a unit normal to the surface 2x2
þ 4yz 5z2
¼ 10 at the point Pð3; 1; 2Þ.
By Problem 7.36, a vector normal to the surface is
rð2x2
þ 4yz 5z2
Þ ¼ 4xi þ 4zj þ ð4y 10zÞk ¼ 12i þ 8j 24k at ð3; 1; 2Þ
Then a unit normal to the surface at P is
12i þ 8j 24k
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð12Þ2
þ ð8Þ2
þ ð24Þ2
q ¼
3i þ 2j 6k
7
:
Another unit normal to the surface at P is
3i þ 2j 6k
7
:
7.38. If ¼ 2x2
y xz3
, find (a) r and (b) r2
.
ðaÞ r ¼
@
@x
i þ
@
@y
j þ
@
@z
k ¼ ð4xy z3
Þi þ 2x2
j 3xz2
k
ðbÞ r2
¼ Laplacian of ¼ r r ¼
@
@x
ð4xy z3
Þ þ
@
@y
ð2x2
Þ þ
@
@z
ð3xz2
Þ ¼ 4y 6xz
CHAP. 7] VECTORS 173](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-182-320.jpg)
![ðdsÞ2
¼ ðdxÞ2
þ ðdyÞ2
þ ðdzÞ2
¼ ðdrÞ2
þ r2
ðdÞ2
þ r2
sin2
ðdÞ2
and
The scale factors are h1 ¼ hr ¼ 1; h2 ¼ h ¼ r; h3 ¼ h ¼ r sin .
7.41. Find the volume element dV in (a) cylindrical and (b) spherical coordinates and sketch.
The volume element in orthogonal curvilinear coordinates u1; u2; u3 is
dV ¼ h1h2h3 du1du2du3 ¼
@ðx; y; zÞ
@ðu1; u2; u3Þ
du1; du2du3
(a) In cylindrical coordinates, u1 ¼ ; u2 ¼ ; u3 ¼ z; h1 ¼ 1; h2 ¼ ; h3 ¼ 1 [see Problem 7.40(a)]. Then
dV ¼ ð1ÞðÞð1Þ d d dz ¼ d d dz
This can also be observed directly from Fig. 7-30(a) below.
(b) In spherical coordinates, u1 ¼ r; u2 ¼ ; u3 ¼ ; h1 ¼ 1; h2 ¼ r; h3 ¼ r sin [see Problem 7.40(b)]. Then
dV ¼ ð1ÞðrÞðr sin Þ dr d d ¼ r2
sin dr d d
This can also be observed directly from Fig. 7-30(b) above.
7.42. Express in cylindrical coordinates: (a) grad ; ðbÞ div A; ðcÞ r2
.
Let u1 ¼ ; u2 ¼ ; u3 ¼ z; h1 ¼ 1; h2 ¼ ; h3 ¼ 1 [see Problem 7.40(a)] in the results 1, 2, and 4 on Pages
174 and 175. Then
ðaÞ grad ¼ r ¼
1
1
@
@
e1 þ
1
@
@
e2 þ
1
1
@
@z
e3 ¼
@
@
e1 þ
1
@
@
e2 þ
@
@z
e3
where e1; e2; e3 are the unit vectors in the directions of increasing ; ; z, respectively.
ðbÞ div A ¼ r A ¼
1
ð1ÞðÞð1Þ
@
@
ðÞð1ÞA1
ð Þ þ
@
@
ðð1Þð1ÞA2Þ þ
@
@z
ðð1ÞðÞA3Þ
¼
1
@
@
ðA1Þ þ
@A2
@
þ
@A3
@z
CHAP. 7] VECTORS 175
Fig. 7-30](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-184-320.jpg)
![(b) From the result in part (a), we have
r2 1
r
¼
d2
dr2
1
r
þ
2
r
d
dr
1
r
¼
2
r3
2
r3
¼ 0
showing that 1=r is a solution of Laplace’s equation.
7.45. A particle moves along a space curve r ¼ rðtÞ, where t is the time measured from some initial time.
If v ¼ jdr=dtj ¼ ds=dt is the magnitude of the velocity of the particle (s is the arc length along the
space curve measured from the initial position), prove that the acceleration a of the particle is
given by
a ¼
dv
dt
T þ
v2
N
where T and N are unit tangent and normal vectors to the space curve and
¼
d2
r
ds2
1
¼
d2
x
ds2
!2
þ
d2
y
ds2
!2
þ
d2
z
ds2
!2
8
:
9
=
;
1=2
The velocity of the particle is given by v ¼ vT. Then the acceleration is given by
a ¼
dv
dt
¼
d
dt
ðvTÞ ¼
dv
dt
T þ v
dT
dt
¼
dv
dt
T þ v
dT
ds
ds
dt
¼
dv
dt
T þ v2 dT
ds
ð1Þ
Since T has a unit magnitude, we have T T ¼ 1. Then differentiating with respect to s,
T
dT
ds
þ
dT
ds
T ¼ 0; 2T
dT
ds
¼ 0 or T
dT
ds
¼ 0
from which it follows that dT=ds is perpendicular to T. Denoting by N the unit vector in the direction of
dT=ds, and called the principal normal to the space curve, we have
dT
ds
¼ N ð2Þ
where is the magnitude of dT=ds. Now since T ¼ dr=ds [see equation (7), Page 157], we have
dT=ds ¼ d2
r=ds2
. Hence
¼
d2
r
ds2
¼
d2
x
ds2
!2
þ
d2
y
ds2
!2
þ
d2
z
ds2
!2
8
:
9
=
;
1=2
Defining ¼ 1= , (2) becomes dT=ds ¼ N=. Thus from (1) we have, as required,
a ¼
dv
dt
T þ
v2
N
The components dv=dt and v2
= in the direction of T and N are called the tangential and normal
components of the acceleration, the latter being sometimes called the centripetal acceleration. The quantities
and are respectively the radius of curvature and curvature of the space curve.
CHAP. 7] VECTORS 177](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-186-320.jpg)
![Supplementary Problems
VECTOR ALGEBRA
7.46. Given any two vectors A and B, illustrate geometrically the equality 4A þ 3ðB AÞ ¼ A þ 3B.
7.47. A man travels 25 miles northeast, 15 miles due east, and 10 miles due south. By using an appropriate scale,
determine graphically (a) how far and (b) in what direction he is from his starting position. Is it possible
to determine the answer analytically? Ans. 33.6 miles, 13.28 north of east.
7.48. If A and B are any two non-zero vectors which do not have the same direction, prove that mA þ nB is a
vector lying in the plane determined by A and B.
7.49. If A, B, and C are non-coplanar vectors (vectors which do not all lie in the same plane) and
x1A þ y1B þ z1C ¼ x2A þ y2B þ z2C, prove that necessarily x1 ¼ x2; y1 ¼ y2; z1 ¼ z2.
7.50. Let ABCD be any quadrilateral and points P; Q; R; and S the midpoints of successive sides. Prove (a) that
PQRS is a parallelogram and (b) that the perimeter of PQRS is equal to the sum of the lengths of the
diagonals of ABCD.
7.51. Prove that the medians of a triangle intersect at a point which is a trisection point of each median.
7.52. Find a unit vector in the direction of the resultant of vectors A ¼ 2i j þ k, B ¼ i þ j þ 2k, C ¼ 3i 2j þ 4k.
Ans. ð6i 2j þ 7kÞ=
ffiffiffiffiffi
89
p
THE DOT OR SCALAR PRODUCT
7.53. Evaluate jðA þ BÞ ðA BÞj if A ¼ 2i 3j þ 5k and B ¼ 3i þ j 2k. Ans. 24
7.54. Verify the consistency of the law of cosines for a triangle. [Hint: Take the sides of A; B; C where C ¼ A B.
Then use C C ¼ ðA BÞ ðA BÞ.]
7.55. Find a so that 2i 3j þ 5k and 3i þ aj 2k are perpendicular. Ans. a ¼ 4=3
7.56. If A ¼ 2i þ j þ k; B ¼ i 2j þ 2k and C ¼ 3i 4j þ 2k, find the projection of A þ C in the direction of B.
Ans. 17/3
7.57. A triangle has vertices at Að2; 3; 1Þ; Bð1; 1; 2Þ; Cð1; 2; 3Þ. Find (a) the length of the median drawn from
B to side AC and (b) the acute angle which this median makes with side BC.
Ans. (a) 1
2
ffiffiffiffiffi
26
p
; ðbÞ cos1
ffiffiffiffiffi
91
p
=14
7.58. Prove that the diagonals of a rhombus are perpendicular to each other.
7.59. Prove that the vector ðAB þ BAÞ=ðA þ BÞ represents the bisector of the angle between A and B.
THE CROSS OR VECTOR PRODUCT
7.60. If A ¼ 2i j þ k and B ¼ i þ 2j 3k, find jð2A þ BÞ ðA 2BÞj: Ans. 5
ffiffiffi
3
p
7.61. Find a unit vector perpendicular to the plane of the vectors A ¼ 3i 2j þ 4k and B ¼ i þ j 2k.
Ans. ð2j þ kÞ=
ffiffiffi
5
p
7.62. If A B ¼ A C, does B ¼ C necessarily?
7.63. Find the area of the triangle with vertices ð2; 3; 1Þ; ð1; 1; 2Þ; ð1; 2; 3Þ. Ans. 1
2
ffiffiffi
3
p
178 VECTORS [CHAP. 7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-187-320.jpg)
![7.64. Find the shortest distance from the point ð3; 2; 1Þ to the plane determine by ð1; 1; 0Þ; ð3; 1; 1Þ; ð1; 0; 2Þ.
Ans. 2
TRIPLE PRODUCTS
7.65. If A ¼ 2i þ j 3k; B ¼ i 2j þ k; C ¼ i þ j 4, find (a) A ðB CÞ, (b) C ðA BÞ, (c) A ðB CÞ,
(d) ðA BÞ C. Ans. (a) 20, (b) 20, (c) 8i 19j k; ðdÞ 25i 15j 10k
7.66. Prove that ðaÞ A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ
ðbÞ A ðB CÞ ¼ BðA CÞ CðA BÞ.
7.67. Find an equation for the plane passing through ð2; 1; 2Þ; ð1; 2; 3Þ; ð4; 1; 0Þ.
Ans. 2x þ y 3z ¼ 9
7.68. Find the volume of the tetrahedron with vertices at ð2; 1; 1Þ; ð1; 1; 2Þ; ð0; 1; 1Þ; ð1; 2; 1Þ.
Ans. 4
3
7.69. Prove that ðA BÞ ðC DÞ þ ðB CÞ ðA DÞ þ ðC AÞ ðB DÞ ¼ 0:
DERIVATIVES
7.70. A particle moves along the space curve r ¼ et
cos t i þ et
sin t j þ et
k. Find the magnitude of the
(a) velocity and (b) acceleration at any time t. Ans. (a)
ffiffiffi
3
p
et
; ðbÞ
ffiffiffi
5
p
et
7.71. Prove that
d
du
ðA BÞ ¼ A
dB
du
þ
dA
du
B where A and B are differentiable functions of u.
7.72. Find a unit vector tangent to the space curve x ¼ t; y ¼ t2
; z ¼ t3
at the point where t ¼ 1.
Ans. ði þ 2j þ 3kÞ=
ffiffiffiffiffi
14
p
7.73. If r ¼ a cos !t þ b sin !t, where a and b are any constant non-collinear vectors and ! is a constant scalar,
prove that (a) r ¼
dr
dr
¼ !ða bÞ; ðbÞ;
d2
r
dt2
þ !2
r ¼ 0.
7.74. If A ¼ x2
i yj þ xzk, B ¼ yi þ xj xyzk and C ¼ i yj þ x3
zk, find (a)
@2
@x @y
ðA BÞ and
(b) d½A ðB CÞ at the point ð1; 1; 2Þ: Ans. (a) 4i þ 8j; ðbÞ 8 dx
7.75. If R ¼ x2
yi 2y2
zj þ xy2
z2
k, find
@2
B
@x2
@2
R
@y2
at the point ð2; 1; 2Þ. Ans. 16
ffiffiffi
5
p
GRADIENT, DIVERGENCE, AND CURL
7.76. If U; V; A; B have continuous partial derivatives prove that:
(a) rðU þ VÞ ¼ rU þ rV; ðbÞ r ðA þ BÞ ¼ r A þ r B; ðcÞ r ðA þ BÞ ¼ r A þ r B.
7.77. If ¼ xy þ yz þ zx and A ¼ x2
yi þ y2
zj þ z2
xk, find (a) A r; ðbÞ r A; and (c) ðrÞ A at the
point ð3; 1; 2Þ. Ans: ðaÞ 25; ðbÞ 2; ðcÞ 56i 30j þ 47k
7.78. Show that r ðr2
rÞ ¼ 0 where r ¼ xi þ yj þ zk and r ¼ jrj.
7.79. Prove: (a) r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ; ðbÞ r ðA BÞ ¼ B ðr AÞ A ðr BÞ.
7.80. Prove that curl grad u ¼ 0, stating appropriate conditions on U.
7.81. Find a unit normal to the surface x2
y 2xz þ 2y2
z4
¼ 10 at the point ð2; 1; 1Þ.
Ans: ð3i þ 4j 6kÞ=
ffiffiffiffiffi
61
p
7.82. If A ¼ 3xz2
i yzj þ ðx þ 2zÞk, find curl curl A. Ans: 6xi þ ð6z 1Þk
7.83. (a) Prove that r ðr AÞ ¼ r2
A þ rðr AÞ. (b) Verify the result in (a) if A is given as in Problem 7.82.
CHAP. 7] VECTORS 179](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-188-320.jpg)
![JACOBIANS AND CURVINLINEAR COORDINATES
7.84. Prove that
@ðx; y; zÞ
@ðu1; u2; u3Þ
¼
@r
@u1
@r
@u2
@r
@u3
.
7.85. Express (a) grad ; ðbÞ div A; ðcÞ r2
in spherical coordinates.
Ans: ðaÞ
@
@r
e1 þ
1
r
@
@
e2 þ
1
r sin
@
@
e3
ðbÞ
1
r2
@
@r
ðr2
A1Þ þ
1
r sin
@
@
ðsin A2Þ þ
1
r sin
@A3
@
where A ¼ A1e1 þ A2e2 þ A3e3
ðcÞ
1
r2
@
@r
r2 @
@r
þ
1
r2 sin
@
sin
@
@
þ
1
r2
sin2
@2
@2
7.86. The transformation from rectangular to parabolic cylindrical coordinates is defined by the equations
x ¼ 1
2 ðu2
v2
Þ, y ¼ uv, z ¼ z. (a) Prove that the system is orthogonal. (b) Find ds2
and the scale
factors. (c) Find the Jacobian of the transformation and the volume element.
Ans. ðbÞ ds2
¼ ðu2
þ v2
Þ du2
þ ðu2
þ v2
Þ dv2
þ dz2
; h1 ¼ h2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 þ v2
p
; h3 ¼ 1
ðcÞ u2
þ v2
; ðu2
þ v2
Þ du dv dz
7.87. Write (a) r2
and (b) div A in parabolic cylindrical coordinates.
Ans: ðaÞ r2
¼
1
u2 þ v2
@2
@u2
þ
@2
@v2
!
þ
@2
@z2
ðbÞ div A ¼
1
u2 þ v2
@
@u
ð
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 þ v2
p
A1Þ þ
@
@v
ð
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 þ v2
p
A2Þ
þ
@A3
@z
7.88. Prove that for orthogonal curvilinear coordinates,
r ¼
e1
h1
@
@u1
þ
e2
h2
@
@u2
þ
e3
h3
@
@u3
[Hint: Let r ¼ a1e1 þ a2e2 þ a3e3 and use the fact that d ¼ r dr must be the same in both rectangular
and the curvilinear coordinates.]
7.89. Give a vector interpretation to the theorem in Problem 6.35 of Chapter 6.
MISCELLANEOUS PROBLEMS
7.90. If A is a differentiable function of u and jAðuÞj ¼ 1, prove that dA=du is perpendicular to A.
7.91. Prove formulas 6, 7, and 8 on Page 159.
7.92. If and are polar coordinates and A; B; n are any constants, prove that U ¼ n
ðA cos n þ B sin nÞ
satisfies Laplace’s equation.
7.93. If V ¼
2 cos þ 3 sin3
cos
r2
, find r2
V. Ans.
6 sin cos ð4 5 sin2
Þ
r4
7.94. Find the most general function of (a) the cylindrical coordinate , (b) the spherical coordinate r, (c) the
spherical coordinate which satisfies Laplace’s equation.
Ans. (a) A þ B ln ; ðbÞ A þ B=r; ðcÞ A þ B lnðcsc cot Þ where A and B are any constants.
7.95. Let T and N denote respectively the unit tangent vector and unit principal normal vector to a space curve
r ¼ rðuÞ, where rðuÞ is assumed differentiable. Define a vector B ¼ T N called the unit binormal vector to
the space curve. Prove that
180 VECTORS [CHAP. 7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-189-320.jpg)
![dT
ds
¼ N;
dB
ds
¼ N;
dN
ds
¼ B T
These are called the Frenet-Serret formulas and are of fundamental importance in differential geometry. In
these formulas is called the curvature, is called the torsion; and the reciprocals of these, ¼ 1= and
¼ 1= , are called the radius of curvature and radius of torsion, respectively.
7.96. (a) Prove that the radius of curvature at any point of the plane curve y ¼ f ðxÞ; z ¼ 0 where f ðxÞ is differ-
entiable, is given by
¼
ð1 þ y02
Þ3=2
y00
(b) Find the radius of curvature at the point ð=2; 1; 0Þ of the curve y ¼ sin x; z ¼ 0.
Ans. (b) 2
ffiffiffi
2
p
7.97. Prove that the acceleration of a particle along a space curve is given respectively in (a) cylindrical,
(b) spherical coordinates by
ð €
_
2
Þe þ ð €
þ 2 _
_
Þe þ €
z
zez
ð€
r
r r _
2
r _
2
sin2
Þer þ ðr €
þ 2_
r
r _
r _
2
sin cos Þe þ ð2_
r
r _
sin þ 2r _
_
cos þ r €
sin Þe
where dots denote time derivatives and e; e; ez; er; e; e are unit vectors in the directions of increasing
; ; z; r; ; , respectively.
7.98. Let E and H be two vectors assumed to have continuous partial derivatives (of second order at least) with
respect to position and time. Suppose further that E and H satisfy the equations
r E ¼ 0; r H ¼ 0; r E ¼
1
c
@H
@t
; r H ¼
1
c
@E
@t
ð1Þ
prove that E and H satisfy the equation
r2
¼
1
c2
@2
@t2
ð2Þ
where is a generic meaning, and in particular can represent any component of E or H.
[The vectors E and H are called electric and magnetic field vectors in electromagnetic theory. Equations (1)
are a special case of Maxwell’s equations. The result (2) led Maxwell to the conclusion that light was an
electromagnetic phenomena. The constant c is the velocity of light.]
7.99. Use the relations in Problem 7.98 to show that
@
@t
f1
2 ðE2
þ H2
Þg þ cr ðE HÞ ¼ 0
7.100. Let A1; A2; A3 be the components of vector A in an xyz rectangular coordinate system with unit vectors
i1; i2; i3 (the usual i; j; k vectors), and A0
1; A0
2; A0
3 the components of A in an x0
y0
z0
rectangular coordinate
system which has the same origin as the xyz system but is rotated with respect to it and has the unit vectors
i0
1; i0
2; i0
3. Prove that the following relations (often called invariance relations) must hold:
An ¼ l1nA0
1 þ l2nA0
2 þ l3nA0
3 n ¼ 1; 2; 3
where i0
m in ¼ lmn.
7.101. If A is the vector of Problem 7.100, prove that the divergence of A, i.e., r A, is an invariant (often called a
scalar invariant), i.e., prove that
@A0
1
@x0 þ
@A0
2
@y0 þ
@A0
3
@z0 ¼
@A1
@x
þ
@A2
@y
þ
@A3
@z
CHAP. 7] VECTORS 181](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-190-320.jpg)
![If R ¼ f ðuÞi þ gðuÞj þ hðuÞk, a vector tangent to C at the point P is given by T0 ¼
dR
du P
. If r0 and r
denote the vectors drawn respectively from O to Pðx0; y0; z0Þ and Qðx; y; zÞ on the tangent line, then since
r r0 is collinear with T0 we have
ðr r0Þ T0 ¼ ðr r0Þ
dR
du P
¼ 0 ð5Þ
In rectangular form this becomes
x x0
f 0ðu0Þ
¼
y y0
g0ðu0Þ
¼
z z0
h0ðu0Þ
ð6Þ
The parametric form is obtained by setting each ratio equal to u.
If the curve C is given as the intersection of two surfaces with equations Fðx; y; zÞ ¼ 0 and
Gðx; y; zÞ ¼ 0 observe that rF rG has the direction of the line of intersection of the tangent planes;
therefore, the corresponding equations of the tangent line are
x x0
Fy Fz
Gy Gz P
¼
y y0
Fz Fx
Gz Gx P
¼
z z0
Fx Fy
Gx Gy P
ð7Þ
Note that the determinants in (7) are Jacobians. A similar result can be found when the surfaces are
given in terms of orthogonal curvilinear coordinates.
4. Normal Plane to a Curve. Suppose we wish to find an equation for the normal plane to curve C
at Pðx0; y0; z0Þ of Fig. 8-2 (i.e., the plane perpendicular to the tangent line to C at this point). Letting r be
the vector from O to any point ðx; y; zÞ on this plane, it follows that r r0 is perpendicular to T0. Then
the required equation is
ðr r0Þ T0 ¼ ðr r0Þ
dR
du P
¼ 0 ð8Þ
When the curve has parametric equations x ¼ f ðuÞ; y ¼ gðuÞ; z ¼ hðuÞ this becomes
f 0
ðu0Þðx x0Þ þ g0
ðu0Þð y y0Þ þ h0
ðu0Þðz z0Þ ¼ 0 ð9Þ
Furthermore, when the curve is the intersection of the implicitly defined surfaces
Fðx; y; zÞ ¼ 0 and Gðx; y; zÞ ¼ 0
then
Fy Fz
Gy Gz P
ðx x0Þ þ
Fz Fx
Gz Gx P
ð y y0Þ þ
Fx Fy
Gx Gy P
ðz z0Þ ¼ 0 ð10Þ
5. Envelopes. Solutions of differential equations in two variables are geometrically represented by
one-parameter families of curves. Sometimes such a family characterizes a curve called an envelope.
For example, the family of all lines (see Problem 8.9) one unit from the origin may be represented by
x sin y cos 1 ¼ 0, where is a parameter. The envelope of this family is the circle x2
þ y2
¼ 1.
If ðx; y; zÞ ¼ 0 is a one-parameter family of curves in the xy plane, there may be a curve E which is
tangent at each point to some member of the family and such that each member of the family is tangent
to E. If E exists, its equation can be found by solving simultaneously the equations
ðx; y; Þ ¼ 0; ðx; y; Þ ¼ 0 ð11Þ
and E is called the envelope of the family.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 185](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-194-320.jpg)
![ðb
a
ð Þ d ¼
ðb
a
ðu2
u1
f ðx; Þ dx
d ¼
ðu2
u1
ðb
a
f ðx; Þ d
dx ð18Þ
The result is known as interchange of the order of integration or integration under the integral sign. (See
Problem 8.18.)
MAXIMA AND MINIMA
In Chapter 4 we briefly examined relative extrema for functions of one variable. The general idea
was that for points of the graph of y ¼ gðxÞ that were locally highest or lowest, the condition g0
ðxÞ ¼ 0
was necessary. Such points P0ðx0Þ were called critical points. (See Fig. 8-3a,b.) The condition g0
ðxÞ ¼ 0
was useful in searching for relative maxima and minima but it was not decisive. (See Fig. 8-3(c).)
To determine the exact nature of the function at a critical point P0, g00
ðx0Þ had to be examined.
0 counterclockwise rotation (rel. min.)
g00
ðx0Þ 0 implied a clockwise rotation (rel. max)
¼ 0 need for further investigation.
This section describes the necessary and sufficient conditions for relative extrema of functions of two
variables. Geometrically we think of surfaces, S, represented by z ¼ f ðx; yÞ. If at a point P0ðx0; y0Þ
then fxðx; y0Þ ¼ 0, means that the curve of intersection of S and the plane y ¼ y0 has a tangent parallel to
the x-axis. Similarly fyðx0; y0Þ ¼ 0 indicates that the curve of intersection of S and the cross section
x ¼ x0 has a tangent parallel the y-axis. (See Problem 8.20.)
Thus
fxðx; y0Þ ¼ 0; fyðx0; yÞ ¼ 0
are necessary conditions for a relative extrema of z ¼ f ðx; yÞ at P0; however, they are not sufficient
because there are directions associated with a rotation through 3608 that have not been examined. Of
course, no differentiation between relative maxima and relative minima has been made. (See Fig. 8-4.)
A very special form, fxy fxfy invariant under plane rotation, and capable of characterizing the
roots of a quadratic equation, Ax2
þ 2Bx þ C ¼ 0, allows us to form sufficient conditions for
relative extrema. (See Problem 8.21.)
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 187
Fig. 8-3 Fig. 8-4](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-196-320.jpg)
![Gðx1; x2; . . . ; xnÞ F þ 11 þ 22 þ þ kk ð23Þ
subject to the (necessary) conditions
@G
@x1
¼ 0;
@G
@x2
¼ 0; . . . ;
@G
@xn
0 ð24Þ
where 1; 2; . . . ; k, which are independent of x1; x2; . . . ; xn, are the Lagrange multipliers.
APPLICATIONS TO ERRORS
The theory of differentials can be applied to obtain errors in a function of x; y; z, etc., when the
errors in x; y; z, etc., are known. See Problem 8.28.
Solved Problems
TANGENT PLANE AND NORMAL LINE TO A SURFACE
8.1. Find equations for the (a) tangent plane and (b) normal line to the surface x2
yz þ 3y2
¼
2xz2
8z at the point ð1; 2; 1Þ.
(a) The equation of the surface is F ¼ x2
yz þ 3y2
2xz2
þ 8z ¼ 0. A normal to the surface at ð1; 2; 1Þ is
N0 ¼ rFjð1;2;1Þ ¼ ð2xyz 2z2
Þi þ ðx2
z þ 6yÞj þ ðx2
y 4xz þ 8Þkjð1;2;1Þ
¼ 6i þ 11j þ 14k
Referring to Fig. 8-1, Page 183:
The vector from O to any point ðx; y; zÞ on the tangent plane is r ¼ xi þ yj þ zk.
The vector from O to the point ð1; 2; 1Þ on the tangent plane is r0 ¼ i þ 2j k.
The vector r r0 ¼ ðx 1Þi þ ð y 2Þj þ ðz þ 1Þk lies in the tangent plane and is thus perpen-
dicular to N0.
Then the required equation is
ðr r0Þ N0 ¼ 0 i:e:; fðx 1Þi þ ð y 2Þj þ ðz þ 1Þkg f6i þ 11j þ 14kg ¼ 0
6ðx 1Þ þ 11ð y 2Þ þ 14ðz þ 1Þ ¼ 0 or 6x 11y 14z þ 2 ¼ 0
(b) Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal N0. The vector from O to
the point ð1; 2; 1Þ on the normal is r0 ¼ i þ 2j k. The vector r r0 ¼ ðx 1Þi þ ð y 2Þj þ ðz þ 1Þk
is collinear with N0. Then
ðr r0Þ N0 ¼ 0 i:e:;
i j k
x 1 y 2 z þ 1
6 11 14
¼ 0
which is equivalent to the equations
11ðx 1Þ ¼ 6ð y 2Þ; 14ð y 2Þ ¼ 11ðz þ 1Þ; 14ðx 1Þ ¼ 6ðz þ 1Þ
These can be written as
x 1
6
¼
y 2
11
¼
z þ 1
14
often called the standard form for the equations of a line. By setting each of these ratios equal to the
parameter t, we have
x ¼ 1 6t; y ¼ 2 þ 11t; z ¼ 14t 1
called the parametric equations for the line.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 189](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-198-320.jpg)
![Denoting the expressions in braces by A; B; C respectively so that rFjP ¼ Ai þ Bj þ Ck, we see
that the required equation is Aðx x0Þ þ Bð y y0Þ þ Cðz z0Þ ¼ 0. This can be written in spherical
coordinates by using the transformation equations for x, y; and z in these coordinates.
(b) We have F ¼ r 4 cos ¼ 0. Then @F=@r ¼ 1, @F=@ ¼ 4 sin , @F=@ ¼ 0.
Since r0 ¼ 2
ffiffiffi
2
p
; 0 ¼ =4; 0 ¼ 3=4, we have from part (a), rFjP ¼ Ai þ Bj þ Ck ¼ i þ j.
From the transformation equations the given point has rectangular coordinates ð
ffiffiffi
2
p
;
ffiffiffi
2
p
; 2Þ, and
so r r0 ¼ ðx þ
ffiffiffi
2
p
Þi þ ð y
ffiffiffi
2
p
Þj þ ðz 2Þk.
The required equation of the plane is thus ðx þ
ffiffiffi
2
p
Þ þ ð y
ffiffiffi
2
p
Þ ¼ 0 or y x ¼ 2
ffiffiffi
2
p
. In sphe-
rical coordinates this becomes r sin sin r sin cos ¼ 2
ffiffiffi
2
p
.
In rectangular coordinates the equation r ¼ 4 cos becomes x2
þ y2
þ ðz 2Þ2
¼ 4 and the tangent
plane can be determined from this as in Problem 8.1. In other cases, however, it may not be so easy to
obtain the equation in rectangular form, and in such cases the method of part (a) is simpler to use.
(c) The equations of the normal line can be represented by
x þ
ffiffiffi
2
p
1
¼
y
ffiffiffi
2
p
1
¼
z 2
0
the significance of the right-hand member being that the line lies in the plane z ¼ 2. Thus, the required
line is given by
x þ
ffiffiffi
2
p
1
¼
y
ffiffiffi
2
p
1
; z ¼ 0 or x þ y ¼ 0; z ¼ 0
TANGENT LINE AND NORMAL PLANE TO A CURVE
8.5. Find equations for the (a) tangent line and (b) normal plane to the curve x ¼ t cos t,
y ¼ 3 þ sin 2t, z ¼ 1 þ cos 3t at the point where t ¼ 1
2 .
(a) The vector from origin O (see Fig. 8-2, Page 183) to any point of curve C is R ¼ ðt cos tÞiþ
ð3 þ sin 2tÞj þ ð1 þ cos 3tÞk. Then a vector tangent to C at the point where t ¼ 1
2 is
T0 ¼
dR
dt t¼1=2
¼ ð1 þ sin tÞi þ 2 cos 2t j 3 sin 3t kjt¼1=2 ¼ 2i 2j þ 3k
The vector from O to the point where t ¼ 1
2 is r0 ¼ 1
2 i þ 3j þ k.
The vector from O to any point ðx; y; zÞ on the tangent line is r ¼ xi þ yj þ zk.
Then r r0 ¼ ðx 1
2 Þi þ y 3Þj þ ðz 1Þk is collinear with T0, so that the required equation is
ðr r0Þ T0 ¼ 0; i:e:;
i j k
x 1
2 y 3 z 1
2 2 3
¼ 0
and the required equations are
x 1
2
2
¼
y 3
2
¼
z 1
3
or in parametric form x ¼ 2t þ 1
2 , y ¼ 3 2t,
z ¼ 3t þ 1:
(b) Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal plane. The vector from O
to the point where t ¼ 1
2 is r0 ¼ 1
2 i þ 3j þ k. The vector r r0 ¼ ðx 1
2 Þi þ ð y 3Þj þ ðz 1Þk lies
in the normal plane and hence is perpendicular to T0. Then the required equation is ðr r0Þ T0 ¼ 0 or
2ðx 1
2 Þ 2ð y 3Þ þ 3ðz 1Þ ¼ 0.
8.6. Find equations for the (a) tangent line and (b) normal plane to the curve 3x2
y þ y2
z ¼ 2,
2xz x2
y ¼ 3 at the point ð1; 1; 1Þ.
(a) The equations of the surfaces intersecting in the curve are
F ¼ 3x2
y þ y2
z þ 2 ¼ 0; G ¼ 2xz x2
y 3 ¼ 0
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 191](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-200-320.jpg)
![192 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
The normals to each surface at the point Pð1; 1; 1Þ are, respectively,
N1 ¼ rFjP ¼ 6xyi þ ð3x2
þ 2yzÞj þ y2
k ¼ 6 þ j þ k
N2 ¼ rGjP ¼ ð2z 2xyÞi x2
j þ 2xk ¼ 4i j þ 2k
Then a tangent vector to the curve at P is
T0 ¼ N1 N2 ¼ ð6i þ j þ kÞ ð4 j þ 2kÞ ¼ 3i þ 16j þ 2k
Thus, as in Problem 8.5(a), the tangent line is given by
ðr r0Þ T0 ¼ 0 or fðx 1Þi þ ð y þ 1Þj þ ðz 1Þkg f3i þ 16j þ 2kg ¼ 0
x 1
3
¼
y þ 1
16
¼
z 1
2
or x ¼ 1 þ 3t; y ¼ 16t 1; z ¼ 2t þ 1
i.e.,
(b) As in Problem 8.5(b) the normal plane is given by
ðr r0Þ T0 ¼ 0 or fðx 1Þi þ ð y þ 1Þj þ ðz 1Þkg f3i þ 16j þ 2kg ¼ 0
3ðx 1Þ þ 16ð y þ 1Þ þ 2ðz 1Þ ¼ 0 or 3x þ 16y þ 2z ¼ 11
i.e.,
The results in (a) and (b) can also be obtained by using equations (7) and (10), respectively, on Page
185.
8.7. Establish equation (10), Page 185.
Suppose the curve is defined by the intersection of two surfaces whose equations are Fðx; y; zÞ ¼ 0,
Gðx; y; zÞ ¼ 0, where we assume F and G continuously differentiable.
The normals to each surface at point P are given respectively by N1 ¼ rFjP and N2 ¼ rGjP. Then a
tangent vector to the curve at P is T0 ¼ N1 N2 ¼ rFjP rGjP. Thus, the equation of the normal plane is
ðr r0Þ T0 ¼ 0. Now
T0 ¼ rFjP rGjP ¼ fðFxi þ Fyj þ FzkÞ ðGxi þ Gyj þ GzkÞgjP
¼
i j k
Fx Fy Fz
Gx Gy Gz P
¼
Fy Fz
Gy Gz P
i þ
Fx Fx
Gx Gx P
j þ
Fx Fy
Gx Gy P
k
and so the required equation is
ðr r0Þ rFjP ¼ 0 or
Fy Fz
Gy Gz P
ðx x0Þ þ
Fz Fx
Gz Gx P
ð y y0Þ þ
Fx Fy
Gx Gy P
ðz z0Þ ¼ 0
ENVELOPES
8.8. Prove that the envelope of the family ðx; y; Þ ¼ 0, if it exists, can be obtained by solving
simultaneously the equations ¼ 0 and ¼ 0.
Assume parametric equations of the envelope to be x ¼ f ð Þ; y ¼ gð Þ. Then ð f ð Þ; gð Þ; Þ ¼ 0
identically, and so upon differentiating with respect to [assuming that , f and g have continuous deriva-
tives], we have
x f 0
ð Þ þ yg0
ð Þ þ ¼ 0 ð1Þ
The slope of any member of the family ðx; y; Þ ¼ 0 at ðx; yÞ is given by x dx þ y dy ¼ 0 or
dy
dx
¼
x
y
. The slope of the envelope at ðx; yÞ is
dy
dx
¼
dy=d
dx=d
¼
g0
ð Þ
f 0
ð Þ
. Then at any point where the envelope and
a member of the family are tangent, we must have
x
y
¼
g0
ð Þ
f 0ð Þ
or x f 0
ð Þ þ yg0
ð Þ ¼ 0 ð2Þ
Comparing (2) with (1) we see that ¼ 0 and the required result follows.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-201-320.jpg)
![8.9. (a) Find the envelope of the family x sin þ y cos ¼ 1. (b) Illus-
trate the results geometrically.
(a) By Problem 8 the envelope, if it exists, is obtained by solving simulta-
neously the equations ðx; y; Þ ¼ x sin þ y cos 1 ¼ 0 and
ðx; y; Þ ¼ x cos y cos ¼ 0. From these equations we find
x ¼ sin ; y ¼ cos or x2
þ y2
¼ 1.
(b) The given family is a family of straight lines, some members of which
are indicated in Fig. 8-5. The envelope is the circle x2
þ y2
¼ 1.
8.10. Find the envelope of the family of surfaces z ¼ 2 x a2
y.
By a generalization of Problem 8.8 the required envelope, if it exists, is obtained by solving simulta-
neously the equations
ð1Þ ¼ 2 x 2
y z ¼ 0 and ð2Þ ¼ 2x 2 y ¼ 0
From (2) ¼ x=y. Then substitution in (1) yields x2
¼ yz, the required envelope.
8.11. Find the envelope of the two-parameter family of surfaces z ¼ x þ y .
The envelope of the family Fðx; y; z; ; Þ ¼ 0, if it exists, is obtained by eliminating and between the
equations F ¼ 0; F ¼ 0; F ¼ 0 (see Problem, 8.43). Now
F ¼ z x y þ ¼ 0; F ¼ x þ ¼ 0; F ¼ y þ ¼ 0
Then ¼ x, ¼ y; and we have z ¼ xy.
DIRECTIONAL DERIVATIVES
8.12. Find the directional derivative of F ¼ x2
yz3
along the curve x ¼ eu
, y ¼ 2 sin u þ 1, z ¼ u cos u
at the point P where u ¼ 0.
The point P corresponding to u ¼ 0 is ð1; 1; 1Þ. Then
rF ¼ 2xyz3
i þ x2
z3
j þ 3x2
yz2
k ¼ 2i j þ 3k at P
A tangent vector to the curve is
dr
du
¼
d
du
feu
i þ ð2 sin u þ 1Þj þ ðu cos uÞkg
¼ eu
i þ 2 cos uj þ ð1 þ sin uÞk ¼ i þ 2j þ k at P
and the unit tangent vector in this direction is T0 ¼
i þ 2j þ k
ffiffiffi
6
p :
Then
Directional derivative ¼ rF T0 ¼ ð2i j þ 3kÞ
i þ 2j þ k
ffiffiffi
6
p
¼
3
ffiffiffi
6
p ¼
1
2
ffiffiffi
6
p
:
Since this is positive, F is increasing in this direction.
8.13. Prove that the greatest rate of change of F, i.e., the maximum directional derivative, takes place in
the direction of, and has the magnitude of, the vector rF.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 193
y
x
Fig. 8-5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-202-320.jpg)
![Then
¼
ðu2ð Þ
u1ð Þ
f ðx; Þ dx þ f ð2; þ Þ
u2
f ð1; þ Þ
u1
Taking the limit as ! 0, making use of the fact that the functions are assumed to have continuous
derivatives, we obtain
d
d
¼
ðu2ð Þ
u1ð Þ
f ðx; Þ dx þ f ½u2ð Þ;
du2
d
f ½u1ð Þ;
du1
d
8.16. If ð Þ ¼
ð 2
sin x
x
dx, find 0
ð Þ where 6¼ 0.
By Leibnitz’s rule,
0
ð Þ ¼
ð 2
@
@
sin x
x
dx þ
sinð 2
Þ
2
d
d
ð 2
Þ
sinð Þ d
d
ð Þ
¼
ð 2
cos x dx þ
2 sin 3
sin 2
¼
sin x
2
þ
2 sin 3
sin 2
¼
3 sin 3
2 sin 2
8.17. If
ð
0
dx
cos x
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1
p ; 1 find
ð
0
dx
ð2 cos xÞ2
. (See Problem 5.58, Chapter 5.)
By Leibnitz’s rule, if ð Þ ¼
ð
0
dx
cos x
¼ ð 2
1Þ1=2
; then
0
ð Þ ¼
ð
0
dx
ð cos xÞ2
¼
1
2
ð 2
1Þ3=2
2 ¼
ð 2
1Þ3=2
Thus
ð
0
dx
ð cos xÞ2
¼
ð 2 1Þ3=2
from which
ð
0
dx
ð2 cos xÞ2
¼
2
3
ffiffiffi
3
p :
INTEGRATION UNDER THE INTEGRAL SIGN
8.18. Prove the result (18), Page 187, for integration under the integral sign.
Consider ð1Þ ð Þ ¼
ðu2
u1
ð
a
f ðx; Þ d
dx
By Leibnitz’s rule,
0
ð Þ ¼
ðu2
u1
@
@
ð
a
f ðx; Þ d
dx ¼
ðu2
u1
f ðx; Þ dx ¼ ð Þ
Then by integration, ð2Þ ð Þ ¼
ð
a
ð Þ d þ c
Since ðaÞ ¼ 0 from (1), we have c ¼ 0 in (2). Thus from (1) and (2) with c ¼ 0, we find
ðu2
u1
ð
a
f ðx; Þ dx
dx ¼
ð
a
ðu2
u1
f ðx; Þ dx
d
Putting ¼ b, the required result follows.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 195](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-204-320.jpg)
![8.19. Prove that
ð
0
ln
b cos x
a cos x
dx ¼ ln
b þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 1
p
a þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
1
p
!
if a; b 1.
From Problem 5.58, Chapter 5,
ð
0
dx
cos x
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1
p ; 1:
Integrating the left side with respect to from a to b yields
ð
0
ðb
a
d
cos x
dx ¼
ð
0
lnð cos xÞ
b
a
dx ¼
ð
0
ln
b cos x
a cos x
dx
Integrating the right side with respect to from a to b yields
ð
0
d
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1
p ¼ lnð þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1
p
Þ
b
a
¼ ln
b þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 1
p
a þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
1
p
!
and the required result follows.
MAXIMA AND MINIMA
8.20. Prove that a necessary condition for f ðx; yÞ to have a relative extremum (maximum or minimum)
at ðx0; y0Þ is that fxðx0; y0Þ ¼ 0, fyðx0; y0Þ ¼ 0.
If f ðx0; y0Þ is to be an extreme value for f ðx; yÞ, then it must be an extreme value for both f ðx; y0Þ and
f ðx0; yÞ. But a necessary condition that these have extreme values at xx ¼ 0 and y ¼ y0, respectively, is
fxðx0; y0Þ ¼ 0, fyðx0; y0Þ ¼ 0 (using results for functions of one variable).
8.21. Let f be continuous and have continuous partial derivatives of order two, at least, in a region R
with the critical point P0ðx0; y0Þ an interior point. Determine the sufficient conditions for relative
extrema at P0.
In the case of one variable, sufficient conditions for a relative extrema were formulated through the
second derivative [if positive then a relative minimum, if negative then a relative maximum, if zero a possible
point of inflection but more investigation is necessary]. In the case of z ¼ f ðx; yÞ that is before us we can
expect the second partial derivatives to supply information. (See Fig. 8-6.)
First observe that solutions of the quadratic equation
At2
þ 2Bt þ C ¼ 0 are t ¼
2B
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4B2
4AC
p
2A
Further observe that the nature of these solutions is determined by B2
AC. If the quantity is positive
the solutions are real and distinct; if negative, they are complex conjugate; and if zero, the two solutions are
coincident.
196 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
Fig. 8-6](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-205-320.jpg)
![The expression B2
AC also has the property of invariance with respect to plane rotations
x ¼
x
x cos
y
y sin
y ¼
x
x sin þ
y
y cos
It has been discovered that with the identifications A ¼ fxx; B ¼ fxy; C ¼ fyy, we have the partial deri-
vative form f 2
xy fxxfyy that characterizes relative extrema.
The demonstration of invariance of this form can be found in analytic geometric books. However, if
you would like to put the problem in the context of the second partial derivative, observe that
f
x
x ¼ fx
@x
@
x
x
þ fy
@y
@
x
x
¼ fx cos þ fy sin
f
y
y ¼ fx
@x
@
y
y
þ fy
@y
@
y
y
¼ fx sin þ fy cos
Then using the chain rule to compute the second partial derivatives and proceeding by straightforward
but tedious calculation one shows that
f 2
xy ¼ fxxfyy ¼ f 2
x
x
y
y f
x
x
x
xf
y
y
y
y:
The following equivalences are a consequence of this invariant form (independently of direction in the
tangent plane at P0):
f 2
xy fxx fyy 0 and fxx fyy 0 ð1Þ
f 2
xy fxx fyy 0 and fxx fyy 0 ð2Þ
The key relation is (1) because in order that this equivalence hold, both fx fy must have the same sign.
We can look to the one variable case (make the same argument for each coordinate direction) and conclude
that there is a relative minimum at P0 if both partial derivatives are positive and a relative maximum if both
are negative. We can make this argument for any pair of coordinate directions because of the invariance
under rotation that was established.
If (2) holds, then the point is called a saddle point. If the quadratic form is zero, no information results.
Observe that this situation is analogous to the one variable extreme value theory in which the nature of
f at x, and with f 0
ðxÞ ¼ 0, is undecided if f 00
ðxÞ ¼ 0.
8.22. Find the relative maxima and minima of f ðx; yÞ ¼ x3
þ y3
3x 12y þ 20.
fx ¼ 3x2
3 ¼ 0 when x ¼ 1; fy ¼ 3y2
12 ¼ 0 when y ¼ 2. Then critical points are Pð1; 2Þ,
Qð1; 2Þ; Rð1; 2Þ; Sð1; 2Þ.
fxx ¼ 6x; fyy ¼ 6y; fxy ¼ 0. Then ¼ fxxfyy f 2
xy ¼ 36xy.
At Pð1; 2Þ; 0 and fxx (or fyyÞ 0; hence P is a relative minimum point.
At Qð1; 2Þ; 0 and Q is neither a relative maximum or minimum point.
At Rð1; 2Þ; 0 and R is neither a relative maximum or minimum point.
At Sð1; 2Þ; 0 and fxx (or fyyÞ 0 so S is a relative maximum point.
Thus, the relative minimum value of f ðx; yÞ occurring at P is 2, while the relative maximum value
occurring at S is 38. Points Q and R are saddle points.
8.23. A rectangular box, open at the top, is to have a volume of 32 cubic feet. What must be the
dimensions so that the total surface is a minimum?
If x, y and z are the edges (see Fig. 8-7), then
ð1Þ Volume of box ¼ V ¼ xyz ¼ 32
ð2Þ Surface area of box ¼ S ¼ xy þ 2yz þ 2xz
or, since z ¼ 32=xy from (1),
S ¼ xy þ
64
x
þ
64
y
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 197
Fig. 8-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-206-320.jpg)
![CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 199
(a) We must find the extrema of F ¼ x2
þ y2
þ z2
subject to the constraint conditions 1 ¼
x2
4
þ
y2
5
þ
z2
25
1 ¼ 0 and 2 ¼ x þ y z ¼ 0. In this case we use two Lagrange multipliers 1; 2 and consider
the function
G ¼ F þ 11 þ 22 ¼ x2
þ y2
þ z2
þ 1
x2
4
þ
y2
5
þ
z2
25
1
!
þ 2ðx þ y zÞ
Taking the partial derivatives of G with respect to x; y; z and setting them equal to zero, we find
Gx ¼ 2x þ
1x
2
þ 2 ¼ 0; Gy ¼ 2y þ
21y
5
þ 2 ¼ 0; Gx ¼ 2z þ
21z
25
2 ¼ 0 ð1Þ
Solving these equations for x; y; z, we find
x ¼
22
1 þ 4
; y ¼
52
21 þ 10
; z ¼
252
21 þ 50
ð2Þ
From the second constraint condition, x þ y z ¼ 0, we obtain on division by 2, assumed dif-
ferent from zero (this is justified since otherwise we would have x ¼ 0; y ¼ 0; z ¼ 0, which would not
satisfy the first constraint condition), the result
2
1 þ 4
þ
5
21 þ 10
þ
25
21 þ 50
¼ 0
Multiplying both sides by 2ð1 þ 4Þð1 þ 5Þð1 þ 25Þ and simplifying yields
172
1 þ 2451 þ 750 ¼ 0 or ð1 þ 10Þð171 þ 75Þ ¼ 0
from which 1 ¼ 10 or 75=17.
Case 1: 1 ¼ 10.
From (2), x ¼ 1
3 2; y ¼ 1
2 2; z ¼ 5
6 2. Substituting in the first constraint condition, x2
=4 þ y2
=5þ
z2
=25 ¼ 1, yields 2
2 ¼ 180=19 or 2 ¼ 6
ffiffiffiffiffiffiffiffiffiffi
5=19
p
. This gives the two critical points
ð2
ffiffiffiffiffiffiffiffiffiffi
5=19
p
; 3
ffiffiffiffiffiffiffiffiffiffi
5=19
p
; 5
ffiffiffiffiffiffiffiffiffiffi
5=19
p
Þ; ð2
ffiffiffiffiffiffiffiffiffiffi
5=19
p
; 3
ffiffiffiffiffiffiffiffiffiffi
5=19
p
; 5
ffiffiffiffiffiffiffiffiffiffi
5=19
p
Þ
The value of x2
þ y2
þ z2
corresponding to these critical points is ð20 þ 45 þ 125Þ=19 ¼ 10.
Case 2: 1 ¼ 75=17:
From (2), x ¼ 34
7 2; y ¼ 17
4 2; z ¼ 17
28 2. Substituting in the first constraint condition,
x2
=4 þ y2
=5 þ z2
=25 ¼ 1, yields 2 ¼ 140=ð17
ffiffiffiffiffiffiffiffi
646
p
Þ which gives the critical points
ð40=
ffiffiffiffiffiffiffiffi
646
p
; 35
ffiffiffiffiffiffiffiffi
646
p
; 5=
ffiffiffiffiffiffiffiffi
646
p
Þ; ð40=
ffiffiffiffiffiffiffiffi
646
p
; 35=
ffiffiffiffiffiffiffiffi
646
p
; 5=
ffiffiffiffiffiffiffiffi
646
p
Þ
The value of x2
þ y2
þ z2
corresponding to these is ð1600 þ 1225 þ 25Þ=646 ¼ 75=17.
Thus, the required maximum value is 10 and the minimum value is 75/17.
(b) Since x2
þ y2
þ z2
represents the square of the distance of ðx; y; zÞ from the origin ð0; 0; 0Þ, the problem
is equivalent to determining the largest and smallest distances from the origin to the curve of intersec-
tion of the ellipsoid x2
=4 þ y2
=5 þ z2
=25 ¼ 1 and the plane z ¼ x þ y. Since this curve is an ellipse, we
have the interpretation that
ffiffiffiffiffi
10
p
and
ffiffiffiffiffiffiffiffiffiffiffiffi
75=17
p
are the lengths of the semi-major and semi-minor axes of
this ellipse.
The fact that the maximum and minimum values happen to be given by 1 in both Case 1 and
Case 2 is more than a coincidence. It follows, in fact, on multiplying equations (1) by x, y, and z in
succession and adding, for we then obtain
2x2
þ
1x2
2
þ 2x þ 2y2
þ
21y2
5
þ 2y þ 2z2
þ
21z2
25
2z ¼ 0
x2
þ y2
þ z2
þ 1
x2
4
þ
y2
5
þ
z2
25
!
þ 2ðx þ y zÞ ¼ 0
i.e.,
Then using the constraint conditions, we find x2
þ y2
þ z2
¼ 1.
For a generalization of this problem, see Problem 8.76.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-208-320.jpg)
![The necessary conditions for a minimum are @F=@a ¼ 0, @F=@b ¼ 0. Performing these differentiations,
we obtain
@F
@a
¼
ð
0
@
@a
fsin x ðax2
þ bxÞg2
dx ¼ 2
ð
0
x2
fsin x ðax2
þ bxÞg dx ¼ 0
@F
@b
¼
ð
0
@
@b
fsin x ðax2
þ bxÞg2
dx ¼ 2
ð
0
xfsin x ðax2
þ bxÞg dx ¼ 0
From these we find
a
ð
0
x4
dx þ b
ð
0
x3
dx ¼
ð
0
x2
sin x dx
a
ð
0
x3
dx þ b
ð
0
x2
dx ¼
ð
0
x sin x dx
8
:
or
5
a
5
þ
4
b
4
¼ 2
4
4
a
4
þ
3
b
3
¼
8
:
Solving for a and b, we find
a ¼
20
3
320
5
0:40065; b
240
4
12
2
1:24798
We can show that for these values, Fða; bÞ is indeed a minimum using the sufficiency conditions on Page
188.
The polynomial ax2
þ bx is said to be a least square approximation of sin x over the interval ð0; Þ. The
ideas involved here are of importance in many branches of mathematics and their applications.
Supplementary Problems
TANGENT PLANE AND NORMAL LINE TO A SURFACE
8.31. Find the equations of the (a) tangent plane and (b) normal line to the surface x2
þ y2
¼ 4z at ð2; 4; 5Þ.
Ans. (a) x 2y z ¼ 5; ðbÞ
x 2
1
¼
y þ 4
2
¼
z 5
1
:
8.32. If z ¼ f ðx; yÞ, prove that the equations for the tangent plane and normal line at point Pðx0; y0; z0Þ are given
respectively by
ðaÞ z z0 ¼ fxjPðx x0Þ þ fyjPð y y0Þ and ðbÞ
x x0
fxjP
¼
y y0
fyjP
¼
z z0
1
8.33. Prove that the acute angle between the z axis and the normal to the surface Fðx; y; zÞ ¼ 0 at any point is
given by sec ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
F2
x þ F2
y þ F2
z
q
=jFzj.
8.34. The equation of a surface is given in cylindrical coordinates by Fð; ; zÞ ¼ 0, where F is continuously
differentiable. Prove that the equations of (a) the tangent plane and (b) the normal line at the point
Pð0; 0; z0Þ are given respectively by
Aðx x0Þ þ Bð y y0Þ þ Cðz z0Þ ¼ 0 and
x x0
A
¼
y y0
B
¼
z z0
C
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 201](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-210-320.jpg)
![where x0 ¼ 0 cos 0, y0 ¼ 0 sin 0 and
A ¼ FjP cos 0
1
FjP sin 0; B ¼ FjP sin 0 þ
1
FjP cos 0; C ¼ FzjP
8.35. Use Problem 8.34 to find the equation of the tangent plane to the surface z ¼ at the point where ¼ 2,
¼ =2, z ¼ 1. To check your answer work the problem using rectangular coordinates.
Ans. 2x y þ 2z ¼ 0
TANGENT LINE AND NORMAL PLANE TO A CURVE
8.36. Find the equations of the (a) tangent line and (b) normal plane to the space curve x ¼ 6 sin t, y ¼ 4 cos 3t,
z ¼ 2 sin 5t at the point where t ¼ =4.
Ans: ðaÞ
x 3
ffiffiffi
2
p
3
¼
y þ 2
ffiffiffi
2
p
6
¼
z þ
ffiffiffi
2
p
5
ðbÞ 3x 6y 5z ¼ 26
ffiffiffi
2
p
8.37. The surfaces x þ y þ z ¼ 3 and x2
y2
þ 2z2
¼ 2 intersect in a space curve. Find the equations of the
(a) tangent line (b) normal plane to this space curve at the point ð1; 1; 1Þ.
Ans: ðaÞ
x 1
3
¼
y 1
1
¼
z 1
2
; ðbÞ 3x y 2z ¼ 0
ENVELOPES
8.38. Find the envelope of each of the following families of curves in the xy plane. In each case construct a graph.
(a) y ¼ x 2
; ðbÞ
x2
þ
y2
1
¼ 1.
Ans. (a) x2
¼ 4y; ðbÞ x þ y ¼ 1; x y ¼ 1
8.39. Find the envelope of a family of lines having the property that the length intercepted between the x and y
axes is a constant a. Ans. x2=3
þ y2=3
¼ a2=3
8.40. Find the envelope of the family of circles having centers on the parabola y ¼ x2
and passing through its
vertex. [Hint: Let ð ; 2
Þ be any point on the parabola.] Ans. x2
¼ y3
=ð2y þ 1Þ
8.41. Find the envelope of the normals (called an evolute) to the parabola y ¼ 1
2 x2
and construct a graph.
Ans. 8ðy 1Þ3
¼ 27x2
8.42. Find the envelope of the following families of surfaces:
ðaÞ ðx yÞ 2
z ¼ 1; ðbÞ ðx Þ2
þ y2
¼ 2 z
Ans. ðaÞ 4z ¼ ðx yÞ2
; ðbÞ y2
¼ z2
þ 2xz
8.43. Prove that the envelope of the two parameter family of surfaces Fðx; y; z; ; Þ ¼ 0, if it exists, is obtained by
eliminating and in the equations F ¼ 0; F ¼ 0; F ¼ 0.
8.44. Find the envelope of the two parameter families (a) z ¼ x þ y 2
2
and (b) x cos þ y cos þ
z cos ¼ a where cos2
þ cos2
þ cos2
¼ 1 and a is a constant.
Ans. ðaÞ 4z ¼ x2
þ y2
; ðbÞ x2
þ y2
þ z2
¼ a2
DIRECTIONAL DERIVATIVES
8.45. (a) Find the directional derivative of U ¼ 2xy z2
at ð2; 1; 1Þ in a direction toward ð3; 1; 1Þ. (b) In what
direction is the directional derivative a maximum? (c) What is the value of this maximum?
Ans. ðaÞ 10=3; ðbÞ 2i þ 4j 2k; ðcÞ 2
ffiffiffi
6
p
202 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-211-320.jpg)
![8.46. The temperature at any point ðx; yÞ in the xy plane is given by T ¼ 100xy=ðx2
þ y2
Þ. (a) Find the direc-
tional derivative at the point ð2; 1Þ in a direction making an angle of 608 with the positive x-axis. (b) In
what direction from ð2; 1Þ would the derivative be a maximum? (c) What is the value of this maximum?
Ans. (a) 12
ffiffiffi
3
p
6; (b) in a direction making an angle of tan1
2 with the positive x-axis, or in the
direction i þ 2j; (c) 12
ffiffiffi
5
p
8.47. Prove that if Fð; ; zÞ is continuously differentiable, the maximum directional derivative of F at any point is
given by
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
@F
@
2
þ
1
2
@F
@
2
þ
@F
@z
2
s
.
DIFFERENTIATION UNDER THE INTEGRAL SIGN
8.48. If ð Þ ¼
ð1=
ffiffi
p
cos x2
dx, find
d
d
. Ans.
ð1=
ffiffi
p
x2
sin x2
dx
1
2
cos
1
1
2
ffiffiffi
p cos 2
8.49. (a) If Fð Þ ¼
ð 2
0
tan1 x
dx, find
dF
d
by Leibnitz’s rule. (b) Check the result in (a) by direct integration.
Ans. ðaÞ 2 tan1
1
2 lnð 2
þ 1Þ
8.50. Given
ð1
0
xp
dx ¼
1
p þ 1
; p 1. Prove that
ð1
0
xp
ðln xÞm
dx ¼
ð1Þm
m!
ðp þ 1Þmþ1
; m ¼ 1; 2; 3; . . . .
8.51. Prove that
ð
0
lnð1 þ cos xÞ dx ¼ ln
1 þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
p
2
!
; j j 1.
8.52. Prove that
ð
0
lnð1 2 cos x þ 2
Þ dx ¼
ln 2
; j j 1
0; j j 1
. Discuss the case j j ¼ 1.
8.53. Show that
ð
0
dx
ð5 3 cos xÞ3
¼
59
2048
:
INTEGRATION UNDER THE INTEGRAL SIGN
8.54. Verify that
ð1
0
ð2
1
ð 2
x2
Þ dx
d ¼
ð2
1
ð1
0
ð 2
x2
Þ d
dx.
8.55. Starting with the result
ð2
0
ð sin xÞ dx ¼ 2 , prove that for all constants a and b,
ð2
0
fðb sin xÞ2
ða sin xÞ2
g dx ¼ 2ðb2
a2
Þ
8.56. Use the result
ð2
0
dx
þ sin x
¼
2
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1
p ; 1 to prove that
ð2
0
ln
5 þ 3 sin x
5 þ 4 sin x
dx ¼ 2 ln
9
8
8.57. (a) Use the result
ð=2
0
dx
1 þ cos x
¼
cos1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
p ; 0 @ 1 to show that for 0 @ a 1; 0 @ b 1
ð=2
0
sec x ln
1 þ b cos x
1 þ a cos x
dx ¼ 1
2 fðcos1
aÞ2
ðcos1
bÞ2
g
(b) Show that
ð=2
0
sec x lnð1 þ 1
2 cos xÞ dx ¼
52
72
.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 203](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-212-320.jpg)
![8.72. (a) Prove that the shortest distance from the point ða; b; cÞ to the plane Ax þ By þ Cz þ D ¼ 0 is
Aa þ Bb þ Cc þ D
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
A2 þ B2 þ C2
p
(b) Find the shortest distance from ð1; 2; 3Þ to the plane 2x 3y þ 6z ¼ 20. Ans. (b) 6
8.73. The potential V due to a charge distribution is given in spherical coordinates ðr; ; Þ by
V ¼
p cos
r2
where p is a constant. Prove that the maximum directional derivative at any point is
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sin2
þ 4 cos2
p
r3
8.74. Prove that
ð1
0
xm
xn
ln x
dx ¼ ln
m þ 1
n þ 1
if m 0; n 0. Can you extend the result to the case
m 1; n 1?
8.75. (a) If b2
4ac 0 and a 0; c 0, prove that the area of the ellipse ax2
þ bxy þ cy2
¼ 1 is 2=
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4ac b2
p
.
[Hint: Find the maximum and minimum values of x2
þ y2
subject to the constraint ax2
þ bxy þ cy2
¼ 1.]
8.76. Prove that the maximum and minimum distances from the origin to the curve of intersection defined by
x2
=a2
þ y2
=b2
þ z2
=c2
¼ 1 and Ax þ By þ Cz ¼ 0 can be obtained by solving for d the equation
A2
a2
a2 d2
þ
B2
b2
b2 d2
þ
C2
c2
c2 d2
¼ 0
8.77. Prove that the last equation in the preceding problem always has two real solutions d2
1 and d2
2 for any real
non-zero constants a; b; c and any real constants A; B; C (not all zero). Discuss the geometrical significance
of this.
8.78. (a) Prove that IM ¼
ðM
0
dx
ðx2 þ 2Þ2
¼
1
2 3
tan1 M
þ
M
2 2
ð 2
þ M2
Þ
ðbÞ Find lim
M!1
IM: This can be denoted by
ðx
0
dx
ðx2 þ 2Þ2
:
ðcÞ Is lim
M!1
d
d
ðM
0
dx
ðx2 þ 2Þ2
¼
d
d
lim
M!1
ðM
0
dx
ðx2 þ 2Þ2
?
8.79. Find the point on the paraboloid z ¼ x2
þ y2
which is closest to the point ð3; 6; 4Þ.
Ans. ð1; 2; 5Þ
8.80. Investigate the maxima and minima of f ðx; yÞ ¼ ðx2
2x þ 4y2
8yÞ2
.
Ans. minimum value ¼ 0
8.81. (a) Prove that
ð=2
0
cos x dx
cos x þ sin x
¼
2ð 2
þ 1Þ
ln
2
þ 1
:
ðbÞ Use ðaÞ to prove that
ð=2
0
cos2
x dx
ð2 cos x þ sin xÞ2
¼
3 þ 5 8 ln 2
50
:
8.82. (a) Find sufficient conditions for a relative maximum or minimum of w ¼ f ðx; y; zÞ.
(b) Examine w ¼ x2
þ y2
þ z2
6xy þ 8xz 10yz for maxima and minima.
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 205](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-214-320.jpg)
![where the integral in braces is to be evaluated first (keeping x constant) and finally integrating with
respect to x from a to b. The result (4) indicates how a double integral can be evaluated by expressing it
in terms of two single integrals called iterated integrals.
The process of iterated integration is visually illustrated in Fig. 9-3a,b and further illustrated as
follows.
The general idea, as demonstrated with respect to a given three-space region, is to establish a plane
section, integrate to determine its area, and then add up all the plane sections through an integration
with respect to the remaining variable. For example, choose a value of x (say, x ¼ x0
Þ. The intersection
of the plane x ¼ x0
with the solid establishes the plane section. In it z ¼ Fðx0
; yÞ is the height function,
and if y ¼ f1ðxÞ and y ¼ f2ðxÞ (for all z) are the bounding cylindrical surfaces of the solid, then the width
is f2ðx0
Þ f1ðx0
Þ, i.e., y2 y1. Thus, the area of the section is A ¼
ðy2
y1
Fðx0
; yÞ dy. Now establish slabs
Ajxj, where for each interval xj ¼ xj xj1, there is an intermediate value x0
j . Then sum these to get
an approximation to the target volume. Adding the slabs and taking the limit yields
V ¼ lim
n!1
X
n
j¼1
Aj xj ¼
ðb
a
ðy2
y1
Fðx; yÞ dy
dx
In some cases the order of integration is dictated by the geometry. For example, if r is such that any
lines parallel to the x-axis meet the boundary of r in at most two points (as in Fig. 9-1), then the
equations of curves CAD and CBD can be written x ¼ g1ðyÞ and x ¼ g2ð yÞ respectively and we find
similarly
ð ð
r
Fðx; yÞ dx dy ¼
ðd
y¼c
ðg2ð yÞ
x¼g1ð yÞ
Fðx; yÞ dx dy ð5Þ
¼
ðd
y¼c
ðg2ð yÞ
x¼g1ð yÞ
Fðx; yÞ dx
dy
If the double integral exists, (4) and (5) yield the same value. (See, however, Problem 9.21.) In writing a
double integral, either of the forms (4) or (5), whichever is appropriate, may be used. We call one form
an interchange of the order of integration with respect to the other form.
CHAP. 9] MULTIPLE INTEGRALS 209
Fig. 9-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-218-320.jpg)
![TRANSFORMATIONS OF MULTIPLE INTEGRALS
In evaluating a multiple integral over a region r, it is often convenient to use coordinates other than
rectangular, such as the curvilinear coordinates considered in Chapters 6 and 7.
If we let ðu; vÞ be curvilinear coordinates of points in a plane, there will be a set of transformation
equations x ¼ f ðu; vÞ; y ¼ gðu; vÞ mapping points ðx; yÞ of the xy plane into points ðu; vÞ of the uv plane.
In such case the region r of the xy plane is mapped into a region r0
of the uv plane. We then have
ð ð
r
Fðx; yÞ dx dy ¼
ð ð
r0
Gðu; vÞ
@ðx; yÞ
@ðu; vÞ
du dv ð9Þ
where Gðu; vÞ Ff f ðu; vÞ; gðu; vÞg and
@ðx; yÞ
@ðu; vÞ
@x
@u
@x
@v
@y
@u
@y
@v
ð10Þ
is the Jacobian of x and y with respect to u and v (see Chapter 6).
Similarly if ðu; v; wÞ are curvilinear coordinates in three dimensions, there will be a set of transfor-
mation equations x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ; z ¼ hðu; v; wÞ and we can write
ð ð ð
r
Fðx; y; zÞ dx dy dz ¼
ð ð ð
r0
Gðu; v; wÞ
@ðx; y; zÞ
@ðu; v; wÞ
du dv dw ð11Þ
where Gðu; v; wÞ Fff ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞg and
@ðx; y; zÞ
@ðu; v; wÞ
@x
@u
@x
@v
@x
@w
@y
@u
@y
@v
@y
@w
@z
@u
@z
@v
@z
@w
ð12Þ
is the Jacobian of x, y, and z with respect to u, v, and w.
The results (9) and (11) correspond to change of variables for double and triple integrals.
Generalizations to higher dimensions are easily made.
THE DIFFERENTIAL ELEMENT OF AREA IN POLAR COORDINATES, DIFFERENTIAL
ELEMENTS OF AREA IN CYLINDRAL AND SPHERICAL COORDINATES
Of special interest is the differential element of area, dA, for polar coordinates in the plane, and the
differential elements of volume, dV, for cylindrical and spherical coordinates in three space. With these
in hand the double and triple integrals as expressed in these systems are seen to take the following forms.
(See Fig. 9-5.)
The transformation equations relating cylindrical coordinates to rectangular Cartesian ones
appeared in Chapter 7, in particular,
x ¼ cos ; y ¼ sin ; z ¼ z
The coordinate surfaces are circular cylinders, planes, and planes. (See Fig. 9-5.)
At any point of the space (other than the origin), the set of vectors
@r
@
;
@r
@
;
@r
@z
constitutes an
orthogonal basis.
CHAP. 9] MULTIPLE INTEGRALS 211](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-220-320.jpg)
![Following the same pattern as with cylindrical coordinates we discover that
dV ¼ r2
sin dr d d
and the iterated triple integral of Fðr; ; Þ has the spherical representation
ðr2
r1
ð2ðÞ
1ðÞ
ð2ðr;Þ
1ðr;Þ
Fðr; ; Þ r2
sin dr d d
Of course, the order of these integrations may be adapted to the geometry.
The coordinate surfaces in spherical coordinates are spheres, cones, and planes. If r is held
constant, say, r ¼ a, then we obtain the differential element of surface area
dA ¼ a2
sin d d
The first octant surface area of a sphere of radius a is
ð=2
0
ð=2
0
a2
sin d d ¼
ð=2
0
a2
ð cos Þ
2
0 d ¼
ð=2
0
a2
d ¼ a2
2
Thus, the surface area of the sphere is 4a2
.
Solved Problems
DOUBLE INTEGRALS
9.1. (a) Sketch the region r in the xy plane bounded by y ¼ x2
; x ¼ 2; y ¼ 1.
(b) Give a physical interpreation to
ð ð
r
ðx2
þ y2
Þ dx dy.
(c) Evaluate the double integral in (b).
(a) The required region r is shown shaded in Fig. 9-6 below.
(b) Since x2
þ y2
is the square of the distance from any point ðx; yÞ to ð0; 0Þ, we can consider the double
integral as representing the polar moment of inertia (i.e., moment of inertia with respect to the origin) of
the region r (assuming unit density).
CHAP. 9] MULTIPLE INTEGRALS 213
Fig. 9-6 Fig. 9-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-222-320.jpg)
![M ¼
ðb
a
ðf2
f1
dy dx ¼
ð1
0
ð ffiffiffiffiffiffiffiffi
2x2
p
x2
yx dy dx ¼
ð1
0
y2
2
# ffiffiffiffiffiffiffiffi
2x2
p
x2
x dx
ðbÞ
¼
ð1
0
1
2
xð2 x2
x4
Þ dx ¼
x2
2
x4
8
x6
12
#1
0
¼
7
24
(c) The coordinates of the center of mass are defined to be
x
x ¼
1
M
ðb
a
ðf2ðxÞ
f1ðxÞ
x dy dx and
y
y ¼
1
M
ðb
a
ðf2ðxÞ
f1ðxÞ
y dy dx
where
M ¼
ðb
a
ðf2ðxÞ
f1ðxÞ
dy dx
Thus,
M
x
x ¼
ð1
0
ð ffiffiffiffiffiffiffiffi
2x2
p
x2
x xy dy dx ¼
ð1
0
x2 y2
2
# ffiffiffiffiffiffiffiffi
2x2
p
x2
dx ¼
ð1
0
x2 1
2
½2 x2
x4
dx
¼
x3
3
x5
10
x7
14
#1
0
¼
1
3
1
10
1
14
¼
17
105
M
y
y ¼
ð1
0
ð ffiffiffiffiffiffiffiffi
2x2
p
x2
yx dy dx ¼
13
120
þ 4
ffiffiffi
2
p
15
9.4. Find the volume of the region common to the intersecting cylinders x2
þ y2
¼ a2
and
x2
þ z2
¼ a2
.
Required volume ¼ 8 times volume of region shown in Fig. 9-9
¼ 8
ða
x¼0
ð ffiffiffiffiffiffiffiffiffiffi
a2x2
p
y¼0
z dy dx
¼ 8
ða
x¼0
ð ffiffiffiffiffiffiffiffiffiffi
a2x2
p
y¼0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x2
p
dy dx
¼ 8
ða
x¼0
ða2
x2
Þ dx ¼
16a3
3
As an aid in setting up this integral, note that z dy dx corresponds to the volume of a column such as
shown darkly shaded in the figure. Keeping x constant and integrating with respect to y from y ¼ 0 to
y ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x2
p
corresponds to adding the volumes of all such columns in a slab parallel to the yz plane, thus
giving the volume of this slab. Finally, integrating with respect to x from x ¼ 0 to x ¼ a corresponds to
adding the volumes of all such slabs in the region, thus giving the required volume.
9.5. Find the volume of the region bounded by
z ¼ x þ y; z ¼ 6; x ¼ 0; y ¼ 0; z ¼ 0
CHAP. 9] MULTIPLE INTEGRALS 215](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-224-320.jpg)
![Let P be any point with coordinates ðx; yÞ or ðu; vÞ, where x ¼ f ðu; vÞ and y ¼ gðu; vÞ. Then the vector r
from O to P is given by r ¼ xi þ yj ¼ f ðu; vÞi þ gðu; vÞj. The tangent vectors to the coordinate curves u ¼ c1
and v ¼ c2, where c1 and c2 are constants, are @r=@v and @r=@u, respectively. Then the area of region r of
Fig. 9-11 is given approximately by
@r
@u
@r
@v
u v.
But
@r
@u
@r
@v
¼
i j k
@x
@u
@y
@u
0
@x
@v
@y
@v
0
¼
@x
@u
@y
@u
@x
@v
@y
@v
k ¼
@ðx; yÞ
@ðu; vÞ
k
@r
@u
@r
@v
u v ¼
@ðx; yÞ
@ðu; vÞ
u v
so that
The double integral is the limit of the sum
X
Ff f ðu; vÞ; gðu; vÞg
@ðx; yÞ
@ðu; vÞ
u v
taken over the entire region r. An investigation reveals that this limit is
ð ð
r0
Ff f ðu; vÞ; gðu; vÞg
@ðx; yÞ
@ðu; vÞ
du dv
where r0
is the region in the uv plane into which the region r is mapped under the transformation
x ¼ f ðu; vÞ; y ¼ gðu; vÞ.
Another method of justifying the above method of change of variables makes use of line integrals and
Green’s theorem in the plane (see Chapter 10, Problem 10.32).
9.7. If u ¼ x2
y2
and v ¼ 2xy, find @ðx; yÞ=@ðu; vÞ in terms of u and v.
@ðu; vÞ
@ðx; yÞ
¼
ux uy
vx vy
¼
2x 2y
2y 2x
¼ 4ðx2
þ y2
Þ
From the identify ðx2
þ y2
Þ2
¼ ðx2
y2
Þ2
þ ð2xyÞ2
we have
ðx2
þ y2
Þ2
¼ u2
þ v2
and x2
þ y2
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
þ v2
p
Then by Problem 6.43, Chapter 6,
@ðx; yÞ
@ðu; vÞ
¼
1
@ðu; vÞ=@ðx; yÞ
¼
1
4ðx2 þ y2Þ
¼
1
4
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2
þ v2
p
Another method: Solve the given equations for x and y in terms of u and v and find the Jacobian directly.
9.8. Find the polar moment of inertia of the region in the xy plane bounded by x2
y2
¼ 1,
x2
y2
¼ 9, xy ¼ 2; xy ¼ 4 assuming unit density.
Under the transformation x2
y2
¼ u, 2xy ¼ v the required region r in the xy plane [shaded in Fig.
9-12(a)] is mapped into region r0
of the uv plane [shaded in Fig. 9-12(b)]. Then:
Required polar moment of inertia ¼
ð ð
r
ðx2
þ y2
Þ dx dy ¼
ð ð
r0
ðx2
þ y2
Þ
@ðx; yÞ
@ðu; vÞ
du dv
¼
ð ð
r0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 þ v2
p du dv
4
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u2 þ v2
p ¼
1
4
ð9
u¼1
ð8
v¼4
du dv ¼ 8
where we have used the results of Problem 9.7.
CHAP. 9] MULTIPLE INTEGRALS 217](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-226-320.jpg)
![Note that the limits of integration for the region r0
can be constructed directly from the region r in the
xy plane without actually constructing the region r0
. In such case we use a grid as in Problem 9.6. The
coordinates ðu; vÞ are curvilinear coordinates, in this case called hyperbolic coordinates.
9.9. Evaluate
ð ð
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
q
dx dy, where r is the region in the xy plane bounded by x2
þ y2
¼ 4 and
x2
þ y2
¼ 9.
The presence of x2
þ y2
suggests the use of polar coordinates ð; Þ, where x ¼ cos ; y ¼ sin (see
Problem 6.39, Chapter 6). Under this transformation the region r [Fig. 9-13(a) below] is mapped into the
region r0
[Fig. 9-13(b) below].
Since
@ðx; yÞ
@ð; Þ
¼ , it follows that
ð ð
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2
q
dx dy ¼
ð ð
r0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2
q
@ðx; yÞ
@ð; Þ
d d ¼
ð ð
r0
d d
¼
ð2
¼0
ð3
¼2
2
d d ¼
ð2
¼0
3
3
3
2
d ¼
ð2
¼0
19
3
d ¼
38
3
218 MULTIPLE INTEGRALS [CHAP. 9
Fig. 9-12
Fig. 9-13](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-227-320.jpg)
![We can also write the integration limits for r0
immediately on observing the region r, since for fixed ,
varies from ¼ 2 to ¼ 3 within the sector shown dashed in Fig. 9-13(a). An integration with respect to
from ¼ 0 to ¼ 2 then gives the contribution from all sectors. Geometrically, d d represents the
area dA as shown in Fig. 9-13(a).
9.10. Find the area of the region in the xy plane bounded by the lemniscate 2
¼ a2
cos 2.
Here the curve is given directly in polar coordinates ð; Þ. By assigning various values to and finding
corresponding values of , we obtain the graph shown in Fig. 9-14. The required area (making use of
symmetry) is
4
ð=4
¼0
ða
ffiffiffiffiffiffiffiffiffi
cos 2
p
¼0
d d ¼ 4
ð=4
¼0
3
2
a
ffiffiffiffiffiffiffiffiffi
cos 2
p
¼0
d
¼ 2
ð=4
¼0
a2
cos 2 d ¼ a2
sin 2
=4
¼0
¼ a2
TRIPLE INTEGRALS
9.11. (a) Sketch the three-dimensional region r bounded by x þ y þ z ¼ a ða 0Þ; x ¼ 0; y ¼ 0; z ¼ 0.
(b) Give a physical interpretation to
ð ð ð
r
ðx2
þ y2
þ z2
Þ dx dy dz
(c) Evaluate the triple integral in (b).
(a) The required region r is shown in Fig. 9-15.
(b) Since x2
þ y2
þ z2
is the square of the distance from any point ðx; y; zÞ to ð0; 0; 0Þ, we can consider the
triple integral as representing the polar moment of inertia (i.e., moment of inertia with respect to the
origin) of the region r (assuming unit density).
We can also consider the triple integral as representing the mass of the region if the density varies
as x2
þ y2
þ z2
.
CHAP. 9] MULTIPLE INTEGRALS 219
Fig. 9-14 Fig. 9-15](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-228-320.jpg)
![ðaÞ Required volume ¼
ð ð ð
r
dx dy dz
¼
ð2
x¼0
ð6
y¼0
ð4x2
z¼0
dz dy dx
¼
ð2
x¼0
ð6
y¼0
ð4 x2
Þ dy dx
¼
ð2
x¼0
ð4 x2
Þy
6
y¼0
dx
¼
ð2
x¼0
ð24 6x2
Þ dx ¼ 32
(b) Total mass ¼
ð2
x¼0
ð6
y¼0
ð4x2
z¼0
dz dy dx ¼ 32 by part (a), since is constant. Then
x
x ¼
Total moment about yz plane
Total mass
¼
ð2
x¼0
ð6
y¼0
ð4x2
z¼0
x dz dy dx
Total mass
¼
24
32
¼
3
4
y
y ¼
Total moment about xz plane
Total mass
¼
Ð2
x¼0
Ð6
y¼0
Ð4x2
z¼0 y dz dy dx
Total mass
¼
96
32
¼ 3
z
z ¼
Total moment about xy plane
Total mass
¼
ð2
x¼0
ð6
y¼0
ð4x2
z¼0
z dz dy dx
Total mass
¼
256 =5
32
¼
8
5
Thus, the centroid has coordinates ð3=4; 3; 8=5Þ.
Note that the value for
y
y could have been predicted because of symmetry.
TRANSFORMATION OF TRIPLE INTEGRALS
9.13. Justify equation (11), Page 211, for changing variables in a triple integral.
By analogy with Problem 9.6, we construct a grid of curvilinear coordinate surfaces which subdivide the
region r into subregions, a typical one of which is r (see Fig. 9-17).
CHAP. 9] MULTIPLE INTEGRALS 221
Fig. 9-17](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-230-320.jpg)
![9.17. Find the volume of the region above the xy plane bounded by the paraboloid z ¼ x2
þ y2
and the
cylinder x2
þ y2
¼ a2
.
The volume is most easily found by using cylindrical coordinates. In these coordinates the equations
for the paraboloid and cylinder are respectively z ¼ 2
and ¼ a. Then
Required volume ¼ 4 times volume shown in Fig. 9-18
¼ 4
ð=2
¼0
ða
¼0
ð2
z¼0
dz d d
¼ 4
ð=2
¼0
ða
¼0
3
d d
¼ 4
ð=2
hi¼0
4
4
a
¼0
d ¼
2
a4
The integration with respect to z (keeping and constant) from z ¼ 0 to z ¼ 2
corresponds to
summing the cubical volumes (indicated by dVÞ in a vertical column extending from the xy plane to the
paraboloid. The subsequent integration with respect to (keeping constant) from ¼ 0 to ¼ a
corresponds to addition of volumes of all columns in the wedge-shaped region. Finally, integration with
respect to corresponds to adding volumes of all such wedge-shaped regions.
The integration can also be performed in other orders to yield the same result.
We can also set up the integral by determining the region r0
in ; ; z space into which r is mapped by
the cylindrical coordinate transformation.
9.18. (a) Find the moment of inertia about the z-axis of the region in Problem 9.17, assuming that the
density is the constant . (b) Find the radius of gyration.
(a) The moment of inertia about the z-axis is
Iz ¼ 4
ð=2
0
ða
¼0
ð2
z¼0
2
dz d d
¼ 4
ð=2
¼0
ða
¼0
5
d d ¼ 4
ð=2
¼0
6
6
a
¼0
d ¼
a6
3
CHAP. 9] MULTIPLE INTEGRALS 223
Fig. 9-18](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-232-320.jpg)
![(b) Letting ¼ , the volume of the sphere thus obtained is
2a3
3
ð1 cos Þ ¼
4
3
a3
9.20. ðaÞ Find the centroid of the region in Problem 9.19.
(b) Use the result in (a) to find the centroid of a hemisphere.
(a) The centroid ð
x
x;
y
y;
z
zÞ is, due to symmetry, given by
x
x ¼
y
y ¼ 0 and
z
z ¼
Total moment about xy plane
Total mass
¼
Ð Ð Ð
z dV
Ð Ð Ð
dV
Since z ¼ r cos and is constant the numerator is
4
ð=2
¼0
ð
¼0
ða
r¼0
r cos r2
sin dr d d ¼ 4
ð=2
¼0
ð
¼0
r4
4
a
r¼0
sin cos d d
¼ a4
ð=2
¼0
ð
¼0
sin cos d d
¼ a4
ð=2
¼0
sin2
2 ¼0
d ¼
a4
sin2
4
The denominator, obtained by multiplying the result of Problem 9.19(a) by , is 2
3 a3
ð1 cos Þ.
Then
z
z ¼
1
4 a4
sin2
2
3 a3ð1 cos Þ
¼
3
8
að1 þ cos Þ:
(b) Letting ¼ =2;
z
z ¼ 3
8 a.
MISCELLANEOUS PROBLEMS
9.21. Prove that (a)
ð1
0
ð1
0
x y
ðx þ yÞ3
dy
dx ¼
1
2
, (b)
ð1
0
ð1
0
x y
ðx þ yÞ3
dx
dy ¼
1
2
.
ðaÞ
ð1
0
ð1
0
x y
ðx þ yÞ3
dy
dx ¼
ð1
0
ð1
0
2x ðx þ yÞ
ðx þ yÞ3
dy
dx
¼
ð1
0
ð1
0
2x
ðx þ yÞ3
1
ðx þ yÞ2
dy
dx
¼
ð1
0
x
ðx þ yÞ2
þ
1
x þ y
1
y¼0
dx
¼
ð1
0
dx
ðx þ 1Þ2
¼
1
x þ 1
1
0
¼
1
2
(b) This follows at once on formally interchanging x and y in (a) to obtain
ð1
0
ð1
0
y x
ðx þ yÞ3
dx
dy ¼
1
2
and
then multiplying both sides by 1.
This example shows that interchange in order of integration may not always produce equal results.
A sufficient condition under which the order may be interchanged is that the double integral over the
corresponding region exists. In this case
ð ð
r
x y
ðx þ yÞ3
dx dy, where r is the region
0 @ x @ 1; 0 @ y @ 1 fails to exist because of the discontinuity of the integrand at the origin. The
integral is actually an improper double integral (see Chapter 12).
9.22. Prove that
ðx
0
ðt
0
FðuÞ du
dt ¼
ðx
0
ðx uÞFðuÞ du.
CHAP. 9] MULTIPLE INTEGRALS 225](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-234-320.jpg)
![9:31: If r is the region of Problem 9.30, evaluate
ð ð
r
eðx2
þy2
Þ
dx dy: Ans: ð1 ea2
Þ
9.32. By using the transformation x þ y ¼ u; y ¼ uv, show that
ð1
x¼0
ð1x
y¼0
ey=ðxþyÞ
dy dx ¼
e 1
2
9.33. Find the area of the region bounded by xy ¼ 4; xy ¼ 8; xy3
¼ 5; xy3
¼ 15. [Hint: Let xy ¼ u; xy3
¼ v.]
Ans: 2 ln 3
9.34. Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas
y2
¼ x; y2
¼ 8x; x2
¼ y; x2
¼ 8y about the x-axis is 279=2. [Hint: Let y2
¼ ux; x2
¼ vy.]
9.35. Find the area of the region in the first quadrant bounded by y ¼ x3
; y ¼ 4x3
; x ¼ y3
; x ¼ 4y3
.
Ans: 1
8
9.36. Let r be the region bounded by x þ y ¼ 1; x ¼ 0; y ¼ 0. Show that
ð ð
r
cos
x y
x þ y
dx dy ¼
sin 1
2
. [Hint: Let
x y ¼ u; x þ y ¼ v.]
TRIPLE INTEGRALS
9.37. (a) Evaluate
ð1
x¼0
ð1
y¼0
ð2
z¼
ffiffiffiffiffiffiffiffiffiffi
x2þy2
p xyz dz dy dx: ðbÞ Give a physical interpretation to the integral in (a).
Ans: ðaÞ 3
8
9.38. Find the (a) volume and (b) centroid of the region in the first octant bounded by x=a þ y=b þ z=c ¼ 1,
where a; b; c are positive. Ans: ðaÞ abc=6; ðbÞ
x
x ¼ a=4;
y
y ¼ b=4;
z
z ¼ c=4
9.39. Find the (a) moment of inertia and (b) radius of gyration about the z-axis of the region in Problem 9.38.
Ans: ðaÞ Mða2
þ b2
Þ=10; ðbÞ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ða2 þ b2Þ=10
p
9.40. Find the mass of the region corresponding to x2
þ y2
þ z2
@ 4; x A 0; y A 0; z A 0, if the density is equal
to xyz. Ans: 4=3
9.41. Find the volume of the region bounded by z ¼ x2
þ y2
and z ¼ 2x. Ans: =2
TRANSFORMATION OF TRIPLE INTEGRALS
9.42. Find the volume of the region bounded by z ¼ 4 x2
y2
and the xy plane. Ans: 8
9.43. Find the centroid of the region in Problem 9.42, assuming constant density .
Ans:
x
x ¼
y
y ¼ 0;
z
z ¼ 4
3
9.44. (a) Evaluate
ð ð ð
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
þ z2
q
dx dy dz, where r is the region bounded by the plane z ¼ 3 and the cone
z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2
p
. (b) Give a physical interpretation of the integral in (a). [Hint: Perform the integration in
cylindrical coordinates in the order ; z; .] Ans: 27ð2
ffiffiffi
2
p
1Þ=2
9.45. Show that the volume of the region bonded by the cone z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2
p
and the paraboloid z ¼ x2
þ y2
is =6.
9.46. Find the moment of inertia of a right circular cylinder of radius a and height b, about its axis if the density is
proportional to the distance from the axis. Ans: 3
5 Ma2
CHAP. 9] MULTIPLE INTEGRALS 227](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-236-320.jpg)
![9.47. (a) Evaluate
ð ð ð
r
dx dy dz
ðx2 þ y2 þ z2Þ3=2
, where r is the region bounded by the spheres x2
þ y2
þ z2
¼ a2
and
x2
þ y2
þ z2
¼ b2
where a b 0. (b) Give a physical interpretation of the integral in (a).
Ans: ðaÞ 4 lnða=bÞ
9.48. (a) Find the volume of the region bounded above by the sphere r ¼ 2a cos , and below by the cone ¼
where 0 =2. (b) Discuss the case ¼ =2. Ans: 4
3 a3
ð1 cos4
Þ
9.49. Find the centroid of a hemispherical shell having outer radius a and inner radius b if the density (a) is
constant, (b) varies as the square of the distance from the base. Discuss the case a ¼ b.
Ans. Taking the z-axis as axis of symmetry: (a)
x
x ¼
y
y ¼ 0;
z
z ¼ 3
8 ða4
b4
Þ=ða3
b3
Þ; ðbÞ
x
x ¼
y
y ¼ 0,
z
z ¼ 5
8 ða6
b6
Þ=ða5
b5
Þ
MISCELLANEOUS PROBLEMS
9.50. Find the mass of a right circular cylinder of radius a and height b if the density varies as the square of the
distance from a point on the circumference of the base.
Ans: 1
6 a2
bkð9a2
þ 2b2
Þ, where k ¼ constant of proportionality.
9.51. Find the (a) volume and (b) centroid of the region bounded above by the sphere x2
þ y2
þ z2
¼ a2
and
below by the plane z ¼ b where a b 0, assuming constant density.
Ans: ðaÞ 1
3 ð2a3
3a2
b þ b3
Þ; ðbÞ
x
x ¼
y
y ¼ 0;
z
z ¼ 3
4 ða þ bÞ2
=ð2a þ bÞ
9.52. A sphere of radius a has a cylindrical hole of radius b bored from it, the axis of the cylinder coinciding with a
diameter of the sphere. Show that the volume of the sphere which remains is 4
3 ½a3
ða2
b2
Þ3=2
.
9.53. A simple closed curve in a plane is revolved about an axis in the plane which does not intersect the curve.
Prove that the volume generated is equal to the area bounded by the curve multiplied by the distance
traveled by the centroid of the area (Pappus’ theorem).
9.54. Use Problem 9.53 to find the volume generated by revolving the circle x2
þ ðy bÞ2
¼ a2
; b a 0 about
the x-axis. Ans: 22
a2
b
9.55. Find the volume of the region bounded by the hyperbolic cylinders xy ¼ 1; xy ¼ 9; xz ¼ 4; xz ¼ 36, yz ¼ 25,
yz ¼ 49. [Hint: Let xy ¼ u; xz ¼ v; yz ¼ w:] Ans: 64
9.56. Evaluate
ð ð ð
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ðx2=a2 þ y2=b2 þ z2=c2Þ
q
dx dy dz, where r is the region interior to the ellipsoid
x2
=a2
þ y2
=b2
þ z2
=c2
¼ 1. [Hint: Let x ¼ au; y ¼ bv; z ¼ cw. Then use spherical coordinates.]
Ans: 1
4 2
abc
9.57. If r is the region x2
þ xy þ y2
@ 1, prove that
ð ð
r
eðx2
þxyþy2
Þ
dx dy ¼
2
e
ffiffiffi
3
p ðe 1Þ. [Hint: Let
x ¼ u cos v sin , y ¼ u sin þ v cos and choose so as to eliminate the xy term in the integrand.
Then let u ¼ a cos , v ¼ b sin where a and b are appropriately chosen.]
9.58. Prove that
ðx
0
ðx
0
ðx
0
FðxÞ dxn
¼
1
ðn 1Þ!
ðx
0
ðx uÞn1
FðuÞ du for n ¼ 1; 2; 3; . . . (see Problem 9.22).
228 MULTIPLE INTEGRALS [CHAP. 9](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-237-320.jpg)
![For this discussion it is assumed that r is continuously differentiable. While (as we are doing) it is
convenient to refer the Euclidean space to a rectangular Cartesian coordinate system, it is not necessary.
(For example, cylindrical and spherical coordinates sometimes are more useful.) In fact, one of the
objectives of the vector language is to free us from any particular frame of reference. Then, a vector
A½xðtÞ; yðtÞ; zðtÞ or a scalar, , is pictured on the domain C, which according to the parametric repre-
sentation, is referred to the real number interval a @ t @ b.
The Integral
ð
C
A dr ð3Þ
of a vector field A defined on a curve segment C is called a line integral. The integrand has the
representation
A1 dx þ A2 dy þ A3 dz
obtained by expanding the dot product.
The scalar and vector integrals
ð
C
ðtÞ dt ¼ lim
n!1
X
n
k¼1
ðk; k; kÞtk ð4Þ
ð
C
AðtÞdt ¼ lim
n!1
X
n
k¼1
Aðk; k; kÞtÞk ð5Þ
can be interpreted as line integrals; however, they do not play a major role [except for the fact that the
scalar integral (3) takes the form (4)].
The following three basic ways are used to evaluate the line integral (3):
1. The parametric equations are used to express the integrand through the parameter t. Then
ð
C
A dr ¼
ðt2
t1
A
dr
dt
dt
2. If the curve C is a plane curve (for example, in the xy plane) and has one of the representations
y ¼ f ðxÞ or x ¼ gðyÞ, then the two integrals that arise are evaluated with respect to x or y,
whichever is more convenient.
230 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
Fig. 10-1](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-239-320.jpg)
![3. If the integrand is a perfect differential, then it may be evaluated through knowledge of the end
points (that is, without reference to any particular joining curve). (See the section on indepen-
dence of path on Page 232; also see Page 237.)
These techniques are further illustrated below for plane curves and for three space in the problems.
EVALUATION OF LINE INTEGRALS FOR PLANE CURVES
If the equation of a curve C in the plane z ¼ 0 is given as y ¼ f ðxÞ, the line integral (2) is evaluated
by placing y ¼ f ðxÞ; dy ¼ f 0
ðxÞ dx in the integrand to obtain the definite integral
ða2
a1
Pfx; f ðxÞg dx þ Qfx; f ðxÞg f 0
ðxÞ dx ð7Þ
which is then evaluated in the usual manner.
Similarly, if C is given as x ¼ gðyÞ, then dx ¼ g0
ðyÞ dy and the line integral becomes
ðb2
b1
Pfgð yÞ; ygg0
ð yÞ dy þ Qfgð yÞ; yg dy ð8Þ
If C is given in parametric form x ¼ ðtÞ; y ¼ ðtÞ, the line integral becomes
ðt2
t1
PfðtÞ; ðtÞg0
ðtÞ dt þ QfðtÞ; ðtÞg; 0
ðtÞ dt ð9Þ
where t1 and t2 denote the values of t corresponding to points A and B, respectively.
Combinations of the above methods may be used in the evaluation. If the integrand A dr is a
perfect differential, d, then
ð
C
A dr ¼
ððc;dÞ
ða;bÞ
d ¼ ðc; dÞ ða; bÞ ð6Þ
Similar methods are used for evaluating line integrals along space curves.
PROPERTIES OF LINE INTEGRALS EXPRESSED FOR PLANE CURVES
Line integrals have properties which are analogous to those of ordinary integrals. For example:
1:
ð
C
Pðx; yÞ dx þ Qðx; yÞ dy ¼
ð
C
Pðx; yÞ dx þ
ð
C
Qðx; yÞ dy
2:
ðða2;b2Þ
ða1;b1Þ
P dx þ Q dy ¼
ðða1;b1Þ
ða2;b2Þ
P dx þ q dy
Thus, reversal of the path of integration changes the sign of the line integral.
3:
ðða2;b2Þ
ða2;b1Þ
P dx þ Q dy ¼
ðða3;b3Þ
ða1;b1Þ
P dx þ Q dy þ
ðða2;b2Þ
ða3;b3Þ
P dx þ Q dy
where ða3; b3Þ is another point on C.
Similar properties hold for line integrals in space.
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 231](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-240-320.jpg)
![Theorem 2. A necessary and sufficient condition that the line integral,
ð
C
A dr be independent of path
is that r A ¼ 0.
Theorem 3. If r A ¼ 0, then the line integral of A over an allowable closed path is 0, i.e.,
þ
A dr ¼ 0.
If C is a plane curve, then Theorem 3 follows immediately from Green’s theorem, since in the plane
case r A reduces to
@A1
@y
¼
@A2
@x
EXAMPLE. Newton’s second law for forces is F ¼
dðmvÞ
dt
, where m is the mass of an object and v is its velocity.
When F has the representation F ¼ r, it is said to be conservative. The previous theorems tell us that the
integrals of conservative fields of force are independent of path. Furthermore, showing that r F ¼ 0 is the
preferred way of showing that F is conservative, since it involves differentiation, while demonstrating that exists
such that F ¼ r requires integration.
SURFACE INTEGRALS
Our previous double integrals have been related to a very special surface, the plane. Now we
consider other surfaces, yet, the approach is quite similar. Surfaces can be viewed intrinsically, i.e., as
non-Euclidean spaces; however, we do not do that. Rather, the surface is thought of as embedded in a
three-dimensional Euclidean space and expressed through a two-parameter vector representation:
x ¼ rðv1; v2Þ
While the purpose of the vector representation is to be general (that is, interpretable through any
allowable three-space coordinate system), it is convenient to initially think in terms of rectangular
Cartesian coordinates; therefore, assume
r ¼ xi þ yj þ zk
and that there is a parametric representation
x ¼ rðv1; v2Þ; y ¼ rðv1; v2Þ; z ¼ rðv1; v2Þ ð11Þ
The functions are assumed to be continuously differentiable.
The parameter curves v2 ¼ const and v1 ¼ const establish a coordinate system on the surface (just as
y ¼ const, and x ¼ const form such a system in the plane). The key to establishing the surface integral
of a function is the differential element of surface area. (For the plane that element is dA ¼ dx; dy.)
At any point, P, of the surface
dx ¼
@r
@v1
dv1 þ
@r
@v2
dv2
spans the tangent plane to the surface. In particular, the directions of the coordinate curves v2 ¼ const
and v1 ¼ const are designated by dx1 ¼
@r
@v1
dv1 and dx2 ¼
@r
@v2
dv2, respectively (see Fig. 10-3).
The cross product
dx1x dx2 ¼
@r
@v1
@r
@v2
dv1 dv2
is normal to the tangent plane at P, and its magnitude
@r
@v1
@r
@v2
is the area of a differential coordinate
parallelogram.
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 233](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-242-320.jpg)
![rF
jrFj
¼
@r
@x
@r
@y
@r
@v1
@r
@v2
ð15Þ
[Now a conclusion of the theory of implicit functions is that from Fðx; y; zÞ ¼ 0 (and under appro-
priate conditions) there can be produced an explicit representation z ¼ f ðx; yÞ of a portion of the surface.
This is an existence statement. The theorem does not say that this representation can be explicitly
produced.] With this fact in hand, we again let v1 ¼ x; v2 ¼ y; z ¼ f ðv1; v2Þ. Then
rF ¼ Fxi þ fyj þ Fzk
Taking the dot product of both sides of (15) yields
Fz
jrFj
¼
1
@r
@v1
@r
@v2
The ambiguity of sign can be eliminated by taking the absolute value of both sides of the equation.
Then
@r
@v1
@r
@v2
¼
jrFj
jFzj
¼
½ðFxÞ2
þ ðFyÞ2
þ ðFzÞ2
1=2
jFzj
and the surface integral of takes the form
ð ð
S
½ðFxÞ2
þ ðFyÞ2
þ ðFzÞ2
1=2
jFzj
dx dy ð16Þ
The formulas (14) and (16) also can be introduced in the following nonvectorial manner.
Let S be a two-sided surface having projection r on the xy plane as in the adjoining Fig. 10-4.
Assume that an equation for S is z ¼ f ðx; yÞ, where f is single-valued and continous for all x and y in r .
Divide r into n subregions of area Ap; p ¼ 1; 2; . . . ; n, and erect a vertical column on each of these
subregions to intersect S in an area Sp.
Let ðx; y; zÞ be single-valued and continuous at all points of S. Form the sum
X
n
p¼1
ðp; p; pÞ Sp ð17Þ
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 235
Fig. 10-4](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-244-320.jpg)
![ð ð ð
@A1
@x
þ
@A2
@y
þ
@A3
@z
dV ¼
ð ð
S
ðA1 cos þ A2 cos þ A3 cos Þ dS ð24Þ
The rectangular Cartesian component form of (23) is
ð ð ð
V
@A1
@x
þ
@A2
@y
þ
@A3
@z
dV ¼
ð ð
S
ðA1 dy dz þ A2 dz dx þ A3 dx dyÞ ð25Þ
EXAMPLE. If B is the magnetic field vector, then one of Maxwell’s equations of electromagnetic theory is
r B ¼ 0. When this equation is substituted into the left member of (23), the right member tells us that the
magnetic flux through a closed surface containing a magnetic field is zero. A simple interpretation of this fact
results by thinking of a magnet enclosed in a ball. All magnetic lines of force that flow out of the ball must return
(so that the total flux is zero). Thus, the lines of force flow from one pole to the other, and there is no dispersion.
STOKES’ THEOREM
Stokes’ theorem establishes the equality of the double integral of a vector field over a portion of a
surface and the line integral of the field over a simple closed curve bounding the surface portion. (See
Fig. 10-6.)
Suppose a closed curve, C, bounds a smooth surface portion, S. If the component functions of
x ¼ rðv1; v2Þ have continuous mixed partial derivatives, then for a vector field A with continuous partial
derivatives on S
þ
C
A dr ¼
ð ð
S
n r A dS ð26Þ
where n ¼ cos i þ cos j þ cos k with ; , and representing the angles made by the outward normal
n and i; j, and k, respectively.
Then the component form of (26) is
þ
C
ðA1 dx þ A2 dy þ A3 dzÞ ¼
ð ð
S
@A3
@y
@A2
@z
cos þ
@A1
@z
@A3
@x
cos þ
@A2
@x
@A1
@y
cos dS
ð27Þ
If r A ¼ 0, Stokes’ theorem tells us that
þ
C
A dr ¼ 0. This is Theorem 3 on Page 237.
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 237
Fig. 10-5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-246-320.jpg)
![ð
C
A dr ¼
ð1
t¼0
f3t2
6ðt2
Þðt3
Þg dt þ f2t2
þ 3ðtÞðt3
Þg dðt2
Þ þ f1 4ðtÞðt2
Þðt3
Þ2
g dðt3
Þ
¼
ð1
t¼0
ð3t2
6t5
Þ dt þ ð4t3
þ 6t5
Þ dt þ ð3t2
12t11
Þ dt ¼ 2
Another method:
Along C, A ¼ ð3t2
6t5
Þi þ ð2t2
þ 3t4
Þj þ ð1 4t9
Þk and r ¼ xi þ yj þ zk ¼ ti þ t2
j þ t3
k,
dr ¼ ði þ 2tj þ 3t2
kÞ dt. Then
ð
C
A dr ¼
ð1
0
ð3t2
6t5
Þ dt þ ð4t3
þ 6t5
Þ dt þ ð3t2
12t11
Þ dt ¼ 2
(b) Along the straight line from ð0; 0; 0Þ to ð0; 1; 1Þ, x ¼ 0; y ¼ 0; dx ¼ 0; dy ¼ 0, while z varies from 0 to 1.
Then the integral over this part of the path is
ð1
z¼0
f3ð0Þ2
6ð0ÞðzÞg0 þ f2ð0Þ þ 3ð0ÞðzÞg0 þ f1 4ð0Þð0Þðz2
Þg dz ¼
ð1
z¼0
dz ¼ 1
Along the straight line from ð0; 0; 1Þ to ð0; 1; 1Þ, x ¼ 0; z ¼ 1; dx ¼ 0; dz ¼ 0, while y varies from 0
to 1. Then the integral over this part of the path is
ð1
y¼0
f3ð0Þ2
6ð yÞð1Þg0 þ f2y þ 3ð0Þð1Þg dy þ f1 4ð0Þð yÞð1Þ2
g0 ¼
ð1
y¼0
2y dy ¼ 1
Along the straight line from ð0; 1; 1Þ to ð1; 1; 1Þ, y ¼ 1; z ¼ 1; dy ¼ 0; dz ¼ 0, while x varies from 0
to 1. Then the integral over this part of the path is
ð1
x¼0
f3x2
6ð1Þð1Þg dx þ f2ð1Þ þ 3xð1Þg0 þ f1 4xð1Þð1Þ2
g0 ¼
ð1
x¼0
ð3x2
6Þ dx ¼ 5
Adding,
ð
C
A dr ¼ 1 þ 1 5 ¼ 3:
(c) The straight line joining ð0; 0; 0Þ and ð1; 1; 1Þ is given in parametric form by x ¼ t; y ¼ t; z ¼ t. Then
ð
C
A dr ¼
ð1
t¼0
ð3t2
6t2
Þ dt þ ð2t þ 3t2
Þ dt þ ð1 4t4
Þ dt ¼ 6=5
10.3. Find the work done in moving a particle once around an
ellipse C in the xy plane, if the ellipse has center at the
origin with semi-major and semi-minor axes 4 and 3,
respectively, as indicated in Fig. 10-7, and if the force
field is given by
F ¼ ð3x 4y þ 2zÞi þ ð4x þ 2y 3z2
Þj þ ð2xz 4y2
þ z3
Þk
In the plane z ¼ 0; F ¼ ð3x 4yÞi þ ð4x þ 2yÞj 4y2
k and
dr ¼ dxi þ dyj so that the work done is
þ
C
F dr ¼
ð
C
fð3x 4yÞi þ ð4x þ 2yÞj 4y2
kg ðdxi þ dyjÞ
¼
þ
C
ð3x 4yÞ dx þ ð4x þ 2yÞ dy
Choose the parametric equations of the ellipse as x ¼ 4 cos t, y ¼ 3 sin t, where t varies from 0 to 2 (see
Fig. 10-7). Then the line integral equals
ð2
t¼0
f3ð4 cos tÞ 4ð3 sin tÞgf4 sin tg dt þ f4ð4 cos tÞ þ 2ð3 sin tÞgf3 cos tg dt
¼
ð2
t¼0
ð48 30 sin t cos tÞ dt ¼ ð48t 15 sin2
tÞj2
0 ¼ 96
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 239
x
y
t
r
r = xi + yj
= 4 cos t i + 3 sin t j
Fig. 10-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-248-320.jpg)
![The plane curves y ¼ x2
and y2
¼ x intersect at ð0; 0Þ and ð1; 1Þ. The positive direction in traversing C
is as shown in Fig. 10-9.
Along y ¼ x2
, the line integral equals
ð1
x¼0
fð2xÞðx2
Þ x2
g dx þ fx þ ðx2
Þ2
g dðx2
Þ ¼
ð1
0
ð2x3
þ x2
þ 2x5
Þ dx ¼ 7=6
Along y2
¼ x the line integral equals
ð0
y¼1
f2ð y2
Þ ð yÞ ð y2
Þ2
g dð y2
Þ þ f y2
þ y2
g dy ¼
ð0
1
ð4y4
2y5
þ 2y2
Þ dy ¼ 17=15
Then the required line integral ¼ 7=6 17=15 ¼ 1=30.
ð ð
r
@Q
@x
@P
@y
dx dy ¼
ð ð
r
@
@x
ðx þ y2
Þ
@
@y
ð2xy x2
Þ
dx dy
¼
ð ð
r
ð1 2xÞ dx dy ¼
ð1
x¼0
ð ffiffi
x
p
y¼x2
ð1 2xÞ dy dx
¼
ð1
x¼0
ð y 2xyÞj
ffiffi
x
p
y¼x2 dx
¼
ð1
0
ðx1=2
2x3=2
x2
þ 2x3
Þ dx ¼ 1=30
Hence, Green’s theorem is verified.
10.7. Extend the proof of Green’s theorem in the plane given in Problem
10.5 to the curves C for which lines parallel to the coordinate axes
may cut C in more than two points.
Consider a closed curve C such as shown in the adjoining Fig. 10-10,
in which lines parallel to the axes may meet C in more than two points.
By constructing line ST the region is divided into two regions r1 and r2,
which are of the type considered in Problem 10.5 and for which Green’s
theorem applies, i.e.,
ð1Þ
ð
STUS
P dx þ Q dy ¼
ð ð
r1
@Q
@x
@P
@y
dx dy;
ð2Þ
ð
SVTS
P dx þ Q dy ¼
ð ð
r2
@Q
@x
@P
@y
dx dy
Adding the left-hand sides of (1) and (2), we have, omitting the integrand P dx þ Q dy in each case,
ð
STUS
þ
ð
SVTS
¼
ð
ST
þ
ð
TUS
þ
ð
SVT
þ
ð
TS
¼
ð
TUS
þ
ð
SVT
¼
ð
TUSVT
using the fact that
ð
ST
¼
ð
TS
.
Adding the right-hand sides of (1) and (2), omitting the integrand,
ð ð
r1
þ
ð ð
r2
¼
ð ð
r
where r consists of
regions r1 and r2.
Then
ð
TUSVT
P dx þ Q dy ¼
ð ð
r
@Q
@x
@P
@y
dx dy and the theorem is proved.
A region r such as considered here and in Problem 10.5, for which any closed curve lying in r can be
continuously shrunk to a point without leaving r, is called a simply connected region. A region which is not
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 241
Fig. 10-9
Fig. 10-10](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-250-320.jpg)
![INDEPENDENCE OF THE PATH
10.11. Let Pðx; yÞ and Qðx; yÞ be continuous and have continuous first partial derivatives at each point
of a simply connected region r. Prove that a necessary and sufficient condition that
þ
C
P dx þ Q dy ¼ 0 around every closed path C in r is that @P=@y ¼ @Q=@x identically in r.
Sufficiency. Suppose @P=@y ¼ @Q=@x. Then by Green’s theorem,
þ
C
P dx þ Q dy ¼
ð ð
r
@Q
@x
@P
@y
dx dy ¼ 0
where r is the region bounded by C.
Necessity.
Suppose
þ
C
P dx þ Q dy ¼ 0 around every closed path C in r and that @P=@y 6¼ @Q=@x at some point of
r. In particular, suppose @P=@y @Q=@x 0 at the point ðx0; y0Þ.
By hypothesis @P=@y and @Q=@x are continuous in r, so that there must be some region containing
ðx0; y0Þ as an interior point for which @P=@y @Q=@x 0. If is the boundary of , then by Green’s
theorem
þ
P dx þ Q dy ¼
ð ð
@Q
@x
@P
@y
dx dy 0
contradicting the hypothesis that
þ
P dx þ Q dy ¼ 0 for all closed curves in r. Thus @Q=@x @P=@y cannot
be positive.
Similarly, we can show that @Q=@x @P=@y cannot be negative, and it follows that it must be identically
zero, i.e., @P=@y ¼ @Q=@x identically in r.
10.12. Let P and Q be defined as in Problem 10.11. Prove that a
necessary and sufficient condition that
ðB
A
P dx þ Q dy be inde-
pendent of the path in r joining points A and B is that
@P=@y ¼ @Q=@x identically in r.
Sufficiency. If @P=@y ¼ @Q=@x, then by Problem 10.11,
ð
ADBEA
P dx þ Q dy ¼ 0
(see Fig. 10-12). From this, omitting for brevity the integrand P dx þ Q dy, we have
ð
ADB
þ
ð
BEA
¼ 0;
ð
ADB
¼
ð
BEA
¼
ð
AEB
and so
ð
C1
¼
ð
C2
i.e., the integral is independent of the path.
Necessity.
If the integral is independent of the path, then for all paths C1 and C2 in r we have
ð
C1
¼
ð
C2
;
ð
ADB
¼
ð
AEB
and
ð
ADBEA
¼ 0
From this it follows that the line integral around any closed path in r is zero, and hence by Problem 10.11
that @P=@y ¼ @Q=@x.
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 243
A
B
E
D
C1
C2
Fig. 10-12](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-252-320.jpg)
![(b) Let A ¼ ðx1; y1Þ; B ¼ ðx2; y2Þ. From part (a),
ðx; yÞ ¼
ððx;yÞ
ða;bÞ
P dx þ Q dy
Then omitting the integrand P dx þ Q dy, we have
ðB
A
¼
ððx2;y2Þ
ðx1;y1Þ
¼
ððx2;y2Þ
ða;bÞ
ððx1;y1Þ
ða;bÞ
¼ ðx2; y2Þ ðx1; y1Þ ¼ ðBÞ ðAÞ
10.14. (a) Prove that
ðð3;4Þ
ð1;2Þ
ð6xy2
y3
Þ dx þ ð6x2
y 3xy2
Þ dy is independent of the path joining ð1; 2Þ and
ð3; 4Þ. (b) Evaluate the integral in (a).
(a) P ¼ 6xy2
y3
; Q ¼ 6x2
y 3xy2
. Then @P=@y ¼ 12xy 3y2
¼ @Q=@x and by Problem 10.12 the line
integral is independent of the path.
(b) Method 1: Since the line integral is independent of the path, choose any path joining ð1; 2Þ and ð3; 4Þ,
for example that consisting of lines from ð1; 2Þ to ð3; 2Þ [along which y ¼ 2; dy ¼ 0] and then ð3; 2Þ to
ð3; 4Þ [along which x ¼ 3; dx ¼ 0]. Then the required integral equals
ð3
x¼1
ð24x 8Þ dx þ
ð4
y¼2
ð54y 9y2
Þ dy ¼ 80 þ 156 ¼ 236
Method 2: Since
@P
@y
¼
@Q
@x
; we must have ð1Þ
@
@x
¼ 6xy2
y3
; ð2Þ
@
@y
¼ 6x2
y 3xy2
:
From (1), ¼ 3x2
y2
xy3
þ f ð yÞ. From (2), ¼ 3x2
y2
xy3
þ gðxÞ. The only way in which
these two expressions for are equal is if f ð yÞ ¼ gðxÞ ¼ c, a constant. Hence ¼ 3x2
y2
xy3
þ c.
Then by Problem 10.13,
ðð3;4Þ
ð1;2Þ
ð6xy2
y3
Þ dx þ ð6x2
y 3xy2
Þ dy ¼
ðð3;4Þ
ð1;2Þ
dð3x2
y2
xy3
þ cÞ
¼ 3x2
y2
xy3
þ cjð3;4Þ
ð1;2Þ ¼ 236
Note that in this evaluation the arbitrary constant c can be omitted. See also Problem 6.16, Page 131.
We could also have noted by inspection that
ð6xy2
y3
Þ dx þ ð6x2
y 3xy2
Þ dy ¼ ð6xy2
dx þ 6x2
y dyÞ ð y3
dx þ 3xy2
dyÞ
¼ dð3x2
y2
Þ dðxy3
Þ ¼ dð3x2
y2
xy3
Þ
from which it is clear that ¼ 3x2
y2
xy3
þ c.
10.15. Evaluate
þ
ðx2
y cos x þ 2xy sin x y2
ex
Þ dx þ ðx2
sin x 2yex
Þ dy around the hypocycloid
x2=3
þ y2=3
¼ a2=3
:
P ¼ x2
y cos x þ 2xy sin x y2
ex
; Q ¼ x2
sin x 2yex
Then @P=@y ¼ x2
cos x þ 2x sin x 2yex
¼ @Q=@x, so that by Problem 10.11 the line integral around any
closed path, in particular x2=3
þ y2=3
¼ a2=3
is zero.
SURFACE INTEGRALS
10.16. If is the angle between the normal line to any point ðx; y; zÞ of a surface S and the
positive z-axis, prove that
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 245](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-254-320.jpg)
![(c) If Uðx; y; zÞ ¼ 3z, (2) becomes
ð ð
r
3z
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ 4x2 þ 4y2
q
dx dy ¼
ð ð
r
3f2 ðx2
þ y2
Þg
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ 4x2 þ 4y2
q
dx dy
or in polar coordinates,
ð2
¼0
ð ffiffi
2
p
¼0
3ð2 2
Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ 42
p
d d ¼
111
10
Physically this could represent the mass of S assuming a density ¼ 3z, or three times the first
moment of S about the xy plane.
10.18. Find the surface area of a hemisphere of radius a cut off
by a cylinder having this radius as diameter.
Equations for the hemisphere and cylinder (see Fig. 10-15)
are given respectively by x2
þ y2
þ z2
¼ a2
(or z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
x2
y2
p
Þ and ðx a=2Þ2
þ y2
¼ a2
=4 (or x2
þ y2
¼ ax).
Since
zx ¼
x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
x2
y2
p and zy ¼
y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
x2
y2
p
we have
Required surface area ¼ 2
ð ð
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ z2
x þ z2
y
q
dx dy ¼ 2
ð ð
r
a
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x2 y2
p dx dy
Two methods of evaluation are possible.
Method 1: Using polar coordinates.
Since x2
þ y2
¼ ax in polar coordinates is ¼ a cos , the integral becomes
2
ð=2
¼0
ða cos
¼0
a
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 2
p d d ¼ 2a
ð=2
¼0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 2
p a cos
¼0
d
¼ 2a2
ð=2
0
ð1 sin Þ d ¼ ð 2Þa2
Method 2: The integral is equal to
2
ða
x¼0
ð ffiffiffiffiffiffiffiffiffiffi
axx2
p
y¼0
a
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2
x2
y2
p dy dx ¼ 2a
ða
x¼0
sin1 y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ax x2
p
ffiffiffiffiffiffiffiffiffiffi
axx2
p
y¼0
dx
¼ 2a
ða
0
sin1
ffiffiffiffiffiffiffiffiffiffiffi
x
a þ x
r
dx
Letting x ¼ a tan2
, this integral becomes
4a2
ð=4
0
tan sec2
d ¼ 4a2 1
2 tan2
j=4
0 1
2
ð=4
0
tan2
d
¼ 2a2
tan2
j=4
0
ð=4
0
ðsec2
1Þ d
¼ 2a2
=4 ðtan Þj=4
0
n o
¼ ð 2Þa2
Note that the above integrals are actually improper and should be treated by appropriate limiting
procedures (see Problem 5.74, Chapter 5, and also Chapter 12).
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 247
Fig. 10-15](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-256-320.jpg)
![is an example of a one-sided surface. This is sometimes called a nonorientable surface. A two-sided surface
is orientable.
THE DIVERGENCE THEOREM
10.22. Prove the divergence theorem. (See Fig. 10-18.)
Let S be a closed surface which is such that any line parallel to the coordinate axes cuts S in at most two
points. Assume the equations of the lower and upper portions, S1 and S2, to be z ¼ f1ðx; yÞ and z ¼ f2ðx; yÞ,
respectively. Denote the projection of the surface on the xy plane by r. Consider
ð ð ð
V
@A3
@z
dV ¼
ð ð ð
V
@A3
@z
dz dy dx ¼
ð ð
r
ðf2ðx;yÞ
z¼f1ðx;yÞ
@A3
@z
dz dy dx
¼
ð ð
r
A3ðx; y; zÞ
f2
z¼f1
dy dx ¼
ð ð
r
½A3ðx; y; f2Þ A3ðx; y; f1Þ dy dx
For the upper portion S2, dy dx ¼ cos 2 dS2 ¼ k n2 dS2 since the normal n2 to S2 makes an acute angle
2 with k.
For the lower portion S1, dy dx ¼ cos 1 dS1 ¼ k n1 dS1 since the normal n1 to S1 makes an obtuse
angle 1 with k.
ð ð
r
A3ðx; y; f2Þ dy dx ¼
ð ð
S2
A3 k n2 dS2
Then
ð ð
r
A3ðx; y; f1Þ dy dx ¼
ð ð
S1
A3 k n1 dS1
and
ð ð
r
A3ðx; y; f2Þ dy dx
ð ð
r
A3ðx; y; f1Þ dy dx ¼
ð ð
S2
A3 k n2 dS2 þ
ð ð
S1
A3 k n1 dS1
¼
ð ð
S
A3 k n dS
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 249
Fig. 10-18](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-258-320.jpg)
![Face BCDE: n ¼ k; z ¼ 1. Then
ð ð
BCDE
A n dS ¼
ð1
0
ð1
0
fð2x 1Þi þ x2
yj xkg k dx dy ¼
ð1
0
ð1
0
x dx dy 1=2
Face AFGO: n ¼ k; z ¼ 0. Then
ð ð
AFGO
A n dS ¼
ð1
0
ð1
0
f2xi x2
yjg ðkÞ dx dy ¼ 0
Adding,
ð ð
S
A n dS ¼ 3
2 þ 1
2 þ 1
3 þ 0 1
2 þ 0 ¼ 11
6 : Since
ð ð ð
V
r A dV ¼
ð1
0
ð1
0
ð1
0
ð2 þ x2
2xzÞ dx dy dz ¼
11
6
the divergence theorem is verified in this case.
10.24. Evaluate
ð ð
S
r n dS, where S is a closed surface.
By the divergence theorem,
ð ð
S
r n dS ¼
ð ð ð
V
r r dV
¼
ð ð ð
V
@
@x
i þ
@
@y
j þ
@
@z
k
ðxi þ yj þ zkÞ dV
¼
ð ð ð
V
@x
@x
þ
@y
@y
þ
@z
@z
dV ¼ 3
ð ð ð
V
dV ¼ 3V
where V is the volume enclosed by S.
10.25. Evaluate
ð ð
S
xz2
dy dz þ ðx2
y z3
Þ dz dx þ ð2xy þ y2
zÞ dx dy, where S is the entire surface of the
hemispherical region bounded by z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x2 y2
p
and z ¼ 0 (a) by the divergence theorem
(Green’s theorem in space), (b) directly.
(a) Since dy dz ¼ dS cos ; dz dx ¼ dS cos ; dx dy ¼ dS cos , the integral can be written
ð ð
S
fxz2
cos þ ðx2
y z3
Þ cos þ ð2xy þ y2
zÞ cos g dS ¼
ð ð
S
A n dS
where A ¼ xz2
i þ ðx2
y z3
Þj þ ð2xy þ y2
zÞk and n ¼ cos i þ cos j þ cos k, the outward drawn unit
normal.
Then by the divergence theorem the integral equals
ð ð ð
V
r A dV ¼
ð ð ð
V
@
@x
ðxz2
Þ þ
@
@y
ðx2
y z3
Þ þ
@
@z
ð2xy þ y2
zÞ
dV ¼
ð ð ð
V
ðx2
þ y2
þ z2
Þ dV
where V is the region bounded by the hemisphere and the xy plane.
By use of spherical coordinates, as in Problem 9.19, Chapter 9, this integral is equal to
4
ð=2
¼0
ð=2
¼0
ð
r¼0
r2
r2
sin dr d d ¼
2a5
5
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 251](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-260-320.jpg)
![½r ðA1iÞ n dS ¼
@A1
@z
n j
@A1
@y
n k
dS ð1Þ
If z ¼ f ðx; yÞ is taken as the equation of S, then the position vector to any point of S is r ¼ xi þ yj þ zk ¼
xi þ yj þ f ðx; yÞk so that
@r
@y
¼ j þ
@z
@y
k ¼ j þ
@ f
@y
k. But
@r
@y
is a vector tangent to S and thus perpendicular to
n, so that
n
@r
@y
¼ n j þ
@z
@y
n k ¼ 0 or n j ¼
@z
@y
n k
Substitute in (1) to obtain
@A1
@z
n j
@A1
@y
n k
dS ¼
@A1
@z
@z
@y
n k
@A1
@y
n k
dS
or
½r ðA1iÞ n dS ¼
@A1
@y
þ
@A1
@z
@z
@y
n k dS ð2Þ
Now on S, A1ðx; y; zÞ ¼ A1½x; y; f ðx; yÞ ¼ Fðx; yÞ; hence,
@A1
@y
þ
@A1
@z
@z
@y
¼
@F
@y
and (2) becomes
½r ðA1iÞ n dS ¼
@F
@y
n k dS ¼
@F
@y
dx dy
Then
ð ð
S
½r ðA1iÞ n dS ¼
ð ð
r
@F
@y
dx dy
where r is the projection of S on the xy plane. By Green’s theorem for the plane, the last integral equals
þ
F dx where is the boundary of r. Since at each point ðx; yÞ of the value of F is the same as the value
of A1 at each point ðx; y; zÞ of C, and since dx is the same for both curves, we must have
þ
F dx ¼
þ
C
A1 dx
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 253
Fig. 10-20](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-262-320.jpg)
![10.28. Prove that a necessary and sufficient condition that
þ
C
A dr ¼ 0 for every closed curve C is that
r A ¼ 0 identically.
Sufficiency. Suppose r A ¼ 0. Then by Stokes’ theorem
þ
C
A dr ¼
ð ð
S
ðr AÞ n dS ¼ 0
Necessity.
Suppose
þ
C
A dr ¼ 0 around every closed path C, and assume r A 6¼ 0 at some point P. Then
assuming r A is continuous, there will be a region with P as an interior point, where r A 6¼ 0. Let S be
a surface contained in this region whose normal n at each point has the same direction as r A, i.e.,
r A ¼ n where is a positive constant. Let C be the boundary of S. Then by Stokes’ theorem
þ
C
A dr ¼
ð ð
S
ðr AÞ n dS ¼
ð ð
S
n n dS 0
which contradicts the hypothesis that
þ
C
A dr ¼ 0 and shows that r A ¼ 0.
It follows that r A ¼ 0 is also a necessary and sufficient condition for a line integral
ðP2
P1
A dr to be
independent of the path joining points P1 and P2.
10.29. Prove that a necessary and sufficient condition that r A ¼ 0 is that A ¼ r.
Sufficiency. If A ¼ r, then r A ¼ r r ¼ 0 by Problem 7.80, Chap. 7, Page 179.
Necessity.
If r A ¼ 0, then by Problem 10.28,
þ
A dr ¼ 0 around every closed path and
ð
C
A dr is independent
of the path joining two points which we take as ða; b; cÞ and ðx; y; zÞ. Let us define
ðx; y; zÞ ¼
ððx;y;zÞ
ða;b;cÞ
A dr ¼
ððx;y;zÞ
ða;b;cÞ
A1 dx þ A2 dy þ A3 dz
Then
ðx þ x; y; zÞ ðx; y; zÞ ¼
ððxþx;y;zÞ
ðx;y;zÞ
A1 dx þ A2 dy þ A3 dz
Since the last integral is independent of the path joining ðx; y; zÞ and ðx þ x; y; zÞ, we can choose the
path to be a straight line joining these points so that dy and dz are zero. Then
ðx þ x; y; zÞ ðx; y; zÞ
x
¼
1
x
ððxþx;y;zÞ
ðx;y;zÞ
A1 dx ¼ A1ðx þ x; y; zÞ 0 1
where we have applied the law of the mean for integrals.
Taking the limit of both sides as x ! 0 gives @=@x ¼ A1.
Similarly, we can show that @=@y ¼ A2; @=@z ¼ A3:
Thus, A ¼ A1i þ A2j þ A3k ¼
@
@x
i þ
@
@y
j þ
@
@z
k ¼ r:
10.30. (a) Prove that a necessary and sufficient condition that A1 dx þ A2 dy þ A3 dz ¼ d, an exact
differential, is that r A ¼ 0 where A ¼ A1i þ A2j þ A3k.
(b) Show that in such case,
ððx2;y2;z2Þ
ðx1;y1;z1Þ
A1 dx þ A2 dy þ A3 dz ¼
ððx2;y2;z2Þ
ðx1;y1;z1Þ
d ¼ ðx2; y2; z2Þ ðx1; y1; z1Þ
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 255](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-264-320.jpg)
![Alternatively, we may notice by inspection that
F dr ¼ ð2xz3
dx þ 3x2
z2
dzÞ þ ð6y dx þ 6x dyÞ ð2yz dy þ y2
dzÞ
¼ dðx2
z3
Þ þ dð6xyÞ dðy2
zÞ ¼ dðx2
z3
þ 6xy y2
z þ cÞ
from which is determined.
Method 2: Since the integral is independent of the path, we can choose any path to evaluate it; in
particular, we can choose the path consisting of straight lines from ð1; 1; 1Þ to ð2; 1; 1Þ, then to
ð2; 1; 1Þ and then to ð2; 1; 1Þ. The result is
ð2
x¼1
ð2x 6Þ dx þ
ð1
y¼1
ð12 2yÞ dy þ
ð1
z¼1
ð12z2
1Þ dz ¼ 15
where the first integral is obtained from the line integral by placing y ¼ 1; z ¼ 1; dy ¼ 0; dz ¼ 0;
the second integral by placing x ¼ 2; z ¼ 1; dx ¼ 0; dz ¼ 0; and the third integral by placing
x ¼ 2; y ¼ 1; dx ¼ 0; dy ¼ 0.
(c) Physically
ð
C
F dr represents the work done in moving an object from ð1; 1; 1Þ to ð2; 1; 1Þ along C.
In a conservative force field the work done is independent of the path C joining these points.
MISCELLANEOUS PROBLEMS
10.32. (a) If x ¼ f ðu; vÞ; y ¼ gðu; vÞ defines a transformation which maps a region r of the xy plane into
a region r0
of the uv plane, prove that
ð ð
r
dx dy ¼
ð ð
r0
@ðx; yÞ
@ðu; vÞ
du dv
(b) Interpret geometrically the result in (a).
(a) If C (assumed to be a simple closed curve) is the boundary of r, then by Problem 10.8,
ð ð
r
dx dy ¼
1
2
þ
C
x dy y dx ð1Þ
Under the given transformation the integral on the right of (1) becomes
1
2
þ
C0
x
@y
@u
du þ
@y
@v
dv
y
@x
@u
du þ
@x
@v
dv
¼
1
2
ð
C0
x
@y
@u
y
@x
@u
du þ x
@y
@v
y
@x
@v
dv ð2Þ
where C0
is the mapping of C in the uv plane (we suppose the mapping to be such that C0
is a simple
closed curve also).
By Green’s theorem if r0
is the region in the uv plane bounded by C0
, the right side of (2) equals
1
2
ð ð
r0
@
@u
x
@y
@v
y
@x
@v
@
@v
x
@y
@u
y
@x
@u
du dv ¼
ð ð
r0
@x
@u
@y
@v
@x
@v
@y
@u
du dv
¼
ð ð
r0
@ðx; yÞ
@ðu; vÞ
du dv
where we have inserted absolute value signs so as to ensure that the result is non-negative as is
ð ð
r
dx dy
In general, we can show (see Problem 10.83) that
ð ð
r
Fðx; yÞ dx dy ¼
ð ð
r0
Ff f ðu; vÞ; gðu; vÞg
@ðx; yÞ
@ðu; vÞ
du dv ð3Þ
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 257](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-266-320.jpg)
![(b)
ð ð
r
dx dy and
ð ð
r0
@ðx; yÞ
@ðu; vÞ
du dv represent the area of region r, the first expressed in rectangular
coordinates, the second in curvilinear coordinates. See Page 212, and the introduction of the differ-
ential element of surface area for an alternative to Part (a).
10.33. Let F ¼
yi þ xj
x2
þ y2
. (a) Calculate r F: ðbÞ Evaluate
þ
F dr around any closed path and
explain the results.
ðaÞ r F ¼
i j k
@
@x
@
@y
@
@z
y
x2 þ y2
x
x2 þ y2
0
¼ 0 in any region excluding ð0; 0Þ:
(b)
þ
F dr ¼
þ
y dx þ x dy
x2 þ y2
. Let x ¼ cos ; y ¼ sin , where ð; Þ are polar coordinates. Then
dx ¼ sin d þ d cos ; dy ¼ cos d þ d sin
y dx þ x dy
x2
þ y2
¼ d ¼ d arc tan
y
x
and so
For a closed curve ABCDA [see Fig. 10-22(a) below] surrounding the origin, ¼ 0 at A and ¼ 2
after a complete circuit back to A. In this case the line integral equals
ð2
0
d ¼ 2.
For a closed curve PQRSP [see Fig. 10-22(b) above] not surrounding the origin, ¼ 0 at P and
¼ 0 after a complete circuit back to P. In this case the line integral equals
ð0
0
d ¼ 0.
Since F ¼ Pi þ Qj; r F ¼ 0 is equivalent to @P=@y ¼ @Q=@x and the results would seem to con-
tradict those of Problem 10.11. However, no contradiction exists since P ¼
y
x2
þ y2
and Q ¼
x
x2
þ y2
do not have continuous derivatives throughout any region including ð0; 0Þ, and this was assumed in
Problem 10.11.
10.34. If div A denotes the divergence of a vector field A at a point P, show that
div A ¼ lim
V!0
ð ð
s
A n dS
V
258 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
y y
x
x
D
R
S
P
Q
O
O
A
B
C
φ
φ
φ0
(a) (b)
Fig. 10-22](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-267-320.jpg)
![where V is the volume enclosed by the surface S and the limit is obtained by shrinking V to
the point P.
By the divergence theorem,
ð ð ð
V
div A dV ¼
ð ð
S
A n dS
By the mean value theorem for integrals, the left side can be written
div A
ð ð ð
V
dV ¼ div A V
where div A is some value intermediate between the maximum and minimum of div A throughout V.
Then
div A ¼
ð ð
S
A n dS
V
Taking the limit as V ! 0 such that P is always interior to V, div A approaches the value div A at
point P; hence
div A ¼ lim
V!0
ð ð
S
A n dS
V
This result can be taken as a starting point for defining the divergence of A, and from it all the
properties may be derived including proof of the divergence theorem. We can also use this to extend the
concept of divergence to coordinate systems other than rectangular (see Page 159).
Physically,
ð ð ð
S
A n ds
0
@
1
A=V represents the flux or net outflow per unit volume of the vector A from
the surface S. If div A is positive in the neighborhood of a point P, it means that the outflow from P is
positive and we call P a source. Similarly, if div A is negative in the neighborhood of P, the outflow is really
an inflow and P is called a sink. If in a region there are no sources or sinks, then div A ¼ 0 and we call A a
solenoidal vector field.
Supplementary Problems
LINE INTEGRALS
10.35. Evaluate
ðð4;2Þ
ð1;1Þ
ðx þ yÞ dx þ ðy xÞ dy along (a) the parabola y2
¼ x, (b) a straight line, (c) straight lines
from ð1; 1Þ to ð1; 2Þ and then to ð4; 2Þ, (d) the curve x ¼ 2t2
þ t þ 1; y ¼ t2
þ 1.
Ans: ðaÞ 34=3; ðbÞ 11; ðcÞ 14; ðdÞ 32=3
10.36. Evaluate
þ
ð2x y þ 4Þ dx þ ð5y þ 3x 6Þ dy around a triangle in the xy plane with vertices at ð0; 0Þ; ð3; 0Þ,
ð3; 2Þ traversed in a counterclockwise direction. Ans. 12
10.37. Evaluate the line integral in the preceding problem around a circle of radius 4 with center at ð0; 0Þ.
Ans: 64
10.38. (a) If F ¼ ðx2
y2
Þi þ 2xyj, evaluate
ð
C
F dr along the curve C in the xy plane given by y ¼ x2
x from the
point ð1; 0Þ to ð2; 2Þ. (b) Interpret physically the result obtained.
Ans. (a) 124/15
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 259](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-268-320.jpg)
![10.39. Evaluate
ð
C
ð2x þ yÞ ds, where C is the curve in the xy plane given by x2
þ y2
¼ 25 and s is the arc length
parameter, from the point ð3; 4Þ to ð4; 3Þ along the shortest path. Ans: 15
10.40. If F ¼ ð3x 2yÞi þ ð y þ 2zÞj x2
k, evaluate
ð
C
F dr from ð0; 0; 0Þ to ð1; 1; 1Þ, where C is a path consisting
of (a) the curve x ¼ t; y ¼ t2
; z ¼ t3
, (b) a straight line joining these points, (c) the straight lines from
ð0; 0; 0Þ to ð0; 1; 0Þ, then to ð0; 1; 1Þ and then to ð1; 1; 1Þ, (d) the curve x ¼ z2
; z ¼ y2
.
Ans: ðaÞ 23=15; ðbÞ 5=3; ðcÞ 0; ðdÞ 13=15
10.41. If T is the unit tangent vector to a curve C (plane or space curve) and F is a given force field, prove that under
appropriate conditions
ð
C
F dr ¼
ð
C
F T ds where s is the arc length parameter. Interpret the result
physically and geometrically.
GREEN’S THEOREM IN THE PLANE, INDEPENDENCE OF THE PATH
10.42. Verify Green’s theorem in the plane for
þ
C
ðx2
xy3
Þ dx þ ð y2
2xyÞ dy where C is a square with vertices at
ð0; 0Þ; ð2; 0Þ; ð2; 2Þ; ð0; 2Þ and counterclockwise orientation. Ans. common value ¼ 8
10.43. Evaluate the line integrals of (a) Problem 10.36 and (b) Problem 10.37 by Green’s theorem.
10.44. (a) Let C be any simple closed curve bounding a region having area A. Prove that if a1; a2; a3; b1; b2; b3 are
constants,
þ
C
ða1x þ a2y þ a3Þ dx þ ðb1x þ b2y þ b3Þ dy ¼ ðb1 a2ÞA
(b) Under what conditions will the line integral around any path C be zero? Ans. (b) a2 ¼ b1
10.45. Find the area bounded by the hypocycloid x2=3
þ y2=3
¼ a2=3
.
[Hint: Parametric equations are x ¼ a cos3
t; y ¼ a sin3
t; 0 @ t @ 2.] Ans: 3a2
=8
10.46. If x ¼ cos ; y ¼ sin , prove that 1
2
þ
x dy y dx ¼ 1
2
ð
2
d and interpret.
10.47. Verify Green’s theorem in the plane for
þ
C
ðx3
x2
yÞ dx þ xy2
dy, where C is the boundary of the region
enclosed by the circles x2
þ y2
¼ 4 and x2
þ y2
¼ 16. Ans: common value ¼ 120
10.48. (a) Prove that
ðð2;1Þ
ð1;0Þ
ð2xy y4
þ 3Þ dx þ ðx2
4xy3
Þ dy is independent of the path joining ð1; 0Þ and ð2; 1Þ.
(b) Evaluate the integral in (a). Ans: ðbÞ 5
10.49. Evaluate
ð
C
ð2xy3
y2
cos xÞ dx þ ð1 2y sin x þ 3x2
y2
Þ dy along the parabola 2x ¼ y2
from ð0; 0Þ to
ð=2; 1Þ. Ans. 2
=4
10.50. Evaluate the line integral in the preceding problem around a parallelogram with vertices at ð0; 0Þ; ð3; 0Þ,
ð5; 2Þ; ð2; 2Þ. Ans: 0
10.51. (a) Prove that G ¼ ð2x2
þ xy 2y2
Þ dx þ ð3x2
þ 2xyÞ dy is not an exact differential. (b) Prove that e y=x
G=x
is an exact differential of and find . (c) Find a solution of the differential equation ð2x2
þ xy 2y2
Þ dxþ
ð3x2
þ 2xyÞ dy ¼ 0.
Ans: ðbÞ ¼ ey=x
ðx2
þ 2xyÞ þ c; ðcÞ x2
þ 2xy þ cey=x
¼ 0
260 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-269-320.jpg)
![SURFACE INTEGRALS
10.52. (a) Evaluate
ð ð
S
ðx2
þ y2
Þ dS, where S is the surface of the cone z2
¼ 3ðx2
þ y2
Þ bounded by z ¼ 0 and z ¼ 3.
(b) Interpret physically the result in (a). Ans: ðaÞ 9
10.53. Determine the surface area of the plane 2x þ y þ 2z ¼ 16 cut off by (a) x ¼ 0; y ¼ 0; x ¼ 2; y ¼ 3,
(b) x ¼ 0; y ¼ 0, and x2
þ y2
¼ 64. Ans: ðaÞ 9; ðbÞ 24
10.54. Find the surface area of the paraboloid 2z ¼ x2
þ y2
which is outside the cone z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2
p
.
Ans: 2
3 ð5
ffiffiffi
5
p
1Þ
10.55. Find the area of the surface of the cone z2
¼ 3ðx2
þ y2
Þ cut out by the paraboloid z ¼ x2
þ y2
.
Ans: 6
10.56. Find the surface area of the region common to the intersecting cylinders x2
þ y2
¼ a2
and x2
þ z2
¼ a2
.
Ans: 16a2
10.57. (a) Obtain the surface area of the sphere x2
þ y2
þ z2
¼ a2
contained within the cone z tan ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
p
,
0 =2. (b) Use the result in (a) to find the surface area of a hemisphere. (c) Explain why formally
placing ¼ in the result of (a) yields the total surface area of a sphere.
Ans: ðaÞ 2a2
ð1 cos Þ; ðbÞ 2a2
(consider the limit as ! =2Þ
10.58. Determine the moment of inertia of the surface of a sphere of radius a about a point on the surface. Assume
a constant density . Ans: 2Ma2
, where mass M ¼ 4a2
10.59. (a) Find the centroid of the surface of the sphere x2
þ y2
þ z2
¼ a2
contained within the cone
z tan ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
p
, 0 =2. (b) From the result in (a) obtain the centroid of the surface of a hemi-
sphere. Ans: ðaÞ 1
2 að1 þ cos Þ; ðbÞ a=2
THE DIVERGENCE THEOREM
10.60. Verify the divergence theorem for A ¼ ð2xy þ zÞi þ y2
j ðx þ 3yÞk taken over the region bounded by
2x þ 2y þ z ¼ 6; x ¼ 0; y ¼ 0; z ¼ 0. Ans: common value ¼ 27
10.61. Evaluate
ð ð
S
F n dS, where F ¼ ðz2
xÞi xyj þ 3zk and S is the surface of the region bounded by
z ¼ 4 y2
; x ¼ 0; x ¼ 3 and the xy plane. Ans. 16
10.62. Evaluate
ð ð
S
A n dS, where A ¼ ð2x þ 3zÞi ðxz þ yÞj þ ð y2
þ 2zÞk and S is the surface of the sphere having
center at ð3; 1; 2Þ and radius 3. Ans: 108
10.63. Determine the value of
ð ð
S
x dy dz þ y dz dx þ z dx dy, where S is the surface of the region bounded by the
cylinder x2
þ y2
¼ 9 and the planes z ¼ 0 and z ¼ 3, (a) by using the divergence theorem, (b) directly.
Ans: 81
10.64. Evaluate
ð ð
S
4xz dy dz y2
dz dx þ yz dx dy, where S is the surface of the cube bounded by x ¼ 0, y ¼ 0,
z ¼ 0, x ¼ 1; y ¼ 1; z ¼ 1, (a) directly, (b) By Green’s theorem in space (divergence theorem).
Ans. 3/2
10.65. Prove that
ð ð
S
ðr AÞ n dS ¼ 0 for any closed surface S.
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 261](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-270-320.jpg)
![10.78. Verify Green’s theorem for a multiply connected region containing two ‘‘holes’’ (see Problem 10.10).
10.79. If P dx þ Q dy is not an exact differential but ðP dx þ Q dyÞ is an exact differential where is some function
of x and y, then is called an integrating factor. (a) Prove that if F and G are functions of x alone, then
ðFy þ GÞ dx þ dy has an integrating factor which is a function of x alone and find . What must be
assumed about F and G? (b) Use (a) to find solutions of the differential equation xy0
¼ 2x þ 3y.
Ans: ðaÞ ¼ e
Ð
FðxÞ dx
ðbÞ y ¼ cx3
x, where c is any constant
10.80. Find the surface area of the sphere x2
þ y2
þ ðz aÞ2
¼ a2
contained within the paraboloid z ¼ x2
þ y2
.
Ans: 2a
10.81. If f ðrÞ is a continuously differentiable function of r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ y2
þ z2
p
, prove that
ð ð
S
f ðrÞ n dS ¼
ð ð ð
V
f 0
ðrÞ
r
r dV
10.82. Prove that
ð ð
S
r ðnÞ dS ¼ 0 where is any continuously differentiable scalar function of position and n is
a unit outward drawn normal to a closed surface S. (See Problem 10.66.)
10.83. Establish equation (3), Problem 10.32, by using Green’s theorem in the plane.
[Hint: Let the closed region r in the xy plane have boundary C and suppose that under the transformation
x ¼ f ðu; vÞ; y ¼ gðu; vÞ, these are transformed into r0
and C0
in the uv plane, respectively. First prove
that
ð ð
r
Fðx; yÞ dx dy ¼
ð
C
Qðx; yÞ dy where @Q=@y ¼ Fðx; yÞ. Then show that apart from sign this last
integral is equal to
ð
C0
Q½ f ðu; vÞ; gðu; vÞ
@g
@u
du þ
@g
@v
dv . Finally, use Green’s theorem to transform this
into
ð ð
r0
F½ f ðu; vÞ; gðu; vÞ
@ðx; yÞ
@ðu; vÞ
du dv.
10.84. If x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ; z ¼ hðu; v; wÞ defines a transformation which maps a region r of xyz space
into a region r0
of uvw space, prove using Stokes’ theorem that
ð ð ð
r
Fðx; y; zÞ dx dy dz ¼
ð ð ð
r0
Gðu; v; wÞ
@ðx; y; zÞ
@ðu; v; wÞ
du dv dw
where Gðu; v; wÞ F½ f ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞ. State sufficient conditions under which the result
is valid. See Problem 10.83. Alternatively, employ the differential element of volume dV ¼
@r
@u
@r
@v
@r
@w
du dv dw (recall the geometric meaning).
10.85. (a) Show that in general the equation r ¼ rðu; vÞ geometrically represents a surface. (b) Discuss the geo-
metric significance of u ¼ c1; v ¼ c2, where c1 and c2 are constants. (c) Prove that the element of arc length
on this surface is given by
ds2
¼ E du2
þ 2F du dv þ G dv2
where E ¼
@r
@u
@r
@u
; F ¼
@r
@u
@r
@v
; G ¼
@r
@v
@r
@v
:
CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 263](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-272-320.jpg)
![1. Comparison test for series of non-negative terms.
(a) Convergence. Let vn A 0 for all n N and suppose that vn converges. Then if
0 @ un @ vn for all n N, un also converges. Note that n N means from some
term onward. Often, N ¼ 1.
EXAMPLE: Since
1
2n
þ 1
@
1
2n and
X 1
2n converges,
X 1
2n
þ 1
also converges.
(b) Divergence. Let vn A 0 for all n N and suppose that vn diverges. Then if un A vn for
all n N, un also diverges.
EXAMPLE: Since
1
ln n
1
n
and
X
1
n¼2
1
n
diverges,
X
1
n¼2
1
ln n
also diverges.
2. The Limit-Comparison or Quotient Test for series of non-negative terms.
(a) If un A 0 and vn A 0 and if lim
n!1
un
vn
¼ A 6¼ 0 or 1, then un and vn either both converge
or both diverge.
(b) If A ¼ 0 in (a) and vn converges, then un converges.
(c) If A ¼ 1 in (a) and vn diverges, then un diverges.
This test is related to the comparison test and is often a very useful alternative to it. In
particlar, taking vn ¼ 1=np
, we have from known facts about the p series the
Theorem 1. Let lim
n!1
np
un ¼ A. Then
(i) un converges if p 1 and A is finite.
(ii) un diverges if p @ 1 and A 6¼ 0 (A may be infinite).
EXAMPLES: 1:
X n
4n3 2
converges since lim
n!1
n2
n
4n3 2
¼
1
4
:
2:
X ln n
ffiffiffiffiffiffiffiffiffiffiffi
n þ 1
p diverges since lim
n!1
n1=2
ln n
ðn þ 1Þ1=2
¼ 1:
3. Integral test for series of non-negative terms.
If f ðxÞ is positive, continuous, and monotonic decreasing for x A N and is such that
f ðnÞ ¼ un; n ¼ N; N þ 1; N þ 2; . . . , then un converges or diverges according as
ð1
N
f ðxÞ dx ¼ lim
M!1
ðM
n
f ðxÞ dx converges or diverges. In particular we may have N ¼ 1, as
is often true in practice.
This theorem borrows from the next chapter since the integral has an unbounded upper
limit. (It is an improper integral. The convergence or divergence of these integrals is defined in
much the same way as for infinite series.)
EXAMPLE:
X
1
n¼1
1
n2
converges since lim
M!1
ðM
1
dx
x2
¼ lim
M!1
1
1
M
exists.
4. Alternating series test. An alternating series is one whose successive terms are alternately
positive and negative.
An alternating series converges if the following two conditions are satisfied (see Problem
11.15).
(a) junþ1j @ junj for n A N (Since a fixed number of terms does not affect the conver-
gence or divergence of a series, N may be any positive integer. Frequently it is chosen to
be 1.)
(b) lim
n!1
un ¼ 0 or lim
n!1
junj ¼ 0
CHAP. 11] INFINITE SERIES 267](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-276-320.jpg)
![THEOREMS ON ABSOLUTELY CONVERGENT SERIES
Theorem 4. (Rearrangement of Terms) The terms of an absolutely convergent series can be rearranged
in any order, and all such rearranged series will converge to the same sum. However, if the terms of a
conditionally convergent series are suitably rearranged, the resulting series may diverge or converge to
any desired sum (see Problem 11.80).
Theorem 5. (Sums, Differences, and Products) The sum, difference, and product of two absolutely
convergent series is absolutely convergent. The operations can be performed as for finite series.
INFINITE SEQUENCES AND SERIES OF FUNCTIONS, UNIFORM CONVERGENCE
We opened this chapter with the thought that functions could be expressed in series form. Such
representation is illustrated by
sin x ¼ x
x3
3!
þ
x5
5!
þ þ ð1Þn1 x2n1
ð2n 1Þ!
þ
where
sin x ¼ lim
n!1
Sn; with S1 ¼ x; S2 ¼ x
x3
3!
; . . . Sn ¼
X
n
k¼1
ð1Þk1 x2k1
ð2k 1Þ!
:
Observe that until this section the sequences and series depended on one element, n. Now there is
variation with respect to x as well. This complexity requires the introduction of a new concept called
uniform convergence, which, in turn, is fundamental in exploring the continuity, differentiation, and
integrability of series.
Let funðxÞg; n ¼ 1; 2; 3; . . . be a sequence of functions defined in ½a; b. The sequence is said to
converge to FðxÞ, or to have the limit FðxÞ in ½a; b, if for each 0 and each x in ½a; b we can find
N 0 such that junðxÞ FðxÞj for all n N. In such case we write lim
n!1
unðxÞ ¼ FðxÞ. The number
N may depend on x as well as . If it depends only on and not on x, the sequence is said to converge to
FðxÞ uniformly in ½a; b or to be uniformly convergent in ½a; b.
The infinite series of functions
X
1
n¼1
unðxÞ ¼ u1ðxÞ þ u2ðxÞ þ u3ðxÞ þ ð3Þ
is said to be convergent in ½a; b if the sequence of partial sums fSnðxÞg, n ¼ 1; 2; 3; . . . ; where
SnðxÞ ¼ u1ðxÞ þ u2ðxÞ þ þ unðxÞ, is convergent in ½a; b. In such case we write lim
n!1
SnðxÞ ¼ SðxÞ
and call SðxÞ the sum of the series.
It follows that unðxÞ converges to SðxÞ in ½a; b if for each 0 and each x in ½a; b we can find
N 0 such that jSnðxÞ SðxÞj for all n N. If N depends only on and not on x, the series is called
uniformly convergent in ½a; b.
Since SðxÞ SnðxÞ ¼ RnðxÞ, the remainder after n terms, we can equivalently say that unðxÞ is
uniformly convergent in ½a; b if for each 0 we can find N depending on but not on x such that
jRnðxÞj for all n N and all x in ½a; b.
These definitions can be modified to include other intervals besides a @ x @ b, such as a x b,
and so on.
The domain of convergence (absolute or uniform) of a series is the set of values of x for which the
series of functions converges (absolutely or uniformly).
EXAMPLE 1. Suppose un ¼ xn
=n and 1
2 @ x @ 1. Now think of the constant function FðxÞ ¼ 0 on this interval.
For any 0 and any x in the interval, there is N such that for all n Njun FðxÞj , i.e., jxn
=nj . Since the
limit does not depend on x, the sequence is uniformly convergent.
CHAP. 11] INFINITE SERIES 269](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-278-320.jpg)
![THEOREMS ON UNIFORMLY CONVERGENT SERIES
If an infinite series of functions is uniformly convergent, it has many of the properties possessed by
sums of finite series of functions, as indicated in the following theorems.
Theorem 6. If funðxÞg; n ¼ 1; 2; 3; . . . are continuous in ½a; b and if unðxÞ converges uniformly to the
sum SðxÞ in ½a; b, then SðxÞ is continuous in ½a; b.
Briefly, this states that a uniformly convergent series of continuous functions is a continuous
function. This result is often used to demonstrate that a given series is not uniformly convergent by
showing that the sum function SðxÞ is discontinuous at some point (see Problem 11.30).
In particular if x0 is in ½a; b, then the theorem states that
lim
x!x0
X
1
n¼1
unðxÞ ¼
X
1
n¼1
lim
x!x0
unðxÞ ¼
X
1
n¼1
unðx0Þ
where we use right- or left-hand limits in case x0 is an endpoint of ½a; b.
Theorem 7. If funðxÞg; n ¼ 1; 2; 3; . . . ; are continuous in ½a; b and if unðxÞ converges uniformly to the
sum SðxÞ in ½a; b, then
ðb
a
SðxÞ dx ¼
X
1
n¼1
ðb
a
unðxÞ dx ð4Þ
or
ðb
a
X
1
n¼1
unðxÞ
( )
dx ¼
X
1
n¼1
ðb
a
unðxÞ dx ð5Þ
Briefly, a uniformly convergent series of continuous functions can be integrated term by term.
Theorem 8. If funðxÞg; n ¼ 1; 2; 3; . . . ; are continuous and have continuous derivatives in ½a; b and if
unðxÞ converges to SðxÞ while u0
nðxÞ is uniformly convergent in ½a; b, then in ½a; b
S0
ðxÞ ¼
X
1
n¼1
u0
nðxÞ ð6Þ
or
d
dx
X
1
n¼1
unðxÞ
( )
¼
X
1
n¼1
d
dx
unðxÞ ð7Þ
This shows conditions under which a series can be differentiated term by term.
Theorems similar to the above can be formulated for sequences. For example, if funðxÞg,
n ¼ 1; 2; 3; . . . is uniformly convergent in ½a; b, then
lim
n!1
ðb
a
unðxÞ dx ¼
ðb
a
lim
n!1
unðxÞ dx ð8Þ
which is the analog of Theorem 7.
CHAP. 11] INFINITE SERIES 271](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-280-320.jpg)
![Theorem 14. Two power series, for example,
X
1
n¼0
anxn
and
X
1
n¼0
bnxn
, can be multiplied to obtain
X
1
n¼0
cnxn
where
cn ¼ a0bn þ a1bn1 þ a2bn2 þ þ anb0 ð11Þ
the result being valid for each x within the common interval of convergence.
Theorem 15. If the power series
X
1
n¼0
anxn
is divided by the power series bnxn
where b0 6¼ 0, the quotient
can be written as a power series which converges for sufficiently small values of x.
Theorem 16. If y ¼
X
1
n¼0
anxn
, then by substituting x ¼
X
1
n¼0
bnyn
, we can obtain the coefficients bn in
terms of an. This process is often called reversion of series.
EXPANSION OF FUNCTIONS IN POWER SERIES
This section gets at the heart of the use of infinite series in analysis. Functions are represented
through them. Certain forms bear the names of mathematicians of the eighteenth and early nineteenth
century who did so much to develop these ideas.
A simple way (and one often used to gain information in mathematics) to explore series representa-
tion of functions is to assume such a representation exists and then discover the details. Of course,
whatever is found must be confirmed in a rigorous manner. Therefore, assume
f ðxÞ ¼ A0 þ A1ðx cÞ þ A2ðx cÞ2
þ þ Anðx cÞn
þ
Notice that the coefficients An can be identified with derivatives of f . In particular
A0 ¼ f ðcÞ; A1 ¼ f 0
ðcÞ; A2 ¼
1
2!
f 00
ðcÞ; . . . ; An ¼
1
n!
f ðnÞ
ðcÞ; . . .
This suggests that a series representation of f is
f ðxÞ ¼ f ðcÞ þ f 0
ðcÞðx cÞ þ
1
2!
f 00
ðcÞðx cÞ2
þ þ
1
n!
f ðnÞ
ðcÞðx cÞn
þ
A first step in formalizing series representation of a function, f , for which the first n derivatives exist,
is accomplished by introducing Taylor polynomials of the function.
P0ðxÞ ¼ f ðcÞ P1ðxÞ ¼ f ðcÞ þ f 0
ðcÞðx cÞ;
P2ðxÞ ¼ f ðcÞ þ f 0
ðcÞðx cÞ þ
1
2!
f 00
ðcÞðx cÞ2
;
PnðxÞ ¼ f ðcÞ þ f 0
ðcÞðx cÞ þ þ
1
n!
f ðnÞ
ðcÞðx cÞn
ð12Þ
TAYLOR’S THEOREM
Let f and its derivatives f 0
; f 00
; . . . ; f ðnÞ
exist and be continuous in a closed interval a x b and
suppose that f ðnþ1Þ
exists in the open interval a x b. Then for c in ½a; b,
f ðxÞ ¼ PnðxÞ þ RnðxÞ;
where the remainder RnðxÞ may be represented in any of the three following ways.
For each n there exists such that
CHAP. 11] INFINITE SERIES 273](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-282-320.jpg)
![Thus, the approximation P4ð:3Þ for sin :3 is correct to four decimal
places.
Additional insight to the process of approximation of functional
values results by constructing a graph of P4ðxÞ and comparing it to
y ¼ sin x. (See Fig. 11-2.)
P4ðxÞ ¼ x
x3
6
The roots of the equation are 0;
ffiffiffi
6
p
. Examination of the first and
second derivatives reveals a relative maximum at x ¼
ffiffiffi
2
p
and a relative
minimum at x ¼
ffiffiffi
2
p
. The graph is a local approximation of the sin
curve. The reader can show that P6ðxÞ produces an even better approximation.
(For an example of series approximation of an integral see the example below.)
SOME IMPORTANT POWER SERIES
The following series, convergent to the given function in the indicated intervals, are frequently
employed in practice:
1. sin x ¼ x
x3
3!
þ
x5
5!
x7
7!
þ ð1Þn1 x2n1
ð2n 1Þ!
þ 1 x 1
2. cos x ¼ 1
x2
2!
þ
x4
4!
x6
6!
þ ð1Þn1 x2n2
ð2n 2Þ!
þ 1 x 1
3. ex
¼ 1 þ x þ
x2
2!
þ
x3
3!
þ þ
xn1
ðn 1Þ!
þ 1 x 1
4. ln j1 þ xj ¼ x
x2
2
þ
x3
3
x4
4
þ ð1Þn1 xn
n
þ 1 x @ 1
5. 1
2 ln
1 þ x
1 x
¼ x þ
x3
3
þ
x5
5
þ
x7
7
þ þ
x2n1
2n 1
þ 1 x 1
6. tan1
x ¼ x
x3
3
þ
x5
5
x7
7
þ ð1Þn1 x2n1
2n 1
þ 1 @ x @ 1
7. ð1 þ xÞp
¼ 1 þ px þ
pð p 1Þ
2!
x2
þ þ
pð p 1Þ . . . ð p n þ 1Þ
n!
xn
þ
This is the binomial series.
(a) If p is a positive integer or zero, the series terminates.
(b) If p 0 but is not an integer, the series converges (absolutely) for 1 @ x @ 1:
ðcÞ If 1 p 0, the series converges for 1 x @ 1:
(d) If p @ 1, the series converges for 1 x 1.
For all p the series certainly converges if 1 x 1.
EXAMPLE. Taylor’s Theorem applied to the series for ex
enables us to estimate the value of the integral
ð1
0
ex2
dx.
Substituting x2
for x, we obtain
Ð1
0 ex2
dx ¼
Ð1
0 1 þ x þ
x4
2!
þ
x6
3!
þ
x8
4!
þ
e
5!
x10
!
dx
where
P4ðxÞ ¼ 1 þ x þ
1
2!
x4
þ
1
3!
x6
þ
1
4!
x8
and
R4ðxÞ ¼
e
5!
x10
; 0 x
CHAP. 11] INFINITE SERIES 275
Fig. 11-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-284-320.jpg)
![f ðx0 þ h; y0 þ kÞ ¼ f ðx0; y0Þ þ h
@
@x
þ k
@
@y
f ðx0; y0Þ þ
1
2!
h
@
@x
þ k
@
@y
2
f ðx0; y0Þ þ
þ
1
n!
h
@
@x
þ k
@
@y
n
f ðx0; y0Þ þ Rn
ð18Þ
where
Rn ¼
1
ðn þ 1Þ!
h
@
@x
þ k
@
@y
nþ1
f ðx0 þ h; y0 þ kÞ; 0 1
and where the meaning of the operator notation is as follows:
h
@
@x
þ k
@
@y
f ¼ hfx þ kfy;
h
@
@x
þ k
@
@y
2
¼ h2
fxx þ 2hkfxy þ k2
fyy
and we formally expand h
@
@x
þ k
@
@y
n
by the binomial theorem.
Note: In alternate notation h ¼ x ¼ x x0, k ¼ y ¼ y y0.
If Rn ! 0 as n ! 1 then an unending continuation of terms produces the Taylor series for f ðx; yÞ.
Multivariable Taylor series have a similar pattern.
4. Double Series. Consider the array of numbers (or functions)
u11 u12 u13 . . .
u21 u22 u23 . . .
u31 u32 u33 . . .
.
.
. .
.
. .
.
.
0
B
B
B
@
1
C
C
C
A
Let Smn ¼
X
m
p¼1
X
n
q¼1
upq be the sum of the numbers in the first m rows and first n columns of this
array. If there exists a number S such that lim
m!1
n!1
Smn ¼ S, we say that the doubles series
X
1
p¼1
X
1
q¼1
upq converges to the sum S; otherwise, it diverges.
Definitions and theorems for double series are very similar to those for series already
considered.
5. Infinite Products. Let Pn ¼ ð1 þ u1Þð1 þ u2Þð1 þ u3Þ . . . ð1 þ unÞ denoted by
Y
n
k¼1
ð1 þ ukÞ, where
we suppose that uk 6¼ 1; k ¼ 1; 2; 3; . . . . If there exists a number P 6¼ 0 such that lim
n!1
Pn ¼ P,
we say that the the infinite product ðð1 þ u1Þð1 þ u2Þð1 þ u3Þ . . . ¼
Y
1
k¼1
ð1 þ ukÞ, or briefly
ð1 þ ukÞ, converges to P; otherwise, it diverges.
If ð1 þ jukjÞ converges, we call the infinite product ð1 þ ukÞ absolutely convergent. It can
be shown that an absolutely convergent infinite product converges and that factors can in such
cases be rearranged without affecting the result.
Theorems about infinite products can (by taking logarithms) often be made to depend on
theorems for infinite series. Thus, for example, we have the following theorem.
Theorem. A necessary and sufficient condition that ð1 þ ukÞ converge absolutely is that uk converge
absolutely.
CHAP. 11] INFINITE SERIES 277](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-286-320.jpg)
![COMPARISON TEST AND QUOTIENT TEST
11.5. If 0 @ un @ vn; n ¼ 1; 2; 3; . . . and if vn converges, prove that un also converges (i.e., establish
the comparison test for convergence).
Let Sn ¼ u1 þ u2 þ þ un; Tn ¼ v1 þ v2 þ þ vn.
Since vn converges, lim
n!1
Tn exists and equals T, say. Also, since vn A 0; Tn @ T.
Then Sn ¼ u1 þ u2 þ þ un @ v1 þ v2 þ þ vn @ T or 0 @ Sn @ T:
Thus Sn is a bounded monotonic increasing sequence and must have a limit (see Chapter 2), i.e., un
converges.
11.6. Using the comparison test prove that 1 þ 1
2 þ 1
3 þ ¼
X
1
n¼1
1
n
diverges.
1 A 1
2
We have
1
2 þ 1
3 A 1
4 þ 1
4 ¼ 1
2
1
4 þ 1
5 þ 1
6 þ 1
7 A 1
8 þ 1
8 þ 1
8 þ 1
8 ¼ 1
2
1
8 þ 1
9 þ 1
10 þ þ 1
15 A 1
16 þ 1
16 þ 1
16 þ þ 1
16 (8 terms) ¼ 1
2
etc. Thus, to any desired number of terms,
1 þ 1
2 þ 1
3
þ 1
4 þ 1
5 þ 1
6 þ 1
7
þ A 1
2 þ 1
2 þ 1
2 þ
Since the right-hand side can be made larger than any positive number by choosing enough terms, the given
series diverges.
By methods analogous to that used here, we can show that
X
1
n¼1
1
np, where p is a constant, diverges if
p @ 1 and converges if p 1. This can also be shown in other ways [see Problem 11.13(a)].
11.7. Test for convergence or divergence
X
1
n¼1
ln n
2n3 1
.
Since ln n n and
1
2n3
1
@
1
n3
; we have
ln n
2n3
1
@
n
n3
¼
1
n2
:
Then the given series converges, since
X
1
n¼1
1
n2
converges.
11.8. Let un and vn be positive. If lim
n!1
un
vn
¼ constant A 6¼ 0, prove that un converges or diverges
according as vn converges or diverges.
By hypothesis, given 0 we can choose an integer N such that
un
vn
A for all n N. Then for
n ¼ N þ 1; N þ 2; . . .
un
vn
A or ðA Þvn un ðA þ Þvn ð1Þ
Summing from N þ 1 to 1 (more precisely from N þ 1 to M and then letting M ! 1),
ðA Þ
X
1
Nþ1
vn @
X
1
Nþ1
un @ ðA þ Þ
X
1
Nþ1
vn ð2Þ
There is no loss in generality in assuming A 0. Then from the right-hand inequality of (2), un
converges when vn does. From the left-hand inequality of (2), un diverges when vn does. For the cases
A ¼ 0 or A ¼ 1, see Problem 11.66.
CHAP. 11] INFINITE SERIES 279](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-288-320.jpg)
![If lim
M!1
ðM
1
f ðxÞ dx exists and is equal to S, we see from the left-hand inequality in (1) that
u2 þ u3 þ þ uM is monotonic increasing and bounded above by S, so that un converges.
If lim
M!1
ðM
1
f ðxÞ dx is unbounded, we see from the right-hand inequality in (1) that un diverges.
Thus the proof is complete.
11.12. Illustrate geometrically the proof in Problem
11.11.
Geometrically, u2 þ u3 þ þ uM is the total area
of the rectangles shown shaded in Fig. 11-3, while
u1 þ u2 þ þ uM1 is the total area of the rectangles
which are shaded and nonshaded.
The area under the curve y ¼ f ðxÞ from x ¼ 1 to
x ¼ M is intermediate in value between the two areas
given above, thus illustrating the result (1) of Problem
11.11.
11.13. Test for convergence: (a)
X
1
1
1
nP
; p ¼ constant;
ðbÞ
X
1
1
n
n2 þ 1
; ðcÞ
X
1
2
1
n ln n
; ðdÞ
X
1
1
nen2
.
ðaÞ Consider
ðM
1
dx
xp ¼
ðM
1
xp
dx ¼
x1p
1 p
M
1
¼
M1p
1
1 p
where p 6¼ 1:
If p 1; lim
M!1
M1p
1
1 p
¼ 1, so that the integral and thus the series diverges.
If p 1; lim
M!1
M1p
1
1 p
¼
1
p 1
, so that the integral and thus the series converges.
If p ¼ 1,
ðM
1
dx
xp ¼
ðM
1
dx
x
¼ ln M and lim
M!1
ln M ¼ 1, so that the integral and thus the series
diverges.
Thus, the series converges if p 1 and diverges if p @ 1.
ðbÞ lim
M!1
ðM
1
x dx
x2 þ 1
¼ lim
M!1
1
2 lnðx2
þ 1ÞjM
1 ¼ lim
M!1
1
2 lnðM2
þ 1Þ 1
2 ln 2
¼ 1 and the series diverges.
ðcÞ lim
M!1
ðM
2
dx
x ln x
¼ lim
M!1
lnðln xÞjM
2 ¼ lim
M!1
flnðln MÞ lnðln 2Þg ¼ 1 and the series diverges.
ðdÞ lim
M!1
ðM
1
xex2
dx ¼ lim
M!1
1
2 ex2
jM
1 ¼ lim
M!1
1
2 e1
1
2 eM2
n o
¼ 1
2 e1
and the series converges.
Note that when the series converges, the value of the corresponding integral is not (in general) the
same as the sum of the series. However, the approximate sum of a series can often be obtained quite
accurately by using integrals. See Problem 11.74.
11.14. Prove that
4
X
1
n¼1
1
n2
þ 1
1
2
þ
4
.
CHAP. 11] INFINITE SERIES 281
Fig. 11-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-290-320.jpg)
![(c) The absolute value of the error made in stopping after M terms is less than 1=ð2M þ 1Þ. To obtain the
desired accuracy, we must have 1=ð2M þ 1Þ @ :001, from which M A 499:5. Thus, at least 500 terms
are needed.
ABSOLUTE AND CONDITIONAL CONVERGENCE
11.17. Prove that an absolutely convergent series is convergent.
Given that junj converges, we must show that un converges.
Let SM ¼ u1 þ u2 þ þ uM and TM ¼ ju1j þ ju2j þ þ juMj. Then
SM þ TM ¼ ðu1 þ ju1jÞ þ ðu2 þ ju2jÞ þ þ ðuM þ juMjÞ
@ 2ju1j þ 2ju2j þ þ 2juMj
Since junj converges and since un þ junj A 0, for n ¼ 1; 2; 3; . . . ; it follows that SM þ TM is a bounded
monotonic increasing sequence, and so lim
M!1
ðSM þ TMÞ exists.
Also, since lim
M!1
TM exists (since the series is absolutely convergent by hypothesis),
lim
M!1
SM ¼ lim
M!1
ðSM þ TM TMÞ ¼ lim
M!1
ðSM þ TMÞ lim
M!1
TM
must also exist and the result is proved.
11.18. Investigate the convergence of the series
sin
ffiffiffi
1
p
13=2
sin
ffiffiffi
2
p
23=2
þ
sin
ffiffiffi
3
p
33=2
.
Since each term is in absolute value less than or equal to the corresponding term of the series
1
13=2
þ
1
23=2
þ
1
33=2
þ , which converges, it follows that the given series is absolutely convergent and
hence convergent by Problem 11.17.
11.19. Examine for convergence and absolute convergence:
ðaÞ
X
1
n¼1
ð1Þn1
n
n2
þ 1
; ðbÞ
X
1
n¼2
ð1Þn1
n ln2
n
; ðcÞ
X
1
n¼1
ð1Þn1
2n
n2
:
(a) The series of absolute values is
X
1
n¼1
n
n2 þ 1
which is divergent by Problem 11.13(b). Hence, the given
series is not absolutely convergent.
However, if an ¼ junj ¼
n
n2 þ 1
and anþ1 ¼ junþ1j ¼
n þ 1
ðn þ 1Þ2
þ 1
, then anþ1 @ an for all n A 1, and
also lim
n!1
an ¼ lim
n!1
n
n2 þ 1
¼ 0. Hence, by Problem 11.15 the series converges.
Since the series converges but is not absolutely convergent, it is conditionally convergent.
(b) The series of absolute values is
X
1
n¼2
1
n ln2
n
.
By the integral test, this series converges or diverges according as lim
M!1
ðM
2
dx
x ln2
x
exists or does not
exist.
If u ¼ ln x;
ð
dx
x ln2
x
¼
ð
du
u2
¼
1
u
þ c ¼
1
ln x
þ c:
Hence, lim
M!1
ðM
2
dx
x ln2
x
¼ lim
M!1
1
ln 2
1
ln M
¼
1
ln 2
and the integral exists. Thus, the series
converges.
Then
X
1
n¼2
ð1Þn1
n ln2
n
converges absolutely and thus converges.
CHAP. 11] INFINITE SERIES 283](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-292-320.jpg)
![Another method:
Since
1
ðn þ 1Þ ln2
ðn þ 1Þ
@
1
n ln2
n
and lim
n!1
1
n ln2
n
¼ 0, it follows by Problem 11.15(a), that the
given alternating series converges. To examine its absolute convergence, we must proceed as above.
(c) Since lim
n!1
un 6¼ 0 where un ¼
ð1Þn1
2n
n2
, the given series cannot be convergent. To show that
lim
n!1
un 6¼ 0, it suffices to show that lim
n!1
junj ¼ lim
n!1
2n
n2
6¼ 0. This can be accomplished by L’Hospital’s
rule or other methods [see Problem 11.21(b)].
RATIO TEST
11.20. Establish the ratio test for convergence.
Consider first the series u1 þ u2 þ u3 þ where each term is non-negative. We must prove that if
lim
n!1
unþ1
un
¼ L 1, then necessarily un converges.
By hypothesis, we can choose an integer N so large that for all n A N, ðunþ1=unÞ r where L r 1.
Then
uNþ1 r uN
uNþ2 r uNþ1 r2
uN
uNþ3 r uNþ2 r3
uN
etc. By addition,
uNþ1 þ uNþ2 þ uNðr þ r2
þ r3
þ Þ
and so the given series converges by the comparison test, since 0 r 1.
In case the series has terms with mixed signs, we consider ju1j þ ju2j þ ju3j þ . Then by the above
proof and Problem 11.17, it follows that if lim
n!1
unþ1
un
¼ L 1, then un converges (absolutely).
Similarly, we can prove that if lim
n!1
unþ1
un
¼ L 1 the series un diverges, while if lim
n!1
unþ1
un
¼ L ¼ 1
the ratio test fails [see Problem 11.21(c)].
11.21. Investigate the convergence of (a)
X
1
n¼1
n4
en2
; ðbÞ
X
1
n¼1
ð1Þn1
2n
n2
; ðcÞ
X
1
n¼1
ð1Þn1
n
n2
þ 1
.
(a) Here un ¼ n4
en2
. Then
lim
n!1
unþ1
un
¼ lim
n!1
ðn þ 1Þ4
eðnþ1Þ2
n4 en2 ¼ lim
n!1
ðn þ 1Þ4
eðn2
þ2nþ1Þ
n4 en2
¼ lim
n!1
n þ 1
n
4
e2n1
¼ lim
n!1
n þ 1
n
4
lim
n!1
e2n1
¼ 1 0 ¼ 0
Since 0 1, the series converges.
(b) Here un ¼
ð1Þn1
2n
n2
. Then
lim
n!1
unþ1
un
¼ lim
n!1
ð1Þn
2nþ1
ðn þ 1Þ2
n2
ð1Þn1
2n
¼ lim
n!1
2n2
ðn þ 1Þ2
¼ 2
Since s 1, the series diverges. Compare Problem 11.19(c).
(c) Here un ¼
ð1Þn1
n
n2 þ 1
. Then
284 INFINITE SERIES [CHAP. 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-293-320.jpg)
![lim
n!1
unþ1
un
¼ lim
n!1
ð1Þn
ðn þ 1Þ
ðn þ 1Þ2
þ 1
n2
þ 1
ð1Þn1
n
¼ lim
n!1
ðn þ 1Þðn2
þ 1Þ
ðn2
þ 2n þ 2Þn
¼ 1
and the ratio test fails. By using other tests [see Problem 11.19(a)], the series is seen to be convergent.
MISCELLANEOUS TESTS
11.22. Test for convergence 1 þ 2r þ r2
þ 2r3
þ r4
þ 2r5
þ where (a) r ¼ 2=3, (b) r ¼ 2=3,
(c) r ¼ 4=3.
Here the ratio test is inapplicable, since
unþ1
un
¼ 2jrj or 1
2 jrj depending on whether n is odd or even.
However, using the nth root test, we have
ffiffiffiffiffiffiffiffi
junj
n
p
¼
ffiffiffiffiffiffiffiffiffi
2jrn
j
n
p
¼
ffiffiffi
2
n
p
jrj if n is odd
ffiffiffiffiffiffiffi
jrnj
n
p
¼ jrj if n is even
(
Then lim
n!1
ffiffiffiffiffiffiffiffi
junj
n
p
¼ jrj (since lim
n!1
21=n
¼ 1).
Thus, if jrj 1 the series converges, and if jrj 1 the series diverges.
Hence, the series converges for cases (a) and (b), and diverges in case (c).
11.23. Test for convergence
1
3
2
þ
1 4
3 6
2
þ
1 4 7
3 6 9
2
þ þ
1 4 7 . . . ð3n 2Þ
3 6 9 . . . ð3nÞ
2
þ .
The ratio test fails since lim
n!1
unþ1
un
¼ lim
n!1
3n þ 1
3n þ 3
2
¼ 1. However, by Raabe’s test,
lim
n!1
n 1
unþ1
un
¼ lim
n!1
n 1
3n þ 1
3n þ 3
2
( )
¼
4
3
1
and so the series converges.
11.24. Test for convergence
1
2
2
þ
1 3
2 4
2
þ
1 3 5
24t
2
þ þ
1 3 5 . . . ð2n 1Þ
2 4 6 . . . ð2nÞ
2
þ .
The ratio test fails since lim
n!1
unþ1
un
¼ lim
n!1
2n þ 1
2n þ 2
2
¼ 1. Also, Raabe’s test fails since
lim
n!1
n 1
unþ1
un
¼ lim
n!1
n 1
2n þ 1
2n þ 2
2
( )
¼ 1
However, using long division,
unþ1
un
¼
2n þ 1
2n þ 2
2
¼ 1
1
n
þ
5 4=n
4n2
þ 8n þ 4
¼ 1
1
n
þ
cn
n2
where jcnj P
so that the series diverges by Gauss’ test.
CHAP. 11] INFINITE SERIES 285](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-294-320.jpg)
![11.26. For what values of x does (a)
X
1
n¼1
1
2n 1
x þ 2
x 1
n
; ðbÞ
X
1
n¼1
1
ðx þ nÞðx þ n 1Þ
converge?
ðaÞ un ¼
1
2n 1
x þ 2
x 1
n
: Then lim
n!1
unþ1
un
¼ lim
n!1
2n 1
2n þ 1
x þ 2
x 1
¼
x þ 2
x 1
if x 6¼ 1; 2:
Then the series converges if
x þ 2
x 1
1, diverges if
x þ 2
x 1
1, and the test fails if
x þ 2
x 1
¼ 1, i.e.,
x ¼ 1
2.
If x ¼ 1 the series diverges.
If x ¼ 2 the series converges.
If x 1
2 the series is
X
1
n¼1
ð1Þn
2n 1
which converges.
Thus, the series converges for
x þ 2
x 1
1, x ¼ 1
2 and x ¼ 2, i.e., for x @ 1
2.
(b) The ratio test fails since lim
n!1
unþ1
un
¼ 1, where un ¼
1
ðx þ nÞðx þ n 1Þ
: However, noting that
1
ðx þ nÞðx þ n 1Þ
¼
1
x þ n 1
1
x þ n
we see that if x 6¼ 0; 1; 2; . . . ; n,
Sn ¼ u1 þ u2 þ þ un ¼
1
x
1
x þ 1
þ
1
x þ 1
1
x þ 2
þ þ
1
x þ n 1
1
x þ n
¼
1
x
1
x þ n
and lim
n!1
Sn ¼ 1=x, provided x 6¼ 0; 1; 2; 3; . . . .
Then the series converges for all x except x ¼ 0; 1; 2; 3; . . . ; and its sum is 1=x.
UNIFORM CONVERGENCE
11.27. Find the domain of convergence of ð1 xÞ þ xð1 xÞ þ x2
ð1 xÞ þ .
Method 1:
Sum of first n terms ¼ SnðxÞ ¼ ð1 xÞ þ xð1 xÞ þ x2
ð1 xÞ þ þ xn1
ð1 xÞ
¼ 1 x þ x x2
þ x2
þ þ xn1
xn
¼ 1 xn
If jxj 1, lim
n!1
SnðxÞ ¼ lim
n!1
ð1 xn
Þ ¼ 1.
If jxj 1, lim
n!1
SnðxÞ does not exist.
If x ¼ 1; SnðxÞ ¼ 0 and lim
n!1
SnðxÞ ¼ 0.
If x ¼ 1; SnðxÞ ¼ 1 ð1Þn
and lim
n!1
SnðxÞ does not exist.
Thus, the series converges for jxj 1 and x ¼ 1, i.e., for 1 x @ 1.
Method 2, using the ratio test.
The series converges if x ¼ 1. If x 6¼ 1 and un ¼ xn1
ð1 xÞ, then lim
n!1
unþ1
un
¼ lim
n!1
jxj.
Thus, the series converges if jxj 1, diverges if jxj 1. The test fails if jxj ¼ 1. If x ¼ 1, the series
converges; if x ¼ 1, the series diverges. Then the series converges for 1 x @ 1:
11.28. Investigate the uniform convergence of the series of Problem 11.27 in the interval
(a) 1
2 x 1
2, (b) 1
2 @ x @ 1
2, ðcÞ :99 @ x @ :99; ðdÞ 1 x 1,
ðeÞ 0 @ x 2.
CHAP. 11] INFINITE SERIES 287](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-296-320.jpg)
![WEIERSTRASS M TEST
11.31. Prove the Weierstrass M test, i.e., if junðxÞj @ Mn; n ¼ 1; 2; 3; . . . ; where Mn are positive
constants such that Mn converges, then unðxÞ is uniformly (and absolutely) convergent.
The remainder of the series unðxÞ after n terms is RnðxÞ ¼ unþ1ðxÞ þ unþ2ðxÞ þ . Now
jRnðxÞj ¼ junþ1ðxÞ þ unþ2ðxÞ þ j @ junþ1ðxÞj þ junþ2ðxÞj þ @ Mnþ1 þ Mnþ2 þ
But Mnþ1 þ Mnþ2 þ can be made less than by choosing n N, since Mn converges. Since N is clearly
independent of x, we have jRnðxÞj for n N, and the series is uniformly convergent. The absolute
convergence follows at once from the comparison test.
11.32. Test for uniform convergence:
ðaÞ
X
1
n¼1
cos nx
n4
; ðbÞ
X
1
n¼1
xn
n3=2
; ðcÞ
X
1
n¼1
sin nx
n
; ðdÞ
X
1
n¼1
1
n2
þ x2
:
(a)
cos nx
n4
@
1
n4
¼ Mn. Then since Mn converges ð p series with p ¼ 4 1Þ, the series is uniformly (and
absolutely) convergent for all x by the M test.
(b) By the ratio test, the series converges in the interval 1 @ x @ 1, i.e., jxj @ 1.
For all x in this interval,
xn
n3=2
¼
jxjn
n3=2
@
1
n3=2
. Choosing Mn ¼
1
n3=2
, we see that Mn converges.
Thus, the given series converges uniformly for 1 @ x @ 1 by the M test.
(c)
sin nx
n
@
1
n
. However, Mn, where Mn ¼
1
n
, does not converge. The M test cannot be used in this
case and we cannot conclude anything about the uniform convergence by this test (see, however,
Problem 11.125).
(d)
1
n2
þ x2
@
1
n2
, and
1
n2
converges. Then by the M test the given series converges uniformly for all x.
11.33. If a power series anxn
converges for x ¼ x0, prove that it converges (a) absolutely in the
interval jxj jx0j, (b) uniformly in the interval jxj @ jx1j; where jx1j jx0j.
(a) Since anxn
0 converges, lim
n!1
anxn
0 ¼ 0 and so we can make janxn
0j 1 by choosing n large enough, i.e.,
janj
1
jx0jn for n N. Then
X
1
Nþ1
janxn
j ¼
X
1
Nþ1
janjjxjn
X
1
Nþ1
jxjn
jx0jn ð1Þ
Since the last series in (1) converges for jxj jx0j, it follows by the comparison test that the first
series converges, i.e., the given series is absolutely convergent.
(b) Let Mn ¼
jx1jn
jx0jn. Then Mn converges since jx1j jx0j. As in part (a), janxn
j Mn for jxj @ jx1j, so
that by the Weierstrass M test, anxn
is uniformly convergent.
It follows that a power series is uniformly convergent in any interval within its interval of con-
vergence.
THEOREMS ON UNIFORM CONVERGENCE
11.34. Prove Theorem 6, Page 271.
We must show that SðxÞ is continuous in ½a; b.
CHAP. 11] INFINITE SERIES 289](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-298-320.jpg)
![11.37. Let SnðxÞ ¼ nxenx2
; n ¼ 1; 2; 3; . . . ; 0 @ x @ 1.
ðaÞ Determine whether lim
n!1
ð1
0
SnðxÞ dx ¼
ð1
0
lim
n!1
SnðxÞ dx:
ðbÞ Explain the result in (aÞ:
ðaÞ
ð1
0
snðxÞ dx ¼
ð1
0
nxenx2
dx ¼ 1
2 enx2
j1
0 ¼ 1
2 ð1 en
Þ: Then
lim
n!1
ð1
0
SnðxÞ dx ¼ lim
n!1
1
2 ð1 en
Þ ¼ 1
2
SðxÞ ¼ lim
n!1
SnðxÞ ¼ lim
n!1
nxenx2
¼ 0; whether x ¼ 0 or 0 x @ 1: Then,
ð1
0
SðxÞ dx ¼ 0
It follows that lim
n!1
ð1
0
SnðxÞ dx 6¼
ð1
0
lim
n!1
SnðxÞ dx, i.e., the limit cannot be taken under the integral
sign.
(b) The reason for the result in (a) is that although the sequence SnðxÞ converges to 0, it does not converge
uniformly to 0. To show this, observe that the function nxenx2
has a maximum at x ¼ 1=
ffiffiffiffiffi
2n
p
(by the
usual rules of elementary calculus), the value of this maximum being
ffiffiffiffiffi
1
2 n
q
e1=2
. Hence, as n ! 1,
SnðxÞ cannot be made arbitrarily small for all x and so cannot converge uniformly to 0.
11.38. Let f ðxÞ ¼
X
1
n¼1
sin nx
n3
: Prove that
ð
0
f ðxÞ dx ¼ 2
X
1
n¼1
1
ð2n 1Þ4
.
We have
sin nx
n3
@
1
n3
. Then by the Weierstrass M test the series is uniformly convergent for all x, in
particular 0 @ x @ , and can be integrated term by term. Thus
ð
0
f ðxÞ dx ¼
ð
0
X
1
n¼1
sin nx
n3
!
dx ¼
X
1
n¼1
ð
0
sin nx
n3
dx
¼
X
1
n¼1
1 cos n
n4
¼ 2
1
14
þ
1
34
þ
1
54
þ
¼ 2
X
1
n¼1
1
ð2n 1Þ4
POWER SERIES
11.39. Prove that both the power series
X
1
n¼0
anxn
and the corresponding series of derivatives
X
1
n¼0
nanxn1
have the same radius of convergence.
Let R 0 be the radius of convergence of anxn
. Let 0 jx0j R. Then, as in Problem 11.33, we can
choose N as that janj
1
jx0jn for n N.
Thus, the terms of the series jnanxn1
j ¼ njanjjxjn1
can for n N be made less than corresponding
terms of the series n
jxjn1
jx0jn , which converges, by the ratio test, for jxj jx0j R.
Hence, nanxn1
converges absolutely for all points x0 (no matter how close jx0j is to R).
If, however, jxj R, lim
n!1
anxn
6¼ 0 and thus lim
n!1
nanxn1
6¼ 0, so that nanxn1
does not converge.
CHAP. 11] INFINITE SERIES 291](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-300-320.jpg)
![Now
RnðxÞ ¼ ðRn Rnþ1Þxn
þ ðRnþ1 Rnþ2Þxnþ1
þ ðRnþ2 Rnþ3Þxnþ2
þ
¼ Rnxn
þ Rnþ1ðxnþ1
xn
Þ þ Rnþ2ðxnþ2
xnþ1
Þ þ
¼ xn
fRn ð1 xÞðRnþ1 þ Rnþ2x þ Rnþ3x2
þ Þg
Hence, for 0 @ x 1,
jRnðxÞj @ jRnj þ ð1 xÞðjRnþ1j þ jRnþ2jx þ jRnþ3jx2
þ Þ ð1Þ
Since ak converges by hypothesis, it follows that given 0 we can choose N such that jRkj =2 for
all k A n. Then for n N we have from (1),
jRnðxÞj @
2
þ ð1 xÞ
2
þ
2
x þ
2
x2
þ
¼
2
þ
2
¼ ð2Þ
since ð1 xÞð1 þ x þ x2
þ x3
þ Þ ¼ 1 (if 0 @ x 1).
Also, for x ¼ 1; jRnðxÞj ¼ jRnj for n N.
Thus, jRnðxÞj for all n N, where N is independent of the value of x in 0 @ x @ 1, and the
required result follows.
Extensions to other power series are easily made.
11.43. Prove Abel’s limit theorem (see Page 272).
As in Problem 11.42, assume the power series to be
X
1
k¼1
akxk
, convergent for 0 @ x @ 1.
Then we must show that lim
x!1
X
1
k¼0
akxk
¼
X
1
k¼0
ak.
This follows at once from Problem 11.42, which shows that akxk
is uniformly convergent for
0 @ x @ 1, and from Problem 11.34, which shows that akxk
is continuous at x ¼ 1.
Extensions to other power series are easily made.
11.44. (a) Prove that tan1
x ¼ x
x3
3
þ
x5
5
x7
7
þ where the series is uniformly convergent in
1 @ x @ 1.
(b) Prove that
4
¼ 1
1
3
þ
1
5
1
7
þ .
(a) By Problem 2.25 of Chapter 2, with r ¼ x2
and a ¼ 1, we have
1
1 þ x2
¼ 1 x2
þ x4
x6
þ 1 x 1 ð1Þ
Integrating from 0 to x, where 1 x 1, yields
ðx
0
dx
1 þ x2
¼ tan1
x ¼ x
x3
3
þ
x5
5
x7
7
þ ð2Þ
using Problems 11.33 and 11.35.
Since the series on the right of (2) converges for x ¼ 1, it follows by Problem 11.42 that the series
is uniformly convergent in 1 @ x @ 1 and represents tan1
x in this interval.
(b) By Problem 11.43 and part (a), we have
lim
x!1
tan1
x ¼ lim
x!1
x
x3
3
þ
x5
5
x7
7
þ
!
or
4
¼ 1
1
3
þ
1
5
1
7
þ
11.45. Evaluate
ð1
0
1 ex2
x2
dx to 3 decimal place accuracy.
CHAP. 11] INFINITE SERIES 293](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-302-320.jpg)
![We have eu
¼ 1 þ u þ
u2
2!
þ
u3
3!
þ
u4
4!
þ
u5
5!
þ ; 1 u 1:
Then if u ¼ x2
; ex2
¼ 1 x2
þ
x4
2!
x6
3!
þ
x8
3!
¼
x10
5!
þ ; 1 x 1:
Thus
1 ex2
x2
¼ 1
x2
2!
þ
x4
3!
x6
4!
þ
x8
5!
:
Since the series converges for all x and so, in particular, converges uniformly for 0 @ x @ 1, we can
integrate term by term to obtain
ð1
0
1 ex2
x2
dx ¼ x
x3
3 2!
þ
x5
5 3!
x7
7 4!
þ
x9
9 5!
1
0
¼ 1
1
3 2!
þ
1
5 3!
1
7 4!
þ
1
9 5!
¼ 1 0:16666 þ 0:03333 0:00595 þ 0:00092 ¼ 0:862
Note that the error made in adding the first four terms of the alternating series is less than the fifth term,
i.e., less than 0.001 (see Problem 11.15).
MISCELLANEOUS PROBLEMS
11.46. Prove that y ¼ JpðxÞ defined by (16), Page 276, satisfies Bessel’s differential equation
x2
y00
þ xy0
þ ðx2
p2
Þy ¼ 0
The series for JpðxÞ converges for all x [see Problem 11.110(a)]. Since a power series can be differ-
entiated term by term within its interval of convergence, we have for all x,
y ¼
X
1
n¼0
ð1Þn
xpþ2n
2pþ2nn!ðn þ pÞ!
y0
¼
X
1
n¼0
ð1Þn
ð p þ 2nÞxpþ2n1
2pþ2nn!ðn þ pÞ!
y00
¼
X
1
n¼0
ð1Þn
ð p þ 2nÞð p þ 2n 1Þ xpþ2n2
2pþ2nn!ðn þ pÞ!
Then,
ðx2
p2
Þy ¼
X
1
n¼0
ð1Þn
xpþ2nþ2
2pþ2nn!ðn þ pÞ!
X
1
n¼0
ð1Þn
p2
xpþ2n
2pþ2nn!ðn þ pÞ!
xy0
¼
X
1
n¼0
ð1Þn
ðp þ 2nÞxpþ2n
2pþ2nn!ðn þ pÞ!
x2
y00
¼
X
1
n¼0
ð1Þn
ð p þ 2nÞð p þ 2n 1Þxpþ2n
2pþ2nn!ðn þ pÞ!
294 INFINITE SERIES [CHAP. 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-303-320.jpg)
![Adding,
x2
y00
þ xy0
þ ðx2
p2
Þy ¼
X
1
n¼0
ð1Þn
xpþ2nþ2
2pþ2nn!ðn þ pÞ!
þ
X
1
n¼0
ð1Þn
½p2
þ ð p þ 2nÞ þ ð p þ 2nÞð p þ 2n 1Þxpþ2n
2pþ2nn!ðn þ pÞ!
¼
X
1
n¼0
ð1Þn
xpþ2nþ2
2pþ2nn!ðn þ pÞ!
þ
X
1
n¼0
ð1Þn
½4nðn þ pÞxpþ2n
2pþ2nn!ðn þ pÞ!
¼
X
1
n¼1
ð1Þn1
xpþ2n
2pþ2n2ðn 1Þ!ðn 1 þ pÞ!
þ
X
1
n¼1
ð1Þn
4xpþ2n
2pþ2nðn 1Þ!ðn þ p 1Þ!
¼
X
1
n¼1
ð1Þn
4xpþ2n
2pþ2nðn 1Þ!ðn þ p 1Þ!
þ
X
1
n¼1
ð1Þn
4xpþ2n
2pþ2nðn 1Þ!ðn þ p 1Þ!
¼ 0
11.47. Test for convergence the complex power series
X
1
n¼1
zn1
n3 3n1
.
Since lim
n!1
unþ1
un
¼ lim
n!1
zn
ðn þ 1Þ3
3n
n3
3n1
zn1
¼ lim
n!1
n3
3ðn þ 1Þ3
jzj ¼
jzj
3
, the series converges for
jzj
3
1,
i.e., jzj 3, and diverges for jzj 3.
For jzj ¼ 3, the series of absolute values is
X
1
n¼1
jzjn1
n3 3n1
¼
X
1
n¼1
1
n3
, so that the series is absolutely
convergent and thus convergent for jzj ¼ 3.
Thus, the series converges within and on the circle jzj ¼ 3.
11.48. Assuming the power series for ex
holds for complex numbers, show that
eix
¼ cos x þ i sin x
Letting z ¼ ix in ez
¼ 1 þ z þ
z2
2!
þ
z3
3!
þ ; we have
eix
¼ 1 þ ix þ
i2
x2
2!
þ
i3
x3
3!
þ ¼ 1
x2
2!
þ
x4
4!
!
þ i x
x3
3!
þ
x5
5!
!
¼ cos x þ i sin x
Similarly, eix
¼ cos x i sin x. The results are called Euler’s identities.
11.49. Prove that lim
n!1
1 þ
1
2
þ
1
3
þ
1
4
þ þ
1
n
ln n
exists.
Letting f ðxÞ ¼ 1=x in (1), Problem 11.11, we find
1
2
þ
1
3
þ
1
4
þ þ
1
M
@ ln M @ 1 þ
1
2
þ
1
3
þ
1
4
þ þ
1
M 1
from which we have on replacing M by n,
1
n
@ 1 þ
1
2
þ
1
3
þ
1
4
þ þ
1
n
ln n @ 1
Thus, the sequence Sn ¼ 1 þ
1
2
þ
1
3
þ
1
4
þ þ
1
n
ln n is bounded by 0 and 1.
CHAP. 11] INFINITE SERIES 295](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-304-320.jpg)
![To obtain the Lagrange form of the remainder Rn, consider the form
f ðxÞ ¼ f ðcÞ þ f 0
ðcÞðx cÞ þ
1
2!
f 00
ðcÞðx cÞ2
þ þ
K
n!
ðx cÞn
This is the Taylor polynomial Pn1ðxÞ plus
K
n!
ðx cÞn
: Also, it could be looked upon as Pn except that
in the last term, f ðnÞ
ðcÞ is replaced by a number K such that for fixed c and x the representation of f ðxÞ is
exact. Now define a new function
ðtÞ ¼ f ðtÞ f ðxÞ þ
X
n1
j¼1
f ð jÞ
ðtÞ
ðx tÞ j
j!
þ
Kðx tÞn
n!
The function satisfies the hypothesis of Rolle’s Theorem in that ðcÞ ¼ ðxÞ ¼ 0, the function is
continuous on the interval bound by c and x, and 0
exists at each point of the interval. Therefore, there
exists in the interval such that 0
ðÞ ¼ 0. We proceed to compute 0
and set it equal to zero.
0
ðtÞ ¼ f 0
ðtÞ þ
X
n1
j¼1
f ð jþ1Þ
ðtÞ
ðx tÞ j
j!
X
n1
j¼1
f ð jÞ
ðtÞ
ðx tÞj1
ð j 1Þ!
Kðx tÞn1
ðn 1Þ!
This reduces to
0
ðtÞ ¼
f ðnÞ
ðtÞ
ðn 1Þ!
ðx tÞn1
K
ðn 1Þ!
ðx tÞn1
According to hypothesis: for each n there is n such that
ðnÞ ¼ 0
Thus
K ¼ f ðnÞ
ðnÞ
and the Lagrange remainder is
Rn1 ¼
f ðnÞ
ðnÞ
n!
ðx cÞn
or equivalently
Rn ¼
1
ðn þ 1Þ!
f ðnþ1Þ
ðnþ1Þðx cÞnþ1
The Cauchy form of the remainder follows immediately by applying the mean value theorem for
integrals. (See Page 274.)
11.53. Extend Taylor’s theorem to functions of two variables x and y.
Define FðtÞ ¼ f ðx0 þ ht; y0 þ ktÞ, then applying Taylor’s theorem for one variable (about t ¼ 0Þ
FðtÞ ¼ Fð0Þ þ F 0
ð0Þ þ
1
2!
F 00
ð0Þt2
þ þ
1
n!
FðnÞ
ð0Þtn
þ
1
ðn þ 1Þ!
Fðnþ1Þ
ðÞtnþ1
; 0 t
Now let t ¼ 1
Fð1Þ ¼ f ðx0 þ h; y0 þ kÞ ¼ Fð0Þ þ F 0
ð0Þ þ
1
2!
F 00
ð0Þ þ þ
1
n!
FðnÞ
ð0Þ þ
1
ðn þ 1Þ!
Fðnþ1Þ
ðÞ
When the derivatives F 0
ðtÞ; . . . ; FðnÞ
ðtÞ; Fðnþ1Þ
ðÞ are computed and substituted into the previous expres-
sion, the two variable version of Taylor’s formula results. (See Page 277, where this form and notational
details can be found.)
11.54. Expand x2
þ 3y 2 in powers of x 1 and y þ 2. Use Taylor’s formula with h ¼ x x0,
k ¼ y y0, where x0 ¼ 1 and y0 ¼ 2.
CHAP. 11] INFINITE SERIES 297](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-306-320.jpg)
![11.65. Use the comparison test to prove that
ðaÞ
X
1
n¼1
@
1
np converges if p 1 and diverges if p @ 1; ðbÞ
X
1
n¼1
tan1
n
n
diverges, ðcÞ
X
1
n¼1
n2
2n converges.
11.66. Establish the results (b) and (c) of the quotient test, Page 267.
11.67. Test for convergence:
ðaÞ
X
1
n¼1
ðln nÞ2
n2
; ðbÞ
X
1
n¼1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n tan1
ð1=n3
Þ
q
; ðcÞ
X
1
n¼1
3 þ sin n
nð1 þ enÞ
; ðdÞ
X
1
n¼1
n sin2
ð1=nÞ:
Ans. (a) conv., (b) div., (c) div., (d) div.
11.68. If un converges, where un A 0 for n N, and if lim
n!1
nun exists, prove that lim
n!1
nun ¼ 0.
11.69. (a) Test for convergence
X
1
n¼1
1
n1þ1=n
. (b) Does your answer to (a) contradict the statement about the p
series made on Page 266 that 1=np
converges for p 1?
Ans. (a) div.
INTEGRAL TEST
11.70. Test for convergence: (a)
X
1
n¼1
n2
2n3 1
; ðbÞ
X
1
n¼2
1
nðln nÞ3
; ðcÞ
X
1
n¼1
n
2n ; ðdÞ
X
1
n¼1
e
ffiffi
n
p
ffiffiffi
n
p ðeÞ
X
1
n¼2
ln n
n
;
ð f Þ
X
1
n¼10
2lnðln nÞ
n ln n
:
Ans: ðaÞ div., ðbÞ conv., ðcÞ conv., ðdÞ conv., ðeÞ div., ð f Þ div.
11.71. Prove that
X
1
n¼2
1
nðln nÞp, where p is a constant, (a) converges if p 1 and (b) diverges if p @ 1.
11.72. Prove that
9
8
X
1
n¼1
1
n3
5
4
.
11.73. Investigate the convergence of
X
1
n¼1
etan1
n
n2 þ 1
:
Ans: conv.
11.74. (a) Prove that 2
3 n3=2
þ 1
3 @
ffiffiffi
1
p
þ
ffiffiffi
2
p
þ
ffiffiffi
3
p
þ þ
ffiffiffi
n
p
@ 2
3 n3=2
þ n1=2
2
3.
(b) Use (a) to estimate the value of
ffiffiffi
1
p
þ
ffiffiffi
2
p
þ
ffiffiffi
3
p
þ þ
ffiffiffiffiffiffiffiffi
100
p
, giving the maximum error.
(c) Show how the accuracy in (b) can be improved by estimating, for example,
ffiffiffiffiffi
10
p
þ
ffiffiffiffiffi
11
p
þ þ
ffiffiffiffiffiffiffiffi
100
p
and adding on the value of
ffiffiffi
1
p
þ
ffiffiffi
2
p
þ þ
ffiffiffi
9
p
computed to some desired degree of accuracy.
Ans: ðbÞ 671:5 4:5
ALTERNATING SERIES
11.75. Test for convergence: (a)
X
1
n¼1
ð1Þnþ1
2n ; ðbÞ
X
1
n¼1
ð1Þn
n2 þ 2n þ 2
; ðcÞ
X
1
n¼1
ð1Þnþ1
n
3n 1
;
ðdÞ
X
1
n¼1
ð1Þn
sin1 1
n
; ðeÞ
X
1
n¼2
ð1Þn ffiffiffi
n
p
ln n
:
Ans. (a) conv., (b) conv., (c) div., (d) conv., (e) div.
CHAP. 11] INFINITE SERIES 299](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-308-320.jpg)
![11.76. (a) What is the largest absolute error made in approximating the sum of the series
X
1
n¼1
ð1Þn
2n
ðn þ 1Þ
by the sum
of the first 5 terms?
Ans. 1/192
(b) What is the least number of terms which must be taken in order that 3 decimal place accuracy will
result?
Ans. 8 terms
11.77. (a) Prove that S ¼
1
13
þ
1
23
þ
1
33
þ ¼
4
3
1
13
1
23
þ
1
33
.
(b) How many terms of the series on the right are needed in order to calculate S to six decimal place
accuracy?
Ans. (b) at least 100 terms
ABSOLUTE AND CONDITIONAL CONVERGENCE
11.78. Test for absolute or conditional convergence:
ðaÞ
X
1
n¼1
ð1Þn1
n2 þ 1
ðcÞ
X
1
n¼2
ð1Þn
n ln n
ðeÞ
X
1
n¼1
ð1Þn1
2n 1
sin
1
ffiffiffi
n
p
ðbÞ
X
1
n¼1
ð1Þn1
n
n2
þ 1
ðdÞ
X
1
n¼1
ð1Þn
n3
ðn2 þ 1Þ4=3
ð f Þ
X
1
n¼1
ð1Þn1
n3
2n 1
Ans. (a) abs. conv., (b) cond. conv., (c) cond. conv., (d) div., (e) abs. conv., ( f ) abs. conv.
11.79. Prove that
X
1
n¼1
cos na
x2 þ n2
converges absolutely for all real x and a.
11.80. If 1 1
2 þ 1
3 1
4 þ converges to S, prove that the rearranged series 1 þ 1
3 1
2 þ 1
5 þ 1
7 1
4 þ 1
9 þ 1
11 1
6 þ
¼ 3
2 S. Explain.
[Hint: Take 1/2 of the first series and write it as 0 þ 1
2 þ 0 1
4 þ 0 þ 1
6 þ ; then add term by term to the first
series. Note that S ¼ ln 2, as shown in Problem 11.100.]
11.81. Prove that the terms of an absolutely convergent series can always be rearranged without altering the sum.
RATIO TEST
11.82. Test for convergence:
ðaÞ
X
1
n¼1
ð1Þn
n
ðn þ 1Þen ; ðbÞ
X
1
n¼1
102n
ð2n 1Þ!
; ðcÞ
X
1
n¼1
3n
n3
; ðdÞ
X
1
n¼1
ð1Þn
23n
32n
; ðeÞ
X
1
n¼1
ð
ffiffiffi
5
p
1Þn
n2 þ 1
:
Ans. (a) conv. (abs.), (b) conv., (c) div., (d) conv. (abs.), (e) div.
11.83. Show that the ratio test cannot be used to establish the conditional convergence of a series.
11.84. Prove that (a)
X
1
n¼1
n!
nn converges and (b) lim
n!1
n!
nn ¼ 0.
MISCELLANEOUS TESTS
11.85. Establish the validity of the nth root test on Page 268.
11.86. Apply the nth root test to work Problems 11.82ðaÞ, (c), (d), and (e).
11.87. Prove that 1
3 þ ð2
3Þ2
þ ð1
3Þ3
þ ð2
3Þ4
þ ð1
3Þ5
þ ð2
3Þ6
þ converges.
300 INFINITE SERIES [CHAP. 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-309-320.jpg)
![11.88. Test for convergence: (a)
1
3
þ
1 4
3 6
þ
1 4 7
3 6 9
þ , (b)
2
9
þ
2 5
9 12
þ
2 5 8
9 12 15
þ .
Ans. (a) div., (b) conv.
11.89. If a; b, and d are positive numbers and b a, prove that
a
b
þ
aða þ dÞ
bðb þ dÞ
þ
aða þ dÞða þ 2dÞ
bðb þ dÞðb þ 2dÞ
þ
converges if b a d, and diverges if b a @ d.
SERIES OF FUNCTIONS
11.90. Find the domain of convergence of the series:
ðaÞ
X
1
n¼1
xn
n3
; ðbÞ
X
1
n¼1
ð1Þn
ðx 1Þn
2nð3n 1Þ
; ðcÞ
X
1
n¼1
1
nð1 þ x2Þn ; ðdÞ
X
1
n¼1
n2 1 x
1 þ x
n
; ðeÞ
X
1
n¼1
enx
n2 n þ 1
Ans: ðaÞ 1 @ x @ 1; ðbÞ 1 x @ 3; ðcÞ all x 6¼ 0; ðdÞ x 0; ðeÞ x @ 0
11.91. Prove that
X
1
n¼1
1 3 5 ð2n 1Þ
2 4 6 ð2nÞ
xn
converges for 1 @ x 1.
UNIFORM CONVERGENCE
11.92. By use of the definition, investigate the uniform convergence of the series
X
1
n¼1
x
½1 þ ðn 1Þx½1 þ nx
Hint: Resolve the nth term into partial fractions and show that the nth partial sum is SnðxÞ ¼ 1
1
1 þ nx
:
Ans. Not uniformly convergent in any interval which includes x ¼ 0; uniformly convergent in any other
interval.
11.93. Work Problem 11.30 directly by first obtaining SnðxÞ.
11.94. Investigate by any method the convergence and uniform convergence of the series:
ðaÞ
X
1
n¼1
x
3
n
; ðbÞ
X
1
n¼1
sin2
nx
2n
1
; ðcÞ
X
1
n¼1
x
ð1 þ xÞn ; x A 0:
Ans. (a) conv. for jxj 3; unif. conv. for jxj @ r 3. (b) unif. conv. for all x. (c) conv. for x A 0; not
unif. conv. for x A 0, but unif. conv. for x A r 0.
11.95. If FðxÞ ¼
X
1
n¼1
sin nx
n3
, prove that:
(a) FðxÞ is continuous for all x, (b) lim
x!0
FðxÞ ¼ 0; ðcÞ F 0
ðxÞ ¼
X
1
n¼1
cos nx
n2
is continous everywhere.
11.96. Prove that
ð
0
cos 2x
1 3
þ
cos 4x
3 5
þ
cos 6x
5 7
þ
dx ¼ 0.
11.97. Prove that FðxÞ ¼
X
1
n¼1
sin nx
sinh n
has derivatives of all orders for any real x.
11.98. Examine the sequence unðxÞ ¼
1
1 þ x2n
; n ¼ 1; 2; 3; . . . ; for uniform convergence.
11.99. Prove that lim
n!1
ð1
0
dx
ð1 þ x=nÞn ¼ 1 e1
.
CHAP. 11] INFINITE SERIES 301](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-310-320.jpg)
![POWER SERIES
11.100. (a) Prove that lnð1 þ xÞ ¼ x
x2
2
þ
x3
3
x4
4
þ .
ðbÞ Prove that ln 2 ¼ 1 1
2 þ 1
3 1
4 þ :
Hint: Use the fact that
1
1 þ x
¼ 1 x þ x2
x3
þ and integrate.
11.101. Prove that sin1
x ¼ x þ
1
2
x3
3
þ
1 3
2 4
x5
5
þ
1 3 5
2 4 6
x7
7
þ , 1 @ x @ 1.
11.102. Evaluate (a)
ð1=2
0
ex2
dx; ðdÞ
ð1
0
1 cos x
x
dx to 3 decimal places, justifying all steps.
Ans. ðaÞ 0:461; ðbÞ 0:486
11.103. Evaluate (a) sin 408; ðbÞ cos 658; ðcÞ tan 128 correct to 3 decimal places.
Ans: ðaÞ 0:643; ðbÞ 0:423; ðcÞ 0:213
11.104. Verify the expansions 4, 5, and 6 on Page 275.
11.105. By multiplying the series for sin x and cos x, verify that 2 sin x cos x ¼ sin 2x.
11.106. Show that ecos x
¼ e 1
x2
2!
þ
4x4
4!
31x6
6!
þ
!
; 1 x 1.
11.107. Obtain the expansions
ðaÞ tanh1
x ¼ x þ
x3
3
þ
x5
5
þ
x7
7
þ 1 x 1
ðbÞ lnðx þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ 1
p
Þ ¼ x
1
2
x3
3
þ
1 3
2 4
x5
5
1 3 5
2 4 6
x7
7
þ 1 @ x @ 1
11.108. Let f ðxÞ ¼ e1=x2
x 6¼ 0
0 x ¼ 0
. Prove that the formal Taylor series about x ¼ 0 corresponding to f ðxÞ exists
but that it does not converge to the given function for any x 6¼ 0.
11.109. Prove that
ðaÞ
lnð1 þ xÞ
1 þ x
¼ x 1 þ
1
2
x2
þ 1 þ
1
2
þ
1
3
x3
for 1 x 1
ðbÞ flnð1 þ xÞg2
¼ x2
1 þ
1
2
2x3
3
þ 1 þ
1
2
þ
1
3
2x4
4
for 1 x @ 1
MISCELLANEOUS PROBLEMS
11.110. Prove that the series for JpðxÞ converges (a) for all x, (b) absolutely and uniformly in any finite interval.
11.111. Prove that (a)
d
dx
fJ0ðxÞg ¼ J1ðxÞ; ðbÞ
d
dx
fxp
JpðxÞg ¼ xp
Jp1ðxÞ; ðcÞ Jpþ1ðxÞ ¼
2p
x
JpðxÞ Jp1ðxÞ.
11.112. Assuming that the result of Problem 11.111(c) holds for p ¼ 0; 1; 2; . . . ; prove that
(a) J1ðxÞ ¼ J1ðxÞ; ðbÞ J2ðxÞ ¼ J2ðxÞ; ðcÞ JnðxÞ ¼ ð1Þn
JnðxÞ; n ¼ 1; 2; 3; . . . :
11.113. Prove that e1=2xðt1=tÞ
¼
X
1
p¼1
JpðxÞ tp
.
[Hint: Write the left side as ext=2
ex=2t
, expand and use Problem 11.112.]
302 INFINITE SERIES [CHAP. 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-311-320.jpg)
![11.114. Prove that
X
1
n¼1
ðn þ 1Þzn
nðn þ 2Þ2
is absolutely and uniformly convergent at all points within and on the circle jzj ¼ 1.
11.115. (a) If
X
1
n¼1
anxn
¼
X
1
n¼1
bnxn
for all x in the common interval of convergence jxj R where R 0, prove that
an ¼ bn for n ¼ 0; 1; 2; . . . . (b) Use (a) to show that the Taylor expansion of a function exists, the
expansion is unique.
11.116. Suppose that lim
ffiffiffiffiffiffiffiffi
junj
n
p
¼ L. Prove that un converges or diverges according as L 1 or L 1. If L ¼ 1
the test fails.
11.117. Prove that the radius of convergence of the series anxn
can be determined by the following limits, when
they exist, and give examples: (a) lim
n!1
an
anþ1
; ðbÞ lim
n!1
1
ffiffiffiffiffiffiffiffi
janj
n
p ; ðcÞ lim
n!1
1
ffiffiffiffiffiffiffiffi
janj
n
p :
11.118. Use Problem 11.117 to find the radius of convergence of the series in Problem 11.22.
11.119. (a) Prove that a necessary and sufficient condition that the series un converge is that, given any 0, we
can find N 0 depending on such that jSp Sqj whenever p N and q N, where
Sk ¼ u1 þ u2 þ þ uk.
ðbÞ Use ðaÞ to prove that the series
X
1
n¼1
n
ðn þ 1Þ3n converges.
ðcÞ How could you use ðaÞ to prove that the series
X
1
n¼1
1
n
diverges?
[Hint: Use the Cauchy convergence criterion, Page 25.]
11.120. Prove that the hypergeometric series (Page 276) (a) is absolutely convergent for jxj 1, (b) is divergent
for jxj 1, (c) is absolutely divergent for jxj ¼ 1 if a þ b c 0; ðdÞ satisfies the differential equation
xð1 xÞy00
þ fc ða þ b þ 1Þxgy0
aby ¼ 0.
11.121. If Fða; b; c; xÞ is the hypergeometric function defined by the series on Page 276, prove that
(a) Fðp; 1; 1; xÞ ¼ ð1 þ xÞp
; ðbÞ xFð1; 1; 2; xÞ ¼ lnð1 þ xÞ; ðcÞ Fð1
2 ; 1
2 ; 3
2 ; x2
Þ ¼ ðsin1
xÞ=x.
11.122. Find the sum of the series SðxÞ ¼ x þ
x3
1 3
þ
x5
1 3 5
þ .
[Hint: Show that S0
ðxÞ 1 þ xSðxÞ and solve.]
Ans: ex2
=2
ðx
0
ex2
=2
dx
11.123. Prove that
1 þ
1
1 3
þ
1
1 3 5
þ
1
1 3 5 7
þ ¼
ffiffiffi
e
p
1
1
2 3
þ
1
22
2! 5
1
23
3! 7
þ
1
24
4! 9
11.124. Establish the Dirichlet test on Page 270.
11.125. Prove that
X
1
n¼1
sin nx
n
is uniformly convergent in any interval which does not include 0; ; 2; . . . .
[Hint: use the Dirichlet test, Page 270, and Problem 1.94, Chapter 1.]
11.126. Establish the results on Page 275 concerning the binomial series.
[Hint: Examine the Lagrange and Cauchy forms of the remainder in Taylor’s theorem.]
CHAP. 11] INFINITE SERIES 303](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-312-320.jpg)
![11.127. Prove that
X
1
n¼1
ð1Þn1
n þ x2
converges uniformly for all x, but not absolutely.
11.128. Prove that 1
1
4
þ
1
7
1
10
þ ¼
3
ffiffiffi
3
p þ
1
3
ln 2
11.129. If x ¼ yey
, prove that y ¼
X
1
n¼1
ð1Þn1
nn1
n!
xn
for 1=e x @ 1=e.
11.130. Prove that the equation e
¼ 1 has only one real root and show that it is given by
¼ 1 þ
X
1
n¼1
ð1Þn1
nn1
en
n!
11.131. Let
x
ex 1
¼ 1 þ B1x þ
B2x2
2!
þ
B3x3
3!
þ . (a) Show that the numbers Bn, called the Bernoulli numbers,
satisfy the recursion formula ðB þ 1Þn
Bn
¼ 0 where Bk
is formally replaced by Bk after expanding.
(b) Using (a) or otherwise, determine B1; . . . ; B6.
Ans: ðbÞ B1 ¼ 1
2 ; B2 ¼ 1
6 ; B3 ¼ 0; B4 ¼ 1
30 ; B5 ¼ 0; B6 ¼ 1
42.
11.132. (a) Prove that
x
ex
1
¼
x
2
coth
x
2
1
: ðbÞ Use Problem 11.127 and part (a) to show that B2kþ1 ¼ 0 if
k ¼ 1; 2; 3; . . . :
11.133. Derive the series expansions:
ðaÞ coth x ¼
1
x
þ
x
3
x3
45
þ þ
B2nð2xÞ2n
ð2nÞ!x
þ
ðbÞ cot x ¼
1
x
x
3
x3
45
þ ð1Þn B2nð2xÞ2n
ð2nÞ!x
þ
ðcÞ tan x ¼ x þ
x3
3
þ
2x5
15
þ ð1Þn1 2ð22n
1ÞB2nð2xÞ2n1
ð2nÞ!
þ
ðdÞ csc x ¼
1
x
þ
x
6
þ
7
360
x3
þ ð1Þn1 2ð22n1
1ÞB2nx2n1
ð2nÞ!
þ
[Hint: For (a) use Problem 11.132; for (b) replace x by ix in (a); for (c) use tan x ¼ cot x 2 cot 2x; for (d) use
csc x ¼ cot x þ tan x=2.]
11.134. Prove that
Y
1
n¼1
1 þ
1
n3
converges.
11.135. Use the definition to prove that
Y
1
n¼1
1 þ
1
n
diverges.
11.136. Prove that
Y
1
n¼1
ð1 unÞ, where 0 un 1, converges if and only if un converges.
11.137. (a) Prove that
Y
1
n¼2
1
1
n2
converges to 1
2. (b) Evaluate the infinite product in (a) to 2 decimal places and
compare with the true value.
11.138. Prove that the series 1 þ 0 1 þ 1 þ 0 1 þ 1 þ 0 1 þ is the C 1 summable to zero.
304 INFINITE SERIES [CHAP. 11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-313-320.jpg)
![11.139. Prove that the Césaro method of summability is regular. [Hint: See Page 278.]
11.140. Prove that the series 1 þ 2x þ 3x2
þ 4x3
þ þ nxn1
þ converges to 1=ð1 xÞ2
for jxj 1.
11.141. A series
X
1
n¼0
an is called Abel summable to S if S ¼ lim
x!1
X
1
n¼0
anxn
exists. Prove that
ðaÞ
X
1
n¼0
ð1Þn
ðn þ 1Þ is Abel summable to 1/4 and
ðbÞ
X
1
n¼0
ð1Þn
ðn þ 1Þðn þ 2Þ
2
is Abel summable to 1/8.
11.142. Prove that the double series
X
1
m¼1
X
1
n¼1
1
ðm2 þ n2Þp, where p is a constant, converges or diverges according as
p 1 or p @ 1, respectively.
11.143. (a) Prove that
ð1
x
exu
u
du ¼
1
x
1
x2
þ
2!
x3
3!
x4
þ
ð1Þn1
ðn 1Þ!
xn þ ð1Þn
n!
ð1
x
exu
unþ1
du.
ðbÞ Use ðaÞ to prove that
ð1
x
exu
u
du
1
x
1
x2
þ
2!
x3
3!
x4
þ
CHAP. 11] INFINITE SERIES 305](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-314-320.jpg)
![(c) If f is integrable on 1 x 1, then
ð1
1
f ðxÞ dx ¼
ða
1
f ðxÞ dx þ
ð1
a
f ðxÞ dx
¼ lim
x!1
ða
x
f ðtÞ dt þ lim
x!1
ðx
a
f ðtÞ dt:
In part (c) it is important to observe that
lim
x!1
ða
x
f ðtÞ dt þ lim
x!1
ðx
a
f ðtÞ dt:
and
lim
x!1
ða
x
f ðtÞ dt þ
ðx
a
f ðtÞ dt
are not necessarily equal.
This can be illustrated with f ðxÞ ¼ xex2
. The first expression is not defined since neither of the
improper integrals (i.e., limits) is defined while the second form yields the value 0.
EXAMPLE. The function FðxÞ ¼
1
ffiffiffiffiffiffi
2
p eðx2
=2Þ
is called the normal density function and has numerous applications
in probability and statistics. In particular (see the bell-shaped curve in Fig. 12-1)
ð1
1
1
ffiffiffiffiffiffi
2
p e
x2
2
: dx ¼ 1
(See Problem 12.31 for the trick of making this evaluation.)
Perhaps at some point in your academic career you were
‘‘graded on the curve.’’ The infinite region under the curve with
the limiting area of 1 corresponds to the assurance of getting a
grade. C’s are assigned to those whose grades fall in a desig-
nated central section, and so on. (Of course, this grading
procedure is not valid for a small number of students, but as
the number increases it takes on statistical meaning.)
In this chapter we formulate tests for convergence or diver-
gence of improper integrals. It will be found that such tests and
proofs of theorems bear close analogy to convergence and
divergence tests and corresponding theorems for infinite series
(See Chapter 11).
CONVERGENCE OR DIVERGENCE OF IMPROPER INTEGRALS OF THE FIRST KIND
Let f ðxÞ be bounded and integrable in every finite interval a @ x @ b. Then we define
ð1
a
f ðxÞ dx ¼ lim
b!1
ðb
a
f ðxÞ dx ð1Þ
where b is a variable on the positive real numbers.
The integral on the left is called convergent or divergent according as the limit on the right does or
does not exist. Note that
ð1
a
f ðxÞ dx bears close analogy to the infinite series
X
1
n¼1
un, where un ¼ f ðnÞ,
while
ðb
a
f ðxÞ dx corresponds to the partial sums of such infinite series. We often write M in place of
b in (1).
CHAP. 12] IMPROPER INTEGRALS 307
Fig. 12-1](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-316-320.jpg)
![2. Quotient test for integrals with non-negative integrands.
(a) If f ðxÞ A 0 and gðxÞ A 0, and if lim
x!1
f ðxÞ
gðxÞ
¼ A 6¼ 0 or 1, then
ð1
a
f ðxÞ dx and
ð1
a
gðxÞ dx
either both converge or both diverge.
(b) If A ¼ 0 in (a) and
ð1
a
gðxÞ dx converges, then
ð1
a
f ðxÞ dx converges.
(c) If A ¼ 1 in (a) and
ð1
a
gðxÞ dx diverges, then
ð1
a
f ðxÞ dx diverges.
This test is related to the comparison test and is often a very useful alternative to it. In particular,
taking gðxÞ ¼ 1=xp
, we have from known facts about the p integral, the following theorem.
Theorem 1. Let lim
x!1
xp
f ðxÞ ¼ A. Then
(i)
ð1
a
f ðxÞ dx converges if p 1 and A is finite
(ii)
ð1
a
f ðxÞ dx diverges if p @ 1 and A 6¼ 0 (A may be infinite).
EXAMPLE 1.
ð1
0
x2
dx
4x4
þ 25
converges since lim
x!1
x2
x2
4x4
þ 25
¼
1
4
.
EXAMPLE 2.
ð1
0
x dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 þ x2 þ 1
p diverges since lim
x!1
x
x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 þ x2 þ 1
p ¼ 1.
Similar test can be devised using gðxÞ ¼ etx
.
3. Series test for integrals with non-negative integrands.
ð1
a
f ðxÞ dx converges or diverges accord-
ing as un, where un ¼ f ðnÞ, converges or diverges.
4. Absolute and conditional convergence.
ð1
a
f ðxÞ dx is called absolutely convergent if
ð1
a
j f ðxÞj dx
converges. If
ð1
a
f ðxÞ dx converges but
ð1
a
j f ðxÞj dx diverges, then
ð1
a
f ðxÞ dx is called con-
ditionally convergent.
Theorem 2. If
ð1
a
j f ðxÞj dx converges, then
ð1
a
f ðxÞ dx converges. In words, an absolutely convergent
integral converges.
EXAMPLE 1.
ð1
a
cos x
x2
þ 1
dx is absolutely convergent and thus convergent since
ð1
0
cos x
x2
þ 1
dx @
ð1
0
dx
x2
þ 1
and
ð1
0
dx
x2
þ 1
converges.
EXAMPLE 2.
ð1
0
sin x
x
dx converges (see Problem 12.11), but
ð1
0
sin x
x
dx does not converge (see
Problem 12.12). Thus,
ð1
0
sin x
x
dx is conditionally convergent.
Any of the tests used for integrals with non-negative integrands can be used to test for absolute
convergence.
CHAP. 12] IMPROPER INTEGRALS 309](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-318-320.jpg)
![SPECIAL IMPROPER INTEGRALS OF THE SECOND KIND
1.
ðb
a
dx
ðx aÞp converges if p 1 and diverges if p A 1.
2.
ðb
a
dx
ðb xÞp converges if p 1 and diverges if p A 1.
These can be called p integrals of the second kind. Note that when p @ 0 the integrals are proper.
CONVERGENCE TESTS FOR IMPROPER INTEGRALS OF THE SECOND KIND
The following tests are given for the case where f ðxÞ is unbounded only at x ¼ a in the interval
a @ x @ b. Similar tests are available if f ðxÞ is unbounded at x ¼ b or at x ¼ x0 where a x0 b.
1. Comparison test for integrals with non-negative integrands.
(a) Convergence. Let gðxÞ A 0 for a x @ b, and suppose that
ðb
a
gðxÞ dx converges. Then if
0 @ f ðxÞ @ gðxÞ for a x @ b,
ðb
a
f ðxÞ dx also converges.
EXAMPLE.
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 1
p
1
ffiffiffiffiffiffiffiffiffiffiffi
x 1
p for x 1. Then since
ð5
1
dx
ffiffiffiffiffiffiffiffiffiffiffi
x 1
p converges ( p integral with a ¼ 1,
p ¼ 1
2),
ð5
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 1
p also converges.
(b) Divergence. Let gðxÞ A 0 for a x @ b, and suppose that
ðb
a
gðxÞ dx diverges. Then if
f ðxÞ A gðxÞ for a x @ b,
ðb
a
f ðxÞ dx also diverges.
EXAMPLE.
ln x
ðx 3Þ4
1
ðx 3Þ4
for x 3. Then since
ð6
3
dx
ðx 3Þ4
diverges ( p integral with a ¼ 3, p ¼ 4),
ð6
3
ln x
ðx 3Þ4
dx also diverges.
2. Quotient test for integrals with non-negative integrands.
(a) If f ðxÞ A 0 and gðxÞ A 0 for a x @ b, and if lim
x!a
f ðxÞ
gðxÞ
¼ A 6¼ 0 or 1, then
ðb
a
f ðxÞ dx and
ðb
a
gðxÞ dx either both converge or both diverge.
CHAP. 12] IMPROPER INTEGRALS 311
Fig. 12-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-320-320.jpg)
![IMPROPER INTEGRALS OF THE THIRD KIND
Improper integrals of the third kind can be expressed in terms of improper integrals of the first and
second kinds, and hence the question of their convergence or divergence is answered by using results
already established.
IMPROPER INTEGRALS CONTAINING A PARAMETER, UNIFORM CONVERGENCE
Let
ð Þ ¼
ð1
a
f ðx; Þ dx ð8Þ
This integral is analogous to an infinite series of functions. In seeking conditions under which we
may differentiate or integrate ð Þ with respect to , it is convenient to introduce the concept of uniform
convergence for integrals by analogy with infinite series.
We shall suppose that the integral (8) converges for 1 @ @ 2, or briefly ½ 1; 2.
Definition.
The integral (8) is said to be uniformly convergent in ½ 1; 2 if for each 0, we can find a number N
depending on but not on , such that
ð Þ
ðu
a
f ðx; Þ dx for all u N and all in ½ 1; 2
This can be restated by nothing that ð Þ
ðu
a
f ðx; Þ dx ¼
ð1
u
f ðx; Þ dx , which is analogous in
an infinite series to the absolute value of the remainder after N terms.
The above definition and the properties of uniform convergence to be developed are formulated in
terms of improper integrals of the first kind. However, analogous results can be given for improper
integrals of the second and third kinds.
SPECIAL TESTS FOR UNIFORM CONVERGENCE OF INTEGRALS
1. Weierstrass M test. If we can find a function MðxÞ A 0 such that
(a) j f ðx; Þj @ MðxÞ 1 @ @ 2; x a
(b)
ð1
a
MðxÞ dx converges,
then
ð1
a
f ðx; Þ dx is uniformly and absolutely convergent in 1 @ @ 2.
EXAMPLE. Since
cos x
x2 þ 1
@
1
x2 þ 1
and
ð1
0
dx
x2 þ 1
converges, it follows that
ð1
0
cos x
x2 þ 1
dx is uniformly
and absolutely convergent for all real values of .
As in the case of infinite series, it is possible for integrals to be uniformly convergent
without being absolutely convergent, and conversely.
CHAP. 12] IMPROPER INTEGRALS 313](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-322-320.jpg)
![The Laplace transform of a function FðxÞ is
defined as
f ðsÞ ¼ lfFðxÞg ¼
ð1
0
esx
FðxÞ dx ð12Þ
and is analogous to power series as seen by replacing
es
by t so that esx
¼ tx
. Many properties of power
series also apply to Laplace transforms. The adjacent
short table of Laplace transforms is useful. In each
case a is a real constant.
LINEARITY
The Laplace transform is a linear operator, i.e.,
fFðxÞ þ GðxÞg ¼ fFðxÞg þ fGðxÞg:
This property is essential for returning to the solution after having calculated in the setting of the
transforms. (See the following example and the previously cited problems.)
CONVERGENCE
The exponential est
contributes to the convergence of the improper integral. What is required is
that FðxÞ does not approach infinity too rapidly as x ! 1. This is formally stated as follows:
If there is some constant a such that jFðxÞj eax
for all sufficiently large values of x, then
f ðsÞ ¼
ð1
0
esx
FðxÞ dx converges when s a and f has derivatives of all orders. (The differentiations
of f can occur under the integral sign .)
APPLICATION
The feature of the Laplace transform that (when combined with linearity) establishes it as a tool for
solving differential equations is revealed by applying integration by parts to f ðsÞ ¼
ðx
0
est
FðtÞ dt. By
letting u ¼ FðtÞ and dv ¼ est
dt, we obtain after letting x ! 1
ðx
0
est
FðtÞ dt ¼
1
s
Fð0Þ þ
1
s
ð1
0
est
F 0
ðtÞ dt:
Conditions must be satisfied that guarantee the convergence of the integrals (for example, est
FðtÞ ! 0
as t ! 1).
This result of integration by parts may be put in the form
(a) fF 0
ðtÞg ¼ sfFðtÞg þ F 0
ð0Þ.
Repetition of the procedure combined with a little algebra yields
(b) fF 00
ðtÞg ¼ s2
fFðtÞg sFð0Þ F 0
ð0Þ.
The Laplace representation of derivatives of the order needed can be obtained by repeating the
process.
To illustrate application, consider the differential equation
d2
y
dt2
þ 4y ¼ 3 sin t;
where y ¼ FðtÞ and Fð0Þ ¼ 1, F 0
ð0Þ ¼ 0. We use
CHAP. 12] IMPROPER INTEGRALS 315
FðxÞ lfFðxÞg
a
a
8
8 0
eax 1
8 a
8 a
sin ax
a
82 þ a2
8 0
cos ax
8
82 þ a2
8 0
xn
n ¼ 1; 2; 3; . . .
n!
8nþ1
8 0
Y 0
ðxÞ 8lfYðxÞg Yð0Þ
Y 00
ðxÞ 82
lfYðxÞg 8Yð0Þ Y 0
ð0Þ](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-324-320.jpg)
![(e) This is a proper integral
since lim
x!0þ
1 cos x
x2
¼
1
2
by applying L’Hospital’s rule
.
12.2. Show how to transform the improper integral of the second kind,
ð2
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xð2 xÞ
p , into
(a) an improper integral of the first kind, (b) a proper integral.
(a) Consider
ð2
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xð2 xÞ
p where 0 1, say. Let 2 x ¼
1
y
. Then the integral becomes
ð1=
1
dy
y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2y 1
p . As ! 0þ, we see that consideration of the given integral is equivalent to considera-
tion of
ð1
1
dy
y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2y 1
p , which is an improper integral of the first kind.
(b) Letting 2 x ¼ v2
in the integral of (a), it becomes 2
ð1
ffiffi
p
dv
ffiffiffiffiffiffiffiffiffiffiffiffiffi
v2 þ 2
p . We are thus led to consideration of
2
ð1
0
dv
ffiffiffiffiffiffiffiffiffiffiffiffiffi
v2
þ 1
p , which is a proper integral.
From the above we see that an improper integral of the first kind may be transformed into an
improper integral of the second kind, and conversely (actually this can always be done).
We also see that an improper integral may be transformed into a proper integral (this can only
sometimes be done).
IMPROPER INTEGRALS OF THE FIRST KIND
12.3. Prove the comparison test (Page 308) for convergence of improper integrals of the first kind.
Since 0 @ f ðxÞ @ gðxÞ for x A a, we have using Property 7, Page 92,
0 @
ðb
a
f ðxÞ dx @
ðb
a
gðxÞ dx @
ð1
a
gðxÞ dx
But by hypothesis the last integral exists. Thus
lim
b!1
ðb
a
f ðxÞ dx exists, and hence
ð1
a
f ðxÞ dx converges
12.4. Prove the quotient test (a) on Page 309.
By hypothesis, lim
x!1
f ðxÞ
gðxÞ
¼ A 0. Then given any 0, we can find N such that
f ðxÞ
gðxÞ
A when
x A N. Thus for x A N, we have
A @
f ðxÞ
gðxÞ
@ A þ or ðA ÞgðxÞ @ f ðxÞ @ ðA þ ÞgðxÞ
Then
ðA Þ
ðb
N
gðxÞ dx @
ðb
N
f ðxÞ dx @ ðA þ Þ
ðb
N
gðxÞ dx ð1Þ
There is no loss of generality in choosing A 0.
If
ð1
a
gðxÞ dx converges, then by the inequality on the right of (1),
lim
b!1
ðb
N
f ðxÞ dx exists, and so
ð1
a
f ðxÞ dx converges
If
ð1
a
gðxÞ dx diverges, then by the inequality on the left of (1),
CHAP. 12] IMPROPER INTEGRALS 317](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-326-320.jpg)
![ðbÞ
ð1
0
1 cos x
x2
dx ¼
ð
0
1 cos x
x2
dx þ
ð1
1 cos x
x2
dx
The first integral on the right converges [see Problem 12.1(e)].
Since lim
x!1
x3=2 1 cos x
x2
¼ 0, the second integral on the right converges by Theorem 1, Page 309,
with A ¼ 0 and p ¼ 3=2.
Thus, the given integral converges.
12.8. Test for convergence: (a)
ð1
1
ex
x
dx; ðbÞ
ð1
1
x3
þ x2
x6 þ 1
dx:
(a) Let x ¼ y. Then the integral becomes
ð1
1
ey
y
dy.
Method 1:
ey
y
@ ey
for y @ 1. Then since
ð1
1
ey
dy converges,
ð1
1
ey
y
dy converges; hence the
given integral converges.
Method 2: lim
y!1
y2 ey
y
¼ lim
y!1
yey
¼ 0. Then the given integral converges by Theorem 1, Page
309, with A ¼ 0 and p ¼ 2.
(b) Write the given integral as
ð0
1
x3
þ x2
x6
þ 1
dx þ
ð1
0
x3
þ x2
x6
þ 1
dx. Letting x ¼ y in the first integral, it
becomes
ð1
0
y3
y2
y6
þ 1
dy. Since lim
y!1
y3 y3
y2
y6
þ 1
!
¼ 1, this integral converges.
Since lim
x!1
x3 x3
þ x2
x6 þ 1
!
¼ 1, the second integral converges.
Thus the given integral converges.
ABSOLUTE AND CONDITIONAL CONVERGENCE FOR IMPROPER INTEGRALS OF THE
FIRST KIND
12.9. Prove that
ð1
a
f ðxÞ dx converges if
ð1
a
j f ðxÞj dx converges, i.e., an absolutely convergent integral is
convergent.
We have j f ðxÞj @ f ðxÞ @ j f ðxÞj, i.e., 0 @ f ðxÞ þ j f ðxÞj @ 2j f ðxÞj. Then
0 @
ðb
a
½ f ðxÞ þ j f ðxÞj dx @ 2
ðb
a
j f ðxÞj dx
If
ð1
a
j f ðxÞj dx converges, it follows that
ð1
a
½ f ðxÞ þ j f ðxÞj dx converges. Hence, by subtracting
ð1
a
j f ðxÞj dx, which converges, we see that
ð1
a
f ðxÞ dx converges.
12.10. Prove that
ð1
1
cos x
x2
dx converges.
Method 1:
cos x
x2
@
1
x2
for x A 1. Then by the comparison test, since
ð1
1
dx
x2
converges, it follows that
ð1
1
cos x
x2
dx converges, i.e.,
ð1
1
cos x
x2
dx converges absolutely, and so converges by Problem 12.9.
CHAP. 12] IMPROPER INTEGRALS 319](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-328-320.jpg)
![ð1
0
sin x
x
dx ¼
X
1
n¼0
ððnþ1Þ
n
sin x
x
dx ¼
X
1
n¼0
ð
0
sin v
v þ n
dv ð1Þ
Now
1
v þ n
A
1
ðn þ 1Þ
for 0 @ v @ : Hence,
ð
0
sin v
v þ n
dv A
1
ðn þ 1Þ
ð
9
sin v dv ¼
2
ðn þ 1Þ
ð2Þ
Since
X
1
n¼0
2
ðn þ 1Þ
diverges, the series on the right of (1) diverges by the comparison test. Hence,
ð1
0
sin x
x
dx diverges and the required result follows.
IMPROPER INTEGRALS OF THE SECOND KIND, CAUCHY PRINCIPAL VALUE
12.13. (a) Prove that
ð7
1
dx
ffiffiffiffiffiffiffiffiffiffiffi
x þ 1
3
p converges and (b) find its value.
The integrand is unbounded at x ¼ 1. Then we define the integral as
lim
!0þ
ð7
1þ
dx
ffiffiffiffiffiffiffiffiffiffiffi
x þ 1
3
p ¼ lim
!0þ
ðx þ 1Þ2=3
2=3
7
1þ
¼ lim
!0þ
6
3
2
2=3
¼ 6
This shows that the integral converges to 6.
12.14. Determine whether
ð5
1
dx
ðx 1Þ3
converges (a) in the usual sense, (b) in the Cauchy principal
value sense.
(a) By definition,
ð5
1
dx
ðx 1Þ3
¼ lim
1!0þ
ð11
1
dx
ðx 1Þ3
þ lim
2!0þ
ð5
1þ2
dx
ðx 1Þ3
¼ lim
1!0þ
1
8
1
22
1
þ lim
2!0þ
1
22
2
1
32
and since the limits do not exist, the integral does not converge in the usual sense.
(b) Since
lim
!0þ
ð1
1
dx
ðx 1Þ3
þ
ð5
1þ
dx
ðx 1Þ3
¼ lim
!0þ
1
8
1
22
þ
1
22
1
32
¼
3
32
the integral exists in the Cauchy principal value sense. The principal value is 3/32.
12.15. Investigate the convergence of:
(a)
ð3
2
dx
x2
ðx3
8Þ2=3
(c)
ð5
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð5 xÞðx 1Þ
p (e)
ð=2
0
dx
ðcos xÞ1=n
; n 1:
(b
ð
0
sin x
x3
dx (d)
ð1
1
2sin1
x
1 x
dx
(a) lim
x!2þ
ðx 2Þ2=3
1
x2
ðx3
8Þ2=3
¼ lim
x!2þ
1
x2
1
x2 þ 2x þ 4
2=3
¼
1
8
ffiffiffiffiffi
18
3
p . Hence, the integral converges by
Theorem 3(i), Page 312.
CHAP. 12] IMPROPER INTEGRALS 321](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-330-320.jpg)
![Write the integral as
ð1
0
xn1
ex
dx þ
ð1
1
xn1
ex
dx ð1Þ
(a) If n A 1, the first integral in (1) converges since the integrand is continuous in 0 @ x @ 1.
If 0 n 1, the first integral in (1) is an improper integral of the second kind at x ¼ 0. Since
lim
x!0þ
x1n
xn1
ex
¼ 1, the integral converges by Theorem 3(i), Page 312, with p ¼ 1 n and a ¼ 0.
Thus, the first integral converges for n 0.
If n 0, the second integral in (1) is an improper integral of the first kind. Since
lim
x!1
x2
xn1
ex
¼ 0 (by L’Hospital’s rule or otherwise), this integral converges by Theorem 1ðiÞ,
Page 309, with p ¼ 2.
Thus, the second integral also converges for n 0, and so the given integral converges for n 0.
(b) If n @ 0, the first integral of (1) diverges since lim
x!0þ
x xn1
ex
¼ 1 [Theorem 3(ii), Page 312].
If n @ 0, the second integral of (1) converges since lim
x!1
x xn1
ex
¼ 0 [Theorem 1(i), Page 309].
Since the first integral in (1) diverges while the second integral converges, their sum also diverges,
i.e., the given integral diverges if n @ 0.
Some interesting properties of the given integral, called the gamma function, are considered in
Chapter 15.
UNIFORM CONVERGENCE OF IMPROPER INTEGRALS
12.19. (a) Evaluate ð Þ ¼
ð1
0
e x
dx for 0.
(b) Prove that the integral in (a) converges uniformly to 1 for A 1 0.
(c) Explain why the integral does not converge uniformly to 1 for 0.
ðaÞ ð Þ ¼ lim
b!1
ðb
a
e x
dx ¼ lim
b!1
e x
b
x¼0
¼ lim
b!1
1 e b
¼ 1 if 0
.
Thus, the integral converges to 1 for all 0.
(b) Method 1, using definition:
The integral converges uniformly to 1 in A 1 0 if for each 0 we can find N, depending on
but not on , such that 1
ðu
0
e x
dx for all u N.
Since 1
ðu
0
e x
dx ¼ j1 ð1 e u
Þj ¼ e u
@ e 1u
for u
1
1
ln
1
¼ N, the result fol-
lows.
Method 2, using the Weierstrass M test:
Since lim
x!1
x2
e x
¼ 0 for A 1 0, we can choose j e x
j
1
x2
for sufficiently large x, say
x A x0. Taking MðxÞ ¼
1
x2
and noting that
ð1
x0
dx
x2
converges, it follows that the given integral is
uniformly convergent to 1 for A 1 0.
(c) As 1 ! 0, the number N in the first method of (b) increases without limit, so that the integral cannot
be uniformly convergent for 0.
12.20. If ð Þ ¼
ð1
0
f ðx; Þ dx is uniformly convergent for 1 @ @ 2, prove that ð Þ is continuous in
this interval.
CHAP. 12] IMPROPER INTEGRALS 323](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-332-320.jpg)
![EVALUATION OF DEFINITE INTEGRALS
12.23. Prove that
ð=2
0
ln sin x dx ¼
2
ln 2.
The given integral converges [Problem 12.42( f )]. Letting x ¼ =2 y,
I ¼
ð=2
0
ln sin x dx ¼
ð=2
0
ln cos y dy ¼
ð=2
0
ln cos x dx
Then
2I ¼
ð=2
0
ðln sin x þ ln cos xÞ dx ¼
ð=2
0
ln
sin 2x
2
dx
¼
ð=2
0
ln sin 2x dx
ð=2
0
ln 2 dx ¼
ð=2
0
ln sin 2x dx
2
ln 2 ð1Þ
Letting 2x ¼ v,
ð=2
0
ln sin 2x dx ¼
1
2
ð
0
ln sin v dv ¼
1
2
ð=2
0
ln sin v dv þ
ð
=2
ln sin v dv
¼
1
2
ðI þ IÞ ¼ I (letting v ¼ u in the last integral)
Hence, (1) becomes 2I ¼ I
2
ln 2 or I ¼
2
ln 2.
12.24. Prove that
ð
0
x ln sin x dx ¼
2
2
ln 2.
Let x ¼ y. Then, using the results in the preceding problem,
J ¼
ð
0
x ln sin x dx ¼
ð
0
ð uÞ ln sin u du ¼
ð
0
ð xÞ ln sin x dx
¼
ð
0
ln sin x dx
ð
0
x ln sin x dx
¼ 2
ln 2 J
or J ¼
2
2
ln 2:
12.25. (a) Prove that ð Þ ¼
ð1
0
dx
x2 þ
is uniformly convergent for A 1.
ðbÞ Show that ð Þ ¼
2
ffiffiffi
p . ðcÞ Evaluate
ð1
0
dx
ðx2 þ 1Þ2
:
ðdÞ Prove that
ð1
0
dx
ðx2
þ 1Þnþ1
¼
ð=2
0
cos2n
d ¼
1 3 5 ð2n 1Þ
2 4 6 ð2nÞ
2
:
(a) The result follows from the Weierestrass test, since
1
x2
þ
@
1
x2
þ 1
for a A 1 and
ð1
0
dx
x2
þ 1
converges.
ðbÞ ð Þ ¼ lim
b!1
ðb
0
dx
x2
þ
¼ lim
b!1
1
ffiffiffi
p tan1 x
ffiffiffi
p
b
0
¼ lim
b!1
1
ffiffiffi
p tan1 b
ffiffiffi
p ¼
2
ffiffiffi
p :
CHAP. 12] IMPROPER INTEGRALS 325](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-334-320.jpg)
![12.28. Prove that
ð1
0
1 cos x
x2
dx ¼
2
.
Since e x 1 cos x
x2
@
1 cos x
x2
for A 0; x A 0 and
ð1
0
1 cos x
x2
dx converges [see Problem
12.7(b)], it follows by the Weierstrass test that
ð1
0
e x 1 cos x
x2
dx is uniformly convergent and represents
a continuous function of for A 0 (Theorem 6, Page 314). Then letting ! 0þ, using Problem 12.27, we
have
lim
!0þ
ð1
0
e x 1 cos x
x2
dx ¼
ð1
0
1 cos x
x2
dx ¼ lim
!0
tan1 1
2
lnð 2
þ 1Þ
¼
2
12.29. Prove that
ð1
0
sin x
x
¼
ð1
0
sin2
x
x2
dx ¼
2
.
Integrating by parts, we have
ðM
1 cos x
x2
dx ¼
1
x
ð1 cos xÞ
M
þ
ðM
sin x
x
dx ¼
1 cos
1 cos M
M
þ
ðM
sin x
x
dx
Taking the limit as ! 0þ and M ! 1 shows that
ð1
0
sin x
x
dx ¼
ð1
0
1 cos x
x
dx ¼
2
Since
ð1
0
1 cos x
x2
dx ¼ 2
ð1
0
sin2
ðx=2Þ
x2
dx ¼
ð1
0
sin2
u
u2
du on letting u ¼ x=2, we also have
ð1
0
sin2
x
x2
dx ¼
2
.
12.30. Prove that
ð1
0
sin3
x
x
dx ¼
4
.
sin3
x ¼
eix
eix
2i
2
¼
ðeix
Þ3
3ðeix
Þ2
ðeix
Þ þ 3ðeix
Þðeix
Þ2
ðeix
Þ3
ð2iÞ3
¼
1
4
e3ix
e3ix
2i
!
þ
3
4
eix
eix
2i
¼
1
4
sin 3x þ
3
4
sin x
Then
ð1
0
sin3
x
x
dx ¼
3
4
ð1
0
sin x
x
dx
1
4
ð1
0
sin 3x
x
dx ¼
3
4
ð1
0
sin x
x
dx
1
4
ð1
0
sin u
u
du
¼
3
4
2
1
4
2
¼
4
MISCELLANEOUS PROBLEMS
12.31. Prove that
ð1
0
ex2
dx ¼
ffiffiffi
p
=2.
By Problem 12.6, the integral converges. Let IM ¼
ðM
0
ex2
dx ¼
ðM
0
ey2
dy and let lim
M!1
IM ¼ I, the
required value of the integral. Then
CHAP. 12] IMPROPER INTEGRALS 327](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-336-320.jpg)
![eðx =xÞ2
ð1 =x2
Þ ¼ ex2
þ2 2
=x2
ð1 =x2
Þ @ e2
ex2
so that I 0
ð Þ converges uniformly for A 0 by the Weierstrass test, since
ð1
0
ex2
dx converges. Now
I 0
ð Þ ¼ 2
ð1
0
eðx =xÞ2
dx 2
ð1
0
eðx =xÞ2
x2
dx ¼ 0
as seen by letting =x ¼ y in the second integral. Thus Ið Þ ¼ c, a constant. To determine c, let ! 0þ in
the required integral and use Problem 12.31 to obtain c ¼
ffiffiffi
p
=2.
(b) From (a),
ð1
0
eðx =xÞ2
dx ¼
ð1
0
eðx2
2 þ 2
x2
Þ
dx ¼ e2
ð1
0
eðx2
þ 2
x2
Þ
dx ¼
ffiffiffi
p
2
.
Then
ð1
0
eðx2
þ 2
x2
Þ
dx ¼
ffiffiffi
p
2
e2
: Putting ¼ 1;
ð1
0
eðx2
þx2
Þ
dx ¼
ffiffiffi
p
2
e2
:
12.34. Verify the results: (a) lfeax
g ¼
1
s a
; s a; ðbÞ lfcos axg ¼
s
s2
þ a2
; s 0.
lfeax
g ¼
ð1
0
esx
eax
dx ¼ lim
M!1
ðM
0
eðsaÞx
dx
ðaÞ
¼ lim
M!1
1 eðsaÞM
s a
¼
1
s a
if s a
ðbÞ lfcos axg ¼
ð1
0
esx
cos ax dx ¼
s
s2 þ a2
by Problem 12.22 with ¼ s; r ¼ a:
Another method, using complex numbers.
From part (a), lfeax
g ¼
1
s a
. Replace a by ai. Then
lfeaix
g ¼ lfcos ax þ i sin axg ¼ lfcos axg þ ilfsin axg
¼
1
s ai
¼
s þ ai
s2 þ a2
¼
s
s2 þ a2
þ i
a
s2 þ a2
Equating real and imaginary parts: lfcos axg ¼
s
s2 þ a2
, lfsin axg ¼
a
s2 þ a2
.
The above formal method can be justified using methods of Chapter 16.
12.35. Prove that (a) lfY 0
ðxÞg ¼ slfYðxÞg Yð0Þ; ðbÞ lfY 00
ðxÞg ¼ s2
lfYðxÞg sYð0Þ Y 0
ð0Þ
under suitable conditions on YðxÞ.
(a) By definition (and with the aid of integration by parts)
lfY 0
ðxÞg ¼
ð1
0
esx
Y 0
ðxÞ dx ¼ lim
M!0
ðM
0
esx
Y 0
ðxÞ dx
¼ lim
M!1
esx
YðxÞ
M
0
þ s
ðM
0
esx
YðxÞ dx
( )
¼ s
ð1
0
esx
YðxÞ dx Yð0Þ ¼ slfYðxÞg Yð0Þ
assuming that s is such that lim
M!1
esM
YðMÞ ¼ 0.
(b) Let UðxÞ ¼ Y 0
ðxÞ. Then by part (a), lfU 0
ðxÞg ¼ slfUðxÞg Uð0Þ. Thus
lfY 00
ðxÞg ¼ slfY 0
ðxÞg Y 0
ð0Þ ¼ s½slfYðxÞg Yð0Þ Y 0
ð0Þ
¼ s2
lfYðxÞg sYð0Þ Y 0
ð0Þ
CHAP. 12] IMPROPER INTEGRALS 329](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-338-320.jpg)
![12.40. Test for convergence, indicating absolute or conditional convergence where possible: (a)
ð1
0
sin 2x
x3 þ 1
dx;
(b)
ð1
1
eax2
cos bx dx, where a; b are positive constants; (c)
ð1
0
cos x
ffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ 1
p dx; ðdÞ
ð1
0
x sin x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
þ a2
p dx;
(e)
ð1
0
cos x
cosh x
dx.
Ans. (a) abs. conv., (b) abs. conv., (c) cond. conv., (d) div., (e) abs. conv.
12.41. Prove the quotient tests (b) and (c) on Page 309.
IMPROPER INTEGRALS OF THE SECOND KIND
12.42. Test for convergence:
ðaÞ
ð1
0
dx
ðx þ 1Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
p ðdÞ
ð2
1
ln x
ffiffiffiffiffiffiffiffiffiffiffiffiffi
8 x3
3
p dx ðgÞ
ð3
0
x2
ð3 xÞ2
dx ð jÞ
ð1
0
dx
xx
ðbÞ
ð1
0
cos x
x2
dx ðeÞ
ð1
0
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lnð1=xÞ
p ðhÞ
ð=2
0
ex
cos x
x
dx
ðcÞ
ð1
1
etan1
x
x
dx ð f Þ
ð=2
0
ln sin x dx ðiÞ
ð1
0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 k2x2
1 x2
s
dx; jkj 1
Ans. (a) conv., (b) div., (c) div., (d) conv., (e) conv., ( f Þ conv., (g) div., (h) div., (i) conv.,
( jÞ conv.
12.43. (a) Prove that
ð5
0
dx
4 x
diverges in the usual sense but converges in the Cauchy principal value senses.
(b) Find the Cauchy principal value of the integral in (a) and give a geometric interpretation.
Ans. (b) ln 4
12.44. Test for convergence, indicating absolute or conditional convergence where possible:
ðaÞ
ð1
0
cos
1
x
dx; ðbÞ
ð1
0
1
x
cos
1
x
dx; ðcÞ
ð1
0
1
x2
cos
1
x
dx:
Ans. (a) abs. conv., (b) cond. conv., (c) div.
12.45. Prove that
ð4
0
3x2
sin
1
x
x cos
1
x
dx ¼
32
ffiffiffi
2
p
3
.
IMPROPER INTEGRALS OF THE THIRD KIND
12.46. Test for convergence: (a)
ð1
0
ex
ln x dx; ðbÞ
ð1
0
ex
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x lnðx þ 1Þ
p ; ðcÞ
ð1
0
ex
dx
ffiffiffi
x
3
p
ð3 þ 2 sin xÞ
.
Ans. (a) conv., (b) div., (c) conv.
12.47. Test for convergence: (a)
ð1
0
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 þ x2
3
p ; ðbÞ
ð1
0
ex
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sinh ðaxÞ
p ; a 0.
Ans. (a) conv., (b) conv. if a 2, div. if 0 a @ 2.
12.48. Prove that
ð1
0
sinh ðaxÞ
sinh ðxÞ
dx converges if 0 @ jaj and diverges if jaj @ .
12.49. Test for convergence, indicating absolute or conditional convergence where possible:
CHAP. 12] IMPROPER INTEGRALS 331](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-340-320.jpg)
![12.65. (a) Prove that
ð1
0
e x cos ax cos bx
x
dx ¼
1
2
ln
2
þ b2
2 þ a2
!
; A 0.
(b) Use (a) to prove that
ð1
0
cos ax cos bx
x
dx ¼ ln
b
a
.
The results of (b) and Problem 12.60 are special cases of Frullani’s integral,
ð1
0
FðaxÞ FðbxÞ
x
dx ¼
Fð0Þ ln
b
a
, where FðtÞ is continuous for t 0, F 0
ð0Þ exists and
ð1
1
FðtÞ
t
dt converges.
12.66. Given
ð1
0
e x2
dx ¼ 1
2
ffiffiffiffiffiffiffiffi
=
p
, 0. Prove that for p ¼ 1; 2; 3; . . .,
ð1
0
x2p
e x2
dx ¼
1
2
3
2
5
2
ð2p 1Þ
2
ffiffiffi
p
2 ð2pþ1Þ=2
12.67. If a 0; b 0, prove that
ð1
0
ðea=x2
eb=x2
Þ dx ¼
ffiffiffiffiffiffi
b
p
ffiffiffiffiffiffi
a
p
.
12.68. Prove that
ð1
0
tan1
ðx=aÞ tan1
ðx=bÞ
x
dx ¼
2
ln
b
a
where a 0; b 0.
12.69. Prove that
ð1
1
dx
ðx2 þ x þ 1Þ3
¼
4
3
ffiffiffi
3
p . [Hint: Use Problem 12.38.]
MISCELLANEOUS PROBLEMS
12.70. Prove that
ð1
0
lnð1 þ xÞ
x
2
dx converges.
12.71. Prove that
ð1
0
dx
1 þ x3
sin2
x
converges. Hint: Consider
X
1
n¼0
ððnþ1Þ
n
dx
1 þ x3
sin2
x
and use the fact that
ððnþ1Þ
n
dx
1 þ x3
sin2
x
@
ððnþ1Þ
n
dx
1 þ ðnÞ3
sin2
x
:
12.72. Prove that
ð1
0
x dx
1 þ x3 sin2
x
diverges.
12.73. (a) Prove that
ð1
0
lnð1 þ 2
x2
Þ
1 þ x2
dx ¼ lnð1 þ Þ; A 0.
(b) Use (a) to show that
ð=2
0
ln sin d ¼
2
ln 2:
12.74. Prove that
ð1
0
sin4
x
x4
dx ¼
3
.
12.75. Evaluate (a) lf1=
ffiffiffi
x
p
g; ðbÞ lfcosh axg; ðcÞ lfðsin xÞ=xg.
Ans: ðaÞ
ffiffiffiffiffiffiffi
=s
p
; s 0 ðbÞ
s
s2 a2
; s jaj ðcÞ tan1 1
s
; s 0:
12.76. (a) If lfFðxÞg ¼ f ðsÞ, prove that lfeax
FðxÞg ¼ f ðs aÞ; ðbÞ Evaluate lfeax
sin bxg.
Ans: ðbÞ
b
ðs aÞ2
þ b2
; s a
CHAP. 12] IMPROPER INTEGRALS 333](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-342-320.jpg)
![12.77. (a) If lfFðxÞg ¼ f ðsÞ, prove that lfxn
FðxÞg ¼ ð1Þn
f ðnÞ
ðsÞ, giving suitable restrictions on FðxÞ.
(b) Evaluate lfx cos xg. Ans: ðbÞ
s2
1
ðs2 þ 1Þ2
; s 0
12.78. Prove that l1
ff ðsÞ þ gðsÞg ¼ l1
f f ðsÞg þ l1
fgðsÞg, stating any restrictions.
12.79. Solve using Laplace transforms, the following differential equations subject to the given conditions.
(a) Y 00
ðxÞ þ 3Y 0
ðxÞ þ 2YðxÞ ¼ 0; Yð0Þ ¼ 3; Y 0
ð0Þ ¼ 0
(b) Y 00
ðxÞ Y 0
ðxÞ ¼ x; Yð0Þ ¼ 2; Y 0
ð0Þ ¼ 3
(c) Y 00
ðxÞ þ 2Y 0
ðxÞ þ 2YðxÞ ¼ 4; Yð0Þ ¼ 0; Y 0
ð0Þ ¼ 0
Ans. ðaÞ YðxÞ ¼ 6ex
3e2x
; ðbÞ YðxÞ ¼ 4 2ex
1
2 x2
x; ðcÞ YðxÞ ¼ 1 ex
ðsin x þ cos xÞ
12.80. Prove that lfFðxÞg exists if FðxÞ is piecewise continuous in every finite interval ½0; b where b 0 and if FðxÞ
is of exponential order as x ! 1, i.e., there exists a constant such that je x
FðxÞj P (a constant) for all
x b.
12.81. If f ðsÞ ¼ lfFðxÞg and gðsÞ ¼ lfGðxÞg, prove that f ðsÞgðsÞ ¼ lfHðxÞg where
HðxÞ ¼
ðx
0
FðuÞGðx uÞ du
is called the convolution of F and G, written F
G.
Hint: Write f ðsÞgðsÞ ¼ lim
M!1
ðM
0
esu
FðuÞ du
ðM
0
esv
GðvÞ dv
¼ lim
M!1
ðM
0
ðM
0
esðuþvÞ
FðuÞ GðvÞ du dv and then let u þ v ¼ t:
12.82. (a) Find l1 1
ðs2 þ 1Þ2
: ðbÞ Solve Y 00
ðxÞ þ YðxÞ ¼ RðxÞ; Yð0Þ ¼ Y 0
ð0Þ ¼ 0.
(c) Solve the integral equation YðxÞ ¼ x þ
ðx
0
YðuÞ sinðx uÞ du. [Hint: Use Problem 12.81.]
Ans. (a) 1
2 ðsin x x cos xÞ; ðbÞ YðxÞ ¼
ðx
0
RðuÞ sinðx uÞ du; ðcÞ YðxÞ ¼ x þ x3
=6
12.83. Let f ðxÞ; gðxÞ, and g0
ðxÞ be continuous in every finite interval a @ x @ b and suppose that g0
ðxÞ @ 0.
Suppose also that hðxÞ ¼
ðx
a
f ðxÞ dx is bounded for all x A a and lim
x!0
gðxÞ ¼ 0.
(a) Prove that
ð1
a
f ðxÞ gðxÞ dx ¼
ð1
a
g0
ðxÞ hðxÞ dx.
(b) Prove that the integral on the right, and hence the integral on the left, is convergent. The result is that
under the give conditions on f ðxÞ and gðxÞ,
ð1
a
f ðxÞ gðxÞ dx converges and is sometimes called Abel’s
integral test.
Hint: For (a), consider lim
b!1
ðb
a
f ðxÞ gðxÞ dx after replacing f ðxÞ by h0
ðxÞ and integrating by parts. For (b),
first prove that if jhðxÞj H (a constant), then
ðb
a
g0
ðxÞ hðxÞ dx @ HfgðaÞ gðbÞg; and then let b ! 1.
12.84. Use Problem 12.83 to prove that (a)
ð1
0
sin x
x
dx and (b)
ð1
0
sin xp
dx; p 1, converge.
334 IMPROPER INTEGRALS [CHAP. 12](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-343-320.jpg)
![12.85. (a) Given that
ð1
0
sin x2
dx ¼
ð1
0
cos x2
dx ¼
1
2
ffiffiffi
2
r
[see Problems 15.27 and 15.68(a), Chapter 15], evaluate
ð1
0
ð1
0
sinðx2
þ y2
Þ dx dy
(b) Explain why the method of Problem 12.31 cannot be used to evaluate the multiple integral in (a).
Ans. =4
CHAP. 12] IMPROPER INTEGRALS 335](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-344-320.jpg)
![FOURIER SERIES
Let f ðxÞ be defined in the interval ðL; LÞ and outside of this interval by f ðx þ 2LÞ ¼ f ðxÞ, i.e., f ðxÞ
is 2L-periodic. It is through this avenue that a new function on an infinite set of real numbers is created
from the image on ðL; LÞ. The Fourier series or Fourier expansion corresponding to f ðxÞ is given by
a0
2
þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
ð1Þ
where the Fourier coefficients an and bn are
an ¼
1
L
ðL
L
f ðxÞ cos
nx
L
dx
n ¼ 0; 1; 2; . . .
bn ¼
1
L
ðL
L
f ðxÞ sin
nx
L
dx
8
:
ð2Þ
ORTHOGONALITY CONDITIONS FOR THE SINE AND COSINE FUNCTIONS
Notice that the Fourier coefficients are integrals. These are obtained by starting with the series, (1),
and employing the following properties called orthogonality conditions:
(a)
ðL
L
cos
mx
L
cos
nx
L
dx ¼ 0 if m 6¼ n and L if m ¼ n
(b)
ðL
L
sin
mx
L
sin
nx
L
dx ¼ 0 if m 6¼ n and L if m ¼ n (3)
(c)
ðL
L
sin
mx
L
cos
nx
L
dx ¼ 0. Where m and n can assume any positive integer values.
An explanation for calling these orthogonality conditions is given on Page 342. Their application in
determining the Fourier coefficients is illustrated in the following pair of examples and then demon-
strated in detail in Problem 13.4.
EXAMPLE 1. To determine the Fourier coefficient a0, integrate both sides of the Fourier series (1), i.e.,
ðL
L
f ðxÞ dx ¼
ðL
L
a0
2
dx þ
ðL
L
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
n o
dx
Now
ðL
L
a0
2
dx ¼ a0L;
ðL
l
sin
nx
L
dx ¼ 0;
ðL
L
cos
nx
L
dx ¼ 0, therefore, a0 ¼
1
L
ðL
L
f ðxÞ dx
EXAMPLE 2. To determine a1, multiply both sides of (1) by cos
x
L
and then integrate. Using the orthogonality
conditions (3)a and (3)c, we obtain a1 ¼
1
L
ðL
L
f ðxÞ cos
x
L
dx. Now see Problem 13.4.
If L ¼ , the series (1) and the coefficients (2) or (3) are particularly simple. The function in this
case has the period 2.
DIRICHLET CONDITIONS
Suppose that
(1) f ðxÞ is defined except possibly at a finite number of points in ðL; LÞ
(2) f ðxÞ is periodic outside ðL; LÞ with period 2L
CHAP. 13] FOURIER SERIES 337](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-346-320.jpg)
![(3) f ðxÞ and f 0
ðxÞ are piecewise continuous in ðL; LÞ.
Then the series (1) with Fourier coefficients converges to
ðaÞ f ðxÞ if x is a point of continuity
ðbÞ
f ðx þ 0Þ þ f ðx 0Þ
2
if x is a point of discontinuity
Here f ðx þ 0Þ and f ðx 0Þ are the right- and left-hand limits of f ðxÞ at x and represent lim
!0þ
f ðx þ Þ and
lim
!0þ
f ðx Þ, respectively. For a proof see Problems 13.18 through 13.23.
The conditions (1), (2), and (3) imposed on f ðxÞ are sufficient but not necessary, and are generally
satisfied in practice. There are at present no known necessary and sufficient conditions for convergence
of Fourier series. It is of interest that continuity of f ðxÞ does not alone ensure convergence of a Fourier
series.
ODD AND EVEN FUNCTIONS
A function f ðxÞ is called odd if f ðxÞ ¼ f ðxÞ. Thus, x3
; x5
3x3
þ 2x; sin x; tan 3x are odd
functions.
A function f ðxÞ is called even if f ðxÞ ¼ f ðxÞ. Thus, x4
; 2x6
4x2
þ 5; cos x; ex
þ ex
are even
functions.
The functions portrayed graphically in Figures 13-1(a) and 13-1ðbÞ are odd and even respectively,
but that of Fig. 13-1(c) is neither odd nor even.
In the Fourier series corresponding to an odd function, only sine terms can be present. In the
Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we
shall consider a cosine term) can be present.
HALF RANGE FOURIER SINE OR COSINE SERIES
A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are
present, respectively. When a half range series corresponding to a given function is desired, the function
is generally defined in the interval ð0; LÞ [which is half of the interval ðL; LÞ, thus accounting for the
name half range] and then the function is specified as odd or even, so that it is clearly defined in the other
half of the interval, namely, ðL; 0Þ. In such case, we have
an ¼ 0; bn ¼
2
L
ðL
0
f ðxÞ sin
nx
L
dx for half range sine series
bn ¼ 0; an ¼
2
L
ðL
0
f ðxÞ cos
nx
L
dx for half range cosine series
8
:
ð4Þ
PARSEVAL’S IDENTITY
If an and bn are the Fourier coefficients corresponding to f ðxÞ and if f ðxÞ satisfies the Dirichlet
conditions.
Then
1
L
ðL
L
f f ðxÞg2
dx ¼
a2
0
2
þ
X
1
n¼1
ða2
n þ b2
nÞ (5)
(See Problem 13.13.)
338 FOURIER SERIES [CHAP. 13](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-347-320.jpg)
![DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES
Differentiation and integration of Fourier series can be justified by using the theorems on Pages 271
and 272, which hold for series in general. It must be emphasized, however, that those theorems provide
sufficient conditions and are not necessary. The following theorem for integration is especially useful.
Theorem. The Fourier series corresponding to f ðxÞ may be integrated term by term from a to x, and the
resulting series will converge uniformly to
ðx
a
f ðxÞ dx provided that f ðxÞ is piecewise continuous in
L @ x @ L and both a and x are in this interval.
COMPLEX NOTATION FOR FOURIER SERIES
Using Euler’s identities,
ei
¼ cos þ i sin ; ei
¼ cos i sin ð6Þ
where i ¼
ffiffiffiffiffiffiffi
1
p
(see Problem 11.48, Chapter 11, Page 295), the Fourier series for f ðxÞ can be written as
f ðxÞ ¼
X
1
n¼1
cn einx=L
ð7Þ
where
cn ¼
1
2L
ðL
L
f ðxÞeinx=L
dx ð8Þ
In writing the equality (7), we are supposing that the Dirichlet conditions are satisfied and further
that f ðxÞ is continuous at x. If f ðxÞ is discontinuous at x, the left side of (7) should be replaced by
ðf ðx þ 0Þ þ f ðx 0Þ
2
:
BOUNDARY-VALUE PROBLEMS
Boundary-value problems seek to determine solutions of partial differential equations satisfying
certain prescribed conditions called boundary conditions. Some of these problems can be solved by
use of Fourier series (see Problem 13.24).
EXAMPLE. The classical problem of a vibrating string may be idealized in the following way. See Fig. 13-2.
Suppose a string is tautly stretched between points ð0; 0Þ and ðL; 0Þ. Suppose the tension, F, is the
same at every point of the string. The string is made to
vibrate in the xy plane by pulling it to the parabolic
position gðxÞ ¼ mðLx x2
Þ and releasing it. (m is a
numerically small positive constant.) Its equation will
be of the form y ¼ f ðx; tÞ. The problem of establishing
this equation is idealized by (a) assuming that the con-
stant tension, F, is so large as compared to the weight wL
of the string that the gravitational force can be neglected,
(b) the displacement at any point of the string is so small
that the length of the string may be taken as L for any of
its positions, and (c) the vibrations are purely transverse.
The force acting on a segment PQ is
w
g
x
@2
y
@t2
;
x x1 x þ x; g 32 ft per sec:2
. If and are the
angles that F makes with the horizontal, then the vertical
CHAP. 13] FOURIER SERIES 339
Fig. 13-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-348-320.jpg)
![With this equation in place the boundary condition
@y
@t
ðx; 0Þ ¼ 0, 0 x L can be considered.
@y
@t
¼ c1 sin
n
L
x
h i
c3
n
L
cos
n
L
t c4
n
L
sin
n
L
t
h i
At t ¼ 0
0 ¼ c1 sin
n
L
x
h i
c3
n
L
Since c1 6¼ 0 and sin
n
L
x is not identically zero, it follows that c3 ¼ 0 and that
y ¼ c1 sin
n
L
x
h i
c4
n
L
cos
n
L
t
h i
The remaining initial condition is
yðx; 0Þ ¼ mðLx x2
Þ; 0 x L
When it is imposed
mðLx x2
Þ ¼ c1c4
n
L
sin
n
L
x
However, this relation cannot be satisfied for all x on the interval ð0; LÞ. Thus, the preceding
extensive analysis of the problem of the vibrating string has led us to an inadequate form
y ¼ c1c4
n
L
sin
n
L
x cos
n
L
t
and an initial condition that is not satisfied. At this point the power of Fourier series is employed. In
particular, a theorem of differential equations states that any finite sum of a particular solution also is a
solution. Generalize this to infinite sum and consider
y ¼
X
1
n¼1
bn sin
n
L
x cos
n
L
t
with the initial condition expressed through a half range sine series, i.e.,
X
1
n¼1
bn sin
n
L
x ¼ mðLx x2
Þ; t ¼ 0
According to the formula of Page 338 for coefficient of a half range sine series
L
2m
bn ¼
ðL
0
ðLx x2
Þ sin
nx
L
dx
That is
L
2m
bn ¼
ðL
0
Lx sin
nx
L
dx
ðL
0
x2
sin
nx
L
dx
Application of integration by parts to the second integral yields
L
2m
bn ¼ L
ðL
0
x sin
nx
L
dx þ
L3
n
cos n þ
ðL
0
L
n
cos
nx
L
2x dx
When integration by parts is applied to the two integrals of this expression and a little algebra is
employed the result is
bn ¼
4L2
ðnÞ3
ð1 cos nÞ
CHAP. 13] FOURIER SERIES 341](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-350-320.jpg)
![Solved Problems
FOURIER SERIES
13.1. Graph each of the following functions.
ðaÞ f ðxÞ ¼
3 0 x 5
3 5 x 0
Period ¼ 10
Since the period is 10, that portion of the graph in 5 x 5 (indicated heavy in Fig. 13-3 above) is
extended periodically outside this range (indicated dashed). Note that f ðxÞ is not defined at
x ¼ 0; 5; 5; 10; 10; 15; 15, and so on. These values are the discontinuities of f ðxÞ.
ðbÞ f ðxÞ ¼
sin x 0 @ x @
0 x 2
Period ¼ 2
Refer to Fig. 13-4 above. Note that f ðxÞ is defined for all x and is continuous everywhere.
ðcÞ f ðxÞ ¼
0 0 @ x 2
1 2 @ x 4
0 4 @ x 6
Period ¼ 6
8
:
Refer to Fig. 13-5 above. Note that f ðxÞ is defined for all x and is discontinuous at x ¼ 2; 4; 8;
10; 14; . . . .
CHAP. 13] FOURIER SERIES 343
Period
f (x)
_25 _20 _15 _10 _5 5
3
3
0 10 15 20 25
x
Fig. 13-3
Period
f (x)
4p
x
3p
2p
p
0
_p
_2p
_3p
Fig. 13-4
14
x
12
10
8
6
4
2
0
_2
_4
_6
_8
_10
_12
1
Period
f (x)
Fig. 13-5](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-352-320.jpg)
![(a) Multiplying
f ðxÞ ¼ A þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
ð1Þ
by cos
mx
L
and integrating from L to L, using Problem 13.3, we have
ðL
L
f ðxÞ cos
mx
L
dx ¼ A
ðL
L
cos
mx
L
dx
þ
X
1
n¼1
an
ðL
L
cos
mx
L
cos
nx
L
dx þ bn
ðL
L
cos
mx
L
sin
nx
L
dx
¼ amL if m 6¼ 0
am ¼
1
L
ðL
L
f ðxÞ cos
mx
L
dx if m ¼ 1; 2; 3; . . .
Thus
(b) Multiplying (1) by sin
mx
L
and integrating from L to L, using Problem 13.3, we have
ðL
L
f ðxÞ sin
mx
L
dx ¼ A
ðL
L
sin
mx
L
dx
þ
X
1
n¼1
an
ðL
L
sin
mx
L
cos
nx
L
dx þ bn
ðL
L
sin
mx
L
sin
nx
L
dx
¼ bmL
bm ¼
1
L
ðL
L
f ðxÞ sin
mx
L
dx if m ¼ 1; 2; 3; . . .
Thus
(c) Integrating of (1) from L to L, using Problem 13.2, gives
ðL
L
f ðxÞ dx ¼ 2AL or A ¼
1
2L
ðL
L
f ðxÞ dx
Putting m ¼ 0 in the result of part (a), we find a0 ¼
1
L
ðL
L
f ðxÞ dx and so A ¼
a0
2
.
The above results also hold when the integration limits L; L are replaced by c; c þ 2L:
Note that in all parts above, interchange of summation and integration is valid because the series is
assumed to converge uniformly to f ðxÞ in ðL; LÞ. Even when this assumption is not warranted, the
coefficients am and bm as obtained above are called Fourier coefficients corresponding to f ðxÞ, and the
corresponding series with these values of am and bm is called the Fourier series corresponding to f ðxÞ.
An important problem in this case is to investigate conditions under which this series actually converges
to f ðxÞ. Sufficient conditions for this convergence are the Dirichlet conditions established in Problems
13.18 through 13.23.
13.5. (a) Find the Fourier coefficients corresponding to the function
f ðxÞ ¼
0 5 x 0
3 0 x 5
Period ¼ 10
(b) Write the corresponding Fourier series.
(c) How should f ðxÞ be defined at x ¼ 5; x ¼ 0; and x ¼ 5 in order that the Fourier series will
converge to f ðxÞ for 5 @ x @ 5?
The graph of f ðxÞ is shown in Fig. 13-6.
CHAP. 13] FOURIER SERIES 345](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-354-320.jpg)
![Period ¼ 2L ¼ 2 and L ¼ . Choosing c ¼ 0, we have
an ¼
1
L
ðcþ2L
c
f ðxÞ cos
nx
L
dx ¼
1
ð2
0
x2
cos nx dx
¼
1
ðx2
Þ
sin nx
n
ð2xÞ
cos nx
n2
þ 2
sin nx
n3
2
0
¼
4
n2
; n 6¼ 0
If n ¼ 0; a0 ¼
1
ð2
0
x2
dx ¼
82
3
:
bn ¼
1
L
ðcþ2L
c
f ðxÞ sin
nx
L
dx ¼
1
ð2
0
x2
sin nx dx
¼
1
ðx2
Þ
cos nx
n
ð2xÞ
sin nx
n2
þ ð2Þ
cos nx
n3
2
0
¼
4
n
Then f ðxÞ ¼ x2
¼
42
3
þ
X
1
n¼1
4
n2
cos nx
4
n
sin nx
:
This is valid for 0 x 2. At x ¼ 0 and x ¼ 2 the series converges to 22
.
(b) If the period is not specified, the Fourier series cannot be determined uniquely in general.
13.7. Using the results of Problem 13.6, prove that
1
12
þ
1
22
þ
1
32
þ ¼
2
6
.
At x ¼ 0 the Fourier series of Problem 13.6 reduces to
42
3
þ
X
1
n¼1
4
n2
.
By the Dirichlet conditions, the series converges at x ¼ 0 to 1
2 ð0 þ 42
Þ ¼ 22
.
Then
42
3
þ
X
1
n¼1
4
n2
¼ 22
, and so
X
1
n¼1
1
n2
¼
2
6
.
ODD AND EVEN FUNCTIONS, HALF RANGE FOURIER SERIES
13.8. Classify each of the following functions according as they are even, odd, or neither even nor odd.
ðaÞ f ðxÞ ¼
2 0 x 3
2 3 x 0
Period ¼ 6
From Fig. 13-8 below it is seen that f ðxÞ ¼ f ðxÞ, so that the function is odd.
ðbÞ f ðxÞ ¼
cos x 0 x
0 x 2
Period ¼ 2
CHAP. 13] FOURIER SERIES 347
_6p _4p _2p O 2p 4p 6p
x
f (x)
4p2
Fig. 13-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-356-320.jpg)
![bn ¼
1
L
ðL
0
f ðxÞ sin
nx
L
dx þ
1
L
ðL
0
f ðxÞ sin
nx
L
dx ¼ 0
f ðxÞ ¼
a0
2
þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
:
Method 2: Assume
f ðxÞ ¼
a0
2
þ
X
1
n¼1
an cos
nx
L
bN sin
nx
L
:
Then
If f ðxÞ is even, f ðxÞ ¼ f ðxÞ. Hence,
a0
2
þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
¼
a0
2
þ
X
1
n¼1
an cos
nx
L
bn sin
nx
L
X
1
n¼1
bn sin
nx
L
¼ 0; i.e., f ðxÞ ¼
a0
2
þ
X
1
n¼1
an cos
nx
L
and so
and no sine terms appear.
In a similar manner we can show that an odd function has no cosine terms (or constant term) in its
Fourier expansion.
13.10. If f ðxÞ is even, show that (a) an ¼
2
L
ðL
0
f ðxÞ cos
nx
L
dx; ðbÞ bn ¼ 0.
an ¼
1
L
ðL
L
f ðxÞ cos
nx
L
dx ¼
1
L
ð0
L
f ðxÞ cos
nx
L
dx þ
1
L
ðL
0
f ðxÞ cos
nx
L
dx
ðaÞ
Letting x ¼ u,
1
L
ð0
L
f ðxÞ cos
nx
L
dx ¼
1
L
ðL
0
f ðuÞ cos
nu
L
du ¼
1
L
ðL
0
f ðuÞ cos
nu
L
du
since by definition of an even function f ðuÞ ¼ f ðuÞ. Then
an ¼
1
L
ðL
0
f ðuÞ cos
nu
L
du þ
1
L
ðL
0
f ðxÞ cos
nx
L
dx ¼
2
L
ðL
0
f ðxÞ cos
nx
L
dx
(b) This follows by Method 1 of Problem 13.9.
13.11. Expand f ðxÞ ¼ sin x; 0 x , in a Fourier cosine series.
A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence, we
extend the definition of f ðxÞ so that it becomes even (dashed part of Fig. 13-11 below). With this extension,
f ðxÞ is then defined in an interval of length 2. Taking the period as 2, we have 2L ¼ 2 so that L ¼ .
By Problem 13.10, bn ¼ 0 and
an ¼
2
L
ðL
0
f ðxÞ cos
nx
L
dx ¼
2
ð
0
sin x cos nx dx
CHAP. 13] FOURIER SERIES 349
f (x)
O
_2p _p p 2p
x
Fig. 13-11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-358-320.jpg)
![Thus bn ¼ 0,
an ¼
2
L
ðL
0
f ðxÞ cos
nx
L
dx ¼
2
2
ð2
0
x cos
nx
2
dx
¼ ðxÞ
2
n
sin
nx
2
ð1Þ
4
n2
2
cos
nx
2
2
0
¼
4
n22
ðcos n 1Þ If n 6¼ 0
If n ¼ 0; a0 ¼
ð2
0
x dx ¼ 2:
f ðxÞ ¼ 1 þ
X
1
n¼1
4
n22
ðcos n 1Þ cos
nx
2
Then
¼ 1
8
2
cos
x
2
þ
1
32
cos
3x
2
þ
1
52
cos
5x
2
þ
It should be noted that the given function f ðxÞ ¼ x, 0 x 2, is represented equally well by the
two different series in (a) and (b).
PARSEVAL’S IDENTITY
13.13. Assuming that the Fourier series corresponding to f ðxÞ converges uniformly to f ðxÞ in ðL; LÞ,
prove Parseval’s identity
1
L
ðL
L
f f ðxÞg2
dx ¼
a2
0
2
þ ða2
n þ b2
nÞ
where the integral is assumed to exist.
If f ðxÞ ¼
a0
2
þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
, then multiplying by f ðxÞ and integrating term by term
from L to L (which is justified since the series is uniformly convergent) we obtain
ðL
L
f f ðxÞg2
dx ¼
a0
2
ðL
L
f ðxÞ dx þ
X
1
n¼1
an
ðL
L
f ðxÞ cos
nx
L
dx þ bn
ðL
L
f ðxÞ sin
nx
L
dx
¼
a2
0
2
L þ L
X
1
n¼1
ða2
n þ b2
nÞ ð1Þ
where we have used the results
ðL
L
f ðxÞ cos
nx
L
dx ¼ Lan;
ðL
L
f ðxÞ sin
nx
L
dx ¼ Lbn;
ðL
L
f ðxÞ dx ¼ La0 ð2Þ
obtained from the Fourier coefficients.
CHAP. 13] FOURIER SERIES 351
f (x)
O
_6 _4 _2 2 4 6
x
Fig. 13-13](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-360-320.jpg)
![Taking the limit as M ! 1, we obtain Bessel’s inequality
a2
0
2
þ
X
1
n¼1
ða2
n þ b2
nÞ @
1
L
ðL
L
f f ðxÞg2
dx ð6Þ
If the equality holds, we have Parseval’s identity (Problem 13.13).
We can think of SMðxÞ as representing an approximation to f ðxÞ, while the left-hand side of (2), divided
by 2L, represents the mean square error of the approximation. Parseval’s identity indicates that as M ! 1
the mean square error approaches zero, while Bessels’ inequality indicates the possibility that this mean
square error does not approach zero.
The results are connected with the idea of completeness of an orthonormal set. If, for example, we were
to leave out one or more terms in a Fourier series (say cos 4x=L, for example), we could never get the mean
square error to approach zero no matter how many terms we took. For an analogy with three-dimensional
vectors, see Problem 13.60.
DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES
13.16. (a) Find a Fourier series for f ðxÞ ¼ x2
; 0 x 2, by integrating the series of Problem 13.12(a).
(b) Use (a) to evaluate the series
X
1
n¼1
ð1Þn1
n2
.
(a) From Problem 13.12(a),
x ¼
4
sin
x
2
1
2
sin
2x
2
þ
1
3
sin
3x
2
ð1Þ
Integrating both sides from 0 to x (applying the theorem of Page 339) and multiplying by 2, we find
x2
¼ C
16
2
cos
x
2
1
22
cos
2x
2
þ
1
32
cos
3x
2
ð2Þ
where C ¼
16
2
1
1
22
þ
1
32
1
42
þ
:
(b) To determine C in another way, note that (2) represents the Fourier cosine series for x2
in 0 x 2.
Then since L ¼ 2 in this case,
C ¼
a0
2
¼
1
L
ðL
0
f ðxÞ ¼
1
2
ð2
0
x2
dx ¼
4
3
Then from the value of C in (a), we have
X
1
n¼1
ð1Þn1
n2
¼ 1
1
22
þ
1
32
¼
1
42
þ ¼
2
16
4
3
¼
2
12
13.17. Show that term by term differentiation of the series in Problem 13.12(a) is not valid.
Term by term differentiation yields 2 cos
x
2
cos
2x
2
þ cos
3x
2
:
Since the nth term of this series does not approach 0, the series does not converge for any value of x.
CHAP. 13] FOURIER SERIES 353](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-362-320.jpg)
![SMðxÞ ¼
a0
2
þ
X
M
n¼1
ðan cos nx þ bn sin nxÞ
Then
¼
1
2
ð
f ðuÞ du þ
1
X
M
n¼1
ð
f ðuÞ cos nðu xÞ du
¼
1
ð
f ðuÞ
1
2
þ
X
M
n¼1
cos nðu xÞ
( )
du
¼
1
ð
f ðuÞ
sinðM þ 1
2Þðu xÞ
2 sin 1
2 ðu xÞ
du
using Problem 13.18. Letting u x ¼ t, we have
SMðxÞ ¼
1
ðx
x
f ðt þ xÞ
sinðM þ 1
2Þt
2 sin 1
2 t
dt
Since the integrand has period 2, we can replace the interval x; x by any other interval of
length 2, in particular, ; . Thus, we obtain the required result.
13.22. Prove that
SMðxÞ
f ðx þ 0Þ þ f ðx 0Þ
2
¼
1
ð0
f ðt þ xÞ f ðx 0Þ
2 sin 1
2 t
!
sinðM þ 1
2Þt dt
þ
1
ð
0
f ðt þ xÞ f ðx þ 0Þ
2 sin 1
2 t
!
sinðM þ 1
2Þt dt
From Problem 13.21,
SMðxÞ ¼
1
ð0
f ðt þ xÞ
sinðM þ 1
2Þt
2 sin 1
2 t
dt þ
1
ð
0
f ðt þ xÞ
sinðM þ 1
2Þt
2 sin 1
2 t
dt ð1Þ
Multiplying the integrals of Problem 13.18(b) by f ðx 0Þ and f ðx þ 0Þ, respectively,
f ðx þ 0Þ þ f ðx 0Þ
2
¼
1
ð0
f ðx 0Þ
sinðM þ 1
2Þt
2 sin 1
2 t
dt þ
1
ð
0
f ðx þ 0Þ
sinðM þ 1
2Þt
2 sin 1
2 t
dt ð2Þ
Subtracting (2) from (1) yields the required result.
13.23. If f ðxÞ and f 0
ðxÞ are piecewise continuous in ð; Þ, prove that
lim
M!1
SMðxÞ ¼
f ðx þ 0Þ þ f ðx 0Þ
2
The function
f ðt þ xÞ f ðx þ 0Þ
2 sin 1
2 t
is piecewise continuous in 0 t @ because f ðxÞ is piecewise con-
tinous.
Also, lim
t!0þ
f ðt þ xÞ f ðx þ 0Þ
2 sin 1
2 t
¼ lim
t!0þ
f ðt þ xÞ f ðx þ 0Þ
t
t
2 sin 1
2 t
¼ lim
t!0þ
f ðt þ xÞ f ðx þ 0Þ
t
exists,
since by hypothesis f 0
ðxÞ is piecewise continuous so that the right-hand derivative of f ðxÞ at each x exists.
Thus,
f ðt þ xÞ f ðx 0Þ
2 sin 1
2 t
is piecewise continous in 0 @ t @ .
Similarly,
f ðt þ xÞ f ðx 0Þ
2 sin 1
2 t
is piecewise continous in @ t @ 0.
CHAP. 13] FOURIER SERIES 355](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-364-320.jpg)
![Since B ¼ 0 makes the solution (8) identically zero, we avoid this choice and instead take
sin 2 ¼ 0; i.e., 2 ¼ m or ¼
m
2
ð10Þ
where m ¼ 0; 1; 2; . . . .
Substitution in (8) now shows that a solution satisfying the first two boundary conditions is
Uðx; tÞ ¼ Bme3m2
2
t=4
sin
mx
2
ð11Þ
where we have replaced B by Bm, indicating that different constants can be used for different values of m.
If we now attempt to satisfy the last boundary condition Uðx; 0Þ ¼ x; 0 x 2, we find it to be
impossible using (11). However, upon recognizing the fact that sums of solutions having the form (11)
are also solutions (called the principle of superposition), we are led to the possible solution
Uðx; tÞ ¼
X
1
m¼1
Bme3m2
2
t=4
sin
mx
2
ð12Þ
From the condition Uðx; 0Þ ¼ x; 0 x 2, we see, on placing t ¼ 0, that (12) becomes
x ¼
X
1
m¼1
Bm sin
mx
2
0 x 2 ð13Þ
This, however, is equivalent to the problem of expanding the function f ðxÞ ¼ x for 0 x 2 into a sine
series. The solution to this is given in Problem 13.12(a), from which we see that Bm ¼
4
m
cos m so that
(12) becomes
Uðx; tÞ ¼
X
1
m¼1
4
m
cos m
e3m2
2
t=4
sin
mx
2
ð14Þ
which is a formal solution. To check that (14) is actually a solution, we must show that it satisfies the partial
differential equation and the boundary conditions. The proof consists in justification of term by term
differentiation and use of limiting procedures for infinite series and may be accomplished by methods of
Chapter 11.
The boundary value problem considered here has an interpretation in the theory of heat conduction.
The equation
@U
@t
¼ k
@2
U
@x2
is the equation for heat conduction in a thin rod or wire located on the x-axis
between x ¼ 0 and x ¼ L if the surface of the wire is insulated so that heat cannot enter or escape. Uðx; tÞ is
the temperature at any place x in the rod at time t. The constant k ¼ K=s (where K is the thermal
conductivity, s is the specific heat, and is the density of the conducting material) is called the diffusivity.
The boundary conditions Uð0; tÞ ¼ 0 and UðL; tÞ ¼ 0 indicate that the end temperatures of the rod are kept
at zero units for all time t 0, while Uðx; 0Þ indicates the initial temperature at any point x of the rod. In
this problem the length of the rod is L ¼ 2 units, while the diffusivity is k ¼ 3 units.
ORTHOGONAL FUNCTIONS
13.25. (a) Show that the set of functions
1; sin
x
L
; cos
x
L
; sin
2x
L
; cos
2x
L
; sin
3x
L
; cos
3x
L
; . . .
forms an orthogonal set in the interval ðL; LÞ.
(b) Determine the corresponding normalizing constants for the set in (a) so that the set is
orthonormal in ðL; LÞ.
(a) This follows at once from the results of Problems 13.2 and 13.3.
(b) By Problem 13.3,
ðL
L
sin2 mx
L
dx ¼ L;
ðL
L
cos2 mx
L
dx ¼ L
CHAP. 13] FOURIER SERIES 357](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-366-320.jpg)
![ln
sin x
x
¼ lim
n!1
ln 1
x2
12
!
þ ln 1
x2
22
!
þ þ ln 1
x2
n2
!
i.e.,
¼ lim
n!1
ln 1
x2
12
!
1
x2
22
!
1
x2
n2
!
( )
¼ ln lim
n!1
1
x2
12
!
1
x2
22
!
1
x2
n2
!
( )
so that
sin x
x
¼ lim
n!1
1
x2
12
!
1
x2
22
!
1
x2
n2
!
¼ 1
x2
12
!
1
x2
22
!
ð2Þ
Replacing x by x=, we obtain
sin x ¼ x 1
x2
2
!
1
x2
ð2Þ2
!
ð3Þ
called the infinite product for sin x, which can be shown valid for all x. The result is of interest since it
corresponds to a factorization of sin x in a manner analogous to factorization of a polynomial.
13.28. Prove that
2
¼
2 2 4 4 6 6 8 8 . . .
1 3 3 5 5 7 7 9 . . .
.
Let x ¼ 1=2 in equation (2) of Problem 13.27. Then,
2
¼ 1
1
22
1
1
42
1
1
62
¼
1
2
3
2
3
4
5
4
5
6
7
6
Taking reciprocals of both sides, we obtain the required result, which is often called Wallis’ product.
Supplementary Problems
FOURIER SERIES
13.29. Graph each of the following functions and find their corresponding Fourier series using properties of even
and odd functions wherever applicable.
ðaÞ f ðxÞ ¼
8 0 x 2
8 2 x 4
Period 4 ðbÞ f ðxÞ ¼
x 4 @ x @ 0
x 0 @ x @ 4
Period 8
ðcÞ f ðxÞ ¼ 4x; 0 x 10; Period 10 ðdÞ f ðxÞ ¼
2x 0 @ x 3
0 3 x 0
Period 6
Ans: ðaÞ
16
X
1
n¼1
ð1 cos nÞ
n
sin
nx
2
ðbÞ 2
8
2
X
1
n¼1
ð1 cos nÞ
n2
cos
nx
4
ðcÞ 20
40
X
1
n¼1
1
n
sin
nx
5
ðdÞ
3
2
þ
X
1
n¼1
6ðcos n 1Þ
n2
2
cos
nx
3
6 cos n
n
sin
nx
3
13.30. In each part of Problem 13.29, tell where the discontinuities of f ðxÞ are located and to what value the series
converges at the discontunities.
Ans. (a) x ¼ 0; 2; 4; . . . ; 0 ðbÞ no discontinuities (c) x ¼ 0; 10; 20; . . . ; 20
(d) x ¼ 3; 9; 15; . . . ; 3
CHAP. 13] FOURIER SERIES 359](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-368-320.jpg)
![CHAP. 13] FOURIER SERIES 361
13.40. By differentiating the result of Problem 13.35(b), prove that for 0 @ x @ ,
x ¼
2
4
cos x
12
þ
cos 3x
32
þ
cos 5x
52
þ
PARSEVAL’S IDENTITY
13.41. By using Problem 13.35 and Parseval’s identity, show that
ðaÞ
X
1
n¼1
1
n4
¼
4
90
ðbÞ
X
1
n¼1
1
n6
¼
6
945
13.42. Show that
1
12 32
þ
1
32 52
þ
1
52 72
þ ¼
2
8
16
. [Hint: Use Problem 13.11.]
13.43. Show that (a)
X
1
n¼1
1
ð2n 1Þ4
¼
4
96
; ðbÞ
X
1
n¼1
1
ð2n 1Þ6
¼
6
960
.
13.44. Show that
1
12
22
32
þ
1
22
32
42
þ
1
32
þ 42
þ 52
þ ¼
42
39
16
.
BOUNDARY-VALUE PROBLEMS
13.45. (a) Solve
@U
@t
¼ 2
@2
U
@x2
subject to the conditions Uð0; tÞ ¼ 0; Uð4; tÞ ¼ 0; Uðx; 0Þ ¼ 3 sin x 2 sin 5x, where
0 x 4; t 0.
(b) Give a possible physical interpretation of the problem and solution.
Ans: ðaÞ Uðx; tÞ ¼ 3e22
t
sin x 2e502
t
sin 5x.
13.46. Solve
@U
@t
¼
@2
U
@x2
subject to the conditions Uð0; tÞ ¼ 0; Uð6; tÞ ¼ 0; Uðx; 0Þ ¼
1 0 x 3
0 3 x 6
and interpret
physically.
Ans: Uðx; tÞ ¼
X
1
m¼1
2
1 cosðm=3Þ
m
em2
2
t=36
sin
mx
6
13.47. Show that if each side of equation (3), Page 356, is a constant c where c A 0, then there is no solution
satisfying the boundary-value problem.
13.48. A flexible string of length is tightly stretched between points x ¼ 0 and x ¼ on the x-axis, its ends are
fixed at these points. When set into small transverse vibration, the displacement Yðx; tÞ from the x-axis of
any point x at time t is given by
@2
Y
@t2
¼ a2 @2
Y
@x2
, where a2
¼ T=; T ¼ tension, ¼ mass per unit length.
(a) Find a solution of this equation (sometimes called the wave equation) with a2
¼ 4 which satisfies the
conditions Yð0; tÞ ¼ 0; Yð; tÞ ¼ 0; Yðx; 0Þ ¼ 0:1 sin x þ 0:01 sin 4x; Ytðx; 0Þ ¼ 0 for 0 x ; t 0.
(b) Interpret physically the boundary conditions in (a) and the solution.
Ans. (a) Yðx; tÞ ¼ 0:1 sin x cos 2t þ 0:01 sin 4x cos 8t
13.49. (a) Solve the boundary-value problem
@2
Y
@t2
¼ 9
@2
Y
@x2
subject to the conditions Yð0; tÞ ¼ 0; Yð2; tÞ ¼ 0,
Yðx; 0Þ ¼ 0:05xð2 xÞ; Ytðx; 0Þ ¼ 0, where 0 x 2; t 0. (b) Interpret physically.
Ans: ðaÞ Yðx; tÞ ¼
1:6
3
X
1
n¼1
1
ð2n 1Þ3
sin
ð2n 1Þx
2
cos
3ð2n 1Þt
2
13.50. Solve the boundary-value problem
@U
@t
¼
@2
U
@x2
; Uð0; tÞ ¼ 1; Uð; tÞ ¼ 3; Uðx; 0Þ ¼ 2.
[Hint: Let Uðx; tÞ ¼ Vðx; tÞ þ FðxÞ and choose FðxÞ so as to simplify the differential equation and boundary
conditions for Vðx; tÞ:](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-370-320.jpg)
![EXAMPLE. Determine the Fourier transform of f if f ðxÞ ¼ ex
for x 0 and e2x
when x 0.
Fð Þ ¼
1
2
ffiffiffiffiffiffi
2
p
ð1
1
ei x
f ðxÞ dx ¼
1
ffiffiffiffiffiffi
2
p
ð0
1
ei x
e2x
dx þ
ð1
0
ei x
ex
dx
¼
1
ffiffiffiffiffiffi
2
p
ei þ2
i þ 2
x!0
x!1
þ
ei 1
i 1
x!1
x!0þ
( )
¼
1
ffiffiffiffiffiffi
2
p
1
2 þ i
þ
1
1 i
If f ðxÞ is an even function, equation (5) yields
Fcð Þ ¼
ffiffiffi
2
r ð1
0
f ðuÞ cos u du
f ðxÞ ¼
ffiffiffi
2
r ð1
0
Fcð Þ cos x d
8
:
ð9Þ
and we call Fcð Þ and f ðxÞ Fourier cosine transforms of each other.
If f ðxÞ is an odd function, equation (6) yields
Fsð Þ ¼
ffiffiffi
2
r ð1
0
f ðuÞ sin u du
f ðxÞ ¼
ffiffiffi
2
r ð1
0
Fsð Þ sin x d
8
:
ð10Þ
and we call Fsð Þ and f ðxÞ Fourier sine transforms of each other.
Note: The Fourier transforms Fc and Fs are (up to a constant) of the same form as Að Þ and Bð Þ.
Since f is even for Fc and odd for Fs, the domains can be shown to be 0 1.
When the product of Fourier transforms is considered, a new concept called convolution comes into
being, and in conjunction with it, a new pair (function and its Fourier transform) arises. In particular, if
Fð Þ and Gð Þ are the Fourier transforms of f and g, respectively, and the convolution of f and g is
defined to be
f g ¼
1
ffiffiffi
p
ð1
1
f ðuÞ gðx uÞ du ð11Þ
then
Fð Þ Gð Þ ¼
1
ffiffiffi
p
ð1
1
ei u
f g du ð12Þ
f g ¼
1
ffiffiffi
p
ð1
1
ei x
Fð Þ Gð Þ d ð13Þ
where in both (11) and (13) the convolution f g is a function of x.
It may be said that multiplication is exchanged with convolution. Also ‘‘the Fourier transform of
the convolution of two functions, f and g is the product of their Fourier transforms,’’ i.e.,
Tðf gÞ ¼ Gð f Þ TðgÞ:
ðFð Þ Gð Þ and f g) are demonstrated to be a Fourier transform pair in Problem 14.29.)
Now equate the representations of f g expressed in (11) and (13), i.e.,
1
ffiffiffi
p
ð1
1
f ðuÞ gðx uÞ du ¼
1
ffiffiffi
p
ð1
1
ei x
Fð Þ Gð Þ d ð14Þ
and let the parameter x be zero, then
ð1
1
f ðuÞ gðuÞ du ¼
ð1
1
Fð Þ Gð Þ d ð15Þ
CHAP. 14] FOURIER INTEGRALS 365](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-374-320.jpg)
![14.2. (a) Use the result of Problem 14.1 to evaluate
ð1
1
sin a cos x
d
ðbÞ Deduce the value of
ð1
0
sin u
u
du:
(a) From Fourier’s integral theorem, if
Fð Þ ¼
1
ffiffiffiffiffiffi
2
p
ð1
1
f ðuÞ ei u
du then f ðxÞ ¼
1
ffiffiffiffiffiffi
2
p
ð1
1
Fð Þ ei x
d
Then from Problem 14.1,
1
ffiffiffiffiffiffi
2
p
ð1
1
ffiffiffi
2
r
sin a
ei x
d ¼
1 jxj a
1=2 jxj ¼ a
0 jxj a
8
:
ð1Þ
The left side of (1) is equal to
1
ð1
1
sin a cos x
d
i
ð1
1
sin a sin x
d ð2Þ
The integrand in the second integral of (2) is odd and so the integral is zero. Then from (1) and
(2), we have
ð1
1
sin a cos x
d ¼
jxj a
=2 jxj ¼ a
0 jxj a
8
:
ð3Þ
Alternative solution: Since the function, f , in Problem 14.1 is an even function, the result follows
immediately from the Fourier cosine transform (9).
(b) If x ¼ 0 and a ¼ 1 in the result of (a), we have
ð1
1
sin
d ¼ or
ð1
0
sin
d ¼
2
since the integrand is even.
14.3. If f ðxÞ is an even function show that:
ðaÞ Fð Þ ¼
ffiffiffi
2
r ð1
0
f ðuÞ cos u du; ðbÞ f ðxÞ ¼
ffiffiffi
2
r ð1
0
Fð Þ cos x d :
We have
Fð Þ ¼
1
ffiffiffiffiffiffi
2
p
ð1
1
f ðuÞ ei u
du ¼
1
ffiffiffiffiffiffi
2
p
ð1
1
f ðuÞ cos u du þ
i
ffiffiffiffiffiffi
2
p
ð1
1
f ðuÞ sin u du ð1Þ
(a) If f ðuÞ is even, f ðuÞ cos u is even and f ðuÞ sin u is odd. Then the second integral on the right of (1) is
zero and the result can be written
Fð Þ ¼
2
ffiffiffiffiffiffi
2
p
ð1
0
f ðuÞ cos u du ¼
ffiffiffi
2
r ð1
0
f ðuÞ cos u du
(b) From (a), Fð Þ ¼ Fð Þ so that Fð Þ is an even function. Then by using a proof exactly analogous to
that in (a), the required result follows.
A similar result holds for odd functions and can be obtained by replacing the cosine by the sine.
14.4. Solve the integral equation
ð1
0
f ðxÞ cos x dx ¼
1 0 @ @ 1
0 1
CHAP. 14] FOURIER INTEGRALS 367](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-376-320.jpg)
![CHAP. 14] FOURIER INTEGRALS 369
This is equivalent to
ða
a
ð1Þ2
dx ¼
ð1
1
2
sin2
a
2
d
ð1
1
sin2
a
2
d ¼ 2
ð1
0
sin2
a
2
d ¼ a
or
ð1
0
sin2
a
2
d ¼
a
2
i.e.,
By letting a ¼ u and using Problem 14.5, it is seen that this is correct. The method can also be used to
find
ð1
0
sin2
u
u2
du directly.
PROOF OF THE FOURIER INTEGRAL THEOREM
14.8. Present a heuristic demonstration of Fourier’s integral theorem by use of a limiting form of
Fourier series.
Let
f ðxÞ ¼
a0
2
þ
X
1
n¼1
an cos
nx
L
þ bn sin
nx
L
ð1Þ
where an ¼
1
L
ðL
L
f ðuÞ cos
nu
L
du and bn ¼
1
L
ðL
L
f ðuÞ sin
nu
L
du:
Then by substitution (see Problem 13.21, Chapter 13),
f ðxÞ ¼
1
2L
ðL
L
f ðuÞ du þ
1
L
X
1
n¼1
ðL
L
f ðuÞ cos
n
L
ðu xÞ du ð2Þ
If we assume that
ð1
1
j f ðuÞj du converges, the first term on the right of (2) approaches zero as L ! 1,
while the remaining part appears to approach
lim
L!1
1
L
X
1
n¼1
ð1
1
f ðuÞ cos
n
L
ðu xÞ du ð3Þ
This last step is not rigorous and makes the demonstration heuristic.
Calling ¼ =L, (3) can be written
f ðxÞ ¼ lim
!0
X
1
n¼1
Fðn Þ ð4Þ
where we have written
Fð Þ ¼
1
ð1
1
f ðuÞ cos ðu xÞ du ð5Þ
But the limit (4) is equal to
f ðxÞ ¼
ð1
0
Fð Þ d ¼
1
ð1
0
d
ð1
1
f ðuÞ cos ðu xÞ du
which is Fourier’s integral formula.
This demonstration serves only to provide a possible result. To be rigorous, we start with the integral
1
ð1
0
d
ð1
1
f ðuÞ cos ðu xÞ dx
and examine the convergence. This method is considered in Problems 14.9 through 14.12.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-378-320.jpg)
![jI3j @ j f ðx þ 0Þj
ð1
L
sin v
v
dv
Also
Since
ð1
0
j f ðxÞj dx and
ð1
0
sin v
v
dv both converge, we can choose L so large that jI2j @ =3, jI3j @ =3.
Also, we can choose so large that jI1j @ =3. Then from (4) we have jIj for and L sufficiently large,
so that the required result follows.
This result follows by reasoning exactly analogous to that in part (a).
14.12. Prove Fourier’s integral formula where f ðxÞ satisfies the conditions stated on Page 364.
We must prove that lim
L!1
1
ðL
¼0
ð1
u¼1
f ðuÞ cos ðx uÞ du d ¼
f ðx þ 0Þ þ f ðx 0Þ
2
Since
ð1
1
f ðuÞ cos ðx uÞ du @
ð1
1
j f ðuÞj du, which converges, it follows by the Weierstrass test
that
ð1
1
f ðuÞ cos ðx uÞ du converges absolutely and uniformly for all . Thus, we can reverse the
order of integration to obtain
1
ðL
¼0
d
ð1
u¼1
f ðuÞ cos ðx uÞ du ¼
1
ð1
u¼1
f ðuÞ du
ðL
¼0
cos ðx uÞ d
¼
1
ð1
u¼1
f ðuÞ
sin Lðu xÞ
u x
du
¼
1
ð1
u¼1
f ðx þ vÞ
sin Lv
v
dv
¼
1
ð0
1
f ðx þ vÞ
sin Lv
v
dv þ
1
ð1
0
f ðx þ vÞ
sin Lv
v
dv
where we have let u ¼ x þ v.
Letting L ! 1, we see by Problem 14.11 that the given integral converges to
f ðx þ 0Þ þ f ðx 0Þ
2
as
required.
MISCELLANEOUS PROBLEMS
14.13. Solve
@U
@t
¼
@2
U
@x2
subject to the conditions Uð0; tÞ ¼ 0; Uðx; 0Þ ¼
1 0 x 1
0 x A 1
, Uðx; tÞ is
bounded where x 0; t 0.
We proceed as in Problem 13.24, Chapter 13. A solution satisfying the partial differential equation and
the first boundary condition is given by Be2
t
sin x. Unlike Problem 13.24, Chapter 13, the boundary
conditions do not prescribe the specific values for , so we must assume that all values of are possible. By
analogy with that problem we sum over all possible values of , which corresponds to an integration in this
case, and are led to the possible solution
Uðx; tÞ ¼
ð1
0
BðÞ e2
t
sin x d ð1Þ
where BðÞ is undetermined. By the second condition, we have
ð1
0
BðÞ sin x d ¼
1 0 x 1
0 x A 1
¼ f ðxÞ
ð2Þ
from which we have by Fourier’s integral formula
BðÞ ¼
2
ð1
0
f ðxÞ sin x dx ¼
2
ð1
0
sin x dx ¼
2ð1 cos Þ
ð3Þ
CHAP. 14] FOURIER INTEGRALS 371](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-380-320.jpg)
![14.18. If f ðxÞ ¼
1 0 @ x 1
0 x A 1
find the (a) Fourier sine transform, (b) Fourier cosine transform of f ðxÞ. In
each case obtain the graph of f ðxÞ and its transform.
Ans: ðaÞ
ffiffiffi
2
r
1 cos
; ðbÞ
ffiffiffi
2
r
sin
:
14.19. (a) Find the Fourier sine transform of ex
, x A 0
ðbÞ Show that
ð1
0
x sin mx
x2
þ 1
dx ¼
2
em
; m 0 by using the result in ðaÞ:
(c) Explain from the viewpoint of Fourier’s integral theorem why the result in (b) does not hold for m ¼ 0.
Ans. (a)
ffiffiffiffiffiffiffiffi
2=
p
½ =ð1 þ 2
Þ
14.20. Solve for YðxÞ the integral equation
ð1
0
YðxÞ sin xt dx ¼
1 0 @ t 1
2 1 @ t 2
0 t A 2
8
:
and verify the solution by direction substitution.
Ans. YðxÞ ¼ ð2 þ 2 cos x 4 cos 2xÞ=x
PARSEVAL’S IDENTITY
14.21. Evaluate (a)
ð1
0
dx
ðx2 þ 1Þ2
; ðbÞ
ð1
0
x2
dx
ðx2 þ 1Þ2
by use of Parseval’s identity.
[Hint: Use the Fourier sine and cosine transforms of ex
, x 0.]
Ans: ðaÞ =4; ðbÞ =4
14.22. Use Problem 14.18 to show that (a)
ð1
0
1 cos x
x
2
dx ¼
2
; ðbÞ
ð1
0
sin4
x
x2
dx ¼
2
.
14.23. Show that
ð1
0
ðx cos x sin xÞ2
x6
dx ¼
15
.
MISCELLANEOUS PROBLEMS
14.24. (a) Solve
@U
@t
¼ 2
@2
U
@x2
, Uð0; tÞ ¼ 0; Uðx; 0Þ ¼ ex
; x 0; Uðx; tÞ is bounded where x 0; t 0.
(b) Give a physical interpretation.
Ans: Uðx; tÞ ¼
2
ð1
0
e22
t
sin x
2 þ 1
d
14.25. Solve
@U
@t
¼
@2
U
@x2
; Uxð0; tÞ ¼ 0; Uðx; 0Þ ¼
x 0 @ x @ 1
0 x 1
, Uðx; tÞ is bounded where x 0; t 0.
Ans: Uðx; tÞ ¼
2
ð1
0
sin
þ
cos 1
2
e2
t
cos x d
14.26. (a) Show that the solution to Problem 14.13 can be written
Uðx; tÞ ¼
2
ffiffiffi
p
ðx=2
ffiffi
t
p
0
ev2
dv
1
ffiffiffi
p
ðð1þxÞ=2
ffiffi
t
p
ð1xÞ=2
ffiffi
t
p ev2
dv
CHAP. 14] FOURIER INTEGRALS 373](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-382-320.jpg)
![This factorial representation for positive integers suggests the possibility that
ðx þ 1Þ ¼ x! ¼ lim
k!1
k! kx
ðx þ 1Þ . . . ðx þ kÞ
x 6¼ 1; 2; k ð6Þ
Gauss verified this identification back in the nineteenth century.
This infinite product is symbolized by ðx; kÞ, i.e., ðx; kÞ ¼
k! kx
ðx þ 1Þ ðx þ kÞ
. It is called Gauss’s
function and through this symbolism,
ðx þ 1Þ ¼ lim
k!1
ðx; kÞ ð7Þ
The expression for
1
ðxÞ
(which has some advantage in developing the derivative of ðxÞ) results as
follows. Put (6a) in the form
lim
k!1
kx
ð1 þ xÞð1 þ x=2Þ . . . ð1 þ x=kÞ
x 6¼
1
2
;
1
3
; . . . ;
1
k
Next, introduce
k ¼ 1 þ
1
2
þ
1
3
þ
1
k
ln k
Then
¼ lim
k!1
k
is Euler’s constant. This constant has been calculated to many places, a few of which are
0:57721566 . . . .
By letting kx
¼ ex ln k
¼ ex½kþ1þ1=2þþ1=k
, the representation (6) can be further modified so that
ðx þ 1Þ ¼ ex
lim
k!1
ex
1 þ x
ex=2
1 þ x=2
ex=k
1 þ x=k
¼ ex
Y
1
k¼1
ex
ex ln k
= 1 þ
x
k
¼
Y
1
¼1
kx
k!ðk þ xÞ ¼ lim
k!1
1 2 3 k
ðx þ 1Þðx þ 2Þ ðx þ kÞ
xx
¼ lim
k!1
ðx; kÞ ð8Þ
Since ðx þ 1Þ ¼ xðxÞ;
1
ðxÞ
¼ xex
lim
k!1
1 þ x
ex
1 þ x=2
ex=2
1 þ x=k
ex=k
¼ xex
Y
1
¼1
ð1 þ x=kÞ ex=k
ð9Þ
Another result of special interest emanates from a comparison of ðxÞð1 xÞ with the ‘‘famous’’
formula
x
sin x
¼ lim
k!1
1
1 x2
1
1 ðx=2Þ2
1
ð1 x=kÞ2
¼
Y
1
¼1
f1 ðx=kÞ2
g ð10Þ
(See Differential and Integral Calculus, by R. Courant (translated by E. J. McShane), Blackie Son
Limited.)
ð1 xÞ is obtained from ð yÞ ¼
1
y
ð y þ 1Þ by letting y ¼ x, i.e.,
ðxÞ ¼
1
x
ð1 xÞ or ð1 xÞ ¼ xðxÞ
CHAP. 15] GAMMA AND BETA FUNCTIONS 377](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-386-320.jpg)
![CHAP. 15] GAMMA AND BETA FUNCTIONS 379
Bðx; yÞ ¼
ð1
0
tx1
ð1 tÞy1
dt ð16Þ
If x 1 and y 1, this is a proper integral. If x 0; y 0 and either or both x 1 or y 1, the
integral is improper but convergent.
It is shown in Problem 15.11 that the beta function can be expressed through gamma functions in the
following way
Bðx; yÞ ¼
ðxÞ ð yÞ
ðx þ yÞ
ð17Þ
Many integrals can be expressed through beta and gamma functions. Two of special interest are
ð=2
0
sin2x1
cos2y1
d ¼
1
2
Bðx; yÞ ¼
1
2
ðxÞ ð yÞ
ðx þ yÞ
ð18Þ
ð1
0
xp1
1 þ x
dx ¼ ð pÞ ð p 1Þ ¼
sin p
0 p 1 ð19Þ
See Problem 15.17. Also see Page 377 where a classical reference is given. Finally, see Chapter 16,
Problem 16.38 where an elegant complex variable resolution of the integral is presented.
DIRICHLET INTEGRALS
If V denotes the closed region in the first octant bounded by the surface
x
a
p
þ
y
b
q
þ
z
c
r
¼ 1 and
the coordinate planes, then if all the constants are positive,
ð ð ð
V
x 1
y 1
z1
dx dy dz ¼
a b c
pqr
p
q
r
1 þ
p
þ
q
þ
r
ð20Þ
Integrals of this type are called Dirichlet integrals and are often useful in evaluating multiple
integrals (see Problem 15.21).
Solved Problems
THE GAMMA FUNCTION
15.1. Prove: (a) ðx þ 1Þ ¼ xðxÞ; x 0; ðbÞ ðn þ 1Þ ¼ n!; n ¼ 1; 2; 3; . . . .
ðaÞ ðv þ 1Þ ¼
ð1
0
xv
ex
dx ¼ lim
M!1
ðM
0
xv
ex
dx
¼ lim
M!1
ðxv
Þðex
ÞjM
0
ðM
0
ðex
Þðvxv1
Þ dx
¼ lim
M!1
Mv
eM
þ v
ðM
0
xv1
ex
dx
¼ vðvÞ if v 0
ðbÞ ð1Þ ¼
ð1
0
ex
dx ¼ lim
M!1
ðM
0
ex
dx ¼ lim
M!1
ð1 eM
Þ ¼ 1:
Put n ¼ 1; 2; 3; . . . in ðn þ 1Þ ¼ nðnÞ. Then
ð2Þ ¼ 1ð1Þ ¼ 1; ð3Þ ¼ 2ð2Þ ¼ 2 1 ¼ 2!; ð4Þ ¼ 3ð3Þ ¼ 3 2! ¼ 3!
In general, ðn þ 1Þ ¼ n! if n is a positive integer.](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-388-320.jpg)
![Letting axn
¼ y, the integral becomes
ð1
0
y
a
1=n
m
ey
d
y
a
1=n
¼
1
naðmþ1Þ=n
ð1
0
yðmþ1Þ=n1
ey
dy ¼
1
naðmþ1Þ=n
m þ 1
n
15.7. Evaluate (a) ð1=2Þ; ðbÞ ð5=2Þ.
We use the generalization to negative values defined by ðxÞ ¼
ðx þ 1Þ
x
.
ðaÞ Letting x ¼ 1
2 ; ð1=2Þ ¼
ð1=2Þ
1=2
¼ 2
ffiffiffi
p
:
ðbÞ Letting x ¼ 3=2; ð3=2Þ ¼
ð1=2Þ
3=2
¼
2
ffiffiffi
p
3=2
¼
4
ffiffiffi
p
3
; using ðaÞ:
Then ð5=2Þ ¼
ð3=2Þ
5=2
¼
8
15
ffiffiffi
p
:
15.8. Prove that
ð1
0
xm
ðln xÞn
dx ¼
ð1Þn
n!
ðm þ 1Þnþ1
, where n is a positive integer and m 1.
Letting x ¼ ey
, the integral becomes ð1Þn
ð1
0
yn
eðmþ1Þy
dy. If ðm þ 1Þy ¼ u, this last integral becomes
ð1Þn
ð1
0
un
ðm þ 1Þn eu du
m þ 1
¼
ð1Þn
ðm þ 1Þnþ1
ð1
0
un
eu
du ¼
ð1Þn
ðm þ 1Þnþ1
ðn þ 1Þ ¼
ð1Þn
n!
ðm þ 1Þnþ1
Compare with Problem 8.50, Chapter 8, page 203.
15.9. A particle is attracted toward a fixed point O with a force inversely proportional to its instanta-
neous distance from O. If the particle is released from rest, find the time for it to reach O.
At time t ¼ 0 let the particle be located on the x-axis at x ¼ a 0 and let O be the origin. Then by
Newton’s law
m
d2
x
dt2
¼
k
x
ð1Þ
where m is the mass of the particle and k 0 is a constant of proportionality.
Let
dx
dt
¼ v, the velocity of the particle. Then
d2
x
dt2
¼
dv
dt
¼
dv
dx
dx
dt
¼ v
dv
dx
and (1) becomes
mv
dv
dx
¼
k
x
or
mv2
2
¼ k ln x þ c ð2Þ
upon integrating. Since v ¼ 0 at x ¼ a, we find c ¼ k ln a. Then
mv2
2
¼ k ln
a
x
or v ¼
dx
dt
¼
ffiffiffiffiffi
2k
m
r ffiffiffiffiffiffiffiffiffi
ln
a
x
r
ð3Þ
where the negative sign is chosen since x is decreasing as t increases. We thus find that the time T taken for
the particle to go from x ¼ a to x ¼ 0 is given by
T ¼
ffiffiffiffiffi
m
2k
r ða
0
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffi
ln a=x
p ð4Þ
CHAP. 15] GAMMA AND BETA FUNCTIONS 381](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-390-320.jpg)
![ðbÞ
ð2
0
x2
dx
ffiffiffiffiffiffiffiffiffiffiffi
2 x
p : Letting x ¼ 2v; the integral becomes
4
ffiffiffi
2
p ð1
0
v2
ffiffiffiffiffiffiffiffiffiffiffi
1 v
p dv ¼ 4
ffiffiffi
2
p ð1
0
v2
ð1 vÞ1=2
dv ¼ 4
ffiffiffi
2
p
Bð3; 1
2Þ ¼
4
ffiffiffi
2
p
ð3Þ ð1=2Þ
ð7=2Þ
¼
64
ffiffiffi
2
p
15
ðcÞ
ða
0
y4
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 y2
q
dy: Letting y2
¼ a2
x or y ¼
ffiffiffi
x
p
; the integral becomes
a6
ð1
0
x3=2
ð1 xÞ1=2
dx ¼ a6
Bð5=2; 3=2Þ ¼
a6
ð5=2Þ ð3=2Þ
ð4Þ
¼
a6
16
15.13. Show that
ð=2
0
sin2u1
cos2v1
d ¼
ðuÞ ðvÞ
2 ðu þ vÞ
u; v 0.
This follows at once from Problems 15.10 and 15.11.
15.14. Evaluate (a)
ð=2
0
sin6
d; ðbÞ
ð=2
0
sin4
cos5
d; ðcÞ
ð
0
cos4
d.
ðaÞ Let 2u 1 ¼ 6; 2v 1 ¼ 0; i.e., u ¼ 7=2; v ¼ 1=2; in Problem 15.13:
Then the required integral has the value
ð7=2Þ ð1=2Þ
2 ð4Þ
¼
5
32
:
ðbÞ Letting 2u 1 ¼ 4; 2v 1 ¼ 5; the required integral has the value
ð5=2Þ ð3Þ
2 ð11=2Þ
¼
8
315
:
ðcÞ The given integral ¼ 2
ð=2
0
cos4
d:
Thus letting 2u 1 ¼ 0; 2v 1 ¼ 4 in Problem 15.13, the value is
2 ð1=2Þ ð5=2Þ
2 ð3Þ
¼
3
8
.
15.15. Prove
ð=2
0
sinp
d ¼
ð=2
0
cosp
d ¼ ðaÞ
1 3 5 ð p 1Þ
2 4 6 p
2
if p is an even positive integer,
(b)
2 4 6 ð p 1Þ
1 3 5 p
is p is an odd positive integer.
From Problem 15.13 with 2u 1 ¼ p; 2v 1 ¼ 0, we have
ð=2
0
sinp
d ¼
½1
2 ð p þ 1Þ ð1
2Þ
2 ½1
2 ð p þ 2Þ
(a) If p ¼ 2r, the integral equals
ðr þ 1
2Þ ð1
2Þ
2 ðr þ 1Þ
¼
ðr 1
2Þðr 3
2Þ 1
2 ð1
2Þ ð1
2Þ
2rðr 1Þ 1
¼
ð2r 1Þð2r 3Þ 1
2rð2r 2Þ 2
2
¼
1 3 5 ð2r 1Þ
2 4 6 2r
2
(b) If p ¼ 2r þ 1, the integral equals
ðr þ 1Þ ð1
2Þ
2 ðr þ 3
2Þ
¼
rðr 1Þ 1
ffiffiffi
p
2ðr þ 1
2Þðr 1
2Þ 1
2
ffiffiffi
p ¼
2 4 6 2r
1 3 5 ð2r þ 1Þ
In both cases
ð=2
0
sinp
d ¼
ð=2
0
cosp
d, as seen by letting ¼ =2 .
CHAP. 15] GAMMA AND BETA FUNCTIONS 383](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-392-320.jpg)
![15.16. Evaluate (a)
ð=2
0
cos6
d; ðbÞ
ð=2
0
sin3
cos2
d; ðcÞ
ð2
0
sin8
d.
(a) From Problem 15.15 the integral equals
1 3 5
2 4 6
¼
5
32
[compare Problem 15.14(a)].
(b) The integral equals
ð=2
0
sin3
ð1 sin2
Þ d ¼
ð=2
0
sin3
d
ð=2
0
sin5
d ¼
2
1 3
2 4
1 3 5
¼
2
15
The method of Problem 15.14(b) can also be used.
ðcÞ The given integral equals 4
ð=2
0
sin8
d ¼ 4
1 3 5 7
2 4 6 8
2
¼
35
64
:
15.17. Given
ð1
0
xp1
1 þ x
dx ¼
sin p
, show that ð pÞ ð1 pÞ ¼
sin p
, where 0 p 1.
Letting
x
1 þ x
¼ y or x ¼
y
1 y
, the given integral becomes
ð1
0
yp1
ð1 yÞp
dy ¼ Bð p; 1 pÞ ¼ ð pÞ ð1 pÞ
and the result follows.
15.18. Evaluate
ð1
0
dy
1 þ y4
.
Let y4
¼ x. Then the integral becomes
1
4
ð1
0
x3=4
1 þ x
dx ¼
4 sinð=4Þ
¼
ffiffiffi
2
p
4
by Problem 15.17 with p ¼ 1
4.
The result can also be obtained by letting y2
¼ tan .
15.19. Show that
ð2
0
x
ffiffiffiffiffiffiffiffiffiffiffiffiffi
8 x3
3
p
dx ¼
16
9
ffiffiffi
3
p .
Letting x3
8y or x ¼ 2y1=3
, the integral comes
ð1
0
2y1=3
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8ð1 yÞ
3
p
2
3 y2=3
dy ¼ 8
3
ð1
0
y1=3
ð1 yÞ1=3
dy ¼ 8
3 Bð2
3 ; 4
3Þ
¼
8
3
ð2
3 ð4
3Þ
ð2Þ
¼
8
9
ð1
3Þ ð2
3Þ ¼
8
9
sin =3
¼
16
9
ffiffiffi
3
p
STIRLING’S FORMULA
15.20. Show that for large positive integers n; n! ¼
ffiffiffiffiffiffiffiffi
2n
p
nn
en
approximately.
By definition ðzÞ ¼
ð1
0
tz1
et
dt. Let lfz ¼ x þ 1 then
ðx þ 1Þ ¼
ð1
0
tx
et
dt ¼
ð1
0
etþln tx
dt ¼
ð1
0
etþx ln t
dt ð1Þ
For a fixed value of x the function x ln t t has a relative maximum for t ¼ x (as is demonstrated by
elementary ideas of calculus). The substutition t ¼ x þ y yields
ðx þ 1Þ ¼ ex
ð1
x
ex lnðxþyÞy
dy ¼ xx
ex
ð1
x
ex lnð1þy
xÞy
dy ð2Þ
384 GAMMA AND BETA FUNCTIONS [CHAP. 15](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-393-320.jpg)
![To this point the analysis has been rigorous. The following formal steps can be made rigorous by
incorporating appropriate limiting procedures; however, because of the difficulty of the proofs, they shall be
omitted.
In (2) introduce the logarithmic expansion
ln 1 þ
y
x
¼
y
x
y2
2x2
þ
y3
3x3
þ ð3Þ
and also let
y ¼
ffiffiffi
x
p
v; dy ¼
ffiffiffi
x
p
dv
Then
ðx þ 1Þ ¼ xx
ex ffiffiffi
x
p
ð1
x
ev2
=2þðv3
=3Þ
ffiffi
x
p
dv ð4Þ
For large values of x
ðx þ 1Þ xx
ex ffiffiffi
x
p
ð1
x
ev2
=2
dv ¼ xx
ex
ffiffiffiffiffiffiffiffi
2x
p
When x is replaced by integer values n, then the Stirling relation
n! ¼ ðx þ 1Þ
ffiffiffiffiffiffiffiffi
2x
p
xx
ex
ð5Þ
is obtained.
It is of interest that from (4) we can also obtain the result (12) on Page 378. See Problem 15.72.
DIRICHLET INTEGRALS
15.21. Evaluate I ¼
ð ð ð
V
x 1
y 1
z1
dx dy dz where V is
the region in the first octant bounded by the sphere
x2
þ y2
þ z2
¼ 1 and the coordinate planes.
Let x2
¼ u; y2
¼ v; z2
¼ w. Then
I ¼
ð ð ð
r
uð 1Þ=2
vð 1Þ=2
wð1Þ=2 du
2
ffiffiffi
u
p
dv
2
ffiffiffi
v
p
dw
2
ffiffiffiffi
w
p
¼
1
8
ð ð ð
r
uð =2Þ1
vð =2Þ1
wð=2Þ1
du dv dw ð1Þ
where r is the region in the uvw space bounded by the plane
u þ v þ w ¼ 1 and the uv; vw, and uw planes as in Fig. 15-2.
Thus,
I ¼
1
8
ð1
u¼0
ð1u
v¼0
ð1uv
w¼0
uð =2Þ1
vð =2Þ1
wð=2Þ1
du dv dw ð2Þ
¼
1
4
ð1
u¼0
ð1u
v¼0
uð =2Þ1
vð =2Þ1
ð1 u vÞ=2
du dv
¼
1
4
ð1
u¼0
uð =2Þ1
ð1u
v¼0
vð =2Þ1
ð1 u vÞ=2
dv
du
Letting v ¼ ð1 uÞt, we have
ð1u
v¼0
vð =2Þ1
ð1 u vÞ=2
dv ¼ ð1 uÞð þÞ=2
ð1
t¼0
tð =2Þ1
ð1 tÞ=2
dt
¼ ð1 uÞð þÞ=2 ð =2Þ ð=2 þ 1Þ
½ð þ Þ=2 þ 1
CHAP. 15] GAMMA AND BETA FUNCTIONS 385
Fig. 15-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-394-320.jpg)
![and the required result follows. (See Problem 15.74, where the duplication formula is developed for the
simpler case of integers.)
15.25. Show that
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1
2 sin2
q ¼
fð1=4Þg2
4
ffiffiffi
p .
Consider
I ¼
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffi
cos
p ¼
ð=2
0
cos1=2
d ¼ 1
2 Bð1
4 ; 1
2Þ ¼
ð1
4Þ
ffiffiffi
p
2 ð3
4Þ
¼
fð1
4Þg2
2
ffiffiffiffiffiffi
2
p
as in Problem 15.23.
But I ¼
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffi
cos
p ¼
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
cos2 =2 sin2
=2
p ¼
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2 sin2
=2
p :
Letting
ffiffiffi
2
p
sin =2 ¼ sin in this last integral, it becomes
ffiffiffi
2
p
ð=2
0
d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
1 1
2 sin2
q , from which the result
follows.
15.26. Prove that
ð1
0
cos x
xp dx ¼
2 ð pÞ cosð p=2Þ
; 0 p 1.
We have
1
xp ¼
1
ð pÞ
ð1
0
up1
exu
du. Then
ð1
0
cos x
xp dx ¼
1
ð pÞ
ð1
0
ð1
0
up1
exu
cos x du dx
¼
1
ð pÞ
ð1
0
up
1 þ u2
du ð1Þ
where we have reversed the order of integration and used Problem 12.22, Chapter 12.
Letting u2
¼ v in the last integral, we have by Problem 15.17
ð1
0
up
1 þ u2
du ¼
1
2
ð1
0
vð p1Þ=2
1 þ v
dv ¼
2 sinð p þ 1Þ=2
¼
2 cos p=2
ð2Þ
Substitution of (2) in (1) yields the required result.
15.27. Evaluate
ð1
0
cos x2
dx.
Letting x2
¼ y, the integral becomes
1
2
ð1
0
cos y
ffiffiffi
y
p dy ¼
1
2
2 ð1
2Þ cos =4
!
¼ 1
2
ffiffiffiffiffiffiffiffi
=2
p
by Problem 15.26.
This integral and the corresponding one for the sine [see Problem 15.68(a)] are called Fresnel integrals.
Supplementary Problems
THE GAMMA FUNCTION
15.28. Evaluate (a)
ð7Þ
2 ð4Þ ð3Þ
; ðbÞ
ð3Þ ð3=2Þ
ð9=2Þ
; ðcÞ ð1=2Þ ð3=2Þ ð5=2Þ.
Ans. ðaÞ 30; ðbÞ 16=105; ðcÞ 3
8 3=2
CHAP. 15] GAMMA AND BETA FUNCTIONS 387](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-396-320.jpg)
![15.45. Prove that (a)
ð1
0
x dx
1 þ x6
¼
3
ffiffiffi
3
p ; ðbÞ
ð1
0
y2
dy
1 þ y4
¼
2
ffiffiffi
2
p .
15.46. Prove that
ð1
1
e2x
ae3x þ b
dx ¼
2
3
ffiffiffi
3
p
a2=3b1=3
where a; b 0.
15.47. Prove that
ð1
1
e2x
ðe3x þ 1Þ
dx ¼
2
9
ffiffiffi
3
p
[Hint: Differentiate with respect to b in Problem 15.46.]
15.48. Use the method of Problem 12.31, Chapter 12, to justify the procedure used in Problem 15.11.
DIRICHLET INTEGRALS
15.49. Find the mass of the region in the xy plane bounded by x þ y ¼ 1; x ¼ 0; y ¼ 0 if the density is ¼
ffiffiffiffiffiffi
xy
p
.
Ans: =24
15.50. Find the mass of the region bounded by the ellipsoid
x2
a2
þ
y2
b2
þ
z2
c2
¼ 1 if the density varies as the square of
the distance from its center.
Ans:
abck
30
ða2
þ b2
þ c2
Þ; k ¼ constant of proportionality
15.51. Find the volume of the region bounded by x2=3
þ y2=3
þ z2=3
¼ 1.
Ans: 4=35
15.52. Find the centroid of the region in the first octant bounded by x2=3
þ y2=3
þ z2=3
¼ 1.
Ans:
x
x ¼
y
y ¼
z
z ¼ 21=128
15.53. Show that the volume of the region bounded by xm
þ ym
þ zm
¼ am
, where m 0, is given by
8fð1=mÞg3
3m2 ð3=mÞ
a3
.
15.54. Show that the centroid of the region in the first octant bounded by xm
þ ym
þ zm
¼ am
, where m 0, is given
by
x
x ¼
y
y ¼
z
z ¼
3 ð2=mÞ ð3=mÞ
4 ð1=mÞ ð4=mÞ
a
MISCELLANEOUS PROBLEMS
15.55. Prove that
ðb
a
ðx aÞ p
ðb xÞq
dx ¼ ðb aÞ pþqþ1
Bð p þ 1; q þ 1Þ where p 1; q 1 and b a.
[Hint: Let x a ¼ ðb aÞy:]
15.56. Evaluate (a)
ð3
1
dx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 1Þð3 xÞ
p ; ðbÞ
ð7
3
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð7 xÞðx 3Þ
4
p
dx.
Ans: ðaÞ ; ðbÞ
2 fð1=4Þg2
3
ffiffiffi
p
15.57. Show that
fð1=3Þg2
ð1=6Þ
¼
ffiffiffi
p ffiffiffi
2
3
p
ffiffiffi
3
p .
15.58. Prove that Bðu; vÞ ¼
1
2
ð1
0
xu1
þ xv1
ð1 þ xÞuþv dx where u; v 0.
[Hint: Let y ¼ x=ð1 þ xÞ:
CHAP. 15] GAMMA AND BETA FUNCTIONS 389](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-398-320.jpg)
![15.59. If 0 p 1 prove that
ð=2
0
tan p
d ¼
2
sec
p
2
.
15.60. Prove that
ð1
0
xu1
ð1 xÞv1
ðx þ rÞuþv ¼
Bðu; vÞ
ruð1 þ rÞuþv where u; v, and r are positive constants.
[Hint: Let x ¼ ðr þ 1Þy=ðr þ yÞ.]
15.61. Prove that
ð=2
0
sin2u1
cos2v1
d
ða sin2
þ b cos2
Þuþv
¼
Bðu; vÞ
2av
bu where u; v 0.
[Hint: Let x ¼ sin2
in Problem 15.60 and choose r appropriately.]
15.62. Prove that
ð1
0
dx
xx ¼
1
11
þ
1
22
þ
1
33
þ
15.63. Prove that for m ¼ 2; 3; 4; . . .
sin
m
sin
2
m
sin
3
m
sin
ðm 1Þ
m
¼
m
2m1
[Hint: Use the factored form xm
1 ¼ ðx 1Þðx 1Þðx 2Þ ðx n1Þ, divide both sides by x 1, and
consider the limit as x ! 1.]
15.64. Prove that
ð=2
0
ln sin x dx ¼ =2 ln 2 using Problem 15.63.
[Hint: Take logarithms of the result in Problem 15.63 and write the limit as m ! 1 as a definite integral.]
15.65. Prove that
1
m
2
m
3
m
m 1
m
¼
ð2Þðm1Þ=2
ffiffiffiffi
m
p :
[Hint: Square the left hand side and use Problem 15.63 and equation (11a), Page 378.]
15.66. Prove that
ð1
0
ln ðxÞ dx ¼ 1
2 lnð2Þ.
[Hint: Take logarithms of the result in Problem 15.65 and let m ! 1.]
15.67. (a) Prove that
ð1
0
sin x
x p dx ¼
2 ð pÞ sinð p=2Þ
; 0 p 1.
(b) Discuss the cases p ¼ 0 and p ¼ 1.
15.68. Evaluate (a)
ð1
0
sin x2
dx; ðbÞ
ð1
0
x cos x3
dx.
Ans: ðaÞ 1
2
ffiffiffiffiffiffiffiffi
=2
p
; ðbÞ
3
ffiffiffi
3
p
ð1=3Þ
15.69. Prove that
ð1
0
x p1
ln x
1 þ x
dx ¼ 2
csc p cot p; 0 p 1.
15.70. Show that
ð1
0
ln x
x4 þ 1
dx ¼
2
ffiffiffi
2
p
16
.
15.71. If a 0; b 0, and 4ac b2
, prove that
ð1
1
ð1
1
eðax2
þbxyþcy2
Þ
dx dy ¼
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4ac b2
p
390 GAMMA AND BETA FUNCTIONS [CHAP. 15](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-399-320.jpg)
![15.72. Obtain (12) on Page 378 from the result (4) of Problem 15.20.
[Hint: Expand ev3
=ð3
ffiffi
n
p
Þ
þ in a power series and replace the lower limit of the integral by 1.]
15.73. Obtain the result (15) on Page 378.
[Hint: Observe that ðxÞ ¼
1
x
ðx þ !Þ, thus ln ðxÞ ¼ ln ðx þ 1Þ ln x, and
0
ðxÞ
ðxÞ
¼
0
ðx þ 1Þ
ðx þ 1Þ
1
x
Furthermore, according to (6) page 377.
ðx þ !Þ ¼ lim
k!1
k! kx
ðx þ 1Þ ðx þ kÞ
Now take the logarithm of this expression and then differentiate. Also recall the definition of the Euler
constant, .
15.74. The duplication formula (13a) Page 378 is proved in Problem 15.24. For further insight, develop it for
positive integers, i.e., show that
22n1
ðn þ 1
2Þ ðnÞ ¼ ð2nÞ
ffiffiffi
p
Hint: Recall that ð1
2Þ ¼ , then show that
ðn þ 1
2Þ ¼
2n þ 1Þ
2
¼
ð2n 1Þ 5 3 1
2n
ffiffiffi
p
:
Observe that
ð2n þ 1Þ
2nðn þ 1Þ
¼
ð2nÞ!
2nn!
¼ ð2n 1Þ 5 3 1
Now substitute and refine.
CHAP. 15] GAMMA AND BETA FUNCTIONS 391](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-400-320.jpg)
![LIMITS AND CONTINUITY
Definitions of limits and continuity for functions of a complex variable are analogous to those for a
real variable. Thus, f ðzÞ is said to have the limit l as z approaches z0 if, given any 0, there exists a
0 such that j f ðzÞ lj whenever 0 jz z0j .
Similarly, f ðzÞ is said to be continuous at z0 if, given any 0, there exists a 0 such that
j f ðzÞ f ðz0Þj whenever jz z0j . Alternatively, f ðzÞ is continuous at z0 if lim
z!z0
f ðzÞ ¼ f ðz0Þ.
Note: While these definitions have the same appearance as in the real variable setting, remember that
jz z0j means
jðx x0j þ ið y y0Þj ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx x0Þ2
þ ð y y0Þ2
q
:
Thus there are two degrees of freedom as ðx; yÞ ! ðx0; y0Þ:
DERIVATIVES
If f ðzÞ is single-valued in some region of the z plane the derivative of f ðzÞ, denoted by f 0
ðzÞ, is defined
as
lim
z!0
ðf ðz þ zÞ f ðzÞ
z
ð1Þ
provided the limit exists independent of the manner in which z ! 0. If the limit (1) exists for z ¼ z0,
then f ðzÞ is called analytic at z0. If the limit exists for all z in a region r, then f ðzÞ is called analytic in r.
In order to be analytic, f ðzÞ must be single-valued and continuous. The converse, however, is not
necessarily true.
We define elementary functions of a complex variable by a natural extension of the corresponding
functions of a real variable. Where series expansions for real functions f ðxÞ exists, we can use as
definition the series with x replaced by z. The convergence of such complex series has already been
considered in Chapter 11.
EXAMPLE 1. We define ex
¼ 1 þ z þ
z2
2!
þ
z3
3!
þ ; sin z ¼ z
z3
3!
þ
z5
5!
z7
7!
þ ; cos z ¼ 1
z2
2!
þ
z4
4!
z6
6!
þ .
From these we can show that ex
¼ exþiy
¼ ex
ðcos y þ i sin yÞ, as well as numerous other relations.
Rules for differentiating functions of a complex variable are much the same as for those of real
variables. Thus,
d
dz
ðzn
Þ ¼ nzn1
;
d
dz
ðsin zÞ ¼ cos z, and so on.
CAUCHY-RIEMANN EQUATIONS
A necessary condition that w ¼ f ðzÞ ¼ uðx; yÞ þ ivðx; yÞ be analytic in a region r is that u and v
satisfy the Cauchy-Riemann equations
@u
@x
¼
@v
@y
;
@u
@y
¼
@v
@x
ð2Þ
(see Problem 16.7). If the partial derivatives in (2) are continuous in r, the equations are sufficient
conditions that f ðzÞ be analytic in r.
If the second derivatives of u and v with respect to x and y exist and are continuous, we find by
differentiating (2) that
@2
u
@x2
þ
@2
u
@y2
¼ 0;
@2
v
@x2
þ
@2
v
@y2
¼ 0 ð3Þ
Thus, the real and imaginary parts satisfy Laplace’s equation in two dimensions. Functions satisfying
Laplace’s equation are called harmonic functions.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 393](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-402-320.jpg)
![derivatives at points within C can be calculated. Thus, if a function of a complex variable has a first
derivative, it has all higher derivatives as well. This, of course, is not necessarily true for functions of
real variables.
TAYLOR’S SERIES
Let f ðzÞ be analytic inside and on a circle having its center at z ¼ a. Then for all points z in the circle
we have the Taylor series representation of f ðzÞ given by
f ðzÞ ¼ f ðaÞ þ f 0
ðaÞðz aÞ þ
f 00
ðaÞ
2!
ðz aÞ2
þ
f 000
ðaÞ
3!
ðz aÞ3
þ ð8Þ
See Problem 16.21.
SINGULAR POINTS
A singular point of a function f ðzÞ is a value of z at which f ðzÞ fails to be analytic. If f ðzÞ is analytic
everywhere in some region except at an interior point z ¼ a, we call z ¼ a an isolated singularity of f ðzÞ.
EXAMPLE. If f ðzÞ ¼
1
ðz 3Þ2
, then z ¼ 3 is an isolated singularity of f ðzÞ.
EXAMPLE. The function f ðzÞ ¼
sin z
z
has a singularity at z ¼ 0. Because lim
z!0
is finite, this singularity is called a
removable singularity.
POLES
If f ðzÞ ¼
ðzÞ
ðz aÞn ; ðaÞ 6¼ 0, where ðzÞ is analytic everywhere in a region including z ¼ a, and if n is a
positive integer, then f ðzÞ has an isolated singularity at z ¼ a, which is called a pole of order n. If n ¼ 1,
the pole is often called a simple pole; if n ¼ 2, it is called a double pole, and so on.
LAURENT’S SERIES
If f ðzÞ has a pole of order n at z ¼ a but is analytic at every other point inside and on a circle C with
center at a, then ðz aÞn
f ðzÞ is analytic at all points inside and on C and has a Taylor series about z ¼ a
so that
f ðzÞ ¼
an
ðz aÞn þ
anþ1
ðz aÞn1
þ þ
a1
z a
þ a0 þ a1ðz aÞ þ a2ðz aÞ2
þ ð9Þ
This is called a Laurent series for f ðzÞ. The part a0 þ a1ðz aÞ þ a2ðz aÞ2
þ is called the analytic
part, while the remainder consisting of inverse powers of z a is called the principal part. More
generally, we refer to the series
X
1
k¼1
akðz aÞk
as a Laurent series, where the terms with k 0 constitute
the principal part. A function which is analytic in a region bounded by two concentric circles having
center at z ¼ a can always be expanded into such a Laurent series (see Problem 16.92).
It is possible to define various types of singularities of a function f ðzÞ from its Laurent series. For
example, when the principal part of a Laurent series has a finite number of terms and an 6¼ 0 while
an1; an2; . . . are all zero, then z ¼ a is a pole of order n. If the principal part has infinitely many
terms, z ¼ a is called an essential singularity or sometimes a pole of infinite order.
EXAMPLE. The function e1=z
¼ 1 þ
1
z
þ
1
2! z2
þ has an essential singularity at z ¼ 0.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 395](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-404-320.jpg)
![and the connecting point constitute a Riemann surface, and w1 and w2 are the branches of the function
each defined in one of the planes. (Since the space of complex variables is the complex plane, this
Riemann surface may be thought of as a flight of fancy that supports a rigorous analytic construction.)
To visualize this Riemann surface and perceive the single-valued character of the new function in it,
first think of duplicates, C1 and C2 of the domain circle, C: jzj ¼ in the planes P1 and P2, respectively.
Start at ¼ on C1, and proceed counterclockwise to the edge U2 of the cut of P1. (This edge
corresponds to ¼ 2). Paste U2 to L1, the initial edge of the cut on P2. Transfer to P2 through
this join and continue on C2. Now after a complete counterclockwise circuit of C2 we reach the edge L2
of the cut. Pasting L2 to U1 provides passage back to P1 and makes it possible to close the curve in the
Riemann plane. See Fig. 16-2.
Note that the function is not continuous on the positive x-axis. Also the cut is somewhat arbitrary.
Other rays and even curves extending from the origin to infinity can be employed. In many integration
applications the cut ¼ i proves valuable. On the other hand, the branch point (0 in this example) is
special. If another point, z0 6¼ 0 were chosen as the center of a small circle with radius less than jz0j, then
the origin would lie outside it. As a point z traversed its circumference, its argument would return to the
original value as would the value of w. However, for any circle that has the branch point as an interior
point, a similar traversal of the circumference will change the value of the argument by 2, and the
values of w1 and w2 will be interchanged. (See Problem 16.37.)
RESIDUES
The coefficients in (9) can be obtained in the customary manner by writing the coefficients for the
Taylor series corresponding to ðz aÞn
f ðzÞ. In further developments, the coefficient a1, called the
residue of f ðzÞ at the pole z ¼ a, is of considerable importance. It can be found from the formula
a1 ¼ lim
z!a
1
ðn 1Þ!
dn1
dzn1
fðz aÞn
f ðzÞg ð10Þ
where n is the order of the pole. For simple poles the calculation of the residue is of particular simplicity
since it reduces to
a1 ¼ lim
z!a
ðz aÞ f ðzÞ ð11Þ
RESIDUE THEOREM
If f ðzÞ is analytic in a region r except for a pole of order n at z ¼ a and if C is any simple closed
curve in r containing z ¼ a, then f ðzÞ has the form (9). Integrating (9), using the fact that
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 397
Fig. 16-2](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-406-320.jpg)
![Solved Problems
FUNCTIONS, LIMITS, CONTINUITY
16.1. Determine the locus represented by
(a) jz 2j ¼ 3; ðbÞ jz 2j ¼ jz þ 4j; ðcÞ jz 3j þ jz þ 3j ¼ 10.
(a) Method 1: jz 2j ¼ jx þ iy 2j ¼ jx 2 þ iyj ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 2Þ2
þ y2
q
¼ 3 or ðx 2Þ2
þ y2
¼ 9, a circle with
center at ð2; 0Þ and radius 3.
Method 2: jz 2j is the distance between the complex numbers z ¼ x þ iy and 2 þ 0i. If this distance is
always 3, the locus is a circle of radius 3 with center at 2 þ 0i or ð2; 0Þ.
(b) Method 1: jx þ iy 2j ¼ jx þ iy þ 4j or
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 2Þ2
þ y2
q
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 4Þ2
þ y
2
q
. Squaring, we find x ¼ 1, a
straight line.
Method 2: The locus is such that the distance from any point on it to ð2; 0Þ and ð4; 0Þ are equal. Thus,
the locus is the perpendicular besector of the line joining ð2; 0Þ and ð4; 0Þ, or x ¼ 1.
(c) Method 2: The locus is given by
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 3Þ2
þ y2
q
þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 3Þ2
þ y2
q
¼ 10 or
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 3Þ2
þ y2
q
¼ 10
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 3Þ2
þ y2
q
. Squaring and simplifying, 25 þ 3x ¼ 5
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx þ 3Þ2
þ y2
q
. Squaring and simplifying
again yields
x2
25
þ
y2
16
¼ 1, an ellipse with semi-major and semi-minor axes of lengths 5 and 4, respec-
tively.
Method 2: The locus is such that the sum of the distances from any point on it to ð3; 0Þ and ð3; 0Þ is 10.
Thus the locus is an ellipse whose foci are at ð3; 0Þ and ð3; 0Þ and whose major axis has length 10.
16.2. Determine the region in the z plane represented by each of the following.
(a) jzj 1.
Interior of a circle of radius 1. See Fig. 16-3(a) below.
(b) 1 jz þ 2ij @ 2.
jz þ 2ij is the distance from z to 2i, so that jz þ 2ij ¼ 1 is a circle of radius 1 with center at 2i,
i.e., ð0; 2Þ; and jz þ 2ij ¼ 2 is a circle of radius 2 with center at 2i. Then 1 jz þ 2ij @ 2 represents
the region exterior to jz þ 2ij ¼ 1 but interior to or on jz þ 2ij ¼ 2. See Fig. 16-3(b) below.
(c) =3 @ arg z @ =2.
Note that arg z ¼ , where z ¼ ei
. The required region is the infinite region bounded by the lines
¼ =3 and ¼ =2, including these lines. See Fig. 16-3(c) below.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 399
Fig. 16-3](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-408-320.jpg)
![d
dz
z
z ¼ lim
z!0
z þ z
z
z
z
¼ lim
x!0
y!0
x þ iy þ x þ i y x þ iy
x þ i y
¼ lim
x!0
y!0
x iy þ x þ i y ðx iyÞ
x þ i y
¼ lim
x!0
y!0
x i y
x þ i y
If y ¼ 0, the required limit is lim
x!0
x
x
¼ 1:
If x ¼ 0, the required limit is lim
y!0
i y
i y
¼ 1:
These two possible approaches show that the limit depends on the manner in which z ! 0, so that the
derivative does not exist; i.e.,
z
z is nonanalytic anywhere.
16.6. (a) If w ¼ f ðzÞ ¼
1 þ z
1 z
, find
dw
dz
. (b) Determine where w is nonanalytic.
ðaÞ Method 1:
dw
dz
¼ lim
z!0
1 þ ðz þ zÞ
1 ðz þ zÞ
1 þ z
1 z
z
¼ lim
z!0
2
ð1 z zÞð1 zÞ
¼
2
ð1 zÞ2
provided z 6¼ 1, independent of the manner in which z ! 0:
Method 2. The usual rules of differentiation apply provided z 6¼ 1. Thus, by the quotient rule for
differentiation,
d
dz
1 þ z
1 z
¼
ð1 zÞ
d
dz
ð1 þ zÞ ð1 þ zÞ
d
dz
ð1 zÞ
ð1 zÞ2
¼
ð1 zÞð1Þ ð1 þ zÞð1Þ
ð1 zÞ2
¼
2
ð1 zÞ2
(b) The function is analytic everywhere except at z ¼ 1, where the derivative does not exist; i.e., the function
is nonanalytic at z ¼ 1.
16.7. Prove that a necessary condition for w ¼ f ðzÞ ¼ uðx; yÞ þ i vðx; yÞ to be analytic in a region is that
the Cauchy-Riemann equations
@u
@x
¼
@v
@y
,
@u
@y
¼
@v
@x
be satisfied in the region.
Since f ðzÞ ¼ f ðx þ iyÞ ¼ uðx; yÞ þ i vðx; yÞ, we have
f ðz þ zÞ ¼ f ½x þ x þ ið y þ yÞ ¼ uðx þ x; y þ yÞ þ i vðx þ x; y þ yÞ
Then
lim
z!0
f ðz þ zÞ f ðzÞ
z
¼ lim
x!0
y!0
uðx þ x; y þ yÞ uðx; yÞ þ ifvðx þ x; y þ yÞ vðx; yÞg
x þ i y
If y ¼ 0, the required limit is
lim
x!0
uðx þ x; yÞ uðx; yÞ
x
þ i
vðx þ x; yÞ vðx; yÞ
x
¼
@u
@x
þ i
@v
@x
If x ¼ 0, the required limit is
lim
y!0
uðx; y þ yÞ uðx; yÞ
i y
þ
vðx; y þ yÞ vðx; yÞ
y
¼
1
i
@u
@y
þ
@v
@y
If the derivative is to exist, these two special limits must be equal, i.e.,
@u
@x
þ i
@v
@x
¼
1
i
@u
@y
þ
@v
@y
¼ i
@u
@y
þ
@v
@y
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 401](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-410-320.jpg)
![This can also be accomplished by nothing that z ¼ x þ iy;
z
z ¼ x iy so that x ¼
z þ
z
z
2
, y ¼
z
z
z
2i
.
The result is then obtained by substitution; the terms involving
z
z drop out.
INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS
16.10. Evaluate
ð2þ4i
1þi
z2
dz
(a) along the parabola x ¼ t; y ¼ t2
where 1 @ t @ 2,
(b) along the straight line joining 1 þ i and 2 þ 4i,
(c) along straight lines from 1 þ i to 2 þ i and then to 2 þ 4i.
We have
ð2þ4i
1þi
z2
dz ¼
ðð2;4Þ
ð1;1Þ
ðx þ iyÞ2
ðdx þ i dyÞ ¼
ðð2;4Þ
ð1;1Þ
ðx2
y2
þ 2ixyÞðdx þ i dyÞ
¼
ðð2;4Þ
ð1;1
ðx2
y2
Þ dx 2xy dy þ i
ðð2;4Þ
ð1;1Þ
2xy dx þ ðx2
y2
Þ dy
Method 1. (a) The points ð1; 1Þ and ð2; 4Þ correspond to t ¼ 1 and t ¼ 2, respectively. Then the above
line integrals become
ð2
t¼1
fðt2
t4
Þ dt 2ðtÞðt2
Þ2t dtg þ i
ð2
t¼1
f2ðtÞðt2
Þ dt þ ðt2
t4
Þð2tÞ dtg ¼
86
3
6i
(b) The line joining ð1; 1Þ and ð2; 4Þ has the equation y 1 ¼
4 1
2 1
ðx 1Þ or y ¼ 3x 2. Then we find
ð2
x¼1
½x2
ð3x 2Þ2
dx 2xð3x 2Þ3 dx
þ i
ð2
x¼1
2xð3x 2Þ dx þ ½x2
ð3x 2Þ2
3 dx
¼
86
3
6i
(c) From 1 þ i to 2 þ i [or ð1; 1Þ to ð2; 1Þ], y ¼ 1; dy ¼ 0 and we have
ð2
x¼1
ðx2
1Þ dx þ i
ð2
x¼1
2x dx ¼
4
3
þ 3i
From 2 þ i to 2 þ 4i [or ð2; 1Þ to ð2; 4Þ], x ¼ 2; dx ¼ 0 and we have
ð4
y¼1
4y dy þ i
ð4
y¼1
ð4 y2
Þ dy ¼ 30 9i
Adding, ð4
3 þ 3iÞ þ ð30 91Þ ¼ 86
3 6i.
Method 2. By the methods of Chapter 10 it is seen that the line integrals are independent of the path, thus
accounting for the same values obtained in (a), (b), and (c) above. In such case the integral can be evaluated
directly, as for real variables, as follows:
ð2þ4i
1þi
z2
dz ¼
z3
3
2þ4i
1i
¼
ð2 þ 4iÞ3
3
ð1 þ iÞ3
3
¼
86
3
6i
16.11. (a) Prove Cauchy’s theorem: If f ðzÞ is analytic inside and on a simple closed curve C, then
þ
C
f ðzÞ dz ¼ 0.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 403](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-412-320.jpg)
![(b) Under these conditions prove that
ðP2
P1
f ðzÞ dz is independent of the path joining P1 and P2.
ðaÞ
þ
C
f ðzÞ dz ¼
þ
C
ðu þ ivÞðdx þ i dyÞ ¼
þ
C
u dx v dy þ i
þ
C
v dx þ u dy
By Green’s theorem (Chapter 10),
þ
C
u dx v dy ¼
ð ð
r
@v
@x
@u
@y
dx dy;
þ
C
v dx þ u dy ¼
ð ð
r
@u
@x
@v
@y
dx dy
where r is the region (simply-connected) bounded by C.
Since f ðzÞ is analytic,
@u
@x
¼
@v
@y
;
@v
@x
¼
@u
@y
(Problem 16.7), and so the above integrals are zero.
Then
þ
C
f ðzÞ dz ¼ 0, assuming f 0
ðzÞ [and thus the partial derivatives] to be continuous.
(b) Consider any two paths joining points P1 and P2 (see Fig. 16-5). By Cauchy’s theorem,
ð
P1AP2BP1
f ðzÞ dz ¼ 0
Then
ð
P1AP2
f ðzÞ dz þ
ð
P2BP1
f ðzÞ dz ¼ 0
or
ð
P1AP2
f ðzÞ dz ¼
ð
P2BP1
f ðzÞ dz ¼
ð
P1BP2
f ðzÞ dz
i.e., the integral along P1AP2 (path 1) ¼ integral along P1BP2
(path 2), and so the integral is independent of the path joining P1
and P2.
This explains the results of Problem 16.10, since f ðzÞ ¼ z2
is analytic.
16.12. If f ðzÞ is analytic within and on the boundary of a region bounded by two closed curves C1 and C2
(see Fig. 16-6), prove that
þ
C1
f ðzÞ dz ¼
þ
C2
f ðzÞ dz
As in Fig. 16-6, construct line AB (called a cross-cut) connecting any point on C2 and a point on C1. By
Cauchy’s theorem (Problem 16.11),
ð
AQPABRSTBA
f ðzÞ dz ¼ 0
since f ðzÞ is analytic within the region shaded and also on the
boundary. Then
ð
AQPA
f ðzÞ dz þ
ð
AB
f ðzÞ dz þ
ð
BRSTB
f ðzÞ dz þ
ð
BA
f ðzÞ dz ¼ 0 ð1Þ
But
ð
AB
f ðzÞ dz ¼
ð
BA
f ðzÞ dz. Hence, (1) gives
ð
AQPA
f ðzÞ dz ¼
ð
BRSTB
f ðzÞ dz ¼
ð
BTSRB
f ðzÞ dz
404 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
Path 1
Path 2
P1
P2
A
B
Fig. 16-5
Fig. 16-6](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-413-320.jpg)
![þ
C1
f ðzÞ dz ¼
þ
C2
f ðzÞ dz
i.e.,
Note that f ðzÞ need not be analytic within curve C2.
16.13. (a) Prove that
þ
C
dz
ðz aÞn ¼
2i if n ¼ 1
0 if n ¼ 2; 3; 4; . . .
, where C is a simple closed curve bounding
a region having z ¼ a as interior point.
(b) What is the value of the integral if n ¼ 0;
1; 2; 3; . . . ?
(a) Let C1 be a circle of radius having center at z ¼ a (see Fig.
16-7). Since ðz aÞn
is analytic within and on the boundary
of the region bounded by C and C1, we have by Problem
16.12,
þ
C
dz
ðz aÞn ¼
þ
C1
dz
ðz aÞn
To evaluate this last integral, note that on C1, jz aj ¼ or z a ¼ ei
and dz ¼ iei
d. The
integral equals
ð2
0
iei
d
n ein
¼
i
n1
ð2
0
eð1nÞi
d ¼
i
n1
eð1nÞi
ð1 nÞi
2
0
¼ 0 if n 6¼ 1
If n ¼ 1, the integral equals i
ð2
0
d ¼ 2i.
(b) For n ¼ 0; 1; 2; . . . the integrand is 1; ðz aÞ; ðz aÞ2
; . . . and is analytic everywhere inside C1,
including z ¼ a. Hence, by Cauchy’s theorem the integral is zero.
16.14. Evaluate
þ
C
dz
z 3
, where C is (a) the circle jzj ¼ 1; ðbÞ the circle jz þ ij ¼ 4.
(a) Since z ¼ 3 is not interior to jzj ¼ 1, the integral equals zero (Problem 16.11).
(b) Since z ¼ 3 is interior to jz þ ij ¼ 4, the integral equals 2i (Problem 16.13).
16.15. If f ðzÞ is analytic inside and on a simple closed curve C, and a is any point within C, prove that
f ðaÞ ¼
1
2i
þ
C
f ðzÞ
z a
dz
Referring to Problem 16.12 and the figure of Problem 16.13, we have
þ
C
f ðzÞ
z a
dz ¼
þ
C1
f ðzÞ
z a
dz
Letting z a ¼ ei
, the last integral becomes i
ð2
0
f ða þ ei
Þ d. But since f ðzÞ is analytic, it is
continuous. Hence,
lim
!0
i
ð2
0
f ða þ ei
Þ d ¼ i
ð2
0
lim
!0
f ða þ ei
Þ d ¼ i
ð2
0
f ðaÞ d ¼ 2i f ðaÞ
and the required result follows.
16.16. Evaluate (a)
þ
C
cos z
z
dz; ðbÞ
þ
C
ex
zðz þ 1Þ
dz, where C is the circle jz 1j ¼ 3.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 405
Fig. 16-7](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-414-320.jpg)
![ðcÞ
X
1
n¼1
ðz iÞn
3n : We have lim
n!1
unþ1
un
¼ lim
n!1
ðz iÞnþ1
3nþ1
3n
ðz iÞn ¼
jz ij
3
:
The series converges if jz ij 3, and diverges if jz ij 3.
If jz ij ¼ 3, then z i ¼ 3ei
and the series becomes
X
1
n¼1
ein
. This series diverges since the nth
term does not approach zero as n ! 1.
Thus, the series converges within the circle jz ij ¼ 3 but not on the boundary.
16.19. If
X
1
n¼0
anzn
is absolutely convergent for jzj @ R, show that it is uniformly convergent for these
values of z.
The definitions, theorems, and proofs for series of complex numbers are analogous to those for real
series.
In this case we have janzn
j @ janjRn
¼ Mn. Since by hypothesis
X
1
n¼1
Mn converges, it follows by the
Weierstrass M test that
X
1
n¼0
anzn
converges uniformly for jzj @ R.
16.20. Locate in the finite z plane all the singularities, if any, of each function and name them.
ðaÞ
z2
ðz þ 1Þ3
: z ¼ 1 is a pole of order 3.
(b)
2z3
z þ 1
ðz 4Þ2
ðz iÞðz 1 þ 2iÞ
. z ¼ 4 is a pole of order 2 (double pole); z ¼ i and z ¼ 1 2i are poles of
order 1 (simple poles).
(c)
sin mz
z2
þ 2z þ 2
, m 6¼ 0. Since z2
þ 2z þ 2 ¼ 0 when z ¼
2
ffiffiffiffiffiffiffiffiffiffiffi
4 8
p
2
¼
2 2i
2
¼ 1 i, we can write
z2
þ 2z þ 2 ¼ fz ð1 þ iÞgfz ð1 iÞg ¼ ðz þ 1 iÞðz þ 1 þ iÞ.
The function has the two simple poles: z ¼ 1 þ i and z ¼ 1 i.
(d)
1 cos z
z
. z ¼ 0 appears to be a singularity. However, since lim
x!0
1 cos z
z
¼ 0, it is a removable
singularity.
Another method:
Since
1 cos z
z
¼
1
z
1 1
z2
2!
þ
z4
4!
z6
6!
þ
!
( )
¼
z
2!
z3
4!
þ , we see that z ¼ 0 is a remova-
ble singularity.
ðeÞ e1=ðx1Þ2
¼ 1
1
ðz 1Þ2
þ
1
2!ðz 1Þ4
:
This is a Laurent series where the principal part has an infinite number of non-zero terms. Then
z ¼ 1 is an essential singularity.
( f ) ez
.
This function has no finite singularity. However, letting z ¼ 1=u, we obtain e1=u
, which has an
essential singularity at u ¼ 0. We conclude that z ¼ 1 is an essential singularity of ez
.
In general, to determine the nature of a possible singularity of f ðzÞ at z ¼ 1, we let z ¼ 1=u and
then examine the behavior of the new function at u ¼ 0.
16.21. If f ðzÞ is analytic at all points inside and on a circle of radius R with center at a, and if a þ h is any
point inside C, prove Taylor’s theorem that
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 407](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-416-320.jpg)
![ðcÞ
sin z
z
; z ¼ : Let z ¼ u: Then z ¼ þ u and
sin z
z
¼
sinðu þ Þ
u
¼
sin u
u
¼
1
u
u
u3
3!
þ
u5
5!
!
¼ 1 þ
u2
3!
u4
5!
þ ¼ 1 þ
ðz Þ2
3!
ðz Þ4
5!
þ
z ¼ is a removable singularity.
The series converges for all values of z.
ðdÞ
z
ðz þ 1Þðz þ 2Þ
; z ¼ 1: Let z þ 1 ¼ u. Then
z
ðz þ 1Þðz þ 2Þ
¼
u 1
uðu þ 1Þ
¼
u 1
u
ð1 u þ u2
u3
þ u4
Þ
¼
1
u
þ 2 2u þ 2u2
2u3
þ
¼
1
z þ 1
þ 2 2ðz þ 1Þ þ 2ðz þ 1Þ2
z ¼ 1 is a pole of order 1, or simple pole.
The series converges for values of z such that 0 jz þ 1j 1.
ðeÞ
1
zðz þ 2Þ3
; z ¼ 0; 2:
Case 1, z ¼ 0. Using the binomial theorem,
1
zðz þ 2Þ3
¼
1
8zð1 þ z=2Þ3
¼
1
8z
1 þ ð3Þ
z
2
þ
ð3Þð4Þ
2!
z
2
2
þ
ð3Þð4Þð5Þ
3!
z
2
3
þ
¼
1
8z
3
16
þ
3
16
z
5
32
z2
þ
z ¼ 0 is a pole of order 1, or simple pole.
The series converges for 0 jzj 2.
Case 2, z ¼ 2. Let z þ 2 ¼ u. Then
1
zðz þ 2Þ3
¼
1
ðu 2Þu3
¼
1
2u3ð1 u=2Þ
¼
1
2u3
1 þ
u
2
þ
u
2
2
þ
u
2
3
þ
u
2
4
þ
¼
1
2u3
1
4u2
1
8u
1
16
1
32
u
¼
1
2ðz þ 2Þ3
1
4ðz þ 2Þ2
1
8ðz þ 2Þ
1
16
1
32
ðz þ 2Þ
z ¼ 2 is a pole of order 3.
The series converges for 0 jz þ 2j 2.
RESIDUES AND THE RESIDUE THEOREM
16.23. Suppose f ðzÞ is analytic everywhere inside and on a simple closed curve C except at z ¼ a which is
a pole of order n. Then
f ðzÞ ¼
an
ðz aÞn þ
anþ1
ðz aÞn1
þ þ a0 þ a1ðz aÞ þ a2ðz aÞ2
þ
where an 6¼ 0. Prove that
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 409](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-418-320.jpg)
![ðaÞ
þ
C
f ðzÞ dz ¼ 2ia1
ðbÞ a1 ¼ lim
z!a
1
ðn 1Þ!
dn1
dzn1
fðz aÞn
f ðzÞg:
(a) By integration, we have on using Problem 16.13
þ
C
f ðzÞ dz ¼
þ
C
an
ðz aÞn dz þ þ
þ
C
a1
z a
dz þ
þ
C
fa0 þ a1ðz aÞ þ a2ðz aÞ2
þ g dz
¼ 2ia1
Since only the term involving a1 remains, we call a1 the residue of f ðzÞ at the pole z ¼ a.
(b) Multiplication by ðz aÞn
gives the Taylor series
ðz aÞn
f ðzÞ ¼ an þ anþ1ðz aÞ þ þ a1ðz aÞn1
þ
Taking the ðn 1Þst derivative of both sides and letting z ! a, we find
ðn 1Þ!a1 ¼ lim
z!a
dn1
dzn1
fðz aÞn
f ðzÞg
from which the required result follows.
16.24. Determine the residues of each function at the indicated poles.
ðaÞ
z2
ðz 2Þðz2 þ 1Þ
; z ¼ 2; i; i: These are simple poles. Then:
Residue at z ¼ 2 is lim
z!2
ðz 2Þ
z2
ðz 2Þðz2
þ 1Þ
( )
¼
4
5
:
Residue at z ¼ i is lim
z!i
ðz iÞ
z2
ðz 2Þðz iÞðz þ iÞ
( )
¼
i2
ði 2Þð2iÞ
¼
1 2i
10
:
Residue at z ¼ i is lim
z!i
ðz þ iÞ
z2
ðz 2Þðz iÞðz þ iÞ
( )
¼
i2
ði 2Þð2iÞ
¼
1 þ 2i
10
:
ðbÞ
1
zðz þ 2Þ3
; z ¼ 0; 2: z ¼ 0 is a simple pole, z ¼ 2 is a pole of order 3. Then:
Residue at z ¼ 0 is lim
z!0
z
1
zðz þ 2Þ3
¼
1
8
:
Residue at z ¼ 2 is lim
z!2
1
2!
d2
dz2
ðz þ 2Þ3
1
zðz þ 2Þ3
¼ lim
z!2
1
2
d2
dz2
1
z
¼ lim
z!2
1
2
2
z3
¼
1
8
:
Note that these residues can also be obtained from the coefficients of 1=z and 1=ðz þ 2Þ in the
respective Laurent series [see Problem 16.22(e)].
ðcÞ
zezt
ðz 3Þ2
; z ¼ 3; a pole of order 2 or double pole. Then:
Residue is lim
z!3
d
dz
ðz 3Þ2
zezt
ðz 3Þ2
¼ lim
z!3
d
dz
ðzezt
Þ ¼ lim
z!3
ðezt
þ ztezt
Þ
¼ e3t
þ 3te3t
410 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-419-320.jpg)
![(d) cot z; z ¼ 5, a pole of order 1. Then:
Residue is lim
z!5
ðz 5Þ
cos z
sin z
¼ lim
z!5
z 5
sin z
lim
z!5
cos z
¼ lim
z!5
1
cos z
ð1Þ
¼ ð1Þð1Þ ¼ 1
where we have used L’Hospital’s rule, which can be shown applicable for functions of a complex
variable.
16.25. If f ðzÞ is analytic within and on a simple closed curve C except at a number of poles a; b; c; . . .
interior to C, prove that
þ
C
f ðzÞ dz ¼ 2i fsum of residues of f ðzÞ at poles a; b; c; etc.g
Refer to Fig. 16-8.
By reasoning similar to that of Problem 16.12 (i.e., by con-
structing cross cuts from C to C1; C2; C3; etc.), we have
þ
C
f ðzÞ dz ¼
þ
C1
f ðzÞ dz þ
þ
C2
f ðzÞ dz þ
For pole a,
f ðzÞ ¼
am
ðz aÞm þ þ
a1
ðz aÞ
þ a0 þ a1ðz aÞ þ
hence, as in Problem 16.23,
þ
C1
f ðzÞ dz ¼ 2i a1:
Similarly for pole b; f ðzÞ ¼
bn
ðz bÞn þ þ
b1
ðz bÞ
þ b0 þ b1ðz bÞ þ
þ
C2
f ðzÞ dz ¼ 2i b1
so that
Continuing in this manner, we see that
þ
C
f ðzÞ dz ¼ 2iða1 þ b1 þ Þ ¼ 2i (sum of residues)
16.26. Evaluate
þ
C
ez
dz
ðz 1Þðz þ 3Þ2
where C is given by (a) jzj ¼ 3=2; ðbÞ jzj ¼ 10.
Residue at simple pole z ¼ 1 is lim
z!1
ðz 1Þ
ez
ðz 1Þðz þ 3Þ2
¼
e
16
Residue at double pole z ¼ 3 is
lim
z!3
d
dz
ðz þ 3Þ2 ez
ðz 1Þðz þ 3Þ2
¼ lim
z!3
ðz 1Þez
ez
ðz 1Þ2
¼
5e3
16
(a) Since jzj ¼ 3=2 encloses only the pole z ¼ 1,
the required integral ¼ 2i
e
16
¼
ie
8
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 411
Fig. 16-8](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-420-320.jpg)
![i.e.,
ðR
R
dx
x4 þ 1
þ
ð
dz
z4 þ 1
¼
ffiffiffi
2
p
2
ð2Þ
Taking the limit of both sides of (2) as R ! 1 and using the results of Problem 16.28, we have
lim
R!1
ðR
R
dx
x4 þ 1
¼
ð1
1
dx
x4 þ 1
¼
ffiffiffi
2
p
2
Since
ð1
1
dx
x4
þ 1
¼ 2
ð1
0
dx
x4
þ 1
; the required integral has the value
ffiffiffi
2
p
4
:
16.30. Show that
ð1
1
x2
dx
ðx2
þ 1Þ2
ðx2
þ 2x þ 2Þ
¼
7
50
:
The poles of
z2
ðz2 þ 1Þ2
ðz2 þ 2z þ 2Þ
enclosed by the contour C of Problem 16.27 are z ¼ i of order 2 and
z ¼ 1 þ i of order 1.
Residue at z ¼ i is lim
z!i
d
dz
ðz iÞ2 z2
ðz þ iÞ2
ðz iÞ2
ðz2 þ 2z þ 2Þ
( )
¼
9i 12
100
:
Residue at z ¼ 1 þ i is lim
z!1þi
ðz þ 1 iÞ
z2
ðz2 þ 1Þ2
ðz þ 1 iÞðz þ 1 þ iÞ
¼
3 4i
25
þ
C
z2
dz
ðz2
þ 1Þ2
ðz2
þ 2z þ 2Þ
¼ 2i
9i 12
100
þ
3 4i
25
¼
7
50
Then
ðR
R
x2
dx
ðx2 þ 1Þ2
ðx2 þ 2x þ 2Þ
þ
ð
z2
dz
ðz2 þ 1Þ2
ðz2 þ 2z þ 2Þ
¼
7
50
or
Taking the limit as R ! 1 and noting that the second integral approaches zero by Problem 16.27, we
obtain the required result.
16.31. Evaluate
ð2
0
d
5 þ 3 sin
.
Let z ¼ ei
. Then sin ¼
ei
ei
2i
¼
z z1
2i
, dz ¼ iei
d ¼ iz d so that
ð2
0
d
5 þ 3 sin
¼
þ
C
dz=iz
5 þ 3
z z1
2i
! ¼
þ
C
2 dz
3z2 þ 10iz 3
where C is the circle of unit radius with center at the origin, as shown in Fig. 16-10 below.
The poles of
2
3z2 þ 10iz 3
are the simple poles
z ¼
10i
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100 þ 36
p
6
¼
10i 8i
6
¼ 3i; i=3:
Only i=3 lies inside C.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 413
Fig. 16-10](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-422-320.jpg)
![16.34. Show that
ð1
0
cos mx
x2 þ 1
dx ¼
2
em
; m 0.
Consider
þ
C
eimz
z2 þ 1
dz where C is the contour of Problem 16.27.
The integrand has simple poles at z ¼ i, but only z ¼ i lies within C.
Residue at z ¼ i is lim
z!i
ðz iÞ
eimz
ðz iÞðz þ iÞ
¼
em
2i
:
þ
C
eimz
z2 þ 1
dz ¼ 2i
em
2i
¼ em
Then
ðR
R
eimx
x2 þ 1
dx þ
ð
eimz
z2 þ 1
dz ¼ em
or
ðR
R
cos mx
x2 þ 1
dx þ i
ðR
R
sin mx
x2 þ 1
dx þ
ð
eimz
z2 þ 1
dz ¼ em
i.e.,
and so
2
ðR
0
cos mx
x2
þ 1
dx þ
ð
eimz
z2
þ 1
dz ¼ em
Taking the limit as R ! 1 and using Problem 16.33 to show that the integral around approaches
zero, we obtain the required result.
16.35. Show that
ð1
0
sin x
x
dx ¼
2
.
The method of Problem 16.34 leads us to consider the integral of eiz
=z around the contour of Problem
16.27. However, since z ¼ 0 lies on this path of integration and since we cannot integrate through a
singularity, we modify that contour by indenting the path at z ¼ 0, as shown in Fig. 16-11, which we call
contour C0
or ABDEFGHJA.
Since z ¼ 0 is outside C0
, we have
ð
C0
eiz
z
dz ¼ 0
or
ðr
R
eix
x
dx þ
ð
HJA
eiz
z
dz þ
ðR
r
eix
x
dx þ
ð
BDEFG
eiz
z
dz ¼ 0
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 415
Fig. 16-11](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-424-320.jpg)
![(b) The slope of the tangent to the curve v2
¼ 4ð1 þ uÞ at ð3; 4Þ is m1 ¼
dv
du ð3;4Þ
¼
2
v ð3;4Þ
¼
1
2
.
The slope of the tangent to the curve u2
¼ 2v þ 1 at ð3; 4Þ is m2 ¼
dv
du ð3;4Þ
¼ u ¼ 3.
Then the angle between the two curves at A0
is given by
tan ¼
m2 m1
1 þ m1m2
¼
3 1
2
1 þ ð3Þð1
2Þ
¼ 1; and ¼ =4
Similarly, we can show that the angle between A0
C0
and B0
C0
is =4, while the angle between A0
B0
and B0
C0
is =2. Therefore, the angles of the curvilinear triangle are equal to the corresponding ones of
the given triangle. In general, if w ¼ f ðzÞ is a transformation where f ðzÞ is analytic, the angle between
two curves in the z plane intersecting at z ¼ z0 has the same magnitude and sense (orientation) as the
angle between the images of the two curves, so long as f 0
ðz0Þ 6¼ 0. This property is called the conformal
property of analytic functions, and for this reason, the transformation w ¼ f ðzÞ is often called a con-
formal transformation or conformal mapping function.
16.37. Let w ¼
ffiffiffi
z
p
define a transformation from the z plane to the w plane. A point moves counter-
clockwise along the circle jzj ¼ 1. Show that when it has returned to its starting position for the
first time, its image point has not yet returned, but that when it has returned for the second time,
its image point returns for the first time.
Let z ¼ ei
. Then w ¼
ffiffiffi
z
p
¼ ei=2
. Let ¼ 0 correspond to the starting position. Then z ¼ 1 and
w ¼ 1 [corresponding to A and P in Figures 16-13(a) and (b)].
When one complete revolution in the z plane has been made, ¼ 2; z ¼ 1, but w ¼ ei=2
¼ ei
¼ 1, so
the image point has not yet returned to its starting position.
However, after two complete revolutions in the z plane have been made, ¼ 4; z ¼ 1 and
w ¼ ei=2
¼ e2i
¼ 1, so the image point has returned for the first time.
It follows from the above that w is not a single-valued function of z but is a double-valued function of z;
i.e., given z, there are two values of w. If we wish to consider it a single-valued function, we must restrict .
We can, for example, choose 0 @ 2, although other possibilities exist. This represents one branch of
the double-valued function w ¼
ffiffiffi
z
p
. In continuing beyond this interval we are on the second branch, e.g.,
2 @ 4. The point z ¼ 0 about which the rotation is taking place is called a branch point. Equiva-
lently, we can insure that f ðzÞ ¼
ffiffiffi
z
p
will be single-valued by agreeing not to cross the line Ox, called a branch
line.
16.38. Show that
ð1
0
xp1
1 þ x
dx ¼
sin p
; 0 p 1.
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 417
Fig. 16-13](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-426-320.jpg)
![(b) Region in the first quadrant bounded by x2
þ y2
¼ 9, the x-axis and the line y ¼ x.
(c) Interior of ellipse x2
=25 þ y2
=16 ¼ 1.
16.41. Express each function in the form uðx; yÞ þ ivðx; yÞ, where u and v are real.
(a) z2
þ 2iz; ðbÞ z=ð3 þ zÞ; ðcÞ ez2
; ðdÞ lnð1 þ zÞ.
Ans. (a) u ¼ x3
3xy2
2y; v ¼ 3x2
y y2
þ 2x
(b) u ¼
x2
þ 3x þ y2
x2
þ 6x þ y2
þ 9
; v ¼
3y
x2
þ 6x þ y2
þ 9
(c) u ¼ ex2
y2
cos 2xy; v ¼ ex2
y2
sin 2xy
(d) u ¼ 1
2 lnfð1 þ xÞ2
þ y2
g; v ¼ tan1 y
1 þ x
þ 2k; k ¼ 0; 1; 2; . . .
16.42. Prove that (a) lim
z!x0
z2
¼ z2
0; ðbÞ f ðzÞ ¼ z2
is continuous at z ¼ z0 directly from the definition.
16.43. (a) If z ¼ ! is any root of z5
¼ 1 different from 1, prove that all the roots are 1; !; !2
; !3
; !4
.
(b) Show that 1 þ ! þ !2
þ !3
þ !4
¼ 0.
(c) Generalize the results in (a) and (b) to the equation zn
¼ 1.
DERIVATIVES, CAUCHY-RIEMANN EQUATIONS
16.44. (a) If w ¼ f ðzÞ ¼ z þ
1
z
, find
dw
dz
directly from the definition.
(b) For what finite values of z is f ðzÞ nonanalytic?
Ans. ðaÞ 1 1=z2
; ðbÞ z ¼ 0
16.45. Given the function w ¼ z4
. (a) Find real functions u and v such that w ¼ u þ iv. (b) Show that the
Cauchy-Riemann equations hold at all points in the finite z plane. (c) Prove that u and v are harmonic
functions. (d) Determine dw=dz.
Ans: ðaÞ u ¼ x4
6x2
y2
þ y4
; v ¼ 4x3
y 4xy2
ðdÞ 4z3
16.46. Prove that f ðzÞ ¼ zjzj is not analytic anywhere.
16.47. Prove that f ðzÞ ¼
1
z 2
is analytic in any region not including z ¼ 2.
16.48. If the imaginary part of an analytic function is 2xð1 yÞ, determine (a) the real part, (b) the function.
Ans: ðaÞ y2
x2
2y þ c; ðbÞ 2iz z2
þ c, where c is real
16.49. Construct an analytic function f ðzÞ whose real part is ex
ðx cos y þ y sin yÞ and for which f ð0Þ ¼ 1.
Ans: zez
þ 1
16.50. Prove that there is no analytic function whose imaginary part is x2
2y.
16.51. Find f ðzÞ such that f 0
ðzÞ ¼ 4z 3 and f ð1 þ iÞ ¼ 3i.
Ans: f ðzÞ ¼ 2z2
3z þ 3 4i
INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS
16.52. Evaluate
ð3þi
12i
ð2z þ 3Þ dz:
(a) along the path x ¼ 2t þ 1; y ¼ 4t2
t 2 0 @ t @ 1.
(b) along the straight line joining 1 2i and 3 þ i.
(c) along straight lines from 1 2i to 1 þ i and then to 3 þ i.
Ans: 17 þ 19i in all cases
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 419](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-428-320.jpg)
![16.53. Evaluate
ð
C
ðz2
z þ 2Þ dz, where C is the upper half of the circle jzj ¼ 1 tranversed in the positive sense.
Ans: 14=3
16.54. Evaluate
þ
C
z;
2z 5
, where C is the circle (a) jzj ¼ 2; ðbÞ jz 3j ¼ 2:
Ans: ðaÞ 0; ðbÞ 5i=2
16.55. Evaluate
þ
C
z2
ðz þ 2Þðz 1Þ
dz, where C is: (a) a square with vertices at 1 i; 1 þ i; 3 þ i; 3 i;
(b) the circle jz þ ij ¼ 3; (c) the circle jzj ¼
ffiffiffi
2
p
.
Ans: ðaÞ 8i=3 ðbÞ 2i ðcÞ 2i=3
16.56. Evaluate (a)
þ
C
cos z
z 1
dz; ðbÞ
þ
C
ez
þ z
ðz 1Þ4
dz where C is any simple closed curve enclosing z ¼ 1.
Ans: ðaÞ 2i ðbÞ ie=3
16.57. Prove Cauchy’s integral formulas.
[Hint: Use the definition of derivative and then apply mathematical induction.]
SERIES AND SINGULARITIES
16.58. For what values of z does each series converge?
ðaÞ
X
1
n¼1
ðz þ 2Þn
n!
; ðbÞ
X
1
n¼1
nðz iÞn
n þ 1
; ðcÞ
X
1
n¼1
ð1Þn
ðz2
þ 2z þ 2Þ2n
:
Ans: ðaÞ all z (b) jz ij 1 ðcÞ z ¼ 1 i
16.59. Prove that the series
X
1
n¼1
zn
nðn þ 1Þ
is (a) absolutely convergent, (b) uniformly convergent for jzj @ 1.
16.60. Prove that the series
X
1
n¼0
ðz þ iÞn
2n converges uniformly within any circle of radius R such that jz þ ij R 2.
16.61. Locate in the finite z plane all the singularities, if any, of each function and name them:
ðaÞ
z 2
ð2z þ 1Þ4
; ðbÞ
z
ðz 1Þðz þ 2Þ2
; ðcÞ
z2
þ 1
z2 þ 2z þ 2
; ðdÞ cos
1
z
; ðeÞ
sinðz =3Þ
3z
; ð f Þ
cos z
ðz2 þ 4Þ2
:
Ans. (a) z ¼ 1
2, pole of order 4 (d) z ¼ 0, essential singularity
(b) z ¼ 1, simple pole; z ¼ 2, double pole (e) z ¼ =3, removable singularity
(c) simple poles z ¼ 1 i ( f ) z ¼ 2i, double poles
16.62. Find Laurent series about the indicated singularity for each of the following functions, naming the singu-
larity in each case. Indicate the region of convergence of each series.
ðaÞ
cos z
z
; z ¼ ðbÞ z2
e1=z
; z ¼ 0 ðcÞ
z2
ðz 1Þ2
ðz þ 3Þ
; z ¼ 1
Ans: ðaÞ
1
z
þ
z
2!
ðz Þ3
4!
þ
ðz Þ5
6!
; simple pole, all z 6¼
ðbÞ z2
z þ
1
2!
1
3! z
þ
1
4! z2
1
5! z3
þ ; essential singularity, all z 6¼ 0
420 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-429-320.jpg)
![ðcÞ
1
4ðz 1Þ2
þ
7
16ðz 1Þ
þ
9
64
9ðz 1Þ
256
þ ; double pole, 0 jz 1j 4
RESIDUES AND THE RESIDUE THEOREM
16.63. Determine the residues of each function at its poles:
ðaÞ
2z þ 3
z2 4
; ðbÞ
z 3
z3 þ 5z2
; ðcÞ
ezt
ðz 2Þ3
; ðdÞ
z
ðz2
þ 1Þ2
:
Ans. (a) z ¼ 2; 7=4; z ¼ 2; 1=4 (c) z ¼ 2; 1
2 t2
e2t
(b) z ¼ 0; 8=25; z ¼ 5; 8=25 (d) z ¼ i; 0; z ¼ i; 0
16.64. Find the residue of ezt
tan z at the simple pole z ¼ 3=2.
Ans: e3t=2
16.65. Evaluate
þ
C
z2
dz
ðz þ 1Þðz þ 3Þ
, where C is a simple closed curve enclosing all the poles.
Ans: 8i
16.66. If C is a simple closed curve enclosing z ¼ i, show that
þ
C
zezt
ðz2 þ 1Þ2
dz ¼ 1
2 t sin t
16.67. If f ðzÞ ¼ PðzÞ=QðzÞ, where PðzÞ and QðzÞ are polynomials such that the degree of PðzÞ is at least two less than
the degree of QðzÞ, prove that
þ
C
f ðzÞ dz ¼ 0, where C encloses all the poles of f ðzÞ.
EVALUATION OF DEFINITE INTEGRALS
Use contour integration to verify each of the following
16.68.
ð1
0
x2
dx
x4 þ 1
¼
2
ffiffiffi
2
p 16.75.
ð2
0
d
ð2 þ cos Þ2
¼
4
ffiffiffi
3
p
9
16.69.
ð1
1
dx
x6 þ a6
¼
2
3a5
; a 0 16.76.
ð
0
sin2
5 4 cos
d ¼
8
16.70.
ð1
0
dx
ðx2 þ 4Þ2
¼
32
16.77.
ð2
0
d
ð1 þ sin2
Þ2
¼
3
2
ffiffiffi
2
p
16.71.
ð1
0
ffiffiffi
x
p
x3
þ 1
dx ¼
3
16.78.
ð2
0
cos n d
1 2a cos þ a2
¼
2an
1 a2
; n ¼ 0; 1; 2; 3; . . . ; 0 a 1
16.72.
ð1
0
dx
ðx4 þ a4Þ2
¼
3
8
ffiffiffi
2
p a7
; a 0 16.79.
ð2
0
d
ða þ b cos Þ3
¼
ð2a2
þ b2
Þ
ða2 b2Þ5=2
; a jbj
16.73.
ð1
1
dx
ðx2
þ 1Þ2
ðx2
þ 4Þ
¼
9
16.80.
ð1
0
x sin 2x
x2 þ 4
dx ¼
e4
4
16.74.
ð2
0
d
2 cos
¼
2
ffiffiffi
3
p 16.81.
ð1
0
cos 2x
x4 þ 4
dx ¼
e
8
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 421](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-430-320.jpg)
![16.82.
ð1
0
x sin x
ðx2 þ 1Þ2
dx ¼
2
e
4
16.84.
ð1
0
sin2
x
x2
dx ¼
2
16.83.
ð1
0
sin x
xðx2 þ 1Þ2
dx ¼
ð2e 3Þ
4e
16.85.
ð1
0
sin3
x
x3
dx ¼
3
8
16.86.
ð1
0
cos x
cosh x
dx ¼
2 coshð=2Þ
. Hint: Consider
þ
C
eiz
cosh z
dz, where C is a rectangle with vertices at ðR; 0Þ,
ðR; 0Þ; ðR; ÞrðR; Þ. Then let R ! 1:
MISCELLANEOUS PROBLEMS
16.87. If z ¼ ei
and f ðzÞ ¼ uð; Þ þ i vð; Þ, where and are polar coordinates, show that the Cauchy-Rie-
mann equations are
@u
@
¼
1
@v
@
;
@v
@
¼
1
@u
@
16.88. If w ¼ f ðzÞ, where f ðzÞ is analytic, defines a transformation from the z plane to the w plane where z ¼ x þ iy
and w ¼ u þ iv, prove that the Jacobian of the transformation is given by
@ðu; vÞ
@ðx; yÞ
¼ j f 0
ðzÞj2
16.89. Let Fðx; yÞ be transformed to Gðu; vÞ by the transformation w ¼ f ðzÞ. Show that if
@2
F
@x2
þ
@2
F
@y2
¼ 0, then at
all points where f 0
ðzÞ 6¼ 0,
@2
G
@u2
þ
@2
G
@v2
¼ 0.
16.90. Show that by the bilinear transformation w ¼
az þ b
cz þ d
, where ad bc 6¼ 0, circles in the z plane are trans-
formed into circles of the w plane.
16.91. If f ðzÞ is analytic inside and on the circle jz aj ¼ R, prove Cauchy’s inequality, namely
j f ðnÞ
ðaÞj @
n!M
Rn
where j f ðzÞj @ M on the circle. [Hint: Use Cauchy’s integral formulas.]
16.92. Let C1 and C2 be concentric circles having center a and radii r1 and r2, respectively, where r1 r2. If a þ h is
any point in the annular region bounded by C1 and C2, and f ðzÞ is analytic in this region, prove Laurent’s
theorem that
f ða þ hÞ ¼
X
1
1
anhn
an ¼
1
2i
þ
C
f ðzÞ dz
ðz aÞnþ1
where
C being any closed curve in the angular region surrounding C1.
Hint: Write f ða þ hÞ ¼
1
2i
þ
C2
f ðzÞ dz
z ða þ hÞ
1
2i
þ
C1
f ðzÞ dz
z ða þ hÞ
and expand
1
z a h
in two different
ways.
16.93. Find a Laurent series expansion for the function f ðzÞ ¼
z
ðz þ 1Þðz þ 2Þ
which converges for 1 jzj 2 and
diverges elsewhere.
Hint: Write
z
ðz þ 1Þðz þ 2Þ
¼
1
z þ 1
þ
2
z þ 2
¼
1
zð1 þ 1=zÞ
þ
1
1 þ z=2
:
422 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-431-320.jpg)
![Ans:
1
z5
þ
1
z4
1
z3
þ
1
z2
1
z
þ 1
z
2
þ
z2
4
z3
8
þ
16.94. Let
ð1
0
est
FðtÞ dt ¼ f ðsÞ where f ðsÞ is a given rational function with numerator of degree less than that of the
denominator. If C is a simple closed curve enclosing all the poles of f ðsÞ, we can show that
FðtÞ ¼
1
2i
þ
C
ezt
f ðzÞ dz ¼ sum of residues of ezt
f ðzÞ at its poles
Use this result to find FðtÞ if f ðsÞ is (a)
s
s2 þ 1
; ðbÞ
1
s2 þ 2s þ 5
; ðcÞ
s2
þ 1
sðs 1Þ2
; ðdÞ
1
ðs2
þ 1Þ2
and
check results in each case.
[Note that f ðsÞ is the Laplace transform of FðtÞ, and FðtÞ is the inverse Laplace transform of f ðsÞ (see Chapter
12). Extensions to other functions f ðxÞ are possible.]
Ans. (a) cos t; ðbÞ 1
2 et
sin 2t; ðcÞ 1
4 þ 5
2 te2t
þ 3
4 e2t
; ðdÞ 1
2 ðsin t t cos tÞ
CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 423](https://image.slidesharecdn.com/advancedcalculus-schaumsoutlineseries-240401123208-3a41c341/85/ADVANCED-CALCULUS-SCHAUMSOUTLINE-SERIES-pdf-432-320.jpg)
ADVANCED CALCULUS-SCHAUMSOUTLINE SERIES.pdf
- 2. Theory and Problems of ADVANCED CALCULUS Second Edition ROBERT WREDE, Ph.D. MURRAY R. SPIEGEL, Ph.D. Former Professor and Chairman of Mathematics Rensselaer Polytechnic Institute Hartford Graduate Center Schaum’s Outline Series McGRAW-HILL New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto
- 3. Copyright © 2002, 1963 by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. 0-07-139834-1 The material in this eBook also appears in the print version of this title: 0-07-137567-8 All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in cor- porate training programs. For more information, please contact George Hoare, Special Sales, at george_hoare@mcgraw- hill.com or (212) 904-4069. TERMS OF USE This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS”. McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WAR- RANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PAR- TICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any dam- ages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, con- sequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise. DOI: 10.1036/0071398341
- 4. iii A key ingredient in learning mathematics is problem solving. This is the strength, and no doubt the reason for the longevity of Professor Spiegel’s advanced calculus. His collection of solved and unsolved problems remains a part of this second edition. Advanced calculus is not a single theory. However, the various sub-theories, including vector analysis, infinite series, and special functions, have in common a dependency on the fundamental notions of the calculus. An important objective of this second edition has been to modernize terminology and concepts, so that the interrelationships become clearer. For exam- ple, in keeping with present usage fuctions of a real variable are automatically single valued; differentials are defined as linear functions, and the universal character of vector notation and theory are given greater emphasis. Further explanations have been included and, on occasion, the appropriate terminology to support them. The order of chapters is modestly rearranged to provide what may be a more logical structure. A brief introduction is provided for most chapters. Occasionally, a historical note is included; however, for the most part the purpose of the introductions is to orient the reader to the content of the chapters. I thank the staff of McGraw-Hill. Former editor, Glenn Mott, suggested that I take on the project. Peter McCurdy guided me in the process. Barbara Gilson, Jennifer Chong, and Elizabeth Shannon made valuable contributions to the finished product. Joanne Slike and Maureen Walker accomplished the very difficult task of combining the old with the new and, in the process, corrected my errors. The reviewer, Glenn Ledder, was especially helpful in the choice of material and with comments on various topics. ROBERT C. WREDE Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
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- 6. v CHAPTER 1 NUMBERS 1 Sets. Real numbers. Decimal representation of real numbers. Geometric representation of real numbers. Operations with real numbers. Inequal- ities. Absolute value of real numbers. Exponents and roots. Logarithms. Axiomatic foundations of the real number system. Point sets, intervals. Countability. Neighborhoods. Limit points. Bounds. Bolzano- Weierstrass theorem. Algebraic and transcendental numbers. The com- plex number system. Polar form of complex numbers. Mathematical induction. CHAPTER 2 SEQUENCES 23 Definition of a sequence. Limit of a sequence. Theorems on limits of sequences. Infinity. Bounded, monotonic sequences. Least upper bound and greatest lower bound of a sequence. Limit superior, limit inferior. Nested intervals. Cauchy’s convergence criterion. Infinite series. CHAPTER 3 FUNCTIONS, LIMITS, AND CONTINUITY 39 Functions. Graph of a function. Bounded functions. Montonic func- tions. Inverse functions. Principal values. Maxima and minima. Types of functions. Transcendental functions. Limits of functions. Right- and left-hand limits. Theorems on limits. Infinity. Special limits. Continuity. Right- and left-hand continuity. Continuity in an interval. Theorems on continuity. Piecewise continuity. Uniform continuity. CHAPTER 4 DERIVATIVES 65 The concept and definition of a derivative. Right- and left-hand deriva- tives. Differentiability in an interval. Piecewise differentiability. Differ- entials. The differentiation of composite functions. Implicit differentiation. Rules for differentiation. Derivatives of elementary func- tions. Higher order derivatives. Mean value theorems. L’Hospital’s rules. Applications. For more information about this title, click here. Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 7. CHAPTER 5 INTEGRALS 90 Introduction of the definite integral. Measure zero. Properties of definite integrals. Mean value theorems for integrals. Connecting integral and differential calculus. The fundamental theorem of the calculus. General- ization of the limits of integration. Change of variable of integration. Integrals of elementary functions. Special methods of integration. Improper integrals. Numerical methods for evaluating definite integrals. Applications. Arc length. Area. Volumes of revolution. CHAPTER 6 PARTIAL DERIVATIVES 116 Functions of two or more variables. Three-dimensional rectangular coordinate systems. Neighborhoods. Regions. Limits. Iterated limits. Continuity. Uniform continuity. Partial derivatives. Higher order par- tial derivatives. Differentials. Theorems on differentials. Differentiation of composite functions. Euler’s theorem on homogeneous functions. Implicit functions. Jacobians. Partial derivatives using Jacobians. The- orems on Jacobians. Transformation. Curvilinear coordinates. Mean value theorems. CHAPTER 7 VECTORS 150 Vectors. Geometric properties. Algebraic properties of vectors. Linear independence and linear dependence of a set of vectors. Unit vectors. Rectangular (orthogonal unit) vectors. Components of a vector. Dot or scalar product. Cross or vector product. Triple products. Axiomatic approach to vector analysis. Vector functions. Limits, continuity, and derivatives of vector functions. Geometric interpretation of a vector derivative. Gradient, divergence, and curl. Formulas involving r. Vec- tor interpretation of Jacobians, Orthogonal curvilinear coordinates. Gradient, divergence, curl, and Laplacian in orthogonal curvilinear coordinates. Special curvilinear coordinates. CHAPTER 8 APPLICATIONS OF PARTIAL DERIVATIVES 183 Applications to geometry. Directional derivatives. Differentiation under the integral sign. Integration under the integral sign. Maxima and minima. Method of Lagrange multipliers for maxima and minima. Applications to errors. CHAPTER 9 MULTIPLE INTEGRALS 207 Double integrals. Iterated integrals. Triple integrals. Transformations of multiple integrals. The differential element of area in polar coordinates, differential elements of area in cylindrical and spherical coordinates. vi CONTENTS
- 8. CHAPTER 10 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 229 Line integrals. Evaluation of line integrals for plane curves. Properties of line integrals expressed for plane curves. Simple closed curves, simply and multiply connected regions. Green’s theorem in the plane. Condi- tions for a line integral to be independent of the path. Surface integrals. The divergence theorem. Stoke’s theorem. CHAPTER 11 INFINITE SERIES 265 Definitions of infinite series and their convergence and divergence. Fun- damental facts concerning infinite series. Special series. Tests for con- vergence and divergence of series of constants. Theorems on absolutely convergent series. Infinite sequences and series of functions, uniform convergence. Special tests for uniform convergence of series. Theorems on uniformly convergent series. Power series. Theorems on power series. Operations with power series. Expansion of functions in power series. Taylor’s theorem. Some important power series. Special topics. Taylor’s theorem (for two variables). CHAPTER 12 IMPROPER INTEGRALS 306 Definition of an improper integral. Improper integrals of the first kind (unbounded intervals). Convergence or divergence of improper integrals of the first kind. Special improper integers of the first kind. Convergence tests for improper integrals of the first kind. Improper integrals of the second kind. Cauchy principal value. Special improper integrals of the second kind. Convergence tests for improper integrals of the second kind. Improper integrals of the third kind. Improper integrals containing a parameter, uniform convergence. Special tests for uniform convergence of integrals. Theorems on uniformly conver- gent integrals. Evaluation of definite integrals. Laplace transforms. Linearity. Convergence. Application. Improper multiple integrals. CHAPTER 13 FOURIER SERIES 336 Periodic functions. Fourier series. Orthogonality conditions for the sine and cosine functions. Dirichlet conditions. Odd and even functions. Half range Fourier sine or cosine series. Parseval’s identity. Differentia- tion and integration of Fourier series. Complex notation for Fourier series. Boundary-value problems. Orthogonal functions. CONTENTS vii
- 9. CHAPTER 14 FOURIER INTEGRALS 363 The Fourier integral. Equivalent forms of Fourier’s integral theorem. Fourier transforms. CHAPTER 15 GAMMA AND BETA FUNCTIONS 375 The gamma function. Table of values and graph of the gamma function. The beta function. Dirichlet integrals. CHAPTER 16 FUNCTIONS OF A COMPLEX VARIABLE 392 Functions. Limits and continuity. Derivatives. Cauchy-Riemann equa- tions. Integrals. Cauchy’s theorem. Cauchy’s integral formulas. Taylor’s series. Singular points. Poles. Laurent’s series. Branches and branch points. Residues. Residue theorem. Evaluation of definite integrals. INDEX 425 viii CONTENTS
- 10. 1 Numbers Mathematics has its own language with numbers as the alphabet. The language is given structure with the aid of connective symbols, rules of operation, and a rigorous mode of thought (logic). These concepts, which previously were explored in elementary mathematics courses such as geometry, algebra, and calculus, are reviewed in the following paragraphs. SETS Fundamental in mathematics is the concept of a set, class, or collection of objects having specified characteristics. For example, we speak of the set of all university professors, the set of all letters A; B; C; D; . . . ; Z of the English alphabet, and so on. The individual objects of the set are called members or elements. Any part of a set is called a subset of the given set, e.g., A, B, C is a subset of A; B; C; D; . . . ; Z. The set consisting of no elements is called the empty set or null set. REAL NUMBERS The following types of numbers are already familiar to the student: 1. Natural numbers 1; 2; 3; 4; . . . ; also called positive integers, are used in counting members of a set. The symbols varied with the times, e.g., the Romans used I, II, III, IV, . . . The sum a þ b and product a b or ab of any two natural numbers a and b is also a natural number. This is often expressed by saying that the set of natural numbers is closed under the operations of addition and multiplication, or satisfies the closure property with respect to these operations. 2. Negative integers and zero denoted by 1; 2; 3; . . . and 0, respectively, arose to permit solu- tions of equations such as x þ b ¼ a, where a and b are any natural numbers. This leads to the operation of subtraction, or inverse of addition, and we write x ¼ a b. The set of positive and negative integers and zero is called the set of integers. 3. Rational numbers or fractions such as 2 3, 5 4, . . . arose to permit solutions of equations such as bx ¼ a for all integers a and b, where b 6¼ 0. This leads to the operation of division, or inverse of multiplication, and we write x ¼ a=b or a b where a is the numerator and b the denominator. The set of integers is a subset of the rational numbers, since integers correspond to rational numbers where b ¼ 1. 4. Irrational numbers such as ffiffiffi 2 p and are numbers which are not rational, i.e., they cannot be expressed as a=b (called the quotient of a and b), where a and b are integers and b 6¼ 0. The set of rational and irrational numbers is called the set of real numbers. Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 11. DECIMAL REPRESENTATION OF REAL NUMBERS Any real number can be expressed in decimal form, e.g., 17=10 ¼ 1:7, 9=100 ¼ 0:09, 1=6 ¼ 0:16666 . . . . In the case of a rational number the decimal exapnsion either terminates, or if it does not terminate, one or a group of digits in the expansion will ultimately repeat, as for example, in 1 7 ¼ 0:142857 142857 142 . . . . In the case of an irrational number such as ffiffiffi 2 p ¼ 1:41423 . . . or ¼ 3:14159 . . . no such repetition can occur. We can always consider a decimal expansion as unending, e.g., 1.375 is the same as 1.37500000 . . . or 1.3749999 . . . . To indicate recurring decimals we some- times place dots over the repeating cycle of digits, e.g., 1 7 ¼ 0:_ 1 1_ 4 4_ 2 2_ 8 8_ 5 5_ 7 7, 19 6 ¼ 3:1_ 6 6. The decimal system uses the ten digits 0; 1; 2; . . . ; 9. (These symbols were the gift of the Hindus. They were in use in India by 600 A.D. and then in ensuing centuries were transmitted to the western world by Arab traders.) It is possible to design number systems with fewer or more digits, e.g. the binary system uses only two digits 0 and 1 (see Problems 32 and 33). GEOMETRIC REPRESENTATION OF REAL NUMBERS The geometric representation of real numbers as points on a line called the real axis, as in the figure below, is also well known to the student. For each real number there corresponds one and only one point on the line and conversely, i.e., there is a one-to-one (see Fig. 1-1) correspondence between the set of real numbers and the set of points on the line. Because of this we often use point and number interchangeably. (The interchangeability of point and number is by no means self-evident; in fact, axioms supporting the relation of geometry and numbers are necessary. The Cantor–Dedekind Theorem is fundamental.) The set of real numbers to the right of 0 is called the set of positive numbers; the set to the left of 0 is the set of negative numbers, while 0 itself is neither positive nor negative. (Both the horizontal position of the line and the placement of positive and negative numbers to the right and left, respectively, are conventions.) Between any two rational numbers (or irrational numbers) on the line there are infinitely many rational (and irrational) numbers. This leads us to call the set of rational (or irrational) numbers an everywhere dense set. OPERATIONS WITH REAL NUMBERS If a, b, c belong to the set R of real numbers, then: 1. a þ b and ab belong to R Closure law 2. a þ b ¼ b þ a Commutative law of addition 3. a þ ðb þ cÞ ¼ ða þ bÞ þ c Associative law of addition 4. ab ¼ ba Commutative law of multiplication 5. aðbcÞ ¼ ðabÞc Associative law of multiplication 6. aðb þ cÞ ¼ ab þ ac Distributive law 7. a þ 0 ¼ 0 þ a ¼ a, 1 a ¼ a 1 ¼ a 0 is called the identity with respect to addition, 1 is called the identity with respect to multi- plication. 2 NUMBERS [CHAP. 1 _5 _4 _3 _2 _1 0 1 3 4 5 2 1 2 4 3 _ _p p e √2 Fig. 1-1
- 12. 8. For any a there is a number x in R such that x þ a ¼ 0. x is called the inverse of a with respect to addition and is denoted by a. 9. For any a 6¼ 0 there is a number x in R such that ax ¼ 1. x is called the inverse of a with respect to multiplication and is denoted by a1 or 1=a. Convention: For convenience, operations called subtraction and division are defined by a b ¼ a þ ðbÞ and a b ¼ ab1 , respectively. These enable us to operate according to the usual rules of algebra. In general any set, such as R, whose members satisfy the above is called a field. INEQUALITIES If a b is a nonnegative number, we say that a is greater than or equal to b or b is less than or equal to a, and write, respectively, a A b or b % a. If there is no possibility that a ¼ b, we write a b or b a. Geometrically, a b if the point on the real axis corresponding to a lies to the right of the point corresponding to b. EXAMPLES. 3 5 or 5 3; 2 1 or 1 2; x @ 3 means that x is a real number which may be 3 or less than 3. If a, b; and c are any given real numbers, then: 1. Either a b, a ¼ b or a b Law of trichotomy 2. If a b and b c, then a c Law of transitivity 3. If a b, then a þ c b þ c 4. If a b and c 0, then ac bc 5. If a b and c 0, then ac bc ABSOLUTE VALUE OF REAL NUMBERS The absolute value of a real number a, denoted by jaj, is defined as a if a 0, a if a 0, and 0 if a ¼ 0. EXAMPLES. j 5j ¼ 5, j þ 2j ¼ 2, j 3 4 j ¼ 3 4, j ffiffiffi 2 p j ¼ ffiffiffi 2 p , j0j ¼ 0. 1. jabj ¼ jajjbj or jabc . . . mj ¼ jajjbjjcj . . . jmj 2. ja þ bj @ jaj þ jbj or ja þ b þ c þ þ mj @ jaj þ jbj þ jcj þ jmj 3. ja bj A jaj jbj The distance between any two points (real numbers) a and b on the real axis is ja bj ¼ jb aj. EXPONENTS AND ROOTS The product a a . . . a of a real number a by itself p times is denoted by ap , where p is called the exponent and a is called the base. The following rules hold: 1. ap aq ¼ apþq 3. ðap Þr ¼ apr 2. ap aq ¼ apq 4. a b p ¼ ap bp CHAP. 1] NUMBERS 3
- 13. These and extensions to any real numbers are possible so long as division by zero is excluded. In particular, by using 2, with p ¼ q and p ¼ 0, respectively, we are lead to the definitions a0 ¼ 1, aq ¼ 1=aq . If ap ¼ N, where p is a positive integer, we call a a pth root of N written ffiffiffiffi N p p . There may be more than one real pth root of N. For example, since 22 ¼ 4 and ð2Þ2 ¼ 4, there are two real square roots of 4, namely 2 and 2. For square roots it is customary to define ffiffiffiffi N p as positive, thus ffiffiffi 4 p ¼ 2 and then ffiffiffi 4 p ¼ 2. If p and q are positive integers, we define ap=q ¼ ffiffiffiffiffi ap q p . LOGARITHMS If ap ¼ N, p is called the logarithm of N to the base a, written p ¼ loga N. If a and N are positive and a 6¼ 1, there is only one real value for p. The following rules hold: 1. loga MN ¼ loga M þ loga N 2. loga M N ¼ loga M loga N 3. loga Mr ¼ r loga M In practice, two bases are used, base a ¼ 10, and the natural base a ¼ e ¼ 2:71828 . . . . The logarithmic systems associated with these bases are called common and natural, respectively. The common loga- rithm system is signified by log N, i.e., the subscript 10 is not used. For natural logarithms the usual notation is ln N. Common logarithms (base 10) traditionally have been used for computation. Their application replaces multiplication with addition and powers with multiplication. In the age of calculators and computers, this process is outmoded; however, common logarithms remain useful in theory and application. For example, the Richter scale used to measure the intensity of earthquakes is a logarith- mic scale. Natural logarithms were introduced to simplify formulas in calculus, and they remain effective for this purpose. AXIOMATIC FOUNDATIONS OF THE REAL NUMBER SYSTEM The number system can be built up logically, starting from a basic set of axioms or ‘‘self-evident’’ truths, usually taken from experience, such as statements 1–9, Page 2. If we assume as given the natural numbers and the operations of addition and multiplication (although it is possible to start even further back with the concept of sets), we find that statements 1 through 6, Page 2, with R as the set of natural numbers, hold, while 7 through 9 do not hold. Taking 7 and 8 as additional requirements, we introduce the numbers 1; 2; 3; . . . and 0. Then by taking 9 we introduce the rational numbers. Operations with these newly obtained numbers can be defined by adopting axioms 1 through 6, where R is now the set of integers. These lead to proofs of statements such as ð2Þð3Þ ¼ 6, ð4Þ ¼ 4, ð0Þð5Þ ¼ 0, and so on, which are usually taken for granted in elementary mathematics. We can also introduce the concept of order or inequality for integers, and from these inequalities for rational numbers. For example, if a, b, c, d are positive integers, we define a=b c=d if and only if ad bc, with similar extensions to negative integers. Once we have the set of rational numbers and the rules of inequality concerning them, we can order them geometrically as points on the real axis, as already indicated. We can then show that there are points on the line which do not represent rational numbers (such as ffiffiffi 2 p , , etc.). These irrational numbers can be defined in various ways, one of which uses the idea of Dedekind cuts (see Problem 1.34). From this we can show that the usual rules of algebra apply to irrational numbers and that no further real numbers are possible. 4 NUMBERS [CHAP. 1
- 14. POINT SETS, INTERVALS A set of points (real numbers) located on the real axis is called a one-dimensional point set. The set of points x such that a @ x @ b is called a closed interval and is denoted by ½a; b. The set a x b is called an open interval, denoted by ða; bÞ. The sets a x @ b and a @ x b, denoted by ða; b and ½a; bÞ, respectively, are called half open or half closed intervals. The symbol x, which can represent any number or point of a set, is called a variable. The given numbers a or b are called constants. Letters were introduced to construct algebraic formulas around 1600. Not long thereafter, the philosopher-mathematician Rene Descartes suggested that the letters at the end of the alphabet be used to represent variables and those at the beginning to represent constants. This was such a good idea that it remains the custom. EXAMPLE. The set of all x such that jxj 4, i.e., 4 x 4, is represented by ð4; 4Þ, an open interval. The set x a can also be represented by a x 1. Such a set is called an infinite or unbounded interval. Similarly, 1 x 1 represents all real numbers x. COUNTABILITY A set is called countable or denumerable if its elements can be placed in 1-1 correspondence with the natural numbers. EXAMPLE. The even natural numbers 2; 4; 6; 8; . . . is a countable set because of the 1-1 correspondence shown. Given set Natural numbers 2 4 6 8 . . . l l l l 1 2 3 4 . . . A set is infinite if it can be placed in 1-1 correspondence with a subset of itself. An infinite set which is countable is called countable infinite. The set of rational numbers is countable infinite, while the set of irrational numbers or all real numbers is non-countably infinite (see Problems 1.17 through 1.20). The number of elements in a set is called its cardinal number. A set which is countably infinite is assigned the cardinal number Fo (the Hebrew letter aleph-null). The set of real numbers (or any sets which can be placed into 1-1 correspondence with this set) is given the cardinal number C, called the cardinality of the continuuum. NEIGHBORHOODS The set of all points x such that jx aj where 0, is called a neighborhood of the point a. The set of all points x such that 0 jx aj in which x ¼ a is excluded, is called a deleted neighborhood of a or an open ball of radius about a. LIMIT POINTS A limit point, point of accumulation, or cluster point of a set of numbers is a number l such that every deleted neighborhood of l contains members of the set; that is, no matter how small the radius of a ball about l there are points of the set within it. In other words for any 0, however small, we can always find a member x of the set which is not equal to l but which is such that jx lj . By considering smaller and smaller values of we see that there must be infinitely many such values of x. A finite set cannot have a limit point. An infinite set may or may not have a limit point. Thus the natural numbers have no limit point while the set of rational numbers has infinitely many limit points. CHAP. 1] NUMBERS 5
- 15. A set containing all its limit points is called a closed set. The set of rational numbers is not a closed set since, for example, the limit point ffiffiffi 2 p is not a member of the set (Problem 1.5). However, the set of all real numbers x such that 0 @ x @ 1 is a closed set. BOUNDS If for all numbers x of a set there is a number M such that x @ M, the set is bounded above and M is called an upper bound. Similarly if x A m, the set is bounded below and m is called a lower bound. If for all x we have m @ x @ M, the set is called bounded. If M is a number such that no member of the set is greater than M but there is at least one member which exceeds M for every 0, then M is called the least upper bound (l.u.b.) of the set. Similarly if no member of the set is smaller than m m but at least one member is smaller than m m þ for every 0, then m m is called the greatest lower bound (g.l.b.) of the set. BOLZANO–WEIERSTRASS THEOREM The Bolzano–Weierstrass theorem states that every bounded infinite set has at least one limit point. A proof of this is given in Problem 2.23, Chapter 2. ALGEBRAIC AND TRANSCENDENTAL NUMBERS A number x which is a solution to the polynomial equation a0xn þ a1xn1 þ a2xn2 þ þ an1x þ an ¼ 0 ð1Þ where a0 6¼ 0, a1; a2; . . . ; an are integers and n is a positive integer, called the degree of the equation, is called an algebraic number. A number which cannot be expressed as a solution of any polynomial equation with integer coefficients is called a transcendental number. EXAMPLES. 2 3 and ffiffiffi 2 p which are solutions of 3x 2 ¼ 0 and x2 2 ¼ 0, respectively, are algebraic numbers. The numbers and e can be shown to be transcendental numbers. Mathematicians have yet to determine whether some numbers such as e or e þ are algebraic or not. The set of algebraic numbers is a countably infinite set (see Problem 1.23), but the set of transcen- dental numbers is non-countably infinite. THE COMPLEX NUMBER SYSTEM Equations such as x2 þ 1 ¼ 0 have no solution within the real number system. Because these equations were found to have a meaningful place in the mathematical structures being built, various mathematicians of the late nineteenth and early twentieth centuries developed an extended system of numbers in which there were solutions. The new system became known as the complex number system. It includes the real number system as a subset. We can consider a complex number as having the form a þ bi, where a and b are real numbers called the real and imaginary parts, and i ¼ ffiffiffiffiffiffiffi 1 p is called the imaginary unit. Two complex numbers a þ bi and c þ di are equal if and only if a ¼ c and b ¼ d. We can consider real numbers as a subset of the set of complex numbers with b ¼ 0. The complex number 0 þ 0i corresponds to the real number 0. The absolute value or modulus of a þ bi is defined as ja þ bij ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2 p . The complex conjugate of a þ bi is defined as a bi. The complex conjugate of the complex number z is often indicated by z z or z . The set of complex numbers obeys rules 1 through 9 of Page 2, and thus constitutes a field. In performing operations with complex numbers, we can operate as in the algebra of real numbers, replac- ing i2 by 1 when it occurs. Inequalities for complex numbers are not defined. 6 NUMBERS [CHAP. 1
- 16. From the point of view of an axiomatic foundation of complex numbers, it is desirable to treat a complex number as an ordered pair ða; bÞ of real numbers a and b subject to certain operational rules which turn out to be equivalent to those above. For example, we define ða; bÞ þ ðc; dÞ ¼ ða þ c; b þ dÞ, ða; bÞðc; dÞ ¼ ðac bd; ad þ bcÞ, mða; bÞ ¼ ðma; mbÞ, and so on. We then find that ða; bÞ ¼ að1; 0Þ þ bð0; 1Þ and we associate this with a þ bi, where i is the symbol for ð0; 1Þ. POLAR FORM OF COMPLEX NUMBERS If real scales are chosen on two mutually perpendicular axes X 0 OX and Y 0 OY (the x and y axes) as in Fig. 1-2 below, we can locate any point in the plane determined by these lines by the ordered pair of numbers ðx; yÞ called rectangular coordinates of the point. Examples of the location of such points are indicated by P, Q, R, S, and T in Fig. 1-2. Since a complex number x þ iy can be considered as an ordered pair ðx; yÞ, we can represent such numbers by points in an xy plane called the complex plane or Argand diagram. Referring to Fig. 1-3 above we see that x ¼ cos , y ¼ sin where ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p ¼ jx þ iyj and , called the amplitude or argument, is the angle which line OP makes with the positive x axis OX. It follows that z ¼ x þ iy ¼ ðcos þ i sin Þ ð2Þ called the polar form of the complex number, where and are called polar coordintes. It is sometimes convenient to write cis instead of cos þ i sin . If z1 ¼ x1 þ iyi ¼ 1ðcos 1 þ i sin 1Þ and z2 ¼ x2 þ iy2 ¼ 2ðcos 2 þ i sin 2Þ and by using the addition formulas for sine and cosine, we can show that z1z2 ¼ 12fcosð1 þ 2Þ þ i sinð1 þ 2Þg ð3Þ z1 z2 ¼ 1 2 fcosð1 2Þ þ i sinð1 2Þg ð4Þ zn ¼ fðcos þ i sin Þgn ¼ n ðcos n þ i sin nÞ ð5Þ where n is any real number. Equation (5) is sometimes called De Moivre’s theorem. We can use this to determine roots of complex numbers. For example, if n is a positive integer, z1=n ¼ fðcos þ i sin Þg1=n ð6Þ ¼ 1=n cos þ 2k n þ i sin þ 2k n k ¼ 0; 1; 2; 3; . . . ; n 1 CHAP. 1] NUMBERS 7 _4 X¢ X _3 _2 _1 1 2 3 4 4 Y Y¢ 3 2 1 _1 _2 _3 O Q(_3, 3) S(2, _2) P(3, 4) T(2.5, 0) R(_2.5, _1.5) Fig. 1-2 X′ X O Y Y′ ρ φ y x P(x, y) Fig. 1-3
- 17. from which it follows that there are in general n different values of z1=n . Later (Chap. 11) we will show that ei ¼ cos þ i sin where e ¼ 2:71828 . . . . This is called Euler’s formula. MATHEMATICAL INDUCTION The principle of mathematical induction is an important property of the positive integers. It is especially useful in proving statements involving all positive integers when it is known for example that the statements are valid for n ¼ 1; 2; 3 but it is suspected or conjectured that they hold for all positive integers. The method of proof consists of the following steps: 1. Prove the statement for n ¼ 1 (or some other positive integer). 2. Assume the statement true for n ¼ k; where k is any positive integer. 3. From the assumption in 2 prove that the statement must be true for n ¼ k þ 1. This is part of the proof establishing the induction and may be difficult or impossible. 4. Since the statement is true for n ¼ 1 [from step 1] it must [from step 3] be true for n ¼ 1 þ 1 ¼ 2 and from this for n ¼ 2 þ 1 ¼ 3, and so on, and so must be true for all positive integers. (This assumption, which provides the link for the truth of a statement for a finite number of cases to the truth of that statement for the infinite set, is called ‘‘The Axiom of Mathematical Induc- tion.’’) Solved Problems OPERATIONS WITH NUMBERS 1.1. If x ¼ 4, y ¼ 15, z ¼ 3, p ¼ 2 3, q ¼ 1 6, and r ¼ 3 4, evaluate (a) x þ ðy þ zÞ, (b) ðx þ yÞ þ z, (c) pðqrÞ, (d) ðpqÞr, (e) xðp þ qÞ (a) x þ ðy þ zÞ ¼ 4 þ ½15 þ ð3Þ ¼ 4 þ 12 ¼ 16 (b) ðx þ yÞ þ z ¼ ð4 þ 15Þ þ ð3Þ ¼ 19 3 ¼ 16 The fact that (a) and (b) are equal illustrates the associative law of addition. (c) pðqrÞ ¼ 2 3 fð 1 6Þð3 4Þg ¼ ð2 3Þð 3 24Þ ¼ ð2 3Þð 1 8Þ ¼ 2 24 ¼ 1 12 (d) ðpqÞr ¼ fð2 3Þð 1 6Þgð3 4Þ ¼ ð 2 18Þð3 4Þ ¼ ð 1 9Þð3 4Þ ¼ 3 36 ¼ 1 12 The fact that (c) and (d) are equal illustrates the associative law of multiplication. (e) xðp þ qÞ ¼ 4ð2 3 1 6Þ ¼ 4ð4 6 1 6Þ ¼ 4ð3 6Þ ¼ 12 6 ¼ 2 Another method: xðp þ qÞ ¼ xp þ xq ¼ ð4Þð2 3Þ þ ð4Þð 1 6Þ ¼ 8 3 4 6 ¼ 8 3 2 3 ¼ 6 3 ¼ 2 using the distributive law. 1.2. Explain why we do not consider (a) 0 0 (b) 1 0 as numbers. (a) If we define a=b as that number (if it exists) such that bx ¼ a, then 0=0 is that number x such that 0x ¼ 0. However, this is true for all numbers. Since there is no unique number which 0/0 can represent, we consider it undefined. (b) As in (a), if we define 1/0 as that number x (if it exists) such that 0x ¼ 1, we conclude that there is no such number. Because of these facts we must look upon division by zero as meaningless. 8 NUMBERS [CHAP. 1
- 18. 1.3. Simplify x2 5x þ 6 x2 2x 3 . x2 5x þ 6 x2 2x 3 ¼ ðx 3Þðx 2Þ ðx 3Þðx þ 1Þ ¼ x 2 x þ 1 provided that the cancelled factor ðx 3Þ is not zero, i.e., x 6¼ 3. For x ¼ 3 the given fraction is undefined. RATIONAL AND IRRATIONAL NUMBERS 1.4. Prove that the square of any odd integer is odd. Any odd integer has the form 2m þ 1. Since ð2m þ 1Þ2 ¼ 4m2 þ 4m þ 1 is 1 more than the even integer 4m2 þ 4m ¼ 2ð2m2 þ 2mÞ, the result follows. 1.5. Prove that there is no rational number whose square is 2. Let p=q be a rational number whose square is 2, where we assume that p=q is in lowest terms, i.e., p and q have no common integer factors except 1 (we sometimes call such integers relatively prime). Then ðp=qÞ2 ¼ 2, p2 ¼ 2q2 and p2 is even. From Problem 1.4, p is even since if p were odd, p2 would be odd. Thus p ¼ 2m: Substituting p ¼ 2m in p2 ¼ 2q2 yields q2 ¼ 2m2 , so that q2 is even and q is even. Thus p and q have the common factor 2, contradicting the original assumption that they had no common factors other than 1. By virtue of this contradiction there can be no rational number whose square is 2. 1.6. Show how to find rational numbers whose squares can be arbitrarily close to 2. We restrict ourselves to positive rational numbers. Since ð1Þ2 ¼ 1 and ð2Þ2 ¼ 4, we are led to choose rational numbers between 1 and 2, e.g., 1:1; 1:2; 1:3; . . . ; 1:9. Since ð1:4Þ2 ¼ 1:96 and ð1:5Þ2 ¼ 2:25, we consider rational numbers between 1.4 and 1.5, e.g., 1:41; 1:42; . . . ; 1:49: Continuing in this manner we can obtain closer and closer rational approximations, e.g. ð1:414213562Þ2 is less than 2 while ð1:414213563Þ2 is greater than 2. 1.7. Given the equation a0xn þ a1xn1 þ þ an ¼ 0, where a0; a1; . . . ; an are integers and a0 and an 6¼ 0. Show that if the equation is to have a rational root p=q, then p must divide an and q must divide a0 exactly. Since p=q is a root we have, on substituting in the given equation and multiplying by qn , the result a0pn þ a1pn1 q þ a2pn2 q2 þ þ an1pqn1 þ anqn ¼ 0 ð1Þ or dividing by p, a0pn1 þ a1pn2 q þ þ an1qn1 ¼ anqn p ð2Þ Since the left side of (2) is an integer, the right side must also be an integer. Then since p and q are relatively prime, p does not divide qn exactly and so must divide an. In a similar manner, by transposing the first term of (1) and dividing by q, we can show that q must divide a0. 1.8. Prove that ffiffiffi 2 p þ ffiffiffi 3 p cannot be a rational number. If x ¼ ffiffiffi 2 p þ ffiffiffi 3 p , then x2 ¼ 5 þ 2 ffiffiffi 6 p , x2 5 ¼ 2 ffiffiffi 6 p and squaring, x4 10x2 þ 1 ¼ 0. The only possible rational roots of this equation are 1 by Problem 1.7, and these do not satisfy the equation. It follows that ffiffiffi 2 p þ ffiffiffi 3 p , which satisfies the equation, cannot be a rational number. CHAP. 1] NUMBERS 9
- 19. 1.9. Prove that between any two rational numbers there is another rational number. The set of rational numbers is closed under the operations of addition and division (non-zero denominator). Therefore, a þ b 2 is rational. The next step is to guarantee that this value is between a and b. To this purpose, assume a b. (The proof would proceed similarly under the assumption b a.) Then 2a a þ b, thus a a þ b 2 and a þ b 2b, therefore a þ b 2 b. INEQUALITIES 1.10. For what values of x is x þ 3ð2 xÞ A 4 x? x þ 3ð2 xÞ A 4 x when x þ 6 3x A 4 x, 6 2x A 4 x, 6 4 A 2x x, 2 A x, i.e. x @ 2. 1.11. For what values of x is x2 3x 2 10 2x? The required inequality holds when x2 3x 2 10 þ 2x 0; x2 x 12 0 or ðx 4Þðx þ 3Þ 0 This last inequality holds only in the following cases. Case 1: x 4 0 and x þ 3 0, i.e., x 4 and x 3. This is impossible, since x cannot be both greater than 4 and less than 3. Case 2: x 4 0 and x þ 3 0, i.e. x 4 and x 3. This is possible when 3 x 4. Thus the inequality holds for the set of all x such that 3 x 4. 1.12. If a A 0 and b A 0, prove that 1 2 ða þ bÞ A ffiffiffiffiffi ab p . The statement is self-evident in the following cases (1) a ¼ b, and (2) either or both of a and b zero. For both a and b positive and a 6¼ b, the proof is by contradiction. Assume to the contrary of the supposition that 1 2 ða þ bÞ ffiffiffiffiffi ab p then 1 4 ða2 þ 2ab þ b2 Þ ab. That is, a2 2ab þ b2 ¼ ða bÞ2 0. Since the left member of this equation is a square, it cannot be less than zero, as is indicated. Having reached this contradiction, we may conclude that our assumption is incorrect and that the original assertion is true. 1.13. If a1; a2; . . . ; an and b1; b2; . . . ; bn are any real numbers, prove Schwarz’s inequality ða1b1 þ a2b2 þ þ anbnÞ2 @ ða2 1 þ a2 2 þ þ a2 nÞðb2 1 þ b2 2 þ þ b2 nÞ For all real numbers , we have ða1 þ b1Þ2 þ ða2 þ b2Þ2 þ þ ðan þ bnÞ2 A 0 Expanding and collecting terms yields A2 2 þ 2C þ B2 A 0 ð1Þ where A2 ¼ a2 1 þ a2 2 þ þ a2 n; B2 ¼ b2 1 þ b2 2 þ þ b2 n; C ¼ a1b1 þ a2b2 þ þ anbn ð2Þ The left member of (1) is a quadratic form in . Since it never is negative, its discriminant, 4C2 4A2 B2 , cannot be positive. Thus C2 A2 B2 0 or C2 A2 B2 This is the inequality that was to be proved. 1.14. Prove that 1 2 þ 1 4 þ 1 8 þ þ 1 2n1 1 for all positive integers n 1. 10 NUMBERS [CHAP. 1
- 20. Sn ¼ 1 2 þ 1 4 þ 1 8 þ þ 1 2n1 Let 1 2 Sn ¼ 1 4 þ 1 8 þ þ 1 2n1 þ 1 2n Then 1 2 Sn ¼ 1 2 1 2n : Thus Sn ¼ 1 1 2n1 1 for all n: Subtracting, EXPONENTS, ROOTS, AND LOGARITHMS 1.15. Evaluate each of the following: ðaÞ 34 38 314 ¼ 34þ8 314 ¼ 34þ814 ¼ 32 ¼ 1 32 ¼ 1 9 ðbÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 106 Þð4 102 Þ 8 105 s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 4 8 106 102 105 s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2:5 109 p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25 1010 p ¼ 5 105 or 0:00005 ðcÞ log2=3 27 8 ¼ x: Then 2 3 x ¼ 27 8 ¼ 3 2 3 ¼ 2 3 3 or x ¼ 3 ðdÞ ðloga bÞðlogb aÞ ¼ u: Then loga b ¼ x; logb a ¼ y assuming a; b 0 and a; b 6¼ 1: Then ax ¼ b, by ¼ a and u ¼ xy. Since ðax Þy ¼ axy ¼ by ¼ a we have axy ¼ a1 or xy ¼ 1 the required value. 1.16. If M 0, N 0; and a 0 but a 6¼ 1, prove that loga M N ¼ loga M loga N. Let loga M ¼ x, loga N ¼ y. Then ax ¼ M, ay ¼ N and so M N ¼ ax ay ¼ axy or loga M N ¼ x y ¼ loga M loga N COUNTABILITY 1.17. Prove that the set of all rational numbers between 0 and 1 inclusive is countable. Write all fractions with denominator 2, then 3; . . . considering equivalent fractions such as 1 2 ; 2 4 ; 3 6 ; . . . no more than once. Then the 1-1 correspondence with the natural numbers can be accomplished as follows: Rational numbers Natural numbers 0 1 1 2 1 3 2 3 1 4 3 4 1 5 2 5 . . . l l l l l l l l l 1 2 3 4 5 6 7 8 9 . . . Thus the set of all rational numbers between 0 and 1 inclusive is countable and has cardinal number Fo (see Page 5). 1.18. If A and B are two countable sets, prove that the set consisting of all elements from A or B (or both) is also countable. Since A is countable, there is a 1-1 correspondence between elements of A and the natural numbers so that we can denote these elements by a1; a2; a3; . . . . Similarly, we can denote the elements of B by b1; b2; b3; . . . . Case 1: Suppose elements of A are all distinct from elements of B. Then the set consisting of elements from A or B is countable, since we can establish the following 1-1 correspondence. CHAP. 1] NUMBERS 11
- 21. A or B Natural numbers a1 b1 a2 b2 a3 b3 . . . l l l l l l 1 2 3 4 5 6 . . . Case 2: If some elements of A and B are the same, we count them only once as in Problem 1.17. Then the set of elements belonging to A or B (or both) is countable. The set consisting of all elements which belong to A or B (or both) is often called the union of A and B, denoted by A [ B or A þ B. The set consisting of all elements which are contained in both A and B is called the intersection of A and B, denoted by A B or AB. If A and B are countable, so is A B. The set consisting of all elements in A but not in B is written A B. If we let B B be the set of elements which are not in B, we can also write A B ¼ A B B. If A and B are countable, so is A B. 1.19. Prove that the set of all positive rational numbers is countable. Consider all rational numbers x 1. With each such rational number we can associate one and only one rational number 1=x in ð0; 1Þ, i.e., there is a one-to-one correspondence between all rational numbers 1 and all rational numbers in ð0; 1Þ. Since these last are countable by Problem 1.17, it follows that the set of all rational numbers 1 is also countable. From Problem 1.18 it then follows that the set consisting of all positive rational numbers is countable, since this is composed of the two countable sets of rationals between 0 and 1 and those greater than or equal to 1. From this we can show that the set of all rational numbers is countable (see Problem 1.59). 1.20. Prove that the set of all real numbers in ½0; 1 is non-countable. Every real number in ½0; 1 has a decimal expansion :a1a2a3 . . . where a1; a2; . . . are any of the digits 0; 1; 2; . . . ; 9. We assume that numbers whose decimal expansions terminate such as 0.7324 are written 0:73240000 . . . and that this is the same as 0:73239999 . . . . If all real numbers in ½0; 1 are countable we can place them in 1-1 correspondence with the natural numbers as in the following list: 1 2 3 . . . $ $ $ 0:a11a12a13a14 . . . 0:a21a22a23a24 . . . 0:a31a32a33a34 . . . . . . We now form a number 0:b1b2b3b4 . . . where b1 6¼ a11; b2 6¼ a22; b3 6¼ a33; b4 6¼ a44; . . . and where all b’s beyond some position are not all 9’s. This number, which is in ½0; 1 is different from all numbers in the above list and is thus not in the list, contradicting the assumption that all numbers in ½0; 1 were included. Because of this contradiction it follows that the real numbers in ½0; 1 cannot be placed in 1-1 corre- spondence with the natural numbers, i.e., the set of real numbers in ½0; 1 is non-countable. LIMIT POINTS, BOUNDS, BOLZANO–WEIERSTRASS THEOREM 1.21. (a) Prove that the infinite sets of numbers 1; 1 2 ; 1 3 ; 1 4 ; . . . is bounded. (b) Determine the least upper bound (l.u.b.) and greatest lower bound (g.l.b.) of the set. (c) Prove that 0 is a limit point of the set. (d) Is the set a closed set? (e) How does this set illustrate the Bolzano–Weierstrass theorem? (a) Since all members of the set are less than 2 and greater than 1 (for example), the set is bounded; 2 is an upper bound, 1 is a lower bound. We can find smaller upper bounds (e.g., 3 2) and larger lower bounds (e.g., 1 2). 12 NUMBERS [CHAP. 1
- 22. (b) Since no member of the set is greater than 1 and since there is at least one member of the set (namely 1) which exceeds 1 for every positive number , we see that 1 is the l.u.b. of the set. Since no member of the set is less than 0 and since there is at least one member of the set which is less than 0 þ for every positive (we can always choose for this purpose the number 1=n where n is a positive integer greater than 1=), we see that 0 is the g.l.b. of the set. (c) Let x be any member of the set. Since we can always find a number x such that 0 jxj for any positive number (e.g. we can always pick x to be the number 1=n where n is a positive integer greater than 1=), we see that 0 is a limit point of the set. To put this another way, we see that any deleted neighborhood of 0 always includes members of the set, no matter how small we take 0. (d) The set is not a closed set since the limit point 0 does not belong to the given set. (e) Since the set is bounded and infinite it must, by the Bolzano–Weierstrass theorem, have at least one limit point. We have found this to be the case, so that the theorem is illustrated. ALGEBRAIC AND TRANSCENDENTAL NUMBERS 1.22. Prove that ffiffiffi 2 3 p þ ffiffiffi 3 p is an algebraic number. Let x ¼ ffiffiffi 2 3 p þ ffiffiffi 3 p . Then x ffiffiffi 3 p ¼ ffiffiffi 2 3 p . Cubing both sides and simplifying, we find x3 þ 9x 2 ¼ 3 ffiffiffi 3 p ðx2 þ 1Þ. Then squaring both sides and simplifying we find x6 9x4 4x3 þ 27x2 þ 36x 23 ¼ 0. Since this is a polynomial equation with integral coefficients it follows that ffiffiffi 2 3 p þ ffiffiffi 3 p , which is a solution, is an algebraic number. 1.23. Prove that the set of all algebraic numbers is a countable set. Algebraic numbers are solutions to polynomial equations of the form a0xn þ a1xn1 þ þ an ¼ 0 where a0; a1; . . . ; an are integers. Let P ¼ ja0j þ ja1j þ þ janj þ n. For any given value of P there are only a finite number of possible polynomial equations and thus only a finite number of possible algebraic numbers. Write all algebraic numbers corresponding to P ¼ 1; 2; 3; 4; . . . avoiding repetitions. Thus, all algebraic numbers can be placed into 1-1 correspondence with the natural numbers and so are countable. COMPLEX NUMBERS 1.24. Perform the indicated operations. (a) ð4 2iÞ þ ð6 þ 5iÞ ¼ 4 2i 6 þ 5i ¼ 4 6 þ ð2 þ 5Þi ¼ 2 þ 3i (b) ð7 þ 3iÞ ð2 4iÞ ¼ 7 þ 3i 2 þ 4i ¼ 9 þ 7i (c) ð3 2iÞð1 þ 3iÞ ¼ 3ð1 þ 3iÞ 2ið1 þ 3iÞ ¼ 3 þ 9i 2i 6i2 ¼ 3 þ 9i 2i þ 6 ¼ 9 þ 7i ðdÞ 5 þ 5i 4 3i ¼ 5 þ 5i 4 3i 4 þ 3i 4 þ 3i ¼ ð5 þ 5iÞð4 þ 3iÞ 16 9i2 ¼ 20 15i þ 20i þ 15i2 16 þ 9 ¼ 35 þ 5i 25 ¼ 5ð7 þ iÞ 25 ¼ 7 5 þ 1 5 i ðeÞ i þ i2 þ i3 þ i4 þ i5 1 þ i ¼ i 1 þ ði2 ÞðiÞ þ ði2 Þ2 þ ði2 Þ2 i 1 þ i ¼ i 1 i þ 1 þ i 1 þ i ¼ i 1 þ i 1 i 1 i ¼ i i2 1 i2 ¼ i þ 1 2 ¼ 1 2 þ 1 2 i ð f Þ j3 4ijj4 þ 3ij ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð3Þ2 þ ð4Þ2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð4Þ2 þ ð3Þ2 q ¼ ð5Þð5Þ ¼ 25 CHAP. 1] NUMBERS 13
- 23. ðgÞ 1 1 þ 3i 1 1 3i ¼ 1 3i 1 9i2 1 þ 3i 1 9i2 ¼ 6i 10 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð0Þ2 þ 6 10 2 s ¼ 3 5 1.25. If z1 and z2 are two complex numbers, prove that jz1z2j ¼ jz1jjz2j. Let z1 ¼ x1 þ iy1 , z2 ¼ x2 þ iy2. Then jz1z2j ¼ jðx1 þ iy1Þðx2 þ iy2Þj ¼ jx1x2 y1y2 þ iðx1y2 þ x2y1Þj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx1x2 y1y2Þ2 þ ðx1y2 þ x2y1Þ2 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1x2 2 þ y2 1y2 2 þ x2 1y2 2 þ x2 2y2 1 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx2 1 þ y2 1Þðx2 2 þ y2 2Þ q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 þ y2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 2 þ y2 2 q ¼ jx1 þ iy1jjx2 þ iy2j ¼ jz1jjz2j: 1.26. Solve x3 2x 4 ¼ 0. The possible rational roots using Problem 1.7 are 1, 2, 4. By trial we find x ¼ 2 is a root. Then the given equation can be written ðx 2Þðx2 þ 2x þ 2Þ ¼ 0. The solutions to the quadratic equation ax2 þ bx þ c ¼ 0 are x ¼ b ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2 4ac p 2a . For a ¼ 1, b ¼ 2, c ¼ 2 this gives x ¼ 2 ffiffiffiffiffiffiffiffiffiffiffi 4 8 p 2 ¼ 2 ffiffiffiffiffiffiffi 4 p 2 ¼ 2 2i 2 ¼ 1 i. The set of solutions is 2, 1 þ i, 1 i. POLAR FORM OF COMPLEX NUMBERS 1.27. Express in polar form (a) 3 þ 3i, (b) 1 þ ffiffiffi 3 p i, (c) 1, (d) 2 2 ffiffiffi 3 p i. See Fig. 1-4. (a) Amplitude ¼ 458 ¼ =4 radians. Modulus ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 32 þ 32 p ¼ 3 ffiffiffi 2 p . Then 3 þ 3i ¼ ðcos þ i sin Þ ¼ 3 ffiffiffi 2 p ðcos =4 þ i sin =4Þ ¼ 3 ffiffiffi 2 p cis =4 ¼ 3 ffiffiffi 2 p ei=4 (b) Amplitude ¼ 1208 ¼ 2=3 radians. Modulus ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1Þ2 þ ð ffiffiffi 3 p Þ2 q ¼ ffiffiffi 4 p ¼ 2. Then 1 þ 3 ffiffiffi 3 p i ¼ 2ðcos 2=3 þ i sin 2=3Þ ¼ 2 cis 2=3 ¼ 2e2i=3 (c) Amplitude ¼ 1808 ¼ radians. Modulus ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1Þ2 þ ð0Þ2 q ¼ 1. Then 1 ¼ 1ðcos þ i sin Þ ¼ cis ¼ ei (d) Amplitude ¼ 2408 ¼ 4=3 radians. Modulus ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þ2 þ ð2 ffiffiffi 3 p Þ2 q ¼ 4. Then 2 2 ffiffiffi 3 p ¼ 4ðcos 4=3 þ i sin 4=3Þ ¼ 4 cis 4=3 ¼ 4e4i=3 14 NUMBERS [CHAP. 1 45° 120° 180° 240° 3 3 3 √ 2 _2√3 √3 (a) (b) (c) (d) 2 _1 _1 _2 4 Fig. 1-4
- 24. 1.28. Evaluate (a) ð1 þ ffiffiffi 3 p iÞ10 , (b) ð1 þ iÞ1=3 . (a) By Problem 1.27(b) and De Moivre’s theorem, ð1 þ ffiffiffi 3 p iÞ10 ¼ ½2ðcos 2=3 þ i sin 2=3Þ10 ¼ 210 ðcos 20=3 þ i sin 20=3Þ ¼ 1024½cosð2=3 þ 6Þ þ i sinð2=3 þ 6Þ ¼ 1024ðcos 2=3 þ i sin 2=3Þ ¼ 1024 1 2 þ 1 2 ffiffiffi 3 p i ¼ 512 þ 512 ffiffiffi 3 p i (b) 1 þ i ¼ ffiffiffi 2 p ðcos 1358 þ i sin 1358Þ ¼ ffiffiffi 2 p ½cosð1358 þ k 3608Þ þ i sinð1358 þ k 3608Þ. Then ð1 þ iÞ1=3 ¼ ð ffiffiffi 2 p Þ1=3 cos 1358 þ k 3608 3 þ i sin 1358 þ k 3608 3 The results for k ¼ 0; 1; 2 are ffiffiffi 2 6 p ðcos 458 þ i sin 458Þ; ffiffiffi 2 6 p ðcos 1658 þ i sin 1658Þ; ffiffiffi 2 6 p ðcos 2858 þ i sin 2858Þ The results for k ¼ 3; 4; 5; 6; 7; . . . give repetitions of these. These complex roots are represented geometrically in the complex plane by points P1; P2; P3 on the circle of Fig. 1-5. MATHEMATICAL INDUCTION 1.29. Prove that 12 þ 22 þ 33 þ 42 þ þ n2 ¼ 1 6 nðn þ 1Þð2n þ 1Þ. The statement is true for n ¼ 1 since 12 ¼ 1 6 ð1Þð1 þ 1Þð2 1 þ 1Þ ¼ 1. Assume the statement true for n ¼ k. Then 12 þ 22 þ 32 þ þ k2 ¼ 1 6 kðk þ 1Þð2k þ 1Þ Adding ðk þ 1Þ2 to both sides, 12 þ 22 þ 32 þ þ k2 þ ðk þ 1Þ2 ¼ 1 6 kðk þ 1Þð2k þ 1Þ þ ðk þ 1Þ2 ¼ ðk þ 1Þ½1 6 kð2k þ 1Þ þ k þ 1 ¼ 1 6 ðk þ 1Þð2k2 þ 7k þ 6Þ ¼ 1 6 ðk þ 1Þðk þ 2Þð2k þ 3Þ which shows that the statement is true for n ¼ k þ 1 if it is true for n ¼ k. But since it is true for n ¼ 1, it follows that it is true for n ¼ 1 þ 1 ¼ 2 and for n ¼ 2 þ 1 ¼ 3; . . . ; i.e., it is true for all positive integers n. 1.30. Prove that xn yn has x y as a factor for all positive integers n. The statement is true for n ¼ 1 since x1 y1 ¼ x y. Assume the statement true for n ¼ k, i.e., assume that xk yk has x y as a factor. Consider xkþ1 ykþ1 ¼ xkþ1 xk y þ xk y ykþ1 ¼ xk ðx yÞ þ yðxk yk Þ The first term on the right has x y as a factor, and the second term on the right also has x y as a factor because of the above assumption. Thus xkþ1 ykþ1 has x y as a factor if xk yk does. Then since x1 y1 has x y as factor, it follows that x2 y2 has x y as a factor, x3 y3 has x y as a factor, etc. CHAP. 1] NUMBERS 15 P2 P3 P1 165° 285° 45° √2 6 Fig. 1-5
- 25. 1.31. Prove Bernoulli’s inequality ð1 þ xÞn 1 þ nx for n ¼ 2; 3; . . . if x 1, x 6¼ 0. The statement is true for n ¼ 2 since ð1 þ xÞ2 ¼ 1 þ 2x þ x2 1 þ 2x. Assume the statement true for n ¼ k, i.e., ð1 þ xÞk 1 þ kx. Multiply both sides by 1 þ x (which is positive since x 1). Then we have ð1 þ xÞkþ1 ð1 þ xÞð1 þ kxÞ ¼ 1 þ ðk þ 1Þx þ kx2 1 þ ðk þ 1Þx Thus the statement is true for n ¼ k þ 1 if it is true for n ¼ k. But since the statement is true for n ¼ 2, it must be true for n ¼ 2 þ 1 ¼ 3; . . . and is thus true for all integers greater than or equal to 2. Note that the result is not true for n ¼ 1. However, the modified result ð1 þ xÞn A 1 þ nx is true for n ¼ 1; 2; 3; . . . . MISCELLANEOUS PROBLEMS 1.32. Prove that every positive integer P can be expressed uniquely in the form P ¼ a02n þ a12n1 þ a22n2 þ þ an where the a’s are 0’s or 1’s. Dividing P by 2, we have P=2 ¼ a02n1 þ a12n2 þ þ an1 þ an=2. Then an is the remainder, 0 or 1, obtained when P is divided by 2 and is unique. Let P1 be the integer part of P=2. Then P1 ¼ a02n1 þ a12n2 þ þ an1. Dividing P1 by 2 we see that an1 is the remainder, 0 or 1, obtained when P1 is divided by 2 and is unique. By continuing in this manner, all the a’s can be determined as 0’s or 1’s and are unique. 1.33. Express the number 23 in the form of Problem 1.32. The determination of the coefficients can be arranged as follows: 2Þ23 2Þ11 Remainder 1 2Þ5 Remainder 1 2Þ2 Remainder 1 2Þ1 Remainder 0 0 Remainder 1 The coefficients are 1 0 1 1 1. Check: 23 ¼ 1 24 þ 0 23 þ 1 22 þ 1 2 þ 1. The number 10111 is said to represent 23 in the scale of two or binary scale. 1.34. Dedekind defined a cut, section, or partition in the rational number system as a separation of all rational numbers into two classes or sets called L (the left-hand class) and R (the right-hand class) having the following properties: I. The classes are non-empty (i.e. at least one number belongs to each class). II. Every rational number is in one class or the other. III. Every number in L is less than every number in R. Prove each of the following statements: (a) There cannot be a largest number in L and a smallest number in R. (b) It is possible for L to have a largest number and for R to have no smallest number. What type of number does the cut define in this case? (c) It is possible for L to have no largest number and for R to have a smallest number. What type of number does the cut define in this case? 16 NUMBERS [CHAP. 1
- 26. (d) It is possible for L to have no largest number and for R to have no smallest number. What type of number does the cut define in this case? (a) Let a be the largest rational number in L, and b the smallest rational number in R. Then either a ¼ b or a b. We cannot have a ¼ b since by definition of the cut every number in L is less than every number in R. We cannot have a b since by Problem 1.9, 1 2 ða þ bÞ is a rational number which would be greater than a (and so would have to be in R) but less than b (and so would have to be in L), and by definition a rational number cannot belong to both L and R. (b) As an indication of the possibility, let L contain the number 2 3 and all rational numbers less than 2 3, while R contains all rational numbers greater than 2 3. In this case the cut defines the rational number 2 3. A similar argument replacing 2 3 by any other rational number shows that in such case the cut defines a rational number. (c) As an indication of the possibility, let L contain all rational numbers less than 2 3, while R contains all rational numbers greaters than 2 3. This cut also defines the rational number 2 3. A similar argument shows that this cut always defines a rational number. (d) As an indication of the possibility let L consist of all negative rational numbers and all positive rational numbers whose squares are less than 2, while R consists of all positive numbers whose squares are greater than 2. We can show that if a is any number of the L class, there is always a larger number of the L class, while if b is any number of the R class, there is always a smaller number of the R class (see Problem 1.106). A cut of this type defines an irrational number. From (b), (c), (d) it follows that every cut in the rational number system, called a Dedekind cut, defines either a rational or an irrational number. By use of Dedekind cuts we can define operations (such as addition, multiplication, etc.) with irrational numbers. Supplementary Problems OPERATIONS WITH NUMBERS 1.35. Given x ¼ 3, y ¼ 2, z ¼ 5, a ¼ 3 2, and b ¼ 1 4, evaluate: ðaÞ ð2x yÞð3y þ zÞð5x 2zÞ; ðbÞ xy 2z2 2ab 1 ; ðcÞ 3a2 b þ ab2 2a22b2 þ 1 ; ðdÞ ðax þ byÞ2 þ ðay bxÞ2 ðay þ bxÞ2 þ ðax byÞ2 : Ans. (a) 2200, (b) 32, (c) 51=41, (d) 1 1.36. Find the set of values of x for which the following equations are true. Justify all steps in each case. ðaÞ 4fðx 2Þ þ 3ð2x 1Þg þ 2ð2x þ 1Þ ¼ 12ðx þ 2Þ 2 ðcÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 8x þ 7 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x þ 2 p ¼ x þ 1 ðbÞ 1 8 x 1 x 2 ¼ 1 4 ðdÞ 1 x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 2x þ 5 p ¼ 3 5 Ans. (a) 2, (b) 6; 4 (c) 1; 1 (d) 1 2 1.37. Prove that x ðz xÞðx yÞ þ y ðx yÞðy zÞ þ z ðy zÞðz xÞ ¼ 0 giving restrictions if any. RATIONAL AND IRRATIONAL NUMBERS 1.38. Find decimal expansions for (a) 3 7, (b) ffiffiffi 5 p . Ans. (a) 0:_ 4 4_ 2 2_ 8 8_ 5 5_ 7 7_ 1 1, (b) 2.2360679 . . . CHAP. 1] NUMBERS 17
- 27. 18 NUMBERS [CHAP. 1 1.39. Show that a fraction with denominator 17 and with numerator 1; 2; 3; . . . ; 16 has 16 digits in the repeating portion of its decimal expansion. Is there any relation between the orders of the digits in these expansions? 1.40. Prove that (a) ffiffiffi 3 p , (b) ffiffiffi 2 3 p are irrational numbers. 1.41. Prove that (a) ffiffiffi 5 3 p ffiffiffi 3 4 p , (b) ffiffiffi 2 p þ ffiffiffi 3 p þ ffiffiffi 5 p are irrational numbers. 1.42. Determine a positive rational number whose square differs from 7 by less than .000001. 1.43. Prove that every rational number can be expressed as a repeating decimal. 1.44. Find the values of x for which (a) 2x3 5x2 9x þ 18 ¼ 0, (b) 3x3 þ 4x2 35x þ 8 ¼ 0, (c) x4 21x2 þ 4 ¼ 0. Ans. (a) 3; 2; 3=2 (b) 8=3; 2 ffiffiffi 5 p (c) 1 2 ð5 ffiffiffiffiffi 17 p Þ; 1 2 ð5 ffiffiffiffiffi 17 p Þ 1.45. If a, b, c, d are rational and m is not a perfect square, prove that a þ b ffiffiffiffi m p ¼ c þ d ffiffiffiffi m p if and only if a ¼ c and b ¼ d. 1.46. Prove that 1 þ ffiffiffi 3 p þ ffiffiffi 5 p 1 ffiffiffi 3 p þ ffiffiffi 5 p ¼ 12 ffiffiffi 5 p 2 ffiffiffiffiffi 15 p þ 14 ffiffiffi 3 p 7 11 : INEQUALITIES 1.47. Find the set of values of x for which each of the following inequalities holds: ðaÞ 1 x þ 3 2x A 5; ðbÞ xðx þ 2Þ @ 24; ðcÞ jx þ 2j jx 5j; ðdÞ x x þ 2 x þ 3 3x þ 1 : Ans. (a) 0 x @ 1 2, (b) 6 @ x @ 4, (c) x 3=2, (d) x 3; 1 x 1 3, or x 2 1.48. Prove (a) jx þ yj @ jxj þ jyj, (b) jx þ y þ zj @ jxj þ jyj þ jzj, (c) jx yj A jxj jyj. 1.49. Prove that for all real x; y; z, x2 þ y2 þ z2 A xy þ yz þ zx: 1.50. If a2 þ b2 ¼ 1 and c2 þ d2 ¼ 1, prove that ac þ bd @ 1. 1.51. If x 0, prove that xnþ1 þ 1 xnþ1 xn þ 1 xn where n is any positive integer. 1.52. Prove that for all real a 6¼ 0, ja þ 1=aj A 2: 1.53. Show that in Schwarz’s inequality (Problem 13) the equality holds if and only if ap ¼ kbp, p ¼ 1; 2; 3; . . . ; n where k is any constant. 1.54. If a1; a2; a3 are positive, prove that 1 3 ða1 þ a2 þ a3Þ A ffiffiffiffiffiffiffiffiffiffiffiffiffiffi a1a2a3 3 p . EXPONENTS, ROOTS, AND LOGARITHMS 1.55. Evaluate (a) 4log2 8 , (b) 3 4 log1=8ð 1 128Þ, (c) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð0:00004Þð25,000Þ ð0:02Þ5 ð0:125Þ s , (d) 32 log3 5 , (e) ð 1 8Þ4=3 ð27Þ2=3 Ans. (a) 64, (b) 7/4, (c) 50,000, (d) 1/25, (e) 7=144 1.56. Prove (a) loga MN ¼ loga M þ loga N, (b) loga Mr ¼ r loga M indicating restrictions, if any. 1.57. Prove blogb a ¼ a giving restrictions, if any.
- 28. CHAP. 1] NUMBERS 19 COUNTABILITY 1.58. (a) Prove that there is a one to one correspondence between the points of the interval 0 @ x @ 1 and 5 @ x @ 3. (b) What is the cardinal number of the sets in (a)? Ans. (b) C, the cardinal number of the continuum. 1.59. (a) Prove that the set of all rational numbers is countable. (b) What is the cardinal number of the set in (a)? Ans. (b) Fo 1.60. Prove that the set of (a) all real numbers, (b) all irrational numbers is non-countable. 1.61. The intersection of two sets A and B, denoted by A B or AB, is the set consisting of all elements belonging to both A and B. Prove that if A and B are countable, so is their intersection. 1.62. Prove that a countable set of countable sets is countable. 1.63. Prove that the cardinal number of the set of points inside a square is equal to the cardinal number of the sets of points on (a) one side, (b) all four sides. (c) What is the cardinal number in this case? (d) Does a corresponding result hold for a cube? Ans. (c) C LIMIT POINTS, BOUNDS, BOLZANO–WEIERSTRASS THEOREM 1.64. Given the set of numbers 1; 1:1; :9; 1:01; :99; 1:001; :999; . . . . (a) Is the set bounded? (b) Does the set have a l.u.b. and g.l.b.? If so, determine them. (c) Does the set have any limit points? If so, determine them. (d) Is the set a closed set? Ans. (a) Yes (b) l:u:b: ¼ 1:1; g:l:b: ¼ :9 (c) 1 (d) Yes 1.65. Give the set :9; :9; :99; :99; :999; :999 answer the questions of Problem 64. Ans. (a) Yes (b) l:u:b: ¼ 1; g:l:b: ¼ 1 (c) 1; 1 (d) No 1.66. Give an example of a set which has (a) 3 limit points, (b) no limit points. 1.67. (a) Prove that every point of the interval 0 x 1 is a limit point. (b) Are there are limit points which do not belong to the set in (a)? Justify your answer. 1.68. Let S be the set of all rational numbers in ð0; 1Þ having denominator 2n , n ¼ 1; 2; 3; . . . . (a) Does S have any limit points? (b) Is S closed? 1.69. (a) Give an example of a set which has limit points but which is not bounded. (b) Does this contradict the Bolzano–Weierstrass theorem? Explain. ALGEBRAIC AND TRANSCENDENTAL NUMBERS 1.70. Prove that (a) ffiffiffi 3 p ffiffiffi 2 p ffiffiffi 3 p þ ffiffiffi 2 p , (b) ffiffiffi 2 p þ ffiffiffi 3 p þ ffiffiffi 5 p are algebraic numbers. 1.71. Prove that the set of transcendental numbers in ð0; 1Þ is not countable. 1.72. Prove that every rational number is algebraic but every irrational number is not necessarily algebraic. COMPLEX NUMBERS, POLAR FORM 1.73. Perform each of the indicated operations: (a) 2ð5 3iÞ 3ð2 þ iÞ þ 5ði 3Þ, (b) ð3 2iÞ3 ðcÞ 5 3 4i þ 10 4 þ 3i ; ðdÞ 1 i 1 þ i 10 ; ðeÞ 2 4i 5 þ 7i 2 ; ð f Þ ð1 þ iÞð2 þ 3iÞð4 2iÞ ð1 þ 2iÞ2 ð1 iÞ : Ans. (a) 1 4i, (b) 9 46i, (c) 11 5 2 5 i, (d) 1, (e) 10 37, ( f ) 16 5 2 5 i.
- 29. 1.74. If z1 and z2 are complex numbers, prove (a) z1 z2 ¼ jz1j jz2j , (b) jz2 1j ¼ jz1j2 giving any restrictions. 1.75. Prove (a) jz1 þ z2j @ jz1j þ jz2j, (b) jz1 þ z2 þ z3j @ jz1j þ jz2j þ jz3j, (c) jz1 z2j A jz1j jz2j. 1.76. Find all solutions of 2x4 3x3 7x2 8x þ 6 ¼ 0. Ans. 3, 1 2, 1 i 1.77. Let z1 and z2 be represented by points P1 and P2 in the Argand diagram. Construct lines OP1 and OP2, where O is the origin. Show that z1 þ z2 can be represented by the point P3, where OP3 is the diagonal of a parallelogram having sides OP1 and OP2. This is called the parallelogram law of addition of complex numbers. Because of this and other properties, complex numbers can be considered as vectors in two dimensions. 1.78. Interpret geometrically the inequalities of Problem 1.75. 1.79. Express in polar form (a) 3 ffiffiffi 3 p þ 3i, (b) 2 2i, (c) 1 ffiffiffi 3 p i, (d) 5, (e) 5i. Ans. (a) 6 cis =6 ðbÞ 2 ffiffiffi 2 p cis 5=4 ðcÞ 2 cis 5=3 ðdÞ 5 cis 0 ðeÞ 5 cis 3=2 1.80. Evaluate (a) ½2ðcos 258 þ i sin 258Þ½5ðcos 1108 þ i sin 1108Þ, (b) 12 cis 168 ð3 cis 448Þð2 cis 628Þ : Ans. (a) 5 ffiffiffi 2 p þ 5 ffiffiffi 2 p i; ðbÞ 2i 1.81. Determine all the indicated roots and represent them graphically: (a) ð4 ffiffiffi 2 p þ 4 ffiffiffi 2 p iÞ1=3 ; ðbÞ ð1Þ1=5 ; ðcÞ ð ffiffiffi 3 p iÞ1=3 ; ðdÞ i1=4 . Ans. (a) 2 cis 158; 2 cis 1358; 2 cis 2558 (b) cis 368; cis 1088; cis 1808 ¼ 1; cis 2528; cis 3248 (c) ffiffiffi 2 3 p cis 1108; ffiffiffi 2 3 p cis 2308; ffiffiffi 2 3 p cis 3508 (d) cis 22:58; cis 112:58; cis 202:58; cis 292:58 1.82. Prove that 1 þ ffiffiffi 3 p i is an algebraic number. 1.83. If z1 ¼ 1 cis 1 and z2 ¼ 2 cis 2, prove (a) z1z2 ¼ 12 cisð1 þ 2Þ, (b) z1=z2 ¼ ð1=2Þ cisð1 2Þ. Interpret geometrically. MATHEMATICAL INDUCTION Prove each of the following. 1.84. 1 þ 3 þ 5 þ þ ð2n 1Þ ¼ n2 1.85. 1 1 3 þ 1 3 5 þ 1 5 7 þ þ 1 ð2n 1Þð2n þ 1Þ ¼ n 2n þ 1 1.86. a þ ða þ dÞ þ ða þ 2dÞ þ þ ½a þ ðn 1Þd ¼ 1 2 n½2a þ ðn 1Þd 1.87. 1 1 2 3 þ 1 2 3 4 þ 1 3 4 5 þ þ 1 nðn þ 1Þðn þ 2Þ ¼ nðn þ 3Þ 4ðn þ 1Þðn þ 2Þ 1.88. a þ ar þ ar2 þ þ arn1 ¼ aðrn 1Þ r 1 ; r 6¼ 1 1.89. 13 þ 23 þ 33 þ þ n3 ¼ 1 4 n2 ðn þ 1Þ2 1.90. 1ð5Þ þ 2ð5Þ2 þ 3ð5Þ3 þ þ nð5Þn1 ¼ 5 þ ð4n 1Þ5nþ1 16 1.91. x2n1 þ y2n1 is divisible by x þ y for n ¼ 1; 2; 3; . . . . 20 NUMBERS [CHAP. 1
- 30. 1.92. ðcos þ i sin Þn ¼ cos n þ i sin n. Can this be proved if n is a rational number? 1.93. 1 2 þ cos x þ cos 2x þ þ cos nx ¼ sinðn þ 1 2Þx 2 sin 1 2 x , x 6¼ 0; 2; 4; . . . 1.94. sin x þ sin 2x þ þ sin nx ¼ cos 1 2 x cosðn þ 1 2Þx 2 sin 1 2 x ; x 6¼ 0; 2; 4; . . . 1.95. ða þ bÞn ¼ an þ nC1an1 b þ nC2an2 b2 þ þ nCn1abn1 þ bn where nCr ¼ nðn 1Þðn 2Þ . . . ðn r þ 1Þ r! ¼ n! r!ðn rÞ! ¼ n Cnr. Here p! ¼ pðp 1Þ . . . 1 and 0! is defined as 1. This is called the binomial theorem. The coefficients nC0 ¼ 1, nC1 ¼ n, nC2 ¼ nðn 1Þ 2! ; . . . ; nCn ¼ 1 are called the binomial coefficients. nCr is also written n r . MISCELLANEOUS PROBLEMS 1.96. Express each of the following integers (scale of 10) in the scale of notation indicated: (a) 87 (two), (b) 64 (three), (c) 1736 (nine). Check each answer. Ans. (a) 1010111, (b) 2101, (c) 2338 1.97. If a number is 144 in the scale of 5, what is the number in the scale of (a) 2, (b) 8? 1.98. Prove that every rational number p=q between 0 and 1 can be expressed in the form p q ¼ a1 2 þ a2 22 þ þ an 2n þ where the a’s can be determined uniquely as 0’s or 1’s and where the process may or may not terminate. The representation 0:a1a2 . . . an . . . is then called the binary form of the rational number. [Hint: Multiply both sides successively by 2 and consider remainders.} 1.99. Express 2 3 in the scale of (a) 2, (b) 3, (c) 8, (d) 10. Ans. (a) 0:1010101 . . . ; (b) 0.2 or 0:2000 . . . ; (c) 0:5252 . . . ; (d) 0:6666 . . . 1.100. A number in the scale of 2 is 11.01001. What is the number in the scale of 10. Ans. 3.28125 1.101. In what scale of notation is 3 þ 4 ¼ 12? Ans. 5 1.102. In the scale of 12, two additional symbols t and e must be used to designate the ‘‘digits’’ 10 and 11, respectively. Using these symbols, represent the integer 5110 (scale of 10) in the scale of 12. Ans. 2e5t 1.103. Find a rational number whose decimal expansion is 1:636363 . . . . Ans. 18/11 1.104. A number in the scale of 10 consists of six digits. If the last digit is removed and placed before the first digit, the new number is one-third as large. Find the original number. Ans. 428571 1.105. Show that the rational numbers form a field. 1.106. Using as axioms the relations 1–9 on Pages 2 and 3, prove that (a) ð3Þð0Þ ¼ 0, (b) ð2Þðþ3Þ ¼ 6, (c) ð2Þð3Þ ¼ 6. CHAP. 1] NUMBERS 21
- 31. 1.107. (a) If x is a rational number whose square is less than 2, show that x þ ð2 x2 Þ=10 is a larger such number. (b) If x is a rational number whose square is greater than 2, find in terms of x a smaller rational number whose square is greater than 2. 1.108. Illustrate how you would use Dedekind cuts to define (a) ffiffiffi 5 p þ ffiffiffi 3 p ; ðbÞ ffiffiffi 3 p ffiffiffi 2 p ; ðcÞ ð ffiffiffi 3 p Þð ffiffiffi 2 p Þ; ðdÞ ffiffiffi 2 p = ffiffiffi 3 p . 22 NUMBERS [CHAP. 1
- 32. 23 Sequences DEFINITION OF A SEQUENCE A sequence is a set of numbers u1; u2; u3; . . . in a definite order of arrangement (i.e., a correspondence with the natural numbers) and formed according to a definite rule. Each number in the sequence is called a term; un is called the nth term. The sequence is called finite or infinite according as there are or are not a finite number of terms. The sequence u1; u2; u3; . . . is also designated briefly by fung. EXAMPLES. 1. The set of numbers 2; 7; 12; 17; . . . ; 32 is a finite sequence; the nth term is given by un ¼ 2 þ 5ðn 1Þ ¼ 5n 3, n ¼ 1; 2; . . . ; 7. 2. The set of numbers 1; 1=3; 1=5; 1=7; . . . is an infinite sequence with nth term un ¼ 1=ð2n 1Þ, n ¼ 1; 2; 3; . . . . Unless otherwise specified, we shall consider infinite sequences only. LIMIT OF A SEQUENCE A number l is called the limit of an infinite sequence u1; u2; u3; . . . if for any positive number we can find a positive number N depending on such that jun lj for all integers n N. In such case we write lim n!1 un ¼ l. EXAMPLE . If un ¼ 3 þ 1=n ¼ ð3n þ 1Þ=n, the sequence is 4; 7=2; 10=3; . . . and we can show that lim n!1 un ¼ 3. If the limit of a sequence exists, the sequence is called convergent; otherwise, it is called divergent. A sequence can converge to only one limit, i.e., if a limit exists, it is unique. See Problem 2.8. A more intuitive but unrigorous way of expressing this concept of limit is to say that a sequence u1; u2; u3; . . . has a limit l if the successive terms get ‘‘closer and closer’’ to l. This is often used to provide a ‘‘guess’’ as to the value of the limit, after which the definition is applied to see if the guess is really correct. THEOREMS ON LIMITS OF SEQUENCES If lim n!1 an ¼ A and lim n!1 bn ¼ B, then 1. lim n!1 ðan þ bnÞ ¼ lim n!1 an þ lim n!1 bn ¼ A þ B 2. lim n!1 ðan bnÞ ¼ lim n!1 an lim n!1 bn ¼ A B 3. lim n!1 ðan bnÞ ¼ ð lim n!1 anÞð lim n!1 bnÞ ¼ AB Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 33. 4. lim n!1 an bn ¼ lim n!1 an lim n!1 bn ¼ A B if lim n!1 bn ¼ B 6¼ 0 If B ¼ 0 and A 6¼ 0, lim n!1 an bn does not exist. If B ¼ 0 and A ¼ 0, lim n!1 an bn may or may not exist. 5. lim n!1 ap n ¼ ð lim n!1 anÞp ¼ Ap , for p ¼ any real number if Ap exists. 6. lim n!1 pan ¼ p liman n!1 ¼ pA , for p ¼ any real number if pA exists. INFINITY We write lim n!1 an ¼ 1 if for each positive number M we can find a positive number N (depending on M) such that an M for all n N. Similarly, we write lim n!1 an ¼ 1 if for each positive number M we can find a positive number N such that an M for all n N. It should be emphasized that 1 and 1 are not numbers and the sequences are not convergent. The terminology employed merely indicates that the sequences diverge in a certain manner. That is, no matter how large a number in absolute value that one chooses there is an n such that the absolute value of an is greater than that quantity. BOUNDED, MONOTONIC SEQUENCES If un @ M for n ¼ 1; 2; 3; . . . ; where M is a constant (independent of n), we say that the sequence fung is bounded above and M is called an upper bound. If un A m, the sequence is bounded below and m is called a lower bound. If m @ un @ M the sequence is called bounded. Often this is indicated by junj @ P. Every convergent sequence is bounded, but the converse is not necessarily true. If unþ1 A un the sequence is called monotonic increasing; if unþ1 un it is called strictly increasing. Similarly, if unþ1 @ un the sequence is called monotonic decreasing, while if unþ1 un it is strictly decreasing. EXAMPLES. 1. The sequence 1; 1:1; 1:11; 1:111; . . . is bounded and monotonic increasing. It is also strictly increasing. 2. The sequence 1; 1; 1; 1; 1; . . . is bounded but not monotonic increasing or decreasing. 3. The sequence 1; 1:5; 2; 2:5; 3; . . . is monotonic decreasing and not bounded. However, it is bounded above. The following theorem is fundamental and is related to the Bolzano–Weierstrass theorem (Chapter 1, Page 6) which is proved in Problem 2.23. Theorem. Every bounded monotonic (increasing or decreasing) sequence has a limit. LEAST UPPER BOUND AND GREATEST LOWER BOUND OF A SEQUENCE A number M is called the least upper bound (l.u.b.) of the sequence fung if un @ M, n ¼ 1; 2; 3; . . . while at least one term is greater than M for any 0. A number m m is called the greatest lower bound (g.l.b.) of the sequence fung if un A m m, n ¼ 1; 2; 3; . . . while at least one term is less than m m þ for any 0. Compare with the definition of l.u.b. and g.l.b. for sets of numbers in general (see Page 6). 24 SEQUENCES [CHAP. 2
- 34. LIMIT SUPERIOR, LIMIT INFERIOR A number l l is called the limit superior, greatest limit or upper limit (lim sup or lim) of the sequence fung if infinitely many terms of the sequence are greater than l l while only a finite number of terms are greater than l l þ , where is any positive number. A number l is called the limit inferior, least limit or lower limit (lim inf or lim) of the sequence fung if infintely many terms of the sequence are less than l þ while only a finite number of terms are less than l , where is any positive number. These correspond to least and greatest limiting points of general sets of numbers. If infintely many terms of fung exceed any positive number M, we define lim sup fung ¼ 1. If infinitely many terms are less than M, where M is any positive number, we define lim inf fung ¼ 1. If lim n!1 un ¼ 1, we define lim sup fung ¼ lim inf fung ¼ 1. If lim n!1 un ¼ 1, we define lim sup fung ¼ lim inf fung ¼ 1. Although every bounded sequence is not necessarily convergent, it always has a finite lim sup and lim inf. A sequence fung converges if and only if lim sup un ¼ lim inf un is finite. NESTED INTERVALS Consider a set of intervals ½an; bn, n ¼ 1; 2; 3; . . . ; where each interval is contained in the preceding one and lim n!1 ðan bnÞ ¼ 0. Such intervals are called nested intervals. We can prove that to every set of nested intervals there corresponds one and only one real number. This can be used to establish the Bolzano–Weierstrass theorem of Chapter 1. (See Problems 2.22 and 2.23.) CAUCHY’S CONVERGENCE CRITERION Cauchy’s convergence criterion states that a sequence fung converges if and only if for each 0 we can find a number N such that jup uqj for all p; q N. This criterion has the advantage that one need not know the limit l in order to demonstrate convergence. INFINITE SERIES Let u1; u2; u3; . . . be a given sequence. Form a new sequence S1; S2; S3; . . . where S1 ¼ u1; S2 ¼ u1 þ u2; S3 ¼ u1 þ u2 þ u3; . . . ; Sn ¼ u1 þ u2 þ u3 þ þ un; . . . where Sn, called the nth partial sum, is the sum of the first n terms of the sequence fung. The sequence S1; S2; S3; . . . is symbolized by u1 þ u2 þ u3 þ ¼ X 1 n¼1 un which is called an infinite series. If lim n!1 Sn ¼ S exists, the series is called convergent and S is its sum, otherwise the series is called divergent. Further discussion of infinite series and other topics related to sequences is given in Chapter 11. CHAP. 2] SEQUENCES 25
- 35. Solved Problems SEQUENCES 2.1. Write the first five terms of each of the following sequences. ðaÞ 2n 1 3n þ 2 ðbÞ 1 ð1Þn n3 ðcÞ ð1Þn1 2 4 6 2n ( ) ðdÞ 1 2 þ 1 4 þ 1 8 þ þ 1 2n ðeÞ ð1Þn1 x2n1 ð2n 1Þ! ( ) ðaÞ 1 5 ; 3 8 ; 5 11 ; 7 14 ; 9 17 ðbÞ 2 13 ; 0; 2 33 ; 0; 2 53 ðcÞ 1 2 ; 1 2 4 ; 1 2 4 6 ; 1 2 4 6 8 ; 1 2 4 6 8 10 ðdÞ 1 2 ; 1 2 þ 1 4 ; 1 2 þ 1 4 þ 1 8 ; 1 2 þ 1 4 þ 1 8 þ 1 16 ; 1 2 þ 1 4 þ 1 8 þ 1 16 þ 1 32 ðeÞ x 1! ; x3 3! ; x5 5! ; x7 7! ; x9 9! Note that n! ¼ 1 2 3 4 n. Thus 1! ¼ 1, 3! ¼ 1 2 3 ¼ 6, 5! ¼ 1 2 3 4 5 ¼ 120, etc. We define 0! ¼ 1. 2.2. Two students were asked to write an nth term for the sequence 1; 16; 81; 256; . . . and to write the 5th term of the sequence. One student gave the nth term as un ¼ n4 . The other student, who did not recognize this simple law of formation, wrote un ¼ 10n3 35n2 þ 50n 24. Which student gave the correct 5th term? If un ¼ n4 , then u1 ¼ 14 ¼ 1, u2 ¼ 24 ¼ 16, u3 ¼ 34 ¼ 81, u4 ¼ 44 ¼ 256, which agrees with the first four terms of the sequence. Hence the first student gave the 5th term as u5 ¼ 54 ¼ 625: If un ¼ 10n3 35n2 þ 50n 24, then u1 ¼ 1; u2 ¼ 16; u3 ¼ 81; u4 ¼ 256, which also agrees with the first four terms given. Hence, the second student gave the 5th term as u5 ¼ 601: Both students were correct. Merely giving a finite number of terms of a sequence does not define a unique nth term. In fact, an infinite number of nth terms is possible. 26 SEQUENCES [CHAP. 2
- 36. LIMIT OF A SEQUENCE 2.3. A sequence has its nth term given by un ¼ 3n 1 4n þ 5 . (a) Write the 1st, 5th, 10th, 100th, 1000th, 10,000th and 100,000th terms of the sequence in decimal form. Make a guess as to the limit of this sequence as n ! 1. (b) Using the definition of limit verify that the guess in (a) is actually correct. ðaÞ n ¼ 1 n ¼ 5 n ¼ 10 n ¼ 100 n ¼ 1000 n ¼ 10,000 n ¼ 100,000 :22222 . . . :56000 . . . :64444 . . . :73827 . . . :74881 . . . :74988 . . . :74998 . . . A good guess is that the limit is :75000 . . . ¼ 3 4. Note that it is only for large enough values of n that a possible limit may become apparent. (b) We must show that for any given 0 (no matter how small) there is a number N (depending on ) such that jun 3 4 j for all n N. Now 3n 1 4n þ 5 3 4 ¼ 19 4ð4n þ 5Þ when 19 4ð4n þ 5Þ or 4ð4n þ 5Þ 19 1 ; 4n þ 5 19 4 ; n 1 4 19 4 5 Choosing N ¼ 1 4 ð19=4 5Þ, we see that jun 3 4 j for all n N, so that lim n!1 ¼ 3 4 and the proof is complete. Note that if ¼ :001 (for example), N ¼ 1 4 ð19000=4 5Þ ¼ 1186 1 4. This means that all terms of the sequence beyond the 1186th term differ from 3 4 in absolute value by less than .001. 2.4. Prove that lim n!1 c np ¼ 0 where c 6¼ 0 and p 0 are constants (independent of n). We must show that for any 0 there is a number N such that jc=np 0j for all n N. Now c np when jcj np , i.e., np jcj or n jcj 1=p . Choosing N ¼ jcj 1=p (depending on ), we see that jc=np j for all n N, proving that lim n!1 ðc=np Þ ¼ 0. 2.5. Prove that lim n!1 1 þ 2 10n 5 þ 3 10n ¼ 2 3 . We must show that for any 0 there is a number N such that 1 þ 2 10n 5 þ 3 10n 2 3 for all n N. Now 1 þ 2 10n 5 þ 3 10n 2 3 ¼ 7 3ð5 þ 3 10nÞ when 7 3ð5 þ 3 10nÞ , i.e. when 3 7 ð5 þ 3 10n Þ 1=, 3 10n 7=3 5, 10n 1 8 ð7=3 5Þ or n log10f1 3 ð7=3 5Þg ¼ N, proving the existence of N and thus establishing the required result. Note that the above value of N is real only if 7=3 5 0, i.e., 0 7=15. If A 7=15, we see that 1 þ 2 10n 5 þ 3 10n 2 3 for all n 0. 2.6. Explain exactly what is meant by the statements (a) lim n!1 32n1 ¼ 1, (b) lim n!1 ð1 2nÞ ¼ 1. (a) If for each positive number M we can find a positive number N (depending on M) such that an M for all n N, then we write lim n!1 an ¼ 1. In this case, 32n1 M when ð2n 1Þ log 3 log M; i.e., n 1 2 log M log 3 þ 1 ¼ N. (b) If for each positive number M we can find a positive number N (depending on M) such that an M for all n N, then we write lim n!1 ¼ 1. In this case, 1 2n M when 2n 1 M or n 1 2 ðM þ 1Þ ¼ N. CHAP. 2] SEQUENCES 27
- 37. It should be emphasized that the use of the notations 1 and 1 for limits does not in any way imply convergence of the given sequences, since 1 and 1 are not numbers. Instead, these are notations used to describe that the sequences diverge in specific ways. 2.7. Prove that lim n!1 xn ¼ 0 if jxj 1. Method 1: We can restrict ourselves to x 6¼ 0, since if x ¼ 0, the result is clearly true. Given 0, we must show that there exists N such that jxn j for n N. Now jxn j ¼ jxjn when n log10 jxj log10 . Dividing by log10 jxj, which is negative, yields n log10 log10 jxj ¼ N, proving the required result. Method 2: Let jxj ¼ 1=ð1 þ pÞ, where p 0. By Bernoulli’s inequality (Problem 1.31, Chapter 1), we have jxn j ¼ jxjn ¼ 1=ð1 þ pÞn 1=ð1 þ npÞ for all n N. Thus lim n!1 xn ¼ 0. THEOREMS ON LIMITS OF SEQUENCES 2.8. Prove that if lim n!1 un exists, it must be unique. We must show that if lim n!1 un ¼ l1 and lim n!1 un ¼ l2, then l1 ¼ l2. By hypothesis, given any 0 we can find N such that jun l1j 1 2 when n N; jun l2j 1 2 when n N Then jl1 l2j ¼ jl1 un þ un l2j @ jl1 unj þ jun l2j 1 2 þ 1 2 ¼ i.e., jl1 l2j is less than any positive (however small) and so must be zero. Thus, l1 ¼ l2. 2.9. If lim n!1 an ¼ A and lim n!1 bn ¼ B, prove that lim n!1 ðan þ bnÞ ¼ A þ B. We must show that for any 0, we can find N 0 such that jðan þ bnÞ ðA þ BÞj for all n N. From absolute value property 2, Page 3 we have jðan þ bnÞ ðA þ BÞj ¼ jðan AÞ þ ðbn BÞj @ jan Aj þ jbn Bj ð1Þ By hypothesis, given 0 we can find N1 and N2 such that jan Aj 1 2 for all n N1 ð2Þ jbn Bj 1 2 for all n N2 ð3Þ Then from (1), (2), and (3), jðan þ bnÞ ðA þ BÞj 1 2 þ 1 2 ¼ for all n N where N is chosen as the larger of N1 and N2. Thus, the required result follows. 2.10. Prove that a convergent sequence is bounded. Given lim n!1 an ¼ A, we must show that there exists a positive number P such that janj P for all n. Now janj ¼ jan A þ Aj @ jan Aj þ jAj But by hypothesis we can find N such that jan Aj for all n N, i.e., janj þ jAj for all n N It follows that janj P for all n if we choose P as the largest one of the numbers a1; a2; . . . ; aN, þ jAj. 28 SEQUENCES [CHAP. 2
- 38. 2.11. If lim n!1 bn ¼ B 6¼ 0, prove there exists a number N such that jbnj 1 2 jBj for all n N. Since B ¼ B bn þ bn, we have: (1) jBj @ jB bnj þ jbnj. Now we can choose N so that jB bnj ¼ jbn Bj 1 2 jBj for all n N, since lim n!1 bn ¼ B by hypothesis. Hence, from (1), jBj 1 2 jBj þ jbnj or jbnj 1 2 jBj for all n N. 2.12. If lim n!1 an ¼ A and lim n!1 bn ¼ B, prove that lim n!1 anbn ¼ AB. We have, using Problem 2.10, janbn ABj ¼ janðbn BÞ þ Bðan AÞj @ janjjbn Bj þ jBjjan Aj ð1Þ @ Pjbn Bj þ ðjBj þ 1Þjan Aj But since lim n!1 an ¼ A and lim n!1 bn ¼ B, given any 0 we can find N1 and N2 such that jbn Bj 2P for all n N1 jan Aj 2ðjBj þ 1Þ for all n N2 Hence, from (1), janbn ABj 1 2 þ 1 2 ¼ for all n N, where N is the larger of N1 and N2. Thus, the result is proved. 2.13. If lim n!1 an ¼ A and lim n!1 bn ¼ B 6¼ 0, prove (a) lim n!1 1 bn ¼ 1 B , (b) lim n!1 an bn ¼ A B . (a) We must show that for any given 0, we can find N such that 1 bn 1 B ¼ jB bnj jBjjbnj for all n N ð1Þ By hypothesis, given any 0, we can find N1, such that jbn Bj 1 2 B2 for all n N1. Also, since lim n!1 bn ¼ B 6¼ 0, we can find N2 such that jbnj 1 2 jBj for all n N2 (see Problem 11). Then if N is the larger of N1 and N2, we can write (1) as 1 bn 1 B ¼ jbn Bj jBjjbnj 1 2 B2 jBj 1 2 jBj ¼ for all n N and the proof is complete. (b) From part (a) and Problem 2.12, we have lim n!1 an bn ¼ lim n!1 an 1 bn ¼ lim n!1 an lim n!1 1 bn ¼ A 1 B ¼ A B This can also be proved directly (see Problem 41). 2.14. Evaluate each of the following, using theorems on limits. ðaÞ lim n!1 3n2 5n 5n2 þ 2n 6 ¼ lim n!1 3 5=n 5 þ 2=n 6=n2 ¼ 3 þ 0 5 þ 0 þ 0 ¼ 3 5 ðbÞ lim n!1 nðn þ 2Þ n þ 1 n3 n2 þ 1 ( ) ¼ lim n!1 n3 þ n2 þ 2n ðn þ 1Þðn2 þ 1Þ ( ) ¼ lim n!1 1 þ 1=n þ 2=n2 ð1 þ 1=nÞð1 þ 1=n2Þ ( ) ¼ 1 þ 0 þ 0 ð1 þ 0Þ ð1 þ 0Þ ¼ 1 ðcÞ lim n!1 ð ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p ffiffiffi n p Þ ¼ lim n!1 ð ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p ffiffiffi n p Þ ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p þ ffiffiffi n p ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p þ ffiffiffi n p ¼ lim n!1 1 ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p þ ffiffiffi n p ¼ 0 CHAP. 2] SEQUENCES 29
- 39. ðdÞ lim n!1 3n2 þ 4n 2n 1 ¼ lim n!1 3 þ 4=n 2=n 1=n2 Since the limits of the numerator and denominator are 3 and 0, respectively, the limit does not exist. Since 3n2 þ 4n 2n 1 3n2 2n ¼ 3n 2 can be made larger than any positive number M by choosing n N, we can write, if desired, lim n!1 3n2 þ 4n 2n 1 ¼ 1. ðeÞ lim n!1 2n 3 2n þ 7 4 ¼ lim n!1 2 3=n 3 þ 7=n 4 ¼ 2 3 4 ¼ 16 81 ð f Þ lim n!1 2n5 4n2 3n7 þ n3 10 ¼ lim n!1 2=n2 4=n5 3 þ 1=n4 10=n7 ¼ 0 3 ¼ 0 ðgÞ lim n!1 1 þ 2 10n 5 þ 3 10n ¼ lim n!1 10n þ 2 5 10n þ 3 ¼ 2 3 (Compare with Problem 2.5.) BOUNDED MONOTONIC SEQUENCES 2.15. Prove that the sequence with nth un ¼ 2n 7 3n þ 2 (a) is monotonic increasing, (b) is bounded above, (c) is bounded below, (d) is bounded, (e) has a limit. (a) fung is monotonic increasing if unþ1 A un, n ¼ 1; 2; 3; . . . . Now 2ðn þ 1Þ 7 3ðn þ 1Þ þ 2 A 2n 7 3n þ 2 if and only if 2n 5 2n þ 5 A 2n 7 3n þ 2 or ð2n 5Þð3n þ 2Þ A ð2n 7Þð3n þ 5Þ, 6n2 11n 10 A 6n2 11n 35, i.e. 10A 35, which is true. Thus, by reversal of steps in the inequalities, we see that fung is monotonic increasing. Actually, since 10 35, the sequence is strictly increasing. (b) By writing some terms of the sequence, we may guess that an upper bound is 2 (for example). To prove this we must show that un @ 2. If ð2n 7Þ=ð3n þ 2Þ @ 2 then 2n 7 @ 6n þ 4 or 4n 11, which is true. Reversal of steps proves that 2 is an upper bound. (c) Since this particular sequence is monotonic increasing, the first term 1 is a lower bound, i.e., un A 1, n ¼ 1; 2; 3; . . . . Any number less than 1 is also a lower bound. (d) Since the sequence has an upper and lower bound, it is bounded. Thus, for example, we can write junj @ 2 for all n. (e) Since every bounded monotonic (increasing or decreasing) sequence has a limit, the given sequence has a limit. In fact, lim n!1 2n 7 3n þ 2 ¼ lim n!1 2 7=n 3 þ 2=n ¼ 2 3 . 2.16. A sequence fung is defined by the recursion formula unþ1 ¼ ffiffiffiffiffiffiffi 3un p , u1 ¼ 1. (a) Prove that lim n!1 un exists. (b) Find the limit in (a). (a) The terms of the sequence are u1 ¼ 1, u2 ¼ ffiffiffiffiffiffiffi 3u1 p ¼ 31=2 , u3 ¼ ffiffiffiffiffiffiffi 3u2 p ¼ 31=2þ1=4 ; . . . . The nth term is given by un ¼ 31=2þ1=4þþ1=2n1 as can be proved by mathematical induction (Chapter 1). Clearly, unþ1 A un. Then the sequence is monotone increasing. By Problem 1.14, Chapter 1, un @ 31 ¼ 3, i.e. un is bounded above. Hence, un is bounded (since a lower bound is zero). Thus, a limit exists, since the sequence is bounded and monotonic increasing. 30 SEQUENCES [CHAP. 2
- 40. CHAP. 2] SEQUENCES 31 (b) Let x ¼ required limit. Since lim n!1 unþ1 ¼ lim n!1 ffiffiffiffiffiffiffi 3un p , we have x ¼ ffiffiffiffiffiffi 3x p and x ¼ 3. (The other possibility, x ¼ 0, is excluded since un A 1:Þ Another method: lim n!1 31=2þ1=4þþ1=2n1 ¼ lim n!1 311=2n ¼ 3 limð11=2nÞ n!1 ¼ 31 ¼ 3 2.17. Verify the validity of the entries in the following table. Sequence Bounded Monotonic Increasing Monotonic Decreasing Limit Exists 2; 1:9; 1:8; 1:7; . . . ; 2 ðn 1Þ=10 . . . No No Yes No 1; 1; 1; 1; . . . ; ð1Þn1 ; . . . Yes No No No 1 2 ; 1 3 ; 1 4 ; 1 5 ; . . . ; ð1Þn1 =ðn þ 1Þ; . . . Yes No No Yes (0) :6; :66; :666; . . . ; 2 3 ð1 1=10n Þ; . . . Yes Yes No Yes (2 3) 1; þ2; 3; þ4; 5; . . . ; ð1Þn n; . . . No No No No 2.18. Prove that the sequence with the nth term un ¼ 1 þ 1 n n is monotonic, increasing, and bounded, and thus a limit exists. The limit is denoted by the symbol e. Note: lim n!1 1 þ 1 n n ¼ e, where e ffi 2:71828 . . . was introduced in the eighteenth century by Leonhart Euler as the base for a system of logarithms in order to simplify certain differentiation and integration formulas. By the binomial theorem, if n is a positive integer (see Problem 1.95, Chapter 1), ð1 þ xÞn ¼ 1 þ nx þ nðn 1Þ 2! x2 þ nðn 1Þðn 2Þ 3! x3 þ þ nðn 1Þ ðn n þ 1Þ n! xn Letting x ¼ 1=n, un ¼ 1 þ 1 n n ¼ 1 þ n 1 n þ nðn 1Þ 2! 1 n2 þ þ nðn 1Þ ðn n þ 1Þ n! 1 nn ¼ 1 þ 1 þ 1 2! 1 1 n þ 1 3! 1 1 n 1 2 n þ þ 1 n! 1 1 n 1 2 n 1 n 1 n Since each term beyond the first two terms in the last expression is an increasing function of n, it follows that the sequence un is a monotonic increasing sequence. It is also clear that 1 þ 1 n n 1 þ 1 þ 1 2! þ 1 3! þ þ 1 n! 1 þ 1 þ 1 2 þ 1 22 þ þ 1 2n1 3 by Problem 1.14, Chapter 1. Thus, un is bounded and monotonic increasing, and so has a limit which we denote by e. The value of e ¼ 2:71828 . . . . 2.19. Prove that lim x!1 1 þ 1 x x ¼ e, where x ! 1 in any manner whatsoever (i.e., not necessarily along the positive integers, as in Problem 2.18). If n ¼ largest integer @ x, then n @ x @ n þ 1 and 1 þ 1 n þ 1 n @ 1 þ 1 x x @ 1 þ 1 n nþ1 . Since lim n!1 1 þ 1 n þ 1 n ¼ lim n!1 1 þ 1 n þ 1 nþ1 , 1 þ 1 n þ 1 ¼ e
- 41. and lim n!1 1 þ 1 n nþ1 ¼ lim n!1 1 þ 1 n n 1 þ 1 n ¼ e it follows that lim x!1 1 þ 1 x x ¼ e: LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR 2.20. Find the (a) l.u.b., (b) g.l.b., (c) lim sup ðlimÞ, and (d) lim inf (limÞ for the sequence 2; 2; 1; 1; 1; 1; 1; 1; . . . . (a) l:u:b: ¼ 2, since all terms are less than equal to 2, while at least one term (the 1st) is greater than 2 for any 0. (b) g:l:b: ¼ 2, since all terms are greater than or equal to 2, while at least one term (the 2nd) is less than 2 þ for any 0. (c) lim sup or lim ¼ 1, since infinitely many terms of the sequence are greater than 1 for any 0 (namely, all 1’s in the sequence), while only a finite number of terms are greater than 1 þ for any 0 (namely, the 1st term). (d) lim inf or lim ¼ 1, since infinitely many terms of the sequence are less than 1 þ for any 0 (namely, all 1’s in the sequence), while only a finite number of terms are less than 1 for any 0 (namely the 2nd term). 2.21. Find the (a) l.u.b., (b) g.l.b., (c) lim sup (lim), and (d) lim inf (lim) for the sequences in Problem 2.17. The results are shown in the following table. Sequence l.u.b. g.l.b. lim sup or lim lim inf or lim 2; 1:9; 1:8; 1:7; . . . ; 2 ðn 1Þ=10 . . . 2 none 1 1 1; 1; 1; 1; . . . ; ð1Þn1 ; . . . 1 1 1 1 1 2 ; 1 3 ; 1 4 1 5 ; . . . ; ð1Þn1 =ðn þ 1Þ; . . . 1 2 1 3 0 0 :6; :66; :666; . . . ; 2 3 ð1 1=10n Þ; . . . 2 3 6 2 3 2 3 1; þ2; 3; þ4; 5; . . . ; ð1Þn n; . . . none none þ1 1 NESTED INTERVALS 2.22. Prove that to every set of nested intervals ½an; bn, n ¼ 1; 2; 3; . . . ; there corresponds one and only one real number. By definition of nested intervals, anþ1 A an; bnþ1 @ bn; n ¼ 1; 2; 3; . . . and lim n!1 ðan bnÞ ¼ 0. Then a1 @ an @ bn @ b1, and the sequences fang and fbng are bounded and respectively monotonic increasing and decreasing sequences and so converge to a and b. To show that a ¼ b and thus prove the required result, we note that b a ¼ ðb bnÞ þ ðbn anÞ þ ðan aÞ ð1Þ jb aj @ jb bnj þ jbn anj þ jan aj ð2Þ Now given any 0, we can find N such that for all n N jb bnj =3; jbn anj =3; jan aj =3 ð3Þ so that from (2), jb aj . Since is any positive number, we must have b a ¼ 0 or a ¼ b. 32 SEQUENCES [CHAP. 2
- 42. 2.23. Prove the Bolzano–Weierstrass theorem (see Page 6). Suppose the given bounded infinite set is contained in the finite interval ½a; b. Divide this interval into two equal intervals. Then at least one of these, denoted by ½a1; b1, contains infinitely many points. Dividing ½a1; b1 into two equal intervals, we obtain another interval, say, ½a2; b2, containing infinitely many points. Continuing this process, we obtain a set of intervals ½an; bn, n ¼ 1; 2; 3; . . . ; each interval contained in the preceding one and such that b1 a1 ¼ ðb aÞ=2; b2 a2 ¼ ðb1 a1Þ=2 ¼ ðb aÞ=22 ; . . . ; bn an ¼ ðb aÞ=2n from which we see that lim n!1 ðbn anÞ ¼ 0. This set of nested intervals, by Problem 2.22, corresponds to a real number which represents a limit point and so proves the theorem. CAUCHY’S CONVERGENCE CRITERION 2.24. Prove Cauchy’s convergence criterion as stated on Page 25. Necessity. Suppose the sequence fung converges to l. Then given any 0, we can find N such that jup lj =2 for all p N and juq lj =2 for all q N Then for both p N and q N, we have jup uqj ¼ jðup lÞ þ ðl uqÞj @ jup lj þ jl uqj =2 þ =2 ¼ Sufficiency. Suppose jup uqj for all p; q N and any 0. Then all the numbers uN; uNþ1; . . . lie in a finite interval, i.e., the set is bounded and infinite. Hence, by the Bolzano–Weierstrass theorem there is at least one limit point, say a. If a is the only limit point, we have the desired proof and lim n!1 un ¼ a. Suppose there are two distinct limit points, say a and b, and suppose b a (see Fig. 2-1). By definition of limit points, we have jup aj ðb aÞ=3 for infinnitely many values of p ð1Þ juq bj ðb aÞ=3 for infinitely many values of q ð2Þ Then since b a ¼ ðb uqÞ þ ðuq upÞ þ ðup aÞ, we have jb aj ¼ b a @ jb uqj þ jup uqj þ jup aj ð3Þ Using (1) and (2) in (3), we see that jup uqj ðb aÞ=3 for infinitely many values of p and q, thus contradicting the hypothesis that jup uqj for p; q N and any 0. Hence, there is only one limit point and the theorem is proved. INFINITE SERIES 2.25. Prove that the infinite series (sometimes called the geometric series) a þ ar þ ar2 þ ¼ X 1 n¼1 arn1 (a) converges to a=ð1 rÞ if jrj 1, (b) diverges if jrj A 1. Sn ¼ a þ ar þ ar2 þ þ arn1 Let rSn ¼ ar þ ar2 þ þ arn1 þ arn Then ð1 rÞSn ¼ a arn Subtract, CHAP. 2] SEQUENCES 33 a b b _ a 3 b _ a 3 Fig. 2-1
- 43. Sn ¼ að1 rn Þ 1 r or ðaÞ If jrj 1; lim n!1 Sn ¼ lim n!1 að1 rn Þ 1 r ¼ a 1 r by Problem 7: (b) If jrj 1, lim n!1 Sn does not exist (see Problem 44). 2.26. Prove that if a series converges, its nth term must necessarily approach zero. Since Sn ¼ u1 þ u2 þ þ un, Sn1 ¼ u1 þ u2 þ þ un1 we have un ¼ Sn Sn1. If the series converges to S, then lim n!1 un ¼ lim n!1 ðSn Sn1Þ ¼ lim n!1 Sn lim n!1 Sn1 ¼ S S ¼ 0 2.27. Prove that the series 1 1 þ 1 1 þ 1 1 þ ¼ X 1 n¼1 ð1Þn1 diverges. Method 1: lim n!1 ð1Þn 6¼ 0, in fact it doesn’t exist. Then by Problem 2.26 the series cannot converge, i.e., it diverges. Method 2: The sequence of partial sums is 1; 1 1; 1 1 þ 1; 1 1 þ 1 1; . . . i.e., 1; 0; 1; 0; 1; 0; 1; . . . . Since this sequence has no limit, the series diverges. MISCELLANEOUS PROBLEMS 2.28. If lim n!1 un ¼ l, prove that lim n!1 u1 þ u2 þ þ un n ¼ l. Let un ¼ vn þ l. We must show that lim n!1 v1 þ v2 þ þ vn n ¼ 0 if lim n!1 vn ¼ 0. Now v1 þ v2 þ þ vn n ¼ v1 þ v2 þ þ vP n þ vPþ1 þ vpþ2 þ þ vn n so that v1 þ v2 þ þ vn n @ jv1 þ v2 þ þ vPj n þ jvPþ1j þ jvPþ2j þ þ jvnj n ð1Þ Since lim n!1 vn ¼ 0, we can choose P so that jvnj =2 for n P. Then jvPþ1j þ jvPþ2j þ þ jvnj n =2 þ =2 þ þ =2 n ¼ ðn PÞ=2 n 2 ð2Þ After choosing P we can choose N so that for n N P, jv1 þ v2 þ þ vPj n 2 ð3Þ Then using (2) and (3), (1) becomes v1 þ v2 þ þ vn n 2 þ 2 ¼ for n N thus proving the required result. 2.29. Prove that lim n!1 ð1 þ n þ n2 Þ1=n ¼ 1. Let ð1 þ n þ n2 Þ1=n ¼ 1 þ un where un A 0. Now by the binomial theorem, 34 SEQUENCES [CHAP. 2
- 44. 1 þ n þ n2 ¼ ð1 þ unÞn ¼ 1 þ nun þ nðn 1Þ 2! u2 n þ nðn 1Þðn 2Þ 3! u3 n þ þ un n Then 1 þ n þ n2 1 þ nðn 1Þðn 2Þ 3! u3 n or 0 u3 n 6ðn2 þ nÞ nðn 1Þðn 2Þ : Hence, lim n!1 u3 n ¼ 0 and lim n!1 un ¼ 0: Thus lim n!1 ð1 þ n þ n2 Þ1=n ¼ lim n!1 ð1 þ unÞ ¼ 1: 2.30. Prove that lim n!1 an n! ¼ 0 for all constants a. The result follows if we can prove that lim n!1 jajn n! ¼ 0 (see Problem 2.38). We can assume a 6¼ 0. Let un ¼ jajn n! . Then un un1 ¼ jaj n . If n is large enough, say, n 2jaj, and if we call N ¼ ½2jaj þ 1, i.e., the greatest integer @ 2jaj þ 1, then uNþ1 uN 1 2 ; uNþ2 uNþ1 1 2 ; . . . ; un un1 1 2 Multiplying these inequalities yields un uN 1 2 nN or un 1 2 nN uN: Since lim n!1 1 2 nN ¼ 0 (using Problem 2.7), it follows that lim n!1 un ¼ 0. Supplementary Problems SEQUENCES 2.31. Write the first four terms of each of the following sequences: ðaÞ ffiffiffi n p n þ 1 ; ðbÞ ð1Þnþ1 n! ( ) ; ðcÞ ð2xÞn1 ð2n 1Þ5 ( ) ; ðdÞ ð1Þn x2n1 1 3 5 ð2n 1Þ ( ) ; ðeÞ cos nx x2 þ n2 : Ans: ðaÞ ffiffiffi 1 p 2 ; ffiffiffi 2 p 3 ; ffiffiffi 3 p 4 ; ffiffiffi 4 p 5 ðcÞ 1 15 ; 2x 35 ; 4x2 55 ; 8x3 75 ðeÞ cos x x2 þ 12 ; cos 2x x2 þ 22 ; cos 3x x2 þ 32 ; cos 4x x2 þ 42 ðbÞ 1 1! ; 1 2! ; 1 3! ; 1 4! ðdÞ x 1 ; x3 1 3 ; x5 1 3 5 ; x7 1 3 5 7 2.32. Find a possible nth term for the sequences whose first 5 terms are indicated and find the 6th term: ðaÞ 1 5 ; 3 8 ; 5 11 ; 7 14 ; 9 17 ; . . . ðbÞ 1; 0; 1; 0; 1; . . . ðcÞ 2 3 ; 0; 3 4 ; 0; 4 5 ; . . . Ans: ðaÞ ð1Þn ð2n 1Þ ð3n þ 2Þ ðbÞ 1 ð1Þn 2 ðcÞ ðn þ 3Þ ðn þ 5Þ 1 ð1Þn 2 2.33. The Fibonacci sequence is the sequence fung where unþ2 ¼ unþ1 þ un and u1 ¼ 1, u2 ¼ 1. (a) Find the first 6 terms of the sequence. (b) Show that the nth term is given by un ¼ ðan bn Þ= ffiffiffi 5 p , where a ¼ 1 2 ð1 þ ffiffiffi 5 p Þ, b ¼ 1 2 ð1 ffiffiffi 5 p Þ. Ans. (a) 1; 1; 2; 3; 5; 8 LIMITS OF SEQUENCES 2.34. Using the definition of limit, prove that: CHAP. 2] SEQUENCES 35
- 45. ðaÞ lim n!1 4 2n 3n þ 2 ¼ 2 3 ; ðbÞ lim n!1 21= ffiffi n p ¼ 1; ðcÞ lim n!1 n4 þ 1 n2 ¼ 1; ðdÞ lim n!1 sin n n ¼ 0: 2.35. Find the least positive integer N such that jð3n þ 2Þ=ðn 1Þ 3j for all n N if (a) ¼ :01, (b) ¼ :001, (c) ¼ :0001. Ans. (a) 502, (b) 5002, (c) 50,002 2.36. Using the definition of limit, prove that lim n!1 ð2n 1Þ=ð3n þ 4Þ cannot be 1 2. 2.37. Prove that lim n!1 ð1Þn n does not exist. 2.38. Prove that if lim n!1 junj ¼ 0 then lim n!1 un ¼ 0. Is the converse true? 2.39. If lim n!1 un ¼ l, prove that (a) lim n!1 cun ¼ cl where c is any constant, (b) lim n!1 u2 n ¼ l2 , (c) lim n!1 up n ¼ l p where p is a positive integer, (d) lim n!1 ffiffiffiffiffi un p ¼ ffiffi l p ; l A 0. 2.40. Give a direct proof that lim n!1 an=bn ¼ A=B if lim n!1 an ¼ A and lim n!1 bn ¼ B 6¼ 0. 2.41. Prove that (a) lim n!1 31=n ¼ 1, (b) lim n!1 2 3 1=n ¼ 1, (c) lim n!1 3 4 n ¼ 0. 2.42. If r 1, prove that lim n!1 rn ¼ 1, carefully explaining the significance of this statement. 2.43. If jrj 1, prove that lim n!1 rn does not exist. 2.44. Evaluate each of the following, using theorems on limits: ðaÞ lim n!1 4 2n 3n2 2n2 þ n ðcÞ lim n!1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3n2 5n þ 4 p 2n 7 ðeÞ lim n!1 ð ffiffiffiffiffiffiffiffiffiffiffiffiffi n2 þ n p nÞ ðbÞ lim n!1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð3 ffiffiffi n p Þð ffiffiffi n p þ 2Þ 8n 4 3 r ðdÞ lim n!1 4 10n 3 102n 3 10n1 þ 2 102n1 ð f Þ lim n!1 ð2n þ 3n Þ1=n Ans: ðaÞ 3=2; ðbÞ 1=2; ðcÞ ffiffiffi 3 p =2; ðdÞ 15; ðeÞ 1=2; ð f Þ 3 BOUNDED MONOTONIC SEQUENCES 2.45. Prove that the sequence with nth term un ¼ ffiffiffi n p =ðn þ 1Þ (a) is monotonic decreasing, (b) is bounded below, (c) is bounded above, (d) has a limit. 2.46. If un ¼ 1 1 þ n þ 1 2 þ n þ 1 3 þ n þ þ 1 n þ n , prove that lim n!1 un exists and lies between 0 and 1. 2.47. If unþ1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi un þ 1 p , u1 ¼ 1, prove that lim n!1 un ¼ 1 2 ð1 þ ffiffiffi 5 p Þ. 2.48. If unþ1 ¼ 1 2 ðun þ p=unÞ where p 0 and u1 0, prove that lim n!1 un ¼ ffiffiffi p p . Show how this can be used to determine ffiffiffi 2 p . 2.49. If un is monotonic increasing (or monotonic decreasing), prove that Sn=n, where Sn ¼ u1 þ u2 þ þ un, is also monotonic increasing (or monotonic decreasing). LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR 2.50. Find the l.u.b., g.l.b., lim sup (lim), lim inf (lim) for each sequence: (a) 1; 1 3 ; 1 5 ; 1 7 ; . . . ; ð1Þn =ð2n 1Þ; . . . ðcÞ 1; 3; 5; 7; . . . ; ð1Þn1 ð2n 1Þ; . . . (b) 2 3 ; 3 4 ; 4 5 ; 5 6 ; . . . ; ð1Þnþ1 ðn þ 1Þ=ðn þ 2Þ; . . . ðdÞ 1; 4; 1; 16; 1; 36; . . . ; n1þð1Þn ; . . . 36 SEQUENCES [CHAP. 2
- 46. Ans. (a) 1 3 ; 1; 0; 0 ðbÞ 1; 1; 1; 1 ðcÞ none, none, þ1, 1 (d) none, 1; þ1; 1 2.51. Prove that a bounded sequence fung is convergent if and only if lim un ¼ lim un. INFINITE SERIES 2.52. Find the sum of the series X 1 n¼1 2 3 n . Ans. 2 2.53. Evaluate X 1 n¼1 ð1Þn1 =5n . Ans. 1 6 2.54. Prove that 1 1 2 þ 1 2 3 þ 1 3 4 þ 1 4 5 þ ¼ X 1 n¼1 1 nðn þ 1Þ ¼ 1. Hint: 1 nðn þ 1Þ ¼ 1 n 1 n þ 1 2.55. Prove that multiplication of each term of an infinite series by a constant (not zero) does not affect the convergence or divergence. 2.56. Prove that the series 1 þ 1 2 þ 1 3 þ þ 1 n þ diverges. Hint: Let Sn ¼ 1 þ 1 2 þ 1 3 þ þ 1 n . Then prove that jS2n Snj 1 2, giving a contradiction with Cauchy’s convergence criterion. MISCELLANEOUS PROBLEMS 2.57. If an @ un @ bn for all n N, and lim n!1 an ¼ lim n!1 bn ¼ l, prove that lim n!1 un ¼ l. 2.58. If lim n!1 an ¼ lim n!1 bn ¼ 0, and is independent of n, prove that lim n!1 ðan cos n þ bn sin nÞ ¼ 0. Is the result true when depends on n? 2.59. Let un ¼ 1 2 f1 þ ð1Þn g, n ¼ 1; 2; 3; . . . . If Sn ¼ u1 þ u2 þ þ un, prove that lim n!1 Sn=n ¼ 1 2. 2.60. Prove that (a) lim n!1 n1=n ¼ 1, (b) lim n!1 ða þ nÞp=n ¼ 1 where a and p are constants. 2.61. If lim n!1 junþ1=unj ¼ jaj 1, prove that lim n!1 un ¼ 0. 2.62. If jaj 1, prove that lim n!1 np an ¼ 0 where the constant p 0. 2.63. Prove that lim 2n n! nn ¼ 0. 2.64. Prove that lim n!1 n sin 1=n ¼ 1. Hint: Let the central angle, , of a circle be measured in radians. Geome- trically illustrate that sin tan , 0 . Let ¼ 1=n. Observe that since n is restricted to positive integers, the angle is restricted to the first quadrant. 2.65. If fung is the Fibonacci sequence (Problem 2.33), prove that lim n!1 unþ1=un ¼ 1 2 ð1 þ ffiffiffi 5 p Þ. 2.66. Prove that the sequence un ¼ ð1 þ 1=nÞnþ1 , n ¼ 1; 2; 3; . . . is a monotonic decreasing sequence whose limit is e. [Hint: Show that un=un1 @ 1: 2.67. If an A bn for all n N and lim n!1 an ¼ A, lim n!1 bn ¼ B, prove that A A B. 2.68. If junj @ jvnj and lim n!1 vn ¼ 0, prove that lim n!1 un ¼ 0. 2.69. Prove that lim n!1 1 n 1 þ 1 2 þ 1 3 þ þ 1 n ¼ 0. CHAP. 2] SEQUENCES 37
- 47. 2.70. Prove that ½an; bn, where an ¼ ð1 þ 1=nÞn and bn ¼ ð1 þ 1=nÞnþ1 , is a set of nested intervals defining the number e. 2.71. Prove that every bounded monotonic (increasing or decreasing) sequence has a limit. 2.72. Let fung be a sequence such that unþ2 ¼ aunþ1 þ bun where a and b are constants. This is called a second order difference equation for un. (a) Assuming a solution of the form un ¼ rn where r is a constant, prove that r must satisfy the equation r2 ar b ¼ 0. (b) Use (a) to show that a solution of the difference equation (called a general solution) is un ¼ Arn 1 þ Brn 2, where A and B are arbitrary constants and r1 and r2 are the two solutions of r2 ar b ¼ 0 assumed different. (c) In case r1 ¼ r2 in (b), show that a (general) solution is un ¼ ðA þ BnÞrn 1. 2.73. Solve the following difference equations subject to the given conditions: (a) unþ2 ¼ unþ1 þ un, u1 ¼ 1, u2 ¼ 1 (compare Prob. 34); (b) unþ2 ¼ 2unþ1 þ 3un, u1 ¼ 3, u2 ¼ 5; (c) unþ2 ¼ 4unþ1 4un, u1 ¼ 2, u2 ¼ 8. Ans. (a) Same as in Prob. 34, (b) un ¼ 2ð3Þn1 þ ð1Þn1 ðcÞ un ¼ n 2n 38 SEQUENCES [CHAP. 2
- 48. 39 Functions, Limits, and Continuity FUNCTIONS A function is composed of a domain set, a range set, and a rule of correspondence that assigns exactly one element of the range to each element of the domain. This definition of a function places no restrictions on the nature of the elements of the two sets. However, in our early exploration of the calculus, these elements will be real numbers. The rule of correspondence can take various forms, but in advanced calculus it most often is an equation or a set of equations. If the elements of the domain and range are represented by x and y, respectively, and f symbolizes the function, then the rule of correspondence takes the form y ¼ f ðxÞ. The distinction between f and f ðxÞ should be kept in mind. f denotes the function as defined in the first paragraph. y and f ðxÞ are different symbols for the range (or image) values corresponding to domain values x. However a ‘‘common practice’’ that provides an expediency in presentation is to read f ðxÞ as, ‘‘the image of x with respect to the function f ’’ and then use it when referring to the function. (For example, it is simpler to write sin x than ‘‘the sine function, the image value of which is sin x.’’) This deviation from precise notation will appear in the text because of its value in exhibiting the ideas. The domain variable x is called the independent variable. The variable y representing the corre- sponding set of values in the range, is the dependent variable. Note: There is nothing exclusive about the use of x, y, and f to represent domain, range, and function. Many other letters will be employed. There are many ways to relate the elements of two sets. [Not all of them correspond a unique range value to a given domain value.] For example, given the equation y2 ¼ x, there are two choices of y for each positive value of x. As another example, the pairs ða; bÞ, ða; cÞ, ða; dÞ, and ða; eÞ can be formed and again the correspondence to a domain value is not unique. Because of such possibilities, some texts, especially older ones, distinguish between multiple-valued and single-valued functions. This viewpoint is not consistent with our definition or modern presentations. In order that there be no ambiguity, the calculus and its applications require a single image associated with each domain value. A multiple- valued rule of correspondence gives rise to a collection of functions (i.e., single-valued). Thus, the rule y2 ¼ x is replaced by the pair of rules y ¼ x1=2 and y ¼ x1=2 and the functions they generate through the establishment of domains. (See the following section on graphs for pictorial illustrations.) Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 49. EXAMPLES. 1. If to each number in 1 @ x @ 1 we associate a number y given by x2 , then the interval 1 @ x @ 1 is the domain. The rule y ¼ x2 generates the range 1 @ y @ 1. The totality is a function f . The functional image of x is given by y ¼ f ðxÞ ¼ x2 . For example, f ð 1 3Þ ¼ ð 1 3Þ2 ¼ 1 9 is the image of 1 3 with respect to the function f . 2. The sequences of Chapter 2 may be interpreted as functions. For infinite sequences consider the domain as the set of positive integers. The rule is the definition of un, and the range is generated by this rule. To illustrate, let un ¼ 1 n with n ¼ 1; 2; . . . . Then the range contains the elements 1; 1 2 ; 1 3 ; 1 4 ; . . . . If the function is denoted by f , then we may write f ðnÞ ¼ 1 n. As you read this chapter, reviewing Chapter 2 will be very useful, and in particular com- paring the corresponding sections. 3. With each time t after the year 1800 we can associate a value P for the population of the United States. The correspondence between P and t defines a function, say F, and we can write P ¼ FðtÞ. 4. For the present, both the domain and the range of a function have been restricted to sets of real numbers. Eventually this limitation will be removed. To get the flavor for greater generality, think of a map of the world on a globe with circles of latitude and longitude as coordinate curves. Assume there is a rule that corresponds this domain to a range that is a region of a plane endowed with a rectangular Cartesian coordinate system. (Thus, a flat map usable for navigation and other purposes is created.) The points of the domain are expressed as pairs of numbers ð; Þ and those of the range by pairs ðx; yÞ. These sets and a rule of correspondence constitute a function whose independent and dependent variables are not single real numbers; rather, they are pairs of real numbers. GRAPH OF A FUNCTION A function f establishes a set of ordered pairs ðx; yÞ of real numbers. The plot of these pairs ðx; f ðxÞÞ in a coordinate system is the graph of f . The result can be thought of as a pictorial representa- tion of the function. For example, the graphs of the functions described by y ¼ x2 , 1 @ x @ 1, and y2 ¼ x, 0 @ x @ 1, y A 0 appear in Fig. 3-1. BOUNDED FUNCTIONS If there is a constant M such that f ðxÞ @ M for all x in an interval (or other set of numbers), we say that f is bounded above in the interval (or the set) and call M an upper bound of the function. If a constant m exists such that f ðxÞ A m for all x in an interval, we say that f ðxÞ is bounded below in the interval and call m a lower bound. 40 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3 Fig. 3-1
- 50. If m @ f ðxÞ @ M in an interval, we call f ðxÞ bounded. Frequencly, when we wish to indicate that a function is bounded, we shall write j f ðxÞj P. EXAMPLES. 1. f ðxÞ ¼ 3 þ x is bounded in 1 @ x @ 1. An upper bound is 4 (or any number greater than 4). A lower bound is 2 (or any number less than 2). 2. f ðxÞ ¼ 1=x is not bounded in 0 x 4 since by choosing x sufficiently close to zero, f ðxÞ can be made as large as we wish, so that there is no upper bound. However, a lower bound is given by 1 4 (or any number less than 1 4). If f ðxÞ has an upper bound it has a least upper bound (l.u.b.); if it has a lower bound it has a greatest lower bound (g.l.b.). (See Chapter 1 for these definitions.) MONOTONIC FUNCTIONS A function is called monotonic increasing in an interval if for any two points x1 and x2 in the interval such that x1 x2, f ðx1Þ @ f ðx2Þ. If f ðx1Þ f ðx2Þ the function is called strictly increasing. Similarly if f ðx1Þ A f ðx2Þ whenever x1 x2, then f ðxÞ is monotonic decreasing; while if f ðx1Þ f ðx2Þ, it is strictly decreasing. INVERSE FUNCTIONS. PRINCIPAL VALUES Suppose y is the range variable of a function f with domain variable x. Furthermore, let the correspondence between the domain and range values be one-to-one. Then a new function f 1 , called the inverse function of f , can be created by interchanging the domain and range of f . This information is contained in the form x ¼ f 1 ðyÞ. As you work with the inverse function, it often is convenient to rename the domain variable as x and use y to symbolize the images, then the notation is y ¼ f 1 ðxÞ. In particular, this allows graphical expression of the inverse function with its domain on the horizontal axis. Note: f 1 does not mean f to the negative one power. When used with functions the notation f 1 always designates the inverse function to f . If the domain and range elements of f are not in one-to-one correspondence (this would mean that distinct domain elements have the same image), then a collection of one-to-one functions may be created. Each of them is called a branch. It is often convenient to choose one of these branches, called the principal branch, and denote it as the inverse function, f 1 . The range values of f that compose the principal branch, and hence the domain of f 1 , are called the principal values. (As will be seen in the section of elementary functions, it is common practice to specify these principal values for that class of functions.) EXAMPLE. Suppose f is generated by y ¼ sin x and the domain is 1 @ x @ 1. Then there are an infinite number of domain values that have the same image. (A finite portion of the graph is illustrated below in Fig. 3-2(a.) In Fig. 3-2(b) the graph is rotated about a line at 458 so that the x-axis rotates into the y-axis. Then the variables are interchanged so that the x-axis is once again the horizontal one. We see that the image of an x value is not unique. Therefore, a set of principal values must be chosen to establish an inverse function. A choice of a branch is accomplished by restricting the domain of the starting function, sin x. For example, choose 2 @ x @ 2 . Then there is a one-to-one correspondence between the elements of this domain and the images in 1 @ x @ 1. Thus, f 1 may be defined with this interval as its domain. This idea is illustrated in Fig. 3-2(c) and Fig. 3-2(d). With the domain of f 1 represented on the horizontal axis and by the variable x, we write y ¼ sin1 x, 1 @ x @ 1. If x ¼ 1 2, then the corresponding range value is y ¼ 6 . Note: In algebra, b1 means 1 b and the fact that bb1 produces the identity element 1 is simply a rule of algebra generalized from arithmetic. Use of a similar exponential notation for inverse functions is justified in that corre- sponding algebraic characteristics are displayed by f 1 ½ f ðxÞ ¼ x and f ½ f 1 ðxÞ ¼ x. CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 41
- 51. MAXIMA AND MINIMA The seventeenth-century development of the calculus was strongly motivated by questions concern- ing extreme values of functions. Of most importance to the calculus and its applications were the notions of local extrema, called relative maximums and relative minimums. If the graph of a function were compared to a path over hills and through valleys, the local extrema would be the high and low points along the way. This intuitive view is given mathematical precision by the following definition. Definition: If there exists an open interval ða; bÞ containing c such that f ðxÞ f ðcÞ for all x other than c in the interval, then f ðcÞ is a relative maximum of f . If f ðxÞ f ðcÞ for all x in ða; bÞ other than c, then f ðcÞ is a relative minimum of f . (See Fig. 3-3.) Functions may have any number of relative extrema. On the other hand, they may have none, as in the case of the strictly increasing and decreasing functions previously defined. Definition: If c is in the domain of f and for all x in the domain of the function f ðxÞ @ f ðcÞ, then f ðcÞ is an absolute maximum of the function f . If for all x in the domain f ðxÞ A f ðcÞ then f ðcÞ is an absolute minimum of f . (See Fig. 3-3.) Note: If defined on closed intervals the strictly increasing and decreasing functions possess absolute extrema. 42 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3 Fig. 3-2
- 52. Absolute extrema are not necessarily unique. For example, if the graph of a function is a horizontal line, then every point is an absolute maximum and an absolute minimum. Note: A point of inflection also is represented in Fig. 3-3. There is an overlap with relative extrema in representation of such points through derivatives that will be addressed in the problem set of Chapter 4. TYPES OF FUNCTIONS It is worth realizing that there is a fundamental pool of functions at the foundation of calculus and advanced calculus. These are called elementary functions. Either they are generated from a real variable x by the fundamental operations of algebra, including powers and roots, or they have relatively simple geometric interpretations. As the title ‘‘elementary functions’’ suggests, there is a more general category of functions (which, in fact, are dependent on the elementary ones). Some of these will be explored later in the book. The elementary functions are described below. 1. Polynomial functions have the form f ðxÞ ¼ a0xn þ a1xn1 þ þ an1x þ an ð1Þ where a0; . . . ; an are constants and n is a positive integer called the degree of the polynomial if a0 6¼ 0. The fundamental theorem of algebra states that in the field of complex numbers every polynomial equation has at least one root. As a consequence of this theorem, it can be proved that every nth degree polynomial has n roots in the complex field. When complex numbers are admitted, the polynomial theoretically may be expressed as the product of n linear factors; with our restriction to real numbers, it is possible that 2k of the roots may be complex. In this case, the k factors generating them will be quadratic. (The corresponding roots are in complex conjugate pairs.) The polynomial x3 5x2 þ 11x 15 ¼ ðx 3Þðx2 2x þ 5Þ illustrates this thought. 2. Algebraic functions are functions y ¼ f ðxÞ satisfying an equation of the form p0ðxÞyn þ p1ðxÞyn1 þ þ pn1ðxÞy þ pnðxÞ ¼ 0 ð2Þ where p0ðxÞ; . . . ; pnðxÞ are polynomials in x. If the function can be expressed as the quotient of two polynomials, i.e., PðxÞ=QðxÞ where PðxÞ and QðxÞ are polynomials, it is called a rational algebraic function; otherwise it is an irrational algebraic function. 3. Transcendental functions are functions which are not algebraic, i.e., they do not satisfy equations of the form (2). CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 43 Fig. 3-3
- 53. Note the analogy with real numbers, polynomials corresponding to integers, rational functions to rational numbers, and so on. TRANSCENDENTAL FUNCTIONS The following are sometimes called elementary transcendental functions. 1. Exponential function: f ðxÞ ¼ ax , a 6¼ 0; 1. For properties, see Page 3. 2. Logarithmic function: f ðxÞ ¼ loga x, a 6¼ 0; 1. This and the exponential function are inverse functions. If a ¼ e ¼ 2:71828 . . . ; called the natural base of logarithms, we write f ðxÞ ¼ loge x ¼ ln x, called the natural logarithm of x. For properties, see Page 4. 3. Trigonometric functions (Also called circular functions because of their geometric interpreta- tion with respect to the unit circle): sin x; cos x; tan x ¼ sin x cos x ; csc x ¼ 1 sin x ; sec x ¼ 1 cos x ; cot x ¼ 1 tan x ¼ cos x sin x The variable x is generally expressed in radians ( radians ¼ 1808). For real values of x, sin x and cos x lie between 1 and 1 inclusive. The following are some properties of these functions: sin2 x þ cos2 x ¼ 1 1 þ tan2 x ¼ sec2 x 1 þ cot2 x ¼ csc2 x sinðx yÞ ¼ sin x cos y cos x sin y sinðxÞ ¼ sin x cosðx yÞ ¼ cos x cos y sin x sin y cosðxÞ ¼ cos x tanðx yÞ ¼ tan x tan y 1 tan x tan y tanðxÞ ¼ tan x 4. Inverse trigonometric functions. The following is a list of the inverse trigonometric functions and their principal values: ðaÞ y ¼ sin1 x; ð=2 @ y @ =2Þ ðdÞ y ¼ csc1 x ¼ sin1 1=x; ð=2 @ y @ =2Þ ðbÞ y ¼ cos1 x; ð0 @ y @ Þ ðeÞ y ¼ sec1 x ¼ cos1 1=x; ð0 @ y @ Þ ðcÞ y ¼ tan1 x; ð=2 y =2Þ ð f Þ y ¼ cot1 x ¼ =2 tan1 x; ð0 y Þ 5. Hyperbolic functions are defined in terms of exponential functions as follows. These functions may be interpreted geometrically, much as the trigonometric functions but with respect to the unit hyperbola. ðaÞ sinh x ¼ ex ex 2 ðdÞ csch x ¼ 1 sinh x ¼ 2 ex ex ðbÞ cosh x ¼ ex þ ex 2 ðeÞ sech x ¼ 1 cosh x ¼ 2 ex þ ex ðcÞ tanh x ¼ sinh x cosh x ¼ ex ex ex þ ex ð f Þ coth x ¼ cosh x sinh x ¼ ex þ ex ex ex The following are some properties of these functions: cosh2 x sinh2 x ¼ 1 1 tanh2 x ¼ sech2 x coth2 x 1 ¼ csch2 x sinhðx yÞ ¼ sinh x cosh y cosh x sinh y sinhðxÞ ¼ sinh x coshðx yÞ ¼ cosh x cosh y sinh x sinh y coshðxÞ ¼ cosh x tanhðx yÞ ¼ tanh x tanh y 1 tanh x tanh y tanhðxÞ ¼ tanh x 44 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 54. 6. Inverse hyperbolic functions. If x ¼ sinh y then y ¼ sinh1 x is the inverse hyperbolic sine of x. The following list gives the principal values of the inverse hyperbolic functions in terms of natural logarithms and the domains for which they are real. ðaÞ sinh1 x ¼ lnðx þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p Þ; all x ðdÞ csch1 x ¼ ln 1 x þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p jxj ! ; x 6¼ 0 ðbÞ cosh1 x ¼ lnðx þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 p Þ; x A 1 ðeÞ sech1 x ¼ ln 1 þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p x ! ; 0 x @ 1 ðcÞ tanh1 x ¼ 1 2 ln 1 þ x 1 x ; jxj 1 ð f Þ coth1 x ¼ 1 2 ln x þ 1 x 1 ; jxj 1 LIMITS OF FUNCTIONS Let f ðxÞ be defined and single-valued for all values of x near x ¼ x0 with the possible exception of x ¼ x0 itslef (i.e., in a deleted neighborhood of x0). We say that the number l is the limit of f ðxÞ as x approaches x0 and write lim x!x0 f ðxÞ ¼ l if for any positive number (however small) we can find some positive number (usually depending on ) such that j f ðxÞ lj whenever 0 jx x0j . In such case we also say that f ðxÞ approaches l as x approaches x0 and write f ðxÞ ! l as x ! x0. In words, this means that we can make f ðxÞ arbitrarily close to l by choosing x sufficiently close to x0. EXAMPLE. Let f ðxÞ ¼ x2 if x 6¼ 2 0 if x ¼ 2 . Then as x gets closer to 2 (i.e., x approaches 2), f ðxÞ gets closer to 4. We thus suspect that lim x!2 f ðxÞ ¼ 4. To prove this we must see whether the above definition of limit (with l ¼ 4) is satisfied. For this proof see Problem 3.10. Note that lim x!2 f ðxÞ 6¼ f ð2Þ, i.e., the limit of f ðxÞ as x ! 2 is not the same as the value of f ðxÞ at x ¼ 2 since f ð2Þ ¼ 0 by definition. The limit would in fact be 4 even if f ðxÞ were not defined at x ¼ 2. When the limit of a function exists it is unique, i.e., it is the only one (see Problem 3.17). RIGHT- AND LEFT-HAND LIMITS In the definition of limit no restriction was made as to how x should approach x0. It is sometimes found convenient to restrict this approach. Considering x and x0 as points on the real axis where x0 is fixed and x is moving, then x can approach x0 from the right or from the left. We indicate these respective approaches by writing x ! x0þ and x ! x0. If lim x!x0þ f ðxÞ ¼ l1 and lim x!x0 f ðxÞ ¼ l2, we call l1 and l2, respectively, the right- and left-hand limits of f at x0 and denote them by f ðx0þÞ or f ðx0 þ 0Þ and f ðx0Þ or f ðx0 0Þ. The ; definitions of limit of f ðxÞ as x ! x0þ or x ! x0 are the same as those for x ! x0 except for the fact that values of x are restricted to x x0 or x x0, respectively. We have lim x!x0 f ðxÞ ¼ l if and only if lim x!x0þ f ðxÞ ¼ lim x!x0 f ðxÞ ¼ l. THEOREMS ON LIMITS If lim x!x0 f ðxÞ ¼ A and lim x!x0 gðxÞ ¼ B, then 1: lim x!x0 ð f ðxÞ þ gðxÞÞ ¼ lim x!x0 f ðxÞ þ lim x!x0 gðxÞ ¼ A þ B CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 45
- 55. 2: lim x!x0 ð f ðxÞ gðxÞÞ ¼ lim x!x0 f ðxÞ lim x!x0 gðxÞ ¼ A B 3: lim x!x0 ð f ðxÞgðxÞÞ ¼ lim x!x0 f ðxÞ lim x!x0 gðxÞ ¼ AB 4: lim x!x0 f ðxÞ gðxÞ ¼ lim x!x0 f ðxÞ lim x!x0 gðxÞ ¼ A B if B 6¼ 0 Similar results hold for right- and left-hand limits. INFINITY It sometimes happens that as x ! x0, f ðxÞ increases or decreases without bound. In such case it is customary to write lim x!x0 f ðxÞ ¼ þ1 or lim x!x0 f ðxÞ ¼ 1, respectively. The symbols þ1 (also written 1) and 1 are read plus infinity (or infinity) and minus infinity, respectively, but it must be emphasized that they are not numbers. In precise language, we say that lim x!x0 f ðxÞ ¼ 1 if for each positive number M we can find a positive number (depending on M in general) such that f ðxÞ M whenever 0 jx x0j . Similarly, we say that lim x!x0 f ðxÞ ¼ 1 if for each positive number M we can find a positive number such that f ðxÞ M whenever 0 jx x0j . Analogous remarks apply in case x ! x0þ or x ! x0. Frequently we wish to examine the behavior of a function as x increases or decreases without bound. In such cases it is customary to write x ! þ1 (or 1) or x ! 1, respectively. We say that lim x!þ1 f ðxÞ ¼ l, or f ðxÞ ! l as x ! þ1, if for any positive number we can find a positive number N (depending on in general) such that j f ðxÞ lj whenever x N. A similar definition can be formulated for lim x!1 f ðxÞ. SPECIAL LIMITS 1. lim x!0 sin x x ¼ 1; lim x!0 1 cos x x ¼ 0 2. lim x!1 1 þ 1 x x ¼ e, lim x!0þ ð1 þ xÞ1=x ¼ e 3. lim x!0 ex 1 x ¼ 1, lim x!1 x 1 ln x ¼ 1 CONTINUITY Let f be defined for all values of x near x ¼ x0 as well as at x ¼ x0 (i.e., in a neighborhood of x0). The function f is called continuous at x ¼ x0 if lim x!x0 f ðxÞ ¼ f ðx0Þ. Note that this implies three conditions which must be met in order that f ðxÞ be continuous at x ¼ x0. 1. lim x!x0 f ðxÞ ¼ l must exist. 2. f ðx0Þ must exist, i.e., f ðxÞ is defined at x0. 3. l ¼ f ðx0Þ. In summary, lim x!x0 f ðxÞ is the value suggested for f at x ¼ x0 by the behavior of f in arbitrarily small neighborhoods of x0. If in fact this limit is the actual value, f ðx0Þ, of the function at x0, then f is continuous there. Equivalently, if f is continuous at x0, we can write this in the suggestive form lim x!x0 f ðxÞ ¼ f ð lim x!x0 xÞ. 46 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 56. EXAMPLES. 1. If f ðxÞ ¼ x2 ; x 6¼ 2 0; x ¼ 2 then from the example on Page 45 lim x!2 f ðxÞ ¼ 4. But f ð2Þ ¼ 0. Hence lim x!2 f ðxÞ 6¼ f ð2Þ and the function is not continuous at x ¼ 2. 2. If f ðxÞ ¼ x2 for all x, then lim x!2 f ðxÞ ¼ f ð2Þ ¼ 4 and f ðxÞ is continuous at x ¼ 2. Points where f fails to be continuous are called discontinuities of f and f is said to be discontinuous at these points. In constructing a graph of a continuous function the pencil need never leave the paper, while for a discontinuous function this is not true since there is generally a jump taking place. This is of course merely a characteristic property and not a definition of continuity or discontinuity. Alternative to the above definition of continuity, we can define f as continuous at x ¼ x0 if for any 0 we can find 0 such that j f ðxÞ f ðx0Þj whenever jx x0j . Note that this is simply the definition of limit with l ¼ f ðx0Þ and removal of the restriction that x 6¼ x0. RIGHT- AND LEFT-HAND CONTINUITY If f is defined only for x A x0, the above definition does not apply. In such case we call f continuous (on the right) at x ¼ x0 if lim x!x0þ f ðxÞ ¼ f ðx0Þ, i.e., if f ðx0þÞ ¼ f ðx0Þ. Similarly, f is continuous (on the left) at x ¼ x0 if lim x!x0 f ðxÞ ¼ f ðx0Þ, i.e., f ðx0Þ ¼ f ðx0Þ. Definitions in terms of and can be given. CONTINUITY IN AN INTERVAL A function f is said to be continuous in an interval if it is continuous at all points of the interval. In particular, if f is defined in the closed interval a @ x @ b or ½a; b, then f is continuous in the interval if and only if lim x!x0 f ðxÞ ¼ f ðx0Þ for a x0 b, lim x!aþ f ðxÞ ¼ f ðaÞ and lim x!b f ðxÞ ¼ f ðbÞ. THEOREMS ON CONTINUITY Theorem 1. If f and g are continuous at x ¼ x0, so also are the functions whose image values satisfy the relations f ðxÞ þ gðxÞ, f ðxÞ gðxÞ, f ðxÞgðxÞ and f ðxÞ gðxÞ , the last only if gðx0Þ 6¼ 0. Similar results hold for continuity in an interval. Theorem 2. Functions described as follows are continuous in every finite interval: (a) all polynomials; (b) sin x and cos x; (c) ax ; a 0 Theorem 3. Let the function f be continuous at the domain value x ¼ x0. Also suppose that a function g, represented by z ¼ gðyÞ, is continuous at y0, where y ¼ f ðxÞ (i.e., the range value of f corresponding to x0 is a domain value of g). Then a new function, called a composite function, f ðgÞ, represented by z ¼ g½ f ðxÞ, may be created which is continuous at its domain point x ¼ x0. [One says that a continuous function of a continuous function is continuous.] Theorem 4. If f ðxÞ is continuous in a closed interval, it is bounded in the interval. Theorem 5. If f ðxÞ is continuous at x ¼ x0 and f ðx0Þ 0 [or f ðx0Þ 0], there exists an interval about x ¼ x0 in which f ðxÞ 0 [or f ðxÞ 0]. Theorem 6. If a function f ðxÞ is continuous in an interval and either strictly increasing or strictly decreasing, the inverse function f 1 ðxÞ is single-valued, continuous, and either strictly increasing or strictly decreasing. CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 47
- 57. Theorem 7. If f ðxÞ is continuous in ½a; b and if f ðaÞ ¼ A and f ðbÞ ¼ B, then corresponding to any number C between A and B there exists at least one number c in ½a; b such that f ðcÞ ¼ C. This is sometimes called the intermediate value theorem. Theorem 8. If f ðxÞ is continuous in ½a; b and if f ðaÞ and f ðbÞ have opposite signs, there is at least one number c for which f ðcÞ ¼ 0 where a c b. This is related to Theorem 7. Theorem 9. If f ðxÞ is continuous in a closed interval, then f ðxÞ has a maximum value M for at least one value of x in the interval and a minimum value m for at least one value of x in the interval. Further- more, f ðxÞ assumes all values between m and M for one or more values of x in the interval. Theorem 10. If f ðxÞ is continuous in a closed interval and if M and m are respectively the least upper bound (l.u.b.) and greatest lower bound (g.l.b.) of f ðxÞ, there exists at least one value of x in the interval for which f ðxÞ ¼ M or f ðxÞ ¼ m. This is related to Theorem 9. PIECEWISE CONTINUITY A function is called piecewise continuous in an interval a @ x @ b if the interval can be subdivided into a finite number of intervals in each of which the function is continuous and has finite right- and left- hand limits. Such a function has only a finite number of discontinuities. An example of a function which is piecewise continuous in a @ x @ b is shown graphically in Fig. 3-4 below. This function has discontinuities at x1, x2, x3, and x4. UNIFORM CONTINUITY Let f be continuous in an interval. Then by definition at each point x0 of the interval and for any 0, we can find 0 (which will in general depend on both and the particular point x0) such that j f ðxÞ f ðx0Þj whenever jx x0j . If we can find for each which holds for all points of the interval (i.e., if depends only on and not on x0), we say that f is uniformly continuous in the interval. Alternatively, f is uniformly continuous in an interval if for any 0 we can find 0 such that j f ðx1Þ f ðx2Þj whenever jx1 x2j where x1 and x2 are any two points in the interval. Theorem. If f is continuous in a closed interval, it is uniformly continuous in the interval. 48 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3 f (x) a x1 x2 x3 x4 b x Fig. 3-4
- 58. Solved Problems FUNCTIONS 3.1. Let f ðxÞ ¼ ðx 2Þð8 xÞ for 2 @ x @ 8. (a) Find f ð6Þ and f ð1Þ. (b) What is the domain of definition of f ðxÞ? (c) Find f ð1 2tÞ and give the domain of definition. (d) Find f ½ f ð3Þ, f ½ f ð5Þ. (e) Graph f ðxÞ. (a) f ð6Þ ¼ ð6 2Þð8 6Þ ¼ 4 2 ¼ 8 f ð1Þ is not defined since f ðxÞ is defined only for 2 @ x @ 8. (b) The set of all x such that 2 @ x @ 8. (c) f ð1 2tÞ ¼ fð1 2tÞ 2gf8 ð1 2tÞg ¼ ð1 þ 2tÞð7 þ 2tÞ where t is such that 2 @ 1 2t @ 8, i.e., 7=2 @ t @ 1=2. (d) f ð3Þ ¼ ð3 2Þð8 3Þ ¼ 5, f ½ f ð3Þ ¼ f ð5Þ ¼ ð5 2Þð8 5Þ ¼ 9. f ð5Þ ¼ 9 so that f ½ f ð5Þ ¼ f ð9Þ is not defined. (e) The following table shows f ðxÞ for various values of x. Plot points ð2; 0Þ; ð3; 5Þ; ð4; 8Þ; ð5; 9Þ; ð6; 8Þ; ð7; 5Þ; ð8; 0Þ; ð2:5; 2:75Þ; ð7:5; 2:75Þ. These points are only a few of the infinitely many points on the required graph shown in the adjoining Fig. 3-5. This set of points defines a curve which is part of a parabola. 3.2. Let gðxÞ ¼ ðx 2Þð8 xÞ for 2 x 8. (a) Discuss the difference between the graph of gðxÞ and that of f ðxÞ in Problem 3.1. (b) What is the l.u.b. and g.l.b. of gðxÞ? (c) Does gðxÞ attain its l.u.b. and g.l.b. for any value of x in the domain of definition? (d) Answer parts (b) and (c) for the function f ðxÞ of Problem 3.1. (a) The graph of gðxÞ is the same as that in Problem 3.1 except that the two points ð2; 0Þ and ð8; 0Þ are missing, since gðxÞ is not defined at x ¼ 2 and x ¼ 8. (b) The l.u.b. of gðxÞ is 9. The g.l.b. of gðxÞ is 0. (c) The l.u.b. of gðxÞ is attained for the value of x ¼ 5. The g.l.b. of gðxÞ is not attained, since there is no value of x in the domain of definition such that gðxÞ ¼ 0. (d) As in (b), the l.u.b. of f ðxÞ is 9 and the g.l.b. of f ðxÞ is 0. The l.u.b. of f ðxÞ is attained for the value x ¼ 5 and the g.l.b. of f ðxÞ is attained at x ¼ 2 and x ¼ 8. Note that a function, such as f ðxÞ, which is continuous in a closed interval attains its l.u.b. and g.l.b. at some point of the interval. However, a function, such as gðxÞ, which is not continuous in a closed interval need not attain its l.u.b. and g.l.b. See Problem 3.34. 3.3. Let f ðxÞ ¼ 1; if x is a rational number 0; if x is an irrational number . (a) Find f ð2 3Þ, f ð5Þ, f ð1:41423Þ, f ð ffiffiffi 2 p Þ, (b) Construct a graph of f ðxÞ and explain why it is misleading by itself. (a) f ð2 3Þ ¼ 1 since 2 3 is a rational number f ð5Þ ¼ 1 since 5 is a rational number f ð1:41423Þ ¼ 1 since 1.41423 is a rational number f ð ffiffiffi 2 p Þ ¼ 0 since ffiffiffi 2 p is an irrational number CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 49 2 4 6 8 x 2 4 6 8 f (x) Fig. 3-5 x 2 3 4 5 6 7 8 2.5 7.5 f ðxÞ 0 5 8 9 8 5 0 2.75 2.75
- 59. (b) The graph is shown in the adjoining Fig. 3-6. Because both the sets of rational numbers and irrational numbers are dense, the visual impression is that there are two images corresponding to each domain value. In actuality, each domain value has only one corresponding range value. 3.4. Referring to Problem 3.1: (a) Draw the graph with axes interchanged, thus illustrating the two possible choices avail- able for definition of f 1 . (b) Solve for x in terms of y to determine the equations describing the two branches, and then interchange the variables. (a) The graph of y ¼ f ðxÞ is shown in Fig. 3-5 of Problem 3.1(a). By interchanging the axes (and the variables), we obtain the graphical form of Fig. 3-7. This figure illustrates that there are two values of y corresponding to each value of x, and hence two branches. Either may be employed to define f 1 . (b) We have y ¼ ðx 2Þð8 xÞ or x2 10x þ 16 þ y ¼ 0. The solu- tion of this quadratic equation is x ¼ 5 ffiffiffiffiffiffiffiffiffiffiffiffi 9 y: p After interchanging variables y ¼ 5 ffiffiffiffiffiffiffiffiffiffiffi 9 x p : In the graph, AP represents y ¼ 5 þ ffiffiffiffiffiffiffiffiffiffiffi 9 x p , and BP designates y ¼ 5 ffiffiffiffiffiffiffiffiffiffiffi 9 x p . Either branch may represent f 1 . Note: The point at which the two branches meet is called a branch point. 3.5. (a) Prove that gðxÞ ¼ 5 þ ffiffiffiffiffiffiffiffiffiffiffi 9 x p is strictly decreasing in 0 @ x @ 9. (b) Is it monotonic decreasing in this interval? (c) Does gðxÞ have a single-valued inverse? (a) gðxÞ is strictly decreasing if gðx1Þ gðx2Þ whenever x1 x2. If x1 x2 then 9 x1 9 x2, ffiffiffiffiffiffiffiffiffiffiffiffiffi 9 x1 p ffiffiffiffiffiffiffiffiffiffiffiffiffi 9 x2 p , 5 þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 9 x1 p 5 þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 9 x2 p showing that gðxÞ is strictly decreasing. (b) Yes, any strictly decreasing function is also monotonic decreasing, since if gðx1Þ gðx2Þ it is also true that gðx1Þ A gðx2Þ. However, if gðxÞ is monotonic decreasing, it is not necessarily strictly decreasing. (c) If y ¼ 5 þ ffiffiffiffiffiffiffiffiffiffiffi 9 x p , then y 5 ¼ ffiffiffiffiffiffiffiffiffiffiffi 9 x p or squaring, x ¼ 16 þ 10y y2 ¼ ðy 2Þð8 yÞ and x is a single-valued function of y, i.e., the inverse function is single-valued. In general, any strictly decreasing (or increasing) function has a single-valued inverse (see Theorem 6, Page 47). The results of this problem can be interpreted graphically using the figure of Problem 3.4. 3.6. Construct graphs for the functions (a) f ðxÞ ¼ x sin 1=x; x 0 0; x ¼ 0 , (b) f ðxÞ ¼ ½x ¼ greatest integer @ x. (a) The required graph is shown in Fig. 3-8. Since jx sin 1=xj @ jxj, the graph is included between y ¼ x and y ¼ x. Note that f ðxÞ ¼ 0 when sin 1=x ¼ 0 or 1=x ¼; m, m ¼ 1; 2; 3; 4; . . . ; i.e., where x ¼ 1=; 1=2; 1=3; . . . . The curve oscillates infinitely often between x ¼ 1= and x ¼ 0. (b) The required graph is shown in Fig. 3-9. If 1 @ x 2, then ½x ¼ 1. Thus ½1:8 ¼ 1, ½ ffiffiffi 2 p ¼ 1, ½1:99999 ¼ 1. However, ½2 ¼ 2. Similarly for 2 @ x 3, ½x ¼ 2, etc. Thus there are jumps at the integers. The function is sometimes called the staircase function or step function. 3.7. (a) Construct the graph of f ðxÞ ¼ tan x. (b) Construct the graph of some of the infinite number of branches available for a definition of tan1 x. (c) Show graphically why the relationship of x 50 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3 0 x 1 f (x) Fig. 3-6 y = f _1(x) 8 6 4 2 2 4 6 8 x B A P Fig. 3-7
- 60. to y is multivalued. (d) Indicate possible principal values for tan1 x. (e) Using your choice, evaluate tan1 ð1Þ. (a) The graph of f ðxÞ ¼ tan x appears in Fig. 3-10 below. (b) The required graph is obtained by interchanging the x and y axes in the graph of (a). The result, with axes oriented as usual, appears in Fig. 3-11 above. (c) In Fig. 3-11 of (b), any vertical line meets the graph in infinitely many points. Thus, the relation of y to x is multivalued and infinitely many branches are available for the purpose of defining tan1 x. (d) To define tan1 x as a single-valued function, it is clear from the graph that we can only do so by restricting its value to any of the following: =2 tan1 x =2; =2 tan1 x 3=2, etc. We shall agree to take the first as defining the principal value. Note that no matter which branch is used to define tan1 x, the resulting function is strictly increasing. (e) tan1 ð1Þ ¼ =4 is the only value lying between =2 and =2, i.e., it is the principal value according to our choice in ðdÞ. 3.8. Show that f ðxÞ ¼ ffiffiffi x p þ 1 x þ 1 , x 6¼ 1, describes an irrational algebraic function. If y ¼ ffiffiffi x p þ 1 x þ 1 then ðx þ 1Þy 1 ¼ ffiffiffi x p or squaring, ðx þ 1Þ2 y2 2ðx þ 1Þy þ 1 x ¼ 0, a polynomial equation in y whose coefficients are polynomials in x. Thus f ðxÞ is an algebraic function. However, it is not the quotient of two polynomials, so that it is an irrational algebraic function. CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 51 _p/2 _p p/2 p 3p/2 2p x y = f (x) = tan x Fig. 3-10 _p p _p/2 p/2 3p/2 x f _1(x) = tan _1x Fig. 3-11 f (x) x y = x y = _ x 1/2p 1/p Fig. 3-8 _3 _2 _1 1 2 3 4 5 x f (x) Fig. 3-9
- 61. 3.9. If f ðxÞ ¼ cosh x ¼ 1 2 ðex þ ex Þ, prove that we can choose as the principal value of the inverse function, cosh1 x ¼ lnðx þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 p Þ, x A 1. If y ¼ 1 2 ðex þ ex Þ, e2x 2yex þ 1 ¼ 0. Then using the quadratic formula, ex ¼ 2y ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4y2 4 p 2 ¼ y ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p . Thus x ¼ lnðy ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p Þ. Since y ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p ¼ ðy ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p Þ y þ ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p y þ ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p ! ¼ 1 y þ ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 p , we can also write x ¼ lnðy þ ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 q Þ or cosh1 y ¼ lnðy þ ffiffiffiffiffiffiffiffiffiffiffiffiffi y2 1 q Þ Choosing the þ sign as defining the principal value and replacing y by x, we have cosh1 x ¼ lnðx þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 p Þ. The choice x A 1 is made so that the inverse function is real. LIMITS 3.10. If (a) f ðxÞ ¼ x2 , (b) f ðxÞ ¼ x2 ; x 6¼ 2 0; x ¼ 2 , prove that lim x!2 f ðxÞ ¼ 4. (a) We must show that given any 0 we can find 0 (depending on in general) such that jx2 4j when 0 jx 2j . Choose @ 1 so that 0 jx 2j 1 or 1 x 3, x 6¼ 2. Then jx2 4j ¼ jðx 2Þðx þ 2Þj ¼ jx 2jjx þ 2j jx þ 2j 5. Take as 1 or =5, whichever is smaller. Then we have jx2 4j whenever 0 jx 2j and the required result is proved. It is of interest to consider some numerical values. If for example we wish to make jx2 4j :05, we can choose ¼ =5 ¼ :05=5 ¼ :01. To see that this is actually the case, note that if 0 jx 2j :01 then 1:99 x 2:01 ðx 6¼ 2Þ and so 3:9601 x2 4:0401, :0399 x2 4 :0401 and certainly jx2 4j :05 ðx2 6¼ 4Þ. The fact that these inequalities also happen to hold at x ¼ 2 is merely coin- cidental. If we wish to make jx2 4j 6, we can choose ¼ 1 and this will be satisfied. (b) There is no difference between the proof for this case and the proof in (a), since in both cases we exclude x ¼ 2. 3.11. Prove that lim x!1 2x4 6x3 þ x2 þ 3 x 1 ¼ 8. We must show that for any 0 we can find 0 such that 2x4 6x3 þ x2 þ 3 x 1 ð8Þ when 0 jx 1j . Since x 6¼ 1, we can write 2x4 6x3 þ x2 þ 3 x 1 ¼ ð2x3 4x2 3x 3Þðx 1Þ x 1 ¼ 2x3 4x2 3x 3 on cancelling the common factor x 1 6¼ 0. Then we must show that for any 0, we can find 0 such that j2x3 4x2 3x þ 5j when 0 jx 1j . Choosing @ 1, we have 0 x 2, x 6¼ 1. Now j2x3 4x2 3x þ 5j ¼ jx 1jj2x2 2x 5j j2x2 2x 5j ðj2x2 j þ j2xj þ 5Þ ð8 þ 4 þ 5Þ ¼ 17. Taking as the smaller of 1 and =17, the required result follows. 3.12. Let f ðxÞ ¼ jx 3j x 3 ; x 6¼ 3 0; x ¼ 3 8 : , (a) Graph the function. (b) Find lim x!3þ f ðxÞ. (c) Find lim x!3 f ðxÞ. (d) Find lim x!3 f ðxÞ. (a) For x 3, jx 3j x 3 ¼ x 3 x 3 ¼ 1. For x 3, jx 3j x 3 ¼ ðx 3Þ x 3 ¼ 1. 52 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 62. Then the graph, shown in the adjoining Fig. 3-12, consists of the lines y ¼ 1, x 3; y ¼ 1, x 3 and the point ð3; 0Þ. (b) As x ! 3 from the right, f ðxÞ ! 1, i.e., lim x!3þ f ðxÞ ¼ 1, as seems clear from the graph. To prove this we must show that given any 0, we can find 0 such that j f ðxÞ 1j whenever 0 x 1 . Now since x 1, f ðxÞ ¼ 1 and so the proof con- sists in the triviality that j1 1j whenever 0 x 1 . (c) As x ! 3 from the left, f ðxÞ ! 1, i.e., lim x!3 f ðxÞ ¼ 1. A proof can be formulated as in (b). (d) Since lim x!3þ f ðxÞ 6¼ lim x!3 f ðxÞ, lim x!3 f ðxÞ does not exist. 3.13. Prove that lim x!0 x sin 1=x ¼ 0. We must show that given any 0, we can find 0 such that jx sin 1=x 0j when 0 jx 0j . If 0 jxj , then jx sin 1=xj ¼ jxjj sin 1=xj @ jxj since j sin 1=xj @ 1 for all x 6¼ 0. Making the choice ¼ , we see that jx sin 1=xj when 0 jxj , completing the proof. 3.14. Evaluate lim x!0þ 2 1 þ e1=x . As x ! 0þ we suspect that 1=x increases indefinitely, e1=x increases indefinitely, e1=x approaches 0, 1 þ e1=x approaches 1; thus the required limit is 2. To prove this conjecture we must show that, given 0, we can find 0 such that 2 1 þ e1=x 2 when 0 x 2 1 þ e1=x 2 ¼ 2 2 2e1=x 1 þ e1=x ¼ 2 e1=x þ 1 Now Since the function on the right is smaller than 1 for all x 0, any 0 will work when e 1. If 0 1, then 2 e1=x þ 1 when e1=x þ 1 2 1 , e1=x 2 1, 1 x ln 2 1 ; or 0 x 1 lnð2= 1Þ ¼ . 3.15. Explain exactly what is meant by the statement lim x!1 1 ðx 1Þ4 ¼ 1 and prove the validity of this statement. The statement means that for each positive number M, we can find a positive number (depending on M in general) such that 1 ðx 1Þ4 4 when 0 jx 1j To prove this note that 1 ðx 1Þ4 M when 0 ðx 1Þ4 1 M or 0 jx 1j 1 ffiffiffiffiffi M 4 p . Choosing ¼ 1= ffiffiffiffiffi M 4 p , the required results follows. 3.16. Present a geometric proof that lim !0 sin ¼ 1. Construct a circle with center at O and radius OA ¼ OD ¼ 1, as in Fig. 3-13 below. Choose point B on OA extended and point C on OD so that lines BD and AC are perpendicular to OD. It is geometrically evident that CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 53 f (x) x (3, 0) 1 1 Fig. 3-12
- 63. 54 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3 Area of triangle OAC Area of sector OAD Area of triangle OBD 1 2 sin cos 1 2 1 2 tan i.e., Dividing by 1 2 sin , cos sin 1 cos cos sin 1 cos or As ! 0, cos ! 1 and it follows that lim !0 sin ¼ 1. THEOREMS ON LIMITS 3.17. If lim x!x0 f ðxÞ exists, prove that it must be unique. We must show that if lim x!x0 f ðxÞ ¼ l1 and lim x!x0 f ðxÞ ¼ l2, then l1 ¼ l2. By hypothesis, given any 0 we can find 0 such that j f ðxÞ l1j =2 when 0 jx x0j j f ðxÞ l2j =2 when 0 jx x0j Then by the absolute value property 2 on Page 3, jl1 l2j ¼ jl1 f ðxÞ þ f ðxÞ l2j @ jl1 f ðxÞj þ j f ðxÞ l2j =2 þ =2 ¼ i.e., jl1 l2j is less than any positive number (however small) and so must be zero. Thus l1 ¼ l2. 3.18. If lim x!x0 gðxÞ ¼ B 6¼ 0, prove that there exists 0 such that jgðxÞj 1 2 jBj for 0 jx x0j Since lim x!x0 gðxÞ ¼ B, we can find 0 such that jgðxÞ Bj 1 2 jBj for 0 jx x0j . Writing B ¼ B gðxÞ þ gðxÞ, we have jBj @ jB gðxÞj þ jgðxÞj 1 2 jBj þ jgðxÞj i.e., jBj 1 2 jBj þ jgðxÞj, from which jgðxÞj 1 2 jBj. 3.19. Given lim x!x0 f ðxÞ ¼ A and lim x!x0 gðxÞ ¼ B, prove (a) lim x!x0 ½ f ðxÞ þ gðxÞ ¼ A þ B, (b) lim x!x0 f ðxÞgðxÞ ¼ AB, (c) lim x!x0 1 gðxÞ ¼ 1 B if B 6¼ 0, (d) lim x!x0 f ðxÞ gðxÞ ¼ A B if B 6¼ 0. (a) We must show that for any 0 we can find 0 such that j½ f ðxÞ þ gðxÞ ðA þ BÞj when 0 jx x0j Using absolute value property 2, Page 3, we have j½ f ðxÞ þ gðxÞ ðA þ BÞj ¼ j½ f ðxÞ A þ ½gðxÞ Bj @ j f ðxÞ Aj þ jgðxÞ Bj ð1Þ By hypothesis, given 0 we can find 1 0 and 2 0 such that j f ðxÞ Aj =2 when 0 jx x0j 1 ð2Þ jgðxÞ Bj =2 when 0 jx x0j 2 ð3Þ Then from (1), (2), and (3), j½ f ðxÞ þ gðxÞ ðA þ BÞj =2 þ =2 ¼ when 0 jx x0j where is chosen as the smaller of 1 and 2. B A C cos G sin G tan G D O G Fig. 3-13
- 64. (b) We have j f ðxÞgðxÞ ABj ¼ j f ðxÞ½gðxÞ B þ B½ f ðxÞ Aj ð4Þ @ j f ðxÞjjgðxÞ Bj þ jBjj f ðxÞ Aj @ j f ðxÞjjgðxÞ Bj þ ðjBj þ 1Þj f ðxÞ Aj Since lim x!x0 f ðxÞ ¼ A, we can find 1 such j f ðxÞ Aj 1 for 0 jx x0j 1, i.e., A 1 f ðxÞ A þ 1, so that f ðxÞ is bounded, i.e., j f ðxÞj P where P is a positive constant. Since lim x!x0 gðxÞ ¼ B, given 0 we can find 2 0 such that jgðxÞ Bj =2P for 0 jx x0j 2. Since lim x!x0 f ðxÞ ¼ A, given 0 we can find 3 0 such that j f ðxÞ Aj 2ðjBj þ 1Þ for 0 jx x0j 2. Using these in (4), we have j f ðxÞgðxÞ ABj P 2P þ ðjBj þ 1Þ 2ðjBj þ 1Þ ¼ for 0 jx x0j where is the smaller of 1; 2; 3 and the proof is complete. (c) We must show that for any 0 we can find 0 such that 1 gðxÞ 1 B ¼ jgðxÞ Bj jBjjgðxÞj when 0 jx x0j ð5Þ By hypothesis, given 0 we can find 1 0 such that jgðxÞ Bj 1 2 B2 when 0 jx x0j 1 By Problem 3.18, since lim x!x0 gðxÞ ¼ B 6¼ 0, we can find 2 0 such that jgðxÞj 1 2 jBj when 0 jx x0j 2 Then if is the smaller of 1 and 2, we can write 1 gðxÞ 1 B ¼ jgðxÞ Bj jBjjgðxÞj 1 2 B2 jBj 1 2 jBj ¼ whenever 0 jx x0j and the required result is proved. (d) From parts (b) and (c), lim x!x0 f ðxÞ gðxÞ ¼ lim x!x0 f ðxÞ 1 gðxÞ ¼ lim x!x0 f ðxÞ lim x!x0 1 gðxÞ ¼ A 1 B ¼ A B This can also be proved directly (see Problem 3.69). The above results can also be proved in the cases x ! x0þ, x ! x0, x ! 1, x ! 1. Note: In the proof of (a) we have used the results j f ðxÞ Aj =2 and jgðxÞ Bj =2, so that the final result would come out to be j f ðxÞ þ gðxÞ ðA þ BÞj . Of course the proof would be just as valid if we had used 2 (or any other positive multiple of ) in place of . A similar remark holds for the proofs of ðbÞ, (c), and (d). 3.20. Evaluate each of the following, using theorems on limits. ðaÞ lim x!2 ðx2 6x þ 4Þ ¼ lim x!2 x2 þ lim x!2 ð6xÞ þ lim x!2 4 ¼ ðlim x!2 xÞðlim x!2 xÞ þ ðlim x!2 6Þðlim x!2 xÞ þ lim x!2 4 ¼ ð2Þð2Þ þ ð6Þð2Þ þ 4 ¼ 4 In practice the intermediate steps are omitted. CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 55
- 65. ðbÞ lim x!1 ðx þ 3Þð2x 1Þ x2 þ 3x 2 ¼ lim x!1 ðx þ 3Þ lim x!1 ð2x 1Þ lim x!1 ðx2 þ 3x 2Þ ¼ 2 ð3Þ 4 ¼ 3 2 ðcÞ lim x!1 2x4 3x2 þ 1 6x4 þ x3 3x ¼ lim x!1 2 3 x2 þ 1 x4 6 þ 1 x 3 x3 ¼ lim x!1 2 þ lim x!1 3 x2 þ lim x!1 1 x4 lim x!1 6 þ lim x!1 1 x þ lim x!1 3 x3 ¼ 2 6 ¼ 1 3 by Problem 3.19. ðdÞ lim h!0 ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p 2 h ¼ lim h!0 ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p 2 h ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p þ 2 ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p þ 2 ¼ lim h!0 4 þ h 4 hð ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p þ 2Þ ¼ lim h!0 1 ffiffiffiffiffiffiffiffiffiffiffi 4 þ h p þ 2 ¼ 1 2 þ 2 ¼ 1 4 ðeÞ lim x!0þ sin x ffiffiffi x p ¼ lim x!0þ sin x x ffiffiffi x p ¼ lim x!0þ sin x x lim x!0þ ffiffiffi x p ¼ 1 0 ¼ 0: Note that in (c), (d), and (e) if we use the theorems on limits indiscriminately we obtain the so called indeterminate forms 1=1 and 0/0. To avoid such predicaments, note that in each case the form of the limit is suitably modified. For other methods of evaluating limits, see Chapter 4. CONTINUITY (Assume that values at which continuity is to be demonstrated, are interior domain values unless otherwise stated.) 3.21. Prove that f ðxÞ ¼ x2 is continuous at x ¼ 2. Method 1: By Problem 3.10, lim x!2 f ðxÞ ¼ f ð2Þ ¼ 4 and so f ðxÞ is continuous at x ¼ 2. Method 2: We must show that given any 0, we can find 0 (depending on ) such that j f ðxÞ f ð2Þj ¼ jx2 4j when jx 2j . The proof patterns that are given in Problem 3.10. 3.22. (a) Prove that f ðxÞ ¼ x sin 1=x; x 6¼ 0 5; x ¼ 0 is not continuous at x ¼ 0. (b) Can one redefine f ð0Þ so that f ðxÞ is continuous at x ¼ 0? (a) From Problem 3.13, lim x!0 f ðxÞ ¼ 0. But this limit is not equal to f ð0Þ ¼ 5, so that f ðxÞ is discontinuous at x ¼ 0. (b) By redefining f ðxÞ so that f ð0Þ ¼ 0, the function becomes continuous. Because the function can be made continuous at a point simply by redefining the function at the point, we call the point a removable discontinuity. 3.23. Is the function f ðxÞ ¼ 2x4 6x3 þ x2 þ 3 x 1 continuous at x ¼ 1? f ð1Þ does not exist, so that f ðxÞ is not continuous at x ¼ 1. By redefining f ðxÞ so that f ð1Þ ¼ lim x!1 f ðxÞ ¼ 8 (see Problem 3.11), it becomes continuous at x ¼ 1, i.e., x ¼ 1 is a removable discontinuity. 3.24. Prove that if f ðxÞ and gðxÞ are continuous at x ¼ x0, so also are (a) f ðxÞ þ gðxÞ, (b) f ðxÞgðxÞ, (c) f ðxÞ gðxÞ if f ðx0Þ 6¼ 0. 56 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 66. These results follow at once from the proofs given in Problem 3.19 by taking A ¼ f ðx0Þ and B ¼ gðx0Þ and rewriting 0 jx x0j as jx x0j , i.e., including x ¼ x0. 3.25. Prove that f ðxÞ ¼ x is continuous at any point x ¼ x0. We must show that, given any 0, we can find 0 such that j f ðxÞ f ðx0Þj ¼ jx x0j when jx x0j . By choosing ¼ , the result follows at once. 3.26. Prove that f ðxÞ ¼ 2x3 þ x is continuous at any point x ¼ x0. Since x is continuous at any point x ¼ x0 (Problem 3.25) so also is x x ¼ x2 , x2 x ¼ x3 , 2x3 , and finally 2x3 þ x, using the theorem (Problem 3.24) that sums and products of continuous functions are continuous. 3.27. Prove that if f ðxÞ ¼ ffiffiffiffiffiffiffiffiffiffiffi x 5 p for 5 @ x @ 9, then f ðxÞ is continuous in this interval. If x0 is any point such that 5 x0 9, then lim x!x0 f ðxÞ ¼ lim x!x0 ffiffiffiffiffiffiffiffiffiffiffi x 5 p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi x0 5 p ¼ f ðx0Þ. Also, lim x!5þ ffiffiffiffiffiffiffiffiffiffiffi x 5 p ¼ 0 ¼ f ð5Þ and lim x!9 ffiffiffiffiffiffiffiffiffiffiffi x 5 p ¼ 2 ¼ f ð9Þ. Thus the result follows. Here we have used the result that lim x!x0 ffiffiffiffiffiffiffiffiffi f ðxÞ p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lim x!x0 f ðxÞ q ¼ ffiffiffiffiffiffiffiffiffiffiffi f ðx0Þ p if f ðxÞ is continuous at x0. An , proof, directly from the definition, can also be employed. 3.28. For what values of x in the domain of definition is each of the following functions continuous? (a) f ðxÞ ¼ x x2 1 Ans. all x except x ¼ 1 (where the denominator is zero) (b) f ðxÞ ¼ 1 þ cos x 3 þ sin x Ans. all x (c) f ðxÞ ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 þ 4 4 p Ans. All x 10 (d) f ðxÞ ¼ 101=ðx3Þ2 Ans. all x 6¼ 3 (see Problem 3.55) (e) f ðxÞ ¼ 101=ðx3Þ2 ; x 6¼ 3 0; x ¼ 3 Ans. all x, since lim x!3 f ðxÞ ¼ f ð3Þ ( f ) f ðxÞ ¼ x jxj x If x 0, f ðxÞ ¼ x x x ¼ 0. If x 0, f ðxÞ ¼ x þ x x ¼ 2. At x ¼ 0, f ðxÞ is undefined. Then f ðxÞ is continuous for all x except x ¼ 0. ðgÞ f ðxÞ ¼ x jxj x ; x 0 2; x ¼ 0 8 : As in ð f Þ, f ðxÞ is continuous for x 0. Then since lim x!0 x jxj x ¼ lim x!0 x þ x x ¼ lim x!0 2 ¼ 2 ¼ f ð0Þ if follows that f ðxÞ is continuous (from the left) at x ¼ 0. Thus, f ðxÞ is continuous for all x @ 0, i.e., everywhere in its domain of definition. ðhÞ f ðxÞ ¼ x csc x ¼ x sin x : Ans: all x except 0; ; 2; 3; . . . : (i) f ðxÞ ¼ x csc x, f ð0Þ ¼ 1. Since lim x!0 x csc x ¼ lim x!0 x sin x ¼ 1 ¼ f ð0Þ, we see that f ðxÞ is continuous for all x except ; 2; 3; . . . [compare (h)]. CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 57
- 67. UNIFORM CONTINUITY 3.29. Prove that f ðxÞ ¼ x2 is uniformly continuous in 0 x 1. Method 1: Using definition. We must show that given any 0 we can find 0 such that jx2 x2 0j when jx x0j , where depends only on and not on x0 where 0 x0 1. If x and x0 are any points in 0 x 1, then jx2 x2 0j ¼ jx þ x0jjx x0j j1 þ 1jjx x0j ¼ 2jx x0j Thus if jx x0j it follows that jx2 x2 0j 2. Choosing ¼ =2, we see that jx2 x2 0j when jx x0j , where depends only on and not on x0. Hence, f ðxÞ ¼ x2 is uniformly continuous in 0 x 1. The above can be used to prove that f ðxÞ ¼ x2 is uniformly continuous in 0 @ x @1. Method 2: The function f ðxÞ ¼ x2 is continuous in the closed interval 0 @ x @ 1. Hence, by the theorem on Page 48 is uniformly continuous in 0 @ x @ 1 and thus in 0 x 1. 3.30. Prove that f ðxÞ ¼ 1=x is not uniformly continuous in 0 x 1. Method 1: Suppose f ðxÞ is uniformly continuous in the given interval. Then for any 0 we should be able to find , say, between 0 and 1, such that j f ðxÞ f ðx0Þj when jx x0j for all x and x0 in the interval. Let x ¼ and x0 ¼ 1 þ : Then jx x0j ¼ 1 þ ¼ 1 þ : However, 1 x 1 x0 ¼ 1 1 þ ¼ (since 0 1Þ: Thus, we have a contradiction and it follows that f ðxÞ ¼ 1=x cannot be uniformly continuous in 0 x 1. Method 2: Let x0 and x0 þ be any two points in ð0; 1Þ. Then j f ðx0Þ f ðx0 þ Þj ¼ 1 x0 1 x0 þ ¼ x0ðx0 þ Þ can be made larger than any positive number by choosing x0 sufficiently close to 0. Hence, the function cannot be uniformly continuous. MISCELLANEOUS PROBLEMS 3.31. If y ¼ f ðxÞ is continuous at x ¼ x0, and z ¼ gðyÞ is continuous at y ¼ y0 where y0 ¼ f ðx0Þ, prove that z ¼ gf f ðxÞg is continuous at x ¼ x0. Let hðxÞ ¼ gf f ðxÞg. Since by hypothesis f ðxÞ and gð yÞ are continuous at x0 and y0, respectively, we have lim x!x0 f ðxÞ ¼ f ð lim x!x0 xÞ ¼ f ðx0Þ lim y!y0 gðyÞ ¼ gð lim y!y0 yÞ ¼ gðy0Þ ¼ gf f ðx0Þg Then lim x!x0 hðxÞ ¼ lim x!x0 gf f ðxÞg ¼ gf lim x!x0 f ðxÞg ¼ gf f ðx0Þg ¼ hðx0Þ which proves that hðxÞ ¼ gf f ðxÞg is continuous at x ¼ x0. 3.32. Prove Theorem 8, Page 48. 58 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 68. Suppose that f ðaÞ 0 and f ðbÞ 0. Since f ðxÞ is continuous there must be an interval ða; a þ hÞ, h 0, for which f ðxÞ 0. The set of points ða; a þ hÞ has an upper bound and so has a least upper bound which we call c. Then f ðcÞ @ 0. Now we cannot have f ðcÞ 0, because if f ðcÞ were negative we would be able to find an interval about c (including values greater than c) for which f ðxÞ 0; but since c is the least upper bound, this is impossible, and so we must have f ðcÞ ¼ 0 as required. If f ðaÞ 0 and f ðbÞ 0, a similar argument can be used. 3.33. (a) Given f ðxÞ ¼ 2x3 3x2 þ 7x 10, evaluate f ð1Þ and f ð2Þ. (b) Prove that f ðxÞ ¼ 0 for some real number x such that 1 x 2. (c) Show how to calculate the value of x in (b). (a) f ð1Þ ¼ 2ð1Þ3 3ð1Þ2 þ 7ð1Þ 10 ¼ 4, f ð2Þ ¼ 2ð2Þ3 3ð2Þ2 þ 7ð2Þ 10 ¼ 8. (b) If f ðxÞ is continuous in a @ x @ b and if f ðaÞ and f ðbÞ have opposite signs, then there is a value of x between a and b such that f ðxÞ ¼ 0 (Problem 3.32). To apply this theorem we need only realize that the given polynomial is continuous in 1 @ x @ 2, since we have already shown in (a) that f ð1Þ 0 and f ð2Þ 0. Thus there exists a number c between 1 and 2 such that f ðcÞ ¼ 0. (c) f ð1:5Þ ¼ 2ð1:5Þ3 3ð1:5Þ2 þ 7ð1:5Þ 10 ¼ 0:5. Then applying the theorem of (b) again, we see that the required root lies between 1 and 1.5 and is ‘‘most likely’’ closer to 1.5 than to 1, since f ð1:5Þ ¼ 0:5 has a value closer to 0 than f ð1Þ ¼ 4 (this is not always a valid conclusion but is worth pursuing in practice). Thus we consider x ¼ 1:4. Since f ð1:4Þ ¼ 2ð1:4Þ3 3ð1:4Þ2 þ 7ð1:4Þ 10 ¼ 0:592, we conclude that there is a root between 1.4 and 1.5 which is most likely closer to 1.5 than to 1.4. Continuing in this manner, we find that the root is 1.46 to 2 decimal places. 3.34. Prove Theorem 10, Page 48. Given any 0, we can find x such that M f ðxÞ by definition of the l.u.b. M. Then 1 M f ðxÞ 1 , so that 1 M f ðxÞ is not bounded and hence cannot be continuous in view of Theorem 4, Page 47. However, if we suppose that f ðxÞ 6¼ M, then since M f ðxÞ is continuous, by hypothesis, we must have 1 M f ðxÞ also continuous. In view of this contradiction, we must have f ðxÞ ¼ M for at least one value of x in the interval. Similarly, we can show that there exists an x in the interval such that f ðxÞ ¼ m (Problem 3.93). Supplementary Problems FUNCTIONS 3.35. Give the largest domain of definition for which each of the following rules of correspondence support the construction of a function. (a) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð3 xÞð2x þ 4Þ p , (b) ðx 2Þ=ðx2 4Þ, (c) ffiffiffiffiffiffiffiffiffiffiffiffi sin 3x p , (d) log10ðx3 3x2 4x þ 12Þ. Ans. (a) 2 @ x @ 3, (b) all x 6¼ 2, (c) 2m=3 @ x @ ð2m þ 1Þ=3, m ¼ 0; 1; 2; . . . ; (d) x 3, 2 x 2. 3.36. If f ðxÞ ¼ 3x þ 1 x 2 , x 6¼ 2, find: (a) 5f ð1Þ 2f ð0Þ þ 3f ð5Þ 6 ; (b) f f ð 1 2Þg2 ; (c) f ð2x 3Þ; (d) f ðxÞ þ f ð4=xÞ, x 6¼ 0; (e) f ðhÞ f ð0Þ h , h 6¼ 0; ( f ) f ðf f ðxÞg. Ans. (a) 61 18 (b) 1 25 (c) 6x 8 2x 5 , x 6¼ 0, 5 2, 2 (d) 5 2, x 6¼ 0; 2 (e) 7 2h 4 , h 6¼ 0; 2 ( f ) 10x þ 1 x þ 5 , x 6¼ 5; 2 CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 59
- 69. 3.37. If f ðxÞ ¼ 2x2 , 0 x @ 2, find (a) the l.u.b. and (b) the g.l.b. of f ðxÞ. Determine whether f ðxÞ attains its l.u.b. and g.l.b. Ans. (a) 8, (b) 0 3.38. Construct a graph for each of the following functions. ðaÞ f ðxÞ ¼ jxj; 3 @ x @ 3 ð f Þ x ½x x where ½x ¼ greatest integer @ x ðbÞ f ðxÞ ¼ 2 jxj x ; 2 @ x @ 2 ðgÞ f ðxÞ ¼ cosh x ðcÞ f ðxÞ ¼ 0; x 0 1 2 ; x ¼ 0 1; x 0 8 : ðhÞ f ðxÞ ¼ sin x x ðdÞ f ðxÞ ¼ x; 2 @ x @ 0 x; 0 @ x @ 2 ðiÞ f ðxÞ ¼ x ðx 1Þðx 2Þðx 3Þ ðeÞ f ðxÞ ¼ x2 sin 1=x; x 6¼ 0 ð jÞ f ðxÞ ¼ sin2 x x2 3.39. Construct graphs for (a) x2 =a2 þ y2 =b2 ¼ 1, (b) x2 =a2 y2 =b2 ¼ 1, (c) y2 ¼ 2px, and (d) y ¼ 2ax x2 , where a; b; p are given constants. In which cases when solved for y is there exactly one value of y assigned to each value of x, thus making possible definitions of functions f , and enabling us to write y ¼ f ðxÞ? In which cases must branches be defined? 3.40. (a) From the graph of y ¼ cos x construct the graph obtained by interchanging the variables, and from which cos1 x will result by choosing an appropriate branch. Indicate possible choices of a principal value of cos1 x. Using this choice, find cos1 ð1=2Þ cos1 ð1=2Þ. Does the value of this depend on the choice? Explain. 3.41. Work parts (a) and (b) of Problem 40 for (a) y ¼ sec1 x, (b) y ¼ cot1 x. 3.42. Given the graph for y ¼ f ðxÞ, show how to obtain the graph for y ¼ f ðax þ bÞ, where a and b are given constants. Illustrate the procedure by obtaining the graphs of (a) y ¼ cos 3x; ðbÞ y ¼ sinð5x þ =3Þ; ðcÞ y ¼ tanð=6 2xÞ. 3.43. Construct graphs for (a) y ¼ ejxj , (b) y ¼ ln jxj, (c) y ¼ ejxj sin x. 3.44. Using the conventional principal values on Pages 44 and 45, evaluate: (a) sin1 ð ffiffiffi 3 p =2Þ ( f ) sin1 x þ cos1 x; 1 @ x @ 1 (b) tan1 ð1Þ tan1 ð1Þ (g) sin1 ðcos 2xÞ; 0 @ x @ =2 (c) cot1 ð1= ffiffiffi 3 p Þ cot1 ð1= ffiffiffi 3 p Þ (h) sin1 ðcos 2xÞ; =2 @ x @ 3=2 (d) cosh1 ffiffiffi 2 p (i) tanh ðcsch1 3xÞ; x 6¼ 0 (e) e coth1 ð25=7Þ ( j) cosð2 tan1 x2 Þ Ans. (a) =3 (c) =3 (e) 3 4 (g) =2 2x (i) jxj x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9x2 þ 1 p ð jÞ 1 x4 1 þ x4 (b) =2 (d) lnð1 þ ffiffiffi 2 p Þ ( f ) =2 (h) 2x 3=2 3.45. Evaluate (a) cosf sinhðln 2Þg, (b) cosh1 fcothðln 3Þg. Ans. (a) ffiffiffi 2 p =2; ðbÞ ln 2 60 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 70. 3.46. (a) Prove that tan1 x þ cot1 x ¼ =2 if the conventional principal values on Page 44 are taken. (b) Is tan1 x þ tan1 ð1=xÞ ¼ =2 also? Explain. 3.47. If f ðxÞ ¼ tan1 x, prove that f ðxÞ þ f ðyÞ ¼ f x þ y 1 xy , discussing the case xy ¼ 1. 3.48. Prove that tan1 a tan1 b ¼ cot1 b cot1 a. 3.49. Prove the identities: (a) 1 tanh2 x ¼ sech2 x, (b) sin 3x ¼ 3 sin x 4 sin3 x, (c) cos 3x ¼ 4 cos3 x 3 cos x, (d) tanh 1 2 x ¼ ðsinh xÞ=ð1 þ cosh xÞ, (e) ln jcsc x cot xj ¼ ln j tan 1 2 xj. 3.50. Find the relative and absolute maxima and minima of: (a) f ðxÞ ¼ ðsin xÞ=x, f ð0Þ ¼ 1; (b) f ðxÞ ¼ ðsin2 xÞ= x2 , f ð0Þ ¼ 1. Discuss the cases when f ð0Þ is undefined or f ð0Þ is defined but 6¼ 1. LIMITS 3.51. Evaluate the following limits, first by using the definition and then using theorems on limits. ðaÞ lim x!3 ðx2 3x þ 2Þ; ðbÞ lim x!1 1 2x 5 ; ðcÞ lim x!2 x2 4 x 2 ; ðdÞ lim x!4 ffiffiffi x p 2 4 x ; ðeÞ lim h!0 ð2 þ hÞ4 16 h ; ð f Þ lim x!1 ffiffiffi x p x þ 1 : Ans. ðaÞ 2; ðbÞ 1 7 ; ðcÞ 4; ðdÞ 1 4 ; ðeÞ 32; ð f Þ 1 2 3.52. Let f ðxÞ ¼ 3x 1; x 0 0; x ¼ 0 2x þ 5; x 0 8 : : ðaÞ Construct a graph of f ðxÞ. Evaluate (b) lim x!2 f ðxÞ; ðcÞ lim x!3 f ðxÞ; ðdÞ lim x!0þ f ðxÞ; ðeÞ lim x!0 f ðxÞ; ð f Þ lim x!0 f ðxÞ, justifying your answer in each case. Ans. (b) 9, (c) 10, (d) 5, (e) 1, ( f ) does not exist 3.53. Evaluate (a) lim h!0þ f ðhÞ f ð0þÞ h and (b) lim h!0 f ðhÞ f ð0Þ h , where f ðxÞ is the function of Prob. 3.52. Ans. (a) 2, (b) 3 3.54. (a) If f ðxÞ ¼ x2 cos 1=x, evaluate lim x!0 f ðxÞ, justifying your answer. (b) Does your answer to (a) still remain the same if we consider f ðxÞ ¼ x2 cos 1=x, x 6¼ 0, f ð0Þ ¼ 2? Explain. 3.55. Prove that lim x!3 101=ðx3Þ2 ¼ 0 using the definition. 3.56. Let f ðxÞ ¼ 1 þ 101=x 2 101=x , x 6¼ 0, f ð0Þ ¼ 1 2. Evaluate (a) lim x!0þ f ðxÞ, (b) lim x!0 f ðxÞ, (c) lim x!0 f ðxÞ, justifying answers in all cases. Ans. (a) 1 2, (b) 1; ðcÞ does not exist. 3.57. Find (a) lim x!0þ jxj x ; ðbÞ lim x!0 jxj x . Illustrate your answers graphically. Ans. (a) 1, (b) 1 3.58. If f ðxÞ is the function defined in Problem 3.56, does lim x!0 f ðjxjÞ exist? Explain. 3.59. Explain exactly what is meant when one writes: ðaÞ lim x!3 2 x ðx 3Þ2 ¼ 1; ðbÞ lim x!0þ ð1 e1=x Þ ¼ 1; ðcÞ lim x!1 2x þ 5 3x 2 ¼ 2 3 : CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 61
- 71. 3.60. Prove that (a) lim x!1 10x ¼ 0; ðbÞ lim x!1 cos x x þ ¼ 0: 3.61. Explain why (a) lim x!1 sin x does not exist, (b) lim x!1 ex sin x does not exist. 3.62. If f ðxÞ ¼ 3x þ jxj 7x 5jxj , evaluate (a) lim x!1 f ðxÞ; ðbÞ lim x!1 f ðxÞ; ðcÞ lim x!0þ f ðxÞ; ðdÞ lim x!0 f ðxÞ; ðeÞ lim x!0 f ðxÞ. Ans. (a) 2, (b) 1/6, (c) 2, (d) 1/6, (e) does not exist. 3.63. If ½x ¼ largest integer @ x, evaluate (a) lim x!2þ fx ½xg; ðbÞ lim x!2 fx ½xg. Ans. (a) 0, (b) 1 3.64. If lim x!x0 f ðxÞ ¼ A, prove that (a) lim x!x0 f f ðxÞg2 ¼ A2 , (b) lim x!x0 ffiffiffiffiffiffiffiffiffi f ðxÞ 3 p ¼ ffiffiffiffi A 3 p . What generalizations of these do you suspect are true? Can you prove them? 3.65. If lim x!x0 f ðxÞ ¼ A and lim x!x0 gðxÞ ¼ B, prove that ðaÞ lim x!x0 f f ðxÞ gðxÞg ¼ A B; ðbÞ lim x!x0 faf ðxÞ þ bgðxÞg ¼ aA þ bB where a; b ¼ any constants. 3.66. If the limits of f ðxÞ, gðxÞ; and hðxÞ are A, B; and C respectively, prove that: (a) lim x!x0 f f ðxÞ þ gðxÞ þ hðxÞg ¼ A þ B þ C, (b) lim x!x0 f ðxÞgðxÞhðxÞ ¼ ABC. Generalize these results. 3.67. Evaluate each of the following using the theorems on limits. ðaÞ lim x!1=2 2x2 1 ð3x þ 2Þð5x 3Þ 2 3x x2 5x þ 3 ( ) Ans: ðaÞ 8=21 ðbÞ lim x!1 ð3x 1Þð2x þ 3Þ ð5x 3Þð4x þ 5Þ ðbÞ 3=10 ðcÞ lim x!1 3x x 1 2x x þ 1 ðcÞ 1 ðdÞ lim x!1 1 x 1 1 x þ 3 2x 3x þ 5 ðdÞ 1=32 3.68. Evaluate lim h!0 ffiffiffiffiffiffiffiffiffiffiffi 8 þ h 3 p 2 h . (Hint: Let 8 þ h ¼ x3 Þ. Ans. 1/12 3.69. If lim x!x0 f ðxÞ ¼ A and lim x!x0 gðxÞ ¼ B 6¼ 0, prove directly that lim x!x0 f ðxÞ gðxÞ ¼ A B . 3.70. Given lim x!0 sin x x ¼ 1, evaluate: ðaÞ lim x!0 sin 3x x ðcÞ lim x!0 1 cos x x2 ðeÞ lim x!0 6x sin 2x 2x þ 3 sin 4x ðgÞ lim x!0 1 2 cos x þ cos 2x x2 ðbÞ lim x!0 1 cos x x ðdÞ lim x!3 ðx 3Þ csc x ð f Þ lim x!0 cos ax cos bx x2 ðhÞ lim x!1 3 sin x sin 3x x3 Ans. (a) 3, (b) 0, (c) 1/2, (d) 1=, (e) 2/7, ( f ) 1 2 ðb2 a2 Þ, (g) 1, (h) 43 3.71. If lim x!0 ex 1 x ¼ 1, prove that: ðaÞ lim x!0 eax ebx x ¼ b a; ðbÞ lim x!0 ax bx x ¼ ln a b ; a; b 0; ðcÞ lim x!0 tanh ax x ¼ a: 62 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 72. 3.72. Prove that lim x!x0 f ðxÞ ¼ l if and only if lim x!x0þ f ðxÞ ¼ lim x!x0 f ðxÞ ¼ l. CONTINUITY In the following problems assume the largest possible domain unless otherwise stated. 3.73. Prove that f ðxÞ ¼ x2 3x þ 2 is continuous at x ¼ 4. 3.74. Prove that f ðxÞ ¼ 1=x is continuous (a) at x ¼ 2, (b) in 1 @ x @ 3. 3.75. Investigate the continuity of each of the following functions at the indicated points: ðaÞ f ðxÞ ¼ sin x x ; x 6¼ 0; f ð0Þ ¼ 0; x ¼ 0 ðcÞ f ðxÞ ¼ x3 8 x2 4 ; x 6¼ 2; f ð2Þ ¼ 3; x ¼ 2 ðbÞ f ðxÞ ¼ x jxj; x ¼ 0 ðdÞ f ðxÞ ¼ sin x; 0 x 1 ln x 1 x 2 ; x ¼ 1: Ans. (a) discontinuous, (b) continuous, (c) continuous, (d) discontinuous 3.76. If ½x ¼ greatest integer @ x, investigate the continuity of f ðxÞ ¼ x ½x in the interval (a) 1 x 2, (b) 1 @ x @ 2. 3.77. Prove that f ðxÞ ¼ x3 is continuous in every finite interval. 3.78. If f ðxÞ=gðxÞ and gðxÞ are continuous at x ¼ x0, prove that f ðxÞ must be continuous at x ¼ x0. 3.79. Prove that f ðxÞ ¼ ðtan1 xÞ=x, f ð0Þ ¼ 1 is continuous at x ¼ 0. 3.80. Prove that a polynomial is continuous in every finite interval. 3.81. If f ðxÞ and gðxÞ are polynomials, prove that f ðxÞ=gðxÞ is continuous at each point x ¼ x0 for which gðx0Þ 6¼ 0. 3.82. Give the points of discontinuity of each of the following functions. ðaÞ f ðxÞ ¼ x ðx 2Þðx 4Þ ðcÞ f ðxÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 3Þð6 xÞ p ; 3 @ x @ 6 ðbÞ f ðxÞ ¼ x2 sin 1=x; x 6¼ 0; f ð0Þ ¼ 0 ðdÞ f ðxÞ ¼ 1 1 þ 2 sin x : Ans. (a) x ¼ 2; 4, (b) none, (c) none, (d) x ¼ 7=6 2m; 11=6 2m; m ¼ 0; 1; 2; . . . UNIFORM CONTINUITY 3.83. Prove that f ðxÞ ¼ x3 is uniformly continuous in (a) 0 x 2, (b) 0 @ x @ 2, (c) any finite interval. 3.84. Prove that f ðxÞ ¼ x2 is not uniformly continuous in 0 x 1. 3.85. If a is a constant, prove that f ðxÞ ¼ 1=x2 is (a) continuous in a x 1 if a A 0, (b) uniformly continuous in a x 1 if a 0, (c) not uniformly continuous in 0 x 1. 3.86. If f ðxÞ and gðxÞ are uniformly continuous in the same interval, prove that (a) f ðxÞ gðxÞ and (b) f ðxÞgðxÞ are uniformly continuous in the interval. State and prove an analogous theorem for f ðxÞ=gðxÞ. MISCELLANEOUS PROBLEMS 3.87. Give an ‘‘; ’’ proof of the theorem of Problem 3.31. 3.88. (a) Prove that the equation tan x ¼ x has a real positive root in each of the intervals =2 x 3=2, 3=2 x 5=2, 5=2 x 7=2; . . . . CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 63
- 73. (b) Illustrate the result in (a) graphically by constructing the graphs of y ¼ tan x and y ¼ x and locating their points of intersection. (c) Determine the value of the smallest positive root of tan x ¼ x. Ans. ðcÞ 4.49 approximately 3.89. Prove that the only real solution of sin x ¼ x is x ¼ 0. 3.90. (a) Prove that cos x cosh x þ 1 ¼ 0 has infinitely many real roots. (b) Prove that for large values of x the roots approximate those of cos x ¼ 0. 3.91. Prove that lim x!0 x2 sinð1=xÞ sin x ¼ 0. 3.92. Suppose f ðxÞ is continuous at x ¼ x0 and assume f ðx0Þ 0. Prove that there exists an interval ðx0 h; x0 þ hÞ, where h 0, in which f ðxÞ 0. (See Theorem 5, page 47.) [Hint: Show that we can make j f ðxÞ f ðx0Þj 1 2 f ðx0Þ. Then show that f ðxÞ A f ðx0Þ j f ðxÞ f ðx0Þj 1 2 f ðx0Þ 0.] 3.93. (a) Prove Theorem 10, Page 48, for the greatest lower bound m (see Problem 3.34). (b) Prove Theorem 9, Page 48, and explain its relationship to Theorem 10. 64 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3
- 74. 65 Derivatives THE CONCEPT AND DEFINITION OF A DERIVATIVE Concepts that shape the course of mathematics are few and far between. The derivative, the fundamental element of the differential calculus, is such a concept. That branch of mathematics called analysis, of which advanced calculus is a part, is the end result. There were two problems that led to the discovery of the derivative. The older one of defining and representing the tangent line to a curve at one of its points had concerned early Greek philosophers. The other problem of representing the instanta- neous velocity of an object whose motion was not constant was much more a problem of the seventeenth century. At the end of that century, these problems and their relationship were resolved. As is usually the case, many mathematicians contributed, but it was Isaac Newton and Gottfried Wilhelm Leibniz who independently put together organized bodies of thought upon which others could build. The tangent problem provides a visual interpretation of the derivative and can be brought to mind no matter what the complexity of a particular application. It leads to the definition of the derivative as the limit of a difference quotient in the following way. (See Fig. 4-1.) Let Poðx0Þ be a point on the graph of y ¼ f ðxÞ. Let PðxÞ be a nearby point on this same graph of the function f . Then the line through these two points is called a secant line. Its slope, ms, is the difference quotient ms ¼ f ðxÞ f ðx0Þ x x0 ¼ y x Fig. 4-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 75. where x and y are called the increments in x and y, respectively. Also this slope may be written ms ¼ f ðx0 þ hÞ f ðx0Þ h where h ¼ x x0 ¼ x. See Fig. 4-2. We can imagine a sequence of lines formed as h ! 0. It is the limiting line of this sequence that is the natural one to be the tangent line to the graph at P0. To make this mode of reasoning precise, the limit (when it exists), is formed as follows: f 0 ðxÞ ¼ lim h!0 f ðx0 þ hÞ f ðx0Þ h As indicated, this limit is given the name f 0 ðx0Þ. It is called the derivative of the function f at its domain value x0. If this limit can be formed at each point of a subdomain of the domain of f , then f is said to be differentiable on that subdomain and a new function f 0 has been constructed. This limit concept was not understood until the middle of the nineteenth century. A simple example illustrates the conceptual problem that faced mathematicians from 1700 until that time. Let the graph of f be the parabola y ¼ x2 , then a little algebraic manipulation yields ms ¼ 2x0h þ h2 h ¼ 2x0 þ h Newton, Leibniz, and their contemporaries simply let h ¼ 0 and said that 2x0 was the slope of the tangent line at P0. However, this raises the ghost of a 0 0 form in the middle term. True understanding of the calculus is in the comprehension of how the introduction of something new (the derivative, i.e., the limit of a difference quotient) resolves this dilemma. Note 1: The creation of new functions from difference quotients is not limited to f 0 . If, starting with f 0 , the limit of the difference quotient exists, then f 00 may be constructed and so on and so on. Note 2: Since the continuity of a function is such a strong property, one might think that differ- entiability followed. This is not necessarily true, as is illustrated in Fig. 4-3. The following theorem puts the matter in proper perspective: Theorem: If f is differentiable at a domain value, then it is continuous at that value. As indicated above, the converse of this theorem is not true. 66 DERIVATIVES [CHAP. 4 y x x0 M P N Fig. 4-3 y a x0 x0 + h b x A B Q S R P α θ y = f (x) f (x0 + h) _ f (x0) h = Dx f (x0) Fig. 4-2
- 76. RIGHT- AND LEFT-HAND DERIVATIVES The status of the derivative at end points of the domain of f , and in other special circumstances, is clarified by the following definitions. The right-hand derivative of f ðxÞ at x ¼ x0 is defined as f 0 þðx0Þ ¼ lim h!0þ f ðx0 þ hÞ f ðx0Þ h ð3Þ if this limit exists. Note that in this case hð¼ xÞ is restricted only to positive values as it approaches zero. Similarly, the left-hand derivative of f ðxÞ at x ¼ x0 is defined as f 0 ðx0Þ ¼ lim h!0 f ðx0 þ hÞ f ðx0Þ h ð4Þ if this limit exists. In this case h is restricted to negative values as it approaches zero. A function f has a derivative at x ¼ x0 if and only if f 0 þðx0Þ ¼ f 0 ðx0Þ. DIFFERENTIABILITY IN AN INTERVAL If a function has a derivative at all points of an interval, it is said to be differentiable in the interval. In particular if f is defined in the closed interval a @ x @ b, i.e. ½a; b, then f is differentiable in the interval if and only if f 0 ðx0Þ exists for each x0 such that a x0 b and if f 0 þðaÞ and f 0 ðbÞ both exist. If a function has a continuous derivative, it is sometimes called continuously differentiable. PIECEWISE DIFFERENTIABILITY A function is called piecewise differentiable or piecewise smooth in an interval a @ x @ b if f 0 ðxÞ is piecewise continuous. An example of a piecewise continuous function is shown graphically on Page 48. An equation for the tangent line to the curve y ¼ f ðxÞ at the point where x ¼ x0 is given by y f ðx0Þ ¼ f 0 ðx0Þðx x0Þ ð7Þ The fact that a function can be continuous at a point and yet not be differentiable there is shown graphically in Fig. 4-3. In this case there are two tangent lines at P represented by PM and PN. The slopes of these tangent lines are f 0 ðx0Þ and f 0 þðx0Þ respectively. DIFFERENTIALS Let x ¼ dx be an increment given to x. Then y ¼ f ðx þ xÞ f ðxÞ ð8Þ is called the increment in y ¼ f ðxÞ. If f ðxÞ is continuous and has a continuous first derivative in an interval, then y ¼ f 0 ðxÞx þ x ¼ f 0 ðxÞdx þ dx ð9Þ where ! 0 as x ! 0. The expression dy ¼ f 0 ðxÞdx ð10Þ is called the differential of y or f(x) or the principal part of y. Note that y 6¼ dy in general. However if x ¼ dx is small, then dy is a close approximation of y (see Problem 11). The quantity dx, called the differential of x, and dy need not be small. CHAP. 4] DERIVATIVES 67
- 77. Because of the definitions (8) and (10), we often write dy dx ¼ f 0 ðxÞ ¼ lim x!0 f ðx þ xÞ f ðxÞ x ¼ lim x!0 y x ð11Þ It is emphasized that dx and dy are not the limits of x and y as x ! 0, since these limits are zero whereas dx and dy are not necessarily zero. Instead, given dx we determine dy from (10), i.e., dy is a dependent variable determined from the independent variable dx for a given x. Geometrically, dy is represented in Fig. 4-1, for the particular value x ¼ x0, by the line segment SR, whereas y is represented by QR. The geometric interpretation of the derivative as the slope of the tangent line to a curve at one of its points is fundamental to its application. Also of importance is its use as representative of instantaneous velocity in the construction of physical models. In particular, this physical viewpoint may be used to introduce the notion of differentials. Newton’s Second and First Laws of Motion imply that the path of an object is determined by the forces acting on it, and that if those forces suddenly disappear, the object takes on the tangential direction of the path at the point of release. Thus, the nature of the path in a small neighborhood of the point of release becomes of interest. With this thought in mind, consider the following idea. Suppose the graph of a function f is represented by y ¼ f ðxÞ. Let x ¼ x0 be a domain value at which f 0 exists (i.e., the function is differentiable at that value). Construct a new linear function dy ¼ f 0 ðx0Þ dx with dx as the (independent) domain variable and dy the range variable generated by this rule. This linear function has the graphical interpretation illustrated in Fig. 4-4. That is, a coordinate system may be constructed with its origin at P0 and the dx and dy axes parallel to the x and y axes, respectively. In this system our linear equation is the equation of the tangent line to the graph at P0. It is representative of the path in a small neighborhood of the point; and if the path is that of an object, the linear equation represents its new path when all forces are released. dx and dy are called differentials of x and y, respectively. Because the above linear equation is valid at every point in the domain of f at which the function has a derivative, the subscript may be dropped and we can write dy ¼ f 0 ðxÞ dx The following important observations should be made. dy dx ¼ f 0 ðxÞ ¼ lim x!0 f ðx þ xÞ f ðxÞ x ¼ lim x!0 y x , thus dy dx is not the same thing as y x . 68 DERIVATIVES [CHAP. 4 Fig. 4-4
- 78. On the other hand, dy and y are related. In particular, lim x!0 y x ¼ f 0 ðxÞ means that for any 0 there exists 0 such that y x dy dx whenever jxj . Now dx is an independent variable and the axes of x and dx are parallel; therefore, dx may be chosen equal to x. With this choice x y dy x or dy x y dy þ x From this relation we see that dy is an approximation to y in small neighborhoods of x. dy is called the principal part of y. The representation of f 0 by dy dx has an algebraic suggestiveness that is very appealing and will appear in much of what follows. In fact, this notation was introduced by Leibniz (without the justification provided by knowledge of the limit idea) and was the primary reason his approach to the calculus, rather than Newton’s was followed. THE DIFFERENTIATION OF COMPOSITE FUNCTIONS Many functions are a composition of simpler ones. For example, if f and g have the rules of correspondence u ¼ x3 and y ¼ sin u, respectively, then y ¼ sin x3 is the rule for a composite function F ¼ gð f Þ. The domain of F is that subset of the domain of F whose corresponding range values are in the domain of g. The rule of composite function differentiation is called the chain rule and is represented by dy dx ¼ dy du du dx ½F 0 ðxÞ ¼ g0 ðuÞf 0 ðxÞ. In the example dy dx dðsin x3 Þ dx ¼ cos x3 ð3x2 dxÞ The importance of the chain rule cannot be too greatly stressed. Its proper application is essential in the differentiation of functions, and it plays a fundamental role in changing the variable of integration, as well as in changing variables in mathematical models involving differential equations. IMPLICIT DIFFERENTIATION The rule of correspondence for a function may not be explicit. For example, the rule y ¼ f ðxÞ is implicit to the equation x2 þ 4xy5 þ 7xy þ 8 ¼ 0. Furthermore, there is no reason to believe that this equation can be solved for y in terms of x. However, assuming a common domain (described by the independent variable x) the left-hand member of the equation can be construed as a composition of functions and differentiated accordingly. (The rules of differentiation are listed below for your review.) In this example, differentiation with respect to x yields 2x þ 4 y5 þ 5xy4 dy dx þ 7 y þ x dy dx ¼ 0 Observe that this equation can be solved for dy dx as a function of x and y (but not of x alone). CHAP. 4] DERIVATIVES 69
- 79. RULES FOR DIFFERENTIATION If f , g; and h are differentiable functions, the following differentiation rules are valid. 1: d dx f f ðxÞ þ gðxÞg ¼ d dx f ðxÞ þ d dx gðxÞ ¼ f 0 ðxÞ þ g0 ðxÞ (Addition Rule) 2: d dx f f ðxÞ gðxÞg ¼ d dx f ðxÞ d dx gðxÞ ¼ f 0 ðxÞ g0 ðxÞ 3: d dx fC f ðxÞg ¼ C d dx f ðxÞ ¼ C f 0 ðxÞ where C is any constant 4: d dx f f ðxÞgðxÞg ¼ f ðxÞ d dx gðxÞ þ gðxÞ d dx f ðxÞ ¼ f ðxÞg0 ðxÞ þ gðxÞ f 0 ðxÞ (Product Rule) 5: d dx f ðxÞ gðxÞ ¼ gðxÞ d dx f ðxÞ f ðxÞ d dx gðxÞ ½gðxÞ2 ¼ gðxÞ f 0 ðxÞ f ðxÞg0 ðxÞ ½gðxÞ2 if gðxÞ 6¼ 0 (Quotient Rule) 6: If y ¼ f ðuÞ where u ¼ gðxÞ; then dy dx ¼ dy du du dx ¼ f 0 ðuÞ du dx ¼ f 0 fgðxÞgg0 ðxÞ ð12Þ Similarly if y ¼ f ðuÞ where u ¼ gðvÞ and v ¼ hðxÞ, then dy dx ¼ dy du du dv dv dx ð13Þ The results (12) and (13) are often called chain rules for differentiation of composite functions. 7: If y ¼ f ðxÞ; and x ¼ f 1 ðyÞ; then dy=dx and dx=dy are related by dy dx ¼ 1 dx=dy ð14Þ 8: If x ¼ f ðtÞ and y ¼ gðtÞ; then dy dx ¼ dy=dt dx=dt ¼ g0 ðtÞ f 0 ðtÞ ð15Þ Similar rules can be formulated for differentials. For example, df f ðxÞ þ gðxÞg ¼ d f ðxÞ þ dgðxÞ ¼ f 0 ðxÞdx þ g0 ðxÞdx ¼ f f 0 ðxÞ þ g0 ðxÞgdx df f ðxÞgðxÞg ¼ f ðxÞdgðxÞ þ gðxÞd f ðxÞ ¼ f f ðxÞg0 ðxÞ þ gðxÞ f 0 ðxÞgdx 70 DERIVATIVES [CHAP. 4
- 80. DERIVATIVES OF ELEMENTARY FUNCTIONS In the following we assume that u is a differentiable function of x; if u ¼ x, du=dx ¼ 1. The inverse functions are defined according to the principal values given in Chapter 3. 1. d dx ðCÞ ¼ 0 16. d dx cot1 u ¼ 1 1 þ u2 du dx 2. d dx un ¼ nun1 du dx 17. d dx sec1 u ¼ 1 u ffiffiffiffiffiffiffiffiffiffiffiffiffi u2 1 p du dx þ if u 1 if u 1 3. d dx sin u ¼ cos u du dx 18. d dx csc1 u ¼ 1 u ffiffiffiffiffiffiffiffiffiffiffiffiffi u2 1 p du dx if u 1 þ if u 1 4. d dx cos u ¼ sin u du dx 19. d dx sinh u ¼ cosh u du dx 5. d dx tan u ¼ sec2 u du dx 20. d dx cosh u ¼ sinh u du dx 6. d dx cot u ¼ csc2 u du dx 21. d dx tanh u ¼ sech2 u du dx 7. d dx sec u ¼ sec u tan u du dx 22. d dx coth u ¼ csch2 u du dx 8. d dx csc u ¼ csc u cot u du dx 23. d dx sech u ¼ sech u tanh u du dx 9. d dx loga u ¼ loga e u du dx a 0; a 6¼ 1 24. d dx csch u ¼ csch u coth u du dx 10. d dx loge u ¼ d dx ln u ¼ 1 u du dx 25. d dx sinh1 u ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2 p du dx 11. d dx au ¼ au ln a du dx 26. d dx cosh1 u ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi u2 1 p du dx 12. d dx eu ¼ eu du dx 27. d dx tanh1 u ¼ 1 1 u2 du dx ; juj 1 13. d dx sin1 u ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u2 p du dx 28. d dx coth1 u ¼ 1 1 u2 du dx ; juj 1 14. d dx cos1 u ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u2 p du dx 29. d dx sech1 u ¼ 1 u ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u2 p du dx 15. d dx tan1 u ¼ 1 1 þ u2 du dx 30. d dx csch1 u ¼ 1 u ffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ 1 p du dx HIGHER ORDER DERIVATIVES If f ðxÞ is differentiable in an interval, its derivative is given by f 0 ðxÞ, y0 or dy=dx, where y ¼ f ðxÞ. If f 0 ðxÞ is also differentiable in the interval, its derivative is denoted by f 00 ðxÞ, y00 or d dx dy dx ¼ d2 y dx2 . Similarly, the nth derivative of f ðxÞ, if it exists, is denoted by f ðnÞ ðxÞ, yðnÞ or dn y dxn, where n is called the order of the derivative. Thus derivatives of the first, second, third, . . . orders are given by f 0 ðxÞ, f 00 ðxÞ, f 000 ðxÞ; . . . . Computation of higher order derivatives follows by repeated application of the differentiation rules given above. CHAP. 4] DERIVATIVES 71
- 81. MEAN VALUE THEOREMS These theorems are fundamental to the rigorous establishment of numerous theorems and formulas. (See Fig. 4-5.) 1. Rolle’s theorem. If f ðxÞ is continuous in ½a; b and differentiable in ða; bÞ and if f ðaÞ ¼ f ðbÞ ¼ 0, then there exists a point in ða; bÞ such that f 0 ðÞ ¼ 0. Rolle’s theorem is employed in the proof of the mean value theorem. It then becomes a special case of that theorem. 2. The mean value theorem. If f ðxÞ is continuous in ½a; b and differentiable in ða; bÞ, then there exists a point in ða; bÞ such that f ðbÞ f ðaÞ b a ¼ f 0 ðÞ a b ð16Þ Rolle’s theorem is the special case of this where f ðaÞ ¼ f ðbÞ ¼ 0. The result (16) can be written in various alternative forms; for example, if x and x0 are in ða; bÞ, then f ðxÞ ¼ f ðx0Þ þ f 0 ðÞðx x0Þ between x0 and x ð17Þ We can also write (16) with b ¼ a þ h, in which case ¼ a þ h, where 0 1. The mean value theorem is also called the law of the mean. 3. Cauchy’s generalized mean value theorem. If f ðxÞ and gðxÞ are continuous in ½a; b and differ- entiable in ða; bÞ, then there exists a point in ða; bÞ such that f ðbÞ f ðaÞ gðbÞ gðaÞ ¼ f 0 ðÞ g0ðÞ a b ð18Þ where we assume gðaÞ 6¼ gðbÞ and f 0 ðxÞ, g0 ðxÞ are not simultaneously zero. Note that the special case gðxÞ ¼ x yields (16). L’HOSPITAL’S RULES If lim x!x0 f ðxÞ ¼ A and lim x!x0 gðxÞ ¼ B, where A and B are either both zero or both infinite, lim x!x0 f ðxÞ gðxÞ is often called an indeterminate of the form 0/0 or 1=1, respectively, although such terminology is somewhat misleading since there is usually nothing indeterminate involved. The following theorems, called L’Hospital’s rules, facilitate evaluation of such limits. 1. If f ðxÞ and gðxÞ are differentiable in the interval ða; bÞ except possibly at a point x0 in this interval, and if g0 ðxÞ 6¼ 0 for x 6¼ x0, then 72 DERIVATIVES [CHAP. 4 y x b ξ a f (a) f (b) B C A D E Fig. 4-5
- 82. lim x!x0 f ðxÞ gðxÞ ¼ lim x!x0 f 0 ðxÞ g0 ðxÞ ð19Þ whenever the limit on the right can be found. In case f 0 ðxÞ and g0 ðxÞ satisfy the same conditions as f ðxÞ and gðxÞ given above, the process can be repeated. 2. If lim x!x0 f ðxÞ ¼ 1 and lim x!x0 gðxÞ ¼ 1, the result (19) is also valid. These can be extended to cases where x ! 1 or 1, and to cases where x0 ¼ a or x0 ¼ b in which only one sided limits, such as x ! aþ or x ! b, are involved. Limits represented by the so-called indeterminate forms 0 1, 10 , 00 , 11 ; and 1 1 can be evaluated on replacing them by equivalent limits for which the above rules are applicable (see Problem 4.29). APPLICATIONS 1. Relative Extrema and Points of Inflection See Chapter 3 where relative extrema and points of inflection were described and a diagram is presented. In this chapter such points are characterized by the variation of the tangent line, and then by the derivative, which represents the slope of that line. Assume that f has a derivative at each point of an open interval and that P1 is a point of the graph of f associated with this interval. Let a varying tangent line to the graph move from left to right through P1. If the point is a relative minimum, then the tangent line rotates counterclockwise. The slope is negative to the left of P1 and positive to the right. At P1 the slope is zero. At a relative maximum a similar analysis can be made except that the rotation is clockwise and the slope varies from positive to negative. Because f 00 designates the change of f 0 , we can state the following theorem. (See Fig. 4-6.) Theorem. Assume that x1 is a number in an open set of the domain of f at which f 0 is continuous and f 00 is defined. If f 0 ðx1Þ ¼ 0 and f 00 ðx1Þ 6¼ 0, then f ðx1Þ is a relative extreme of f . Specifically: (a) If f 00 ðx1Þ 0, then f ðx1Þ is a relative minimum, (b) If f 00 ðx1Þ 0; then f ðx1Þ is a relative maximum. (The domain value x1 is called a critical value.) This theorem may be generalized in the following way. Assume existence and continuity of derivatives as needed and suppose that f 0 ðx1Þ ¼ f 00 ðx1Þ ¼ f ð2p1Þ ðx1Þ ¼ 0 and f ð2pÞ ðx1Þ 6¼ 0 ( p a posi- tive integer). Then: (a) f has a relative minimum at x1 if f ð2pÞ ðx1Þ 0, (b) f has a relative maximum at x1 if f ð2pÞ ðx1Þ 0. CHAP. 4] DERIVATIVES 73 Fig. 4-6
- 83. (Notice that the order of differentiation in each succeeding case is two greater. The nature of the intermediate possibilities is suggested in the next paragraph.) It is possible that the slope of the tangent line to the graph of f is positive to the left of P1, zero at the point, and again positive to the right. Then P1 is called a point of inflection. In the simplest case this point of inflection is characterized by f 0 ðx1Þ ¼ 0, f 00 ðx1Þ ¼ 0, and f 000 ðx1Þ 6¼ 0. 2. Particle motion The fundamental theories of modern physics are relativity, electromagnetism, and quantum mechanics. Yet Newtonian physics must be studied because it is basic to many of the concepts in these other theories, and because it is most easily applied to many of the circumstances found in every- day life. The simplest aspect of Newtonian mechanics is called kinematics, or the geometry of motion. In this model of reality, objects are idealized as points and their paths are represented by curves. In the simplest (one-dimensional) case, the curve is a straight line, and it is the speeding up and slowing down of the object that is of importance. The calculus applies to the study in the following way. If x represents the distance of a particle from the origin and t signifies time, then x ¼ f ðtÞ designates the position of a particle at time t. Instantaneous velocity (or speed in the one-dimensional case) is represented by dx dt ¼ lim t!0 f ðt þ tÞ t (the limiting case of the formula change in distance change in time for speed when the motion is constant). Furthermore, the instantaneous change in velocity is called acceleration and represented by d2 x dt2 . Path, velocity, and acceleration of a particle will be represented in three dimensions in Chapter 7 on vectors. 3. Newton’s method It is difficult or impossible to solve algebraic equations of higher degree than two. In fact, it has been proved that there are no general formulas representing the roots of algebraic equations of degree five and higher in terms of radicals. However, the graph y ¼ f ðxÞ of an algebraic equation f ðxÞ ¼ 0 crosses the x- axis at each single-valued real root. Thus, by trial and error, consecutive integers can be found between which a root lies. Newton’s method is a systematic way of using tangents to obtain a better approx- imation of a specific real root. The procedure is as follows. (See Fig. 4-7.) Suppose that f has as many derivatives as required. Let r be a real root of f ðxÞ ¼ 0, i.e., f ðrÞ ¼ 0. Let x0 be a value of x near r. For example, the integer preceding or following r. Let f 0 ðx0Þ be the slope of the graph of y ¼ f ðxÞ at P0½x0; f ðx0Þ. Let Q1ðx1; 0Þ be the x-axis intercept of the tangent line at P0 then 0 f ðx0Þ x x0 ¼ f 0 ðx0Þ 74 DERIVATIVES [CHAP. 4 Fig. 4-7
- 84. where the two representations of the slope of the tangent line have been equated. The solution of this relation for x1 is x1 ¼ x0 f ðx0Þ f 0 ðx0Þ Starting with the tangent line to the graph at P1½x1; f ðx1Þ and repeating the process, we get x2 ¼ x1 f ðx1Þ f 0ðx1Þ ¼ x0 f ðx0Þ f 0ðx0Þ f ðx1Þ f 0ðx1Þ and in general xn ¼ x0 X n k¼0 f ðxkÞ f 0ðxkÞ Under appropriate circumstances, the approximation xn to the root r can be made as good as desired. Note: Success with Newton’s method depends on the shape of the function’s graph in the neighbor- hood of the root. There are various cases which have not been explored here. Solved Problems DERIVATIVES 4.1. (a) Let f ðxÞ ¼ 3 þ x 3 x , x 6¼ 3. Evaluate f 0 ð2Þ from the definition. f 0 ð2Þ ¼ lim h!0 f ð2 þ hÞ f ð2Þ h ¼ lim h!0 1 h 5 þ h 1 h 5 ¼ lim h!0 1 h 6h 1 h ¼ lim h!0 6 1 h ¼ 6 Note: By using rules of differentiation we find f 0 ðxÞ ¼ ð3 xÞ d dx ð3 þ xÞ ð3 þ xÞ d dx ð3 xÞ ð3 xÞ2 ¼ ð3 xÞð1Þ ð3 þ xÞð1Þ ð3 xÞ2 ¼ 6 ð3 xÞ2 at all points x where the derivative exists. Putting x ¼ 2, we find f 0 ð2Þ ¼ 6. Although such rules are often useful, one must be careful not to apply them indiscriminately (see Problem 4.5). (b) Let f ðxÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x 1 p . Evaluate f 0 ð5Þ from the definition. f 0 ð5Þ ¼ lim h!0 f ð5 þ hÞ f ð5Þ h ¼ lim h!0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 2h p 3 h ¼ lim h!0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 2h p 3 h ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 2h p þ 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 2h p þ 3 ¼ lim h!0 9 þ 2h 9 hð ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 9 þ 2h p þ 3Þ ¼ lim h!0 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 þ 2h p þ 3 ¼ 1 3 By using rules of differentiation we find f 0 ðxÞ ¼ d dx ð2x 1Þ1=2 ¼ 1 2 ð2x 1Þ1=2 d dx ð2x 1Þ ¼ ð2x 1Þ1=2 . Then f 0 ð5Þ ¼ 91=2 ¼ 1 3. 4.2. (a) Show directly from definition that the derivative of f ðxÞ ¼ x3 is 3x2 . (b) Show from definition that d dx ffiffiffi x p Þ ¼ 1 2 ffiffiffi x p . CHAP. 4] DERIVATIVES 75
- 85. ðaÞ f ðx þ hÞ f ðxÞ h ¼ 1 h ½ðx þ hÞ3 x3 ¼ 1 h ½x3 þ 3x2 h þ 3xh2 þ h3 x3 ¼ 3x2 þ 3xh þ h2 Then f 0 ðxÞ ¼ lim h!0 f ðx þ hÞ f ðxÞ h ¼ 3x2 ðbÞ lim h!0 f ðx þ hÞ f ðxÞ h ¼ lim h!0 ffiffiffiffiffiffiffiffiffiffiffi x þ h p ffiffiffi x p h The result follows by multiplying numerator and denominator by ffiffiffiffiffiffiffiffiffiffiffi x þ h p ffiffiffi x p and then letting h ! 0. 4.3. If f ðxÞ has a derivative at x ¼ x0, prove that f ðxÞ must be continuous at x ¼ x0. f ðx0 þ hÞ f ðx0Þ ¼ f ðx0 þ hÞ f ðx0Þ h h; h 6¼ 0 lim h!0 f ðx0 þ hÞ f ðx0Þ ¼ lim h!0 f ðx0 þ hÞ f ðx0Þ h lim h!0 h ¼ f 0 ðx0Þ 0 ¼ 0 Then since f 0 ðx0Þ exists by hypothesis. Thus lim h!0 f ðx0 þ hÞ f ðx0Þ ¼ 0 or lim h!0 f ðx0 þ hÞ ¼ f ðx0Þ showing that f ðxÞ is continuous at x ¼ x0. 4.4. Let f ðxÞ ¼ x sin 1=x; x 6¼ 0 0; x ¼ 0 . (a) Is f ðxÞ continuous at x ¼ 0? (b) Does f ðxÞ have a derivative at x ¼ 0? (a) By Problem 3.22(b) of Chapter 3, f ðxÞ is continuous at x ¼ 0. ðbÞ f 0 ð0Þ ¼ lim h!0 f ð0 þ hÞ f ð0Þ h ¼ lim h!0 f ðhÞ f ð0Þ h ¼ lim h!0 h sin 1=h 0 h ¼ lim h!0 sin 1 h which does not exist. This example shows that even though a function is continuous at a point, it need not have a derivative at the point, i.e., the converse of the theorem in Problem 4.3 is not necessarily true. It is possible to construct a function which is continuous at every point of an interval but has a derivative nowhere. 4.5. Let f ðxÞ ¼ x2 sin 1=x; x 6¼ 0 0; x ¼ 0 . (a) Is f ðxÞ differentiable at x ¼ 0? (b) Is f 0 ðxÞ continuous at x ¼ 0? ðaÞ f 0 ð0Þ ¼ lim h!0 f ðhÞ f ð0Þ h ¼ lim h!0 h2 sin 1=h 0 h ¼ lim h!0 h sin 1 h ¼ 0 by Problem 3.13, Chapter 3. Then f ðxÞ has a derivative (is differentiable) at x ¼ 0 and its value is 0. (b) From elementary calculus differentiation rules, if x 6¼ 0, f 0 ðxÞ ¼ d dx x2 sin 1 x ¼ x2 d dx sin 1 x þ sin 1 x d dx ðx2 Þ ¼ x2 cos 1 x 1 x2 þ sin 1 x ð2xÞ ¼ cos 1 x þ 2x sin 1 x 76 DERIVATIVES [CHAP. 4
- 86. Since lim x!0 f 0 ðxÞ ¼ lim x!0 cos 1 x þ 2x sin 1 x does not exist (because lim x!0 cos 1=x does not exist), f 0 ðxÞ cannot be continuous at x ¼ 0 in spite of the fact that f 0 ð0Þ exists. This shows that we cannot calculate f 0 ð0Þ in this case by simply calculating f 0 ðxÞ and putting x ¼ 0, as is frequently supposed in elementary calculus. It is only when the derivative of a function is continuous at a point that this procedure gives the right answer. This happens to be true for most functions arising in elementary calculus. 4.6. Present an ‘‘; ’’ definition of the derivative of f ðxÞ at x ¼ x0. f ðxÞ has a derivative f 0 ðx0Þ at x ¼ x0 if, given any 0, we can find 0 such that f ðx0 þ hÞ f ðx0Þ h f 0 ðx0Þ when 0 jhj RIGHT- AND LEFT-HAND DERIVATIVES 4.7. Let f ðxÞ ¼ jxj. (a) Calculate the right-hand derivatives of f ðxÞ at x ¼ 0. (b) Calculate the left- hand derivative of f ðxÞ at x ¼ 0. (c) Does f ðxÞ have a derivative at x ¼ 0? (d) Illustrate the conclusions in (a), (b), and (c) from a graph. ðaÞ f 0 þð0Þ ¼ lim h!0þ f ðhÞ f ð0Þ h ¼ lim h!0þ jhj 0 h ¼ lim h!0þ h h ¼ 1 since jhj ¼ h for h 0. ðbÞ f 0 ð0Þ ¼ lim h!0 f ðhÞ f ð0Þ h ¼ lim h!0 jhj 0 h ¼ lim h!0 h h ¼ 1 since jhj ¼ h for h 0. (c) No. The derivative at 0 does not exist if the right and left hand derivatives are unequal. (d) The required graph is shown in the adjoining Fig. 4-8. Note that the slopes of the lines y ¼ x and y ¼ x are 1 and 1 respectively, representing the right and left hand derivatives at x ¼ 0. However, the derivative at x ¼ 0 does not exist. 4.8. Prove that f ðxÞ ¼ x2 is differentiable in 0 @ x @ 1. Let x0 be any value such that 0 x0 1. Then f 0 ðx0Þ ¼ lim h!0 f ðx0 þ hÞ f ðx0Þ h ¼ lim h!0 ðx0 þ hÞ2 x2 0 h ¼ lim h!0 ð2x0 þ hÞ ¼ 2x0 At the end point x ¼ 0, f 0 þð0Þ ¼ lim h!0þ f ð0 þ hÞ f ð0Þ h ¼ lim h!0þ h2 0 h ¼ lim h!0þ h ¼ 0 At the end point x ¼ 1, f 0 ð1Þ ¼ lim h!0 f ð1 þ hÞ f ð1Þ h ¼ lim h!0 ð1 þ hÞ2 1 h ¼ lim h!0 ð2 þ hÞ ¼ 2 Then f ðxÞ is differentiable in 0 @ x @ 1. We may write f 0 ðxÞ ¼ 2x for any x in this interval. It is customary to write f 0 þð0Þ ¼ f 0 ð0Þ and f 0 ð1Þ ¼ f 0 ð1Þ in this case. CHAP. 4] DERIVATIVES 77 y x y = x y = _ x Fig. 4-8
- 87. 4.9. Find an equation for the tangent line to y ¼ x2 at the point where (a) x ¼ 1=3; ðbÞ x ¼ 1. (a) From Problem 4.8, f 0 ðx0Þ ¼ 2x0 so that f 0 ð1=3Þ ¼ 2=3. Then the equation of the tangent line is y f ðx0Þ ¼ f 0 ðx0Þðx x0Þ or y 1 9 ¼ 2 3 ðx 1 3Þ; i:e:; y ¼ 2 3 x 1 9 (b) As in part (a), y f ð1Þ ¼ f 0 ð1Þðx 1Þ or y 1 ¼ 2ðx 1Þ, i.e., y ¼ 2x 1. DIFFERENTIALS 4.10. If y ¼ f ðxÞ ¼ x3 6x, find (a) y; ðbÞ dy; ðcÞ y dy. ðaÞ y ¼ f ðx þ xÞ f ðxÞ ¼ fðx þ xÞ3 6ðx þ xÞg fx3 6xg ¼ x3 þ 3x2 x þ 3xðxÞ2 þ ðxÞ3 6x 6x x3 þ 6x ¼ ð3x2 6Þx þ 3xðxÞ2 þ ðxÞ3 (b) dy ¼ principal part of y ¼ ð3x2 6Þx ¼ ð3x2 6Þdx, since by definition x ¼ dx. Note that f 0 ðxÞ ¼ 3x2 6 and dy ¼ ð3x2 6Þdx, i.e., dy=dx ¼ 3x2 6. It must be emphasized that dy and dx are not necessarily small. (c) From (a) and (b), y dy ¼ 3xðxÞ2 þ ðxÞ3 ¼ x, where ¼ 3xx þ ðxÞ2 . Note that ! 0 as x ! 0, i.e., y dy x ! 0 as x ! 0. Hence y dy is an infinitesimal of higher order than x (see Problem 4.83). In case x is small, dy and y are approximately equal. 4.11. Evaluate ffiffiffiffiffi 25 3 p approximately by use of differentials. If x is small, y ¼ f ðx þ xÞ f ðxÞ ¼ f 0 ðxÞx approximately. Let f ðxÞ ¼ ffiffiffi x 3 p . Then ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ x 3 p ffiffiffi x 3 p 1 3 x2=3 x (where denotes approximately equal to). If x ¼ 27 and x ¼ 2, we have ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 27 2 3 p ffiffiffiffiffi 27 3 p 1 3 ð27Þ2=3 ð2Þ; i.e., ffiffiffiffiffi 25 3 p 3 2=27 Then ffiffiffiffiffi 25 3 p 3 2=27 or 2.926. If is interesting to observe that ð2:926Þ3 ¼ 25:05, so that the approximation is fairly good. DIFFERENTIATION RULES: DIFFERENTIATION OF ELEMENTARY FUNCTIONS 4.12. Prove the formula d dx f f ðxÞgðxÞg ¼ f ðxÞ d dx gðxÞ þ gðxÞ d dx f ðxÞ, assuming f and g are differentiable. By definition, d dx f f ðxÞgðxÞg ¼ lim x!0 f ðx þ xÞgðx þ xÞ f ðxÞgðxÞ x ¼ lim x!0 f ðx þ xÞfgðx þ xÞ gðxÞg þ gðxÞf f ðx þ xÞ f ðxÞg x ¼ lim x!0 f ðx þ xÞ gðx þ xÞ gðxÞ x þ lim x!0 gðxÞ f ðx þ xÞ f ðxÞ x ¼ f ðxÞ d dx gðxÞ þ gðxÞ d dx f ðxÞ Another method: Let u ¼ f ðxÞ, v ¼ gðxÞ. Then u ¼ f ðx þ xÞ f ðxÞ and v ¼ gðx þ xÞ gðxÞ, i.e., f ðx þ xÞ ¼ u þ u, gðx þ xÞ ¼ v þ v. Thus 78 DERIVATIVES [CHAP. 4
- 88. d dx uv ¼ lim x!0 ðu þ uÞðv þ vÞ uv x ¼ lim x!0 uv þ vu þ uv x ¼ lim x!0 u v x þ v u x þ u x v ¼ u dv dx þ v du dx where it is noted that v ! 0 as x ! 0, since v is supposed differentiable and thus continuous. 4.13. If y ¼ f ðuÞ where u ¼ gðxÞ, prove that dy dx ¼ dy du du dx assuming that f and g are differentiable. Let x be given an increment x 6¼ 0. Then as a consequence u and y take on increments u and y respectively, where y ¼ f ðu þ uÞ f ðuÞ; u ¼ gðx þ xÞ gðxÞ ð1Þ Note that as x ! 0, y ! 0 and u ! 0. If u 6¼ 0, let us write ¼ y u dy du so that ! 0 as u ! 0 and y ¼ dy du u þ u ð2Þ If u ¼ 0 for values of x, then (1) shows that y ¼ 0 for these values of x. For such cases, we define ¼ 0. It follows that in both cases, u 6¼ 0 or u ¼ 0, (2) holds. Dividing (2) by x 6¼ 0 and taking the limit as x ! 0, we have dy dx ¼ lim x!0 y x ¼ lim x!0 dy du u x þ u x ¼ dy du lim x!0 u x þ lim x!0 lim x!0 u x ¼ dy du du dx þ 0 du dx ¼ dy du du dx ð3Þ 4.14. Given d dx ðsin xÞ ¼ cos x and d dx ðcos xÞ ¼ sin x, derive the formulas ðaÞ d dx ðtan xÞ ¼ sec2 x; ðbÞ d dx ðsin1 xÞ ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p ðaÞ d dx ðtan xÞ ¼ d dx sin x cos x ¼ cos x d dx ðsin xÞ sin x d dx ðcos xÞ cos2 x ¼ ðcos xÞðcos xÞ ðsin xÞð sin xÞ cos2 x ¼ 1 cos2 x ¼2 x (b) If y ¼ sin1 x, then x ¼ sin y. Taking the derivative with respect to x, 1 ¼ cos y dy dx or dy dx ¼ 1 cos y ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 y q ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p We have supposed here that the principal value =2 @ sin1 x @ =2, is chosen so that cos y is positive, thus accounting for our writing cos y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 y q rather than cos y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 y q . 4.15. Derive the formula d dx ðloga uÞ ¼ loga e u du dx ða 0; a 6¼ 1Þ, where u is a differentiable function of x. Consider y ¼ f ðuÞ ¼ loga u. By definition, dy du ¼ lim u!0 f ðu þ uÞ f ðuÞ u ¼ lim u!0 logaðu þ uÞ loga u u ¼ lim u!0 1 u loga u þ u u ¼ lim u!0 1 u loga 1 þ u u u=u CHAP. 4] DERIVATIVES 79
- 89. Since the logarithm is a continuous function, this can be written 1 u loga lim u!0 1 þ u u u=u ( ) ¼ 1 u loga e by Problem 2.19, Chapter 2, with x ¼ u=u. Then by Problem 4.13, d dx ðloga uÞ ¼ loga e u du dx . 4.16. Calculate dy=dx if (a) xy3 3x2 ¼ xy þ 5, (b) exy þ y ln x ¼ cos 2x. (a) Differentiate with respect to x, considering y as a function of x. (We sometimes say that y is an implicit function of x, since we cannot solve explicitly for y in terms of x.) Then d dx ðxy3 Þ d dx ð3x2 Þ ¼ d dx ðxyÞ þ d dx ð5Þ or ðxÞð3y2 y0 Þ þ ðy3 Þð1Þ 6x ¼ ðxÞðy0 Þ þ ðyÞð1Þ þ 0 where y0 ¼ dy=dx. Solving, y0 ¼ ð6x y3 þ yÞ=ð3xy2 xÞ. ðbÞ d dx ðexy Þ þ d dx ðy ln xÞ ¼ d dx ðcos 2xÞ; exy ðxy0 þ yÞ þ y x þ ðln xÞy0 ¼ 2 sin 2x: y0 ¼ 2x sin 2x þ xyexy þ y x2exy þ x ln x Solving; 4.17. If y ¼ coshðx2 3x þ 1Þ, find (a) dy=dx; ðbÞ d2 y=dx2 . (a) Let y ¼ cosh u, where u ¼ x2 3x þ 1. Then dy=du ¼ sinh u, du=dx ¼ 2x 3, and dy dx ¼ dy du du dx ¼ ðsinh uÞð2x 3Þ ¼ ð2x 3Þ sinhðx2 3x þ 1Þ ðbÞ d2 y dx2 ¼ d dx dy dx ¼ d dx sinh u du dx ¼ sinh u d2 u dx2 þ cosh u du dx 2 ¼ ðsinh uÞð2Þ þ ðcosh uÞð2x 3Þ2 ¼ 2 sinhðx2 3x þ 1Þ þ ð2x 3Þ2 coshðx2 3x þ 1Þ 4.18. If x2 y þ y3 ¼ 2, find (a) y0 ; ðbÞ y00 at the point ð1; 1Þ. (a) Differentiating with respect to x, x2 y0 þ 2xy þ 3y2 y0 ¼ 0 and y0 ¼ 2xy x2 þ 3xy2 ¼ 1 2 at ð1; 1Þ ðbÞ y00 ¼ d dx ðy0 Þ ¼ d dx 2xy x2 þ 3y2 ¼ ðx2 þ 3y2 Þð2xy0 þ 2yÞ ð2xyÞð2x þ 6yy0 Þ ðx2 þ 3y2Þ2 Substituting x ¼ 1, y ¼ 1; and y0 ¼ 1 2, we find y00 ¼ 3 8. MEAN VALUE THEOREMS 4.19. Prove Rolle’s theorem. Case 1: f ðxÞ 0 in ½a; b. Then f 0 ðxÞ ¼ 0 for all x in ða; bÞ. Case 2: f ðxÞ 6 0 in ½a; b. Since f ðxÞ is continuous there are points at which f ðxÞ attains its maximum and minimum values, denoted by M and m respectively (see Problem 3.34, Chapter 3). Since f ðxÞ 6 0, at least one of the values M; m is not zero. Suppose, for example, M 6¼ 0 and that f ðÞ ¼ M (see Fig. 4-9). For this case, f ð þ hÞ @ f ðÞ. 80 DERIVATIVES [CHAP. 4
- 90. If h 0, then f ð þ hÞ f ðÞ h @ 0 and lim h!0þ f ð þ hÞ f ðÞ h @ 0 ð1Þ If h 0, then f ð þ hÞ f ðÞ h A 0 and lim h!0 f ð þ hÞ f ðÞ h A 0 ð2Þ But by hypothesis f ðxÞ has a derivative at all points in ða; bÞ. Then the right-hand derivative (1) must be equal to the left-hand derivative (2). This can happen only if they are both equal to zero, in which case f 0 ðÞ ¼ 0 as required. A similar argument can be used in case M ¼ 0 and m 6¼ 0. 4.20. Prove the mean value theorem. Define FðxÞ ¼ f ðxÞ f ðaÞ ðx aÞ f ðbÞ f ðaÞ b a . Then FðaÞ ¼ 0 and FðbÞ ¼ 0. Also, if f ðxÞ satisfies the conditions on continuity and differentiability specified in Rolle’s theorem, then FðxÞ satisfies them also. Then applying Rolle’s theorem to the function FðxÞ, we obtain F 0 ðÞ ¼ f 0 ðÞ f ðbÞ f ðaÞ b a ¼ 0; a b or f 0 ðÞ ¼ f ðbÞ f ðaÞ b a ; a b 4.21. Verify the mean value theorem for f ðxÞ ¼ 2x2 7x þ 10, a ¼ 2, b ¼ 5. f ð2Þ ¼ 4, f ð5Þ ¼ 25, f 0 ðÞ ¼ 4 7. Then the mean value theorem states that 4 7 ¼ ð25 4Þ=ð5 2Þ or ¼ 3:5. Since 2 5, the theorem is verified. 4.22. If f 0 ðxÞ ¼ 0 at all points of the interval ða; bÞ, prove that f ðxÞ must be a constant in the interval. Let x1 x2 be any two different points in ða; bÞ. By the mean value theorem for x1 x2, f ðx2Þ f ðx1Þ x2 x1 ¼ f 0 ðÞ ¼ 0 Thus, f ðx1Þ ¼ f ðx2Þ ¼ constant. From this it follows that if two functions have the same derivative at all points of ða; bÞ, the functions can only differ by a constant. 4.23. If f 0 ðxÞ 0 at all points of the interval ða; bÞ, prove that f ðxÞ is strictly increasing. Let x1 x2 be any two different points in ða; bÞ. By the mean value theorem for x1 x2, f ðx2Þ f ðx1Þ x2 x1 ¼ f 0 ðÞ 0 Then f ðx2Þ f ðx1Þ for x2 x1, and so f ðxÞ is strictly increasing. 4.24. (a) Prove that b a 1 þ b2 tan1 b tan1 a b a 1 þ a2 if a b. (b) Show that 4 þ 3 25 tan1 4 3 4 þ 1 6 . (a) Let f ðxÞ ¼ tan1 x. Since f 0 ðxÞ ¼ 1=ð1 þ x2 Þ and f 0 ðÞ ¼ 1=ð1 þ 2 Þ, we have by the mean value theorem CHAP. 4] DERIVATIVES 81 f (x) a ξ b x M Fig. 4-9
- 91. tan1 b tan1 a b a ¼ 1 1 þ 2 a b Since a, 1=ð1 þ 2 Þ 1=ð1 þ a2 Þ. Since b, 1=ð1 þ 2 Þ 1=ð1 þ b2 Þ. Then 1 1 þ b2 tan1 b tan1 a b a 1 1 þ a2 and the required result follows on multiplying by b a. (b) Let b ¼ 4=3 and a ¼ 1 in the result of part (a). Then since tan1 1 ¼ =4, we have 3 25 tan1 4 3 tan1 1 1 6 or 4 þ 3 25 tan1 4 3 4 þ 1 6 4.25. Prove Cauchy’s generalized mean value theorem. Consider GðxÞ ¼ f ðxÞ f ðaÞ fgðxÞ gðaÞg, where is a constant. Then GðxÞ satisfies the conditions of Rolle’s theorem, provided f ðxÞ and gðxÞ satisfy the continuity and differentiability conditions of Rolle’s theorem and if GðaÞ ¼ GðbÞ ¼ 0. Both latter conditions are satisfied if the constant ¼ f ðbÞ f ðaÞ gðbÞ gðaÞ . Applying Rolle’s theorem, G0 ðÞ ¼ 0 for a b, we have f 0 ðÞ g0 ðÞ ¼ 0 or f 0 ðÞ g0 ðÞ ¼ f ðbÞ f ðaÞ gðbÞ gðaÞ ; a b as required. L’HOSPITAL’S RULE 4.26. Prove L’Hospital’s rule for the case of the ‘‘indeterminate forms’’ (a) 0/0, (b) 1=1. (a) We shall suppose that f ðxÞ and gðxÞ are differentiable in a x b and f ðx0Þ ¼ 0, gðx0Þ ¼ 0, where a x0 b. By Cauchy’s generalized mean value theorem (Problem 25), f ðxÞ gðxÞ ¼ f ðxÞ f ðx0Þ gðxÞ gðx0Þ ¼ f 0 ðÞ g0 ðÞ x0 x Then lim x!x0þ f ðxÞ gðxÞ ¼ lim x!x0þ f 0 ðÞ g0ðÞ ¼ lim x!x0þ f 0 ðxÞ g0ðxÞ ¼ L since as x ! x0þ, ! x0þ. Modification of the above procedure can be used to establish the result if x ! x0, x ! x0, x ! 1, x ! 1. (b) We suppose that f ðxÞ and gðxÞ are differentiable in a x b, and lim x!x0þ f ðxÞ ¼ 1, lim x!x0þ gðxÞ ¼ 1 where a x0 b. Assume x1 is such that a x0 x x1 b. By Cauchy’s generalized mean value theorem, f ðxÞ f ðx1Þ gðxÞ gðx1Þ ¼ f 0 ðÞ g0 ðÞ x x1 Hence f ðxÞ f ðx1Þ gðxÞ gðx1Þ ¼ f ðxÞ gðxÞ 1 f ðx1Þ=f ðxÞ 1 gðx1Þ=gðxÞ ¼ f 0 ðÞ g0ðÞ 82 DERIVATIVES [CHAP. 4
- 92. from which we see that f ðxÞ gðxÞ ¼ f 0 ðÞ g0ðÞ 1 gðx1Þ=gðxÞ 1 f ðx1Þ=f ðxÞ ð1Þ Let us now suppose that lim x!x0þ f 0 ðxÞ g0ðxÞ ¼ L and write (1) as f ðxÞ gðxÞ ¼ f 0 ðÞ g0ðÞ L 1 gðx1Þ=gðxÞ 1 f ðx1Þ=f ðxÞ þ L 1 gðx1Þ=gðxÞ 1 f ðx1Þ=f ðxÞ ð2Þ We can choose x1 so close to x0 that j f 0 ðÞ=g0 ðÞ Lj . Keeping x1 fixed, we see that lim x!x0þ 1 gðx1Þ=gðxÞ 1 f ðx1Þ=f ðxÞ ¼ 1 since lim x!x0þ f ðxÞ ¼ 1 and lim x!x0 gðxÞ ¼ 1 Then taking the limit as x ! x0þ on both sides of (2), we see that, as required, lim x!x0þ f ðxÞ gðxÞ ¼ L ¼ lim x!x0þ f 0 ðxÞ g0ðxÞ Appropriate modifications of the above procedure establish the result if x ! x0, x ! x0, x ! 1, x ! 1. 4.27. Evaluate (a) lim x!0 e2x 1 x ðbÞ lim x!1 1 þ cos x x2 2x þ 1 All of these have the ‘‘indeterminate form’’ 0/0. ðaÞ lim x!0 e2x 1 x ¼ lim x!0 2e2x 1 ¼ 2 ðbÞ lim x!1 1 þ cos x x2 2x þ 1 ¼ lim x!1 sin x 2x 2 ¼ lim x!1 2 cos x 2 ¼ 2 2 Note: Here L’Hospital’s rule is applied twice, since the first application again yields the ‘‘indeter- minate form’’ 0/0 and the conditions for L’Hospital’s rule are satisfied once more. 4.28. Evaluate (a) lim x!1 3x2 x þ 5 5x2 þ 6x 3 ðbÞ lim x!1 x2 ex All of these have or can be arranged to have the ‘‘indeterminate form’’ 1=1. ðaÞ lim x!1 3x2 x þ 5 5x2 þ 6x 3 ¼ lim x!1 6x 1 10x þ 6 ¼ lim x!1 6 10 ¼ 3 5 ðbÞ lim x!1 x2 ex ¼ lim x!1 x2 ex ¼ lim x!1 2x ex ¼ lim x!1 2 ex ¼ 0 4.29. Evaluate lim x!0þ x2 ln x. lim x!0þ x2 ln x ¼ lim x!0þ ln x 1=x2 ¼ lim x!0þ 1=x 2=x3 ¼ lim x!0þ x2 2 ¼ 0 The given limit has the ‘‘indeterminate form’’ 0 1. In the second step the form is altered so as to give the indeterminate form 1=1 and L’Hospital’s rule is then applied. 4.30. Find lim x!0 ðcos xÞ1=x2 . Since lim x!0 cos x ¼ 1 and lim x!0 1=x2 ¼ 1, the limit takes the ‘‘indeterminate form’’ 11 . CHAP. 4] DERIVATIVES 83
- 93. Let FðxÞ ¼ ðcos xÞ1=x2 . Then ln FðxÞ ¼ ðln cos xÞ=x2 to which L’Hospital’s rule can be applied. We have lim x!0 ln cos x x2 ¼ lim x!0 ð sin xÞ=ðcos xÞ 2x ¼ lim x!0 sin x 2x cos x ¼ lim x!0 cos x 2x sin x þ 2 cos x ¼ 1 2 Thus, lim x!0 ln FðxÞ ¼ 1 2. But since the logarithm is a continuous function, lim x!0 ln FðxÞ ¼ lnðlim x!0 FðxÞÞ. Then lnðlim x!0 FðxÞÞ ¼ 1 2 or lim x!0 FðxÞ ¼ lim x!0 ðcos xÞ1=x2 ¼ e1=2 4.31. If FðxÞ ¼ ðe3x 5xÞ1=x , find (a) lim x!0 FðxÞ and (b) lim x!0 FðxÞ. The respective indeterminate forms in (a) and (b) are 10 and 11 . Let GðxÞ ¼ ln FðxÞ ¼ lnðe3x 5xÞ x . Then lim x!1 GðxÞ and lim x!0 GðxÞ assume the indeterminate forms 1=1 and 0/0 respectively, and L’Hospital’s rule applies. We have ðaÞ lim x!1 lnðe3x 5xÞ x ¼ lim x!1 3e3x 5 e3x 5x ¼ lim x!0 9e3x 3e3x 5 ¼ lim x!1 27e3x 9e3x ¼ 3 Then, as in Problem 4.30, lim x!1 ðe3x 5xÞ1=x ¼ e3 . ðbÞ lim x!0 lnðe3x 5xÞ x ¼ lim x!0 3e3x 5 e3x 5x ¼ 2 and lim x!0 ðe3x 5xÞ1=x ¼ e2 4.32. Suppose the equation of motion of a particle is x ¼ sinðc1t þ c2Þ, where c1 and c2 are constants. (Simple harmonic motion.) (a) Show that the acceleration of the particle is proportional to its distance from the origin. (b) If c1 ¼ 1, c2 ¼ , and t 0, determine the velocity and acceleration at the end points and at the midpoint of the motion. ðaÞ dx dt ¼ c1 cosðc1t þ c2Þ; d2 x dt2 ¼ c2 1 sinðc1t þ c2Þ ¼ c2 1x: This relation demonstrates the proportionality of acceleration and distance. (b) The motion starts at 0 and moves to 1. Then it oscillates between this value and 1. The absolute value of the velocity is zero at the end points, and that of the acceleration is maximum there. The particle coasts through the origin (zero acceleration), while the absolute value of the velocity is maximum there. 4.33. Use Newton’s method to determine ffiffiffi 3 p to three decimal points of accuracy. ffiffiffi 3 p is a solution of x2 3 ¼ 0, which lies between 1 and 2. Consider f ðxÞ ¼ x2 3 then f 0 ðxÞ ¼ 2x. The graph of f crosses the x-axis between 1 and 2. Let x0 ¼ 2. Then f ðx0Þ ¼ 1 and f 0 ðx0Þ ¼ 1:75. According to the Newton formula, x1 ¼ x0 f ðx0Þ f 0 ðx0Þ ¼ 2 :25 ¼ 1:75. Then x2 ¼ x1 f ðx1Þ f 0 ðx1Þ ¼ 1:732. To verify the three decimal point accuracy, note that ð1:732Þ2 ¼ 2:9998 and ð1:7333Þ2 ¼ 3:0033. 84 DERIVATIVES [CHAP. 4
- 94. MISCELLANEOUS PROBLEMS 4.34. If x ¼ gðtÞ and y ¼ f ðtÞ are twice differentiable, find (a) dy=dx; ðbÞ d2 y=dx2 . (a) Letting primes denote derivatives with respect to t, we have dy dx ¼ dy=dt dx=dt ¼ f 0 ðtÞ g0ðtÞ if g0 ðtÞ 6¼ 0 ðbÞ d2 y dx2 ¼ d dx dy dx ¼ d dx f 0 ðtÞ g0ðtÞ ¼ d dt f 0 ðtÞ g0 ðtÞ dx=dt ¼ d dt f 0 ðtÞ g0 ðtÞ g0ðtÞ ¼ 1 g0ðtÞ g0 ðtÞf 00 ðtÞ f 0 ðtÞg00 ðtÞ ½g0ðtÞ2 ¼ g0 ðtÞf 00 ðtÞ f 0 ðtÞg00 ðtÞ ½g0ðtÞ3 if g0 ðtÞ 6¼ 0 4.35. Let f ðxÞ ¼ e1=x2 ; x 6¼ 0 0; x ¼ 0 . Prove that (a) f 0 ð0Þ ¼ 0; ðbÞ f 00 ð0Þ ¼ 0. ðaÞ f 0 þð0Þ ¼ lim h!0þ f ðhÞ f ð0Þ h ¼ lim h!0þ e1=h2 0 h ¼ lim h!0þ e1=h2 h If h ¼ 1=u, using L’Hospital’s rule this limit equals lim u!1 ueu2 ¼ lim u!1 u=eu2 ¼ lim u!1 1=2ueu2 ¼ 0 Similarly, replacing h ! 0þ by h ! 0 and u ! 1 by u ! 1, we find f 0 ð0Þ ¼ 0. Thus f 0 þð0Þ ¼ f 0 ð0Þ ¼ 0, and so f 0 ð0Þ ¼ 0. ðbÞ f 00 þ ð0Þ ¼ lim h!0þ f 0 ðhÞ f 0 ð0Þ h ¼ lim h!0þ e1=h2 2h3 0 h ¼ lim h!0þ 2e1=h2 h4 ¼ lim u!1 2u4 eu2 ¼ 0 by successive applications of L’Hospital’s rule. Similarly, f 00 ð0Þ ¼ 0 and so f 00 ð0Þ ¼ 0. In general, f ðnÞ ð0Þ ¼ 0 for n ¼ 1; 2; 3; . . . 4.36. Find the length of the longest ladder which can be carried around the corner of a corridor, whose dimensions are indicated in the figure below, if it is assumed that the ladder is carried parallel to the floor. The length of the longest ladder is the same as the shortest straight line segment AB [Fig. 4-10], which touches both outer walls and the corner formed by the inner walls. As seen from Fig. 4-10, the length of the ladder AB is L ¼ a sec þ b csc L is a minimum when dL=d ¼ a sec tan b csc cot ¼ 0 a sin3 ¼ b cos3 or tan ¼ ffiffiffiffiffiffiffiffi b=a 3 p i.e.; sec ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2=3 þ b2=3 p a1=3 ; csc ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2=3 þ b2=3 p b1=3 Then L ¼ a sec þ b csc ¼ ða2=3 þ b2=3 Þ3=2 so that Although it is geometrically evident that this gives the minimum length, we can prove this analytically by showing that d2 L=d2 for ¼ tan1 ffiffiffiffiffiffiffiffi b=a 3 p is positive (see Problem 4.78). CHAP. 4] DERIVATIVES 85 a b B O A a s e c G b c s c G G G Fig. 4-10
- 95. Supplementary Problems DERIVATIVES 4.37. Use the definition to compute the derivatives of each of the following functions at the indicated point: (a) ð3x 4Þ=ð2x þ 3Þ; x ¼ 1; ðbÞ x3 3x2 þ 2x 5; x ¼ 2; ðcÞ ffiffiffi x p ; x ¼ 4; ðdÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6x 4 3 p ; x ¼ 2: Ans. (a) 17/25, (b) 2, (c) 1 4, (d) 1 2 4.38. Show from definition that (a) d dx x4 ¼ 4x3 ; ðbÞ d dx 3 þ x 3 x ¼ 6 ð3 xÞ2 ; x 6¼ 3 4.39. Let f ðxÞ ¼ x3 sin 1=x; x 6¼ 0 0; x ¼ 0 . Prove that (a) f ðxÞ is continuous at x ¼ 0, (b) f ðxÞ has a derivative at x ¼ 0, (c) f 0 ðxÞ is continuous at x ¼ 0. 4.40. Let f ðxÞ ¼ xe1=x2 ; x 6¼ 0 0; x ¼ 0 . Determine whether f ðxÞ (a) is continuous at x ¼ 0, (b) has a derivative at x ¼ 0: Ans. (a) Yes; (b) Yes, 0 4.41. Give an alternative proof of the theorem in Problem 4.3, Page 76, using ‘‘; definitions’’. 4.42. If f ðxÞ ¼ ex , show that f 0 ðx0Þ ¼ ex0 depends on the result lim h!0 ðeh 1Þ=h ¼ 1. 4.43. Use the results lim h!0 ðsin hÞ=h ¼ 1, lim h!0 ð1 cos hÞ=h ¼ 0 to prove that if f ðxÞ ¼ sin x, f 0 ðx0Þ ¼ cos x0. RIGHT- AND LEFT-HAND DERIVATIVES 4.44. Let f ðxÞ ¼ xjxj. (a) Calculate the right-hand derivative of f ðxÞ at x ¼ 0. (b) Calculate the left-hand derivative of f ðxÞ at x ¼ 0. (c) Does f ðxÞ have a derivative at x ¼ 0? (d) Illustrate the conclusions in ðaÞ, (b), and (c) from a graph. Ans. (a) 0; (b) 0; (c) Yes, 0 4.45. Discuss the (a) continuity and (b) differentiability of f ðxÞ ¼ xp sin 1=x, f ð0Þ ¼ 0, where p is any positive number. What happens in case p is any real number? 4.46. Let f ðxÞ ¼ 2x 3; 0 @ x @ 2 x2 3; 2 x @ 4 . Discuss the (a) continuity and (b) differentiability of f ðxÞ in 0 @ x @ 4. 4.47. Prove that the derivative of f ðxÞ at x ¼ x0 exists if and only if f 0 þðx0Þ ¼ f 0 ðx0Þ. 4.48. (a) Prove that f ðxÞ ¼ x3 x2 þ 5x 6 is differentiable in a @ x @ b, where a and b are any constants. (b) Find equations for the tangent lines to the curve y ¼ x3 x2 þ 5x 6 at x ¼ 0 and x ¼ 1. Illustrate by means of a graph. (c) Determine the point of intersection of the tangent lines in (b). (d) Find f 0 ðxÞ; f 00 ðxÞ; f 000 ðxÞ; f ðIVÞ ðxÞ; . . . . Ans. (b) y ¼ 5x 6; y ¼ 6x 7; ðcÞ ð1; 1Þ; ðdÞ 3x2 2x þ 5; 6x 2; 6; 0; 0; 0; . . . 4.49. If f ðxÞ ¼ x2 jxj, discuss the existence of successive derivatives of f ðxÞ at x ¼ 0. DIFFERENTIALS 4.50. If y ¼ f ðxÞ ¼ x þ 1=x, find (a) y; ðbÞ dy; ðcÞ y dy; ðdÞ ðy dyÞ=x; ðeÞ dy=dx. Ans: ðaÞ x x xðx þ xÞ ; ðbÞ 1 1 x2 x; ðcÞ ðxÞ2 x2ðx þ xÞ ; ðdÞ x x2ðx þ xÞ ; ðeÞ 1 1 x2 : Note: x ¼ dx. 86 DERIVATIVES [CHAP. 4
- 96. 4.51. If f ðxÞ ¼ x2 þ 3x, find (a) y; ðbÞ dy; ðcÞ y=x; ðdÞ dy=dx; and (e) ðy dyÞ=x, if x ¼ 1 and x ¼ :01. Ans. (a) .0501, (b) .05, (c) 5.01, (d) 5, (e) .01 4.52. Using differentials, compute approximate values for each of the following: (a) sin 318; ðbÞ lnð1:12Þ, (c) ffiffiffiffiffi 36 5 p . Ans. (a) 0.515, (b) 0.12, (c) 2.0125 4.53. If y ¼ sin x, evaluate (a) y; ðbÞ dy. (c) Prove that ðy dyÞ=x ! 0 as x ! 0. DIFFERENTIATION RULES AND ELEMENTARY FUNCTIONS 4.54. Prove: (a) d dx f f ðxÞ þ gðxÞg ¼ d dx f ðxÞ þ d dx gðxÞ; ðbÞ d dx f f ðxÞ gðxÞg ¼ d dx f ðxÞ d dx gðxÞ, ðcÞ d dx f ðxÞ gðxÞ ¼ gðxÞ f 0 ðxÞ f ðxÞg0 ðxÞ ½gðxÞ2 ; gðxÞ 6¼ 0: 4.55. Evaluate (a) d dx fx3 lnðx2 2x þ 5Þg at x ¼ 1; ðbÞ d dx fsin2 ð3x þ =6Þg at x ¼ 0. Ans. (a) 3 ln 4; ðbÞ 3 2 ffiffiffi 3 p 4.56. Derive the formulas: (a) d dx au ¼ au ln a du dx ; a 0; a 6¼ 1; ðbÞ d dx csc u ¼ csc u cot u du dx ; ðcÞ d dx tanh u ¼ sech2 u du dx where u is a differentiable function of x: 4.57. Compute (a) d dx tan1 x; ðbÞ d dx csc1 x; ðcÞ d dx sinh1 x; ðdÞ d dx coth1 x, paying attention to the use of principal values. 4.58. If y ¼ xx , computer dy=dx. [Hint: Take logarithms before differentiating.] Ans. xx ð1 þ ln xÞ 4.59. If y ¼ flnð3x þ 2Þgsin1 ð2xþ:5Þ , find dy=dx at x ¼ 0: Ans: 4 ln 2 þ 2 ln ln 2 ffiffiffi 3 p ðln 2Þ=6 4.60. If y ¼ f ðuÞ, where u ¼ gðvÞ and v ¼ hðxÞ, prove that dy dx ¼ dy du du dv dv dx assuming f , g; and h are differentiable. 4.61. Calculate (a) dy=dx and (b) d2 y=dx2 if xy ln y ¼ 1. Ans. (a) y2 =ð1 xyÞ; ðbÞ ð3y3 2xy4 Þ=ð1 xyÞ3 provided xy 6¼ 1 4.62. If y ¼ tan x, prove that y000 ¼ 2ð1 þ y2 Þð1 þ 3y2 Þ. 4.63. If x ¼ sec t and y ¼ tan t, evaluate (a) dy=dx; ðbÞ d2 y=dx2 ; ðcÞ d3 y=dx3 , at t ¼ =4. Ans. (a) ffiffiffi 2 p ; ðbÞ 1; ðcÞ 3 ffiffiffi 2 p 4.64. Prove that d2 y dx2 ¼ d2 x dy2 dx dy 3 , stating precise conditions under which it holds. 4.65. Establish formulas (a) 7, (b) 18, and (c) 27, on Page 71. MEAN VALUE THEOREMS 4.66. Let f ðxÞ ¼ 1 ðx 1Þ2=3 , 0 @ x @ 2. (a) Construct the graph of f ðxÞ. (b) Explain why Rolle’s theorem is not applicable to this function, i.e., there is no value for which f 0 ðÞ ¼ 0, 0 2. CHAP. 4] DERIVATIVES 87
- 97. 4.67. Verify Rolle’s theorem for f ðxÞ ¼ x2 ð1 xÞ2 , 0 @ x @ 1. 4.68. Prove that between any two real roots of ex sin x ¼ 1 there is at least one real root of ex cos x ¼ 1. [Hint: Apply Rolle’s theorem to the function ex sin x: 4.69. (a) If 0 a b, prove that ð1 a=bÞ ln b=a ðb=a 1Þ (b) Use the result of (a) to show that 1 6 ln 1:2 1 5. 4.70. Prove that ð=6 þ ffiffiffi 3 p =15Þ sin1 :6 ð=6 þ 1=8Þ by using the mean value theorem. 4.71. Show that the function FðxÞ in Problem 4.20(a) represents the difference in ordinants of curve ACB and line AB at any point x in ða; bÞ. 4.72. (a) If f 0 ðxÞ @ 0 at all points of ða; bÞ, prove that f ðxÞ is monotonic decreasing in ða; bÞ. (b) Under what conditions is f ðxÞ strictly decreasing in ða; bÞ? 4.73. (a) Prove that ðsin xÞ=x is strictly decreasing in ð0; =2Þ. (b) Prove that 0 @ sin x @ 2x= for 0 @ x @ =2. 4.74. (a) Prove that sin b sin a cos a cos b ¼ cot , where is between a and b. (b) By placing a ¼ 0 and b ¼ x in (a), show that ¼ x=2. Does the result hold if x 0? L’HOSPITAL’S RULE 4.75. Evaluate each of the following limits. (a) lim x!0 x sin x x3 (e) lim x!0þ x3 ln x (i) lim x!0 ð1=x csc xÞ (m) lim x!1 x ln x þ 3 x 3 (b) lim x!0 e2x 2ex þ 1 cos 3x 2 cos 2x þ cos x ( f ) lim x!0 ð3x 2x Þ=x ( j) lim x!0 xsin x (n) lim x!0 sin x x 1=x2 (c) lim x!1þ ðx2 1Þ tan x=2 (g) lim x!1 ð1 3=xÞ2x ðkÞ lim x!0 ð1=x2 cot2 xÞ (o) lim x!1 ðx þ ex þ e2x Þ1=x (d) lim x!1 x3 e2x (h) lim x!1 ð1 þ 2xÞ1=3x (l) lim x!0 tan1 x sin1 x xð1 cos xÞ (p) lim x!0þ ðsin xÞ1= ln x Ans. (a) 1 6 ; ðbÞ 1; ðcÞ 4=; ðdÞ 0; ðeÞ 0; ð f Þ ln 3=2; ðgÞ e6 ; ðhÞ 1; ðiÞ 0; ð jÞ 1, (k) 2 3 ; ðlÞ 1 3 ; ðmÞ 6; ðnÞ e1=6 ; ðoÞ e2 ; ð pÞ e MISCELLANEOUS PROBLEMS 4.76. Prove that ffiffiffiffiffiffiffiffiffiffiffi 1 x 1 þ x r lnð1 þ xÞ sin1 x 1 if 0 x 1. 4.77. If f ðxÞ ¼ f ðx þ xÞ f ðxÞ, (a) Prove that ff ðxÞg ¼ 2 f ðxÞ ¼ f ðx þ 2xÞ 2f ðx þ xÞ þ f ðxÞ, (b) derive an expression for n f ðxÞ where n is any positive integer, (c) show that lim x!0 n f ðxÞ ðxÞn ¼ f ðnÞ ðxÞ if this limit exists. 4.78. Complete the analytic proof mentioned at the end of Problem 4.36. 4.79. Find the relative maximum and minima of f ðxÞ ¼ x2 , x 0. Ans. f ðxÞ has a relative minimum when x ¼ e1 . 4.80. A train moves according to the rule x ¼ 5t3 þ 30t, where t and x are measured in hours and miles, respectively. (a) What is the acceleration after 1 minute? (b) What is the speed after 2 hours? 4.81. A stone thrown vertically upward has the law of motion x ¼ 16t2 þ 96t. (Assume that the stone is at ground level at t ¼ 0, that t is measured in seconds, and that x is measured in feet.) (a) What is the height of the stone at t ¼ 2 seconds? (b) To what height does the stone rise? (c) What is the initial velocity, and what is the maximum speed attained? 88 DERIVATIVES [CHAP. 4
- 98. 4.82. A particle travels with constant velocities v1 and v2 in mediums I and II, respectively (see adjoining Fig. 4-11). Show that in order to go from point P to point Q in the least time, it must follow path PAQ where A is such that ðsin 1Þ=ðsin 2Þ ¼ v1=v2 Note: This is Snell’s Law; a fundamental law of optics first discovered experimentally and then derived mathematically. 4.83. A variable is called an infinitesimal if it has zero as a limit. Given two infinitesimals and , we say that is an infinitesimal of higher order (or the same order) if lim = ¼ 0 (or lim = ¼ l 6¼ 0). Prove that as x ! 0, (a) sin2 2x and ð1 cos 3xÞ are infinitesimals of the same order, (b) ðx3 sin3 xÞ is an infinitesimal of higher order than fx lnð1 þ xÞ 1 þ cos xg. 4.84. Why can we not use L’Hospital’s rule to prove that lim x!0 x2 sin 1=x sin x ¼ 0 (see Problem 3.91, Chap. 3)? 4.85. Can we use L’Hospital’s rule to evaluate the limit of the sequence un ¼ n3 en2 , n ¼ 1; 2; 3; . . . ? Explain. 4.86 (1) Determine decimal approximations with at least three places of accuracy for each of the following irrational numbers. (a) ffiffiffi 2 p ; ðbÞ ffiffiffi 5 p ; ðcÞ 71=3 (2) The cubic equation x3 3x2 þ x 4 ¼ 0 has a root between 3 and 4. Use Newton’s Method to determine it to at least three places of accuracy. 4.87. Using successive applications of Newton’s method obtain the positive root of (a) x3 2x2 2x 7 ¼ 0, (b) 5 sin x ¼ 4x to 3 decimal places. Ans. (a) 3.268, (b) 1.131 4.88. If D denotes the operator d=dx so that Dy dy=dx while Dk y dk y=dxk , prove Leibnitz’s formula Dn ðuvÞ ¼ ðDn uÞv þ nC1ðDn1 uÞðDvÞ þ nC2ðDn2 uÞðD2 vÞ þ þ nCrðDnr uÞðDr vÞ þ þ uDn v where nCr ¼ ðn rÞ are the binomial coefficients (see Problem 1.95, Chapter 1). 4.89. Prove that dn dxn ðx2 sin xÞ ¼ fx2 nðn 1Þg sinðx þ n=2Þ 2nx cosðx þ n=2Þ. 4.90. If f 0 ðx0Þ ¼ f 00 ðx0Þ ¼ ¼ f ð2nÞ ðx0Þ ¼ 0 but f ð2nþ1Þ ðx0Þ 6¼ 0, discuss the behavior of f ðxÞ in the neighborhood of x ¼ x0. The point x0 in such case is often called a point of inflection. This is a generalization of the previously discussed case corresponding to n ¼ 1. 4.91. Let f ðxÞ be twice differentiable in ða; bÞ and suppose that f 0 ðaÞ ¼ f 0 ðbÞ ¼ 0. Prove that there exists at least one point in ða; bÞ such that j f 00 ðÞj A 4 ðb aÞ2 f f ðbÞ f ðaÞg. Give a physical interpretation involving velocity and acceration of a particle. CHAP. 4] DERIVATIVES 89 P Q A G1 G2 Medium I velocity L1 Medium II velocity L2 Fig. 4-11
- 99. 90 Integrals INTRODUCTION OF THE DEFINITE INTEGRAL The geometric problems that motivated the development of the integral calculus (determination of lengths, areas, and volumes) arose in the ancient civilizations of Northern Africa. Where solutions were found, they related to concrete problems such as the measurement of a quantity of grain. Greek philosophers took a more abstract approach. In fact, Eudoxus (around 400 B.C.) and Archimedes (250 B.C.) formulated ideas of integration as we know it today. Integral calculus developed independently, and without an obvious connection to differential calculus. The calculus became a ‘‘whole’’ in the last part of the seventeenth century when Isaac Barrow, Isaac Newton, and Gottfried Wilhelm Leibniz (with help from others) discovered that the integral of a function could be found by asking what was differentiated to obtain that function. The following introduction of integration is the usual one. It displays the concept geometrically and then defines the integral in the nineteenth-century language of limits. This form of definition establishes the basis for a wide variety of applications. Consider the area of the region bound by y ¼ f ðxÞ, the x-axis, and the joining vertical segments (ordinates) x ¼ a and x ¼ b. (See Fig. 5-1.) y a ξ1 ξ2 ξn _ 1 ξ3 ξn x1 x2 xn _ 2 xn _ 1 b x x3 y = f (x) Fig. 5-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 100. Subdivide the interval a @ x @ b into n sub-intervals by means of the points x1; x2; . . . ; xn1 chosen arbitrarily. In each of the new intervals ða; x1Þ; ðx1; x2Þ; . . . ; ðxn1; bÞ choose points 1; 2; . . . ; n arbitrarily. Form the sum f ð1Þðx1 aÞ þ f ð2Þðx2 x1Þ þ f ð3Þðx3 x2Þ þ þ f ðnÞðb xn1Þ ð1Þ By writing x0 ¼ a, xn ¼ b; and xk xk1 ¼ xk, this can be written X n k¼1 f ðkÞðxk xk1Þ ¼ X n k¼1 f ðkÞxk ð2Þ Geometrically, this sum represents the total area of all rectangles in the above figure. We now let the number of subdivisions n increase in such a way that each xk ! 0. If as a result the sum (1) or (2) approaches a limit which does not depend on the mode of subdivision, we denote this limit by ðb a f ðxÞ dx ¼ lim n!1 X n k¼1 f ðkÞxk ð3Þ This is called the definite integral of f ðxÞ between a and b. In this symbol f ðxÞ dx is called the integrand, and ½a; b is called the range of integration. We call a and b the limits of integration, a being the lower limit of integration and b the upper limit. The limit (3) exists whenever f ðxÞ is continuous (or piecewise continuous) in a @ x @ b (see Problem 5.31). When this limit exists we say that f is Riemann integrable or simply integrable in ½a; b. The definition of the definite integral as the limit of a sum was established by Cauchy around 1825. It was named for Riemann because he made extensive use of it in this 1850 exposition of integration. Geometrically the value of this definite integral represents the area bounded by the curve y ¼ f ðxÞ, the x-axis and the ordinates at x ¼ a and x ¼ b only if f ðxÞ A 0. If f ðxÞ is sometimes positive and sometimes negative, the definite integral represents the algebraic sum of the areas above and below the x- axis, treating areas above the x-axis as positive and areas below the x-axis as negative. MEASURE ZERO A set of points on the x-axis is said to have measure zero if the sum of the lengths of intervals enclosing all the points can be made arbitrary small (less than any given positive number ). We can show (see Problem 5.6) that any countable set of points on the real axis has measure zero. In particular, the set of rational numbers which is countable (see Problems 1.17 and 1.59, Chapter 1), has measure zero. An important theorem in the theory of Riemann integration is the following: Theorem. If f ðxÞ is bounded in ½a; b, then a necessary and sufficient condition for the existence of Ðb a f ðxÞ dx is that the set of discontinuities of f ðxÞ have measure zero. PROPERTIES OF DEFINITE INTEGRALS If f ðxÞ and gðxÞ are integrable in ½a; b then 1. ðb a f f ðxÞ gðxÞg dx ¼ ðb a f ðxÞ dx ðb a gðxÞ dx 2. ðb a A f ðxÞ dx ¼ A ðb a f ðxÞ dx where A is any constant CHAP. 5] INTEGRALS 91
- 101. 3. ðb a f ðxÞ dx ¼ ðc a f ðxÞ dx þ ðb c f ðxÞ dx provided f ðxÞ is integrable in ½a; c and ½c; b. 4. ðb a f ðxÞ dx ¼ ða b f ðxÞ dx 5. ða a f ðxÞ dx ¼ 0 6. If in a @ x @ b, m @ f ðxÞ @ M where m and M are constants, then mðb aÞ @ ðb a f ðxÞ dx @ Mðb aÞ 7. If in a @ x @ b, f ðxÞ @ gðxÞ then ðb a f ðxÞ dx @ ðb a gðxÞ dx 8. ðb a f ðxÞ dx @ ðb a j f ðxÞj dx if a b MEAN VALUE THEOREMS FOR INTEGRALS As in differential calculus the mean value theorems listed below are existence theorems. The first one generalizes the idea of finding an arithmetic mean (i.e., an average value of a given set of values) to a continuous function over an interval. The second mean value theorem is an extension of the first one that defines a weighted average of a continuous function. By analogy, consider determining the arithmetic mean (i.e., average value) of temperatures at noon for a given week. This question is resolved by recording the 7 temperatures, adding them, and dividing by 7. To generalize from the notion of arithmetic mean and ask for the average temperature for the week is much more complicated because the spectrum of temperatures is now continuous. However, it is reasonable to believe that there exists a time at which the average temperature takes place. The manner in which the integral can be employed to resolve the question is suggested by the following example. Let f be continuous on the closed interval a @ x @ b. Assume the function is represented by the correspondence y ¼ f ðxÞ, with f ðxÞ 0. Insert points of equal subdivision, a ¼ x0; x1; . . . ; xn ¼ b. Then all xk ¼ xk xk1 are equal and each can be designated by x. Observe that b a ¼ nx. Let k be the midpoint of the interval xk and f ðkÞ the value of f there. Then the average of these functional values is f ð1Þ þ þ f ðnÞ n ¼ ½ f ð1Þ þ þ f ðnÞx b a ¼ 1 b a X n k¼1 f ð Þ This sum specifies the average value of the n functions at the midpoints of the intervals. However, we may abstract the last member of the string of equalities (dropping the special conditions) and define lim n!1 1 b a X n k¼1 f ð Þ ¼ 1 b a ðb a f ðxÞ dx as the average value of f on ½a; b. 92 INTEGRALS [CHAP. 5
- 102. Of course, the question of for what value x ¼ the average is attained is not answered; and, in fact, in general, only existence not the value can be demonstrated. To see that there is a point x ¼ such that f ðÞ represents the average value of f on ½a; b, recall that a continuous function on a closed interval has maximum and minimum values, M and m, respectively. Thus (think of the integral as representing the area under the curve). (See Fig. 5-2.) mðb aÞ @ ðb a f ðxÞ dx @ Mðb aÞ or m @ 1 b a ðb a f ðxÞ dx @ M Since f is a continuous function on a closed interval, there exists a point x ¼ in ða; bÞ intermediate to m and M such that f ðÞ ¼ 1 b a ðb a f ðxÞ dx While this example is not a rigorous proof of the first mean value theorem, it motivates it and provides an interpretation. (See Chapter 3, Theorem 10.) 1. First mean value theorem. If f ðxÞ is continuous in ½a; b, there is a point in ða; bÞ such that ðb a f ðxÞ dx ¼ ðb aÞ f ðÞ ð4Þ 2. Generalized first mean value theorem. If f ðxÞ and gðxÞ are continuous in ½a; b, and gðxÞ does not change sign in the interval, then there is a point in ða; bÞ such that ðb a f ðxÞgðxÞ dx ¼ f ðÞ ðb a gðxÞ dx ð5Þ This reduces to (4) if gðxÞ ¼ 1. CHAP. 5] INTEGRALS 93 y b a x b _ a A D F E M m B C y = f (x) Fig. 5-2
- 103. CONNECTING INTEGRAL AND DIFFERENTIAL CALCULUS In the late seventeenth century the key relationship between the derivative and the integral was established. The connection which is embodied in the fundamental theorem of calculus was responsible for the creation of a whole new branch of mathematics called analysis. Definition: Any function F such that F 0 ðxÞ ¼ f ðxÞ is called an antiderivative, primitive, or indefinite integral of f . The antiderivative of a function is not unique. This is clear from the observation that for any constant c ðFðxÞ þ cÞ0 ¼ F 0 ðxÞ ¼ f ðxÞ The following theorem is an even stronger statement. Theorem. Any two primitives (i.e., antiderivatives), F and G of f differ at most by a constant, i.e., FðxÞ GðxÞ ¼ C. (See the problem set for the proof of this theorem.) EXAMPLE. If F 0 ðxÞ ¼ x2 , then FðxÞ ¼ ð x2 dx ¼ x3 3 þ c is an indefinite integral (antiderivative or primitive) of x2 . The indefinite integral (which is a function) may be expressed as a definite integral by writing ð f ðxÞ dx ¼ ðx c f ðtÞ dt The functional character is expressed through the upper limit of the definite integral which appears on the right-hand side of the equation. This notation also emphasizes that the definite integral of a given function only depends on the limits of integration, and thus any symbol may be used as the variable of integration. For this reason, that variable is often called a dummy variable. The indefinite integral notation on the left depends on continuity of f on a domain that is not described. One can visualize the definite integral on the right by thinking of the dummy variable t as ranging over a subinterval ½c; x. (There is nothing unique about the letter t; any other convenient letter may represent the dummy variable.) The previous terminology and explanation set the stage for the fundamental theorem. It is stated in two parts. The first states that the antiderivative of f is a new function, the integrand of which is the derivative of that function. Part two demonstrates how that primitive function (antiderivative) enables us to evaluate definite integrals. THE FUNDAMENTAL THEOREM OF THE CALCULUS Part 1 Let f be integrable on a closed interval ½a; b. Let c satisfy the condition a @ c @ b, and define a new function FðxÞ ¼ ðx c f ðtÞ dt if a @ x @ b Then the derivative F 0 ðxÞ exists at each point x in the open interval ða; bÞ, where f is continuous and F 0 ðxÞ ¼ f ðxÞ. (See Problem 5.10 for proof of this theorem.) Part 2 As in Part 1, assume that f is integrable on the closed interval ½a; b and continuous in the open interval ða; bÞ. Let F be any antiderivative so that F 0 ðxÞ ¼ f ðxÞ for each x in ða; bÞ. If a c b, then for any x in ða; bÞ ðx c f ðtÞ dt ¼ FðxÞ FðcÞ 94 INTEGRALS [CHAP. 5
- 104. If the open interval on which f is continuous includes a and b, then we may write ðb a f ðxÞ dx ¼ FðbÞ FðaÞ: (See Problem 5.11) This is the usual form in which the theorem is used. EXAMPLE. To evaluate ð2 1 x2 dx we observe that F 0 ðxÞ ¼ x2 , FðxÞ ¼ x3 3 þ c and ð2 1 x2 dx ¼ 23 3 þ c 13 3 þ c ¼ 7 3. Since c subtracts out of this evaluation it is convenient to exclude it and simply write 23 3 13 3 . GENERALIZATION OF THE LIMITS OF INTEGRATION The upper and lower limits of integration may be variables. For example: ðcos x sin x t dt ¼ t2 2 #cos x sin x ¼ ðcos2 x sin2 xÞ=2 In general, if F 0 ðxÞ ¼ f ðxÞ then ðvðxÞ uðxÞ f ðtÞ dt ¼ F½vðxÞ ¼ F½uðxÞ CHANGE OF VARIABLE OF INTEGRATION If a determination of Ð f ðxÞ dx is not immediately obvious in terms of elementary functions, useful results may be obtained by changing the variable from x to t according to the transformation x ¼ gðtÞ. (This change of integrand that follows is suggested by the differential relation dx ¼ g0 ðtÞ dt.) The funda- mental theorem enabling us to do this is summarized in the statement ð f ðxÞ dx ¼ ð f fgðtÞgg0 ðtÞ dt ð6Þ where after obtaining the indefinite integral on the right we replace t by its value in terms of x, i.e., t ¼ g1 ðxÞ. This result is analogous to the chain rule for differentiation (see Page 69). The corresponding theorem for definite integrals is ðb a f ðxÞ dx ¼ ð f fgðtÞgg0 ðtÞ dt ð7Þ where gð Þ ¼ a and gð Þ ¼ b, i.e., ¼ g1 ðaÞ, ¼ g1 ðbÞ. This result is certainly valid if f ðxÞ is con- tinuous in ½a; b and if gðtÞ is continuous and has a continuous derivative in @ t @ . INTEGRALS OF ELEMENTARY FUNCTIONS The following results can be demonstrated by differentiating both sides to produce an identity. In each case an arbitrary constant c (which has been omitted here) should be added. CHAP. 5] INTEGRALS 95
- 105. 1. ð un du ¼ unþ1 n þ 1 n 6¼ 1 18. ð coth u du ¼ ln j sinh uj 2. ð du u ¼ ln juj 19. ð sech u du ¼ tan1 ðsinh uÞ 3. ð sin u du ¼ cos u 20. ð csch u du ¼ coth1 ðcosh uÞ 4. ð cos u du ¼ sin u 21. ð sech2 u du ¼ tanh u 5. ð tan u du ¼ ln j sec uj 22. ð csch2 u du ¼ coth u ¼ ln j cos uj 6. ð cot u du ¼ ln j sin uj 23. ð sech u tanh u du ¼ sech u 7. ð sec u du ¼ ln j sec u þ tan uj 24. ð csch u coth u du ¼ csch u ¼ ln j tanðu=2 þ =4Þj 8. ð csc u du ¼ ln jcsc u cot uj 25. ð du ffiffiffiffiffiffiffiffiffiffiffiffiffiffi s2 u2 p ¼ sin1 u a or cos1 u a ¼ ln j tan u=2j 9. ð sec2 u du ¼ tan u 26. ð du ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p ¼ ln ju þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p j 10. ð csc2 u du ¼ cot u 27. ð du u2 þ a2 ¼ 1 a tan1 u a or 1 a cot1 u a 11. ð sec u tan u du ¼ sec u 28. ð du u2 a2 ¼ 1 2a ln u a u þ a 12. ð csc u cot u du ¼ csc u 29. ð du u ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p ¼ 1 a ln u a þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p 13. ð au du ¼ au ln a a 0; a 6¼ 1 30. ð du u ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p ¼ 1 a cos1 a u or 1 a sec1 u a 14. ð eu du ¼ eu 31. ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p du ¼ u 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p a2 2 ln ju þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p j 15. ð sinh u du ¼ cosh u 32. ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p du ¼ u 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p þ a2 2 sin1 u a 16. ð cosh u du ¼ sinh u 33. ð eau sin bu du ¼ eau ða sin bu b cos buÞ a2 þ b2 17. ð tanh u du ¼ ln cosh u 34. ð eau cos bu du ¼ eau ða cos bu þ b sin buÞ a2 þ b2 96 INTEGRALS [CHAP. 5
- 106. SPECIAL METHODS OF INTEGRATION 1. Integration by parts. Let u and v be differentiable functions. According to the product rule for differentials dðuvÞ ¼ u dv þ v du Upon taking the antiderivative of both sides of the equation, we obtain uv ¼ ð u dv þ ð v du This is the formula for integration by parts when written in the form ð u dv ¼ uv ð v du or ð f ðxÞg0 ðxÞ dx ¼ f ðxÞgðxÞ ð f 0 ðxÞgðxÞ dx where u ¼ f ðxÞ and v ¼ gðxÞ. The corresponding result for definite integrals over the interval ½a; b is certainly valid if f ðxÞ and gðxÞ are continuous and have continuous derivatives in ½a; b. See Problems 5.17 to 5.19. 2. Partial fractions. Any rational function PðxÞ QðxÞ where PðxÞ and QðxÞ are polynomials, with the degree of PðxÞ less than that of QðxÞ, can be written as the sum of rational functions having the form A ðax þ bÞr, Ax þ B ðax2 þ bx þ cÞr where r ¼ 1; 2; 3; . . . which can always be integrated in terms of elementary functions. EXAMPLE 1. 3x 2 ð4x 3Þð2x þ 5Þ3 ¼ A 4x 3 þ B ð2x þ 5Þ3 þ C ð2x þ 5Þ2 þ D 2x þ 5 EXAMPLE 2. 5x2 x þ 2 ðx2 þ 2x þ 4Þ2 ðx 1Þ ¼ Ax þ B ðx2 þ 2x þ 4Þ2 þ Cx þ D x2 þ 2x þ 4 þ E x 1 The constants, A, B, C, etc., can be found by clearing of fractions and equating coefficients of like powers of x on both sides of the equation or by using special methods (see Problem 5.20). 3. Rational functions of sin x and cos x can always be integrated in terms of elementary functions by the substitution tan x=2 ¼ u (see Problem 5.21). 4. Special devices depending on the particular form of the integrand are often employed (see Problems 5.22 and 5.23). IMPROPER INTEGRALS If the range of integration ½a; b is not finite or if f ðxÞ is not defined or not bounded at one or more points of ½a; b, then the integral of f ðxÞ over this range is called an improper integral. By use of appropriate limiting operations, we may define the integrals in such cases. EXAMPLE 1. ð1 0 dx 1 þ x2 ¼ lim M!1 ðM 0 dx 1 þ x2 ¼ lim M!1 tan1 x M 0 ¼ lim M!1 tan1 M ¼ =2 EXAMPLE 2. ð1 0 dx ffiffiffi x p ¼ lim !0þ ð1 dx ffiffiffi x p ¼ lim !0þ 2 ffiffiffi x p 1 ¼ lim !0þ ð2 2 ffiffi ffi p Þ ¼ 2 EXAMPLE 3. ð1 0 dx x ¼ lim !0þ ð1 dx x ¼ lim !0þ ln x 1 ¼ lim !0þ ð ln Þ Since this limit does not exist we say that the integral diverges (i.e., does not converge). CHAP. 5] INTEGRALS 97
- 107. For further examples, see Problems 5.29 and 5.74 through 5.76. For further discussion of improper integrals, see Chapter 12. NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS Numerical methods for evaluating definite integrals are available in case the integrals cannot be evaluated exactly. The following special numerical methods are based on subdividing the interval ½a; b into n equal parts of length x ¼ ðb aÞ=n. For simplicity we denote f ða þ kxÞ ¼ f ðxkÞ by yk, where k ¼ 0; 1; 2; . . . ; n. The symbol means ‘‘approximately equal.’’ In general, the approximation improves as n increases. 1. Rectangular rule. ðb a f ðxÞ dx xfy0 þ y1 þ y2 þ þ yn1g or xf y1 þ y2 þ y3 þ þ yng ð8Þ The geometric interpretation is evident from the figure on Page 90. When left endpoint function values y0; y1; . . . ; yn1 are used, the rule is called ‘‘the left-hand rule.’’ Similarly, when right endpoint evaluations are employed, it is called ‘‘the right-hand rule.’’ 2. Trapezoidal rule. ðb a f ðxÞ dx x 2 f y0 þ 2y1 þ 2y2 þ þ 2yn1 þ yng ð9Þ This is obtained by taking the mean of the approximations in (8). Geometrically this replaces the curve y ¼ f ðxÞ by a set of approximating line segments. 3. Simpson’s rule. ðb a f ðxÞ dx x 3 f y0 þ 4y1 þ 2y2 þ 4y3 þ 2y4 þ 4y5 þ þ 2yn2 þ 4yn1 þ yng ð10Þ The above formula is obtained by approximating the graph of y ¼ gðxÞ by a set of parabolic arcs of the form y ¼ ax2 þ bx þ c. The correlation of two observations lead to 10. First, ðh h ½ax2 þ bx þ c dx ¼ h 3 ½2ah2 þ 6c The second observation is related to the fact that the vertical parabolas employed here are determined by three nonlinear points. In particular, consider ðh; y0Þ, ð0; y1Þ, ðh; y2Þ then y0 ¼ aðhÞ2 þ bðhÞ þ c, y1 ¼ c, y2 ¼ ah2 þ bh þ c. Consequently, y0 þ 4y1 þ y2 ¼ 2ah2 þ 6c. Thus, this combination of ordinate values (corresponding to equally space domain values) yields the area bound by the parabola, vertical segments, and the x-axis. Now these ordinates may be interpreted as those of the function, f , whose integral is to be approximated. Then, as illu- strated in Fig. 5-3: X n k¼1 h 3 ½yk1 þ 4yk þ ykþ1 ¼ x 3 ½ y0 þ 4y1 þ 2y2 þ 4y3 þ 2y4 þ 4y5 þ þ 2yn2 þ 4yn1 þ yn The Simpson rule is likely to give a better approximation than the others for smooth curves. APPLICATIONS The use of the integral as a limit of a sum enables us to solve many physical or geometrical problems such as determination of areas, volumes, arc lengths, moments of intertia, centroids, etc. 98 INTEGRALS [CHAP. 5
- 108. ARC LENGTH As you walk a twisting mountain trail, it is possible to determine the distance covered by using a pedometer. To create a geometric model of this event, it is necessary to describe the trail and a method of measuring distance along it. The trail might be referred to as a path, but in more exacting geometric terminology the word, curve is appropriate. That segment to be measured is an arc of the curve. The arc is subject to the following restrictions: 1. It does not intersect itself (i.e., it is a simple arc). 2. There is a tangent line at each point. 3. The tangent line varies continuously over the arc. These conditions are satisfied with a parametric representation x ¼ f ðtÞ; y ¼ gðtÞ; z ¼ hðtÞ; a @ t @ b, where the functions f , g, and h have continuous derivatives that do not simultaneously vanish at any point. This arc is in Euclidean three space and will be discussed in Chapter 10. In this introduction to curves and their arc length, we let z ¼ 0, thereby restricting the discussion to the plane. A careful examination of your walk would reveal movement on a sequence of straight segments, each changed in direction from the previous one. This suggests that the length of the arc of a curve is obtained as the limit of a sequence of lengths of polygonal approximations. (The polygonal approx- imations are characterized by the number of divisions n ! 1 and no subdivision is bound from zero. (See Fig. 5-4.) Geometrically, the measurement of the kth segment of the arc, 0 @ t @ s, is accomplished by employing the Pythagorean theorem, and thus, the measure is defined by CHAP. 5] INTEGRALS 99 Fig. 5-4 Fig. 5-3
- 109. lim n!1 X n k¼1 fðxkÞ2 þ ðykÞ2 g1=2 or equivalently lim n!1 X n k¼1 1 þ yk xk 2 ( )1=2 ðxkÞ where xk ¼ xk xk1 and yk ¼ yk yk1. Thus, the length of the arc of a curve in rectangular Cartesian coordinates is L ¼ ðb a f½ f 0 ðtÞ2 þ ½g0 ðtÞ2 g1=2 dt ¼ ð dx dt 2 þ dy dt 2 ( )1=2 dt (This form may be generalized to any number of dimensions.) Upon changing the variable of integration from t to x we obtain the planar form L ¼ ðf ðbÞ f ðaÞ 1 þ dy dx 2 ( )1=2 (This form is only appropriate in the plane.) The generic differential formula ds2 ¼ dx2 þ dy2 is useful, in that various representations algebrai- cally arise from it. For example, ds dt expresses instantaneous speed. AREA Area was a motivating concept in introducing the integral. Since many applications of the integral are geometrically interpretable in the context of area, an extended formula is listed and illustrated below. Let f and g be continuous functions whose graphs intersect at the graphical points corresponding to x ¼ a and x ¼ b, a b. If gðxÞ A f ðxÞ on ½a; b, then the area bounded by f ðxÞ and gðxÞ is A ¼ ðb a fgðxÞ f ðxÞg dx If the functions intersect in ða; bÞ, then the integral yields an algebraic sum. For example, if gðxÞ ¼ sin x and f ðxÞ ¼ 0 then: ð2 0 sin x dx ¼ cos x 2 0 ¼ 0 VOLUMES OF REVOLUTION Disk Method Assume that f is continuous on a closed interval a @ x @ b and that f ðxÞ A 0. Then the solid realized through the revolution of a plane region R (bound by f ðxÞ, the x-axis, and x ¼ a and x ¼ b) about the x-axis has the volume V ¼ ðb a ½ f ðxÞ2 dx 100 INTEGRALS [CHAP. 5
- 110. This method of generating a volume is called the disk method because the cross sections of revolution are circular disks. (See Fig. 5-5(a).) EXAMPLE. A solid cone is generated by revolving the graph of y ¼ kx, k 0 and 0 @ x @ b, about the x-axis. Its volume is V ¼ ðb 0 k2 x2 dx ¼ k3 x3 3 b 0 ¼ k3 b3 3 Shell Method Suppose f is a continuous function on ½a; b, a A 0, satisfying the condition f ðxÞ A 0. Let R be a plane region bound by f ðxÞ, x ¼ a, x ¼ b, and the x-axis. The volume obtained by orbiting R about the y-axis is V ¼ ðb a 2x f ðxÞ dx This method of generating a volume is called the shell method because of the cylindrical nature of the vertical lines of revolution. (See Fig. 5-5(b).) EXAMPLE. If the region bounded by y ¼ kx, 0 @ x @ b and x ¼ b (with the same conditions as in the previous example) is orbited about the y-axis the volume obtained is V ¼ 2 ðb 0 xðkxÞ dx ¼ 2k x3 3 b 0 ¼ 2k b3 3 By comparing this example with that in the section on the disk method, it is clear that for the same plane region the disk method and the shell method produce different solids and hence different volumes. Moment of Inertia Moment of inertia is an important physical concept that can be studied through its idealized geo- metric form. This form is abstracted in the following way from the physical notions of kinetic energy, K ¼ 1 2 mv2 , and angular velocity, v ¼ !r. (m represents mass and v signifies linear velocity). Upon substituting for v K ¼ 1 2 m!2 r2 ¼ 1 2 ðmr2 Þ!2 When this form is compared to the original representation of kinetic energy, it is reasonable to identify mr2 as rotational mass. It is this quantity, l ¼ mr2 that we call the moment of inertia. Then in a purely geometric sense, we denote a plane region R described through continuous func- tions f and g on ½a; b, where a 0 and f ðxÞ and gðxÞ intersect at a and b only. For simplicity, assume gðxÞ A f ðxÞ 0. Then CHAP. 5] INTEGRALS 101 Fig. 5-5
- 111. l ¼ ðb a x2 ½gðxÞ f ðxÞ dx By idealizing the plane region, R, as a volume with uniform density one, the expression ½ f ðxÞ gðxÞ dx stands in for mass and r2 has the coordinate representation x2 . (See Problem 5.25(b) for more details.) Solved Problems DEFINITION OF A DEFINITE INTEGRAL 5.1. If f ðxÞ is continuous in ½a; b prove that lim n!1 b a n X n k¼1 f a þ kðb aÞ n ¼ ðb a f ðxÞ dx Since f ðxÞ is continuous, the limit exists independent of the mode of subdivision (see Problem 5.31). Choose the subdivision of ½a; b into n equal parts of equal length x ¼ ðb aÞ=n (see Fig. 5-1, Page 90). Let k ¼ a þ kðb aÞ=n, k ¼ 1; 2; . . . ; n. Then lim n!1 X n k¼1 f ðkÞxk ¼ lim n!1 b a n X n k¼1 f a þ kðb aÞ n ¼ ðb a f ðxÞ dx 5.2. Express lim n!1 1 n X n k¼1 f k n as a definite integral. Let a ¼ 0, b ¼ 1 in Problem 1. Then lim n!1 1 n X n k¼1 f k n ¼ ð1 0 f ðxÞ dx 5.3. (a) Express ð1 0 x2 dx as a limit of a sum, and use the result to evaluate the given definite integral. (b) Interpret the result geometrically. (a) If f ðxÞ ¼ x2 , then f ðk=nÞ ¼ ðk=nÞ2 ¼ k2 =n2 . Thus by Problem 5.2, lim n!1 1 n X n k¼1 k2 n2 ¼ ð1 0 x2 dx This can be written, using Problem 1.29 of Chapter 1, ð1 0 x2 dx ¼ lim n!1 1 n 12 n2 þ 22 n2 þ þ n2 n2 ! ¼ lim n!1 12 þ 22 þ þ n2 n3 ¼ lim n!1 nðn þ 1Þð2n þ 1Þ 6n3 ¼ lim n!1 ð1 þ 1=nÞð2 þ 1=nÞ 6 ¼ 1 3 which is the required limit. Note: By using the fundamental theorem of the calculus, we observe that Ð1 0 x2 dx ¼ ðx3 =3Þj1 0 ¼ 13 =3 03 =3 ¼ 1=3. (b) The area bounded by the curve y ¼ x2 , the x-axis and the line x ¼ 1 is equal to 1 3. 102 INTEGRALS [CHAP. 5
- 112. 5.4. Evaluate lim n!1 1 n þ 1 þ 1 n þ 2 þ þ 1 n þ n . The required limit can be written lim n!1 1 n 1 1 þ 1=n þ 1 1 þ 2=n þ þ 1 1 þ n=n ¼ lim n!1 1 n X n k¼1 1 1 þ k=n ¼ ð1 0 dx 1 þ x ¼ lnð1 þ xÞj1 0 ¼ ln 2 using Problem 5.2 and the fundamental theorem of the calculus. 5.5. Prove that lim n!1 1 n sin t n þ sin 2t n þ þ sin ðn 1Þt n ¼ 1 cos t t . Let a ¼ 0; b ¼ t; f ðxÞ ¼ sin x in Problem 1. Then lim n!1 t n X n k¼1 sin kt n ¼ ðt 0 sin x dx ¼ 1 cos t and so lim n!1 1 n X n1 k¼1 sin kt n ¼ 1 cos t t using the fact that lim n!1 sin t n ¼ 0. MEASURE ZERO 5.6. Prove that a countable point set has measure zero. Let the point set be denoted by x1; x2; x3; x4; . . . and suppose that intervals of lengths less than =2; =4; =8; =16; . . . respectively enclose the points, where is any positive number. Then the sum of the lengths of the intervals is less than =2 þ =4 þ =8 þ ¼ (let a ¼ =2 and r ¼ 1 2 in Problem 2.25(a) of Chapter 2), showing that the set has measure zero. PROPERTIES OF DEFINITE INTEGRALS 5.7. Prove that ðb a f ðxÞ dx @ ðb a j f ðxÞj dx if a b. By absolute value property 2, Page 3, X n k¼1 f ðkÞxk @ X n k¼1 j f ðkÞxkj ¼ X n k¼1 j f ðkÞjxk Taking the limit as n ! 1 and each xk ! 0, we have the required result. 5.8. Prove that lim n!1 ð2 0 sin nx x2 þ n2 dx ¼ 0. ð2 0 sin nx x2 þ n2 dx @ ð2 0 sin nx x2 þ n2 dx @ ð2 0 dx n2 ¼ 2 n2 Then lim n!1 ð2 0 sin nx x2 þ n2 dx ¼ 0, and so the required result follows. CHAP. 5] INTEGRALS 103
- 113. MEAN VALUE THEOREMS FOR INTEGRALS 5.9. Given the right triangle pictured in Fig. 5-6: (a) Find the average value of h. (b) At what point does this average value occur? (c) Determine the average value of f ðxÞ ¼ sin1 x; 0 @ x @ 1 2. (Use integration by parts.) (d) Determine the average value of f ðxÞ ¼ cos2 x; 0 @ x @ 2 . (a) hðxÞ ¼ H B x. According to the mean value theorem for integrals, the average value of the function h on the interval ½0; B is A ¼ 1 B ðB 0 H B x dx ¼ H 2 (b) The point, , at which the average value of h occurs may be obtained by equating f ðÞ with that average value, i.e., H B ¼ H 2 . Thus, ¼ B 2 . FUNDAMENTAL THEOREM OF THE CALCULUS 5.10. If FðxÞ ¼ ðx a f ðtÞ dt where f ðxÞ is continuous in ½a; b, prove that F 0 ðxÞ ¼ f ðxÞ. Fðx þ hÞ FðxÞ h ¼ 1 h ðxþh a f ðtÞ dt ðx a f ðtÞ dt ¼ 1 h ðxþh x f ðtÞ dt ¼ f ðÞ between x and x þ h by the first mean value theorem for integrals (Page 93). Then if x is any point interior to ½a; b, F 0 ðxÞ ¼ lim h!0 Fðx þ hÞ FðxÞ h ¼ lim h!0 f ðÞ ¼ f ðxÞ since f is continuous. If x ¼ a or x ¼ b, we use right- or left-hand limits, respectively, and the result holds in these cases as well. 5.11. Prove the fundamental theorem of the calculus, Part 2 (Pages 94 and 95). By Problem 5.10, if FðxÞ is any function whose derivative is f ðxÞ, we can write FðxÞ ¼ ðx a f ðtÞ dt þ c where c is any constant (see last line of Problem 22, Chapter 4). Since FðaÞ ¼ c, it follows that FðbÞ ¼ ðb a f ðtÞ dt þ FðaÞ or ðb a f ðtÞ dt ¼ FðbÞ FðaÞ. 5.12. If f ðxÞ is continuous in ½a; b, prove that FðxÞ ¼ ðx a f ðtÞ dt is continuous in ½a; b. If x is any point interior to ½a; b, then as in Problem 5.10, lim h!0 Fðx þ hÞ FðxÞ ¼ lim h!0 h f ðÞ ¼ 0 and FðxÞ is continuous. If x ¼ a and x ¼ b, we use right- and left-hand limits, respectively, to show that FðxÞ is continuous at x ¼ a and x ¼ b. 104 INTEGRALS [CHAP. 5 Fig. 5-6
- 114. Another method: By Problem 5.10 and Problem 4.3, Chapter 4, it follows that F 0 ðxÞ exists and so FðxÞ must be con- tinuous. CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION 5.13. Prove the result (7), Page 95, for changing the variable of integration. Let FðxÞ ¼ ðx a f ðxÞ dx and GðtÞ ¼ ðt f fgðtÞg g0 ðtÞ dt, where x ¼ gðtÞ. Then dF ¼ f ðxÞ dx, dG ¼ f fgðtÞg g0 ðtÞ dt. Since dx ¼ g0 ðtÞ dt, it follows that f ðxÞ dx ¼ f fgðtÞg g0 ðtÞ dt so that dFðxÞ ¼ dGðtÞ, from which FðxÞ ¼ GðtÞ þ c. Now when x ¼ a, t ¼ or FðaÞ ¼ Gð Þ þ c. But FðaÞ ¼ Gð Þ ¼ 0, so that c ¼ 0. Hence FðxÞ ¼ GðtÞ. Since x ¼ b when t ¼ , we have ðb a f ðxÞ dx ¼ ð f fgðtÞg g0 ðtÞ dt as required. 5.14. Evaluate: ðaÞ ð ðx þ 2Þ sinðx2 þ 4x 6Þ dx ðcÞ ð1 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 2Þð3 xÞ p ðeÞ ð1= ffiffi 2 p 0 x sin1 x2 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x4 p dx ðbÞ ð cotðln xÞ x dx ðdÞ ð 2x tanh 21x dx ð f Þ ð x dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p (a) Method 1: Let x2 þ 4x 6 ¼ u. Then ð2x þ 4Þ dx ¼ du, ðx þ 2Þ dx ¼ 1 2 du and the integral becomes 1 2 ð sin u du ¼ 1 2 cos u þ c ¼ 1 2 cosðx2 þ 4x 6Þ þ c Method 2: ð ðx þ 2Þ sinðx2 þ 4x 6Þ dx ¼ 1 2 ð sinðx2 þ 4x 6Þdðx2 þ 4x 6Þ ¼ 1 2 cosðx2 þ 4x 6Þ þ c (b) Let ln x ¼ u. Then ðdxÞ=x ¼ du and the integral becomes ð cot u du ¼ ln j sin uj þ c ¼ ln j sinðln xÞj þ c ðcÞ Method 1: ð dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 2Þð3 xÞ p ¼ ð dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 þ x x2 p ¼ ð dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 ðx2 xÞ p ¼ ð dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25=4 ðx 1 2Þ2 q Letting x 1 2 ¼ u, this becomes ð du ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25=4 u2 p ¼ sin1 u 5=2 þ c ¼ sin1 2x 1 5 þ c Then ð1 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 2Þð3 xÞ p ¼ sin1 2x 1 5 1 1 ¼ sin1 1 5 sin1 3 5 ¼ sin1 :2 þ sin1 :6 CHAP. 5] INTEGRALS 105
- 115. Method 2: Let x 1 2 ¼ u as in Method 1. Now when x ¼ 1, u ¼ 3 2; and when x ¼ 1, u ¼ 1 2. Thus by Formula 25, Page 96. ð1 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 2Þð3 xÞ p ¼ ð1 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25=4 ðx 1 2Þ2 q ¼ ð1=2 3=2 du ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25=4 u2 p ¼ sin1 u 5=2 1=2 3=2 ¼ sin1 :2 þ sin1 :6 (d) Let 21x ¼ u. Then 21x ðln 2Þdx ¼ du and 2x dx ¼ du 2 ln 2 , so that the integral becomes 1 2 ln 2 ð tanh u du ¼ 1 2 ln 2 ln cosh 21x þ c (e) Let sin1 x2 ¼ u. Then du ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðx2Þ2 q 2x dx ¼ 2x dx ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x4 p and the integral becomes 1 2 ð u du ¼ 1 4 u2 þ c ¼ 1 4 ðsin1 x2 Þ2 þ c Thus ð1= ffiffi 2 p 0 x sin1 x2 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x4 p dx ¼ 1 4 ðsin1 x2 Þ2 1= ffiffi 2 p 0 ¼ 1 4 sin1 1 2 2 ¼ 2 144 : ð f Þ ð x dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p ¼ 1 2 ð 2x þ 1 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p dx ¼ 1 2 ð 2x þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p dx 1 2 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p ¼ 1 2 ð ðx2 þ x þ 1Þ1=2 dðx2 þ x þ 1Þ 1 2 ð dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 1 2Þ2 þ 3 4 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p 1 2 ln jx þ 1 2 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 1 2Þ2 þ 3 4 q j þ c 5.15. Show that ð2 1 dx ðx2 2x þ 4Þ3=2 ¼ 1 6 . Write the integral as ð2 1 dx ½ðx 1Þ2 þ 33=2 . Let x 1 ¼ ffiffiffi 3 p tan u, dx ¼ ffiffiffi 3 p sec2 u du. When x ¼ 1, u ¼ tan1 0 ¼ 0; when x ¼ 2, u ¼ tan1 1= ffiffiffi 3 p ¼ =6. Then the integral becomes ð=6 0 ffiffiffi 3 p sec2 u du ½3 þ 3 tan2 u3=2 ¼ ð=6 0 ffiffiffi 3 p sec2 u du ½3 sec2 u3=2 ¼ 1 3 ð=6 0 cos u du ¼ 1 3 sin u =6 0 ¼ 1 6 5.16. Determine ðe2 e dx xðln xÞ3 . Let ln x ¼ y, ðdxÞ=x ¼ dy. When x ¼ e, y ¼ 1; when x ¼ e2 , y ¼ 2. Then the integral becomes ð2 1 dy y3 ¼ y2 2 2 1 ¼ 3 8 5.17. Find ð xn ln x dx if (a) n 6¼ 1, (b) n ¼ 1. 106 INTEGRALS [CHAP. 5
- 116. (a) Use integration by parts, letting u ¼ ln x, dv ¼ xn dx, so that du ¼ ðdxÞ=x, v ¼ xnþ1 =ðn þ 1Þ. Then ð xn ln x dx ¼ ð u dv ¼ uv ð v du ¼ xnþ1 n þ 1 ln x ð xnþ1 n þ 1 dx x ¼ xnþ1 n þ 1 ln x xnþ1 ðn þ 1Þ2 þ c ðbÞ ð x1 ln x dx ¼ ð ln x dðln xÞ ¼ 1 2 ðln xÞ2 þ c: 5.18. Find ð 3 ffiffiffiffiffiffiffiffi 2xþ1 p dx. Let ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2x þ 1 p ¼ y, 2x þ 1 ¼ y2 . Then dx ¼ y dy and the integral becomes ð 3y y dy. Integrate by parts, letting u ¼ y, dv ¼ 3y dy; then du ¼ dy, v ¼ 3y =ðln 3Þ, and we have ð 3y y dy ¼ ð u dv ¼ uv ð v du ¼ y 3y ln 3 ð 3y ln 3 dy ¼ y 3y ln 3 3y ðln 3Þ2 þ c 5.19. Find ð1 0 x lnðx þ 3Þ dx. Let u ¼ lnðx þ 3Þ, dv ¼ x dx. Then du ¼ dx x þ 3 , v ¼ x2 2 . Hence on integrating by parts, ð x lnðx þ 3Þ dx ¼ x2 2 lnðx þ 3Þ 1 2 ð x2 dx x þ 3 ¼ x2 2 lnðx þ 3Þ 1 2 ð x 3 þ 9 x þ 3 dx ¼ x2 2 lnðx þ 3Þ 1 2 x2 2 3x þ 9 lnðx þ 3Þ ( ) þ c ð1 0 x lnðx þ 3Þ dx ¼ 5 4 4 ln 4 þ 9 2 ln 3 Then 5.20. Determine ð 6 x ðx 3Þð2x þ 5Þ dx. Use the method of partial fractions. Let 6 x ðx 3Þð2x þ 5Þ ¼ A x 3 þ B 2x þ 5 . Method 1: To determine the constants A and B, multiply both sides by ðx 3Þð2x þ 5Þ to obtain 6 x ¼ Að2x þ 5Þ þ Bðx 3Þ or 6 x ¼ 5A 3B þ ð2A þ BÞx ð1Þ Since this is an identity, 5A 3B ¼ 6, 2A þ B ¼ 1 and A ¼ 3=11, B ¼ 17=11. Then ð 6 x ðx 3Þð2x þ 5Þ dx ¼ ð 3=11 x 3 dx þ ð 17=11 2x þ 5 dx ¼ 3 11 ln jx 3j 17 22 ln j2x þ 5j þ c Method 2: Substitute suitable values for x in the identity (1). For example, letting x ¼ 3 and x ¼ 5=2 in (1), we find at once A ¼ 3=11, B ¼ 17=11. 5.21. Evaluate ð dx 5 þ 3 cos x by using the substitution tan x=2 ¼ u. From Fig. 5-7 we see that sin x=2 ¼ u ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2 p ; cos x=2 ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2 p CHAP. 5] INTEGRALS 107 x/2 u √1 + u 2 1 Fig. 5-7
- 117. Then cos x ¼ cos2 x=2 sin2 x=2 ¼ 1 u2 1 þ u2 : Also du ¼ 1 2 sec2 x=2 dx or dx ¼ 2 cos2 x=2 du ¼ 2 du 1 þ u2 : Thus the integral becomes ð du u2 þ 4 ¼ 1 2 tan1 u=2 þ c ¼ 1 2 tan1 1 2 tan x=2 þ c: 5.22. Evaluate ð 0 x sin x 1 þ cos2 x dx. Let x ¼ y. Then I ¼ ð 0 x sin x 1 þ cos2 x dx ¼ ð 0 ð yÞ sin y 1 þ cos2 y dy ¼ ð 0 sin y 1 þ cos2 y dy ð 0 y sin y 1 þ cos2 y dy ¼ ð 0 dðcos yÞ 1 þ cos2 y I ¼ tan1 ðcos yÞj 0 I ¼ 2 =2 I i.e.; I ¼ 2 =2 I or I ¼ 2 =4: 5.23. Prove that ð=2 0 ffiffiffiffiffiffiffiffiffiffi sin x p ffiffiffiffiffiffiffiffiffiffi sin x p þ ffiffiffiffiffiffiffiffiffiffi ffi cos x p dx ¼ 4 . Letting x ¼ =2 y, we have I ¼ ð=2 0 ffiffiffiffiffiffiffiffiffiffi sin x p ffiffiffiffiffiffiffiffiffiffi sin x p þ ffiffiffiffiffiffiffiffiffiffiffi cos x p dx ¼ ð=2 0 ffiffiffiffiffiffiffiffiffiffi cos y p ffiffiffiffiffiffiffiffiffiffi cos y p þ ffiffiffiffiffiffiffiffiffi sin y p dy ¼ ð=2 0 ffiffiffiffiffiffiffiffiffiffiffi cos x p ffiffiffiffiffiffiffiffiffiffiffi cos x p þ ffiffiffiffiffiffiffiffiffiffi sin x p dx Then I þ I ¼ ð=2 0 ffiffiffiffiffiffiffiffiffiffi sin x p ffiffiffiffiffiffiffiffiffiffi sin x p þ ffiffiffiffiffiffiffiffiffiffiffi cos x p dx þ ð=2 0 ffiffiffiffiffiffiffiffiffiffiffi cos x p ffiffiffiffiffiffiffiffiffiffiffi cos x p þ ffiffiffiffiffiffiffiffiffiffi sin x p dx ¼ ð=2 0 ffiffiffiffiffiffiffiffiffiffi sin x p þ ffiffiffiffiffiffiffiffiffiffiffi cos x p ffiffiffiffiffiffiffiffiffiffi sin x p þ ffiffiffiffiffiffiffiffiffiffiffi cos x p dx ¼ ð=2 0 dx ¼ 2 from which 2I ¼ =2 and I ¼ =4. The same method can be used to prove that for all real values of m, ð=2 0 sinm x sinm x þ cosm x dx ¼ 4 (see Problem 5.89). Note: This problem and Problem 5.22 show that some definite integrals can be evaluated without first finding the corresponding indefinite integrals. NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS 5.24. Evaluate ð1 0 dx 1 þ x2 approximately, using (a) the trapezoidal rule, (b) Simpson’s rule, where the interval ½0; 1 is divided into n ¼ 4 equal parts. Let f ðxÞ ¼ 1=ð1 þ x2 Þ. Using the notation on Page 98, we find x ¼ ðb aÞ=n ¼ ð1 0Þ=4 ¼ 0:25. Then keeping 4 decimal places, we have: y0 ¼ f ð0Þ ¼ 1:0000, y1 ¼ f ð0:25Þ ¼ 0:9412, y2 ¼ f ð0:50Þ ¼ 0:8000, y3 ¼ f ð0:75Þ ¼ 0:6400, y4 ¼ f ð1Þ ¼ 0:50000. (a) The trapezoidal rule gives 108 INTEGRALS [CHAP. 5
- 118. x 2 fy0 þ 2y1 þ 2y2 þ 2y3 þ y4g ¼ 0:25 2 f1:0000 þ 2ð0:9412Þ þ 2ð0:8000Þ þ 2ð0:6400Þ þ 0:500g ¼ 0:7828: (b) Simpson’s rule gives x 3 fy0 þ 4y1 þ 2y2 þ 4y3 þ y4g ¼ 0:25 3 f1:0000 þ 4ð0:9412Þ þ 2ð0:8000Þ þ 4ð0:6400Þ þ 0:5000g ¼ 0:7854: The true value is =4 0:7854: APPLICATIONS (AREA, ARC LENGTH, VOLUME, MOMENT OF INTERTIA) 5.25. Find the (a) area and (b) moment of inertia about the y-axis of the region in the xy plane bounded by y ¼ 4 x2 and the x-axis. (a) Subdivide the region into rectangles as in the figure on Page 90. A typical rectangle is shown in the adjoining Fig. 5-8. Then Required area ¼ lim n!1 X n k¼1 f ðkÞ xk ¼ lim n!1 X n k¼1 ð4 2 kÞ xk ¼ ð2 2 ð4 x2 Þ dx ¼ 32 3 (b) Assuming unit density, the moment of inertia about the y- axis of the typical rectangle shown above is 2 k f ðkÞ xk. Then Required moment of inertia ¼ lim n!1 X n k¼1 2 k f ðkÞ xk ¼ lim n!1 X n k¼1 2 kð4 2 kÞ xk ¼ ð2 2 x2 ð4 x2 Þ dx ¼ 128 15 5.26. Find the length of arc of the parabola y ¼ x2 from x ¼ 0 to x ¼ 1. Required arc length ¼ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðdy=dxÞ2 q dx ¼ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ð2xÞ2 q dx ¼ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 p dx ¼ 1 2 ð2 0 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2 p du ¼ 1 2 f1 2 u ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2 p þ 1 2 lnðu þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ u2Þ p gj2 0 ¼ 1 2 ffiffiffi 5 p þ 1 4 lnð2 þ ffiffiffi 5 p Þ 5.27. (a) (Disk Method) Find the volume generated by revolving the region of Problem 5.25 about the x-axis. Required volume ¼ lim n!1 X n k¼1 y2 kxk ¼ ð2 2 ð4 x2 Þ2 dx ¼ 512=15: (b) (Disk Method) Find the volume of the frustrum of a paraboloid obtained by revolving f ðxÞ ¼ ffiffiffiffiffiffi kx p , 0 a @ x @ b about the x-axis. CHAP. 5] INTEGRALS 109 Fig. 5-8
- 119. V ¼ ðb a kx dx ¼ k 2 ðb2 a2 Þ: (c) (Shell Method) Find the volume obtained by orbiting the region of part (b) about the y-axis. Compare this volume with that obtained in part (b). V ¼ 2 ðb 0 xðkxÞ dx ¼ 2kb3 =3 The solids generated by the two regions are different, as are the volumes. MISCELLANEOUS PROBLEMS 5.28 If f ðxÞ and gðxÞ are continuous in ½a; b, prove Schwarz’s inequality for integrals: ðb a f ðxÞ gðxÞ dx 2 @ ðb a f f ðxÞg2 dx ðb a fgðxÞg2 dx We have ðb a f f ðxÞ þ gðxÞg2 dx ¼ ðb a f f ðxÞg2 dx þ 2 ðb a f ðxÞ gðxÞ dx þ 2 ðb a fgðxÞg2 dx A 0 for all real values of . Hence by Problem 1.13 of Chapter 1, using (1) with A2 ¼ ðb a gðxÞg2 dx; B2 ¼ ðb a f f ðxÞg2 dx; C ¼ ðb a f ðxÞ gðxÞ dx we find C2 @ A2 B2 , which gives the required result. 5.29. Prove that lim M!1 ðM 0 dx x4 þ 4 ¼ 8 . We have x4 þ 4 ¼ x4 þ 4x2 þ 4 4x2 ¼ ðx2 þ 2Þ2 ð2xÞ2 ¼ ðx2 þ 2 þ 2xÞðx2 þ 2 2xÞ: According to the method of partial fractions, assume 1 x4 þ 4 ¼ Ax þ B x2 þ 2x þ 2 þ Cx þ D x2 2x þ 2 Then 1 ¼ ðA þ CÞx3 þ ðB 2A þ 2C þ DÞx2 þ ð2A 2B þ 2C þ 2DÞx þ 2B þ 2D so that A þ C ¼ 0, B 2A þ 2C þ D ¼ 0, 2A 2B þ 2C þ 2D ¼ 0, 2B þ 2D ¼ 1 Solving simultaneously, A ¼ 1 8, B ¼ 1 4, C ¼ 1 8, D ¼ 1 4. Thus ð dx x4 þ 4 ¼ 1 8 ð x þ 2 x2 þ 2x þ 2 dx 1 8 ð x 2 x2 2x þ 2 dx ¼ 1 8 ð x þ 1 ðx þ 1Þ2 þ 1 dx þ 1 8 ð dx ðx þ 1Þ2 þ 1 1 8 ð x 1 ðx 1Þ2 þ 1 dx þ 1 8 ð dx ðx 1Þ2 þ 1 ¼ 1 16 lnðx2 þ 2x þ 2Þ þ 1 8 tan1 ðx þ 1Þ 1 16 lnðx2 2x þ 2Þ þ 1 8 tan1 ðx 1Þ þ C Then lim M!1 ðM 0 dx x4 þ 4 ¼ lim M!1 1 16 ln M2 þ 2M þ 2 M2 2M þ 2 ! þ 1 8 tan1 ðM þ 1Þ þ 1 8 tan1 ðM 1Þ ( ) ¼ 8 We denote this limit by ð1 0 dx x4 þ 4 , called an improper integral of the first kind. Such integrals are considered further in Chapter 12. See also Problem 5.74. 110 INTEGRALS [CHAP. 5
- 120. 5.30. Evaluate lim x!0 Ðx 0 sin t3 dt x4 . The conditions of L’Hospital’s rule are satisfied, so that the required limit is lim x!0 d dx ðx 0 sin t3 dt d dx ðx4 Þ ¼ lim x!0 sin x3 4x3 ¼ lim x!0 d dx ðsin x3 Þ d dx ð4x3 Þ ¼ lim x!0 3x2 cos x3 12x2 ¼ 1 4 5.31. Prove that if f ðxÞ is continuous in ½a; b then ðb a f ðxÞ dx exists. Let ¼ X n k¼1 f ðkÞ xk, using the notation of Page 91. Since f ðxÞ is continuous we can find numbers Mk and mk representing the l.u.b. and g.l.b. of f ðxÞ in the interval ½xk1; xk, i.e., such that mk @ f ðxÞ @ Mk. We then have mðb aÞ @ s ¼ X n k¼1 mkxk @ @ X n k¼1 Mkxk ¼ S @ Mðb aÞ ð1Þ where m and M are the g.l.b. and l.u.b. of f ðxÞ in ½a; b. The sums s and S are sometimes called the lower and upper sums, respectively. Now choose a second mode of subdivision of ½a; b and consider the corresponding lower and upper sums denoted by s0 and S0 respectively. We have must s0 @ S and S0 A s ð2Þ To prove this we choose a third mode of subdivision obtained by using the division points of both the first and second modes of subdivision and consider the corresponding lower and upper sums, denoted by t and T, respectively. By Problem 5.84, we have s @ t @ T @ S0 and s0 @ t @ T @ S ð3Þ which proves (2). From (2) it is also clear that as the number of subdivisions is increased, the upper sums are monotonic decreasing and the lower sums are monotonic increasing. Since according to (1) these sums are also bounded, it follows that they have limiting values which we shall call s s and S respectively. By Problem 5.85, s s @ S. In order to prove that the integral exists, we must show that s s ¼ S. Since f ðxÞ is continuous in the closed interval ½a; b, it is uniformly continuous. Then given any 0, we can take each xk so small that Mk mk =ðb aÞ. It follows that S s ¼ X n k¼1 ðMk mkÞxk b a X n k¼1 xk ¼ ð4Þ Now S s ¼ ðS SÞ þ ðS s sÞ þ ð s s sÞ and it follows that each term in parentheses is positive and so is less than by (4). In particular, since S s s is a definite number it must be zero, i.e., S ¼ s s. Thus, the limits of the upper and lower sums are equal and the proof is complete. Supplementary Problems DEFINITION OF A DEFINITE INTEGRAL 5.32. (a) Express ð1 0 x3 dx as a limit of a sum. (b) Use the result of (a) to evaluate the given definite integral. (c) Interpret the result geometrically. Ans. (b) 1 4 5.33. Using the definition, evaluate (a) ð2 0 ð3x þ 1Þ dx; ðbÞ ð6 3 ðx2 4xÞ dx. Ans. (a) 8, (b) 9 CHAP. 5] INTEGRALS 111
- 121. 5.34. Prove that lim n!1 n n2 þ 12 þ n n2 þ 22 þ þ n n2 þ n2 ¼ 4 . 5.35. Prove that lim n!1 1p þ 2p þ 3p þ þ np npþ1 ¼ 1 p þ 1 if p 1. 5.36. Using the definition, prove that ðb a ex dx ¼ eb ea . 5.37. Work Problem 5.5 directly, using Problem 1.94 of Chapter 1. 5.38. Prove that lim n!1 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 þ 12 p þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 þ 22 p þ þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 þ n2 p ( ) ¼ lnð1 þ ffiffiffi 2 p Þ. 5.39. Prove that lim n!1 X n k¼1 n n2 þ k2 x2 ¼ tan1 x x if x 6¼ 0. PROPERTIES OF DEFINITE INTEGRALS 5.40. Prove (a) Property 2, (b) Property 3 on Pages 91 and 92. 5.41. If f ðxÞ is integrable in ða; cÞ and ðc; bÞ, prove that ðb a f ðxÞ dx ¼ ðc a f ðxÞ dx þ ðb c f ðxÞ dx. 5.42. If f ðxÞ and gðxÞ are integrable in ½a; b and f ðxÞ @ gðxÞ, prove that ðb a f ðxÞ dx @ ðb a gðxÞ dx. 5.43. Prove that 1 cos x A x2 = for 0 @ x @ =2. 5.44. Prove that ð1 0 cos nx x þ 1 dx @ ln 2 for all n. 5.45. Prove that ð ffiffi 3 p 1 ex sin x x2 þ 1 dx @ 12e . MEAN VALUE THEOREMS FOR INTEGRALS 5.46. Prove the result (5), Page 92. [Hint: If m @ f ðxÞ @ M, then mgðxÞ @ f ðxÞgðxÞ @ MgðxÞ. Now integrate and divide by ðb a gðxÞ dx. Then apply Theorem 9 in Chapter 3. 5.47. Prove that there exist values 1 and 2 in 0 @ x @ 1 such that ð1 0 sin x x2 þ 1 dx ¼ 2 ð2 1 þ 1Þ ¼ 4 sin 2 Hint: Apply the first mean value theorem. 5.48. (a) Prove that there is a value in 0 @ x @ such that ð 0 ex cos x dx ¼ sin . (b) Suppose a wedge in the shape of a right triangle is idealized by the region bound by the x-axis, f ðxÞ ¼ x, and x ¼ L. Let the weight distribution for the wedge be defined by WðxÞ ¼ x2 þ 1. Use the generalized mean value theorem to show that the point at which the weighted value occurs is 3L 4 L2 þ 2 L2 þ 3 . 112 INTEGRALS [CHAP. 5
- 122. CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION 5.49. Evaluate: (a) ð x2 esin x3 cos x3 dx; ðbÞ ð1 0 tan1 t 1 þ t2 dt; ðcÞ ð3 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4x x2 p ; ðdÞ ð csch2 ffiffiffi u p ffiffiffi u p du, (e) ð2 2 dx 16 x2 . Ans. (a) 1 3 esin x3 þ c; ðbÞ 2 =32; ðcÞ =3; ðdÞ 2 coth ffiffiffi u p þ c; ðeÞ 1 4 ln 3. 5.50. Show that (a) ð1 0 dx ð3 þ 2x x2Þ3=2 ¼ ffiffiffi 3 p 12 ; ðbÞ ð dx x2 ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 p x þ c. 5.51. Prove that (a) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p du ¼ 1 2 u ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p 1 2 a2 ln ju þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 a2 p j (b) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p du ¼ 1 2 u ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 u2 p þ 1 2 a2 sin1 u=a þ c; a 0. 5.52. Find ð x dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 2x þ 5 p : Ans. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 2x þ 5 p ln jx þ 1 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 2x þ 5 p j þ c. 5.53. Establish the validity of the method of integration by parts. 5.54. Evaluate (a) ð 0 x cos 3x dx; ðbÞ ð x3 e2x dx: Ans. (a) 2=9; ðbÞ 1 3 e2x ð4x3 þ 6x2 þ 6x þ 3Þ þ c 5.55. Show that (a) ð1 0 x2 tan1 x dx ¼ 1 12 1 6 þ 1 6 ln 2 ðbÞ ð2 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 1 p dx ¼ 5 ffiffiffi 7 p 4 þ 3 ffiffiffi 3 p 4 þ 3 8 ln 5 þ 2 ffiffiffi 7 p 2 ffiffiffi 3 p 3 . 5.56. (a) If u ¼ f ðxÞ and v ¼ gðxÞ have continuous nth derivatives, prove that ð uvðnÞ dx ¼ uvðn1Þ u0 vðn2Þ þ u00 vðn3Þ ð1Þn ð uðnÞ v dx called generalized integration by parts. (b) What simplifications occur if uðnÞ ¼ 0? Discuss. (c) Use (a) to evaluate ð 0 x4 sin x dx. Ans. (c) 4 122 þ 48 5.57. Show that ð1 0 x dx ðx þ 1Þ2 ðx2 þ 1Þ ¼ 2 8 . [Hint: Use partial fractions, i.e., assume x ðx þ 1Þ2 ðx2 þ 1Þ ¼ A ðx þ 1Þ2 þ B x þ 1 þ Cx þ D x2 þ 1 and find A; B; C; D.] 5.58. Prove that ð 0 dx cos x ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p ; 1. NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS 5.59. Evaluate ð1 0 dx 1 þ x approximately, using (a) the trapezoidal rule, (b) Simpson’s rule, taking n ¼ 4. Compare with the exact value, ln 2 ¼ 0:6931. 5.60. Using (a) the trapezoidal rule, (b) Simpson’s rule evaluate ð=2 0 sin2 x dx by obtaining the values of sin2 x at x ¼ 08; 108; . . . ; 908 and compare with the exact value =4. 5.61. Prove the (a) rectangular rule, (b) trapezoidal rule, i.e., (16) and (17) of Page 98. 5.62. Prove Simpson’s rule. CHAP. 5] INTEGRALS 113
- 123. 114 INTEGRALS [CHAP. 5 5.63. Evaluate to 3 decimal places using numerical integration: (a) ð2 1 dx 1 þ x2 ; ðbÞ ð1 0 cosh x2 dx. Ans. (a) 0.322, (b) 1.105. APPLICATIONS 5.64. Find the (a) area and (b) moment of inertia about the y-axis of the region in the xy plane bounded by y ¼ sin x, 0 @ x @ and the x-axis, assuming unit density. Ans. (a) 2, (b) 2 4 5.65. Find the moment of inertia about the x-axis of the region bounded by y ¼ x2 and y ¼ x, if the density is proportional to the distance from the x-axis. Ans. 1 8 M, where M ¼ mass of the region. 5.66. (a) Show that the arc length of the catenary y ¼ cosh x from x ¼ 0 to x ¼ ln 2 is 3 4. (b) Show that the length of arc of y ¼ x3=2 , 2 @ x @ 5 is 343 27 2 ffiffiffi 2 p 113=2 . 5.67. Show that the length of one arc of the cycloid x ¼ að sin Þ, y ¼ að1 cos Þ, ð0 @ @ 2Þ is 8a. 5.68. Prove that the area bounded by the ellipse x2 =a2 þ y2 =b2 ¼ 1 is ab. 5.69. (a) (Disk Method) Find the volume of the region obtained by revolving the curve y ¼ sin x, 0 @ x @ , about the x-axis. Ans: ðaÞ 2 =2 (b) (Disk Method) Show that the volume of the frustrum of a paraboloid obtained by revolving f ðxÞ ¼ ffiffiffiffiffiffi kx p , 0 a @ x @ b, about the x-axis is ðb a kx dx ¼ k 2 ðb2 a2 Þ. (c) Determine the volume obtained by rotating the region bound by f ðxÞ ¼ 3, gðxÞ ¼ 5 x2 on ffiffiffi 2 p @ x @ ffiffiffi 2 p . (d) (Shell Method) A spherical bead of radius a has a circular cylindrical hole of radius b, b a, through the center. Find the volume of the remaining solid by the shell method. (e) (Shell Method) Find the volume of a solid whose outer boundary is a torus (i.e., the solid is generated by orbiting a circle ðx aÞ2 þ y2 ¼ b2 about the y-axis (a b). 5.70. Prove that the centroid of the region bounded by y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 p , a @ x @ a and the x-axis is located at ð0; 4a=3Þ. 5.71. (a) If ¼ f ðÞ is the equation of a curve in polar coordinates, show that the area bounded by this curve and the lines ¼ 1 and ¼ 2 is 1 2 ð2 1 2 d. (b) Find the area bounded by one loop of the lemniscate 2 ¼ a2 cos 2. Ans. (b) a2 5.72. (a) Prove that the arc length of the curve in Problem 5.71(a) is ð2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ ðd=dÞ2 q d. (b) Find the length of arc of the cardioid ¼ að1 cos Þ. Ans. (b) 8a MISCELLANEOUS PROBLEMS 5.73. Establish the mean value theorem for derivatives from the first mean value theorem for integrals. [Hint: Let f ðxÞ ¼ F 0 ðxÞ in (4), Page 93.] 5.74. Prove that (a) lim !0þ ð4 0 dx ffiffiffiffiffiffiffiffiffiffiffi 4 x p ¼ 4; ðbÞ lim !0þ ð3 dx ffiffiffi x 3 p ¼ 6; ðcÞ lim !0þ ð1 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p ¼ 2 and give a geo- metric interpretation of the results. [These limits, denoted usually by ð4 0 dx ffiffiffiffiffiffiffiffiffiffiffi 4 x p ; ð3 0 dx ffiffiffi x 3 p and ð1 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p respectively, are called impro- per integrals of the second kind (see Problem 5.29) since the integrands are not bounded in the range of integration. For further discussion of improper integrals, see Chapter 12.] 5.75. Prove that (a) lim M!1 ðM 0 x5 ex dx ¼ 4! ¼ 24; ðbÞ lim !0þ ð2 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xð2 xÞ p ¼ 2 .
- 124. CHAP. 5] INTEGRALS 115 5.76. Evaluate (a) ð1 0 dx 1 þ x3 ; ðbÞ ð=2 0 sin 2x ðsin xÞ4=3 dx; ðcÞ ð1 0 dx x þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p . Ans. (a) 2 3 ffiffiffi 3 p ðbÞ 3 ðcÞ does not exist 5.77. Evaluate lim x!=2 ex2 = e=4 þ Ð=2 x esin t dt 1 þ cos 2x . Ans. e=2 5.78. Prove: (a) d dx ðx3 x2 ðt2 þ t þ 1Þ dt ¼ 3x3 þ x5 2x3 þ 3x2 2x; ðb d dx ðx2 x cos t2 dt ¼ 2x cos x4 cos x2 . 5.79. Prove that (a) ð 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ sin x p dx ¼ 4; ðbÞ ð=2 0 dx sin x þ cos x ¼ ffiffiffi 2 p lnð ffiffiffi 2 p þ 1Þ. 5.80. Explain the fallacy: I ¼ ð1 1 dx 1 þ x2 ¼ ð1 1 dy 1 þ y2 ¼ I, using the transformation x ¼ 1=y. Hence I ¼ 0. But I ¼ tan1 ð1Þ tan1 ð1Þ ¼ =4 ð=4Þ ¼ =2. Thus =2 ¼ 0. 5.81. Prove that ð1=2 0 cos x ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ x2 p dx @ 1 4 tan1 1 2 . 5.82. Evaluate lim n!1 ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p þ ffiffiffiffiffiffiffiffiffiffiffi n þ 2 p þ þ ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2n 1 p n3=2 ( ) . Ans. 2 3 ð2 ffiffiffi 2 p 1Þ 5.83. Prove that f ðxÞ ¼ 1 if x is irrational 0 if x is rational is not Riemann integrable in ½0; 1. [Hint: In (2), Page 91, let k, k ¼ 1; 2; 3; . . . ; n be first rational and then irrational points of subdivision and examine the lower and upper sums of Problem 5.31.] 5.84. Prove the result (3) of Problem 5.31. [Hint: First consider the effect of only one additional point of subdivision.] 5.85. In Problem 5.31, prove that s s @ S. [Hint: Assume the contrary and obtain a contradiction.] 5.86. If f ðxÞ is sectionally continuous in ½a; b, prove that ðb a f ðxÞ dx exists. [Hint: Enclose each point of disconti- nuity in an interval, noting that the sum of the lengths of such intervals can be made arbitrarily small. Then consider the difference between the upper and lower sums. 5.87. If f ðxÞ ¼ 2x 0 x 1 3 x ¼ 1 6x 1 1 x 2 8 : , find ð2 0 f ðxÞ dx. Interpret the result graphically. Ans. 9 5.88. Evaluate ð3 0 fx ½x þ 1 2g dx where ½x denotes the greatest integer less than or equal to x. Interpret the result graphically. Ans. 3 5.89. (a) Prove that ð=2 0 sinm x sinm x þ cosm x dx ¼ 4 for all real values of m. (b) Prove that ð2 0 dx 1 þ tan4 x ¼ . 5.90. Prove that ð=2 0 sin x x dx exists. 5.91. Show that ð0:5 0 tan1 x x dx ¼ 0:4872 approximately. 5.92. Show that ð 0 x dx 1 þ cos2 x ¼ 2 2 ffiffiffi 2 p :
- 125. 116 Partial Derivatives FUNCTIONS OF TWO OR MORE VARIABLES The definition of a function was given in Chapter 3 (page 39). For us the distinction for functions of two or more variables is that the domain is a set of n-tuples of numbers. The range remains one dimensional and is referred to an interval of numbers. If n ¼ 2, the domain is pictured as a two- dimensional region. The region is referred to a rectangular Cartesian coordinate system described through number pairs ðx; yÞ, and the range variable is usually denoted by z. The domain variables are independent while the range variable is dependent. We use the notation f ðx; yÞ, Fðx; yÞ, etc., to denote the value of the function at ðx; yÞ and write z ¼ f ðx; yÞ, z ¼ Fðx; yÞ, etc. We shall also sometimes use the notation z ¼ zðx; yÞ although it should be understood that in this case z is used in two senses, namely as a function and as a variable. EXAMPLE. If f ðx; yÞ ¼ x2 þ 2y3 , then f ð3; 1Þ ¼ ð3Þ2 þ 2ð1Þ3 ¼ 7: The concept is easily extended. Thus w ¼ Fðx; y; zÞ denotes the value of a function at ðx; y; zÞ [a point in three-dimensional space], etc. EXAMPLE. If z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðx2 þ y2Þ p , the domain for which z is real consists of the set of points ðx; yÞ such that x2 þ y2 @ 1, i.e., the set of points inside and on a circle in the xy plane having center at ð0; 0Þ and radius 1. THREE-DIMENSIONAL RECTANGULAR COORDINATE SYSTEMS A three-dimensional rectangular coordinate system, as referred to in the previous paragraph, obtained by constructing three mutually perpendicular axes (the x-, y-, and z-axes) intersecting in point O (the origin). It forms a natural extension of the usual xy plane for representing functions of two variables graphically. A point in three dimensions is represented by the triplet ðx; y; zÞ called coordinates of the point. In this coordinate system z ¼ f ðx; yÞ [or Fðx; y; zÞ ¼ 0] represents a surface, in general. EXAMPLE. The set of points ðx; y; zÞ such that z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðx2 þ y2 Þ p comprises the surface of a hemisphere of radius 1 and center at ð0; 0; 0Þ. For functions of more than two variables such geometric interpretation fails, although the termi- nology is still employed. For example, ðx; y; z; wÞ is a point in four-dimensional space, and w ¼ f ðx; y; zÞ [or Fðx; y; z; wÞ ¼ 0] represents a hypersurface in four dimensions; thus x2 þ y2 þ z2 þ w2 ¼ a2 represents a hypersphere in four dimensions with radius a 0 and center at ð0; 0; 0; 0Þ. w ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 ðx2 þ y2 þ z2Þ p , x2 þ y2 þ z2 @ a2 describes a function generated from the hypersphere. Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 126. NEIGHBORHOODS The set of all points ðx; yÞ such that jx x0j , j y y0j where 0, is called a rectangular neighborhood of ðx0; y0Þ; the set 0 jx x0j , 0 j y y0j which excludes ðx0; y0Þ is called a rectangular deleted neighborhood of ðx0; y0Þ. Similar remarks can be made for other neighborhoods, e.g., ðx x0Þ2 þ ð y y0Þ2 2 is a circular neighborhood of ðx0; y0Þ. The term ‘‘open ball’’ is used to designate this circular neighborhood. This terminology is appropriate for generalization to more dimensions. Whether neighborhoods are viewed as circular or square is immaterial, since the descrip- tions are interchangeable. Simply notice that given an open ball (circular neighborhood) of radius there is a centered square whose side is of length less than ffiffiffi 2 p that is interior to the open ball, and conversely for a square of side there is an interior centered of radius of radius less than =2. (See Fig. 6-1.) A point ðx0; y0Þ is called a limit point, accumulation point, or cluster point of a point set S if every deleted neighborhood of ðx0; y0Þ contains points of S. As in the case of one-dimensional point sets, every bounded infinite set has at least one limit point (the Bolzano–Weierstrass theorem, see Pages 6 and 12). A set containing all its limit points is called a closed set. REGIONS A point P belonging to a point set S is called an interior point of S if there exists a deleted neighborhood of P all of whose points belong to S. A point P not belonging to S is called an exterior point of S if there exists a deleted neighborhood of P all of whose points do not belong to S. A point P is called a boundary point of S if every deleted neighborhood of P contains points belonging to S and also points not belonging to S. If any two points of a set S can be joined by a path consisting of a finite number of broken line segments all of whose points belong to S, then S is called a connected set. A region is a connected set which consists of interior points or interior and boundary points. A closed region is a region containing all its boundary points. An open region consists only of interior points. The complement of a set, S, in the xy plane is the set of all points in the plane not belonging to S. (See Fig. 6-2.) Examples of some regions are shown graphically in Figs 6-3(a), (b), and (c) below. The rectangular region of Fig. 6-1(a), including the boundary, represents the sets of points a @ x @ b, c @ y @ d which is a natural extension of the closed interval a @ x @ b for one dimension. The set a x b, c y d corresponds to the boundary being excluded. In the regions of Figs 6-3(a) and 6-3(b), any simple closed curve (one which does not intersect itself anywhere) lying inside the region can be shrunk to a point which also lies in the region. Such regions are called simply-connected regions. In Fig. 6-3(c) however, a simple closed curve ABCD surrounding one of the ‘‘holes’’ in the region cannot be shrunk to a point without leaving the region. Such regions are called multiply-connected regions. CHAP. 6] PARTIAL DERIVATIVES 117 Fig. 6-1 Fig. 6-2
- 127. LIMITS Let f ðx; yÞ be defined in a deleted neighborhood of ðx0; y0Þ [i.e.; f ðx; yÞ may be undefined at ðx0; y0Þ]. We say that l is the limit of f ðx; yÞ as x approaches x0 and y approaches y0 [or ðx; yÞ approaches ðx0; y0Þ] and write lim x!x0 y!y0 f ðx; yÞ ¼ l [or lim ðx;yÞ!ðx0;y0Þ f ðx; yÞ ¼ l] if for any positive number we can find some positive number [depending on and ðx0; y0Þ, in general] such that j f ðx; yÞ lj whenever 0 jx x0j and 0 j y y0j . If desired we can use the deleted circular neighborhood open ball 0 ðx x0Þ2 þ ð y y0Þ2 2 instead of the deleted rectangular neighborhood. EXAMPLE. Let f ðx; yÞ ¼ 3xy if ðx; yÞ 6¼ ð1; 2Þ 0 if ðx; yÞ ¼ ð1; 2Þ . As x ! 1 and y ! 2 [or ðx; yÞ ! ð1; 2Þ], f ðx; yÞ gets closer to 3ð1Þð2Þ ¼ 6 and we suspect that lim x!1 y!2 f ðx; yÞ ¼ 6. To prove this we must show that the above definition of limit with l ¼ 6 is satisfied. Such a proof can be supplied by a method similar to that of Problem 6.4. Note that lim x!1 y!2 f ðx; yÞ 6¼ f ð1; 2Þ since f ð1; 2Þ ¼ 0. The limit would in fact be 6 even if f ðx; yÞ were not defined at ð1; 2Þ. Thus the existence of the limit of f ðx; yÞ as ðx; yÞ ! ðx0; y0Þ is in no way dependent on the existence of a value of f ðx; yÞ at ðx0; y0Þ. Note that in order for lim ðx;yÞ!ðx0;y0Þ f ðx; yÞ to exist, it must have the same value regardless of the approach of ðx; yÞ to ðx0; y0Þ. It follows that if two different approaches give different values, the limit cannot exist (see Problem 6.7). This implies, as in the case of functions of one variable, that if a limit exists it is unique. The concept of one-sided limits for functions of one variable is easily extended to functions of more than one variable. EXAMPLE 1. lim x!0þ y!1 tan1 ð y=xÞ ¼ =2, lim x!0 y!1 tan1 ð y=xÞ ¼ =2. EXAMPLE 2. lim x!0 y!1 tan1 ð y=xÞ does not exist, as is clear from the fact that the two different approaches of Example 1 give different results. In general the theorems on limits, concepts of infinity, etc., for functions of one variable (see Page 21) apply as well, with appropriate modifications, to functions of two or more variables. 118 PARTIAL DERIVATIVES [CHAP. 6 Fig. 6-3
- 128. ITERATED LIMITS The iterated limits lim x!x0 lim y!y0 f ðx; yÞ and lim y!y0 lim x!x0 f ðx; yÞ , [also denoted by lim x!x0 lim y!y0 f ðx; yÞ and lim y!y0 lim x!x0 f ðx; yÞ respectively] are not necessarily equal. Although they must be equal if lim x!x0 y!y0 f ðx; yÞ is to exist, their equality does not guarantee the existence of this last limit. EXAMPLE. If f ðx; yÞ ¼ x y x þ y , then lim x!0 lim y!0 x y x þ y ¼ lim x!0 ð1Þ ¼ 1 and lim y!0 lim x!0 x y x þ y ¼ lim y!0 ð1Þ ¼ 1. Thus the iterated limits are not equal and so lim x!0 y!0 f ðx; yÞ cannot exist. CONTINUITY Let f ðx; yÞ be defined in a neighborhood of ðx0; y0Þ [i.e.; f ðx; yÞ must be defined at ðx0; y0Þ as well as near it]. We say that f ðx; yÞ is continuous at ðx0; y0Þ if for any positive number we can find some positive number [depending on and ðx0; y0Þ in general] such that j f ðx; yÞ f ðx0; y0Þj whenever jx x0j and jy y0j , or alternatively ðx x0Þ2 þ ð y y0Þ2 2 . Note that three conditions must be satisfied in order that f ðx; yÞ be continuous at ðx0; y0Þ. 1. lim ðx;yÞ!ðx0;y0Þ f ðx; yÞ ¼ l, i.e., the limit exists as ðx; yÞ ! ðx0; y0Þ 2. f ðx0; y0Þ must exist, i.e., f ðx; yÞ is defined at ðx0; y0Þ 3. l ¼ f ðx0; y0Þ If desired we can write this in the suggestive form lim x!x0 y!y0 f ðx; yÞ ¼ f ð lim x!x0 x; lim y!y0 yÞ. EXAMPLE. If f ðx; yÞ ¼ 3xy ðx; yÞ 6¼ ð1; 2Þ 0 ðx; yÞ ¼ ð1; 2Þ , then lim ðx;yÞ!ð1;2Þ f ðx; yÞ ¼ 6 6¼ f ð1; 2Þ. Hence, f ðx; yÞ is not contin- uous at ð1; 2Þ. If we redefine the function so that f ðx; yÞ ¼ 6 for ðx; yÞ ¼ ð1; 2Þ, then the function is continuous at ð1; 2Þ. If a function is not continuous at a point ðx0; y0Þ, it is said to be discontinuous at ðx0; y0Þ which is then called a point of discontinuity. If, as in the above example, it is possible to redefine the value of a function at a point of discontinuity so that the new function is continuous, we say that the point is a removable discontinuity of the old function. A function is said to be continuous in a region r of the xy plane if it is continuous at every point of r. Many of the theorems on continuity for functions of a single variable can, with suitable modifica- tion, be extended to functions of two more variables. UNIFORM CONTINUITY In the definition of continuity of f ðx; yÞ at ðx0; y0Þ, depends on and also ðx0; y0Þ in general. If in a region r we can find a which depends only on but not on any particular point ðx0; y0Þ in r [i.e., the same will work for all points in r], then f ðx; yÞ is said to be uniformly continuous in r. As in the case of functions of one variable, it can be proved that a function which is continuous in a closed and bounded region is uniformly continuous in the region. PARTIAL DERIVATIVES The ordinary derivative of a function of several variables with respect to one of the independent variables, keeping all other independent variables constant, is called the partial derivative of the function with respect to the variable. Partial derivatives of f ðx; yÞ with respect to x and y are denoted by CHAP. 6] PARTIAL DERIVATIVES 119
- 129. @f @x or fx; fxðx; yÞ; @f x y # and @f @y or fy; fyðx; yÞ; @f @y x , respectively, the latter notations being used when it is needed to emphasize which variables are held constant. By definition, @f @x ¼ lim x!0 f ðx þ x; yÞ f ðx; yÞ x ; @f @y ¼ lim y!0 f ðx; y þ yÞ f ðx; yÞ y ð1Þ when these limits exist. The derivatives evaluated at the particular point ðx0; y0Þ are often indicated by @f @x ðx0;y0Þ ¼ fxðx0; y0Þ and @f @y ðx0;y0Þ ¼ fyðx0; y0Þ, respectively. EXAMPLE. If f ðx; yÞ ¼ 2x3 þ 3xy2 , then fx ¼ @f =@x ¼ 6x2 þ 3y2 and fy ¼ @f =@y ¼ 6xy. Also, fxð1; 2Þ ¼ 6ð1Þ2 þ 3ð2Þ2 ¼ 18, fyð1; 2Þ ¼ 6ð1Þð2Þ ¼ 12. If a function f has continuous partial derivatives @f =@x, @f =@y in a region, then f must be continuous in the region. However, the existence of these partial derivatives alone is not enough to guarantee the continuity of f (see Problem 6.9). HIGHER ORDER PARTIAL DERIVATIVES If f ðx; yÞ has partial derivatives at each point ðx; yÞ in a region, then @f =@x and @f =@y are themselves functions of x and y, which may also have partial derivatives. These second derivatives are denoted by @ @x @f @x ¼ @2 f @x2 ¼ fxx; @ @y @f @y ¼ @2 f @y2 ¼ fyy; @ @x @f @y ¼ @2 f @x @y ¼ fyx; @ @y @f @x ¼ @2 f @y @x ¼ fxy ð2Þ If fxy and fyx are continuous, then fxy ¼ fyx and the order of differentiation is immaterial; otherwise they may not be equal (see Problems 6.13 and 6.41). EXAMPLE. If f ðx; yÞ ¼ 2x3 þ 3xy2 (see preceding example), then fxx ¼ 12x, fyy ¼ 6x, fxy ¼ 6y ¼ fyx. In such case fxxð1; 2Þ ¼ 12, fyyð1; 2Þ ¼ 6, fxyð1; 2Þ ¼ fyxð1; 2Þ ¼ 12. In a similar manner, higher order derivatives are defined. For example @3 f @x2@y ¼ fyxx is the derivative of f taken once with respect to y and twice with respect to x. DIFFERENTIALS (The section of differentials in Chapter 4 should be read before beginning this one.) Let x ¼ dx and y ¼ dy be increments given to x and y, respectively. Then z ¼ f ðx þ x; y þ yÞ f ðx; yÞ ¼ f ð3Þ is called the increment in z ¼ f ðx; yÞ. If f ðx; yÞ has continuous first partial derivatives in a region, then z ¼ @f @x x þ @f @y y þ 1x þ 2y ¼ @z @x dx þ @z @y dy þ 1 dx þ 2 dy ¼ f ð4Þ where 1 and 2 approach zero as x and y approach zero (see Problem 6.14). The expression dz ¼ @z @x dx þ @z @y dy or df ¼ @f @x dx þ @f @y dy ð5Þ is called the total differential or simply differential of z or f , or the principal part of z or f . Note that z 6¼ dz in general. However, if x ¼ dx and y ¼ dy are ‘‘small,’’ then dz is a close approximation of z (see Problem 6.15). The quantities dx and dy, called differentials of x and y respectively, need not be small. 120 PARTIAL DERIVATIVES [CHAP. 6
- 130. The form dz ¼ fxðx0; y0Þdx þ fyðx0; y0Þdy signifies a linear function with the independent variables dx and dy and the dependent range variable dz. In the one variable case, the corresponding linear function represents the tangent line to the underlying curve. In this case, the underlying entity is a surface and the linear function generates the tangent plane at P0. In a small enough neighborhood, this tangent plane is an approximation of the surface (i.e., the linear representation of the surface at P0). If y is held constant, then one obtains the curve of intersection of the surface and the coordinate plane y ¼ y0. The differential form reduces to dz ¼ fxðx0; y0Þdx (i.e., the one variable case). A similar statement follows when x is held constant. See Fig. 6-4. If f is such that f (or zÞ can be expressed in the form (4) where 1 and 2 approach zero as x and y approach zero, we call f differentiable at ðx; yÞ. The mere existence of fx and fy does not in itself guarantee differentiability; however, continuity of fx and fy does (although this condition happens to be slightly stronger than necessary). In case fx and fy are continuous in a region r, we shall say that f is continuously differentiable in r. THEOREMS ON DIFFERENTIALS In the following we shall assume that all functions have continuous first partial derivatives in a region r, i.e., the functions are continuously differentiable in r. 1. If z ¼ f ðx1; x2; . . . ; xnÞ, then df ¼ @f @x1 dx1 þ @f @x2 dx2 þ þ @f @xn dxn ð6Þ regardless of whether the variables x1; x2; . . . ; xn are independent or dependent on other vari- ables (see Problem 6.20). This is a generalization of the result (5). In (6) we often use z in place of f . 2. If f ðx1; x2; . . . ; xnÞ ¼ c, a constant, then df ¼ 0. Note that in this case x1; x2; . . . ; xn cannot all be independent variables. CHAP. 6] PARTIAL DERIVATIVES 121 Fig. 6-4
- 131. 3. The expression Pðx; yÞdx þ Qðx; yÞdy or briefly P dx þ Q dy is the differential of f ðx; yÞ if and only if @P @y ¼ @Q @x . In such case P dx þ Q dy is called an exact differential. Note: Observe that @P @y ¼ @Q @x implies that @2 f @y @x ¼ @2 f @x @y . 4. The expression Pðx; y; zÞ dx þ Qðx; y; zÞ dy þ Rðx; y; zÞ dz or briefly P dx þ Q dy þ R dz is the differential of f ðx; y; zÞ if and only if @P @y ¼ @Q @x ; @Q @z ¼ @R @y ; @R @x ¼ @P @z . In such case P dx þ Q dy þ R dz is called an exact differential. Proofs of Theorems 3 and 4 are best supplied by methods of later chapters (see Chapter 10, Problems 10.13 and 10.30). DIFFERENTIATION OF COMPOSITE FUNCTIONS Let z ¼ f ðx; yÞ where x ¼ gðr; sÞ, y ¼ hðr; sÞ so that z is a function of r and s. Then @z @r ¼ @z @x @x @r þ @z @y @y @r ; @z @s ¼ @z @x @x @s þ @z @y @y @s ð7Þ In general, if u ¼ Fðx1; . . . ; xnÞ where x1 ¼ f1ðr1; . . . ; rpÞ; . . . ; xn ¼ fnðr1; . . . ; rpÞ, then @u @rk ¼ @u @x1 @x1 @rk þ @u @x2 @x2 @rk þ þ @u @xn @xn @rk k ¼ 1; 2; . . . ; p ð8Þ If in particular x1; x2; . . . ; xn depend on only one variable s, then du ds ¼ @u @x1 dx1 ds þ @u @x2 dx2 ds þ þ @u @xn dxn ds ð9Þ These results, often called chain rules, are useful in transforming derivatives from one set of variables to another. Higher derivatives are obtained by repeated application of the chain rules. EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS A function represented by Fðx1; x2; . . . ; xnÞ is called homogeneous of degree p if, for all values of the parameter and some constant p, we have the identity Fðx1; x2; . . . ; xnÞ ¼ p Fðx1; x2; . . . ; xnÞ ð10Þ EXAMPLE. Fðx; yÞ ¼ x4 þ 2xy3 5y4 is homogeneous of degree 4, since Fðx; yÞ ¼ ðxÞ4 þ 2ðxÞðyÞ3 5ðyÞ4 ¼ 4 ðx4 þ 2xy3 5y4 Þ ¼ 4 Fðx; yÞ Euler’s theorem on homogeneous functions states that if Fðx1; x2; . . . ; xnÞ is homogeneous of degree p then (see Problem 6.25) x1 @F @x1 þ x2 @F @x2 þ þ xn @F @xn ¼ pF ð11Þ 122 PARTIAL DERIVATIVES [CHAP. 6
- 132. IMPLICIT FUNCTIONS In general, an equation such as Fðx; y; zÞ ¼ 0 defines one variable, say z, as a function of the other two variables x and y. Then z is sometimes called an implicit function of x and y, as distinguished from a so-called explicit function f, where z ¼ f ðx; yÞ, which is such that F½x; y; f ðx; yÞ 0. Differentiation of implicit functions requires considerable discipline in interpreting the independent and dependent character of the variables and in distinguishing the intent of one’s notation. For example, suppose that in the implicit equation F½x; y; f ðx; zÞ ¼ 0, the independent variables are x and y and that z ¼ f ðx; yÞ. In order to find @f @x and @f @y , we initially write (observe that Fðx; t; zÞ is zero for all domain pairs ðx; yÞ, in other words it is a constant): 0 ¼ dF ¼ Fx dx þ Fy dy þ Fz dz and then compute the partial derivatives Fx; Fy; Fz as though y; y; z constituted an independent set of variables. At this stage we invoke the dependence of z on x and y to obtain the differential form dz ¼ @f @x dx þ @f @y dy. Upon substitution and some algebra (see Problem 6.30) the following results are obtained: @f @x ¼ Fx Fz ; @f @y ¼ Fy Fz EXAMPLE. If 0 ¼ Fðx; y; zÞ ¼ x2 z þ yz2 þ 2xy2 z3 and z ¼ f ðx; yÞ then Fx ¼ 2xz þ 2y2 , Fy ¼ z2 þ 4xy. Fz ¼ x2 þ 2yz 3z2 . Then @f @x ¼ ð2xz þ 2y2 Þ x2 þ 2yz 3z2 ; @f @y ¼ ðz2 þ 4xyÞ x2 þ 2yz 3x2 Observe that f need not be known to obtain these results. If that information is available then (at least theoretically) the partial derivatives may be expressed through the independent variables x and y. JACOBIANS If Fðu; vÞ and Gðu; vÞ are differentiable in a region, the Jacobian determinant, or briefly the Jacobian, of F and G with respect to u and v is the second order functional determinant defined by @ðF; GÞ @ðu; vÞ ¼ @F @u @F @v @G @u @G @v ¼ Fu Fv Gu Gv ð7Þ Similarly, the third order determinant @ðF; G; HÞ @ðu; v; wÞ ¼ Fu Fv Fw Gu Gv Gw Hu Hv Hw is called the Jacobian of F, G, and H with respect to u, v, and w. Extensions are easily made. PARTIAL DERIVATIVES USING JACOBIANS Jacobians often prove useful in obtaining partial derivatives of implicit functions. Thus, for example, given the simultaneous equations Fðx; y; u; vÞ ¼ 0; Gðx; y; u; vÞ ¼ 0 CHAP. 6] PARTIAL DERIVATIVES 123
- 133. we may, in general, consider u and v as functions of x and y. In this case, we have (see Problem 6.31) @u @x ¼ @ðF; GÞ @ðx; vÞ @ðF; GÞ @ðu; vÞ ; @u @y ¼ @ðF; GÞ @ðy; vÞ @ðF; GÞ @ðu; vÞ ; @v @x ¼ @ðF; GÞ @ðu; xÞ @ðF; GÞ @ðu; vÞ ; @v @y ¼ @ðF; GÞ @ðu; yÞ @ðF; GÞ @ðu; vÞ The ideas are easily extended. Thus if we consider the simultaneous equations Fðu; v; w; x; yÞ ¼ 0; Gðu; v; w; x; yÞ ¼ 0; Hðu; v; w; x; yÞ ¼ 0 we may, for example, consider u, v, and w as functions of x and y. In this case, @u @x ¼ @ðF; G; HÞ @ðx; v; wÞ @ðF; G; HÞ @ðu; v; wÞ ; @w @y ¼ @ðF; G; HÞ @ðu; v; yÞ @ðF; G; HÞ @ðu; v; wÞ with similar results for the remaining partial derivatives (see Problem 6.33). THEOREMS ON JACOBIANS In the following we assume that all functions are continuously differentiable. 1. A necessary and sufficient condition that the equations Fðu; v; x; y; zÞ ¼ 0, Gðu; v; x; y; zÞ ¼ 0 can be solved for u and v (for example) is that @ðF; GÞ @ðu; vÞ is not identically zero in a region r. Similar results are valid for m equations in n variables, where m n. 2. If x and y are functions of u and v while u and v are functions of r and s, then (see Problem 6.43) @ðx; yÞ @ðr; sÞ ¼ @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðr; sÞ ð9Þ This is an example of a chain rule for Jacobians. These ideas are capable of generalization (see Problems 6.107 and 6.109, for example). 3. If u ¼ f ðx; yÞ and v ¼ gðx; yÞ, then a necessary and sufficient condition that a functional relation of the form ðu; vÞ ¼ 0 exists between u and v is that @ðu; vÞ @ðx; yÞ be identically zero. Similar results hold for n functions of n variables. Further discussion of Jacobians appears in Chapter 7 where vector interpretations are employed. TRANSFORMATIONS The set of equations x ¼ Fðu; vÞ y ¼ Gðu; vÞ ð10Þ defines, in general, a transformation or mapping which establishes a correspondence between points in the uv and xy planes. If to each point in the uv plane there corresponds one and only one point in the xy plane, and conversely, we speak of a one-to-one transformation or mapping. This will be so if F and G are continuously differentiable with Jacobian not identically zero in a region. In such case (which we shall assume unless otherwise stated) equations (10) are said to define a continuously differentiable transformation or mapping. 124 PARTIAL DERIVATIVES [CHAP. 6
- 134. Under the transformation (10) a closed region r of the xy plane is, in general, mapped into a closed region r0 of the uv plane. Then if Axy and Auv denote respectively the areas of these regions, we can show that lim Axy Auv ¼ @ðx; yÞ @ðu; vÞ ð11Þ where lim denotes the limit as Axy (or Auv) approaches zero. The Jacobian on the right of (11) is often called the Jacobian of the transformation (10). If we solve (10) for u and v in terms of x and y, we obtain the transformation u ¼ f ðx; yÞ, v ¼ gðx; yÞ often called the inverse transformation corresponding to (10). The Jacobians @ðu; vÞ @ðx; yÞ and @ðx; yÞ @ðu; vÞ of these transformations are reciprocals of each other (see Problem 6.43). Hence, if one Jacobian is different from zero in a region, so also is the other. The above ideas can be extended to transformations in three or higher dimensions. We shall deal further with these topics in Chapter 7, where use is made of the simplicity of vector notation and interpretation. CURVILINEAR COORDINATES If ðx; yÞ are the rectangular coordinates of a point in the xy plane, we can think of ðu; vÞ as also specifying coordinates of the same point, since by knowing ðu; vÞ we can determine ðx; yÞ from (10). The coordinates ðu; vÞ are called curvilinear coordinates of the point. EXAMPLE. The polar coordinates ð; Þ of a point correspond to the case u ¼ , v ¼ . In this case the transformation equations (10) are x ¼ cos , y ¼ sin . For curvilinear coordinates in higher dimensional spaces, see Chapter 7. MEAN VALUE THEOREM If f ðx; yÞ is continuous in a closed region and if the first partial derivatives exist in the open region (i.e., excluding boundary points), then f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ 0 1 ð12Þ This is sometimes written in a form in which h ¼ x ¼ x x0 and k ¼ y ¼ y y0. Solved Problems FUNCTIONS AND GRAPHS 6.1. If f ðx; yÞ ¼ x3 2xy þ 3y2 , find: (a) f ð2; 3Þ; ðbÞ f 1 x ; 2 y ; ðcÞ f ðx; y þ kÞ f ðx; yÞ k ; k 6¼ 0. ðaÞ f ð2; 3Þ ¼ ð2Þ3 2ð2Þð3Þ þ 3ð3Þ2 ¼ 8 þ 12 þ 27 ¼ 31 ðbÞ f 1 x ; 2 y ¼ 1 x 3 2 1 x 2 y þ 3 2 y 2 ¼ 1 x3 4 xy þ 12 y2 CHAP. 6] PARTIAL DERIVATIVES 125
- 135. ðcÞ f ðx; y þ kÞ f ðx; yÞ k ¼ 1 k f½x3 2xðy þ kÞ þ 3ðy þ kÞ2 ½x3 2xy þ 3y2 g ¼ 1 k ðx3 2xy 2kx þ 3y2 þ 6ky þ 3k2 x2 þ 2xy 3y2 Þ ¼ 1 k ð2kx þ 6ky þ 3k2 Þ ¼ 2x þ 6y þ 3k: 6.2. Give the domain of definition for which each of the following functions are defined and real, and indicate this domain graphically. (a) f ðx; yÞ ¼ lnfð16 x2 y2 Þðx2 þ y2 4Þg The function is defined and real for all points ðx; yÞ such that ð16 x2 y2 Þðx2 þ y2 4Þ 0; i.e., 4 x2 þ y2 16 which is the required domain of definition. This point set consists of all points interior to the circle of radius 4 with center at the origin and exterior to the circle of radius 2 with center at the origin, as in the figure. The corresponding region, shown shaded in Fig. 6-5 below, is an open region. (b) f ðx; yÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 ð2x þ 3yÞ p The function is defined and real for all points ðx; yÞ such that 2x þ 3y @ 6, which is the required domain of definition. The corresponding (unbounded) region of the xy plane is shown shaded in Fig. 6-6 above. 6.3. Sketch and name the surface in three-dimensional space represented by each of the following. What are the traces on the coordinate planes? (a) 2x þ 4y þ 3z ¼ 12. Trace on xy plane ðz ¼ 0Þ is the straight line x þ 2y ¼ 6, z ¼ 0: Trace on yz plane ðx ¼ 0Þ is the straight line 4y þ 3z ¼ 12, x ¼ 0. Trace on xz plane ðy ¼ 0Þ is the straight line 2x þ 3z ¼ 12, y ¼ 0. These are represented by AB, BC; and AC in Fig. 6-7. The surface is a plane intersecting the x-, y-, and z-axes in the points Að6; 0; 0Þ, Bð0; 3; 0Þ, Cð0; 0; 4Þ. The lengths OA ¼ 6, OB ¼ 3, OC ¼ 4 are called the x, y, and z intercepts, respectively. ðbÞ x2 a2 þ y2 b2 z2 c2 ¼ 1 Trace on xy plane ðz ¼ 0Þ is the ellipse x2 a2 þ y2 b2 ¼ 1, z ¼ 0. Trace on yz plane ðx ¼ 0Þ is the hyperbola y2 b2 z2 c2 ¼ 1, x ¼ 0. 126 PARTIAL DERIVATIVES [CHAP. 6 Fig. 6-5 Fig. 6-6 Fig. 6-7
- 136. Trace on xz plane ðy ¼ 0Þ is the hyperbola x2 a2 z2 c2 ¼ 1, y ¼ 0. Trace on any plane z ¼ p parallel to the xy plane is the ellipse x2 a2ð1 þ p2=c2Þ þ y2 b2ð1 þ p2=c2Þ ¼ 1 As j pj increases from zero, the elliptic cross section increases in size. The surface is a hyperboloid of one sheet (see Fig. 6-8). LIMITS AND CONTINUITY 6.4. Prove that lim x!1 y!2 ðx2 þ 2yÞ ¼ 5. Method 1, using definition of limit. We must show that given any 0, we can find 0 such that jx2 þ 2y 5j when 0 jx 1j , 0 j y 2j . If 0 jx 1j and 0 j y 2j , then 1 x 1 þ and 2 y 2 þ , excluding x ¼ 1; y ¼ 2. Thus, 1 2 þ 2 x2 1 þ 2 þ 2 and 4 2 2y 4 þ 2. Adding, 5 4 þ 2 x2 þ 2y 5 þ 4 þ 2 or 4 þ 2 x2 þ 2y 5 4 þ 2 Now if @ 1, it certainly follows that 5 x2 þ 2y 5 5, i.e., jx2 þ 2y 5j 5 whenever 0 jx 1j , 0 j y 2j . Then choosing 5 ¼ , i.e., ¼ =5 (or ¼ 1, whichever is smaller), it follows that jx2 þ 2y 5j when 0 jx 1j , 0 j y 2j , i.e., lim x!1 y!2 ðx2 þ 2yÞ ¼ 5. Method 2, using theorems on limits. lim x!1 y!2 ðx2 þ 2yÞ ¼ lim x!1 y!2 x2 þ lim x!1 y!2 2y ¼ 1 þ 4 ¼ 5 6.5. Prove that f ðx; yÞ ¼ x2 þ 2y is continuous at ð1; 2Þ. By Problem 6.4, lim x!1 y!2 f ðx; yÞ ¼ 5. Also, f ð1; 2Þ ¼ 12 þ 2ð2Þ ¼ 5. Then lim x!1 y!2 f ðx; yÞ ¼ f ð1; 2Þ and the function is continuous at ð1; 2Þ. Alternatively, we can show, in much the same manner as in the first method of Problem 6.4, that given any 0 we can find 0 such that j f ðx; yÞ f ð1; 2Þj when jx 1j ; j y 2j . 6.6. Determine whether f ðx; yÞ ¼ x2 þ 2y; ðx; yÞ 6¼ ð1; 2Þ 0; ðx; yÞ ¼ ð1; 2Þ . (a) has a limit as x ! 1 and y ! 2, (b) is continuous at ð1; 2Þ. (a) By Problem 6.4, it follows that lim x!1 y!2 f ðx; yÞ ¼ 5, since the limit has nothing to do with the value at ð1; 2Þ. (b) Since lim x!1 y!2 f ðx; yÞ ¼ 5 and f ð1; 2Þ ¼ 0, it follows that lim x!1 y!2 f ðx; yÞ 6¼ f ð1; 2Þ. Hence, the function is discontinuous at ð1; 2Þ: 6.7. Investigate the continuity of f ðx; yÞ ¼ x2 y2 x2 þ y2 ðx; yÞ 6¼ ð0; 0Þ 0 ðx; yÞ ¼ ð0; 0Þ 8 : at ð0; 0Þ. Let x ! 0 and y ! 0 in such a way that y ¼ mx (a line in the xy plane). Then along this line, lim x!0 y!0 x2 y2 x2 þ y2 ¼ lim x!0 x2 m2 x2 x2 þ m2 x2 ¼ lim x!0 x2 ð1 m2 Þ x2 ð1 þ m2 Þ ¼ 1 m2 1 þ m2 CHAP. 6] PARTIAL DERIVATIVES 127 Fig. 6-8
- 137. Since the limit of the function depends on the manner of approach to ð0; 0Þ (i.e., the slope m of the line), the function cannot be continuous at ð0; 0Þ. Another method: Since lim x!0 lim y!0 x2 y2 x2 þ y2 ( ) ¼ lim x!0 x2 x2 ¼ 1 and lim y¼0 lim x!0 x2 y2 x2 þ y2 ( ) ¼ 1 are not equal, lim x!0 y!0 f ðx; yÞ cannot exist. Hence, f ðx; yÞ cannot be continuous at ð0; 0Þ. PARTIAL DERIVATIVES 6.8. If f ðx; yÞ ¼ 2x2 xy þ y2 , find (a) @f =@x, and (b) @f =@y at ðx0; y0Þ directly from the definition. ðaÞ @f @x ðx0:y0Þ ¼ fxðx0; y0Þ ¼ lim h!0 f ðx0 þ h; y0Þ f ðx0; y0Þ h ¼ lim h!0 ½2ðx0 þ hÞ2 ðx0 þ hÞy0 þ y2 0 ¼ ½2x2 0 x0y0 þ y2 0 h ¼ lim h!0 4hx0 þ 2h2 hy0 h ¼ lim h!0 ð4x0 þ 2h y0Þ ¼ 4x0 y0 ðbÞ @f @y ðx0;y0Þ ¼ fyðx0; y0Þ ¼ lim k!0 f ðx0; y0 þ kÞ f ðx0; y0Þ k ¼ lim k!0 ½2x2 0 x0ðy0 þ kÞ þ ð y0 þ kÞ2 ½2x2 0 x0y0 þ y2 0 k ¼ lim k!0 kx0 þ 2ky0 þ k2 k ¼ lim k!0 ðx0 þ 2y0 þ kÞ ¼ x0 þ 2y0 Since the limits exist for all points ðx0; y0Þ, we can write fxðx; yÞ ¼ fx ¼ 4x y, fyðx; yÞ ¼ fy ¼ x þ 2y which are themselves functions of x and y. Note that formally fxðx0; y0Þ is obtained from f ðx; yÞ by differentiating with respect to x, keeping y constant and then putting x ¼ x0; y ¼ y0. Similarly, fyðx0; y0Þ is obtained by differentiating f with respect to y, keeping x constant. This procedure, while often lucrative in practice, need not always yield correct results (see Problem 6.9). It will work if the partial derivatives are continuous. 6.9. Let f ðx; yÞ ¼ xy=ðx2 þ y2 Þ ðx; yÞ 6¼ ð0; 0Þ 0 otherwise : Prove that (a) fxð0; 0Þ and fyð0; 0Þ both exist but that (b) f ðx; yÞ is discontinuous at ð0; 0Þ. ðaÞ fxð0; 0Þ ¼ lim h!0 f ðh; 0Þ f ð0; 0Þ h ¼ lim h!0 0 h ¼ 0 fyð0; 0Þ ¼ lim k!0 f ð0; 0Þ f ð0; 0Þ k ¼ lim k!0 0 k ¼ 0 (b) Let ðx; yÞ ! ð0; 0Þ along the line y ¼ mx in the xy plane. Then lim x!0 y!0 f ðx; yÞ ¼ lim x!0 mx2 x2 þ m2 x2 ¼ m 1 þ m2 so that the limit depends on m and hence on the approach and therefore does not exist. Hence, f ðx; yÞ is not continuous at ð0; 0Þ: Note that unlike the situation for functions of one variable, the existence of the first partial derivatives at a point does not imply continuity at the point. Note also that if ðx; yÞ 6¼ ð0; 0Þ, fx ¼ y2 x2 y ðx2 þ y2 Þ2 , fy ¼ x3 xy2 ðx2 þ y2 Þ2 and fxð0; 0Þ, fyð0; 0Þ cannot be computed from them by merely letting x ¼ 0 and y ¼ 0. See remark at the end of Problem 4.5(b) Chapter 4. 6.10. If ðx; yÞ ¼ x3 y þ exy2 , find (a) x; ðbÞ y; ðcÞ xx; ðdÞ yy; ðeÞ xy; ð f Þ yx. 128 PARTIAL DERIVATIVES [CHAP. 6
- 138. ðaÞ x ¼ @ @x ¼ @ @x ðx3 y þ exy2 Þ ¼ 3x2 y þ exy2 y2 ¼ 3x2 y þ y2 exy2 ðbÞ y ¼ @ @y ¼ @ @y ðx3 y þ exy2 Þ ¼ x3 þ exy2 2xy ¼ x3 þ 2xy exy2 ðcÞ xx ¼ @2 @x2 ¼ @ @x @ @x ¼ @ @x ð3x2 y þ y2 exy2 Þ ¼ 6xy þ y2 ðexy2 y2 Þ ¼ 6xy þ y4 exy2 ðdÞ yy ¼ @2 @y2 ¼ @ @y ðx3 þ 2xy exy2 Þ ¼ 0 þ 2xy @ @y ðexy2 Þ þ exy2 @ @y ð2xyÞ ¼ 2xy exy2 2xy þ exy2 2x ¼ 4x2 y2 exy2 þ 2x exy2 ðeÞ xy ¼ @2 @y @x ¼ @ @y @ @x ¼ @ @y ð3x2 y þ y2 exy2 Þ ¼ 3x2 þ y2 exy2 2xy þ exy2 2y ¼ 3x2 þ 2xy3 exy2 þ 2y exy2 ð f Þ yx ¼ @2 @x @y ¼ @ @x @ @y ¼ @ @x ðx3 þ 2xy exy2 Þ ¼ 3x2 þ 2xy exy2 y2 þ exy2 2y ¼ 3x2 þ 2xy3 exy2 þ 2y exy2 Note that xy ¼ yx in this case. This is because the second partial derivatives exist and are continuous for all ðx; yÞ in a region r. When this is not true we may have xy 6¼ yx (see Problem 6.41, for example). 6.11. Show that Uðx; y; zÞ ¼ ðx2 þ y2 þ z2 Þ1=2 satisfies Laplace’s partial differential equation @2 U @x2 þ @2 U @y2 þ @2 U @z2 ¼ 0. We assume here that ðx; y; zÞ 6¼ ð0; 0; 0Þ. Then @U @x ¼ 1 2 ðx2 þ y2 þ z2 Þ3=2 2x ¼ xðx2 þ y2 þ z2 Þ3=2 @2 U @x2 ¼ @ @x ½xðx2 þ y2 þ z2 Þ3=2 ¼ ðxÞ½ 3 2 ðx2 þ y2 þ z2 Þ5=2 2x þ ðx2 þ y2 þ z2 Þ3=2 ð1Þ ¼ 3x2 ðx2 þ y2 þ z2Þ5=2 ðx2 þ y2 þ z2 Þ ðx2 þ y2 þ z2Þ5=2 ¼ 2x2 y2 z2 ðx2 þ y2 þ z2Þ5=2 @2 U @y2 ¼ 2y2 x2 z2 ðx2 þ y2 þ z2Þ5=2 ; @2 U @x2 ¼ 2z2 x2 y2 ðx2 þ y2 þ z2Þ5=2 : Similarly @2 U @x2 þ @2 U @y2 þ @2 U @z2 ¼ 0: Adding, 6.12. If z ¼ x2 tan1 y x , find @2 z @x @y at ð1; 1Þ. @z @y ¼ x2 1 1 þ ð y=xÞ2 @ @y y x ¼ x2 x2 x2 þ y2 1 x ¼ x3 x2 þ y2 CHAP. 6] PARTIAL DERIVATIVES 129
- 139. @2 z @x @y ¼ @ @x @z @y ¼ @ @x x3 x2 þ y2 ! ¼ ðx2 þ y2 Þð3x2 Þ ðx3 Þð2xÞ ðx2 þ y2Þ2 ¼ 2 3 1 2 22 ¼ 1 at ð1; 1Þ The result can be written zxyð1; 1Þ ¼ 1: Note: In this calculation we are using the fact that zxy is continuous at ð1; 1Þ (see remark at the end of Problem 6.9). 6.13. If f ðx; yÞ is defined in a region r and if fxy and fyx exist and are continuous at a point of r, prove that fxy ¼ fyx at this point. Let ðx0; y0Þ be the point of r. Consider G ¼ f ðx0 þ h; y0 þ kÞ f ðx0; y0 þ kÞ f ðx0 þ h; y0Þ þ f ðx0; y0Þ Define (1) ðx; yÞ ¼ f ðx þ h; yÞ f ðx; yÞ (2) ðx; yÞ ¼ f ðx; y þ kÞ f ðx; yÞ Then (3) G ¼ ðx0; y0 þ kÞ ðx0; y0Þ (4) G ¼ ðx0 þ h; y0Þ ðx0; y0Þ Applying the mean value theorem for functions of one variable (see Page 72) to (3) and (4), we have (5) G ¼ kyðx0; y0 þ 1kÞ ¼ kf fyðx0 þ h; y0 þ 1kÞ fyðx0; y0 þ 1kÞg 0 1 1 (6) G ¼ h xðx0 þ 2h; y0Þ ¼ hf fxðx0 þ 2h; y0 þ kÞ fxðx0 þ 2h; y0Þg 0 2 1 Applying the mean value theorem again to (5) and (6), we have (7) G ¼ hk fyxðx0 þ 3h; y0 þ 1kÞ 0 1 1; 0 3 1 (8) G ¼ hk fxyðx0 þ 2h; y0 þ 4kÞ 0 2 1; 0 4 1 From (7) and (8) we have ð9Þ fyxðx0 þ 3h; y0 þ 1kÞ ¼ fxyðx0 þ 2h; y0 þ 4kÞ Letting h ! 0 and k ! 0 in (9) we have, since fxy and fyx are assumed continuous at ðx0; y0Þ, fyxðx0; y0Þ ¼ fxyðx0; y0Þ as required. For example where this fails to hold, see Problem 6.41. DIFFERENTIALS 6.14. Let f ðx; yÞ have continuous first partial derivatives in a region r of the xy plane. Prove that f ¼ f ðx þ x; y þ yÞ f ðx; yÞ ¼ fxx þ fyy þ 1x þ 2y where 1 and 2 approach zero as x and y approach zero. Applying the mean value theorem for functions of one variable (see Page 72), we have ð1Þ f ¼ f f ðx þ x; y þ yÞ f ðx; y þ yÞg þ f f ðx; y þ yÞ f ðx; yÞg ¼ x fxðx þ 1x; y þ yÞ þ y fyðx; y þ 2yÞ 0 1 1; 0 2 1 Since, by hypothesis, fx and fy are continuous, it follows that fxðx þ 1x; y þ yÞ ¼ fxðx; yÞ þ 1; fyðx; y þ 2yÞ ¼ fyðx; yÞ þ 2 where 1 ! 0, 2 ! 0 as x ! 0 and y ! 0. Thus, f ¼ fxx þ fyy þ 1x þ 2y as required. Defining x ¼ dx; y ¼ dy, we have f ¼ fx dx þ fy dy þ 1 dx þ 2 dy: We call df ¼ fx dx þ fy dy the differential of f (or z) or the principal part of f (or z). 6.15. If z ¼ f ðx; yÞ ¼ x2 y 3y, find (a) z; ðbÞ dz: ðcÞ Determine z and dz if x ¼ 4, y ¼ 3, x ¼ 0:01, y ¼ 0:02. (d) How might you determine f ð5:12; 6:85Þ without direct computa- tion? 130 PARTIAL DERIVATIVES [CHAP. 6
- 140. Solution: ðaÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ ¼ fðx þ xÞ2 ð y þ yÞ 3ð y þ yÞg fx2 y 3yg ¼ 2xy x þ ðx2 3Þy |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðAÞ þ ðxÞ2 y þ 2x x y þ ðxÞ2 y |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðBÞ The sum (A) is the principal part of z and is the differential of z, i.e., dz. Thus, ðbÞ dz ¼ 2xy x þ ðx2 3Þy ¼ 2xy dx þ ðx2 3Þ dy Another method: dz ¼ @z @x dx þ @z @y dy ¼ 2xy dx þ ðx2 3Þ dy ðcÞ z ¼ f ðx þ x; y þ yÞ f ðx; yÞ ¼ f ð4 0:01; 3 þ 0:02Þ f ð4; 3Þ ¼ fð3:99Þ2 ð3:02Þ 3ð3:02Þg fð4Þ2 ð3Þ 3ð3Þg ¼ 0:018702 dz ¼ 2xy dx þ ðx2 3Þ dy ¼ 2ð4Þð3Þð0:01Þ þ ð43 3Þð0:02Þ ¼ 0:02 Note that in this case z and dz are approximately equal, because x ¼ dx and y ¼ dy are sufficiently small. (d) We must find f ðx þ x; y þ yÞ when x þ x ¼ 5:12 and y ¼ y ¼ 6:85. We can accomplish this by choosing x ¼ 5, x ¼ 0:12, y ¼ 7, y ¼ 0:15. Since x and y are small, we use the fact that f ðx þ x; y þ yÞ ¼ f ðx; yÞ þ z is approximately equal to f ðx; yÞ þ dz, i.e., z þ dz. Now z ¼ f ðx; yÞ ¼ f ð5; 7Þ ¼ ð5Þ2 ð7Þ 3ð7Þ ¼ 154 dz ¼ 2xy dx þ ðx2 3Þ dy ¼ 2ð5Þð7Þð0:12Þ þ ð52 3Þð0:15Þ ¼ 5:1: Then the required value is 154 þ 5:1 ¼ 159:1 approximately. The value obtained by direct com- putation is 159.01864. 6.16. (a) Let U ¼ x2 ey=x . Find dU. (b) Show that ð3x2 y 2y2 Þ dx þ ðx3 4xy þ 6y2 Þ dy can be written as an exact differential of a function ðx; yÞ and find this function. (a) Method 1: @U @x ¼ x2 ey=x y x2 þ 2xey=x ; @U @y ¼ x2 ey=x 1 x dU ¼ @U @x dx þ @U @y dy ¼ ð2xey=x yey=x Þ dx þ xey=x dy Then Method 2: dU ¼ x2 dðey=x Þ þ ey=x dðx2 Þ ¼ x2 ey=x dðy=xÞ þ 2xey=x dx ¼ x2 ey=x x dy y dx x2 þ 2xey=x dx ¼ ð2xey=x yey=x Þ dx þ xey=x dy (b) Method 1: Suppose that ð3x2 y 2y2 Þ dx þ ðx3 4xy þ 6y2 Þ dy ¼ d ¼ @ @x dx þ @ @y dy: Then (1) @ @x ¼ 3x2 y 2y2 ; (2) @ @y ¼ x3 4xy þ 6y2 From (1), integrating with respect to x keeping y constant, we have ¼ x3 y ¼ 2xy2 þ FðyÞ CHAP. 6] PARTIAL DERIVATIVES 131
- 141. where FðyÞ is the ‘‘constant’’ of integration. Substituting this into (2) yields x3 4xy þ F 0 ð yÞ ¼ x3 4xy þ 6y2 from which F 0 ð yÞ ¼ 6y2 ; i.e., Fð yÞ ¼ 2y3 þ c Hence, the required function is ¼ x3 y 2xy2 þ 2y3 þ c, where c is an arbitrary constant. Note that by Theorem 3, Page 122, the existence of such a function is guaranteed, since if P ¼ 3x2 y 2y2 and Q ¼ x3 4xy þ 6y2 , then @P=@y ¼ 3x2 4y ¼ @Q=@x identically. If @P=@y 6¼ @Q=@x this function would not exist and the given expression would not be an exact differential. Method 2: ð3x2 y 2y2 Þ dx þ ðx3 4xy þ 6y2 Þ dy ¼ ð3x2 y dx þ x3 dyÞ ð2y2 dx þ 4xy dyÞ þ 6y2 dy ¼ dðx3 yÞ dð2xy2 Þ þ dð2y3 Þ ¼ dðx3 y 2xy2 þ 2y3 Þ ¼ dðx3 y 2xy2 þ 2y3 þ cÞ Then the required function is x3 y 2xy2 þ 2y3 þ c. This method, called the grouping method, is based on one’s ability to recognize exact differential combinations and is less than Method 1. Naturally, before attempting to apply any method, one should determine whether the given expression is an exact differential by using Theorem 3, Page 122. See Theorem 4, Page 122. DIFFERENTIATION OF COMPOSITE FUNCTIONS 6.17. Let z ¼ f ðx; yÞ and x ¼ ðtÞ, y ¼ ðtÞ where f ; ; are assumed differentiable. Prove dz dt ¼ @z @x dx dt þ @z @y @y dt Using the results of Problem 6.14, we have dz dt ¼ lim t!0 z t ¼ lim t!0 @z @x x t þ @z @y y t þ 1 x t þ 2 y t ¼ @z @x dx dt þ @z @y dy dt since as t ! 0 we have x ! 0; y ! 0; 1 ! 0; 2 ! 0; x t ! dx dt ; y t ! dy dt : 6.18. If z ¼ exy2 , x ¼ t cos t, y ¼ t sin t, computer dz=dt at t ¼ =2. dz dt ¼ @z @x dx dt þ @z @y dy dt ¼ ð y2 exy2 Þðt sin t þ cos tÞ þ ð2xyexy2 Þðt cos t þ sin tÞ: At t ¼ =2; x ¼ 0; y ¼ =2: Then dz dt t¼=2 ¼ ð2 =4Þð=2Þ þ ð0Þð1Þ ¼ 3 =8: Another method. Substitute x and y to obtain z ¼ et3 sin2 t cos t and then differentiate. 6.19. If z ¼ f ðx; yÞ where x ¼ ðu; vÞ and y ¼ ðu; vÞ, prove that ðaÞ @z @u ¼ @z @x @x @u þ @z @y @y @u ; ðbÞ @z @v ¼ @z @x @x @v þ @z @y @y @v : (a) From Problem 6.14, assuming the differentiability of f ; ; , we have @z @u ¼ lim u!0 z u ¼ lim u!0 @z @x x u þ @z @y y u þ 1 x u þ 2 y u ¼ @z @x @x @u þ @z @y @y @u (b) The result is proved as in (a) by replacing u by v and letting v ! 0. 132 PARTIAL DERIVATIVES [CHAP. 6
- 142. 6.20. Prove that dz ¼ @z @x dx þ @z @y dy even if x and y are dependent variables. Suppose x and y depend on three variables u; v; w, for example. Then ð1Þ dx ¼ xu du þ xv dv þ xw dw ð2Þ dy ¼ yu du þ yv dv þ yw dw zx dx þ zy dy ¼ ðzxxu þ zyyuÞ du þ ðzxxv þ zyyvÞ dv þ ðzxxw þ zyywÞ dw Thus, ¼ zu du þ zv dv þ zw dw ¼ dz using obvious generalizations of Problem 6.19. 6.21. If T ¼ x3 xy þ y3 , x ¼ cos , y ¼ sin , find (a) @T=@, (b) @T=@. @T @ ¼ @T @x @x @ þ @T @y @y @ ¼ ð3x2 yÞðcos Þ þ ð3y2 xÞðsin Þ @T @ ¼ @T @x @x @ þ @T @y @y @ ¼ ð3x2 yÞð sin Þ þ ð3y2 xÞð cos Þ This may also be worked by direct substitution of x and y in T. 6.22. If U ¼ z sin y=x where x ¼ 3r2 þ 2s, y ¼ 4r 2s3 , z ¼ 2r2 3s2 , find (a) @U=@r; ðbÞ @U=@s. ðaÞ @U @r ¼ @U @x @x @r þ @U @y @y @r þ @U @z @z @r ¼ z cos y x y x2 ð6rÞ þ z cos y x 1 x ð4Þ þ sin y x ð4rÞ ¼ 6ryz x2 cos y x þ 4z x cos y x þ 4r sin y x ðbÞ @U @s ¼ @U @x @x @s þ @U @y @y @s þ @U @z @z @s ¼ z cos y x y x2 ð2Þ þ z cos y x 1 x ð6s2 Þ þ sin y x ð6sÞ ¼ 2yz x2 cos y x 6s2 z x cos y x 6s sin y x 6.23. If x ¼ cos , y ¼ sin , show that @V @x 2 þ @V @y 2 ¼ @V @ 2 þ 1 2 @V @ 2 . Using the subscript notation for partial derivatives, we have V ¼ Vxx þ Vyy ¼ Vx cos þ Vy sin ð1Þ V ¼ Vxx þ Vyy ¼ Vxð sin Þ þ Vyð cos Þ ð2Þ Dividing both sides of (2) by , we have 1 V ¼ Vx sin þ Vy cos ð3Þ Then from (1) and (3), we have V2 þ 1 2 V2 ¼ ðVx cos þ Vy sin Þ2 þ ðVx sin þ Vy cos Þ2 ¼ V2 x þ V2 y 6.24. Show that z ¼ f ðx2 yÞ, where f is differentiable, satisfies xð@z=@xÞ ¼ 2yð@z=@yÞ. Let x2 y ¼ u. Then z ¼ f ðuÞ. Thus CHAP. 6] PARTIAL DERIVATIVES 133
- 143. @z @x ¼ @z @u @u @x ¼ f 0 ðuÞ 2xy; @z @y ¼ @z @u @u @y ¼ f 0 ðuÞ x2 Then x @z @x ¼ f 0 ðuÞ 2x2 y; 2y @z @y ¼ f 0 ðuÞ 2x2 y and so x @z @x ¼ 2y @z @y : Another method: We have dz ¼ f 0 ðx2 yÞ dðx2 yÞ ¼ f 0 ðx2 yÞð2xy dx þ x2 dyÞ: Also, dz ¼ @z @x dx þ @z @y dy: Then @z @x ¼ 2xy f 0 ðx2 yÞ; @z @y ¼ x3 f 0 ðx2 yÞ. Elimination of f 0 ðx2 yÞ yields x @z @x ¼ 2y @z @y . 6.25. If for all values of the parameter and for some constant p, Fðx; yÞ ¼ p Fðx; yÞ identically, where F is assumed differentiable, prove that xð@F=@xÞ þ yð@F=@yÞ ¼ pF. Let x ¼ u, y ¼ v. Then Fðu; vÞ ¼ p Fðx; yÞ ð1Þ The derivative with respect to of the left side of (1) is @F @ ¼ @F @u @u @ þ @F @v dv @ ¼ @F @u x þ @F @v y The derivative with respect to of the right side of (1) is pp1 F. Then x @F @u þ y @F @v ¼ pp1 F ð2Þ Letting ¼ 1 in (2), so that u ¼ x; v ¼ y, we have xð@F=@xÞ þ yð@F=@yÞ ¼ pF. 6.26. If Fðx; yÞ ¼ x4 y2 sin1 y=x, show that xð@F=@xÞ þ yð@F=@yÞ ¼ 6F. Since Fðx; yÞ ¼ ðxÞ4 ðyÞ2 sin1 y=x ¼ 6 x4 y2 sin1 y=x ¼ 6 Fðx; yÞ, the result follows from Pro- blem 6.25 with p ¼ 6. It can of course also be shown by direct differentiation. 6.27. Prove that Y ¼ f ðx þ atÞ þ gðx atÞ satisfies @2 Y=@t2 ¼ a2 ð@2 Y=@x2 Þ, where f and g are assumed to be at least twice differentiable and a is any constant. Let u ¼ x þ at; v ¼ x at so that Y ¼ f ðuÞ þ gðvÞ. Then if f 0 ðuÞ df =du, g0 ðvÞ dg=dv, @Y @t ¼ @Y @u @u @t þ @Y @v @v @t ¼ a f 0 ðuÞ ag0 ðvÞ; @Y @x ¼ @Y @x @u @x þ @Y @v @v @x ¼ f 0 ðuÞ þ g0 ðvÞ By further differentiation, using the notation f 00 ðuÞ d2 f =du2 , g00 ðvÞ d2 g=dv2 , we have ð1Þ @2 Y @t2 ¼ @Yt @t ¼ @Yt @u @u @t þ @Yt @v @v @t ¼ @ @u fa f 0 ðuÞ a g0 ðvÞgðaÞ þ @ @v fa f 0 ðuÞ a g0 ðvÞg ðaÞ ¼ a2 f 00 ðuÞ þ a2 g00 ðvÞ ð2Þ @2 Y @x2 ¼ @Yx @x ¼ @Yx @u @u @x þ @Yx @v @v @x ¼ @ @u f f 0 ðuÞ þ g0 ðvÞg þ @ @v f f 0 ðuÞ þ g0 ÞðvÞg ¼ f 00 ðuÞ þ g00 ðvÞ Then from (1) and (2), @2 Y=@t2 ¼ a2 ð@2 Y=@x2 Þ. 134 PARTIAL DERIVATIVES [CHAP. 6
- 144. 6.28. If x ¼ 2r s and y ¼ r þ 2s, find @2 U @y @x in terms of derivatives with respect to r and s. Solving x ¼ 2r s, y ¼ r þ 2s for r and s: r ¼ ð2x þ yÞ=5, s ¼ ð2y xÞ=5. Then @r=@x ¼ 2=5, @s=@x ¼ 1=5, @r=@y ¼ 1=5, @s=@y ¼ 2=5. Hence we have @U @x ¼ @U @r @r @x þ @U @s @s @x ¼ 2 5 @U @r 1 5 @U @s @2 U @y @x ¼ @ @y @U @x ¼ @ @r 2 5 @U @r 1 5 @U @s @r @y þ @ @s 2 5 @U @r 1 5 @U @s @s @y ¼ 2 5 @2 U @r2 1 5 @2 U @r @s ! 1 5 þ 2 5 @2 U @s @r 1 5 @2 U @s2 ! 2 5 ¼ 1 25 2 @2 U @r2 þ 3 @2 U @r @s 2 @2 U @s2 ! assuming U has continuous second partial derivatives. IMPLICIT FUNCTIONS AND JACOBIANS 6.29. If U ¼ x3 y, find dU=dt if (1) x5 þ y ¼ t; ð2) x2 þ y3 ¼ t2 . Equations (1) and (2) define x and y as (implicit) functions of t. Then differentiating with respect to t, we have ð3Þ 5x4 ðdx=dtÞ þ dy=t ¼ 1 ð4Þ 2xðdx=dtÞ þ 3y2 ðdy=dtÞ ¼ 2t Solving (3) and (4) simultaneously for dx=dt and dy=dt, dx dt ¼ 1 1 2t 3y2 5x4 1 2x 3y2 ¼ 3y2 2t 15x4y2 2x ; dy dt ¼ 5x4 1 2x 2t 5x4 1 2x 3y2 ¼ 10x4 t 2x 15x4y2 2x Then dU dt ¼ @U @x dx dt þ @U @y dy dt ¼ ð3x2 yÞ 3y2 2t 15x4 y2 2x ! þ ðx3 Þ 10x4 t 2x 15x4 y2 2x ! : 6.30. If Fðx; y; zÞ ¼ 0 defines z as an implicit function of x and y in a region r of the xy plane, prove that (a) @z=@x ¼ Fx=Fz and (b) @z=@y ¼ Fy=Fz, where Fz 6¼ 0. Since z is a function of x and y, dz ¼ @z @x dx þ @z @y dy. Then dF ¼ @F @x dx þ @F @y dy þ @F @z dz ¼ @F @x þ @F @z @z @x dx þ @F @y þ @F @z @z @y dy ¼ 0. Since x and y are independent, we have ð1Þ @F @x þ @F @z @z @x ¼ 0 ð2Þ @F @y þ @F @z @z @y ¼ 0 from which the required results are obtained. If desired, equations (1) and (2) can be written directly. 6.31. If Fðx; y; u; vÞ ¼ 0 and Gðx; y; u; vÞ ¼ 0, find (a) @u=@x; ðbÞ @u=@y; ðcÞ @v=@x; ðdÞ @v=@y. The two equations in general define the dependent variables u and v as (implicit) functions of the independent variables x and y. Using the subscript notation, we have CHAP. 6] PARTIAL DERIVATIVES 135
- 145. ð1Þ dF ¼ Fx dx þ Fy dy þ Fu du þ Fv dv ¼ 0 ð2Þ dG ¼ Gx dx þ Gy dy þ Gu du þ Gv dv ¼ 0 Also, since u and v are functions of x and y, ð3Þ du ¼ ux dx þ uy dy ð4Þ dv ¼ vx dx þ vy dy: Substituting (3) and (4) in (1) and (2) yields ð5Þ dF ¼ ðFx þ Fu ux þ Fv vxÞ dx þ ðFy þ Fu uy þ Fv vyÞ dy ¼ 0 ð6Þ dG ¼ ðGx þ Gu ux þ Gv vxÞ dx þ ðGy þ Gu uy þ Gv vyÞ dy ¼ 0 Since x and y are independent, the coefficients of dx and dy in (5) and (6) are zero. Hence we obtain ð7Þ Fu ux þ Fv vx ¼ Fx Gu ux þ Gv vx ¼ Gx ð8Þ Fu uy þ Fv vy ¼ Fy Gu uy þ Gv vy ¼ Gy Solving (7) and (8) gives ðaÞ ux ¼ @u @x ¼ Fx Fv Gx Gv Fu Fv Gu Gv ¼ @ðF; GÞ @ðx; vÞ @ðF; GÞ @ðu; vÞ ðbÞ vx ¼ @v @x ¼ Fu Fx Gu Gx Fu Fv Gu Gv ¼ @ðF; GÞ @ðu; xÞ @ðF; GÞ @ðu; vÞ ðcÞ uy ¼ @u @y ¼ Fy Fv Gy Gv Fu Fv Gu Gv ¼ @ðF; GÞ @ðy; vÞ @ðF; GÞ @ðu; vÞ ðdÞ vy ¼ @v @y ¼ Fu Fy Gu Gy Fu Fv Gu Gv ¼ @ðF; GÞ @ðu; yÞ @ðF; GÞ @ðu; vÞ The functional determinant Fu Fv Gu Gv , denoted by @ðF; GÞ @ðu; vÞ or J F; G u; v , is the Jacobian of F and G with respect to u and v and is supposed 6¼ 0. Note that it is possible to devise mnemonic rules for writing at once the required partial derivatives in terms of Jacobians (see also Problem 6.33). 6.32. If u2 v ¼ 3x þ y and u 2v2 ¼ x 2y, find (a) @u=@x; ðbÞ @v=@x; ðcÞ @u=@y; ðdÞ @v=@y. Method 1: Differentiate the given equations with respect to x, considering u and v as functions of x and y. Then ð1Þ 2u @u @x @v @x ¼ 3 ð2Þ @u @x 4v @v @x ¼ 1 Solving, @u @x ¼ 1 12v 1 8uv ; @v @x ¼ 2u 3 1 8uv : Differentiating with respect to y, we have ð3Þ 2u @u @y @v @y ¼ 1 ð4Þ @u @y 4v @v @y ¼ 2 Solving, @u @y ¼ 2 4v 1 8uv ; @v @y ¼ 4u 1 1 8uv : We have, of course, assumed that 1 8uv 6¼ 0. Method 2: The given equations are F ¼ u2 v 3x y ¼ 0, G ¼ u 2v2 x þ 2y ¼ 0. Then by Problem 6.31, 136 PARTIAL DERIVATIVES [CHAP. 6
- 146. @u @x ¼ @ðF; GÞ @ðx; vÞ @ðF; GÞ @ðu; vÞ ¼ Fx Fv Gx Gv Fu Fv Gu Gv ¼ 3 1 1 4v 2u 1 1 4v ¼ 1 12v 1 8uv provided 1 8uv 6¼ 0. Similarly, the other partial derivatives are obtained. 6.33. If Fðu; v; w; x; yÞ ¼ 0, Gðu; v; w; x; yÞ ¼ 0, Hðu; v; w; x; yÞ ¼ 0, find ðaÞ @v @y x ; ðbÞ @x @v w ; ðcÞ @w @u y : From 3 equations in 5 variables, we can (theoretically at least) determine 3 variables in terms of the remaining 2. Thus, 3 variables are dependent and 2 are independent. If we were asked to determine @v=@y, we would know that v is a dependent variable and y is an independent variable, but would not know the remaining independent variable. However, the particular notation @v @y x serves to indicate that we are to obtain @v=@y keeping x constant, i.e., x is the other independent variable. (a) Differentiating the given equations with respect to y, keeping x constant, gives ð1Þ Fu uy þ Fv vy þ Fw wy þ Fy ¼ 0 ð2Þ Gu uy þ Gv vy þ Gw wy þ Gy ¼ 0 ð3Þ Hu uy þ Hv vy þ Hw wy þ Hy ¼ 0 Solving simultaneously for vy, we have vy ¼ @v @y x ¼ Fu Fy Fw Gu Gy Gw Hu Hy Hw Fu Fv Fw Gu Gv Gw Hu Hv Hw ¼ @ðF; G; HÞ @ðu; y; wÞ @ðF; G; HÞ @ðu; v; wÞ Equations (1), (2), and (3) can also be obtained by using differentials as in Problem 6.31. The Jacobian method is very suggestive for writing results immediately, as seen in this problem and Problem 6.31. Thus, observe that in calculating @v @y x the result is the negative of the quotient of two Jacobians, the numerator containing the independent variable y, the denominator containing the dependent variable v in the same relative positions. Using this scheme, we have ðbÞ @x @v w ¼ @ðF; G; HÞ @ðv; y; uÞ @ðF; G; HÞ @ðx; y; uÞ ðcÞ @w @u y ¼ @ðF; G; HÞ @ðu; x; vÞ @ðF; G; HÞ @ðw; x; vÞ 6.34. If z3 xz y ¼ 0, prove that @2 z @x @y ¼ 3z2 þ x ð3z2 xÞ3 . Differentiating with respect to x, keeping y constant and remembering that z is the dependent variable depending on the independent variables x and y, we find 3z2 @z @x x @z @x z ¼ 0 and ð1Þ @z @x ¼ z 3z2 x Differentiating with respect to y, keeping x constant, we find 3z2 @z @y x @z @y 1 ¼ 0 and ð2Þ @z @y ¼ 1 3z2 x CHAP. 6] PARTIAL DERIVATIVES 137
- 147. Differentiating (2) with respect to x and using (1), we have @2 z @x @y ¼ 1 ð3z2 xÞ2 6z @z @x 1 ¼ 1 6z½z=ð3z2 xÞ ð3z2 xÞ2 ¼ 3z2 þ x ð3z2 xÞ3 The result can also be obtained by differentiating (1) with respect to y and using (2). 6.35. Let u ¼ f ðx; yÞ and v ¼ gðx; yÞ, where f and g are continuously differentiable in some region r. Prove that a necessary and sufficient condition that there exists a functional relation between u and v of the form ðu; vÞ ¼ 0 is the vanishing of the Jacobian, i.e., @ðu; vÞ @ðx; yÞ ¼ 0 identically. Necessity. We have to prove that if the functional relation ðu; vÞ ¼ 0 exists, then the Jacobian @ðu; vÞ @ðx; yÞ ¼ 0 identically. To do this, we note that d ¼ u du þ v dv ¼ uðux dx þ uy dyÞ þ vðvx dx þ vy dyÞ ¼ ðu ux þ v vxÞ dx þ ðu uy þ v vyÞ dy ¼ 0 ð1Þ u ux þ v vx ¼ 0 ð2Þ u uy þ v vy ¼ 0 Then Now u and v cannot be identically zero since if they were, there would be no functional relation, contrary to hypothesis. Hence it follows from (1) and (2) that ux vx uy vy ¼ @ðu; vÞ @ðx; yÞ ¼ 0 identically. Sufficiency. We have to prove that if the Jacobian @ðu; vÞ @ðx; yÞ ¼ 0 identically, then there exists a functional relation between u and v, i.e., ðu; vÞ ¼ 0. Let us first suppose that both ux ¼ 0 and uy ¼ 0. In this case the Jacobian is identically zero and u is a constant c1, so that the trival functional relation u ¼ c1 is obtained. Let us now assume that we do not have both ux ¼ 0 and uy ¼ 0; for definiteness, assume ux 6¼ 0. We may then, according to Theorem 1, Page 124, solve for x in the equation u ¼ f ðx; yÞ to obtain x ¼ Fðu; yÞ, from which it follows that ð1Þ u ¼ f fFðu; yÞ; yg ð2Þ v ¼ gfFðu; yÞ; yg From these we have respectively, ð3Þ du ¼ ux dx þ uy dy ¼ uxðFu du þ Fy dyÞ þ uy dy ¼ uxFu du þ ðuxFy þ uyÞ dy ð4Þ dv ¼ vx dx þ vy dy ¼ vxðFu du þ Fy dyÞ þ vy dy ¼ vxFu du þ ðvxFy þ vyÞ dy From (3), uxFu ¼ 1 and uxFy þ uy ¼ 0 or (5) Fy ¼ uy=ux. Using this, (4) becomes dv ¼ vxFu du þ fvxðuy=uxÞ þ vyg dy ¼ vxFu du þ uxvy uyvx ux dy: ð6Þ But by hypothesis @ðu; vÞ @ðx; yÞ ¼ ux uy vx vy ¼ uxvy uyvx ¼ 0 identically, so that (6) becomes dv ¼ vxFu du. This means essentially that referring to (2), @v=@y ¼ 0 which means that v is not dependent on y but depends only on u, i.e., v is a function of u, which is the same as saying that the functional relation ðu; vÞ ¼ 0 exists. 6.36. (a) If u ¼ x þ y 1 xy and v ¼ tan1 x þ tan1 y, find @ðu; vÞ @ðx; yÞ . (b) Are u and v functionally related? If so, find the relationship. 138 PARTIAL DERIVATIVES [CHAP. 6
- 148. ðaÞ @ðu; vÞ @ðx; yÞ ¼ ux uy vx vy ¼ 1 þ y2 ð1 xyÞ2 1 þ x2 ð1 xyÞ2 1 1 þ x2 1 1 þ y2 ¼ 0 if xy 6¼ 1: (b) By Problem 6.35, since the Jacobian is identically zero in a region, there must be a functional relation- ship between u and v. This is seen to be tan v ¼ u, i.e., ðu; vÞ ¼ u tan v ¼ 0. We can show this directly by solving for x (say) in one of the equations and then substituting in the other. Thus, for example, from v ¼ tan1 x þ tan1 y we find tan1 x ¼ v tan1 y and so x ¼ tanðv tan1 yÞ ¼ tan v tanðtan1 yÞ 1 þ tan v tanðtan1 yÞ ¼ tan v y 1 þ y tan v Then substituting this in u ¼ ðx þ yÞ=ð1 xyÞ and simplifying, we find u ¼ tan v. 6.37. (a) If x ¼ u v þ w, y ¼ u2 v2 w2 and z ¼ u3 þ v, evaluate the Jacobian @ðx; y; zÞ @ðu; v; wÞ and (b) explain the significance of the non-vanishing of this Jacobian. ðaÞ @ðx; y; zÞ @ðu; v; wÞ ¼ xu xv xw yu yv yw zu zv zw ¼ 1 1 1 2u 2v 2w 3u2 1 0 ¼ 6wu2 þ 2u þ 6u2 v þ 2w (b) The given equations can be solved simultaneously for u; v; w in terms of x; y; z in a region r if the Jacobian is not zero in r. TRANSFORMATIONS, CURVILINEAR COORDINATES 6.38. A region r in the xy plane is bounded by x þ y ¼ 6, x y ¼ 2; and y ¼ 0. (a) Determine the region r0 in the uv plane into which r is mapped under the transformation x ¼ u þ v, y ¼ u v. (b) Compute @ðx; yÞ @ðu; vÞ . (c) Compare the result of (b) with the ratio of the areas of r and r0 . (a) The region r shown shaded in Fig. 6-9(a) below is a triangle bounded by the lines x þ y ¼ 6, x y ¼ 2, and y ¼ 0 which for distinguishing purposes are shown dotted, dashed, and heavy respectively. Under the given transformation the line x þ y ¼ 6 is transformed into ðu þ vÞ þ ðu vÞ ¼ 6, i.e., 2u ¼ 6 or u ¼ 3, which is a line (shown dotted) in the uv plane of Fig. 6-9(b) above. CHAP. 6] PARTIAL DERIVATIVES 139 Fig. 6-9
- 149. Similarly, x y ¼ 2 becomes ðu þ vÞ ðu vÞ ¼ 2 or v ¼ 1, which is a line (shown dashed) in the uv plane. In like manner, y ¼ 0 becomes u v ¼ 0 or u ¼ v, which is a line shown heavy in the uv plane. Then the required region is bounded by u ¼ 3, v ¼ 1 and u ¼ v, and is shown shaded in Fig. 6- 9(b). ðbÞ @ðx; yÞ @ðu; vÞ ¼ @x @u @x @v @y @u @y @v ¼ @ @u ðu þ vÞ @ @v ðu þ vÞ @ @u ðu vÞ @ @v ðu vÞ ¼ 1 1 1 1 ¼ 2 (c) The area of triangular region r is 4, whereas the area of triangular region r0 is 2. Hence, the ratio is 4=2 ¼ 2, agreeing with the value of the Jacobian in (b). Since the Jacobian is constant in this case, the areas of any regions r in the xy plane are twice the areas of corresponding mapped regions r0 in the uv plane. 6.39. A region r in the xy plane is bounded by x2 þ y2 ¼ a2 , x2 þ y2 ¼ b2 , x ¼ 0 and y ¼ 0, where 0 a b. (a) Determine the region r0 into which r is mapped under the transformation x ¼ cos , y ¼ sin , where 0, 0 @ 2. (b) Discuss what happens when a ¼ 0. (c) Compute @ðx; yÞ @ð; Þ . (d) Compute @ð; Þ @ðx; yÞ . (a) The region r [shaded in Fig. 6-10(a) above] is bounded by x ¼ 0 (dotted), y ¼ 0 (dotted and dashed), x2 þ y2 ¼ a2 (dashed), x2 þ y2 ¼ b2 (heavy). Under the given transformation, x2 þ y2 ¼ a2 and x2 þ y2 ¼ b2 become 2 ¼ a2 and 2 ¼ b2 or ¼ a and ¼ b respectively. Also, x ¼ 0, a @ y @ b becomes ¼ =2, a @ @ b; y ¼ 0, a @ x @ b becomes ¼ 0, a @ @ b. The required region r0 is shown shaded in Fig. 6-10(b) above. Another method: Using the fact that is the distance from the origin O of the xy plane and is the angle measured from the positive x-axis, it is clear that the required region is given by a @ @ b, 0 @ @ =2 as indicated in Fig. 6-10(b). (b) If a ¼ 0, the region r becomes one-fourth of a circular region of radius b (bounded by 3 sides) while r0 remains a rectangle. The reason for this is that the point x ¼ 0, y ¼ 0 is mapped into ¼ 0, ¼ an indeterminate and the transformation is not one to one at this point which is sometimes called a singular point. 140 PARTIAL DERIVATIVES [CHAP. 6 Fig. 6-10
- 150. ðcÞ @ðx; yÞ @ð; Þ ¼ @ @ ð cos Þ @ @ ð cos Þ @ @ ð sin Þ @ @ ð sin Þ ¼ cos sin sin cos ¼ ðcos2 þ sin2 Þ ¼ (d) From Problem 6.43(b) we have, letting u ¼ , v ¼ , @ðx; yÞ @ð; Þ @ð; Þ @ðx; yÞ ¼ 1 so that, using ðcÞ; @ð; Þ @ðx; yÞ ¼ 1 This can also be obtained by direct differentiation. Note that from the Jacobians of these transformations it is clear why ¼ 0 (i.e., x ¼ 0, y ¼ 0) is a singular point. MEAN VALUE THEOREMS 6.40. Prove the mean value theorem for functions of two variables. Let f ðtÞ ¼ f ðx0 þ ht; y0 þ ktÞ. By the mean value theorem for functions of one variable, Fð1Þ ¼ Fð0Þ ¼ F 0 ðÞ 0 1 ð1Þ If x ¼ x0 þ ht, y ¼ y0 þ kt, then FðtÞ ¼ f ðx; yÞ, so that by Problem 6.17, F 0 ðtÞ ¼ fxðdx=dtÞ þ fyðdy=dtÞ ¼ hfx þ kfy and F 0 ðÞ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ where 0 1. Thus, (1) becomes f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ ¼ h fxðx0 þ h; y0 þ kÞ þ k fyðx0 þ h; y0 þ kÞ ð2Þ where 0 1 as required. Note that (2), which is analogous to (1) of Problem 6.14 where h ¼ x, has the advantage of being more symmetric (and also more useful), since only a single number is involved. MISCELLANEOUS PROBLEMS 6.41. Let f ðx; yÞ ¼ xy x2 y2 x2 þ y2 ! ðx; yÞ 6¼ ð0; 0Þ 0 ðx; yÞ ¼ ð0; 0Þ 8 : . Compute (a) fxð0; 0Þ; ðbÞ fyð0; 0Þ; ðcÞ fxxð0; 0Þ; ðdÞ fyyð0; 0Þ; ðeÞ fxyð0; 0Þ; ð f Þ fyxð0; 0Þ: ðaÞ fxð0; 0Þ ¼ lim h!0 f ðh; 0Þ f ð0; 0Þ h ¼ lim h!0 0 h ¼ 0 ðbÞ fyð0; 0Þ ¼ lim h!0 f ð0; kÞ f ð0; 0Þ k ¼ lim k!0 0 k ¼ 0 If ðx; yÞ 6¼ ð0; 0Þ, fxðx; yÞ ¼ @ @x xy x2 y2 x2 þ y2 ! ( ) ¼ xy 4xy2 ðx2 þ y2Þ2 ! þ y x2 y2 x2 þ y2 ! fyðx; yÞ ¼ @ @y xy x2 y2 x2 þ y2 ! ( ) ¼ xy 4xy2 ðx2 þ y2Þ2 ! þ x x2 y2 x2 þ y2 ! CHAP. 6] PARTIAL DERIVATIVES 141
- 151. Then ðcÞ fxxð0; 0Þ ¼ lim h!0 fxðh; 0Þ fxð0; 0Þ h ¼ lim h!0 0 h ¼ 0 ðdÞ fyyð0; 0Þ ¼ lim k!0 fyð0; kÞ fyð0; 0Þ k ¼ lim k!0 0 k ¼ 0 ðeÞ fxyð0; 0Þ ¼ lim k!0 fxð0; kÞ fxð0; 0Þ k ¼ lim k!0 k k ¼ 1 ð f Þ fyxð0; 0Þ ¼ lim h!0 fyðh; 0Þ fyð0; 0Þ h ¼ lim h!0 h h ¼ 1 Note that fxy 6¼ fyx at ð0; 0Þ. See Problem 6.13. 6.42. Show that under the transformation x ¼ cos , y ¼ sin the equation @2 V @x2 þ @2 V @y2 ¼ 0 becomes @2 V @2 þ 1 @V @ þ 1 2 @2 V @2 ¼ 0. We have ð1Þ @V @x ¼ @V @ @ @x þ @V @ @ @x ð2Þ @V @y ¼ @V @ @ @y þ @V @ @ @y Differentiate x ¼ cos , y ¼ sin with respect to x, remembering that and are functions of x and y 1 ¼ sin @ @x þ cos @ @x ; 0 ¼ cos @ @x þ sin @ @x Solving simultaneously, @ @x ¼ cos ; @ @x ¼ sin ð3Þ Similarly, differentiate with respect to y. Then 0 ¼ sin @ @y þ cos @ @y ; 1 ¼ cos @ @y þ sin @ @y Solving simultaneously, @ @y ¼ sin ; @ @y ¼ cos ð4Þ Then from (1) and (2), ð5Þ @V @x ¼ cos @V @ sin @V @ ð6Þ @V @y ¼ sin @V @ þ cos @V @ Hence @2 V @x2 ¼ @ @x @V @x ¼ @ @ @V @x @ @x þ @ @ @V @x @ @x ¼ @ @ cos @V @ sin @V @ @ @x þ @ @ cos @V @ sin @V @ @ @x ¼ cos @2 V @2 þ sin 2 @V @ sin @2 V @ @ ! ðcos Þ þ sin @V @ þ cos @2 V @ @ cos @V @ sin @2 V @2 ! sin 142 PARTIAL DERIVATIVES [CHAP. 6
- 152. which simplifies to @2 V @x2 ¼ cos2 @2 V @2 þ 2 sin cos 2 @V @ 2 sin cos @2 V @ @ þ sin2 @V @ þ sin2 2 @2 V @2 ð7Þ Similarly, @2 V @y2 ¼ sin2 @2 V @2 2 sin cos 2 @V @ þ 2 sin cos @2 V @ @ þ cos2 @V @ þ cos2 2 @2 V @2 ð8Þ Adding ð7Þ and ð8Þ we find, as required, @2 V @x2 þ @2 V @y2 ¼ @2 V @2 þ 1 @V @ þ 1 2 @2 V @2 ¼ 0: 6.43. (a) If x ¼ f ðu; vÞ and y ¼ gðu; vÞ, where u ¼ ðr; sÞ and v ¼ ðr; sÞ, prove that @ðx; yÞ @ðr; sÞ ¼ @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðr; sÞ . (b) Prove that @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðx; yÞ ¼ 1 provided @ðx; yÞ @ðu; vÞ 6¼ 0, and interpret geometrically. ðaÞ @ðx; yÞ @ðr; sÞ ¼ xr xs yr ys ¼ xuur þ xvvr xuus þ xvvs yuur þ yvvr yuus þ yvvs ¼ xu xv yu yv ur us vr vs ¼ @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðr; sÞ using a theorem on multiplication of determinants (see Problem 6.108). We have assumed here, of course, the existence of the partial derivatives involved. ðbÞ Place r ¼ x; s ¼ y in the result of ðaÞ: Then @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðx; yÞ ¼ @ðx; yÞ @ðx; yÞ ¼ 1: The equations x ¼ f ðu; vÞ, y ¼ gðu; vÞ defines a transformation between points ðx; yÞ in the xy plane and points ðu; vÞ in the uv plane. The inverse transformation is given by u ¼ ðx; yÞ, v ¼ ðx; yÞ. The result obtained states that the Jacobians of these transformations are reciprocals of each other. 6.44. Show that Fðxy; z 2xÞ ¼ 0 satisfies under suitable conditions the equation xð@z=@xÞ yð@z=@yÞ ¼ 2x. What are these conditions? Let u ¼ xy, v ¼ z 2x. Then Fðu; vÞ ¼ 0 and dF ¼ Fu du þ Fv dv ¼ Fuðx dy þ y dxÞ þ Fvðdz 2 dxÞ ¼ 0 ð1Þ Taking z as dependent variable and x and y as independent variables, we have dz ¼ zx dx þ zy dy. Then substituting in (1), we find ð yFu þ Fvzx 2Þ dx þ ðxFu þ FvzyÞ dy ¼ 0 Hence, we have, since x and y are independent, ð2Þ yFu þ Fvzx 2 ¼ 0 ð3Þ xFu þ Fvzy ¼ 0 Solve for Fu in (3) and substitute in (2). Then we obtain the required result xzx yzy ¼ 2x upon dividing by Fv (supposed not equal to zero). The result will certainly be valid if we assume that Fðu; vÞ is continuously differentiable and that Fv 6¼ 0. CHAP. 6] PARTIAL DERIVATIVES 143
- 153. Supplementary Problems FUNCTIONS AND GRAPHS 6.45. If f ðx; yÞ ¼ 2x þ y 1 xy , find (a) f ð1; 3Þ; ðbÞ f ð2 þ h; 3Þ f ð2; 3Þ h ; ðcÞ f ðx þ y; xyÞ. Ans. ðaÞ 1 4 ; ðbÞ 11 5ð3h þ 5Þ ; ðcÞ 2x þ 2y þ xy 1 x2 y xy2 6.46. If gðx; y; zÞ ¼ x2 yz þ 3xy, find (a) gð1; 2; 2Þ; ðbÞ gðx þ 1; y 1; z2 Þ; ðcÞ gðxy; xz; x þ yÞ. Ans. ðaÞ 1; ðbÞ x2 x 2 yz2 þ z2 þ 3xy þ 3y; ðcÞ x2 y2 x2 z xyz þ 3x2 yz 6.47. Give the domain of definition for which each of the following functional rules are defined and real, and indicate this domain graphically. ðaÞ f ðx; yÞ ¼ 1 x2 þ y2 1 ; ðbÞ f ðx; yÞ ¼ lnðx þ yÞ; ðcÞ f ðx; yÞ ¼ sin1 2x y x þ y : Ans: ðaÞ x2 þ y2 6¼ 1; ðbÞ x þ y 0; ðcÞ 2x y x þ y @ 1 6.48. (a) What is the domain of definition for which f ðx; y; zÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ y þ z 1 x2 þ y2 þ z2 1 s is defined and real? (b) Indi- cate this domain graphically. Ans. (a) x þ y þ z @ 1; x2 þ y2 þ z2 1 and x þ y þ z A 1; x2 þ y2 þ z2 1 6.49. Sketch and name the surface in three-dimensional space represented by each of the following. ðaÞ 3x þ 2z ¼ 12; ðdÞ x2 þ z2 ¼ y2 ; ðgÞ x2 þ y2 ¼ 2y; ðbÞ 4z ¼ x2 þ y2 ; ðeÞ x2 þ y2 þ z2 ¼ 16; ðhÞ z ¼ x þ y; ðcÞ z ¼ x2 4y2 ; ð f Þ x2 4y2 4z2 ¼ 36; ðiÞ y2 ¼ 4z; ð jÞ x2 þ y2 þ z2 4x þ 6y þ 2z 2 ¼ 0: Ans. ðaÞ plane, (b) paraboloid of revolution, (c) hyperbolic paraboloid, (d) right circular cone, (e) sphere, ( f ) hyperboloid of two sheets, (g) right circular cylinder, (h) plane, (i) parabolic cylinder, ( j) sphere, center at ð2; 3; 1Þ and radius 4. 6.50. Construct a graph of the region bounded by x2 þ y2 ¼ a2 and x2 þ z2 ¼ a2 , where a is a constant. 6.51. Describe graphically the set of points ðx; y; zÞ such that: (a) x2 þ y2 þ z2 ¼ 1; x2 þ y2 ¼ z2 ; ðbÞ x2 þ y2 z x þ y. 52. The level curves for a function z ¼ f ðx; yÞ are curves in the xy plane defined by f ðx; yÞ ¼ c, where c is any constant. They provide a way of representing the function graphically. Similarly, the level surfaces of w ¼ f ðx; y; zÞ are the surfaces in a rectangular ðxyz) coordinate system defined by f ðx; y; zÞ ¼ c, where c is any constant. Describe and graph the level curves and surfaces for each of the following functions: (a) f ðx; yÞ ¼ lnðx2 þ y2 1Þ; ðbÞ f ðx; yÞ ¼ 4xy; ðcÞ f ðx; yÞ ¼ tan1 y=ðx þ 1Þ; ðdÞ f ðx; yÞ ¼ x2=3 þ y2=3 ; ðeÞ f ðx; y; zÞ ¼ x2 þ 4y2 þ 16z2 ; ð f Þ sinðx þ zÞ=ð1 yÞ: LIMITS AND CONTINUITY 6.53. Prove that (a) lim x!4 y!1 ð3x 2yÞ ¼ 14 and (b) lim ðx;yÞ!ð2;1Þ ðxy 3x þ 4Þ ¼ 0 by using the definition. 6.54. If lim f ðx; yÞ ¼ A and lim gðx; yÞ ¼ B, where lim denotes limit as ðx; yÞ ! ðx0; y0Þ, prove that: (a) lim f f ðx; yÞ þ gðx; yÞg ¼ A þ B; ðbÞ lim f f ðx; yÞ gðx; yÞg ¼ AB. 6.55. Under what conditions is the limit of the quotient of two functions equal to the quotient of their limits? Prove your answer. 144 PARTIAL DERIVATIVES [CHAP. 6
- 154. 6.56. Evaluate each of the following limits where they exist. (a) lim x!1 y!2 3 x þ y 4 þ x 2y ðcÞ lim x!4 y! x2 sin y x ðeÞ lim x!0 y!1 e1=x2 ðy1Þ2 ðgÞ lim x!0þ y!1 x þ y 1 ffiffiffi x p ffiffiffiffiffiffiffiffiffiffiffi 1 y p ðbÞ lim x!0 y!0 3x 2y 2x 3y ðdÞ lim x!0 y!0 x sinðx2 þ y2 Þ x2 þ y2 ð f Þ lim x!0 y!0 2x y x2 þ y2 ðhÞ lim x!2 y!1 sin1 ðxy 2Þ tan1ð3xy 6Þ Ans. (a) 4, (b) does not exist, (c) 8 ffiffiffi 2 p ; ðdÞ 0; ðeÞ 0; ð f Þ does not exist, (g) 0, (h) 1/3 6.57. Formulate a definition of limit for functions of (a) 3, (b) n variables. 6.58. Does lim 4x þ y 3z 2x 5y þ 2z as ðx; y; zÞ ! ð0; 0; 0Þ exist? Justify your answer. 6.59. Investigate the continuity of each of the following functions at the indicated points: ðaÞ x2 þ y2 ; ðx0; y0Þ: ðbÞ x 3x þ 5y ; ð0; 0Þ: ðcÞ ðx2 þ y2 Þ sin 1 x2 þ y2 if ðx; yÞ 6¼ ð0; 0Þ, 0 if ðx; yÞ ¼ ð0; 0Þ; ð0; 0Þ: Ans. (a) continuous, (b) discontinuous, (c) continuous 6.60. Using the definition, prove that f ðx; yÞ ¼ xy þ 6x is continuous at (a) ð1; 2Þ; ðbÞ ðx0; y0Þ. 6.61. Prove that the function of Problem 6.60 is uniformly continuous in the square region defined by 0 @ x @ 1, 0 @ y @ 1. PARTIAL DERIVATIVES 6.62. If f ðx; yÞ ¼ x y x þ y , find (a) @f =@x and (b) @f =@y at ð2; 1Þ from the definition and verify your answer by differentiation rules. Ans. (a) 2; ðbÞ 4 6.63. If f ðx; yÞ ¼ ðx2 xyÞ=ðx þ yÞ for ðx; yÞ 6¼ ð0; 0Þ 0 for ðx; yÞ ¼ ð0; 0Þ , find (a) fxð0; 0Þ; ðbÞ fyð0; 0Þ. Ans. (a) 1, (b) 0 6.64. Investigate lim ðx;yÞ!ð0;0Þ fxðx; yÞ for the function in the preceding problem and explain why this limit (if it exists) is or is not equal to fxð0; 0Þ: 6.65. If f ðx; yÞ ¼ ðx yÞ sinð3x þ 2yÞ, compute (a) fx; ðbÞ fy; ðcÞ fxx; ðdÞ fyy; ð f Þ fyx at ð0; =3Þ. Ans. (a) 1 2 ð þ ffiffiffi 3 p Þ; ðbÞ 1 6 ð2 3 ffiffiffi 3 p Þ; ðcÞ 3 2 ð ffiffiffi 3 p 2Þ; ðdÞ 2 3 ði ffiffiffi 3 p þ 3Þ; ðeÞ 1 2 ð2 ffiffiffi 3 p þ 1Þ, ( f ) 1 2 ð2 ffiffiffi 3 p þ 1Þ 6.66. (a) Prove by direct differentiation that z ¼ xy tanðy=xÞ satisfies the equation xð@z=@xÞ þ yð@z=@yÞ ¼ 2z if ðx; yÞ 6¼ ð0; 0Þ. (b) Discuss part (a) for all other points ðx; yÞ assuming z ¼ 0 at ð0; 0Þ. 6.67. Verify that fxy ¼ fyx for the functions (a) ð2x yÞ=ðx þ yÞ, (b) x tan xy; and (c) coshðy þ cos xÞ, indicating possible exceptional points and investigate these points. 6.68. Show that z ¼ lnfðx aÞ2 þ ðy bÞ2 g satisfies @2 z=@x2 þ @2 z=@y2 ¼ 0 except at ða; bÞ. 6.69. Show that z ¼ x cosðy=xÞ þ tanðy=xÞ satisfies x2 zxx þ 2xyzxy þ y2 zyy ¼ 0 except at points for which x ¼ 0. 6.70. Show that if w ¼ x y þ z x þ y z n , then: CHAP. 6] PARTIAL DERIVATIVES 145
- 155. ðaÞ x @w @x þ y @w @y þ z @w @z ¼ 0; ðbÞ x2 @2 w @x2 þ y2 @2 w @y2 þ z2 @2 w @z2 þ 2xy @2 w @x @y þ 2xz @2 w @x @z þ 2yz @2 w @y @z ¼ 0: Indicate possible exceptional points. DIFFERENTIALS 6.71. If z ¼ x3 xy þ 3y2 , compute (a) z and (b) dz where x ¼ 5, y ¼ 4, x ¼ 0:2, y ¼ 0:1. Explain why z and dz are approximately equal. (c) Find z and dz if x ¼ 5, y ¼ 4, x ¼ 2, y ¼ 1. Ans. (a) 11:658; ðbÞ 12:3; ðcÞ z ¼ 66; dz ¼ 123 6.72. Computer ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð3:8Þ2 þ 2ð2:1Þ3 5 q approximately, using differentials. Ans. 2.01 6.73. Find dF and dG if (a) Fðx; yÞ ¼ x3 y 4xy2 þ 8y3 ; ðbÞ Gðx; y; zÞ ¼ 8xy2 z3 3x2 yz, (c) Fðx; yÞ ¼ xy2 lnðy=xÞ. Ans. ðaÞ ð3x2 y 4y2 Þ dx þ ðx3 8xy þ 24y2 Þ dy ðbÞ ð8y2 z3 6xyzÞ dx þ ð16xyz3 3x2 zÞ dy þ ð24xy2 z2 3x2 yÞ dz ðcÞ f y2 lnðy=xÞ y2 g dx þ f2xy lnðy=xÞ þ xyg dy 6.74. Prove that (a) dðUVÞ ¼ U dV þ V dU; ðbÞ dðU=VÞ ¼ ðV dU U dVÞ=V2 ; ðcÞ dðln UÞ ¼ ðdUÞ=U, (d) dðtan1 VÞ ¼ ðdVÞ=ð1 þ v2 Þ where U and V are differentiable functions of two or more variables. 6.75. Determine whether each of the following are exact differentials of a function and if so, find the function. ðaÞ ð2xy2 þ 3y cos 3xÞ dx þ ð2x2 y þ sin 3xÞ dy ðbÞ ð6xy y2 Þ dx þ ð2xey x2 Þ dy ðcÞ ðz3 3yÞ dx þ ð12y2 3xÞ dy þ 3xz2 dz Ans. ðaÞ x2 y2 þ y sin 3x þ c; ðbÞ not exact, ðcÞ xz2 þ 4y3 3xy þ c DIFFERENTIATION OF COMPOSITE FUNCTIONS 6.76. (a) If Uðx; y; zÞ ¼ 2x2 yz þ xz2 , x ¼ 2 sin t, y ¼ t2 t þ 1, z ¼ 3et , find dU=dt at t ¼ 0. (b) if Hðx; yÞ ¼ sinð3x yÞ, x3 þ 2y ¼ 2t3 , x y2 ¼ t2 þ 3t, find dH=dt. Ans: ðaÞ 24; ðbÞ 36t2 y þ 12t þ 9x2 6t2 þ 6x2 t þ 18 6x2y þ 2 ! cosð3x yÞ 6.77. If Fðx; yÞ ¼ ð2x þ yÞ=ðy 2xÞ, x ¼ 2u 3v, y ¼ u þ 2v, find (a) @F=@u; ðbÞ @F=@v; ðcÞ @2 F=@u2 , (d) @2 F=@v2 ; ðeÞ @2 F=@u @v, where u ¼ 2, v ¼ 1. Ans. (a) 7, (b) 14; ðcÞ 21; ðdÞ 112; ðeÞ 49 6.78. If U ¼ x2 Fðy=xÞ, show that under suitable restrictions on F, xð@U=@xÞ þ yð@U=@yÞ ¼ 2U. 6.79. If x ¼ u cos v sin and y ¼ u sin þ v cos , where is a constant, show that ð@V=@xÞ2 þ ð@V=@yÞ2 ¼ ð@V=@uÞ2 þ ð@V=@vÞ2 6.80. Show that if x ¼ cos , y ¼ sin , the equations @u @x ¼ @v @y ; @u @y ¼ @v @x becomes @u @ ¼ 1 @v @ ; @v @ ¼ 1 @u @ 6.81. Use Problem 6.80 to show that under the transformation x ¼ cos , y ¼ sin , the equation @2 u @x2 þ @2 u @y2 ¼ 0 becomes @2 u @2 þ 1 @u @ þ 1 2 @2 u @2 ¼ 0 IMPLICIT FUNCTIONS AND JACOBIANS 6.82. If Fðx; yÞ ¼ 0, prove that dy=dx ¼ Fx=Fy. 146 PARTIAL DERIVATIVES [CHAP. 6
- 156. 6.83. Find (a) dy=dx and (b) d2 y=dx2 if x3 þ y3 3xy ¼ 0. Ans. (a) ðy x2 Þ=ðy2 xÞ; ðbÞ 2xy=ðy2 xÞ3 6.84. If xu2 þ v ¼ y3 , 2yu xv3 ¼ 4x, find (a) @u @x ; ðbÞ @v @y . Ans: ðaÞ v3 3xu2 v2 þ 4 6x2 uv2 þ 2y ; ðbÞ 2xu2 þ 3y3 3x2 uv2 þ y 6.85. If u ¼ f ðx; yÞ, v ¼ gðx; yÞ are differentiable, prove that @u @x @x @u þ @v @x @x @v ¼ 1. Explain clearly which variables are considered independent in each partial derivative. 6.86. If f ðx; y; r; sÞ ¼ 0, gðx; y; r; sÞ ¼ 0, prove that @y @r @r @x þ @y @s @s @x ¼ 0, explaining which variables are independent. What notation could you use to indicate the independent variables considered? 6.87. If Fðx; yÞ ¼ 0, show that d2 y dx2 ¼ FxxF2 y 2FxyFxFy þ FyyF2 x F3 y 6.88. Evaluate @ðF; GÞ @ðu; vÞ if Fðu; vÞ ¼ 3u2 uv, Gðu; vÞ ¼ 2uv2 þ v3 . Ans. 24u2 v þ 16uv2 3v3 6.89. If F ¼ x þ 3y2 z3 , G ¼ 2x2 yz, and H ¼ 2z2 xy, evaluate @ðF; G; HÞ @ðx; y; zÞ at ð1; 1; 0Þ. Ans. 10 6.90. If u ¼ sin1 x þ sin1 y and v ¼ x ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 y2 p þ y ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p , determine whether there is a functional relationship between u and v, and if so find it. 6.91. If F ¼ xy þ yz þ zx, G ¼ x2 þ y2 þ z2 , and H ¼ x þ y þ z, determine whether there is a functional relation- ship connecting F, G, and H, and if so find it. Ans. H2 G 2F ¼ 0. 6.92. (a) If x ¼ f ðu; v; wÞ, y ¼ gðu; v; wÞ, and z ¼ hðu; v; wÞ, prove that @ðx; y; zÞ @ðu; v; wÞ @ðu; v; wÞ @ðx; y; wÞ ¼ 1 provided @ðx; y; zÞ @ðu; v; wÞ 6¼ 0. (b) Give an interpretation of the result of (a) in terms of transformations. 6.93. If f ðx; y; zÞ ¼ 0 and gðx; y; zÞ ¼ 0, show that dx @ð f ; gÞ @ðy; zÞ ¼ dy @ð f ; gÞ @ðz; xÞ ¼ dz @ð f ; gÞ @ðx; yÞ giving conditions under which the result is valid. 6.94. If x þ y2 ¼ u, y þ z2 ¼ v, z þ x2 ¼ w, find (a) @x @u ; ðbÞ @2 x @u2 ; ðcÞ @2 x @u @v assuming that the equations define x; y; and z as twice differentiable functions of u, v; and w. Ans: ðaÞ 1 1 þ 8xyz ; ðbÞ 16x2 y 8yz 32x2 z2 ð1 þ 8xyzÞ3 ; ðcÞ 16y2 z 8xz 32x2 y2 ð1 þ 8xyzÞ3 6.95. State and prove a theorem similar to that in Problem 6.35, for the case where u ¼ f ðx; y; zÞ, v ¼ gðx; y; zÞ, w ¼ hðx; y; zÞ. TRANSFORMATIONS, CURVILINEAR COORDINATES 6.96. Given the transformation x ¼ 2u þ v, y ¼ u 3v. (a) Sketch the region r0 of the uv plane into which the region r of the xy plane bounded by x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 1 is mapped under the transformation. (b) Compute @ðx; yÞ @ðu; vÞ . (c) Compare the result of (b) with the ratios of the areas of r and r0 . Ans. (b) 7 CHAP. 6] PARTIAL DERIVATIVES 147
- 157. 6.97. (a) Prove that under a linear transformation x ¼ a1u þ a2v, y ¼ b1u þ b2v (a1b2 a2b1 6¼ 0Þ lines and circles in the xy plane are mapped respectively into lines and circles in the uv plane. (b) Compute the Jacobian J of the transformation and discuss the significance of J ¼ 0. 6.98. Given x ¼ cos u cosh v, y ¼ sin u sinh v. (a) Show that in general the coordinate curves u ¼ a and v ¼ b in the uv plane are mapped into hyperbolas and ellipses, respectively, in the xy plane. (b) Compute @ðx; yÞ @ðu; vÞ . (c) Compute @ðu; vÞ @ðx; yÞ . Ans. (b) sin2 u cosh2 v þ cos2 u sinh2 v; ðcÞ ðsin2 u cosh2 v þ cos2 u sinh2 vÞ1 6.99. Given the transformation x ¼ 2u þ 3v w, y ¼ 2v þ w, z ¼ 2u 2v þ w. (a) Sketch the region r0 of the uvw space into which the region r of the xyz space bounded by x ¼ 0; x ¼ 8; y ¼ 0; y ¼ 4; z ¼ 0; z ¼ 6 is mapped. (b) Compute @ðx; y; zÞ @ðu; v; wÞ . (c) Compare the result of (b) with the ratios of the volumes of r and r0 . Ans. (b) 1 6.100. Given the spherical coordinate transformation x ¼ r sin cos , y ¼ r sin sin , z ¼ r cos , where r A 0, 0 @ @ , 0 @ 2. Describe the coordinate surfaces (a) r ¼ a; ðbÞ ¼ b, and (c) ¼ c, where a; b; c are any constants. Ans. (a) spheres, (b) cones, (c) planes 6.101. (a) Verify that for the spherical coordinate transformation of Problem 6.100, J ¼ @ðx; y; zÞ @ðr; ; Þ ¼ r2 sin . (b) Discuss the case where J ¼ 0. MISCELLANEOUS PROBLEMS 6.102. If FðP; V; TÞ ¼ 0, prove that (a) @P @T V @T @V P ¼ @P @V T ; ðbÞ @P @T V @T @V P @V @P T ¼ 1. These results are useful in thermodynamics, where P; V; T correspond to pressure, volume, and temperature of a physical system. 6.103. Show that Fðx=y; z=yÞ ¼ 0 satisfies xð@z=@xÞ þ yð@z=@yÞ ¼ z. 6.104. Show that Fðx þ y z; x2 þ y2 Þ ¼ 0 satisfies xð@z=@yÞ yð@z=@xÞ ¼ x y. 6.105. If x ¼ f ðu; vÞ and y ¼ gðu; vÞ, prove that @v @x ¼ 1 J @y @u where J ¼ @ðx; yÞ @ðu; vÞ . 6.106. If x ¼ f ðu; vÞ, y ¼ gðu; vÞ, z ¼ hðu; vÞ and Fðx; y; zÞ ¼ 0, prove that @ðy; zÞ @ðu; vÞ dx þ @ðz; xÞ @ðu; vÞ dy þ @ðx; yÞ @ðu; vÞ dz ¼ 0 6.107. If x ¼ ðu; v; wÞ, y ¼ ðu; v; wÞ and u ¼ f ðr; sÞ, v ¼ gðr; sÞ, w ¼ hðr; sÞ, prove that @ðx; yÞ @ðr; sÞ ¼ @ðx; yÞ @ðu; vÞ @ðu; vÞ @ðr; sÞ þ @ðx; yÞ @ðv; wÞ @ðv; wÞ @ðr; sÞ þ @ðx; yÞ @ðw; uÞ @ðw; uÞ @ðr; sÞ 6.108. (a) Prove that a b c d e f g h ¼ ae þ bg af þ bh ce þ dg cf þ dh , thus establishing the rule for the product of two second order determinants referred to in Problem 6.43. (b) Generalize the result of ðaÞ to determinants of 3; 4 . . . . 6.109. If x; y; and z are functions of u; v; and w, while u; v; and w are functions of r; s; and t, prove that 148 PARTIAL DERIVATIVES [CHAP. 6
- 158. @ðx; y; zÞ @ðr; s; tÞ ¼ @ðx; y; zÞ @ðu; v; wÞ @ðu; v; wÞ @ðr; s; tÞ 6.110. Given the equations Fjðx1; . . . ; xm; y1; . . . ; ynÞ ¼ 0 where j ¼ 1; 2; . . . ; n. Prove that under suitable condi- tions on Fj, @yr @xs ¼ @ðF1; F2; . . . ; Fr; . . . ; FnÞ @ð y1; y2; . . . ; xs; . . . ; ynÞ , @ðF1; F2; . . . ; FnÞ @ð y1; y2; . . . ; ynÞ 6.111. (a) If Fðx; yÞ is homogeneous of degree 2, prove that x2 @2 F @x2 þ 2xy @2 F @x @y þ y2 @2 F @y2 ¼ 2F: (b) Illustrate by using the special case Fðx; yÞ ¼ x2 lnðy=xÞ: Note that the result can be written in operator form, using Dx @=@x and Dy @=@y, as ðx Dx þ y DyÞ2 F ¼ 2F. [Hint: Differentiate both sides of equation (1), Problem 6.25, twice with respect to .] 6.112. Generalize the result of Problem 6.11 as follows. If Fðx1; x2; . . . ; xnÞ is homogeneous of degree p, then for any positive integer r, if Dxj @=@xj, ðx1Dx1 þ x2Dx2 þ þ xnDxn Þr F ¼ pðp 1Þ . . . ðp r þ 1ÞF 6.113. (a) Let x and y be determined from u and v according to x þ iy ¼ ðu þ ivÞ3 . Prove that under this transformation the equation @2 @x2 þ @2 @y2 ¼ 0 is transformed into @2 @u2 þ @2 @v2 ¼ 0 (b) Is the result in ða) true if x þ iy ¼ Fðu þ ivÞ? Prove your statements. CHAP. 6] PARTIAL DERIVATIVES 149
- 159. 150 Vectors VECTORS The foundational ideas for vector analysis were formed indepen- dently in the nineteenth century by William Rowen Hamilton and Herman Grassmann. We are indebted to the physicist John Willard Gibbs, who formulated the classical presentation of the Hamilton viewpoint in his Yale lectures, and his student E. B. Wilson, who considered the mathematical material presented in class worthy of organizing as a book (published in 1901). Hamilton was searching for a mathematical language appropriate to a comprehensive exposition of the physical knowledge of the day. His geometric presentation emphasizing magnitude and direction, and compact notation for the entities of the calculus, was refined in the following years to the benefit of expressing Newtonian mechanics, electromagnetic theory, and so on. Grassmann developed an algebraic and more philo- sophic mathematical structure which was not appreciated until it was needed for Riemanian (non- Euclidean) geometry and the special and general theories of relativity. Our introduction to vectors is geometric. We conceive of a vector as a directed line segment PQ ! from one point P called the initial point to another point Q called the terminal point. We denote vectors by boldfaced letters or letters with an arrow over them. Thus PQ ! is denoted by A or ~ A A as in Fig. 7-1. The magnitude or length of the vector is then denoted by jPQ ! j, PQ, jAj or j ~ A Aj. Vectors are defined to satisfy the following geometric properties. GEOMETRIC PROPERTIES 1. Two vectors A and B are equal if they have the same magnitude and direction regardless of their initial points. Thus A ¼ B in Fig. 7-1 above. In other words, a vector is geometrically represented by any one of a class of commonly directed line segments of equal magnitude. Since any one of the class of line segments may be chosen to represent it, the vector is said to be free. In certain circumstances (tangent vectors, forces bound to a point), the initial point is fixed, then the vector is bound. Unless specifically stated, the vectors in this discussion are free vectors. 2. A vector having direction opposite to that of vector A but with the same magnitude is denoted by A [see Fig. 7-2]. Q P A or A B Fig. 7-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 160. 3. The sum or resultant of vectors A and B of Fig. 7-3(a) below is a vector C formed by placing the initial point of B on the terminal point of A and joining the initial point of A to the terminal point of B [see Fig. 7-3(b) below]. The sum C is written C ¼ A þ B. The definition here is equivalent to the parallelogram law for vector addition as indicated in Fig.7-3(c) below. Extensions to sums of more than two vectors are immediate. For example, Fig. 7-4 below shows how to obtain the sum or resultant E of the vectors A, B, C, and D. 4. The difference of vectors A and B, represented by A B, is that vector C which added to B gives A. Equivalently, A B may be defined as A þ ðBÞ. If A ¼ B, then A B is defined as the null or zero vector and is represented by the symbol 0. This has a magnitude of zero but its direction is not defined. The expression of vector equations and related concepts is facilitated by the use of real numbers and functions. In this context, these are called scalars. This special designation arises from application where the scalars represent object that do not have direction, such as mass, length, and temperature. 5. Multiplication of a vector A by a scalar m produces a vector mA with magnitude jmj times the magnitude of A and direction the same as or opposite to that of A according as m is positive or negative. If m ¼ 0, mA ¼ 0, the null vector. ALGEBRAIC PROPERTIES OF VECTORS The following algebraic properties are consequences of the geometric definition of a vector. (See Problems 7.1 and 7.2.) CHAP. 7] VECTORS 151 A _ A Fig. 7-2 B A A A B B C = A + B C = A + B (a) (b) (c) Fig. 7-3 A A C C B B D D E = A + B + C + D Fig. 7-4
- 161. If A, B and C are vectors, and m and n are scalars, then 1. A þ B ¼ B þ A Commutative Law for Addition 2. A þ ðB þ CÞ ¼ ðA þ BÞ þ C Associative Law for Addition 3. mðnAÞ ¼ ðmnÞA ¼ nðmAÞ Associative Law for Multiplication 4. ðm þ nÞA ¼ mA þ nA Distributive Law 5. mðA þ BÞ ¼ mA þ mB Distributive Law Note that in these laws only multiplication of a vector by one or more scalars is defined. On Pages 153 and 154 we define products of vectors. LINEAR INDEPENDENCE AND LINEAR DEPENDENCE OF A SET OF VECTORS A set of vectors, A1; A2; . . . ; Ap, is linearly independent means that a1A1 þ a2A2 þ þ apAp þ þ apAp ¼ 0 if and only if a1 ¼ a2 ¼ ¼ ap ¼ 0 (i.e., the algebraic sum is zero if and only if all the coefficients are zero). The set of vectors is linearly dependent when it is not linearly independent. UNIT VECTORS Unit vectors are vectors having unit length. If A is any vector with length A 0, then A=A is a unit vector, denoted by a, having the same direction as A. Then A ¼ Aa. RECTANGULAR (ORTHOGONAL) UNIT VECTORS The rectangular unit vectors i, j, and k are unit vectors having the direction of the positive x, y, and z axes of a rectangular coordinate system [see Fig. 7-5]. We use right-handed rectangular coordinate systems unless otherwise specified. Such systems derive their name from the fact that a right-threaded screw rotated through 908 from Ox to Oy will advance in the positive z direction. In general, three vectors A, B, and C which have coincident initial points and are not coplanar are said to form a right- handed system or dextral system if a right-threaded screw rotated through an angle less than 1808 from A to B will advance in the direction C [see Fig. 7-6 below]. 152 VECTORS [CHAP. 7 Fig. 7-5 Fig. 7-6
- 162. COMPONENTS OF A VECTOR Any vector A in 3 dimensions can be represented with initial point at the origin O of a rectangular coordinate system [see Fig. 7-7]. Let ðA1; A2; A3Þ be the rectangular coordinates of the terminal point of vector A with initial point at O. The vectors A1i; A2j; and A3k are called the rectangular component vectors, or simply component vec- tors, of A in the x, y; and z directions respectively. A1; A2; and A3 are called the rectangular components, or simply components, of A in the x, y; and z directions respectively. The vectors of the set fi; j; kg are perpendicular to one another, and they are unit vectors. The words orthogonal and normal, respectively, are used to describe these charac- teristics; hence, the set is what is called an orthonormal basis. It is easily shown to be linearly independent. In an n-dimensional space, any set of n linearly indepen- dent vectors is a basis. The further characteristic of a basis is that any vector of the space can be expressed through it. It is the basis representation that provides the link between the geometric and algebraic expressions of vectors and vector concepts. The sum or resultant of A1i; A2j; and A3k is the vector A, so that we can write A ¼ A1i þ A2j þ A3k ð1Þ The magnitude of A is A ¼ jAj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q ð2Þ In particular, the position vector or radius vector r from O to the point ðx; y; zÞ is written r ¼ xi þ yj þ zk ð3Þ and has magnitude r ¼ jrj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 þ z2 p : DOT OR SCALAR PRODUCT The dot or scalar product of two vectors A and B, denoted by A B (read A dot B) is defined as the product of the magnitudes of A and B and the cosine of the angle between them. In symbols, A B ¼ AB cos ; 0 @ @ ð4Þ Assuming that neither A nor B is the zero vector, an immediate consequence of the definition is that A B ¼ 0 if and only if A and B are perpendicular. Note that A B is a scalar and not a vector. The following laws are valid: 1. A B ¼ B A Commutative Law for Dot Products 2. A ðB þ CÞ ¼ A B þ A C Distributive Law 3. mðA BÞ ¼ ðmAÞ B ¼ A ðmBÞ ¼ ðA BÞm, where m is a scalar. 4. i i ¼ j j ¼ k k ¼ 1; i j ¼ j k ¼ k i ¼ 0 5. If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, then A B ¼ A1B1 þ A2B2 þ A3B3 The equivalence of this component form the dot product with the geometric definition 4 follows from the law of cosines. (See Fig. 7-8.) CHAP. 7] VECTORS 153 Fig. 7-7
- 163. In particular, jCj2 ¼ jAj2 þ jBj2 2jAjjBj cos Since C ¼ B A its components are B1 A1; B2 A2; B3 A3 and the square of its magnitude is ðB2 1 þ B2 2 þ B2 3Þ þ ðA2 1 þ A2 2 þ A2 3Þ 2ðA1B1Þ þ A2B2 þ A3B3Þ or jBj2 þ jAj2 2ðA1B1 þ A2B2 þ A3B3Þ When this representation for jC2 j is placed in the original equation and cancellations are made, we obtain A1B1 þ A2B2 þ A3B3 ¼ jAj jBj cos : CROSS OR VECTOR PRODUCT The cross or vector product of A and B is a vector C ¼ A B (read A cross B). The magnitude of A B is defined as the product of the magnitudes of A and B and the sine of the angle between them. The direction of the vector C ¼ A B is perpendicular to the plane of A and B and such that A, B, and C form a right-handed system. In symbols, A B ¼ AB sin u; 0 @ @ ð5Þ where u is a unit vector indicating the direction of A B. If A ¼ B or if A is parallel to B, then sin ¼ 0 and A B ¼ 0. The following laws are valid: 1. A B ¼ B A (Commutative Law for Cross Products Fails) 2. A ðB þ CÞ ¼ A B þ A C Distributive Law 3. mðA BÞ ¼ ðmAÞ B ¼ A ðmBÞ ¼ ðA BÞm, where m is a scalar. Also the following consequences of the definition are important: 4. i i ¼ j j ¼ k k ¼ 0, i j ¼ k; j k ¼ i; k i ¼ j 5. If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, then A B ¼ i j k A1 A2 A3 B1 B2 B3 The equivalence of this component representation (5) and the geometric definition may be seen as follows. Choose a coodinate system such that the direction of the x-axis is that of A and the xy plane is the plane of the vectors A and B. (See Fig. 7-9.) A B ¼ i j k A1 0 0 B1 B2 0 ¼ A1B2k ¼ jAjjBjsine K Then Since this choice of coordinate system places no restrictions on the vectors A and B, the result is general and thus establishes the equivalence. 154 VECTORS [CHAP. 7 Fig. 7-8 Fig. 7-9
- 164. 6. jA Bj ¼ the area of a parallelogram with sides A and B. 7. If A B ¼ 0 and neither A nor B is a null vector, then A and B are parallel. TRIPLE PRODUCTS Dot and cross multiplication of three vectors, A, B, and C may produce meaningful products of the form ðA BÞC; A ðB CÞ; and A ðB CÞ. The following laws are valid: 1. ðA BÞC 6¼ AðB CÞ in general 2. A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ ¼ volume of a parallelepiped having A, B, and C as edges, or the negative of this volume according as A, B, and C do or do not form a right- handed system. If A ¼ A1i þ A2j þ A3k, B ¼ B1i þ B2j þ B3k and C ¼ C1i þ C2j þ C3k, then A ðB CÞ ¼ A1 A2 A3 B1 B2 B3 C1 C2 C3 ð6Þ 3. A ðB CÞ 6¼ ðA BÞ C (Associative Law for Cross Products Fails) 4. A ðB CÞ ¼ ðA CÞB ðA BÞC ðA BÞ C ¼ ðA CÞB ðB CÞA The product A ðB CÞ is called the scalar triple product or box product and may be denoted by ½ABC. The product A ðB CÞ is called the vector triple product. In A ðB CÞ parentheses are sometimes omitted and we write A B C. However, parentheses must be used in A ðB CÞ (see Problem 7.29). Note that A ðB CÞ ¼ ðA BÞ C. This is often expressed by stating that in a scalar triple product the dot and the cross can be interchanged without affecting the result (see Problem 7.26). AXIOMATIC APPROACH TO VECTOR ANALYSIS From the above remarks it is seen that a vector r ¼ xi þ yj þ zk is determined when its 3 components ðx; y; zÞ relative to some coordinate system are known. In adopting an axiomatic approach, it is thus quite natural for us to make the following Definition. A three-dimensional vector is an ordered triplet of real numbers with the following properties. If A ¼ ðA1; A2; A3Þ and B ¼ ðB1; B2; B3Þ then 1. A ¼ B if and only if A1 ¼ B1; A2 ¼ B2; A3 ¼ B3 2. A þ B ¼ ðA1 þ B1; A2 þ B2; A3 þ B3Þ 3. A B ¼ ðA1 B1; A2 B2; A3 B3Þ 4. 0 ¼ ð0; 0; 0Þ 5. mA ¼ mðA1; A2; A3Þ ¼ ðmA1; mA2; mA3Þ In addition, two forms of multiplication are established. 6. A B ¼ A1B1 þ A2B2 þ A3B3 7. Length or magnitude of A ¼ jAj ¼ ffiffiffiffiffiffiffiffiffiffiffi A A p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q 8. A B ¼ ðA2B3 A3B2; A3B1 A1B3; A1B2 A2B1Þ Unit vectors are defined to be ð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þ and then designated by i; j; k, respectively, thereby identifying the components axiomatically introduced with the geometric orthonormal basis elements. If one wishes, this axiomatic formulation (which provides a component representation for vectors) can be used to reestablish the fundamental laws previously introduced geometrically; however, the CHAP. 7] VECTORS 155
- 165. primary reason for introducing this approach was to formalize a component representation of the vectors. It is that concept that will be used in the remainder of this chapter. Note 1: One of the advantages of component representation of vectors is the easy extension of the ideas to all dimensions. In an n-dimensional space, the component representation is AðA1; A2; . . . ; AnÞ An exception is the cross-product which is specifi- cally restricted to three-dimensional space. There are generalizations of the cross-product to higher dimen- sional spaces, but there is no direct extension.) Note 2: The geometric interpretation of a vector endows it with an absolute meaning at any point of space. The component representation (as an ordered triple of numbers) in Euclidean three space is not unique, rather, it is attached to the coordinate system employed. This follows because the components are geometrically interpreted as the projections of the arrow representation on the coordinate directions. Therefore, the projections on the axes of a second coordinate system rotated (for example) from the first one will be different. (See Fig. 7-10.) Therefore, for theories where groups of coordinate systems play a role, a more adequate component defini- tion of a vector is as a collection of ordered triples of numbers, each one identified with a coordinate system of the group, and any two related by a coordinate transformation. This viewpoint is indispensable in New- tonian mechanics, electromagnetic theory, special relativ- ity, and so on. VECTOR FUNCTIONS If corresponding to each value of a scalar u we associate a vector A, then A is called a function of u denoted by AðuÞ. In three dimensions we can write AðuÞ ¼ A1ðuÞi þ A2ðuÞj þ A3ðuÞk. The function concept is easily extended. Thus, if to each point ðx; y; zÞ there corresponds a vector A, then A is a function of ðx; y; zÞ, indicated by Aðx; y; zÞ ¼ A1ðx; y; zÞi þ A2ðx; y; zÞj þ A3ðx; y; zÞk. We sometimes say that a vector function A defines a vector field since it associates a vector with each point of a region. Similarly, ðx; y; zÞ defines a scalar field since it associates a scalar with each point of a region. LIMITS, CONTINUITY, AND DERIVATIVES OF VECTOR FUNCTIONS Limits, continuity, and derivatives of vector functions follow rules similar to those for scalar func- tions already considered. The following statements show the analogy which exists. 1. The vector function represented by AðuÞ is said to be continuous at u0 if given any positive number , we can find some positive number such that jAðuÞ Aðu0Þj whenever ju u0j . This is equivalent to the statement lim u!u0 AðuÞ ¼ Aðu0Þ. 2. The derivative of AðuÞ is defined as dA du ¼ lim u!0 Aðu þ uÞ AðuÞ u provided this limit exists. In case AðuÞ ¼ A1ðuÞi þ A2ðuÞj þ A3ðuÞk; then 156 VECTORS [CHAP. 7 Fig. 7-10
- 166. dA du ¼ dA1 du i þ dA2 du j þ dA3 du k Higher derivatives such as d2 A=du2 , etc., can be similarly defined. 3. If Aðx; y; zÞ ¼ A1ðx; y; zÞi þ A2ðx; y; zÞj þ A3ðx; y; zÞk, then dA ¼ @A @x dx þ @A @y dy þ @A @z dz is the differential of A. 4. Derivatives of products obey rules similar to those for scalar functions. However, when cross products are involved the order may be important. Some examples are ðaÞ d du ðAÞ ¼ dA du þ d du A; ðbÞ @ @y ðA BÞ ¼ A @B @y þ @A @y B; ðcÞ @ @z ðA BÞ ¼ A @B @z þ @A @z B (Maintain the order of A and BÞ GEOMETRIC INTERPRETATION OF A VECTOR DERIVATIVE If r is the vector joining the origin O of a coordinate system and the point ðx; y; zÞ, then specification of the vector function rðuÞ defines x, y; and z as functions of u (r is called a position vector). As u changes, the terminal point of r describes a space curve (see Fig. 7-11) having parametric equations x ¼ xðuÞ; y ¼ yðuÞ; z ¼ zðuÞ. If the parameter u is the arc length s measured from some fixed point on the curve, then recall from the discussion of arc length that ds2 ¼ dr dr. Thus dr ds ¼ T ð7Þ is a unit vector in the direction of the tangent to the curve and is called the unit tangent vector. If u is the time t, then dr dt ¼ v ð8Þ is the velocity with which the terminal point of r describes the curve. We have v ¼ dr dt ¼ dr ds ds dt ¼ ds dt T ¼ vT ð9Þ CHAP. 7] VECTORS 157 Fig. 7-11
- 167. from which we see that the magnitude of v is v ¼ ds=dt. Similarly, d2 r dt2 ¼ a ð10Þ is the acceleration with which the terminal point of r describes the curve. These concepts have important applications in mechanics and differential geometry. A primary objective of vector calculus is to express concepts in an intuitive and compact form. Success is nowhere more apparent than in applications involving the partial differentiation of scalar and vector fields. [Illustrations of such fields include implicit surface representation, fx; y; zðx; yÞ ¼ 0, the electromagnetic potential function ðx; y; zÞ, and the electromagnetic vector field Fðx; y; zÞ.] To give mathematics the capability of addressing theories involving such functions, William Rowen Hamilton and others of the nineteenth century introduced derivative concepts called gradient, divergence, and curl, and then developed an analytic structure around them. An intuitive understanding of these entities begins with examination of the differential of a scalar field, i.e., d ¼ @ @x dx þ @ @y dy þ @ @z dz Now suppose the function is constant on a surface S and that C; x ¼ f1ðtÞ; y ¼ f2 ðtÞ; z ¼ f3ðtÞ is a curve on S. At any point of this curve dr dt ¼ dx dt i þ dy dt j þ dz dt k lies in the tangent plane to the surface. Since this statement is true for every surface curve through a given point, the differential dr spans the tangent plane. Thus, the triple @ @x , @ @y , @ @z represents a vector perpendicualr to S. With this special geometric characteristic in mind we define r ¼ @ @x i þ @ @y j þ @ @z k to be the gradient of the scalar field . Furthermore, we give the symbol r a special significance by naming it del. EXAMPLE 1. If ðx; y; zÞ ¼ 0 is an implicity defined surface, then, because the function always has the value zero for points on it, the condition of constancy is satisfied and r is normal to the surface at any of its points. This allows us to form an equation for the tangent plane to the surface at any one of its points. See Problem 7.36. EXAMPLE 2. For certain purposes, surfaces on which is constant are called level surfaces. In meteorology, surfaces of equal temperature or of equal atmospheric pressure fall into this category. From the previous devel- opment, we see that r is perpendicular to the level surface at any one of its points and hence has the direction of maximum change at that point. The introduction of the vector operator r and the interaction of it with the multiplicative properties of dot and cross come to mind. Indeed, this line of thought does lead to new concepts called divergence and curl. A summary follows. GRADIENT, DIVERGENCE, AND CURL Consider the vector operator r (del) defined by r i @ @x þ j @ @y þ k @ @z ð11Þ Then if ðx; y; zÞ and Aðx; y; zÞ have continuous first partial derivatives in a region (a condition which is in many cases stronger than necessary), we can define the following. 158 VECTORS [CHAP. 7
- 168. 1. Gradient. The gradient of is defined by grad ¼ r ¼ i @ @x þ j @ @y þ k @ @z ¼ i @ @x þ j @ @y þ k @ @z ð12Þ ¼ @ @x i þ @ @y j þ @ @z k 2. Divergence. The divergence of A is defined by div A ¼ r A ¼ i @ @x þ j @ @y þ k @ @z ðA1i þ A2j þ A3kÞ ð13Þ ¼ @A1 @x þ @A2 @y þ @A3 @z 3. Curl. The curl of A is defined by curl A ¼ r A ¼ i @ @x þ j @ @y þ k @ @z ðA1i þ A2j þ A3kÞ ð14Þ ¼ i j k @ @x @ @y @ @z A1 A2 A3 ¼ i @ @y @ @z A2 A3 j @ @x @ @z A1 A2 þ k @ @x @ @y A1 A2 ¼ @A3 @y @A2 @z i þ @A1 @z @A3 @x j þ @A2 @x @A1 @y k Note that in the expansion of the determinant, the operators @=@x; @=@y; @=@z must precede A1; A2; A3. In other words, r is a vector operator, not a vector. When employing it the laws of vector algebra either do not apply or at the very least must be validated. In particular, r A is a new vector obtained by the specified partial differentiation on A, while A r is an operator waiting to act upon a vector or a scalar. FORMULAS INVOLVING r If the partial derivatives of A, B, U, and V are assumed to exist, then 1. rðU þ VÞ ¼ rU þ rV or grad ðU þ VÞ ¼ grad u þ grad V 2. r ðA þ BÞ ¼ r A þ r B or div ðA þ BÞ þ div A þ div B 3. r ðA þ BÞ ¼ r A þ r B or curl ðA þ BÞ ¼ curl A þ curl B 4. r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ 5. r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ 6. r ðA BÞ ¼ B ðr AÞ A ðr BÞ 7. r ðA BÞ ¼ ðB rÞA Bðr AÞ ðA rÞB þ Aðr BÞ 8. rðA BÞ ¼ ðB rÞA þ ðA rÞB þ B ðr AÞ þ A ðr BÞ 9: r ðrUÞ r2 U @2 U @x2 þ @2 U @y2 þ @2 U @z2 is called the Laplacian of U and r2 @2 @x2 þ @2 @y2 þ @2 @z2 is called the Laplacian operator: 10. r ðrUÞ ¼ 0. The curl of the gradient of U is zero. CHAP. 7] VECTORS 159
- 169. 11. r ðr AÞ ¼ 0. The divergence of the curl of A is zero. 12. r ðr AÞ ¼ rðr AÞ r2 A VECTOR INTERPRETATION OF JACOBIANS, ORTHOGONAL CURVILINEAR COORDINATES The transformation equations x ¼ f ðu1; u2; u3Þ; y ¼ gðu1; u2; u3Þ; z ¼ hðu1; u2; u3Þ ð15Þ [where we assume that f ; g; h are continuous, have continuous partial derivatives, and have a single- valued inverse] establish a one-to-one correspondence between points in an xyz and u1u2u3 rectangular coordinate system. In vector notation the transformation (17) can be written r ¼ xi þ yj þ zk ¼ f ðu1; u2; u3Þi þ gðu1; u2; u3Þj þ hðu1; u2; u3Þk ð16Þ A point P in Fig. 7-12 can then be defined not only by rectangular coordinates ðx; y; zÞ but by coordinates ðu1; u2; u3Þ as well. We call ðu1; u2; u3Þ the curvilinear coordinates of the point. If u2 and u3 are constant, then as u1 varies, r describes a curve which we call the u1 coordinate curve. Similarly, we define the u2 and u3 coordinate curves through P. From (16), we have dr ¼ @r @u1 du1 þ @r @u2 du2 þ @r @u3 du3 ð17Þ The collection of vectors @r @x ; @r @y ; @r @z is a basis for the vector structure associated with the curvilinear system. If the curvilinear system is orthogonal, then so is this set; however, in general, the vectors are not unit vectors. he differential form for arc length may be written ds2 ¼ g11ðdu1Þ2 þ g22ðdu2Þ2 þ g33ðdu3 Þ2 where g11 ¼ @r @x @r @x ; g22 ¼ @r @y @r @y ; g33 ¼ @r @z @r @z The vector @r=@u1 is tangent to the u1 coordinate curve at P. If e1 is a unit vector at P in this direction, we can write @r=@u1 ¼ h1e1 where h1 ¼ j@r=@u1j. Similarly we can write @r=@u2 ¼ h2e2 and @r=@u3 ¼ h3e3, where h2 ¼ j@r=@u2j and h3 ¼ j@r=@u3j respectively. Then (17) can be written 160 VECTORS [CHAP. 7 Fig. 7-12
- 170. dr ¼ h1 du1e1 þ h2 du2e2 þ h3 du3e3 ð18Þ The quantities h1; h2; h3 are sometimes caleld scale factors. If e1; e2; e3 are mutually perpendicular at any point P, the curvilinear coordinates are called orthogonal. Since the basis elements are unit vectors as well as orthogonal this is an orthonormal basis. In such case the element of arc length ds is given by ds2 ¼ dr dr ¼ h2 1 du2 1 þ h2 2 du2 2 þ h2 3 du2 3 ð19Þ and corresponds to the square of the length of the diagonal in the above parallelepiped. Also, in the case of othogonal coordinates, referred to the orthonormal basis e1; e2; e3, the volume of the parallelepiped is given by dV ¼ jgjkjdu1du2du3 ¼ jðh1 du1e1Þ ðh2 du2e2Þ ðh3 du3e3Þj ¼ h1h2h3 du1du2du3 ð20Þ which can be written as dV ¼ @r @u1 @r @u2 @r @u3 du1du2du3 ¼ @ðx; y; zÞ @ðu1; u2; u3Þ du1du2du3 ð21Þ where @ðx; y; zÞ=@ðu1; u2; u3Þ is the Jacobian of the transformation. It is clear that when the Jacobian vanishes there is no parallelepiped and explains geometrically the significance of the vanishing of a Jacobian as treated in Chapter 6. Note: The further significance of the Jacobian vanishing is that the transformation degenerates at the point. GRADIENT DIVERGENCE, CURL, AND LAPLACIAN IN ORTHOGONAL CURVILINEAR COORDINATES If is a scalar function and A ¼ A1e1 þ A2e2 þ A3e3 a vector function of orthogonal curvilinear coordinates u1; u2; u3, we have the following results. 1: r ¼ grad ¼ 1 h1 @ @u1 e1 þ 1 h2 @ @u2 e2 þ 1 h3 @ @u3 e3 2: r A ¼ div A ¼ 1 h1h2h3 @ @u1 ðh2; h3A1Þ þ @ @u2 ðh3h1A2Þ þ @ @u3 ðh1h2A3Þ 3: r A ¼ curl A ¼ 1 h1h2h3 h1e1 h2e2 h3e3 @ @u1 @ @u2 @ @u3 h1A1 h2A2 h3A3 4: r2 ¼ Laplacian of ¼ 1 h1h2h3 @ @u1 h2h3 h1 @ @u1 þ @ @u2 h3h1 h2 @ @u2 þ @ @u3 h1h2 h3 @ @u3 These reduce to the usual expressions in rectangular coordinates if we replace ðu1; u2; u3Þ by ðx; y; zÞ, in which case e1; e2; and e3 are replaced by i, j, and k and h1 ¼ h2 ¼ h3 ¼ 1. SPECIAL CURVILINEAR COORDINATES 1. Cylindrical Coordinates (; ; z). See Fig. 7-13. Transformation equations: x ¼ cos ; y ¼ sin ; z ¼ z CHAP. 7] VECTORS 161
- 171. where A 0; 0 @ 2; 1 z 1. Scale factors: h1 ¼ 1; h2 ¼ ; h3 ¼ 1 Element of arc length: ds2 ¼ d2 þ 2 d2 þ dz2 Jacobian : @ðx; y; zÞ @ð; ; zÞ ¼ Element of volume: dV ¼ d d dz Laplacian: r2 U ¼ 1 @ @ @U @ þ 1 2 @2 U @2 þ @2 U @z2 ¼ @2 U @2 þ 1 @U @ þ 1 2 @2 U @2 þ @2 U @z2 Note that corresponding results can be obtained for polar coordinates in the plane by omit- ting z dependence. In such case for example, ds2 ¼ d2 þ 2 d2 , while the element of volume is replaced by the element of area, dA ¼ d d. 2. Spherical Coordinates (r; ; Þ. See Fig. 7-14. Transformation equations: x ¼ r sin cos ; y ¼ r sin sin ; z ¼ r cos where r A 0; 0 @ @ ; 0 @ 2. Scale factors: h1 ¼ 1; h2 ¼ r; h3 ¼ r sin Element of arc length: ds2 ¼ dr2 þ r2 d2 þ r2 sin2 d2 Jacobian : @ðx; y; zÞ @ðr; ; Þ ¼ r2 sin Element of volume: dV ¼ r2 sin dr d d Laplacian: r2 U ¼ 1 r2 @ @r r2 @U @r þ 1 r2 sin @ @ sin @U @ þ 1 r2 sin2 @2 U @2 Other types of coordinate systems are possible. 162 VECTORS [CHAP. 7 Fig. 7-13 Fig. 7-14
- 172. Solved Problems VECTOR ALGEBRA 7.1. Show that addition of vectors is commutative, i.e., A þ B ¼ B þ A. See Fig. 7-15 below. OP þ PQ ¼ OQ or A þ B ¼ C; OR þ RQ ¼ OQ or B þ A ¼ C: and Then A þ B ¼ B þ A. 7.2. Show that the addition of vectors is associative, i.e., A þ ðB þ CÞ ¼ ðA þ BÞ þ C. See Fig. 7-16 above. OP þ PQ ¼ OQ ¼ ðA þ BÞ and PQ þ QR ¼ PR ¼ ðB þ CÞ OP þ PR ¼ OR ¼ D; i:e:; A þ ðB þ CÞ ¼ D Since OQ þ QR ¼ OR ¼ D; i:e:; ðA þ BÞ þ C ¼ D we have A þ ðB þ CÞ ¼ ðA þ BÞ þ C. Extensions of the results of Problems 7.1 and 7.2 show that the order of addition of any number of vectors is immaterial. 7.3. An automobile travels 3 miles due north, then 5 miles northeast as shown in Fig. 7-17. Represent these displace- ments graphically and determine the resultant displacement (a) graphically, (b) analytically. Vector OP or A represents displacement of 3 mi due north. Vector PQ or B represents displacement of 5 mi northeast. Vector OQ or C represents the resultant displacement or sum of vectors A and B, i.e., C ¼ A þ B. This is the triangle law of vector addition. The resultant vector OQ can also be obtained by constructing the diagonal of the parallelogram OPQR having vectors OP ¼ A and OR (equal to vector PQ or B) as sides. This is the parallelo- gram law of vector addition. (a) Graphical Determination of Resultant. Lay off the 1 mile unit on vector OQ to find the magnitude 7.4 mi (approximately). CHAP. 7] VECTORS 163 A A B B P R O Q C = A + B C = B + A Fig. 7-15 (A + B) ( B + C ) A B C D O P Q R Fig. 7-16 N S O P R Q A A B B W E Unit = 1 mile 45° 135° C = A + B Fig. 7-17
- 173. Angle EOQ ¼ 61:58, using a protractor. Then vector OQ has magnitude 7.4 mi and direction 61.58 north of east. (b) Analytical Determination of Resultant. From triangle OPQ, denoting the magnitudes of A; B; C by A; B; C, we have by the law of cosines C2 ¼ A2 þ B2 2AB cos ff OPQ ¼ 32 þ 52 2ð3Þð5Þ cos 1358 ¼ 34 þ 15 ffiffiffi 2 p ¼ 55:21 and C ¼ 7:43 (approximately). By the law of sines, A sin ff OQP ¼ C sin ff OPQ : Then sin ff OQP ¼ A sin ff OPQ C ¼ 3ð0:707Þ 7:43 ¼ 0:2855 and ff OQP ¼ 168350 Thus vector OQ has magnitude 7.43 mi and direction ð458 þ 168350 Þ ¼ 618350 north of east. 7.4. Prove that if a and b are non-collinear, then xa þ yb ¼ 0 implies x ¼ y ¼ 0. Is the set fa; bg linearly independent or linearly dependent? Suppose x 6¼ 0. Then xa þ yb ¼ 0 implies xa ¼ yb or a ¼ ðy=xÞb, i.e., a and b must be parallel to the same line (collinear), contrary to hypothesis. Thus, x ¼ 0; then yb ¼ 0, from which y ¼ 0. The set is linearly independent. 7.5. If x1a þ y1b ¼ x2a þ y2b, where a and b are non-collinear, then x1 ¼ x2 and y1 ¼ y2. x1a þ y1b ¼ x2a þ y2b can be written x1a þ y1b ðx2a þ y2bÞ ¼ 0 or ðx1 x2Þa þ ðy1 y2Þb ¼ 0 Hence, by Problem 7.4, x1 x2 ¼ 0; y1 y2 ¼ 0; or x1 ¼ x2; y1 ¼ y2: Extensions are possible (see Problem 7.49). 7.6. Prove that the diagonals of a parallelogram bisect each other. Let ABCD be the given parallelogram with diagonals intersect- ing at P as shown in Fig. 7-18. Since BD þ a ¼ b; BD ¼ b a. Then BP ¼ xðb aÞ. Since AC ¼ a þ b, AP ¼ yða þ bÞ. But AB ¼ AP þ PB ¼ AP BP, i.e., a ¼ yða þ bÞ xðb aÞ ¼ ðx þ yÞa þ ðy xÞb. Since a and b are non-collinear, we have by Problem 7.5, x þ y ¼ 1 and y x ¼ 0, i.e., x ¼ y ¼ 1 2 and P is the midpoint of both diagonals. 7.7. Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and has half its length. From Fig. 7-19, AC þ CB ¼ AB or b þ a ¼ c. Let DE ¼ d be the line joining the midpoints of sides AC and CB. Then d ¼ DC þ CE ¼ 1 2 b þ 1 2 a ¼ 1 2 ðb þ aÞ ¼ 1 2 c Thus, d is parallel to c and has half its length. 7.8. Prove that the magnitude A of the vector A ¼ A1i þ A2j þ A3k is A ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q . See Fig. 7-20. 164 VECTORS [CHAP. 7 b b A D a a B P C Fig. 7-18
- 174. By the Pythagorean theorem, ðOPÞ2 ¼ ðOQÞ2 þ ðQPÞ2 where OP denotes the magnitude of vector OP, etc. Similarly, ðOQÞ2 ¼ ðORÞ2 þ ðRQÞ2 . Then ðOPÞ2 ¼ ðORÞ2 þ ðRQÞ2 þ ðQPÞ2 or A2 ¼ A2 1 þ A2 2 þ A2 3, i.e., A ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q . 7.9. Determine the vector having initial point Pðx1; y1; z1Þ and terminal point Qðx2; y2; z2Þ and find its magnitude. See Fig. 7-21. The position vector of P is r1 ¼ x1i þ y1j þ z1k. The position vector of Q is r2 ¼ x2i þ y2j þ z2k. r1 ¼ PQ ¼ r2 or PQ ¼ r2 r1 ¼ ðx2i þ y2j þ z2kÞ ðx1i þ y1j þ z1kÞ ¼ ðx2 x1Þi þ ðy2 y1Þj þ ðz2 z1Þk Magnitude of PQ ¼ PQ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx2 x1Þ2 þ ðy2 y1Þ2 þ ðz2 z1Þ2 q : Note that this is the distance between points P and Q. THE DOT OR SCALAR PRODUCT 7.10. Prove that the projection of A on B is equal to A b, where b is a unit vector in the direction of B. Through the initial and terminal points of A pass planes perpen- dicular to B at G and H respectively, as in the adjacent Fig. 7-22: then Projection of A on B ¼ GH ¼ EF ¼ A cos ¼ A b 7.11. Prove A ðB þ CÞ ¼ A B þ A C. See Fig. 7-23. Let a be a unit vector in the direction of A; then Projection of ðB þ CÞ on A ¼ projection of B on A þ projection of C on A ðB þ CÞ a ¼ B a þ C a CHAP. 7] VECTORS 165 C A B D d E a c b b 1 2 a 1 2 Fig. 7-19 Fig. 7-20 A B H G E F G Fig. 7-22 Fig. 7-21 E F G A B C (B + C) Fig. 7-23
- 175. Multiplying by A, ðB þ CÞ Aa ¼ B Aa þ C Aa ðB þ CÞ A ¼ B A þ C A and Then by the commutative law for dot products, A ðB þ CÞ ¼ A B þ A C and the distributive law is valid. 7.12. Prove that ðA þ BÞ ðC þ DÞ ¼ A C þ A D þ B C þ B D. By Problem 7.11, ðA þ BÞ ðC þ DÞ ¼ A ðC þ DÞ þ B ðC þ DÞ ¼ A C þ A D þ B C þ B D. The ordinary laws of algebra are valid for dot products where the operations are defined. 7.13. Evaluate each of the following. ðaÞ i i ¼ jijjij cos 08 ¼ ð1Þð1Þð1Þ ¼ 1 ðbÞ i k ¼ jijjkj cos 908 ¼ ð1Þð1Þð0Þ ¼ 0 ðcÞ k j ¼ jkjjjj cos 908 ¼ ð1Þð1Þð0Þ ¼ 0 ðdÞ j ð2i 3j þ kÞ ¼ 2j i 3j j þ j k ¼ 0 3 þ 0 ¼ 3 ðeÞ ð2i jÞ ð3i þ kÞ ¼ 2i ð3i þ kÞ j ð3i þ kÞ ¼ 6i i þ 2i k 3j i j k ¼ 6 þ 0 0 0 ¼ 6 7.14. If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, prove that A B ¼ A1B1 þ A2B2 þ A3B3. A B ¼ ðA1i þ A2j þ A3kÞ ðB1i þ B2j þ B3kÞ ¼ A1i ðB1i þ B2j þ B3kÞ þ A2j ðB1i þ B2j þ B3kÞ þ A3k ðB1i þ B2j þ B3kÞ ¼ A1B1i i þ A1B2i j þ A1B3i k þ A2B1j i þ A2B2j j þ A2B3j k þ A3B1k i þ A3B2k j þ A3B3k k ¼ A1B1 þ A2B2 þ A3B3 since i j ¼ k k ¼ 1 and all other dot products are zero. 7.15. If A ¼ A1i þ A2j þ A3k, show that A ¼ ffiffiffiffiffiffiffiffiffiffiffi A A p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q . A A ¼ ðAÞðAÞ cos 08 ¼ A2 . Then A ¼ ffiffiffiffiffiffiffiffiffiffiffi A A p . Also, A A ¼ ðA1i þ A2j þ A3kÞ ðA1i þ A2j þ A3kÞ ¼ ðA1ÞðA1Þ þ ðA2ÞðA2Þ þ ðA3ÞðA3Þ ¼ A2 1 þ A2 2 þ A2 3 By Problem 7.14, taking B ¼ A. Then A ¼ ffiffiffiffiffiffiffiffiffiffiffi A A p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 1 þ A2 2 þ A2 3 q is the magnitude of A. Sometimes A A is written A2 . THE CROSS OR VECTOR PRODUCT 7.16. Prove A B ¼ B A. A B ¼ C has magnitude AB sin and direction such that A, B, and C form a right-handed system [Fig. 7-24(a)]. B A ¼ D has magnitude BA sin and direction such that B, A, and D form a right-handed system [Fig. 7-24(b)]. Then D has the same magnitude as C but is opposite in direction, i.e., C ¼ D or A B ¼ B A. The commutative law for cross products is not valid. 7.17. Prove that A ðB þ CÞ ¼ A B þ A C for the case where A is perpendicular to B and also to C. 166 VECTORS [CHAP. 7
- 176. Since A is perpendicular to B, A B is a vector perpendicular to the plane of A and B and having magnitude AB sin 908 ¼ AB or magnitude of AB. This is equivalent to multiplying vector B by A and rotating the resultant vector through 908 to the position shown in Fig. 7-25. Similarly, A C is the vector obtained by multiplying C by A and rotating the resultant vector through 908 to the position shown. In like manner, A ðB þ CÞ is the vector obtained by multiplying B þ C by A and rotating the resultant vector through 908 to the position shown. Since A ðB þ CÞ is the diagonal of the parallelogram with A B and A C as sides, we have A ðB þ CÞ ¼ A B þ A C. 7.18. Prove that A ðB þ CÞ ¼ A B þ A C in the general case where A, B, and C are non- coplanar. See Fig. 7-26. Resolve B into two component vectors, one perpendicular to A and the other parallel to A, and denote them by B? and Bk respectively. Then B ¼ B? þ Bk. If is the angle between A and B, then B? ¼ B sin . Thus the magnitude of A B? is AB sin , the same as the magnitude of A B. Also, the direction of A B? is the same as the direction of A B. Hence A B? ¼ A B. Similarly, if C is resolved into two component vectors Ck and C?, parallel and perpendicular respec- tively to A, then A C? ¼ A C. Also, since B þ C ¼ B? þ Bk þ C? þ Ck ¼ ðB? þ C?Þ þ ðBk þ CkÞ it follows that A ðB? þ C?Þ ¼ A ðB þ CÞ Now B? and C? are vectors perpendicular to A and so by Problem 7.17, A ðB? þ C?Þ ¼ A B? þ A C? A ðB þ CÞ ¼ A B þ A C Then CHAP. 7] VECTORS 167 Fig. 7-24 Fig. 7-25 Fig. 7-26
- 177. and the distributive law holds. Multiplying by 1, using Problem 7.16, this becomes ðB þ CÞ A ¼ B A þ C A. Note that the order of factors in cross products is important. The usual laws of algebra apply only if proper order is maintained. 7.19. (a) If A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k, prove that A B ¼ i j k A1 A2 A3 B1 B2 B3 . A B ¼ ðA1i þ A2j þ A3kÞ ðB1i þ B2j þ B3kÞ ¼ A1i ðB1i þ B2j þ B3kÞ þ A2j ðB1i þ B2j þ B3kÞ þ A3k ðB1i þ B2j þ B3kÞ ¼ A1B1i i þ A1B2i j þ A1B3i k þ A2B1j i þ A2B2j j þ A2B3j k þ A3B1k i þ A3B2k j þ A3B3k k ¼ ðA2B3 A3B2Þi þ ðA3B1 A1B3Þj þ ðA1B2 A2B1Þk ¼ i j k A1 A2 A3 B1 B2 B3 (b) Use the determinant representation to prove the result of Problem 7.18. 7.20. If A ¼ 3i j þ 2k and B ¼ 2i þ 3j k, find A B. A B ¼ i j k 3 1 2 2 3 1 ¼ i 1 2 3 1 j 3 2 2 1 þ k 3 1 2 3 ¼ 5i þ 7j þ 11k 7.21. Prove that the area of a parallelogram with sides A and B is jA Bj. See Fig. 7-27. Area of parallelogram ¼ hjBj ¼ jAj sin jBj ¼ jA Bj Note that the area of the triangle with sides A and B ¼ 1 2 jA Bj. 7.22. Find the area of the triangle with vertices at Pð2; 3; 5Þ; Qð4; 2; 1Þ; Rð3; 6; 4Þ. PQ ¼ ð4 2Þi þ ð2 3Þj þ ð1 5Þk ¼ 2i j 6k PR ¼ ð3 2Þi þ ð6 3Þj þ ð4 5Þk ¼ i þ 3j k Area of triangle ¼ 1 2 jPQ PRj ¼ 1 2 jð2i j6kÞ ði þ 3j kÞj ¼ 1 2 i j k 2 1 6 1 3 1 ¼ 1 2 j19i 4j þ 7kj ¼ 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð19Þ2 þ ð4Þ2 þ ð7Þ2 q ¼ 1 2 ffiffiffiffiffiffiffiffi 426 p 168 VECTORS [CHAP. 7 G A B h Fig. 7-27
- 178. TRIPLE PRODUCTS 7.23. Show that A ðB CÞ is in absolute value equal to the volume of a parallelepiped with sides A, B, and C. See Fig. 7-28. Let n be a unit normal to parallelogram I, having the direction of B C, and let h be the height of the terminal point of A above the parallelogram I. Volume of a parallelepiped ¼ ðheight hÞðarea of parallelogram IÞ ¼ ðA nÞðjB CjÞ ¼ A fjB Cjng ¼ A ðB CÞ If A, B and C do not form a right-handed system, A n 0 and the volume =jA ðB CÞj. 7.24. If A ¼ A1i þ A2j þ A3k, B ¼ B1i þ B2j þ B3k, C ¼ C1i þ C2j þ C3k show that A ðB CÞ ¼ A1 A2 A3 B1 B2 B3 C1 C2 C3 A ðB CÞ ¼ A i j k B1 B2 B3 C1 C2 C3 ¼ ðA1i þ A2j þ A3kÞ ½ðB2C3 B3C2Þi þ ðB3C1 B1C3Þj þ ðB1C2 B2C1Þk ¼ A1ðB2C3 B3C2Þ þ A2ðB3C1 B1C3Þ þ A3ðB1C2 B2C1Þ ¼ A1 A2 A3 B1 B2 B3 C1 C2 C3 : 7.25. Find the volume of a parallelepiped with sides A ¼ 3i j; B ¼ j þ 2k; C ¼ i þ 5j þ 4k. By Problems 7.23 and 7.24, volume of parallelepiped ¼ jA ðB CÞj ¼ j 3 1 0 0 1 2 1 5 4 j ¼ j 20j ¼ 20: 7.26. Prove that A ðB CÞ ¼ ðA BÞ C, i.e., the dot and cross can be interchanged. By Problem 7.24: A ðB CÞ ¼ A1 A2 A3 B1 B2 B3 C1 C2 C3 ; ðA BÞ C ¼ C ðA BÞ ¼ C1 C2 C3 A1 A2 A3 B1 B2 B3 Since the two determinants are equal, the required result follows. 7.27. Let r1 ¼ x1i þ y1j þ z1k, r2 ¼ x2i þ y2j þ z2k and r3 ¼ x3i þ y3j þ z3k be the position vectors of points P1ðx1; y1; z1Þ, P2ðx2; yx; z2Þ and P3ðx3; y3; z3Þ. Find an equation for the plane passing through P1, P2; and P3. See Fig. 7-29. We assume that P1, P2, and P3 do not lie in the same straight line; hence, they determine a plane. Let r ¼ xi þ yj þ zk denote the position vectors of any point Pðx; y; zÞ in the plane. Consider vectors P1P2 ¼ r2 r1, P1P3 ¼ r3 r1 and P1P ¼ r r1 which all lie in the plane. Then P1P P1P2 P1P3 ¼ 0 CHAP. 7] VECTORS 169 Fig. 7-28
- 179. ðr r1Þ ðr2 r1Þ ðr3 r1Þ ¼ 0 or In terms of rectangular coordinates this becomes ½ðx x1Þi þ ðy y1Þj þ ðz z1Þk ½ðx2 x1Þi þ ðy2 y1Þj þ ðz2 z1Þk ½ðx3 x1Þi þ ðy3 y1Þj þ ðz3 z1Þk ¼ 0 or, using Problem 7.24, x x1 y y1 z z1 x2 x1 y2 y1 z2 z1 x3 x1 y3 y1 z3 z1 ¼ 0 7.28. Find an equation for the plane passing through the points P1ð3; 1; 2Þ, P2ð1; 2; 4Þ, P3ð2; 1; 1Þ. The positions vectors of P1; P2; P3 and any point Pðx; y; zÞ on the plane are respectively r1 ¼ 3i þ j 2k; r2 ¼ i þ 2j þ 4k; r3 ¼ 2i j þ k; r ¼ xi þ jj þ zk Then PP1 ¼ r r1, P2P1 ¼ r2 r1, P3P1 ¼ r3 r1, all lie in the required plane and so the required equation is ðr r1Þ ðr2 r1Þ ðr3 r1Þ ¼ 0, i.e., fðx 3Þi þ ðy 1Þj þ ðz þ 2Þkg f4i þ j þ 6kg fi 2j þ 3kg ¼ 0 fðx 3Þi þ ðy 1Þj þ ðz þ 2Þkg f15i þ 6j þ 9kg ¼ 0 15ðx 3Þ þ 6ðy 1Þ þ 9ðz þ 2Þ ¼ 0 or 5x 2y þ 3z ¼ 11 Another method: By Problem 7.27, the required equation is x 3 y 1 z þ 2 1 3 2 1 4 þ 2 2 3 1 1 1 þ 2 ¼ 0 or 5x þ 2y þ 3z ¼ 11 170 VECTORS [CHAP. 7 Fig. 7-29
- 180. 7.29. If A ¼ i þ j, B ¼ 2i 3j þ k, C ¼ 4j 3k, find (a) ðA BÞ C, (b) A ðB CÞ. ðaÞ A B ¼ i j k 1 1 0 2 3 1 ¼ i j 5k. Then ðA BÞ C ¼ i j k 1 1 5 0 4 3 ¼ 23i þ 3j þ 4k: ðbÞ B C ¼ i j k 2 3 1 0 4 3 ¼ 5i þ 6j þ 8k. Then A ðB CÞ ¼ i j k 1 1 0 5 6 8 ¼ 8i 8j þ k: It can be proved that, in general, ðA BÞ C 6¼ A ðB CÞ. DERIVATIVES 7.30. If r ¼ ðt3 þ 2tÞi 3e2t j þ 2 sin 5tk, find (a) dr dt ; ðbÞ dr dt ; ðcÞ d2 r dt2 ; ðdÞ d2 r dt2 at t ¼ 0 and give a possible physical significance. ðaÞ dr dt ¼ d dt ðt3 þ 2tÞi þ d dt ð3e2t Þj þ d dt ð2 sin 5tÞk ¼ ð3t2 þ 2Þi þ 6e2t j þ 10 cos 5tk At t ¼ 0, dr=dt ¼ 2i þ 6j þ 10k ðbÞ From ðaÞ; jdr=dtj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þ2 þ ð6Þ2 þ ð10Þ2 q ¼ ffiffiffiffiffiffiffiffi 140 p ¼ 2 ffiffiffiffiffi 35 p at t ¼ 0: ðcÞ d2 r dt2 ¼ d dt dr dt ¼ d dt fð3t2 þ 2Þi þ 6e2t j þ 10 cos 5tkg ¼ 6ti 12e2t j 50 sin 5tk At t ¼ 0, d2 r=dt2 ¼ 12j. ðdÞ From ðcÞ; jd2 r=dt2 j ¼ 12 at t ¼ 0: If t represents time, these represent respectively the velocity, magnitude of the velocity, acceleration, and magnitude of the acceleration at t ¼ 0 of a particle moving along the space curve x ¼ t3 þ 2t, y ¼ 3e2t , z ¼ 2 sin 5t. 7.31. Prove that d du ðA BÞ ¼ A dB du þ dA du B, where A and B are differentiable functions of u. Method 1: d du ðA BÞ ¼ lim u!0 ðA þ AÞ ðB þ BÞ A B u ¼ lim u!0 A B þ A B þ A B u ¼ lim u!0 A B u þ A u B þ A u B ¼ A dB du þ dA du B Method 2: Let A ¼ A1i þ A2j þ A3k, B þ B1i þ B2j þ B3k. Then d du ðA BÞ ¼ d du ðA1B1 þ A2B2 þ A3B3Þ ¼ A1 dB1 du þ A2 dB2 du þ A3 dB3 du þ dA1 du B1 þ dA2 du B2 þ dA3 du B3 ¼ A dB du þ dA du B CHAP. 7] VECTORS 171
- 181. 7.32. If ðx; y; zÞ ¼ x2 yz and A ¼ 3x2 yi þ yz2 j xzk, find @2 @y @z ðAÞ at the point ð1; 2; 1Þ. A ¼ ðx2 yzÞð3x2 yi þ yz2 j xzkÞ ¼ 3x4 y2 zi þ x2 y2 z3 j x3 yz2 k @ @z ðAÞ ¼ @ @z ð3x4 y2 zi þ x2 y2 z3 j x3 yz2 kÞ ¼ 3x4 y2 i þ 3x2 y2 z2 j 2x3 yzk @2 @y @z ðAÞ ¼ @ @y ð3x4 y2 i þ 3x2 y2 z2 j 2x3 yzkÞ ¼ 6x4 yi þ 6x2 yz2 j 2x3 zk If x ¼ 1, y ¼ 2, z ¼ 1, this becomes 12i 12j þ 2k. 7.33. If A ¼ x2 sin yi þ z2 cos yj xy2 k, find dA. Method 1: @A @x ¼ 2x sin yi y2 k; @A @y ¼ x2 cos yi z2 sin yj 2xyk; @A @z ¼ 2z cos yj dA ¼ @A @x dx þ @A @y dy þ @A @z dz ¼ ð2x sin yi y2 kÞ dx þ ðx2 cos yi z2 sin yj 2xykÞ dy þ ð2z cos yjÞ dz ¼ ð2x sin y dx þ x2 cos y dyÞi þ ð2z cos y dz z2 sin y dyÞj ðy2 dx þ 2xy dyÞk Method 2: dA ¼ dðx2 sin yÞi þ dðz2 cos yÞj dðxy2 Þk ¼ ð2x sin y dx þ x2 cos y dyÞi þ ð2z cos y dz z2 sin y dyÞj ðy2 dx þ 2xy dyÞk GRADIENT, DIVERGENCE, AND CURL 7.34. If ¼ x2 yz3 and A ¼ xzi y2 j þ 2x2 yk, find (a) r; ðbÞ r A; ðcÞ r A; ðdÞ div ðAÞ, (e) curl ðAÞ. ðaÞ r ¼ i @ @x þ j @ @y þ k @ @z ¼ @ @x i þ @ @y j þ @ @z k ¼ @ @x ðx2 yz3 Þi þ @ @y ðx2 yz3 Þj þ @ @z ðx2 yz3 Þk ¼ 2xyz3 i þ x2 z3 j þ 3x2 yz2 k ðbÞ r A ¼ i @ @x þ j @ @y þ k @ @z ðxzi y2 j þ 2x2 ykÞ ¼ @ @x ðxzÞ þ @ @y ðy2 Þ þ @ @z ð2x2 yÞ ¼ z 2y ðcÞ r A ¼ i @ @x þ j @ @y þ k @ @z ðxzi y2 j þ 2x2 ykÞ ¼ i j k @=@x @=@y @=@z xz y2 2x2 y ¼ @ @y ð2x2 yÞ @ @z ðy2 Þ i þ @ @z ðxzÞ @ @x ð2x2 yÞ j þ @ @x ðy2 Þ @ @y ðxzÞ k ¼ 2x2 i þ ðx 4xyÞj 172 VECTORS [CHAP. 7
- 182. ðdÞ div ðAÞ ¼ r ðAÞ ¼ r ðx3 yz4 i x2 y3 z3 j þ 2x4 y2 z3 kÞ ¼ @ @x ðx3 yz4 Þ þ @ @y ðx2 y3 z3 Þ þ @ @z ð2x4 y2 z3 Þ ¼ 3x2 yz4 3x2 y2 z3 þ 6x4 y2 z2 ðeÞ curl ðAÞ ¼ r ðAÞ ¼ r ðx3 yz4 i x2 y3 z3 j þ 2x4 y2 z3 kÞ ¼ i j k @=@x @=@y @=@z x3 yz4 x2 y3 z3 2x4 y2 z3 ¼ ð4x4 yz3 3x2 y3 z2 Þi þ ð4x3 yz3 8x3 y2 z3 Þj ð2xy3 z3 þ x3 z4 Þk 7.35. Prove r ðAÞ ¼ ðrÞ A þ ðr AÞ. r ðAÞ ¼ r ðA1i þ A2j þ A3kÞ ¼ @ @x ðA1Þ þ @ @y ðA2Þ þ @ @z ðA3Þ ¼ @ @x A1 þ @ @y A2 þ @ @z A3 þ @A1 @x þ @A2 @y þ @A3 @z ¼ @ @x i þ @ @y j þ @ @z k ðA1i þ A2j þ A3kÞ þ @ @x i þ @ @y j þ @ @z k ðA1i þ A2j þ A3kÞ ¼ ðrÞ A þ ðr AÞ 7.36. Express a formula for the tangent plane to the surface ðx; y; zÞ ¼ 0 at one of its points P0ðx0; y0; z0Þ. Ans: ðrÞ0 ðr r0Þ ¼ 0 7.37. Find a unit normal to the surface 2x2 þ 4yz 5z2 ¼ 10 at the point Pð3; 1; 2Þ. By Problem 7.36, a vector normal to the surface is rð2x2 þ 4yz 5z2 Þ ¼ 4xi þ 4zj þ ð4y 10zÞk ¼ 12i þ 8j 24k at ð3; 1; 2Þ Then a unit normal to the surface at P is 12i þ 8j 24k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð12Þ2 þ ð8Þ2 þ ð24Þ2 q ¼ 3i þ 2j 6k 7 : Another unit normal to the surface at P is 3i þ 2j 6k 7 : 7.38. If ¼ 2x2 y xz3 , find (a) r and (b) r2 . ðaÞ r ¼ @ @x i þ @ @y j þ @ @z k ¼ ð4xy z3 Þi þ 2x2 j 3xz2 k ðbÞ r2 ¼ Laplacian of ¼ r r ¼ @ @x ð4xy z3 Þ þ @ @y ð2x2 Þ þ @ @z ð3xz2 Þ ¼ 4y 6xz CHAP. 7] VECTORS 173
- 183. Another method: r2 ¼ @2 @x2 þ @2 @y2 þ @2 @z2 ¼ @2 @x2 ð2x2 y xz3 Þ þ @2 @y2 ð2x2 y xz3 Þ þ @2 @z2 ð2x2 y xz3 Þ ¼ 4y 6xz 7.39. Prove div curl A ¼ 0. div curl A ¼ r ðr AÞ ¼ r i j k @=@x @=@y @=@z A1 A2 A3 ¼ r @A3 @y @A2 @z i þ @A1 @z @A3 @x j þ @A2 @x @A1 @y k ¼ @ @x @A3 @y @A2 @z þ @ @y @A1 @z @A3 @x þ @ @z @A2 @x @A1 @y ¼ @2 A3 @x @y @2 A2 @x @z þ @2 A1 @y @z @2 A3 @y @x þ @2 A2 @z @x @2 A1 @z @y ¼ 0 assuming that A has continuous second partial derivatives so that the order of differentiation is immaterial. JACOBIANS AND CURVLINEAR COORDINATES 7.40. Find ds2 in (a) cylindrical and (b) spherical coordinates and determine the scale factors. (a) Method 1: x ¼ cos ; y ¼ sin ; ¼ z dx ¼ sin d þ cos d; dy ¼ cos d þ sin d; dz ¼ dz ds2 ¼ dx2 þ dy2 þ dz2 ¼ ð sin d þ cos dÞ2 Then þ ð cos d þ sin dÞ2 þ ðdzÞ2 ¼ ðdÞ2 þ 2 ðdÞ2 þ ðdzÞ2 ¼ h2 1ðdÞ2 þ h2 2ðdÞ2 þ d2 3 ðdzÞ2 and h1 ¼ h ¼ 1, h2 ¼ h ¼ , h3 ¼ hz ¼ 1 are the scale factors. Method 2: The position vector is r ¼ cos i þ sin j þ zk. Then dr ¼ @r @ d þ @r @ d þ @r @z dz ¼ ðcos i þ sin jÞ d þ ð sin i þ cos jÞ d þ k dz ¼ ðcos d sin dÞi þ ðsin d þ cos dÞj þ k dz Thus ds2 ¼ dr dr ¼ ðcos d sin dÞ2 þ ðsin d þ cos dÞ2 þ ðdzÞ2 ¼ ðdÞ2 þ 2 ðdÞ2 þ ðdzÞ2 x ¼ r sin cos ; y ¼ r sin sin ; z ¼ r cos ðbÞ dx ¼ r sin sin d þ r cos cos d þ sin cos dr Then dy ¼ r sin cos d þ r cos sin d þ sin sin dr dz ¼ r sin d þ cos dr 174 VECTORS [CHAP. 7
- 184. ðdsÞ2 ¼ ðdxÞ2 þ ðdyÞ2 þ ðdzÞ2 ¼ ðdrÞ2 þ r2 ðdÞ2 þ r2 sin2 ðdÞ2 and The scale factors are h1 ¼ hr ¼ 1; h2 ¼ h ¼ r; h3 ¼ h ¼ r sin . 7.41. Find the volume element dV in (a) cylindrical and (b) spherical coordinates and sketch. The volume element in orthogonal curvilinear coordinates u1; u2; u3 is dV ¼ h1h2h3 du1du2du3 ¼ @ðx; y; zÞ @ðu1; u2; u3Þ du1; du2du3 (a) In cylindrical coordinates, u1 ¼ ; u2 ¼ ; u3 ¼ z; h1 ¼ 1; h2 ¼ ; h3 ¼ 1 [see Problem 7.40(a)]. Then dV ¼ ð1ÞðÞð1Þ d d dz ¼ d d dz This can also be observed directly from Fig. 7-30(a) below. (b) In spherical coordinates, u1 ¼ r; u2 ¼ ; u3 ¼ ; h1 ¼ 1; h2 ¼ r; h3 ¼ r sin [see Problem 7.40(b)]. Then dV ¼ ð1ÞðrÞðr sin Þ dr d d ¼ r2 sin dr d d This can also be observed directly from Fig. 7-30(b) above. 7.42. Express in cylindrical coordinates: (a) grad ; ðbÞ div A; ðcÞ r2 . Let u1 ¼ ; u2 ¼ ; u3 ¼ z; h1 ¼ 1; h2 ¼ ; h3 ¼ 1 [see Problem 7.40(a)] in the results 1, 2, and 4 on Pages 174 and 175. Then ðaÞ grad ¼ r ¼ 1 1 @ @ e1 þ 1 @ @ e2 þ 1 1 @ @z e3 ¼ @ @ e1 þ 1 @ @ e2 þ @ @z e3 where e1; e2; e3 are the unit vectors in the directions of increasing ; ; z, respectively. ðbÞ div A ¼ r A ¼ 1 ð1ÞðÞð1Þ @ @ ðÞð1ÞA1 ð Þ þ @ @ ðð1Þð1ÞA2Þ þ @ @z ðð1ÞðÞA3Þ ¼ 1 @ @ ðA1Þ þ @A2 @ þ @A3 @z CHAP. 7] VECTORS 175 Fig. 7-30
- 185. where A ¼ A1e1 þ A2e2 þ A3e3. ðcÞ r2 ¼ 1 ð1ÞðÞð1Þ @ @ ðÞð1Þ ð1Þ @ @ þ @ @ ð1Þð1Þ ðÞ @ @ þ @ @z ð1ÞðÞ ð1Þ @ @z ¼ 1 @ @ @ @ þ 1 2 @2 @2 þ @2 @z2 MISCELLANEOUS PROBLEMS 7.43. Prove that grad f ðrÞ ¼ f 0 ðrÞ r r, where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 þ z2 p and f 0 ðrÞ ¼ df =dr is assumed to exist. grad f ðrÞ ¼ r f ðrÞ ¼ @ @x f ðrÞ i þ @ @y f ðrÞ j þ @ @z f ðrÞ k ¼ f 0 ðrÞ @r @x i þ f 0 ðrÞ @r @y j þ f 0 ðrÞ @r @z k ¼ f 0 ðrÞ x r i þ f 0 ðrÞ y r j þ f 0 ðrÞ z r k ¼ f 0 ðrÞ r ðxi þ yj þ zkÞ ¼ f 0 ðrÞ r r Another method: In orthogonal curvilinear coordinates u1; u2; u3, we have r ¼ 1 h1 @ @u1 e1 þ 1 h2 @ @u2 e2 þ 1 h3 @ @u3 e3 ð1Þ If, in particular, we use spherical coordinates, we have u1 ¼ r; u2 ¼ ; u3 ¼ . Then letting ¼ f ðrÞ, a function of r alone, the last two terms on the right of (1) are zero. Hence, we have, on observing that e1 ¼ r=r and h1 ¼ 1, the result r f ðrÞ ¼ 1 1 @f ðrÞ @r r r ¼ f 0 ðrÞ r r ð2Þ 7.44. (a) Find the Laplacian of ¼ f ðrÞ. (b) Prove that ¼ 1=r is a solution of Laplace’s equation r2 ¼ 0. (a) By Problem 7.43, r ¼ r f ðrÞ ¼ f 0 ðrÞ r r By Problem 7.35, assuming that f ðrÞ has continuous second partial derivatives, we have Laplacian of ¼ r2 ¼ r ðrÞ ¼ r f 0 ðrÞ r r ¼ r f 0 ðrÞ r r þ f 0 ðrÞ r ðr rÞ ¼ 1 r d dr f 0 ðrÞ r r r þ f 0 ðrÞ r ð3Þ ¼ r f 00 ðrÞ f 0 ðrÞ r3 r2 þ 3 f 0 ðrÞ r ¼ f 00 ðrÞ þ 2 r f 0 ðrÞ Another method: In spherical coordinates, we have r2 U ¼ 1 r2 @ @r r2 @U @r þ 1 r2 sin @ @ sin @U @ þ 1 r2 sin2 @2 U @2 If U ¼ f ðrÞ, the last two terms on the right are zero and we find r2 f ðrÞ ¼ 1 r2 d dr ðr2 f 0 ðrÞÞ ¼ f 00 ðrÞ þ 2 r f 0 ðrÞ 176 VECTORS [CHAP. 7
- 186. (b) From the result in part (a), we have r2 1 r ¼ d2 dr2 1 r þ 2 r d dr 1 r ¼ 2 r3 2 r3 ¼ 0 showing that 1=r is a solution of Laplace’s equation. 7.45. A particle moves along a space curve r ¼ rðtÞ, where t is the time measured from some initial time. If v ¼ jdr=dtj ¼ ds=dt is the magnitude of the velocity of the particle (s is the arc length along the space curve measured from the initial position), prove that the acceleration a of the particle is given by a ¼ dv dt T þ v2 N where T and N are unit tangent and normal vectors to the space curve and ¼ d2 r ds2 1 ¼ d2 x ds2 !2 þ d2 y ds2 !2 þ d2 z ds2 !2 8 : 9 = ; 1=2 The velocity of the particle is given by v ¼ vT. Then the acceleration is given by a ¼ dv dt ¼ d dt ðvTÞ ¼ dv dt T þ v dT dt ¼ dv dt T þ v dT ds ds dt ¼ dv dt T þ v2 dT ds ð1Þ Since T has a unit magnitude, we have T T ¼ 1. Then differentiating with respect to s, T dT ds þ dT ds T ¼ 0; 2T dT ds ¼ 0 or T dT ds ¼ 0 from which it follows that dT=ds is perpendicular to T. Denoting by N the unit vector in the direction of dT=ds, and called the principal normal to the space curve, we have dT ds ¼ N ð2Þ where is the magnitude of dT=ds. Now since T ¼ dr=ds [see equation (7), Page 157], we have dT=ds ¼ d2 r=ds2 . Hence ¼ d2 r ds2 ¼ d2 x ds2 !2 þ d2 y ds2 !2 þ d2 z ds2 !2 8 : 9 = ; 1=2 Defining ¼ 1= , (2) becomes dT=ds ¼ N=. Thus from (1) we have, as required, a ¼ dv dt T þ v2 N The components dv=dt and v2 = in the direction of T and N are called the tangential and normal components of the acceleration, the latter being sometimes called the centripetal acceleration. The quantities and are respectively the radius of curvature and curvature of the space curve. CHAP. 7] VECTORS 177
- 187. Supplementary Problems VECTOR ALGEBRA 7.46. Given any two vectors A and B, illustrate geometrically the equality 4A þ 3ðB AÞ ¼ A þ 3B. 7.47. A man travels 25 miles northeast, 15 miles due east, and 10 miles due south. By using an appropriate scale, determine graphically (a) how far and (b) in what direction he is from his starting position. Is it possible to determine the answer analytically? Ans. 33.6 miles, 13.28 north of east. 7.48. If A and B are any two non-zero vectors which do not have the same direction, prove that mA þ nB is a vector lying in the plane determined by A and B. 7.49. If A, B, and C are non-coplanar vectors (vectors which do not all lie in the same plane) and x1A þ y1B þ z1C ¼ x2A þ y2B þ z2C, prove that necessarily x1 ¼ x2; y1 ¼ y2; z1 ¼ z2. 7.50. Let ABCD be any quadrilateral and points P; Q; R; and S the midpoints of successive sides. Prove (a) that PQRS is a parallelogram and (b) that the perimeter of PQRS is equal to the sum of the lengths of the diagonals of ABCD. 7.51. Prove that the medians of a triangle intersect at a point which is a trisection point of each median. 7.52. Find a unit vector in the direction of the resultant of vectors A ¼ 2i j þ k, B ¼ i þ j þ 2k, C ¼ 3i 2j þ 4k. Ans. ð6i 2j þ 7kÞ= ffiffiffiffiffi 89 p THE DOT OR SCALAR PRODUCT 7.53. Evaluate jðA þ BÞ ðA BÞj if A ¼ 2i 3j þ 5k and B ¼ 3i þ j 2k. Ans. 24 7.54. Verify the consistency of the law of cosines for a triangle. [Hint: Take the sides of A; B; C where C ¼ A B. Then use C C ¼ ðA BÞ ðA BÞ.] 7.55. Find a so that 2i 3j þ 5k and 3i þ aj 2k are perpendicular. Ans. a ¼ 4=3 7.56. If A ¼ 2i þ j þ k; B ¼ i 2j þ 2k and C ¼ 3i 4j þ 2k, find the projection of A þ C in the direction of B. Ans. 17/3 7.57. A triangle has vertices at Að2; 3; 1Þ; Bð1; 1; 2Þ; Cð1; 2; 3Þ. Find (a) the length of the median drawn from B to side AC and (b) the acute angle which this median makes with side BC. Ans. (a) 1 2 ffiffiffiffiffi 26 p ; ðbÞ cos1 ffiffiffiffiffi 91 p =14 7.58. Prove that the diagonals of a rhombus are perpendicular to each other. 7.59. Prove that the vector ðAB þ BAÞ=ðA þ BÞ represents the bisector of the angle between A and B. THE CROSS OR VECTOR PRODUCT 7.60. If A ¼ 2i j þ k and B ¼ i þ 2j 3k, find jð2A þ BÞ ðA 2BÞj: Ans. 5 ffiffiffi 3 p 7.61. Find a unit vector perpendicular to the plane of the vectors A ¼ 3i 2j þ 4k and B ¼ i þ j 2k. Ans. ð2j þ kÞ= ffiffiffi 5 p 7.62. If A B ¼ A C, does B ¼ C necessarily? 7.63. Find the area of the triangle with vertices ð2; 3; 1Þ; ð1; 1; 2Þ; ð1; 2; 3Þ. Ans. 1 2 ffiffiffi 3 p 178 VECTORS [CHAP. 7
- 188. 7.64. Find the shortest distance from the point ð3; 2; 1Þ to the plane determine by ð1; 1; 0Þ; ð3; 1; 1Þ; ð1; 0; 2Þ. Ans. 2 TRIPLE PRODUCTS 7.65. If A ¼ 2i þ j 3k; B ¼ i 2j þ k; C ¼ i þ j 4, find (a) A ðB CÞ, (b) C ðA BÞ, (c) A ðB CÞ, (d) ðA BÞ C. Ans. (a) 20, (b) 20, (c) 8i 19j k; ðdÞ 25i 15j 10k 7.66. Prove that ðaÞ A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ ðbÞ A ðB CÞ ¼ BðA CÞ CðA BÞ. 7.67. Find an equation for the plane passing through ð2; 1; 2Þ; ð1; 2; 3Þ; ð4; 1; 0Þ. Ans. 2x þ y 3z ¼ 9 7.68. Find the volume of the tetrahedron with vertices at ð2; 1; 1Þ; ð1; 1; 2Þ; ð0; 1; 1Þ; ð1; 2; 1Þ. Ans. 4 3 7.69. Prove that ðA BÞ ðC DÞ þ ðB CÞ ðA DÞ þ ðC AÞ ðB DÞ ¼ 0: DERIVATIVES 7.70. A particle moves along the space curve r ¼ et cos t i þ et sin t j þ et k. Find the magnitude of the (a) velocity and (b) acceleration at any time t. Ans. (a) ffiffiffi 3 p et ; ðbÞ ffiffiffi 5 p et 7.71. Prove that d du ðA BÞ ¼ A dB du þ dA du B where A and B are differentiable functions of u. 7.72. Find a unit vector tangent to the space curve x ¼ t; y ¼ t2 ; z ¼ t3 at the point where t ¼ 1. Ans. ði þ 2j þ 3kÞ= ffiffiffiffiffi 14 p 7.73. If r ¼ a cos !t þ b sin !t, where a and b are any constant non-collinear vectors and ! is a constant scalar, prove that (a) r ¼ dr dr ¼ !ða bÞ; ðbÞ; d2 r dt2 þ !2 r ¼ 0. 7.74. If A ¼ x2 i yj þ xzk, B ¼ yi þ xj xyzk and C ¼ i yj þ x3 zk, find (a) @2 @x @y ðA BÞ and (b) d½A ðB CÞ at the point ð1; 1; 2Þ: Ans. (a) 4i þ 8j; ðbÞ 8 dx 7.75. If R ¼ x2 yi 2y2 zj þ xy2 z2 k, find @2 B @x2 @2 R @y2 at the point ð2; 1; 2Þ. Ans. 16 ffiffiffi 5 p GRADIENT, DIVERGENCE, AND CURL 7.76. If U; V; A; B have continuous partial derivatives prove that: (a) rðU þ VÞ ¼ rU þ rV; ðbÞ r ðA þ BÞ ¼ r A þ r B; ðcÞ r ðA þ BÞ ¼ r A þ r B. 7.77. If ¼ xy þ yz þ zx and A ¼ x2 yi þ y2 zj þ z2 xk, find (a) A r; ðbÞ r A; and (c) ðrÞ A at the point ð3; 1; 2Þ. Ans: ðaÞ 25; ðbÞ 2; ðcÞ 56i 30j þ 47k 7.78. Show that r ðr2 rÞ ¼ 0 where r ¼ xi þ yj þ zk and r ¼ jrj. 7.79. Prove: (a) r ðUAÞ ¼ ðrUÞ A þ Uðr AÞ; ðbÞ r ðA BÞ ¼ B ðr AÞ A ðr BÞ. 7.80. Prove that curl grad u ¼ 0, stating appropriate conditions on U. 7.81. Find a unit normal to the surface x2 y 2xz þ 2y2 z4 ¼ 10 at the point ð2; 1; 1Þ. Ans: ð3i þ 4j 6kÞ= ffiffiffiffiffi 61 p 7.82. If A ¼ 3xz2 i yzj þ ðx þ 2zÞk, find curl curl A. Ans: 6xi þ ð6z 1Þk 7.83. (a) Prove that r ðr AÞ ¼ r2 A þ rðr AÞ. (b) Verify the result in (a) if A is given as in Problem 7.82. CHAP. 7] VECTORS 179
- 189. JACOBIANS AND CURVINLINEAR COORDINATES 7.84. Prove that @ðx; y; zÞ @ðu1; u2; u3Þ ¼ @r @u1 @r @u2 @r @u3 . 7.85. Express (a) grad ; ðbÞ div A; ðcÞ r2 in spherical coordinates. Ans: ðaÞ @ @r e1 þ 1 r @ @ e2 þ 1 r sin @ @ e3 ðbÞ 1 r2 @ @r ðr2 A1Þ þ 1 r sin @ @ ðsin A2Þ þ 1 r sin @A3 @ where A ¼ A1e1 þ A2e2 þ A3e3 ðcÞ 1 r2 @ @r r2 @ @r þ 1 r2 sin @ sin @ @ þ 1 r2 sin2 @2 @2 7.86. The transformation from rectangular to parabolic cylindrical coordinates is defined by the equations x ¼ 1 2 ðu2 v2 Þ, y ¼ uv, z ¼ z. (a) Prove that the system is orthogonal. (b) Find ds2 and the scale factors. (c) Find the Jacobian of the transformation and the volume element. Ans. ðbÞ ds2 ¼ ðu2 þ v2 Þ du2 þ ðu2 þ v2 Þ dv2 þ dz2 ; h1 ¼ h2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p ; h3 ¼ 1 ðcÞ u2 þ v2 ; ðu2 þ v2 Þ du dv dz 7.87. Write (a) r2 and (b) div A in parabolic cylindrical coordinates. Ans: ðaÞ r2 ¼ 1 u2 þ v2 @2 @u2 þ @2 @v2 ! þ @2 @z2 ðbÞ div A ¼ 1 u2 þ v2 @ @u ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p A1Þ þ @ @v ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p A2Þ þ @A3 @z 7.88. Prove that for orthogonal curvilinear coordinates, r ¼ e1 h1 @ @u1 þ e2 h2 @ @u2 þ e3 h3 @ @u3 [Hint: Let r ¼ a1e1 þ a2e2 þ a3e3 and use the fact that d ¼ r dr must be the same in both rectangular and the curvilinear coordinates.] 7.89. Give a vector interpretation to the theorem in Problem 6.35 of Chapter 6. MISCELLANEOUS PROBLEMS 7.90. If A is a differentiable function of u and jAðuÞj ¼ 1, prove that dA=du is perpendicular to A. 7.91. Prove formulas 6, 7, and 8 on Page 159. 7.92. If and are polar coordinates and A; B; n are any constants, prove that U ¼ n ðA cos n þ B sin nÞ satisfies Laplace’s equation. 7.93. If V ¼ 2 cos þ 3 sin3 cos r2 , find r2 V. Ans. 6 sin cos ð4 5 sin2 Þ r4 7.94. Find the most general function of (a) the cylindrical coordinate , (b) the spherical coordinate r, (c) the spherical coordinate which satisfies Laplace’s equation. Ans. (a) A þ B ln ; ðbÞ A þ B=r; ðcÞ A þ B lnðcsc cot Þ where A and B are any constants. 7.95. Let T and N denote respectively the unit tangent vector and unit principal normal vector to a space curve r ¼ rðuÞ, where rðuÞ is assumed differentiable. Define a vector B ¼ T N called the unit binormal vector to the space curve. Prove that 180 VECTORS [CHAP. 7
- 190. dT ds ¼ N; dB ds ¼ N; dN ds ¼ B T These are called the Frenet-Serret formulas and are of fundamental importance in differential geometry. In these formulas is called the curvature, is called the torsion; and the reciprocals of these, ¼ 1= and ¼ 1= , are called the radius of curvature and radius of torsion, respectively. 7.96. (a) Prove that the radius of curvature at any point of the plane curve y ¼ f ðxÞ; z ¼ 0 where f ðxÞ is differ- entiable, is given by ¼ ð1 þ y02 Þ3=2 y00 (b) Find the radius of curvature at the point ð=2; 1; 0Þ of the curve y ¼ sin x; z ¼ 0. Ans. (b) 2 ffiffiffi 2 p 7.97. Prove that the acceleration of a particle along a space curve is given respectively in (a) cylindrical, (b) spherical coordinates by ð € _ 2 Þe þ ð € þ 2 _ _ Þe þ € z zez ð€ r r r _ 2 r _ 2 sin2 Þer þ ðr € þ 2_ r r _ r _ 2 sin cos Þe þ ð2_ r r _ sin þ 2r _ _ cos þ r € sin Þe where dots denote time derivatives and e; e; ez; er; e; e are unit vectors in the directions of increasing ; ; z; r; ; , respectively. 7.98. Let E and H be two vectors assumed to have continuous partial derivatives (of second order at least) with respect to position and time. Suppose further that E and H satisfy the equations r E ¼ 0; r H ¼ 0; r E ¼ 1 c @H @t ; r H ¼ 1 c @E @t ð1Þ prove that E and H satisfy the equation r2 ¼ 1 c2 @2 @t2 ð2Þ where is a generic meaning, and in particular can represent any component of E or H. [The vectors E and H are called electric and magnetic field vectors in electromagnetic theory. Equations (1) are a special case of Maxwell’s equations. The result (2) led Maxwell to the conclusion that light was an electromagnetic phenomena. The constant c is the velocity of light.] 7.99. Use the relations in Problem 7.98 to show that @ @t f1 2 ðE2 þ H2 Þg þ cr ðE HÞ ¼ 0 7.100. Let A1; A2; A3 be the components of vector A in an xyz rectangular coordinate system with unit vectors i1; i2; i3 (the usual i; j; k vectors), and A0 1; A0 2; A0 3 the components of A in an x0 y0 z0 rectangular coordinate system which has the same origin as the xyz system but is rotated with respect to it and has the unit vectors i0 1; i0 2; i0 3. Prove that the following relations (often called invariance relations) must hold: An ¼ l1nA0 1 þ l2nA0 2 þ l3nA0 3 n ¼ 1; 2; 3 where i0 m in ¼ lmn. 7.101. If A is the vector of Problem 7.100, prove that the divergence of A, i.e., r A, is an invariant (often called a scalar invariant), i.e., prove that @A0 1 @x0 þ @A0 2 @y0 þ @A0 3 @z0 ¼ @A1 @x þ @A2 @y þ @A3 @z CHAP. 7] VECTORS 181
- 191. The results of this and the preceding problem express an obvious requirement that physical quantities must not depend on coordinate systems in which they are observed. Such ideas when generalized lead to an important subject called tensor analysis, which is basic to the theory of relativity. 7.102. Prove that (a) A B; ðbÞ A B; ðcÞ r A are invariant under the transformation of Problem 7.100. 7.103. If u1; u2; u3 are orthogonal curvilinear coordinates, prove that ðaÞ @ðu1; u2; u3Þ @ðx; y; zÞ ¼ ru1 ru2 ru3 ðbÞ @r @u1 @r @u2 @r @u3 ðru1 ru2 ru3Þ ¼ 1 and give the significance of these in terms of Jacobians. 7.104. Use the axiomatic approach to vectors to prove relation (8) on Page 155. 7.105. A set of n vectors A1; A2; ; An is called linearly dependent if there exists a set of scalars c1; c2; . . . ; cn not all zero such that c1A1 þ c2A2 þ þ cnAn ¼ 0 identically; otherwise, the set is called linearly independent. (a) Prove that the vectors A1 ¼ 2i 3j þ 5k, A2 ¼ i þ j 2k; A3 ¼ 3i 7j þ 12k are linearly dependent. (b) Prove that any four three-dimensional vectors are linearly dependent. (c) Prove that a necessary and sufficient condition that the vectors A1 ¼ a1i þ b1j þ c1k, A2 ¼ a2i þ b2j þ c2k; A3 ¼ a3i þ b3j þ c3k be linearly independent is that A1 A2 A3 6¼ 0. Give a geometrical interpretation of this. 7.106. A complex number can be defined as an ordered pair ða; bÞ of real numbers a and b subject to certain rules of operation for addition and multiplication. (a) What are these rules? (b) How can the rules in (a) be used to define subtraction and division? (c) Explain why complex numbers can be considered as two-dimen- sional vectors. (d) Describe similarities and differences between various operations involving complex numbers and the vectors considered in this chapter. 182 VECTORS [CHAP. 7
- 192. 183 Applications of Partial Derivatives APPLICATIONS TO GEOMETRY The theoretical study of curves and surfaces began more than two thousand years ago when Greek phi- losopher-mathematicians explored the properties of conic sections, helixes, spirals, and surfaces of revolu- tion generated from them. While applications were not on their minds, many practical consequences evolved. These included representation of the ellipti- cal paths of planets about the sun, employment of the focal properties of paraboloids, and use of the special properties of helixes to construct the double helical model of DNA. The analytic tool for studying functions of more than one variable is the partial derivative. Surfaces are a geometric starting point, since they are represented by functions of two independent variables. Vector forms of many of these these concepts were introduced in the previous chapter. In this one, corresponding coordinate equations are exhibited. 1. Tangent Plane to a Surface. Let Fðx; y; zÞ ¼ 0 be the equation of a surface S such as shown in Fig. 8-1. We shall assume that F, and all other functions in this chapter, is continuously differentiable unless otherwise indicated. Suppose we wish to find the equation of a tangent plane to S at the point Pðx0; y0; z0Þ. A vector normal to S at this point is N0 ¼ rFjP, the subscript P indicating that the gradient is to be evaluated at the point Pðx0; y0; z0Þ. If r0 and r are the vectors drawn respectively from O to Pðx0; y0; z0Þ and Qðx; y; zÞ on the plane, the equation of the plane is ðr r0Þ N0 ¼ ðr r0Þ rFjP ¼ 0 ð1Þ since r r0 is perpendicular to N0. Fig. 8-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 193. In rectangular form this is @F @x P ðx x0Þ þ @F @y P ðy y0Þ þ @F @z P ðz z0Þ ¼ 0 ð2Þ In case the equation of the surface is given in orthogonal curvilinear coordinates in the form Fðu1; u2; u3Þ ¼ 0, the equation of the tangent plane can be obtained using the result on Page 162 for the gradient in these coordinates. See Problem 8.4. 2. Normal Line to a Surface. Suppose we require equations for the normal line to the surface S at Pðx0; y0; z0Þ i.e., the line perpendicular to the tangent plane of the surface at P. If we now let r be the vector drawn from O in Fig. 8-1 to any point ðx; y; zÞ on the normal N0, we see that r r0 is collinear with N0 and so the required condition is ðr r0Þ N0 ¼ ðr r0Þ rFjP ¼ 0 ð3Þ By expressing the cross product in the determinant form i j k x x0 y y0 z z0 FxjP FyjP FzjP we find that x x0 @F @x P ¼ y y0 @F @y P ¼ z z0 @F @z P ð4Þ Setting each of these ratios equal to a parameter (such as t or u) and solving for x, y; and z yields the parametric equations of the normal line. The equations for the normal line can also be written when the equation of the surface is expressed in orthogonal curvilinear coordinates. (See Problem 8.1(b).) 3. Tangent Line to a Curve. Let the parametric equations of curve C of Fig. 8-2 be x ¼ f ðuÞ; y ¼ gðuÞ; z ¼ hðuÞ; where we shall suppose, unless otherwise indicated, that f , g; and h are continuously differentiable. We wish to find equations for the tangent line to C at the point Pðx0; y0; z0Þ where u ¼ u0. 184 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8 Fig. 8-2
- 194. If R ¼ f ðuÞi þ gðuÞj þ hðuÞk, a vector tangent to C at the point P is given by T0 ¼ dR du P . If r0 and r denote the vectors drawn respectively from O to Pðx0; y0; z0Þ and Qðx; y; zÞ on the tangent line, then since r r0 is collinear with T0 we have ðr r0Þ T0 ¼ ðr r0Þ dR du P ¼ 0 ð5Þ In rectangular form this becomes x x0 f 0ðu0Þ ¼ y y0 g0ðu0Þ ¼ z z0 h0ðu0Þ ð6Þ The parametric form is obtained by setting each ratio equal to u. If the curve C is given as the intersection of two surfaces with equations Fðx; y; zÞ ¼ 0 and Gðx; y; zÞ ¼ 0 observe that rF rG has the direction of the line of intersection of the tangent planes; therefore, the corresponding equations of the tangent line are x x0 Fy Fz Gy Gz P ¼ y y0 Fz Fx Gz Gx P ¼ z z0 Fx Fy Gx Gy P ð7Þ Note that the determinants in (7) are Jacobians. A similar result can be found when the surfaces are given in terms of orthogonal curvilinear coordinates. 4. Normal Plane to a Curve. Suppose we wish to find an equation for the normal plane to curve C at Pðx0; y0; z0Þ of Fig. 8-2 (i.e., the plane perpendicular to the tangent line to C at this point). Letting r be the vector from O to any point ðx; y; zÞ on this plane, it follows that r r0 is perpendicular to T0. Then the required equation is ðr r0Þ T0 ¼ ðr r0Þ dR du P ¼ 0 ð8Þ When the curve has parametric equations x ¼ f ðuÞ; y ¼ gðuÞ; z ¼ hðuÞ this becomes f 0 ðu0Þðx x0Þ þ g0 ðu0Þð y y0Þ þ h0 ðu0Þðz z0Þ ¼ 0 ð9Þ Furthermore, when the curve is the intersection of the implicitly defined surfaces Fðx; y; zÞ ¼ 0 and Gðx; y; zÞ ¼ 0 then Fy Fz Gy Gz P ðx x0Þ þ Fz Fx Gz Gx P ð y y0Þ þ Fx Fy Gx Gy P ðz z0Þ ¼ 0 ð10Þ 5. Envelopes. Solutions of differential equations in two variables are geometrically represented by one-parameter families of curves. Sometimes such a family characterizes a curve called an envelope. For example, the family of all lines (see Problem 8.9) one unit from the origin may be represented by x sin y cos 1 ¼ 0, where is a parameter. The envelope of this family is the circle x2 þ y2 ¼ 1. If ðx; y; zÞ ¼ 0 is a one-parameter family of curves in the xy plane, there may be a curve E which is tangent at each point to some member of the family and such that each member of the family is tangent to E. If E exists, its equation can be found by solving simultaneously the equations ðx; y; Þ ¼ 0; ðx; y; Þ ¼ 0 ð11Þ and E is called the envelope of the family. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 185
- 195. The result can be extended to determine the envelope of a one-parameter family of surfaces ðx; y; z; Þ. This envelope can be found from ðx; y; z; Þ ¼ 0; ðx; y; z; Þ ¼ 0 ð12Þ Extensions to two- (or more) parameter families can be made. DIRECTIONAL DERIVATIVES Suppose Fðx; y; zÞ is defined at a point ðx; y; zÞ on a given space curve C. Let Fðx þ x; y þ y; z þ zÞ be the value of the function at a neighboring point on C and let s denote the length of arc of the curve between those points. Then lim s!0 F s ¼ lim s!0 Fðx þ x; y þ y; z þ zÞ Fðx; y; zÞ s ð13Þ if it exists, is called the directional derivative of F at the point ðx; y; zÞ along the curve C and is given by dF ds ¼ @F @x dx ds þ @F @y dy ds þ @F @z dz ds ð14Þ In vector form this can be written dF ds ¼ @F @x i þ @F @y j þ @F @z k dx ds i þ dy ds j þ dz ds k ¼ rF dr ds ¼ rF T ð15Þ from which it follows that the directional derivative is given by the component of rF in the direction of the tangent to C. In the previous chapter we observed the following fact: The maximum value of the directional derivative is given by jrFj. These maxima occur in directions normal to the surfaces Fðx; y; zÞ ¼ c (where c is any constant) which are sometimes called equipotential surfaces or level surfaces. DIFFERENTIATION UNDER THE INTEGRAL SIGN Let ð Þ ¼ ðu2 u1 f ðx; Þ dx a @ @ b ð16Þ where u1 and u2 may depend on the parameter . Then d d ¼ ðu2 u1 @ f @ dx þ f ðu2; Þ du2 d f ðu1; Þ du1 d ð17Þ for a @ @ b, if f ðx; Þ and @ f =@ are continuous in both x and in some region of the x plane including u1 @ x @ u2, a @ @ b and if u1 and u2 are continuous and have continuous derivatives for a @ @ b. In case u1 and u2 are constants, the last two terms of (17) are zero. The result (17), called Leibnitz’s rule, is often useful in evaluating definite integrals (see Problems 8.15, 8.29). INTEGRATION UNDER THE INTEGRAL SIGN If ð Þ is defined by (16) and f ðx; Þ is continuous in x and in a region including u1 @ x @ u2; a @ x @ b, then if u1 and u2 are constants, 186 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 196. ðb a ð Þ d ¼ ðb a ðu2 u1 f ðx; Þ dx d ¼ ðu2 u1 ðb a f ðx; Þ d dx ð18Þ The result is known as interchange of the order of integration or integration under the integral sign. (See Problem 8.18.) MAXIMA AND MINIMA In Chapter 4 we briefly examined relative extrema for functions of one variable. The general idea was that for points of the graph of y ¼ gðxÞ that were locally highest or lowest, the condition g0 ðxÞ ¼ 0 was necessary. Such points P0ðx0Þ were called critical points. (See Fig. 8-3a,b.) The condition g0 ðxÞ ¼ 0 was useful in searching for relative maxima and minima but it was not decisive. (See Fig. 8-3(c).) To determine the exact nature of the function at a critical point P0, g00 ðx0Þ had to be examined. 0 counterclockwise rotation (rel. min.) g00 ðx0Þ 0 implied a clockwise rotation (rel. max) ¼ 0 need for further investigation. This section describes the necessary and sufficient conditions for relative extrema of functions of two variables. Geometrically we think of surfaces, S, represented by z ¼ f ðx; yÞ. If at a point P0ðx0; y0Þ then fxðx; y0Þ ¼ 0, means that the curve of intersection of S and the plane y ¼ y0 has a tangent parallel to the x-axis. Similarly fyðx0; y0Þ ¼ 0 indicates that the curve of intersection of S and the cross section x ¼ x0 has a tangent parallel the y-axis. (See Problem 8.20.) Thus fxðx; y0Þ ¼ 0; fyðx0; yÞ ¼ 0 are necessary conditions for a relative extrema of z ¼ f ðx; yÞ at P0; however, they are not sufficient because there are directions associated with a rotation through 3608 that have not been examined. Of course, no differentiation between relative maxima and relative minima has been made. (See Fig. 8-4.) A very special form, fxy fxfy invariant under plane rotation, and capable of characterizing the roots of a quadratic equation, Ax2 þ 2Bx þ C ¼ 0, allows us to form sufficient conditions for relative extrema. (See Problem 8.21.) CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 187 Fig. 8-3 Fig. 8-4
- 197. A point ðx0; y0Þ is called a relative maximum point or relative minimum point of f ðx; yÞ respectively according as f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ or f ðx0 þ h; y0 þ kÞ f ðx0; y0Þ for all h and k such that 0 jhj ; 0 jkj where is a sufficiently small positive number. A necessary condition that a differentiable function f ðx; yÞ have a relative maximum or minimum is @ f @x ¼ 0; @ f @y ¼ 0 ð19Þ If ðx0; y0Þ is a point (called a critical point) satisfying equations (19) and if is defined by ¼ @2 f @x2 ! @2 f @y2 ! @2 f @x @y !2 8 : 9 = ; ðx0;y0Þ ð20Þ then 1. ðx0; y0Þ is a relative maximum point if 0 and @2 f @x2 ðx0;y0Þ 0 or @2 f @y2 ðx0;y0Þ 0 ! 2. ðx0; y0Þ is a relative minimum point if 0 and @2 f @x2 ðx0;y0Þ 0 or @2 f @y2 ðx0;y0Þ 0 ! 3. ðx0; y0Þ is neither a relative maximum or minimum point if 0. If 0, ðx0; y0Þ is some- times called a saddle point. 4. No information is obtained if ¼ 0 (in such case further investigation is necessary). METHOD OF LAGRANGE MULTIPLIERS FOR MAXIMA AND MINIMA A method for obtaining the relative maximum or minimum values of a function Fðx; y; zÞ subject to a constraint condition ðx; y; zÞ ¼ 0, consists of the formation of the auxiliary function Gðx; y; zÞ Fðx; y; zÞ þ ðx; y; zÞ ð21Þ subject to the conditions @G @x ¼ 0; @G @y ¼ 0; @G @z ¼ 0 ð22Þ which are necessary conditions for a relative maximum or minimum. The parameter , which is independent of x; y; z, is called a Lagrange multiplier. The conditions (22) are equivalent to rG ¼ 0, and hence, 0 ¼ rF þ r Geometrically, this means that rF and r are parallel. This fact gives rise to the method of Lagrange multipliers in the following way. Let the maximum value of F on ðx; y; zÞ ¼ 0 be A and suppose it occurs at P0ðx0; y0; z0Þ. (A similar argument can be made for a minimum value of F.) Now consider a family of surfaces Fðx; y; zÞ ¼ C. The member Fðx; y; zÞ ¼ A passes through P0, while those surfaces Fðx; y; zÞ ¼ B with B A do not. (This choice of a surface, i.e., f ðx; y; zÞ ¼ A, geometrically imposes the condition ðx; y; zÞ ¼ 0 on F.) Since at P0 the condition 0 ¼ rF þ r tells us that the gradients of Fðx; y; zÞ ¼ A and ðx; y; zÞ are parallel, we know that the surfaces have a common tangent plane at a point that is maximum for F. Thus, rG ¼ 0 is a necessary condition for a relative maximum of F at P0. Of course, the condition is not sufficient. The critical point so determined may not be unique and it may not produce a relative extremum. The method can be generalized. If we wish to find the relative maximum or minimum values of a function Fðx1; x2; x3; . . . ; xnÞ subject to the constraint conditions ðx1; . . . ; xnÞ ¼ 0; 2ðx1; . . . ; xnÞ ¼ 0; . . . ; kðx1; . . . ; xnÞ ¼ 0, we form the auxiliary function 188 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 198. Gðx1; x2; . . . ; xnÞ F þ 11 þ 22 þ þ kk ð23Þ subject to the (necessary) conditions @G @x1 ¼ 0; @G @x2 ¼ 0; . . . ; @G @xn 0 ð24Þ where 1; 2; . . . ; k, which are independent of x1; x2; . . . ; xn, are the Lagrange multipliers. APPLICATIONS TO ERRORS The theory of differentials can be applied to obtain errors in a function of x; y; z, etc., when the errors in x; y; z, etc., are known. See Problem 8.28. Solved Problems TANGENT PLANE AND NORMAL LINE TO A SURFACE 8.1. Find equations for the (a) tangent plane and (b) normal line to the surface x2 yz þ 3y2 ¼ 2xz2 8z at the point ð1; 2; 1Þ. (a) The equation of the surface is F ¼ x2 yz þ 3y2 2xz2 þ 8z ¼ 0. A normal to the surface at ð1; 2; 1Þ is N0 ¼ rFjð1;2;1Þ ¼ ð2xyz 2z2 Þi þ ðx2 z þ 6yÞj þ ðx2 y 4xz þ 8Þkjð1;2;1Þ ¼ 6i þ 11j þ 14k Referring to Fig. 8-1, Page 183: The vector from O to any point ðx; y; zÞ on the tangent plane is r ¼ xi þ yj þ zk. The vector from O to the point ð1; 2; 1Þ on the tangent plane is r0 ¼ i þ 2j k. The vector r r0 ¼ ðx 1Þi þ ð y 2Þj þ ðz þ 1Þk lies in the tangent plane and is thus perpen- dicular to N0. Then the required equation is ðr r0Þ N0 ¼ 0 i:e:; fðx 1Þi þ ð y 2Þj þ ðz þ 1Þkg f6i þ 11j þ 14kg ¼ 0 6ðx 1Þ þ 11ð y 2Þ þ 14ðz þ 1Þ ¼ 0 or 6x 11y 14z þ 2 ¼ 0 (b) Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal N0. The vector from O to the point ð1; 2; 1Þ on the normal is r0 ¼ i þ 2j k. The vector r r0 ¼ ðx 1Þi þ ð y 2Þj þ ðz þ 1Þk is collinear with N0. Then ðr r0Þ N0 ¼ 0 i:e:; i j k x 1 y 2 z þ 1 6 11 14 ¼ 0 which is equivalent to the equations 11ðx 1Þ ¼ 6ð y 2Þ; 14ð y 2Þ ¼ 11ðz þ 1Þ; 14ðx 1Þ ¼ 6ðz þ 1Þ These can be written as x 1 6 ¼ y 2 11 ¼ z þ 1 14 often called the standard form for the equations of a line. By setting each of these ratios equal to the parameter t, we have x ¼ 1 6t; y ¼ 2 þ 11t; z ¼ 14t 1 called the parametric equations for the line. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 189
- 199. 8.2. In what point does the normal line of Problem 8.1(b) meet the plane x þ 3y 2z ¼ 10? Substituting the parametric equations of Problem 8.1(b), we have 1 6t þ 3ð2 þ 11tÞ 2ð14t 1Þ ¼ 10 or t ¼ 1 Then x ¼ 1 6t ¼ 7; y ¼ 2 þ 11t ¼ 9; z ¼ 14t 1 ¼ 15 and the required point is ð7; 9; 15Þ. 8.3. Show that the surface x2 2yz þ y3 ¼ 4 is perpendicular to any member of the family of surfaces x2 þ 1 ¼ ð2 4aÞy2 þ az2 at the point of intersection ð1; 1; 2Þ: Let the equations of the two surfaces be written in the form F ¼ x2 2yz þ y3 4 ¼ 0 and G ¼ x2 þ 1 ð2 4aÞy2 az2 ¼ 0 Then rF ¼ 2xi þ ð3y2 2zÞj 2yk; rG ¼ 2xi 2ð2 4aÞyj 2azk Thus, the normals to the two surfaces at ð1; 1; 2Þ are given by N1 ¼ 2i j þ 2k; N2 ¼ 2i þ 2ð2 4aÞj 4ak Since N1 N2 ¼ ð2Þð2Þ 2ð2 4aÞ ð2Þð4aÞ 0, it follows that N1 and N2 are perpendicular for all a, and so the required result follows. 8.4. The equation of a surface is given in spherical coordinates by Fðr; ; Þ ¼ 0, where we suppose that F is continuously differentiable. (a) Find an equation for the tangent plane to the surface at the point ðr0; 0; 0Þ. (b) Find an equation for the tangent plane to the surface r ¼ 4 cos at the point ð2 ffiffiffi 2 p ; =4; 3=4Þ. (c) Find a set of equations for the normal line to the surface in (b) at the indicated point. (a) The gradient of in orthogonal curvilinear coordinates is r ¼ 1 h1 @ @u1 e1 þ 1 h2 @ @u2 e2 þ 1 h3 @ @u3 e3 e1 ¼ 1 h1 @r @u1 ; e2 ¼ 1 h2 @r @u2 ; e3 ¼ 1 h3 @r @u3 where (see Pages 161, 175). In spherical coordinates u1 ¼ r; u2 ¼ ; u3 ¼ ; h1 ¼ 1; h2 ¼ r; h3 ¼ r sin and r ¼ xi þ yjþ zk ¼ r sin cos i þ r sin sin j þ r cos k. Then e1 ¼ sin cos i þ sin sin j þ cos k e2 ¼ cos cos i þ cos sin j sin k e3 ¼ sin i þ cos j 8 : ð1Þ and rF ¼ @F @r e1 þ 1 r @F @ e2 þ 1 r sin @F @ e3 ð2Þ As on Page 183 the required equation is ðr r0Þ rFjP ¼ 0. Now substituting (1) and (2), we have rFjP ¼ @F @r P sin 0 cos 0 þ 1 r0 @F @ P cos 0 cos 0 sin 0 r0 sin 0 @F @ P i þ @F @r P sin 0 sin 0 þ 1 r0 @F @ P cos 0 sin 0 þ cos 0 r0 sin 0 @F @ P j þ @F @r P cos 0 1 r0 @F @ P sin 0 k 190 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 200. Denoting the expressions in braces by A; B; C respectively so that rFjP ¼ Ai þ Bj þ Ck, we see that the required equation is Aðx x0Þ þ Bð y y0Þ þ Cðz z0Þ ¼ 0. This can be written in spherical coordinates by using the transformation equations for x, y; and z in these coordinates. (b) We have F ¼ r 4 cos ¼ 0. Then @F=@r ¼ 1, @F=@ ¼ 4 sin , @F=@ ¼ 0. Since r0 ¼ 2 ffiffiffi 2 p ; 0 ¼ =4; 0 ¼ 3=4, we have from part (a), rFjP ¼ Ai þ Bj þ Ck ¼ i þ j. From the transformation equations the given point has rectangular coordinates ð ffiffiffi 2 p ; ffiffiffi 2 p ; 2Þ, and so r r0 ¼ ðx þ ffiffiffi 2 p Þi þ ð y ffiffiffi 2 p Þj þ ðz 2Þk. The required equation of the plane is thus ðx þ ffiffiffi 2 p Þ þ ð y ffiffiffi 2 p Þ ¼ 0 or y x ¼ 2 ffiffiffi 2 p . In sphe- rical coordinates this becomes r sin sin r sin cos ¼ 2 ffiffiffi 2 p . In rectangular coordinates the equation r ¼ 4 cos becomes x2 þ y2 þ ðz 2Þ2 ¼ 4 and the tangent plane can be determined from this as in Problem 8.1. In other cases, however, it may not be so easy to obtain the equation in rectangular form, and in such cases the method of part (a) is simpler to use. (c) The equations of the normal line can be represented by x þ ffiffiffi 2 p 1 ¼ y ffiffiffi 2 p 1 ¼ z 2 0 the significance of the right-hand member being that the line lies in the plane z ¼ 2. Thus, the required line is given by x þ ffiffiffi 2 p 1 ¼ y ffiffiffi 2 p 1 ; z ¼ 0 or x þ y ¼ 0; z ¼ 0 TANGENT LINE AND NORMAL PLANE TO A CURVE 8.5. Find equations for the (a) tangent line and (b) normal plane to the curve x ¼ t cos t, y ¼ 3 þ sin 2t, z ¼ 1 þ cos 3t at the point where t ¼ 1 2 . (a) The vector from origin O (see Fig. 8-2, Page 183) to any point of curve C is R ¼ ðt cos tÞiþ ð3 þ sin 2tÞj þ ð1 þ cos 3tÞk. Then a vector tangent to C at the point where t ¼ 1 2 is T0 ¼ dR dt t¼1=2 ¼ ð1 þ sin tÞi þ 2 cos 2t j 3 sin 3t kjt¼1=2 ¼ 2i 2j þ 3k The vector from O to the point where t ¼ 1 2 is r0 ¼ 1 2 i þ 3j þ k. The vector from O to any point ðx; y; zÞ on the tangent line is r ¼ xi þ yj þ zk. Then r r0 ¼ ðx 1 2 Þi þ y 3Þj þ ðz 1Þk is collinear with T0, so that the required equation is ðr r0Þ T0 ¼ 0; i:e:; i j k x 1 2 y 3 z 1 2 2 3 ¼ 0 and the required equations are x 1 2 2 ¼ y 3 2 ¼ z 1 3 or in parametric form x ¼ 2t þ 1 2 , y ¼ 3 2t, z ¼ 3t þ 1: (b) Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal plane. The vector from O to the point where t ¼ 1 2 is r0 ¼ 1 2 i þ 3j þ k. The vector r r0 ¼ ðx 1 2 Þi þ ð y 3Þj þ ðz 1Þk lies in the normal plane and hence is perpendicular to T0. Then the required equation is ðr r0Þ T0 ¼ 0 or 2ðx 1 2 Þ 2ð y 3Þ þ 3ðz 1Þ ¼ 0. 8.6. Find equations for the (a) tangent line and (b) normal plane to the curve 3x2 y þ y2 z ¼ 2, 2xz x2 y ¼ 3 at the point ð1; 1; 1Þ. (a) The equations of the surfaces intersecting in the curve are F ¼ 3x2 y þ y2 z þ 2 ¼ 0; G ¼ 2xz x2 y 3 ¼ 0 CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 191
- 201. 192 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8 The normals to each surface at the point Pð1; 1; 1Þ are, respectively, N1 ¼ rFjP ¼ 6xyi þ ð3x2 þ 2yzÞj þ y2 k ¼ 6 þ j þ k N2 ¼ rGjP ¼ ð2z 2xyÞi x2 j þ 2xk ¼ 4i j þ 2k Then a tangent vector to the curve at P is T0 ¼ N1 N2 ¼ ð6i þ j þ kÞ ð4 j þ 2kÞ ¼ 3i þ 16j þ 2k Thus, as in Problem 8.5(a), the tangent line is given by ðr r0Þ T0 ¼ 0 or fðx 1Þi þ ð y þ 1Þj þ ðz 1Þkg f3i þ 16j þ 2kg ¼ 0 x 1 3 ¼ y þ 1 16 ¼ z 1 2 or x ¼ 1 þ 3t; y ¼ 16t 1; z ¼ 2t þ 1 i.e., (b) As in Problem 8.5(b) the normal plane is given by ðr r0Þ T0 ¼ 0 or fðx 1Þi þ ð y þ 1Þj þ ðz 1Þkg f3i þ 16j þ 2kg ¼ 0 3ðx 1Þ þ 16ð y þ 1Þ þ 2ðz 1Þ ¼ 0 or 3x þ 16y þ 2z ¼ 11 i.e., The results in (a) and (b) can also be obtained by using equations (7) and (10), respectively, on Page 185. 8.7. Establish equation (10), Page 185. Suppose the curve is defined by the intersection of two surfaces whose equations are Fðx; y; zÞ ¼ 0, Gðx; y; zÞ ¼ 0, where we assume F and G continuously differentiable. The normals to each surface at point P are given respectively by N1 ¼ rFjP and N2 ¼ rGjP. Then a tangent vector to the curve at P is T0 ¼ N1 N2 ¼ rFjP rGjP. Thus, the equation of the normal plane is ðr r0Þ T0 ¼ 0. Now T0 ¼ rFjP rGjP ¼ fðFxi þ Fyj þ FzkÞ ðGxi þ Gyj þ GzkÞgjP ¼ i j k Fx Fy Fz Gx Gy Gz P ¼ Fy Fz Gy Gz P i þ Fx Fx Gx Gx P j þ Fx Fy Gx Gy P k and so the required equation is ðr r0Þ rFjP ¼ 0 or Fy Fz Gy Gz P ðx x0Þ þ Fz Fx Gz Gx P ð y y0Þ þ Fx Fy Gx Gy P ðz z0Þ ¼ 0 ENVELOPES 8.8. Prove that the envelope of the family ðx; y; Þ ¼ 0, if it exists, can be obtained by solving simultaneously the equations ¼ 0 and ¼ 0. Assume parametric equations of the envelope to be x ¼ f ð Þ; y ¼ gð Þ. Then ð f ð Þ; gð Þ; Þ ¼ 0 identically, and so upon differentiating with respect to [assuming that , f and g have continuous deriva- tives], we have x f 0 ð Þ þ yg0 ð Þ þ ¼ 0 ð1Þ The slope of any member of the family ðx; y; Þ ¼ 0 at ðx; yÞ is given by x dx þ y dy ¼ 0 or dy dx ¼ x y . The slope of the envelope at ðx; yÞ is dy dx ¼ dy=d dx=d ¼ g0 ð Þ f 0 ð Þ . Then at any point where the envelope and a member of the family are tangent, we must have x y ¼ g0 ð Þ f 0ð Þ or x f 0 ð Þ þ yg0 ð Þ ¼ 0 ð2Þ Comparing (2) with (1) we see that ¼ 0 and the required result follows.
- 202. 8.9. (a) Find the envelope of the family x sin þ y cos ¼ 1. (b) Illus- trate the results geometrically. (a) By Problem 8 the envelope, if it exists, is obtained by solving simulta- neously the equations ðx; y; Þ ¼ x sin þ y cos 1 ¼ 0 and ðx; y; Þ ¼ x cos y cos ¼ 0. From these equations we find x ¼ sin ; y ¼ cos or x2 þ y2 ¼ 1. (b) The given family is a family of straight lines, some members of which are indicated in Fig. 8-5. The envelope is the circle x2 þ y2 ¼ 1. 8.10. Find the envelope of the family of surfaces z ¼ 2 x a2 y. By a generalization of Problem 8.8 the required envelope, if it exists, is obtained by solving simulta- neously the equations ð1Þ ¼ 2 x 2 y z ¼ 0 and ð2Þ ¼ 2x 2 y ¼ 0 From (2) ¼ x=y. Then substitution in (1) yields x2 ¼ yz, the required envelope. 8.11. Find the envelope of the two-parameter family of surfaces z ¼ x þ y . The envelope of the family Fðx; y; z; ; Þ ¼ 0, if it exists, is obtained by eliminating and between the equations F ¼ 0; F ¼ 0; F ¼ 0 (see Problem, 8.43). Now F ¼ z x y þ ¼ 0; F ¼ x þ ¼ 0; F ¼ y þ ¼ 0 Then ¼ x, ¼ y; and we have z ¼ xy. DIRECTIONAL DERIVATIVES 8.12. Find the directional derivative of F ¼ x2 yz3 along the curve x ¼ eu , y ¼ 2 sin u þ 1, z ¼ u cos u at the point P where u ¼ 0. The point P corresponding to u ¼ 0 is ð1; 1; 1Þ. Then rF ¼ 2xyz3 i þ x2 z3 j þ 3x2 yz2 k ¼ 2i j þ 3k at P A tangent vector to the curve is dr du ¼ d du feu i þ ð2 sin u þ 1Þj þ ðu cos uÞkg ¼ eu i þ 2 cos uj þ ð1 þ sin uÞk ¼ i þ 2j þ k at P and the unit tangent vector in this direction is T0 ¼ i þ 2j þ k ffiffiffi 6 p : Then Directional derivative ¼ rF T0 ¼ ð2i j þ 3kÞ i þ 2j þ k ffiffiffi 6 p ¼ 3 ffiffiffi 6 p ¼ 1 2 ffiffiffi 6 p : Since this is positive, F is increasing in this direction. 8.13. Prove that the greatest rate of change of F, i.e., the maximum directional derivative, takes place in the direction of, and has the magnitude of, the vector rF. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 193 y x Fig. 8-5
- 203. dF ds ¼ rF dr ds is the projection of rF in the direction dr ds . This projection is a maximum when rF and dr=ds have the same direction. Then the maximum value of dF=ds takes place in the direction of rF, and the magnitude is jrFj. 8.14. (a) Find the directional derivative of U ¼ 2x3 y 3y2 z at Pð1; 2; 1Þ in a direction toward Qð3; 1; 5Þ. (b) In what direction from P is the directional derivative a maximum? (c) What is the magnitude of the maximum directional derivative? ðaÞ rU ¼ 6x2 yi þ ð2x3 6yzÞj 3y2 k ¼ 12i þ 14j 12k at P: The vector from P to Q ¼ ð3 1Þi þ ð1 2Þj þ ½5 ð1Þk ¼ 2i 3j þ 6k. The unit vector from P to Q ¼ T ¼ 2i 3j þ 6k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ð2Þ2 þ ð3Þ2 þ ð6Þ2 q ¼ 2i 3j þ 6k 7 : Then Directional derivative at P ¼ ð12i þ 14j 12kÞ 2i 3j þ 6k 7 ¼ 90 7 i.e., U is decreasing in this direction. (b) From Problem 8.13, the directional derivative is a maximum in the direction 12i þ 14j 12k. (c) From Problem 8.13, the value of the maximum directional derivative is j12i þ 14j 12kj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 144 þ 196 þ 144 p ¼ 22: DIFFERENTIATION UNDER THE INTEGRAL SIGN 8.15. Prove Leibnitz’s rule for differentiating under the integral sign. Let ð Þ ¼ ðu2ð Þ u1ð Þ f ðx; Þ dx: Then ¼ ð þ Þ ð Þ ¼ ðu2ð þ Þ u1ð þ Þ f ðx; þ Þ dx ðu2ð Þ u1ð Þ f ðx; Þ dx ¼ ðu1ð Þ u1ð þ Þ f ðx; þ Þ dx þ ðu2ð Þ u1ð Þ f ðx; þ Þ dx þ ðu2ð þ Þ u2ð Þ f ðx; þ Þ dx ðu2ð Þ u1ð Þ f ðx; Þ dx ¼ ðu2ð Þ u1ð Þ ½ f ðx; þ Þ f ðx; Þ dx þ ðu2ð þ Þ u2ð Þ f ðx; þ Þ dx ðu1ð þ Þ u1ð Þ f ðx; þ Þ dx By the mean value theorems for integrals, we have ðu2ð Þ u1ð Þ ½ f ðx; þ Þ f ðx; Þ dx ¼ ðu2ð Þ u1ð Þ f ðx; Þ dx ð1Þ ðu1ð þ Þ u1ð Þ f ðx; þ Þ dx ¼ f ð1; þ Þ½u1ð þ Þ u1ð Þ ð2Þ ðu2ð þ Þ u2ð Þ f ðx; þ Þ dx ¼ f ð2; þ Þ½u2ð þ Þ u2ð Þ ð3Þ where is between and þ , 1 is between u1ð Þ and u1ð þ Þ and 2 is between u2ð Þ and u2ð þ Þ. 194 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 204. Then ¼ ðu2ð Þ u1ð Þ f ðx; Þ dx þ f ð2; þ Þ u2 f ð1; þ Þ u1 Taking the limit as ! 0, making use of the fact that the functions are assumed to have continuous derivatives, we obtain d d ¼ ðu2ð Þ u1ð Þ f ðx; Þ dx þ f ½u2ð Þ; du2 d f ½u1ð Þ; du1 d 8.16. If ð Þ ¼ ð 2 sin x x dx, find 0 ð Þ where 6¼ 0. By Leibnitz’s rule, 0 ð Þ ¼ ð 2 @ @ sin x x dx þ sinð 2 Þ 2 d d ð 2 Þ sinð Þ d d ð Þ ¼ ð 2 cos x dx þ 2 sin 3 sin 2 ¼ sin x 2 þ 2 sin 3 sin 2 ¼ 3 sin 3 2 sin 2 8.17. If ð 0 dx cos x ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p ; 1 find ð 0 dx ð2 cos xÞ2 . (See Problem 5.58, Chapter 5.) By Leibnitz’s rule, if ð Þ ¼ ð 0 dx cos x ¼ ð 2 1Þ1=2 ; then 0 ð Þ ¼ ð 0 dx ð cos xÞ2 ¼ 1 2 ð 2 1Þ3=2 2 ¼ ð 2 1Þ3=2 Thus ð 0 dx ð cos xÞ2 ¼ ð 2 1Þ3=2 from which ð 0 dx ð2 cos xÞ2 ¼ 2 3 ffiffiffi 3 p : INTEGRATION UNDER THE INTEGRAL SIGN 8.18. Prove the result (18), Page 187, for integration under the integral sign. Consider ð1Þ ð Þ ¼ ðu2 u1 ð a f ðx; Þ d dx By Leibnitz’s rule, 0 ð Þ ¼ ðu2 u1 @ @ ð a f ðx; Þ d dx ¼ ðu2 u1 f ðx; Þ dx ¼ ð Þ Then by integration, ð2Þ ð Þ ¼ ð a ð Þ d þ c Since ðaÞ ¼ 0 from (1), we have c ¼ 0 in (2). Thus from (1) and (2) with c ¼ 0, we find ðu2 u1 ð a f ðx; Þ dx dx ¼ ð a ðu2 u1 f ðx; Þ dx d Putting ¼ b, the required result follows. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 195
- 205. 8.19. Prove that ð 0 ln b cos x a cos x dx ¼ ln b þ ffiffiffiffiffiffiffiffiffiffiffiffiffi b2 1 p a þ ffiffiffiffiffiffiffiffiffiffiffiffiffi a2 1 p ! if a; b 1. From Problem 5.58, Chapter 5, ð 0 dx cos x ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p ; 1: Integrating the left side with respect to from a to b yields ð 0 ðb a d cos x dx ¼ ð 0 lnð cos xÞ b a dx ¼ ð 0 ln b cos x a cos x dx Integrating the right side with respect to from a to b yields ð 0 d ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p ¼ lnð þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p Þ b a ¼ ln b þ ffiffiffiffiffiffiffiffiffiffiffiffiffi b2 1 p a þ ffiffiffiffiffiffiffiffiffiffiffiffiffi a2 1 p ! and the required result follows. MAXIMA AND MINIMA 8.20. Prove that a necessary condition for f ðx; yÞ to have a relative extremum (maximum or minimum) at ðx0; y0Þ is that fxðx0; y0Þ ¼ 0, fyðx0; y0Þ ¼ 0. If f ðx0; y0Þ is to be an extreme value for f ðx; yÞ, then it must be an extreme value for both f ðx; y0Þ and f ðx0; yÞ. But a necessary condition that these have extreme values at xx ¼ 0 and y ¼ y0, respectively, is fxðx0; y0Þ ¼ 0, fyðx0; y0Þ ¼ 0 (using results for functions of one variable). 8.21. Let f be continuous and have continuous partial derivatives of order two, at least, in a region R with the critical point P0ðx0; y0Þ an interior point. Determine the sufficient conditions for relative extrema at P0. In the case of one variable, sufficient conditions for a relative extrema were formulated through the second derivative [if positive then a relative minimum, if negative then a relative maximum, if zero a possible point of inflection but more investigation is necessary]. In the case of z ¼ f ðx; yÞ that is before us we can expect the second partial derivatives to supply information. (See Fig. 8-6.) First observe that solutions of the quadratic equation At2 þ 2Bt þ C ¼ 0 are t ¼ 2B ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4B2 4AC p 2A Further observe that the nature of these solutions is determined by B2 AC. If the quantity is positive the solutions are real and distinct; if negative, they are complex conjugate; and if zero, the two solutions are coincident. 196 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8 Fig. 8-6
- 206. The expression B2 AC also has the property of invariance with respect to plane rotations x ¼ x x cos y y sin y ¼ x x sin þ y y cos It has been discovered that with the identifications A ¼ fxx; B ¼ fxy; C ¼ fyy, we have the partial deri- vative form f 2 xy fxxfyy that characterizes relative extrema. The demonstration of invariance of this form can be found in analytic geometric books. However, if you would like to put the problem in the context of the second partial derivative, observe that f x x ¼ fx @x @ x x þ fy @y @ x x ¼ fx cos þ fy sin f y y ¼ fx @x @ y y þ fy @y @ y y ¼ fx sin þ fy cos Then using the chain rule to compute the second partial derivatives and proceeding by straightforward but tedious calculation one shows that f 2 xy ¼ fxxfyy ¼ f 2 x x y y f x x x xf y y y y: The following equivalences are a consequence of this invariant form (independently of direction in the tangent plane at P0): f 2 xy fxx fyy 0 and fxx fyy 0 ð1Þ f 2 xy fxx fyy 0 and fxx fyy 0 ð2Þ The key relation is (1) because in order that this equivalence hold, both fx fy must have the same sign. We can look to the one variable case (make the same argument for each coordinate direction) and conclude that there is a relative minimum at P0 if both partial derivatives are positive and a relative maximum if both are negative. We can make this argument for any pair of coordinate directions because of the invariance under rotation that was established. If (2) holds, then the point is called a saddle point. If the quadratic form is zero, no information results. Observe that this situation is analogous to the one variable extreme value theory in which the nature of f at x, and with f 0 ðxÞ ¼ 0, is undecided if f 00 ðxÞ ¼ 0. 8.22. Find the relative maxima and minima of f ðx; yÞ ¼ x3 þ y3 3x 12y þ 20. fx ¼ 3x2 3 ¼ 0 when x ¼ 1; fy ¼ 3y2 12 ¼ 0 when y ¼ 2. Then critical points are Pð1; 2Þ, Qð1; 2Þ; Rð1; 2Þ; Sð1; 2Þ. fxx ¼ 6x; fyy ¼ 6y; fxy ¼ 0. Then ¼ fxxfyy f 2 xy ¼ 36xy. At Pð1; 2Þ; 0 and fxx (or fyyÞ 0; hence P is a relative minimum point. At Qð1; 2Þ; 0 and Q is neither a relative maximum or minimum point. At Rð1; 2Þ; 0 and R is neither a relative maximum or minimum point. At Sð1; 2Þ; 0 and fxx (or fyyÞ 0 so S is a relative maximum point. Thus, the relative minimum value of f ðx; yÞ occurring at P is 2, while the relative maximum value occurring at S is 38. Points Q and R are saddle points. 8.23. A rectangular box, open at the top, is to have a volume of 32 cubic feet. What must be the dimensions so that the total surface is a minimum? If x, y and z are the edges (see Fig. 8-7), then ð1Þ Volume of box ¼ V ¼ xyz ¼ 32 ð2Þ Surface area of box ¼ S ¼ xy þ 2yz þ 2xz or, since z ¼ 32=xy from (1), S ¼ xy þ 64 x þ 64 y CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 197 Fig. 8-7
- 207. @S @x ¼ y 64 x2 ¼ 0 when ð3Þ x2 y ¼ 64; @S @y ¼ x 64 y2 ¼ 0 when ð4Þ xy2 ¼ 64 Dividing equations (3) and (4), we find y ¼ x so that x3 ¼ 64 or x ¼ y ¼ 4 and z ¼ 2. For x ¼ y ¼ 4, ¼ SxxSyy S2 xy ¼ 128 x3 128 y3 1 0 and sxx ¼ 128 x3 0. Hence, it follows that the dimensions 4 ft 4 ft 2 ft give the minimum surface. LAGRANGE MULTIPLIERS FOR MAXIMA AND MINIMA 8.24. Consider Fðx; y; zÞ subject to the constraint condition Gðx; y; zÞ ¼ 0. Prove that a necessary condition that Fðx; y; zÞ have an extreme value is that FxGy FyGx ¼ 0. Since Gðx; y; zÞ ¼ 0, we can consider z as a function of x and y, say z ¼ f ðx; yÞ. A necessary condition that F½x; y; f ðx; yÞ have an extreme value is that the partial derivatives with respect to x and y be zero. This gives ð1Þ Fx þ Fzzx ¼ 0 ð2Þ Fy þ FzZy ¼ 0 Since Gðx; y; zÞ ¼ 0, we also have ð3Þ Gx þ Gxzx ¼ 0 ð4Þ Gy þ Gzzy ¼ 0 From (1) and (3) we have (5) FxGx FxGx ¼ 0, and from (2) and (4) we have (6) FyGz FzGy ¼ 0. Then from (5) and (6) we find FxGy FyGx ¼ 0: The above results hold only if Fz 6¼ 0; Gz 6¼ 0. 8.25. Referring to the preceding problem, show that the stated condition is equivalent to the conditions x ¼ 0; y ¼ 0 where ¼ F þ G and is a constant. If x ¼ 0; Fx þ Gx ¼ 0. If y ¼ 0; Fy þ Gy ¼ 0. Elimination of between these equations yields FxGy FyGx ¼ 0. The multiplier is the Lagrange multiplier. If desired we can consider equivalently ¼ F þ G where x ¼ 0; y ¼ 0. 8.26. Find the shortest distance from the origin to the hyperbola x2 þ 8xy þ 7y2 ¼ 225, z ¼ 0. We must find the minimum value of x2 þ y2 (the square of the distance from the origin to any point in the xy plane) subject to the constraint x2 þ 8xy þ 7y2 ¼ 225. According to the method of Lagrange multipliers, we consider ¼ x2 þ 8xy þ 7y2 225 þ ðx2 þ y2 Þ. Then x ¼ 2x þ 8y þ 2x ¼ 0 or ð1Þ ð þ 1Þx þ 4y ¼ 0 y ¼ 8x þ 14y þ 2y ¼ 0 or ð2Þ 4x þ ð þ 7Þy ¼ 0 From (1) and (2), since ðx; yÞ 6¼ ð0; 0Þ, we must have þ 1 4 4 þ 7 ¼ 0; i:e:; 2 þ 8 9 ¼ 0 or ¼ 1; 9 Case 1: ¼ 1. From (1) or (2), x ¼ 2y and substitution in x2 þ 8xy þ 7y2 ¼ 225 yields 5y2 ¼ 225, for which no real solution exists. Case 2: ¼ 9. From (1) or (2), y ¼ 2x and substitution in x2 þ 8xy þ 7y2 ¼ 225 yields 45x2 ¼ 225. Then x2 ¼ 5; y2 ¼ 4x2 ¼ 20 and so x2 þ y2 ¼ 25. Thus the required shortest distance is ffiffiffiffiffi 25 p ¼ 5. 8.27 (a) Find the maximum and minimum values of x2 þ y2 þ z2 subject to the constraint conditions x2 =4 þ y2 =5 þ z2 =25 ¼ 1 and z ¼ x þ y. (b) Give a geometric interpretation of the result in (a). 198 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 208. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 199 (a) We must find the extrema of F ¼ x2 þ y2 þ z2 subject to the constraint conditions 1 ¼ x2 4 þ y2 5 þ z2 25 1 ¼ 0 and 2 ¼ x þ y z ¼ 0. In this case we use two Lagrange multipliers 1; 2 and consider the function G ¼ F þ 11 þ 22 ¼ x2 þ y2 þ z2 þ 1 x2 4 þ y2 5 þ z2 25 1 ! þ 2ðx þ y zÞ Taking the partial derivatives of G with respect to x; y; z and setting them equal to zero, we find Gx ¼ 2x þ 1x 2 þ 2 ¼ 0; Gy ¼ 2y þ 21y 5 þ 2 ¼ 0; Gx ¼ 2z þ 21z 25 2 ¼ 0 ð1Þ Solving these equations for x; y; z, we find x ¼ 22 1 þ 4 ; y ¼ 52 21 þ 10 ; z ¼ 252 21 þ 50 ð2Þ From the second constraint condition, x þ y z ¼ 0, we obtain on division by 2, assumed dif- ferent from zero (this is justified since otherwise we would have x ¼ 0; y ¼ 0; z ¼ 0, which would not satisfy the first constraint condition), the result 2 1 þ 4 þ 5 21 þ 10 þ 25 21 þ 50 ¼ 0 Multiplying both sides by 2ð1 þ 4Þð1 þ 5Þð1 þ 25Þ and simplifying yields 172 1 þ 2451 þ 750 ¼ 0 or ð1 þ 10Þð171 þ 75Þ ¼ 0 from which 1 ¼ 10 or 75=17. Case 1: 1 ¼ 10. From (2), x ¼ 1 3 2; y ¼ 1 2 2; z ¼ 5 6 2. Substituting in the first constraint condition, x2 =4 þ y2 =5þ z2 =25 ¼ 1, yields 2 2 ¼ 180=19 or 2 ¼ 6 ffiffiffiffiffiffiffiffiffiffi 5=19 p . This gives the two critical points ð2 ffiffiffiffiffiffiffiffiffiffi 5=19 p ; 3 ffiffiffiffiffiffiffiffiffiffi 5=19 p ; 5 ffiffiffiffiffiffiffiffiffiffi 5=19 p Þ; ð2 ffiffiffiffiffiffiffiffiffiffi 5=19 p ; 3 ffiffiffiffiffiffiffiffiffiffi 5=19 p ; 5 ffiffiffiffiffiffiffiffiffiffi 5=19 p Þ The value of x2 þ y2 þ z2 corresponding to these critical points is ð20 þ 45 þ 125Þ=19 ¼ 10. Case 2: 1 ¼ 75=17: From (2), x ¼ 34 7 2; y ¼ 17 4 2; z ¼ 17 28 2. Substituting in the first constraint condition, x2 =4 þ y2 =5 þ z2 =25 ¼ 1, yields 2 ¼ 140=ð17 ffiffiffiffiffiffiffiffi 646 p Þ which gives the critical points ð40= ffiffiffiffiffiffiffiffi 646 p ; 35 ffiffiffiffiffiffiffiffi 646 p ; 5= ffiffiffiffiffiffiffiffi 646 p Þ; ð40= ffiffiffiffiffiffiffiffi 646 p ; 35= ffiffiffiffiffiffiffiffi 646 p ; 5= ffiffiffiffiffiffiffiffi 646 p Þ The value of x2 þ y2 þ z2 corresponding to these is ð1600 þ 1225 þ 25Þ=646 ¼ 75=17. Thus, the required maximum value is 10 and the minimum value is 75/17. (b) Since x2 þ y2 þ z2 represents the square of the distance of ðx; y; zÞ from the origin ð0; 0; 0Þ, the problem is equivalent to determining the largest and smallest distances from the origin to the curve of intersec- tion of the ellipsoid x2 =4 þ y2 =5 þ z2 =25 ¼ 1 and the plane z ¼ x þ y. Since this curve is an ellipse, we have the interpretation that ffiffiffiffiffi 10 p and ffiffiffiffiffiffiffiffiffiffiffiffi 75=17 p are the lengths of the semi-major and semi-minor axes of this ellipse. The fact that the maximum and minimum values happen to be given by 1 in both Case 1 and Case 2 is more than a coincidence. It follows, in fact, on multiplying equations (1) by x, y, and z in succession and adding, for we then obtain 2x2 þ 1x2 2 þ 2x þ 2y2 þ 21y2 5 þ 2y þ 2z2 þ 21z2 25 2z ¼ 0 x2 þ y2 þ z2 þ 1 x2 4 þ y2 5 þ z2 25 ! þ 2ðx þ y zÞ ¼ 0 i.e., Then using the constraint conditions, we find x2 þ y2 þ z2 ¼ 1. For a generalization of this problem, see Problem 8.76.
- 209. 200 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8 APPLICATIONS TO ERRORS 8.28. The period T of a simple pendulum of length l is given by T ¼ 2 ffiffiffiffiffiffiffi l=g p . Find the (a) error and (b) percent error made in computing T by using l ¼ 2 m and g ¼ 9:75 m=sec2 , if the true values are l ¼ 19:5 m and g ¼ 9:81 m=sec2 . (a) T ¼ 2l1=2 g1=2 . Then dT ¼ ð2g1=2 ð1 2 l1=2 dlÞ þ ð2l1=2 Þð 1 2 g3=2 dgÞ ¼ ffiffiffiffi lg p dl ffiffiffiffiffi l g3 s dg ð1Þ Error in g ¼ g ¼ dg ¼ þ0:06; error in l ¼ l ¼ dl ¼ 0:5 The error in T is actually T, which is in this case approximately equal to dT. Thus, we have from (1), Error in T ¼ dT ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:75Þ p ð0:05Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð9:75Þ3 s ðþ0:06Þ ¼ 0:0444 sec (approx.) The value of T for l ¼ 2; g ¼ 9:75 is T ¼ 2 ffiffiffiffiffiffiffiffiffi 2 9:75 r ¼ 2:846 sec (approx.) ðbÞ Percent error (or relative error) in T ¼ dT T ¼ 0:0444 2:846 ¼ 1:56%: Another method: Since ln T ¼ ln 2 þ 1 2 ln l 1 2 ln g, dT T ¼ 1 2 dl l 1 2 dg g ¼ 1 2 0:05 2 1 2 þ0:06 9:75 ¼ 1:56% ð2Þ as before. Note that (2) can be written Percent error in T ¼ 1 2 Percent error in l 1 2 Percent error in g MISCELLANEOUS PROBLEMS 8.29. Evaluate ð1 0 x 1 ln x dx. In order to evaluate this integral, we resort to the following device. Define ð Þ ¼ ð1 0 x 1 ln x dx 0 Then by Leibnitz’s rule 0 ð Þ ¼ ð1 0 @ @ x 1 ln x dx ¼ ð1 0 x ln x ln x dx ¼ ð1 0 x dx ¼ 1 þ 1 Integrating with respect to , ð Þ ¼ lnð þ 1Þ þ c. But since ð0Þ ¼ 0; c ¼ 0; and so ð Þ ¼ lnð þ 1Þ. Then the value of the required integral is ð1Þ ¼ ln 2. The applicability of Leibnitz’s rule can be justified here, since if we define Fðx; Þ ¼ ðx 1Þ= ln x, 0 x 1, Fð0; Þ ¼ 0; Fð1; Þ ¼ , then Fðx; Þ is continuous in both x and for 0 @ x @ 1 and all finite 0. 8.30. Find constants a and b for which Fða; bÞ ¼ ð 0 fsin x ðax2 þ bxÞg2 dx is a minimum.
- 210. The necessary conditions for a minimum are @F=@a ¼ 0, @F=@b ¼ 0. Performing these differentiations, we obtain @F @a ¼ ð 0 @ @a fsin x ðax2 þ bxÞg2 dx ¼ 2 ð 0 x2 fsin x ðax2 þ bxÞg dx ¼ 0 @F @b ¼ ð 0 @ @b fsin x ðax2 þ bxÞg2 dx ¼ 2 ð 0 xfsin x ðax2 þ bxÞg dx ¼ 0 From these we find a ð 0 x4 dx þ b ð 0 x3 dx ¼ ð 0 x2 sin x dx a ð 0 x3 dx þ b ð 0 x2 dx ¼ ð 0 x sin x dx 8 : or 5 a 5 þ 4 b 4 ¼ 2 4 4 a 4 þ 3 b 3 ¼ 8 : Solving for a and b, we find a ¼ 20 3 320 5 0:40065; b 240 4 12 2 1:24798 We can show that for these values, Fða; bÞ is indeed a minimum using the sufficiency conditions on Page 188. The polynomial ax2 þ bx is said to be a least square approximation of sin x over the interval ð0; Þ. The ideas involved here are of importance in many branches of mathematics and their applications. Supplementary Problems TANGENT PLANE AND NORMAL LINE TO A SURFACE 8.31. Find the equations of the (a) tangent plane and (b) normal line to the surface x2 þ y2 ¼ 4z at ð2; 4; 5Þ. Ans. (a) x 2y z ¼ 5; ðbÞ x 2 1 ¼ y þ 4 2 ¼ z 5 1 : 8.32. If z ¼ f ðx; yÞ, prove that the equations for the tangent plane and normal line at point Pðx0; y0; z0Þ are given respectively by ðaÞ z z0 ¼ fxjPðx x0Þ þ fyjPð y y0Þ and ðbÞ x x0 fxjP ¼ y y0 fyjP ¼ z z0 1 8.33. Prove that the acute angle between the z axis and the normal to the surface Fðx; y; zÞ ¼ 0 at any point is given by sec ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F2 x þ F2 y þ F2 z q =jFzj. 8.34. The equation of a surface is given in cylindrical coordinates by Fð; ; zÞ ¼ 0, where F is continuously differentiable. Prove that the equations of (a) the tangent plane and (b) the normal line at the point Pð0; 0; z0Þ are given respectively by Aðx x0Þ þ Bð y y0Þ þ Cðz z0Þ ¼ 0 and x x0 A ¼ y y0 B ¼ z z0 C CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 201
- 211. where x0 ¼ 0 cos 0, y0 ¼ 0 sin 0 and A ¼ FjP cos 0 1 FjP sin 0; B ¼ FjP sin 0 þ 1 FjP cos 0; C ¼ FzjP 8.35. Use Problem 8.34 to find the equation of the tangent plane to the surface z ¼ at the point where ¼ 2, ¼ =2, z ¼ 1. To check your answer work the problem using rectangular coordinates. Ans. 2x y þ 2z ¼ 0 TANGENT LINE AND NORMAL PLANE TO A CURVE 8.36. Find the equations of the (a) tangent line and (b) normal plane to the space curve x ¼ 6 sin t, y ¼ 4 cos 3t, z ¼ 2 sin 5t at the point where t ¼ =4. Ans: ðaÞ x 3 ffiffiffi 2 p 3 ¼ y þ 2 ffiffiffi 2 p 6 ¼ z þ ffiffiffi 2 p 5 ðbÞ 3x 6y 5z ¼ 26 ffiffiffi 2 p 8.37. The surfaces x þ y þ z ¼ 3 and x2 y2 þ 2z2 ¼ 2 intersect in a space curve. Find the equations of the (a) tangent line (b) normal plane to this space curve at the point ð1; 1; 1Þ. Ans: ðaÞ x 1 3 ¼ y 1 1 ¼ z 1 2 ; ðbÞ 3x y 2z ¼ 0 ENVELOPES 8.38. Find the envelope of each of the following families of curves in the xy plane. In each case construct a graph. (a) y ¼ x 2 ; ðbÞ x2 þ y2 1 ¼ 1. Ans. (a) x2 ¼ 4y; ðbÞ x þ y ¼ 1; x y ¼ 1 8.39. Find the envelope of a family of lines having the property that the length intercepted between the x and y axes is a constant a. Ans. x2=3 þ y2=3 ¼ a2=3 8.40. Find the envelope of the family of circles having centers on the parabola y ¼ x2 and passing through its vertex. [Hint: Let ð ; 2 Þ be any point on the parabola.] Ans. x2 ¼ y3 =ð2y þ 1Þ 8.41. Find the envelope of the normals (called an evolute) to the parabola y ¼ 1 2 x2 and construct a graph. Ans. 8ðy 1Þ3 ¼ 27x2 8.42. Find the envelope of the following families of surfaces: ðaÞ ðx yÞ 2 z ¼ 1; ðbÞ ðx Þ2 þ y2 ¼ 2 z Ans. ðaÞ 4z ¼ ðx yÞ2 ; ðbÞ y2 ¼ z2 þ 2xz 8.43. Prove that the envelope of the two parameter family of surfaces Fðx; y; z; ; Þ ¼ 0, if it exists, is obtained by eliminating and in the equations F ¼ 0; F ¼ 0; F ¼ 0. 8.44. Find the envelope of the two parameter families (a) z ¼ x þ y 2 2 and (b) x cos þ y cos þ z cos ¼ a where cos2 þ cos2 þ cos2 ¼ 1 and a is a constant. Ans. ðaÞ 4z ¼ x2 þ y2 ; ðbÞ x2 þ y2 þ z2 ¼ a2 DIRECTIONAL DERIVATIVES 8.45. (a) Find the directional derivative of U ¼ 2xy z2 at ð2; 1; 1Þ in a direction toward ð3; 1; 1Þ. (b) In what direction is the directional derivative a maximum? (c) What is the value of this maximum? Ans. ðaÞ 10=3; ðbÞ 2i þ 4j 2k; ðcÞ 2 ffiffiffi 6 p 202 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 212. 8.46. The temperature at any point ðx; yÞ in the xy plane is given by T ¼ 100xy=ðx2 þ y2 Þ. (a) Find the direc- tional derivative at the point ð2; 1Þ in a direction making an angle of 608 with the positive x-axis. (b) In what direction from ð2; 1Þ would the derivative be a maximum? (c) What is the value of this maximum? Ans. (a) 12 ffiffiffi 3 p 6; (b) in a direction making an angle of tan1 2 with the positive x-axis, or in the direction i þ 2j; (c) 12 ffiffiffi 5 p 8.47. Prove that if Fð; ; zÞ is continuously differentiable, the maximum directional derivative of F at any point is given by ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @F @ 2 þ 1 2 @F @ 2 þ @F @z 2 s . DIFFERENTIATION UNDER THE INTEGRAL SIGN 8.48. If ð Þ ¼ ð1= ffiffi p cos x2 dx, find d d . Ans. ð1= ffiffi p x2 sin x2 dx 1 2 cos 1 1 2 ffiffiffi p cos 2 8.49. (a) If Fð Þ ¼ ð 2 0 tan1 x dx, find dF d by Leibnitz’s rule. (b) Check the result in (a) by direct integration. Ans. ðaÞ 2 tan1 1 2 lnð 2 þ 1Þ 8.50. Given ð1 0 xp dx ¼ 1 p þ 1 ; p 1. Prove that ð1 0 xp ðln xÞm dx ¼ ð1Þm m! ðp þ 1Þmþ1 ; m ¼ 1; 2; 3; . . . . 8.51. Prove that ð 0 lnð1 þ cos xÞ dx ¼ ln 1 þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 p 2 ! ; j j 1. 8.52. Prove that ð 0 lnð1 2 cos x þ 2 Þ dx ¼ ln 2 ; j j 1 0; j j 1 . Discuss the case j j ¼ 1. 8.53. Show that ð 0 dx ð5 3 cos xÞ3 ¼ 59 2048 : INTEGRATION UNDER THE INTEGRAL SIGN 8.54. Verify that ð1 0 ð2 1 ð 2 x2 Þ dx d ¼ ð2 1 ð1 0 ð 2 x2 Þ d dx. 8.55. Starting with the result ð2 0 ð sin xÞ dx ¼ 2 , prove that for all constants a and b, ð2 0 fðb sin xÞ2 ða sin xÞ2 g dx ¼ 2ðb2 a2 Þ 8.56. Use the result ð2 0 dx þ sin x ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 p ; 1 to prove that ð2 0 ln 5 þ 3 sin x 5 þ 4 sin x dx ¼ 2 ln 9 8 8.57. (a) Use the result ð=2 0 dx 1 þ cos x ¼ cos1 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 p ; 0 @ 1 to show that for 0 @ a 1; 0 @ b 1 ð=2 0 sec x ln 1 þ b cos x 1 þ a cos x dx ¼ 1 2 fðcos1 aÞ2 ðcos1 bÞ2 g (b) Show that ð=2 0 sec x lnð1 þ 1 2 cos xÞ dx ¼ 52 72 . CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 203
- 213. MAXIMA AND MINIMA, LAGRANGE MULTIPLIERS 8.58. Find the maxima and minima of Fðx; y; zÞ ¼ xy2 z3 subject to the conditions x þ y þ z ¼ 6, x 0; y 0, z 0. Ans. maximum value ¼ 108 at x ¼ 1; y ¼ 2; z ¼ 3 8.59. What is the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid x2 =9 þ y2 =16 þ z2 =36 ¼ 1? Ans. 64 ffiffiffi 3 p 8.60. (a) Find the maximum and minimum values of x2 þ y2 subject to the condition 3x2 þ 4xy þ 6y2 ¼ 140. (b) Give a geometrical interpretation of the results in (a). Ans. maximum value ¼ 70, minimum value ¼ 20 8.61. Solve Problem 8.23 using Lagrange multipliers. 8.62. Prove that in any triangle ABC there is a point P such that PA 2 þ PB 2 þ PC 2 is a minimum and that P is the intersection of the medians. 8.63. (a) Prove that the maximum and minimum values of f ðx; yÞ ¼ x2 þ xy þ y2 in the unit square 0 @ x @ 1, 0 @ y @ 1 are 3 and 0, respectively. (b) Can the result of (a) be obtained by setting the partial derivatives of f ðx; yÞ with respect to x and y equal to zero. Explain. 8.64. Find the extreme values of z on the surface 2x2 þ 3y2 þ z2 12xy þ 4xz ¼ 35. Ans. maximum ¼ 5, minimum ¼ 5 8.65. Establish the method of Lagrange multipliers in the case where we wish to find the extreme values of Fðx; y; zÞ subject to the two constraint conditions Gðx; y; zÞ ¼ 0, Hðx; y; zÞ ¼ 0. 8.66. Prove that the shortest distance from the origin to the curve of intersection of the surfaces xyz ¼ a and y ¼ bx where a 0; b 0, is 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aðb2 þ 1Þ=2b p . 8.67. Find the volume of the ellipsoid 11x2 þ 9y2 þ 15z2 4xy þ 10yz 20xz ¼ 80. Ans. 64 ffiffiffi 2 p =3 APPLICATIONS TO ERRORS 8.68. The diameter of a right circular cylinder is measured as 6:0 0:03 inches, while its height is measured as 4:0 0:02 inches. What is the largest possible (a) error and (b) percent error made in computing the volume? Ans. (a) 1.70 in3 , (b) 1.5% 8.69. The sides of a triangle are measured to be 12.0 and 15.0 feet, and the included angle 60.08. If the lengths can be measured to within 1% accuracy, while the angle can be measured to within 2% accuracy, find the maximum error and percent error in determining the (a) area and (b) opposite side of the triangle. Ans. (a) 2.501 ft2 , 3.21%; (b) 0.287 ft, 2.08% MISCELLANEOUS PROBLEMS 8.70. If and are cylindrical coordinates, a and b are any positive constants, and n is a positive integer, prove that the surfaces n sin n ¼ a and n cos n ¼ b are mutually perpendicular along their curves of intersec- tion. 8.71. Find an equation for the (a) tangent plane and (b) normal line to the surface 8r ¼ 2 at the point where r ¼ 1, ¼ =4; ¼ =2; ðr; ; Þ being spherical coordinates. Ans: ðaÞ 4x ð2 þ 4Þ y þ ð4 2 Þz ¼ 2 ffiffiffi 2 p ; ðbÞ x 4 ¼ y ffiffiffi 2 p =2 2 þ 4 ¼ z ffiffiffi 2 p =2 2 4 204 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 214. 8.72. (a) Prove that the shortest distance from the point ða; b; cÞ to the plane Ax þ By þ Cz þ D ¼ 0 is Aa þ Bb þ Cc þ D ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 þ B2 þ C2 p (b) Find the shortest distance from ð1; 2; 3Þ to the plane 2x 3y þ 6z ¼ 20. Ans. (b) 6 8.73. The potential V due to a charge distribution is given in spherical coordinates ðr; ; Þ by V ¼ p cos r2 where p is a constant. Prove that the maximum directional derivative at any point is p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin2 þ 4 cos2 p r3 8.74. Prove that ð1 0 xm xn ln x dx ¼ ln m þ 1 n þ 1 if m 0; n 0. Can you extend the result to the case m 1; n 1? 8.75. (a) If b2 4ac 0 and a 0; c 0, prove that the area of the ellipse ax2 þ bxy þ cy2 ¼ 1 is 2= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4ac b2 p . [Hint: Find the maximum and minimum values of x2 þ y2 subject to the constraint ax2 þ bxy þ cy2 ¼ 1.] 8.76. Prove that the maximum and minimum distances from the origin to the curve of intersection defined by x2 =a2 þ y2 =b2 þ z2 =c2 ¼ 1 and Ax þ By þ Cz ¼ 0 can be obtained by solving for d the equation A2 a2 a2 d2 þ B2 b2 b2 d2 þ C2 c2 c2 d2 ¼ 0 8.77. Prove that the last equation in the preceding problem always has two real solutions d2 1 and d2 2 for any real non-zero constants a; b; c and any real constants A; B; C (not all zero). Discuss the geometrical significance of this. 8.78. (a) Prove that IM ¼ ðM 0 dx ðx2 þ 2Þ2 ¼ 1 2 3 tan1 M þ M 2 2 ð 2 þ M2 Þ ðbÞ Find lim M!1 IM: This can be denoted by ðx 0 dx ðx2 þ 2Þ2 : ðcÞ Is lim M!1 d d ðM 0 dx ðx2 þ 2Þ2 ¼ d d lim M!1 ðM 0 dx ðx2 þ 2Þ2 ? 8.79. Find the point on the paraboloid z ¼ x2 þ y2 which is closest to the point ð3; 6; 4Þ. Ans. ð1; 2; 5Þ 8.80. Investigate the maxima and minima of f ðx; yÞ ¼ ðx2 2x þ 4y2 8yÞ2 . Ans. minimum value ¼ 0 8.81. (a) Prove that ð=2 0 cos x dx cos x þ sin x ¼ 2ð 2 þ 1Þ ln 2 þ 1 : ðbÞ Use ðaÞ to prove that ð=2 0 cos2 x dx ð2 cos x þ sin xÞ2 ¼ 3 þ 5 8 ln 2 50 : 8.82. (a) Find sufficient conditions for a relative maximum or minimum of w ¼ f ðx; y; zÞ. (b) Examine w ¼ x2 þ y2 þ z2 6xy þ 8xz 10yz for maxima and minima. CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 205
- 215. [Hint: For (a) use the fact that the quadratic form A 2 þ B 2 þ C2 þ 2D þ 2E þ 2F 0 (i.e., is positive definite) if A 0; A D D B 0; A D F D B E F E C 0 206 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8
- 216. 207 Multiple Integrals Much of the procedure for double and triple integrals may be thought of as a reversal of partial differentiation and otherwise is analogous to that for single integrals. However, one complexity that must be addressed relates to the domain of definition. With single integrals, the functions of one variable were defined on intervals of real numbers. Thus, the integrals only depended on the properties of the functions. The integrands of double and triple integrals are functions of two and three variables, respec- tively, and as such are defined on two- and three-dimensional regions. These regions have a flexibility in shape not possible in the single-variable cases. For example, with functions of two variables, and the corresponding double integrals, rectangular regions, a @ x @ b, c @ y @ d are common. However, in many problems the domains are regions bound above and below by segments of plane curves. In the case of functions of three variables, and the corresponding triple integrals other than the regions a @ x @ b; c @ y @ d; e @ z @ f , there are those bound above and below by portions of surfaces. In very special cases, double and triple integrals can be directly evaluated. However, the systematic technique of iterated integration is the usual procedure. It is here that the reversal of partial differentia- tion comes into play. Definitions of double and triple integrals are given below. Also, the method of iterated integration is described. DOUBLE INTEGRALS Let Fðx; yÞ be defined in a closed region r of the xy plane (see Fig. 9-1). Subdivide r into n subregions rk of area Ak, k ¼ 1; 2; . . . ; n. Let ðk; kÞ be some point of Ak. Form the sum X n k¼1 Fðk; kÞ Ak ð1Þ Consider lim n!1 X n k¼1 Fðk; kÞ Ak ð2Þ Fig. 9-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 217. where the limit is taken so that the number n of subdivisions increases without limit and such that the largest linear dimension of each Ak approaches zero. See Fig. 9-2(a). If this limit exists, it is denoted by ð r ð Fðx; yÞ dA ð3Þ and is called the double integral of Fðx; yÞ over the region r. It can be proved that the limit does exist if Fðx; yÞ is continuous (or sectionally continuous) in r. The double integral has a great variety of interpretations with any individual one dependent on the form of the integrand. For example, if Fðx; yÞ ¼ ðx; yÞ represents the variable density of a flat iron plate then the double integral, Ð A dA, of this function over a same shaped plane region, A, is the mass of the plate. In Fig. 9-2(b) we assume that Fðx; yÞ is a height function (established by a portion of a surface z ¼ Fðx; yÞÞ for a cylindrically shaped object. In this case the double integral represents a volume. ITERATED INTEGRALS If r is such that any lines parallel to the y-axis meet the boundary of r in at most two points (as is true in Fig. 9-1), then we can write the equations of the curves ACB and ADB bounding r as y ¼ f1ðxÞ and y ¼ f2ðxÞ, respectively, where f1ðxÞ and f2ðxÞ are single-valued and continuous in a @ x @ b. In this case we can evaluate the double integral (3) by choosing the regions rk as rectangles formed by constructing a grid of lines parallel to the x- and y-axes and Ak as the corresponding areas. Then (3) can be written ð ð r Fðx; yÞ dx dy ¼ ðb x¼a ðf2ðxÞ y¼f1ðxÞ Fðx; yÞ dy dx ð4Þ ¼ ðb x¼a ðf2ðxÞ y¼f1ðxÞ Fðx; yÞ dy dx 208 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-2
- 218. where the integral in braces is to be evaluated first (keeping x constant) and finally integrating with respect to x from a to b. The result (4) indicates how a double integral can be evaluated by expressing it in terms of two single integrals called iterated integrals. The process of iterated integration is visually illustrated in Fig. 9-3a,b and further illustrated as follows. The general idea, as demonstrated with respect to a given three-space region, is to establish a plane section, integrate to determine its area, and then add up all the plane sections through an integration with respect to the remaining variable. For example, choose a value of x (say, x ¼ x0 Þ. The intersection of the plane x ¼ x0 with the solid establishes the plane section. In it z ¼ Fðx0 ; yÞ is the height function, and if y ¼ f1ðxÞ and y ¼ f2ðxÞ (for all z) are the bounding cylindrical surfaces of the solid, then the width is f2ðx0 Þ f1ðx0 Þ, i.e., y2 y1. Thus, the area of the section is A ¼ ðy2 y1 Fðx0 ; yÞ dy. Now establish slabs Ajxj, where for each interval xj ¼ xj xj1, there is an intermediate value x0 j . Then sum these to get an approximation to the target volume. Adding the slabs and taking the limit yields V ¼ lim n!1 X n j¼1 Aj xj ¼ ðb a ðy2 y1 Fðx; yÞ dy dx In some cases the order of integration is dictated by the geometry. For example, if r is such that any lines parallel to the x-axis meet the boundary of r in at most two points (as in Fig. 9-1), then the equations of curves CAD and CBD can be written x ¼ g1ðyÞ and x ¼ g2ð yÞ respectively and we find similarly ð ð r Fðx; yÞ dx dy ¼ ðd y¼c ðg2ð yÞ x¼g1ð yÞ Fðx; yÞ dx dy ð5Þ ¼ ðd y¼c ðg2ð yÞ x¼g1ð yÞ Fðx; yÞ dx dy If the double integral exists, (4) and (5) yield the same value. (See, however, Problem 9.21.) In writing a double integral, either of the forms (4) or (5), whichever is appropriate, may be used. We call one form an interchange of the order of integration with respect to the other form. CHAP. 9] MULTIPLE INTEGRALS 209 Fig. 9-3
- 219. In case r is not of the type shown in the above figure, it can generally be subdivided into regions r1; r2; . . . which are of this type. Then the double integral over r is found by taking the sum of the double integrals over r1; r2; . . . . TRIPLE INTEGRALS The above results are easily generalized to closed regions in three dimensions. For example, consider a function Fðx; y; zÞ defined in a closed three-dimensional region r. Subdivide the region into n subregions of volume Vk, k ¼ 1; 2; . . . ; n. Letting ðk; k; kÞ be some point in each subregion, we form lim n!1 X n k¼1 Fðk; k; kÞ Vk ð6Þ where the number n of subdivisions approaches infinity in such a way that the largest linear dimension of each subregion approaches zero. If this limit exists, we denote it by ð ð r ð Fðx; y; zÞ dV ð7Þ called the triple integral of Fðx; y; zÞ over r. The limit does exist if Fð; x; y; zÞ is continuous (or piecemeal continuous) in r. If we construct a grid consisting of planes parallel to the xy, yz, and xz planes, the region r is subdivided into subregions which are rectangular parallelepipeds. In such case we can express the triple integral over r given by (7) as an iterated integral of the form ðb x¼a ðg2ðaÞ y¼g1ðxÞ ðf2ðx;yÞ z¼f1ðx;yÞ Fðx; y; zÞ dx dy dz ¼ ðb x¼a ðg2ðxÞ y¼g1ðxÞ ðf2ðx;yÞ z¼f1ðx;yÞ Fðx; y; zÞ dz dy dx ð8Þ (where the innermost integral is to be evaluated first) or the sum of such integrals. The integration can also be performed in any other order to give an equivalent result. The iterated triple integral is a sequence of integrations; first from surface portion to surface portion, then from curve segment to curve segment, and finally from point to point. (See Fig. 9-4.) Extensions to higher dimensions are also possible. 210 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-4
- 220. TRANSFORMATIONS OF MULTIPLE INTEGRALS In evaluating a multiple integral over a region r, it is often convenient to use coordinates other than rectangular, such as the curvilinear coordinates considered in Chapters 6 and 7. If we let ðu; vÞ be curvilinear coordinates of points in a plane, there will be a set of transformation equations x ¼ f ðu; vÞ; y ¼ gðu; vÞ mapping points ðx; yÞ of the xy plane into points ðu; vÞ of the uv plane. In such case the region r of the xy plane is mapped into a region r0 of the uv plane. We then have ð ð r Fðx; yÞ dx dy ¼ ð ð r0 Gðu; vÞ @ðx; yÞ @ðu; vÞ du dv ð9Þ where Gðu; vÞ Ff f ðu; vÞ; gðu; vÞg and @ðx; yÞ @ðu; vÞ @x @u @x @v @y @u @y @v ð10Þ is the Jacobian of x and y with respect to u and v (see Chapter 6). Similarly if ðu; v; wÞ are curvilinear coordinates in three dimensions, there will be a set of transfor- mation equations x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ; z ¼ hðu; v; wÞ and we can write ð ð ð r Fðx; y; zÞ dx dy dz ¼ ð ð ð r0 Gðu; v; wÞ @ðx; y; zÞ @ðu; v; wÞ du dv dw ð11Þ where Gðu; v; wÞ Fff ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞg and @ðx; y; zÞ @ðu; v; wÞ @x @u @x @v @x @w @y @u @y @v @y @w @z @u @z @v @z @w ð12Þ is the Jacobian of x, y, and z with respect to u, v, and w. The results (9) and (11) correspond to change of variables for double and triple integrals. Generalizations to higher dimensions are easily made. THE DIFFERENTIAL ELEMENT OF AREA IN POLAR COORDINATES, DIFFERENTIAL ELEMENTS OF AREA IN CYLINDRAL AND SPHERICAL COORDINATES Of special interest is the differential element of area, dA, for polar coordinates in the plane, and the differential elements of volume, dV, for cylindrical and spherical coordinates in three space. With these in hand the double and triple integrals as expressed in these systems are seen to take the following forms. (See Fig. 9-5.) The transformation equations relating cylindrical coordinates to rectangular Cartesian ones appeared in Chapter 7, in particular, x ¼ cos ; y ¼ sin ; z ¼ z The coordinate surfaces are circular cylinders, planes, and planes. (See Fig. 9-5.) At any point of the space (other than the origin), the set of vectors @r @ ; @r @ ; @r @z constitutes an orthogonal basis. CHAP. 9] MULTIPLE INTEGRALS 211
- 221. In the cylindrical case r ¼ cos i þ sin j þ zk and the set is @r @ ¼ cos i þ sin j; @r @ ¼ sin i þ cos j; @r @z ¼ k Therefore @r @ @r @ @r @z ¼ . That the geometric interpretation of @r @ @r @ @r @z d d dz is an infinitesimal rectangular parallele- piped suggests the differential element of volume in cylindrical coordinates is dV ¼ d d dz Thus, for an integrable but otherwise arbitrary function, Fð; ; zÞ, of cylindrical coordinates, the iterated triple integral takes the form ðz2 z1 ðg2ðzÞ g1ðzÞ ðf2ð;zÞ f1ð;zÞ Fð; ; zÞ d d dz The differential element of area for polar coordinates in the plane results by suppressing the z coordinate. It is dA ¼ @r @ @r @ d d and the iterated form of the double integral is ð2 1 ð2ðÞ 1ðÞ Fð; Þ d d The transformation equations relating spherical and rectangular Cartesian coordinates are x ¼ r sin cos ; y ¼ r sin sin ; z ¼ r cos In this case the coordinate surfaces are spheres, cones, and planes. (See Fig. 9-5.) 212 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-5
- 222. Following the same pattern as with cylindrical coordinates we discover that dV ¼ r2 sin dr d d and the iterated triple integral of Fðr; ; Þ has the spherical representation ðr2 r1 ð2ðÞ 1ðÞ ð2ðr;Þ 1ðr;Þ Fðr; ; Þ r2 sin dr d d Of course, the order of these integrations may be adapted to the geometry. The coordinate surfaces in spherical coordinates are spheres, cones, and planes. If r is held constant, say, r ¼ a, then we obtain the differential element of surface area dA ¼ a2 sin d d The first octant surface area of a sphere of radius a is ð=2 0 ð=2 0 a2 sin d d ¼ ð=2 0 a2 ð cos Þ 2 0 d ¼ ð=2 0 a2 d ¼ a2 2 Thus, the surface area of the sphere is 4a2 . Solved Problems DOUBLE INTEGRALS 9.1. (a) Sketch the region r in the xy plane bounded by y ¼ x2 ; x ¼ 2; y ¼ 1. (b) Give a physical interpreation to ð ð r ðx2 þ y2 Þ dx dy. (c) Evaluate the double integral in (b). (a) The required region r is shown shaded in Fig. 9-6 below. (b) Since x2 þ y2 is the square of the distance from any point ðx; yÞ to ð0; 0Þ, we can consider the double integral as representing the polar moment of inertia (i.e., moment of inertia with respect to the origin) of the region r (assuming unit density). CHAP. 9] MULTIPLE INTEGRALS 213 Fig. 9-6 Fig. 9-7
- 223. We can also consider the double integral as representing the mass of the region r assuming a density varying as x2 þ y2 . (c) Method 1: The double integral can be expressed as the iterated integral ð2 x¼1 ðx2 y¼1 ðx2 þ y2 Þ dy dx ¼ ð2 x¼1 ðx2 y¼1 ðx2 þ y2 Þ dy ( ) dx ¼ ð2 x¼1 x2 y þ y3 3 x2 y¼1 dx ¼ ð2 x¼1 x4 þ x6 3 x2 1 3 ! dx ¼ 1006 105 The integration with respect to y (keeping x constant) from y ¼ 1 to y ¼ x2 corresponds formally to summing in a vertical column (see Fig. 9-6). The subsequent integration with respect to x from x ¼ 1 to x ¼ 2 corresponds to addition of contributions from all such vertical columns between x ¼ 1 and x ¼ 2. Method 2: The double integral can also be expressed as the iterated integral ð4 y¼1 ð2 x¼ ffiffi y p ðx2 þ y2 Þ dx dy ¼ ð4 y¼1 ð2 x¼ ffiffi y p ðx2 þ y2 Þ dx ( ) dy ¼ ð4 y¼1 x3 3 þ xy2 2 x¼ ffiffi y p dy ¼ ð4 y¼1 8 3 þ 2y2 y3=2 3 y5=2 ! dy ¼ 1006 105 In this case the vertical column of region r in Fig. 9-6 above is replaced by a horizontal column as in Fig. 9-7 above. Then the integration with respect to x (keeping y constant) from x ¼ ffiffiffi y p to x ¼ 2 corresponds to summing in this horizontal column. Subsequent integration with respect to y from y ¼ 1 to y ¼ 4 corresponds to addition of contributions for all such horizontal columns between y ¼ 1 and y ¼ 4. 9.2. Find the volume of the region bound by the elliptic paraboloid z ¼ 4 x2 1 4 y2 and the plane z ¼ 0. Because of the symmetry of the elliptic paraboloid, the result can be obtained by multiplying the first octant volume by 4. Letting z ¼ 0 yields 4x2 þ y2 ¼ 16. The limits of integration are determined from this equation. The required volume is 4 ð2 0 ð2 ffiffiffiffiffiffiffiffi 4x2 p 0 4 x2 1 4 y2 dy dx ¼ 4 ð2 0 4y x2 y 1 4 y3 3 !2 ffiffiffiffiffiffiffiffi 4x2 p 0 dx ¼ 16 Hint: Use trigonometric substitutions to complete the integrations. 9.3. The geometric model of a material body is a plane region R bound by y ¼ x2 and y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 p on the interval 0 @ x @ 1, and with a density function ¼ xy (a) Draw the graph of the region. (b) Find the mass of the body. (c) Find the coordinates of the center of mass. (See Fig. 9-8.) (a) 214 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-8
- 224. M ¼ ðb a ðf2 f1 dy dx ¼ ð1 0 ð ffiffiffiffiffiffiffiffi 2x2 p x2 yx dy dx ¼ ð1 0 y2 2 # ffiffiffiffiffiffiffiffi 2x2 p x2 x dx ðbÞ ¼ ð1 0 1 2 xð2 x2 x4 Þ dx ¼ x2 2 x4 8 x6 12 #1 0 ¼ 7 24 (c) The coordinates of the center of mass are defined to be x x ¼ 1 M ðb a ðf2ðxÞ f1ðxÞ x dy dx and y y ¼ 1 M ðb a ðf2ðxÞ f1ðxÞ y dy dx where M ¼ ðb a ðf2ðxÞ f1ðxÞ dy dx Thus, M x x ¼ ð1 0 ð ffiffiffiffiffiffiffiffi 2x2 p x2 x xy dy dx ¼ ð1 0 x2 y2 2 # ffiffiffiffiffiffiffiffi 2x2 p x2 dx ¼ ð1 0 x2 1 2 ½2 x2 x4 dx ¼ x3 3 x5 10 x7 14 #1 0 ¼ 1 3 1 10 1 14 ¼ 17 105 M y y ¼ ð1 0 ð ffiffiffiffiffiffiffiffi 2x2 p x2 yx dy dx ¼ 13 120 þ 4 ffiffiffi 2 p 15 9.4. Find the volume of the region common to the intersecting cylinders x2 þ y2 ¼ a2 and x2 þ z2 ¼ a2 . Required volume ¼ 8 times volume of region shown in Fig. 9-9 ¼ 8 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼0 z dy dx ¼ 8 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 p dy dx ¼ 8 ða x¼0 ða2 x2 Þ dx ¼ 16a3 3 As an aid in setting up this integral, note that z dy dx corresponds to the volume of a column such as shown darkly shaded in the figure. Keeping x constant and integrating with respect to y from y ¼ 0 to y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 p corresponds to adding the volumes of all such columns in a slab parallel to the yz plane, thus giving the volume of this slab. Finally, integrating with respect to x from x ¼ 0 to x ¼ a corresponds to adding the volumes of all such slabs in the region, thus giving the required volume. 9.5. Find the volume of the region bounded by z ¼ x þ y; z ¼ 6; x ¼ 0; y ¼ 0; z ¼ 0 CHAP. 9] MULTIPLE INTEGRALS 215
- 225. Required volume ¼ volume of region shown in Fig. 9-10 ¼ ð6 x¼0 ð6x y¼0 f6 ðx þ yÞg dy dx ¼ ð6 x¼0 ð6 xÞy 1 2 y2 6x y¼0 dx ¼ ð6 x¼0 1 2 ð6 xÞ2 dx ¼ 36 In this case the volume of a typical column (shown darkly shaded) corresponds to f6 ðx þ yÞg dy dx. The limits of integration are then obtained by integrating over the region r of the figure. Keeping x constant and integrating with respect to y from y ¼ 0 to y ¼ 6 x (obtained from z ¼ 6 and z ¼ x þ yÞ corresponds to summing all columns in a slab parallel to the yz plane. Finally, integrating with respect to x from x ¼ 0 to x ¼ 6 corresponds to adding the volumes of all such slabs and gives the required volume. TRANSFORMATION OF DOUBLE INTEGRALS 9.6. Justify equation (9), Page 211, for changing variables in a double integral. In rectangular coordinates, the double integral of Fðx; yÞ over the region r (shaded in Fig. 9-11) is ð ð r Fðx; yÞ dx dy. We can also evaluate this double integral by considering a grid formed by a family of u and v curvilinear coordinate curves constructed on the region r as shown in the figure. 216 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-9 Fig. 9-10 Fig. 9-11
- 226. Let P be any point with coordinates ðx; yÞ or ðu; vÞ, where x ¼ f ðu; vÞ and y ¼ gðu; vÞ. Then the vector r from O to P is given by r ¼ xi þ yj ¼ f ðu; vÞi þ gðu; vÞj. The tangent vectors to the coordinate curves u ¼ c1 and v ¼ c2, where c1 and c2 are constants, are @r=@v and @r=@u, respectively. Then the area of region r of Fig. 9-11 is given approximately by @r @u @r @v u v. But @r @u @r @v ¼ i j k @x @u @y @u 0 @x @v @y @v 0 ¼ @x @u @y @u @x @v @y @v k ¼ @ðx; yÞ @ðu; vÞ k @r @u @r @v u v ¼ @ðx; yÞ @ðu; vÞ u v so that The double integral is the limit of the sum X Ff f ðu; vÞ; gðu; vÞg @ðx; yÞ @ðu; vÞ u v taken over the entire region r. An investigation reveals that this limit is ð ð r0 Ff f ðu; vÞ; gðu; vÞg @ðx; yÞ @ðu; vÞ du dv where r0 is the region in the uv plane into which the region r is mapped under the transformation x ¼ f ðu; vÞ; y ¼ gðu; vÞ. Another method of justifying the above method of change of variables makes use of line integrals and Green’s theorem in the plane (see Chapter 10, Problem 10.32). 9.7. If u ¼ x2 y2 and v ¼ 2xy, find @ðx; yÞ=@ðu; vÞ in terms of u and v. @ðu; vÞ @ðx; yÞ ¼ ux uy vx vy ¼ 2x 2y 2y 2x ¼ 4ðx2 þ y2 Þ From the identify ðx2 þ y2 Þ2 ¼ ðx2 y2 Þ2 þ ð2xyÞ2 we have ðx2 þ y2 Þ2 ¼ u2 þ v2 and x2 þ y2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p Then by Problem 6.43, Chapter 6, @ðx; yÞ @ðu; vÞ ¼ 1 @ðu; vÞ=@ðx; yÞ ¼ 1 4ðx2 þ y2Þ ¼ 1 4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p Another method: Solve the given equations for x and y in terms of u and v and find the Jacobian directly. 9.8. Find the polar moment of inertia of the region in the xy plane bounded by x2 y2 ¼ 1, x2 y2 ¼ 9, xy ¼ 2; xy ¼ 4 assuming unit density. Under the transformation x2 y2 ¼ u, 2xy ¼ v the required region r in the xy plane [shaded in Fig. 9-12(a)] is mapped into region r0 of the uv plane [shaded in Fig. 9-12(b)]. Then: Required polar moment of inertia ¼ ð ð r ðx2 þ y2 Þ dx dy ¼ ð ð r0 ðx2 þ y2 Þ @ðx; yÞ @ðu; vÞ du dv ¼ ð ð r0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p du dv 4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 þ v2 p ¼ 1 4 ð9 u¼1 ð8 v¼4 du dv ¼ 8 where we have used the results of Problem 9.7. CHAP. 9] MULTIPLE INTEGRALS 217
- 227. Note that the limits of integration for the region r0 can be constructed directly from the region r in the xy plane without actually constructing the region r0 . In such case we use a grid as in Problem 9.6. The coordinates ðu; vÞ are curvilinear coordinates, in this case called hyperbolic coordinates. 9.9. Evaluate ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 q dx dy, where r is the region in the xy plane bounded by x2 þ y2 ¼ 4 and x2 þ y2 ¼ 9. The presence of x2 þ y2 suggests the use of polar coordinates ð; Þ, where x ¼ cos ; y ¼ sin (see Problem 6.39, Chapter 6). Under this transformation the region r [Fig. 9-13(a) below] is mapped into the region r0 [Fig. 9-13(b) below]. Since @ðx; yÞ @ð; Þ ¼ , it follows that ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 q dx dy ¼ ð ð r0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 q @ðx; yÞ @ð; Þ d d ¼ ð ð r0 d d ¼ ð2 ¼0 ð3 ¼2 2 d d ¼ ð2 ¼0 3 3 3 2 d ¼ ð2 ¼0 19 3 d ¼ 38 3 218 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-12 Fig. 9-13
- 228. We can also write the integration limits for r0 immediately on observing the region r, since for fixed , varies from ¼ 2 to ¼ 3 within the sector shown dashed in Fig. 9-13(a). An integration with respect to from ¼ 0 to ¼ 2 then gives the contribution from all sectors. Geometrically, d d represents the area dA as shown in Fig. 9-13(a). 9.10. Find the area of the region in the xy plane bounded by the lemniscate 2 ¼ a2 cos 2. Here the curve is given directly in polar coordinates ð; Þ. By assigning various values to and finding corresponding values of , we obtain the graph shown in Fig. 9-14. The required area (making use of symmetry) is 4 ð=4 ¼0 ða ffiffiffiffiffiffiffiffiffi cos 2 p ¼0 d d ¼ 4 ð=4 ¼0 3 2 a ffiffiffiffiffiffiffiffiffi cos 2 p ¼0 d ¼ 2 ð=4 ¼0 a2 cos 2 d ¼ a2 sin 2 =4 ¼0 ¼ a2 TRIPLE INTEGRALS 9.11. (a) Sketch the three-dimensional region r bounded by x þ y þ z ¼ a ða 0Þ; x ¼ 0; y ¼ 0; z ¼ 0. (b) Give a physical interpretation to ð ð ð r ðx2 þ y2 þ z2 Þ dx dy dz (c) Evaluate the triple integral in (b). (a) The required region r is shown in Fig. 9-15. (b) Since x2 þ y2 þ z2 is the square of the distance from any point ðx; y; zÞ to ð0; 0; 0Þ, we can consider the triple integral as representing the polar moment of inertia (i.e., moment of inertia with respect to the origin) of the region r (assuming unit density). We can also consider the triple integral as representing the mass of the region if the density varies as x2 þ y2 þ z2 . CHAP. 9] MULTIPLE INTEGRALS 219 Fig. 9-14 Fig. 9-15
- 229. (c) The triple integral can be expressed as the iterated integral ða x¼0 ðax y¼0 ðaxy z¼0 ðx2 þ y2 þ z2 Þ dz dy dx ¼ ða x¼0 ðax y¼0 x2 z þ y2 z þ z3 3 axy z¼0 dy dx ¼ ða x¼0 ðax y¼0 x2 ða xÞ x2 y þ ða xÞy2 y3 þ ða x yÞ3 3 ( ) dy dx ¼ ða x¼0 x2 ða xÞy x2 y2 2 þ ða xÞy3 3 y4 4 ða x yÞ4 12 ax y¼0 dx ¼ ða 0 x2 ða xÞ2 x2 ða xÞ2 2 þ ða xÞ4 3 ða xÞ4 4 þ ða xÞ4 12 ( ) dx ¼ ða 0 x2 ða xÞ2 2 þ ða xÞ4 6 ( ) dx ¼ a5 20 The integration with respect to z (keeping x and y constant) from z ¼ 0 to z ¼ a x y corre- sponds to summing the polar moments of inertia (or masses) corresponding to each cube in a vertical column. The subsequent integration with respect to y from y ¼ 0 to y ¼ a x (keeping x constant) corresponds to addition of contributions from all vertical columns contained in a slab parallel to the yz plane. Finally, integration with respect to x from x ¼ 0 to x ¼ a adds up contributions from all slabs parallel to the yz plane. Although the above integration has been accomplished in the order z; y; x, any other order is clearly possible and the final answer should be the same. 9.12. Find the (a) volume and (b) centroid of the region r bounded by the parabolic cylinder z ¼ 4 x2 and the planes x ¼ 0, y ¼ 0, y ¼ 6, z ¼ 0 assuming the density to be a constant . The region r is shown in Fig. 9-16. 220 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-16
- 230. ðaÞ Required volume ¼ ð ð ð r dx dy dz ¼ ð2 x¼0 ð6 y¼0 ð4x2 z¼0 dz dy dx ¼ ð2 x¼0 ð6 y¼0 ð4 x2 Þ dy dx ¼ ð2 x¼0 ð4 x2 Þy 6 y¼0 dx ¼ ð2 x¼0 ð24 6x2 Þ dx ¼ 32 (b) Total mass ¼ ð2 x¼0 ð6 y¼0 ð4x2 z¼0 dz dy dx ¼ 32 by part (a), since is constant. Then x x ¼ Total moment about yz plane Total mass ¼ ð2 x¼0 ð6 y¼0 ð4x2 z¼0 x dz dy dx Total mass ¼ 24 32 ¼ 3 4 y y ¼ Total moment about xz plane Total mass ¼ Ð2 x¼0 Ð6 y¼0 Ð4x2 z¼0 y dz dy dx Total mass ¼ 96 32 ¼ 3 z z ¼ Total moment about xy plane Total mass ¼ ð2 x¼0 ð6 y¼0 ð4x2 z¼0 z dz dy dx Total mass ¼ 256 =5 32 ¼ 8 5 Thus, the centroid has coordinates ð3=4; 3; 8=5Þ. Note that the value for y y could have been predicted because of symmetry. TRANSFORMATION OF TRIPLE INTEGRALS 9.13. Justify equation (11), Page 211, for changing variables in a triple integral. By analogy with Problem 9.6, we construct a grid of curvilinear coordinate surfaces which subdivide the region r into subregions, a typical one of which is r (see Fig. 9-17). CHAP. 9] MULTIPLE INTEGRALS 221 Fig. 9-17
- 231. The vector r from the origin O to point P is r ¼ xi þ yj þ zk ¼ f ðu; v; wÞi þ gðu; v; wÞj þ hðu; v; wÞk assuming that the transformation equations are x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ, and z ¼ hðu; v; wÞ. Tangent vectors to the coordinate curves corresponding to the intersection of pairs of coordinate surfaces are given by @r=@u; @r=@v; @r=@w. Then the volume of the region r of Fig. 9-17 is given approxi- mately by @r @u @r @v @r @w u v w ¼ @ðx; y; zÞ @ðu; v; wÞ u v w The triple integral of Fðx; y; zÞ over the region is the limit of the sum X Ff f ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞg @ðx; y; zÞ @ðu; v; wÞ u v w An investigation reveals that this limit is ð ð ð r0 F f f ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞg @ðx; y; zÞ @ðu; v; wÞ du dv dw where r0 is the region in the uvw space into which the region r is mapped under the transformation. Another method for justifying the above change of variables in triple integrals makes use of Stokes’ theorem (see Problem 10.84, Chapter 10). 9.14. What is the mass of a circular cylindrical body represented by the region 0 @ @ c; 0 @ @ 2; 0 @ z @ h, and with the density function ¼ z sin2 ? M ¼ ðh 0 ð2 0 ðc 0 z sin2 d d dz ¼ 9.15. Use spherical coordinates to calculate the volume of a sphere of radius a. V ¼ 8 ða 0 ð=2 0 ð=2 0 a2 sin dr d d ¼ 4 3 a3 9.16. Express ð ð ð r Fðx; y; zÞ dx dy dz in (a) cylindrical and (b) spherical coordinates. (a) The transformation equations in cylindrical coordinates are x ¼ cos ; y ¼ sin ; z ¼ z. As in Problem 6.39, Chapter 6, @ðx; y; zÞ=@ð; ; zÞ ¼ . Then by Problem 9.13 the triple integral becomes ð ð ð r0 Gð; ; zÞ d d dz where r0 is the region in the ; ; z space corresponding to r and where Gð; ; z Fð cos ; sin ; zÞ. (b) The transformation equations in spherical coordinates are x ¼ r sin cos ; y ¼ r sin sin ; z ¼ r cos . By Problem 6.101, Chapter 6, @ðx; y; zÞ=@ðr; ; Þ ¼ r2 sin . Then by Problem 9.13 the triple integral becomes ð ð ð r0 Hðr; ; Þr2 sin dr d d where r0 is the region in the r; ; space corresponding to r, and where Hðr; ; Þ Fðr sin cos , r sin sin ; r cos Þ. 222 MULTIPLE INTEGRALS [CHAP. 9
- 232. 9.17. Find the volume of the region above the xy plane bounded by the paraboloid z ¼ x2 þ y2 and the cylinder x2 þ y2 ¼ a2 . The volume is most easily found by using cylindrical coordinates. In these coordinates the equations for the paraboloid and cylinder are respectively z ¼ 2 and ¼ a. Then Required volume ¼ 4 times volume shown in Fig. 9-18 ¼ 4 ð=2 ¼0 ða ¼0 ð2 z¼0 dz d d ¼ 4 ð=2 ¼0 ða ¼0 3 d d ¼ 4 ð=2 hi¼0 4 4 a ¼0 d ¼ 2 a4 The integration with respect to z (keeping and constant) from z ¼ 0 to z ¼ 2 corresponds to summing the cubical volumes (indicated by dVÞ in a vertical column extending from the xy plane to the paraboloid. The subsequent integration with respect to (keeping constant) from ¼ 0 to ¼ a corresponds to addition of volumes of all columns in the wedge-shaped region. Finally, integration with respect to corresponds to adding volumes of all such wedge-shaped regions. The integration can also be performed in other orders to yield the same result. We can also set up the integral by determining the region r0 in ; ; z space into which r is mapped by the cylindrical coordinate transformation. 9.18. (a) Find the moment of inertia about the z-axis of the region in Problem 9.17, assuming that the density is the constant . (b) Find the radius of gyration. (a) The moment of inertia about the z-axis is Iz ¼ 4 ð=2 0 ða ¼0 ð2 z¼0 2 dz d d ¼ 4 ð=2 ¼0 ða ¼0 5 d d ¼ 4 ð=2 ¼0 6 6 a ¼0 d ¼ a6 3 CHAP. 9] MULTIPLE INTEGRALS 223 Fig. 9-18
- 233. The result can be expressed in terms of the mass M of the region, since by Problem 9.17, M ¼ volume density ¼ 2 a4 so that Iz ¼ a6 3 ¼ a6 3 2M a4 ¼ 2 3 Ma2 Note that in setting up the integral for Iz we can think of dz d d as being the mass of the cubical volume element, 2 dz d d, as the moment of inertia of this mass with respect to the z-axis and ð ð ð r 2 dz d d as the total moment of inertia about the z-axis. The limits of integration are determined as in Problem 9.17. (b) The radius of gyration is the value K such that MK2 ¼ 2 3 Ma2 , i.e., K2 ¼ 2 3 a2 or K ¼ a ffiffiffiffiffiffiffiffi 2=3 p . The physical significance of K is that if all the mass M were concentrated in a thin cylindrical shell of radius K, then the moment of inertia of this shell about the axis of the cylinder would be Iz. 9.19. (a) Find the volume of the region bounded above by the sphere x2 þ y2 þ z2 ¼ a2 and below by the cone z2 sin2 ¼ ðx2 þ y2 Þ cos2 , where is a constant such that 0 @ @ . (b) From the result in (a), find the volume of a sphere of radius a. In spherical coordinates the equation of the sphere is r ¼ a and that of the cone is ¼ . This can be seen directly or by using the transformation equations x ¼ r sin cos ; y ¼ r sin sin , z ¼ r cos . For example, z2 sin2 ¼ ðx2 þ y2 Þ cos2 becomes, on using these equations, r2 cos2 sin2 ¼ ðr2 sin2 cos2 þ r2 sin2 sin2 Þ cos2 i.e., r2 cos2 sin2 ¼ r2 sin2 cos2 from which tan ¼ tan and so ¼ or ¼ . It is sufficient to consider one of these, say, ¼ . ðaÞ Required volume ¼ 4 times volume (shaded) in Fig. 9-19 ¼ 4 ð=2 ¼0 ð ¼0 ða r¼0 r2 sin dr d d ¼ 4 ð=2 ¼0 ð ¼0 r3 3 sin r¼0 d d ¼ 4a3 3 ð=2 ¼0 ð ¼0 sin d d ¼ 4a3 3 ð=2 ¼0 cos ¼0 d ¼ 2a3 3 ð1 cos Þ The integration with respect to r (keeping and constant) from r ¼ 0 to r ¼ a corresponds to summing the volumes of all cubical elements (such as indicated by dV) in a column extending from r ¼ 0 to r ¼ a. The subsequent integration with respect to (keeping constant) from ¼ 0 to ¼ =4 corresponds to summing the volumes of all columns in the wedge-shaped region. Finally, integration with respect to corresponds to adding volumes of all such wedge-shaped regions. 224 MULTIPLE INTEGRALS [CHAP. 9 Fig. 9-19
- 234. (b) Letting ¼ , the volume of the sphere thus obtained is 2a3 3 ð1 cos Þ ¼ 4 3 a3 9.20. ðaÞ Find the centroid of the region in Problem 9.19. (b) Use the result in (a) to find the centroid of a hemisphere. (a) The centroid ð x x; y y; z zÞ is, due to symmetry, given by x x ¼ y y ¼ 0 and z z ¼ Total moment about xy plane Total mass ¼ Ð Ð Ð z dV Ð Ð Ð dV Since z ¼ r cos and is constant the numerator is 4 ð=2 ¼0 ð ¼0 ða r¼0 r cos r2 sin dr d d ¼ 4 ð=2 ¼0 ð ¼0 r4 4 a r¼0 sin cos d d ¼ a4 ð=2 ¼0 ð ¼0 sin cos d d ¼ a4 ð=2 ¼0 sin2 2 ¼0 d ¼ a4 sin2 4 The denominator, obtained by multiplying the result of Problem 9.19(a) by , is 2 3 a3 ð1 cos Þ. Then z z ¼ 1 4 a4 sin2 2 3 a3ð1 cos Þ ¼ 3 8 að1 þ cos Þ: (b) Letting ¼ =2; z z ¼ 3 8 a. MISCELLANEOUS PROBLEMS 9.21. Prove that (a) ð1 0 ð1 0 x y ðx þ yÞ3 dy dx ¼ 1 2 , (b) ð1 0 ð1 0 x y ðx þ yÞ3 dx dy ¼ 1 2 . ðaÞ ð1 0 ð1 0 x y ðx þ yÞ3 dy dx ¼ ð1 0 ð1 0 2x ðx þ yÞ ðx þ yÞ3 dy dx ¼ ð1 0 ð1 0 2x ðx þ yÞ3 1 ðx þ yÞ2 dy dx ¼ ð1 0 x ðx þ yÞ2 þ 1 x þ y 1 y¼0 dx ¼ ð1 0 dx ðx þ 1Þ2 ¼ 1 x þ 1 1 0 ¼ 1 2 (b) This follows at once on formally interchanging x and y in (a) to obtain ð1 0 ð1 0 y x ðx þ yÞ3 dx dy ¼ 1 2 and then multiplying both sides by 1. This example shows that interchange in order of integration may not always produce equal results. A sufficient condition under which the order may be interchanged is that the double integral over the corresponding region exists. In this case ð ð r x y ðx þ yÞ3 dx dy, where r is the region 0 @ x @ 1; 0 @ y @ 1 fails to exist because of the discontinuity of the integrand at the origin. The integral is actually an improper double integral (see Chapter 12). 9.22. Prove that ðx 0 ðt 0 FðuÞ du dt ¼ ðx 0 ðx uÞFðuÞ du. CHAP. 9] MULTIPLE INTEGRALS 225
- 235. Let IðxÞ ¼ ðx 0 ðt 0 FðuÞ du dt; JðxÞ ¼ ðz 0 ðx uÞFðuÞ du: Then I 0 ðxÞ ¼ ðz 0 FðuÞ du; J 0 ðxÞ ¼ ðz 0 FðuÞ du using Leibnitz’s rule, Page 186. Thus, I 0 ðxÞ ¼ J 0 ðxÞ, and so IðxÞ JðxÞ ¼ c, where c is a constant. Since Ið0Þ ¼ Jð0Þ ¼ 0, c ¼ 0, and so IðxÞ ¼ JðxÞ. The result is sometimes written in the form ðx 0 ðx 0 FðxÞ dx2 ¼ ðx 0 ðx uÞFðuÞ du The result can be generalized to give (see Problem 9.58) ðx 0 ðx 0 ðx 0 FðxÞ dxn ¼ 1 ðn 1Þ! ðx 0 ðx uÞn1 FðuÞ du Supplementary Problems DOUBLE INTEGRALS 9.23. (a) Sketch the region r in the xy plane bounded by y2 ¼ 2x and y ¼ x. (b) Find the area of r. (c) Find the polar moment of inertia of r assuming constant density . Ans. (b) 2 3 ; ðcÞ 48 =35 ¼ 72M=35, where M is the mass of r. 9.24. Find the centroid of the region in the preceding problem. Ans. x x ¼ 4 5 ; y y ¼ 1 9.25. Given ð3 y¼0 ð ffiffiffiffiffiffi 4y p x¼1 ðx þ yÞ dx dy. (a) Sketch the region and give a possible physical interpretation of the double integral. (b) Interchange the order of integration. (c) Evaluate the double integral. Ans: ðbÞ ð2 x¼1 ð4x2 y¼0 ðx þ yÞ dy dx; ðcÞ 241=60 9:26: Show that ð2 x¼1 ðx y¼ ffiffi x p sin x 2y dy dx þ ð4 x¼2 ð2 y¼ ffiffi x p sin x 2y dy dx ¼ 4ð þ 2Þ 3 : 9.27. Find the volume of the tetrahedron bounded by x=a þ y=b þ z=c ¼ 1 and the coordinate planes. Ans. abc=6 9.28. Find the volume of the region bounded by z ¼ x3 þ y2 ; z ¼ 0; x ¼ a; x ¼ a; y ¼ a; y ¼ a. Ans. 8a4 =3 9.29. Find (a) the moment of inertia about the z-axis and (b) the centroid of the region in Problem 9.28 assuming a constant density . Ans. (a) 112 45 a6 ¼ 14 15 Ma2 , where M ¼ mass; (b) x x ¼ y y ¼ 0; z z ¼ 7 15 a2 TRANSFORMATION OF DOUBLE INTEGRALS 9.30. Evaluate ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 q dx dy, where r is the region x2 þ y2 @ a2 . Ans. 2 3 a3 226 MULTIPLE INTEGRALS [CHAP. 9
- 236. 9:31: If r is the region of Problem 9.30, evaluate ð ð r eðx2 þy2 Þ dx dy: Ans: ð1 ea2 Þ 9.32. By using the transformation x þ y ¼ u; y ¼ uv, show that ð1 x¼0 ð1x y¼0 ey=ðxþyÞ dy dx ¼ e 1 2 9.33. Find the area of the region bounded by xy ¼ 4; xy ¼ 8; xy3 ¼ 5; xy3 ¼ 15. [Hint: Let xy ¼ u; xy3 ¼ v.] Ans: 2 ln 3 9.34. Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas y2 ¼ x; y2 ¼ 8x; x2 ¼ y; x2 ¼ 8y about the x-axis is 279=2. [Hint: Let y2 ¼ ux; x2 ¼ vy.] 9.35. Find the area of the region in the first quadrant bounded by y ¼ x3 ; y ¼ 4x3 ; x ¼ y3 ; x ¼ 4y3 . Ans: 1 8 9.36. Let r be the region bounded by x þ y ¼ 1; x ¼ 0; y ¼ 0. Show that ð ð r cos x y x þ y dx dy ¼ sin 1 2 . [Hint: Let x y ¼ u; x þ y ¼ v.] TRIPLE INTEGRALS 9.37. (a) Evaluate ð1 x¼0 ð1 y¼0 ð2 z¼ ffiffiffiffiffiffiffiffiffiffi x2þy2 p xyz dz dy dx: ðbÞ Give a physical interpretation to the integral in (a). Ans: ðaÞ 3 8 9.38. Find the (a) volume and (b) centroid of the region in the first octant bounded by x=a þ y=b þ z=c ¼ 1, where a; b; c are positive. Ans: ðaÞ abc=6; ðbÞ x x ¼ a=4; y y ¼ b=4; z z ¼ c=4 9.39. Find the (a) moment of inertia and (b) radius of gyration about the z-axis of the region in Problem 9.38. Ans: ðaÞ Mða2 þ b2 Þ=10; ðbÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða2 þ b2Þ=10 p 9.40. Find the mass of the region corresponding to x2 þ y2 þ z2 @ 4; x A 0; y A 0; z A 0, if the density is equal to xyz. Ans: 4=3 9.41. Find the volume of the region bounded by z ¼ x2 þ y2 and z ¼ 2x. Ans: =2 TRANSFORMATION OF TRIPLE INTEGRALS 9.42. Find the volume of the region bounded by z ¼ 4 x2 y2 and the xy plane. Ans: 8 9.43. Find the centroid of the region in Problem 9.42, assuming constant density . Ans: x x ¼ y y ¼ 0; z z ¼ 4 3 9.44. (a) Evaluate ð ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 þ z2 q dx dy dz, where r is the region bounded by the plane z ¼ 3 and the cone z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p . (b) Give a physical interpretation of the integral in (a). [Hint: Perform the integration in cylindrical coordinates in the order ; z; .] Ans: 27ð2 ffiffiffi 2 p 1Þ=2 9.45. Show that the volume of the region bonded by the cone z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p and the paraboloid z ¼ x2 þ y2 is =6. 9.46. Find the moment of inertia of a right circular cylinder of radius a and height b, about its axis if the density is proportional to the distance from the axis. Ans: 3 5 Ma2 CHAP. 9] MULTIPLE INTEGRALS 227
- 237. 9.47. (a) Evaluate ð ð ð r dx dy dz ðx2 þ y2 þ z2Þ3=2 , where r is the region bounded by the spheres x2 þ y2 þ z2 ¼ a2 and x2 þ y2 þ z2 ¼ b2 where a b 0. (b) Give a physical interpretation of the integral in (a). Ans: ðaÞ 4 lnða=bÞ 9.48. (a) Find the volume of the region bounded above by the sphere r ¼ 2a cos , and below by the cone ¼ where 0 =2. (b) Discuss the case ¼ =2. Ans: 4 3 a3 ð1 cos4 Þ 9.49. Find the centroid of a hemispherical shell having outer radius a and inner radius b if the density (a) is constant, (b) varies as the square of the distance from the base. Discuss the case a ¼ b. Ans. Taking the z-axis as axis of symmetry: (a) x x ¼ y y ¼ 0; z z ¼ 3 8 ða4 b4 Þ=ða3 b3 Þ; ðbÞ x x ¼ y y ¼ 0, z z ¼ 5 8 ða6 b6 Þ=ða5 b5 Þ MISCELLANEOUS PROBLEMS 9.50. Find the mass of a right circular cylinder of radius a and height b if the density varies as the square of the distance from a point on the circumference of the base. Ans: 1 6 a2 bkð9a2 þ 2b2 Þ, where k ¼ constant of proportionality. 9.51. Find the (a) volume and (b) centroid of the region bounded above by the sphere x2 þ y2 þ z2 ¼ a2 and below by the plane z ¼ b where a b 0, assuming constant density. Ans: ðaÞ 1 3 ð2a3 3a2 b þ b3 Þ; ðbÞ x x ¼ y y ¼ 0; z z ¼ 3 4 ða þ bÞ2 =ð2a þ bÞ 9.52. A sphere of radius a has a cylindrical hole of radius b bored from it, the axis of the cylinder coinciding with a diameter of the sphere. Show that the volume of the sphere which remains is 4 3 ½a3 ða2 b2 Þ3=2 . 9.53. A simple closed curve in a plane is revolved about an axis in the plane which does not intersect the curve. Prove that the volume generated is equal to the area bounded by the curve multiplied by the distance traveled by the centroid of the area (Pappus’ theorem). 9.54. Use Problem 9.53 to find the volume generated by revolving the circle x2 þ ðy bÞ2 ¼ a2 ; b a 0 about the x-axis. Ans: 22 a2 b 9.55. Find the volume of the region bounded by the hyperbolic cylinders xy ¼ 1; xy ¼ 9; xz ¼ 4; xz ¼ 36, yz ¼ 25, yz ¼ 49. [Hint: Let xy ¼ u; xz ¼ v; yz ¼ w:] Ans: 64 9.56. Evaluate ð ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðx2=a2 þ y2=b2 þ z2=c2Þ q dx dy dz, where r is the region interior to the ellipsoid x2 =a2 þ y2 =b2 þ z2 =c2 ¼ 1. [Hint: Let x ¼ au; y ¼ bv; z ¼ cw. Then use spherical coordinates.] Ans: 1 4 2 abc 9.57. If r is the region x2 þ xy þ y2 @ 1, prove that ð ð r eðx2 þxyþy2 Þ dx dy ¼ 2 e ffiffiffi 3 p ðe 1Þ. [Hint: Let x ¼ u cos v sin , y ¼ u sin þ v cos and choose so as to eliminate the xy term in the integrand. Then let u ¼ a cos , v ¼ b sin where a and b are appropriately chosen.] 9.58. Prove that ðx 0 ðx 0 ðx 0 FðxÞ dxn ¼ 1 ðn 1Þ! ðx 0 ðx uÞn1 FðuÞ du for n ¼ 1; 2; 3; . . . (see Problem 9.22). 228 MULTIPLE INTEGRALS [CHAP. 9
- 238. 229 Line Integrals, Surface Integrals, and Integral Theorems Construction of mathematical models of physical phenomena requires functional domains of greater complexity than the previously employed line segments and plane regions. This section makes progress in meeting that need by enriching integral theory with the introduction of segments of curves and portions of surfaces as domains. Thus, single integrals as functions defined on curve segments take on new meaning and are then called line integrals. Stokes’s theorem exhibits a striking relation between the line integral of a function on a closed curve and the double integral of the surface portion that is enclosed. The divergence theorem relates the triple integral of a function on a three-dimensional region of space to its double integral on the bounding surface. The elegant language of vectors best describes these concepts; therefore, it would be useful to reread the introduction to Chapter 7, where the impor- tance of vectors is emphasized. (The integral theorems also are expressed in coordinate form.) LINE INTEGRALS The objective of this section is to geometrically view the domain of a vector or scalar function as a segment of a curve. Since the curve is defined on an interval of real numbers, it is possible to refer the function to this primitive domain, but to do so would suppress much geometric insight. A curve, C, in three-dimensional space may be represented by parametric equations: x ¼ f1ðtÞ; y ¼ f2ðtÞ; z ¼ f3ðtÞ; a @ t @ b ð1Þ or in vector notation: x ¼ rðtÞ ð2Þ where rðtÞ ¼ xi þ yj þ zk (see Fig. 10-1). Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 239. For this discussion it is assumed that r is continuously differentiable. While (as we are doing) it is convenient to refer the Euclidean space to a rectangular Cartesian coordinate system, it is not necessary. (For example, cylindrical and spherical coordinates sometimes are more useful.) In fact, one of the objectives of the vector language is to free us from any particular frame of reference. Then, a vector A½xðtÞ; yðtÞ; zðtÞ or a scalar, , is pictured on the domain C, which according to the parametric repre- sentation, is referred to the real number interval a @ t @ b. The Integral ð C A dr ð3Þ of a vector field A defined on a curve segment C is called a line integral. The integrand has the representation A1 dx þ A2 dy þ A3 dz obtained by expanding the dot product. The scalar and vector integrals ð C ðtÞ dt ¼ lim n!1 X n k¼1 ðk; k; kÞtk ð4Þ ð C AðtÞdt ¼ lim n!1 X n k¼1 Aðk; k; kÞtÞk ð5Þ can be interpreted as line integrals; however, they do not play a major role [except for the fact that the scalar integral (3) takes the form (4)]. The following three basic ways are used to evaluate the line integral (3): 1. The parametric equations are used to express the integrand through the parameter t. Then ð C A dr ¼ ðt2 t1 A dr dt dt 2. If the curve C is a plane curve (for example, in the xy plane) and has one of the representations y ¼ f ðxÞ or x ¼ gðyÞ, then the two integrals that arise are evaluated with respect to x or y, whichever is more convenient. 230 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-1
- 240. 3. If the integrand is a perfect differential, then it may be evaluated through knowledge of the end points (that is, without reference to any particular joining curve). (See the section on indepen- dence of path on Page 232; also see Page 237.) These techniques are further illustrated below for plane curves and for three space in the problems. EVALUATION OF LINE INTEGRALS FOR PLANE CURVES If the equation of a curve C in the plane z ¼ 0 is given as y ¼ f ðxÞ, the line integral (2) is evaluated by placing y ¼ f ðxÞ; dy ¼ f 0 ðxÞ dx in the integrand to obtain the definite integral ða2 a1 Pfx; f ðxÞg dx þ Qfx; f ðxÞg f 0 ðxÞ dx ð7Þ which is then evaluated in the usual manner. Similarly, if C is given as x ¼ gðyÞ, then dx ¼ g0 ðyÞ dy and the line integral becomes ðb2 b1 Pfgð yÞ; ygg0 ð yÞ dy þ Qfgð yÞ; yg dy ð8Þ If C is given in parametric form x ¼ ðtÞ; y ¼ ðtÞ, the line integral becomes ðt2 t1 PfðtÞ; ðtÞg0 ðtÞ dt þ QfðtÞ; ðtÞg; 0 ðtÞ dt ð9Þ where t1 and t2 denote the values of t corresponding to points A and B, respectively. Combinations of the above methods may be used in the evaluation. If the integrand A dr is a perfect differential, d, then ð C A dr ¼ ððc;dÞ ða;bÞ d ¼ ðc; dÞ ða; bÞ ð6Þ Similar methods are used for evaluating line integrals along space curves. PROPERTIES OF LINE INTEGRALS EXPRESSED FOR PLANE CURVES Line integrals have properties which are analogous to those of ordinary integrals. For example: 1: ð C Pðx; yÞ dx þ Qðx; yÞ dy ¼ ð C Pðx; yÞ dx þ ð C Qðx; yÞ dy 2: ðða2;b2Þ ða1;b1Þ P dx þ Q dy ¼ ðða1;b1Þ ða2;b2Þ P dx þ q dy Thus, reversal of the path of integration changes the sign of the line integral. 3: ðða2;b2Þ ða2;b1Þ P dx þ Q dy ¼ ðða3;b3Þ ða1;b1Þ P dx þ Q dy þ ðða2;b2Þ ða3;b3Þ P dx þ Q dy where ða3; b3Þ is another point on C. Similar properties hold for line integrals in space. CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 231
- 241. SIMPLE CLOSED CURVES, SIMPLY AND MULTIPLY CONNECTED REGIONS A simple closed curve is a closed curve which does not intersect itself anywhere. Mathematically, a curve in the xy plane is defined by the parametric equations x ¼ ðtÞ; y ¼ ðtÞ where and are single- valued and continuous in an interval t1 @ t @ t2. If ðt1Þ ¼ ðt2Þ and ðt1Þ ¼ ðt2Þ, the curve is said to be closed. If ðuÞ ¼ ðvÞ and ðuÞ ¼ ðvÞ only when u ¼ v (except in the special case where u ¼ t1 and v ¼ t2), the curve is closed and does not intersect itself and so is a simple closed curve. We shall also assume, unless otherwise stated, that and are piecewise differentiable in t1 @ t @ t2. If a plane region has the property that any closed curve in it can be continuously shrunk to a point without leaving the region, then the region is called simply connected; otherwise, it is called multiply con- nected (see Fig. 10-2 and Page 118 of Chapter 6). As the parameter t varies from t1 to t2, the plane curve is described in a certain sense or direction. For curves in the xy plane, we arbitrarily describe this direction as positive or negative according as a person traversing the curve in this direction with his head pointing in the positive z direction has the region enclosed by the curve always toward his left or right, respectively. If we look down upon a simple closed curve in the xy plane, this amounts to saying that traversal of the curve in the counterclockwise direction is taken as positive while traversal in the clockwise direction is taken as negative. GREEN’S THEOREM IN THE PLANE This theorem is needed to prove Stokes’ theorem (Page 237). Then it becomes a special case of that theorem. Let P, Q, @P=@y; @Q=@x be single-valued and continuous in a simply connected region r bounded by a simple closed curve C. Then þ C P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy ð10Þ where þ C is used to emphasize that C is closed and that it is described in the positive direction. This theorem is also true for regions bounded by two or more closed curves (i.e., multiply connected regions). See Problem 10.10. CONDITIONS FOR A LINE INTEGRAL TO BE INDEPENDENT OF THE PATH The line integral of a vector field A is independent of path if its value is the same regardless of the (allowable) path from initial to terminal point. (Thus, the integral is evaluated from knowledge of the coordinates of these two points.) For example, the integral of the vector field A ¼ yi þ xj is independent of path since ð C A dr ¼ ð C y dx þ x dy ¼ ðx2y2 x1y1 dðxyÞ ¼ x2y2 x1y1 Thus, the value of the integral is obtained without reference to the curve joining P1 and P2. This notion of the independence of path of line integrals of certain vector fields, important to theory and application, is characterized by the following three theorems: Theorem 1. A necessary and sufficient condition that ð C A dr be independent of path is that there exists a scalar function such that A ¼ r. 232 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-2
- 242. Theorem 2. A necessary and sufficient condition that the line integral, ð C A dr be independent of path is that r A ¼ 0. Theorem 3. If r A ¼ 0, then the line integral of A over an allowable closed path is 0, i.e., þ A dr ¼ 0. If C is a plane curve, then Theorem 3 follows immediately from Green’s theorem, since in the plane case r A reduces to @A1 @y ¼ @A2 @x EXAMPLE. Newton’s second law for forces is F ¼ dðmvÞ dt , where m is the mass of an object and v is its velocity. When F has the representation F ¼ r, it is said to be conservative. The previous theorems tell us that the integrals of conservative fields of force are independent of path. Furthermore, showing that r F ¼ 0 is the preferred way of showing that F is conservative, since it involves differentiation, while demonstrating that exists such that F ¼ r requires integration. SURFACE INTEGRALS Our previous double integrals have been related to a very special surface, the plane. Now we consider other surfaces, yet, the approach is quite similar. Surfaces can be viewed intrinsically, i.e., as non-Euclidean spaces; however, we do not do that. Rather, the surface is thought of as embedded in a three-dimensional Euclidean space and expressed through a two-parameter vector representation: x ¼ rðv1; v2Þ While the purpose of the vector representation is to be general (that is, interpretable through any allowable three-space coordinate system), it is convenient to initially think in terms of rectangular Cartesian coordinates; therefore, assume r ¼ xi þ yj þ zk and that there is a parametric representation x ¼ rðv1; v2Þ; y ¼ rðv1; v2Þ; z ¼ rðv1; v2Þ ð11Þ The functions are assumed to be continuously differentiable. The parameter curves v2 ¼ const and v1 ¼ const establish a coordinate system on the surface (just as y ¼ const, and x ¼ const form such a system in the plane). The key to establishing the surface integral of a function is the differential element of surface area. (For the plane that element is dA ¼ dx; dy.) At any point, P, of the surface dx ¼ @r @v1 dv1 þ @r @v2 dv2 spans the tangent plane to the surface. In particular, the directions of the coordinate curves v2 ¼ const and v1 ¼ const are designated by dx1 ¼ @r @v1 dv1 and dx2 ¼ @r @v2 dv2, respectively (see Fig. 10-3). The cross product dx1x dx2 ¼ @r @v1 @r @v2 dv1 dv2 is normal to the tangent plane at P, and its magnitude @r @v1 @r @v2 is the area of a differential coordinate parallelogram. CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 233
- 243. (This is the usual geometric interpretation of the cross product abstracted to the differential level.) This strongly suggests the following definition: Definition. The differential element of surface area is dS ¼ @r @v1 @r @v2 dv1 dv2 ð12Þ For a function ðv1; v2Þ that is everywhere integrable on S ð ð S dS ¼ ð ð S ðv1; v2Þ @r @v1 @r @v2 dv1 dv2 ð13Þ is the surface integral of the function : In general, the surface integral must be referred to three-space coordinates to be evaluated. If the surface has the Cartesian representation z ¼ f ðx; yÞ and the identifications v1 ¼ x; v2 ¼ y; z ¼ f ðv1; v2Þ are made then @r @v1 ¼ i þ @z @x k; @r @v2 ¼ j þ @z @y k and @r @v2 @r @v2 ¼ k @z @y j @z @x i Therefore, @r @v1 @r @v2 ¼ 1 þ @z @x 2 þ @z @y 2 #1=2 Thus, the surface integral of has the special representation S ¼ ð ð S ðx; y; zÞ 1 þ @z @x 2 þ @z @y 2 #1=2 dx dy ð14Þ If the surface is given in the implicit form Fðx; y; zÞ ¼ 0, then the gradient may be employed to obtain another representation. To establish it, recall that at any surface point P the gradient, rF is perpendicular (normal) to the tangent plane (and hence to S). Therefore, the following equality of the unit vectors holds (up to sign): 234 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-3
- 244. rF jrFj ¼ @r @x @r @y @r @v1 @r @v2 ð15Þ [Now a conclusion of the theory of implicit functions is that from Fðx; y; zÞ ¼ 0 (and under appro- priate conditions) there can be produced an explicit representation z ¼ f ðx; yÞ of a portion of the surface. This is an existence statement. The theorem does not say that this representation can be explicitly produced.] With this fact in hand, we again let v1 ¼ x; v2 ¼ y; z ¼ f ðv1; v2Þ. Then rF ¼ Fxi þ fyj þ Fzk Taking the dot product of both sides of (15) yields Fz jrFj ¼ 1 @r @v1 @r @v2 The ambiguity of sign can be eliminated by taking the absolute value of both sides of the equation. Then @r @v1 @r @v2 ¼ jrFj jFzj ¼ ½ðFxÞ2 þ ðFyÞ2 þ ðFzÞ2 1=2 jFzj and the surface integral of takes the form ð ð S ½ðFxÞ2 þ ðFyÞ2 þ ðFzÞ2 1=2 jFzj dx dy ð16Þ The formulas (14) and (16) also can be introduced in the following nonvectorial manner. Let S be a two-sided surface having projection r on the xy plane as in the adjoining Fig. 10-4. Assume that an equation for S is z ¼ f ðx; yÞ, where f is single-valued and continous for all x and y in r . Divide r into n subregions of area Ap; p ¼ 1; 2; . . . ; n, and erect a vertical column on each of these subregions to intersect S in an area Sp. Let ðx; y; zÞ be single-valued and continuous at all points of S. Form the sum X n p¼1 ðp; p; pÞ Sp ð17Þ CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 235 Fig. 10-4
- 245. where ðp; p; pÞ is some point of Sp. If the limit of this sum as n ! 1 in such a way that each Sp ! 0 exists, the resulting limit is called the surface integral of ðx; y; zÞ over S and is designated by ð ð S ðx; y; zÞ dS ð18Þ Since Sp ¼ j sec pj Ap approximately, where p is the angle between the normal line to S and the positive z-axis, the limit of the sum (17) can be written ð ð r ðx; y; zÞj sec j dA ð19Þ The quantity j sec j is given by j sec j ¼ 1 jnp kj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ @z @x 2 þ @z @y 2 s ð20Þ Then assuming that z ¼ f ðx; yÞ has continuous (or sectionally continuous) derivatives in r, (19) can be written in rectangular form as ð ð r ðx; y; zÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ @z @x 2 þ @z @y 2 s dx dy ð21Þ In case the equation for S is given as Fðx; y; zÞ ¼ 0, (21) can also be written ð ð r ðx; y; zÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðFxÞ2 þ ðFyÞ2 þ ðFzÞ2 q jFzj dx dy ð22Þ The results (21) or (22) can be used to evaluate (18). In the above we have assumed that S is such that any line parallel to the z-axis intersects S in only one point. In case S is not of this type, we can usually subdivide S into surfaces S1; S2; . . . ; which are of this type. Then the surface integral over S is defined as the sum of the surface integrals over S1; S2; . . . . The results stated hold when S is projected on to a region r on the xy plane. In some cases it is better to project S on to the yz or xz planes. For such cases (18) can be evaluated by appropriately modifying (21) and (22). THE DIVERGENCE THEOREM The divergence theorem establishes equality between triple integral (volume integral) of a function over a region of three-dimensional space and the double integral of the function over the surface that bounds that region. This relation is very important in the expression of physical theory. (See Fig. 10-5.) Divergence (or Gauss) Theorem Let A be a vector field that is continuously differentiable on a closed-space region, V, bound by a smooth surface, S. Then ð ð ð V r A dV ¼ ð ð S A n dS ð23Þ where n is an outwardly drawn normal. If n is expressed through direction cosines, i.e., n ¼ i cos þ j cos þ k cos , then (23) may be written 236 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 246. ð ð ð @A1 @x þ @A2 @y þ @A3 @z dV ¼ ð ð S ðA1 cos þ A2 cos þ A3 cos Þ dS ð24Þ The rectangular Cartesian component form of (23) is ð ð ð V @A1 @x þ @A2 @y þ @A3 @z dV ¼ ð ð S ðA1 dy dz þ A2 dz dx þ A3 dx dyÞ ð25Þ EXAMPLE. If B is the magnetic field vector, then one of Maxwell’s equations of electromagnetic theory is r B ¼ 0. When this equation is substituted into the left member of (23), the right member tells us that the magnetic flux through a closed surface containing a magnetic field is zero. A simple interpretation of this fact results by thinking of a magnet enclosed in a ball. All magnetic lines of force that flow out of the ball must return (so that the total flux is zero). Thus, the lines of force flow from one pole to the other, and there is no dispersion. STOKES’ THEOREM Stokes’ theorem establishes the equality of the double integral of a vector field over a portion of a surface and the line integral of the field over a simple closed curve bounding the surface portion. (See Fig. 10-6.) Suppose a closed curve, C, bounds a smooth surface portion, S. If the component functions of x ¼ rðv1; v2Þ have continuous mixed partial derivatives, then for a vector field A with continuous partial derivatives on S þ C A dr ¼ ð ð S n r A dS ð26Þ where n ¼ cos i þ cos j þ cos k with ; , and representing the angles made by the outward normal n and i; j, and k, respectively. Then the component form of (26) is þ C ðA1 dx þ A2 dy þ A3 dzÞ ¼ ð ð S @A3 @y @A2 @z cos þ @A1 @z @A3 @x cos þ @A2 @x @A1 @y cos dS ð27Þ If r A ¼ 0, Stokes’ theorem tells us that þ C A dr ¼ 0. This is Theorem 3 on Page 237. CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 237 Fig. 10-5
- 247. Solved Problems LINE INTEGRALS 10.1. Evaluate ðð1;2Þ ð0;1Þ ðx2 yÞ dx þ ðy2 þ xÞ dy along (a) a straight line from ð0; 1Þ to ð1; 2Þ, (b) straight lines from ð0; 1Þ to ð1; 1Þ and then from ð1; 1Þ to ð1; 2Þ, (c) the parabola x ¼ t, y ¼ t2 þ 1. (a) An equation for the line joining ð0; 1Þ and ð1; 2Þ in the xy plane is y ¼ x þ 1. Then dy ¼ dx and the line integral equals ð1 x¼0 fx2 ðx þ 1Þg dx þ fðx þ 1Þ2 þ xg dx ¼ ð1 0 ð2x2 þ 2xÞ dx ¼ 5=3 (b) Along the straight line from ð0; 1Þ to ð1; 1Þ, y ¼ 1; dy ¼ 0 and the line integral equals ð1 x¼0 ðx2 1Þ dx þ ð1 þ xÞð0Þ ¼ ð1 0 ðx2 1Þ dx ¼ 2=3 Along the straight line from ð1; 1Þ to ð1; 2Þ, x ¼ 1; dx ¼ 0 and the line integral equals ð2 y¼1 ð1 yÞð0Þ þ ð y2 þ 1Þ dy ¼ ð2 1 ð y2 þ 1Þ dy ¼ 10=3 Then the required value ¼ 2=3 þ 10=3 ¼ 8:3. (c) Since t ¼ 0 at ð0; 1Þ and t ¼ 1 at ð1; 2Þ, the line integral equals ð1 t¼0 ft2 ðt2 þ 1Þg dt þ fðt2 þ 1Þ2 þ tg 2t dt ¼ ð1 0 ð2t5 þ 4t3 þ 2t2 þ 2t 1Þ dt ¼ 2 10.2. If A ¼ ð3x2 6yzÞi þ ð2y þ 3xzÞj þ ð1 4xyz2 Þk, evaluate ð C A dr from ð0; 0; 0Þ to ð1; 1; 1Þ along the following paths C: ðaÞ x ¼ t; y ¼ t2 ; z ¼ t3 ðbÞ The straight lines from ð0; 0; 0Þ to ð0; 0; 1Þ, then to ð0; 1; 1Þ, and then to ð1; 1; 1Þ ðcÞ The straight line joining ð0; 0; 0Þ and ð1; 1; 1Þ ð C A dr ¼ ð C fð3x2 6yzÞi þ ð2y þ 3xzÞj þ ð1 4xyz2 Þkg ðdxi þ dyj þ dzkÞ ¼ ð C ð3x2 6yzÞ dx þ ð2y þ 3xzÞ dy þ ð1 4xyz2 Þ dz (a) If x ¼ t; y ¼ t2 ; z ¼ t3 , points ð0; 0; 0Þ and ð1; 1; 1Þ correspond to t ¼ 0 and t ¼ 1, respectively. Then 238 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-6
- 248. ð C A dr ¼ ð1 t¼0 f3t2 6ðt2 Þðt3 Þg dt þ f2t2 þ 3ðtÞðt3 Þg dðt2 Þ þ f1 4ðtÞðt2 Þðt3 Þ2 g dðt3 Þ ¼ ð1 t¼0 ð3t2 6t5 Þ dt þ ð4t3 þ 6t5 Þ dt þ ð3t2 12t11 Þ dt ¼ 2 Another method: Along C, A ¼ ð3t2 6t5 Þi þ ð2t2 þ 3t4 Þj þ ð1 4t9 Þk and r ¼ xi þ yj þ zk ¼ ti þ t2 j þ t3 k, dr ¼ ði þ 2tj þ 3t2 kÞ dt. Then ð C A dr ¼ ð1 0 ð3t2 6t5 Þ dt þ ð4t3 þ 6t5 Þ dt þ ð3t2 12t11 Þ dt ¼ 2 (b) Along the straight line from ð0; 0; 0Þ to ð0; 1; 1Þ, x ¼ 0; y ¼ 0; dx ¼ 0; dy ¼ 0, while z varies from 0 to 1. Then the integral over this part of the path is ð1 z¼0 f3ð0Þ2 6ð0ÞðzÞg0 þ f2ð0Þ þ 3ð0ÞðzÞg0 þ f1 4ð0Þð0Þðz2 Þg dz ¼ ð1 z¼0 dz ¼ 1 Along the straight line from ð0; 0; 1Þ to ð0; 1; 1Þ, x ¼ 0; z ¼ 1; dx ¼ 0; dz ¼ 0, while y varies from 0 to 1. Then the integral over this part of the path is ð1 y¼0 f3ð0Þ2 6ð yÞð1Þg0 þ f2y þ 3ð0Þð1Þg dy þ f1 4ð0Þð yÞð1Þ2 g0 ¼ ð1 y¼0 2y dy ¼ 1 Along the straight line from ð0; 1; 1Þ to ð1; 1; 1Þ, y ¼ 1; z ¼ 1; dy ¼ 0; dz ¼ 0, while x varies from 0 to 1. Then the integral over this part of the path is ð1 x¼0 f3x2 6ð1Þð1Þg dx þ f2ð1Þ þ 3xð1Þg0 þ f1 4xð1Þð1Þ2 g0 ¼ ð1 x¼0 ð3x2 6Þ dx ¼ 5 Adding, ð C A dr ¼ 1 þ 1 5 ¼ 3: (c) The straight line joining ð0; 0; 0Þ and ð1; 1; 1Þ is given in parametric form by x ¼ t; y ¼ t; z ¼ t. Then ð C A dr ¼ ð1 t¼0 ð3t2 6t2 Þ dt þ ð2t þ 3t2 Þ dt þ ð1 4t4 Þ dt ¼ 6=5 10.3. Find the work done in moving a particle once around an ellipse C in the xy plane, if the ellipse has center at the origin with semi-major and semi-minor axes 4 and 3, respectively, as indicated in Fig. 10-7, and if the force field is given by F ¼ ð3x 4y þ 2zÞi þ ð4x þ 2y 3z2 Þj þ ð2xz 4y2 þ z3 Þk In the plane z ¼ 0; F ¼ ð3x 4yÞi þ ð4x þ 2yÞj 4y2 k and dr ¼ dxi þ dyj so that the work done is þ C F dr ¼ ð C fð3x 4yÞi þ ð4x þ 2yÞj 4y2 kg ðdxi þ dyjÞ ¼ þ C ð3x 4yÞ dx þ ð4x þ 2yÞ dy Choose the parametric equations of the ellipse as x ¼ 4 cos t, y ¼ 3 sin t, where t varies from 0 to 2 (see Fig. 10-7). Then the line integral equals ð2 t¼0 f3ð4 cos tÞ 4ð3 sin tÞgf4 sin tg dt þ f4ð4 cos tÞ þ 2ð3 sin tÞgf3 cos tg dt ¼ ð2 t¼0 ð48 30 sin t cos tÞ dt ¼ ð48t 15 sin2 tÞj2 0 ¼ 96 CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 239 x y t r r = xi + yj = 4 cos t i + 3 sin t j Fig. 10-7
- 249. In traversing C we have chosen the counterclockwise direction indicated in Fig. 10-7. We call this the positive direction, or say that C has been traversed in the positive sense. If C were tranversed in the clockwise (negative) direction, the value of the integral would be 96. 10.4. Evaluate ð C y ds along the curve C given by y ¼ 2 ffiffiffi x p from x ¼ 3 to x ¼ 24. Since ds ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx2 þ dy2 p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ð y0Þ2 q dx ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 1=x p dx, we have ð C y ds ¼ ð24 2 2 ffiffiffi x p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 1=x p dx ¼ 2 ð24 3 ffiffiffiffiffiffiffiffiffiffiffi x þ 1 p dx ¼ 4 3 ðx þ 1Þ3=2 24 3 ¼ 156 GREEN’S THEOREM IN THE PLANE 10.5. Prove Green’s theorem in the plane if C is a closed curve which has the property that any straight line parallel to the coordinate axes cuts C in at most two points. Let the equations of the curves AEB and AFB (see adjoin- ing Fig. 10-8) be y ¼ Y1ðxÞ and y ¼ Y2ðxÞ, respectively. If r is the region bounded by C, we have ð ð r @P @y dx dy ¼ ðb x¼a ðY2ðxÞ y¼Y1ðxÞ @P @y dy dx ¼ ðb x¼a Pðx; yÞjY2ðxÞ y¼Y1ðxÞ dx ¼ ðb a ½Pðx; Y2Þ Pðx; Y1Þ dx ¼ ðb a Pðx; Y1Þ dx ða b Pðx; Y2Þ dx ¼ þ C P dx Then ð1Þ þ C P dx ¼ ð ð r @P @y dx dy Similarly let the equations of curves EAF and EBF be x ¼ X1ð yÞ and x ¼ X2ð yÞ respectively. Then ð ð r @Q @x dx dy ¼ ðf y¼c ðX2ð yÞ x¼x1ð yÞ @Q @x dx dy ¼ ðf c ½QðX2; yÞ QðX1; yÞ dy ¼ ðc f QðX1; yÞ dy þ ðf c QðX2; yÞ dy ¼ þ C Q dy ð2Þ þ C Q dy ¼ ð ð r @Q @x dx dy Then Adding (1) and (2), þ C P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy 10.6. Verify Green’s theorem in the plane for þ C ð2xy x2 Þ dx þ ðx þ y2 Þ dy where C is the closed curve of the region bounded by y ¼ x2 and y2 ¼ x. 240 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-8
- 250. The plane curves y ¼ x2 and y2 ¼ x intersect at ð0; 0Þ and ð1; 1Þ. The positive direction in traversing C is as shown in Fig. 10-9. Along y ¼ x2 , the line integral equals ð1 x¼0 fð2xÞðx2 Þ x2 g dx þ fx þ ðx2 Þ2 g dðx2 Þ ¼ ð1 0 ð2x3 þ x2 þ 2x5 Þ dx ¼ 7=6 Along y2 ¼ x the line integral equals ð0 y¼1 f2ð y2 Þ ð yÞ ð y2 Þ2 g dð y2 Þ þ f y2 þ y2 g dy ¼ ð0 1 ð4y4 2y5 þ 2y2 Þ dy ¼ 17=15 Then the required line integral ¼ 7=6 17=15 ¼ 1=30. ð ð r @Q @x @P @y dx dy ¼ ð ð r @ @x ðx þ y2 Þ @ @y ð2xy x2 Þ dx dy ¼ ð ð r ð1 2xÞ dx dy ¼ ð1 x¼0 ð ffiffi x p y¼x2 ð1 2xÞ dy dx ¼ ð1 x¼0 ð y 2xyÞj ffiffi x p y¼x2 dx ¼ ð1 0 ðx1=2 2x3=2 x2 þ 2x3 Þ dx ¼ 1=30 Hence, Green’s theorem is verified. 10.7. Extend the proof of Green’s theorem in the plane given in Problem 10.5 to the curves C for which lines parallel to the coordinate axes may cut C in more than two points. Consider a closed curve C such as shown in the adjoining Fig. 10-10, in which lines parallel to the axes may meet C in more than two points. By constructing line ST the region is divided into two regions r1 and r2, which are of the type considered in Problem 10.5 and for which Green’s theorem applies, i.e., ð1Þ ð STUS P dx þ Q dy ¼ ð ð r1 @Q @x @P @y dx dy; ð2Þ ð SVTS P dx þ Q dy ¼ ð ð r2 @Q @x @P @y dx dy Adding the left-hand sides of (1) and (2), we have, omitting the integrand P dx þ Q dy in each case, ð STUS þ ð SVTS ¼ ð ST þ ð TUS þ ð SVT þ ð TS ¼ ð TUS þ ð SVT ¼ ð TUSVT using the fact that ð ST ¼ ð TS . Adding the right-hand sides of (1) and (2), omitting the integrand, ð ð r1 þ ð ð r2 ¼ ð ð r where r consists of regions r1 and r2. Then ð TUSVT P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy and the theorem is proved. A region r such as considered here and in Problem 10.5, for which any closed curve lying in r can be continuously shrunk to a point without leaving r, is called a simply connected region. A region which is not CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 241 Fig. 10-9 Fig. 10-10
- 251. simply connected is called multiply connected. We have shown here that Green’s theorem in the plane applies to simply connected regions bounded by closed curves. In Problem 10.10 the theorem is extended to multiply connected regions. For more complicated simply connected regions, it may be necessary to construct more lines, such as ST, to establish the theorem. 10.8. Show that the area bounded by a simple closed curve C is given by 1 2 þ C x dy y dx. In Green’s theorem, put P ¼ y; Q ¼ x. Then þ C x dy y dx ¼ ð ð r @ @x ðxÞ @ @y ðyÞ dx dy ¼ 2 ð ð r dx dy ¼ 2A where A is the required area. Thus, A ¼ 1 2 þ C x dy y dx. 10.9. Find the area of the ellipse x ¼ a cos ; y ¼ b sin . Area ¼ 1 2 þ C x dy y dx ¼ 1 2 ð2 0 ða cos Þðb cos Þ d ðb sin Þða sin Þ d ¼ 1 2 ð2 0 abðcos2 þ sin2 Þ d ¼ 1 2 ð2 0 ab d ¼ ab 10.10. Show that Green’s theorem in the plane is also valid for a multiply connected region r such as shown in Fig. 10-11. The shaded region r, shown in the figure, is multiply connected since not every closed curve lying in r can be shrunk to a point without leaving r, as is observed by con- sidering a curve surrounding DEFGD, for example. The boundary of r, which consists of the exterior boundary AHJKLA and the interior boundary DEFGD, is to be tra- versed in the positive direction, so that a person traveling in this direction always has the region on his left. It is seen that the positive directions are those indicated in the adjoining figure. In order to establish the theorem, construct a line, such as AD, called a cross-cut, connecting the exterior and interior boundaries. The region bounded by ADEFGDALKJHA is simply connected, and so Green’s theorem is valid. Then þ ADEFGDALKJHA P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy But the integral on the left, leaving out the integrand, is equal to ð AD þ ð DEFGD þ ð DA þ ð ALKJHA ¼ ð DEFGD þ ð ALKJHA since ð AD ¼ ð DA . Thus, if C1 is the curve ALKJHA, C2 is the curve DEFGD and C is the boundary of r consisting of C1 and C2 (traversed in the positive directions), then ð C1 þ ð C2 ¼ ð C and so þ C P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy 242 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-11
- 252. INDEPENDENCE OF THE PATH 10.11. Let Pðx; yÞ and Qðx; yÞ be continuous and have continuous first partial derivatives at each point of a simply connected region r. Prove that a necessary and sufficient condition that þ C P dx þ Q dy ¼ 0 around every closed path C in r is that @P=@y ¼ @Q=@x identically in r. Sufficiency. Suppose @P=@y ¼ @Q=@x. Then by Green’s theorem, þ C P dx þ Q dy ¼ ð ð r @Q @x @P @y dx dy ¼ 0 where r is the region bounded by C. Necessity. Suppose þ C P dx þ Q dy ¼ 0 around every closed path C in r and that @P=@y 6¼ @Q=@x at some point of r. In particular, suppose @P=@y @Q=@x 0 at the point ðx0; y0Þ. By hypothesis @P=@y and @Q=@x are continuous in r, so that there must be some region containing ðx0; y0Þ as an interior point for which @P=@y @Q=@x 0. If is the boundary of , then by Green’s theorem þ P dx þ Q dy ¼ ð ð @Q @x @P @y dx dy 0 contradicting the hypothesis that þ P dx þ Q dy ¼ 0 for all closed curves in r. Thus @Q=@x @P=@y cannot be positive. Similarly, we can show that @Q=@x @P=@y cannot be negative, and it follows that it must be identically zero, i.e., @P=@y ¼ @Q=@x identically in r. 10.12. Let P and Q be defined as in Problem 10.11. Prove that a necessary and sufficient condition that ðB A P dx þ Q dy be inde- pendent of the path in r joining points A and B is that @P=@y ¼ @Q=@x identically in r. Sufficiency. If @P=@y ¼ @Q=@x, then by Problem 10.11, ð ADBEA P dx þ Q dy ¼ 0 (see Fig. 10-12). From this, omitting for brevity the integrand P dx þ Q dy, we have ð ADB þ ð BEA ¼ 0; ð ADB ¼ ð BEA ¼ ð AEB and so ð C1 ¼ ð C2 i.e., the integral is independent of the path. Necessity. If the integral is independent of the path, then for all paths C1 and C2 in r we have ð C1 ¼ ð C2 ; ð ADB ¼ ð AEB and ð ADBEA ¼ 0 From this it follows that the line integral around any closed path in r is zero, and hence by Problem 10.11 that @P=@y ¼ @Q=@x. CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 243 A B E D C1 C2 Fig. 10-12
- 253. 10.13. Let P and Q be as in Problem 10.11. (a) Prove that a necessary and sufficient condition that P dx þ Q dy be an exact differential of a function ðx; yÞ is that @P=@y ¼ @Q=@x. (b) Show that in such case ðB A P dx þ Q dy ¼ ðB A d ¼ ðBÞ ðAÞ where A and B are any two points. (a) Necessity. If P dx þ Q dy ¼ d ¼ @ @x dx þ @ @y dy, an exact differential, then (1) @=@x ¼ P, (2) @=@y ¼ 0. Thus, by differentiating (1) and (2) with respect to y and x, respectively, @P=@y ¼ @Q=@x since we are assuming continuity of the partial derivatives. Sufficiency. By Problem 10.12, if @P=@y ¼ @Q=@x, then ð P dx þ Q dy is independent of the path joining two points. In particular, let the two points be ða; bÞ and ðx; yÞ and define ðx; yÞ ¼ ððx;yÞ ða;bÞ P dx þ Q dy Then ðx þ x; yÞ ðx; yÞ ¼ ðxþx;y ða;bÞ P dx þ Q dy ððx;yÞ ða;bÞ P dx þ Q dy ¼ ððxþx;yÞ ðx;yÞ P dx þ Q dy Since the last integral is independent of the path joining ðx; yÞ and ðx þ x; yÞ, we can choose the path to be a straight line joining these points (see Fig. 10-13) so that dy ¼ 0. Then by the mean value theorem for integrals, ðx þ x; yÞ ðx; yÞ x ¼ 1 x ððxþx;yÞ ðx;yÞ P dx ¼ Pðx þ x; yÞ 0 1 Taking the limit as x ! 0, we have @=@x ¼ P. Similarly we can show that @=@y ¼ Q. Thus it follows that P dx þ Q dy ¼ @ @x dx þ @ @y dy ¼ d: 244 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 x y (a, b) (x, y) (x + Dx, y) Fig. 10-13
- 254. (b) Let A ¼ ðx1; y1Þ; B ¼ ðx2; y2Þ. From part (a), ðx; yÞ ¼ ððx;yÞ ða;bÞ P dx þ Q dy Then omitting the integrand P dx þ Q dy, we have ðB A ¼ ððx2;y2Þ ðx1;y1Þ ¼ ððx2;y2Þ ða;bÞ ððx1;y1Þ ða;bÞ ¼ ðx2; y2Þ ðx1; y1Þ ¼ ðBÞ ðAÞ 10.14. (a) Prove that ðð3;4Þ ð1;2Þ ð6xy2 y3 Þ dx þ ð6x2 y 3xy2 Þ dy is independent of the path joining ð1; 2Þ and ð3; 4Þ. (b) Evaluate the integral in (a). (a) P ¼ 6xy2 y3 ; Q ¼ 6x2 y 3xy2 . Then @P=@y ¼ 12xy 3y2 ¼ @Q=@x and by Problem 10.12 the line integral is independent of the path. (b) Method 1: Since the line integral is independent of the path, choose any path joining ð1; 2Þ and ð3; 4Þ, for example that consisting of lines from ð1; 2Þ to ð3; 2Þ [along which y ¼ 2; dy ¼ 0] and then ð3; 2Þ to ð3; 4Þ [along which x ¼ 3; dx ¼ 0]. Then the required integral equals ð3 x¼1 ð24x 8Þ dx þ ð4 y¼2 ð54y 9y2 Þ dy ¼ 80 þ 156 ¼ 236 Method 2: Since @P @y ¼ @Q @x ; we must have ð1Þ @ @x ¼ 6xy2 y3 ; ð2Þ @ @y ¼ 6x2 y 3xy2 : From (1), ¼ 3x2 y2 xy3 þ f ð yÞ. From (2), ¼ 3x2 y2 xy3 þ gðxÞ. The only way in which these two expressions for are equal is if f ð yÞ ¼ gðxÞ ¼ c, a constant. Hence ¼ 3x2 y2 xy3 þ c. Then by Problem 10.13, ðð3;4Þ ð1;2Þ ð6xy2 y3 Þ dx þ ð6x2 y 3xy2 Þ dy ¼ ðð3;4Þ ð1;2Þ dð3x2 y2 xy3 þ cÞ ¼ 3x2 y2 xy3 þ cjð3;4Þ ð1;2Þ ¼ 236 Note that in this evaluation the arbitrary constant c can be omitted. See also Problem 6.16, Page 131. We could also have noted by inspection that ð6xy2 y3 Þ dx þ ð6x2 y 3xy2 Þ dy ¼ ð6xy2 dx þ 6x2 y dyÞ ð y3 dx þ 3xy2 dyÞ ¼ dð3x2 y2 Þ dðxy3 Þ ¼ dð3x2 y2 xy3 Þ from which it is clear that ¼ 3x2 y2 xy3 þ c. 10.15. Evaluate þ ðx2 y cos x þ 2xy sin x y2 ex Þ dx þ ðx2 sin x 2yex Þ dy around the hypocycloid x2=3 þ y2=3 ¼ a2=3 : P ¼ x2 y cos x þ 2xy sin x y2 ex ; Q ¼ x2 sin x 2yex Then @P=@y ¼ x2 cos x þ 2x sin x 2yex ¼ @Q=@x, so that by Problem 10.11 the line integral around any closed path, in particular x2=3 þ y2=3 ¼ a2=3 is zero. SURFACE INTEGRALS 10.16. If is the angle between the normal line to any point ðx; y; zÞ of a surface S and the positive z-axis, prove that CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 245
- 255. j sec j ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ z2 x þ z2 y q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F2 x þ F2 y þ F2 z q jFzj according as the equation for S is z ¼ f ðx; yÞ or Fðx; y; zÞ ¼ 0. If the equation of S is Fðx; y; zÞ ¼ 0, a normal to S at ðx; y; zÞ is rF ¼ Fxi þ Fyj þ Fzk. Then rF k ¼ jrFjjkj cos or Fz ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F2 x þ F2 y þ F2 z q cos from which j sec j ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F2 x þ F2 y þ F2 z q jFzj as required. In case the equation is z ¼ f ðx; yÞ, we can write Fðx; y; zÞ ¼ z f ðx; yÞ ¼ 0, from which Fx ¼ zx; Fy zy; Fz ¼ 1 and we find j sec j ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ z2 x þ z2 y q . 10.17. Evaluate ð ð S Uðx; y; zÞ dS where S is the surface of the paraboloid z ¼ 2 ðx2 þ y2 Þ above the xy plane and Uðx; y; zÞ is equal to (a) 1, (b) x2 þ y2 , (c) 3z. Give a physical interpretation in each case. (See Fig. 10-14.) The required integral is equal to ð ð r Uðx; y; zÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ z2 x þ z2 y q dx dy ð1Þ where r is the projection of S on the xy plane given by x2 þ y2 ¼ 2; z ¼ 0. Since zx ¼ 2x; zy ¼ 2y, (1) can be written ð ð r Uðx; y; zÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy ð2Þ (a) If Uðx; y; zÞ ¼ 1, (2) becomes ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy To evaluate this, transform to polar coordinates ð; Þ. Then the integral becomes ð2 ¼0 ð ffiffi 2 p ¼0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 42 p d d ¼ ð2 ¼0 1 12 ð1 þ 42 Þ3=2 ffiffi 2 p ¼0 d ¼ 13 3 Physically this could represent the surface area of S, or the mass of S assuming unit density. (b) If Uðx; y; zÞ ¼ x2 þ y2 , (2) becomes ð ð r ðx2 þ y2 Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy or in polar coordinates ð2 ¼0 ð ffiffi 2 p ¼0 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 42 p d d ¼ 149 30 where the integration with respect to is accomplished by the substitution ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 42 p ¼ u. Physically this could represent the moment of inertia of S about the z-axis assuming unit density, or the mass of S assuming a density ¼ x2 þ y2 . 246 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-14
- 256. (c) If Uðx; y; zÞ ¼ 3z, (2) becomes ð ð r 3z ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy ¼ ð ð r 3f2 ðx2 þ y2 Þg ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy or in polar coordinates, ð2 ¼0 ð ffiffi 2 p ¼0 3ð2 2 Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 42 p d d ¼ 111 10 Physically this could represent the mass of S assuming a density ¼ 3z, or three times the first moment of S about the xy plane. 10.18. Find the surface area of a hemisphere of radius a cut off by a cylinder having this radius as diameter. Equations for the hemisphere and cylinder (see Fig. 10-15) are given respectively by x2 þ y2 þ z2 ¼ a2 (or z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p Þ and ðx a=2Þ2 þ y2 ¼ a2 =4 (or x2 þ y2 ¼ ax). Since zx ¼ x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p and zy ¼ y ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p we have Required surface area ¼ 2 ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ z2 x þ z2 y q dx dy ¼ 2 ð ð r a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p dx dy Two methods of evaluation are possible. Method 1: Using polar coordinates. Since x2 þ y2 ¼ ax in polar coordinates is ¼ a cos , the integral becomes 2 ð=2 ¼0 ða cos ¼0 a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 2 p d d ¼ 2a ð=2 ¼0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 2 p a cos ¼0 d ¼ 2a2 ð=2 0 ð1 sin Þ d ¼ ð 2Þa2 Method 2: The integral is equal to 2 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi axx2 p y¼0 a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p dy dx ¼ 2a ða x¼0 sin1 y ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ax x2 p ffiffiffiffiffiffiffiffiffiffi axx2 p y¼0 dx ¼ 2a ða 0 sin1 ffiffiffiffiffiffiffiffiffiffiffi x a þ x r dx Letting x ¼ a tan2 , this integral becomes 4a2 ð=4 0 tan sec2 d ¼ 4a2 1 2 tan2 j=4 0 1 2 ð=4 0 tan2 d ¼ 2a2 tan2 j=4 0 ð=4 0 ðsec2 1Þ d ¼ 2a2 =4 ðtan Þj=4 0 n o ¼ ð 2Þa2 Note that the above integrals are actually improper and should be treated by appropriate limiting procedures (see Problem 5.74, Chapter 5, and also Chapter 12). CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 247 Fig. 10-15
- 257. 10.19. Find the centroid of the surface in Problem 10.17. By symmetry, x x ¼ y y ¼ 0 and z z ¼ ð ð S z dS ð ð S dS ¼ ð ð r z ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy ð ð r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y2 q dx dy The numerator and denominator can be obtained from the results of Problems 10.17(c) and 10.17(a), respectively, and we thus have z z ¼ 37=10 13=3 ¼ 111 130 . 10.20. Evaluate ð ð S A n dS, where A ¼ xyi x2 j þ ðx þ zÞk, S is that portion of the plane 2x þ 2y þ z ¼ 6 included in the first octant, and n is a unit normal to S. (See Fig. 10-16.) A normal to S is rð2x þ 2y þ z 6Þ ¼ 2i þ 2j þ k, and so n ¼ 2i þ 2j þ k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 22 þ 22 þ 12 p ¼ 2i þ 2j þ k 3 . Then A n ¼ fxyi x2 j þ ðx þ zÞkg 2i þ 2j þ k 3 ¼ 2xy 2x2 þ ðx þ zÞ 3 ¼ 2xy 2x2 þ ðx þ 6 2x 2yÞ 3 ¼ 2xy 2x2 x 2y þ 6 3 The required surface integral is therefore ð ð S 2xy 2x2 x 2y þ 6 3 ! dS ¼ ð ð r 2xy 2x2 x 2y þ 6 3 ! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ z2 x þ z2 y q dx dy ¼ ð ð r 2xy 2x2 x 2y þ 6 3 ! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 þ 22 þ 22 p dx dy ¼ ð3 x¼0 ð3x y¼0 ð2xy 2x2 x 2y þ 6Þ dy dx ¼ ð3 x¼0 ðxy2 2x2 y xy y2 þ 6yÞj3x 0 dx ¼ 27=4 10.21. In dealing with surface integrals we have restricted ourselves to surfaces which are two-sided. Give an example of a surface which is not two-sided. Take a strip of paper such as ABCD as shown in the adjoining Fig. 10-17. Twist the strip so that points A and B fall on D and C, respectively, as in the adjoining figure. If n is the positive normal at point P of the surface, we find that as n moves around the surface, it reverses its original direction when it reaches P again. If we tried to color only one side of the surface, we would find the whole thing colored. This surface, called a Möbius strip, 248 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-16 Fig. 10-17
- 258. is an example of a one-sided surface. This is sometimes called a nonorientable surface. A two-sided surface is orientable. THE DIVERGENCE THEOREM 10.22. Prove the divergence theorem. (See Fig. 10-18.) Let S be a closed surface which is such that any line parallel to the coordinate axes cuts S in at most two points. Assume the equations of the lower and upper portions, S1 and S2, to be z ¼ f1ðx; yÞ and z ¼ f2ðx; yÞ, respectively. Denote the projection of the surface on the xy plane by r. Consider ð ð ð V @A3 @z dV ¼ ð ð ð V @A3 @z dz dy dx ¼ ð ð r ðf2ðx;yÞ z¼f1ðx;yÞ @A3 @z dz dy dx ¼ ð ð r A3ðx; y; zÞ f2 z¼f1 dy dx ¼ ð ð r ½A3ðx; y; f2Þ A3ðx; y; f1Þ dy dx For the upper portion S2, dy dx ¼ cos 2 dS2 ¼ k n2 dS2 since the normal n2 to S2 makes an acute angle 2 with k. For the lower portion S1, dy dx ¼ cos 1 dS1 ¼ k n1 dS1 since the normal n1 to S1 makes an obtuse angle 1 with k. ð ð r A3ðx; y; f2Þ dy dx ¼ ð ð S2 A3 k n2 dS2 Then ð ð r A3ðx; y; f1Þ dy dx ¼ ð ð S1 A3 k n1 dS1 and ð ð r A3ðx; y; f2Þ dy dx ð ð r A3ðx; y; f1Þ dy dx ¼ ð ð S2 A3 k n2 dS2 þ ð ð S1 A3 k n1 dS1 ¼ ð ð S A3 k n dS CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 249 Fig. 10-18
- 259. so that ð1Þ ð ð ð V @A3 @z dV ¼ ð ð S A3k n dS Similarly, by projecting S on the other coordinate planes, ð2Þ ð ð ð V @A1 @x dV ¼ ð ð S A1 i n dS ð3Þ ð ð ð V @A2 @y dV ¼ ð ð S A2 j n dS Adding (1), (2), and (3), ð ð ð V @A1 @x þ @A2 @y þ @A3 @z dV ¼ ð ð S ðA1i þ A2j þ A3kÞ n dS ð ð ð V r A dV ¼ ð ð S A n dS or The theorem can be extended to surfaces which are such that lines parallel to the coordinate axes meet them in more than two points. To establish this extension, subdivide the region bounded by S into subregions whose surfaces do satisfy this condition. The procedure is analogous to that used in Green’s theorem for the plane. 10.23. Verify the divergence theorem for A ¼ ð2x zÞi þ x2 yj xz2 k taken over the region bounded by x ¼ 0; x ¼ 1; y ¼ 0; y ¼ 1; z ¼ 0; z ¼ 1. We first evaluate ð ð S A n dS where S is the surface of the cube in Fig. 10-19. Face DEFG: n ¼ i; x ¼ 1. Then ð ð DEFG A n dS ¼ ð1 0 ð1 0 fð2 zÞi þ j z2 kg i dy dz ¼ ð1 0 ð1 0 ð2 zÞ dy dz ¼ 3=2 Face ABCO: n ¼ i; x ¼ 0. Then ð ð ABCO A n dS ¼ ð1 0 ð1 0 ðziÞ ðiÞ dy dz ¼ ð1 0 ð1 0 z dy dz ¼ 1=2 Face ABEF: n ¼ j; y ¼ 1. Then ð ð ABEF A n dS ¼ ð1 0 ð1 0 fð2x zÞi þ x2 j xz2 kg j dx dz ¼ ð1 0 ð1 0 x2 dx dz ¼ 1=3 Face OGDC: n ¼ j; y ¼ 0. Then ð ð OGDC A n dS ¼ ð1 0 ð1 0 fð2x zÞi xz2 kg ðjÞ dx dz ¼ 0 250 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-19
- 260. Face BCDE: n ¼ k; z ¼ 1. Then ð ð BCDE A n dS ¼ ð1 0 ð1 0 fð2x 1Þi þ x2 yj xkg k dx dy ¼ ð1 0 ð1 0 x dx dy 1=2 Face AFGO: n ¼ k; z ¼ 0. Then ð ð AFGO A n dS ¼ ð1 0 ð1 0 f2xi x2 yjg ðkÞ dx dy ¼ 0 Adding, ð ð S A n dS ¼ 3 2 þ 1 2 þ 1 3 þ 0 1 2 þ 0 ¼ 11 6 : Since ð ð ð V r A dV ¼ ð1 0 ð1 0 ð1 0 ð2 þ x2 2xzÞ dx dy dz ¼ 11 6 the divergence theorem is verified in this case. 10.24. Evaluate ð ð S r n dS, where S is a closed surface. By the divergence theorem, ð ð S r n dS ¼ ð ð ð V r r dV ¼ ð ð ð V @ @x i þ @ @y j þ @ @z k ðxi þ yj þ zkÞ dV ¼ ð ð ð V @x @x þ @y @y þ @z @z dV ¼ 3 ð ð ð V dV ¼ 3V where V is the volume enclosed by S. 10.25. Evaluate ð ð S xz2 dy dz þ ðx2 y z3 Þ dz dx þ ð2xy þ y2 zÞ dx dy, where S is the entire surface of the hemispherical region bounded by z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 p and z ¼ 0 (a) by the divergence theorem (Green’s theorem in space), (b) directly. (a) Since dy dz ¼ dS cos ; dz dx ¼ dS cos ; dx dy ¼ dS cos , the integral can be written ð ð S fxz2 cos þ ðx2 y z3 Þ cos þ ð2xy þ y2 zÞ cos g dS ¼ ð ð S A n dS where A ¼ xz2 i þ ðx2 y z3 Þj þ ð2xy þ y2 zÞk and n ¼ cos i þ cos j þ cos k, the outward drawn unit normal. Then by the divergence theorem the integral equals ð ð ð V r A dV ¼ ð ð ð V @ @x ðxz2 Þ þ @ @y ðx2 y z3 Þ þ @ @z ð2xy þ y2 zÞ dV ¼ ð ð ð V ðx2 þ y2 þ z2 Þ dV where V is the region bounded by the hemisphere and the xy plane. By use of spherical coordinates, as in Problem 9.19, Chapter 9, this integral is equal to 4 ð=2 ¼0 ð=2 ¼0 ð r¼0 r2 r2 sin dr d d ¼ 2a5 5 CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 251
- 261. (b) If S1 is the convex surface of the hemispherical region and S2 is the base ðz ¼ 0Þ, then ð ð S1 xz2 dy dz ¼ ða y¼a ð ffiffiffiffiffiffiffiffiffi a2y2 p z¼0 z2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 y2 z2 q dz dy ða y¼a ð ffiffiffiffiffiffiffiffiffi a2y2 p z¼0 z2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 y2 z2 q dz dy ð ð S1 ðx2 y z3 Þ dz dx ¼ ða x¼a ð ffiffiffiffiffiffiffiffiffiffi a2x2 p x¼0 fx2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 z2 p z3 g dz dx ða x¼a ð ffiffiffiffiffiffiffiffiffiffi a2x2 p z¼0 fx2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 z2 p z3 g dz dx ð ð S1 ð2xy þ y2 zÞ dx dy ¼ ða x¼a ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼ ffiffiffiffiffiffiffiffiffiffi a2x2 p f2xy þ y2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 q g dy dx ð ð S2 xz2 dy dz ¼ 0; ð ð S2 ðx2 y z3 Þ dz dx ¼ 0; ð ð S2 ð2xy þ y2 zÞ dx dy ¼ ð ð S2 f2xy þ y2 ð0Þg dx dy ¼ ða x¼a ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼ ffiffiffiffiffiffiffiffiffiffi a2x2 p 2xy dy dx ¼ 0 By addition of the above, we obtain 4 ða y¼0 ð ffiffiffiffiffiffiffiffiffiffi a 2 y2 p x¼0 z2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 y2 z2 q dz dy þ 4 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi a2x2 p z¼0 x2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 z2 p dz dx þ 4 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼0 y2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 q dy dx Since by symmetry all these integrals are equal, the result is, on using polar coordinates, 12 ða x¼0 ð ffiffiffiffiffiffiffiffiffiffi a2x2 p y¼0 y2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 x2 y2 q dy dx ¼ 12 ð=2 ¼0 ða ¼0 2 sin2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 2 p d d ¼ 2a5 5 STOKES’ THEOREM 10.26. Prove Stokes’ theorem. Let S be a surface which is such that its projections on the xy, yz, and xz planes are regions bounded by simple closed curves, as indicated in Fig. 10-20. Assume S to have representation z ¼ f ðx; yÞ or x ¼ gðy; zÞ or y ¼ hðx; zÞ, where f ; g; h are single-valued, continuous, and differentiable functions. We must show that ð ð S ðr AÞ n dS ¼ ð ð S ½r ðA1i þ A2j þ A3kÞ n dS ¼ ð C A dr where C is the boundary of S. Consider first ð ð S ½r ðA1iÞ n dS: Since r ðA1iÞ ¼ i j k @ @x @ @y @ @z A1 0 0 ¼ @A1 @z j @A1 @y k; 252 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 262. ½r ðA1iÞ n dS ¼ @A1 @z n j @A1 @y n k dS ð1Þ If z ¼ f ðx; yÞ is taken as the equation of S, then the position vector to any point of S is r ¼ xi þ yj þ zk ¼ xi þ yj þ f ðx; yÞk so that @r @y ¼ j þ @z @y k ¼ j þ @ f @y k. But @r @y is a vector tangent to S and thus perpendicular to n, so that n @r @y ¼ n j þ @z @y n k ¼ 0 or n j ¼ @z @y n k Substitute in (1) to obtain @A1 @z n j @A1 @y n k dS ¼ @A1 @z @z @y n k @A1 @y n k dS or ½r ðA1iÞ n dS ¼ @A1 @y þ @A1 @z @z @y n k dS ð2Þ Now on S, A1ðx; y; zÞ ¼ A1½x; y; f ðx; yÞ ¼ Fðx; yÞ; hence, @A1 @y þ @A1 @z @z @y ¼ @F @y and (2) becomes ½r ðA1iÞ n dS ¼ @F @y n k dS ¼ @F @y dx dy Then ð ð S ½r ðA1iÞ n dS ¼ ð ð r @F @y dx dy where r is the projection of S on the xy plane. By Green’s theorem for the plane, the last integral equals þ F dx where is the boundary of r. Since at each point ðx; yÞ of the value of F is the same as the value of A1 at each point ðx; y; zÞ of C, and since dx is the same for both curves, we must have þ F dx ¼ þ C A1 dx CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 253 Fig. 10-20
- 263. or ð ð S ½r ðA1iÞ n dS ¼ þ C A1 dx Similarly, by projections on the other coordinate planes, ð ð S ½r ðA2jÞ n dS ¼ þ C A2 dy; ð ð S ½r ðA3kÞ n dS ¼ þ C A3 dz Thus, by addition, ð ð S ðr AÞ n dS ¼ þ C A dr The theorem is also valid for surfaces S which may not satisfy the restrictions imposed above. For assume that S can be subdivided into surfaces S1; S2; . . . ; Sk with boundaries C1; C2; . . . ; Ck which do satisfy the restrictions. Then Stokes’ theorem holds for each such surface. Adding these surface integrals, the total surface integral over S is obtained. Adding the corresponding line integrals over C1; C2; . . . ; Ck, the line integral over C is obtained. 10.27. Verify Stoke’s theorem for A ¼ 3yi xzj þ yz2 k, where S is the surface of the paraboloid 2z ¼ x2 þ y2 bounded by z ¼ 2 and C is its boundary. See Fig. 10-21. The boundary C of S is a circle with equations x2 þ y2 ¼ 4; z ¼ 2 and parametric equations x ¼ 2 cos t; y ¼ 2 sin t; z ¼ 2, where 0 @ t 2. Then þ C A dr ¼ þ C 3y dx xz dy þ yz2 dz ¼ ð0 2 3ð2 sin tÞð2 sin tÞ dt ð2 cos tÞð2Þð2 cos tÞ dt ¼ ð2 0 ð12 sin2 t þ 8 cos2 tÞ dt ¼ 20 r A ¼ i j k @ @x @ @y @ z 3y xz yz2 ¼ ðz2 þ xÞi ðz þ 3Þk Also, n ¼ rðx2 þ y2 2zÞ jrðx2 þ y2 2zÞj ¼ xi þ yj k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi x2 þ y2 þ 1 p : and Then ð ð S ðr AÞ n dS ¼ ð ð r ðr AÞ n dx dy jn kj ¼ ð ð r ðxz2 þ x2 þ z þ 3Þ dx dy ¼ ð ð r x x2 þ y2 2 !2 þx2 þ x2 þ y2 2 þ 3 8 : 9 = ; dx dy In polar coordinates this becomes ð2 ¼0 ð2 ¼0 fð cos Þð4 =2Þ þ 2 cos2 þ 2 =2 þ 3g d d ¼ 20 254 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 Fig. 10-21
- 264. 10.28. Prove that a necessary and sufficient condition that þ C A dr ¼ 0 for every closed curve C is that r A ¼ 0 identically. Sufficiency. Suppose r A ¼ 0. Then by Stokes’ theorem þ C A dr ¼ ð ð S ðr AÞ n dS ¼ 0 Necessity. Suppose þ C A dr ¼ 0 around every closed path C, and assume r A 6¼ 0 at some point P. Then assuming r A is continuous, there will be a region with P as an interior point, where r A 6¼ 0. Let S be a surface contained in this region whose normal n at each point has the same direction as r A, i.e., r A ¼ n where is a positive constant. Let C be the boundary of S. Then by Stokes’ theorem þ C A dr ¼ ð ð S ðr AÞ n dS ¼ ð ð S n n dS 0 which contradicts the hypothesis that þ C A dr ¼ 0 and shows that r A ¼ 0. It follows that r A ¼ 0 is also a necessary and sufficient condition for a line integral ðP2 P1 A dr to be independent of the path joining points P1 and P2. 10.29. Prove that a necessary and sufficient condition that r A ¼ 0 is that A ¼ r. Sufficiency. If A ¼ r, then r A ¼ r r ¼ 0 by Problem 7.80, Chap. 7, Page 179. Necessity. If r A ¼ 0, then by Problem 10.28, þ A dr ¼ 0 around every closed path and ð C A dr is independent of the path joining two points which we take as ða; b; cÞ and ðx; y; zÞ. Let us define ðx; y; zÞ ¼ ððx;y;zÞ ða;b;cÞ A dr ¼ ððx;y;zÞ ða;b;cÞ A1 dx þ A2 dy þ A3 dz Then ðx þ x; y; zÞ ðx; y; zÞ ¼ ððxþx;y;zÞ ðx;y;zÞ A1 dx þ A2 dy þ A3 dz Since the last integral is independent of the path joining ðx; y; zÞ and ðx þ x; y; zÞ, we can choose the path to be a straight line joining these points so that dy and dz are zero. Then ðx þ x; y; zÞ ðx; y; zÞ x ¼ 1 x ððxþx;y;zÞ ðx;y;zÞ A1 dx ¼ A1ðx þ x; y; zÞ 0 1 where we have applied the law of the mean for integrals. Taking the limit of both sides as x ! 0 gives @=@x ¼ A1. Similarly, we can show that @=@y ¼ A2; @=@z ¼ A3: Thus, A ¼ A1i þ A2j þ A3k ¼ @ @x i þ @ @y j þ @ @z k ¼ r: 10.30. (a) Prove that a necessary and sufficient condition that A1 dx þ A2 dy þ A3 dz ¼ d, an exact differential, is that r A ¼ 0 where A ¼ A1i þ A2j þ A3k. (b) Show that in such case, ððx2;y2;z2Þ ðx1;y1;z1Þ A1 dx þ A2 dy þ A3 dz ¼ ððx2;y2;z2Þ ðx1;y1;z1Þ d ¼ ðx2; y2; z2Þ ðx1; y1; z1Þ CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 255
- 265. (a) Necessity. If A1 dx þ A2 dy þ A3 dz ¼ d ¼ @ @x dx þ @ @y dy þ @ @z dz, then ð1Þ @ @x ¼ A1 ð2Þ @ @y ¼ A2; ð3Þ @ @z ¼ A3 Then by differentiating we have, assuming continuity of the partial derivatives, @A1 @y ¼ @A2 @x ; @A2 @z ¼ @A3 @y ; @A1 @z ¼ @A3 @x which is precisely the condition r A ¼ 0. Another method: If A1 dx þ A2 dy þ A3 dz ¼ d, then A ¼ A1i þ A2j þ A3k ¼ @ @x i þ @ @y j þ @ @z k ¼ r from which r A ¼ r r ¼ 0. Sufficiency. If r A ¼ 0, then by Problem 10.29, A ¼ r and A1 dx þ A2 dy þ A3 dz ¼ A dr ¼ r dr ¼ @ @x dx þ @ @y dy þ @ @z dz ¼ d (b) From part (a), ðx; y; zÞ ¼ ððx;y;zÞ ða;b;cÞ A1 dx þ A2 dy þ A3 dz. Then omitting the integrand A1 dx þ A2 dy þ A3 dz, we have ððx2;y2;z2Þ ðx1;y1;z1Þ ¼ ððx2;y2;z2Þ ða;b;cÞ ððx1;y1;z1Þ ða;b;cÞ ¼ ðx2; y2; z2Þ ðx1; y1; z1Þ 10.31. (a) Prove that F ¼ ð2xz3 þ 6yÞi þ ð6x 2yzÞj þ ð3x2 z2 y2 Þk is a conservative force field. (b) Evaluate ð C F dr where C is any path from ð1; 1; 1Þ to ð2; 1; 1Þ. (c) Give a physical interpretation of the results. (a) A force field F is conservative if the line integral ð C F dr is independent of the path C joining any two points. A necessary and sufficient condition that F be conservative is that r F ¼ 0. Since here r F ¼ i j k @ @x @ @y @ @z 2xz3 þ 6y 6x 2yz 3x2 z2 y2 ¼ 0; F is conservative (b) Method 1: By Problem 10.30, F dr ¼ ð2xz3 þ 6yÞ dx þ ð6x 2yzÞ dy þ ð3x2 z2 y2 Þ dz is an exact dif- ferential d, where is such that ð1Þ @ @x ¼ 2xz3 þ 6y ð2Þ @ @y ¼ 6x 2yz ð3Þ @ @z ¼ 3x2 z2 y2 From these we obtain, respectively, ¼ x2 z3 þ 6xy þ f1ð y; zÞ ¼ 6xy y2 z þ f2ðx; zÞ ¼ x2 z3 y2 z þ f3ðx; yÞ These are consistent if f1ðy; zÞ ¼ y2 z þ c; f2ðx; zÞ ¼ x2 z3 þ c; f3ðx; yÞ ¼ 6xy þ c, in which case ¼ x2 z3 þ 6xy y2 z þ c. Thus, by Problem 10.30, ðð2;1;1Þ ð1;1;1Þ F dr ¼ x2 z3 þ 6xy y2 z þ cjð2;1;1Þ ð1;1;1Þ ¼ 15 256 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 266. Alternatively, we may notice by inspection that F dr ¼ ð2xz3 dx þ 3x2 z2 dzÞ þ ð6y dx þ 6x dyÞ ð2yz dy þ y2 dzÞ ¼ dðx2 z3 Þ þ dð6xyÞ dðy2 zÞ ¼ dðx2 z3 þ 6xy y2 z þ cÞ from which is determined. Method 2: Since the integral is independent of the path, we can choose any path to evaluate it; in particular, we can choose the path consisting of straight lines from ð1; 1; 1Þ to ð2; 1; 1Þ, then to ð2; 1; 1Þ and then to ð2; 1; 1Þ. The result is ð2 x¼1 ð2x 6Þ dx þ ð1 y¼1 ð12 2yÞ dy þ ð1 z¼1 ð12z2 1Þ dz ¼ 15 where the first integral is obtained from the line integral by placing y ¼ 1; z ¼ 1; dy ¼ 0; dz ¼ 0; the second integral by placing x ¼ 2; z ¼ 1; dx ¼ 0; dz ¼ 0; and the third integral by placing x ¼ 2; y ¼ 1; dx ¼ 0; dy ¼ 0. (c) Physically ð C F dr represents the work done in moving an object from ð1; 1; 1Þ to ð2; 1; 1Þ along C. In a conservative force field the work done is independent of the path C joining these points. MISCELLANEOUS PROBLEMS 10.32. (a) If x ¼ f ðu; vÞ; y ¼ gðu; vÞ defines a transformation which maps a region r of the xy plane into a region r0 of the uv plane, prove that ð ð r dx dy ¼ ð ð r0 @ðx; yÞ @ðu; vÞ du dv (b) Interpret geometrically the result in (a). (a) If C (assumed to be a simple closed curve) is the boundary of r, then by Problem 10.8, ð ð r dx dy ¼ 1 2 þ C x dy y dx ð1Þ Under the given transformation the integral on the right of (1) becomes 1 2 þ C0 x @y @u du þ @y @v dv y @x @u du þ @x @v dv ¼ 1 2 ð C0 x @y @u y @x @u du þ x @y @v y @x @v dv ð2Þ where C0 is the mapping of C in the uv plane (we suppose the mapping to be such that C0 is a simple closed curve also). By Green’s theorem if r0 is the region in the uv plane bounded by C0 , the right side of (2) equals 1 2 ð ð r0 @ @u x @y @v y @x @v @ @v x @y @u y @x @u du dv ¼ ð ð r0 @x @u @y @v @x @v @y @u du dv ¼ ð ð r0 @ðx; yÞ @ðu; vÞ du dv where we have inserted absolute value signs so as to ensure that the result is non-negative as is ð ð r dx dy In general, we can show (see Problem 10.83) that ð ð r Fðx; yÞ dx dy ¼ ð ð r0 Ff f ðu; vÞ; gðu; vÞg @ðx; yÞ @ðu; vÞ du dv ð3Þ CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 257
- 267. (b) ð ð r dx dy and ð ð r0 @ðx; yÞ @ðu; vÞ du dv represent the area of region r, the first expressed in rectangular coordinates, the second in curvilinear coordinates. See Page 212, and the introduction of the differ- ential element of surface area for an alternative to Part (a). 10.33. Let F ¼ yi þ xj x2 þ y2 . (a) Calculate r F: ðbÞ Evaluate þ F dr around any closed path and explain the results. ðaÞ r F ¼ i j k @ @x @ @y @ @z y x2 þ y2 x x2 þ y2 0 ¼ 0 in any region excluding ð0; 0Þ: (b) þ F dr ¼ þ y dx þ x dy x2 þ y2 . Let x ¼ cos ; y ¼ sin , where ð; Þ are polar coordinates. Then dx ¼ sin d þ d cos ; dy ¼ cos d þ d sin y dx þ x dy x2 þ y2 ¼ d ¼ d arc tan y x and so For a closed curve ABCDA [see Fig. 10-22(a) below] surrounding the origin, ¼ 0 at A and ¼ 2 after a complete circuit back to A. In this case the line integral equals ð2 0 d ¼ 2. For a closed curve PQRSP [see Fig. 10-22(b) above] not surrounding the origin, ¼ 0 at P and ¼ 0 after a complete circuit back to P. In this case the line integral equals ð0 0 d ¼ 0. Since F ¼ Pi þ Qj; r F ¼ 0 is equivalent to @P=@y ¼ @Q=@x and the results would seem to con- tradict those of Problem 10.11. However, no contradiction exists since P ¼ y x2 þ y2 and Q ¼ x x2 þ y2 do not have continuous derivatives throughout any region including ð0; 0Þ, and this was assumed in Problem 10.11. 10.34. If div A denotes the divergence of a vector field A at a point P, show that div A ¼ lim V!0 ð ð s A n dS V 258 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10 y y x x D R S P Q O O A B C φ φ φ0 (a) (b) Fig. 10-22
- 268. where V is the volume enclosed by the surface S and the limit is obtained by shrinking V to the point P. By the divergence theorem, ð ð ð V div A dV ¼ ð ð S A n dS By the mean value theorem for integrals, the left side can be written div A ð ð ð V dV ¼ div A V where div A is some value intermediate between the maximum and minimum of div A throughout V. Then div A ¼ ð ð S A n dS V Taking the limit as V ! 0 such that P is always interior to V, div A approaches the value div A at point P; hence div A ¼ lim V!0 ð ð S A n dS V This result can be taken as a starting point for defining the divergence of A, and from it all the properties may be derived including proof of the divergence theorem. We can also use this to extend the concept of divergence to coordinate systems other than rectangular (see Page 159). Physically, ð ð ð S A n ds 0 @ 1 A=V represents the flux or net outflow per unit volume of the vector A from the surface S. If div A is positive in the neighborhood of a point P, it means that the outflow from P is positive and we call P a source. Similarly, if div A is negative in the neighborhood of P, the outflow is really an inflow and P is called a sink. If in a region there are no sources or sinks, then div A ¼ 0 and we call A a solenoidal vector field. Supplementary Problems LINE INTEGRALS 10.35. Evaluate ðð4;2Þ ð1;1Þ ðx þ yÞ dx þ ðy xÞ dy along (a) the parabola y2 ¼ x, (b) a straight line, (c) straight lines from ð1; 1Þ to ð1; 2Þ and then to ð4; 2Þ, (d) the curve x ¼ 2t2 þ t þ 1; y ¼ t2 þ 1. Ans: ðaÞ 34=3; ðbÞ 11; ðcÞ 14; ðdÞ 32=3 10.36. Evaluate þ ð2x y þ 4Þ dx þ ð5y þ 3x 6Þ dy around a triangle in the xy plane with vertices at ð0; 0Þ; ð3; 0Þ, ð3; 2Þ traversed in a counterclockwise direction. Ans. 12 10.37. Evaluate the line integral in the preceding problem around a circle of radius 4 with center at ð0; 0Þ. Ans: 64 10.38. (a) If F ¼ ðx2 y2 Þi þ 2xyj, evaluate ð C F dr along the curve C in the xy plane given by y ¼ x2 x from the point ð1; 0Þ to ð2; 2Þ. (b) Interpret physically the result obtained. Ans. (a) 124/15 CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 259
- 269. 10.39. Evaluate ð C ð2x þ yÞ ds, where C is the curve in the xy plane given by x2 þ y2 ¼ 25 and s is the arc length parameter, from the point ð3; 4Þ to ð4; 3Þ along the shortest path. Ans: 15 10.40. If F ¼ ð3x 2yÞi þ ð y þ 2zÞj x2 k, evaluate ð C F dr from ð0; 0; 0Þ to ð1; 1; 1Þ, where C is a path consisting of (a) the curve x ¼ t; y ¼ t2 ; z ¼ t3 , (b) a straight line joining these points, (c) the straight lines from ð0; 0; 0Þ to ð0; 1; 0Þ, then to ð0; 1; 1Þ and then to ð1; 1; 1Þ, (d) the curve x ¼ z2 ; z ¼ y2 . Ans: ðaÞ 23=15; ðbÞ 5=3; ðcÞ 0; ðdÞ 13=15 10.41. If T is the unit tangent vector to a curve C (plane or space curve) and F is a given force field, prove that under appropriate conditions ð C F dr ¼ ð C F T ds where s is the arc length parameter. Interpret the result physically and geometrically. GREEN’S THEOREM IN THE PLANE, INDEPENDENCE OF THE PATH 10.42. Verify Green’s theorem in the plane for þ C ðx2 xy3 Þ dx þ ð y2 2xyÞ dy where C is a square with vertices at ð0; 0Þ; ð2; 0Þ; ð2; 2Þ; ð0; 2Þ and counterclockwise orientation. Ans. common value ¼ 8 10.43. Evaluate the line integrals of (a) Problem 10.36 and (b) Problem 10.37 by Green’s theorem. 10.44. (a) Let C be any simple closed curve bounding a region having area A. Prove that if a1; a2; a3; b1; b2; b3 are constants, þ C ða1x þ a2y þ a3Þ dx þ ðb1x þ b2y þ b3Þ dy ¼ ðb1 a2ÞA (b) Under what conditions will the line integral around any path C be zero? Ans. (b) a2 ¼ b1 10.45. Find the area bounded by the hypocycloid x2=3 þ y2=3 ¼ a2=3 . [Hint: Parametric equations are x ¼ a cos3 t; y ¼ a sin3 t; 0 @ t @ 2.] Ans: 3a2 =8 10.46. If x ¼ cos ; y ¼ sin , prove that 1 2 þ x dy y dx ¼ 1 2 ð 2 d and interpret. 10.47. Verify Green’s theorem in the plane for þ C ðx3 x2 yÞ dx þ xy2 dy, where C is the boundary of the region enclosed by the circles x2 þ y2 ¼ 4 and x2 þ y2 ¼ 16. Ans: common value ¼ 120 10.48. (a) Prove that ðð2;1Þ ð1;0Þ ð2xy y4 þ 3Þ dx þ ðx2 4xy3 Þ dy is independent of the path joining ð1; 0Þ and ð2; 1Þ. (b) Evaluate the integral in (a). Ans: ðbÞ 5 10.49. Evaluate ð C ð2xy3 y2 cos xÞ dx þ ð1 2y sin x þ 3x2 y2 Þ dy along the parabola 2x ¼ y2 from ð0; 0Þ to ð=2; 1Þ. Ans. 2 =4 10.50. Evaluate the line integral in the preceding problem around a parallelogram with vertices at ð0; 0Þ; ð3; 0Þ, ð5; 2Þ; ð2; 2Þ. Ans: 0 10.51. (a) Prove that G ¼ ð2x2 þ xy 2y2 Þ dx þ ð3x2 þ 2xyÞ dy is not an exact differential. (b) Prove that e y=x G=x is an exact differential of and find . (c) Find a solution of the differential equation ð2x2 þ xy 2y2 Þ dxþ ð3x2 þ 2xyÞ dy ¼ 0. Ans: ðbÞ ¼ ey=x ðx2 þ 2xyÞ þ c; ðcÞ x2 þ 2xy þ cey=x ¼ 0 260 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 270. SURFACE INTEGRALS 10.52. (a) Evaluate ð ð S ðx2 þ y2 Þ dS, where S is the surface of the cone z2 ¼ 3ðx2 þ y2 Þ bounded by z ¼ 0 and z ¼ 3. (b) Interpret physically the result in (a). Ans: ðaÞ 9 10.53. Determine the surface area of the plane 2x þ y þ 2z ¼ 16 cut off by (a) x ¼ 0; y ¼ 0; x ¼ 2; y ¼ 3, (b) x ¼ 0; y ¼ 0, and x2 þ y2 ¼ 64. Ans: ðaÞ 9; ðbÞ 24 10.54. Find the surface area of the paraboloid 2z ¼ x2 þ y2 which is outside the cone z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p . Ans: 2 3 ð5 ffiffiffi 5 p 1Þ 10.55. Find the area of the surface of the cone z2 ¼ 3ðx2 þ y2 Þ cut out by the paraboloid z ¼ x2 þ y2 . Ans: 6 10.56. Find the surface area of the region common to the intersecting cylinders x2 þ y2 ¼ a2 and x2 þ z2 ¼ a2 . Ans: 16a2 10.57. (a) Obtain the surface area of the sphere x2 þ y2 þ z2 ¼ a2 contained within the cone z tan ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p , 0 =2. (b) Use the result in (a) to find the surface area of a hemisphere. (c) Explain why formally placing ¼ in the result of (a) yields the total surface area of a sphere. Ans: ðaÞ 2a2 ð1 cos Þ; ðbÞ 2a2 (consider the limit as ! =2Þ 10.58. Determine the moment of inertia of the surface of a sphere of radius a about a point on the surface. Assume a constant density . Ans: 2Ma2 , where mass M ¼ 4a2 10.59. (a) Find the centroid of the surface of the sphere x2 þ y2 þ z2 ¼ a2 contained within the cone z tan ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p , 0 =2. (b) From the result in (a) obtain the centroid of the surface of a hemi- sphere. Ans: ðaÞ 1 2 að1 þ cos Þ; ðbÞ a=2 THE DIVERGENCE THEOREM 10.60. Verify the divergence theorem for A ¼ ð2xy þ zÞi þ y2 j ðx þ 3yÞk taken over the region bounded by 2x þ 2y þ z ¼ 6; x ¼ 0; y ¼ 0; z ¼ 0. Ans: common value ¼ 27 10.61. Evaluate ð ð S F n dS, where F ¼ ðz2 xÞi xyj þ 3zk and S is the surface of the region bounded by z ¼ 4 y2 ; x ¼ 0; x ¼ 3 and the xy plane. Ans. 16 10.62. Evaluate ð ð S A n dS, where A ¼ ð2x þ 3zÞi ðxz þ yÞj þ ð y2 þ 2zÞk and S is the surface of the sphere having center at ð3; 1; 2Þ and radius 3. Ans: 108 10.63. Determine the value of ð ð S x dy dz þ y dz dx þ z dx dy, where S is the surface of the region bounded by the cylinder x2 þ y2 ¼ 9 and the planes z ¼ 0 and z ¼ 3, (a) by using the divergence theorem, (b) directly. Ans: 81 10.64. Evaluate ð ð S 4xz dy dz y2 dz dx þ yz dx dy, where S is the surface of the cube bounded by x ¼ 0, y ¼ 0, z ¼ 0, x ¼ 1; y ¼ 1; z ¼ 1, (a) directly, (b) By Green’s theorem in space (divergence theorem). Ans. 3/2 10.65. Prove that ð ð S ðr AÞ n dS ¼ 0 for any closed surface S. CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 261
- 271. 10.66. Prove that ð ð S n dS ¼ 0, where n is the outward drawn normal to any closed surface S. [Hint: Let A ¼ c, where c is an arbitrary vector constant. Express the divergence theorem in this special case. Use the arbitrary property of c. 10.67. If n is the unit outward drawn normal to any closed surface S bounding the region V, prove that ð ð ð V div n dV ¼ S STOKES’ THEOREM 10.68. Verify Stokes’ theorem for A ¼ 2yi þ 3xj z2 k, where S is the upper half surface of the sphere x2 þ y2 þ z2 ¼ 9 and C is its boundary. Ans. common value ¼ 9 10.69. Verify Stokes’ theorem for A ¼ ð y þ zÞi xzj þ y2 k, where S is the surface of the region in the first octant bounded by 2x þ z ¼ 6 and y ¼ 2 which is not included in the (a) xy plane, (b) plane y ¼ 2, (c) plane 2x þ z ¼ 6 and C is the corresponding boundary. Ans. The common value is (a) 6; ðbÞ 9; ðcÞ 18 10.70. Evaluate ð ð S ðr AÞ n dS, where A ¼ ðx zÞi þ ðx3 þ yzÞj 3xy2 k and S is the surface of the cone z ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 p above the xy plane. Ans: 12 10.71. If V is a region bounded by a closed surface S and B ¼ r A, prove that ð ð S B n dS ¼ 0. 10.72. (a) Prove that F ¼ ð2xy þ 3Þi þ ðx2 4zÞj 4yk is a conservative force field. (b) Find such that F ¼ r. (c) Evaluate ð C F dr, where C is any path from ð3; 1; 2Þ to ð2; 1; 1Þ. Ans: ðbÞ ¼ x2 y 4yz þ 3x þ constant; (c) 6 10.73. Let C be any path joining any point on the sphere x2 þ y2 þ z2 ¼ a2 to any point on the sphere x2 þ y2 þ z2 ¼ b2 . Show that if F ¼ 5r3 r, where r ¼ xi þ yj þ zk, then ð C F dr ¼ b5 a5 . 10.74. In Problem 10.73 evaluate ð C F dr is F ¼ f ðrÞr, where f ðrÞ is assumed to be continuous. Ans: ðb a r f ðrÞ dr 10.75. Determine whether there is a function such that F ¼ r, where: (a) F ¼ ðxz yÞi þ ðx2 y þ z3 Þj þ ð3xz2 xyÞk: (b) F ¼ 2xey i þ ðcos z x2 ey Þj y sin zk. If so, find it. Ans: ðaÞ does not exist. ðbÞ ¼ x2 ey þ y cos z þ constant 10.76. Solve the differential equation ðz3 4xyÞ dx þ ð6y 2x2 Þ dy þ ð3xz2 þ 1Þ dz ¼ 0. Ans: xz3 2x2 y þ 3y2 þ z ¼ constant MISCELLANEOUS PROBLEMS 10.77. Prove that a necessary and sufficient condition that þ C @U @x dy @U @y dx be zero around every simple closed path C in a region r (where U is continuous and has continuous partial derivatives of order two, at least) is that @2 U @x2 þ @2 U @y2 ¼ 0. 262 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 272. 10.78. Verify Green’s theorem for a multiply connected region containing two ‘‘holes’’ (see Problem 10.10). 10.79. If P dx þ Q dy is not an exact differential but ðP dx þ Q dyÞ is an exact differential where is some function of x and y, then is called an integrating factor. (a) Prove that if F and G are functions of x alone, then ðFy þ GÞ dx þ dy has an integrating factor which is a function of x alone and find . What must be assumed about F and G? (b) Use (a) to find solutions of the differential equation xy0 ¼ 2x þ 3y. Ans: ðaÞ ¼ e Ð FðxÞ dx ðbÞ y ¼ cx3 x, where c is any constant 10.80. Find the surface area of the sphere x2 þ y2 þ ðz aÞ2 ¼ a2 contained within the paraboloid z ¼ x2 þ y2 . Ans: 2a 10.81. If f ðrÞ is a continuously differentiable function of r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 þ z2 p , prove that ð ð S f ðrÞ n dS ¼ ð ð ð V f 0 ðrÞ r r dV 10.82. Prove that ð ð S r ðnÞ dS ¼ 0 where is any continuously differentiable scalar function of position and n is a unit outward drawn normal to a closed surface S. (See Problem 10.66.) 10.83. Establish equation (3), Problem 10.32, by using Green’s theorem in the plane. [Hint: Let the closed region r in the xy plane have boundary C and suppose that under the transformation x ¼ f ðu; vÞ; y ¼ gðu; vÞ, these are transformed into r0 and C0 in the uv plane, respectively. First prove that ð ð r Fðx; yÞ dx dy ¼ ð C Qðx; yÞ dy where @Q=@y ¼ Fðx; yÞ. Then show that apart from sign this last integral is equal to ð C0 Q½ f ðu; vÞ; gðu; vÞ @g @u du þ @g @v dv . Finally, use Green’s theorem to transform this into ð ð r0 F½ f ðu; vÞ; gðu; vÞ @ðx; yÞ @ðu; vÞ du dv. 10.84. If x ¼ f ðu; v; wÞ; y ¼ gðu; v; wÞ; z ¼ hðu; v; wÞ defines a transformation which maps a region r of xyz space into a region r0 of uvw space, prove using Stokes’ theorem that ð ð ð r Fðx; y; zÞ dx dy dz ¼ ð ð ð r0 Gðu; v; wÞ @ðx; y; zÞ @ðu; v; wÞ du dv dw where Gðu; v; wÞ F½ f ðu; v; wÞ; gðu; v; wÞ; hðu; v; wÞ. State sufficient conditions under which the result is valid. See Problem 10.83. Alternatively, employ the differential element of volume dV ¼ @r @u @r @v @r @w du dv dw (recall the geometric meaning). 10.85. (a) Show that in general the equation r ¼ rðu; vÞ geometrically represents a surface. (b) Discuss the geo- metric significance of u ¼ c1; v ¼ c2, where c1 and c2 are constants. (c) Prove that the element of arc length on this surface is given by ds2 ¼ E du2 þ 2F du dv þ G dv2 where E ¼ @r @u @r @u ; F ¼ @r @u @r @v ; G ¼ @r @v @r @v : CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 263
- 273. 10.86. (a) Referring to Problem 10.85, show that the element of surface area is given by dS ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi EG F2 p du dv. (b) Deduce from (a) that the area of a surface r ¼ rðu; vÞ is ð ð S ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi EG F2 p du dv. [Hint: Use the fact that @r @u @r @v ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @r @u @r @v @r @u @r @v s and then use the identity ðA BÞ ðC DÞ ¼ ðA CÞðB DÞ ðA DÞðB CÞ. 10.87. (a) Prove that r ¼ a sin u cos v i þ a sin u sin v j þ a cos u, 0 @ u @ ; 0 @ v 2 represents a sphere of radius a. (b) Use Problem 10.86 to show that the surface area of this sphere is 4a2 . 10.88. Use the result of Problem 10.34 to obtain div A in (a) cylindrical and (b) spherical coordinates. See Page 161. 264 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10
- 274. 265 Infinite Series The early developers of the calculus, including Newton and Leibniz, were well aware of the importance of infinite series. The values of many functions such as sine and cosine were geometrically obtainable only in special cases. Infinite series provided a way of developing extensive tables of values for them. This chapter begins with a statement of what is meant by infinite series, then the question of when these sums can be assigned values is addressed. Much information can be obtained by exploring infinite sums of constant terms; however, the eventual objective in analysis is to introduce series that depend on variables. This presents the possibility of representing functions by series. Afterward, the question of how continuity, differentiability, and integrability play a role can be examined. The question of dividing a line segment into infinitesimal parts has stimulated the imaginations of philosophers for a very long time. In a corruption of a paradox introduce by Zeno of Elea (in the fifth century B.C.) a dimensionless frog sits on the end of a one-dimensional log of unit length. The frog jumps halfway, and then halfway and halfway ad infinitum. The question is whether the frog ever reaches the other end. Mathematically, an unending sum, 1 2 þ 1 4 þ þ 1 2n þ is suggested. ‘‘Common sense’’ tells us that the sum must approach one even though that value is never attained. We can form sequences of partial sums S1 ¼ 1 2 ; S2 ¼ 1 2 þ 1 4 ; . . . ; Sn ¼ 1 2 þ 1 4 þ þ 1 2n þ and then examine the limit. This returns us to Chapter 2 and the modern manner of thinking about the infinitesimal. In this chapter consideration of such sums launches us on the road to the theory of infinite series. DEFINITIONS OF INFINITE SERIES AND THEIR CONVERGENCE AND DIVERGENCE Definition: The sum S ¼ X 1 n¼1 un ¼ u1 þ u2 þ þ un þ ð1Þ Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 275. is an infinite series. Its value, if one exists, is the limit of the sequence of partial sums fSng S ¼ lim n!1 Sn ð2Þ If there is a unique value, the series is said to converge to that sum, S. If there is not a unique sum the series is said to diverge. Sometimes the character of a series is obvious. For example, the series X 1 n¼1 1 2n generated by the frog on the log surely converges, while X 1 n¼1 n is divergent. On the other hand, the variable series 1 x þ x2 x3 þ x4 x5 þ raises questions. This series may be obtained by carrying out the division 1 1 x . If 1 x 1, the sums Sn yields an approximations to 1 1 x and (2) is the exact value. The indecision arises for x ¼ 1. Some very great mathematicians, including Leonard Euler, thought that S should be equal to 1 2, as is obtained by substituting 1 into 1 1 x . The problem with this conclusion arises with examination of 1 1 þ 1 1þ 1 1 þ and observation that appropriate associations can produce values of 1 or 0. Imposition of the condition of uniqueness for convergence put this series in the category of divergent and eliminated such possibility of ambiguity in other cases. FUNDAMENTAL FACTS CONCERNING INFINITE SERIES 1. If un converges, then lim n!1 un ¼ 0 (see Problem 2.26, Chap. 2). The converse, however, is not necessarily true, i.e., if lim n!1 un ¼ 0, un may or may not converge. It follows that if the nth term of a series does not approach zero the series is divergent. 2. Multiplication of each term of a series by a constant different from zero does not affect the convergence or divergence. 3. Removal (or addition) of a finite number of terms from (or to) a series does not affect the convergence or divergence. SPECIAL SERIES 1. Geometric series X 1 n¼1 arn1 ¼ a þ ar þ ar2 þ , where a and r are constants, converges to S ¼ a 1 r if jrj 1 and diverges if jrj A 1. The sum of the first n terms is Sn ¼ að1 rn Þ 1 r (see Problem 2.25, Chap. 2). 2. The p series X 1 n¼1 1 n p ¼ 1 1p þ 1 2p þ 1 3p þ ; where p is a constant, converges for p 1 and diverges for p @ 1. The series with p ¼ 1 is called the harmonic series. TESTS FOR CONVERGENCE AND DIVERGENCE OF SERIES OF CONSTANTS More often than not, exact values of infinite series cannot be obtained. Thus, the search turns toward information about the series. In particular, its convergence or divergence comes in question. The following tests aid in discovering this information. 266 INFINITE SERIES [CHAP. 11
- 276. 1. Comparison test for series of non-negative terms. (a) Convergence. Let vn A 0 for all n N and suppose that vn converges. Then if 0 @ un @ vn for all n N, un also converges. Note that n N means from some term onward. Often, N ¼ 1. EXAMPLE: Since 1 2n þ 1 @ 1 2n and X 1 2n converges, X 1 2n þ 1 also converges. (b) Divergence. Let vn A 0 for all n N and suppose that vn diverges. Then if un A vn for all n N, un also diverges. EXAMPLE: Since 1 ln n 1 n and X 1 n¼2 1 n diverges, X 1 n¼2 1 ln n also diverges. 2. The Limit-Comparison or Quotient Test for series of non-negative terms. (a) If un A 0 and vn A 0 and if lim n!1 un vn ¼ A 6¼ 0 or 1, then un and vn either both converge or both diverge. (b) If A ¼ 0 in (a) and vn converges, then un converges. (c) If A ¼ 1 in (a) and vn diverges, then un diverges. This test is related to the comparison test and is often a very useful alternative to it. In particlar, taking vn ¼ 1=np , we have from known facts about the p series the Theorem 1. Let lim n!1 np un ¼ A. Then (i) un converges if p 1 and A is finite. (ii) un diverges if p @ 1 and A 6¼ 0 (A may be infinite). EXAMPLES: 1: X n 4n3 2 converges since lim n!1 n2 n 4n3 2 ¼ 1 4 : 2: X ln n ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p diverges since lim n!1 n1=2 ln n ðn þ 1Þ1=2 ¼ 1: 3. Integral test for series of non-negative terms. If f ðxÞ is positive, continuous, and monotonic decreasing for x A N and is such that f ðnÞ ¼ un; n ¼ N; N þ 1; N þ 2; . . . , then un converges or diverges according as ð1 N f ðxÞ dx ¼ lim M!1 ðM n f ðxÞ dx converges or diverges. In particular we may have N ¼ 1, as is often true in practice. This theorem borrows from the next chapter since the integral has an unbounded upper limit. (It is an improper integral. The convergence or divergence of these integrals is defined in much the same way as for infinite series.) EXAMPLE: X 1 n¼1 1 n2 converges since lim M!1 ðM 1 dx x2 ¼ lim M!1 1 1 M exists. 4. Alternating series test. An alternating series is one whose successive terms are alternately positive and negative. An alternating series converges if the following two conditions are satisfied (see Problem 11.15). (a) junþ1j @ junj for n A N (Since a fixed number of terms does not affect the conver- gence or divergence of a series, N may be any positive integer. Frequently it is chosen to be 1.) (b) lim n!1 un ¼ 0 or lim n!1 junj ¼ 0 CHAP. 11] INFINITE SERIES 267
- 277. EXAMPLE. For the series 1 1 2 þ 1 3 1 4 þ 1 5 ¼ X 1 n¼1 ð1Þn1 n , we have un ¼ ð1Þn1 n , junj ¼ 1 n , junþ1j ¼ 1 n þ 1 . Then for n A 1, junþ1j @ junj. Also lim n!1 junj ¼ 0. Hence, the series converges. Theorem 2. The numerical error made in stopping at any particular term of a convergent alternating series which satisfies conditions (a) and (b) is less than the absolute value of the next term. EXAMPLE. If we stop at the 4th term of the series 1 1 2 þ 1 3 1 4 þ 1 5 , the error made is less than 1 5 ¼ 0:2. 5. Absolute and conditional convergence. The series un is called absolutely convergent if junj converges. If un converges but junj diverges, then un is called conditionally convergent. Theorem 3. If junj converges, then un converges. In words, an absolutely convergent series is convergent (see Problem 11.17). EXAMPLE 1. 1 12 þ 1 22 1 32 1 42 þ 1 52 þ 1 62 is absolutely convergent and thus convergent, since the series of absolute values 1 12 þ 1 22 þ 1 32 þ 1 42 þ converges. EXAMPLE 2. 1 1 2 þ 1 3 1 4 þ converges, but 1 þ 1 2 þ 1 3 þ 1 4 þ diverges. Thus, 1 1 2 þ 1 3 1 4 þ is conditionally convergent. Any of the tests used for series with non-negative terms can be used to test for absolute convergence. Also, tests that compare successive terms are common. Tests 6, 8, and 9 are of this type. 6. Ratio test. Let lim n!1 unþ1 un ¼ L. Then the series un (a) converges (absolutely) if L 1 (b) diverges if L 1. If L ¼ 1 the test fails. 7. The nth root test. Let lim n!1 ffiffiffiffiffiffiffiffi junj n p ¼ L. Then the series un (a) converges (absolutely) if L 1 (b) diverges if L 1: If L ¼ 1 the test fails. 8. Raabe’s test. Let lim n!1 n 1 un þ 1 un ¼ L. Then the series un (a) converges (absolutely) if L 1 (b) diverges or converges conditionally if L 1. If L ¼ 1 the test fails. This test is often used when the ratio tests fails. 9. Gauss’ test. If unþ1 un ¼ 1 L n þ cn n2 , where jcnj P for all n N, then the series un (a) converges (absolutely) if L 1 (b) diverges or converges conditionally if L @ 1. This test is often used when Raabe’s test fails. 268 INFINITE SERIES [CHAP. 11
- 278. THEOREMS ON ABSOLUTELY CONVERGENT SERIES Theorem 4. (Rearrangement of Terms) The terms of an absolutely convergent series can be rearranged in any order, and all such rearranged series will converge to the same sum. However, if the terms of a conditionally convergent series are suitably rearranged, the resulting series may diverge or converge to any desired sum (see Problem 11.80). Theorem 5. (Sums, Differences, and Products) The sum, difference, and product of two absolutely convergent series is absolutely convergent. The operations can be performed as for finite series. INFINITE SEQUENCES AND SERIES OF FUNCTIONS, UNIFORM CONVERGENCE We opened this chapter with the thought that functions could be expressed in series form. Such representation is illustrated by sin x ¼ x x3 3! þ x5 5! þ þ ð1Þn1 x2n1 ð2n 1Þ! þ where sin x ¼ lim n!1 Sn; with S1 ¼ x; S2 ¼ x x3 3! ; . . . Sn ¼ X n k¼1 ð1Þk1 x2k1 ð2k 1Þ! : Observe that until this section the sequences and series depended on one element, n. Now there is variation with respect to x as well. This complexity requires the introduction of a new concept called uniform convergence, which, in turn, is fundamental in exploring the continuity, differentiation, and integrability of series. Let funðxÞg; n ¼ 1; 2; 3; . . . be a sequence of functions defined in ½a; b. The sequence is said to converge to FðxÞ, or to have the limit FðxÞ in ½a; b, if for each 0 and each x in ½a; b we can find N 0 such that junðxÞ FðxÞj for all n N. In such case we write lim n!1 unðxÞ ¼ FðxÞ. The number N may depend on x as well as . If it depends only on and not on x, the sequence is said to converge to FðxÞ uniformly in ½a; b or to be uniformly convergent in ½a; b. The infinite series of functions X 1 n¼1 unðxÞ ¼ u1ðxÞ þ u2ðxÞ þ u3ðxÞ þ ð3Þ is said to be convergent in ½a; b if the sequence of partial sums fSnðxÞg, n ¼ 1; 2; 3; . . . ; where SnðxÞ ¼ u1ðxÞ þ u2ðxÞ þ þ unðxÞ, is convergent in ½a; b. In such case we write lim n!1 SnðxÞ ¼ SðxÞ and call SðxÞ the sum of the series. It follows that unðxÞ converges to SðxÞ in ½a; b if for each 0 and each x in ½a; b we can find N 0 such that jSnðxÞ SðxÞj for all n N. If N depends only on and not on x, the series is called uniformly convergent in ½a; b. Since SðxÞ SnðxÞ ¼ RnðxÞ, the remainder after n terms, we can equivalently say that unðxÞ is uniformly convergent in ½a; b if for each 0 we can find N depending on but not on x such that jRnðxÞj for all n N and all x in ½a; b. These definitions can be modified to include other intervals besides a @ x @ b, such as a x b, and so on. The domain of convergence (absolute or uniform) of a series is the set of values of x for which the series of functions converges (absolutely or uniformly). EXAMPLE 1. Suppose un ¼ xn =n and 1 2 @ x @ 1. Now think of the constant function FðxÞ ¼ 0 on this interval. For any 0 and any x in the interval, there is N such that for all n Njun FðxÞj , i.e., jxn =nj . Since the limit does not depend on x, the sequence is uniformly convergent. CHAP. 11] INFINITE SERIES 269
- 279. EXAMPLE 2. If un ¼ xn and 0 @ x @ 1, the sequence is not uniformly convergent because (think of the function FðxÞ ¼ 0, 0 @ x 1, Fð1Þ ¼ 1Þ jxn 0j when xn ; thus n ln x ln : On the interval 0 @ x 1, and for 0 1, both members of the inequality are negative, therefore, n ln ln x : Since ln ln x ¼ ln 1 ln ln 1 nn x ¼ lnð=Þ lnð1=xÞ , it follows that we must choose N such that n N ln 1= ln 1=x From this expression we see that ! 0 then ln 1 ! 1 and also as x ! 1 from the left ln 1 x ! 0 from the right; thus, in either case, N must increase without bound. This dependency on both and x demonstrations that the sequence is not uniformly convergent. For a pictorial view of this example, see Fig. 11-1. SPECIAL TESTS FOR UNIFORM CONVERGENCE OF SERIES 1. Weierstrass M test. If sequence of positive constants M1; M2; M3; . . . can be found such that in some interval (a) junðxÞj @ Mn n ¼ 1; 2; 3; . . . (b) Mn converges then unðxÞ is uniformly and absolutely convergent in the interval. EXAMPLE. X 1 n¼1 cos nx n2 is uniformly and absolutely convergent in ½0; 2 since cos nx n2 @ 1 n2 and X 1 n2 converges. This test supplies a sufficient but not a necessary condition for uniform convergence, i.e., a series may be uniformly convergent even when the test cannot be made to apply. One may be led because of this test to believe that uniformly convergent series must be absolutely convergent, and conversely. However, the two properties are independent, i.e., a series can be uniformly convergent without being absolutely convergent, and conversely. See Problems 11.30, 11.127. 2. Dirichlet’s test. Suppose that (a) the sequence fang is a monotonic decreasing sequence of positive constants having limit zero, (b) there exists a constant P such that for a @ x @ b ju1ðxÞ þ u2ðxÞ þ þ unðxÞj P for all n N: Then the series a1u1ðxÞ þ a2u2ðxÞ þ ¼ X 1 n¼1 anunðxÞ is uniformly convergent in a @ x @ b. 270 INFINITE SERIES [CHAP. 11 Fig. 11-1
- 280. THEOREMS ON UNIFORMLY CONVERGENT SERIES If an infinite series of functions is uniformly convergent, it has many of the properties possessed by sums of finite series of functions, as indicated in the following theorems. Theorem 6. If funðxÞg; n ¼ 1; 2; 3; . . . are continuous in ½a; b and if unðxÞ converges uniformly to the sum SðxÞ in ½a; b, then SðxÞ is continuous in ½a; b. Briefly, this states that a uniformly convergent series of continuous functions is a continuous function. This result is often used to demonstrate that a given series is not uniformly convergent by showing that the sum function SðxÞ is discontinuous at some point (see Problem 11.30). In particular if x0 is in ½a; b, then the theorem states that lim x!x0 X 1 n¼1 unðxÞ ¼ X 1 n¼1 lim x!x0 unðxÞ ¼ X 1 n¼1 unðx0Þ where we use right- or left-hand limits in case x0 is an endpoint of ½a; b. Theorem 7. If funðxÞg; n ¼ 1; 2; 3; . . . ; are continuous in ½a; b and if unðxÞ converges uniformly to the sum SðxÞ in ½a; b, then ðb a SðxÞ dx ¼ X 1 n¼1 ðb a unðxÞ dx ð4Þ or ðb a X 1 n¼1 unðxÞ ( ) dx ¼ X 1 n¼1 ðb a unðxÞ dx ð5Þ Briefly, a uniformly convergent series of continuous functions can be integrated term by term. Theorem 8. If funðxÞg; n ¼ 1; 2; 3; . . . ; are continuous and have continuous derivatives in ½a; b and if unðxÞ converges to SðxÞ while u0 nðxÞ is uniformly convergent in ½a; b, then in ½a; b S0 ðxÞ ¼ X 1 n¼1 u0 nðxÞ ð6Þ or d dx X 1 n¼1 unðxÞ ( ) ¼ X 1 n¼1 d dx unðxÞ ð7Þ This shows conditions under which a series can be differentiated term by term. Theorems similar to the above can be formulated for sequences. For example, if funðxÞg, n ¼ 1; 2; 3; . . . is uniformly convergent in ½a; b, then lim n!1 ðb a unðxÞ dx ¼ ðb a lim n!1 unðxÞ dx ð8Þ which is the analog of Theorem 7. CHAP. 11] INFINITE SERIES 271
- 281. POWER SERIES A series having the form a0 þ a1x þ a2x2 þ ¼ X 1 n¼0 anxn ð9Þ where a0; a1; a2; . . . are constants, is called a power series in x. It is often convenient to abbreviate the series (9) as anxn . In general a power series converges for jxj R and diverges for jxj R, where the constant R is called the radius of convergence of the series. For jxj ¼ R, the series may or may not converge. The interval jxj R or R x R, with possible inclusion of endpoints, is called the interval of convergence of the series. Although the ratio test is often successful in obtaining this interval, it may fail and in such cases, other tests may be used (see Problem 11.22). The two special cases R ¼ 0 and R ¼ 1 can arise. In the first case the series converges only for x ¼ 0; in the second case it converges for all x, sometimes written 1 x 1 (see Problem 11.25). When we speak of a convergent power series, we shall assume, unless otherwise indicated, that R 0. Similar remarks hold for a power series of the form (9), where x is replaced by ðx aÞ. THEOREMS ON POWER SERIES Theorem 9. A power series converges uniformly and absolutely in any interval which lies entirely within its interval of convergence. Theorem 10. A power series can be differentiated or integrated term by term over any interval lying entirely within the interval of convergence. Also, the sum of a convergent power series is continuous in any interval lying entirely within its interval of convergence. This follows at once from Theorem 9 and the theorems on uniformly convergent series on Pages 270 and 271. The results can be extended to include end points of the interval of convergence by the following theorems. Theorem 11. Abel’s theorem. When a power series converges up to and including an endpoint of its interval of convergence, the interval of uniform convergence also extends so far as to include this endpoint. See Problem 11.42. Theorem 12. Abel’s limit theorem. If X 1 n¼0 anxn converges at x ¼ x0, which may be an interior point or an endpoint of the interval of convergence, then lim x!x0 X 1 n¼0 anxn ( ) ¼ X 1 n¼0 lim x!x0 anxn ¼ X 1 n¼0 anxn 0 ð10Þ If x0 is an end point, we must use x ! x0þ or x ! x0 in (10) according as x0 is a left- or right-hand end point. This follows at once from Theorem 11 and Theorem 6 on the continuity of sums of uniformly convergent series. OPERATIONS WITH POWER SERIES In the following theorems we assume that all power series are convergent in some interval. Theorem 13. Two power series can be added or subtracted term by term for each value of x common to their intervals of convergence. 272 INFINITE SERIES [CHAP. 11
- 282. Theorem 14. Two power series, for example, X 1 n¼0 anxn and X 1 n¼0 bnxn , can be multiplied to obtain X 1 n¼0 cnxn where cn ¼ a0bn þ a1bn1 þ a2bn2 þ þ anb0 ð11Þ the result being valid for each x within the common interval of convergence. Theorem 15. If the power series X 1 n¼0 anxn is divided by the power series bnxn where b0 6¼ 0, the quotient can be written as a power series which converges for sufficiently small values of x. Theorem 16. If y ¼ X 1 n¼0 anxn , then by substituting x ¼ X 1 n¼0 bnyn , we can obtain the coefficients bn in terms of an. This process is often called reversion of series. EXPANSION OF FUNCTIONS IN POWER SERIES This section gets at the heart of the use of infinite series in analysis. Functions are represented through them. Certain forms bear the names of mathematicians of the eighteenth and early nineteenth century who did so much to develop these ideas. A simple way (and one often used to gain information in mathematics) to explore series representa- tion of functions is to assume such a representation exists and then discover the details. Of course, whatever is found must be confirmed in a rigorous manner. Therefore, assume f ðxÞ ¼ A0 þ A1ðx cÞ þ A2ðx cÞ2 þ þ Anðx cÞn þ Notice that the coefficients An can be identified with derivatives of f . In particular A0 ¼ f ðcÞ; A1 ¼ f 0 ðcÞ; A2 ¼ 1 2! f 00 ðcÞ; . . . ; An ¼ 1 n! f ðnÞ ðcÞ; . . . This suggests that a series representation of f is f ðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ þ 1 2! f 00 ðcÞðx cÞ2 þ þ 1 n! f ðnÞ ðcÞðx cÞn þ A first step in formalizing series representation of a function, f , for which the first n derivatives exist, is accomplished by introducing Taylor polynomials of the function. P0ðxÞ ¼ f ðcÞ P1ðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ; P2ðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ þ 1 2! f 00 ðcÞðx cÞ2 ; PnðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ þ þ 1 n! f ðnÞ ðcÞðx cÞn ð12Þ TAYLOR’S THEOREM Let f and its derivatives f 0 ; f 00 ; . . . ; f ðnÞ exist and be continuous in a closed interval a x b and suppose that f ðnþ1Þ exists in the open interval a x b. Then for c in ½a; b, f ðxÞ ¼ PnðxÞ þ RnðxÞ; where the remainder RnðxÞ may be represented in any of the three following ways. For each n there exists such that CHAP. 11] INFINITE SERIES 273
- 283. RnðxÞ ¼ 1 ðn þ 1Þ! f ðnþ1Þ ðÞðx cÞnþ1 (Lagrange form) ð13Þ ( is between c and x.) (The theorem with this remainder is a mean value theorem. Also, it is called Taylor’s formula.) For each n there exists such that RnðxÞ ¼ 1 n! f ðnþ1Þ ðÞðx Þn ðx cÞ (Cauchy form) ð14Þ RnðxÞ ¼ 1 n! ðx c ðx tÞn f ðnþ1Þ ðtÞ dt (Integral form) ð15Þ If all the derivatives of f exist, then f ðxÞ ¼ X 1 n¼0 1 n! f ðnÞ ðcÞðx cÞn ð16Þ This infinite series is called a Taylor series, although when c ¼ 0, it can also be referred to as a MacLaurin series or expansion. One might be tempted to believe that if all derivatives of f ðxÞ exist at x ¼ c, the expansion (16) would be valid. This, however, is not necessarily the case, for although one can then formally obtain the series on the right of (16), the resulting series may not converge to f ðxÞ. For an example of this see Problem 11.108. Precise conditions under which the series converges to f ðxÞ are best obtained by means of the theory of functions of a complex variable. See Chapter 16. The determination of values of functions at desired arguments is conveniently approached through Taylor polynomials. EXAMPLE. The value of sin x may be determined geometrically for 0; 6 , and an infinite number of other arguments. To obtain values for other real number arguments, a Taylor series may be expanded about any of these points. For example, let c ¼ 0 and evaluate several derivatives there, i.e., f ð0Þ ¼ sin 0 ¼ 0; f 0 ð0Þ ¼ cos 0 ¼ 1, f 00 ð0Þ ¼ sin 0 ¼ 0; f 000 ð0Þ ¼ cos 0 ¼ 1; f 1v ð0Þ ¼ sin 0 ¼ 0; f v ð0Þ ¼ cos 0 ¼ 1. Thus, the MacLaurin expansion to five terms is sin x ¼ 0 þ x 0 1 3! x3 þ 0 1 51 x5 þ Since the fourth term is 0 the Taylor polynomials P3 and P4 are equal, i.e., P3ðxÞ ¼ P4ðxÞ ¼ x x3 3! and the Lagrange remainder is R4ðxÞ ¼ 1 5! cos x5 Suppose an approximation of the value of sin :3 is required. Then P4ð:3Þ ¼ :3 1 6 ð:3Þ3 :2945: The accuracy of this approximation can be determined from examination of the remainder. In particular, (remember j cos j 1) jR4j ¼ 1 5! cos ð:3Þ5 1 120 243 105 :000021 274 INFINITE SERIES [CHAP. 11
- 284. Thus, the approximation P4ð:3Þ for sin :3 is correct to four decimal places. Additional insight to the process of approximation of functional values results by constructing a graph of P4ðxÞ and comparing it to y ¼ sin x. (See Fig. 11-2.) P4ðxÞ ¼ x x3 6 The roots of the equation are 0; ffiffiffi 6 p . Examination of the first and second derivatives reveals a relative maximum at x ¼ ffiffiffi 2 p and a relative minimum at x ¼ ffiffiffi 2 p . The graph is a local approximation of the sin curve. The reader can show that P6ðxÞ produces an even better approximation. (For an example of series approximation of an integral see the example below.) SOME IMPORTANT POWER SERIES The following series, convergent to the given function in the indicated intervals, are frequently employed in practice: 1. sin x ¼ x x3 3! þ x5 5! x7 7! þ ð1Þn1 x2n1 ð2n 1Þ! þ 1 x 1 2. cos x ¼ 1 x2 2! þ x4 4! x6 6! þ ð1Þn1 x2n2 ð2n 2Þ! þ 1 x 1 3. ex ¼ 1 þ x þ x2 2! þ x3 3! þ þ xn1 ðn 1Þ! þ 1 x 1 4. ln j1 þ xj ¼ x x2 2 þ x3 3 x4 4 þ ð1Þn1 xn n þ 1 x @ 1 5. 1 2 ln 1 þ x 1 x ¼ x þ x3 3 þ x5 5 þ x7 7 þ þ x2n1 2n 1 þ 1 x 1 6. tan1 x ¼ x x3 3 þ x5 5 x7 7 þ ð1Þn1 x2n1 2n 1 þ 1 @ x @ 1 7. ð1 þ xÞp ¼ 1 þ px þ pð p 1Þ 2! x2 þ þ pð p 1Þ . . . ð p n þ 1Þ n! xn þ This is the binomial series. (a) If p is a positive integer or zero, the series terminates. (b) If p 0 but is not an integer, the series converges (absolutely) for 1 @ x @ 1: ðcÞ If 1 p 0, the series converges for 1 x @ 1: (d) If p @ 1, the series converges for 1 x 1. For all p the series certainly converges if 1 x 1. EXAMPLE. Taylor’s Theorem applied to the series for ex enables us to estimate the value of the integral ð1 0 ex2 dx. Substituting x2 for x, we obtain Ð1 0 ex2 dx ¼ Ð1 0 1 þ x þ x4 2! þ x6 3! þ x8 4! þ e 5! x10 ! dx where P4ðxÞ ¼ 1 þ x þ 1 2! x4 þ 1 3! x6 þ 1 4! x8 and R4ðxÞ ¼ e 5! x10 ; 0 x CHAP. 11] INFINITE SERIES 275 Fig. 11-2
- 285. Then ð1 0 P4ðxÞ dx ¼ 1 þ 1 3 þ 1 5ð2!Þ þ 1 7ð3!Þ þ 1 9ð4!Þ 1:4618 ð1 0 R4ðxÞ dx ð1 0 e 5! x10 dx e ð1 0 x10 5! dx ¼ e 11:5 :0021 Thus, the maximum error is less than .0021 and the value of the integral is accurate to two decimal places. SPECIAL TOPICS 1. Functions defined by series are often useful in applications and frequently arise as solutions of differential equations. For example, the function defined by JpðxÞ ¼ xp 2p p! 1 x2 2ð2p þ 2Þ þ x4 2 4ð2p þ 2Þð2p þ 4Þ ( ) ¼ X 1 n¼0 ð1Þn ðx=2Þpþ2n n!ðn þ pÞ! ð16Þ is a solution of Bessel’s differential equation x2 y00 þ xy0 þ ðx2 p2 Þy ¼ 0 and is thus called a Bessel function of order p. See Problems 11.46, 11.110 through 11.113. Similarly, the hypergeometric function Fða; b; c; xÞ ¼ 1 þ a B 1 c x þ aða þ 1Þbðb þ 1Þ 1 2 cðc þ 1Þ x2 þ ð17Þ is a solution of Gauss’ differential equation xð1 xÞy00 þ fc ða þ b þ 1Þxgy0 aby ¼ 0. These functions have many important properties. 2. Infinite series of complex terms, in particular power series of the form X 1 n¼0 anzn , where z ¼ x þ iy and an may be complex, can be handled in a manner similar to real series. Such power series converge for jzj R, i.e., interior to a circle of convergence x2 þ y2 ¼ R2 , where R is the radius of convergence (if the series converges only for z ¼ 0, we say that the radius of convergence R is zero; if it converges for all z, we say that the radius of convergence is infinite). On the boundary of this circle, i.e., jzj ¼ R, the series may or may not converge, depending on the particular z. Note that for y ¼ 0 the circle of convergence reduces to the interval of convergence for real power series. Greater insight into the behavior of power series is obtained by use of the theory of functions of a complex variable (see Chapter 16). 3. Infinite series of functions of two (or more) variables, such as X 1 n¼1 unðx; yÞ can be treated in a manner analogous to series in one variable. In particular, we can discuss power series in x and y having the form a00 þ ða10x þ a01yÞ þ ða20x2 þ a11xy þ a02y2 Þ þ ð18Þ using double subscripts for the constants. As for one variable, we can expand suitable functions of x and y in such power series. In particular, the Taylor theroem may be extended as follows. TAYLOR’S THEOREM (FOR TWO VARIABLES) Let f be a function of two variables x and y. If all partial derivatives of order n are continuous in a closed region and if all the ðn þ 1Þ partial derivatives exist in the open region, then 276 INFINITE SERIES [CHAP. 11
- 286. f ðx0 þ h; y0 þ kÞ ¼ f ðx0; y0Þ þ h @ @x þ k @ @y f ðx0; y0Þ þ 1 2! h @ @x þ k @ @y 2 f ðx0; y0Þ þ þ 1 n! h @ @x þ k @ @y n f ðx0; y0Þ þ Rn ð18Þ where Rn ¼ 1 ðn þ 1Þ! h @ @x þ k @ @y nþ1 f ðx0 þ h; y0 þ kÞ; 0 1 and where the meaning of the operator notation is as follows: h @ @x þ k @ @y f ¼ hfx þ kfy; h @ @x þ k @ @y 2 ¼ h2 fxx þ 2hkfxy þ k2 fyy and we formally expand h @ @x þ k @ @y n by the binomial theorem. Note: In alternate notation h ¼ x ¼ x x0, k ¼ y ¼ y y0. If Rn ! 0 as n ! 1 then an unending continuation of terms produces the Taylor series for f ðx; yÞ. Multivariable Taylor series have a similar pattern. 4. Double Series. Consider the array of numbers (or functions) u11 u12 u13 . . . u21 u22 u23 . . . u31 u32 u33 . . . . . . . . . . . . 0 B B B @ 1 C C C A Let Smn ¼ X m p¼1 X n q¼1 upq be the sum of the numbers in the first m rows and first n columns of this array. If there exists a number S such that lim m!1 n!1 Smn ¼ S, we say that the doubles series X 1 p¼1 X 1 q¼1 upq converges to the sum S; otherwise, it diverges. Definitions and theorems for double series are very similar to those for series already considered. 5. Infinite Products. Let Pn ¼ ð1 þ u1Þð1 þ u2Þð1 þ u3Þ . . . ð1 þ unÞ denoted by Y n k¼1 ð1 þ ukÞ, where we suppose that uk 6¼ 1; k ¼ 1; 2; 3; . . . . If there exists a number P 6¼ 0 such that lim n!1 Pn ¼ P, we say that the the infinite product ðð1 þ u1Þð1 þ u2Þð1 þ u3Þ . . . ¼ Y 1 k¼1 ð1 þ ukÞ, or briefly ð1 þ ukÞ, converges to P; otherwise, it diverges. If ð1 þ jukjÞ converges, we call the infinite product ð1 þ ukÞ absolutely convergent. It can be shown that an absolutely convergent infinite product converges and that factors can in such cases be rearranged without affecting the result. Theorems about infinite products can (by taking logarithms) often be made to depend on theorems for infinite series. Thus, for example, we have the following theorem. Theorem. A necessary and sufficient condition that ð1 þ ukÞ converge absolutely is that uk converge absolutely. CHAP. 11] INFINITE SERIES 277
- 287. 6. Summability. Let S1; S2; S3; . . . be the partial sums of a divergent series un. If the sequence S1; S1; S2 2 ; S1 þ S2 þ S3 3 ; . . . (formed by taking arithmetic means of the first n terms of S1; S2; S3; . . .) converges to S, we say that the series un is summable in the Ce´saro sense, or C-1 summable to S (see Problem 11.51). If un converges to S, the Césaro method also yields the result S. For this reason the Césaro method is said to be a regular method of summability. In case the Césaro limit does not exist, we can apply the same technique to the sequence S1; S1 þ S2 3 ; S1 þ S2 þ S3 3 ; . . . : If the C-1 limit for this sequence exists and equals S, we say that uk converges to S in the C-2 sense. The process can be continued indefinitely. Solved Problems CONVERGENCE AND DIVERGENCE OF SERIES OF CONSTANTS 11.1. (a) Prove that 1 1 3 þ 1 3 5 þ 1 5 7 þ ¼ X 1 n¼1 1 ð2n 1Þð2n þ 1Þ converges and (b) find its sum. un ¼ 1 ð2n 1Þð2n þ 1Þ ¼ 1 2 1 2n 1 1 2n þ 1 : Then Sn ¼ u1 þ u2 þ þ un ¼ 1 2 1 1 1 3 þ 1 2 1 3 1 5 þ þ 1 2 1 2n 1 1 2n þ 1 ¼ 1 2 1 1 1 3 þ 1 3 1 5 þ 1 5 þ 1 2n 1 1 2n þ 1 ¼ 1 2 1 1 2n þ 1 Since lim n!1 Sn ¼ lim n!1 1 2 1 1 2n þ 1 ¼ 1 2 ; the series converges and its sum is 1 2 : The series is sometimes called a telescoping series, since the terms of Sn, other than the first and last, cancel out in pairs. 11.2. (a) Prove that 2 3 þ ð2 3Þ2 þ ð2 3Þ3 þ ¼ X 1 n¼1 ð2 3Þn converges and (b) find its sum. This is a geometric series; therefore, the partial sums are of the form Sn ¼ að1 rn Þ 1 r . Since jrj 1 S ¼ lim n!1 Sn ¼ a 1 r and in particular with r ¼ 2 3 and a ¼ 2 3, we obtain S ¼ 2. 11.3. Prove that the series 1 2 þ 2 3 þ 3 4 þ 4 5 þ ¼ X 1 n¼1 n n þ 1 diverges. lim n!1 un ¼ lim n!1 n n þ 1 ¼ 1. Hence by Problem 2.26, Chapter 2, the series is divergent. 11.4. Show that the series whose nth term is un ¼ ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p ffiffiffi n p diverges although lim n!1 un ¼ 0. The fact that lim n!1 un ¼ 0 follows from Problem 2.14(c), Chapter 2. Now Sn ¼ u1 þ u2 þ þ un ¼ ð ffiffiffi 2 p ffiffiffi 1 p Þ þ ð ffiffiffi 3 p ffiffiffi 2 p Þ þ þ ð ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p ffiffiffi n p Þ ¼ ffiffiffiffiffiffiffiffiffiffiffi n þ 1 p ffiffiffi 1 p . Then Sn increases without bound and the series diverges. This problem shows that lim n!1 ¼ 0 is a necessary but not sufficient condition for the convergence of un. See also Problem 11.6. 278 INFINITE SERIES [CHAP. 11
- 288. COMPARISON TEST AND QUOTIENT TEST 11.5. If 0 @ un @ vn; n ¼ 1; 2; 3; . . . and if vn converges, prove that un also converges (i.e., establish the comparison test for convergence). Let Sn ¼ u1 þ u2 þ þ un; Tn ¼ v1 þ v2 þ þ vn. Since vn converges, lim n!1 Tn exists and equals T, say. Also, since vn A 0; Tn @ T. Then Sn ¼ u1 þ u2 þ þ un @ v1 þ v2 þ þ vn @ T or 0 @ Sn @ T: Thus Sn is a bounded monotonic increasing sequence and must have a limit (see Chapter 2), i.e., un converges. 11.6. Using the comparison test prove that 1 þ 1 2 þ 1 3 þ ¼ X 1 n¼1 1 n diverges. 1 A 1 2 We have 1 2 þ 1 3 A 1 4 þ 1 4 ¼ 1 2 1 4 þ 1 5 þ 1 6 þ 1 7 A 1 8 þ 1 8 þ 1 8 þ 1 8 ¼ 1 2 1 8 þ 1 9 þ 1 10 þ þ 1 15 A 1 16 þ 1 16 þ 1 16 þ þ 1 16 (8 terms) ¼ 1 2 etc. Thus, to any desired number of terms, 1 þ 1 2 þ 1 3 þ 1 4 þ 1 5 þ 1 6 þ 1 7 þ A 1 2 þ 1 2 þ 1 2 þ Since the right-hand side can be made larger than any positive number by choosing enough terms, the given series diverges. By methods analogous to that used here, we can show that X 1 n¼1 1 np, where p is a constant, diverges if p @ 1 and converges if p 1. This can also be shown in other ways [see Problem 11.13(a)]. 11.7. Test for convergence or divergence X 1 n¼1 ln n 2n3 1 . Since ln n n and 1 2n3 1 @ 1 n3 ; we have ln n 2n3 1 @ n n3 ¼ 1 n2 : Then the given series converges, since X 1 n¼1 1 n2 converges. 11.8. Let un and vn be positive. If lim n!1 un vn ¼ constant A 6¼ 0, prove that un converges or diverges according as vn converges or diverges. By hypothesis, given 0 we can choose an integer N such that un vn A for all n N. Then for n ¼ N þ 1; N þ 2; . . . un vn A or ðA Þvn un ðA þ Þvn ð1Þ Summing from N þ 1 to 1 (more precisely from N þ 1 to M and then letting M ! 1), ðA Þ X 1 Nþ1 vn @ X 1 Nþ1 un @ ðA þ Þ X 1 Nþ1 vn ð2Þ There is no loss in generality in assuming A 0. Then from the right-hand inequality of (2), un converges when vn does. From the left-hand inequality of (2), un diverges when vn does. For the cases A ¼ 0 or A ¼ 1, see Problem 11.66. CHAP. 11] INFINITE SERIES 279
- 289. 11.9. Test for convergence; (a) X 1 n¼1 4n2 n þ 3 n3 þ 2n ; ðbÞ X 1 n¼1 n þ ffiffiffi n p 2n3 1 ; ðcÞ X 1 n¼1 ln n n2 þ 3 . (a) For large n, 4n2 n þ 3 n3 þ 2n is approximately 4n2 n3 ¼ 4 n . Taking un ¼ 4n2 n þ 3 n3 þ 2n and vn ¼ 4 n , we have lim n!1 un vn ¼ 1. Since vn ¼ 41=n diverges, un also diverges by Problem 11.8. Note that the purpose of considering the behavior of un for large n is to obtain an appropriate comparison series vn. In the above we could just as well have taken vn ¼ 1=n. Another method: lim n!1 n 4n2 n þ 3 n3 þ 2n ! ¼ 4. Then by Theorem 1, Page 267, the series converges. (b) For large n, un ¼ n þ ffiffiffi n p 2n3 1 is approximately vn ¼ n 2n3 ¼ 1 2n2 . Since lim n!1 un vn ¼ 1 and X vn ¼ 1 2 X 1 n2 converges ( p series with p ¼ 2), the given series converges. Another method: lim n!1 n2 n þ ffiffiffi n p 2n3 1 ¼ 1 2 . Then by Theorem 1, Page 267, the series converges. (c) lim n!1 n3=2 ln n n2 þ 3 @ lim n!1 n3=2 ln n n2 ¼ lim n!1 ln n ffiffiffi n p ¼ 0 (by L’Hospital’s rule or otherwise). Then by Theorem 1 with p ¼ 3=2, the series converges. Note that the method of Problem 11.6(a) yields ln n n2 þ 3 n n2 ¼ 1 n , but nothing can be deduced since 1=n diverges. 11.10. Examine for convergence: (a) X 1 n¼1 en2 ; ðbÞ X 1 n¼1 sin3 1 n . (a) lim n!1 n2 en2 ¼ 0 (by L’Hospital’s rule or otherwise). Then by Theorem 1 with p ¼ 2, the series con- verges. (b) For large n, sinð1=nÞ is approximately 1=n. This leads to consideration of lim n!1 n3 sin3 1 n ¼ lim n!1 sinð1=nÞ 1=n 3 ¼ 1 from which we deduce, by Theorem 1 with p ¼ 3, that the given series converges. INTEGRAL TEST 11.11. Establish the integral test (see Page 267). We perform the proof taking N ¼ 1. Modifications are easily made if N 1. From the monotonicity of f ðxÞ, we have unþ1 ¼ f ðn þ 1Þ @ f ðxÞ @ f ðnÞ ¼ un n ¼ 1; 2; 3; . . . Integrating from x ¼ n to x ¼ n þ 1, using Property 7, Page 92, unþ1 @ ðnþ1 n f ðxÞ dx @ un n ¼ 1; 2; 3 . . . Summing from n ¼ 1 to M 1, u2 þ u3 þ þ uM @ ðM 1 f ðxÞ dx @ u1 þ u2 þ þ uM1 ð1Þ If f ðxÞ is strictly decreasing, the equality signs in (1) can be omitted. 280 INFINITE SERIES [CHAP. 11
- 290. If lim M!1 ðM 1 f ðxÞ dx exists and is equal to S, we see from the left-hand inequality in (1) that u2 þ u3 þ þ uM is monotonic increasing and bounded above by S, so that un converges. If lim M!1 ðM 1 f ðxÞ dx is unbounded, we see from the right-hand inequality in (1) that un diverges. Thus the proof is complete. 11.12. Illustrate geometrically the proof in Problem 11.11. Geometrically, u2 þ u3 þ þ uM is the total area of the rectangles shown shaded in Fig. 11-3, while u1 þ u2 þ þ uM1 is the total area of the rectangles which are shaded and nonshaded. The area under the curve y ¼ f ðxÞ from x ¼ 1 to x ¼ M is intermediate in value between the two areas given above, thus illustrating the result (1) of Problem 11.11. 11.13. Test for convergence: (a) X 1 1 1 nP ; p ¼ constant; ðbÞ X 1 1 n n2 þ 1 ; ðcÞ X 1 2 1 n ln n ; ðdÞ X 1 1 nen2 . ðaÞ Consider ðM 1 dx xp ¼ ðM 1 xp dx ¼ x1p 1 p M 1 ¼ M1p 1 1 p where p 6¼ 1: If p 1; lim M!1 M1p 1 1 p ¼ 1, so that the integral and thus the series diverges. If p 1; lim M!1 M1p 1 1 p ¼ 1 p 1 , so that the integral and thus the series converges. If p ¼ 1, ðM 1 dx xp ¼ ðM 1 dx x ¼ ln M and lim M!1 ln M ¼ 1, so that the integral and thus the series diverges. Thus, the series converges if p 1 and diverges if p @ 1. ðbÞ lim M!1 ðM 1 x dx x2 þ 1 ¼ lim M!1 1 2 lnðx2 þ 1ÞjM 1 ¼ lim M!1 1 2 lnðM2 þ 1Þ 1 2 ln 2 ¼ 1 and the series diverges. ðcÞ lim M!1 ðM 2 dx x ln x ¼ lim M!1 lnðln xÞjM 2 ¼ lim M!1 flnðln MÞ lnðln 2Þg ¼ 1 and the series diverges. ðdÞ lim M!1 ðM 1 xex2 dx ¼ lim M!1 1 2 ex2 jM 1 ¼ lim M!1 1 2 e1 1 2 eM2 n o ¼ 1 2 e1 and the series converges. Note that when the series converges, the value of the corresponding integral is not (in general) the same as the sum of the series. However, the approximate sum of a series can often be obtained quite accurately by using integrals. See Problem 11.74. 11.14. Prove that 4 X 1 n¼1 1 n2 þ 1 1 2 þ 4 . CHAP. 11] INFINITE SERIES 281 Fig. 11-3
- 291. From Problem 11.11 it follows that lim M!1 X M n¼2 1 n2 þ 1 lim M!1 ðM 1 dx x2 þ 1 lim M!1 X M1 n¼1 1 n2 þ 1 i.e., X 1 n¼2 1 n2 þ 1 4 X 1 n¼1 1 n2 þ 1 , from which 4 X 1 n¼1 1 n2 þ 1 as required. Since X 1 n¼2 1 n2 þ 1 4 , we obtain, on adding 1 2 to each side, X 1 n¼1 1 n2 þ 1 1 2 þ 4 : The required result is therefore proved. ALTERNATING SERIES 11.15. Given the alternating series a1 a2 þ a3 a4 þ where 0 @ anþ1 @ an and where lim n!1 an ¼ 0. Prove that (a) the series converges, (b) the error made in stopping at any term is not greater than the absolute value of the next term. (a) The sum of the series to 2M terms is S2M ¼ ða1 a2Þ þ ða3 a4Þ þ þ ða2M1 a2MÞ ¼ a1 ða2 a3Þ ða4 a5Þ ða2M2 a2M1Þ a2M Since the quantities in parentheses are non-negative, we have S2M A 0; S2 @ S4 @ S6 @ S8 @ @ S2M @ a1 Therefore, fS2Mg is a bounded monotonic increasing sequence and thus has limit S. Also, S2Mþ1 ¼ S2M þ a2Mþ1. Since lim M!1 S2M ¼ S and lim M!1 a2Mþ1 ¼ 0 (for, by hypothesis, lim n!1 an ¼ 0), it follows that lim M!1 S2Mþ1 ¼ lim M!1 S2M þ lim M!1 a2Mþ1 ¼ S þ 0 ¼ S. Thus, the partial sums of the series approach the limit S and the series converges. (b) The error made in stopping after 2M terms is ða2Mþ1 a2Mþ2Þ þ ða2Mþ3 a2Mþ4Þ þ ¼ a2Mþ1 ða2Mþ2 a2Mþ3Þ and is thus non-negative and less than or equal to a2Mþ1, the first term which is omitted. Similarly, the error made in stopping after 2M þ 1 terms is a2Mþ2 þ ða2Mþ3 a2Mþ4Þ þ ¼ ða2Mþ2 a2Mþ3Þ ða2Mþ4 a2Mþ5Þ which is non-positive and greater than a2Mþ2. 11.16. (a) Prove that the series X 1 n¼1 ð1Þnþ1 2n 1 converges. (b) Find the maximum error made in approx- imating the sum by the first 8 terms and the first 9 terms of the series. (c) How many terms of the series are needed in order to obtain an error which does not exceed .001 in absolute value? (a) The series is 1 1 3 þ 1 5 1 7 þ 1 9 . If un ¼ ð1Þnþ1 2n 1 , then an ¼ junj ¼ 1 2n 1 , anþ1 ¼ junþ1j ¼ 1 2n þ 1 . Since 1 2n þ 1 @ 1 2n 1 and since lim n!1 1 2n 1 ¼ 0, it follows by Problem 11.5(a) that the series converges. (b) Use the results of Problem 11.15(b). Then the first 8 terms give 1 1 3 þ 1 5 1 7 þ 1 9 1 11 þ 1 13 1 15 and the error is positive and does not exceed 1 17. Similarly, the first 9 terms are 1 1 3 þ 1 5 1 7 þ 1 9 1 11 þ 1 13 1 15 þ 1 17 and the error is negative and greater than or equal to 1 19, i.e., the error does not exceed 1 19 in absolute value. 282 INFINITE SERIES [CHAP. 11
- 292. (c) The absolute value of the error made in stopping after M terms is less than 1=ð2M þ 1Þ. To obtain the desired accuracy, we must have 1=ð2M þ 1Þ @ :001, from which M A 499:5. Thus, at least 500 terms are needed. ABSOLUTE AND CONDITIONAL CONVERGENCE 11.17. Prove that an absolutely convergent series is convergent. Given that junj converges, we must show that un converges. Let SM ¼ u1 þ u2 þ þ uM and TM ¼ ju1j þ ju2j þ þ juMj. Then SM þ TM ¼ ðu1 þ ju1jÞ þ ðu2 þ ju2jÞ þ þ ðuM þ juMjÞ @ 2ju1j þ 2ju2j þ þ 2juMj Since junj converges and since un þ junj A 0, for n ¼ 1; 2; 3; . . . ; it follows that SM þ TM is a bounded monotonic increasing sequence, and so lim M!1 ðSM þ TMÞ exists. Also, since lim M!1 TM exists (since the series is absolutely convergent by hypothesis), lim M!1 SM ¼ lim M!1 ðSM þ TM TMÞ ¼ lim M!1 ðSM þ TMÞ lim M!1 TM must also exist and the result is proved. 11.18. Investigate the convergence of the series sin ffiffiffi 1 p 13=2 sin ffiffiffi 2 p 23=2 þ sin ffiffiffi 3 p 33=2 . Since each term is in absolute value less than or equal to the corresponding term of the series 1 13=2 þ 1 23=2 þ 1 33=2 þ , which converges, it follows that the given series is absolutely convergent and hence convergent by Problem 11.17. 11.19. Examine for convergence and absolute convergence: ðaÞ X 1 n¼1 ð1Þn1 n n2 þ 1 ; ðbÞ X 1 n¼2 ð1Þn1 n ln2 n ; ðcÞ X 1 n¼1 ð1Þn1 2n n2 : (a) The series of absolute values is X 1 n¼1 n n2 þ 1 which is divergent by Problem 11.13(b). Hence, the given series is not absolutely convergent. However, if an ¼ junj ¼ n n2 þ 1 and anþ1 ¼ junþ1j ¼ n þ 1 ðn þ 1Þ2 þ 1 , then anþ1 @ an for all n A 1, and also lim n!1 an ¼ lim n!1 n n2 þ 1 ¼ 0. Hence, by Problem 11.15 the series converges. Since the series converges but is not absolutely convergent, it is conditionally convergent. (b) The series of absolute values is X 1 n¼2 1 n ln2 n . By the integral test, this series converges or diverges according as lim M!1 ðM 2 dx x ln2 x exists or does not exist. If u ¼ ln x; ð dx x ln2 x ¼ ð du u2 ¼ 1 u þ c ¼ 1 ln x þ c: Hence, lim M!1 ðM 2 dx x ln2 x ¼ lim M!1 1 ln 2 1 ln M ¼ 1 ln 2 and the integral exists. Thus, the series converges. Then X 1 n¼2 ð1Þn1 n ln2 n converges absolutely and thus converges. CHAP. 11] INFINITE SERIES 283
- 293. Another method: Since 1 ðn þ 1Þ ln2 ðn þ 1Þ @ 1 n ln2 n and lim n!1 1 n ln2 n ¼ 0, it follows by Problem 11.15(a), that the given alternating series converges. To examine its absolute convergence, we must proceed as above. (c) Since lim n!1 un 6¼ 0 where un ¼ ð1Þn1 2n n2 , the given series cannot be convergent. To show that lim n!1 un 6¼ 0, it suffices to show that lim n!1 junj ¼ lim n!1 2n n2 6¼ 0. This can be accomplished by L’Hospital’s rule or other methods [see Problem 11.21(b)]. RATIO TEST 11.20. Establish the ratio test for convergence. Consider first the series u1 þ u2 þ u3 þ where each term is non-negative. We must prove that if lim n!1 unþ1 un ¼ L 1, then necessarily un converges. By hypothesis, we can choose an integer N so large that for all n A N, ðunþ1=unÞ r where L r 1. Then uNþ1 r uN uNþ2 r uNþ1 r2 uN uNþ3 r uNþ2 r3 uN etc. By addition, uNþ1 þ uNþ2 þ uNðr þ r2 þ r3 þ Þ and so the given series converges by the comparison test, since 0 r 1. In case the series has terms with mixed signs, we consider ju1j þ ju2j þ ju3j þ . Then by the above proof and Problem 11.17, it follows that if lim n!1 unþ1 un ¼ L 1, then un converges (absolutely). Similarly, we can prove that if lim n!1 unþ1 un ¼ L 1 the series un diverges, while if lim n!1 unþ1 un ¼ L ¼ 1 the ratio test fails [see Problem 11.21(c)]. 11.21. Investigate the convergence of (a) X 1 n¼1 n4 en2 ; ðbÞ X 1 n¼1 ð1Þn1 2n n2 ; ðcÞ X 1 n¼1 ð1Þn1 n n2 þ 1 . (a) Here un ¼ n4 en2 . Then lim n!1 unþ1 un ¼ lim n!1 ðn þ 1Þ4 eðnþ1Þ2 n4 en2 ¼ lim n!1 ðn þ 1Þ4 eðn2 þ2nþ1Þ n4 en2 ¼ lim n!1 n þ 1 n 4 e2n1 ¼ lim n!1 n þ 1 n 4 lim n!1 e2n1 ¼ 1 0 ¼ 0 Since 0 1, the series converges. (b) Here un ¼ ð1Þn1 2n n2 . Then lim n!1 unþ1 un ¼ lim n!1 ð1Þn 2nþ1 ðn þ 1Þ2 n2 ð1Þn1 2n ¼ lim n!1 2n2 ðn þ 1Þ2 ¼ 2 Since s 1, the series diverges. Compare Problem 11.19(c). (c) Here un ¼ ð1Þn1 n n2 þ 1 . Then 284 INFINITE SERIES [CHAP. 11
- 294. lim n!1 unþ1 un ¼ lim n!1 ð1Þn ðn þ 1Þ ðn þ 1Þ2 þ 1 n2 þ 1 ð1Þn1 n ¼ lim n!1 ðn þ 1Þðn2 þ 1Þ ðn2 þ 2n þ 2Þn ¼ 1 and the ratio test fails. By using other tests [see Problem 11.19(a)], the series is seen to be convergent. MISCELLANEOUS TESTS 11.22. Test for convergence 1 þ 2r þ r2 þ 2r3 þ r4 þ 2r5 þ where (a) r ¼ 2=3, (b) r ¼ 2=3, (c) r ¼ 4=3. Here the ratio test is inapplicable, since unþ1 un ¼ 2jrj or 1 2 jrj depending on whether n is odd or even. However, using the nth root test, we have ffiffiffiffiffiffiffiffi junj n p ¼ ffiffiffiffiffiffiffiffiffi 2jrn j n p ¼ ffiffiffi 2 n p jrj if n is odd ffiffiffiffiffiffiffi jrnj n p ¼ jrj if n is even ( Then lim n!1 ffiffiffiffiffiffiffiffi junj n p ¼ jrj (since lim n!1 21=n ¼ 1). Thus, if jrj 1 the series converges, and if jrj 1 the series diverges. Hence, the series converges for cases (a) and (b), and diverges in case (c). 11.23. Test for convergence 1 3 2 þ 1 4 3 6 2 þ 1 4 7 3 6 9 2 þ þ 1 4 7 . . . ð3n 2Þ 3 6 9 . . . ð3nÞ 2 þ . The ratio test fails since lim n!1 unþ1 un ¼ lim n!1 3n þ 1 3n þ 3 2 ¼ 1. However, by Raabe’s test, lim n!1 n 1 unþ1 un ¼ lim n!1 n 1 3n þ 1 3n þ 3 2 ( ) ¼ 4 3 1 and so the series converges. 11.24. Test for convergence 1 2 2 þ 1 3 2 4 2 þ 1 3 5 24t 2 þ þ 1 3 5 . . . ð2n 1Þ 2 4 6 . . . ð2nÞ 2 þ . The ratio test fails since lim n!1 unþ1 un ¼ lim n!1 2n þ 1 2n þ 2 2 ¼ 1. Also, Raabe’s test fails since lim n!1 n 1 unþ1 un ¼ lim n!1 n 1 2n þ 1 2n þ 2 2 ( ) ¼ 1 However, using long division, unþ1 un ¼ 2n þ 1 2n þ 2 2 ¼ 1 1 n þ 5 4=n 4n2 þ 8n þ 4 ¼ 1 1 n þ cn n2 where jcnj P so that the series diverges by Gauss’ test. CHAP. 11] INFINITE SERIES 285
- 295. SERIES OF FUNCTIONS 11.25. For what values of x do the following series converge? ðaÞ X 1 n¼1 xn1 n 3n ; ðbÞ X 1 n¼1 ð1Þn1 x2n1 ð2n 1Þ! ; ðcÞ X 1 n¼1 n!ðx aÞn ; ðdÞ X 1 n¼1 nðx 1Þn 2n ð3n 1Þ : (a) un ¼ xn1 n 3n. Assuming x 6¼ 0 (if x ¼ 0 the series converges), we have lim n!1 unþ1 un ¼ lim n!1 xn ðn þ 1Þ 3nþ1 n 3n xn1 ¼ lim n!1 n 3ðn þ 1Þ jxj ¼ jxj 3 Then the series converges if jxj 3 1, and diverges if jxj 3 1. If jxj 3 ¼ 1, i.e., x ¼ 3, the test fails. If x ¼ 3 the series becomes X 1 n¼1 1 3n ¼ 1 3 X 1 n¼1 1 n , which diverges. If x ¼ 3 the series becomes X 1 n¼1 ð1Þn1 3n ¼ 1 3 X 1 n¼1 ð1Þn1 n , which converges. Then the interval of convergence is 3 @ x 3. The series diverges outisde this interval. Note that the series converges absolutely for 3 x 3. At x ¼ 3 the series converges con- ditionally. (b) Proceed as in part (a) with un ¼ ð1Þn1 x2n1 ð2n 1Þ! . Then lim n!1 unþ1 un ¼ lim n!1 ð1Þn x2nþ1 ð2n þ 1Þ! ð2n 1Þ! ð1Þn1 x2n1 ¼ lim n!1 ð2n 1Þ! ð2n þ 1Þ! x2 ¼ lim n!1 ð2n 1Þ! ð2n þ 1Þð2nÞð2n 1Þ! x2 ¼ lim n!1 x2 ð2n þ 1Þð2nÞ ¼ 0 Then the series converges (absolutely) for all x, i.e., the interval of (absolute) convergence is 1 x 1. ðcÞ un ¼ n!ðx aÞn ; lim n!1 unþ1 un ¼ lim n!1 ðn þ 1Þ!ðx aÞnþ1 n!ðx aÞn ¼ lim n!1 ðn þ 1Þjx aj: This limit is infinite if x 6¼ a. Then the series converges only for x ¼ a. ðdÞ un ¼ nðx 1Þn 2nð3n 1Þ ; unþ1 ¼ ðn þ 1Þðx 1Þnþ1 2nþ1ð3n þ 2Þ : Then lim n!1 unþ1 un ¼ lim n!1 ðn þ 1Þð3n 1Þðx 1Þ 2nð3n þ 2Þ ¼ x 1 2 ¼ jx 1j 2 Thus, the series converges for jx 1j 2 and diverges for jx 1j 2. The test fails for jx 1j ¼ 2, i.e., x 1 ¼ 2 or x ¼ 3 and x ¼ 1. For x ¼ 3 the series becomes X 1 n¼1 n 3n 1 , which diverges since the nth term does not approach zero. For x ¼ 1 the series becomes X 1 n¼1 ð1Þn n 3n 1 , which also diverges since the nth term does not approach zero. Then the series converges only for jx 1j 2, i.e., 2 x 1 2 or 1 x 3. 286 INFINITE SERIES [CHAP. 11
- 296. 11.26. For what values of x does (a) X 1 n¼1 1 2n 1 x þ 2 x 1 n ; ðbÞ X 1 n¼1 1 ðx þ nÞðx þ n 1Þ converge? ðaÞ un ¼ 1 2n 1 x þ 2 x 1 n : Then lim n!1 unþ1 un ¼ lim n!1 2n 1 2n þ 1 x þ 2 x 1 ¼ x þ 2 x 1 if x 6¼ 1; 2: Then the series converges if x þ 2 x 1 1, diverges if x þ 2 x 1 1, and the test fails if x þ 2 x 1 ¼ 1, i.e., x ¼ 1 2. If x ¼ 1 the series diverges. If x ¼ 2 the series converges. If x 1 2 the series is X 1 n¼1 ð1Þn 2n 1 which converges. Thus, the series converges for x þ 2 x 1 1, x ¼ 1 2 and x ¼ 2, i.e., for x @ 1 2. (b) The ratio test fails since lim n!1 unþ1 un ¼ 1, where un ¼ 1 ðx þ nÞðx þ n 1Þ : However, noting that 1 ðx þ nÞðx þ n 1Þ ¼ 1 x þ n 1 1 x þ n we see that if x 6¼ 0; 1; 2; . . . ; n, Sn ¼ u1 þ u2 þ þ un ¼ 1 x 1 x þ 1 þ 1 x þ 1 1 x þ 2 þ þ 1 x þ n 1 1 x þ n ¼ 1 x 1 x þ n and lim n!1 Sn ¼ 1=x, provided x 6¼ 0; 1; 2; 3; . . . . Then the series converges for all x except x ¼ 0; 1; 2; 3; . . . ; and its sum is 1=x. UNIFORM CONVERGENCE 11.27. Find the domain of convergence of ð1 xÞ þ xð1 xÞ þ x2 ð1 xÞ þ . Method 1: Sum of first n terms ¼ SnðxÞ ¼ ð1 xÞ þ xð1 xÞ þ x2 ð1 xÞ þ þ xn1 ð1 xÞ ¼ 1 x þ x x2 þ x2 þ þ xn1 xn ¼ 1 xn If jxj 1, lim n!1 SnðxÞ ¼ lim n!1 ð1 xn Þ ¼ 1. If jxj 1, lim n!1 SnðxÞ does not exist. If x ¼ 1; SnðxÞ ¼ 0 and lim n!1 SnðxÞ ¼ 0. If x ¼ 1; SnðxÞ ¼ 1 ð1Þn and lim n!1 SnðxÞ does not exist. Thus, the series converges for jxj 1 and x ¼ 1, i.e., for 1 x @ 1. Method 2, using the ratio test. The series converges if x ¼ 1. If x 6¼ 1 and un ¼ xn1 ð1 xÞ, then lim n!1 unþ1 un ¼ lim n!1 jxj. Thus, the series converges if jxj 1, diverges if jxj 1. The test fails if jxj ¼ 1. If x ¼ 1, the series converges; if x ¼ 1, the series diverges. Then the series converges for 1 x @ 1: 11.28. Investigate the uniform convergence of the series of Problem 11.27 in the interval (a) 1 2 x 1 2, (b) 1 2 @ x @ 1 2, ðcÞ :99 @ x @ :99; ðdÞ 1 x 1, ðeÞ 0 @ x 2. CHAP. 11] INFINITE SERIES 287
- 297. (a) By Problem 11.27, SnðxÞ ¼ 1 xn ; SðxÞ ¼ lim n!1 SnðxÞ ¼ 1 if 1 2 x 1 2; thus, the series converges in this interval. We have Remainder after n terms ¼ RnðxÞ ¼ SðxÞ SnðxÞ ¼ 1 ð1 xn Þ ¼ xn The series is uniformly convergent in the interval if given any 0 we can find N dependent on , but not on x, such that jRnðxÞj for all n N. Now jRnðxÞj ¼ jxn j ¼ jxjn when n ln jxj ln or n ln ln jxj since division by ln jxj (which is negative since jxj 1 2) reverses the sense of the inequality. But if jxj 1 2 ; ln jxj ln ð1 2Þ, and n ln ln jxj ln lnð1 2Þ ¼ N. Thus, since N is independent of x, the series is uniformly convergent in the interval. (b) In this case jxj @ 1 2 ; ln jxj @ ln ð1 2Þ; and n ln ln jxj A ln lnð1 2Þ ¼ N, so that the series is also uniformly convergent in 1 2 @ x @ 1 2 : (c) Reasoning similar to the above, with 1 2 replaced by .99, shows that the series is uniformly convergent in :99 @ x @ :99. (d) The arguments used above break down in this case, since ln ln jxj can be made larger than any positive number by choosing jxj sufficiently close to 1. Thus, no N exists and it follows that the series is not uniformly convergent in 1 x 1. (e) Since the series does not even converge at all points in this interval, it cannot converge uniformly in the interval. 11.29. Discuss the continuity of the sum function SðxÞ ¼ lim n!1 SnðxÞ of Problem 11.27 for the interval 0 @ x @ 1. If 0 @ x 1; SðxÞ ¼ lim n!1 SnðxÞ ¼ lim n!1 ð1 xn Þ ¼ 1. If x ¼ 1; SnðxÞ ¼ 0 and SðxÞ ¼ 0. Thus, SðxÞ ¼ 1 if 0 @ x 1 0 if x ¼ 1 and SðxÞ is discontinuous at x ¼ 1 but continuous at all other points in 0 @ x 1. In Problem 11.34 it is shown that if a series is uniformly convergent in an interval, the sum function SðxÞ must be continuous in the interval. It follows that if the sum function is not continuous in an interval, the series cannot be uniformly convergent. This fact is often used to demonstrate the nonuniform convergence of a series (or sequence). 11.30. Investigate the uniform convergence of x2 þ x2 1 þ x2 þ x2 ð1 þ x2 Þ2 þ þ x2 ð1 þ x2Þn þ . Suppose x 6¼ 0. Then the series is a geometric series with ratio 1=ð1 þ x2 Þ whose sum is (see Problem 2.25, Chap. 2). SðxÞ ¼ x2 1 1=ð1 þ x2Þ ¼ 1 þ x2 If x ¼ 0 the sum of the first n terms is Snð0Þ ¼ 0; hence Sð0Þ ¼ lim n!1 Snð0Þ ¼ 0. Since lim x!0 SðxÞ ¼ 1 6¼ Sð0Þ, SðxÞ is discontinuous at x ¼ 0. Then by Problem 11.34, the series cannot be uniformly convergent in any interval which includes x ¼ 0, although it is (absolutely) convergent in any interval. However, it is uniformly convergent in any interval which excludes x ¼ 0. This can also be shown directly (see Problem 11.93). 288 INFINITE SERIES [CHAP. 11
- 298. WEIERSTRASS M TEST 11.31. Prove the Weierstrass M test, i.e., if junðxÞj @ Mn; n ¼ 1; 2; 3; . . . ; where Mn are positive constants such that Mn converges, then unðxÞ is uniformly (and absolutely) convergent. The remainder of the series unðxÞ after n terms is RnðxÞ ¼ unþ1ðxÞ þ unþ2ðxÞ þ . Now jRnðxÞj ¼ junþ1ðxÞ þ unþ2ðxÞ þ j @ junþ1ðxÞj þ junþ2ðxÞj þ @ Mnþ1 þ Mnþ2 þ But Mnþ1 þ Mnþ2 þ can be made less than by choosing n N, since Mn converges. Since N is clearly independent of x, we have jRnðxÞj for n N, and the series is uniformly convergent. The absolute convergence follows at once from the comparison test. 11.32. Test for uniform convergence: ðaÞ X 1 n¼1 cos nx n4 ; ðbÞ X 1 n¼1 xn n3=2 ; ðcÞ X 1 n¼1 sin nx n ; ðdÞ X 1 n¼1 1 n2 þ x2 : (a) cos nx n4 @ 1 n4 ¼ Mn. Then since Mn converges ð p series with p ¼ 4 1Þ, the series is uniformly (and absolutely) convergent for all x by the M test. (b) By the ratio test, the series converges in the interval 1 @ x @ 1, i.e., jxj @ 1. For all x in this interval, xn n3=2 ¼ jxjn n3=2 @ 1 n3=2 . Choosing Mn ¼ 1 n3=2 , we see that Mn converges. Thus, the given series converges uniformly for 1 @ x @ 1 by the M test. (c) sin nx n @ 1 n . However, Mn, where Mn ¼ 1 n , does not converge. The M test cannot be used in this case and we cannot conclude anything about the uniform convergence by this test (see, however, Problem 11.125). (d) 1 n2 þ x2 @ 1 n2 , and 1 n2 converges. Then by the M test the given series converges uniformly for all x. 11.33. If a power series anxn converges for x ¼ x0, prove that it converges (a) absolutely in the interval jxj jx0j, (b) uniformly in the interval jxj @ jx1j; where jx1j jx0j. (a) Since anxn 0 converges, lim n!1 anxn 0 ¼ 0 and so we can make janxn 0j 1 by choosing n large enough, i.e., janj 1 jx0jn for n N. Then X 1 Nþ1 janxn j ¼ X 1 Nþ1 janjjxjn X 1 Nþ1 jxjn jx0jn ð1Þ Since the last series in (1) converges for jxj jx0j, it follows by the comparison test that the first series converges, i.e., the given series is absolutely convergent. (b) Let Mn ¼ jx1jn jx0jn. Then Mn converges since jx1j jx0j. As in part (a), janxn j Mn for jxj @ jx1j, so that by the Weierstrass M test, anxn is uniformly convergent. It follows that a power series is uniformly convergent in any interval within its interval of con- vergence. THEOREMS ON UNIFORM CONVERGENCE 11.34. Prove Theorem 6, Page 271. We must show that SðxÞ is continuous in ½a; b. CHAP. 11] INFINITE SERIES 289
- 299. Now SðxÞ ¼ SnðxÞ þ RnðxÞ, so that Sðx þ hÞ ¼ Snðx þ hÞ þ Rnðx þ hÞ and thus Sðx þ hÞ SðxÞ ¼ Snðx þ hÞ SnðxÞ þ Rnðx þ hÞ RnðxÞ ð1Þ where we choose h so that both x and x þ h lie in ½a; b (if x ¼ b, for example, this will require h 0). Since SnðxÞ is a sum of finite number of continuous functions, it must also be continuous. Then given 0, we can find so that jSnðx þ hÞ SnðxÞj =3 whenever jhj ð2Þ Since the series, by hypothesis, is uniformly convergent, we can choose N so that jRnðxÞj =3 and jRnðx þ hÞj =3 for n N ð3Þ Then from (1), (2), and (3), jSðx þ hÞ SðxÞj @ jSnðx þ hÞ SnðxÞj þ jRnðx þ hÞj þ jRnðxÞj for jhj , and so the continuity is established. 11.35. Prove Theorem 7, Page 271. If a function is continuous in ½a; b, its integral exists. Then since SðxÞ; SnðxÞ, and RnðxÞ are continuous, ðb a SðxÞ ¼ ðb a SnðxÞ dx þ ðb a RnðxÞ dx To prove the theorem we must show that ðb a SðxÞ dx ðb a SnðxÞ dx ¼ ðb a RnðxÞ dx can be made arbitrarily small by choosing n large enough. This, however, follows at once, since by the uniform convergence of the series we can make jRnðxÞj =ðb aÞ for n N independent of x in ½a; b, and so ðb a RnðxÞ dx @ ðb a jRnðxÞj dx ðb a b a dx ¼ This is equivalent to the statements ðb a SðxÞ dx ¼ lim n!1 ðb a SnðxÞ dx or lim n!1 ðb a SnðxÞ dx ¼ ðb a lim n!1 SnðxÞ n o dx 11.36. Prove Theorem 8, Page 271. Let gðxÞ ¼ X 1 n¼1 u0 nðxÞ. Since, by hypothesis, this series converges uniformly in ½a; b, we can integrate term by term (by Problem 11.35) to obtain ðx a gðxÞ dx ¼ X 1 n¼1 ðx a u0 nðxÞ dx ¼ X 1 n¼1 funðxÞ unðaÞg ¼ X 1 n¼1 unðxÞ X 1 n¼1 unðaÞ ¼ SðxÞ SðaÞ because, by hypothesis, X 1 n¼1 unðxÞ converges to SðxÞ in ½a; b. Differentiating both sides of ðx a gðxÞ dx ¼ SðxÞ SðaÞ then shows that gðxÞ ¼ S0 ðxÞ, which proves the theorem. 290 INFINITE SERIES [CHAP. 11
- 300. 11.37. Let SnðxÞ ¼ nxenx2 ; n ¼ 1; 2; 3; . . . ; 0 @ x @ 1. ðaÞ Determine whether lim n!1 ð1 0 SnðxÞ dx ¼ ð1 0 lim n!1 SnðxÞ dx: ðbÞ Explain the result in (aÞ: ðaÞ ð1 0 snðxÞ dx ¼ ð1 0 nxenx2 dx ¼ 1 2 enx2 j1 0 ¼ 1 2 ð1 en Þ: Then lim n!1 ð1 0 SnðxÞ dx ¼ lim n!1 1 2 ð1 en Þ ¼ 1 2 SðxÞ ¼ lim n!1 SnðxÞ ¼ lim n!1 nxenx2 ¼ 0; whether x ¼ 0 or 0 x @ 1: Then, ð1 0 SðxÞ dx ¼ 0 It follows that lim n!1 ð1 0 SnðxÞ dx 6¼ ð1 0 lim n!1 SnðxÞ dx, i.e., the limit cannot be taken under the integral sign. (b) The reason for the result in (a) is that although the sequence SnðxÞ converges to 0, it does not converge uniformly to 0. To show this, observe that the function nxenx2 has a maximum at x ¼ 1= ffiffiffiffiffi 2n p (by the usual rules of elementary calculus), the value of this maximum being ffiffiffiffiffi 1 2 n q e1=2 . Hence, as n ! 1, SnðxÞ cannot be made arbitrarily small for all x and so cannot converge uniformly to 0. 11.38. Let f ðxÞ ¼ X 1 n¼1 sin nx n3 : Prove that ð 0 f ðxÞ dx ¼ 2 X 1 n¼1 1 ð2n 1Þ4 . We have sin nx n3 @ 1 n3 . Then by the Weierstrass M test the series is uniformly convergent for all x, in particular 0 @ x @ , and can be integrated term by term. Thus ð 0 f ðxÞ dx ¼ ð 0 X 1 n¼1 sin nx n3 ! dx ¼ X 1 n¼1 ð 0 sin nx n3 dx ¼ X 1 n¼1 1 cos n n4 ¼ 2 1 14 þ 1 34 þ 1 54 þ ¼ 2 X 1 n¼1 1 ð2n 1Þ4 POWER SERIES 11.39. Prove that both the power series X 1 n¼0 anxn and the corresponding series of derivatives X 1 n¼0 nanxn1 have the same radius of convergence. Let R 0 be the radius of convergence of anxn . Let 0 jx0j R. Then, as in Problem 11.33, we can choose N as that janj 1 jx0jn for n N. Thus, the terms of the series jnanxn1 j ¼ njanjjxjn1 can for n N be made less than corresponding terms of the series n jxjn1 jx0jn , which converges, by the ratio test, for jxj jx0j R. Hence, nanxn1 converges absolutely for all points x0 (no matter how close jx0j is to R). If, however, jxj R, lim n!1 anxn 6¼ 0 and thus lim n!1 nanxn1 6¼ 0, so that nanxn1 does not converge. CHAP. 11] INFINITE SERIES 291
- 301. Thus, R is the radius of convergence of nanxn1 . Note that the series of derivatives may or may not converge for values of x such that jxj ¼ R. 11.40. Illustrate Problem 11.39 by using the series X 1 n¼1 xn n2 3n . lim n!1 unþ1 un ¼ lim n!1 xnþ1 ðn þ 1Þ2 3nþ1 n2 3n xn ¼ lim n!1 n2 3ðn þ 1Þ2 jxj ¼ jxj 3 so that the series converges for jxj 3. At x ¼ 3 the series also converges, so that the interval of convergence is 3 @ x @ 3. The series of derivatives is X 1 n¼1 nxn1 n2 3n ¼ X 1 n¼1 xn1 n 3n By Problem 11.25(a) this has the interval of convergence 3 @ x 3. The two series have the same radius of convergence, i.e., R ¼ 3, although they do not have the same interval of convergence. Note that the result of Problem 11.39 can also be proved by the ratio test if this test is applicable. The proof given there, however, applies even when the test is not applicable, as in the series of Problem 11.22. 11.41. Prove that in any interval within its interval of convergence a power series ðaÞ represents a continuous function, say, f ðxÞ, ðbÞ can be integrated term by term to yield the integral of f ðxÞ, ðcÞ can be differentiated term by term to yield the derivative of f ðxÞ. We consider the power series anxn , although analogous results hold for anðx aÞn . (a) This follows from Problem 11.33 and 11.34, and the fact that each term anxn of the series is continuous. (b) This follows from Problems 11.33 and 11.35, and the fact that each term anxn of the series is continuous and thus integrable. (c) From Problem 11.39, the series of derivatives of a power series always converges within the interval of convergence of the original power series and therefore is uniformly convergent within this interval. Thus, the required result follows from Problems 11.33 and 11.36. If a power series converges at one (or both) end points of the interval of convergence, it is possible to establish (a) and (b) to include the end point (or end points). See Problem 11.42. 11.42. Prove Abel’s theroem that if a power series converges at an end point of its interval of conver- gence, then the interval of uniform convergence includes this end point. For simplicity in the proof, we assume the power series to be X 1 k¼0 akxk with the end point of its interval of convergence at x ¼ 1, so that the series surely converges for 0 @ x @ 1. Then we must show that the series converges uniformly in this interval. Let RnðxÞ ¼ anxn þ anþ1xnþ1 þ anþ2xnþ2 þ ; Rn ¼ an þ anþ1 þ anþ2 þ To prove the required result we must show that given any 0, we can find N such that jRnðxÞj for all n N, where N is independent of the particular x in 0 @ x @ 1. 292 INFINITE SERIES [CHAP. 11
- 302. Now RnðxÞ ¼ ðRn Rnþ1Þxn þ ðRnþ1 Rnþ2Þxnþ1 þ ðRnþ2 Rnþ3Þxnþ2 þ ¼ Rnxn þ Rnþ1ðxnþ1 xn Þ þ Rnþ2ðxnþ2 xnþ1 Þ þ ¼ xn fRn ð1 xÞðRnþ1 þ Rnþ2x þ Rnþ3x2 þ Þg Hence, for 0 @ x 1, jRnðxÞj @ jRnj þ ð1 xÞðjRnþ1j þ jRnþ2jx þ jRnþ3jx2 þ Þ ð1Þ Since ak converges by hypothesis, it follows that given 0 we can choose N such that jRkj =2 for all k A n. Then for n N we have from (1), jRnðxÞj @ 2 þ ð1 xÞ 2 þ 2 x þ 2 x2 þ ¼ 2 þ 2 ¼ ð2Þ since ð1 xÞð1 þ x þ x2 þ x3 þ Þ ¼ 1 (if 0 @ x 1). Also, for x ¼ 1; jRnðxÞj ¼ jRnj for n N. Thus, jRnðxÞj for all n N, where N is independent of the value of x in 0 @ x @ 1, and the required result follows. Extensions to other power series are easily made. 11.43. Prove Abel’s limit theorem (see Page 272). As in Problem 11.42, assume the power series to be X 1 k¼1 akxk , convergent for 0 @ x @ 1. Then we must show that lim x!1 X 1 k¼0 akxk ¼ X 1 k¼0 ak. This follows at once from Problem 11.42, which shows that akxk is uniformly convergent for 0 @ x @ 1, and from Problem 11.34, which shows that akxk is continuous at x ¼ 1. Extensions to other power series are easily made. 11.44. (a) Prove that tan1 x ¼ x x3 3 þ x5 5 x7 7 þ where the series is uniformly convergent in 1 @ x @ 1. (b) Prove that 4 ¼ 1 1 3 þ 1 5 1 7 þ . (a) By Problem 2.25 of Chapter 2, with r ¼ x2 and a ¼ 1, we have 1 1 þ x2 ¼ 1 x2 þ x4 x6 þ 1 x 1 ð1Þ Integrating from 0 to x, where 1 x 1, yields ðx 0 dx 1 þ x2 ¼ tan1 x ¼ x x3 3 þ x5 5 x7 7 þ ð2Þ using Problems 11.33 and 11.35. Since the series on the right of (2) converges for x ¼ 1, it follows by Problem 11.42 that the series is uniformly convergent in 1 @ x @ 1 and represents tan1 x in this interval. (b) By Problem 11.43 and part (a), we have lim x!1 tan1 x ¼ lim x!1 x x3 3 þ x5 5 x7 7 þ ! or 4 ¼ 1 1 3 þ 1 5 1 7 þ 11.45. Evaluate ð1 0 1 ex2 x2 dx to 3 decimal place accuracy. CHAP. 11] INFINITE SERIES 293
- 303. We have eu ¼ 1 þ u þ u2 2! þ u3 3! þ u4 4! þ u5 5! þ ; 1 u 1: Then if u ¼ x2 ; ex2 ¼ 1 x2 þ x4 2! x6 3! þ x8 3! ¼ x10 5! þ ; 1 x 1: Thus 1 ex2 x2 ¼ 1 x2 2! þ x4 3! x6 4! þ x8 5! : Since the series converges for all x and so, in particular, converges uniformly for 0 @ x @ 1, we can integrate term by term to obtain ð1 0 1 ex2 x2 dx ¼ x x3 3 2! þ x5 5 3! x7 7 4! þ x9 9 5! 1 0 ¼ 1 1 3 2! þ 1 5 3! 1 7 4! þ 1 9 5! ¼ 1 0:16666 þ 0:03333 0:00595 þ 0:00092 ¼ 0:862 Note that the error made in adding the first four terms of the alternating series is less than the fifth term, i.e., less than 0.001 (see Problem 11.15). MISCELLANEOUS PROBLEMS 11.46. Prove that y ¼ JpðxÞ defined by (16), Page 276, satisfies Bessel’s differential equation x2 y00 þ xy0 þ ðx2 p2 Þy ¼ 0 The series for JpðxÞ converges for all x [see Problem 11.110(a)]. Since a power series can be differ- entiated term by term within its interval of convergence, we have for all x, y ¼ X 1 n¼0 ð1Þn xpþ2n 2pþ2nn!ðn þ pÞ! y0 ¼ X 1 n¼0 ð1Þn ð p þ 2nÞxpþ2n1 2pþ2nn!ðn þ pÞ! y00 ¼ X 1 n¼0 ð1Þn ð p þ 2nÞð p þ 2n 1Þ xpþ2n2 2pþ2nn!ðn þ pÞ! Then, ðx2 p2 Þy ¼ X 1 n¼0 ð1Þn xpþ2nþ2 2pþ2nn!ðn þ pÞ! X 1 n¼0 ð1Þn p2 xpþ2n 2pþ2nn!ðn þ pÞ! xy0 ¼ X 1 n¼0 ð1Þn ðp þ 2nÞxpþ2n 2pþ2nn!ðn þ pÞ! x2 y00 ¼ X 1 n¼0 ð1Þn ð p þ 2nÞð p þ 2n 1Þxpþ2n 2pþ2nn!ðn þ pÞ! 294 INFINITE SERIES [CHAP. 11
- 304. Adding, x2 y00 þ xy0 þ ðx2 p2 Þy ¼ X 1 n¼0 ð1Þn xpþ2nþ2 2pþ2nn!ðn þ pÞ! þ X 1 n¼0 ð1Þn ½p2 þ ð p þ 2nÞ þ ð p þ 2nÞð p þ 2n 1Þxpþ2n 2pþ2nn!ðn þ pÞ! ¼ X 1 n¼0 ð1Þn xpþ2nþ2 2pþ2nn!ðn þ pÞ! þ X 1 n¼0 ð1Þn ½4nðn þ pÞxpþ2n 2pþ2nn!ðn þ pÞ! ¼ X 1 n¼1 ð1Þn1 xpþ2n 2pþ2n2ðn 1Þ!ðn 1 þ pÞ! þ X 1 n¼1 ð1Þn 4xpþ2n 2pþ2nðn 1Þ!ðn þ p 1Þ! ¼ X 1 n¼1 ð1Þn 4xpþ2n 2pþ2nðn 1Þ!ðn þ p 1Þ! þ X 1 n¼1 ð1Þn 4xpþ2n 2pþ2nðn 1Þ!ðn þ p 1Þ! ¼ 0 11.47. Test for convergence the complex power series X 1 n¼1 zn1 n3 3n1 . Since lim n!1 unþ1 un ¼ lim n!1 zn ðn þ 1Þ3 3n n3 3n1 zn1 ¼ lim n!1 n3 3ðn þ 1Þ3 jzj ¼ jzj 3 , the series converges for jzj 3 1, i.e., jzj 3, and diverges for jzj 3. For jzj ¼ 3, the series of absolute values is X 1 n¼1 jzjn1 n3 3n1 ¼ X 1 n¼1 1 n3 , so that the series is absolutely convergent and thus convergent for jzj ¼ 3. Thus, the series converges within and on the circle jzj ¼ 3. 11.48. Assuming the power series for ex holds for complex numbers, show that eix ¼ cos x þ i sin x Letting z ¼ ix in ez ¼ 1 þ z þ z2 2! þ z3 3! þ ; we have eix ¼ 1 þ ix þ i2 x2 2! þ i3 x3 3! þ ¼ 1 x2 2! þ x4 4! ! þ i x x3 3! þ x5 5! ! ¼ cos x þ i sin x Similarly, eix ¼ cos x i sin x. The results are called Euler’s identities. 11.49. Prove that lim n!1 1 þ 1 2 þ 1 3 þ 1 4 þ þ 1 n ln n exists. Letting f ðxÞ ¼ 1=x in (1), Problem 11.11, we find 1 2 þ 1 3 þ 1 4 þ þ 1 M @ ln M @ 1 þ 1 2 þ 1 3 þ 1 4 þ þ 1 M 1 from which we have on replacing M by n, 1 n @ 1 þ 1 2 þ 1 3 þ 1 4 þ þ 1 n ln n @ 1 Thus, the sequence Sn ¼ 1 þ 1 2 þ 1 3 þ 1 4 þ þ 1 n ln n is bounded by 0 and 1. CHAP. 11] INFINITE SERIES 295
- 305. 296 INFINITE SERIES [CHAP. 11 Consider Snþ1 Sn ¼ 1 n þ 1 ln n þ 1 n . By integrating the inequality 1 n þ 1 @ 1 x @ 1 n with respect to x from n to n þ 1, we have 1 n þ 1 @ ln n þ 1 n @ 1 n or 1 n þ 1 1 n @ 1 n þ 1 ln n þ 1 n @ 0 i.e., Snþ1 Sn @ 0, so that Sn is monotonic decreasing. Since Sn is bounded and monotonic decreasing, it has a limit. This limit, denoted by , is equal to 0:577215 . . . and is called Euler’s constant. It is not yet known whether is rational or not. 11.50. Prove that the infinite product Y 1 k¼1 ð1 þ ukÞ, where uk 0, converges if X 1 k¼1 uk converges. According to the Taylor series for ex (Page 275), 1 þ x @ ex for x 0, so that Pn ¼ Y n k¼1 ð1 þ ukÞ ¼ ð1 þ u1Þð1 þ u2Þ ð1 þ unÞ @ eu1 eu2 eun ¼ eu1þu2þþun Since u1 þ u2 þ converges, it follows that Pn is a bounded monotonic increasing sequence and so has a limit, thus proving the required result. 11.51. Prove that the series 1 1 þ 1 1 þ 1 1 þ is C 1 summable to 1/2. The sequence of partial sums is 1; 0; 1; 0; 1; 0; . . . . Then S1 ¼ 1; S1 þ S2 2 ¼ 1 þ 0 2 ¼ 1 2 ; S1 þ S2 þ S3 3 ¼ 1 þ 0 þ 1 3 ¼ 2 3 ; . . . : Continuing in this manner, we obtain the sequence 1; 1 2 ; 2 3 ; 1 2 ; 3 5 ; 1 2 ; . . . ; the nth term being Tn ¼ 1=2 if n is even n=ð2n 1Þ if n is odd . Thus, lim n!1 Tn ¼ 1 2 and the required result follows. 11.52. (a) If f ðnþ1Þ ðxÞ is continuous in ½a; b prove that for c in ½a; b, f ðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ þ 1 2! f 00 ðcÞðx cÞ2 þ þ 1 n! f ðnÞ ðcÞðx cÞn þ 1 n! ðx c ðx tÞn f ðnþ1Þ ðtÞ dt. (b) Obtain the Lagrange and Cauchy forms of the remainder in Taylor’s Formula. (See Page 274.) The proof of (a) is made using mathematical induction. (See Chapter 1.) The result holds for n ¼ 0 since f ðxÞ ¼ f ðcÞ þ ðx C f 0 ðtÞ dt ¼ f ðcÞ þ f ðxÞ f ðcÞ We make the induction assumption that it holds for n ¼ k and then use integration by parts with dv ¼ ðx tÞk k! dt and u ¼ f kþ1 ðtÞ Then v ¼ ðx tÞkþ1 ðk þ 1Þ! and du ¼ f kþ2 ðtÞ dt Thus, 1 k! ðx C ðx tÞk f ðkþ1Þ ðtÞ dt ¼ f kþ1 ðtÞðx tÞkþ1 ðk þ 1Þ! x C þ 1 ðk þ 1Þ! ðx C ðx tÞkþ1 f ðkþ2Þ ðtÞ dt ¼ f kþ1 ðcÞðx cÞkþ1 ðk þ 1Þ! þ 1 ðk þ 1Þ! ðx C ðx tÞkþ1 f ðkþ2Þ ðtÞ dt Having demonstrated that the result holds for k þ 1, we conclude that it holds for all positive integers.
- 306. To obtain the Lagrange form of the remainder Rn, consider the form f ðxÞ ¼ f ðcÞ þ f 0 ðcÞðx cÞ þ 1 2! f 00 ðcÞðx cÞ2 þ þ K n! ðx cÞn This is the Taylor polynomial Pn1ðxÞ plus K n! ðx cÞn : Also, it could be looked upon as Pn except that in the last term, f ðnÞ ðcÞ is replaced by a number K such that for fixed c and x the representation of f ðxÞ is exact. Now define a new function ðtÞ ¼ f ðtÞ f ðxÞ þ X n1 j¼1 f ð jÞ ðtÞ ðx tÞ j j! þ Kðx tÞn n! The function satisfies the hypothesis of Rolle’s Theorem in that ðcÞ ¼ ðxÞ ¼ 0, the function is continuous on the interval bound by c and x, and 0 exists at each point of the interval. Therefore, there exists in the interval such that 0 ðÞ ¼ 0. We proceed to compute 0 and set it equal to zero. 0 ðtÞ ¼ f 0 ðtÞ þ X n1 j¼1 f ð jþ1Þ ðtÞ ðx tÞ j j! X n1 j¼1 f ð jÞ ðtÞ ðx tÞj1 ð j 1Þ! Kðx tÞn1 ðn 1Þ! This reduces to 0 ðtÞ ¼ f ðnÞ ðtÞ ðn 1Þ! ðx tÞn1 K ðn 1Þ! ðx tÞn1 According to hypothesis: for each n there is n such that ðnÞ ¼ 0 Thus K ¼ f ðnÞ ðnÞ and the Lagrange remainder is Rn1 ¼ f ðnÞ ðnÞ n! ðx cÞn or equivalently Rn ¼ 1 ðn þ 1Þ! f ðnþ1Þ ðnþ1Þðx cÞnþ1 The Cauchy form of the remainder follows immediately by applying the mean value theorem for integrals. (See Page 274.) 11.53. Extend Taylor’s theorem to functions of two variables x and y. Define FðtÞ ¼ f ðx0 þ ht; y0 þ ktÞ, then applying Taylor’s theorem for one variable (about t ¼ 0Þ FðtÞ ¼ Fð0Þ þ F 0 ð0Þ þ 1 2! F 00 ð0Þt2 þ þ 1 n! FðnÞ ð0Þtn þ 1 ðn þ 1Þ! Fðnþ1Þ ðÞtnþ1 ; 0 t Now let t ¼ 1 Fð1Þ ¼ f ðx0 þ h; y0 þ kÞ ¼ Fð0Þ þ F 0 ð0Þ þ 1 2! F 00 ð0Þ þ þ 1 n! FðnÞ ð0Þ þ 1 ðn þ 1Þ! Fðnþ1Þ ðÞ When the derivatives F 0 ðtÞ; . . . ; FðnÞ ðtÞ; Fðnþ1Þ ðÞ are computed and substituted into the previous expres- sion, the two variable version of Taylor’s formula results. (See Page 277, where this form and notational details can be found.) 11.54. Expand x2 þ 3y 2 in powers of x 1 and y þ 2. Use Taylor’s formula with h ¼ x x0, k ¼ y y0, where x0 ¼ 1 and y0 ¼ 2. CHAP. 11] INFINITE SERIES 297
- 307. x2 þ 3y 2 ¼ 10 4ðx 1Þ þ 4ð y þ 2Þ 2ðx 1Þ2 þ 2ðx 1Þð y þ 2Þ þ ðx 1Þ2 ð y þ 2Þ (Check this algebraically.) 11.55. Prove that ln x þ y 2 ¼ x þ y 2 2 þ ðx þ y 2Þ ; 0 1; x 0; y 0. Hint: Use the Taylor formula with the linear term as the remainder. 11.56. Expand f ðx; yÞ ¼ sin xy in powers of x 1 and y 2 to second-degree terms. 1 1 8 2 ðx 1Þ2 2 ðx 1Þ y 2 y 2 2 Supplementary Problems CONVERGENCE AND DIVERGENCE OF SERIES OF CONSTANTS 11.57. (a) Prove that the series 1 3 7 þ 1 7 11 þ 1 11 15 þ ¼ X 1 n¼1 1 ð4n 1Þð4n þ 3Þ converges and (b) find its sum. Ans. (b) 1/12 11.58. Prove that the convergence or divergence of a series is not affected by (a) multiplying each term by the same non-zero constant, (b) removing (or adding) a finite number of terms. 11.59. If un and vn converge to A and B, respectively, prove that ðun þ vnÞ converges to A þ B. 11.60. Prove that the series 3 2 þ ð3 2Þ2 þ ð3 2Þ3 þ ¼ ð3 2Þn diverges. 11.61. Find the fallacy: Let S ¼ 1 1 þ 1 1 þ 1 1 þ . Then S ¼ 1 ð1 1Þ ð1 1Þ ¼ 1 and S ¼ ð1 1Þ þ ð1 1Þ þ ð1 1Þ þ ¼ 0. Hence, 1 ¼ 0. COMPARISON TEST AND QUOTIENT TEST 11.62. Test for convergence: ðaÞ X 1 n¼1 1 n2 þ 1 ; ðbÞ X 1 n¼1 n 4n2 3 ; ðcÞ X 1 n¼1 n þ 2 ðn þ 1Þ ffiffiffiffiffiffiffiffiffiffiffi n þ 3 p ; ðdÞ X 1 n¼1 3n n 5n ; ðeÞ X 1 n¼1 1 5n 3 ; ð f Þ X 1 n¼1 2n 1 ð3n þ 2Þn4=3: Ans: ðaÞ conv., ðbÞ div., ðcÞ div., ðdÞ conv., ðeÞ div., ð f Þ conv. 11.63. Investigate the convergence of (a) X 1 n¼1 4n2 þ 5n 2 nðn2 þ 1Þ3=2 ; ðbÞ X 1 n¼1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n ln n n2 þ 10n3 r . Ans. (a) conv., (b) div. 11.64. Establish the comparison test for divergence (see Page 267). 298 INFINITE SERIES [CHAP. 11
- 308. 11.65. Use the comparison test to prove that ðaÞ X 1 n¼1 @ 1 np converges if p 1 and diverges if p @ 1; ðbÞ X 1 n¼1 tan1 n n diverges, ðcÞ X 1 n¼1 n2 2n converges. 11.66. Establish the results (b) and (c) of the quotient test, Page 267. 11.67. Test for convergence: ðaÞ X 1 n¼1 ðln nÞ2 n2 ; ðbÞ X 1 n¼1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n tan1 ð1=n3 Þ q ; ðcÞ X 1 n¼1 3 þ sin n nð1 þ enÞ ; ðdÞ X 1 n¼1 n sin2 ð1=nÞ: Ans. (a) conv., (b) div., (c) div., (d) div. 11.68. If un converges, where un A 0 for n N, and if lim n!1 nun exists, prove that lim n!1 nun ¼ 0. 11.69. (a) Test for convergence X 1 n¼1 1 n1þ1=n . (b) Does your answer to (a) contradict the statement about the p series made on Page 266 that 1=np converges for p 1? Ans. (a) div. INTEGRAL TEST 11.70. Test for convergence: (a) X 1 n¼1 n2 2n3 1 ; ðbÞ X 1 n¼2 1 nðln nÞ3 ; ðcÞ X 1 n¼1 n 2n ; ðdÞ X 1 n¼1 e ffiffi n p ffiffiffi n p ðeÞ X 1 n¼2 ln n n ; ð f Þ X 1 n¼10 2lnðln nÞ n ln n : Ans: ðaÞ div., ðbÞ conv., ðcÞ conv., ðdÞ conv., ðeÞ div., ð f Þ div. 11.71. Prove that X 1 n¼2 1 nðln nÞp, where p is a constant, (a) converges if p 1 and (b) diverges if p @ 1. 11.72. Prove that 9 8 X 1 n¼1 1 n3 5 4 . 11.73. Investigate the convergence of X 1 n¼1 etan1 n n2 þ 1 : Ans: conv. 11.74. (a) Prove that 2 3 n3=2 þ 1 3 @ ffiffiffi 1 p þ ffiffiffi 2 p þ ffiffiffi 3 p þ þ ffiffiffi n p @ 2 3 n3=2 þ n1=2 2 3. (b) Use (a) to estimate the value of ffiffiffi 1 p þ ffiffiffi 2 p þ ffiffiffi 3 p þ þ ffiffiffiffiffiffiffiffi 100 p , giving the maximum error. (c) Show how the accuracy in (b) can be improved by estimating, for example, ffiffiffiffiffi 10 p þ ffiffiffiffiffi 11 p þ þ ffiffiffiffiffiffiffiffi 100 p and adding on the value of ffiffiffi 1 p þ ffiffiffi 2 p þ þ ffiffiffi 9 p computed to some desired degree of accuracy. Ans: ðbÞ 671:5 4:5 ALTERNATING SERIES 11.75. Test for convergence: (a) X 1 n¼1 ð1Þnþ1 2n ; ðbÞ X 1 n¼1 ð1Þn n2 þ 2n þ 2 ; ðcÞ X 1 n¼1 ð1Þnþ1 n 3n 1 ; ðdÞ X 1 n¼1 ð1Þn sin1 1 n ; ðeÞ X 1 n¼2 ð1Þn ffiffiffi n p ln n : Ans. (a) conv., (b) conv., (c) div., (d) conv., (e) div. CHAP. 11] INFINITE SERIES 299
- 309. 11.76. (a) What is the largest absolute error made in approximating the sum of the series X 1 n¼1 ð1Þn 2n ðn þ 1Þ by the sum of the first 5 terms? Ans. 1/192 (b) What is the least number of terms which must be taken in order that 3 decimal place accuracy will result? Ans. 8 terms 11.77. (a) Prove that S ¼ 1 13 þ 1 23 þ 1 33 þ ¼ 4 3 1 13 1 23 þ 1 33 . (b) How many terms of the series on the right are needed in order to calculate S to six decimal place accuracy? Ans. (b) at least 100 terms ABSOLUTE AND CONDITIONAL CONVERGENCE 11.78. Test for absolute or conditional convergence: ðaÞ X 1 n¼1 ð1Þn1 n2 þ 1 ðcÞ X 1 n¼2 ð1Þn n ln n ðeÞ X 1 n¼1 ð1Þn1 2n 1 sin 1 ffiffiffi n p ðbÞ X 1 n¼1 ð1Þn1 n n2 þ 1 ðdÞ X 1 n¼1 ð1Þn n3 ðn2 þ 1Þ4=3 ð f Þ X 1 n¼1 ð1Þn1 n3 2n 1 Ans. (a) abs. conv., (b) cond. conv., (c) cond. conv., (d) div., (e) abs. conv., ( f ) abs. conv. 11.79. Prove that X 1 n¼1 cos na x2 þ n2 converges absolutely for all real x and a. 11.80. If 1 1 2 þ 1 3 1 4 þ converges to S, prove that the rearranged series 1 þ 1 3 1 2 þ 1 5 þ 1 7 1 4 þ 1 9 þ 1 11 1 6 þ ¼ 3 2 S. Explain. [Hint: Take 1/2 of the first series and write it as 0 þ 1 2 þ 0 1 4 þ 0 þ 1 6 þ ; then add term by term to the first series. Note that S ¼ ln 2, as shown in Problem 11.100.] 11.81. Prove that the terms of an absolutely convergent series can always be rearranged without altering the sum. RATIO TEST 11.82. Test for convergence: ðaÞ X 1 n¼1 ð1Þn n ðn þ 1Þen ; ðbÞ X 1 n¼1 102n ð2n 1Þ! ; ðcÞ X 1 n¼1 3n n3 ; ðdÞ X 1 n¼1 ð1Þn 23n 32n ; ðeÞ X 1 n¼1 ð ffiffiffi 5 p 1Þn n2 þ 1 : Ans. (a) conv. (abs.), (b) conv., (c) div., (d) conv. (abs.), (e) div. 11.83. Show that the ratio test cannot be used to establish the conditional convergence of a series. 11.84. Prove that (a) X 1 n¼1 n! nn converges and (b) lim n!1 n! nn ¼ 0. MISCELLANEOUS TESTS 11.85. Establish the validity of the nth root test on Page 268. 11.86. Apply the nth root test to work Problems 11.82ðaÞ, (c), (d), and (e). 11.87. Prove that 1 3 þ ð2 3Þ2 þ ð1 3Þ3 þ ð2 3Þ4 þ ð1 3Þ5 þ ð2 3Þ6 þ converges. 300 INFINITE SERIES [CHAP. 11
- 310. 11.88. Test for convergence: (a) 1 3 þ 1 4 3 6 þ 1 4 7 3 6 9 þ , (b) 2 9 þ 2 5 9 12 þ 2 5 8 9 12 15 þ . Ans. (a) div., (b) conv. 11.89. If a; b, and d are positive numbers and b a, prove that a b þ aða þ dÞ bðb þ dÞ þ aða þ dÞða þ 2dÞ bðb þ dÞðb þ 2dÞ þ converges if b a d, and diverges if b a @ d. SERIES OF FUNCTIONS 11.90. Find the domain of convergence of the series: ðaÞ X 1 n¼1 xn n3 ; ðbÞ X 1 n¼1 ð1Þn ðx 1Þn 2nð3n 1Þ ; ðcÞ X 1 n¼1 1 nð1 þ x2Þn ; ðdÞ X 1 n¼1 n2 1 x 1 þ x n ; ðeÞ X 1 n¼1 enx n2 n þ 1 Ans: ðaÞ 1 @ x @ 1; ðbÞ 1 x @ 3; ðcÞ all x 6¼ 0; ðdÞ x 0; ðeÞ x @ 0 11.91. Prove that X 1 n¼1 1 3 5 ð2n 1Þ 2 4 6 ð2nÞ xn converges for 1 @ x 1. UNIFORM CONVERGENCE 11.92. By use of the definition, investigate the uniform convergence of the series X 1 n¼1 x ½1 þ ðn 1Þx½1 þ nx Hint: Resolve the nth term into partial fractions and show that the nth partial sum is SnðxÞ ¼ 1 1 1 þ nx : Ans. Not uniformly convergent in any interval which includes x ¼ 0; uniformly convergent in any other interval. 11.93. Work Problem 11.30 directly by first obtaining SnðxÞ. 11.94. Investigate by any method the convergence and uniform convergence of the series: ðaÞ X 1 n¼1 x 3 n ; ðbÞ X 1 n¼1 sin2 nx 2n 1 ; ðcÞ X 1 n¼1 x ð1 þ xÞn ; x A 0: Ans. (a) conv. for jxj 3; unif. conv. for jxj @ r 3. (b) unif. conv. for all x. (c) conv. for x A 0; not unif. conv. for x A 0, but unif. conv. for x A r 0. 11.95. If FðxÞ ¼ X 1 n¼1 sin nx n3 , prove that: (a) FðxÞ is continuous for all x, (b) lim x!0 FðxÞ ¼ 0; ðcÞ F 0 ðxÞ ¼ X 1 n¼1 cos nx n2 is continous everywhere. 11.96. Prove that ð 0 cos 2x 1 3 þ cos 4x 3 5 þ cos 6x 5 7 þ dx ¼ 0. 11.97. Prove that FðxÞ ¼ X 1 n¼1 sin nx sinh n has derivatives of all orders for any real x. 11.98. Examine the sequence unðxÞ ¼ 1 1 þ x2n ; n ¼ 1; 2; 3; . . . ; for uniform convergence. 11.99. Prove that lim n!1 ð1 0 dx ð1 þ x=nÞn ¼ 1 e1 . CHAP. 11] INFINITE SERIES 301
- 311. POWER SERIES 11.100. (a) Prove that lnð1 þ xÞ ¼ x x2 2 þ x3 3 x4 4 þ . ðbÞ Prove that ln 2 ¼ 1 1 2 þ 1 3 1 4 þ : Hint: Use the fact that 1 1 þ x ¼ 1 x þ x2 x3 þ and integrate. 11.101. Prove that sin1 x ¼ x þ 1 2 x3 3 þ 1 3 2 4 x5 5 þ 1 3 5 2 4 6 x7 7 þ , 1 @ x @ 1. 11.102. Evaluate (a) ð1=2 0 ex2 dx; ðdÞ ð1 0 1 cos x x dx to 3 decimal places, justifying all steps. Ans. ðaÞ 0:461; ðbÞ 0:486 11.103. Evaluate (a) sin 408; ðbÞ cos 658; ðcÞ tan 128 correct to 3 decimal places. Ans: ðaÞ 0:643; ðbÞ 0:423; ðcÞ 0:213 11.104. Verify the expansions 4, 5, and 6 on Page 275. 11.105. By multiplying the series for sin x and cos x, verify that 2 sin x cos x ¼ sin 2x. 11.106. Show that ecos x ¼ e 1 x2 2! þ 4x4 4! 31x6 6! þ ! ; 1 x 1. 11.107. Obtain the expansions ðaÞ tanh1 x ¼ x þ x3 3 þ x5 5 þ x7 7 þ 1 x 1 ðbÞ lnðx þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p Þ ¼ x 1 2 x3 3 þ 1 3 2 4 x5 5 1 3 5 2 4 6 x7 7 þ 1 @ x @ 1 11.108. Let f ðxÞ ¼ e1=x2 x 6¼ 0 0 x ¼ 0 . Prove that the formal Taylor series about x ¼ 0 corresponding to f ðxÞ exists but that it does not converge to the given function for any x 6¼ 0. 11.109. Prove that ðaÞ lnð1 þ xÞ 1 þ x ¼ x 1 þ 1 2 x2 þ 1 þ 1 2 þ 1 3 x3 for 1 x 1 ðbÞ flnð1 þ xÞg2 ¼ x2 1 þ 1 2 2x3 3 þ 1 þ 1 2 þ 1 3 2x4 4 for 1 x @ 1 MISCELLANEOUS PROBLEMS 11.110. Prove that the series for JpðxÞ converges (a) for all x, (b) absolutely and uniformly in any finite interval. 11.111. Prove that (a) d dx fJ0ðxÞg ¼ J1ðxÞ; ðbÞ d dx fxp JpðxÞg ¼ xp Jp1ðxÞ; ðcÞ Jpþ1ðxÞ ¼ 2p x JpðxÞ Jp1ðxÞ. 11.112. Assuming that the result of Problem 11.111(c) holds for p ¼ 0; 1; 2; . . . ; prove that (a) J1ðxÞ ¼ J1ðxÞ; ðbÞ J2ðxÞ ¼ J2ðxÞ; ðcÞ JnðxÞ ¼ ð1Þn JnðxÞ; n ¼ 1; 2; 3; . . . : 11.113. Prove that e1=2xðt1=tÞ ¼ X 1 p¼1 JpðxÞ tp . [Hint: Write the left side as ext=2 ex=2t , expand and use Problem 11.112.] 302 INFINITE SERIES [CHAP. 11
- 312. 11.114. Prove that X 1 n¼1 ðn þ 1Þzn nðn þ 2Þ2 is absolutely and uniformly convergent at all points within and on the circle jzj ¼ 1. 11.115. (a) If X 1 n¼1 anxn ¼ X 1 n¼1 bnxn for all x in the common interval of convergence jxj R where R 0, prove that an ¼ bn for n ¼ 0; 1; 2; . . . . (b) Use (a) to show that the Taylor expansion of a function exists, the expansion is unique. 11.116. Suppose that lim ffiffiffiffiffiffiffiffi junj n p ¼ L. Prove that un converges or diverges according as L 1 or L 1. If L ¼ 1 the test fails. 11.117. Prove that the radius of convergence of the series anxn can be determined by the following limits, when they exist, and give examples: (a) lim n!1 an anþ1 ; ðbÞ lim n!1 1 ffiffiffiffiffiffiffiffi janj n p ; ðcÞ lim n!1 1 ffiffiffiffiffiffiffiffi janj n p : 11.118. Use Problem 11.117 to find the radius of convergence of the series in Problem 11.22. 11.119. (a) Prove that a necessary and sufficient condition that the series un converge is that, given any 0, we can find N 0 depending on such that jSp Sqj whenever p N and q N, where Sk ¼ u1 þ u2 þ þ uk. ðbÞ Use ðaÞ to prove that the series X 1 n¼1 n ðn þ 1Þ3n converges. ðcÞ How could you use ðaÞ to prove that the series X 1 n¼1 1 n diverges? [Hint: Use the Cauchy convergence criterion, Page 25.] 11.120. Prove that the hypergeometric series (Page 276) (a) is absolutely convergent for jxj 1, (b) is divergent for jxj 1, (c) is absolutely divergent for jxj ¼ 1 if a þ b c 0; ðdÞ satisfies the differential equation xð1 xÞy00 þ fc ða þ b þ 1Þxgy0 aby ¼ 0. 11.121. If Fða; b; c; xÞ is the hypergeometric function defined by the series on Page 276, prove that (a) Fðp; 1; 1; xÞ ¼ ð1 þ xÞp ; ðbÞ xFð1; 1; 2; xÞ ¼ lnð1 þ xÞ; ðcÞ Fð1 2 ; 1 2 ; 3 2 ; x2 Þ ¼ ðsin1 xÞ=x. 11.122. Find the sum of the series SðxÞ ¼ x þ x3 1 3 þ x5 1 3 5 þ . [Hint: Show that S0 ðxÞ 1 þ xSðxÞ and solve.] Ans: ex2 =2 ðx 0 ex2 =2 dx 11.123. Prove that 1 þ 1 1 3 þ 1 1 3 5 þ 1 1 3 5 7 þ ¼ ffiffiffi e p 1 1 2 3 þ 1 22 2! 5 1 23 3! 7 þ 1 24 4! 9 11.124. Establish the Dirichlet test on Page 270. 11.125. Prove that X 1 n¼1 sin nx n is uniformly convergent in any interval which does not include 0; ; 2; . . . . [Hint: use the Dirichlet test, Page 270, and Problem 1.94, Chapter 1.] 11.126. Establish the results on Page 275 concerning the binomial series. [Hint: Examine the Lagrange and Cauchy forms of the remainder in Taylor’s theorem.] CHAP. 11] INFINITE SERIES 303
- 313. 11.127. Prove that X 1 n¼1 ð1Þn1 n þ x2 converges uniformly for all x, but not absolutely. 11.128. Prove that 1 1 4 þ 1 7 1 10 þ ¼ 3 ffiffiffi 3 p þ 1 3 ln 2 11.129. If x ¼ yey , prove that y ¼ X 1 n¼1 ð1Þn1 nn1 n! xn for 1=e x @ 1=e. 11.130. Prove that the equation e ¼ 1 has only one real root and show that it is given by ¼ 1 þ X 1 n¼1 ð1Þn1 nn1 en n! 11.131. Let x ex 1 ¼ 1 þ B1x þ B2x2 2! þ B3x3 3! þ . (a) Show that the numbers Bn, called the Bernoulli numbers, satisfy the recursion formula ðB þ 1Þn Bn ¼ 0 where Bk is formally replaced by Bk after expanding. (b) Using (a) or otherwise, determine B1; . . . ; B6. Ans: ðbÞ B1 ¼ 1 2 ; B2 ¼ 1 6 ; B3 ¼ 0; B4 ¼ 1 30 ; B5 ¼ 0; B6 ¼ 1 42. 11.132. (a) Prove that x ex 1 ¼ x 2 coth x 2 1 : ðbÞ Use Problem 11.127 and part (a) to show that B2kþ1 ¼ 0 if k ¼ 1; 2; 3; . . . : 11.133. Derive the series expansions: ðaÞ coth x ¼ 1 x þ x 3 x3 45 þ þ B2nð2xÞ2n ð2nÞ!x þ ðbÞ cot x ¼ 1 x x 3 x3 45 þ ð1Þn B2nð2xÞ2n ð2nÞ!x þ ðcÞ tan x ¼ x þ x3 3 þ 2x5 15 þ ð1Þn1 2ð22n 1ÞB2nð2xÞ2n1 ð2nÞ! þ ðdÞ csc x ¼ 1 x þ x 6 þ 7 360 x3 þ ð1Þn1 2ð22n1 1ÞB2nx2n1 ð2nÞ! þ [Hint: For (a) use Problem 11.132; for (b) replace x by ix in (a); for (c) use tan x ¼ cot x 2 cot 2x; for (d) use csc x ¼ cot x þ tan x=2.] 11.134. Prove that Y 1 n¼1 1 þ 1 n3 converges. 11.135. Use the definition to prove that Y 1 n¼1 1 þ 1 n diverges. 11.136. Prove that Y 1 n¼1 ð1 unÞ, where 0 un 1, converges if and only if un converges. 11.137. (a) Prove that Y 1 n¼2 1 1 n2 converges to 1 2. (b) Evaluate the infinite product in (a) to 2 decimal places and compare with the true value. 11.138. Prove that the series 1 þ 0 1 þ 1 þ 0 1 þ 1 þ 0 1 þ is the C 1 summable to zero. 304 INFINITE SERIES [CHAP. 11
- 314. 11.139. Prove that the Césaro method of summability is regular. [Hint: See Page 278.] 11.140. Prove that the series 1 þ 2x þ 3x2 þ 4x3 þ þ nxn1 þ converges to 1=ð1 xÞ2 for jxj 1. 11.141. A series X 1 n¼0 an is called Abel summable to S if S ¼ lim x!1 X 1 n¼0 anxn exists. Prove that ðaÞ X 1 n¼0 ð1Þn ðn þ 1Þ is Abel summable to 1/4 and ðbÞ X 1 n¼0 ð1Þn ðn þ 1Þðn þ 2Þ 2 is Abel summable to 1/8. 11.142. Prove that the double series X 1 m¼1 X 1 n¼1 1 ðm2 þ n2Þp, where p is a constant, converges or diverges according as p 1 or p @ 1, respectively. 11.143. (a) Prove that ð1 x exu u du ¼ 1 x 1 x2 þ 2! x3 3! x4 þ ð1Þn1 ðn 1Þ! xn þ ð1Þn n! ð1 x exu unþ1 du. ðbÞ Use ðaÞ to prove that ð1 x exu u du 1 x 1 x2 þ 2! x3 3! x4 þ CHAP. 11] INFINITE SERIES 305
- 315. 306 Improper Integrals DEFINITION OF AN IMPROPER INTEGRAL The functions that generate the Riemann integrals of Chapter 6 are continuous on closed intervals. Thus, the functions are bounded and the intervals are finite. Integrals of functions with these char- acteristics are called proper integrals. When one or more of these restrictions is relaxed, the integrals are said to be improper. Categories of improper integrals are established below. The integral ðb a f ðxÞ dx is called an improper integral if 1. a ¼ 1 or b ¼ 1 or both, i.e., one or both integration limits is infinite, 2. f ðxÞ is unbounded at one or more points of a @ x @ b. Such points are called singularities of f ðxÞ. Integrals corresponding to (1) and (2) are called improper integrals of the first and second kinds, respectively. Integrals with both conditions (1) and (2) are called improper integrals of the third kind. EXAMPLE 1. ð1 0 sin x2 dx is an improper integral of the first kind. EXAMPLE 2. ð4 0 dx x 3 is an improper integral of the second kind. EXAMPLE 3. ð1 0 ex ffiffiffi x p dx is an improper integral of the third kind. EXAMPLE 4. ð1 0 sin x x dx is a proper integral since lim x!0þ sin x x ¼ 1. IMPROPER INTEGRALS OF THE FIRST KIND (Unbounded Intervals) If f is an integrable on the appropriate domains, then the indefinite integrals ðx a f ðtÞ dt and ða x f ðtÞ dt (with variable upper and lower limits, respectively) are functions. Through them we define three forms of the improper integral of the first kind. Definition (a) If f is integrable on a @ x 1, then ð1 a f ðxÞ dx ¼ lim x!1 ðx a f ðtÞ dt. (b) If f is integrable on 1 x @ a, then ða 1 f ðxÞ dx ¼ lim x!1 ða x f ðtÞ dt: Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 316. (c) If f is integrable on 1 x 1, then ð1 1 f ðxÞ dx ¼ ða 1 f ðxÞ dx þ ð1 a f ðxÞ dx ¼ lim x!1 ða x f ðtÞ dt þ lim x!1 ðx a f ðtÞ dt: In part (c) it is important to observe that lim x!1 ða x f ðtÞ dt þ lim x!1 ðx a f ðtÞ dt: and lim x!1 ða x f ðtÞ dt þ ðx a f ðtÞ dt are not necessarily equal. This can be illustrated with f ðxÞ ¼ xex2 . The first expression is not defined since neither of the improper integrals (i.e., limits) is defined while the second form yields the value 0. EXAMPLE. The function FðxÞ ¼ 1 ffiffiffiffiffiffi 2 p eðx2 =2Þ is called the normal density function and has numerous applications in probability and statistics. In particular (see the bell-shaped curve in Fig. 12-1) ð1 1 1 ffiffiffiffiffiffi 2 p e x2 2 : dx ¼ 1 (See Problem 12.31 for the trick of making this evaluation.) Perhaps at some point in your academic career you were ‘‘graded on the curve.’’ The infinite region under the curve with the limiting area of 1 corresponds to the assurance of getting a grade. C’s are assigned to those whose grades fall in a desig- nated central section, and so on. (Of course, this grading procedure is not valid for a small number of students, but as the number increases it takes on statistical meaning.) In this chapter we formulate tests for convergence or diver- gence of improper integrals. It will be found that such tests and proofs of theorems bear close analogy to convergence and divergence tests and corresponding theorems for infinite series (See Chapter 11). CONVERGENCE OR DIVERGENCE OF IMPROPER INTEGRALS OF THE FIRST KIND Let f ðxÞ be bounded and integrable in every finite interval a @ x @ b. Then we define ð1 a f ðxÞ dx ¼ lim b!1 ðb a f ðxÞ dx ð1Þ where b is a variable on the positive real numbers. The integral on the left is called convergent or divergent according as the limit on the right does or does not exist. Note that ð1 a f ðxÞ dx bears close analogy to the infinite series X 1 n¼1 un, where un ¼ f ðnÞ, while ðb a f ðxÞ dx corresponds to the partial sums of such infinite series. We often write M in place of b in (1). CHAP. 12] IMPROPER INTEGRALS 307 Fig. 12-1
- 317. Similarly, we define ðb 1 f ðxÞ dx ¼ lim a!1 ðb a f ðxÞ dx ð2Þ where a is a variable on the negative real numbers. And we call the integral on the left convergent or divergent according as the limit on the right does or does not exist. EXAMPLE 1. ð1 1 dx x2 ¼ lim b!1 ðb 1 dx x2 ¼ lim b!1 1 1 b ¼ 1 so that ð1 1 dx x2 converges to 1. EXAMPLE 2. ðu 1 cos x dx ¼ lim a!1 ðu a cos x dx ¼ lim a!1 ðsin u sin aÞ. Since this limit does not exist, ðu 1 cos x dx is divergent. In like manner, we define ð1 1 f ðxÞ dx ¼ ðx0 1 f ðxÞ dx þ ð1 x0 f ðxÞ dx ð3Þ where x0 is a real number, and call the integral convergent or divergent according as the integrals on the right converge or not as in definitions (1) and (2). (See the previous remarks in part (c) of the definition of improper integrals of the first kind.) SPECIAL IMPROPER INTEGRALS OF THE FIRST KIND 1. Geometric or exponential integral ð1 a etx dx, where t is a constant, converges if t 0 and diverges if t @ 0. Note the analogy with the geometric series if r ¼ et so that etx ¼ rx . 2. The p integral of the first kind ð1 a dx xp, where p is a constant and a 0, converges if p 1 and diverges if p @ 1. Compare with the p series. CONVERGENCE TESTS FOR IMPROPER INTEGRALS OF THE FIRST KIND The following tests are given for cases where an integration limit is 1. Similar tests exist where an integration limit is 1 (a change of variable x ¼ y then makes the integration limit 1). Unless otherwise specified we shall assume that f ðxÞ is continuous and thus integrable in every finite interval a @ x @ b. 1. Comparison test for integrals with non-negative integrands. (a) Convergence. Let gðxÞ A 0 for all x A a, and suppose that ð1 a gðxÞ dx converges. Then if 0 @ f ðxÞ @ gðxÞ for all x A a, ð1 a f ðxÞ dx also converges. EXAMPLE. Since 1 ex þ 1 @ 1 ex ¼ ex and ð1 0 ex dx converges, ð1 0 dx ex þ 1 also converges. (b) Divergence. Let gðxÞ A 0 for all x A a, and suppose that ð1 a gðxÞ dx diverges. Then if f ðxÞ A gðxÞ for all x A a, ð1 a f ðxÞ dx also diverges. EXAMPLE. Since 1 ln x 1 x for x A 2 and ð1 2 dx x diverges ( p integral with p ¼ 1), ð1 2 dx ln x also diverges. 308 IMPROPER INTEGRALS [CHAP. 12
- 318. 2. Quotient test for integrals with non-negative integrands. (a) If f ðxÞ A 0 and gðxÞ A 0, and if lim x!1 f ðxÞ gðxÞ ¼ A 6¼ 0 or 1, then ð1 a f ðxÞ dx and ð1 a gðxÞ dx either both converge or both diverge. (b) If A ¼ 0 in (a) and ð1 a gðxÞ dx converges, then ð1 a f ðxÞ dx converges. (c) If A ¼ 1 in (a) and ð1 a gðxÞ dx diverges, then ð1 a f ðxÞ dx diverges. This test is related to the comparison test and is often a very useful alternative to it. In particular, taking gðxÞ ¼ 1=xp , we have from known facts about the p integral, the following theorem. Theorem 1. Let lim x!1 xp f ðxÞ ¼ A. Then (i) ð1 a f ðxÞ dx converges if p 1 and A is finite (ii) ð1 a f ðxÞ dx diverges if p @ 1 and A 6¼ 0 (A may be infinite). EXAMPLE 1. ð1 0 x2 dx 4x4 þ 25 converges since lim x!1 x2 x2 4x4 þ 25 ¼ 1 4 . EXAMPLE 2. ð1 0 x dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x4 þ x2 þ 1 p diverges since lim x!1 x x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x4 þ x2 þ 1 p ¼ 1. Similar test can be devised using gðxÞ ¼ etx . 3. Series test for integrals with non-negative integrands. ð1 a f ðxÞ dx converges or diverges accord- ing as un, where un ¼ f ðnÞ, converges or diverges. 4. Absolute and conditional convergence. ð1 a f ðxÞ dx is called absolutely convergent if ð1 a j f ðxÞj dx converges. If ð1 a f ðxÞ dx converges but ð1 a j f ðxÞj dx diverges, then ð1 a f ðxÞ dx is called con- ditionally convergent. Theorem 2. If ð1 a j f ðxÞj dx converges, then ð1 a f ðxÞ dx converges. In words, an absolutely convergent integral converges. EXAMPLE 1. ð1 a cos x x2 þ 1 dx is absolutely convergent and thus convergent since ð1 0 cos x x2 þ 1 dx @ ð1 0 dx x2 þ 1 and ð1 0 dx x2 þ 1 converges. EXAMPLE 2. ð1 0 sin x x dx converges (see Problem 12.11), but ð1 0 sin x x dx does not converge (see Problem 12.12). Thus, ð1 0 sin x x dx is conditionally convergent. Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence. CHAP. 12] IMPROPER INTEGRALS 309
- 319. IMPROPER INTEGRALS OF THE SECOND KIND If f ðxÞ becomes unbounded only at the end point x ¼ a of the interval a @ x @ b, then we define ðb a f ðxÞ dx ¼ lim !0þ ðb aþ f ðxÞ dx ð4Þ and define it to be an improper integral of the second kind. If the limit on the right of (4) exists, we call the integral on the left convergent; otherwise, it is divergent. Similarly if f ðxÞ becomes unbounded only at the end point x ¼ b of the interval a @ x @ b, then we extend the category of improper integrals of the second kind. ðb a f ðxÞ dx ¼ lim !0þ ðb a f ðxÞ dx ð5Þ Note: Be alert to the word unbounded. This is distinct from undefined. For example, ð1 0 sin x x dx ¼ lim !0 ð1 sin x x dx is a proper integral, since lim x!0 sin x x ¼ 1 and hence is bounded as x ! 0 even though the function is undefined at x ¼ 0. In such case the integral on the left of (5) is called convergent or divergent according as the limit on the right exists or does not exist. Finally, the category of improper integrals of the second kind also includes the case where f ðxÞ becomes unbounded only at an interior point x ¼ x0 of the interval a @ x @ b, then we define ðb a f ðxÞ dx ¼ lim 1!0þ ðx01 a f ðxÞ dx þ lim 2!0þ ðb x0þ2 f ðxÞ dx ð6Þ The integral on the left of (6) converges or diverges according as the limits on the right exist or do not exist. Extensions of these definitions can be made in case f ðxÞ becomes unbounded at two or more points of the interval a @ x @ b. CAUCHY PRINCIPAL VALUE It may happen that the limits on the right of (6) do not exist when 1 and 2 approach zero independently. In such case it is possible that by choosing 1 ¼ 2 ¼ in (6), i.e., writing ðb a f ðxÞ dx ¼ lim !0þ ðx0 a f ðxÞ dx þ ðb x0þ f ðxÞ dx ð7Þ the limit does exist. If the limit on the right of (7) does exist, we call this limiting value the Cauchy principal value of the integral on the left. See Problem 12.14. EXAMPLE. The natural logarithm (i.e., base e) may be defined as follows: ln x ¼ ðx 1 dt t ; 0 x 1 Since f ðxÞ ¼ 1 x is unbounded as x ! 0, this is an improper integral of the second kind (see Fig. 12-2). Also, ð1 0 dt t is an integral of the third kind, since the interval to the right is unbounded. Now lim !0 ð1 dt t ¼ lim !0 ½ln 1 ln ! 1 as ! 0; therefore, this improper integral of the second kind is divergent. Also, ð1 1 dt t ¼ lim x!1 ðx 1 dt t ¼ lim x!1 ½ln x ln i ! 1; this integral (which is of the first kind) also diverges. 310 IMPROPER INTEGRALS [CHAP. 12
- 320. SPECIAL IMPROPER INTEGRALS OF THE SECOND KIND 1. ðb a dx ðx aÞp converges if p 1 and diverges if p A 1. 2. ðb a dx ðb xÞp converges if p 1 and diverges if p A 1. These can be called p integrals of the second kind. Note that when p @ 0 the integrals are proper. CONVERGENCE TESTS FOR IMPROPER INTEGRALS OF THE SECOND KIND The following tests are given for the case where f ðxÞ is unbounded only at x ¼ a in the interval a @ x @ b. Similar tests are available if f ðxÞ is unbounded at x ¼ b or at x ¼ x0 where a x0 b. 1. Comparison test for integrals with non-negative integrands. (a) Convergence. Let gðxÞ A 0 for a x @ b, and suppose that ðb a gðxÞ dx converges. Then if 0 @ f ðxÞ @ gðxÞ for a x @ b, ðb a f ðxÞ dx also converges. EXAMPLE. 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi x4 1 p 1 ffiffiffiffiffiffiffiffiffiffiffi x 1 p for x 1. Then since ð5 1 dx ffiffiffiffiffiffiffiffiffiffiffi x 1 p converges ( p integral with a ¼ 1, p ¼ 1 2), ð5 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffi x4 1 p also converges. (b) Divergence. Let gðxÞ A 0 for a x @ b, and suppose that ðb a gðxÞ dx diverges. Then if f ðxÞ A gðxÞ for a x @ b, ðb a f ðxÞ dx also diverges. EXAMPLE. ln x ðx 3Þ4 1 ðx 3Þ4 for x 3. Then since ð6 3 dx ðx 3Þ4 diverges ( p integral with a ¼ 3, p ¼ 4), ð6 3 ln x ðx 3Þ4 dx also diverges. 2. Quotient test for integrals with non-negative integrands. (a) If f ðxÞ A 0 and gðxÞ A 0 for a x @ b, and if lim x!a f ðxÞ gðxÞ ¼ A 6¼ 0 or 1, then ðb a f ðxÞ dx and ðb a gðxÞ dx either both converge or both diverge. CHAP. 12] IMPROPER INTEGRALS 311 Fig. 12-2
- 321. (b) If A ¼ 0 in (a), then ðb a gðxÞ dx converges, then ðb a f ðxÞ dx converges. (c) If A ¼ 1 in (a), and ðb a gðxÞ dx diverges, then ðb a f ðxÞ dx diverges. This test is related to the comparison test and is a very useful alternative to it. In particular taking gðxÞ ¼ 1=ðx aÞp we have from known facts about the p integral the following theorems. Theorem 3. Let lim x!aþ ðx aÞp f ðxÞ ¼ A. Then (i) ðb a f ðxÞ dx converges if p 1 and A is finite (ii) ðb a f ðxÞ dx diverges if p A 1 and A 6¼ 0 (A may be infinite). If f ðxÞ becomes unbounded only at the upper limit these conditions are replaced by those in Theorem 4. Let lim x!b ðb xÞp f ðxÞ ¼ B. Then (i) ðb a f ðxÞ dx converges if p 1 and B is finite (ii) ðb a f ðxÞ dx diverges if p A 1 and B 6¼ 0 (B may be infinite). EXAMPLE 1. ð5 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffi x4 1 p converges, since lim x!1þ ðx 1Þ1=2 1 ðx4 1Þ1=2 ¼ lim x!1þ ffiffiffiffiffiffiffiffiffiffiffiffiffi x 1 x4 1 r ¼ 1 2 . EXAMPLE 2. ð3 0 dx ð3 xÞ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p diverges, since lim x!3 ð3 xÞ 1 ð3 xÞ ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p ¼ 1 ffiffiffiffiffi 10 p . 3. Absolute and conditional convergence. ðb a f ðxÞ dx is called absolute convergent if ðb a j f ðxÞj dx converges. If ðb a f ðxÞ dx converges but ðb a j f ðxÞj dx diverges, then ðb a f ðxÞ dx is called condition- ally convergent. Theorem 5. If ðb a j f ðxÞj dx converges, then ðb a f ðxÞ dx converges. In words, an absolutely convergent integral converges. EXAMPLE. Since sin x ffiffiffiffiffiffiffiffiffiffiffiffi x 3 p @ 1 ffiffiffiffiffiffiffiffiffiffiffiffi x 3 p and ð4 dx ffiffiffiffiffiffiffiffiffiffiffiffi x 3 p converges ( p integral with a ¼ ; p ¼ 1 3), it follows that ð4 sin x ffiffiffiffiffiffiffiffiffiffiffiffi x 3 p dx converges and thus ð4 sin x ffiffiffiffiffiffiffiffiffiffiffiffi x 3 p dx converges (absolutely). Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence. 312 IMPROPER INTEGRALS [CHAP. 12
- 322. IMPROPER INTEGRALS OF THE THIRD KIND Improper integrals of the third kind can be expressed in terms of improper integrals of the first and second kinds, and hence the question of their convergence or divergence is answered by using results already established. IMPROPER INTEGRALS CONTAINING A PARAMETER, UNIFORM CONVERGENCE Let ð Þ ¼ ð1 a f ðx; Þ dx ð8Þ This integral is analogous to an infinite series of functions. In seeking conditions under which we may differentiate or integrate ð Þ with respect to , it is convenient to introduce the concept of uniform convergence for integrals by analogy with infinite series. We shall suppose that the integral (8) converges for 1 @ @ 2, or briefly ½ 1; 2. Definition. The integral (8) is said to be uniformly convergent in ½ 1; 2 if for each 0, we can find a number N depending on but not on , such that ð Þ ðu a f ðx; Þ dx for all u N and all in ½ 1; 2 This can be restated by nothing that ð Þ ðu a f ðx; Þ dx ¼ ð1 u f ðx; Þ dx , which is analogous in an infinite series to the absolute value of the remainder after N terms. The above definition and the properties of uniform convergence to be developed are formulated in terms of improper integrals of the first kind. However, analogous results can be given for improper integrals of the second and third kinds. SPECIAL TESTS FOR UNIFORM CONVERGENCE OF INTEGRALS 1. Weierstrass M test. If we can find a function MðxÞ A 0 such that (a) j f ðx; Þj @ MðxÞ 1 @ @ 2; x a (b) ð1 a MðxÞ dx converges, then ð1 a f ðx; Þ dx is uniformly and absolutely convergent in 1 @ @ 2. EXAMPLE. Since cos x x2 þ 1 @ 1 x2 þ 1 and ð1 0 dx x2 þ 1 converges, it follows that ð1 0 cos x x2 þ 1 dx is uniformly and absolutely convergent for all real values of . As in the case of infinite series, it is possible for integrals to be uniformly convergent without being absolutely convergent, and conversely. CHAP. 12] IMPROPER INTEGRALS 313
- 323. 2. Dirichlet’s test. Suppose that (a) ðxÞ is a positive monotonic decreasing function which approaches zero as x ! 1. (b) ðu a f ðx; Þ dx P for all u a and 1 @ @ 2. Then the integral ð1 a f ðx; Þ ðxÞ dx is uniformly convergent for 1 @ @ 2. THEOREMS ON UNIFORMLY CONVERGENT INTEGRALS Theorem 6. If f ðx; Þ is continuous for x A a and 1 @ @ 2, and if ð1 a f ðx; Þ dx is uniformly convergent for 1 @ @ 2, then ð Þ ¼ ð1 a f ðx; Þ dx is continous in 1 @ @ 2. In particular, if 0 is any point of 1 @ @ 2, we can write lim ! 0 ð Þ ¼ lim ! 0 ð1 a f ðx; Þ dx ¼ ð1 a lim ! 0 f ðx; Þ dx ð9Þ If 0 is one of the end points, we use right or left hand limits. Theorem 7. Under the conditions of Theorem 6, we can integrate ð Þ with respect to from 1 to 2 to obtain ð 2 1 ð Þ d ¼ ð 2 1 ð1 a f ðx; Þ dx d ¼ ð1 a ð 2 1 f ðx; Þ d dx ð10Þ which corresponds to a change of the order of integration. Theorem 8. If f ðx; Þ is continuous and has a continuous partial derivative with respect to for x A a and 1 @ @ 2, and if ð1 a @f @ dx converges uniformly in 1 @ @ 2, then if a does not depend on , d d ¼ ð1 a @f @ dx ð11Þ If a depends on , this result is easily modified (see Leibnitz’s rule, Page 186). EVALUATION OF DEFINITE INTEGRALS Evaluation of definite integrals which are improper can be achieved by a variety of techniques. One useful device consists of introducing an appropriately placed parameter in the integral and then differ- entiating or integrating with respect to the parameter, employing the above properties of uniform convergence. LAPLACE TRANSFORMS Operators that transform one set of objects into another are common in mathematics. The derivative and the indefinite integral both are examples. Logarithms provide an immediate arithmetic advantage by replacing multiplication, division, and powers, respectively, by the relatively simpler processes of addition, subtraction, and multiplication. After obtaining a result with logarithms an anti-logarithm procedure is necessary to find its image in the original framework. The Laplace trans- form has a role similar to that of logarithms but in the more sophisticated world of differential equations. (See Problems 12.34 and 12.36.) 314 IMPROPER INTEGRALS [CHAP. 12
- 324. The Laplace transform of a function FðxÞ is defined as f ðsÞ ¼ lfFðxÞg ¼ ð1 0 esx FðxÞ dx ð12Þ and is analogous to power series as seen by replacing es by t so that esx ¼ tx . Many properties of power series also apply to Laplace transforms. The adjacent short table of Laplace transforms is useful. In each case a is a real constant. LINEARITY The Laplace transform is a linear operator, i.e., fFðxÞ þ GðxÞg ¼ fFðxÞg þ fGðxÞg: This property is essential for returning to the solution after having calculated in the setting of the transforms. (See the following example and the previously cited problems.) CONVERGENCE The exponential est contributes to the convergence of the improper integral. What is required is that FðxÞ does not approach infinity too rapidly as x ! 1. This is formally stated as follows: If there is some constant a such that jFðxÞj eax for all sufficiently large values of x, then f ðsÞ ¼ ð1 0 esx FðxÞ dx converges when s a and f has derivatives of all orders. (The differentiations of f can occur under the integral sign .) APPLICATION The feature of the Laplace transform that (when combined with linearity) establishes it as a tool for solving differential equations is revealed by applying integration by parts to f ðsÞ ¼ ðx 0 est FðtÞ dt. By letting u ¼ FðtÞ and dv ¼ est dt, we obtain after letting x ! 1 ðx 0 est FðtÞ dt ¼ 1 s Fð0Þ þ 1 s ð1 0 est F 0 ðtÞ dt: Conditions must be satisfied that guarantee the convergence of the integrals (for example, est FðtÞ ! 0 as t ! 1). This result of integration by parts may be put in the form (a) fF 0 ðtÞg ¼ sfFðtÞg þ F 0 ð0Þ. Repetition of the procedure combined with a little algebra yields (b) fF 00 ðtÞg ¼ s2 fFðtÞg sFð0Þ F 0 ð0Þ. The Laplace representation of derivatives of the order needed can be obtained by repeating the process. To illustrate application, consider the differential equation d2 y dt2 þ 4y ¼ 3 sin t; where y ¼ FðtÞ and Fð0Þ ¼ 1, F 0 ð0Þ ¼ 0. We use CHAP. 12] IMPROPER INTEGRALS 315 FðxÞ lfFðxÞg a a 8 8 0 eax 1 8 a 8 a sin ax a 82 þ a2 8 0 cos ax 8 82 þ a2 8 0 xn n ¼ 1; 2; 3; . . . n! 8nþ1 8 0 Y 0 ðxÞ 8lfYðxÞg Yð0Þ Y 00 ðxÞ 82 lfYðxÞg 8Yð0Þ Y 0 ð0Þ
- 325. fsin atg ¼ a s2 þ a2 ; fcos atg ¼ s s2 þ a2 and recall that f ðsÞ ¼ fFðtÞgfF 00 ðtÞg þ 4fFðtÞg ¼ 3fsin tg Using (b) we obtain s2 f ðsÞ s þ 4f ðsÞ ¼ 3 s2 þ 1 : Solving for f ðsÞ yields f ðsÞ ¼ 3 ðs2 þ 4Þðs2 þ 1Þ þ s s2 þ 4 ¼ 1 s2 þ 1 1 s2 þ 4 þ s s2 þ 4 : (Partial fractions were employed.) Referring to the table of Laplace transforms, we see that this last expression may be written f ðsÞ ¼ fsin tg 1 2 fsin 2tg þ fcos 2tg then using the linearity of the Laplace transform f ðsÞ ¼ fsin t 1 2 sin 2t þ cos 2tg: We find that FðtÞ ¼ sin t 1 2 sin 2t þ cos 2t satisfies the differential equation. IMPROPER MULTIPLE INTEGRALS The definitions and results for improper single integrals can be extended to improper multiple integrals. Solved Problems IMPROPER INTEGRALS 12.1. Classify according to the type of improper integral. (a) ð1 1 dx ffiffiffi x 3 p ðx þ 1Þ (c) ð10 3 x dx ðx 2Þ2 (e) ð 0 1 cos x x2 dx (b) ð1 0 dx 1 þ tan x (d) ð1 1 x2 dx x4 þ x2 þ 1 (a) Second kind (integrand is unbounded at x ¼ 0 and x ¼ 1). (b) Third kind (integration limit is infinite and integrand is unbounded where tan x ¼ 1Þ. (c) This is a proper integral (integrand becomes unbounded at x ¼ 2, but this is outside the range of integration 3 @ x @ 10). (d) First kind (integration limits are infinite but integrand is bounded). 316 IMPROPER INTEGRALS [CHAP. 12
- 326. (e) This is a proper integral since lim x!0þ 1 cos x x2 ¼ 1 2 by applying L’Hospital’s rule . 12.2. Show how to transform the improper integral of the second kind, ð2 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xð2 xÞ p , into (a) an improper integral of the first kind, (b) a proper integral. (a) Consider ð2 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xð2 xÞ p where 0 1, say. Let 2 x ¼ 1 y . Then the integral becomes ð1= 1 dy y ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2y 1 p . As ! 0þ, we see that consideration of the given integral is equivalent to considera- tion of ð1 1 dy y ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2y 1 p , which is an improper integral of the first kind. (b) Letting 2 x ¼ v2 in the integral of (a), it becomes 2 ð1 ffiffi p dv ffiffiffiffiffiffiffiffiffiffiffiffiffi v2 þ 2 p . We are thus led to consideration of 2 ð1 0 dv ffiffiffiffiffiffiffiffiffiffiffiffiffi v2 þ 1 p , which is a proper integral. From the above we see that an improper integral of the first kind may be transformed into an improper integral of the second kind, and conversely (actually this can always be done). We also see that an improper integral may be transformed into a proper integral (this can only sometimes be done). IMPROPER INTEGRALS OF THE FIRST KIND 12.3. Prove the comparison test (Page 308) for convergence of improper integrals of the first kind. Since 0 @ f ðxÞ @ gðxÞ for x A a, we have using Property 7, Page 92, 0 @ ðb a f ðxÞ dx @ ðb a gðxÞ dx @ ð1 a gðxÞ dx But by hypothesis the last integral exists. Thus lim b!1 ðb a f ðxÞ dx exists, and hence ð1 a f ðxÞ dx converges 12.4. Prove the quotient test (a) on Page 309. By hypothesis, lim x!1 f ðxÞ gðxÞ ¼ A 0. Then given any 0, we can find N such that f ðxÞ gðxÞ A when x A N. Thus for x A N, we have A @ f ðxÞ gðxÞ @ A þ or ðA ÞgðxÞ @ f ðxÞ @ ðA þ ÞgðxÞ Then ðA Þ ðb N gðxÞ dx @ ðb N f ðxÞ dx @ ðA þ Þ ðb N gðxÞ dx ð1Þ There is no loss of generality in choosing A 0. If ð1 a gðxÞ dx converges, then by the inequality on the right of (1), lim b!1 ðb N f ðxÞ dx exists, and so ð1 a f ðxÞ dx converges If ð1 a gðxÞ dx diverges, then by the inequality on the left of (1), CHAP. 12] IMPROPER INTEGRALS 317
- 327. lim b!1 ðb N f ðxÞ dx ¼ 1 and so ð1 a f ðxÞ dx diverges For the cases where A ¼ 0 and A ¼ 1, see Problem 12.41. As seen in this and the preceding problem, there is in general a marked similarity between proofs for infinite series and improper integrals. 12.5. Test for convergence: (a) ð1 1 x dx 3x4 þ 5x2 þ 1 ; ðbÞ ð1 2 x2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x6 þ 16 p dx. (a) Method 1: For large x, the integrand is approximately x=3x4 ¼ 1=3x3 . Since x 3x4 þ 5x2 þ 1 @ 1 3x3 , and 1 3 ð1 1 dx x3 converges ( p integral with p ¼ 3), it follows by the comparison test that ð1 1 x dx 3x4 þ 5x2 þ 1 also converges. Note that the purpose of examining the integrand for large x is to obtain a suitable comparison integral. Method 2: Let f ðxÞ ¼ x 3x4 þ 5x2 þ 1 ; gðxÞ ¼ 1 x3 . Since lim x!1 f ðxÞ gðxÞ ¼ 1 3 , and ð1 1 gðxÞ dx converges, ð1 1 f ðxÞ dx also converges by the quotient test. Note that in the comparison function gðxÞ, we have discarded the factor 1 3. It could, however, just as well have been included. Method 3: lim x!1 x3 x 3x4 þ 5x2 þ 1 ¼ 1 3 . Hence, by Theorem 1, Page 309, the required integral converges. (b) Method 1: For large x, the integrand is approximately x2 = ffiffiffiffiffi x6 p ¼ 1=x. For x A 2, x2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi x6 þ 1 p A 1 2 1 x . Since 1 2 ð1 2 dx x diverges, ð1 2 x2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x6 þ 16 p dx also diverges. Method 2: Let f ðxÞ ¼ x2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x6 16 p , gðxÞ ¼ 1 x . Then since lim x!1 f ðxÞ gðxÞ ¼ 1, and ð1 2 gðxÞ dx diverges, ð1 2 f ðxÞ dx also diverges. Method 3: Since lim x!1 x x2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x6 þ 16 p ! ¼ 1, the required integral diverges by Theorem 1, Page 309. Note that Method 1 may (and often does) require one to obtain a suitable inequality factor (in this case 1 2, or any positive constant less than 1 2) before the comparison test can be applied. Methods 2 and 3, however, do not require this. 12.6. Prove that ð1 0 ex2 dx converges. lim x!1 x2 ex2 ¼ 0 (by L’Hospital’s rule or otherwise). Then by Theorem 1, with A ¼ 0; p ¼ 2 the given integral converges. Compare Problem 11.10(a), Chapter 11. 12.7. Examine for convergence: ðaÞ ð1 1 ln x x þ a dx; where a is a positive constant; ðbÞ ð1 0 1 cos x x2 dx: (a) lim x!1 x ln x x þ a ¼ 1. Hence by Theorem 1, Page 309, with A ¼ 1; p ¼ 1, the given integral diverges. 318 IMPROPER INTEGRALS [CHAP. 12
- 328. ðbÞ ð1 0 1 cos x x2 dx ¼ ð 0 1 cos x x2 dx þ ð1 1 cos x x2 dx The first integral on the right converges [see Problem 12.1(e)]. Since lim x!1 x3=2 1 cos x x2 ¼ 0, the second integral on the right converges by Theorem 1, Page 309, with A ¼ 0 and p ¼ 3=2. Thus, the given integral converges. 12.8. Test for convergence: (a) ð1 1 ex x dx; ðbÞ ð1 1 x3 þ x2 x6 þ 1 dx: (a) Let x ¼ y. Then the integral becomes ð1 1 ey y dy. Method 1: ey y @ ey for y @ 1. Then since ð1 1 ey dy converges, ð1 1 ey y dy converges; hence the given integral converges. Method 2: lim y!1 y2 ey y ¼ lim y!1 yey ¼ 0. Then the given integral converges by Theorem 1, Page 309, with A ¼ 0 and p ¼ 2. (b) Write the given integral as ð0 1 x3 þ x2 x6 þ 1 dx þ ð1 0 x3 þ x2 x6 þ 1 dx. Letting x ¼ y in the first integral, it becomes ð1 0 y3 y2 y6 þ 1 dy. Since lim y!1 y3 y3 y2 y6 þ 1 ! ¼ 1, this integral converges. Since lim x!1 x3 x3 þ x2 x6 þ 1 ! ¼ 1, the second integral converges. Thus the given integral converges. ABSOLUTE AND CONDITIONAL CONVERGENCE FOR IMPROPER INTEGRALS OF THE FIRST KIND 12.9. Prove that ð1 a f ðxÞ dx converges if ð1 a j f ðxÞj dx converges, i.e., an absolutely convergent integral is convergent. We have j f ðxÞj @ f ðxÞ @ j f ðxÞj, i.e., 0 @ f ðxÞ þ j f ðxÞj @ 2j f ðxÞj. Then 0 @ ðb a ½ f ðxÞ þ j f ðxÞj dx @ 2 ðb a j f ðxÞj dx If ð1 a j f ðxÞj dx converges, it follows that ð1 a ½ f ðxÞ þ j f ðxÞj dx converges. Hence, by subtracting ð1 a j f ðxÞj dx, which converges, we see that ð1 a f ðxÞ dx converges. 12.10. Prove that ð1 1 cos x x2 dx converges. Method 1: cos x x2 @ 1 x2 for x A 1. Then by the comparison test, since ð1 1 dx x2 converges, it follows that ð1 1 cos x x2 dx converges, i.e., ð1 1 cos x x2 dx converges absolutely, and so converges by Problem 12.9. CHAP. 12] IMPROPER INTEGRALS 319
- 329. Method 2: Since lim x!1 x3=2 cos x x2 ¼ lim x!1 cos x x1=2 ¼ 0, it follows from Theorem 1, Page 309, with A ¼ 0 and p ¼ 3=2, that ð1 1 cos x x2 dx converges, and hence ð1 1 cos x x2 dx converges (absolutely). 12.11. Prove that ð1 0 sin x x dx converges. Since ð1 0 sin x x dx converges because sin x x is continuous in 0 x @ 1 and lim x!0þ sin x x ¼ 1 we need only show that ð1 1 sin x x dx converges. Method 1: Integration by parts yields ðM 1 sin x x dx ¼ cos x x M 1 þ ðM 1 cos x x2 dx ¼ cos 1 cos M M þ ðM 1 cos x x2 dx ð1Þ or on taking the limit on both sides of (1) as M ! 1 and using the fact that lim M!1 cos M M ¼ 0, ð1 1 sin x x dx ¼ cos 1 þ ð1 1 cos x x2 dx ð2Þ Since the integral on the right of (2) converges by Problem 12.10, the required results follows. The technique of integration by parts to establish convergence is often useful in practice. Method 2: ð1 0 sin x x dx ¼ ð 0 sin x x dx þ ð2 sin x x dx þ þ ððnþ1Þ n sin x x dx þ ¼ X 1 n¼0 ððnþ1Þ n sin x x dx Letting x ¼ v þ n, the summation becomes X 1 n¼0 ð1Þn ð 0 sin v n þ n dv ¼ ð 0 sin v v dv ð 0 sin v v þ dv þ ð 0 sin v v þ 2 dv This is an alternating series. Since 1 v þ n @ 1 v þ ðn þ 1Þ and sin v A 0 in ½0; , it follows that ð 0 sin v v þ n dv @ ð 0 sin v v þ ðn þ 1Þ dv lim n!1 ð 0 sin v v þ n dv @ lim n!1 ð 0 dv n ¼ 0 Also, Thus, each term of the alternating series is in absolute value less than or equal to the preceding term, and the nth term approaches zero as n ! 1. Hence, by the alternating series test (Page 267) the series and thus the integral converges. 12.12. Prove that ð1 0 sin x x dx converges conditionally. Since by Problem 12.11 the given integral converges, we must show that it is not absolutely convergent, i.e., ð1 0 sin x x dx diverges. As in Problem 12.11, Method 2, we have 320 IMPROPER INTEGRALS [CHAP. 12
- 330. ð1 0 sin x x dx ¼ X 1 n¼0 ððnþ1Þ n sin x x dx ¼ X 1 n¼0 ð 0 sin v v þ n dv ð1Þ Now 1 v þ n A 1 ðn þ 1Þ for 0 @ v @ : Hence, ð 0 sin v v þ n dv A 1 ðn þ 1Þ ð 9 sin v dv ¼ 2 ðn þ 1Þ ð2Þ Since X 1 n¼0 2 ðn þ 1Þ diverges, the series on the right of (1) diverges by the comparison test. Hence, ð1 0 sin x x dx diverges and the required result follows. IMPROPER INTEGRALS OF THE SECOND KIND, CAUCHY PRINCIPAL VALUE 12.13. (a) Prove that ð7 1 dx ffiffiffiffiffiffiffiffiffiffiffi x þ 1 3 p converges and (b) find its value. The integrand is unbounded at x ¼ 1. Then we define the integral as lim !0þ ð7 1þ dx ffiffiffiffiffiffiffiffiffiffiffi x þ 1 3 p ¼ lim !0þ ðx þ 1Þ2=3 2=3 7 1þ ¼ lim !0þ 6 3 2 2=3 ¼ 6 This shows that the integral converges to 6. 12.14. Determine whether ð5 1 dx ðx 1Þ3 converges (a) in the usual sense, (b) in the Cauchy principal value sense. (a) By definition, ð5 1 dx ðx 1Þ3 ¼ lim 1!0þ ð11 1 dx ðx 1Þ3 þ lim 2!0þ ð5 1þ2 dx ðx 1Þ3 ¼ lim 1!0þ 1 8 1 22 1 þ lim 2!0þ 1 22 2 1 32 and since the limits do not exist, the integral does not converge in the usual sense. (b) Since lim !0þ ð1 1 dx ðx 1Þ3 þ ð5 1þ dx ðx 1Þ3 ¼ lim !0þ 1 8 1 22 þ 1 22 1 32 ¼ 3 32 the integral exists in the Cauchy principal value sense. The principal value is 3/32. 12.15. Investigate the convergence of: (a) ð3 2 dx x2 ðx3 8Þ2=3 (c) ð5 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 xÞðx 1Þ p (e) ð=2 0 dx ðcos xÞ1=n ; n 1: (b ð 0 sin x x3 dx (d) ð1 1 2sin1 x 1 x dx (a) lim x!2þ ðx 2Þ2=3 1 x2 ðx3 8Þ2=3 ¼ lim x!2þ 1 x2 1 x2 þ 2x þ 4 2=3 ¼ 1 8 ffiffiffiffiffi 18 3 p . Hence, the integral converges by Theorem 3(i), Page 312. CHAP. 12] IMPROPER INTEGRALS 321
- 331. (b) lim x!0þ x2 sin x x3 ¼ 1. Hence, the integral diverges by Theorem 3(ii) on Page 312. ðcÞ Write the integral as ð3 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 xÞðx 1Þ p þ ð5 3 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 xÞðx 1Þ p : Since lim x!1þ ðx 1Þ1=2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 xÞðx 1Þ p ¼ 1 2 , the first integral converges. Since lim x!5 ð5 xÞ1=2 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð5 xÞðx 1Þ p ¼ 1 2 , the second integral converges. Thus, the given integral converges. (d) lim x!1 ð1 xÞ 2sin1 x 1 x ¼ 2=2 . Hence, the integral diverges. Another method: 2sin1 x 1 x A 2=2 1 x , and ð1 1 dx 1 x diverges. Hence, the given integral diverges. ðeÞ lim x!1=2 ð=2 xÞ1=n 1 ðcos xÞ1=n ¼ lim x!1=2 =2 x cos x 1=n ¼ 1: Hence the integral converges. 12.16. If m and n are real numbers, prove that ð1 0 xm1 ð1 xÞn1 dx (a) converges if m 0 and n 0 simultaneously and (b) diverges otherwise. (a) For m A 1 and n A 1 simultaneously, the integral converges, since the integrand is continuous in 0 @ x @ 1. Write the integral as ð1=2 0 xm1 ð1 xÞn1 dx þ ð1 1=2 xm1 ð1 xÞn1 dx ð1Þ If 0 m 1 and 0 n 1, the first integral converges, since lim x!0þ x1m xm1 ð1 xÞn1 ¼ 1, using Theorem 3(i), Page 312, with p ¼ 1 m and a ¼ 0. Similarly, the second integral converges since lim x!1 ð1 xÞ1n xm1 ð1 xÞn1 ¼ 1, using Theorem 4(i), Page 312, with p ¼ 1 n and b ¼ 1. Thus, the given integral converges if m 0 and n 0 simultaneously. (b) If m @ 0, lim x!0þ x xm1 ð1 xÞn1 ¼ 1. Hence, the first integral in (1) diverges, regardless of the value of n, by Theorem 3(ii), Page 312, with p ¼ 1 and a ¼ 0. Similarly, the second integral diverges if n @ 0 regardless of the value of m, and the required result follows. Some interesting properties of the given integral, called the beta integral or beta function, are considered in Chapter 15. 12.17. Prove that ð 0 1 x sin 1 x dx converges conditionally. Letting x ¼ 1=y, the integral becomes ð1 1= sin y y dy and the required result follows from Problem 12.12. IMPROPER INTEGRALS OF THE THIRD KIND 12.18. If n is a real number, prove that ð1 0 xn1 ex dx (a) converges if n 0 and (b) diverges if n @ 0. 322 IMPROPER INTEGRALS [CHAP. 12
- 332. Write the integral as ð1 0 xn1 ex dx þ ð1 1 xn1 ex dx ð1Þ (a) If n A 1, the first integral in (1) converges since the integrand is continuous in 0 @ x @ 1. If 0 n 1, the first integral in (1) is an improper integral of the second kind at x ¼ 0. Since lim x!0þ x1n xn1 ex ¼ 1, the integral converges by Theorem 3(i), Page 312, with p ¼ 1 n and a ¼ 0. Thus, the first integral converges for n 0. If n 0, the second integral in (1) is an improper integral of the first kind. Since lim x!1 x2 xn1 ex ¼ 0 (by L’Hospital’s rule or otherwise), this integral converges by Theorem 1ðiÞ, Page 309, with p ¼ 2. Thus, the second integral also converges for n 0, and so the given integral converges for n 0. (b) If n @ 0, the first integral of (1) diverges since lim x!0þ x xn1 ex ¼ 1 [Theorem 3(ii), Page 312]. If n @ 0, the second integral of (1) converges since lim x!1 x xn1 ex ¼ 0 [Theorem 1(i), Page 309]. Since the first integral in (1) diverges while the second integral converges, their sum also diverges, i.e., the given integral diverges if n @ 0. Some interesting properties of the given integral, called the gamma function, are considered in Chapter 15. UNIFORM CONVERGENCE OF IMPROPER INTEGRALS 12.19. (a) Evaluate ð Þ ¼ ð1 0 e x dx for 0. (b) Prove that the integral in (a) converges uniformly to 1 for A 1 0. (c) Explain why the integral does not converge uniformly to 1 for 0. ðaÞ ð Þ ¼ lim b!1 ðb a e x dx ¼ lim b!1 e x b x¼0 ¼ lim b!1 1 e b ¼ 1 if 0 . Thus, the integral converges to 1 for all 0. (b) Method 1, using definition: The integral converges uniformly to 1 in A 1 0 if for each 0 we can find N, depending on but not on , such that 1 ðu 0 e x dx for all u N. Since 1 ðu 0 e x dx ¼ j1 ð1 e u Þj ¼ e u @ e 1u for u 1 1 ln 1 ¼ N, the result fol- lows. Method 2, using the Weierstrass M test: Since lim x!1 x2 e x ¼ 0 for A 1 0, we can choose j e x j 1 x2 for sufficiently large x, say x A x0. Taking MðxÞ ¼ 1 x2 and noting that ð1 x0 dx x2 converges, it follows that the given integral is uniformly convergent to 1 for A 1 0. (c) As 1 ! 0, the number N in the first method of (b) increases without limit, so that the integral cannot be uniformly convergent for 0. 12.20. If ð Þ ¼ ð1 0 f ðx; Þ dx is uniformly convergent for 1 @ @ 2, prove that ð Þ is continuous in this interval. CHAP. 12] IMPROPER INTEGRALS 323
- 333. Let ð Þ ¼ ðu a f ðx; Þ dx þ Rðu; Þ; where Rðu; Þ ¼ ð1 u f ðx; Þ dx: Then ð þ hÞ ¼ ðu a f ðx; þ hÞ dx þ Rðu; þ hÞ and so ð þ hÞ ð Þ ¼ ðu a f f ðx; þ hÞ f ðx; Þg dx þ Rðu; þ hÞ Rðu; Þ Thus jð þ hÞ ð Þj @ ðu a j f ðx; þ hÞ f ðx; Þjdx þ jRðu; þ hÞj þ jRðu; Þj ð1Þ Since the integral is uniformly convergent in 1 @ @ 2, we can, for each 0, find N independent of such that for u N, jRðu; þ hÞj =3; jRðu; Þj =3 ð2Þ Since f ðx; Þ is continuous, we can find 0 corresponding to each 0 such that ðu a j f ðx; þ hÞ f ðx; Þj dx =3 for jhj ð3Þ Using (2) and (3) in (1), we see that jð þ hÞ ð Þj for jhj , so that ð Þ is continuous. Note that in this proof we assume that and þ h are both in the interval 1 @ @ 2. Thus, if ¼ 1, for example, h 0 and right-hand continuity is assumed. Also note the analogy of this proof with that for infinite series. Other properties of uniformly convergent integrals can be proved similarly. 12.21. (a) Show that lim !0þ ð1 0 e x dx 6¼ ð1 0 lim !0þ e x dx: ðbÞ Explain the result in (a). ðaÞ lim !0þ ð1 0 e x dx ¼ lim !0þ ¼ 1 by Problem 12.19ðaÞ: ð1 0 lim !0þ e x dx ¼ ð1 0 0 dx ¼ 0. Thus the required result follows. (b) Since ð Þ ¼ ð1 0 eax dx is not uniformly convergent for A 0 (see Problem 12.19), there is no guarantee that ð Þ will be continuous for A 0. Thus lim !0þ ð Þ may not be equal to ð0Þ. 12.22. (a) Prove that ð1 0 e x cos rx dx ¼ 2 þ r2 for 0 and any real value of r. (b) Prove that the integral in (a) converges uniformly and absolutely for a @ @ b, where 0 a b and any r. (a) From integration formula 34, Page 96, we have lim M!1 ðM 0 e x cos rx dx ¼ lim M!1 e x ðr sin rx cos rxÞ 2 þ r2 M 0 ¼ 2 þ r2 (b) This follows at once from the Weierstrass M test for integrals, by noting that je x cos rxj @ e x and ð1 0 e x dx converges. 324 IMPROPER INTEGRALS [CHAP. 12
- 334. EVALUATION OF DEFINITE INTEGRALS 12.23. Prove that ð=2 0 ln sin x dx ¼ 2 ln 2. The given integral converges [Problem 12.42( f )]. Letting x ¼ =2 y, I ¼ ð=2 0 ln sin x dx ¼ ð=2 0 ln cos y dy ¼ ð=2 0 ln cos x dx Then 2I ¼ ð=2 0 ðln sin x þ ln cos xÞ dx ¼ ð=2 0 ln sin 2x 2 dx ¼ ð=2 0 ln sin 2x dx ð=2 0 ln 2 dx ¼ ð=2 0 ln sin 2x dx 2 ln 2 ð1Þ Letting 2x ¼ v, ð=2 0 ln sin 2x dx ¼ 1 2 ð 0 ln sin v dv ¼ 1 2 ð=2 0 ln sin v dv þ ð =2 ln sin v dv ¼ 1 2 ðI þ IÞ ¼ I (letting v ¼ u in the last integral) Hence, (1) becomes 2I ¼ I 2 ln 2 or I ¼ 2 ln 2. 12.24. Prove that ð 0 x ln sin x dx ¼ 2 2 ln 2. Let x ¼ y. Then, using the results in the preceding problem, J ¼ ð 0 x ln sin x dx ¼ ð 0 ð uÞ ln sin u du ¼ ð 0 ð xÞ ln sin x dx ¼ ð 0 ln sin x dx ð 0 x ln sin x dx ¼ 2 ln 2 J or J ¼ 2 2 ln 2: 12.25. (a) Prove that ð Þ ¼ ð1 0 dx x2 þ is uniformly convergent for A 1. ðbÞ Show that ð Þ ¼ 2 ffiffiffi p . ðcÞ Evaluate ð1 0 dx ðx2 þ 1Þ2 : ðdÞ Prove that ð1 0 dx ðx2 þ 1Þnþ1 ¼ ð=2 0 cos2n d ¼ 1 3 5 ð2n 1Þ 2 4 6 ð2nÞ 2 : (a) The result follows from the Weierestrass test, since 1 x2 þ @ 1 x2 þ 1 for a A 1 and ð1 0 dx x2 þ 1 converges. ðbÞ ð Þ ¼ lim b!1 ðb 0 dx x2 þ ¼ lim b!1 1 ffiffiffi p tan1 x ffiffiffi p b 0 ¼ lim b!1 1 ffiffiffi p tan1 b ffiffiffi p ¼ 2 ffiffiffi p : CHAP. 12] IMPROPER INTEGRALS 325
- 335. (c) From (b), ð1 0 dx x2 þ ¼ 2 ffiffiffi p . Differentiating both sides with respect to , we have ð1 0 @ @ 1 x2 þ dx ¼ ð1 0 dx ðx2 þ Þ2 ¼ 4 3=2 the result being justified by Theorem 8, Page 314, since ð1 0 dx ðx2 þ Þ2 is uniformly convergent for A 1 because 1 ðx2 þ Þ2 @ 1 ðx2 þ 1Þ2 and ð1 0 dx ðx2 þ 1Þ2 converges . Taking the limit as ! 1þ, using Theorem 6, Page 314, we find ð1 0 dx ðx2 þ 1Þ2 ¼ 4 . (d) Differentiating both sides of ð1 0 dx x2 þ ¼ 2 1=2 n times, we find ð1Þð2Þ ðnÞ ð1 0 dx ðx2 þ Þnþ1 ¼ 1 2 3 2 5 2 2n 1 2 2 ð2nþ1Þ=2 where justification proceeds as in part (c). Letting ! 1þ, we find ð1 0 dx ðx2 þ 1Þnþ1 ¼ 1 3 5 ð2n 1Þ 2n n! 2 ¼ 1 3 5 ð2n 1Þ 2 4 6 ð2nÞ 2 Substituting x ¼ tan , the integral becomes ð=2 0 cos2n d and the required result is obtained. 12.26. Prove that ð1 0 eax ebx x sec rx dx ¼ 1 2 ln b2 þ r2 a2 þ r2 where a; b 0. From Problem 12.22 and Theorem 7, Page 314, we have ð1 x¼0 ðb ¼a e x cos rx d dx ¼ ðb ¼a ð1 x¼0 e x cos rx dx d or ð1 x¼0 e x cos rx x b ¼a dx ¼ ðb ¼a 2 þ r2 d ð1 0 eax ebx x sec rx dx ¼ 1 2 ln b2 þ r2 a2 þ r2 i.e., 12.27. Prove that ð1 0 e x 1 cos x x2 dx ¼ tan1 1 2 lnð 2 þ 1Þ, 0. By Problem 12.22 and Theorem 7, Page 314, we have ðr 0 ð1 0 e x cos rx dx dr ¼ ð1 0 ðr 0 e x cos rx dr dx ð1 0 e x sin rx x dx ¼ ðr 0 a 2 þ r2 ¼ tan1 r or Integrating again with respect to r from 0 to r yields ð1 0 e x 1 cos rx x2 dx ¼ ðr 0 tan1 r dr ¼ r tan1 r 2 lnð 2 þ r2 Þ using integration by parts. The required result follows on letting r ¼ 1. 326 IMPROPER INTEGRALS [CHAP. 12
- 336. 12.28. Prove that ð1 0 1 cos x x2 dx ¼ 2 . Since e x 1 cos x x2 @ 1 cos x x2 for A 0; x A 0 and ð1 0 1 cos x x2 dx converges [see Problem 12.7(b)], it follows by the Weierstrass test that ð1 0 e x 1 cos x x2 dx is uniformly convergent and represents a continuous function of for A 0 (Theorem 6, Page 314). Then letting ! 0þ, using Problem 12.27, we have lim !0þ ð1 0 e x 1 cos x x2 dx ¼ ð1 0 1 cos x x2 dx ¼ lim !0 tan1 1 2 lnð 2 þ 1Þ ¼ 2 12.29. Prove that ð1 0 sin x x ¼ ð1 0 sin2 x x2 dx ¼ 2 . Integrating by parts, we have ðM 1 cos x x2 dx ¼ 1 x ð1 cos xÞ M þ ðM sin x x dx ¼ 1 cos 1 cos M M þ ðM sin x x dx Taking the limit as ! 0þ and M ! 1 shows that ð1 0 sin x x dx ¼ ð1 0 1 cos x x dx ¼ 2 Since ð1 0 1 cos x x2 dx ¼ 2 ð1 0 sin2 ðx=2Þ x2 dx ¼ ð1 0 sin2 u u2 du on letting u ¼ x=2, we also have ð1 0 sin2 x x2 dx ¼ 2 . 12.30. Prove that ð1 0 sin3 x x dx ¼ 4 . sin3 x ¼ eix eix 2i 2 ¼ ðeix Þ3 3ðeix Þ2 ðeix Þ þ 3ðeix Þðeix Þ2 ðeix Þ3 ð2iÞ3 ¼ 1 4 e3ix e3ix 2i ! þ 3 4 eix eix 2i ¼ 1 4 sin 3x þ 3 4 sin x Then ð1 0 sin3 x x dx ¼ 3 4 ð1 0 sin x x dx 1 4 ð1 0 sin 3x x dx ¼ 3 4 ð1 0 sin x x dx 1 4 ð1 0 sin u u du ¼ 3 4 2 1 4 2 ¼ 4 MISCELLANEOUS PROBLEMS 12.31. Prove that ð1 0 ex2 dx ¼ ffiffiffi p =2. By Problem 12.6, the integral converges. Let IM ¼ ðM 0 ex2 dx ¼ ðM 0 ey2 dy and let lim M!1 IM ¼ I, the required value of the integral. Then CHAP. 12] IMPROPER INTEGRALS 327
- 337. I2 M ¼ ðM 0 ex2 dx ðM 0 ey2 dy ¼ ðM 0 ðM 0 eðx2 þy2 Þ dx dy ¼ ð ð rM eðx2 þy2 Þ dx dy where rM is the square OACE of side M (see Fig. 12-3). Since integrand is positive, we have ð ð r1 eðx2 þy2 Þ dx dy @ I2 M @ ð ð r2 eðx2 þy2 Þ dx dy ð1Þ where r1 and r2 are the regions in the first quadrant bounded by the circles having radii M and M ffiffiffi 2 p , respectively. Using polar coordinates, we have from (1), ð=2 ¼0 ðM ¼0 e2 d d @ I2 M @ ð=2 ¼0 ðM ffiffi 2 p ¼0 e2 d d ð2Þ or 4 ð1 eM2 Þ @ I2 M @ 4 ð1 e2M2 Þ ð3Þ Then taking the limit as M ! 1 in (3), we find lim M!1 I2 M ¼ I2 ¼ =4 and I ¼ ffiffiffi p =2. 12.32. Evaluate ð1 0 ex2 cos x dx. Let Ið Þ ¼ ð1 0 ex2 cos x dx. Then using integration by parts and appropriate limiting procedures, dI d ¼ ð1 0 xex2 sin x dx ¼ 1 2 ex2 sin xj1 0 1 2 ð1 0 ex2 cos x dx ¼ 2 I The differentiation under the integral sign is justified by Theorem 8, Page 314, and the fact that ð1 0 xex2 sin x dx is uniformly convergent for all (since by the Weierstrass test, jxex2 sin xj @ xex2 and ð1 0 xex2 dx converges). From Problem 12.31 and the uniform convergence, and thus continuity, of the given integral (since jex2 cos xj @ ex2 and ð1 0 ex2 dx converges, so that that Weierstrass test applies), we have Ið0Þ ¼ lim !0 Ið Þ ¼ 1 2 ffiffiffi p . Solving dI d ¼ 2 I subject to Ið0Þ ¼ ffiffiffi p 2 , we find Ið Þ ¼ ffiffiffi p 2 e 2 =4 . 12.33. (a) Prove that Ið Þ ¼ ð1 0 eðx =xÞ2 dx ¼ ffiffiffi p 2 . (b) Evaluate ð1 0 eðx2 þx2 Þ dx. (a) We have I 0 ð Þ ¼ 2 ð1 0 eðx =xÞ2 ð1 =x2 Þ dx. The differentiation is proved valid by observing that the integrand remains bounded as x ! 0 þ and that for sufficiently large x, 328 IMPROPER INTEGRALS [CHAP. 12 y D E C B A O M M√2 x Fig. 12-3
- 338. eðx =xÞ2 ð1 =x2 Þ ¼ ex2 þ2 2 =x2 ð1 =x2 Þ @ e2 ex2 so that I 0 ð Þ converges uniformly for A 0 by the Weierstrass test, since ð1 0 ex2 dx converges. Now I 0 ð Þ ¼ 2 ð1 0 eðx =xÞ2 dx 2 ð1 0 eðx =xÞ2 x2 dx ¼ 0 as seen by letting =x ¼ y in the second integral. Thus Ið Þ ¼ c, a constant. To determine c, let ! 0þ in the required integral and use Problem 12.31 to obtain c ¼ ffiffiffi p =2. (b) From (a), ð1 0 eðx =xÞ2 dx ¼ ð1 0 eðx2 2 þ 2 x2 Þ dx ¼ e2 ð1 0 eðx2 þ 2 x2 Þ dx ¼ ffiffiffi p 2 . Then ð1 0 eðx2 þ 2 x2 Þ dx ¼ ffiffiffi p 2 e2 : Putting ¼ 1; ð1 0 eðx2 þx2 Þ dx ¼ ffiffiffi p 2 e2 : 12.34. Verify the results: (a) lfeax g ¼ 1 s a ; s a; ðbÞ lfcos axg ¼ s s2 þ a2 ; s 0. lfeax g ¼ ð1 0 esx eax dx ¼ lim M!1 ðM 0 eðsaÞx dx ðaÞ ¼ lim M!1 1 eðsaÞM s a ¼ 1 s a if s a ðbÞ lfcos axg ¼ ð1 0 esx cos ax dx ¼ s s2 þ a2 by Problem 12.22 with ¼ s; r ¼ a: Another method, using complex numbers. From part (a), lfeax g ¼ 1 s a . Replace a by ai. Then lfeaix g ¼ lfcos ax þ i sin axg ¼ lfcos axg þ ilfsin axg ¼ 1 s ai ¼ s þ ai s2 þ a2 ¼ s s2 þ a2 þ i a s2 þ a2 Equating real and imaginary parts: lfcos axg ¼ s s2 þ a2 , lfsin axg ¼ a s2 þ a2 . The above formal method can be justified using methods of Chapter 16. 12.35. Prove that (a) lfY 0 ðxÞg ¼ slfYðxÞg Yð0Þ; ðbÞ lfY 00 ðxÞg ¼ s2 lfYðxÞg sYð0Þ Y 0 ð0Þ under suitable conditions on YðxÞ. (a) By definition (and with the aid of integration by parts) lfY 0 ðxÞg ¼ ð1 0 esx Y 0 ðxÞ dx ¼ lim M!0 ðM 0 esx Y 0 ðxÞ dx ¼ lim M!1 esx YðxÞ M 0 þ s ðM 0 esx YðxÞ dx ( ) ¼ s ð1 0 esx YðxÞ dx Yð0Þ ¼ slfYðxÞg Yð0Þ assuming that s is such that lim M!1 esM YðMÞ ¼ 0. (b) Let UðxÞ ¼ Y 0 ðxÞ. Then by part (a), lfU 0 ðxÞg ¼ slfUðxÞg Uð0Þ. Thus lfY 00 ðxÞg ¼ slfY 0 ðxÞg Y 0 ð0Þ ¼ s½slfYðxÞg Yð0Þ Y 0 ð0Þ ¼ s2 lfYðxÞg sYð0Þ Y 0 ð0Þ CHAP. 12] IMPROPER INTEGRALS 329
- 339. 12.36. Solve the differential equation Y 00 ðxÞ þ YðxÞ ¼ x; Yð0Þ ¼ 0; Y 0 ð0Þ ¼ 2. Take the Laplace transform of both sides of the given differential equation. Then by Problem 12.35, lfY 00 ðxÞ þ YðxÞg ¼ lfxg; lfY 00 ðxÞg þ lfYðxÞg ¼ 1=s2 s2 lfYðxÞg sYð0Þ Y 0 ð0Þ þ lfYðxÞg ¼ 1=s2 and so Solving for lfYðxÞg using the given conditions, we find lfYðxÞg ¼ 2s2 s2ðs2 þ 1Þ ¼ 1 s2 þ 1 s2 þ 1 ð1Þ by methods of partial fractions. Since 1 s2 ¼ lfxg and 1 s2 þ 1 ¼ lfsin xg; it follows that 1 s2 þ 1 s2 þ 1 ¼ lfx þ sin xg: Hence, from (1), lfYðxÞg ¼ lfx þ sin xg, from which we can conclude that YðxÞ ¼ x þ sin x which is, in fact, found to be a solution. Another method: If lfFðxÞg ¼ f ðsÞ, we call f ðsÞ the inverse Laplace transform of FðxÞ and write f ðsÞ ¼ l1 fFðxÞg. By Problem 12.78, l1 f f ðsÞ þ gðsÞg ¼ l1 f f ðsÞg þ l1 fgðsÞg. Then from (1), YðxÞ ¼ l1 1 s2 þ 1 s2 þ 1 ¼ l1 1 s2 þ l1 1 s2 þ 1 ¼ x þ sin x Inverse Laplace transforms can be read from the table on Page 315. Supplementary Problems IMPROPER INTEGRALS OF THE FIRST KIND 12.37. Test for convergence: ðaÞ ð1 0 x2 þ 1 x4 þ 1 dx ðdÞ ð1 1 dx x4 þ 4 ðgÞ ð1 1 x2 dx ðx2 þ x þ 1Þ5=2 ðbÞ ð1 2 x dx ffiffiffiffiffiffiffiffiffiffiffiffiffi x3 1 p ðeÞ ð1 1 2 þ sin x x2 þ 1 dx ðhÞ ð1 1 ln x dx x þ ex ðcÞ ð1 1 dx x ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3x þ 2 p ð f Þ ð1 2 x dx ðln xÞ3 ðiÞ ð1 0 sin2 x x2 dx Ans. (a) conv., (b) div., (c) conv., (d) conv., (e) conv., ( f ) div., (g) conv., (h) div., (i) conv. 12.38. Prove that ð1 1 dx x2 þ 2ax þ b2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2 a2 p if b jaj. 12.39. Test for convergence: (a) ð1 1 ex ln x dx; ðbÞ ð1 0 ex lnð1 þ ex Þ dx; ðcÞ ð1 0 ex cosh x2 dx. Ans. (a) conv., (b) conv., (c) div. 330 IMPROPER INTEGRALS [CHAP. 12
- 340. 12.40. Test for convergence, indicating absolute or conditional convergence where possible: (a) ð1 0 sin 2x x3 þ 1 dx; (b) ð1 1 eax2 cos bx dx, where a; b are positive constants; (c) ð1 0 cos x ffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 1 p dx; ðdÞ ð1 0 x sin x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ a2 p dx; (e) ð1 0 cos x cosh x dx. Ans. (a) abs. conv., (b) abs. conv., (c) cond. conv., (d) div., (e) abs. conv. 12.41. Prove the quotient tests (b) and (c) on Page 309. IMPROPER INTEGRALS OF THE SECOND KIND 12.42. Test for convergence: ðaÞ ð1 0 dx ðx þ 1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 p ðdÞ ð2 1 ln x ffiffiffiffiffiffiffiffiffiffiffiffiffi 8 x3 3 p dx ðgÞ ð3 0 x2 ð3 xÞ2 dx ð jÞ ð1 0 dx xx ðbÞ ð1 0 cos x x2 dx ðeÞ ð1 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnð1=xÞ p ðhÞ ð=2 0 ex cos x x dx ðcÞ ð1 1 etan1 x x dx ð f Þ ð=2 0 ln sin x dx ðiÞ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2x2 1 x2 s dx; jkj 1 Ans. (a) conv., (b) div., (c) div., (d) conv., (e) conv., ( f Þ conv., (g) div., (h) div., (i) conv., ( jÞ conv. 12.43. (a) Prove that ð5 0 dx 4 x diverges in the usual sense but converges in the Cauchy principal value senses. (b) Find the Cauchy principal value of the integral in (a) and give a geometric interpretation. Ans. (b) ln 4 12.44. Test for convergence, indicating absolute or conditional convergence where possible: ðaÞ ð1 0 cos 1 x dx; ðbÞ ð1 0 1 x cos 1 x dx; ðcÞ ð1 0 1 x2 cos 1 x dx: Ans. (a) abs. conv., (b) cond. conv., (c) div. 12.45. Prove that ð4 0 3x2 sin 1 x x cos 1 x dx ¼ 32 ffiffiffi 2 p 3 . IMPROPER INTEGRALS OF THE THIRD KIND 12.46. Test for convergence: (a) ð1 0 ex ln x dx; ðbÞ ð1 0 ex dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x lnðx þ 1Þ p ; ðcÞ ð1 0 ex dx ffiffiffi x 3 p ð3 þ 2 sin xÞ . Ans. (a) conv., (b) div., (c) conv. 12.47. Test for convergence: (a) ð1 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x4 þ x2 3 p ; ðbÞ ð1 0 ex dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sinh ðaxÞ p ; a 0. Ans. (a) conv., (b) conv. if a 2, div. if 0 a @ 2. 12.48. Prove that ð1 0 sinh ðaxÞ sinh ðxÞ dx converges if 0 @ jaj and diverges if jaj @ . 12.49. Test for convergence, indicating absolute or conditional convergence where possible: CHAP. 12] IMPROPER INTEGRALS 331
- 341. ðaÞ ð1 0 sin x ffiffiffi x p dx; ðbÞ ð1 0 sin ffiffiffi x p sinh ffiffiffi x p dx: Ans: ðaÞ cond. conv., ðbÞ abs. conv. UNIFORM CONVERGENCE OF IMPROPER INTEGRALS 12.50. (a) Prove that ð Þ ¼ ð1 0 cos x 1 þ x2 dx is uniformly convergent for all . (b) Prove that ð Þ is continuous for all . (c) Find lim !0 ð Þ: Ans. (c) =2: 12.51. Let ð Þ ¼ ð1 0 Fðx; Þ dx, where Fðx; Þ ¼ 2 xe x2 . (a) Show that ð Þ is not continuous at ¼ 0, i.e., lim !0 ð1 0 Fðx; Þ dx 6¼ ð1 0 lim !0 Fðx; Þ dx. (b) Explain the result in (a). 12.52. Work Problem 12.51 if Fðx; Þ ¼ 2 xe x . 12.53. If FðxÞ is bounded and continuous for 1 x 1 and Vðx; yÞ ¼ 1 ð1 1 yFðÞ d y2 þ ð xÞ2 prove that lim y!0 Vðx; yÞ ¼ FðxÞ. 12.54. Prove (a) Theorem 7 and (b) Theorem 8 on Page 314. 12.55. Prove the Weierstrass M test for uniform convergence of integrals. 12.56. Prove that if ð1 0 FðxÞ dx converges, then ð1 0 e x FðxÞ dx converges uniformly for A 0. 12.57. Prove that ðaÞ ðaÞ ¼ ð1 0 eax sin x x dx converges uniformly for a A 0, ðbÞ ðaÞ ¼ 2 tan1 a, (c) ð1 0 sin x x dx ¼ 2 (compare Problems 12.27 through 12.29). 12.58. State the definition of uniform convergence for improper integrals of the second kind. 12.59. State and prove a theorem corresponding to Theorem 8, Page 314, if a is a differentiable function of . EVALUATION OF DEFINITE INTEGRALS Establish each of the following results. Justify all steps in each case. 12.60. ð1 0 eax ebx x dx ¼ lnðb=aÞ; a; b 0 12.61. ð1 0 eax ebx x csc rx dx ¼ tan1 ðb=rÞ tan1 ða=rÞ; a; b; r 0 12.62. ð1 0 sin rx xð1 þ x2 Þ dx ¼ 2 ð1 er Þ; r A 0 12.63. ð1 0 1 cos rx x2 dx ¼ 2 jrj 12.64. ð1 0 x sin rx a2 þ x2 dx ¼ 2 ear ; a; r A 0 332 IMPROPER INTEGRALS [CHAP. 12
- 342. 12.65. (a) Prove that ð1 0 e x cos ax cos bx x dx ¼ 1 2 ln 2 þ b2 2 þ a2 ! ; A 0. (b) Use (a) to prove that ð1 0 cos ax cos bx x dx ¼ ln b a . The results of (b) and Problem 12.60 are special cases of Frullani’s integral, ð1 0 FðaxÞ FðbxÞ x dx ¼ Fð0Þ ln b a , where FðtÞ is continuous for t 0, F 0 ð0Þ exists and ð1 1 FðtÞ t dt converges. 12.66. Given ð1 0 e x2 dx ¼ 1 2 ffiffiffiffiffiffiffiffi = p , 0. Prove that for p ¼ 1; 2; 3; . . ., ð1 0 x2p e x2 dx ¼ 1 2 3 2 5 2 ð2p 1Þ 2 ffiffiffi p 2 ð2pþ1Þ=2 12.67. If a 0; b 0, prove that ð1 0 ðea=x2 eb=x2 Þ dx ¼ ffiffiffiffiffiffi b p ffiffiffiffiffiffi a p . 12.68. Prove that ð1 0 tan1 ðx=aÞ tan1 ðx=bÞ x dx ¼ 2 ln b a where a 0; b 0. 12.69. Prove that ð1 1 dx ðx2 þ x þ 1Þ3 ¼ 4 3 ffiffiffi 3 p . [Hint: Use Problem 12.38.] MISCELLANEOUS PROBLEMS 12.70. Prove that ð1 0 lnð1 þ xÞ x 2 dx converges. 12.71. Prove that ð1 0 dx 1 þ x3 sin2 x converges. Hint: Consider X 1 n¼0 ððnþ1Þ n dx 1 þ x3 sin2 x and use the fact that ððnþ1Þ n dx 1 þ x3 sin2 x @ ððnþ1Þ n dx 1 þ ðnÞ3 sin2 x : 12.72. Prove that ð1 0 x dx 1 þ x3 sin2 x diverges. 12.73. (a) Prove that ð1 0 lnð1 þ 2 x2 Þ 1 þ x2 dx ¼ lnð1 þ Þ; A 0. (b) Use (a) to show that ð=2 0 ln sin d ¼ 2 ln 2: 12.74. Prove that ð1 0 sin4 x x4 dx ¼ 3 . 12.75. Evaluate (a) lf1= ffiffiffi x p g; ðbÞ lfcosh axg; ðcÞ lfðsin xÞ=xg. Ans: ðaÞ ffiffiffiffiffiffiffi =s p ; s 0 ðbÞ s s2 a2 ; s jaj ðcÞ tan1 1 s ; s 0: 12.76. (a) If lfFðxÞg ¼ f ðsÞ, prove that lfeax FðxÞg ¼ f ðs aÞ; ðbÞ Evaluate lfeax sin bxg. Ans: ðbÞ b ðs aÞ2 þ b2 ; s a CHAP. 12] IMPROPER INTEGRALS 333
- 343. 12.77. (a) If lfFðxÞg ¼ f ðsÞ, prove that lfxn FðxÞg ¼ ð1Þn f ðnÞ ðsÞ, giving suitable restrictions on FðxÞ. (b) Evaluate lfx cos xg. Ans: ðbÞ s2 1 ðs2 þ 1Þ2 ; s 0 12.78. Prove that l1 ff ðsÞ þ gðsÞg ¼ l1 f f ðsÞg þ l1 fgðsÞg, stating any restrictions. 12.79. Solve using Laplace transforms, the following differential equations subject to the given conditions. (a) Y 00 ðxÞ þ 3Y 0 ðxÞ þ 2YðxÞ ¼ 0; Yð0Þ ¼ 3; Y 0 ð0Þ ¼ 0 (b) Y 00 ðxÞ Y 0 ðxÞ ¼ x; Yð0Þ ¼ 2; Y 0 ð0Þ ¼ 3 (c) Y 00 ðxÞ þ 2Y 0 ðxÞ þ 2YðxÞ ¼ 4; Yð0Þ ¼ 0; Y 0 ð0Þ ¼ 0 Ans. ðaÞ YðxÞ ¼ 6ex 3e2x ; ðbÞ YðxÞ ¼ 4 2ex 1 2 x2 x; ðcÞ YðxÞ ¼ 1 ex ðsin x þ cos xÞ 12.80. Prove that lfFðxÞg exists if FðxÞ is piecewise continuous in every finite interval ½0; b where b 0 and if FðxÞ is of exponential order as x ! 1, i.e., there exists a constant such that je x FðxÞj P (a constant) for all x b. 12.81. If f ðsÞ ¼ lfFðxÞg and gðsÞ ¼ lfGðxÞg, prove that f ðsÞgðsÞ ¼ lfHðxÞg where HðxÞ ¼ ðx 0 FðuÞGðx uÞ du is called the convolution of F and G, written F G. Hint: Write f ðsÞgðsÞ ¼ lim M!1 ðM 0 esu FðuÞ du ðM 0 esv GðvÞ dv ¼ lim M!1 ðM 0 ðM 0 esðuþvÞ FðuÞ GðvÞ du dv and then let u þ v ¼ t: 12.82. (a) Find l1 1 ðs2 þ 1Þ2 : ðbÞ Solve Y 00 ðxÞ þ YðxÞ ¼ RðxÞ; Yð0Þ ¼ Y 0 ð0Þ ¼ 0. (c) Solve the integral equation YðxÞ ¼ x þ ðx 0 YðuÞ sinðx uÞ du. [Hint: Use Problem 12.81.] Ans. (a) 1 2 ðsin x x cos xÞ; ðbÞ YðxÞ ¼ ðx 0 RðuÞ sinðx uÞ du; ðcÞ YðxÞ ¼ x þ x3 =6 12.83. Let f ðxÞ; gðxÞ, and g0 ðxÞ be continuous in every finite interval a @ x @ b and suppose that g0 ðxÞ @ 0. Suppose also that hðxÞ ¼ ðx a f ðxÞ dx is bounded for all x A a and lim x!0 gðxÞ ¼ 0. (a) Prove that ð1 a f ðxÞ gðxÞ dx ¼ ð1 a g0 ðxÞ hðxÞ dx. (b) Prove that the integral on the right, and hence the integral on the left, is convergent. The result is that under the give conditions on f ðxÞ and gðxÞ, ð1 a f ðxÞ gðxÞ dx converges and is sometimes called Abel’s integral test. Hint: For (a), consider lim b!1 ðb a f ðxÞ gðxÞ dx after replacing f ðxÞ by h0 ðxÞ and integrating by parts. For (b), first prove that if jhðxÞj H (a constant), then ðb a g0 ðxÞ hðxÞ dx @ HfgðaÞ gðbÞg; and then let b ! 1. 12.84. Use Problem 12.83 to prove that (a) ð1 0 sin x x dx and (b) ð1 0 sin xp dx; p 1, converge. 334 IMPROPER INTEGRALS [CHAP. 12
- 344. 12.85. (a) Given that ð1 0 sin x2 dx ¼ ð1 0 cos x2 dx ¼ 1 2 ffiffiffi 2 r [see Problems 15.27 and 15.68(a), Chapter 15], evaluate ð1 0 ð1 0 sinðx2 þ y2 Þ dx dy (b) Explain why the method of Problem 12.31 cannot be used to evaluate the multiple integral in (a). Ans. =4 CHAP. 12] IMPROPER INTEGRALS 335
- 345. 336 Fourier Series Mathematicians of the eighteenth century, including Daniel Bernoulli and Leonard Euler, expressed the problem of the vibratory motion of a stretched string through partial differential equations that had no solutions in terms of ‘‘elementary functions.’’ Their resolution of this difficulty was to introduce infinite series of sine and cosine functions that satisfied the equations. In the early nineteenth century, Joseph Fourier, while studying the problem of heat flow, developed a cohesive theory of such series. Consequently, they were named after him. Fourier series and Fourier integrals are investigated in this and the next chapter. As you explore the ideas, notice the similarities and differences with the chapters on infinite series and improper integrals. PERIODIC FUNCTIONS A function f ðxÞ is said to have a period T or to be periodic with period T if for all x, f ðx þ TÞ ¼ f ðxÞ, where T is a positive constant. The least value of T 0 is called the least period or simply the period of f ðxÞ. EXAMPLE 1. The function sin x has periods 2; 4; 6; . . . ; since sin ðx þ 2Þ; sin ðx þ 4Þ; sin ðx þ 6Þ; . . . all equal sin x. However, 2 is the least period or the period of sin x. EXAMPLE 2. The period of sin nx or cos nx, where n is a positive integer, is 2=n. EXAMPLE 3. The period of tan x is . EXAMPLE 4. A constant has any positive number as period. Other examples of periodic functions are shown in the graphs of Figures 13-1(a), (b), and (c) below. f (x) x Period f (x) f (x) x x Period Period (a) (b) (c) Fig. 13-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 346. FOURIER SERIES Let f ðxÞ be defined in the interval ðL; LÞ and outside of this interval by f ðx þ 2LÞ ¼ f ðxÞ, i.e., f ðxÞ is 2L-periodic. It is through this avenue that a new function on an infinite set of real numbers is created from the image on ðL; LÞ. The Fourier series or Fourier expansion corresponding to f ðxÞ is given by a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L ð1Þ where the Fourier coefficients an and bn are an ¼ 1 L ðL L f ðxÞ cos nx L dx n ¼ 0; 1; 2; . . . bn ¼ 1 L ðL L f ðxÞ sin nx L dx 8 : ð2Þ ORTHOGONALITY CONDITIONS FOR THE SINE AND COSINE FUNCTIONS Notice that the Fourier coefficients are integrals. These are obtained by starting with the series, (1), and employing the following properties called orthogonality conditions: (a) ðL L cos mx L cos nx L dx ¼ 0 if m 6¼ n and L if m ¼ n (b) ðL L sin mx L sin nx L dx ¼ 0 if m 6¼ n and L if m ¼ n (3) (c) ðL L sin mx L cos nx L dx ¼ 0. Where m and n can assume any positive integer values. An explanation for calling these orthogonality conditions is given on Page 342. Their application in determining the Fourier coefficients is illustrated in the following pair of examples and then demon- strated in detail in Problem 13.4. EXAMPLE 1. To determine the Fourier coefficient a0, integrate both sides of the Fourier series (1), i.e., ðL L f ðxÞ dx ¼ ðL L a0 2 dx þ ðL L X 1 n¼1 an cos nx L þ bn sin nx L n o dx Now ðL L a0 2 dx ¼ a0L; ðL l sin nx L dx ¼ 0; ðL L cos nx L dx ¼ 0, therefore, a0 ¼ 1 L ðL L f ðxÞ dx EXAMPLE 2. To determine a1, multiply both sides of (1) by cos x L and then integrate. Using the orthogonality conditions (3)a and (3)c, we obtain a1 ¼ 1 L ðL L f ðxÞ cos x L dx. Now see Problem 13.4. If L ¼ , the series (1) and the coefficients (2) or (3) are particularly simple. The function in this case has the period 2. DIRICHLET CONDITIONS Suppose that (1) f ðxÞ is defined except possibly at a finite number of points in ðL; LÞ (2) f ðxÞ is periodic outside ðL; LÞ with period 2L CHAP. 13] FOURIER SERIES 337
- 347. (3) f ðxÞ and f 0 ðxÞ are piecewise continuous in ðL; LÞ. Then the series (1) with Fourier coefficients converges to ðaÞ f ðxÞ if x is a point of continuity ðbÞ f ðx þ 0Þ þ f ðx 0Þ 2 if x is a point of discontinuity Here f ðx þ 0Þ and f ðx 0Þ are the right- and left-hand limits of f ðxÞ at x and represent lim !0þ f ðx þ Þ and lim !0þ f ðx Þ, respectively. For a proof see Problems 13.18 through 13.23. The conditions (1), (2), and (3) imposed on f ðxÞ are sufficient but not necessary, and are generally satisfied in practice. There are at present no known necessary and sufficient conditions for convergence of Fourier series. It is of interest that continuity of f ðxÞ does not alone ensure convergence of a Fourier series. ODD AND EVEN FUNCTIONS A function f ðxÞ is called odd if f ðxÞ ¼ f ðxÞ. Thus, x3 ; x5 3x3 þ 2x; sin x; tan 3x are odd functions. A function f ðxÞ is called even if f ðxÞ ¼ f ðxÞ. Thus, x4 ; 2x6 4x2 þ 5; cos x; ex þ ex are even functions. The functions portrayed graphically in Figures 13-1(a) and 13-1ðbÞ are odd and even respectively, but that of Fig. 13-1(c) is neither odd nor even. In the Fourier series corresponding to an odd function, only sine terms can be present. In the Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present. HALF RANGE FOURIER SINE OR COSINE SERIES A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are present, respectively. When a half range series corresponding to a given function is desired, the function is generally defined in the interval ð0; LÞ [which is half of the interval ðL; LÞ, thus accounting for the name half range] and then the function is specified as odd or even, so that it is clearly defined in the other half of the interval, namely, ðL; 0Þ. In such case, we have an ¼ 0; bn ¼ 2 L ðL 0 f ðxÞ sin nx L dx for half range sine series bn ¼ 0; an ¼ 2 L ðL 0 f ðxÞ cos nx L dx for half range cosine series 8 : ð4Þ PARSEVAL’S IDENTITY If an and bn are the Fourier coefficients corresponding to f ðxÞ and if f ðxÞ satisfies the Dirichlet conditions. Then 1 L ðL L f f ðxÞg2 dx ¼ a2 0 2 þ X 1 n¼1 ða2 n þ b2 nÞ (5) (See Problem 13.13.) 338 FOURIER SERIES [CHAP. 13
- 348. DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES Differentiation and integration of Fourier series can be justified by using the theorems on Pages 271 and 272, which hold for series in general. It must be emphasized, however, that those theorems provide sufficient conditions and are not necessary. The following theorem for integration is especially useful. Theorem. The Fourier series corresponding to f ðxÞ may be integrated term by term from a to x, and the resulting series will converge uniformly to ðx a f ðxÞ dx provided that f ðxÞ is piecewise continuous in L @ x @ L and both a and x are in this interval. COMPLEX NOTATION FOR FOURIER SERIES Using Euler’s identities, ei ¼ cos þ i sin ; ei ¼ cos i sin ð6Þ where i ¼ ffiffiffiffiffiffiffi 1 p (see Problem 11.48, Chapter 11, Page 295), the Fourier series for f ðxÞ can be written as f ðxÞ ¼ X 1 n¼1 cn einx=L ð7Þ where cn ¼ 1 2L ðL L f ðxÞeinx=L dx ð8Þ In writing the equality (7), we are supposing that the Dirichlet conditions are satisfied and further that f ðxÞ is continuous at x. If f ðxÞ is discontinuous at x, the left side of (7) should be replaced by ðf ðx þ 0Þ þ f ðx 0Þ 2 : BOUNDARY-VALUE PROBLEMS Boundary-value problems seek to determine solutions of partial differential equations satisfying certain prescribed conditions called boundary conditions. Some of these problems can be solved by use of Fourier series (see Problem 13.24). EXAMPLE. The classical problem of a vibrating string may be idealized in the following way. See Fig. 13-2. Suppose a string is tautly stretched between points ð0; 0Þ and ðL; 0Þ. Suppose the tension, F, is the same at every point of the string. The string is made to vibrate in the xy plane by pulling it to the parabolic position gðxÞ ¼ mðLx x2 Þ and releasing it. (m is a numerically small positive constant.) Its equation will be of the form y ¼ f ðx; tÞ. The problem of establishing this equation is idealized by (a) assuming that the con- stant tension, F, is so large as compared to the weight wL of the string that the gravitational force can be neglected, (b) the displacement at any point of the string is so small that the length of the string may be taken as L for any of its positions, and (c) the vibrations are purely transverse. The force acting on a segment PQ is w g x @2 y @t2 ; x x1 x þ x; g 32 ft per sec:2 . If and are the angles that F makes with the horizontal, then the vertical CHAP. 13] FOURIER SERIES 339 Fig. 13-2
- 349. difference in tensions is Fðsin sin Þ. This is the force producing the acceleration that accounts for the vibratory motion. Now Ffsin sin g ¼ F tan ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ tan2 p tan ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ tan2 p ( ) Fftan tan g ¼ F @y @x ðx þ x; tÞ @y @x ðx; tÞ , where the squared terms in the denominator are neglected because the vibrations are small. Next, equate the two forms of the force, i.e., F @y @x ðx þ x; tÞ @y @x ðx; tÞ ¼ w g x @2 y @t2 divide by x, and then let x ! 0. After letting ¼ ffiffiffiffiffiffi Fg w r , the resulting equation is @2 y @t2 ¼ 2 @2 y @x2 This homogeneous second partial derivative equation is the classical equation for the vibrating string. Associated boundary conditions are yð0; tÞ ¼ 0; yðL; tÞ ¼ 0; t 0 The initial conditions are yðx; 0Þ ¼ mðLx x2 Þ; @y @t ðx; 0Þ ¼ 0; 0 x L The method of solution is to separate variables, i.e., assume yðx; tÞ ¼ GðxÞHðtÞ Then upon substituting GðxÞ H 00 ðtÞ ¼ 2 G00 ðxÞ HðtÞ Separating variables yields G00 G ¼ k; H 00 H ¼ 2 k; where k is an arbitrary constant Since the solution must be periodic, trial solutions are GðxÞ ¼ c1 sin ffiffiffiffiffiffiffi k p x þ c2 cos ffiffiffiffiffiffiffi k p x; 0 HðtÞ ¼ c3 sin ffiffiffiffiffiffiffi k p t þ c4 cos ffiffiffiffiffiffiffi k p t Therefore y ¼ GH ¼ ½c1 sin ffiffiffiffiffiffiffi k p x þ c2 cos ffiffiffiffiffiffiffi k p x½c3 sin ffiffiffiffiffiffiffi k p t þ c4 cos ffiffiffiffiffiffiffi k p t The initial condition y ¼ 0 at x ¼ 0 for all t leads to the evaluation c2 ¼ 0. Thus y ¼ ½c1 sin ffiffiffiffiffiffiffi k p x½c3 sin ffiffiffiffiffiffiffi k p t þ c4 cos ffiffiffiffiffiffiffi k p t Now impose the boundary condition y ¼ 0 at x ¼ L, thus 0 ¼ ½c1 sin ffiffiffiffiffiffiffi k p L½c3 sin ffiffiffiffiffiffiffi k p t þ c4 cos ffiffiffiffiffiffiffi k p t: c1 6¼ 0 as that would imply y ¼ 0 and a trivial solution. The next simplest solution results from the choice ffiffiffiffiffiffiffi k p ¼ n L , since y ¼ c1 sin n L x h i c3 sin n l t þ c4 cos n L t h i and the first factor is zero when x ¼ L. 340 FOURIER SERIES [CHAP. 13
- 350. With this equation in place the boundary condition @y @t ðx; 0Þ ¼ 0, 0 x L can be considered. @y @t ¼ c1 sin n L x h i c3 n L cos n L t c4 n L sin n L t h i At t ¼ 0 0 ¼ c1 sin n L x h i c3 n L Since c1 6¼ 0 and sin n L x is not identically zero, it follows that c3 ¼ 0 and that y ¼ c1 sin n L x h i c4 n L cos n L t h i The remaining initial condition is yðx; 0Þ ¼ mðLx x2 Þ; 0 x L When it is imposed mðLx x2 Þ ¼ c1c4 n L sin n L x However, this relation cannot be satisfied for all x on the interval ð0; LÞ. Thus, the preceding extensive analysis of the problem of the vibrating string has led us to an inadequate form y ¼ c1c4 n L sin n L x cos n L t and an initial condition that is not satisfied. At this point the power of Fourier series is employed. In particular, a theorem of differential equations states that any finite sum of a particular solution also is a solution. Generalize this to infinite sum and consider y ¼ X 1 n¼1 bn sin n L x cos n L t with the initial condition expressed through a half range sine series, i.e., X 1 n¼1 bn sin n L x ¼ mðLx x2 Þ; t ¼ 0 According to the formula of Page 338 for coefficient of a half range sine series L 2m bn ¼ ðL 0 ðLx x2 Þ sin nx L dx That is L 2m bn ¼ ðL 0 Lx sin nx L dx ðL 0 x2 sin nx L dx Application of integration by parts to the second integral yields L 2m bn ¼ L ðL 0 x sin nx L dx þ L3 n cos n þ ðL 0 L n cos nx L 2x dx When integration by parts is applied to the two integrals of this expression and a little algebra is employed the result is bn ¼ 4L2 ðnÞ3 ð1 cos nÞ CHAP. 13] FOURIER SERIES 341
- 351. Therefore, y ¼ X 1 n¼1 bn sin n L x cos n L t with the coefficients bn defined above. ORTHOGONAL FUNCTIONS Two vectors A and B are called orthogonal (perpendicular) if A B ¼ 0 or A1B1 þ A2B2 þ A3B3 ¼ 0, where A ¼ A1i þ A2j þ A3k and B ¼ B1i þ B2j þ B3k. Although not geometrically or physically evi- dent, these ideas can be generalized to include vectors with more than three components. In particular, we can think of a function, say, AðxÞ, as being a vector with an infinity of components (i.e., an infinite dimensional vector), the value of each component being specified by substituting a particular value of x in some interval ða; bÞ. It is natural in such case to define two functions, AðxÞ and BðxÞ, as orthogonal in ða; bÞ if ðb a AðxÞ BðxÞ dx ¼ 0 ð9Þ A vector A is called a unit vector or normalized vector if its magnitude is unity, i.e., if A A ¼ A2 ¼ 1. Extending the concept, we say that the function AðxÞ is normal or normalized in ða; bÞ if ðb a fAðxÞg2 dx ¼ 1 ð10Þ From the above it is clear that we can consider a set of functions fkðxÞg; k ¼ 1; 2; 3; . . . ; having the properties ðb a mðxÞnðxÞ dx ¼ 0 m 6¼ n ð11Þ ðb a fmðxÞg2 dx ¼ 1 m ¼ 1; 2; 3; . . . ð12Þ In such case, each member of the set is orthogonal to every other member of the set and is also normalized. We call such a set of functions an orthonormal set. The equations (11) and (12) can be summarized by writing ðb a mðxÞnðxÞ dx ¼ mn ð13Þ where mn, called Kronecker’s symbol, is defined as 0 if m 6¼ n and 1 if m ¼ n. Just as any vector r in three dimensions can be expanded in a set of mutually orthogonal unit vectors i; j; k in the form r ¼ c1i þ c2j þ c3k, so we consider the possibility of expanding a function f ðxÞ in a set of orthonormal functions, i.e., f ðxÞ ¼ X 1 n¼1 cnnðxÞ a @ x @ b ð14Þ As we have seen, Fourier series are constructed from orthogonal functions. Generalizations of Fourier series are of great interest and utility both from theoretical and applied viewpoints. 342 FOURIER SERIES [CHAP. 13
- 352. Solved Problems FOURIER SERIES 13.1. Graph each of the following functions. ðaÞ f ðxÞ ¼ 3 0 x 5 3 5 x 0 Period ¼ 10 Since the period is 10, that portion of the graph in 5 x 5 (indicated heavy in Fig. 13-3 above) is extended periodically outside this range (indicated dashed). Note that f ðxÞ is not defined at x ¼ 0; 5; 5; 10; 10; 15; 15, and so on. These values are the discontinuities of f ðxÞ. ðbÞ f ðxÞ ¼ sin x 0 @ x @ 0 x 2 Period ¼ 2 Refer to Fig. 13-4 above. Note that f ðxÞ is defined for all x and is continuous everywhere. ðcÞ f ðxÞ ¼ 0 0 @ x 2 1 2 @ x 4 0 4 @ x 6 Period ¼ 6 8 : Refer to Fig. 13-5 above. Note that f ðxÞ is defined for all x and is discontinuous at x ¼ 2; 4; 8; 10; 14; . . . . CHAP. 13] FOURIER SERIES 343 Period f (x) _25 _20 _15 _10 _5 5 3 3 0 10 15 20 25 x Fig. 13-3 Period f (x) 4p x 3p 2p p 0 _p _2p _3p Fig. 13-4 14 x 12 10 8 6 4 2 0 _2 _4 _6 _8 _10 _12 1 Period f (x) Fig. 13-5
- 353. 13.2. Prove ðL L sin kx L dx ¼ ðL L cos kx L dx ¼ 0 if k ¼ 1; 2; 3; . . . . ðL L sin kx L dx ¼ L k cos kx L L L ¼ L k cos k þ L k cosðkÞ ¼ 0 ðL L cos kx L dx ¼ L k sin kx L L L ¼ L k sin k L k sinðkÞ ¼ 0 13.3. Prove (a) ðL L cos mx L cos nx L dx ¼ ðL L sin mx L sin nx L dx ¼ 0 m 6¼ n L m ¼ n (b) ðL L sin mx L cos nx L dx ¼ 0 where m and n can assume any of the values 1; 2; 3; . . . . (a) From trigonometry: cos A cos B ¼ 1 2 fcosðA BÞ þ cosðA þ BÞg; sin A sin B ¼ 1 2 fcosðA BÞ cos ðA þ BÞg: Then, if m 6¼ n, by Problem 13.2, ðL L cos mx L cos nx L dx ¼ 1 2 ðL L cos ðm nÞx L þ cos ðm þ nÞx L dx ¼ 0 Similarly, if m 6¼ n, ðL L sin mx L sin nx L dx ¼ 1 2 ðL L cos ðm nÞx L cos ðm þ nÞx L dx ¼ 0 If m ¼ n, we have ðL L cos mx L cos nx L dx ¼ 1 2 ðL L 1 þ cos 2nx L dx ¼ L ðL L sin mx L sin nx L dx ¼ 1 2 ðL L 1 cos 2nx L dx ¼ L Note that if m ¼ n these integrals are equal to 2L and 0 respectively. (b) We have sin A cos B ¼ 1 2 fsinðA BÞ þ sinðA þ BÞg. Then by Problem 13.2, if m 6¼ n, ðL L sin mx L cos nx L dx ¼ 1 2 ðL L sin ðm nÞx L þ sin ðm þ nÞx L dx ¼ 0 If m ¼ n, ðL L sin mx L cos nx L dx ¼ 1 2 ðL L sin 2nx L dx ¼ 0 The results of parts (a) and (b) remain valid even when the limits of integration L; L are replaced by c; c þ 2L, respectively. 13.4. If the series A þ X 1 n¼1 an cos nx L þ bn sin nx L converges uniformly to f ðxÞ in ðL; LÞ, show that for n ¼ 1; 2; 3; . . . ; ðaÞ an ¼ 1 L ðL L f ðxÞ cos nx L dx; ðbÞ bn ¼ 1 L ðL L f ðxÞ sin nx L dx; ðcÞ A ¼ a0 2 : 344 FOURIER SERIES [CHAP. 13
- 354. (a) Multiplying f ðxÞ ¼ A þ X 1 n¼1 an cos nx L þ bn sin nx L ð1Þ by cos mx L and integrating from L to L, using Problem 13.3, we have ðL L f ðxÞ cos mx L dx ¼ A ðL L cos mx L dx þ X 1 n¼1 an ðL L cos mx L cos nx L dx þ bn ðL L cos mx L sin nx L dx ¼ amL if m 6¼ 0 am ¼ 1 L ðL L f ðxÞ cos mx L dx if m ¼ 1; 2; 3; . . . Thus (b) Multiplying (1) by sin mx L and integrating from L to L, using Problem 13.3, we have ðL L f ðxÞ sin mx L dx ¼ A ðL L sin mx L dx þ X 1 n¼1 an ðL L sin mx L cos nx L dx þ bn ðL L sin mx L sin nx L dx ¼ bmL bm ¼ 1 L ðL L f ðxÞ sin mx L dx if m ¼ 1; 2; 3; . . . Thus (c) Integrating of (1) from L to L, using Problem 13.2, gives ðL L f ðxÞ dx ¼ 2AL or A ¼ 1 2L ðL L f ðxÞ dx Putting m ¼ 0 in the result of part (a), we find a0 ¼ 1 L ðL L f ðxÞ dx and so A ¼ a0 2 . The above results also hold when the integration limits L; L are replaced by c; c þ 2L: Note that in all parts above, interchange of summation and integration is valid because the series is assumed to converge uniformly to f ðxÞ in ðL; LÞ. Even when this assumption is not warranted, the coefficients am and bm as obtained above are called Fourier coefficients corresponding to f ðxÞ, and the corresponding series with these values of am and bm is called the Fourier series corresponding to f ðxÞ. An important problem in this case is to investigate conditions under which this series actually converges to f ðxÞ. Sufficient conditions for this convergence are the Dirichlet conditions established in Problems 13.18 through 13.23. 13.5. (a) Find the Fourier coefficients corresponding to the function f ðxÞ ¼ 0 5 x 0 3 0 x 5 Period ¼ 10 (b) Write the corresponding Fourier series. (c) How should f ðxÞ be defined at x ¼ 5; x ¼ 0; and x ¼ 5 in order that the Fourier series will converge to f ðxÞ for 5 @ x @ 5? The graph of f ðxÞ is shown in Fig. 13-6. CHAP. 13] FOURIER SERIES 345
- 355. (a) Period ¼ 2L ¼ 10 and L ¼ 5. Choose the interval c to c þ 2L as 5 to 5, so that c ¼ 5. Then an ¼ 1 L ðcþ2L c f ðxÞ cos nx L dx ¼ 1 5 ð5 5 f ðxÞ cos nx 5 dx ¼ 1 5 ð0 5 ð0Þ cos nx 5 dx þ ð5 0 ð3Þ cos nx 5 dx ¼ 3 5 ð5 0 cos nx 5 dx ¼ 3 5 5 n sin nx 5 5 0 ¼ 0 if n 6¼ 0 If n ¼ 0; an ¼ a0 ¼ 3 5 ð5 0 cos 0x 5 dx ¼ 3 5 ð5 0 dx ¼ 3: bn ¼ 1 L ðcþ2L c f ðxÞ sin nx L dx ¼ 1 5 ð5 5 f ðxÞ sin nx 5 dx ¼ 1 5 ð0 5 ð0Þ sin nx 5 dx þ ð5 0 ð3Þ sin nx 5 dx ¼ 3 5 ð5 0 sin nx 5 dx ¼ 3 5 5 n cos nx 5 5 0 ¼ 3ð1 cos nÞ n (b) The corresponding Fourier series is a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L ¼ 3 2 þ X 1 n¼1 3ð1 cos nÞ n sin nx 5 ¼ 3 2 þ 6 sin x 5 þ 1 3 sin 3x 5 þ 1 5 sin 5x 5 þ (c) Since f ðxÞ satisfies the Dirichlet conditions, we can say that the series converges to f ðxÞ at all points of continuity and to f ðx þ 0Þ þ f ðx 0Þ 2 at points of discontinuity. At x ¼ 5, 0, and 5, which are points of discontinuity, the series converges to ð3 þ 0Þ=2 ¼ 3=2 as seen from the graph. If we redefine f ðxÞ as follows, f ðxÞ ¼ 3=2 x ¼ 5 0 5 x 0 3=2 x ¼ 0 3 0 x 5 3=2 x ¼ 5 Period ¼ 10 8 : then the series will converge to f ðxÞ for 5 @ x @ 5. 13.6. Expand f ðxÞ ¼ x2 ; 0 x 2 in a Fourier series if (a) the period is 2, (b) the period is not specified. (a) The graph of f ðxÞ with period 2 is shown in Fig. 13-7 below. 346 FOURIER SERIES [CHAP. 13 _15 _10 _5 5 3 10 x 15 Period f (x) Fig. 13-6
- 356. Period ¼ 2L ¼ 2 and L ¼ . Choosing c ¼ 0, we have an ¼ 1 L ðcþ2L c f ðxÞ cos nx L dx ¼ 1 ð2 0 x2 cos nx dx ¼ 1 ðx2 Þ sin nx n ð2xÞ cos nx n2 þ 2 sin nx n3 2 0 ¼ 4 n2 ; n 6¼ 0 If n ¼ 0; a0 ¼ 1 ð2 0 x2 dx ¼ 82 3 : bn ¼ 1 L ðcþ2L c f ðxÞ sin nx L dx ¼ 1 ð2 0 x2 sin nx dx ¼ 1 ðx2 Þ cos nx n ð2xÞ sin nx n2 þ ð2Þ cos nx n3 2 0 ¼ 4 n Then f ðxÞ ¼ x2 ¼ 42 3 þ X 1 n¼1 4 n2 cos nx 4 n sin nx : This is valid for 0 x 2. At x ¼ 0 and x ¼ 2 the series converges to 22 . (b) If the period is not specified, the Fourier series cannot be determined uniquely in general. 13.7. Using the results of Problem 13.6, prove that 1 12 þ 1 22 þ 1 32 þ ¼ 2 6 . At x ¼ 0 the Fourier series of Problem 13.6 reduces to 42 3 þ X 1 n¼1 4 n2 . By the Dirichlet conditions, the series converges at x ¼ 0 to 1 2 ð0 þ 42 Þ ¼ 22 . Then 42 3 þ X 1 n¼1 4 n2 ¼ 22 , and so X 1 n¼1 1 n2 ¼ 2 6 . ODD AND EVEN FUNCTIONS, HALF RANGE FOURIER SERIES 13.8. Classify each of the following functions according as they are even, odd, or neither even nor odd. ðaÞ f ðxÞ ¼ 2 0 x 3 2 3 x 0 Period ¼ 6 From Fig. 13-8 below it is seen that f ðxÞ ¼ f ðxÞ, so that the function is odd. ðbÞ f ðxÞ ¼ cos x 0 x 0 x 2 Period ¼ 2 CHAP. 13] FOURIER SERIES 347 _6p _4p _2p O 2p 4p 6p x f (x) 4p2 Fig. 13-7
- 357. From Fig. 13-9 below it is seen that the function is neither even nor odd. ðcÞ f ðxÞ ¼ xð10 xÞ; 0 x 10; Period ¼ 10: From Fig. 13-10 below the function is seen to be even. 13.9. Show that an even function can have no sine terms in its Fourier expansion. Method 1: No sine terms appear if bn ¼ 0; n ¼ 1; 2; 3; . . . . To show this, let us write bn ¼ 1 L ðL L f ðxÞ sin nx L dx ¼ 1 L ð0 L f ðxÞ sin nx L dx þ 1 L ðL 0 f ðxÞ sin nx L dx ð1Þ If we make the transformation x ¼ u in the first integral on the right of (1), we obtain 1 L ð0 L f ðxÞ sin nx L dx ¼ 1 L ðL 0 f ðuÞ sin nu L du ¼ 1 L ðL 0 f ðuÞ sin nu L du ð2Þ ¼ 1 L ðL 0 f ðuÞ sin nu L du ¼ 1 L ðL 0 f ðxÞ sin nx L dx where we have used the fact that for an even function f ðuÞ ¼ f ðuÞ and in the last step that the dummy variable of integration u can be replaced by any other symbol, in particular x. Thus, from (1), using (2), we have 348 FOURIER SERIES [CHAP. 13 f (x) 2 _2 _6 _3 3 6 x Fig. 13-8 f (x) O 1 _2p _p p 2p 3p x Fig. 13-9 f (x) O 25 _10 5 10 x Fig. 13-10
- 358. bn ¼ 1 L ðL 0 f ðxÞ sin nx L dx þ 1 L ðL 0 f ðxÞ sin nx L dx ¼ 0 f ðxÞ ¼ a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L : Method 2: Assume f ðxÞ ¼ a0 2 þ X 1 n¼1 an cos nx L bN sin nx L : Then If f ðxÞ is even, f ðxÞ ¼ f ðxÞ. Hence, a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L ¼ a0 2 þ X 1 n¼1 an cos nx L bn sin nx L X 1 n¼1 bn sin nx L ¼ 0; i.e., f ðxÞ ¼ a0 2 þ X 1 n¼1 an cos nx L and so and no sine terms appear. In a similar manner we can show that an odd function has no cosine terms (or constant term) in its Fourier expansion. 13.10. If f ðxÞ is even, show that (a) an ¼ 2 L ðL 0 f ðxÞ cos nx L dx; ðbÞ bn ¼ 0. an ¼ 1 L ðL L f ðxÞ cos nx L dx ¼ 1 L ð0 L f ðxÞ cos nx L dx þ 1 L ðL 0 f ðxÞ cos nx L dx ðaÞ Letting x ¼ u, 1 L ð0 L f ðxÞ cos nx L dx ¼ 1 L ðL 0 f ðuÞ cos nu L du ¼ 1 L ðL 0 f ðuÞ cos nu L du since by definition of an even function f ðuÞ ¼ f ðuÞ. Then an ¼ 1 L ðL 0 f ðuÞ cos nu L du þ 1 L ðL 0 f ðxÞ cos nx L dx ¼ 2 L ðL 0 f ðxÞ cos nx L dx (b) This follows by Method 1 of Problem 13.9. 13.11. Expand f ðxÞ ¼ sin x; 0 x , in a Fourier cosine series. A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence, we extend the definition of f ðxÞ so that it becomes even (dashed part of Fig. 13-11 below). With this extension, f ðxÞ is then defined in an interval of length 2. Taking the period as 2, we have 2L ¼ 2 so that L ¼ . By Problem 13.10, bn ¼ 0 and an ¼ 2 L ðL 0 f ðxÞ cos nx L dx ¼ 2 ð 0 sin x cos nx dx CHAP. 13] FOURIER SERIES 349 f (x) O _2p _p p 2p x Fig. 13-11
- 359. ¼ 1 ð 0 fsinðx þ nxÞ þ sinðx nxÞg ¼ 1 cosðn þ 1Þx n þ 1 þ cosðn 1Þx n 1 0 ¼ 1 1 cosðn þ 1Þ n þ 1 þ cosðn 1Þ 1 n 1 ¼ 1 1 þ cos n n þ 1 1 þ cos n n 1 ¼ 2ð1 þ cos nÞ ðn2 1Þ if n 6¼ 1: For n ¼ 1; a1 ¼ 2 ð 0 sin x cos x dx ¼ 2 sin2 x 2 0 ¼ 0: For n ¼ 0; a0 ¼ 2 ð 0 sin x dx ¼ 2 ð cos xÞ 0 ¼ 4 : f ðxÞ ¼ 2 2 X 1 n¼2 ð1 þ cos nÞ n2 1 cos nx Then ¼ 2 4 cos 2x 22 1 þ cos 4x 42 1 þ cos 6x 62 1 þ 13.12. Expand f ðxÞ ¼ x; 0 x 2, in a half range (a) sine series, (b) cosine series. (a) Extend the definition of the given function to that of the odd function of period 4 shown in Fig. 13-12 below. This is sometimes called the odd extension of f ðxÞ. Then 2L ¼ 4; L ¼ 2. Thus an ¼ 0 and bn ¼ 2 L ðL 0 f ðxÞ sin nx L dx ¼ 2 2 ð2 0 x sin nx 2 dx ¼ ðxÞ 2 n cos nx 2 ð1Þ 4 n22 sin nx 2 2 0 ¼ 4 n cos n f ðxÞ ¼ X 1 n¼1 4 n cos n sin nx 2 Then ¼ 4 sin x 2 1 2 sin 2x 2 þ 1 3 sin 3x 2 (b) Extend the definition of f ðxÞ to that of the even function of period 4 shown in Fig. 13-13 below. This is the even extension of f ðxÞ. Then 2L ¼ 4; L ¼ 2. 350 FOURIER SERIES [CHAP. 13 f (x) O _6 _4 _2 2 4 6 x Fig. 13-12
- 360. Thus bn ¼ 0, an ¼ 2 L ðL 0 f ðxÞ cos nx L dx ¼ 2 2 ð2 0 x cos nx 2 dx ¼ ðxÞ 2 n sin nx 2 ð1Þ 4 n2 2 cos nx 2 2 0 ¼ 4 n22 ðcos n 1Þ If n 6¼ 0 If n ¼ 0; a0 ¼ ð2 0 x dx ¼ 2: f ðxÞ ¼ 1 þ X 1 n¼1 4 n22 ðcos n 1Þ cos nx 2 Then ¼ 1 8 2 cos x 2 þ 1 32 cos 3x 2 þ 1 52 cos 5x 2 þ It should be noted that the given function f ðxÞ ¼ x, 0 x 2, is represented equally well by the two different series in (a) and (b). PARSEVAL’S IDENTITY 13.13. Assuming that the Fourier series corresponding to f ðxÞ converges uniformly to f ðxÞ in ðL; LÞ, prove Parseval’s identity 1 L ðL L f f ðxÞg2 dx ¼ a2 0 2 þ ða2 n þ b2 nÞ where the integral is assumed to exist. If f ðxÞ ¼ a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L , then multiplying by f ðxÞ and integrating term by term from L to L (which is justified since the series is uniformly convergent) we obtain ðL L f f ðxÞg2 dx ¼ a0 2 ðL L f ðxÞ dx þ X 1 n¼1 an ðL L f ðxÞ cos nx L dx þ bn ðL L f ðxÞ sin nx L dx ¼ a2 0 2 L þ L X 1 n¼1 ða2 n þ b2 nÞ ð1Þ where we have used the results ðL L f ðxÞ cos nx L dx ¼ Lan; ðL L f ðxÞ sin nx L dx ¼ Lbn; ðL L f ðxÞ dx ¼ La0 ð2Þ obtained from the Fourier coefficients. CHAP. 13] FOURIER SERIES 351 f (x) O _6 _4 _2 2 4 6 x Fig. 13-13
- 361. The required result follows on dividing both sides of (1) by L. Parseval’s identity is valid under less restrictive conditions than that imposed here. 13.14. (a) Write Parseval’s identity corresponding to the Fourier series of Problem 13.12(b). (b) Determine from (a) the sum S of the series 1 14 þ 1 24 þ 1 34 þ þ 1 n4 þ . (a) Here L ¼ 2; a0 ¼ 2; an ¼ 4 n2 2 ðcos n 1Þ; n 6¼ 0; bn ¼ 0. Then Parseval’s identity becomes 1 2 ð2 2 f f ðxÞg2 dx ¼ 1 2 ð2 2 x2 dx ¼ ð2Þ2 2 þ X 1 n¼1 16 n44 ðcos n 1Þ2 or 8 3 ¼ 2 þ 64 4 1 14 þ 1 34 þ 1 54 þ ; i.e., 1 14 þ 1 34 þ 1 54 þ ¼ 4 96: ðbÞ S ¼ 1 14 þ 1 24 þ 1 34 þ ¼ 1 14 þ 1 34 þ 1 54 þ þ 1 24 þ 1 44 þ 1 64 þ ¼ 1 14 þ 1 34 þ 1 54 þ þ 1 24 1 14 þ 1 24 þ 1 34 þ ¼ 4 96 þ S 16 ; from which S ¼ 4 90 13.15. Prove that for all positive integers M, a2 0 2 þ X M n¼1 ða2 n þ b2 nÞ @ 1 L ðL L f f ðxÞg2 dx where an and bn are the Fourier coefficients corresponding to f ðxÞ, and f ðxÞ is assumed piecewise continuous in ðL; LÞ. Let SMðxÞ ¼ a0 2 þ X M n¼1 an cos nx L þ bn sin nx L (1) For M ¼ 1; 2; 3; . . . this is the sequence of partial sums of the Fourier series corresponding to f ðxÞ. We have ðL L f f ðxÞ SMðxÞg2 dx A 0 ð2Þ since the integrand is non-negative. Expanding the integrand, we obtain 2 ðL L f ðxÞ SMðxÞ dx ðL L S2 MðxÞ dx @ ðL L f f ðxÞg2 dx ð3Þ Multiplying both sides of (1) by 2 f ðxÞ and integrating from L to L, using equations (2) of Problem 13.13, gives 2 ðL L f ðxÞ SMðxÞ dx ¼ 2L a2 0 2 þ X M n¼1 ða2 n þ b2 nÞ ( ) ð4Þ Also, squaring (1) and integrating from L to L, using Problem 13.3, we find ðL L S2 MðxÞ dx ¼ L a2 0 2 þ X M n¼1 ða2 n þ b2 nÞ ( ) ð5Þ Substitution of (4) and (5) into (3) and dividing by L yields the required result. 352 FOURIER SERIES [CHAP. 13
- 362. Taking the limit as M ! 1, we obtain Bessel’s inequality a2 0 2 þ X 1 n¼1 ða2 n þ b2 nÞ @ 1 L ðL L f f ðxÞg2 dx ð6Þ If the equality holds, we have Parseval’s identity (Problem 13.13). We can think of SMðxÞ as representing an approximation to f ðxÞ, while the left-hand side of (2), divided by 2L, represents the mean square error of the approximation. Parseval’s identity indicates that as M ! 1 the mean square error approaches zero, while Bessels’ inequality indicates the possibility that this mean square error does not approach zero. The results are connected with the idea of completeness of an orthonormal set. If, for example, we were to leave out one or more terms in a Fourier series (say cos 4x=L, for example), we could never get the mean square error to approach zero no matter how many terms we took. For an analogy with three-dimensional vectors, see Problem 13.60. DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 13.16. (a) Find a Fourier series for f ðxÞ ¼ x2 ; 0 x 2, by integrating the series of Problem 13.12(a). (b) Use (a) to evaluate the series X 1 n¼1 ð1Þn1 n2 . (a) From Problem 13.12(a), x ¼ 4 sin x 2 1 2 sin 2x 2 þ 1 3 sin 3x 2 ð1Þ Integrating both sides from 0 to x (applying the theorem of Page 339) and multiplying by 2, we find x2 ¼ C 16 2 cos x 2 1 22 cos 2x 2 þ 1 32 cos 3x 2 ð2Þ where C ¼ 16 2 1 1 22 þ 1 32 1 42 þ : (b) To determine C in another way, note that (2) represents the Fourier cosine series for x2 in 0 x 2. Then since L ¼ 2 in this case, C ¼ a0 2 ¼ 1 L ðL 0 f ðxÞ ¼ 1 2 ð2 0 x2 dx ¼ 4 3 Then from the value of C in (a), we have X 1 n¼1 ð1Þn1 n2 ¼ 1 1 22 þ 1 32 ¼ 1 42 þ ¼ 2 16 4 3 ¼ 2 12 13.17. Show that term by term differentiation of the series in Problem 13.12(a) is not valid. Term by term differentiation yields 2 cos x 2 cos 2x 2 þ cos 3x 2 : Since the nth term of this series does not approach 0, the series does not converge for any value of x. CHAP. 13] FOURIER SERIES 353
- 363. CONVERGENCE OF FOURIER SERIES 13.18. Prove that (a) 1 2 þ cos t þ cos 2t þ þ cos Mt ¼ sinðM þ 1 2Þt 2 sin 1 2 t (b) 1 ð 0 sinðM þ 1 2Þt 2 sin 1 2 t dt ¼ 1 2 ; 1 ð0 sinðM þ 1 2Þt 2 sin 1 2 t dt ¼ 1 2 : (a) We have cos nt sin 1 2 t ¼ 1 2 fsinðn þ 1 2Þt sinðn 1 2Þtg. Then summing from n ¼ 1 to M, sin 1 2 tfcos t þ cos 2t þ þ cos Mtg ¼ ðsin 3 2 t sin 1 2 tÞ þ ðsin 5 2 t sin 3 2 tÞ þ þ sinðM þ 1 2Þt sinðM 1 2Þt ¼ 1 2 fsinðM þ 1 2Þt sin 1 2 tg On dividing by sin 1 2 t and adding 1 2, the required result follows. (b) Integrating the result in (a) from to 0 and 0 to , respectively. This gives the required results, since the integrals of all the cosine terms are zero. 13.19. Prove that lim n!1 ð f ðxÞ sin nx dx ¼ lim n!1 ð f ðxÞ cos nx dx ¼ 0 if f ðxÞ is piecewise continuous. This follows at once from Problem 13.15, since if the series a2 0 2 þ X 1 n¼1 ða2 n þ b2 nÞ is convergent, lim n!1 an ¼ lim n!1 bn ¼ 0. The result is sometimes called Riemann’s theorem. 13.20. Prove that lim M!1 ð f ðxÞ sinðM þ 1 2Þx dx ¼ 0 if f ðxÞ is piecewise continuous. We have ð f ðxÞ sinðM þ 1 2Þx dx ¼ ð f f ðxÞ sin 1 2 xg cos Mx dx þ ð f f ðxÞ cos 1 2 xg sin Mx dx Then the required result follows at once by using the result of Problem 13.19, with f ðxÞ replaced by f ðxÞ sin 1 2 x and f ðxÞ cos 1 2 x respectively, which are piecewise continuous if f ðxÞ is. The result can also be proved when the integration limits are a and b instead of and . 13.21. Assuming that L ¼ , i.e., that the Fourier series corresponding to f ðxÞ has period 2L ¼ 2, show that SMðxÞ ¼ a0 2 þ X M n¼1 ðan cos nx þ bn sin nxÞ ¼ 1 ð f ðt þ xÞ sinðM þ 1 2Þt 2 sin 1 2 t dt Using the formulas for the Fourier coefficients with L ¼ , we have an cos nx þ bn sin nx ¼ 1 ð f ðuÞ cos nu du cos nx þ 1 ð f ðuÞ sin nu du sin nx ¼ 1 ð f ðuÞ cos nu cos nx þ sin nu sin nx ð Þ du ¼ 1 ð f ðuÞ cos nðu xÞ du a0 2 ¼ 1 2 ð f ðuÞ du Also, 354 FOURIER SERIES [CHAP. 13
- 364. SMðxÞ ¼ a0 2 þ X M n¼1 ðan cos nx þ bn sin nxÞ Then ¼ 1 2 ð f ðuÞ du þ 1 X M n¼1 ð f ðuÞ cos nðu xÞ du ¼ 1 ð f ðuÞ 1 2 þ X M n¼1 cos nðu xÞ ( ) du ¼ 1 ð f ðuÞ sinðM þ 1 2Þðu xÞ 2 sin 1 2 ðu xÞ du using Problem 13.18. Letting u x ¼ t, we have SMðxÞ ¼ 1 ðx x f ðt þ xÞ sinðM þ 1 2Þt 2 sin 1 2 t dt Since the integrand has period 2, we can replace the interval x; x by any other interval of length 2, in particular, ; . Thus, we obtain the required result. 13.22. Prove that SMðxÞ f ðx þ 0Þ þ f ðx 0Þ 2 ¼ 1 ð0 f ðt þ xÞ f ðx 0Þ 2 sin 1 2 t ! sinðM þ 1 2Þt dt þ 1 ð 0 f ðt þ xÞ f ðx þ 0Þ 2 sin 1 2 t ! sinðM þ 1 2Þt dt From Problem 13.21, SMðxÞ ¼ 1 ð0 f ðt þ xÞ sinðM þ 1 2Þt 2 sin 1 2 t dt þ 1 ð 0 f ðt þ xÞ sinðM þ 1 2Þt 2 sin 1 2 t dt ð1Þ Multiplying the integrals of Problem 13.18(b) by f ðx 0Þ and f ðx þ 0Þ, respectively, f ðx þ 0Þ þ f ðx 0Þ 2 ¼ 1 ð0 f ðx 0Þ sinðM þ 1 2Þt 2 sin 1 2 t dt þ 1 ð 0 f ðx þ 0Þ sinðM þ 1 2Þt 2 sin 1 2 t dt ð2Þ Subtracting (2) from (1) yields the required result. 13.23. If f ðxÞ and f 0 ðxÞ are piecewise continuous in ð; Þ, prove that lim M!1 SMðxÞ ¼ f ðx þ 0Þ þ f ðx 0Þ 2 The function f ðt þ xÞ f ðx þ 0Þ 2 sin 1 2 t is piecewise continuous in 0 t @ because f ðxÞ is piecewise con- tinous. Also, lim t!0þ f ðt þ xÞ f ðx þ 0Þ 2 sin 1 2 t ¼ lim t!0þ f ðt þ xÞ f ðx þ 0Þ t t 2 sin 1 2 t ¼ lim t!0þ f ðt þ xÞ f ðx þ 0Þ t exists, since by hypothesis f 0 ðxÞ is piecewise continuous so that the right-hand derivative of f ðxÞ at each x exists. Thus, f ðt þ xÞ f ðx 0Þ 2 sin 1 2 t is piecewise continous in 0 @ t @ . Similarly, f ðt þ xÞ f ðx 0Þ 2 sin 1 2 t is piecewise continous in @ t @ 0. CHAP. 13] FOURIER SERIES 355
- 365. Then from Problems 13.20 and 13.22, we have lim M!1 SMðxÞ f ðx þ 0Þ þ f ðx 0Þ 2 ¼ 0 or lim M!1 SMðxÞ ¼ f ðx þ 0Þ þ f ðx 0Þ 2 BOUNDARY-VALUE PROBLEMS 13.24. Find a solution Uðx; tÞ of the boundary-value problem @U @t ¼ 3 @2 U @x2 t 0; 0 x 2 Uð0; tÞ ¼ 0; Uð2; tÞ ¼ 0 t 0 Uðx; 0Þ ¼ x 0 x 2 A method commonly employed in practice is to assume the existence of a solution of the partial differential equation having the particular form Uðx; tÞ ¼ XðxÞ TðtÞ, where XðxÞ and TðtÞ are functions of x and t, respectively, which we shall try to determine. For this reason the method is often called the method of separation of variables. Substitution in the differential equation yields ð1Þ @ @t ðXTÞ ¼ 3 @2 @x2 ðXTÞ or ð2Þ X dT dt ¼ 3T d2 X dx2 where we have written X and T in place of XðxÞ and TðtÞ. Equation (2) can be written as 1 3T dT dt ¼ 1 X d2 X dx2 ð3Þ Since one side depends only on t and the other only on x, and since x and t are independent variables, it is clear that each side must be a constant c. In Problem 13.47 we see that if c A 0, a solution satisfying the given boundary conditions cannot exist. Let us thus assume that c is a negative constant which we write as 2 . Then from (3) we obtain two ordinary differentiation equations dT dt þ 32 T ¼ 0; d2 X dx2 þ 2 X ¼ 0 ð4Þ whose solutions are respectively T ¼ C1e32 t ; X ¼ A1 cos x þ B1 sin x ð5Þ A solution is given by the product of X and T which can be written Uðx; tÞ ¼ e32 t ðA cos x þ B sin xÞ ð6Þ where A and B are constants. We now seek to determine A and B so that (6) satisfies the given boundary conditions. To satisfy the condition Uð0; tÞ ¼ 0, we must have es2 t ðAÞ ¼ 0 or A ¼ 0 ð7Þ so that (6) becomes Uðx; tÞ ¼ Bes2 t sin x ð8Þ To satisfy the condition Uð2; tÞ ¼ 0, we must then have Bes2 t sin 2 ¼ 0 ð9Þ 356 FOURIER SERIES [CHAP. 13
- 366. Since B ¼ 0 makes the solution (8) identically zero, we avoid this choice and instead take sin 2 ¼ 0; i.e., 2 ¼ m or ¼ m 2 ð10Þ where m ¼ 0; 1; 2; . . . . Substitution in (8) now shows that a solution satisfying the first two boundary conditions is Uðx; tÞ ¼ Bme3m2 2 t=4 sin mx 2 ð11Þ where we have replaced B by Bm, indicating that different constants can be used for different values of m. If we now attempt to satisfy the last boundary condition Uðx; 0Þ ¼ x; 0 x 2, we find it to be impossible using (11). However, upon recognizing the fact that sums of solutions having the form (11) are also solutions (called the principle of superposition), we are led to the possible solution Uðx; tÞ ¼ X 1 m¼1 Bme3m2 2 t=4 sin mx 2 ð12Þ From the condition Uðx; 0Þ ¼ x; 0 x 2, we see, on placing t ¼ 0, that (12) becomes x ¼ X 1 m¼1 Bm sin mx 2 0 x 2 ð13Þ This, however, is equivalent to the problem of expanding the function f ðxÞ ¼ x for 0 x 2 into a sine series. The solution to this is given in Problem 13.12(a), from which we see that Bm ¼ 4 m cos m so that (12) becomes Uðx; tÞ ¼ X 1 m¼1 4 m cos m e3m2 2 t=4 sin mx 2 ð14Þ which is a formal solution. To check that (14) is actually a solution, we must show that it satisfies the partial differential equation and the boundary conditions. The proof consists in justification of term by term differentiation and use of limiting procedures for infinite series and may be accomplished by methods of Chapter 11. The boundary value problem considered here has an interpretation in the theory of heat conduction. The equation @U @t ¼ k @2 U @x2 is the equation for heat conduction in a thin rod or wire located on the x-axis between x ¼ 0 and x ¼ L if the surface of the wire is insulated so that heat cannot enter or escape. Uðx; tÞ is the temperature at any place x in the rod at time t. The constant k ¼ K=s (where K is the thermal conductivity, s is the specific heat, and is the density of the conducting material) is called the diffusivity. The boundary conditions Uð0; tÞ ¼ 0 and UðL; tÞ ¼ 0 indicate that the end temperatures of the rod are kept at zero units for all time t 0, while Uðx; 0Þ indicates the initial temperature at any point x of the rod. In this problem the length of the rod is L ¼ 2 units, while the diffusivity is k ¼ 3 units. ORTHOGONAL FUNCTIONS 13.25. (a) Show that the set of functions 1; sin x L ; cos x L ; sin 2x L ; cos 2x L ; sin 3x L ; cos 3x L ; . . . forms an orthogonal set in the interval ðL; LÞ. (b) Determine the corresponding normalizing constants for the set in (a) so that the set is orthonormal in ðL; LÞ. (a) This follows at once from the results of Problems 13.2 and 13.3. (b) By Problem 13.3, ðL L sin2 mx L dx ¼ L; ðL L cos2 mx L dx ¼ L CHAP. 13] FOURIER SERIES 357
- 367. ðL L ffiffiffiffi 1 L r sin mx L !2 dx ¼ 1; ðL L ffiffiffiffi 1 L r cos mx L !2 dx ¼ 1 Then ðL L ð1Þ2 dx ¼ 2L or ðL L 1 ffiffiffiffiffiffi 2L p 2 dx ¼ 1 Also, Thus the required orthonormal set is given by 1 ffiffiffiffiffiffi 2L p ; 1 ffiffiffiffi L p sin x L ; 1 ffiffiffiffi L p cos x L ; 1 ffiffiffiffi L p sin 2x L ; 1 ffiffiffiffi L p cos 2x L ; . . . MISCELLANEOUS PROBLEMS 13.26. Find a Fourier series for f ðxÞ ¼ cos x; @ x @ , where 6¼ 0; 1; 2; 3; . . . . We shall take the period as 2 so that 2L ¼ 2; L ¼ . Since the function is even, bn ¼ 0 and an ¼ 2 L ðL 0 f ðxÞ cos nx dx ¼ 2 ð 0 cos x cos nx dx ¼ 1 ð 0 fcosð nÞx þ cosð þ nÞxg dx ¼ 1 sinð nÞ n þ sinð þ nÞ þ n ¼ 2 sin cos n ð 2 n2Þ 0 ¼ 2 sin Then cos x ¼ sin þ 2 sin X 1 n¼1 cos n 2 n2 cos nx ¼ sin 1 2 2 12 cos x þ 2 2 22 cos 2x 2 2 32 cos 3x þ 13.27. Prove that sin x ¼ x 1 x2 2 ! 1 x2 ð2Þ2 ! 1 x2 ð3Þ2 ! . Let x ¼ in the Fourier series obtained in Problem 13.26. Then cos ¼ sin 1 þ 2 2 12 þ 2 2 22 þ 2 2 32 þ or cot 1 ¼ 2 2 12 þ 2 2 22 þ 2 2 32 þ ð1Þ This result is of interest since it represents an expansion of the contangent into partial fractions. By the Weierstrass M test, the series on the right of (1) converges uniformly for 0 @ j j @ jxj 1 and the left-hand side of (1) approaches zero as ! 0, as is seen by using L’Hospital’s rule. Thus, we can integrate both sides of (1) from 0 to x to obtain ðx 0 cot 1 d ¼ ðx 0 2 2 1 d þ ðx 0 2 2 22 d þ ln sin x 0 ¼ ln 1 x2 12 ! þ ln 1 x2 22 ! þ or 358 FOURIER SERIES [CHAP. 13
- 368. ln sin x x ¼ lim n!1 ln 1 x2 12 ! þ ln 1 x2 22 ! þ þ ln 1 x2 n2 ! i.e., ¼ lim n!1 ln 1 x2 12 ! 1 x2 22 ! 1 x2 n2 ! ( ) ¼ ln lim n!1 1 x2 12 ! 1 x2 22 ! 1 x2 n2 ! ( ) so that sin x x ¼ lim n!1 1 x2 12 ! 1 x2 22 ! 1 x2 n2 ! ¼ 1 x2 12 ! 1 x2 22 ! ð2Þ Replacing x by x=, we obtain sin x ¼ x 1 x2 2 ! 1 x2 ð2Þ2 ! ð3Þ called the infinite product for sin x, which can be shown valid for all x. The result is of interest since it corresponds to a factorization of sin x in a manner analogous to factorization of a polynomial. 13.28. Prove that 2 ¼ 2 2 4 4 6 6 8 8 . . . 1 3 3 5 5 7 7 9 . . . . Let x ¼ 1=2 in equation (2) of Problem 13.27. Then, 2 ¼ 1 1 22 1 1 42 1 1 62 ¼ 1 2 3 2 3 4 5 4 5 6 7 6 Taking reciprocals of both sides, we obtain the required result, which is often called Wallis’ product. Supplementary Problems FOURIER SERIES 13.29. Graph each of the following functions and find their corresponding Fourier series using properties of even and odd functions wherever applicable. ðaÞ f ðxÞ ¼ 8 0 x 2 8 2 x 4 Period 4 ðbÞ f ðxÞ ¼ x 4 @ x @ 0 x 0 @ x @ 4 Period 8 ðcÞ f ðxÞ ¼ 4x; 0 x 10; Period 10 ðdÞ f ðxÞ ¼ 2x 0 @ x 3 0 3 x 0 Period 6 Ans: ðaÞ 16 X 1 n¼1 ð1 cos nÞ n sin nx 2 ðbÞ 2 8 2 X 1 n¼1 ð1 cos nÞ n2 cos nx 4 ðcÞ 20 40 X 1 n¼1 1 n sin nx 5 ðdÞ 3 2 þ X 1 n¼1 6ðcos n 1Þ n2 2 cos nx 3 6 cos n n sin nx 3 13.30. In each part of Problem 13.29, tell where the discontinuities of f ðxÞ are located and to what value the series converges at the discontunities. Ans. (a) x ¼ 0; 2; 4; . . . ; 0 ðbÞ no discontinuities (c) x ¼ 0; 10; 20; . . . ; 20 (d) x ¼ 3; 9; 15; . . . ; 3 CHAP. 13] FOURIER SERIES 359
- 369. 360 FOURIER SERIES [CHAP. 13 13.31. Expand f ðxÞ ¼ 2 x 0 x 4 x 6 4 x 8 in a Fourier series of period 8. Ans: 16 2 cos x 4 þ 1 32 cos 3x 4 þ 1 52 cos 5x 4 þ 13.32. (a) Expand f ðxÞ ¼ cos x; 0 x , in a Fourier sine series. (b) How should f ðxÞ be defined at x ¼ 0 and x ¼ so that the series will converge to f ðxÞ for 0 @ x @ ? Ans: ðaÞ 8 X 1 n¼1 n sin 2nx 4n2 1 ðbÞ f ð0Þ ¼ f ðÞ ¼ 0 13.33. (a) Expand in a Fourier series f ðxÞ ¼ cos x; 0 x if the period is ; and (b) compare with the result of Problem 13.32, explaining the similarities and differences if any. Ans. Answer is the same as in Problem 13.32. 13.34. Expand f ðxÞ ¼ x 0 x 4 8 x 4 x 8 in a series of (a) sines, (b) cosines. Ans: ðaÞ 32 2 X 1 n¼1 1 n2 sin n 2 sin nx 8 ðbÞ 16 2 X 1 n¼1 2 cos n=2 cos n 1 n2 cos nx 8 13.35. Prove that for 0 @ x @ , ðaÞ xð xÞ ¼ 2 6 cos 2x 12 þ cos 4x 22 þ cos 6x 32 þ ðbÞ xð xÞ ¼ 8 sin x 13 þ sin 3x 33 þ sin 5 53 þ 13.36. Use the preceding problem to show that ðaÞ X 1 n¼1 1 n2 ¼ 2 6 ; ðbÞ X 1 n¼1 ð1Þn1 n2 ¼ 2 12 ; ðcÞ X 1 n¼1 ð1Þn1 ð2n 1Þ3 ¼ 3 32 : 13.37. Show that 1 13 þ 1 33 1 53 1 73 þ 1 93 þ 1 113 ¼ 32 ffiffiffi 2 p 16 . DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 13.38. (a) Show that for x , x ¼ 2 sin x 1 sin 2x 2 þ sin 3x 3 (b) By integrating the result of (a), show that for @ x @ , x2 ¼ 2 3 4 cos x 12 cos 2x 22 þ cos 3x 32 (c) By integrating the result of (b), show that for @ x @ , xð xÞð þ xÞ ¼ 12 sin x 13 sin 2x 23 þ sin 3x 33 13.39. (a) Show that for x , x cos x ¼ 1 2 sin x þ 2 2 1 3 sin 2x 3 2 4 sin 3x þ 4 3 5 sin 4x (b) Use (a) to show that for @ x @ , x sin x ¼ 1 1 2 cos x 2 cos 2x 1 3 cos 3x 2 4 þ cos 4x 3 5
- 370. CHAP. 13] FOURIER SERIES 361 13.40. By differentiating the result of Problem 13.35(b), prove that for 0 @ x @ , x ¼ 2 4 cos x 12 þ cos 3x 32 þ cos 5x 52 þ PARSEVAL’S IDENTITY 13.41. By using Problem 13.35 and Parseval’s identity, show that ðaÞ X 1 n¼1 1 n4 ¼ 4 90 ðbÞ X 1 n¼1 1 n6 ¼ 6 945 13.42. Show that 1 12 32 þ 1 32 52 þ 1 52 72 þ ¼ 2 8 16 . [Hint: Use Problem 13.11.] 13.43. Show that (a) X 1 n¼1 1 ð2n 1Þ4 ¼ 4 96 ; ðbÞ X 1 n¼1 1 ð2n 1Þ6 ¼ 6 960 . 13.44. Show that 1 12 22 32 þ 1 22 32 42 þ 1 32 þ 42 þ 52 þ ¼ 42 39 16 . BOUNDARY-VALUE PROBLEMS 13.45. (a) Solve @U @t ¼ 2 @2 U @x2 subject to the conditions Uð0; tÞ ¼ 0; Uð4; tÞ ¼ 0; Uðx; 0Þ ¼ 3 sin x 2 sin 5x, where 0 x 4; t 0. (b) Give a possible physical interpretation of the problem and solution. Ans: ðaÞ Uðx; tÞ ¼ 3e22 t sin x 2e502 t sin 5x. 13.46. Solve @U @t ¼ @2 U @x2 subject to the conditions Uð0; tÞ ¼ 0; Uð6; tÞ ¼ 0; Uðx; 0Þ ¼ 1 0 x 3 0 3 x 6 and interpret physically. Ans: Uðx; tÞ ¼ X 1 m¼1 2 1 cosðm=3Þ m em2 2 t=36 sin mx 6 13.47. Show that if each side of equation (3), Page 356, is a constant c where c A 0, then there is no solution satisfying the boundary-value problem. 13.48. A flexible string of length is tightly stretched between points x ¼ 0 and x ¼ on the x-axis, its ends are fixed at these points. When set into small transverse vibration, the displacement Yðx; tÞ from the x-axis of any point x at time t is given by @2 Y @t2 ¼ a2 @2 Y @x2 , where a2 ¼ T=; T ¼ tension, ¼ mass per unit length. (a) Find a solution of this equation (sometimes called the wave equation) with a2 ¼ 4 which satisfies the conditions Yð0; tÞ ¼ 0; Yð; tÞ ¼ 0; Yðx; 0Þ ¼ 0:1 sin x þ 0:01 sin 4x; Ytðx; 0Þ ¼ 0 for 0 x ; t 0. (b) Interpret physically the boundary conditions in (a) and the solution. Ans. (a) Yðx; tÞ ¼ 0:1 sin x cos 2t þ 0:01 sin 4x cos 8t 13.49. (a) Solve the boundary-value problem @2 Y @t2 ¼ 9 @2 Y @x2 subject to the conditions Yð0; tÞ ¼ 0; Yð2; tÞ ¼ 0, Yðx; 0Þ ¼ 0:05xð2 xÞ; Ytðx; 0Þ ¼ 0, where 0 x 2; t 0. (b) Interpret physically. Ans: ðaÞ Yðx; tÞ ¼ 1:6 3 X 1 n¼1 1 ð2n 1Þ3 sin ð2n 1Þx 2 cos 3ð2n 1Þt 2 13.50. Solve the boundary-value problem @U @t ¼ @2 U @x2 ; Uð0; tÞ ¼ 1; Uð; tÞ ¼ 3; Uðx; 0Þ ¼ 2. [Hint: Let Uðx; tÞ ¼ Vðx; tÞ þ FðxÞ and choose FðxÞ so as to simplify the differential equation and boundary conditions for Vðx; tÞ:
- 371. Ans: Uðx; tÞ ¼ 1 þ 2x þ X 1 m¼1 4 cos m m em2 t sin mx 13.51. Give a physical interpretation to Problem 13.50. 13.52. Solve Problem 13.49 with the boundary conditions for Yðx; 0Þ and Ytðx; 0Þ interchanged, i.e., Yðx; Þ ¼ 0; Ytðx; 0Þ ¼ 0:05xð2 xÞ, and give a physical interpretation. Ans: Yðx; tÞ ¼ 3:2 34 X 1 n¼1 1 ð2n 1Þ4 sin ð2n 1Þx 2 sin 3ð2n 1Þt 2 13.53. Verify that the boundary-value problem of Problem 13.24 actually has the solution (14), Page 357. MISCELLANEOUS PROBLEMS 13.54. If x and 6¼ 0; 1; 2; . . . ; prove that 2 sin x sin ¼ sin x 12 2 2 sin 2x 22 2 þ 3 sin 3x 32 2 13.55. If x , prove that ðaÞ 2 sinh x sinh ¼ sin x 2 þ 12 2 sin 2x 2 þ 23 þ 3 sin 3x 2 þ 32 ðbÞ 2 cosh x sinh ¼ 1 2 cos x 2 þ 12 þ cos 2x 2 þ 22 13.56. Prove that sinh x ¼ x 1 þ x2 2 ! 1 þ x2 ð2Þ2 ! 1 þ x2 ð3Þ2 ! 13.57. Prove that cos x ¼ 1 4x2 2 ! 1 4x2 ð3Þ2 ! 1 4x2 ð5Þ2 ! [Hint: cos x ¼ ðsin 2xÞ=ð2 sin xÞ: 13.58. Show that (a) ffiffiffi 2 p 2 ¼ 1 3 5 7 9 22 13 15 . . . 2 2 6 6 10 10 14 14 . . . (b) ffiffiffi 2 p ¼ 4 4 4 8 8 12 12 16 16 . . . 3 5 7 9 11 13 15 17 . . . 13.59. Let r be any three dimensional vector. Show that (a) ðr iÞ2 þ ðr jÞ2 @ ðrÞ2 ; ðbÞ ðr iÞ2 þ ðr jÞ2 þ ðr kÞ2 ¼ r2 and discusse these with reference to Parseval’s identity. 13.60. If fnðxÞg; n ¼ 1; 2; 3; . . . is orthonormal in (a; b), prove that ðb a f ðxÞ X 1 n¼1 cnnðxÞ ( )2 dx is a minimum when cn ¼ ðb a f ðxÞ nðxÞ dx Discuss the relevance of this result to Fourier series. 362 FOURIER SERIES [CHAP. 13
- 372. 363 Fourier Integrals Fourier integrals are generalizations of Fourier series. The series representation a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L n o of a function is a periodic form on 1 x 1 obtained by gen- erating the coefficients from the function’s definition on the least period ½L; L. If a function defined on the set of all real numbers has no period, then an analogy to Fourier integrals can be envisioned as letting L ! 1 and replacing the integer valued index, n, by a real valued function . The coefficients an and bn then take the form Að Þ and Bð Þ. This mode of thought leads to the following definition. (See Problem 14.8.) THE FOURIER INTEGRAL Let us assume the following conditions on f ðxÞ: 1. f ðxÞ satisfies the Dirichlet conditions (Page 337) in every finite interval ðL; LÞ. 2. ð1 1 j f ðxÞj dx converges, i.e. f ðxÞ is absolutely integrable in ð1; 1Þ. Then Fourier’s integral theorem states that the Fourier integral of a function f is f ðxÞ ¼ ð1 0 fAð Þ cos x þ Bð Þ sin xg d ð1Þ where Að Þ ¼ 1 ð1 1 f ðxÞ cos x dx Bð Þ ¼ 1 ð1 1 f ðxÞ sin x dx 8 : (2) Að Þ and Bð Þ with 1 1 are generalizations of the Fourier coefficients an and bn. The right-hand side of (1) is also called a Fourier integral expansion of f . (Since Fourier integrals are improper integrals, a review of Chapter 12 is a prerequisite to the study of this chapter.) The result (1) holds if x is a point of continuity of f ðxÞ. If x is a point of discontinuity, we must replace f ðxÞ by f ðx þ 0Þ þ f ðx 0Þ 2 as in the case of Fourier series. Note that the above conditions are sufficient but not necessary. Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 373. In the generalization of Fourier coefficients to Fourier integrals, a0 may be neglected, since whenever ð1 1 f ðxÞ dx exists, ja0j ¼ 1 L ðL L f ðxÞ dx ! 0 as L ! 1 EQUIVALENT FORMS OF FOURIER’S INTEGRAL THEOREM Fourier’s integral theorem can also be written in the forms f ðxÞ ¼ 1 ð1 ¼0 ð1 u¼1 f ðuÞ cos ðx uÞ du d ð3Þ f ðxÞ ¼ 1 2 ð1 1 ei x d ð1 1 f ðuÞ ei u du ð4Þ ¼ 1 2 ð1 1 ð1 1 f ðuÞ ei ðuxÞ du d where it is understood that if f ðxÞ is not continuous at x the left side must be replaced by f ðx þ 0Þ þ f ðx 0Þ 2 . These results can be simplified somewhat if f ðxÞ is either an odd or an even function, and we have f ðxÞ ¼ 2 ð1 0 cos x d ð1 0 f ðuÞ cos u du if f ðxÞ is even ð5Þ f ðxÞ ¼ 2 ð1 0 sin x d ð1 0 f ðuÞ sin u du if f ðxÞ is odd ð6Þ An entity of importance in evaluating integrals and solving differential and integral equations is introduced in the next paragraph. It is abstracted from the Fourier integral form of a function, as can be observed by putting (4) in the form f ðxÞ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 ei x 1 ffiffiffiffiffiffi 2 p ð1 1 ei u f ðuÞ du d and observing the parenthetic expression. FOURIER TRANSFORMS From (4) it follows that Fð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ ei u du ð7Þ then f ðxÞ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 Fð Þ ei x d (8) The function Fð Þ is called the Fourier transform of f ðxÞ and is sometimes written Fð Þ ¼ ff f ðxÞg. The function f ðxÞ is the inverse Fourier transform of Fð Þ and is written f ðxÞ ¼ f1 fFð Þg. Note: The constants preceding the integral signs in (7) and (8) were here taken as equal to 1= ffiffiffiffiffiffi 2 p . However, they can be any constants different from zero so long as their product is 1=2. The above is called the symmetric form. The literature is not uniform as to whether the negative exponent appears in (7) or in (8). 364 FOURIER INTEGRALS [CHAP. 14
- 374. EXAMPLE. Determine the Fourier transform of f if f ðxÞ ¼ ex for x 0 and e2x when x 0. Fð Þ ¼ 1 2 ffiffiffiffiffiffi 2 p ð1 1 ei x f ðxÞ dx ¼ 1 ffiffiffiffiffiffi 2 p ð0 1 ei x e2x dx þ ð1 0 ei x ex dx ¼ 1 ffiffiffiffiffiffi 2 p ei þ2 i þ 2 x!0 x!1 þ ei 1 i 1 x!1 x!0þ ( ) ¼ 1 ffiffiffiffiffiffi 2 p 1 2 þ i þ 1 1 i If f ðxÞ is an even function, equation (5) yields Fcð Þ ¼ ffiffiffi 2 r ð1 0 f ðuÞ cos u du f ðxÞ ¼ ffiffiffi 2 r ð1 0 Fcð Þ cos x d 8 : ð9Þ and we call Fcð Þ and f ðxÞ Fourier cosine transforms of each other. If f ðxÞ is an odd function, equation (6) yields Fsð Þ ¼ ffiffiffi 2 r ð1 0 f ðuÞ sin u du f ðxÞ ¼ ffiffiffi 2 r ð1 0 Fsð Þ sin x d 8 : ð10Þ and we call Fsð Þ and f ðxÞ Fourier sine transforms of each other. Note: The Fourier transforms Fc and Fs are (up to a constant) of the same form as Að Þ and Bð Þ. Since f is even for Fc and odd for Fs, the domains can be shown to be 0 1. When the product of Fourier transforms is considered, a new concept called convolution comes into being, and in conjunction with it, a new pair (function and its Fourier transform) arises. In particular, if Fð Þ and Gð Þ are the Fourier transforms of f and g, respectively, and the convolution of f and g is defined to be f g ¼ 1 ffiffiffi p ð1 1 f ðuÞ gðx uÞ du ð11Þ then Fð Þ Gð Þ ¼ 1 ffiffiffi p ð1 1 ei u f g du ð12Þ f g ¼ 1 ffiffiffi p ð1 1 ei x Fð Þ Gð Þ d ð13Þ where in both (11) and (13) the convolution f g is a function of x. It may be said that multiplication is exchanged with convolution. Also ‘‘the Fourier transform of the convolution of two functions, f and g is the product of their Fourier transforms,’’ i.e., Tðf gÞ ¼ Gð f Þ TðgÞ: ðFð Þ Gð Þ and f g) are demonstrated to be a Fourier transform pair in Problem 14.29.) Now equate the representations of f g expressed in (11) and (13), i.e., 1 ffiffiffi p ð1 1 f ðuÞ gðx uÞ du ¼ 1 ffiffiffi p ð1 1 ei x Fð Þ Gð Þ d ð14Þ and let the parameter x be zero, then ð1 1 f ðuÞ gðuÞ du ¼ ð1 1 Fð Þ Gð Þ d ð15Þ CHAP. 14] FOURIER INTEGRALS 365
- 375. Now suppose that g ¼ f f and thus G ¼ F F, where the bar symbolizes the complex conjugate function. Then (15) takes the form ð1 1 j f ðuÞj2 du ¼ ð1 1 jFð Þj2 d ð16Þ This is Parseval’s theorem for Fourier integrals. Furthermore, if f and g are even functions, it can be shown that (15) reduces to the following Parseval identities: ð1 0 f ðuÞ gðuÞ du ¼ ð1 0 Fcð Þ Gcð Þ d ð17Þ where Fc and Gc are the Fourier cosine transforms of f and g. If f and g are odd functions, the (15) takes the form ð1 0 f ðuÞ gðuÞ du ¼ ð1 0 Fsð Þ Gsð Þ d ð18Þ where Fs and Gs are the Fourier sine transforms of f and g. (See Problem 14.3.) Solved Problems THE FOURIER INTEGRAL AND FOURIER TRANSFORMS 14.1. (a) Find the Fourier transform of f ðxÞ ¼ 1 jxj a 0 jxj a . (b) Graph f ðxÞ and its Fourier transform for a ¼ 3. (a) The Fourier transform of f ðxÞ is Fð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ ei u du ¼ 1 ffiffiffiffiffiffi 2 p ða a ð1Þ ei u du ¼ 1 ffiffiffiffiffiffi 2 p ei u i a a ¼ 1 ffiffiffiffiffiffi 2 p ei a ei a i ¼ ffiffiffi 2 r sin a ; 6¼ 0 For ¼ 0, we obtain Fð Þ ¼ ffiffiffiffiffiffiffiffi 2= p a. (b) The graphs of f ðxÞ and Fð Þ for a ¼ 3 are shown in Figures 14-1 and 14-2, respectively. 366 FOURIER INTEGRALS [CHAP. 14 f (x) 3 _3 _2 _1 1 2 3 x 2 1 O 1 Fig. 14-1 F(α) O 3 2 1 _1 _2p/3 2p/3 _p p _p/3 p/3 α Fig. 14-2
- 376. 14.2. (a) Use the result of Problem 14.1 to evaluate ð1 1 sin a cos x d ðbÞ Deduce the value of ð1 0 sin u u du: (a) From Fourier’s integral theorem, if Fð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ ei u du then f ðxÞ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 Fð Þ ei x d Then from Problem 14.1, 1 ffiffiffiffiffiffi 2 p ð1 1 ffiffiffi 2 r sin a ei x d ¼ 1 jxj a 1=2 jxj ¼ a 0 jxj a 8 : ð1Þ The left side of (1) is equal to 1 ð1 1 sin a cos x d i ð1 1 sin a sin x d ð2Þ The integrand in the second integral of (2) is odd and so the integral is zero. Then from (1) and (2), we have ð1 1 sin a cos x d ¼ jxj a =2 jxj ¼ a 0 jxj a 8 : ð3Þ Alternative solution: Since the function, f , in Problem 14.1 is an even function, the result follows immediately from the Fourier cosine transform (9). (b) If x ¼ 0 and a ¼ 1 in the result of (a), we have ð1 1 sin d ¼ or ð1 0 sin d ¼ 2 since the integrand is even. 14.3. If f ðxÞ is an even function show that: ðaÞ Fð Þ ¼ ffiffiffi 2 r ð1 0 f ðuÞ cos u du; ðbÞ f ðxÞ ¼ ffiffiffi 2 r ð1 0 Fð Þ cos x d : We have Fð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ ei u du ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ cos u du þ i ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ sin u du ð1Þ (a) If f ðuÞ is even, f ðuÞ cos u is even and f ðuÞ sin u is odd. Then the second integral on the right of (1) is zero and the result can be written Fð Þ ¼ 2 ffiffiffiffiffiffi 2 p ð1 0 f ðuÞ cos u du ¼ ffiffiffi 2 r ð1 0 f ðuÞ cos u du (b) From (a), Fð Þ ¼ Fð Þ so that Fð Þ is an even function. Then by using a proof exactly analogous to that in (a), the required result follows. A similar result holds for odd functions and can be obtained by replacing the cosine by the sine. 14.4. Solve the integral equation ð1 0 f ðxÞ cos x dx ¼ 1 0 @ @ 1 0 1 CHAP. 14] FOURIER INTEGRALS 367
- 377. Let ffiffiffi 2 r ð1 0 f ðxÞ cos x dx ¼ Fð Þ and choose Fð Þ ¼ ffiffiffiffiffiffiffiffi 2= p ð1 Þ 0 @ @ 1 0 1 . Then by Problem 14.3, f ðxÞ ¼ ffiffiffi 2 r ð1 0 Fð Þ cos x d ¼ ffiffiffi 2 r ð1 0 ffiffiffi 2 r ð1 Þ cos x d ¼ 2 ð1 0 ð1 Þ cos x d ¼ 2ð1 cos xÞ x2 14.5. Use Problem 14.4 to show that ð1 0 sin2 u u2 du ¼ 2 . As obtained in Problem 14.4, 2 ð1 0 1 cos x x2 cos x dx ¼ 1 0 @ @ 1 0 1 Taking the limit as ! 0þ, we find ð1 0 1 cos x x2 dx ¼ 2 But this integral can be written as ð1 0 2 sin2 ðx=2Þ x2 dx which becomes ð1 0 sin2 u u2 du on letting x ¼ 2u, so that the required result follows. 14.6. Show that ð1 0 cos x 2 þ 1 d ¼ 2 ex ; x A 0. Let f ðxÞ ¼ ex in the Fourier integral theorem f ðxÞ ¼ 2 ð1 0 cos x d ð1 0 f ðuÞ cos u du 2 ð1 0 cos x d ð1 0 eu cos u du ¼ ex Then But by Problem 12.22, Chapter 12, we have ð1 0 eu cos u du ¼ 1 2 þ 1 . Then 2 ð1 0 cos x 2 þ 1 d ¼ ex or ð1 0 cos x 2 þ 1 d ¼ 2 ex PARSEVAL’S IDENTITY 14.7. Verify Parseval’s identity for Fourier integrals for the Fourier transforms of Problem 14.1. We must show that ð1 1 f f ðxÞg2 dx ¼ ð1 1 fFð Þg2 d where f ðxÞ ¼ 1 jxj a 0 jxj a and Fð Þ ¼ ffiffiffi 2 r sin a : ( 368 FOURIER INTEGRALS [CHAP. 14
- 378. CHAP. 14] FOURIER INTEGRALS 369 This is equivalent to ða a ð1Þ2 dx ¼ ð1 1 2 sin2 a 2 d ð1 1 sin2 a 2 d ¼ 2 ð1 0 sin2 a 2 d ¼ a or ð1 0 sin2 a 2 d ¼ a 2 i.e., By letting a ¼ u and using Problem 14.5, it is seen that this is correct. The method can also be used to find ð1 0 sin2 u u2 du directly. PROOF OF THE FOURIER INTEGRAL THEOREM 14.8. Present a heuristic demonstration of Fourier’s integral theorem by use of a limiting form of Fourier series. Let f ðxÞ ¼ a0 2 þ X 1 n¼1 an cos nx L þ bn sin nx L ð1Þ where an ¼ 1 L ðL L f ðuÞ cos nu L du and bn ¼ 1 L ðL L f ðuÞ sin nu L du: Then by substitution (see Problem 13.21, Chapter 13), f ðxÞ ¼ 1 2L ðL L f ðuÞ du þ 1 L X 1 n¼1 ðL L f ðuÞ cos n L ðu xÞ du ð2Þ If we assume that ð1 1 j f ðuÞj du converges, the first term on the right of (2) approaches zero as L ! 1, while the remaining part appears to approach lim L!1 1 L X 1 n¼1 ð1 1 f ðuÞ cos n L ðu xÞ du ð3Þ This last step is not rigorous and makes the demonstration heuristic. Calling ¼ =L, (3) can be written f ðxÞ ¼ lim !0 X 1 n¼1 Fðn Þ ð4Þ where we have written Fð Þ ¼ 1 ð1 1 f ðuÞ cos ðu xÞ du ð5Þ But the limit (4) is equal to f ðxÞ ¼ ð1 0 Fð Þ d ¼ 1 ð1 0 d ð1 1 f ðuÞ cos ðu xÞ du which is Fourier’s integral formula. This demonstration serves only to provide a possible result. To be rigorous, we start with the integral 1 ð1 0 d ð1 1 f ðuÞ cos ðu xÞ dx and examine the convergence. This method is considered in Problems 14.9 through 14.12.
- 379. 14.9. Prove that: (a) lim !1 ðL 0 sin v v dv ¼ 2 ; ðbÞ lim !1 ð0 L sin v v dv ¼ 2 . (a) Let v ¼ y. Then lim !1 ðL 0 sin v v dv ¼ lim !1 ð L 0 sin y y dy ¼ ð1 0 sin y y dy ¼ 2 by Problem 12.29, Chap. 12. ðbÞ Let v ¼ y. Then lim !1 ð0 L sin v v dv ¼ lim !1 ð L 0 sin y y dy ¼ 2 : 14.10. Riemann’s theorem states that if FðxÞ is piecewise continuous in ða; bÞ, then lim !1 ðb a FðxÞ sin x dx ¼ 0 with a similar result for the cosine (see Problem 14.32). Use this to prove that ðaÞ lim !1 ðL 0 f ðx þ vÞ sin v v dv ¼ 2 f ðx þ 0Þ ðbÞ lim !1 ð0 L f ðx þ vÞ sin v v dv ¼ 2 f ðx 0Þ where f ðxÞ and f 0 ðxÞ are assumed piecewise continuous in ð0; LÞ and ðL; 0Þ respectively. (a) Using Problem 9(a), it is seen that a proof of the given result amounts to proving that lim !1 ðL 0 f f ðx þ vÞ f ðx þ 0Þg sin v v dv ¼ 0 This follows at once from Riemann’s theorem, because FðvÞ ¼ f ðx þ vÞ f ðx þ 0Þ v is piecewise contin- uous in ð0; LÞ since lim n!0þ FðvÞ exists and f ðxÞ is piecewise continuous. (b) A proof of this is analogous to that in part (a) if we make use of Problem 14.9(b). 14.11. If f ðxÞ satisfies the additional condition that ð1 1 j f ðxÞj dx converges, prove that ðaÞ lim !1 ð1 0 f ðx þ vÞ sin v v dv ¼ 2 f ðx þ 0Þ; ðbÞ lim !1 ð0 1 f ðx þ vÞ sin v v dv ¼ 2 f ðx 0Þ: We have ð1 0 f ðx þ vÞ sin v v dv ¼ ðL 0 f ðx þ vÞ sin v v dv þ ð1 L f ðx þ vÞ sin v v dv ð1Þ ð1 0 f ðx þ 0Þ sin v v dv ¼ ðL 0 f ðx þ 0Þ sin v v dv þ ð1 L f ðx þ 0Þ sin v v dv ð2Þ Subtracting, ð1 0 f f ðx þ vÞ f ðx þ 0Þg sin v v dv ¼ ðL 0 f f ðx þ vÞ f ðx þ 0Þg sin v v dv þ ð1 L f ðx þ vÞ sin v v dv ð1 L f ðx þ 0Þ sin v v dv Denoting the integrals in (3) by I; I1; I2, and I3, respectively, we have I ¼ I1 þ I2 þ I3 so that jIj @ jI1j þ jI2j þ jI3j ð4Þ jI2j @ ð1 L f ðx þ vÞ sin v v dv @ 1 L ð1 L j f ðx þ vÞj dv Now 370 FOURIER INTEGRALS [CHAP. 14
- 380. jI3j @ j f ðx þ 0Þj ð1 L sin v v dv Also Since ð1 0 j f ðxÞj dx and ð1 0 sin v v dv both converge, we can choose L so large that jI2j @ =3, jI3j @ =3. Also, we can choose so large that jI1j @ =3. Then from (4) we have jIj for and L sufficiently large, so that the required result follows. This result follows by reasoning exactly analogous to that in part (a). 14.12. Prove Fourier’s integral formula where f ðxÞ satisfies the conditions stated on Page 364. We must prove that lim L!1 1 ðL ¼0 ð1 u¼1 f ðuÞ cos ðx uÞ du d ¼ f ðx þ 0Þ þ f ðx 0Þ 2 Since ð1 1 f ðuÞ cos ðx uÞ du @ ð1 1 j f ðuÞj du, which converges, it follows by the Weierstrass test that ð1 1 f ðuÞ cos ðx uÞ du converges absolutely and uniformly for all . Thus, we can reverse the order of integration to obtain 1 ðL ¼0 d ð1 u¼1 f ðuÞ cos ðx uÞ du ¼ 1 ð1 u¼1 f ðuÞ du ðL ¼0 cos ðx uÞ d ¼ 1 ð1 u¼1 f ðuÞ sin Lðu xÞ u x du ¼ 1 ð1 u¼1 f ðx þ vÞ sin Lv v dv ¼ 1 ð0 1 f ðx þ vÞ sin Lv v dv þ 1 ð1 0 f ðx þ vÞ sin Lv v dv where we have let u ¼ x þ v. Letting L ! 1, we see by Problem 14.11 that the given integral converges to f ðx þ 0Þ þ f ðx 0Þ 2 as required. MISCELLANEOUS PROBLEMS 14.13. Solve @U @t ¼ @2 U @x2 subject to the conditions Uð0; tÞ ¼ 0; Uðx; 0Þ ¼ 1 0 x 1 0 x A 1 , Uðx; tÞ is bounded where x 0; t 0. We proceed as in Problem 13.24, Chapter 13. A solution satisfying the partial differential equation and the first boundary condition is given by Be2 t sin x. Unlike Problem 13.24, Chapter 13, the boundary conditions do not prescribe the specific values for , so we must assume that all values of are possible. By analogy with that problem we sum over all possible values of , which corresponds to an integration in this case, and are led to the possible solution Uðx; tÞ ¼ ð1 0 BðÞ e2 t sin x d ð1Þ where BðÞ is undetermined. By the second condition, we have ð1 0 BðÞ sin x d ¼ 1 0 x 1 0 x A 1 ¼ f ðxÞ ð2Þ from which we have by Fourier’s integral formula BðÞ ¼ 2 ð1 0 f ðxÞ sin x dx ¼ 2 ð1 0 sin x dx ¼ 2ð1 cos Þ ð3Þ CHAP. 14] FOURIER INTEGRALS 371
- 381. so that, at least formally, the solution is given by Uðx; tÞ ¼ 2 ð1 0 1 cos e2 t sin x dx ð4Þ See Problem 14.26. 14.14. Show that ex2 =2 is its own Fourier transform. Since ex2 =2 is even, its Fourier transform is given by ffiffiffiffiffiffiffiffi 2= p ¼ ð1 0 ex2 =2 cos x dx. Letting x ¼ ffiffiffi 2 p u and using Problem 12.32, Chapter 12, the integral becomes 2 ffiffiffi p ð1 0 eu2 cosð ffiffiffi 2 p uÞ du ¼ 2 ffiffiffi p ffiffiffi p 2 e 2 =2 ¼ e 2 =2 which proves the required result. 14.15. Solve the integral equation yðxÞ ¼ gðxÞ þ ð1 1 yðuÞ rðx uÞ du where gðxÞ and rðxÞ are given. Suppose that the Fourier transforms of yðxÞ; gðxÞ; and rðxÞ exist, and denote them by Yð Þ; Gð Þ; and Rð Þ, respectively. Then taking the Fourier transform of both sides of the given integral equation, we have by the convolution theorem Yð Þ ¼ Gð Þ þ ffiffiffiffiffiffi 2 p Yð Þ Rð Þ or Yð Þ ¼ Gð Þ 1 ffiffiffiffiffiffi 2 p Rð Þ yðxÞ ¼ f1 Gð Þ 1 ffiffiffiffiffiffi 2 p Rð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 Gð Þ 1 ffiffiffiffiffiffi 2 p Rð Þ ei x d Then assuming this integral exists. Supplementary Problems THE FOURIER INTEGRAL AND FOURIER TRANSFORMS 14.16. (a) Find the Fourier transform of f ðxÞ ¼ 1=2 jxj @ 0 jxj (b) Determine the limit of this transform as ! 0þ and discuss the result. Ans: ðaÞ 1 ffiffiffiffiffiffi 2 p sin ; ðbÞ 1 ffiffiffiffiffiffi 2 p 14.17. (a) Find the Fourier transform of f ðxÞ ¼ 1 x2 jxj 1 0 jxj 1 (b) Evaluate ð1 0 x cos x sin x x3 cos x 2 dx. Ans: ðaÞ 2 ffiffiffi 2 r cos sin 3 ; ðbÞ 3 16 372 FOURIER INTEGRALS [CHAP. 14
- 382. 14.18. If f ðxÞ ¼ 1 0 @ x 1 0 x A 1 find the (a) Fourier sine transform, (b) Fourier cosine transform of f ðxÞ. In each case obtain the graph of f ðxÞ and its transform. Ans: ðaÞ ffiffiffi 2 r 1 cos ; ðbÞ ffiffiffi 2 r sin : 14.19. (a) Find the Fourier sine transform of ex , x A 0 ðbÞ Show that ð1 0 x sin mx x2 þ 1 dx ¼ 2 em ; m 0 by using the result in ðaÞ: (c) Explain from the viewpoint of Fourier’s integral theorem why the result in (b) does not hold for m ¼ 0. Ans. (a) ffiffiffiffiffiffiffiffi 2= p ½ =ð1 þ 2 Þ 14.20. Solve for YðxÞ the integral equation ð1 0 YðxÞ sin xt dx ¼ 1 0 @ t 1 2 1 @ t 2 0 t A 2 8 : and verify the solution by direction substitution. Ans. YðxÞ ¼ ð2 þ 2 cos x 4 cos 2xÞ=x PARSEVAL’S IDENTITY 14.21. Evaluate (a) ð1 0 dx ðx2 þ 1Þ2 ; ðbÞ ð1 0 x2 dx ðx2 þ 1Þ2 by use of Parseval’s identity. [Hint: Use the Fourier sine and cosine transforms of ex , x 0.] Ans: ðaÞ =4; ðbÞ =4 14.22. Use Problem 14.18 to show that (a) ð1 0 1 cos x x 2 dx ¼ 2 ; ðbÞ ð1 0 sin4 x x2 dx ¼ 2 . 14.23. Show that ð1 0 ðx cos x sin xÞ2 x6 dx ¼ 15 . MISCELLANEOUS PROBLEMS 14.24. (a) Solve @U @t ¼ 2 @2 U @x2 , Uð0; tÞ ¼ 0; Uðx; 0Þ ¼ ex ; x 0; Uðx; tÞ is bounded where x 0; t 0. (b) Give a physical interpretation. Ans: Uðx; tÞ ¼ 2 ð1 0 e22 t sin x 2 þ 1 d 14.25. Solve @U @t ¼ @2 U @x2 ; Uxð0; tÞ ¼ 0; Uðx; 0Þ ¼ x 0 @ x @ 1 0 x 1 , Uðx; tÞ is bounded where x 0; t 0. Ans: Uðx; tÞ ¼ 2 ð1 0 sin þ cos 1 2 e2 t cos x d 14.26. (a) Show that the solution to Problem 14.13 can be written Uðx; tÞ ¼ 2 ffiffiffi p ðx=2 ffiffi t p 0 ev2 dv 1 ffiffiffi p ðð1þxÞ=2 ffiffi t p ð1xÞ=2 ffiffi t p ev2 dv CHAP. 14] FOURIER INTEGRALS 373
- 383. (b) Prove directly that the function in (a) satisfies @U @t ¼ @2 U @x2 and the conditions of Problem 14.13. 14.27. Verify the convolution theorem for the functions f ðxÞ ¼ gðxÞ ¼ 1 jxj 1 0 jxj 1 . 14.28. Establish equation (4), Page 364, from equation (3), Page 364. 14.29. Prove the result (12), Page 365. Hint: If Fð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 f ðuÞ ei u du and Gð Þ ¼ 1 ffiffiffiffiffiffi 2 p ð1 1 gðvÞ ei v dv, then Fð Þ Gð Þ ¼ 1 2 ð1 1 ð1 1 ei ðuþvÞ f ðuÞ gðvÞ du dv Now make the transformation u þ v ¼ x: Fð Þ Gð Þ ¼ 1 ffiffiffi p ð1 1 ð1 1 ei x f ðuÞ gðx uÞ du dx Define f g ¼ 1 ffiffiffi p ð1 1 f ðuÞ gðx uÞ du ð f g is a function of xÞ then Fð Þ Gð Þ ¼ 1 ffiffiffi p ð1 1 ð1 1 ei x f g dx Thus, Fð Þ Gð Þ is the Fourier transform of the convolution f g and conversely as indicated in (13) f g is the Fourier transform of Fð Þ Gð Þ. 14.30. If Fð Þ and Gð Þ are the Fourier transforms of f ðxÞ and gðxÞ respectively, prove (by repeating the pattern of Problem 14.29) that ð1 1 Fð Þ Gð Þ d ¼ ð1 1 f ðxÞ gðxÞ dx where the bar signifies the complex conjugate. Observe that if G is expressed as in Problem 14.29 then G Gð Þ ¼ 1 ð1 1 ei x f ðuÞ g gðvÞ dv 14.31. Show that the Fourier transform of gðu xÞ is ei x , i.e., ei x Gð Þ ¼ 1 ffiffiffi p ð1 1 ei u f ðuÞ gðu xÞ du Hint: See Problem 14.29. Let v ¼ u x. 14.32. Prove Riemann’s theorem (see Problem 14.10). 374 FOURIER INTEGRALS [CHAP. 14
- 384. 375 Gamma and Beta Functions THE GAMMA FUNCTION The gamma function may be regarded as a generalization of n! (n-factorial), where n is any positive integer to x!, where x is any real number. (With limited exceptions, the discussion that follows will be restricted to positive real numbers.) Such an extension does not seem reasonable, yet, in certain ways, the gamma function defined by the improper integral ðxÞ ¼ ð1 0 tx1 et dt ð1Þ meets the challenge. This integral has proved valuable in applications. However, because it cannot be represented through elementary functions, establishment of its properties take some effort. Some of the important ones are outlined below. The gamma function is convergent for x 0. (See Problem 12.18, Chapter 12.) The fundamental property ðx þ 1Þ ¼ xðxÞ ð2Þ may be obtained by employing the technique of integration by parts to (1). The process is carried out in Problem 15.1. From the form (2) the function ðxÞ can be evaluated for all x 0 when its values in the interval 1 % x 2 are known. (Any other interval of unit length will suffice.) The table and graph in Fig. 15-1 illustrates this idea. TABLES OF VALUES AND GRAPH OF THE GAMMA FUNCTION n ðnÞ 1.00 1.0000 1.10 0.9514 1.20 0.9182 1.30 0.8975 5 4 3 2 1 _1 _2 _3 _4 _5 _5 _4 _3 _2 _1 1 2 3 4 5 n Γ(n) Fig. 15-1 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 385. 1.40 0.8873 1.50 0.8862 1.60 0.8935 1.70 0.9086 1.80 0.9314 1.90 0.9618 2.00 1.0000 The equation (2) is a recurrence relationship that leads to the factorial concept. First observe that if x ¼ 1, then (1) can be evaluated, and in particular, ð1Þ ¼ 1: From (2) ðx þ 1Þ ¼ xðxÞ ¼ xðx 1Þðx 1Þ ¼ xðx 1Þðx 2Þ ðx kÞðx kÞ If x ¼ n, where n is a positive integer, then ðn þ 1Þ ¼ nðn 1Þðn 2Þ . . . 1 ¼ n! ð3Þ If x is a real number, then x! ¼ ðx þ 1Þ is defined by ðx þ 1Þ. The value of this identification is in intuitive guidance. If the recurrence relation (2) is characterized as a differential equation, then the definition of ðxÞ can be extended to negative real numbers by a process called analytic continuation. The key idea is that even though ðxÞ is defined in (1) is not convergent for x 0, the relation ðxÞ ¼ 1 x ðx þ 1Þ allows the meaning to be extended to the interval 1 x 0, and from there to 2 x 1, and so on. A general development of this concept is beyond the scope of this presentation; however, some information is presented in Problem 15.7. The factorial notion guides us to information about ðx þ 1Þ in more than one way. In the eighteenth century, Sterling introduced the formula (for positive integer values n) lim n!1 ffiffiffiffiffiffi 2 p nnþ1 en n! ¼ 1 ð4Þ This is called Sterling’s formula and it indicates that n! asymptotically approaches ffiffiffiffiffiffi 2 p nnþ1 en for large values of n. This information has proved useful, since n! is difficult to calculate for large values of n. There is another consequence of Sterling’s formula. It suggests the possibility that for sufficiently large values of x, x! ¼ ðx þ 1Þ ffiffiffiffiffiffi 2 p xxþ1 ex ð5aÞ (An argument supporting this is made in Problem 15.20.) It is known that ðx þ 1Þ satisfies the inequality ffiffiffiffiffiffi 2 p xxþ1 ex ðx þ 1Þ ffiffiffiffiffiffi 2 p xxþ1 ex e 1 12ðxþ1Þ ð5bÞ Since the factor e 1 12ðxþ1Þ ! 0 for large values of x, the suggested value (5a) of ðx þ 1Þ is consistent with (5b). An exact representation of ðx þ 1Þ is suggested by the following manipulation of n!. (It depends on ðn þ kÞ! ¼ ðk þ nÞ!.) n! ¼ lim k!1 12 . . . nðn þ 1Þ þ ðn þ 2Þ . . . ðn þ kÞ ðn þ 1Þðn þ 2Þ . . . ðn þ kÞ ¼ lim k!1 k! kn ðn þ 1Þ . . . ðn þ kÞ lim k!1 ðk þ 1Þðk þ 2Þ . . . ðk þ nÞ kn : Since n is fixed the second limit is one, therefore, n! ¼ lim k!1 k! kn ðn þ 1Þ . . . ðn þ kÞ : (This must be read as an infinite product.) 376 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 386. This factorial representation for positive integers suggests the possibility that ðx þ 1Þ ¼ x! ¼ lim k!1 k! kx ðx þ 1Þ . . . ðx þ kÞ x 6¼ 1; 2; k ð6Þ Gauss verified this identification back in the nineteenth century. This infinite product is symbolized by ðx; kÞ, i.e., ðx; kÞ ¼ k! kx ðx þ 1Þ ðx þ kÞ . It is called Gauss’s function and through this symbolism, ðx þ 1Þ ¼ lim k!1 ðx; kÞ ð7Þ The expression for 1 ðxÞ (which has some advantage in developing the derivative of ðxÞ) results as follows. Put (6a) in the form lim k!1 kx ð1 þ xÞð1 þ x=2Þ . . . ð1 þ x=kÞ x 6¼ 1 2 ; 1 3 ; . . . ; 1 k Next, introduce k ¼ 1 þ 1 2 þ 1 3 þ 1 k ln k Then ¼ lim k!1 k is Euler’s constant. This constant has been calculated to many places, a few of which are 0:57721566 . . . . By letting kx ¼ ex ln k ¼ ex½kþ1þ1=2þþ1=k , the representation (6) can be further modified so that ðx þ 1Þ ¼ ex lim k!1 ex 1 þ x ex=2 1 þ x=2 ex=k 1 þ x=k ¼ ex Y 1 k¼1 ex ex ln k = 1 þ x k ¼ Y 1 ¼1 kx k!ðk þ xÞ ¼ lim k!1 1 2 3 k ðx þ 1Þðx þ 2Þ ðx þ kÞ xx ¼ lim k!1 ðx; kÞ ð8Þ Since ðx þ 1Þ ¼ xðxÞ; 1 ðxÞ ¼ xex lim k!1 1 þ x ex 1 þ x=2 ex=2 1 þ x=k ex=k ¼ xex Y 1 ¼1 ð1 þ x=kÞ ex=k ð9Þ Another result of special interest emanates from a comparison of ðxÞð1 xÞ with the ‘‘famous’’ formula x sin x ¼ lim k!1 1 1 x2 1 1 ðx=2Þ2 1 ð1 x=kÞ2 ¼ Y 1 ¼1 f1 ðx=kÞ2 g ð10Þ (See Differential and Integral Calculus, by R. Courant (translated by E. J. McShane), Blackie Son Limited.) ð1 xÞ is obtained from ð yÞ ¼ 1 y ð y þ 1Þ by letting y ¼ x, i.e., ðxÞ ¼ 1 x ð1 xÞ or ð1 xÞ ¼ xðxÞ CHAP. 15] GAMMA AND BETA FUNCTIONS 377
- 387. Now use (8) to produce ðxÞð1 xÞ ¼ x1 ex lim k!1 Y 1 ¼1 ð1 þ x=kÞ1 ex=k ( )! ex lim k!1 Y 1 ¼1 ð1 x=kÞ1 ex=k ! ¼ 1 x lim k!1 Y 1 ¼1 ð1 ðx=kÞ2 Þ Thus ðxÞð1 xÞ ¼ sin x ; 0 x 1 ð11aÞ Observe that (11) yields the result ð1 2Þ ¼ ffiffiffi p ð11bÞ Another exact representation of ðx þ 1Þ is ðx þ 1Þ ¼ ffiffiffiffiffiffi 2 p xxþ1 ex 1 þ 1 12x þ 1 288x2 þ 139 51840x3 þ ð12Þ The method of obtaining this result is closely related to Sterling’s asymptotic series for the gamma function. (See Problem 15.20 and Problem 15.74.) The duplication formula 22x1 ðxÞ x þ 1 2 ¼ ffiffiffi p ð2xÞ ð13aÞ also is part of the literature. Its proof is given in Problem 15.24. The duplication formula is a special case ðm ¼ 2Þ of the following product formula: ðxÞ x þ 1 m x þ 2 m X þ m 1 m ¼ m 1 2mx ð2Þ m1 2 ðmxÞ ð13bÞ It can be shown that the gamma function has continuous derivatives of all orders. They are obtained by differentiating (with respect to the parameter) under the integral sign. It helps to recall that ðxÞ ¼ ð1 0 tx1 eyt dt and that if y ¼ tx1 , then ln y ¼ ln tx1 ¼ ðx 1Þ ln t. Therefore, 1 y y0 ¼ ln t. It follows that 0 ðxÞ ¼ ð1 0 tx1 et ln t dt: ð14aÞ This result can be obtained (after making assumptions about the interchange of differentiation with limits) by taking the logarithm of both sides of (9) and then differentiating. In particular, 0 ð1Þ ¼ ð is Euler’s constant.) ð14bÞ It also may be shown that 0 ðxÞ ðxÞ ¼ þ 1 1 1 x þ 1 2 1 x þ 1 þ 1 n 1 x þ n 1 ð15Þ (See Problem 15.73 for further information.) THE BETA FUNCTION The beta function is a two-parameter composition of gamma functions that has been useful enough in application to gain its own name. Its definition is 378 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 388. CHAP. 15] GAMMA AND BETA FUNCTIONS 379 Bðx; yÞ ¼ ð1 0 tx1 ð1 tÞy1 dt ð16Þ If x 1 and y 1, this is a proper integral. If x 0; y 0 and either or both x 1 or y 1, the integral is improper but convergent. It is shown in Problem 15.11 that the beta function can be expressed through gamma functions in the following way Bðx; yÞ ¼ ðxÞ ð yÞ ðx þ yÞ ð17Þ Many integrals can be expressed through beta and gamma functions. Two of special interest are ð=2 0 sin2x1 cos2y1 d ¼ 1 2 Bðx; yÞ ¼ 1 2 ðxÞ ð yÞ ðx þ yÞ ð18Þ ð1 0 xp1 1 þ x dx ¼ ð pÞ ð p 1Þ ¼ sin p 0 p 1 ð19Þ See Problem 15.17. Also see Page 377 where a classical reference is given. Finally, see Chapter 16, Problem 16.38 where an elegant complex variable resolution of the integral is presented. DIRICHLET INTEGRALS If V denotes the closed region in the first octant bounded by the surface x a p þ y b q þ z c r ¼ 1 and the coordinate planes, then if all the constants are positive, ð ð ð V x 1 y 1 z1 dx dy dz ¼ a b c pqr p q r 1 þ p þ q þ r ð20Þ Integrals of this type are called Dirichlet integrals and are often useful in evaluating multiple integrals (see Problem 15.21). Solved Problems THE GAMMA FUNCTION 15.1. Prove: (a) ðx þ 1Þ ¼ xðxÞ; x 0; ðbÞ ðn þ 1Þ ¼ n!; n ¼ 1; 2; 3; . . . . ðaÞ ðv þ 1Þ ¼ ð1 0 xv ex dx ¼ lim M!1 ðM 0 xv ex dx ¼ lim M!1 ðxv Þðex ÞjM 0 ðM 0 ðex Þðvxv1 Þ dx ¼ lim M!1 Mv eM þ v ðM 0 xv1 ex dx ¼ vðvÞ if v 0 ðbÞ ð1Þ ¼ ð1 0 ex dx ¼ lim M!1 ðM 0 ex dx ¼ lim M!1 ð1 eM Þ ¼ 1: Put n ¼ 1; 2; 3; . . . in ðn þ 1Þ ¼ nðnÞ. Then ð2Þ ¼ 1ð1Þ ¼ 1; ð3Þ ¼ 2ð2Þ ¼ 2 1 ¼ 2!; ð4Þ ¼ 3ð3Þ ¼ 3 2! ¼ 3! In general, ðn þ 1Þ ¼ n! if n is a positive integer.
- 389. 15.2. Evaluate each of the following. ðaÞ ð6Þ 2ð3Þ ¼ 5! 2 2! ¼ 5 4 3 2 2 2 ¼ 30 ðbÞ ð5 2Þ ð1 2Þ ¼ 3 2 ð3 2Þ ð1 2Þ ¼ 3 2 1 2 ð1 2Þ ð1 2Þ ¼ 3 4 ðcÞ ð3Þ ð2:5Þ ð5:5Þ ¼ 2!ð1:5Þð0:5Þ ð0:5Þ ð4:5Þð3:5Þð2:5Þð1:5Þð0:5Þ ð0:5Þ ¼ 16 315 ðdÞ 6 ð8 3Þ 5 ð2 3Þ ¼ 6ð5 3Þð2 3Þ ð2 3Þ 5 ð2 3Þ ¼ 4 3 15.3. Evaluate each integral. ðaÞ ð1 0 x3 ex dx ¼ ð4Þ ¼ 3! ¼ 6 ðbÞ ð1 0 x6 e2x dx: Let 2x ¼ 7. Then the integral becomes ð1 0 y 2 6 ey dy 2 ¼ 1 27 ð1 0 y6 ey dy ¼ ð7Þ 27 ¼ 6! 27 ¼ 45 8 15.4. Prove that ð1 2Þ ¼ ffiffiffi p . ð1 2Þ ¼ ð1 0 x1=2 ex dx. Letting x ¼ u2 this integral becomes 2 ð1 0 eu2 du ¼ 2 ffiffiffi p 2 ¼ ffiffiffi p using Problem 12.31, Chapter 12 This result also is described in equation (11a,b) earlier in the chapter. 15.5. Evaluate each integral. ðaÞ ð1 0 ffiffiffi y p ey2 dy. Letting y3 ¼ x, the integral becomes ð1 0 ffiffiffiffiffiffiffiffi x1=3 p ex 1 3 x2=3 dx ¼ 1 3 ð1 0 x1=2 ex dx ¼ 1 3 1 2 ¼ ffiffiffi p 3 ðbÞ ð1 0 34x2 dx ¼ ð1 0 ðeln 3 Þð4x2 Þ dz ¼ ð1 0 eð4 ln 3Þz2 dz. Let ð4 ln 3Þz2 ¼ x and the integral becomes ð1 0 ex d x1=2 ffiffiffiffiffiffiffiffiffiffiffi 4 ln 3 p ! ¼ 1 2 ffiffiffiffiffiffiffiffiffiffiffi 4 ln 3 p ð1 0 x1=2 ex dx ¼ ð1=2Þ 2 ffiffiffiffiffiffiffiffiffiffiffi 4 ln 3 p ¼ ffiffiffi p 4 ffiffiffiffiffiffiffiffi ln 3 p (c) ð1 0 dx ffiffiffiffiffiffiffiffiffiffiffiffi ln x p : Let ln x ¼ u. Then x ¼ eu . When x ¼ 1; u ¼ 0; when x ¼ 0; u ¼ 1. The integral becomes ð1 0 eu ffiffiffi u p du ¼ ð1 0 u1=2 eu du ¼ ð1=2Þ ¼ ffiffiffi p 15.6. Evaluate ð1 0 xm eaxn dx where m; n; a are positive constants. 380 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 390. Letting axn ¼ y, the integral becomes ð1 0 y a 1=n m ey d y a 1=n ¼ 1 naðmþ1Þ=n ð1 0 yðmþ1Þ=n1 ey dy ¼ 1 naðmþ1Þ=n m þ 1 n 15.7. Evaluate (a) ð1=2Þ; ðbÞ ð5=2Þ. We use the generalization to negative values defined by ðxÞ ¼ ðx þ 1Þ x . ðaÞ Letting x ¼ 1 2 ; ð1=2Þ ¼ ð1=2Þ 1=2 ¼ 2 ffiffiffi p : ðbÞ Letting x ¼ 3=2; ð3=2Þ ¼ ð1=2Þ 3=2 ¼ 2 ffiffiffi p 3=2 ¼ 4 ffiffiffi p 3 ; using ðaÞ: Then ð5=2Þ ¼ ð3=2Þ 5=2 ¼ 8 15 ffiffiffi p : 15.8. Prove that ð1 0 xm ðln xÞn dx ¼ ð1Þn n! ðm þ 1Þnþ1 , where n is a positive integer and m 1. Letting x ¼ ey , the integral becomes ð1Þn ð1 0 yn eðmþ1Þy dy. If ðm þ 1Þy ¼ u, this last integral becomes ð1Þn ð1 0 un ðm þ 1Þn eu du m þ 1 ¼ ð1Þn ðm þ 1Þnþ1 ð1 0 un eu du ¼ ð1Þn ðm þ 1Þnþ1 ðn þ 1Þ ¼ ð1Þn n! ðm þ 1Þnþ1 Compare with Problem 8.50, Chapter 8, page 203. 15.9. A particle is attracted toward a fixed point O with a force inversely proportional to its instanta- neous distance from O. If the particle is released from rest, find the time for it to reach O. At time t ¼ 0 let the particle be located on the x-axis at x ¼ a 0 and let O be the origin. Then by Newton’s law m d2 x dt2 ¼ k x ð1Þ where m is the mass of the particle and k 0 is a constant of proportionality. Let dx dt ¼ v, the velocity of the particle. Then d2 x dt2 ¼ dv dt ¼ dv dx dx dt ¼ v dv dx and (1) becomes mv dv dx ¼ k x or mv2 2 ¼ k ln x þ c ð2Þ upon integrating. Since v ¼ 0 at x ¼ a, we find c ¼ k ln a. Then mv2 2 ¼ k ln a x or v ¼ dx dt ¼ ffiffiffiffiffi 2k m r ffiffiffiffiffiffiffiffiffi ln a x r ð3Þ where the negative sign is chosen since x is decreasing as t increases. We thus find that the time T taken for the particle to go from x ¼ a to x ¼ 0 is given by T ¼ ffiffiffiffiffi m 2k r ða 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffi ln a=x p ð4Þ CHAP. 15] GAMMA AND BETA FUNCTIONS 381
- 391. Letting ln a=x ¼ u or x ¼ aeu , this becomes T ¼ a ffiffiffiffiffi m 2k r ð1 0 u1=2 eu du ¼ a ffiffiffiffiffi m 2k r ð1 2Þ ¼ a ffiffiffiffiffiffiffi m 2k r THE BETA FUNCTION 15.10. Prove that (a) Bðu; vÞ ¼ Bðv; uÞ; ðbÞ Bðu; vÞ ¼ 2 ð=2 0 sin2u1 cos2v1 d. (a) Using the transformation x ¼ 1 y, we have Bðu; vÞ ¼ ð1 0 xu1 ð1 xÞv1 dx ¼ ð1 0 ð1 yÞu1 yv1 dy ¼ ð1 0 yv1 ð1 yÞu1 dy ¼ Bðv; uÞ (b) Using the transformation x ¼ sin2 , we have Bðu; vÞ ¼ ð1 0 xu1 ð1 xÞv1 dx ¼ ð=2 0 ðsin2 Þu1 ðcos2 Þv1 2 sin cos d ¼ 2 ð=2 0 sin2u1 cos2v1 d 15.11. Prove that Bðu; vÞ ¼ ðuÞ ðvÞ ðu þ vÞ u; v 0. Letting z2 ¼ x2 ; we have ðuÞ ¼ ð1 0 zu1 ez dx ¼ 2 ð1 0 x2u1 ex2 dx: Similarly, ðvÞ ¼ 2 ð1 0 y2v1 ey2 dy: Then ðuÞ ðvÞ ¼ 4 ð1 0 x2u1 ex2 dx ð1 0 y2v1 ey2 dy ¼ 4 ð1 0 ð1 0 x2u1 y2v1 eðx2 þy2 Þ dx dy Transforming to polar coordiantes, x ¼ cos ; y ¼ sin , ðuÞ ðvÞ ¼ 4 ð=2 ¼0 ð1 ¼0 2ðuþvÞ1 e2 cos2u1 sin2v1 d d ¼ 4 ð1 ¼0 2ðuþvÞ1 e2 d ð=2 ¼0 cos2u1 sin2v1 d ¼ 2ðu þ vÞ ð=2 0 cos2u1 sin2v1 d ¼ ðu þ vÞ Bðv; uÞ ¼ ðu þ vÞ Bðu; vÞ using the results of Problem 15.10. Hence, the required result follows. The above argument can be made rigorous by using a limiting procedure as in Problem 12.31, Chapter 12. 15.12. Evaluate each of the following integrals. ðaÞ ð1 0 x4 ð1 xÞ3 dx ¼ Bð5; 4Þ ¼ ð5Þ ð4Þ ð9Þ ¼ 4!3! 8! ¼ 1 280 382 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 392. ðbÞ ð2 0 x2 dx ffiffiffiffiffiffiffiffiffiffiffi 2 x p : Letting x ¼ 2v; the integral becomes 4 ffiffiffi 2 p ð1 0 v2 ffiffiffiffiffiffiffiffiffiffiffi 1 v p dv ¼ 4 ffiffiffi 2 p ð1 0 v2 ð1 vÞ1=2 dv ¼ 4 ffiffiffi 2 p Bð3; 1 2Þ ¼ 4 ffiffiffi 2 p ð3Þ ð1=2Þ ð7=2Þ ¼ 64 ffiffiffi 2 p 15 ðcÞ ða 0 y4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 y2 q dy: Letting y2 ¼ a2 x or y ¼ ffiffiffi x p ; the integral becomes a6 ð1 0 x3=2 ð1 xÞ1=2 dx ¼ a6 Bð5=2; 3=2Þ ¼ a6 ð5=2Þ ð3=2Þ ð4Þ ¼ a6 16 15.13. Show that ð=2 0 sin2u1 cos2v1 d ¼ ðuÞ ðvÞ 2 ðu þ vÞ u; v 0. This follows at once from Problems 15.10 and 15.11. 15.14. Evaluate (a) ð=2 0 sin6 d; ðbÞ ð=2 0 sin4 cos5 d; ðcÞ ð 0 cos4 d. ðaÞ Let 2u 1 ¼ 6; 2v 1 ¼ 0; i.e., u ¼ 7=2; v ¼ 1=2; in Problem 15.13: Then the required integral has the value ð7=2Þ ð1=2Þ 2 ð4Þ ¼ 5 32 : ðbÞ Letting 2u 1 ¼ 4; 2v 1 ¼ 5; the required integral has the value ð5=2Þ ð3Þ 2 ð11=2Þ ¼ 8 315 : ðcÞ The given integral ¼ 2 ð=2 0 cos4 d: Thus letting 2u 1 ¼ 0; 2v 1 ¼ 4 in Problem 15.13, the value is 2 ð1=2Þ ð5=2Þ 2 ð3Þ ¼ 3 8 . 15.15. Prove ð=2 0 sinp d ¼ ð=2 0 cosp d ¼ ðaÞ 1 3 5 ð p 1Þ 2 4 6 p 2 if p is an even positive integer, (b) 2 4 6 ð p 1Þ 1 3 5 p is p is an odd positive integer. From Problem 15.13 with 2u 1 ¼ p; 2v 1 ¼ 0, we have ð=2 0 sinp d ¼ ½1 2 ð p þ 1Þ ð1 2Þ 2 ½1 2 ð p þ 2Þ (a) If p ¼ 2r, the integral equals ðr þ 1 2Þ ð1 2Þ 2 ðr þ 1Þ ¼ ðr 1 2Þðr 3 2Þ 1 2 ð1 2Þ ð1 2Þ 2rðr 1Þ 1 ¼ ð2r 1Þð2r 3Þ 1 2rð2r 2Þ 2 2 ¼ 1 3 5 ð2r 1Þ 2 4 6 2r 2 (b) If p ¼ 2r þ 1, the integral equals ðr þ 1Þ ð1 2Þ 2 ðr þ 3 2Þ ¼ rðr 1Þ 1 ffiffiffi p 2ðr þ 1 2Þðr 1 2Þ 1 2 ffiffiffi p ¼ 2 4 6 2r 1 3 5 ð2r þ 1Þ In both cases ð=2 0 sinp d ¼ ð=2 0 cosp d, as seen by letting ¼ =2 . CHAP. 15] GAMMA AND BETA FUNCTIONS 383
- 393. 15.16. Evaluate (a) ð=2 0 cos6 d; ðbÞ ð=2 0 sin3 cos2 d; ðcÞ ð2 0 sin8 d. (a) From Problem 15.15 the integral equals 1 3 5 2 4 6 ¼ 5 32 [compare Problem 15.14(a)]. (b) The integral equals ð=2 0 sin3 ð1 sin2 Þ d ¼ ð=2 0 sin3 d ð=2 0 sin5 d ¼ 2 1 3 2 4 1 3 5 ¼ 2 15 The method of Problem 15.14(b) can also be used. ðcÞ The given integral equals 4 ð=2 0 sin8 d ¼ 4 1 3 5 7 2 4 6 8 2 ¼ 35 64 : 15.17. Given ð1 0 xp1 1 þ x dx ¼ sin p , show that ð pÞ ð1 pÞ ¼ sin p , where 0 p 1. Letting x 1 þ x ¼ y or x ¼ y 1 y , the given integral becomes ð1 0 yp1 ð1 yÞp dy ¼ Bð p; 1 pÞ ¼ ð pÞ ð1 pÞ and the result follows. 15.18. Evaluate ð1 0 dy 1 þ y4 . Let y4 ¼ x. Then the integral becomes 1 4 ð1 0 x3=4 1 þ x dx ¼ 4 sinð=4Þ ¼ ffiffiffi 2 p 4 by Problem 15.17 with p ¼ 1 4. The result can also be obtained by letting y2 ¼ tan . 15.19. Show that ð2 0 x ffiffiffiffiffiffiffiffiffiffiffiffiffi 8 x3 3 p dx ¼ 16 9 ffiffiffi 3 p . Letting x3 8y or x ¼ 2y1=3 , the integral comes ð1 0 2y1=3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8ð1 yÞ 3 p 2 3 y2=3 dy ¼ 8 3 ð1 0 y1=3 ð1 yÞ1=3 dy ¼ 8 3 Bð2 3 ; 4 3Þ ¼ 8 3 ð2 3 ð4 3Þ ð2Þ ¼ 8 9 ð1 3Þ ð2 3Þ ¼ 8 9 sin =3 ¼ 16 9 ffiffiffi 3 p STIRLING’S FORMULA 15.20. Show that for large positive integers n; n! ¼ ffiffiffiffiffiffiffiffi 2n p nn en approximately. By definition ðzÞ ¼ ð1 0 tz1 et dt. Let lfz ¼ x þ 1 then ðx þ 1Þ ¼ ð1 0 tx et dt ¼ ð1 0 etþln tx dt ¼ ð1 0 etþx ln t dt ð1Þ For a fixed value of x the function x ln t t has a relative maximum for t ¼ x (as is demonstrated by elementary ideas of calculus). The substutition t ¼ x þ y yields ðx þ 1Þ ¼ ex ð1 x ex lnðxþyÞy dy ¼ xx ex ð1 x ex lnð1þy xÞy dy ð2Þ 384 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 394. To this point the analysis has been rigorous. The following formal steps can be made rigorous by incorporating appropriate limiting procedures; however, because of the difficulty of the proofs, they shall be omitted. In (2) introduce the logarithmic expansion ln 1 þ y x ¼ y x y2 2x2 þ y3 3x3 þ ð3Þ and also let y ¼ ffiffiffi x p v; dy ¼ ffiffiffi x p dv Then ðx þ 1Þ ¼ xx ex ffiffiffi x p ð1 x ev2 =2þðv3 =3Þ ffiffi x p dv ð4Þ For large values of x ðx þ 1Þ xx ex ffiffiffi x p ð1 x ev2 =2 dv ¼ xx ex ffiffiffiffiffiffiffiffi 2x p When x is replaced by integer values n, then the Stirling relation n! ¼ ðx þ 1Þ ffiffiffiffiffiffiffiffi 2x p xx ex ð5Þ is obtained. It is of interest that from (4) we can also obtain the result (12) on Page 378. See Problem 15.72. DIRICHLET INTEGRALS 15.21. Evaluate I ¼ ð ð ð V x 1 y 1 z1 dx dy dz where V is the region in the first octant bounded by the sphere x2 þ y2 þ z2 ¼ 1 and the coordinate planes. Let x2 ¼ u; y2 ¼ v; z2 ¼ w. Then I ¼ ð ð ð r uð 1Þ=2 vð 1Þ=2 wð1Þ=2 du 2 ffiffiffi u p dv 2 ffiffiffi v p dw 2 ffiffiffiffi w p ¼ 1 8 ð ð ð r uð =2Þ1 vð =2Þ1 wð=2Þ1 du dv dw ð1Þ where r is the region in the uvw space bounded by the plane u þ v þ w ¼ 1 and the uv; vw, and uw planes as in Fig. 15-2. Thus, I ¼ 1 8 ð1 u¼0 ð1u v¼0 ð1uv w¼0 uð =2Þ1 vð =2Þ1 wð=2Þ1 du dv dw ð2Þ ¼ 1 4 ð1 u¼0 ð1u v¼0 uð =2Þ1 vð =2Þ1 ð1 u vÞ=2 du dv ¼ 1 4 ð1 u¼0 uð =2Þ1 ð1u v¼0 vð =2Þ1 ð1 u vÞ=2 dv du Letting v ¼ ð1 uÞt, we have ð1u v¼0 vð =2Þ1 ð1 u vÞ=2 dv ¼ ð1 uÞð þÞ=2 ð1 t¼0 tð =2Þ1 ð1 tÞ=2 dt ¼ ð1 uÞð þÞ=2 ð =2Þ ð=2 þ 1Þ ½ð þ Þ=2 þ 1 CHAP. 15] GAMMA AND BETA FUNCTIONS 385 Fig. 15-2
- 395. so that (2) becomes I ¼ 1 4 ð =2Þ ð=2 þ 1Þ ½ð þ Þ=2 þ 1 ð1 u¼0 uð =2Þ1 ð1 uÞð þÞ=2 du ð3Þ ¼ 1 4 ð =2Þ ð=2 þ 1Þ ½ð þ Þ=2 þ 1 ð =2Þ ½ð þ Þ=2 þ 1 ½ð þ þ Þ=2 þ 1 ¼ ð =2Þ ð =2Þ ð=2Þ 8 ½ð þ þÞ=2 þ 1 where we have used ð=2Þ ð=2Þ ¼ ð=2 þ 1Þ. The integral evaluated here is a special case of the Dirichlet integral (20), Page 379. The general case can be evaluated similarly. 15.22. Find the mass of the region bounded by x2 þ y2 þ z2 ¼ a2 if the density is ¼ x2 y2 z2 . The required mass ¼ 8 ð ð ð V x2 y2 z2 dx dy dz, where V is the region in the first octant bounded by the sphere x2 þ y2 þ z2 ¼ a2 and the coordinate planes. In the Dirichlet integral (20), Page 379, let b ¼ c ¼ a; p ¼ q ¼ r ¼ 2 and ¼ ¼ ¼ 3. Then the required result is 8 a3 a3 a3 2 2 2 ð3=2Þ ð3=2Þ ð3=2Þ ð1 þ 3=2 þ 3=2 þ 3=2Þ ¼ 4s9 945 MISCELLANEOUS PROBLEMS 15.23. Show that ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x4 p dx ¼ fð1:4Þg2 6 ffiffiffiffiffiffi 2 p . Let x4 ¼ y. Then the integral becomes 1 4 ð1 0 y3=4 ð1 yÞ1=2 dy ¼ 1 4 ð1=4Þ ð3=2Þ ð7=4Þ ¼ ffiffiffi p 6 fð1=4Þg2 ð1:4Þ ð3=4Þ From Problem 15.17 with p ¼ 1=4; ð1=4Þ ð3=4Þ ¼ ffiffiffi 2 p so that the required result follows. 15.24. Prove the duplication formula 22p1 ð pÞ ð p þ 1 2Þ ¼ ffiffiffi p ð2pÞ. Let I ¼ ð=2 0 sin2p x dx; J ¼ ð=2 0 sin2p 2x dx. Then I ¼ 1 2 Bð p þ 1 2 ; 1 2Þ ¼ ð p þ 1 2Þ ffiffiffi p 2 ð p þ 1Þ Letting 2x ¼ u, we find J ¼ 1 2 ð 0 sin2p u du ¼ ð=2 0 sin2p u du ¼ I J ¼ ð=2 0 ð2 sin x cos xÞ2p dx ¼ 22p ð=2 0 sin2p x cos2p x dx But ¼ 22p1 Bð p þ 1 2 ; p þ 1 2Þ ¼ 22p1 fð p þ 1 2Þg2 ð2p þ 1Þ Then since I ¼ J, ð p þ 1 2Þ ffiffiffi p 2p ð pÞ ¼ 22p1 fð p þ 1 2Þg2 2p ð2pÞ 386 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 396. and the required result follows. (See Problem 15.74, where the duplication formula is developed for the simpler case of integers.) 15.25. Show that ð=2 0 d ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 2 sin2 q ¼ fð1=4Þg2 4 ffiffiffi p . Consider I ¼ ð=2 0 d ffiffiffiffiffiffiffiffiffiffi cos p ¼ ð=2 0 cos1=2 d ¼ 1 2 Bð1 4 ; 1 2Þ ¼ ð1 4Þ ffiffiffi p 2 ð3 4Þ ¼ fð1 4Þg2 2 ffiffiffiffiffiffi 2 p as in Problem 15.23. But I ¼ ð=2 0 d ffiffiffiffiffiffiffiffiffiffi cos p ¼ ð=2 0 d ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos2 =2 sin2 =2 p ¼ ð=2 0 d ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 sin2 =2 p : Letting ffiffiffi 2 p sin =2 ¼ sin in this last integral, it becomes ffiffiffi 2 p ð=2 0 d ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 1 2 sin2 q , from which the result follows. 15.26. Prove that ð1 0 cos x xp dx ¼ 2 ð pÞ cosð p=2Þ ; 0 p 1. We have 1 xp ¼ 1 ð pÞ ð1 0 up1 exu du. Then ð1 0 cos x xp dx ¼ 1 ð pÞ ð1 0 ð1 0 up1 exu cos x du dx ¼ 1 ð pÞ ð1 0 up 1 þ u2 du ð1Þ where we have reversed the order of integration and used Problem 12.22, Chapter 12. Letting u2 ¼ v in the last integral, we have by Problem 15.17 ð1 0 up 1 þ u2 du ¼ 1 2 ð1 0 vð p1Þ=2 1 þ v dv ¼ 2 sinð p þ 1Þ=2 ¼ 2 cos p=2 ð2Þ Substitution of (2) in (1) yields the required result. 15.27. Evaluate ð1 0 cos x2 dx. Letting x2 ¼ y, the integral becomes 1 2 ð1 0 cos y ffiffiffi y p dy ¼ 1 2 2 ð1 2Þ cos =4 ! ¼ 1 2 ffiffiffiffiffiffiffiffi =2 p by Problem 15.26. This integral and the corresponding one for the sine [see Problem 15.68(a)] are called Fresnel integrals. Supplementary Problems THE GAMMA FUNCTION 15.28. Evaluate (a) ð7Þ 2 ð4Þ ð3Þ ; ðbÞ ð3Þ ð3=2Þ ð9=2Þ ; ðcÞ ð1=2Þ ð3=2Þ ð5=2Þ. Ans. ðaÞ 30; ðbÞ 16=105; ðcÞ 3 8 3=2 CHAP. 15] GAMMA AND BETA FUNCTIONS 387
- 397. 15.29. Evaluate (a) ð1 0 x4 ex dx; ðbÞ ð1 0 x6 e3x dx; ðcÞ ð1 0 x2 e2x2 dx: Ans. ðaÞ 24; ðbÞ 80 243 ; ðcÞ ffiffiffiffiffiffi 2 p 16 15.30. Find (a) ð1 0 ex2 dx; ðbÞ ð1 0 ffiffiffi x 4 p e ffiffi x p dx; ðcÞ ð1 0 y3 e2y5 dy. Ans. ðaÞ 1 3 ð1 3Þ; ðbÞ 3 ffiffiffi p 2 ; ðcÞ ð4=5Þ 5 ffiffiffiffiffi 16 5 p 15.31. Show that ð1 0 est ffiffi t p dt ¼ ffiffiffi 8 r ; s 0. 15.32. Prove that ðvÞ ¼ ð1 0 ln 1 x v1 dx; v 0. 15.33. Evaluate (a) ð1 0 ðln xÞ4 dx; ðbÞ ð1 0 ðx ln xÞ3 dx; ðcÞ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnð1=xÞ 3 p dx. Ans: ðaÞ 24; ðbÞ 3=128; ðcÞ 1 3 ð1 3Þ 15.34. Evaluate (a) ð7=2Þ; ðbÞ ð1=3Þ. Ans: ðaÞ ð16 ffiffiffi p Þ=105; ðbÞ 3 ð2=3Þ 15.35. Prove that lim x!m ðxÞ ¼ 1 where m ¼ 0; 1; 2; 3; . . . 15.36. Prove that if m is a positive interger, ðm þ 1 2Þ ¼ ð1Þm 2m ffiffiffi p 1 3 5 ð2m 1Þ 15.37. Prove that 0 ð1Þ ¼ ð1 0 ex ln x dx is a negative number (it is equal to , where ¼ 0:577215 . . . is called Euler’s constant as in Problem 11.49, Page 296). THE BETA FUNCTION 15.38. Evaluate (a) Bð3; 5Þ; ðbÞ Bð3=2; 2Þ; ðcÞ Bð1=3; 2=3Þ: Ans: ðaÞ 1=105; ðbÞ 4=15; ðcÞ 2= ffiffiffi 3 p 15.39. Find (a) ð1 0 x2 ð1 xÞ3 dx; ðbÞ ð1 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 xÞ=x p dx; ðcÞ ð2 0 ð4 x2 Þ3=2 dx. Ans: ðaÞ 1=60; ðbÞ =2; ðcÞ 3 15.40. Evaluate (a) ð4 0 u3=2 ð4 uÞ5=2 du; ðbÞ ð3 0 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3x x2 p : Ans: ðaÞ 12; ðbÞ 15.41. Prove that ða 0 dy ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a4 y4 p ¼ fð1=4Þg2 4a ffiffiffiffiffiffi 2 p : 15.42. Evaluate (a) ð=2 0 sin4 cos4 d; ðbÞ ð2 0 cos6 d: Ans: ðaÞ 3=256; ðbÞ 5=8 15.43. Evaluate (a) ð 0 sin5 d; ðbÞ ð=2 0 cos5 sin2 d: Ans: ðaÞ 16=15; ðbÞ 8=105 15.44. Prove that ð=2 0 ffiffiffiffiffiffiffiffiffiffi tan p d ¼ = ffiffiffi 2 p . 388 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 398. 15.45. Prove that (a) ð1 0 x dx 1 þ x6 ¼ 3 ffiffiffi 3 p ; ðbÞ ð1 0 y2 dy 1 þ y4 ¼ 2 ffiffiffi 2 p . 15.46. Prove that ð1 1 e2x ae3x þ b dx ¼ 2 3 ffiffiffi 3 p a2=3b1=3 where a; b 0. 15.47. Prove that ð1 1 e2x ðe3x þ 1Þ dx ¼ 2 9 ffiffiffi 3 p [Hint: Differentiate with respect to b in Problem 15.46.] 15.48. Use the method of Problem 12.31, Chapter 12, to justify the procedure used in Problem 15.11. DIRICHLET INTEGRALS 15.49. Find the mass of the region in the xy plane bounded by x þ y ¼ 1; x ¼ 0; y ¼ 0 if the density is ¼ ffiffiffiffiffiffi xy p . Ans: =24 15.50. Find the mass of the region bounded by the ellipsoid x2 a2 þ y2 b2 þ z2 c2 ¼ 1 if the density varies as the square of the distance from its center. Ans: abck 30 ða2 þ b2 þ c2 Þ; k ¼ constant of proportionality 15.51. Find the volume of the region bounded by x2=3 þ y2=3 þ z2=3 ¼ 1. Ans: 4=35 15.52. Find the centroid of the region in the first octant bounded by x2=3 þ y2=3 þ z2=3 ¼ 1. Ans: x x ¼ y y ¼ z z ¼ 21=128 15.53. Show that the volume of the region bounded by xm þ ym þ zm ¼ am , where m 0, is given by 8fð1=mÞg3 3m2 ð3=mÞ a3 . 15.54. Show that the centroid of the region in the first octant bounded by xm þ ym þ zm ¼ am , where m 0, is given by x x ¼ y y ¼ z z ¼ 3 ð2=mÞ ð3=mÞ 4 ð1=mÞ ð4=mÞ a MISCELLANEOUS PROBLEMS 15.55. Prove that ðb a ðx aÞ p ðb xÞq dx ¼ ðb aÞ pþqþ1 Bð p þ 1; q þ 1Þ where p 1; q 1 and b a. [Hint: Let x a ¼ ðb aÞy:] 15.56. Evaluate (a) ð3 1 dx ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 1Þð3 xÞ p ; ðbÞ ð7 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð7 xÞðx 3Þ 4 p dx. Ans: ðaÞ ; ðbÞ 2 fð1=4Þg2 3 ffiffiffi p 15.57. Show that fð1=3Þg2 ð1=6Þ ¼ ffiffiffi p ffiffiffi 2 3 p ffiffiffi 3 p . 15.58. Prove that Bðu; vÞ ¼ 1 2 ð1 0 xu1 þ xv1 ð1 þ xÞuþv dx where u; v 0. [Hint: Let y ¼ x=ð1 þ xÞ: CHAP. 15] GAMMA AND BETA FUNCTIONS 389
- 399. 15.59. If 0 p 1 prove that ð=2 0 tan p d ¼ 2 sec p 2 . 15.60. Prove that ð1 0 xu1 ð1 xÞv1 ðx þ rÞuþv ¼ Bðu; vÞ ruð1 þ rÞuþv where u; v, and r are positive constants. [Hint: Let x ¼ ðr þ 1Þy=ðr þ yÞ.] 15.61. Prove that ð=2 0 sin2u1 cos2v1 d ða sin2 þ b cos2 Þuþv ¼ Bðu; vÞ 2av bu where u; v 0. [Hint: Let x ¼ sin2 in Problem 15.60 and choose r appropriately.] 15.62. Prove that ð1 0 dx xx ¼ 1 11 þ 1 22 þ 1 33 þ 15.63. Prove that for m ¼ 2; 3; 4; . . . sin m sin 2 m sin 3 m sin ðm 1Þ m ¼ m 2m1 [Hint: Use the factored form xm 1 ¼ ðx 1Þðx 1Þðx 2Þ ðx n1Þ, divide both sides by x 1, and consider the limit as x ! 1.] 15.64. Prove that ð=2 0 ln sin x dx ¼ =2 ln 2 using Problem 15.63. [Hint: Take logarithms of the result in Problem 15.63 and write the limit as m ! 1 as a definite integral.] 15.65. Prove that 1 m 2 m 3 m m 1 m ¼ ð2Þðm1Þ=2 ffiffiffiffi m p : [Hint: Square the left hand side and use Problem 15.63 and equation (11a), Page 378.] 15.66. Prove that ð1 0 ln ðxÞ dx ¼ 1 2 lnð2Þ. [Hint: Take logarithms of the result in Problem 15.65 and let m ! 1.] 15.67. (a) Prove that ð1 0 sin x x p dx ¼ 2 ð pÞ sinð p=2Þ ; 0 p 1. (b) Discuss the cases p ¼ 0 and p ¼ 1. 15.68. Evaluate (a) ð1 0 sin x2 dx; ðbÞ ð1 0 x cos x3 dx. Ans: ðaÞ 1 2 ffiffiffiffiffiffiffiffi =2 p ; ðbÞ 3 ffiffiffi 3 p ð1=3Þ 15.69. Prove that ð1 0 x p1 ln x 1 þ x dx ¼ 2 csc p cot p; 0 p 1. 15.70. Show that ð1 0 ln x x4 þ 1 dx ¼ 2 ffiffiffi 2 p 16 . 15.71. If a 0; b 0, and 4ac b2 , prove that ð1 1 ð1 1 eðax2 þbxyþcy2 Þ dx dy ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4ac b2 p 390 GAMMA AND BETA FUNCTIONS [CHAP. 15
- 400. 15.72. Obtain (12) on Page 378 from the result (4) of Problem 15.20. [Hint: Expand ev3 =ð3 ffiffi n p Þ þ in a power series and replace the lower limit of the integral by 1.] 15.73. Obtain the result (15) on Page 378. [Hint: Observe that ðxÞ ¼ 1 x ðx þ !Þ, thus ln ðxÞ ¼ ln ðx þ 1Þ ln x, and 0 ðxÞ ðxÞ ¼ 0 ðx þ 1Þ ðx þ 1Þ 1 x Furthermore, according to (6) page 377. ðx þ !Þ ¼ lim k!1 k! kx ðx þ 1Þ ðx þ kÞ Now take the logarithm of this expression and then differentiate. Also recall the definition of the Euler constant, . 15.74. The duplication formula (13a) Page 378 is proved in Problem 15.24. For further insight, develop it for positive integers, i.e., show that 22n1 ðn þ 1 2Þ ðnÞ ¼ ð2nÞ ffiffiffi p Hint: Recall that ð1 2Þ ¼ , then show that ðn þ 1 2Þ ¼ 2n þ 1Þ 2 ¼ ð2n 1Þ 5 3 1 2n ffiffiffi p : Observe that ð2n þ 1Þ 2nðn þ 1Þ ¼ ð2nÞ! 2nn! ¼ ð2n 1Þ 5 3 1 Now substitute and refine. CHAP. 15] GAMMA AND BETA FUNCTIONS 391
- 401. 392 Functions of a Complex Variable Ultimately it was realized that to accept numbers that provided solutions to equations such as x2 þ 1 ¼ 0 was no less meaningful than had been the extension of the real number system to admit a solution for x þ 1 ¼ 0, or roots for x2 2 ¼ 0. The complex number system was in place around 1700, and by the early nineteenth century, mathematicians were comfortable with it. Physical theories took on a completeness not possible without this foundation of complex numbers and the analysis emanating from it. The theorems of the differential and integral calculus of complex functions introduce math- ematical surprises as well as analytic refinement. This chapter is a summary of the basic ideas. FUNCTIONS If to each of a set of complex numbers which a variable z may assume there corresponds one or more values of a variable w, then w is called a function of the complex variable z, written w ¼ f ðzÞ. The fundamental operations with complex numbers have already been considered in Chapter 1. A function is single-valued if for each value of z there corresponds only one value of w; otherwise it is multiple-valued or many-valued. In general, we can write w ¼ f ðzÞ ¼ uðx; yÞ þ ivðx; yÞ, where u and v are real functions of x and y. EXAMPLE. w ¼ z2 ¼ ðx þ iyÞ2 ¼ x2 y2 þ 2ixy ¼ u þ iv so that uðx; yÞ ¼ x2 y2 ; vðx; yÞ ¼ 2xy. These are called the real and imaginary parts of w ¼ z2 respectively. In complex variables, multiple-valued functions often are replaced by a specially constructed single- valued function with branches. This idea is discussed in a later paragraph. EXAMPLE. Since e2ki ¼ 1, the general polar form of z is z ¼ eiðþ2kÞ . This form and the fact that the logarithm and exponential functions are inverse leads to the following definition of ln z ln z ¼ ln þ ð þ 2kÞi k ¼ 0; 1; 2; . . . ; n . . . Each value of k determines a single-valued function from this collection of multiple-valued functions. These are the branches from which (in the realm of complex variables) a single-valued function can be constructed. Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 402. LIMITS AND CONTINUITY Definitions of limits and continuity for functions of a complex variable are analogous to those for a real variable. Thus, f ðzÞ is said to have the limit l as z approaches z0 if, given any 0, there exists a 0 such that j f ðzÞ lj whenever 0 jz z0j . Similarly, f ðzÞ is said to be continuous at z0 if, given any 0, there exists a 0 such that j f ðzÞ f ðz0Þj whenever jz z0j . Alternatively, f ðzÞ is continuous at z0 if lim z!z0 f ðzÞ ¼ f ðz0Þ. Note: While these definitions have the same appearance as in the real variable setting, remember that jz z0j means jðx x0j þ ið y y0Þj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx x0Þ2 þ ð y y0Þ2 q : Thus there are two degrees of freedom as ðx; yÞ ! ðx0; y0Þ: DERIVATIVES If f ðzÞ is single-valued in some region of the z plane the derivative of f ðzÞ, denoted by f 0 ðzÞ, is defined as lim z!0 ðf ðz þ zÞ f ðzÞ z ð1Þ provided the limit exists independent of the manner in which z ! 0. If the limit (1) exists for z ¼ z0, then f ðzÞ is called analytic at z0. If the limit exists for all z in a region r, then f ðzÞ is called analytic in r. In order to be analytic, f ðzÞ must be single-valued and continuous. The converse, however, is not necessarily true. We define elementary functions of a complex variable by a natural extension of the corresponding functions of a real variable. Where series expansions for real functions f ðxÞ exists, we can use as definition the series with x replaced by z. The convergence of such complex series has already been considered in Chapter 11. EXAMPLE 1. We define ex ¼ 1 þ z þ z2 2! þ z3 3! þ ; sin z ¼ z z3 3! þ z5 5! z7 7! þ ; cos z ¼ 1 z2 2! þ z4 4! z6 6! þ . From these we can show that ex ¼ exþiy ¼ ex ðcos y þ i sin yÞ, as well as numerous other relations. Rules for differentiating functions of a complex variable are much the same as for those of real variables. Thus, d dz ðzn Þ ¼ nzn1 ; d dz ðsin zÞ ¼ cos z, and so on. CAUCHY-RIEMANN EQUATIONS A necessary condition that w ¼ f ðzÞ ¼ uðx; yÞ þ ivðx; yÞ be analytic in a region r is that u and v satisfy the Cauchy-Riemann equations @u @x ¼ @v @y ; @u @y ¼ @v @x ð2Þ (see Problem 16.7). If the partial derivatives in (2) are continuous in r, the equations are sufficient conditions that f ðzÞ be analytic in r. If the second derivatives of u and v with respect to x and y exist and are continuous, we find by differentiating (2) that @2 u @x2 þ @2 u @y2 ¼ 0; @2 v @x2 þ @2 v @y2 ¼ 0 ð3Þ Thus, the real and imaginary parts satisfy Laplace’s equation in two dimensions. Functions satisfying Laplace’s equation are called harmonic functions. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 393
- 403. INTEGRALS Let f ðzÞ be defined, single-valued and continuous in a region r. We define the integral of f ðzÞ along some path C in r from point z1 to point z2, where z1 ¼ x1 þ iy1; z2 ¼ x2 þ iy2, as ð C f ðzÞ dz ¼ ððx2;y2Þ ðx1;y1Þ ðu þ ivÞðdx þ i dyÞ ¼ ððx2;y2Þ ðx1;y1Þ u dx v dy þ i ððx2;y2Þ ðx1;y1Þ v dx þ u dy With this definition the integral of a function of a complex variable can be made to depend on line integrals for real functions already considered in Chapter 10. An alternative definition based on the limit of a sum, as for functions of a real variable, can also be formulated and turns out to be equivalent to the one above. The rules for complex integration are similar to those for real integrals. An important result is ð C f ðzÞ dz @ ð C j f ðzÞjjdzj @ M ð C ds ¼ ML ð4Þ where M is an upper bound of j f ðzÞj on C, i.e., j f ðzÞj @ M, and L is the length of the path C. Complex function integral theory is one of the most esthetically pleasing constructions in all of mathematics. Major results are outlined below. CAUCHY’S THEOREM Let C be a simple closed curve. If f ðzÞ is analytic within the region bounded by C as well as on C, then we have Cauchy’s theorem that ð C f ðzÞ dz þ C f ðzÞ dz ¼ 0 ð5Þ where the second integral emphasizes the fact that C is a simple closed curve. Expressed in another way, (5) is equivalent to the statement that ðz2 z1 f ðzÞ dz has a value independent of the path joining z1 and z2. Such integrals can be evaluated as Fðz2Þ Fðz1Þ, where F 0 ðzÞ ¼ f ðzÞ. These results are similar to corresponding results for line integrals developed in Chapter 10. EXAMPLE. Since f ðzÞ ¼ 2z is analytic everywhere, we have for any simple closed curve C þ C 2z dz ¼ 0 Also, ð1þi 2i 2z dz ¼ z2 1þi 2i ¼ ð1 þ iÞ2 ð2iÞ2 ¼ 2i þ 4 CAUCHY’S INTEGRAL FORMULAS If f ðzÞ is analytic within and on a simple closed curve C and a is any point interior to C, then f ðaÞ ¼ 1 2i þ C f ðzÞ z a dz ð6Þ where C is traversed in the positive (counterclockwise) sense. Also, the nth derivative of f ðzÞ at z ¼ a is given by f ðnÞ ðaÞ ¼ n! 2i þ C f ðzÞ ðz aÞnþ1 dz ð7Þ These are called Cauchy’s integral formulas. They are quite remarkable because they show that if the function f ðzÞ is known on the closed curve C then it is also known within C, and the various 394 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 404. derivatives at points within C can be calculated. Thus, if a function of a complex variable has a first derivative, it has all higher derivatives as well. This, of course, is not necessarily true for functions of real variables. TAYLOR’S SERIES Let f ðzÞ be analytic inside and on a circle having its center at z ¼ a. Then for all points z in the circle we have the Taylor series representation of f ðzÞ given by f ðzÞ ¼ f ðaÞ þ f 0 ðaÞðz aÞ þ f 00 ðaÞ 2! ðz aÞ2 þ f 000 ðaÞ 3! ðz aÞ3 þ ð8Þ See Problem 16.21. SINGULAR POINTS A singular point of a function f ðzÞ is a value of z at which f ðzÞ fails to be analytic. If f ðzÞ is analytic everywhere in some region except at an interior point z ¼ a, we call z ¼ a an isolated singularity of f ðzÞ. EXAMPLE. If f ðzÞ ¼ 1 ðz 3Þ2 , then z ¼ 3 is an isolated singularity of f ðzÞ. EXAMPLE. The function f ðzÞ ¼ sin z z has a singularity at z ¼ 0. Because lim z!0 is finite, this singularity is called a removable singularity. POLES If f ðzÞ ¼ ðzÞ ðz aÞn ; ðaÞ 6¼ 0, where ðzÞ is analytic everywhere in a region including z ¼ a, and if n is a positive integer, then f ðzÞ has an isolated singularity at z ¼ a, which is called a pole of order n. If n ¼ 1, the pole is often called a simple pole; if n ¼ 2, it is called a double pole, and so on. LAURENT’S SERIES If f ðzÞ has a pole of order n at z ¼ a but is analytic at every other point inside and on a circle C with center at a, then ðz aÞn f ðzÞ is analytic at all points inside and on C and has a Taylor series about z ¼ a so that f ðzÞ ¼ an ðz aÞn þ anþ1 ðz aÞn1 þ þ a1 z a þ a0 þ a1ðz aÞ þ a2ðz aÞ2 þ ð9Þ This is called a Laurent series for f ðzÞ. The part a0 þ a1ðz aÞ þ a2ðz aÞ2 þ is called the analytic part, while the remainder consisting of inverse powers of z a is called the principal part. More generally, we refer to the series X 1 k¼1 akðz aÞk as a Laurent series, where the terms with k 0 constitute the principal part. A function which is analytic in a region bounded by two concentric circles having center at z ¼ a can always be expanded into such a Laurent series (see Problem 16.92). It is possible to define various types of singularities of a function f ðzÞ from its Laurent series. For example, when the principal part of a Laurent series has a finite number of terms and an 6¼ 0 while an1; an2; . . . are all zero, then z ¼ a is a pole of order n. If the principal part has infinitely many terms, z ¼ a is called an essential singularity or sometimes a pole of infinite order. EXAMPLE. The function e1=z ¼ 1 þ 1 z þ 1 2! z2 þ has an essential singularity at z ¼ 0. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 395
- 405. BRANCHES AND BRANCH POINTS Another type of singularity is a branch point. These points play a vital role in the construction of single-valued functions from ones that are multiple-valued, and they have an important place in the computation of integrals. In the study of functions of a real variable, domains were chosen so that functions were single- valued. This guaranteed inverses and removed any ambiguities from differentiation and integration. The applications of complex variables are best served by the approach illustrated below. It is in the realm of real variables and yet illustrates a pattern appropriate to complex variables. Let y2 ¼ x; x 0, then y ¼ ffiffiffi x p . In real variables two functions f1 and f2 are described by y ¼ þ ffiffiffi x p on x 0, and y ¼ ffiffiffi x p on x 0, respectively. Each of them is single-valued. An approach that can be extended to complex variable results by defining the positive x-axis (not including zero) as a cut in the plane. This creates two branches f1 and f2 of a new function on a domain called the Riemann axis. The only passage joining the spaces in which the branches f1 and f2, respec- tively, are defined is through 0. This connecting point, zero, is given the special name branch point. Observe that two points x in the space of f1 and x in that of f2 can appear to be near each other in the ordinary view but are not from the Riemannian perspective. (See Fig. 16-1.) The above real variables construction suggests one for complex variables illustrated by w ¼ z1=2 . In polar coordinates e2i ¼ 1; therefore, the general representation of w ¼ z1=2 in that system is w ¼ 1=2 eiðþ2kÞ=2 , k ¼ 0; 1. Thus, this function is double-valued. If k ¼ 0, then w1 ¼ 1=2 ei=2 , 0 2; 0 If k ¼ 1, then w2 ¼ 1=2 eiðþ2Þ=2 ¼ 1=2 ei=2 i ¼ 1=2 ei=2 ; 2 4; 0. Thus, the two branches of w are w1 and w2, where w1 ¼ w2. (The double valued characteristic of w is illustrated by noticing that as z traverses a circle, C: jzj ¼ through the values to 2. The functional values run from 1=2 ei=2 to 1=2 ei . In other words, as z navigates the entire circle, the range variable only moves halfway around the corresponding range circle. In order for that variable to complete the circuit, z would have to make a second revolution. Thus, we would have coincident positions of z giving rise to distinct values of w. For example, z1 ¼ eð=2Þ=i and z2 ¼ eð=2þ2Þi are coincident points on the unit circle. The distinct functional values are z1=2 1 ¼ ffiffiffi 2 p 2 ð1 þ iÞ and z1=2 2 ¼ ffiffiffi 2 p 2 ð1 þ iÞ. The following abstract construction replaces the multiple-valued function with a new single-valued one. Make a cut in the complex plane that includes all of the positive x-axis except the origin. Think of two planes, P1 and P2, the first one of infinitesimal distance above the complex plane and the other infinitesimally below it. The point 0 which connects these spaces is called a branch point. The planes 396 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 Fig. 16-1
- 406. and the connecting point constitute a Riemann surface, and w1 and w2 are the branches of the function each defined in one of the planes. (Since the space of complex variables is the complex plane, this Riemann surface may be thought of as a flight of fancy that supports a rigorous analytic construction.) To visualize this Riemann surface and perceive the single-valued character of the new function in it, first think of duplicates, C1 and C2 of the domain circle, C: jzj ¼ in the planes P1 and P2, respectively. Start at ¼ on C1, and proceed counterclockwise to the edge U2 of the cut of P1. (This edge corresponds to ¼ 2). Paste U2 to L1, the initial edge of the cut on P2. Transfer to P2 through this join and continue on C2. Now after a complete counterclockwise circuit of C2 we reach the edge L2 of the cut. Pasting L2 to U1 provides passage back to P1 and makes it possible to close the curve in the Riemann plane. See Fig. 16-2. Note that the function is not continuous on the positive x-axis. Also the cut is somewhat arbitrary. Other rays and even curves extending from the origin to infinity can be employed. In many integration applications the cut ¼ i proves valuable. On the other hand, the branch point (0 in this example) is special. If another point, z0 6¼ 0 were chosen as the center of a small circle with radius less than jz0j, then the origin would lie outside it. As a point z traversed its circumference, its argument would return to the original value as would the value of w. However, for any circle that has the branch point as an interior point, a similar traversal of the circumference will change the value of the argument by 2, and the values of w1 and w2 will be interchanged. (See Problem 16.37.) RESIDUES The coefficients in (9) can be obtained in the customary manner by writing the coefficients for the Taylor series corresponding to ðz aÞn f ðzÞ. In further developments, the coefficient a1, called the residue of f ðzÞ at the pole z ¼ a, is of considerable importance. It can be found from the formula a1 ¼ lim z!a 1 ðn 1Þ! dn1 dzn1 fðz aÞn f ðzÞg ð10Þ where n is the order of the pole. For simple poles the calculation of the residue is of particular simplicity since it reduces to a1 ¼ lim z!a ðz aÞ f ðzÞ ð11Þ RESIDUE THEOREM If f ðzÞ is analytic in a region r except for a pole of order n at z ¼ a and if C is any simple closed curve in r containing z ¼ a, then f ðzÞ has the form (9). Integrating (9), using the fact that CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 397 Fig. 16-2
- 407. þ C dz ðz aÞn ¼ 0 if n 6¼ 1 2i if n ¼ 1 ð12Þ (see Problem 16.13), it follows that þ C f ðzÞ dz ¼ 2ia1 ð13Þ i.e., the integral of f ðzÞ around a closed path enclosing a single pole of f ðzÞ is 2i times the residue at the pole. More generally, we have the following important theorem. Theorem. If f ðzÞ is analytic within and on the boundary C of a region r except at a finite number of poles a; b; c; . . . within r, having residues a1; b1; c1; . . . ; respectively, then þ C f ðzÞ dz ¼ 2iða1 þ b1 þ c1 þ Þ ð14Þ i.e., the integral of f ðzÞ is 2i times the sum of the residues of f ðzÞ at the poles enclosed by C. Cauchy’s theorem and integral formulas are special cases of this result, which we call the residue theorem. EVALUATION OF DEFINITE INTEGRALS The evaluation of various definite integrals can often be achieved by using the residue theorem together with a suitable function f ðzÞ and a suitable path or contour C, the choice of which may reuqire great ingenuity. The following types are most common in practice. 1. ð1 0 FðxÞ dx; FðxÞ is an even function. Consider þ C FðzÞ dz along a contour C consisting of the line along the x-axis from R to þR and the semicircle above the x-axis having this line as diameter. Then let R ! 1. See Problems 16.29 and 16.30. 2. ð2 0 Gðsin ; cos Þ d, G is a rational function of sin and cos . Let z ¼ ei . Then sin ¼ z z1 2i ; cos ¼ z þ z1 2 and dz ¼ iei d or d ¼ dz=iz. The given integral is equivalent to þ C FðzÞ dz, where C is the unit circle with center at the origin. See Problems 16.31 and 16.32. 3. ð1 1 FðxÞ cos mx sin mx dx; FðxÞ is a rational function. Here we consider þ C FðzÞeimz dz where C is the same contour as that in Type 1. See Problem 16.34. 4. Miscellaneous integrals involving particular contours. See Problems 16.35 and 16.38. In particular, Problem 16.38 illustrates a choice of path for an integration about a branch point. 398 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 408. Solved Problems FUNCTIONS, LIMITS, CONTINUITY 16.1. Determine the locus represented by (a) jz 2j ¼ 3; ðbÞ jz 2j ¼ jz þ 4j; ðcÞ jz 3j þ jz þ 3j ¼ 10. (a) Method 1: jz 2j ¼ jx þ iy 2j ¼ jx 2 þ iyj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 2Þ2 þ y2 q ¼ 3 or ðx 2Þ2 þ y2 ¼ 9, a circle with center at ð2; 0Þ and radius 3. Method 2: jz 2j is the distance between the complex numbers z ¼ x þ iy and 2 þ 0i. If this distance is always 3, the locus is a circle of radius 3 with center at 2 þ 0i or ð2; 0Þ. (b) Method 1: jx þ iy 2j ¼ jx þ iy þ 4j or ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 2Þ2 þ y2 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 4Þ2 þ y 2 q . Squaring, we find x ¼ 1, a straight line. Method 2: The locus is such that the distance from any point on it to ð2; 0Þ and ð4; 0Þ are equal. Thus, the locus is the perpendicular besector of the line joining ð2; 0Þ and ð4; 0Þ, or x ¼ 1. (c) Method 2: The locus is given by ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 3Þ2 þ y2 q þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 3Þ2 þ y2 q ¼ 10 or ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx 3Þ2 þ y2 q ¼ 10 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 3Þ2 þ y2 q . Squaring and simplifying, 25 þ 3x ¼ 5 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðx þ 3Þ2 þ y2 q . Squaring and simplifying again yields x2 25 þ y2 16 ¼ 1, an ellipse with semi-major and semi-minor axes of lengths 5 and 4, respec- tively. Method 2: The locus is such that the sum of the distances from any point on it to ð3; 0Þ and ð3; 0Þ is 10. Thus the locus is an ellipse whose foci are at ð3; 0Þ and ð3; 0Þ and whose major axis has length 10. 16.2. Determine the region in the z plane represented by each of the following. (a) jzj 1. Interior of a circle of radius 1. See Fig. 16-3(a) below. (b) 1 jz þ 2ij @ 2. jz þ 2ij is the distance from z to 2i, so that jz þ 2ij ¼ 1 is a circle of radius 1 with center at 2i, i.e., ð0; 2Þ; and jz þ 2ij ¼ 2 is a circle of radius 2 with center at 2i. Then 1 jz þ 2ij @ 2 represents the region exterior to jz þ 2ij ¼ 1 but interior to or on jz þ 2ij ¼ 2. See Fig. 16-3(b) below. (c) =3 @ arg z @ =2. Note that arg z ¼ , where z ¼ ei . The required region is the infinite region bounded by the lines ¼ =3 and ¼ =2, including these lines. See Fig. 16-3(c) below. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 399 Fig. 16-3
- 409. 16.3. Express each function in the form uðx; yÞ þ ivðx; yÞ, where u and v are real: (a) z3 ; ðbÞ 1=ð1 zÞ; ðcÞ e3z ; ðdÞ ln z. ðaÞ w ¼ z3 ¼ ðx þ iyÞ3 ¼ x3 þ 3x2 ðiyÞ þ 3xðiyÞ2 þ ðiyÞ3 ¼ x3 þ 3ix2 y 3xy2 iy2 ¼ x3 3xy2 þ ið3x2 y y3 Þ Then uðx; yÞ ¼ x3 3xy2 ; vðx; yÞ ¼ 3x2 y y3 . ðbÞ w ¼ 1 1 z ¼ 1 1 ðx þ iyÞ ¼ 1 1 x iy 1 x þ iy 1 x þ iy ¼ 1 x þ iy ð1 xÞ2 þ y2 Then uðx; yÞ ¼ 1 x ð1 xÞ2 þ y2 ; vðx; yÞ ¼ y ð1 xÞ2 þ y2 : ðcÞ e3z ¼ e3ðxþiyÞ ¼ e3x e3iy ¼ e3x ðcos 3y þ i sin 3yÞ and u ¼ e3x cos 3y; v ¼ e3x sin 3y ðdÞ ln z ¼ lnðei Þ ¼ ln þ i ¼ ln ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2 q þ i tan1 y=x and u ¼ 1 2 lnðx2 þ y2 Þ; v ¼ tan1 y=x Note that ln z is a multiple-valued function (in this case it is infinitely many-valued), since can be increased by any multiple of 2. The principal value of the logarithm is defined as that value for which 0 @ 2 and is called the principal branch of ln z. 16.4. Prove (a) sinðx þ iyÞ ¼ sin x cosh y þ i cos x sinh y (b) cosðx þ iyÞ ¼ cos x cosh y i sin x sinh y. We use the relations eix ¼ cos z þ i sin z; eix ¼ cos z i sin z, from which sin z ¼ eiz eiz 2i ; cos z ¼ eiz þ eiz 2 Then sin z ¼ sinðx þ iyÞ ¼ eiðxþiyÞ eiðxþiyÞ 2i ¼ eixy eixþy 2i ¼ 1 2i fey ðcos x þ i sin xÞ e y ðcos x i sin xÞg ¼ ðsin xÞ e y þ ey 2 þ iðcos xÞ e y ey 2 ¼ sin x cosh y þ i cos x sinh y Similarly, cos z ¼ cosðx þ iyÞ ¼ eiðxþiyÞ þ eiðxþiyÞ 2 ¼ 1 2 feixy þ eixþy g ¼ 1 2 fey ðcos x þ i sin xÞ þ ey ðcos x i sin xÞg ¼ ðcos xÞ e y þ ey 2 iðsin xÞ e y ey 2 ¼ cos x cosh y i sin x sinh y DERIVATIVES, CAUCHY-RIEMANN EQUATIONS 16.5. Prove that d dz z z, where z z is the conjugate of z, does not exist anywhere. By definition, d dz f ðzÞ ¼ lim z!0 f ðz þ zÞ f ðzÞ z if this limit exists independent of the manner in which z ¼ x þ i y approaches zero. Then 400 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 410. d dz z z ¼ lim z!0 z þ z z z z ¼ lim x!0 y!0 x þ iy þ x þ i y x þ iy x þ i y ¼ lim x!0 y!0 x iy þ x þ i y ðx iyÞ x þ i y ¼ lim x!0 y!0 x i y x þ i y If y ¼ 0, the required limit is lim x!0 x x ¼ 1: If x ¼ 0, the required limit is lim y!0 i y i y ¼ 1: These two possible approaches show that the limit depends on the manner in which z ! 0, so that the derivative does not exist; i.e., z z is nonanalytic anywhere. 16.6. (a) If w ¼ f ðzÞ ¼ 1 þ z 1 z , find dw dz . (b) Determine where w is nonanalytic. ðaÞ Method 1: dw dz ¼ lim z!0 1 þ ðz þ zÞ 1 ðz þ zÞ 1 þ z 1 z z ¼ lim z!0 2 ð1 z zÞð1 zÞ ¼ 2 ð1 zÞ2 provided z 6¼ 1, independent of the manner in which z ! 0: Method 2. The usual rules of differentiation apply provided z 6¼ 1. Thus, by the quotient rule for differentiation, d dz 1 þ z 1 z ¼ ð1 zÞ d dz ð1 þ zÞ ð1 þ zÞ d dz ð1 zÞ ð1 zÞ2 ¼ ð1 zÞð1Þ ð1 þ zÞð1Þ ð1 zÞ2 ¼ 2 ð1 zÞ2 (b) The function is analytic everywhere except at z ¼ 1, where the derivative does not exist; i.e., the function is nonanalytic at z ¼ 1. 16.7. Prove that a necessary condition for w ¼ f ðzÞ ¼ uðx; yÞ þ i vðx; yÞ to be analytic in a region is that the Cauchy-Riemann equations @u @x ¼ @v @y , @u @y ¼ @v @x be satisfied in the region. Since f ðzÞ ¼ f ðx þ iyÞ ¼ uðx; yÞ þ i vðx; yÞ, we have f ðz þ zÞ ¼ f ½x þ x þ ið y þ yÞ ¼ uðx þ x; y þ yÞ þ i vðx þ x; y þ yÞ Then lim z!0 f ðz þ zÞ f ðzÞ z ¼ lim x!0 y!0 uðx þ x; y þ yÞ uðx; yÞ þ ifvðx þ x; y þ yÞ vðx; yÞg x þ i y If y ¼ 0, the required limit is lim x!0 uðx þ x; yÞ uðx; yÞ x þ i vðx þ x; yÞ vðx; yÞ x ¼ @u @x þ i @v @x If x ¼ 0, the required limit is lim y!0 uðx; y þ yÞ uðx; yÞ i y þ vðx; y þ yÞ vðx; yÞ y ¼ 1 i @u @y þ @v @y If the derivative is to exist, these two special limits must be equal, i.e., @u @x þ i @v @x ¼ 1 i @u @y þ @v @y ¼ i @u @y þ @v @y CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 401
- 411. so that we must have @u @x ¼ @v @y and @v @x ¼ @u @y : Conversely, we can prove that if the first partial derivatives of u and v with respect to x and y are continuous in a region, then the Cauchy–Riemann equations provide sufficient conditions for f ðzÞ to be analytic. 16.8. (a) If f ðzÞ ¼ uðx; yÞ þ i vðx; yÞ is analytic in a region r, prove that the one parameter families of curves uðx; yÞ ¼ C1 and vðx; yÞ ¼ C2 are orthogonal families. (b) Illustrate by using f ðzÞ ¼ z2 . (a) Consider any two particular members of these families uðx; yÞ ¼ u0; vðx; yÞ ¼ v0 which intersect at the point ðx0; y0Þ. Since du ¼ ux dx þ uy dy ¼ 0, we have dy dx ¼ ux uy : Also since dv ¼ vx dx þ vy dy ¼ 0; dy dx ¼ vx vy : When evaluated at ðx0; y0Þ, these represent respectively the slopes of the two curves at this point of intersection. By the Cauchy–Riemann equations, ux ¼ vy; uy ¼ vx, we have the product of the slopes at the point ðx0; y0Þ equal to ux uy vx vy ¼ 1 so that any two members of the respective families are orthogonal, and thus the two families are ortho- gonal. (b) If f ðzÞ ¼ z2 , then u ¼ x2 y2 ; v ¼ 2xy. The graphs of several members of x2 y2 ¼ C1, 2xy ¼ C2 are shown in Fig. 16-4. 16.9. In aerodynamics and fluid mechanics, the functions and in f ðzÞ ¼ þ i , where f ðzÞ is analytic, are called the velocity potential and stream function, respectively. If ¼ x2 þ 4x y2 þ 2y, (a) find and (b) find f ðzÞ. (a) By the Cauchy-Riemann equations, @ @x ¼ @ @y ; @ @x ¼ @ @y . Then ð1Þ @ @y ¼ 2x þ 4 ð2Þ @ @x ¼ 2y 2 Method 1. Integrating (1), ¼ 2xy þ 4y þ FðxÞ. Integrating (2), ¼ 2xy 2x þ Gð yÞ. These are identical if FðxÞ ¼ 2x þ c; Gð yÞ ¼ 4y þ c, where c is a real constant. Thus, ¼ 2xy þ 4y 2x þ c. Method 2. Integrating (1), ¼ 2xy þ 4y þ FðxÞ. Then substituting in (2), 2y þ F 0 ðxÞ ¼ 2y 2 or F 0 ðxÞ ¼ 2 and FðxÞ ¼ 2x þ c. Hence, ¼ 2xy þ 4y 2x þ c. ðaÞ From ðaÞ; f ðzÞ ¼ þ i ¼ x2 þ 4x y2 þ 2y þ ið2xy þ 4y 2x þ cÞ ¼ ðx2 y2 þ 2ixyÞ þ 4ðx þ iyÞ 2iðx þ iyÞ þ ic ¼ z2 þ 4z 2iz þ c1 where c1 is a pure imaginary constant. 402 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 y x Fig. 16-4
- 412. This can also be accomplished by nothing that z ¼ x þ iy; z z ¼ x iy so that x ¼ z þ z z 2 , y ¼ z z z 2i . The result is then obtained by substitution; the terms involving z z drop out. INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS 16.10. Evaluate ð2þ4i 1þi z2 dz (a) along the parabola x ¼ t; y ¼ t2 where 1 @ t @ 2, (b) along the straight line joining 1 þ i and 2 þ 4i, (c) along straight lines from 1 þ i to 2 þ i and then to 2 þ 4i. We have ð2þ4i 1þi z2 dz ¼ ðð2;4Þ ð1;1Þ ðx þ iyÞ2 ðdx þ i dyÞ ¼ ðð2;4Þ ð1;1Þ ðx2 y2 þ 2ixyÞðdx þ i dyÞ ¼ ðð2;4Þ ð1;1 ðx2 y2 Þ dx 2xy dy þ i ðð2;4Þ ð1;1Þ 2xy dx þ ðx2 y2 Þ dy Method 1. (a) The points ð1; 1Þ and ð2; 4Þ correspond to t ¼ 1 and t ¼ 2, respectively. Then the above line integrals become ð2 t¼1 fðt2 t4 Þ dt 2ðtÞðt2 Þ2t dtg þ i ð2 t¼1 f2ðtÞðt2 Þ dt þ ðt2 t4 Þð2tÞ dtg ¼ 86 3 6i (b) The line joining ð1; 1Þ and ð2; 4Þ has the equation y 1 ¼ 4 1 2 1 ðx 1Þ or y ¼ 3x 2. Then we find ð2 x¼1 ½x2 ð3x 2Þ2 dx 2xð3x 2Þ3 dx þ i ð2 x¼1 2xð3x 2Þ dx þ ½x2 ð3x 2Þ2 3 dx ¼ 86 3 6i (c) From 1 þ i to 2 þ i [or ð1; 1Þ to ð2; 1Þ], y ¼ 1; dy ¼ 0 and we have ð2 x¼1 ðx2 1Þ dx þ i ð2 x¼1 2x dx ¼ 4 3 þ 3i From 2 þ i to 2 þ 4i [or ð2; 1Þ to ð2; 4Þ], x ¼ 2; dx ¼ 0 and we have ð4 y¼1 4y dy þ i ð4 y¼1 ð4 y2 Þ dy ¼ 30 9i Adding, ð4 3 þ 3iÞ þ ð30 91Þ ¼ 86 3 6i. Method 2. By the methods of Chapter 10 it is seen that the line integrals are independent of the path, thus accounting for the same values obtained in (a), (b), and (c) above. In such case the integral can be evaluated directly, as for real variables, as follows: ð2þ4i 1þi z2 dz ¼ z3 3 2þ4i 1i ¼ ð2 þ 4iÞ3 3 ð1 þ iÞ3 3 ¼ 86 3 6i 16.11. (a) Prove Cauchy’s theorem: If f ðzÞ is analytic inside and on a simple closed curve C, then þ C f ðzÞ dz ¼ 0. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 403
- 413. (b) Under these conditions prove that ðP2 P1 f ðzÞ dz is independent of the path joining P1 and P2. ðaÞ þ C f ðzÞ dz ¼ þ C ðu þ ivÞðdx þ i dyÞ ¼ þ C u dx v dy þ i þ C v dx þ u dy By Green’s theorem (Chapter 10), þ C u dx v dy ¼ ð ð r @v @x @u @y dx dy; þ C v dx þ u dy ¼ ð ð r @u @x @v @y dx dy where r is the region (simply-connected) bounded by C. Since f ðzÞ is analytic, @u @x ¼ @v @y ; @v @x ¼ @u @y (Problem 16.7), and so the above integrals are zero. Then þ C f ðzÞ dz ¼ 0, assuming f 0 ðzÞ [and thus the partial derivatives] to be continuous. (b) Consider any two paths joining points P1 and P2 (see Fig. 16-5). By Cauchy’s theorem, ð P1AP2BP1 f ðzÞ dz ¼ 0 Then ð P1AP2 f ðzÞ dz þ ð P2BP1 f ðzÞ dz ¼ 0 or ð P1AP2 f ðzÞ dz ¼ ð P2BP1 f ðzÞ dz ¼ ð P1BP2 f ðzÞ dz i.e., the integral along P1AP2 (path 1) ¼ integral along P1BP2 (path 2), and so the integral is independent of the path joining P1 and P2. This explains the results of Problem 16.10, since f ðzÞ ¼ z2 is analytic. 16.12. If f ðzÞ is analytic within and on the boundary of a region bounded by two closed curves C1 and C2 (see Fig. 16-6), prove that þ C1 f ðzÞ dz ¼ þ C2 f ðzÞ dz As in Fig. 16-6, construct line AB (called a cross-cut) connecting any point on C2 and a point on C1. By Cauchy’s theorem (Problem 16.11), ð AQPABRSTBA f ðzÞ dz ¼ 0 since f ðzÞ is analytic within the region shaded and also on the boundary. Then ð AQPA f ðzÞ dz þ ð AB f ðzÞ dz þ ð BRSTB f ðzÞ dz þ ð BA f ðzÞ dz ¼ 0 ð1Þ But ð AB f ðzÞ dz ¼ ð BA f ðzÞ dz. Hence, (1) gives ð AQPA f ðzÞ dz ¼ ð BRSTB f ðzÞ dz ¼ ð BTSRB f ðzÞ dz 404 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 Path 1 Path 2 P1 P2 A B Fig. 16-5 Fig. 16-6
- 414. þ C1 f ðzÞ dz ¼ þ C2 f ðzÞ dz i.e., Note that f ðzÞ need not be analytic within curve C2. 16.13. (a) Prove that þ C dz ðz aÞn ¼ 2i if n ¼ 1 0 if n ¼ 2; 3; 4; . . . , where C is a simple closed curve bounding a region having z ¼ a as interior point. (b) What is the value of the integral if n ¼ 0; 1; 2; 3; . . . ? (a) Let C1 be a circle of radius having center at z ¼ a (see Fig. 16-7). Since ðz aÞn is analytic within and on the boundary of the region bounded by C and C1, we have by Problem 16.12, þ C dz ðz aÞn ¼ þ C1 dz ðz aÞn To evaluate this last integral, note that on C1, jz aj ¼ or z a ¼ ei and dz ¼ iei d. The integral equals ð2 0 iei d n ein ¼ i n1 ð2 0 eð1nÞi d ¼ i n1 eð1nÞi ð1 nÞi 2 0 ¼ 0 if n 6¼ 1 If n ¼ 1, the integral equals i ð2 0 d ¼ 2i. (b) For n ¼ 0; 1; 2; . . . the integrand is 1; ðz aÞ; ðz aÞ2 ; . . . and is analytic everywhere inside C1, including z ¼ a. Hence, by Cauchy’s theorem the integral is zero. 16.14. Evaluate þ C dz z 3 , where C is (a) the circle jzj ¼ 1; ðbÞ the circle jz þ ij ¼ 4. (a) Since z ¼ 3 is not interior to jzj ¼ 1, the integral equals zero (Problem 16.11). (b) Since z ¼ 3 is interior to jz þ ij ¼ 4, the integral equals 2i (Problem 16.13). 16.15. If f ðzÞ is analytic inside and on a simple closed curve C, and a is any point within C, prove that f ðaÞ ¼ 1 2i þ C f ðzÞ z a dz Referring to Problem 16.12 and the figure of Problem 16.13, we have þ C f ðzÞ z a dz ¼ þ C1 f ðzÞ z a dz Letting z a ¼ ei , the last integral becomes i ð2 0 f ða þ ei Þ d. But since f ðzÞ is analytic, it is continuous. Hence, lim !0 i ð2 0 f ða þ ei Þ d ¼ i ð2 0 lim !0 f ða þ ei Þ d ¼ i ð2 0 f ðaÞ d ¼ 2i f ðaÞ and the required result follows. 16.16. Evaluate (a) þ C cos z z dz; ðbÞ þ C ex zðz þ 1Þ dz, where C is the circle jz 1j ¼ 3. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 405 Fig. 16-7
- 415. (a) Since z ¼ lies within C, 1 2i þ C cos z z dz ¼ cos ¼ 1 by Problem 16.15 with f ðzÞ ¼ cos z, a ¼ . Then þ C cos z z dz ¼ 2i. þ C ez zðz þ 1Þ dz ¼ þ C ez 1 z 1 z þ 1 dz ¼ þ C ez z dz þ C ez z þ 1 dz ðbÞ ¼ 2ie0 2ie1 ¼ 2ið1 e1 Þ by Problem 16.15, since z ¼ 0 and z ¼ 1 are both interior to C. 16.17. Evaluate þ C 5z2 3z þ 2 ðz 1Þ3 dz where C is any simple closed curve enclosing z ¼ 1. Method 1. By Cauchy’s integral formula, f ðnÞ ðaÞ ¼ n! 2i þ C f ðzÞ ðz aÞnþ1 dz. If n ¼ 2 and f ðzÞ ¼ 5z2 3z þ 2, then f 00 ð1Þ ¼ 10. Hence, 10 ¼ 2! 2i þ C 5z2 3z þ 2 ðz 1Þ3 dz or þ C 5z2 3z þ 2 ðz 1Þ3 dz ¼ 10i Method 2. 5z2 3z þ 2 ¼ 5ðz 1Þ2 þ 7ðz 1Þ þ 4. Then þ C 5z2 3z þ 2 ðz 1Þ3 dz ¼ þ C 5ðz 1Þ2 þ 7ðz 1Þ þ 4 ðz 1Þ3 dz ¼ 5 þ C d z 1 þ 7 þ C dz ðz 1Þ2 þ 4 þ C dz ðz 1Þ3 ¼ 5ð2iÞ þ 7ð0Þ þ 4ð0Þ ¼ 10i by Problem 16.13. SERIES AND SINGULARITIES 16.18. For what values of z does each series converge? ðaÞ X 1 n¼1 zn n2 2n : The nth term ¼ un ¼ zn n2 2n : Then lim n!1 unþ1 un ¼ lim n!1 znþ1 ðn þ 1Þ2 2nþ1 n2 2n zn ¼ jzj 2 By the ratio test the series converges if jzj 2 and diverges if jzj 2. If jzj ¼ 2 the ratio test fails. However, the series of absolute values X 1 n¼1 zn n2 2n ¼ X 1 n¼1 jzjn n2 2n converges if jzj ¼ 2, since X 1 n¼1 1 n2 converges. Thus, the series converges (absolutely) for jzj @ 2, i.e., at all points inside and on the circle jzj ¼ 2. ðbÞ X 1 n¼1 ð1Þn1 z2n1 ð2n 1Þ! ¼ z z3 3! þ z5 5! : We have lim n!1 unþ1 un ¼ lim n!1 ð1Þn z2nþ1 ð2n þ 1Þ! ð2n 1Þ! ð1Þn1 z2n1 ¼ lim n!1 z2 2nð2n þ 1Þ ¼ 0 Then the series, which represents sin z, converges for all values of z. 406 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 416. ðcÞ X 1 n¼1 ðz iÞn 3n : We have lim n!1 unþ1 un ¼ lim n!1 ðz iÞnþ1 3nþ1 3n ðz iÞn ¼ jz ij 3 : The series converges if jz ij 3, and diverges if jz ij 3. If jz ij ¼ 3, then z i ¼ 3ei and the series becomes X 1 n¼1 ein . This series diverges since the nth term does not approach zero as n ! 1. Thus, the series converges within the circle jz ij ¼ 3 but not on the boundary. 16.19. If X 1 n¼0 anzn is absolutely convergent for jzj @ R, show that it is uniformly convergent for these values of z. The definitions, theorems, and proofs for series of complex numbers are analogous to those for real series. In this case we have janzn j @ janjRn ¼ Mn. Since by hypothesis X 1 n¼1 Mn converges, it follows by the Weierstrass M test that X 1 n¼0 anzn converges uniformly for jzj @ R. 16.20. Locate in the finite z plane all the singularities, if any, of each function and name them. ðaÞ z2 ðz þ 1Þ3 : z ¼ 1 is a pole of order 3. (b) 2z3 z þ 1 ðz 4Þ2 ðz iÞðz 1 þ 2iÞ . z ¼ 4 is a pole of order 2 (double pole); z ¼ i and z ¼ 1 2i are poles of order 1 (simple poles). (c) sin mz z2 þ 2z þ 2 , m 6¼ 0. Since z2 þ 2z þ 2 ¼ 0 when z ¼ 2 ffiffiffiffiffiffiffiffiffiffiffi 4 8 p 2 ¼ 2 2i 2 ¼ 1 i, we can write z2 þ 2z þ 2 ¼ fz ð1 þ iÞgfz ð1 iÞg ¼ ðz þ 1 iÞðz þ 1 þ iÞ. The function has the two simple poles: z ¼ 1 þ i and z ¼ 1 i. (d) 1 cos z z . z ¼ 0 appears to be a singularity. However, since lim x!0 1 cos z z ¼ 0, it is a removable singularity. Another method: Since 1 cos z z ¼ 1 z 1 1 z2 2! þ z4 4! z6 6! þ ! ( ) ¼ z 2! z3 4! þ , we see that z ¼ 0 is a remova- ble singularity. ðeÞ e1=ðx1Þ2 ¼ 1 1 ðz 1Þ2 þ 1 2!ðz 1Þ4 : This is a Laurent series where the principal part has an infinite number of non-zero terms. Then z ¼ 1 is an essential singularity. ( f ) ez . This function has no finite singularity. However, letting z ¼ 1=u, we obtain e1=u , which has an essential singularity at u ¼ 0. We conclude that z ¼ 1 is an essential singularity of ez . In general, to determine the nature of a possible singularity of f ðzÞ at z ¼ 1, we let z ¼ 1=u and then examine the behavior of the new function at u ¼ 0. 16.21. If f ðzÞ is analytic at all points inside and on a circle of radius R with center at a, and if a þ h is any point inside C, prove Taylor’s theorem that CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 407
- 417. f ða þ hÞ ¼ f ðaÞ þ h f 0 ðaÞ þ h2 2! f 00 ðaÞ þ h3 3! f 000 ðaÞ þ By Cauchy’s integral formula (Problem 16.15), we have f ða þ hÞ ¼ 1 2i þ C f ðzÞ dz z a h ð1Þ By division 1 z a h ¼ 1 ðz aÞ½1 h=ðz aÞ ¼ 1 ðz aÞ 1 þ h ðz aÞ þ h2 ðz aÞ2 þ þ hn ðz aÞn þ hnþ1 ðz aÞn ðz a hÞ ( ) ð2Þ Substituting (2) in (1) and using Cauchy’s integral formulas, we have f ða þ hÞ ¼ 1 2i þ C f ðzÞ dz z a þ h 2i þ C f ðzÞ dz ðz aÞ2 þ þ hn 2i þ C f ðzÞ dz ðz aÞnþ1 þ Rn ¼ f ðaÞ þ h f 0 ðaÞ þ h2 2! f 00 ðaÞ þ þ hn n! f ðnÞ ðaÞ þ Rn Rn ¼ hnþ1 2i þ C f ðzÞ dz ðz aÞnþ1 ðz a hÞ where Now when z is on C, f ðzÞ z a h @ M and jz aj ¼ R, so that by (4), Page 394, we have, since 2R is the length of C jRnj @ jhjnþ1 M 2Rnþ1 2R As n ! 1; jRnj ! 0. Then Rn ! 0 and the required result follows. If f ðzÞ is analytic in an annular region r1 @ jz aj @ r2, we can generalize the Taylor series to a Laurent series (see Problem 16.92). In some cases, as shown in Problem 16.22, the Laurent series can be obtained by use of known Taylor series. 16.22. Find Laurent series about the indicated singularity for each of the following functions. Name the singularity in each case and give the region of convergence of each series. ðaÞ ez ðz 1Þ2 ; z ¼ 1: Let z 1 ¼ u: Then z ¼ 1 þ u and ez ðz 1Þ2 ¼ e1þu u2 ¼ e eu u2 ¼ e u2 1 þ u þ u2 2! þ u3 3! þ u4 4! þ ( ) ¼ e ðz 1Þ2 þ e z 1 þ e 2! þ eðz 1Þ 3! þ eðz 1Þ2 4! þ z ¼ 1 is a pole of order 2, or double pole. The series converges for all values of z 6¼ 1. ðbÞ z cos 1 z ; z ¼ 0: z cos 1 z ¼ z 1 1 2! z2 þ 1 4! z4 1 6! z6 þ ¼ z 1 2! z þ 1 4! z3 1 6! z5 þ z ¼ 0 is an essential singularity. The series converges for all values of z 6¼ 0. 408 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 418. ðcÞ sin z z ; z ¼ : Let z ¼ u: Then z ¼ þ u and sin z z ¼ sinðu þ Þ u ¼ sin u u ¼ 1 u u u3 3! þ u5 5! ! ¼ 1 þ u2 3! u4 5! þ ¼ 1 þ ðz Þ2 3! ðz Þ4 5! þ z ¼ is a removable singularity. The series converges for all values of z. ðdÞ z ðz þ 1Þðz þ 2Þ ; z ¼ 1: Let z þ 1 ¼ u. Then z ðz þ 1Þðz þ 2Þ ¼ u 1 uðu þ 1Þ ¼ u 1 u ð1 u þ u2 u3 þ u4 Þ ¼ 1 u þ 2 2u þ 2u2 2u3 þ ¼ 1 z þ 1 þ 2 2ðz þ 1Þ þ 2ðz þ 1Þ2 z ¼ 1 is a pole of order 1, or simple pole. The series converges for values of z such that 0 jz þ 1j 1. ðeÞ 1 zðz þ 2Þ3 ; z ¼ 0; 2: Case 1, z ¼ 0. Using the binomial theorem, 1 zðz þ 2Þ3 ¼ 1 8zð1 þ z=2Þ3 ¼ 1 8z 1 þ ð3Þ z 2 þ ð3Þð4Þ 2! z 2 2 þ ð3Þð4Þð5Þ 3! z 2 3 þ ¼ 1 8z 3 16 þ 3 16 z 5 32 z2 þ z ¼ 0 is a pole of order 1, or simple pole. The series converges for 0 jzj 2. Case 2, z ¼ 2. Let z þ 2 ¼ u. Then 1 zðz þ 2Þ3 ¼ 1 ðu 2Þu3 ¼ 1 2u3ð1 u=2Þ ¼ 1 2u3 1 þ u 2 þ u 2 2 þ u 2 3 þ u 2 4 þ ¼ 1 2u3 1 4u2 1 8u 1 16 1 32 u ¼ 1 2ðz þ 2Þ3 1 4ðz þ 2Þ2 1 8ðz þ 2Þ 1 16 1 32 ðz þ 2Þ z ¼ 2 is a pole of order 3. The series converges for 0 jz þ 2j 2. RESIDUES AND THE RESIDUE THEOREM 16.23. Suppose f ðzÞ is analytic everywhere inside and on a simple closed curve C except at z ¼ a which is a pole of order n. Then f ðzÞ ¼ an ðz aÞn þ anþ1 ðz aÞn1 þ þ a0 þ a1ðz aÞ þ a2ðz aÞ2 þ where an 6¼ 0. Prove that CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 409
- 419. ðaÞ þ C f ðzÞ dz ¼ 2ia1 ðbÞ a1 ¼ lim z!a 1 ðn 1Þ! dn1 dzn1 fðz aÞn f ðzÞg: (a) By integration, we have on using Problem 16.13 þ C f ðzÞ dz ¼ þ C an ðz aÞn dz þ þ þ C a1 z a dz þ þ C fa0 þ a1ðz aÞ þ a2ðz aÞ2 þ g dz ¼ 2ia1 Since only the term involving a1 remains, we call a1 the residue of f ðzÞ at the pole z ¼ a. (b) Multiplication by ðz aÞn gives the Taylor series ðz aÞn f ðzÞ ¼ an þ anþ1ðz aÞ þ þ a1ðz aÞn1 þ Taking the ðn 1Þst derivative of both sides and letting z ! a, we find ðn 1Þ!a1 ¼ lim z!a dn1 dzn1 fðz aÞn f ðzÞg from which the required result follows. 16.24. Determine the residues of each function at the indicated poles. ðaÞ z2 ðz 2Þðz2 þ 1Þ ; z ¼ 2; i; i: These are simple poles. Then: Residue at z ¼ 2 is lim z!2 ðz 2Þ z2 ðz 2Þðz2 þ 1Þ ( ) ¼ 4 5 : Residue at z ¼ i is lim z!i ðz iÞ z2 ðz 2Þðz iÞðz þ iÞ ( ) ¼ i2 ði 2Þð2iÞ ¼ 1 2i 10 : Residue at z ¼ i is lim z!i ðz þ iÞ z2 ðz 2Þðz iÞðz þ iÞ ( ) ¼ i2 ði 2Þð2iÞ ¼ 1 þ 2i 10 : ðbÞ 1 zðz þ 2Þ3 ; z ¼ 0; 2: z ¼ 0 is a simple pole, z ¼ 2 is a pole of order 3. Then: Residue at z ¼ 0 is lim z!0 z 1 zðz þ 2Þ3 ¼ 1 8 : Residue at z ¼ 2 is lim z!2 1 2! d2 dz2 ðz þ 2Þ3 1 zðz þ 2Þ3 ¼ lim z!2 1 2 d2 dz2 1 z ¼ lim z!2 1 2 2 z3 ¼ 1 8 : Note that these residues can also be obtained from the coefficients of 1=z and 1=ðz þ 2Þ in the respective Laurent series [see Problem 16.22(e)]. ðcÞ zezt ðz 3Þ2 ; z ¼ 3; a pole of order 2 or double pole. Then: Residue is lim z!3 d dz ðz 3Þ2 zezt ðz 3Þ2 ¼ lim z!3 d dz ðzezt Þ ¼ lim z!3 ðezt þ ztezt Þ ¼ e3t þ 3te3t 410 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 420. (d) cot z; z ¼ 5, a pole of order 1. Then: Residue is lim z!5 ðz 5Þ cos z sin z ¼ lim z!5 z 5 sin z lim z!5 cos z ¼ lim z!5 1 cos z ð1Þ ¼ ð1Þð1Þ ¼ 1 where we have used L’Hospital’s rule, which can be shown applicable for functions of a complex variable. 16.25. If f ðzÞ is analytic within and on a simple closed curve C except at a number of poles a; b; c; . . . interior to C, prove that þ C f ðzÞ dz ¼ 2i fsum of residues of f ðzÞ at poles a; b; c; etc.g Refer to Fig. 16-8. By reasoning similar to that of Problem 16.12 (i.e., by con- structing cross cuts from C to C1; C2; C3; etc.), we have þ C f ðzÞ dz ¼ þ C1 f ðzÞ dz þ þ C2 f ðzÞ dz þ For pole a, f ðzÞ ¼ am ðz aÞm þ þ a1 ðz aÞ þ a0 þ a1ðz aÞ þ hence, as in Problem 16.23, þ C1 f ðzÞ dz ¼ 2i a1: Similarly for pole b; f ðzÞ ¼ bn ðz bÞn þ þ b1 ðz bÞ þ b0 þ b1ðz bÞ þ þ C2 f ðzÞ dz ¼ 2i b1 so that Continuing in this manner, we see that þ C f ðzÞ dz ¼ 2iða1 þ b1 þ Þ ¼ 2i (sum of residues) 16.26. Evaluate þ C ez dz ðz 1Þðz þ 3Þ2 where C is given by (a) jzj ¼ 3=2; ðbÞ jzj ¼ 10. Residue at simple pole z ¼ 1 is lim z!1 ðz 1Þ ez ðz 1Þðz þ 3Þ2 ¼ e 16 Residue at double pole z ¼ 3 is lim z!3 d dz ðz þ 3Þ2 ez ðz 1Þðz þ 3Þ2 ¼ lim z!3 ðz 1Þez ez ðz 1Þ2 ¼ 5e3 16 (a) Since jzj ¼ 3=2 encloses only the pole z ¼ 1, the required integral ¼ 2i e 16 ¼ ie 8 CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 411 Fig. 16-8
- 421. (b) Since jzj ¼ 10 encloses both poles z ¼ 1 and z ¼ 3 the required integral ¼ 2i e 16 5e3 16 ! ¼ iðe 5e3 Þ 8 EVALUATION OF DEFINITE INTEGRALS 16.27. If j f ðzÞj @ M Rk for z ¼ Rei , where k 1 and M are constants, prove that lim R!1 ð f ðzÞ dz ¼ 0 where is the semicircular arc of radius R shown in Fig. 16-9. By the result (4), Page 394, we have ð f ðzÞ dz @ ð j f ðzÞjjdzj @ M Rk R þ M Rk1 since the length of arc L ¼ R. Then lim R!1 ð f ðzÞ dz ¼ 0 and so lim R!1 ð f ðzÞ dz ¼ 0 16.28. Show that for z ¼ Rei , j f ðzÞj @ M Rk ; k 1 if f ðzÞ ¼ 1 1 þ z4 . If z ¼ Rei , j f ðzÞj ¼ 1 1 þ R4e4i @ 1 jR4e4ij 1 ¼ 1 R4 1 @ 2 R4 if R is large enough (say R 2, for example) so that M ¼ 2; k ¼ 4. Note that we have made use of the inequality jz1 þ z2j A jz1j jz2j with z1 ¼ R4 e4i and z2 ¼ 1. 16.29. Evaluate ð1 0 dx x4 þ 1 . Consider þ C dz z4 þ 1 , where C is the closed contour of Problem 16.27 consisting of the line from R to R and the semicircle , traversed in the positive (counterclockwise) sense. Since z4 þ 1 ¼ 0 when z ¼ ei=4 ; e3i=4 ; e5i=4 ; e7i=4 , these are simple poles of 1=ðz4 þ 1Þ. Only the poles ei=4 and e3i=4 lie within C. Then using L’Hospital’s rule, Residue at ei=4 ¼ lim z!ei=4 ðz ei=4 Þ 1 z4 þ 1 ¼ lim z!ei=4 1 4z3 ¼ 1 4 e3i=4 Residue at e3i=4 ¼ lim z!e3i=4 ðz e3i=4 Þ 1 z4 þ 1 ¼ lim z!e3i=4 1 4z3 ¼ 1 4 e9i=4 Thus þ C dz z4 þ 1 ¼ 2i 1 4 e3i=4 þ 1 4 e9i=4 ¼ ffiffiffi 2 p 2 ð1Þ 412 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 Fig. 16-9
- 422. i.e., ðR R dx x4 þ 1 þ ð dz z4 þ 1 ¼ ffiffiffi 2 p 2 ð2Þ Taking the limit of both sides of (2) as R ! 1 and using the results of Problem 16.28, we have lim R!1 ðR R dx x4 þ 1 ¼ ð1 1 dx x4 þ 1 ¼ ffiffiffi 2 p 2 Since ð1 1 dx x4 þ 1 ¼ 2 ð1 0 dx x4 þ 1 ; the required integral has the value ffiffiffi 2 p 4 : 16.30. Show that ð1 1 x2 dx ðx2 þ 1Þ2 ðx2 þ 2x þ 2Þ ¼ 7 50 : The poles of z2 ðz2 þ 1Þ2 ðz2 þ 2z þ 2Þ enclosed by the contour C of Problem 16.27 are z ¼ i of order 2 and z ¼ 1 þ i of order 1. Residue at z ¼ i is lim z!i d dz ðz iÞ2 z2 ðz þ iÞ2 ðz iÞ2 ðz2 þ 2z þ 2Þ ( ) ¼ 9i 12 100 : Residue at z ¼ 1 þ i is lim z!1þi ðz þ 1 iÞ z2 ðz2 þ 1Þ2 ðz þ 1 iÞðz þ 1 þ iÞ ¼ 3 4i 25 þ C z2 dz ðz2 þ 1Þ2 ðz2 þ 2z þ 2Þ ¼ 2i 9i 12 100 þ 3 4i 25 ¼ 7 50 Then ðR R x2 dx ðx2 þ 1Þ2 ðx2 þ 2x þ 2Þ þ ð z2 dz ðz2 þ 1Þ2 ðz2 þ 2z þ 2Þ ¼ 7 50 or Taking the limit as R ! 1 and noting that the second integral approaches zero by Problem 16.27, we obtain the required result. 16.31. Evaluate ð2 0 d 5 þ 3 sin . Let z ¼ ei . Then sin ¼ ei ei 2i ¼ z z1 2i , dz ¼ iei d ¼ iz d so that ð2 0 d 5 þ 3 sin ¼ þ C dz=iz 5 þ 3 z z1 2i ! ¼ þ C 2 dz 3z2 þ 10iz 3 where C is the circle of unit radius with center at the origin, as shown in Fig. 16-10 below. The poles of 2 3z2 þ 10iz 3 are the simple poles z ¼ 10i ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100 þ 36 p 6 ¼ 10i 8i 6 ¼ 3i; i=3: Only i=3 lies inside C. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 413 Fig. 16-10
- 423. Residue at i=3 ¼ lim z!i=2 z þ i 3 2 3z2 þ 10iz 3 ¼ lim z!i=2 2 6z þ 10i ¼ 1 4i by L’Hospital’s rule. Then þ C 2 dz 3z2 þ 10iz 3 ¼ 2i 1 4i ¼ 2 , the required value. 16.32. Show that ð2 0 cos 3 5 4 cos d ¼ 12 . If z ¼ ei , cos ¼ z þ z1 2 ; cos 3 ¼ e3i þ e3i 2 ¼ z3 þ z3 2 ; dz ¼ iz d. ð2 0 cos 3 5 4 cos d ¼ þ C ðz3 þ z3 Þ=2 5 4 z þ z1 2 ! dz iz Then ¼ 1 2i þ C z6 þ 1 z3 ð2z 1Þðz 2Þ dz where C is the contour of Problem 16.31. The integrand has a pole of order 3 at z ¼ 0 and a simple pole z ¼ 1 2 within C. Residue at z ¼ 0 is lim z!0 1 2! d2 dz2 z3 z6 þ 1 z3ð2z 1Þðz 2Þ ( ) ¼ 21 8 : Residue at z ¼ 1 2 is lim z!1=2 ðz 1 2Þ z6 þ 1 z3ð2z 1Þðz 2Þ ( ) ¼ 65 24 : Then 1 2i þ C z6 þ 1 z3ð2z 1Þðz 2Þ dz ¼ 1 2i ð2iÞ 21 8 65 24 ¼ 12 as required. 16.33. If j f ðzÞj @ M Rk for z ¼ Rei , where k 0 and M are constants, prove that lim R!1 ð eimz f ðzÞ dz ¼ 0 where is the semicircular arc of the contour in Problem 16.27 and m is a positive constant. If z ¼ Rei ; ð eimz f ðzÞ dz ¼ ð 0 eimRei f ðRei Þ iRei d: ð 0 eimRei f ðRei Þ iRei d @ ð 0 eimRei f ðRei Þ iRei d Then ¼ ð 0 eimR cos mR sin f ðRei Þ iRei d ¼ ð 0 emR sin j f ðRei Þj R d @ M Rk1 ð 0 emR sin d ¼ 2M Rk1 ð=2 0 emr sin d Now sin A 2= for 0 @ @ =2 (see Problem 4.73, Chapter 4). Then the last integral is less than or equal to 2M Rk1 ð=2 0 e2mR= d ¼ M mRk ð1 emR Þ As R ! 1 this approaches zero, since m and k are positive, and the required result is proved. 414 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 424. 16.34. Show that ð1 0 cos mx x2 þ 1 dx ¼ 2 em ; m 0. Consider þ C eimz z2 þ 1 dz where C is the contour of Problem 16.27. The integrand has simple poles at z ¼ i, but only z ¼ i lies within C. Residue at z ¼ i is lim z!i ðz iÞ eimz ðz iÞðz þ iÞ ¼ em 2i : þ C eimz z2 þ 1 dz ¼ 2i em 2i ¼ em Then ðR R eimx x2 þ 1 dx þ ð eimz z2 þ 1 dz ¼ em or ðR R cos mx x2 þ 1 dx þ i ðR R sin mx x2 þ 1 dx þ ð eimz z2 þ 1 dz ¼ em i.e., and so 2 ðR 0 cos mx x2 þ 1 dx þ ð eimz z2 þ 1 dz ¼ em Taking the limit as R ! 1 and using Problem 16.33 to show that the integral around approaches zero, we obtain the required result. 16.35. Show that ð1 0 sin x x dx ¼ 2 . The method of Problem 16.34 leads us to consider the integral of eiz =z around the contour of Problem 16.27. However, since z ¼ 0 lies on this path of integration and since we cannot integrate through a singularity, we modify that contour by indenting the path at z ¼ 0, as shown in Fig. 16-11, which we call contour C0 or ABDEFGHJA. Since z ¼ 0 is outside C0 , we have ð C0 eiz z dz ¼ 0 or ðr R eix x dx þ ð HJA eiz z dz þ ðR r eix x dx þ ð BDEFG eiz z dz ¼ 0 CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 415 Fig. 16-11
- 425. Replacing x by x in the first integral and combining with the third integral, we find, ðR r eix eix x dx þ ð HJA eiz z dz þ ð BDEFG eiz z dz ¼ 0 or 2i ðR r sin x x dx ¼ ð HJA eiz z dz ð BDEFG eix z dz Let r ! 0 and R ! 1. By Problem 16.33, the second integral on the right approaches zero. The first integral on the right approaches lim r!0 ð0 eirei rei irei d ¼ lim r!0 ð0 ieirei d ¼ i since the limit can be taken under the integral sign. Then we have lim R!1 r!0 2i ðR r sin x x dx ¼ i or ð1 0 sin x x dx ¼ 2 MISCELLANEOUS PROBLEMS 16.36. Let w ¼ z2 define a transformation from the z plane (xy plane) to the w plane ðuv plane). Consider a triangle in the z plane with vertices at Að2; 1Þ; Bð4; 1Þ; Cð4; 3Þ. (a) Show that the image or mapping of this triangle is a curvilinear triangle in the uv plane. (b) Find the angles of this curvilinear triangle and compare with those of the original triangle. (a) Since w ¼ z2 , we have u ¼ x2 y2 ; v ¼ 2xy as the transformation equations. Then point Að2; 1Þ in the xy plane maps into point A0 ð3; 4Þ of the uv plane (see figures below). Similarly, points B and C map into points B0 and C0 respectively. The line segments AC; BC; AB of triangle ABC map respectively into parabolic segments A0 C0 ; B0 C0 ; A0 B0 of curvilinear triangle A0 B0 C0 with equations as shown in Figures 16-12(a) and (b). 416 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 Fig. 16-12
- 426. (b) The slope of the tangent to the curve v2 ¼ 4ð1 þ uÞ at ð3; 4Þ is m1 ¼ dv du ð3;4Þ ¼ 2 v ð3;4Þ ¼ 1 2 . The slope of the tangent to the curve u2 ¼ 2v þ 1 at ð3; 4Þ is m2 ¼ dv du ð3;4Þ ¼ u ¼ 3. Then the angle between the two curves at A0 is given by tan ¼ m2 m1 1 þ m1m2 ¼ 3 1 2 1 þ ð3Þð1 2Þ ¼ 1; and ¼ =4 Similarly, we can show that the angle between A0 C0 and B0 C0 is =4, while the angle between A0 B0 and B0 C0 is =2. Therefore, the angles of the curvilinear triangle are equal to the corresponding ones of the given triangle. In general, if w ¼ f ðzÞ is a transformation where f ðzÞ is analytic, the angle between two curves in the z plane intersecting at z ¼ z0 has the same magnitude and sense (orientation) as the angle between the images of the two curves, so long as f 0 ðz0Þ 6¼ 0. This property is called the conformal property of analytic functions, and for this reason, the transformation w ¼ f ðzÞ is often called a con- formal transformation or conformal mapping function. 16.37. Let w ¼ ffiffiffi z p define a transformation from the z plane to the w plane. A point moves counter- clockwise along the circle jzj ¼ 1. Show that when it has returned to its starting position for the first time, its image point has not yet returned, but that when it has returned for the second time, its image point returns for the first time. Let z ¼ ei . Then w ¼ ffiffiffi z p ¼ ei=2 . Let ¼ 0 correspond to the starting position. Then z ¼ 1 and w ¼ 1 [corresponding to A and P in Figures 16-13(a) and (b)]. When one complete revolution in the z plane has been made, ¼ 2; z ¼ 1, but w ¼ ei=2 ¼ ei ¼ 1, so the image point has not yet returned to its starting position. However, after two complete revolutions in the z plane have been made, ¼ 4; z ¼ 1 and w ¼ ei=2 ¼ e2i ¼ 1, so the image point has returned for the first time. It follows from the above that w is not a single-valued function of z but is a double-valued function of z; i.e., given z, there are two values of w. If we wish to consider it a single-valued function, we must restrict . We can, for example, choose 0 @ 2, although other possibilities exist. This represents one branch of the double-valued function w ¼ ffiffiffi z p . In continuing beyond this interval we are on the second branch, e.g., 2 @ 4. The point z ¼ 0 about which the rotation is taking place is called a branch point. Equiva- lently, we can insure that f ðzÞ ¼ ffiffiffi z p will be single-valued by agreeing not to cross the line Ox, called a branch line. 16.38. Show that ð1 0 xp1 1 þ x dx ¼ sin p ; 0 p 1. CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 417 Fig. 16-13
- 427. Consider þ C zp1 1 þ z dz. Since z ¼ 0 is a branch point, choose C as the contour of Fig. 16-14 where AB and GH are actually coincident with x-axis but are shown separated for visual purposes. The integrand has the pole z ¼ 1 lying within C. Residue at z ¼ 1 ¼ ei is lim z!1 ðz þ 1Þ z p1 1 þ z ¼ ðei Þp1 ¼ eðp1Þi þ C zp1 1 þ z dz ¼ 2i eðp1Þi Then or, omitting the integrand, ð AB þ ð BDEFG þ ð GH þ ð HJA ¼ 2i eðp1Þi We thus have ðR r xp1 1 þ x dx þ ð2 0 ðRei Þp1 iRei d 1 þ Rei þ ðr R ðxe2i Þp1 1 þ xe2i dx þ ð0 2 ðrei Þp1 irei d 1 þ rei ¼ 2i eðp1Þi where we have to use z ¼ xe2i for the integral along GH, since the argument of z is increased by 2 in going round the circle BDEFG. Taking the limit as r ! 0 and R ! 1 and noting that the second and fourth integrals approach zero, we find ð1 0 xp1 1 þ x dx þ ð0 1 e2iðp1Þ xp1 1 þ x dx ¼ 2 eðp1Þi ð1 e2iðp1Þ Þ ð1 0 xp1 1 þ x dx ¼ 2i eðp1Þi or so that ð1 0 xp1 1 þ x dx ¼ 2i eðp1Þi 1 e2iðp1Þ ¼ 2i epi epi ¼ sin p Supplementary Problems FUNCTIONS, LIMITS, CONTINUITY 16.39. Describe the locus represented by (a) jz þ 2 3ij ¼ 5; ðbÞ jz þ 2j ¼ 2jz 1j; ðcÞ jz þ 5j jz 5j ¼ 6. Construct a figure in each case. Ans. ðaÞ Circle ðx þ 2Þ2 þ ð y 3Þ2 ¼ 25, center ð2; 3Þ, radius 5. (b) Circle ðx 2Þ2 þ y2 ¼ 4, center ð2; 0Þ, radius 2. (c) Branch of hyperbola x2 =9 y2 =16 ¼ 1, where x A 3. 16.40. Determine the region in the z plane represented by each of the following: (a) jz 2 þ ij A 4; ðbÞ jzj @ 3; 0 @ arg z @ 4 ; ðcÞ jz 3j þ jz þ 3j 10. Construct a figure in each case. Ans. (a) Boundary and exterior of circle ðx 2Þ2 þ ð y þ 1Þ2 ¼ 16. 418 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16 Fig. 16-14
- 428. (b) Region in the first quadrant bounded by x2 þ y2 ¼ 9, the x-axis and the line y ¼ x. (c) Interior of ellipse x2 =25 þ y2 =16 ¼ 1. 16.41. Express each function in the form uðx; yÞ þ ivðx; yÞ, where u and v are real. (a) z2 þ 2iz; ðbÞ z=ð3 þ zÞ; ðcÞ ez2 ; ðdÞ lnð1 þ zÞ. Ans. (a) u ¼ x3 3xy2 2y; v ¼ 3x2 y y2 þ 2x (b) u ¼ x2 þ 3x þ y2 x2 þ 6x þ y2 þ 9 ; v ¼ 3y x2 þ 6x þ y2 þ 9 (c) u ¼ ex2 y2 cos 2xy; v ¼ ex2 y2 sin 2xy (d) u ¼ 1 2 lnfð1 þ xÞ2 þ y2 g; v ¼ tan1 y 1 þ x þ 2k; k ¼ 0; 1; 2; . . . 16.42. Prove that (a) lim z!x0 z2 ¼ z2 0; ðbÞ f ðzÞ ¼ z2 is continuous at z ¼ z0 directly from the definition. 16.43. (a) If z ¼ ! is any root of z5 ¼ 1 different from 1, prove that all the roots are 1; !; !2 ; !3 ; !4 . (b) Show that 1 þ ! þ !2 þ !3 þ !4 ¼ 0. (c) Generalize the results in (a) and (b) to the equation zn ¼ 1. DERIVATIVES, CAUCHY-RIEMANN EQUATIONS 16.44. (a) If w ¼ f ðzÞ ¼ z þ 1 z , find dw dz directly from the definition. (b) For what finite values of z is f ðzÞ nonanalytic? Ans. ðaÞ 1 1=z2 ; ðbÞ z ¼ 0 16.45. Given the function w ¼ z4 . (a) Find real functions u and v such that w ¼ u þ iv. (b) Show that the Cauchy-Riemann equations hold at all points in the finite z plane. (c) Prove that u and v are harmonic functions. (d) Determine dw=dz. Ans: ðaÞ u ¼ x4 6x2 y2 þ y4 ; v ¼ 4x3 y 4xy2 ðdÞ 4z3 16.46. Prove that f ðzÞ ¼ zjzj is not analytic anywhere. 16.47. Prove that f ðzÞ ¼ 1 z 2 is analytic in any region not including z ¼ 2. 16.48. If the imaginary part of an analytic function is 2xð1 yÞ, determine (a) the real part, (b) the function. Ans: ðaÞ y2 x2 2y þ c; ðbÞ 2iz z2 þ c, where c is real 16.49. Construct an analytic function f ðzÞ whose real part is ex ðx cos y þ y sin yÞ and for which f ð0Þ ¼ 1. Ans: zez þ 1 16.50. Prove that there is no analytic function whose imaginary part is x2 2y. 16.51. Find f ðzÞ such that f 0 ðzÞ ¼ 4z 3 and f ð1 þ iÞ ¼ 3i. Ans: f ðzÞ ¼ 2z2 3z þ 3 4i INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS 16.52. Evaluate ð3þi 12i ð2z þ 3Þ dz: (a) along the path x ¼ 2t þ 1; y ¼ 4t2 t 2 0 @ t @ 1. (b) along the straight line joining 1 2i and 3 þ i. (c) along straight lines from 1 2i to 1 þ i and then to 3 þ i. Ans: 17 þ 19i in all cases CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 419
- 429. 16.53. Evaluate ð C ðz2 z þ 2Þ dz, where C is the upper half of the circle jzj ¼ 1 tranversed in the positive sense. Ans: 14=3 16.54. Evaluate þ C z; 2z 5 , where C is the circle (a) jzj ¼ 2; ðbÞ jz 3j ¼ 2: Ans: ðaÞ 0; ðbÞ 5i=2 16.55. Evaluate þ C z2 ðz þ 2Þðz 1Þ dz, where C is: (a) a square with vertices at 1 i; 1 þ i; 3 þ i; 3 i; (b) the circle jz þ ij ¼ 3; (c) the circle jzj ¼ ffiffiffi 2 p . Ans: ðaÞ 8i=3 ðbÞ 2i ðcÞ 2i=3 16.56. Evaluate (a) þ C cos z z 1 dz; ðbÞ þ C ez þ z ðz 1Þ4 dz where C is any simple closed curve enclosing z ¼ 1. Ans: ðaÞ 2i ðbÞ ie=3 16.57. Prove Cauchy’s integral formulas. [Hint: Use the definition of derivative and then apply mathematical induction.] SERIES AND SINGULARITIES 16.58. For what values of z does each series converge? ðaÞ X 1 n¼1 ðz þ 2Þn n! ; ðbÞ X 1 n¼1 nðz iÞn n þ 1 ; ðcÞ X 1 n¼1 ð1Þn ðz2 þ 2z þ 2Þ2n : Ans: ðaÞ all z (b) jz ij 1 ðcÞ z ¼ 1 i 16.59. Prove that the series X 1 n¼1 zn nðn þ 1Þ is (a) absolutely convergent, (b) uniformly convergent for jzj @ 1. 16.60. Prove that the series X 1 n¼0 ðz þ iÞn 2n converges uniformly within any circle of radius R such that jz þ ij R 2. 16.61. Locate in the finite z plane all the singularities, if any, of each function and name them: ðaÞ z 2 ð2z þ 1Þ4 ; ðbÞ z ðz 1Þðz þ 2Þ2 ; ðcÞ z2 þ 1 z2 þ 2z þ 2 ; ðdÞ cos 1 z ; ðeÞ sinðz =3Þ 3z ; ð f Þ cos z ðz2 þ 4Þ2 : Ans. (a) z ¼ 1 2, pole of order 4 (d) z ¼ 0, essential singularity (b) z ¼ 1, simple pole; z ¼ 2, double pole (e) z ¼ =3, removable singularity (c) simple poles z ¼ 1 i ( f ) z ¼ 2i, double poles 16.62. Find Laurent series about the indicated singularity for each of the following functions, naming the singu- larity in each case. Indicate the region of convergence of each series. ðaÞ cos z z ; z ¼ ðbÞ z2 e1=z ; z ¼ 0 ðcÞ z2 ðz 1Þ2 ðz þ 3Þ ; z ¼ 1 Ans: ðaÞ 1 z þ z 2! ðz Þ3 4! þ ðz Þ5 6! ; simple pole, all z 6¼ ðbÞ z2 z þ 1 2! 1 3! z þ 1 4! z2 1 5! z3 þ ; essential singularity, all z 6¼ 0 420 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 430. ðcÞ 1 4ðz 1Þ2 þ 7 16ðz 1Þ þ 9 64 9ðz 1Þ 256 þ ; double pole, 0 jz 1j 4 RESIDUES AND THE RESIDUE THEOREM 16.63. Determine the residues of each function at its poles: ðaÞ 2z þ 3 z2 4 ; ðbÞ z 3 z3 þ 5z2 ; ðcÞ ezt ðz 2Þ3 ; ðdÞ z ðz2 þ 1Þ2 : Ans. (a) z ¼ 2; 7=4; z ¼ 2; 1=4 (c) z ¼ 2; 1 2 t2 e2t (b) z ¼ 0; 8=25; z ¼ 5; 8=25 (d) z ¼ i; 0; z ¼ i; 0 16.64. Find the residue of ezt tan z at the simple pole z ¼ 3=2. Ans: e3t=2 16.65. Evaluate þ C z2 dz ðz þ 1Þðz þ 3Þ , where C is a simple closed curve enclosing all the poles. Ans: 8i 16.66. If C is a simple closed curve enclosing z ¼ i, show that þ C zezt ðz2 þ 1Þ2 dz ¼ 1 2 t sin t 16.67. If f ðzÞ ¼ PðzÞ=QðzÞ, where PðzÞ and QðzÞ are polynomials such that the degree of PðzÞ is at least two less than the degree of QðzÞ, prove that þ C f ðzÞ dz ¼ 0, where C encloses all the poles of f ðzÞ. EVALUATION OF DEFINITE INTEGRALS Use contour integration to verify each of the following 16.68. ð1 0 x2 dx x4 þ 1 ¼ 2 ffiffiffi 2 p 16.75. ð2 0 d ð2 þ cos Þ2 ¼ 4 ffiffiffi 3 p 9 16.69. ð1 1 dx x6 þ a6 ¼ 2 3a5 ; a 0 16.76. ð 0 sin2 5 4 cos d ¼ 8 16.70. ð1 0 dx ðx2 þ 4Þ2 ¼ 32 16.77. ð2 0 d ð1 þ sin2 Þ2 ¼ 3 2 ffiffiffi 2 p 16.71. ð1 0 ffiffiffi x p x3 þ 1 dx ¼ 3 16.78. ð2 0 cos n d 1 2a cos þ a2 ¼ 2an 1 a2 ; n ¼ 0; 1; 2; 3; . . . ; 0 a 1 16.72. ð1 0 dx ðx4 þ a4Þ2 ¼ 3 8 ffiffiffi 2 p a7 ; a 0 16.79. ð2 0 d ða þ b cos Þ3 ¼ ð2a2 þ b2 Þ ða2 b2Þ5=2 ; a jbj 16.73. ð1 1 dx ðx2 þ 1Þ2 ðx2 þ 4Þ ¼ 9 16.80. ð1 0 x sin 2x x2 þ 4 dx ¼ e4 4 16.74. ð2 0 d 2 cos ¼ 2 ffiffiffi 3 p 16.81. ð1 0 cos 2x x4 þ 4 dx ¼ e 8 CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 421
- 431. 16.82. ð1 0 x sin x ðx2 þ 1Þ2 dx ¼ 2 e 4 16.84. ð1 0 sin2 x x2 dx ¼ 2 16.83. ð1 0 sin x xðx2 þ 1Þ2 dx ¼ ð2e 3Þ 4e 16.85. ð1 0 sin3 x x3 dx ¼ 3 8 16.86. ð1 0 cos x cosh x dx ¼ 2 coshð=2Þ . Hint: Consider þ C eiz cosh z dz, where C is a rectangle with vertices at ðR; 0Þ, ðR; 0Þ; ðR; ÞrðR; Þ. Then let R ! 1: MISCELLANEOUS PROBLEMS 16.87. If z ¼ ei and f ðzÞ ¼ uð; Þ þ i vð; Þ, where and are polar coordinates, show that the Cauchy-Rie- mann equations are @u @ ¼ 1 @v @ ; @v @ ¼ 1 @u @ 16.88. If w ¼ f ðzÞ, where f ðzÞ is analytic, defines a transformation from the z plane to the w plane where z ¼ x þ iy and w ¼ u þ iv, prove that the Jacobian of the transformation is given by @ðu; vÞ @ðx; yÞ ¼ j f 0 ðzÞj2 16.89. Let Fðx; yÞ be transformed to Gðu; vÞ by the transformation w ¼ f ðzÞ. Show that if @2 F @x2 þ @2 F @y2 ¼ 0, then at all points where f 0 ðzÞ 6¼ 0, @2 G @u2 þ @2 G @v2 ¼ 0. 16.90. Show that by the bilinear transformation w ¼ az þ b cz þ d , where ad bc 6¼ 0, circles in the z plane are trans- formed into circles of the w plane. 16.91. If f ðzÞ is analytic inside and on the circle jz aj ¼ R, prove Cauchy’s inequality, namely j f ðnÞ ðaÞj @ n!M Rn where j f ðzÞj @ M on the circle. [Hint: Use Cauchy’s integral formulas.] 16.92. Let C1 and C2 be concentric circles having center a and radii r1 and r2, respectively, where r1 r2. If a þ h is any point in the annular region bounded by C1 and C2, and f ðzÞ is analytic in this region, prove Laurent’s theorem that f ða þ hÞ ¼ X 1 1 anhn an ¼ 1 2i þ C f ðzÞ dz ðz aÞnþ1 where C being any closed curve in the angular region surrounding C1. Hint: Write f ða þ hÞ ¼ 1 2i þ C2 f ðzÞ dz z ða þ hÞ 1 2i þ C1 f ðzÞ dz z ða þ hÞ and expand 1 z a h in two different ways. 16.93. Find a Laurent series expansion for the function f ðzÞ ¼ z ðz þ 1Þðz þ 2Þ which converges for 1 jzj 2 and diverges elsewhere. Hint: Write z ðz þ 1Þðz þ 2Þ ¼ 1 z þ 1 þ 2 z þ 2 ¼ 1 zð1 þ 1=zÞ þ 1 1 þ z=2 : 422 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16
- 432. Ans: 1 z5 þ 1 z4 1 z3 þ 1 z2 1 z þ 1 z 2 þ z2 4 z3 8 þ 16.94. Let ð1 0 est FðtÞ dt ¼ f ðsÞ where f ðsÞ is a given rational function with numerator of degree less than that of the denominator. If C is a simple closed curve enclosing all the poles of f ðsÞ, we can show that FðtÞ ¼ 1 2i þ C ezt f ðzÞ dz ¼ sum of residues of ezt f ðzÞ at its poles Use this result to find FðtÞ if f ðsÞ is (a) s s2 þ 1 ; ðbÞ 1 s2 þ 2s þ 5 ; ðcÞ s2 þ 1 sðs 1Þ2 ; ðdÞ 1 ðs2 þ 1Þ2 and check results in each case. [Note that f ðsÞ is the Laplace transform of FðtÞ, and FðtÞ is the inverse Laplace transform of f ðsÞ (see Chapter 12). Extensions to other functions f ðxÞ are possible.] Ans. (a) cos t; ðbÞ 1 2 et sin 2t; ðcÞ 1 4 þ 5 2 te2t þ 3 4 e2t ; ðdÞ 1 2 ðsin t t cos tÞ CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 423
- 434. 425 Abel, integral test of, 334 summability, 305 theorems of, 282, 293 Absolute convergence: of integrals, 309, 312, 319–321 of series, 268, 283, 300 theorems on, 269, 283 Absolute maximum or minimum, 42 (See also Maxima and minima) Absolute value, 3 of complex numbers, 6 Acceleration, 74, 158 centripetal, 175 in cylindrical and spherical coordinates, 181 normal and tangential components of, 177 Accumulation, point of, 5, 117 (See also Limit points) Addition, 1 associative law of, 2, 8 commutative law of, 2 of complex numbers, 7, 13 of vectors, 151, 152, 163 Aerodynamics, 402 Aleph-null, 5 Algebra: fundamental theorem of, 43 of complex numbers, 6, 7, 13–15 of vectors, 151, 152, 163–171 Algebraic functions, 43 Algebraic numbers, 6, 13 countability of, 13 Alternating series, 267, 268, 282 convergence test for, 267, 268 error in computations using, 268, 282 Amplitude, 7 Analytic continuation, of gamma function, 376 Analytic functions, 393 Analytic part, of a Laurent series, 395 Anti-derivatives, 94 Approximations (see Numerical methods) by partial sums of Fourier series, 352 by use of differentials, 78, 79, 130, 131 least square, 201 to irrational numbers, 9 using Newton’s method, 74 using Taylor’s theorem, 274–275 Archimedes, 90 Arc length, 99, 109 element, 157, 161, 174 Area, 100, 109 expressed as a line integral, 242 of an ellipse, 205 of a parallelogram, 155, 168 Argand diagram, 7 Argument, 7 Arithmetic mean, 10 Associative law, 2, 4 for vectors, 152, 155 Asymptotic series or expansions: for gamma function, 286, 292 Axiomatic foundations: of complex numbers, 6, 7 of real number, 4 of vector analysis, 155 Axis, real, 2 x, y and z, 121 Base, of logarithms, 4 Bases, orthonormal, 152 Bernoulli, Daniel, 336 Bernoulli numbers, 304 Bernoulli’s inequality, 16 Bessel functions, 276 Bessel differential equation, 276 Beta functions, 375, 378, 379, 382, 384 relation, to gamma functions, 379 Bilinear transformation, 422 (See also Fractional linear transformation) Binary scale, 16, 21 system, 2 Binomial coefficients, 21 series, 275 theorem, 21 Bolzano-Weierstrass theorem, 6, 12, 19, 117 Boundary conditions, 339 Boundary point, 117 Boundary-value problems: and Fourier integrals, 371 and Fourier series, 339, 356, 357 in heat conduction, 356, 357 in vibration of strings, 361 separation of variables method for solving, 356 Bounded functions, 40, 41 sequences, 24, 30–32, 36 sets, 6 Bounds, lower and upper, 6, 12, 13 Box product, 155 Branches of a function, 41 Branch line, 417 Branch point, 396, 417 Calculus, fundamental theorem of integral, 94, 95, 104 Cardinal number of the continuum, 5 Cardioid, 114 Catenary, 113 Cauchy principal value, 310, 321 Cauchy-Riemann equations, 393, 400–403 derivation of, 401 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
- 435. Cauchy-Riemann equations (Cont.): in polar form, 422 Cauchy’s: convergence criterion, 25, 33 form of remainder in Taylor’s theorem, 274, 296 generalized theorem of the mean, 72, 82 inequality, 422 integral formulas, 394, 403–406 theorem, 394, 403–406 Centripetal acceleration, 175 Chain rules, 69, 122, 133 for Jacobians, 124 Circle of convergence, 276 Class, 1 (See also Sets) Closed interval, 5 region, 117 set, 6, 12, 13, 117 Closure law or property, 2 Cluster point, 5, 117 (See also Limit points) Collection, 1 (See also Sets) Commutative law, 2 for dot products, 153 for vectors, 154, 166, 167 Comparison test: for integrals, 308, 311, 319 for series, 267, 279, 280 Completeness, of an orthonormal set, 310 Complex numbers, 6, 13, 14 absolute value of, 6 amplitude of, 7 argument of, 7 as ordered pairs of real numbers, 7 as vectors, 20 axiomatic foundations of, 7 conjugate of, 6 equality of, 6 modulus of, 6 operations with, 6, 13, 14 polar form of, 7, 14 real and imaginary parts of, 6 roots of, 7, 13 Complex plane, 7 Complex variable, 392, 393 (See also Functions of a complex variable) Components, of a vector, 153 Composite fuctions, 47 continuity of, 47 differentiation of, 69, 132–135 Conditional convergence: of integrals, 309, 312, 319, 320 of series, 268, 300 Conductivity, thermal, 357 Conformal mapping or transformation, 417 (See also Transformations) Conjugate, complex 6 Connected region, 237 set, 117 Connected region (Cont.): simply-, 117, 232 Conservative field, 233 Constants, 5 Constraints, 188 Continuity, 46–64, 119, 127, 128, 399 and differentiability, 66, 72, 73, 120, 121 definition of, 46, 47 in an interval, 47 in a region, 119 of an infinite series of functions, 271, 272, 288 of functions of a complex variable, 393, 399, 400 of integrals, 99, 314 of vector functions, 156 right- and left-hand, 47 piecewise, 48 theorems on, 47, 48 uniform, 48, 119 Continuous (see Continuity) differentiability, 67, 121 Continuously differentiable functions, 66, 67, 120 Continuum, cardinality of, 5 Contour integration, 398 Convergence: absolute (see Absolute convergence) circle of, 276 conditional (see Conditional convergence) criterion of Cauchy, 33, 37 domain of, 272 interval of, 25, 272 of Fourier integrals (see Fourier’s integral theorem) of Fourier series, 338, 354–356 of improper integrals (see Improper integrals) of infinite series (see Infinite series) of series of constants, 278–285 radius of, 272, 276 region of, 117 uniform (see Uniform convergence) Convergent (see Convergence) integrals, 306–309 (See also Improper integrals) sequences, 23, 269 (See also Sequences) series, 25 (See also Infinite series) Convolution theorem: for Fourier transforms, 365 for Laplace transforms, 334 Coordinate curve, 160 Coordinates: curvilinear, 139, 160 (See also Curvilinear coordinates) cylindrical, 161, 174 hyperbolic, 218 Coordinates (Cont.): polar, 7 rectangular, 152 spherical, 162, 190 Correspondence, 2, 11, 23, 39, 160 one to one, 2, 11 Countability, 5, 11, 12 of algebraic numbers, 13 of rational numbers, 11, 12 Countable set, 5, 11, 12 measure of a, 91, 103 Critical points, 73 Cross products, 154, 166–169 proof of distributive law for, 166 Curl, 158, 159, 172–174 in curvilinear coordinates, 161 Curvature, radius of, 177, 181 Curve, coordinate, 150 simple closed, 117, 232, 242 space, 157 Curvilinear coordinates, 125, 139 curl, divergence, gradient, and Laplacian in, 161, 162 Jacobians and, 161, 162 multiple integrals in, 207–228 orthogonal, 207–228 special, 161, 162 transformations and, 139, 140, 160 vectors and, 161, 162 Cut (see Dedekind cut) Cycloid, 99 Cylindricalcoordinates,161,174,175 arc length element in, 161 divergence in, 175 gradient in, 175 Laplacian in, 161, 173 multiple integrals in, 222 parabolic, 180 volume element in, 161, 175 Decimal representation of real numbers, 2 Decimals, recurring, 2 Decreasing functions, 41, 47 monotonic, 41 strictly, 41, 47 Decreasing sequences, monotonic and strictly, 24 Dedekind cuts, 4, 16 Definite integrals, 90–95, 103 (See also Integrals) change of variable in, 95, 105–108 definition of, 90, 91 mean value theorems for, 92, 93, 104 numerical methods for evaluating, 98, 108, 109 properties of, 91, 92 theorem for existence of, 91 with variable limits, 95, 186, 313, 314 426 INDEX
- 436. Degree, of a polynomial equation, 6 of homogeneous fuctions, 122 Del (r), 159 formulas involving, 159 in curl, gradient, and divergence, 159 Deleted neighborhood, 6, 117 De Moivre’s theorem, 7, 15 Dense, everywhere, 2 Denumerable set (see Countable set) Dependent variable, 39, 116 Derivatives, 65–89, 75–79 (See also Differentiation) chain rules for, 70, 124 continuity and, 66, 72, 121, 130 definition of, 65, 66 directional, 186, 193, 202 evaluation of, 71, 75–89 graphical interpretation of, 66 higher order, 71, 120 of functions of a complex variable, 393, 400–403 of infinite series of functions, 269, 400–403 of elementary functions, 71, 78–80 of vector functions, 157, 171, 172 partial (see Partial derivatives) right- and left-hand, 67, 77, 78, 86 rules for finding, 70 table of, 71 Determinant: for cross product, 154 for curl, 159 for scalar triple product, 155 Jacobian (see Jacobian) Dextral system, 152 Difference equations, 65 Differentiability, 66, 67, 121 and continuity, 66, 72, 73 continuous, 66 piecewise, 66 Differential: as a linear function, 68, 121 elements of area, of volume, 160, 163, 212, 213, 233 Differential equation: Gauss’, 276 solution of, by Laplace transforms, 314, 330 Differential geometry, 158, 181 Differentials, 67, 68, 69, 78, 120–122 approximations by use of, 78, 79, 120 exact, 122, 131, 132 geometric interpretation of, 68, 69, 121 of functions of several variables, 120, 130 of vector functions, 156 total, 120 Differentiation (See also Derivatives) of Fourier series, 339, 353 Differentiation (Cont.): rules for, 70, 78–80, 87 under the integral sign, 186, 194, 203 Diffusivity, 357 Directed line segments, 150 Directional derivatives, 186, 193, 202 Dirichlet conditions, 337, 345 integrals, 379, 385, 389 Dirichlet’s test: for integrals, 314 for series, 270, 303 Discontinuities, 47, 119 removable, 56, 119 Distance between points, 165 Distributive law, 2 for cross products, 154 for dot products, 153 Divergence, 158, 159, 172–174 in curvilinear coordinates, 161 in cylindrical coordinates, 161 of improper integrals, 306–309 (See also Improper integrals) of infinite series (see Infinite series) Divergence theorem, 236, 249–252, 261 proof of, 249, 250 Divergent integrals, 306–335 sequences, 23 (See also Sequences) series, 25 (See also Series) Division, 1 by zero, 8 of complex numbers, 6, 7 Domain, of a function, 39, 116 of convergence, 272 Dot products, 153, 154, 165, 166 commutative law for, 153 distributive law for, 153 laws for, 153, 154 Double series, 277 Dummy variable, 94 Duplication formula for gamma function, 286, 378, 386 e, 4 Electric field vector, 181 Electromagnetic theory, 181 Elementary transcendental functions, 43, 71, 95 Elements, of a set, 1 Ellipse, 114 area of, 114 Empty set, 1 Envelopes, 185, 186, 192 Equality, of complex number, 6 of vectors, 158 Equations: difference, 65 differential (see Differential equation) Equations (Cont.): integral, 364, 369, 370 polynomial, 6, 43 Equipotential surfaces, 186 Errors, applications to, 189, 200, 204 in computing sums of alternating series, 266, 282 mean square, 353 Essential singularity, 395 Eudoxus, 90 Euler, Leonhart, 336 Euler’s, constant, 296, 378, 388 formulas or indentities, 8, 295 theorem on homogeneous functions, 122 Even function, 338, 347–351 Everywhere dense set, 2 Evolute, 185 Exact differentials, 122, 131, 132, 231 (See also Differentials) Expansion of functions: in Fourier series (see Fourier series) in power series, 272 Expansions (see Series) Explicit functions, 123 Exponential function, 42 order, 334 Exponents, 3, 11 Factorial function (see Gamma functions) Fibonacci sequence, 35, 37 Field, 2 conservative, 233 scalar, 153 vector, 153 Fluid mechanics, 402 Fourier coefficients, 337, 345 expansion (see Fourier series) Fourier integrals, 363–374 (See also Fourier transforms) convergence of (see Fourier’s integral theorem) solution of boundary-value problems by, 371 Fourier, Joseph, 336 Fourier series, 336 –362 complex notation for, 339 convergence of, 338, 354–356 differentiation and integration of, 339 Dirichlet conditions for convergence of, 337 half range, 338, 347–351 Parseval’s identity for, 310, 338, 351, 352 solution of boundary-value problems by, 339, 356–358 Fourier’s integral theorem, 363, 364 heuristic demonstration of, 369 proof of, 369 INDEX 427
- 437. Fourier transforms, 364–368 (See also Fourier integrals) convolution theorem for, 365 inverse, 364 Parselval’s identities for, 366, 368, 373 symmetric form for, 364 Fractions, 1 Frenet-Serret formulas, 159 Fresnel integrals, 387 Frullani’s integral, 333 Functional determinant, 123, 136 (See also Jacobians) Functional notation, 39, 116 Functions, 39–64, 116, 132, 392 algebraic, 43 beta (see Beta functions) bounded, 40, 41 branches of, 41 composite (see Composite functions) continuity of (see Continuity) decreasing, 41, 42 definition of, 39, 116 derivatives of (see Derivatives) differential of (see Differentials) domain of, 39, 116 elementary transcendental, 43, 44 even, 338, 347–351 explicit and implicit, 123 gamma (see Gamma functions) harmonic, 393 hyperbolic, 44, 45 hypergeometric, 276, 303 increasing, 41, 42 inverse (see Inverse functions) limits of (see Limits of functions) maxima and minima of (see Maxima and minima) monotonic, 41 multipled-valued (see Multiple- valued function) normalized, 342 odd, 238, 347–351 of a complex variable (see Functions of a complex variable) of a function (see Composite function) of several variables, 116, 123, 126 orthogonal, 342, 357, 358 orthonormal, 342 periodic, 336 polynomial, 43 sequences and series of, 269, 270, 272, 286, 289 single-valued, 39, 392 staircase of step, 51 transcendental, 43, 44 types of, 43, 44 vector (see Vector fuctions) Functions of a complex variable, 392–423 Functions of a complex variable (Cont.): analytic, 393 Cauchy-Riemann equations, 393, 400 (see Cauchy-Riemann equations) continuity of, 393, 399, 400 definition of, 392 derivatives of, 393, 400–403 elementary, 393 imaginary part of, 392, 400 integrals of, 394, 403–406 Jacobians and, 422 Laplace transforms and, 423 limits of, 393, 399, 400 line integrals and, 394 multiple-valued, 392 poles of, 395 real part of, 392, 400 residue theorem for (see Residue theorem) series of, 286, 395, 406–409 single-valued, 392 singular points of, 395 Fundamental theorem: of algebra, 43 of calculus, 94, 104 Gamma functions, 375–391 analytic continuation of, 376 asymptotic formulas for, 376, 378 duplication formula for, 378, 386 infinite product for, 377 recurrence formula for, 375, 376 Stirling’s formulas and asymptotic series for, 378, 384 table and graph of, 375 Gauss’: differential equation, 276 function, 377 test, 268, 283 Geometric integral, 308 Gibbs, Williard, 150 G.l.b (see Greatest lower bound) Gradient, 158, 161, 162, 172 in curvilinear coordinates, 161 in cylindrical coordinates, 162 Graph, of a function of one variable, 41 of a function of two variables, 144, 145 Grassman, Herman, 150 Greater than, 3 Greatest limit (see Limit superior) Greatest lower bound, 6 of a function, 41 of a sequence, 24, 32, 36 Green’s theorem in the plane, 232, 240–243, 260 in space, (see Divergence theorem) Grouping method, for exact differentials, 132 Half range Fourier sine or cosine series, 238, 239, 347–351 Hamilton, William Rowen, 150, 158 Harmonic functions, 393 series, 266 Heat conduction equation, 357 solution of, by Fourier integrals, 369 solution of, by Fourier series, 354, 355 Homogeneous functions, Euler’s theorem on, 122 Hyperbolic coordinates, 218 Hyperbolic functions, 44, 45 inverse, 41 Hyperboloid of one sheet, 127 Hypergeometric function or series, 276 Hypersphere, 116 Hypersurface, 116 Identity, with respect to addition and multiplication, 2 Image or mapping, 124, 416 Imaginary part, of a complex number, 6 of functions of a complex variable, 392, 399, 400 Imaginary unit, 6 Implicit functions, 69, 123 and Jacobians, 135–139 Improper integrals, 97, 110, 114, 306–335 absolute and conditional convergence of, 309, 312, 319–321 comparison test for, 308, 311 containing a parameter, 313 definition of, 306 of the first kind, 306–308, 317–321 of the second kind, 306, 310–312, 321, 322 of the third kind, 306, 313, 322 quotient test for, 304, 311, 315 uniform convergence of, 313, 314, 323, 324 Weierstrass M test for, 313, 324–329 Increasing functions, 41 monotonic, 41 strictly, 41, 47 Increasing sequences, monotonic and strictly, 24 Indefinite integrals, 94 (See also Integrals) Independence of the path, 212, 213, 243–245, 260 Independent variable, 39, 116 Indeterminate forms, 56, 82–84 L’Hospital’s rules for (see L’Hospital’s rules) Induction, mathematical, 8 428 INDEX
- 438. Inequalities, 3, 10 Inequality, 3 Bernoulli’s, 16 Cauchy’s, 422 Schwarz’s, 10, 18, 110 Inferior limit (see Limit inferior) Infinite: countably, 5 interval, 5 Infinite product, 277 for gamma function, 376 Infinite series, 25, 33, 37, 265–305 (See also Series) absolute convergence of, 268, 283, 300 comparison test for, 267, 279, 280 conditional convergence of, 268, 283 convergence tests for, 266–268 functions defined by, 276 Gauss’ test for, 268 integral test for, 267, 280–283 nth root test for, 268 of complex terms, 276 of functions, 269, 270, 276, 277 partial sums of, 25, 266 quotient test for, 267, 278 Raabe’s test for, 268, 285 ratio test for, 268, 284, 300 rearrangement of terms in, 269 special, 270 uniform convergence of, 269, 270 (See also Uniform convergence) Weierstrass M test for, 270, 289 Infinitesimal, 89 Infinity, 25, 46 Inflection, point of, 74 Initial point, of a vector, 150 Integers, positive and negative, 1 Integrable, 91 Integral equations, 367, 372, 373 Integral formulas of Cauchy, 394, 403–406 Integrals, 90–115, 207–228, 306–335, 363–374, 394, 398, 409–423 (See also Integration) definite, 90, 91 (See also Definite integrals) Dirichlet, 379, 385, 389 double, 207, 213–219 evaluation of, 314, 325–327, 398, 412–416 Fresnel, 387 Frullani’s, 333 improper, 97 (see Improper integrals) indefinite, 94 iterated, 208–210 line (see Line integrals) mean value theorems for, 72, 92 multiple (see Multiple integrals) of functions of a complex variable, 392–423 Integrals (Cont.): of infinite series of functions, 272, 275 of elementary functions, 96 Schwarz’s inequality for, 110 table of, 96 transformations of, 95, 103–108, 299 uniform convergence of, 313, 314, 323, 324 Integral test for infinite series, 267 Integrand, 91 Integrating factor, 223 Integration, applications of, 98, 109, 110, 114 (See also Integrals) by parts, 97–102 contour, 398 interchange of order of, 209 limits, of, 91 of Fourier series, 339, 353 of elementary functions, 96, 97, 107 range of, 91 special methods of, 97, 105–108 under integral sign, 186, 195 Intercepts, 126 Interior point, 117 Intermediate value theorem, 48 Intersection of sets, 12 Intervals: closed, 5 continuity in, 47 infinite, 5 nested, 25, 32 of convergence, 25 open, 5 unbounded, 5 Invariance relations, 181, 182 Invariant, scalar, 182 Fourier transforms, 369 (See also Fourier transforms) Laplace transforms, 315, 423, (See also Laplace transforms) Inverse functions, 41 continuity of, 47 hyperbolic, 45 trigonometric, 44 Inverse, of addition and multiplication, 2, 3 Irrantional algebraic functions, 43 Irrationality of ffiffiffi 2 p , proof of, 9 Irrational numbers, 2, 9, 10 approximations to, 9 definition of, 2 (See also Dedekind cut) Isolated singularity, 395 Iterated integrals, 208–210 limits, 119 Jacobian determinant (see Jacobians) Jacobians, 123, 135–139, 161, 162, 174, 175 Jacobians (Cont.): chain rules for, 124 curvilinear coordinates and, 161, 162 functions of a complex variable and, 422 implicit functions and, 135–139 multiple integrals and, 211 of transformations, 124 partial derivatives using, 123 theorems on, 124, 162 vector interpretation of, 160 Kronecker’s symbol, 342 Lagrange multipliers, 188, 198, 199 Lagrange’s form of the remainder, in Taylor series, 274, 297 Laplace’s equation, 129 (See also Laplacian operator) Laplace transforms, 314, 315, 333 convolution theorem for, 334 inverse, 330, 423 relation of functions of a complex variable to, 423 table of, 315 use of, in solving differential equations, 315, 330 Laplacian operator, 161, 162 (See also Laplace’s equation) in curvilinear coordinates, 161 in cylindrical coordinates, 161, 173 in spherical coordinates, 161 Laurent’s series, 395, 407, 408 theorem, 408, 409 Least limit (see Limit inferior) Least square approximations, 201 Least upper bound, 6, 32 of functions, 41 of sequences, 24, 36 Left-hand continuity, 47 derivatives, 67, 77, 78 limits, 45 Leibnitz’s formula for nth derivative of a product, 89 rule for differentiating under the integral sign, 186, 194 Leibniz, Gottfried Wilhelm, 65, 90, 265 Lemniscate, 114 Length, of a vector, 150 Less than, 2 Level curves and surfaces, 144, 186 L’Hospital’s rules, 72, 82–84, 88 proofs of, 82, 83 Limit inferior, 32, 36 Limit points, 5, 12, 117 Bolzano-Weirstrass theorem on (see Bolzano-Weirstrass) Limits of functions, 39–64, 117, 118, 393, 399, 400 definition of, 43, 118, 119 INDEX 429
- 439. Limits of functions (Cont.): iterated, 119, 208 of a complex variable, 393, 399, 400 proofs of theorems on, 54–56 right- and left-hand, 45 special, 46 theorems on, 45 Limits of integration, 91 Limits of sequences, 23, 24, 25, 27 definition of, 23 of functions, 45, 269 theorems of, 23, 24, 28–30 Limits of vector functions, 156 Limit superior, 32, 36 Linear dependence of vectors, 182 Linear transformations, 148 fractional (see Fractional linear transformation) Line integrals, 229–231, 238–240, 259 evaluation of, 231 independence of path of, 232, 238, 243–245 properties of, 231 relation of, to functions of a complex variable, 394 vector notation for, 230 Line, normal (see Normal line) tangent (see Tangent line) Logarithms, 4, 10, 11, 351 as multiple-valued functions, 392 base of, 4 Lower bound, 6, 12, 13 of functions, 40 of sequences, 24 Lower limit (see Limit inferior) L.u.b. (see Least upper bound) Maclaurin series, 274 Magnetic field vector, 181 Magnitude, of a vector, 150 Many-valued function (see Multiple-valued function) Mappings, 124 (See also Transformations) conformal, 417 Mathematical induction, 8, 15 Maxima and minima, 42, 73, 174, 185, 187, 196–198 absolute, 42 Lagrange’s multiplier method for, 188, 198, 199, 204 of functions of several variables, 187, 188, 196–198 relative, 42 Taylor’s theorem and, 276, 277, 297, 298 Maximum (see Maxima and minima) Maxwell’s equations, 181 Mean square error, 353 Mean value theorems: for derivatives, 72, 80–82, 87, 125, 141 for integrals, 93, 104, 112 Measure zero, 91, 103 Mechanics, 158 fluid, 402 Members, of a set, 1 Minimum (see Maxima and minima) Moebius strip, 248 Moment of inertia, 101 polar, 213, 219 Monotonic functions, 41 Monotonic sequences, 24, 30–32 fundamental theorem on, 24 Multiple integrals, 207–228 improper, 316 in curvilinear coordinates, 211, 212, 221, 222 in cylindrical coordinates, 211 in spherical coordinates, 212 Jacobians and, 211 transformations of, 211–213 Multiple-valued functions, 39, 117, 392 logarithm as a, 392 Multiplication, 2 associative law of, 2 involving vectors, 153–155 of complex numbers, 6, 7 Multiply-connected regions, 117 Natural base of logarithms, 3 Natural numbers, 4 Negative integers, 1 numbers, 1, 2 Neighborhoods, 6, 117 Nested intervals, 25, 32 Newton, Isaac, 65, 90, 265 first and second laws, 68 Newton’s methods, 74 Normal component of acceleration, 177 Normalized vectors and functions, 342 Normal line: parametric equations for, 184, 201 principal, 177, 180 to a surface, 184, 189–191 Normal plane, 184, 185, 191, 192 nth root test, 268 Null set, 1 vector, 151 Number, cardinal, 5 Numbers, 1–22 algebraic (see Algebraic number) Bernoulli, 304 complex (see Complex numbers) history, 2, 5 irrational (see Irrational numbers) natural, 1 Numbers (Cont.): negative, 1, 2 operations with, 2–15 positive, 1, 2 rational (see Rational numbers) real (see Real numbers) roots of, 3 transcendental, 6, 13 Numerator, 1 Numerical methods (see Approximations) for evaluating definite integrals, 98, 108–110 Odd functions, 338, 347–351 Open ball, 117 Open interval, 5 region, 117 Operations: with complex numbers, 6, 13, 14 with power series, 372, 373 with real numbers, 2, 8 Ordered pairs of real numbers, 7 triplets of real numbers, 155 Order, exponential, 334 of derivatives, 71 of poles, 395, 396 Orientable surface, 248 Origin, of a coordinate system, 116 Orthogonal curvilinear coordinates (see Curvilinear coordinates) Orthogonal families, 402, 403 functions, 153, 342, 357, 358 Orthonormal functions, 357 Pappus’ theorem, 228 Parabola, 50 Parabolic cylindrical coordinates, 180 Parallelepiped, volume of, 155, 169 Parallelogram, area of, 155, 168 law, 151, 163 Parametric equations, of line, 189 of normal line, 184 of space curve, 157 Parseval’s identity: for Fourier integrals, 366, 368 for Fourier series, 338, 351, 362, 373 Partial derivatives, 116–149 applications of, 183–206 definition of, 120 evaluation of, 120, 128–130 higher order, 120 notations for, 120 order of differentiation of, 120 using Jacobians, 123 Partial sums of infinite series, 25, 265, 266 Period, of a function, 336 Piecewise continous, 48 differentiable, 66 430 INDEX
- 440. p integrals, 308 Plane, complex, 7 Plane, equation of, 170 normal to a curve (see Normal plane) tangent to a surface (see Tangent place) Point: boundary, 117 branch, 396, 397 cluster, 5, 117 (See also Limit points) critical, 73 interior, 117 limit (see Limit points) neighborhood of, 5, 117 of accumulation, 5 (See also Limit points) singular (see Singular points) Point set: one-dimensional, 5 two-dimensional, 117 Polar coordinates, 7 Polar form, of complex numbers, 7, 14 Poles, 395 defined from a Laurent series, 395 of infinite order, 395 residues at, 395 Polynomial functions, 43 degree of, 43 Position vector, 157 Positive definite quadratic form, 206 Positive direction, 232 normal, 236 Positive integers, 1 numbers, 1, 2 Potential, velocity, 402 Power series, 272, 275, 276, 291–294 Abel’s theorem on, 272 expansion of functions in, 273 operations with, 273, 274 radius of convergence of, 272 special, 276, 277 theorems on, 272 uniform covergence of, 272 Prime, relatively, 9 Principal branch: of a function, 41 of a logarithm, 397 Principal normal, to a space curve, 177, 180 Principal part, 67, 120 of a Laurent series, 395 Principal value: of functions, 41, 44, 45 of integrals (see Cauchy principal value) of inverse hyperbolic functions, 44 of inverse trigonometric functions, 44 of logarithms, 392 Product, 1 box, 155 cross or vector (see Cross products) dot or scalar (see Dot products) infinite (see Infinite product) nth derivative of, 89 triple (see Triple products) Wallis’, 359 p series, 266 Quadratic equation, solutions of, 14 Quadratic form, 206 Quotient, 1 Quotient test: for integrals, 309, 311, 317 for series, 267, 279, 280 Raabe’s test, 268, 285 Radius of convergence, 272, 276 of curvature, 177, 181 of torsion, 181 Radius vector, 153 Range, of integration, 91 Rates of change, 74 Rational algebraic functions, 43 Rational numbers, 1, 9, 10 countability of, 11, 12 Ratio test, 268, 284, 285 proof of, 284 Real axis, 2 Real numbers, 1 (See also Numbers) absolute value of, 3 axiomatic foundations of, 3 decimal representation of, 2 geometric representation of, 2 inequalities for (see Inequality) non-countability of, 12 operations with, 2, 8, 9 ordered pairs and triplets of, 7, 155 roots of, 3, 11 Real part: of a complex number, 6 of functions of a complex variable, 392, 399, 400 Rectangular component vectors, 152 Rectangular coordinates, 7, 116, 160 Rectangular neighborhood, 117 rule for integration, 98 Recurring: decimal, 2 Region, 117 closed, 117 connected, 232 multiply-connected, 117 of convergence, 117 open, 117 simply-connected, 117, 232, 241 Regular summability, 278, 304 Relative extrema, 73 Relativity, theory of, 182 Removable discontinuity, 56, 119 singularity, 393, 407 Residues, 397, 409–412 Residue theorem, 397, 398, 409–412 evaluation of integrals by, 398, 403–406 proof of, 409, 410 Resultant of vectors, 151, 163 Reversion of series, 273 Riemann: axis, 396 surface, 397 Riemann integrable, 91 Riemann’s theorem, 354, 370 Right-hand continuity, 47 derivatives, 67, 77, 78 limits, 45 Right-handed rectangular coordinate system, 152, 153 Rolle’s theorem, 72 proof of, 80 Roots: of complex numbers, 7, 14 of real numbers, 3, 11 Roots of equations, 43 computations, 59 Newton’s method for finding, 89 Saddle points, 188 Scalar, 153 field, 153 invariant, 182 product (see Dot products) triple product, 155 Scale factors, 160 Scale of two (see Binary scale) Schwarz’s inequality: for integrals, 110 for real numbers, 10, 18 Section (see Dedekind cut) Separation of variables in boundary- value problems, 356 Sequence, Fibonacci, 35 Sequences, 23–38, 269 bounded, monotonic, 24, 30–32 convergent and divergent, 23, 269 decreasing, 25 definition of, 23 finite and infinite, 269 increasing, 25 limits of, 23, 27, 269 (See also Limits of sequences) of functions, 269 terms of, 26 uniform covergence of, 269 Series (see Infinite series) alternating (see Alternating series) asymptotic (see Asymptotic series) binomial, 275 double, 277 INDEX 431
- 441. Series (Cont.): geometric, 25, 266 harmonic, 266 Laurent’s, 395, 407, 408, 420 Maclaurin, 274 of functions of a complex variable, 406–409 p-, 266 partial sums of, 25, 266 power (see Power series) reversion of, 273 sum of, 25, 266 Taylor (see Taylor series) telescoping, 278 terms of, 266 test for integrals, 280 Sets, 1 bounded, 6 closed, 6, 12, 13 connected, 117 countable or denumerable (see Countable set) elements of, 1 everywhere dense, 2 intersection of, 12 orthonormal, 337, 342 point, 117 union of, 12 Simple closed curves, 117, 232, 241 Simple poles, 395 Simply connected region, 117, 232, 241 Simpson’s rule, 98, 108, 109 Single-valued function, 39, 116, 392 Singular points or singularities, 395–398, 406–409 defined from Laurent series, 395 essential, 395, 407 isolated, 395 removable, 395, 407 Sink, 259 Slope, 66 Smooth function (see Piecewise differentiability) Solenoidal vector fields, 259 Source, 259 Space curve, 157 Specific heat, 356, 357 Spherical coordinates, 162, 174, 175 arc length element in, 162, 174 Laplacian in, 162, 176 multiple integrals in, 222 volume element in, 162, 175 Staircase or step function, 51 Stirling’s asymptotic formula and series, 378, 384 Stokes’ theorem, 237, 252–257 proof of, 252, 253 Stream function, 402 Subset, 1 Subtraction, 2 of complex numbers, 13, 14 of vectors, 151 Sum, 2 of series, 25, 266 of vectors, 151, 163 partial, 25, 266 Summability, 278, 296, 304 Abel, 305 Césaro, 278, 296 regular, 278, 304 Superior limit (see Limit superior) Superposition, principal of, 357 Surface, 116 equipotential, 186 level, 144, 186 normal line to (see Normal line) orientable, 248 tangent place to (see Tangent plane) Surface integrals, 233–236, 245–249, 261 Tangential component of acceleration, 180, 181 Tangent line, to a coordinate curve, 84 to a curve, 65, 184, 202 Tangent plane, 183, 189–191, 200 in curvilinear coordinates, 201, 202 Tangent vector, 157, 177 Taylor polynomials, 273 Taylor series, in one variable, 274 (See also Taylor’s theorem) in several variables, 276 of functions of a complex variable, 395 Taylor’s theorem, 273, 297 (See also Taylor series) for functions of one variable, 273 for functions of several variables, 276, 277 proof of, 297, 407, 408 remainder in, 274 Telescoping series, 278 Tensor analysis, 182 Term, of a sequence, 23 of a series, 266 Terminal point of a vector, 150 Thermal conductivity, 356, 357 Thermodynamics, 148 Torsion, radius of, 181 Total differential, 122 (See also Differentials) Trace, on a place, 127 Transcendental functions, 45, 46 numbers, 6, 13 Transformations, 124, 139, 140 and curvilinear coordinates, 139, 140, 160 conformal, 417 Jacobians of, 125, 160 of integrals, 95, 105–108, 211–213, 216–219 Transforms (see Fourier transforms and Laplace transforms) Transitivity, law of, 2 Trigonometric functions, 46, 96 derivatives of, 71 integrals of, 95, 96 inverse, 44 Triple integrals, 210, 219–221 transformation of, 221–225 Triple products, scalar, 155 vector, 155 Unbounded interval, 5 Uniform continuity, 48, 58, 63, 119 Uniform convergence, 269, 270, 287, 288 of integrals, 313, 314 of power series, 272 of sequences, 269 of series, 269, 270 tests for integrals, 313, 314 tests for series, 270 theorems for integrals, 314 theorems for series, 270, 271, 272 Weirstrass M test for (see Weirstrass M test) Union of sets, 12 Unit tangent vector, 157 Unit vectors, 152, 342 infinite dimensional, 342 rectangular, 152 Upper bound, 6 of functions, 40, 41 of sequences, 24 Upper limit (see Limit superior) Variable, 5, 39 change of, in differentiation, 69, 70 change of, in integration, 95, 105–108, 211 complex, 392, 393 (See also Functions of a complex variable) dependent and independent, 40, 116 dummy, 94 limits of integration, 94, 186, 194, 313 Vector algebra, 151, 152, 161–165 Vector analysis (see Vectors) Vector: bound, 150 free, 150 Vector field, 156 solenoidal, 259 Vector functions, 156 limits, continuity and derivatives of, 156, 171, 172 Vector product (see Cross products) Vectors, 20, 150–182 algebra of, 151, 152, 178 axiomatic foundations for, 155 432 INDEX
- 442. Vectors (Cont.): complex numbers as, 20 components of, 153 curvilinear coordinates and, 161, 162 equality of, 150 infinite dimensional, 342 Jacobians interpreted in terms of, 160 length or magnitude of, 150 normalized, 342 null, 151 position, 153 radius, 153 resultant or sum of, 151, 163 scalar product, 153 tangent, 157, 177, 179 Vectors (Cont.): unit, 152, 153, 342 Vector triple product, 155, 169–171 Velocity, 74, 177 of light, 181 potential, 402 Vibrating string, equation of, 361 Volume, 100 element of, 161, 162, 175 of parallelepiped, 161, 169 Volume of revolution: disk method, 100 shell method, 101 Wallis’ product, 359 Wave equation, 361 Weierstrass M test: for integrals, 313, 324–329 for series, 270, 289 Wilson, E.B., 150 Work, as a line integral, 239 x-axis, 116 z-axis, 116 intercept, 126 Zeno of Elea, 265 Zero, 1 division by, 8 measure, 91, 103 vector, 151 INDEX 433