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Fibonacci Number Proof

by ryou00730
Tags: fibonacci, number, proof
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ryou00730
#1
Nov8-10, 11:05 AM
P: 29
1. The problem statement, all variables and given/known data
The Fibonacci numbers are defined recusively by:
F(0) = 0, F(1) = 1, for n > 1, F(n) = F(n − 1) + F(n − 2).
Use strong induction to show that F(n) < 2^n for all n.


2. Relevant equations
n/a

3. The attempt at a solution
I use my base case as F(2) = F(1) + F(0) = 1 which is less than 2^n = 2^2 = 4. After that I am not sure where to go with strong induction...
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micromass
#2
Nov8-10, 11:21 AM
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The induction hypthesis gives us

[tex]F(n)=F(n-1)+F(n-2)\leq 2^{n-1}+2^{n-2}[/tex]

Can you now prove that this is smaller then [tex]2^n[/tex]?
ryou00730
#3
Nov8-10, 01:04 PM
P: 29
well 2^n-1 + 2^n-2 is equal to 2^n (2^-1 + 2^-2) = 3/4(2^n) but I don't see the logic behind equating the problem to less than or equal to 2^n-1 + 2^n-2

micromass
#4
Nov8-10, 01:12 PM
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Fibonacci Number Proof

If you want to apply induction? Then what is the induction hypothesis?
ryou00730
#5
Nov8-10, 01:15 PM
P: 29
Let k>2, for all integers i, with 2<=i<k, F(i)<2^i would be my induction hypothesis I think?
micromass
#6
Nov8-10, 01:19 PM
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Yes, that is correct. So in particular, we have

[tex]F(n-1)\leq 2^{n-1}~\text{and}~F(n-2)\leq 2^{n-2}[/tex]

So

[tex]F(n-1)+F(n-2)\leq 2^{n-1}+2^{n-2}[/tex]
ryou00730
#7
Nov8-10, 01:20 PM
P: 29
woops, I meant n>2, so just replace all my k's with n's and so if I know that F(i)<2^i, then F(i-1) + F(i-2) < 2^i-1 + 2^i-2 which is F(i) < 3/4 2^i, so therefore F(i) must also be less than 2^i if its less than 0.75 of it, so the statement is true for all F(n)
ryou00730
#8
Nov8-10, 01:21 PM
P: 29
does that look about right?
micromass
#9
Nov8-10, 01:25 PM
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Yeah, that looks fine
ryou00730
#10
Nov8-10, 01:27 PM
P: 29
thank you :), so in general with strong induction, you prove a base, then your induction hypothesis is that it works for all numbers between your base up until some value n, and you have to prove using this, that it also works for the n? thank you for all your help!
micromass
#11
Nov8-10, 01:33 PM
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Yep, that is the idea behind strong induction.

Good luck with your next problems!
ryou00730
#12
Nov8-10, 01:34 PM
P: 29
thank you:)


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