How to compute the coefficient of a generating function?

Question

I would like to find a closed formula for the coefficients of the generating function $$f(x)=-{\frac {{x}^{4}+6\,{x}^{3}-2\,{x}^{2}+6\,x+1}{ \left( x+1 \right) \left( {x}^{2}+1 \right) \left( x-1 \right) ^{3}}} .$$

I am trying to do the following. \begin{align} f(x)=\sum_{i,j,k_1,k_2,k_3=0}^{\infty} (-x)^i (-x^2)^j x^{k_1+k_2+k_3}(1+6x-2x^2+6x^3+x^4). \end{align}

The expression is very complicated. Are there some simpler method (or some software) to find a closed formula for the coefficients? Thank you very much.

Answer 1

$$-{\frac {{x}^{4}+6\,{x}^{3}-2\,{x}^{2}+6\,x+1}{ \left( x+1 \right) \left( {x}^{2}+1 \right) \left( x-1 \right) ^{3}}} =\frac{0.25}{1-x}-\frac{0.75}{1+x}+\frac{x}{1+x^2}-\frac{1.5}{(1-x)^2}+\frac{3}{(1-x)^3}$$ then use the following series and differentiate it two times to replace all terms to series $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$

Answer 2

Split into partial fractions, use the generalized binomial theorem:

$\begin{align*} (1 - u)^{-m} &= \sum_{n \ge 0} (-1)^n \binom{-m}{n} u^n \\\\ &= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} u^n \end{align*}$

and you are all set:

$\begin{align*} [x^n] f(x) &= [x^n] \left( \frac{x}{1 + x^2} - \frac{3}{4} \frac{1}{1 + x} + \frac{1}{4} \frac{1}{1 - x} - \frac{3}{2} \frac{1}{(1 - x)^2} + 3 \frac{1}{(1 - x)^3} \right) \\\\ &= [x^{n - 1}] \frac{1}{1 + x^2} - \frac{3}{4} \cdot (-1)^n + \frac{1}{4} - \frac{3}{2} \binom{n + 2 - 1}{2 - 1} + 3 \binom{n + 3 - 1}{3 - 1} \\\\ &= [n \text{ odd}] - \frac{3}{4} \cdot (-1)^n + \frac{6 n^2 + 12 n + 7}{4} \end{align*}$

Here $[n \text{ odd}]$ is Iverson's convention: 1 if the condition is true, 0 otherwise.