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Fibonnaci function

  • Thread starter talolard
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Homework Statement



I am making a mistake and i can't find it. Pleasde help me


Let [tex] \left\{ a_{n}\right\} [/tex] be the fibonacci sequence and [tex] f(x)=\sum a_{n}x^{n} [/tex]. Prove that [/tex] f(x)=\frac{1}{1-x-s^{2}} [/tex]in its radius of convergence.

Solution

[tex] f(x)=\sum a_{n}x^{n}=1+x+\sum_{n=2}a_{n}x^{n}=1+x+\sum_{n=2}a_{n-1}x^{n}+\sum_{n=2}a_{n-2}x^{n}=1+x+x\sum_{n=2}a_{n-1}x^{n-1}+x^{2}\sum_{n=2}a_{n-2}x^{n-2}=1+x+xf(x)+x^{2}f(x) [/tex]

=[tex] f(x)(x+x^{2})+1+x=f(x)\iff f(x)(1-x-x^{2})=1+x\iff f(x)=\frac{1+x}{(1-x-x^{2})} [/tex]

Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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I don't know if it will solve all your problems, but you claim that
[tex]\sum_{n = 2}^\infty a_{n - 1} x^{n - 1} = f(x)[/tex]
However, I think it is
[tex]\sum_{n = 2}^\infty a_{n - 1} x^{n - 1} = f(x) - 1[/tex]
 
  • #3
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Actually., I think that exactly solves my problem.
[tex] 1+x+x\sum_{n=2}a_{n-1}x^{n-1}+x^{2}\sum_{n=2}a_{n-2}x^{n-2}=1+x+xf(x)-x+x^{2}f(x)=1+f(x)(x+x^{2})=f(x)\iff f(x)(1-x-x^{2})=1\iff=\frac{1}{1-x-x^{2}} [/tex]
THanks
 

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