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Thread: Fibonnacci sequence

  1. #1
    Oct 2009

    Fibonnacci sequence

    Can anyone explain if the Fibonnacci sequence can be placed into a parabolic or hyperbolic function?
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  2. #2
    MHF Contributor
    Sep 2012
    Washington DC USA

    Re: Fibonnacci sequence

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

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

    Parabolas and hyperbolas only have a few "free parameters", and the requirement $\displaystyle \phi(n) = F_n \forall n \in \mathbb{Z}^+$ 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 ($\displaystyle \phi(x) = b\sqrt{1+(x/a)^2}$).

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

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

    Thus $\displaystyle \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

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

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

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

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

    But that obviously doesn't satisfy $\displaystyle \phi(n) = F_n \forall n \in \mathbb{Z}^+$, 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; Sep 14th 2012 at 06:49 PM.
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