@apple_pi
simplest form of fibonacci polynomial
$$\frac{1}{1-x-x^2}=F_0+F_1x+F_2x^2+...+F_kx^k+...$$
$F_k$ is the $k\text{th}$ term of Recursive Fibonacci Sequence

this series will be converge if $|x+x^2|

Question

@apple_pi
simplest form of fibonacci polynomial
$$\frac{1}{1-x-x^2}=F_0+F_1x+F_2x^2+...+F_kx^k+...$$
$F_k$  is the  $k\text{th}$ term of Recursive Fibonacci Sequence

this series will be converge if  $|x+x^2|<1$

anonymous · Answer

let suppose  $F_0$ , $F_1$ , $F_2$ ,..... are the fibonacci terms....

anonymous · Answer

we want to find a function $ f(x)$  such that :$$f(x)=F_0+F_1x+F_2x^2+....+F_kx^k+...$$

experimentx · Answer

i know the generating function!! but don't know how it works.

anonymous · Answer

$$\ \ f(x)=F_0+F_1x+F_2x^2+....$$$$xf(x)= \ \  \ \ \ \ \ F_0x+F_1x^2+F_2x^3+....$$$$x^2f(x)= \ \  \ \ \ \ \ \ \ \ \ \ \ \ \  F_0x^2+F_1x^3+F_2x^4+....$$  subtract  $xf(x)$  and  $x^2 f(x)$  from  $f(x)$ u will get 

$$f(x)-xf(x)-x^2f(x)\ \ \ \ \ \  =F_0+(F_1-F_0)x+(F_2-F_1-F_0)x^2+...+(F_k-F_{k-1}-F_{k-2})x^k+...$$

anonymous · Answer

@apple_pi 
can u tell me $RHS$  of  last equation equals to what ... see the coefficients of polynomial

anonymous · Answer

wait, does it equal F0

anonymous · Answer

F1-F0 = 1-1 = 0
F2-F1-F0 = 2-1-1 = 0

anonymous · Answer

Yep thats right

anonymous · Answer

and because $$F_k=F_{k-1}+F_{k-2}$$

anonymous · Answer

Oh, so 1 = f(x)-xf(x)-x^2f(x)

anonymous · Answer

then divide by f(x) to get
1/f(x) = 1-x-x^2
f(x) = 1/(1-x-x^2)!

anonymous · Answer

OK thank you very much

anonymous · Answer

Right