Recursive Definitions and Induction Proofs

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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 n2
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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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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)!

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Golden Ratio
b
a
c
Try this
Thus a gives the golden ratio.
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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?
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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)

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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

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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)

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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)

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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)

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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)

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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

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

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When n is a positive integer, show that

• Basis step n 1, remember f0 0, f1 1, f2 1

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When n is a positive integer, show that

• Inductive step
• Assume that
• We must show that

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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

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