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Does anybody know the partial fraction decomposition of $$ \prod_{j=1}^{N}\ \frac{1}{(x-a_{j})^{n_{j}}} $$ with all $a_{j}$ different and $n_j$ positive integers? I know you can get it with the residue but the differentiation is awkward.

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I would try $$\sum_{j=1}^N \sum_{i=1}^{n_j} \frac{c_{i,j}}{(x-a_j)^i}$$ and do various things (substituting $x = a_j$ and differentiating some number of times) to determine the $c_{i,j}$. – marty cohen Jan 27 '12 at 3:11
To be honest, I've tried it a little bit and I feel like I can get some information about the fraction decomposition, but I don't feel like the explicit formula is gonna be beautiful. Perhaps if you want to do that for the sole purpose of not needing to compute it in the future, you should just get used to computing. If this is for a proof purpose, then I can't do much. – Patrick Da Silva Jan 27 '12 at 3:40
After looking through this, I don't see how it's possible. (In the sense that you can't move this to the form where each fraction has only one factor) I'm likely wrong. I'd like to see if this could be done and how. – 000 Jan 27 '12 at 23:04
up vote 2 down vote accepted

New answer

If we write the partial fraction decomposition as $$ \prod_{j=1}^N\ \frac{1}{(x-a_j)^{n_j}} =\sum_{j=1}^N\ \sum_{p=0}^{n_j-1}\ \frac{c_{j,p}}{(x-a_j)^{n_j-p}}\quad, $$ then we have $$ c_{j,p}=(-1)^p\sum_{u\in S(j,p)}\ \prod_{k\neq j}\ \binom{n_k-1+u_k}{n_k-1}\ \frac{1}{(a_j-a_k)^{n_k+u_k}}\quad, $$ where $S(j,p)$ is the set of those maps $$ u:\{1,\dots,N\}\setminus\{j\}\to\mathbb N,\quad k\mapsto u_k $$ which satisfy $$ \sum_{k\neq j}\ u_k=p. $$ To prove this, it suffices to observe that, for $1\le j\neq k\le N$, the degree less than $n_j$ Taylor polynomial of $$ \frac{1}{(x-a_k)^{n_k}} $$ at $a_j$ is $$ \sum_{q=0}^{n_j-1}\ \binom{n_k-1+q}{n_k-1}\ \frac{(-1)^q}{(a_j-a_k)^{n_k+q}}\ (x-a_j)^q\quad. $$

Old answer

An expression for the partial fraction decomposition is obtained by combining Robert's answer with Theorem $11$ p. $8$ in this pdf document, accessible from this html page.

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Wow yes, this is what I was looking for: the actual explicit form. But could you write down the expression for my case without me having to become an expert myself? Thanks! – lcv Jan 28 '12 at 3:54
Dear @lcv: I tried to do comply with your kind request. By the way, I upvoted your question. – Pierre-Yves Gaillard Jan 28 '12 at 7:50
many thanks for taking the time to write down the formula explicitly. I must admit in the meantime I got it myself "by differentiating". In any case thank you for teaching me the connection with the Chinese Remainder Theorem. I will cite you in my work. (Btw I accepted and upvoted your answer) – lcv Jan 31 '12 at 2:03
@lcv - Dear Lorenzo: Thank you! I see that you're a new user. Welcome to MSE! – Pierre-Yves Gaillard Jan 31 '12 at 6:17

Let $P(x) = \prod_{j=1}^N (x-a_j)^{n_j}$, and for each $k \in \{1 \ldots N\}$ let $U_k(x) = P(x)/(x-a_k)^{n_k}$ be the product of the terms not involving $x-a_j$. The coefficient of $(x-a_k)^{-m}$, where $1 \le m \le n_k$, is the coefficient of $t^{n_k - m}$ in the Maclaurin series of $1/U_k(t+a_k)$.

EDIT: In the case $N=2$, $1/U_1(t+a_1) = (t+a_1 - a_2)^{-n_2}$ so the coefficient of $(x-a_1)^{-m}$ is ${n_1 + n_2 - m - 1 \choose n_2 - 1} \frac{(-1)^{n_2}}{(a_2 - a_1)^{n_1 + n_2 - m}}$ for $1 \le m \le n_1$ (and similarly for the coefficient of $(x -a_2)^{-m}$ with $1$ and $2$ interchanged). The general case can be reduced to this one, since if $$\frac{1}{\prod_{j=1}^{N-1} (x-a_j)^{n_j}} = \sum_{j=1}^{N-1} \sum_{k=1}^{n_j} \frac{b_{j,k}}{(x - a_j)^k}$$ $$\frac{1}{\prod_{j=1}^N (x-a_j)^{n_j}} = \sum_{j=1}^{N-1} \sum_{k=1}^{n_j} \frac{b_{j,k}}{(x-a_j)^k (x-a_N)^{n_N}}$$

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Sure, but I think that's just moving the problem elsewhere. – Patrick Da Silva Jan 27 '12 at 5:30

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