Task 4

task 4 :
sequences, mathematical induction and
recursion
sequences, mathematical induction and
recursion
TAF 3023 ; Discrete Math

sequences

A)Definition of sequences.
a)-Sequence is a function from a subset of the natural
numbers(usually of the form {0, 1, 2, . . . } to a set S.

-For an example:-

A={0,1,2,3,4…….x} and B={1,2,3,4……..x}
are called initial segments of N.

is a sequence of the form
c) arithmetic progression
a, a + d, a 2d, . . . , a + nd, . . .

where the initial term a and the common difference d are real
numbers.

-An arithmetic progression is a discrete analogue of the linear
function

f (x) = dx + a.

-Example:-

The sequence of sn=-1+4n.We start at n=0. Initial terms and
common differences equal to −1 and 4. The list of terms s0, s1,
s2, s3, . . . begins with
−1, 3, 7, 11, . . . ,

d) recurrence relation
for the sequence {an} is an equation that expresses an in terms of one or
more of the previous terms of the sequence, namely, a0, a1, . . . , an−1, for
all integers n with n ≥ n0, where n0 is a nonnegative integer. A sequence is
called a solution of a recurrencerelation if its terms satisfy the recurrence
relation.

Let {an} be a sequence that satisfies the recurrence relation an = an−1 +
3 for n = 1, 2, 3, . . . , and suppose that a0 = 2. What are a1, a2, and a3?

- Example:-
Let {an} be a sequence that satisfies the recurrence relation an = an−1+ 3
for n = 1, 2, 3, . . . ,
and suppose that a0 = 2. What are a1, a2, and a3?
We see from the recurrence relation that a1 = a0 + 3 = 2 + 3 = 5. It then
follows that
a2 = 5 + 3 = 8 and a3 = 8 + 3 = 11.

= 2, 3, 4, . . . .

-Example:- e) The Fibonacci sequence
Find the Fibonacci numbers f2, f3, f4, f5, and f6.

The recurrence relation for the Fibonacci sequence tells us that
we find successive erms by adding the previous two terms.
Because the initial conditions tell us that f0 = 0 and f1 = 1, using
the recurrence relation in the definition we find that

Types of sequences
B)Types of sequences :

-Geometric progression is a discrete analogue of the
exponential function f(x)=arn.

-Example:-

The sequence bn=(-1)n. We start with n=0. Initial term
and common ratio equal to 1 and −1.The list of the
term begins with b0,b1,b2,b3…

1,−1, 1,−1, 1, . . . ;

2) An arithmetic progression

-An arithmetic progression is a discrete analogue of the linear
function

f (x) = dx + a.

-Example:-
The sequence of sn=-1+4n.We start at n=0. Initial terms and
common differences equal to −1 and 4. The list of terms s0, s1, s2,
s3, . . . begins with

−1, 3, 7, 11, . . . ,

3)The Fibonacci sequence

f0, f1, f2, . . . , is defined by the initial conditions

f0 = 0, f1 = 1,and the recurrence relation fn = fn−1 + fn−2 ,for n = 2, 3, 4,
....

-Example:-
Find the Fibonacci numbers f2, f3, f4, f5, and f6.

The recurrence relation for the Fibonacci sequence tells us that we find
successive erms by adding the previous two terms. Because the initial
conditions tell us that f0 = 0 and f1 = 1, using the recurrence relation in
the definition we find that

b)-Geometric progression

a,ar,ar2,ar3………,arn,….

-Geometric progression is a discrete analogue of the
exponential function f(x)=arn.

-Example:-

The sequence bn=(-1)n. We start with n=0. Initial term
and common ratio equal to 1 and −1.The list of the
term begins with b0,b1,b2,b3…

1,−1, 1,−1, 1, . . . ;

-A recurrence relation for the sequence {an} is an equation that expresses
an in terms of one or more of the previous terms of the sequence, namely,
a0, a1, . . . , an−1, for all integers n with n ≥ n0, where n0 is a nonnegative
integer. A sequence is called a solution of a recurrencerelation if its terms
satisfy the recurrence relation.

- Let {an} be a sequence that satisfies the recurrence relation an = an−1 +
3 for n = 1, 2, 3, . . . , and suppose that a0 = 2. What are a1, a2, and a3?

- Example:-
Let {an} be a sequence that satisfies the recurrence relation an = an−1 + 3
for n = 1, 2, 3, . . . ,

and suppose that a0 = 2. What are a1, a2, and a3?

We see from the recurrence relation that a1 = a0 + 3 = 2 + 3 = 5. It then
follows that

a2 = 5 + 3 = 8 and a3 = 8 + 3 = 11.

Scanner k=new scanner(system.in);
C)Example sequences used in computer
int number;
programming
number=k.nextInt();

while(number<=100){

if(number%2==0){

System.out.println(“even”);}

else{

System.out.println(“odd”);}

The statement above shows that when the modulus 2 of an input number is 0.The
value has no remainder and is divisible by 2.In nutshell, it is an even number. On
the other hand, when the modulus 2 of the number is 1.The number input is odd.
Sequence is applied in a way that each number shows a pattern closely to
sequence and it would be tiresome to calculate every single number to know if it’s
odd or even. Sequence makes it easier to know the even and odd properties of a
number with a much more larger value by tracing out the pattern.

EXAMPLE PROBLEM INVOLVING SEQUENCE

A sequence is a set of numbers such as 1, 2, 3, … or 96, 48, 24, … that follow some
sort of formula.

Example 1 :
Find five consecutive integers that sum up to 405.

Solution :
Instead of implementing a guess and check method, suppose the five integers were
x, x+1, x+2, x+3, and x+4 = 405. Their sum is 405, so we have
x + (x+1) + (x+2) + (x+3) + (x+4) = 405
5x + 10 = 405
x = 79

The five integers are 79, 80, 81, 82, and 83.

Example 2 :
Find the sum of all the integers starting from 1 to 1000.

Solution :
The sequence of integers starting from 1 to 1000 is given by

1, 2, 3, 4, …, 1000

The first term is 1 and the last term is 1000. The common
difference is equal to 1. The formula that gives the sum of the first
and terms of an arithmetic sequence knowing the first and last term
of the sequence and the number of terms.

S1000 = 1000 (1 + 1000) / 2 = 500500

(over) the natural numbers N = {0; 1; 2; …}. If
First Principle of Mathematical
Induction.
(1) P(0), and

(2) P(n) → P(n + 1)

then 8nP(n).

Terminology: The hypothesis P(0) is called the basis
step and the hypothesis,

P(n) → P(n + 1), is called the induction (or inductive)
step.

The Principle of Mathematical Induction is an axiom of the system of
natural

numbers that may be used to prove a quantified statement of the
form 8nP(n), where

the universe of discourse is the set of natural numbers. The principle
of induction has

a number of equivalent forms and is based on the last of the four
Peano Axioms. The axiom of induction states that if S is a set of
natural numbers such that (i) 0 ϵ S and (ii) if n ϵ S, then n + 1 ϵ S, then
S = N. This is a fairly complicated statement: Not only is it an “if ...,
then ..." statement, but its hypotheses also contains an “if ..., then ..."
statement (if n ϵ S, then n + 1 ϵ S). When we apply the axiom to the
truth set of a predicate P(n), we arrive at the first principle of
mathematical induction stated above.

More generally,
we may apply the principle of induction whenever the
universe of discourse is a set of
integers of the form {k; k + 1; k + 2; …} where k is some
fixed integer. In this case
it would be stated as follows:

Let P(n) be a predicate over {k; k + 1; k + 2; k + 3; …},
where k ϵ Z. If

(1) P(k), and

(2) P(n) → P(n + 1)
then 8nP(n).

In this context the “for all n", of course, means for all n ≥ k.

Why does the principle of induction work? This is essentially the domino
effect.
Assume you have shown the premises. In other words you know P(0) is
true and you
know that P(n) implies P(n + 1) for any integer n ≥ 0.
Since you know P(0) from the basis step and P(0) → P(1) from the
inductive
step, we have P(1) (by modus ponens).
Since you now know P(1) and P(1) → P(2) from the inductive step, you
have

P(2).

Since you now know P(2) and P(2) → P(3) from the inductive step, you
have

P(3).

And so on ad infinitum (or ad nauseum).

The Second Principle of
Mathematical Induction.
Let k be an integer, and let P(n) be a predicate whose
universe of discourse is the set of integers {k; k + 1; k
+ 2; …}. Suppose

1. P(k), and

2. P(j) for k ≤ j ≤ n implies P(n + 1).

Then 8nP(n).

The second principle of induction differs from the first
only in the form of the induction hypothesis. Here we
assume not just P(n), but P(j) for all the integers j
between k and n (inclusive). We use this assumption to
show P(n+1). This method
of induction is also called strong mathematical induction.
It is used in computer
science in a variety of settings such as proving recursive
formulas and estimating the
number of operations involved in so-called “divide-and-
conquer" procedures.

i) examples of mathematical induction
1. Version:1.0 StartHTML:0000000167 EndHTML:0000001580
StartFragment:0000000457 EndFragment:0000001564

Scanner k=new scanner(system.in);

int number;

number=k.nextInt();

while(number<=100){

if(number%2==0){

System.out.println(“even”);}

else{

System.out.println(“odd”);}

The statement above shows that when the modulus
2 of an input number is 0.The value has no
remainder and is divisible by 2.In nutshell, it is an
even number. On the other hand, when the
modulus 2 of the number is 1.The number input is
odd. Sequence is applied in a way that each number
shows a pattern closely to sequence and it would
be tiresome to calculate every single number to
know if it’s odd or even. Sequence makes it easier
to know the even and odd properties of a number
with a much more larger value by tracing out the
pattern.

ii. examples of mathematical
induction
2. Express gcd(252, 198) = 18 as a linear combination of 252 and 198. Solution: To show that
gcd(252, 198) = 18, the Euclidean algorithm uses these divisions:

252 = 1 · 198 + 54 198 = 3 · 54 + 36 54 = 1 · 36 + 18

36 = 2 · 18. Using the next-to-last division (the third division), we can express gcd(252, 198)
= 18 as a

linear combination of 54 and 36. We find that 18 = 54 − 1 · 36. The second division tells us
that 36 = 198 − 3 · 54.

Substituting this expression for 36 into the previous equation, we can express 18 as a linear
combination of 54 and 198. We have

18 = 54 − 1 · 36 = 54 − 1 · (198 − 3 · 54) = 4 · 54 − 1 · 198. The first division tells us that
54 = 252 − 1 · 198.

Substituting this expression for 54 into the previous
equation, we can express 18 as a linear combination of
252 and 198. We conclude that

18 = 4 · (252 − 1 · 198) − 1 · 198 = 4 · 252 − 5 · 198,
completing the solution.

We will use Theorem 6 to develop several useful results.
One of our goals will be to prove the part of the
fundamental theorem of arithmetic asserting that a
positive integer has at most one prime factorization. We
will show that if a positive integer has a factorization into
primes, where the primes are written in nondecreasing
order, then this factorization is unique.

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