How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)

Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?

I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$

So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$

However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$

Appreciate any and all help!

Thanks for reading,

A


  Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$$
$$x(\frac{1}{1-(x+x^3)})=x(1+(x+x^3)+(x+x^3)^2+(x+x^3)^3+.....)$$
$$x\left(\frac{1}{1-(x+x^3)}\right)$$ will do it.