recursive definitions and induction proofs rosen 3.4

Recursive Definitions and Induction Proofs

Rosen 3.4

More Fibonacci Numbers

Prove ,

for n 2

nn

nf2n1nn

2n1nn

Basis step, for n=2

1)1(1

f22

2


Inductive Step

Assume ,

for 2 k n

kk

kf2k1kk

2k1kk

Show

1n1n

1nf

1n1n

1nn1nn

1n1nnn

1n1nnn

1nn1n

)(

fff



But we showed before that if ,then = (1+5)/2 .More generally,= (1+5)/2= (1-5)/2and is the golden ratio (often labeled )!

2n1nn

Golden Ratio

bac,c

a

a

b

a b

c

0baba

a)ba(b,abc22

22

0

babb)1a(baba

b)1(a,b2

51ba

2222

22

Try this

Thus gives the golden ratio.

Golden Ratio

What is fn+1/ fn as n gets very large?

Recall

= (1+5)/2 = (1-5)/2

nn

nf

What happens to n and n as n gets very large?

fn+1/ fn approachesthe golden ratio ()as n gets very large!

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) = f2*1 - 1 = f1 = 1 = f2

Inductive Step

Assume g(k) = f1 + f3 + … + f2k-1 = f2k for k n, we must show that this implies g(n+1) = f2(n+1)

g(n+1) = f1 + f3 + … + f2n-1 + f2n+1

g(n+1) = g(n) + f2n+1

g(n+1) = f2n + f2n+1 = f2n+2 = f2(n+1)

Find a closed form solution for the recursive definition: T(n) = T(n/2) + c1, T(1) = c0

where n is 2k for kN.

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

Proof of guessProof by Induction.Basis Step: T(1) = c0 + c1log21 = c0 + c1*0 = 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)

Ackermann’s FunctionA(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) =


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) =


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) =


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

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

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

Basis step: n = 1, remember f0 =0, f1 =1, f2 =1

LetA

1 1

1 0

When n is a positive integer, show that

An fn1 fn

fn fn 1

A1

1 1

1 0

f2 f1

f1 f0

f11 f1

f1 f1 1

Inductive step:

Assume that

We must show that

LetA

1 1

1 0

When n is a positive integer, show that

An fn1 fn

fn fn 1

An

fn1 fn

fn fn 1

An1

fn11 fn1

fn1 fn11

An1 AAn 1 1

1 0

fn1 fn

fn fn 1

fn1 fn fn fn 1

fn1 fn

fn2 fn1

fn1 fn

fn11 fn1

fn1 fn1 1

1

j( j1)

j1

n

n

n1

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

1

j( j1)

j1

n

n

n1

1

j( j1)

j1

n1

n1

n 2

1

j( j1)

j1

n1

1

j( j1)

j1

n

1

(n1)(n 2)

nn1

1

(n 1)(n 2)

n 2 n1

(n1)(n 2)

n2 2n1

n1 n 2

n1 n1 n1 n 2

n1

n 2

recursive definitions and induction proofs rosen 3.4

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