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Published byDonald Webster Modified over 5 years ago

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Problems to Solve Involving Induction

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Proof by Induction Basis Step: Does it work for n=0?

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Inductive Step Assume that for r 1 We must show that

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Prove that is O(n 2 ) First let’s find a closed form solution for the sum. = 3n(n+1)/2 - 2n = (3n 2 + 3n -4n)/2 = (3n 2 -n)/2

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How can we be sure that our arithmetic is correct? Prove that Proof by Induction Basis Step: For n=1

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Inductive Step Assume that Then we must show that:

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Big-Oh (3n 2 -n)/2 is clearly O(n 2 ) since (3n 2 -n)/2 (3n 2 -n) 3n 2 since n is positive. So chose C = 3 and k=1 and we have | (3n 2 -n)/2| (3) |n 2 | when n > 1.

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Does 1*1! + 2*2! + 3*3!+…+n*n! = (n+1)! -1 n 1? Basis Step: n = 1 1*1! = 1, (1+1)!-1 = 2-1=1 Inductive Step: Assume that 1*1! + 2*2! + 3*3!+…+n*n! = (n+1)! -1 We must show that 1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)! =(n+2)!-1

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1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)! = (n+1)! -1 + (n+1)*(n+1)! = (n+1+1)(n+1)! -1 = (n+2)(n+1)! - 1 = (n+2)! - 1

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Prove that every 2 n by 2 n (n > 1) chessboard can be tiled with T-ominos

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Basis Case: 2 2 x 2 2 Board

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Inductive Step Assume that we can tile a board of size 2 n by 2 n. We must show that this implies that we can tile a board of size 2 n+1 by 2 n+1. Proof: Divide the 2 n+1 by 2 n+1 board into 4 parts, each of size 2 n by 2 n. Since we know that each of these boards can be tiled, then we can put them together to tile the 2 n+1 by 2 n+1 board.

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