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Unformatted text preview: MATH 135 Fall 2008 Lectures V/VI/VII Notes Strong Induction Sometimes induction doesnt work where it looks like it should. We then need to change our ap proach a bit. The following example is similar to examples that weve done earlier. Lets try to make regular induction work and see where things go wrong. Example A sequence { x n } is defined by x 1 = 0, x 2 = 30 and x m = x m 1 + 6 x m 2 for m 3. Prove that x n = 2 3 n + 3 ( 2) n for n 1. Proof Lets try to prove this by regular induction. Base Case s n = 1: x 1 = 0 and 2 3 1 + 3 ( 2) 1 = 0 so the result is true for n = 1. TO ADD: n = 2 : x 2 = 30 and 2 3 2 + 3 ( 2) 2 = 30 so the result is true for n = 2. (We use two Base Cases because there are two initial conditions.) Induction Hypothesis Suppose the result is true for n = k , for some k P , k 1. That is, suppose x k = 2 3 k + 3 ( 2) k . CROSS OUT PREVIOUS AND REPLACE WITH: Assume the result is true for n = 1 , 2 ,...,k for some k P , k 2. Induction Conclusion Consider n = k + 1. Then x k +1 = x k + 6 x k 1 (Range for k ?) = (2 3 k + 3 ( 2) k ) + 6 x k 1 (by Induction Hypothesis) PROBLEM What about x k 1 ? We need to know something here = (2 3 k + 3 ( 2) k ) + 6(2 3 k 1 + 3 ( 2) k 1 ) (by Induction Hypothesis) = 3 k 1 [2 3 + 6 2] + ( 2) k 1 [3 ( 2) + 6 3] = 18 3 k 1 + 12 ( 2) k 1 = 2 3 k +1 + 3 ( 2) k +1 Therefore the result is true for n = k + 1, so holds for all n by POSI. The main problem is that our hypothesis needs to include both something about n = k and about n = k 1. Can we just add this? Yes: Principle of Strong Induction (POSI) Let P ( n ) be a statement that depends on n P . If (i) P (1) is true, and (ii) P (1) , P (2) ,...,P ( k ) are all true P ( k + 1) is true then P ( n ) is true for all n P ....
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 Fall '08
 ANDREWCHILDS
 Math, Algebra

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