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Math Help - Fibonnacci sequence

  1. #1
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    Fibonnacci sequence

    Can anyone explain if the Fibonnacci sequence can be placed into a parabolic or hyperbolic function?
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  2. #2
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    Re: Fibonnacci sequence

    If you're asking if there's a function \phi:\mathbb{R}^+ \rightarrow \mathbb{R} such that the graph of \phi is a parabola or hyperbola,

    and that \phi satistfies, for all positive integers n, \ \phi(n) = F_n, the n^{th} Fibonnacci number, then the answer is no.

    Parabolas and hyperbolas only have a few "free parameters", and the requirement \phi(n) = F_n \forall n \in \mathbb{Z}^+ would quickly exhaust them.

    It would be possible that it algebraically worked out, but it doesn't. I'll do the parabola case, and leave to you the hyperbola ( \phi(x) = b\sqrt{1+(x/a)^2}).

    If a parabola is the graph of a function \phi that satisfies  \phi(n) = F_n, then let \phi(x) = ax^2 + bx + c for some  a, b, c \in \mathbb{R}.

    Then \forall n \in \mathbb{Z}^+, \phi(n+2) = \phi(n+1) + \phi(n), so a(n+2)^2 + b(n+2) + c = ( a(n+1)^2 + b (n+1) + c )  + ( an^2 + bn + c ).

    Thus \forall n \in \mathbb{Z}^+, a \[ (n+2)^2 - (n+1)^2 - n^2 \] + b\[ (n+2) - (n+1) - (n) \] + \[ c - c - c \] = 0, so

    a(- n^2 -2n + 3 )+ b(-n + 1) - c = 0, so (-a)n^2 + (-2a-b)n + (3a+b-c) = 0

    The only way that could be true for all n is if each of those n-coefficients was 0. Thus -a=0, -2a-b = 0, 3a+b-c = 0.

    That system of equations (3 equations in the 3 unknowns a, b, and c) has only one solution: a = b = c = 0.

    That means the only quadratic that has a prayer of working is the 0 quadratic \phi(x) = 0x^2+0x+0 = 0.

    But that obviously doesn't satisfy \phi(n) = F_n \forall n \in \mathbb{Z}^+, and it was the only one that had a chance.

    -----

    I suppose I should note that even if you consider rotated parabolas/hyperbolas, you're still going to quickly run out of free parameters.
    (The standard equation for the graph of a generic conic section in the plane has 6 free parameters. That's more than the 3 a quadratic
    polynomial has, but it's still less than the infinity required.).
    Last edited by johnsomeone; September 14th 2012 at 06:49 PM.
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