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Unformatted text preview: MATH 135 Winter 2009 Lectures V/VI/VII Notes Strong Induction Sometimes induction doesn’t 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 we’ve done earlier. Let’s 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” Let’s 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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This note was uploaded on 06/03/2010 for the course MATH 135 taught by Professor Peter during the Winter '07 term at University of the West.
 Winter '07
 peter
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