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Recursive Definitions and Induction Proofs

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Title: Recursive Definitions and Induction Proofs


1
Recursive Definitions and Induction Proofs
  • Rosen 3.4

2
More Fibonacci Numbers
Prove , for n ? 2
Basis step, for n2
3
More Fibonacci Numbers
Inductive Step Assume , for 2 ? k ? n
Show
4
More Fibonacci Numbers
5
More Fibonacci Numbers
  • 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 f)!

6
Golden Ratio
b
a
c
Try this
Thus a gives the golden ratio.
7
Golden Ratio
  • What is fn1/ fn as n gets very large?
  • Recall
  • (1?5)/2
  • b (1-?5)/2

fn1/ fn approaches the golden ratio (a) as n
gets very large!
What happens to an and bn as n gets very large?
8
Prove that the function g(n) f1 f3
f2n-1 (where fi is a Fibonacci number) is equal
to f2n whenever n is a positive integer.
  • Basis Step
  • If n 1, then g(1) f21 - 1 f1 1 f2
  • Inductive Step
  • Assume g(k) f1 f3 f2k-1 f2k for k ?
    n, we must show that this implies g(n1)
    f2(n1)
  • g(n1) f1 f3 f2n-1 f2n1
  • g(n1) g(n) f2n1
  • g(n1) f2n f2n1 f2n2 f2(n1)

9
Find a closed form solution for the recursive
definition T(n) T(n/2) c1, T(1) c0where
n is 2k for k?N.
  • T(1) c0
  • T(2) T(1) c1 c0 c1
  • T(4) T(2) c1 c0 c1 c1 c0 2c1
  • T(8) T(4) c1 c0 2c1 c1 c0 3c1
  • T(16) T(8) c1 c0 3c1 c1 c0 4c1
  • Guess that T(n) c0 c1log2n

10
Proof of guess
  • Proof by Induction.
  • Basis Step T(1) c0 c1log21 c0 c10 c0
  • Inductive Step Assume that T(n) c0 c1log2n,
    then we must show that T(2n) c0 c1log2(2n) .
  • T(2n) T(n) c1
  • c0 c1log2n c1
  • c0 c1log2n c1log22 c0 c1(log2n log22)
  • c0 c1log2(2n)

11
Ackermanns 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
Ackermanns 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
Ackermanns 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
Ackermanns 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) 22 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(k1, 2) 4.
  • A(k1,2) A(k,A(k1,1)) A(k,2) 4

17
When n is a positive integer, show that
  • Basis step n 1, remember f0 0, f1 1, f2 1

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

19
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20
Prove that
  • Proof
  • Basis Step For n 1 we get 1/(1(11)) 1/2 for
    both the sum and the closed form.
  • Inductive Step
  • Assume that
  • We must show that

21
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