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Chapter 4: Sequences and Mathematical Induction

Chapter 4: Sequences and Mathematical Induction. 4.1- Sequences . 4.2 , 4.3 Mathematical Induction 4.4 Strong Mathematical induction and WOP. Instructor: Hayk Melikya melikyan@nccu.edu.

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Chapter 4: Sequences and Mathematical Induction

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  1. Chapter 4: Sequences and Mathematical Induction 4.1- Sequences . 4.2 , 4.3 Mathematical Induction 4.4 Strong Mathematical induction and WOP Instructor: Hayk Melikya melikyan@nccu.edu

  2. A sequence (informally) is a collection of elements (objects or numbers usually infinite number of) indexed by integers. Sequences Examples Each individual element ak is called a term, where k is called an index a1, a2, a3, …, an, … General formula 1, 2, 3, 4, … 1/2, 2/3, 3/4, 4/5,… 1, -1, 1, -1, … 1, -1/4, 1/9, -1/16, … Sequences can be computed using an explicit formula: ak = k / (k + 1) for k > 0 Finding an explicit formula given initial terms of the sequence

  3. Summation

  4. A Telescoping Sum

  5. Product

  6. Factorial Factorial defines a product: How to estimate n!? Turn product into a sum taking logs: ln(n!) = ln(1·2·3 ··· (n – 1)·n) = ln 1 + ln 2 + ··· + ln(n – 1) + ln(n)

  7. Arithmetic Series Given n numbers, a1, a2, …, an with common difference d, i.e. ai+1 - ai =d. What is a simple closed form expression of the sum? Adding the equations together gives: Rearranging and remembering that an = a1 + (n − 1)d, we get:

  8. Geometric Series What is the closed form expression of Gn? xn+1 GnxGn= 1

  9. Infinite Geometric Series Consider infinitesum (series) for |x|<1

  10. Some Examples

  11. Let P(n) be a property that is defined for integers n, and let a be a fixed integer Suppose the following two statements are true: 1. P(a) is true. 2. for all integers k ≥a, if P(k) is true then P(k+1) is true. Then the statement P(n) is true for all integers n ≥a Principle of Mathematics induction • Domino effect • Inductive sets of integer. • A subset of positive integers S is called inductive set • if k  S then k +1  S. • Principle of mathematical induction states that if a  S then all integers greater than or equal to a also are in S. (chapter 4.2-4.4 of the book

  12. Method of Proof by Mathematical Induction Consider a statement of the form ”For all integers n ≥ a, a property P(n) is true.” To prove such a statement, perform the following two steps: Step1 (Basic step): Show tah the property is true for n =a ( P(a) is true.) Step2 ( Inductive step): show that for all integers k ≥ a, If property is true for n = k then it is true for n = k+1 (  k ≥ a) (P(k)P(k + 1)) Let P(n) be the property “ n cent can be obtained using 3 cent and 5 cent coins”. Proposition(4.2.1): P(n) is true for all integers n ≥ 8.

  13. The Induction Rule when a =1 1 and (from n ton +1), proves 1, 2, 3,…. P(1), P(n)P(n +1) (m  N)P(m) Much easier to prove with P (n) as an assumption. Very easy to prove For any n>=1 Like domino effect…

  14. Proof by Induction Let’s prove: Statements in green form a template for inductive proofs. Proof: (by induction on n) The induction hypothesis, P(n), is:

  15. Proof by Induction Base Case (n = 0): Wait: divide by zero bug! This is only true for r 1 Theorem:

  16. Proof by Induction Induction Step: Assume P(n) for some n  0 and prove P(n + 1): Have P (n) by assumption: So let r be any number  1, then from P (n) we have How do we proceed?

  17. Proof by Induction adding r n+1 to both sides, But since r  1 was arbitrary, we conclude (by UG), that which is P (n+1). This completes the induction proof.

  18. Summation Try to prove:

  19. Proving a Property Base Case (n = 1): Induction Step: Assume P(i) for some i  1 and prove P(i + 1):

  20. Proving an Inequality Base Case (n = 3): Induction Step: Assume P(i) for some i  3 and prove P(i + 1):

  21. … … Paradox Proposition: All horses are the same color. Proof: (by induction on n) Induction hypothesis: P(n) ::= any set of n horses have the same color Base case (n =1): true! Inductive case: Assume any n horses have the same color. Prove that any n+1 horses have the same color. Second set of n horses have the same color First set of n horses have the same color

  22. Paradox What is wrong? n =1 Proof that P(n) → P(n+1) is false ifn = 1, because the two horse groups do not overlap. Second set of n=1 horses First set of n=1 horses (But proof works for all n ≠1)

  23. Strong Induction Prove P(1). Then prove P(n+1) assuming all of P(1), …, P(n)(instead of just P(n)). Conclude (n.)P(n) Strong induction equivalent 1  2, 2  3, …, n-1n. So by the time we got to n+1, already know all of P(1), …, P(n) Ordinary induction

  24. Principle of Strong Mathematical Induction Let P(n) be a property that is defined for integers n, and let a and b be a fixed integers with a ≤ b Suppose the following two statements are true: 1. P(a), P(a+1), . . . And P(b) are true. 2. for all integers k ≥b, if P(i) is true for all integers i from a through k then P(k+1) is true. Then the statement P(n) is true for all integers n ≥a

  25. Prime Products (revisited) Every integer > 1 is a product of primes. Proof:(by strong induction) Base case is easy. Suppose the claim is true for all 2 <= i < n. Consider an integer n. In particular, n is not prime. • So n = k·m for integers k, m where n > k,m >1. • Since k,m smaller than n, • By the induction hypothesis, both k and m are product of primes • k = p1 p2ps • m = q1 q2   qt

  26. Prime Products Every integer > 1 is a product of primes. …So n = k m = p1 p2psq1q2 qt is a prime product.  This completes the proof of the induction step.

  27. Postage by Strong Induction Available stamps: 5¢ 3¢ What amount can you form? Theorem: Can form any amount  8¢ Prove by strong induction on n > 0. P(n) ::= can form (n+7)¢.

  28. Postage by Strong Induction Base case (n= 1): (1 +7)¢: Inductive Step: assume (m +7)¢ for 1  mn, then prove ((n +1) + 7)¢ cases: n +1= 1, 9¢: n +1= 2, 10¢:

  29. + 3 Postage by Strong Induction casen +1  3: let m =n 2. now n  m  0, so by induction hypothesis have: = (n +1)+8 (n2)+8 We’re done! In fact, use at most two 5-cent stamps!

  30. Postage by Strong Induction Given an unlimited supply of 5 cent and 7 cent stamps, what postages are possible? Theorem: For all n >= 24, it is possible to produce n cents of postage from 5¢ and 7¢ stamps.

  31. Well Ordering Principle Every nonempty set ofnonnegative integers has a least element. Familiar? Now you mention it, Yes. Obvious?Yes. Trivial? Yes. But watch out: Every nonempty set of nonnegative rationals has a least element. NO! Every nonempty set of nonnegative integers has a least element. NO!

  32. Well Ordering Principle Theorem: is irrational Proof: suppose …can always find such m, n without common factors… why always? By WOP, minimum |m| s.t. where |m0| is minimum. so

  33. Well Ordering Principle but if m0, n0had common factor c > 1, then and contradicting minimality of |m0| • The well ordering principle is usually used in “proof by contradiction”. • Assume the statement is not true, so there is a counterexample. • Choose the “smallest” counterexample, and find a even smaller counterexample. • Conclude that a counterexample does not exist.

  34. Well Ordering Principle in Proofs To prove ``n. P(n)’’ using WOP: • Define the set of counterexamples • C ::= {n | ~ P(n)} • 2. Assume C is not empty. • 3. By WOP, have minimum element m0C. • 4. Reach a contradiction(somehow) – • usually by finding a member of C that is < m0 . • 5. Conclude no counterexamples exist. QED

  35. Induction (Proving equation) For any integer n>=2, Proof: We prove by induction on n . Let P(n) be the proposition that

  36. Base case, n=2: So P(2) is true. Inductive step: Suppose that P(n) is true for some n>=2. So,

  37. Then, for n>=2, By induction, P(n) is true for all integers n>=2. By the inductive hypothesis

  38. Induction (Divisibility) For any integer n>=1, is divisible by 6 Proof: We prove by induction on n . Base case, n=1: is divisible by 6. So it is true for n=1.

  39. Inductive step: Suppose that for some n>=1, is divisible by 6 Then, Either n+1 or n+2 is even, so the last term is divisible by 6. Therefore is divisible by 6. By induction, is divisible by 6 for all integers n>=1 By the inductive assumption

  40. Induction (Proving inequality) For any integer n>=4, Proof: We prove by induction on n . Base case, n=4: So the claim is true for n=4.

  41. Inductive step: Assume that for some n>=4 Then, By induction, for all integer n>=4. By the inductive hypothesis By assumption, n>=2

  42. Chapter 4 (Sections 4.1, 4.2, 4.3, 4.4) 1. (4.1) Expand a sequence/sum/product from sequence/sum/product notation, and vice versa 2. (4.1) Rewrite a sum by separating off and adding on the last term 3. (4.1) Know the definition of a factorial. 4. (4.2) Know the method of proof by induction 5. (4.2) Prove formulas for the sums of sequences using induction 6. (4.3) Use induction to prove inequalities 7. (4.3) Use induction to prove results with divisibility 8. (4.4) Use strong induction on recursively defined sequences. Practice Problems: (Section 4.1) 1, 2, 10, 11, 19-22, 34, 35, 43, 46 (Section 4.2) 4, 6, 10, (11-15) (Section 4.3) 8-10, 16-20 (Section 4.4) 1-8, 14, 17

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