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Finding the formula for the fibunacci sequence

  1. May 16, 2010 #1
    1 If I am given the function [tex]\frac{1}{-x^2-x+1} = \sum_{j=0}^{\infty} F_{j} x^{j}[/tex]

    Which discribes a sequence of fibunacci numbers 1,1,2,3,5,8,13,21......

    Find the formula for the fibunacci sequence for n>=2 and where [tex]F_j = F_{j-1}+F_{j-2}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I know that the recusive relation can be written as [tex]F_j = \alpha_1(r_1)^j + \alpha_2 (r_2)^j[/tex]

    With the inital conditions [tex]F_0 = F_1 = 1[/tex]

    Since the poles of the function are [tex]r_1 = \frac{\sqrt{5}-1}{2}[/tex] and [tex]r_2 = \frac{-(\sqrt{5}+1)}{2}[/tex]

    which gives me the expression [tex]F_j = \alpha_1(\frac{\sqrt{5}-1}{2})^n + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}})^n[/tex]

    so this gives me [tex]F_0 = \alpha_1 + \alpha_2 = 1[/tex]

    and [tex]F_1 = \alpha_1(\frac{\sqrt{5}-1}{2}) + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}}) = 1[/tex] and I end up with

    [tex]\alpha_1, \alpha_2 = \pm \frac{1}{\sqrt{5}}[/tex]

    and this j >= 2 then the formula for the jth fibunacci number must

    [tex]F_j = \frac{1}{\sqrt{5}}(\frac{\sqrt{5}+1}{2})^{j+1} -\frac{1}{\sqrt{5}} (\frac{-(\sqrt{5}+1)}{2}})^{j+1}[/tex]

    How is that HallsoftIvy?

    Last edited: May 16, 2010
  2. jcsd
  3. May 16, 2010 #2


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    Science Advisor

    You mean [tex]F_n= F_{n-1}+ F_{n-2}[/tex]

    Okay, so its just a matter of finding [tex]r_1[/tex] and [tex]r_2[/tex]. If [tex]F_n= r^n[/tex] the equation [tex]F_n= F_{n-1}+ F_{n-2}[/tex] becomes [tex]r^n= r^{n-1}+ r^{n-2}[/tex]. Dividing that equation by [tex]r^{n-1}[/tex] gives [tex]r^2= r+ 1[/tex]. What are the roots of that equation?

    No fair! You edited and added your solution while I was responding!
  4. May 16, 2010 #3
    Sorry HallsoftIvy :)
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