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Math Help - Why is Fibonacci =< golden ration^n-1

  1. #1
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    Why is Fibonacci =< golden ration^n-1

    How can it be proved by induction that the Fibonaaci series is always:

    Fn <= ( (1 +sqrt 5)/2 )^(n−1).





    F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn, for n = 0, 1, 2, . . .



    golden ratio=
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  2. #2
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    I don't know if you can just use the well-known formula for the nth term of the Fibonacci sequence, but if not you can find a derivation here as well as some other identities.
    Last edited by Jameson; September 9th 2009 at 05:11 PM.
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  3. #3
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    Quote Originally Posted by DeRSeD View Post
    How can it be proved by induction that the Fibonaaci series is always:

    Fn <= ( (1 +sqrt 5)/2 )^(n−1).





    F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn, for n = 0, 1, 2, . . .



    golden ratio=
    Start by showing that F_n \leq \phi ^{n-1} for n = 0 and n = 1.

    Then assume F_n \leq \phi ^{n-1} for all n \leq k, where k > 1. If so, then
    F_{k+2} = F_{k} + F_{k+1} \leq \phi ^{k-1} + \phi ^k
    If you can show \phi ^{k-1} + \phi ^k \leq \phi ^{k+1} then you will be done. You might want to start by looking at the case k=1.
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