1. Recursive Definitions and Induction Proofs Rosen 3.4

2. More Fibonacci Numbers Prove, for n  2 Basis step, for n=2

3. More Fibonacci Numbers Inductive Step Assume, for 2  k  n Show

4. More Fibonacci Numbers

5. 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  )!

6. Golden Ratio a b c Try this Thus  gives the golden ratio.

7. 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!

8. 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)

9. 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

10. 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)

11. 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) =

12. 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) =

13. 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) =

14. 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

15. 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

16. 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

17. Basis step: n = 1, remember f 0 =0, f 1 =1, f 2 =1 When n is a positive integer, show that

18. Inductive step: Assume that We must show that When n is a positive integer, show that

20. 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