# Fibonnaci function

## Homework Statement

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

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

Solution

$$f(x)=\sum a_{n}x^{n}=1+x+\sum_{n=2}a_{n}x^{n}=1+x+\sum_{n=2}a_{n-1}x^{n}+\sum_{n=2}a_{n-2}x^{n}=1+x+x\sum_{n=2}a_{n-1}x^{n-1}+x^{2}\sum_{n=2}a_{n-2}x^{n-2}=1+x+xf(x)+x^{2}f(x)$$

=$$f(x)(x+x^{2})+1+x=f(x)\iff f(x)(1-x-x^{2})=1+x\iff f(x)=\frac{1+x}{(1-x-x^{2})}$$

## The Attempt at a Solution

CompuChip
Homework Helper
I don't know if it will solve all your problems, but you claim that
$$\sum_{n = 2}^\infty a_{n - 1} x^{n - 1} = f(x)$$
However, I think it is
$$\sum_{n = 2}^\infty a_{n - 1} x^{n - 1} = f(x) - 1$$

Actually., I think that exactly solves my problem.
$$1+x+x\sum_{n=2}a_{n-1}x^{n-1}+x^{2}\sum_{n=2}a_{n-2}x^{n-2}=1+x+xf(x)-x+x^{2}f(x)=1+f(x)(x+x^{2})=f(x)\iff f(x)(1-x-x^{2})=1\iff=\frac{1}{1-x-x^{2}}$$
THanks