31 Slides--Induction III

31 Slides--Induction III - CS103 HO#31 Slides--Induction...

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Unformatted text preview: CS103 HO#31 Slides--Induction III April 28, 2010 1 Complete Induction Well-Ordering Principle Mathematical Induction MI CI P 1 P 2 __ __ __ . . . __ __ P k P k+1 Suppose the sequence of propositions P n satisfies the conditions for Complete Induction: We want to show nP n MI CI P 1 P 2 __ __ __ . . . __ __ P k P k+1 Suppose the sequence of propositions P n satisfies the conditions for Complete Induction: Let Q n = P 1 P 2 P 3 P n Q 1 is true since we are assuming that P 1 is true Suppose Q k is true. Then P k+1 is true by the diagram above. Q k P k+1 = Q k+1 , so Q k Q k+1 So by MI, we have nQ n But Q n P n , so we have nP n So MI CI Complete Induction Well-Ordering Principle Mathematical Induction Specifying a Sequence Enumerate: 2, 4, 6, 8, ... Explicit formula: a n = 2n, n = 1, 2, ... Recursive formula: a 1 = 2, a n = a n-1 + 2 Recursive Definition of a Sequence 1, 3, 6, 10, 15, 21, ... a 1 a 2 a 3 a 4 a 5 a 6 a 1 = 1 a n = a n-1 + n Solving the recurrence relation to obtain an explicit formula can be difficult, but here, we can see that a n is the sum of the 1 st n integers, so which we proved by induction. a n = n(n + 1) 2 CS103 HO#31 Slides--Induction III April 28, 2010 2 Recursive Definition of Factorial n!=1 2 3 ... n can be defined recursively: n! = 1 if n = 1 n(n-1)! if n > 1 Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 The Explicit Formula (1 + 5) n (1 - 5) n 2 n 5 F n = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = Prove P n : F i = F n F n+1 for n 1 i=1 n 2 0+1+1+4+9+25 = 40 5 8 = 40 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = Prove P n : F i = F n F n+1 for n 1 i=1 n 2 BASE CASE: P 1 asserts that 1 2 = 1 1, which is true. Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = Prove P n : F i = F n F n+1 for n 1 i=1 n 2 INDUCTIVE STEP: Assume P k : F i = F k F k+1 for some k 1 Show P k+1 : F i = F k+1 F k+2 i=1 k 2 2 i=1 k+1 Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = Prove P n : F i = F n F n+1 for n 1 i=1 n 2 INDUCTIVE STEP: 2 Assume P k : F i = F k F k+1 for some k 1 Show P k+1 : F i = F k+1 F k+2 F i + F k+1 = F k F k+1 + F k+1 by Inductive Hypothesis and adding to both sides F i = F k+1 (F k + F k+1 ) def. of and factoring RHS i=1 k 2 2 i=1 k+1 i=1 k 2 2 2 k+1 i=1 CS103 HO#31 Slides--Induction III April 28, 2010 3 Fibonacci Sequence i f n = 1 i f n = 1 F n-1 + F n-2 if n > 1 F n = Prove P n : F i = F n F n+1 for n 1 i=1 n 2 INDUCTIVE STEP: 2 Assume P k :...
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31 Slides--Induction III - CS103 HO#31 Slides--Induction...

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