SUBTOPIC 5               :       MATHEMATICAL INDUCTION        Mathematical induction can be used to prove statements that...
The base case can start with any nonnegative number. If that number is          then you would provethat assertion P(n) ho...
Consider the left hand side(n+1)! – 1 + (n+1) x (n+1) =(n+1)! (1+n+1) -1 = (n+2)! -1Suppose we have a sequence of proposit...
P (k)                                                trueSuppose that the dominos satisfy two constraints.   1. Well posit...
P (1)        P (2)                   P (k)    P                                              …..           (k+1)P (0) true...
P (0) true       P (1) true      P (2) true      true      ….. P (k) true   P (k+1) truePrinciple of Mathematical Inductio...
Prove that:                 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ nBase case (n=1):                                 ...
EXERCISE:  1. For every n ≥ 4, n! >  2. Proof that,     We have seen that for any natural number n ,                      ...
ANSWER:  1. Proof:     In this problem          .     Basis Step: If n = 4, then LHS = 4! = 24, and RHS =               . ...
Now to determine the value of a, b, c, d and e, we compare the value of the both sides ofthe equation (1) for five values ...
Upcoming SlideShare
Loading in...5
×

Chapter 5

148

Published on

Published in: Education, Technology, Business
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
148
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
2
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Transcript of "Chapter 5"

  1. 1. SUBTOPIC 5 : MATHEMATICAL INDUCTION Mathematical induction can be used to prove statements that assert that ∀ n ∈ , P (n) istrue. But, it is important to note that mathematical induction can be used only to prove resultsobtained in some other way. It is not a tool for discovering formula or theorem. Principle of Induction. To prove that P (n) is true for all positive integers n, where P (n)is a propositional function. A proof by mathematical induction has two parts: Basic step: We verify that P (1) is true. Inductive step: We show that the conditional proposition ∀ k ∈ , P (k) ⇒ P (k + 1) is true. In the induction step, the assumption that P (n) holds is called the Induction Hypothesis(IH). In more formal notation, this proof technique can be stated as [P (0) ˄∀ k (P (k) ⇒ P (k + 1) ∀n P (n) You can think of the proof by (mathematical) induction as a kind of recursive proof:Instead of attacking the problem directly, we only explain how to get a proof for P(n + 1) out of aproof for P(n).How would you prove that the proof by induction indeed works??Proof (by contradiction) Assume that for some values of n, P(n) is false. Let be the least suchn that P_n0_ is false. Cannot be 0, because P(0) is true. Thus, must be in the form =1+ . Since < then by P ( is true. Therefore, by inductive hypothesis P ( + 1) must betrue. It follows then that P ( is true. 33
  2. 2. The base case can start with any nonnegative number. If that number is then you would provethat assertion P(n) holds for all n ≥ . [P ( ) ˄∀ k ≥ (P (k) ⇒ P (k + 1) ∀n P (n)The induction step not necessarily should start with n. You can change the step from n -1 to n,where n > 0. Sometimes this yields slightly shorter expressions. However, you cannot make astep from n - 1 to n + 1.Example:Prove for ≥ 1 1 x 1! + 2 x 2! + 3 x 3! + ... + n x n! = (n + 1)! - 1This could be also written by using ∑ notationIf you take 15-355 (http://www.andrew.cmu.edu/course/15-355/) you will learn a more generalapproach for deriving and proving combinatorial identities.ProofBase case: n _ 1The left hand side is 1x1! The right hand side is 2! - 1. They are equal.Inductive hypothesis: Suppose this holdsWe need to prove 34
  3. 3. Consider the left hand side(n+1)! – 1 + (n+1) x (n+1) =(n+1)! (1+n+1) -1 = (n+2)! -1Suppose we have a sequence of propositions which we would like to prove:P (0), P (1), P (2), P (3)… P (n)We can picture each proposition as a domino: P (k)A sequence of propositions is visualized as a sequence of dominos. P (0) P (1) P (2) P (k) P ….. (k+1)When the domino falls (to right), the corresponding proposition is considered true: 35
  4. 4. P (k) trueSuppose that the dominos satisfy two constraints. 1. Well positioned: if any domino falls to right, the next domino to right must fall also. P (k) P true (k+1) true 2. First domino has fallen to right. P (0) trueThan can conclude that all the domino fall. P (0) P (1) P (2) P (k) P ….. (k+1) P (0) P (1) P (2) P (k) P ….. (k+1) true 36
  5. 5. P (1) P (2) P (k) P ….. (k+1)P (0) true true true P (2) P (k) P ….. (k+1)P (0) true P (1) true True true P (k) P ….. (k+1)P (0) true P (1) true P (2) true P (k) P ….. (k+1)P (0) true P (1) true P (2) true true true 37
  6. 6. P (0) true P (1) true P (2) true true ….. P (k) true P (k+1) truePrinciple of Mathematical Induction.If: 1. Basis P(1) is true 2. Induction ∀ n P(k) P(k+1) is true P (0) true P (1) true P (2) true true ….. P (k) true P (k+1) trueThen, ∀ n P (n) is true.Use induction to prove that the sum of the first n odd integers is Yeah!Base case (n=1): the sum of the first 1 odd integer isAssume p (k): the sum of the first k odd integers is =1Where, 1 + 3 + … + (2k+1) =Prove that 1 + 3 + … + + (2k-1) + (2k+1) = 1 + 3 + … + + (2k-1) + (2k+1) = = 38
  7. 7. Prove that: 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ nBase case (n=1): 1 x 1! = (1 x 1)! -1? 1 x 1! = 1Assume P (k): 1 x 1! + 2 x 2! + … + k x k! = (k+1)! -1 2! – 1 = 1Prove that: 1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+2)! -1 1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+1)! -1+ (k+1) (k+1) = (1 + (k+1)) (k+1)! – 1 = (k+2) (k+1)! -1 = (k+2)! - 1 39
  8. 8. EXERCISE: 1. For every n ≥ 4, n! > 2. Proof that, We have seen that for any natural number n , 1 + 2 + ... + n = n( n + 1 )/2 and 12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6 hold. 40
  9. 9. ANSWER: 1. Proof: In this problem . Basis Step: If n = 4, then LHS = 4! = 24, and RHS = . Hence LHS > RHS . Induction: Assume that n! > for an arbitrary n ≥ 4 . -- Induction Hypothesis To prove that this inequality holds for n+1, first try to express LHS for n+1 in terms of LHS for n and try to use the induction hypothesis. Note here (n + 1)! = (n + 1) n! Thus using the induction hypothesis, we get (n + 1)! = (n + 1)n! > (n + 1) . Since n ≥ 4 , (n+1) > 2. Hence, (n + 1) > . Hence, (n + 1)! > . 2. One might then wonder what 13 + 23 + ... + n3 would be equal to. One way to find that out is to make a conjecture (i.e. educated guess) and prove the conjecture. Let us first take a guess. By looking at the sums 1 + 2 + ... + n = n( n + 1 )/2 --- a function of n2 and 12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6 --- a function of n3 , one can guess that 13 + 23 + ... + n3 is equal to some function of n4, that is 13 + 23 + ... + n3 = an4 + bn3 + cn2 + dn + e ---- (1) for some constants a, b, c, d and e. 41
  10. 10. Now to determine the value of a, b, c, d and e, we compare the value of the both sides ofthe equation (1) for five values of n.For n = 0, LHS = 0, and RHS = e. Hence we get e = 0.Similarly for n = 1, 1 = a + b + c + d,for n = 2, 9 = 16a + 8b + 4c + 2d,for n = 3, 36 = 81a + 27b + 9c + 3d , andfor n = 4, 100 = 256a + 64b + 16c + 4d .Solving this system of equations we obtaina = c = 1/4, b = 1/2, and d = 0.Hence our conjecture is13 + 23 + ... + n3 = 1/4 n2 ( n2 + 2n + 1 ) = ( 1/2 n ( n + 1 ) )2 .What we do next is to try to prove it by mathematical induction.Proof by Mathematical Induction:Basis Step: For n = 0, LHS = 03 = 0 , andRHS = ( 1/2* 0 ( 0 + 1 ) )2 = 0 .Hence LHS = RHS.Induction: Assume that 13 + 23 + ... + n3 = ( 1/2 n ( n + 1 ) )2 for an arbitrary n. ---------Induction HypothesisThen for n + 1, LHS = 13 + 23 + ... + n3 + ( n + 1 )3 = (13 + 23 + ... + n3) + ( n + 1 )3 = ( 1/2 n ( n + 1 ) )2 + ( n + 1 )3 by the induction hypothesis. = ( 1/4 ( n + 1 )2( n2 + 4*( n + 1 ) ) = ( 1/2 ( n + 1 )( n + 2 ) )2 ,which is equal to RHS.End of ProofThus we now know for sure that13 + 23 + ... + n3 = ( 1/2 n ( n + 1 ) )2 . 42

×