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.
Partial fraction of $\prod_{j=1}^{N}\frac{1}{(x-a_{j})^{n_{j}}}$
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0After looki$n$g 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. – 2012-01-27
2 Answers
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 (Internet Archive), accessible from this html page (Internet Archive).
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0@lcv - Dear Lorenzo: Thank **you**! I see that you're a new user. Welcome to MSE! – 2012-01-31
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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2Sure, but I think that's just moving the problem elsewhere. – 2012-01-27