Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots.

Question

Let $d \ge 2$ be an integer. Let $\left\{ m_j \right\}_{j=1}^d$ be strictly positive integers and $\left\{ b_j \right\}_{j=1}^d$ be parameters. Define the following quantity: \begin{equation} {\mathfrak F}_d(x) := \frac{1}{\prod\limits_{j=1}^d (x+b_j)^{m_j}} \end{equation} Below we decompose the quantity above into partial fractions for $d=3$ using differentiation with respect to the $b$-parameters. We have: \begin{eqnarray} &&{\mathfrak F}_d(x) = \\ &&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_1 \\ l_1 \le l \le m_1 \end{array}} \binom{m_1+m_2-1-l}{m_2-1} \binom{l+m_3-1-l_1}{m_3-1} \frac{(-1)^{m_2+m_3}}{(x+b_1)^{l_1} (b_1-b_3)^{l+m_3-l_1} (b_1-b_2)^{m_1+m_2-l}} + \\ &&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_2 \\ l_1 \le l \le m_2 \end{array}} \binom{m_1+m_2-1-l}{m_1-1} \binom{l+m_3-1-l_1}{m_3-1} \frac{(-1)^{m_1+m_3}}{(x+b_2)^{l_1} (b_2-b_3)^{l+m_3-l_1} (b_2-b_1)^{m_1+m_2-l}} + \\ &&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_3 \\ l_1 \le l \le m_3 \end{array}} \binom{m_2+m_3-1-l}{m_2-1} \binom{l+m_1-1-l_1}{m_1-1} \frac{(-1)^{m_1+m_2}}{(x+b_3)^{l_1} (b_3-b_1)^{l+m_1-l_1} (b_3-b_2)^{m_2+m_3-l}} \end{eqnarray} Now the question is to provide the result for arbitrary $d$.

Answer 1

The result is given below: \begin{eqnarray} &&{\mathfrak F}_d(x) = \sum\limits_{k=1}^d \sum\limits_{\begin{array}{rrr} 1&\le l_{d-2}& \le m_k \\l_{d-2}& \le l_{d-3} &\le m_k\\&\vdots&\\l_1&\le l_0&\le m_k\end{array}} \prod\limits_{j=-1}^{d-3} \binom{l_j + m_{f_{k,j}}-1-l_{j+1}}{m_{f_{k,j}}-1} \frac{1}{\left(b_k-b_{f_{k,j}}\right)^{l_j + m_{f_{k,j}}-l_{j+1}}} \cdot \frac{(-1)^{M-m_k}}{\left(x+b_k\right)^{l_{d-2}}} \end{eqnarray} subject to $l_{-1}= m_k$.

Here $f_{k,j} := (k-2-j) 1_{j\le k-3} + (j+3) 1_{j>k-3}$ and $M= \sum\limits_{j=1}^d m_j$.

As a sanity check we analyze special cases.

Firstly let us take $m_1=\cdots=m_d=1$ then all the $l$-indices are equal to one and we immediately get: \begin{equation} rhs = \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^1} \cdot \frac{(-1)^{d-1}}{(x+b_k)^1} \end{equation} as it should be.

Now let us take $m_1=\cdots=m_d=2$. In this cases there are two cases. (A) $l_{d-2}=2$ then $(l_{d-3},\cdots,l_0,l_{-1})=(2,\cdots,2)$ or (B) $l_{d-2}=1$ then $(l_{d-3},\cdots,l_0,l_{-1})=(1,\cdots,1,2,\cdots,2)$ where the number of one's can be zero but the number of two's has to be strictly positive. This yields: \begin{eqnarray} rhs&=& \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^2} \cdot \frac{(-1)^{2(d-1)}}{(x+b_k)^2} +\\ &&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^2} \cdot \left\{\sum\limits_{j=-1}^{d-3} \frac{2}{b_k-b_{f_{k,j}})^1}\right\} \frac{(-1)^{2(d-1)}}{(x+b_k)^1} \end{eqnarray} as it should be.

Finally we take the case $m_1=\cdots=m_d=3$. Then there are three cases. (A) $l_{d-2}=3$ then $(l_{d-3},\cdots,l_0,l_{-1})=(3,\cdots,3)$ or (B) $l_{d-2}=2$ then $(l_{d-3},\cdots,l_0,l_{-1})=(2,\cdots,2,3,\cdots,3)$ or (C) $l_{d-2}=1$ then $(l_{d-3},\cdots,l_0,l_{-1})=(1,\cdots,1,2,\cdots,2,3,\cdots,3)$. In the last two cases the number of three's has to be strictly positive. This immediately gives: \begin{eqnarray} rhs&=& \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \frac{(-1)^{3(d-1)}}{(x+b_k)^3} +\\ &&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \left\{\sum\limits_{j=-1}^{d-3} \frac{3}{b_k-b_{f_{k,j}})^1}\right\} \frac{(-1)^{3(d-1)}}{(x+b_k)^2}+\\ &&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \left\{ \sum\limits_{-1\le j_1<j_2\le d-3} \frac{3^2}{\prod\limits_{\xi=1}^2 (b_k-b_{f_{k,j_\xi}})^1}+ \sum\limits_{j=-1}^{d-3} \frac{6}{(b_k-b_{f_{k,j}})^2} \right\} \frac{(-1)^{3(d-1)}}{(x+b_k)^1} \end{eqnarray} again as it should be.