S = 1 + 1x +2x^2 +3x^3 + 5x^4 + 8x^5 +13x^6 + ...

Evaluate S - xS -x^2S

I have the proof at http://en.wikipedia.org/wiki/Fibonacci_number#Power_series but don't know if it matters that my series starts differently.
$$(F _{n-1} + F _{n-2})x ^{n+1}$$

Question

S = 1 + 1x +2x^2 +3x^3 + 5x^4 + 8x^5 +13x^6 + ...

Evaluate S - xS -x^2S

I have the proof at http://en.wikipedia.org/wiki/Fibonacci_number#Power_series but don't know if it matters that my series starts differently.
$$(F _{n-1} + F _{n-2})x ^{n+1}$$

bibby · Answer

F(x) = 1 + 1x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + 13x^6 + 21x^7 + ...
xF(x) =       x + 1x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + ...
x^2F(x) =         1x^2 + 1x^3 + 2x^4 + 3x^5 + 5x^6 + 8x^7 + 13x^8

so they obviously all cancel as the first as 5 = 3 + 2, 8 = 5 + 3 etc

phi · Answer

based on wikipedia  F(x) should be
$$ F(x)= 0 + 1x +1x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + ... $$
as noted,  F(x) - x F(x) - x^2 F(x) = x (everything else cancels)
so F(x) ( 1 - x -x^2) = x
and
$$ F(x) = \frac{x}{1 - x -x^2} $$

bibby · Answer

In my case I'm left with a 1 after the subtraction.
I can't /alter/ the series. I'm just given S.

In my case would it be $$\frac{ 1 }{ 1-x-x^2 }$$

phi · Answer

Ok, in that case, I get the same thing.

bibby · Answer

Sweet, thanks for everything.

phi · Answer

It looks like you get a closed solution except for when x = phi

bibby · Answer

It seems all we have to do is represent S as a rational function and then get a series from that