L6W09 - 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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 8

L6W09 - 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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online