LarsonETF6_ch09_sec10 - Infinite Series.ppt

Infinite Series
Copyright © Cengage Learning. All rights reserved.

Taylor and Maclaurin Series
Copyright © Cengage Learning. All rights reserved.

3
 Find a Taylor or Maclaurin series for a function.
 Find a binomial series.
 Use a basic list of Taylor series to find other Taylor
series.
Objectives

4
Taylor Series and Maclaurin Series

5
The coefficients of the power series in Theorem 9.22 are
precisely the coefficients of the Taylor polynomials for f(x)
at c. For this reason, the series is called the Taylor series
for f(x) at c.
The next theorem gives the form that every convergent
power series must take.

6

7
Use the function f(x) = sin x to form the Maclaurin series
and determine the interval of convergence.
Solution:
Successive differentiation of f(x) yields
f(x) = sin x f(0) = sin 0 = 0
f'(x) = cos x f'(0) = cos 0 = 1
f''(x) = –sin x f''(0) = –sin 0 = 0
f(3)(x) = –cos x f(3)(0) = –cos 0 = –1
Example 1 – Forming a Power Series

8
f(4)(x) = sin x f(4)(0) = sin 0 = 0
f(5)(x) = cos x f(5)(0) = cos 0 = 1
and so on.
The pattern repeats after the third derivative.
Example 1 – Solution cont’d

9
So, the power series is as follows.
By the Ratio Test, you can conclude that this series
converges for all x.

10
You cannot conclude that the power series converges to
sin x for all x.
You can simply conclude that the power series converges
to some function, but you are not sure what function it is.
This is a subtle, but important, point in dealing with Taylor
or Maclaurin series.
To persuade yourself that the series
might converge to a function other than f, remember that
the derivatives are being evaluated at a single point.

11
It can easily happen that another function will agree with
the values of f (n)(x) when x = c and disagree at other
x-values.
If you formed the power series for
the function shown in Figure 9.23,
you would obtain the same series
as in Example 1.
You know that the series converges
for all x, and yet it obviously cannot
converge to both f(x) and sin x
for all x . Figure 9.23

12
Let f have derivatives of all orders in an open interval I
centered at c.
The Taylor series for f may fail to converge for some x in I.
Or, even if it is convergent, it may fail to have f(x) as its
sum.
Nevertheless, Theorem 9.19 tells us that for each n,
where

13
Note that in this remainder formula, the particular value of z
that makes the remainder formula true depends on the values
of x and n. If then the next theorem tells us that the
Taylor series for f actually converges to f (x) for all x in I.

14
Show that the Maclaurin series for f(x) = sin x converges to
sin x for all x.
Solution:
You need to show that
is true for all x.
Example 2 – A Convergent Maclaurin Series

15
Because
or
you know that for every real number z.
Therefore, for any fixed x, you can apply Taylor’s Theorem
(Theorem 9.19) to conclude that

16
From the discussion regarding the relative rates of
convergence of exponential and factorial sequences, it
follows that for a fixed x
Finally, by the Squeeze Theorem, it follows that for all x,
Rn(x)→0 as n→ .
So, by Theorem 9.23, the Maclaurin series for sin x
converges to sin x for all x.
cont’d
Example 2 – Solution

17
Figure 9.24 visually illustrates the convergence of the
Maclaurin series for sin x by comparing the graphs of the
Maclaurin polynomials P1(x), P3(x), P5(x), and P7(x) with the
graph of the sine function. Notice that as the degree of the
polynomial increases, its graph more closely resembles
that of the sine function.
Figure 9.24

18

20
Before presenting the basic list for elementary functions,
you will develop one more series—for a function of the form
f(x) = (1 + x)k. This produces the binomial series.
Binomial Series

21
Find the Maclaurin series for f(x) = (1 + x)k and determine its
radius of convergence.
Assume that k is not a positive integer and k ≠ 0.
Solution:
By successive differentiation, you have
f(x) = (1 + x)k f(0) = 1
f'(x) = k(1 + x)k – 1 f'(0) = k
f''(x) = k(k – 1)(1 + x)k – 2 f''(0) = k(k – 1)
f'''(x) = k(k – 1)(k – 2)(1 + x)k – 3 f'''(0) = k(k – 1)(k – 2)
. .
. .
. .
f(n)(x) = k…(k – n + 1)(1 + x)k – n f(n)(0) = k(k – 1)…(k – n + 1)
Example 4 – Binomial Series

22
which produces the series
Because an + 1/an→1, you can apply the Ratio Test to
conclude that the radius of convergence is R = 1.
So, the series converges to some function in the interval
(–1, 1).
cont’d
Example 4 – Binomial Series

23
Deriving Taylor Series from a Basic
List

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Deriving Taylor Series from a Basic List

25
Find the power series for
Solution:
Using the power series
you can replace x by to obtain the series
This series converges for all x in the domain of —that
is, for x ≥ 0.
Example 6 – Deriving a Power Series from a Basic List

LarsonETF6_ch09_sec10 - Infinite Series.ppt

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