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Consider the following expression: $$\frac{x^n}{x^m+k},$$ for non-negative integers $n$ and $m$, $m>n$, and $k\in\mathbb{C}$. For $k=0$ the expression clearly simplifies to $x^{n-m}$. For $|k|>0$ we have the following identity:

$$\frac{x^n}{x^m+k}=\sum_{p=1}^m\frac{c_p^n}{(x-c_p)\prod_{q\neq p}(c_p-c_q)},$$ where the product runs from $q=1$ to $m$ but skips $p$. We define $c_j$ by $$c_j:=\exp\left[\frac{1}{m}(2\pi \text{i} j+\text{i}\text{Arg}(-k)+\log|k|)\right]. $$ How can we prove the identity?

Addendum:

A simpler decomposition is the following, as suggested and proven by achille hui in a comment below: $$\frac{x^n}{x^m+k}=\frac{-1}{mk}\sum_{p=1}^m\frac{c_p^{n+1}}{x-c_p},$$ enjoy!

Martin Sleziak
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SDiv
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    The identity you tried to prove/use is more complicated than necessary. A much simpler decomposition is $$\frac{x^n}{x^m + k} = \sum_{p=1}^m \left.\frac{y^n}{(y^m + k)'( x - c_p )}\right|{y=c_p} = \sum{p=1}^m \left.\frac{y^n}{m y^{m-1}( x - c_p )}\right|{y=c_p} = -\frac{1}{mk} \sum{p=1}^n \frac{c_p^{n+1}}{x-c_p} $$ In any event, $\prod_{q\ne p} (c_p - c_q)$ is $(y^m+k)' = m y^{m-1}$ evaluated at $y = c_p$. – achille hui Oct 08 '14 at 10:32
  • @achille hui Thanks for a great comment! I was wondering if a simplification was possible and clearly it is! I will add it to the question. – SDiv Oct 08 '14 at 12:03
  • @achillehui. Nice remark. Just a small correction: the exponent of $c_p$ in the result is $n-m+1$ not $n+1$. – Tom-Tom Oct 08 '14 at 12:46
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    @Tom-Tom. achille hui has replaced $c_p^m$ with -k. – SDiv Oct 08 '14 at 12:54
  • @SDIv. Oh... that's true. Thanks ! – Tom-Tom Oct 08 '14 at 13:19

2 Answers2

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First remark that the polynomial $X^m+k$ has simple roots. Let us call the roots $c_j$ for $1\leq j\leq m$. Therefore $x^n/(x^m+k)$ rewrites as a sum of simple elements $$\frac{x^n}{x^m+k}=\sum_{j=1}^m \frac{a_j}{x-c_j}.$$ To compute the values of $a_j$, multiply by $(x-c_j)$ and take the limit $x\to c_j$. The right-hand side goes to $a_j$ as $x$ goes to $c_j$ while the left hand side goes to $$\frac{c_j^n}{\displaystyle\prod_{\substack{1\leq p\leq m\\k\neq j}}(c_j-c_p)}.$$ It remains to prove that $$c_j=|k|^{1/m}\,\exp\left(j\frac{2\pi \mathrm i}m+\mathrm i\frac{\arg(-k)}m\right).$$ It is clear that the $c_j$'s are distinct since they have distinct arguments. Take now $c_j^m$ and you find $-k$.

Tom-Tom
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One method for proving the identity is the following and is based on answers given in the following posts: post 1 and post 2.

  1. Note the following: $$\sum_{p=1}^m\frac{c_p^n}{(x-c_p)\prod_{q\neq p}(c_p-c_q)}\\ = \sum_{p=1}^m\frac{c_p^n\prod_{q\neq p}(x-c_q)}{\prod_{q=1}^m(x-c_q)\prod_{q\neq p}(c_p-c_q)}\\ = \frac{1}{\prod_{p=1}^m(x-c_p)}\sum_{p=1}^m \left ( c_p^n\prod_{q\neq p}\frac{(x-c_q)}{(c_p-c_q)} \right )$$
  2. As shown in post 1, $x^m+k=\prod_{p=1}^m(x-c_p),$ where the set of all $c_j$ are the mth roots of -k.
  3. As shown in post 2, $$x^n=\sum_{p=1}^m \left ( c_p^n\prod_{q\neq p}\frac{(x-c_q)}{(c_p-c_q)} \right ) $$
  4. Then $$\frac{1}{\prod_{p=1}^m(x-c_p)}\sum_{p=1}^m \left ( c_p^n\prod_{q\neq p}\frac{(x-c_q)}{(c_p-c_q)} \right )=\frac{x^n}{\prod_{p=1}^m(x-c_p)}=\frac{x^n}{x^m+k} \Box. $$
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SDiv
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