Partial fraction decomposition of a parameter-dependent rational function

Question

Let $n$ and $m$ be strictly positive integers such that $n< 2 m$. By using mathematical induction in $n$ we have derived the following equality identity: \begin{equation} \frac{x^n}{\left(1-(1+a) x+x^2\right)^m}= \sum\limits_{\xi=0}^{\lfloor \frac{n}{2} \rfloor} \frac{\left\{ P^{(n-1)}_{\xi-1}(a) 1_{\xi \ge 1} - P^{(n-1)}_\xi(a) 1_{\xi \le \lfloor \frac{n-2}{2} \rfloor}\right\} + x \cdot P^{(n)}_\xi(a) 1_{\xi \le \lfloor\frac{n-1}{2}\rfloor}}{\left(1-(1+a) x+x^2\right)^{m-\xi}} \end{equation} where the quantities $P^{(n)}_\xi(a)$ are polynomials in $a$ which are defined as follows: \begin{equation} P^{(n)}_\xi(a) := \sum\limits_{l=0}^{\tilde{n}-\xi} \binom{\tilde{n}+l}{\xi} \binom{\tilde{n}-\xi+l}{2 l+1-r} (-1)^{\tilde{n}-\xi-l-1+r} \cdot (1+a)^{2 l+1-r} \end{equation} for $\xi=0,\cdots,\tilde{n}-1+r$ and $\tilde{n}:=\lfloor n/2 \rfloor$ and $r:= n\% 2$. Now my question is twofold. Firstly can we find a closed form expression for the sum in the definition of $P^{(n)}_\xi(a)$. Secondly, how do we decompose into partial fractions a more generic function as given below: \begin{equation} \frac{x^n}{(x-b)^l\left(1-(1+a) x+x^2\right)^m}= ? \end{equation} where $n<2 m+l$.

Answer 1

Here we provide an answer to the second question. The following decomposition holds: \begin{eqnarray} &&\frac{x^n}{(x-b)^p(1-(1+a) x+x^2)^m}=\\ && \sum\limits_{\xi_1=1}^p\sum\limits_{\xi_2=0}^n Q^{(p-\xi_1)}_{m} \binom{n}{\xi_2} \frac{b^{n-\xi_2}}{(x-b)^{\xi_1-\xi_2}}+\\ && \sum\limits_{\xi=1}^{m+\lfloor\frac{n}{2}\rfloor} \frac{{\mathcal A}^{(n,m,p)}_{1,\xi} + {\mathcal A}^{(n,m,p)}_{2,\xi} x}{(1-(1+a) x+x^2)^{\xi-\lfloor \frac{n}{2} \rfloor}} - \sum\limits_{\xi=1}^{m+\lfloor\frac{n+1}{2}\rfloor} \frac{{\mathcal B}^{(n,m,p)}_{1,\xi} + {\mathcal B}^{(n,m,p)}_{2,\xi} x}{(1-(1+a) x+x^2)^{\xi-\lfloor \frac{n+1}{2} \rfloor}} (I) \end{eqnarray} Here: \begin{eqnarray} &&\left({\mathcal A}^{(n,m,p)}_{\eta,\xi}\right)_{\eta=1}^2 = \sum\limits_{l=1 \vee (\xi - \lfloor \frac{n}{2} \rfloor)}^{m \wedge \xi}((1+a-b) Q^{(p-1)}_{m-l+1} - Q^{(p-2)}_{m-l+1}) \cdot\\ && \left\{P^{(n-1)}_{l-\xi+\lfloor \frac{n}{2}\rfloor}1_{l-1\ge \xi-\lfloor \frac{n}{2} \rfloor} - P^{(n-1)}_{1+l-\xi+\lfloor \frac{n}{2}\rfloor}1_{\xi \ge l+1 }\quad,\quad P^{(n)}_{1+l-\xi+\lfloor \frac{n}{2} \rfloor}1_{l-\lfloor\frac{n-1}{2} \rfloor \le \xi-\lfloor \frac{n}{2} \rfloor}\right\} \end{eqnarray} for $\xi=1,\cdots,m+\lfloor\frac{n}{2}\rfloor$ and \begin{eqnarray} &&\left({\mathcal B}^{(n,m,p)}_{\eta,\xi}\right)_{\eta=1}^2 =\sum\limits_{l=1 \vee (\xi - \lfloor \frac{n+1}{2} \rfloor)}^{m \wedge \xi} Q^{(p-1)}_{m-l+1} \cdot \\ &&\left\{P^{(n)}_{l-\xi+\lfloor \frac{n+1}{2} \rfloor}1_{l-1\ge\xi-\lfloor \frac{n+1}{2} \rfloor} - P^{(n)}_{1+l-\xi+\lfloor \frac{n+1}{2} \rfloor}1_{\xi \ge l+1}\quad,\quad P^{(n+1)}_{1+l-\xi+\lfloor \frac{n+1}{2} \rfloor}1_{l-\lfloor \frac{n}{2}\rfloor \le \xi -\lfloor \frac{n+1}{2}\rfloor }\right\} \end{eqnarray} for $\xi=1,\cdots,m+\lfloor\frac{n+1}{2}\rfloor$. Finally the quantities $Q$ are defined as follows: \begin{equation} Q^{(n)}_m = \frac{(2b-1-a)^n}{(1-(1+a) b+b^2)^{n+m}} \cdot \sum\limits_{l=0}^{\lfloor \frac{n}{2}\rfloor} \binom{n-l}{l} \binom{n-l+m-1}{m-1} (-1)^{n-l} \left(\frac{1-(1+a) b+b^2}{(2b-1-a)^2}\right)^l \end{equation} Note 1: The decomposition holds for any $n\ge 1$, $m\ge 1$ and $p\ge 1$. However if $n < p+2 m$ then of course there cannot be any positive powers of $x$ on the right hand side and therefore in the first sum on the rhs in (I) $\xi_1-\xi_2 \ge 1$ and in the second and in the third sums on the rhs in (I) $\xi \ge 1+\lfloor \frac{n}{2}\rfloor$ and $\xi \ge 1+\lfloor \frac{n+1}{2}\rfloor$.