task 4 : sequences, mathematical induction and

recursionsequences, mathematical induction and

recursionTAF 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.

c) arithmetic progressionis a sequence of the form    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 relationfor 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.

e) The Fibonacci sequence

0, 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             

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 progressionis a sequence of the form 

-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 progressionis a sequence of the form

             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.

C)Example sequences used in computer programming

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.

  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 havex + (x+1) + (x+2) + (x+3) + (x+4) = 4055x + 10 = 405x = 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

First Principle of Mathematical Induction.

Let P(n) be a predicate with domain of discourse (over) the natural numbers N = {0; 1; 2; …}. If 

          (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 S and (ii) if n S, then n + 1 S, then ϵ ϵ ϵS = 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 S, then n + 1 S). When we apply the axiom to the ϵ ϵtruth 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 ofintegers of the form {k; k + 1; k + 2; …} where k is some fixed integer. In this caseit would be stated as follows:

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

(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 youknow 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 →inductivestep, 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 jbetween k and n (inclusive). We use this assumption to show P(n+1). This methodof induction is also called strong mathematical induction. It is used in computerscience in a variety of settings such as proving recursive formulas and estimating thenumber of operations involved in so-called “divide-and-conquer" procedures.

i) examples of mathematical induction1.  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.