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Recursive Definitions and Induction Proofs Rosen 3.4
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More Fibonacci Numbers Prove, for n 2 Basis step, for n=2
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More Fibonacci Numbers Inductive Step Assume, for 2 k n Show
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More Fibonacci Numbers
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But we showed before that if, then = (1+ 5)/2. More generally, = (1+ 5)/2 = (1- 5)/2 and is the golden ratio (often labeled )!
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Golden Ratio a b c Try this Thus gives the golden ratio.
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Golden Ratio What is f n+1 / f n as n gets very large? Recall = (1+ 5)/2 = (1- 5)/2 What happens to n and n as n gets very large? f n+1 / f n approaches the golden ratio ( ) as n gets very large!
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Prove that the function g(n) = f 1 + f 3 + … + f 2n-1 (where f i is a Fibonacci number) is equal to f 2n whenever n is a positive integer. Basis Step If n = 1, then g(1) = f 2*1 - 1 = f 1 = 1 = f 2 Inductive Step Assume g(k) = f 1 + f 3 + … + f 2k-1 = f 2k for k n, we must show that this implies g(n+1) = f 2(n+1) g(n+1) = f 1 + f 3 + … + f 2n-1 + f 2n+1 g(n+1) = g(n) + f 2n+1 g(n+1) = f 2n + f 2n+1 = f 2n+2 = f 2(n+1)
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Find a closed form solution for the recursive definition: T(n) = T(n/2) + c1, T(1) = c0 where n is 2 k for k N. T(1) = c 0 T(2) = T(1) + c 1 = c 0 + c 1 T(4) = T(2) + c 1 = c 0 + c 1 + c 1 = c 0 + 2c 1 T(8) = T(4) + c 1 = c 0 + 2c 1 = c 1 = c 0 + 3c 1 T(16) = T(8) + c 1 = c 0 + 3c 1 + c 1 = c 0 + 4c 1 Guess that T(n) = c 0 + c 1 log 2 n
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Proof of guess Proof by Induction. Basis Step: T(1) = c 0 + c 1 log 2 1 = c 0 + c 1 *0 = c 0 Inductive Step: Assume that T(n) = c 0 + c 1 log 2 n, then we must show that T(2n) = c 0 + c 1 log 2 (2n). T(2n) = T(n) + c 1 = c 0 + c 1 log 2 n + c 1 = c 0 + c 1 log 2 n + c 1 log 2 2 = c 0 + c 1 (log 2 n + log 2 2) = c 0 + c 1 log 2 (2n)
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Ackermann’s Function A(m,n) = 2n if m = 0 0 if m 1 and n = 0 2 if m 1 and n = 1 A(m-1, A(m,n-1))if m 1 and n 2 A(1,0) = 0 A(0,1) = A(2,2) = A(1,1) =
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Ackermann’s Function A(m,n) = 2n if m = 0 0 if m 1 and n = 0 2 if m 1 and n = 1 A(m-1, A(m,n-1))if m 1 and n 2 A(1,0) = 0 A(0,1) = 2 A(2,2) = A(1,1) =
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Ackermann’s Function A(m,n) = 2n if m = 0 0 if m 1 and n = 0 2 if m 1 and n = 1 A(m-1, A(m,n-1))if m 1 and n 2 A(1,0) = 0 A(0,1) = 2 A(2,2) = A(1,A(2,1)) = A(1,2) = A(0,A(1,1)) =A(0, 2) = 4 A(1,1) =
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Ackermann’s Function A(m,n) = 2n if m = 0 0 if m 1 and n = 0 2 if m 1 and n = 1 A(m-1, A(m,n-1))if m 1 and n 2 A(1,0) = 0 A(0,1) = 2 A(2,2) = A(1,A(2,1)) = A(1,2) = A(0,A(1,1)) =A(0, 2) = 4 A(1,1) = 2
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Show that A(m,2) = 4 whenever m 1 A(m,n) = 2n if m = 0 = 0 if m 1 and n = 0 = 2 if m 1 and n = 1 = A(m-1, A(m,n-1))if m 1 and n 2 Basis step: When m = 1, A(1,2) = A(0,A(1,1)) = A(0, 2) = 2*2 = 4
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Show that A(m,2) = 4 whenever m 1 A(m,n) = 2n if m = 0 = 0 if m 1 and n = 0 = 2 if m 1 and n = 1 = A(m-1, A(m,n-1))if m 1 and n 2 Inductive Step: Assume that A(j,2) = 4 for 1 j k. We must show that A(k+1, 2) = 4. A(k+1,2) = A(k,A(k+1,1)) = A(k,2) = 4
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Basis step: n = 1, remember f 0 =0, f 1 =1, f 2 =1 When n is a positive integer, show that
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Inductive step: Assume that We must show that When n is a positive integer, show that
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Proof: Basis Step: For n = 1 we get 1/(1(1+1)) = 1/2 for both the sum and the closed form. Inductive Step: Assume that We must show that Prove that
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