# The partial fraction expansion of $\frac{1}{x^n - 1}$

If $n$ is an integer, is there a nice way to write the partial fraction expansion of $\frac{1}{x^n - 1}$? I figure that if $\zeta$ is the $n$-th root of unity, then for some coefficients $a_0, a_1, \ldots, a_{n-1}$ we may write

$$\frac{1}{x^n - 1} = \frac{a_0}{x - 1} + \frac{a_1}{x - \zeta} + \frac{a_2}{x - \zeta^2} + \ldots + \frac{a_{n-1}}{x - \zeta^{n-1}}.$$

Then for $0 \leq i \leq n -1$, $$a_i = \lim_{x \to \zeta^{i}} \frac{x - \zeta^{i}}{x^n - 1} = \frac{1}{(\zeta^{i} - 1) \ldots (\zeta^{i} - \zeta^{i-1}) (\zeta^{i} - \zeta^{i + 1}) \ldots (\zeta^{i} - \zeta^{n - 1})}.$$ Is there a simpler expression for this and if so, how could I see it easily?

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@bla: by the way, if you want to actually start from the expression you've got and turn it into what it should be, divide the denominator by (zeta^i)^{n-1} and note that the result is just the evaluation of the factorization of (x^n-1)/(x-1) at 1. The real lesson to take away here is that since the LHS is invariant under multiplying x by zeta, the RHS must be also. – Qiaochu Yuan Sep 17 '10 at 0:19
More generally, if a rational fraction $f(X)$ is defined at a point $a$, then the polar part of $f(X)/(X-a)^n$ at $a$ is obtained by dividing the order $n-1$ Taylor approximation of $f(X)$ at $a$ by $(X-a)^n$. [The "polar part at $a$" is by definition the contribution of $a$ in the partial fraction decomposition.] – Pierre-Yves Gaillard Sep 17 '10 at 3:48

Use l'Hopital's rule. $a_i = \lim_{x \to \zeta^i} \frac{1}{nx^{n-1}} = \frac{\zeta^i}{n}$. Note that the identity I am claiming is true is equivalent to the identity

$$\sum_{i=0}^{n-1} (\zeta^i)^k = \begin{cases} n \text{ if } n | k \\ 0 \text{ otherwise} \end{cases}$$

by taking the generating function of both sides. More generally, the coefficient of $\frac{1}{x - a}$ in the partial fraction decomposition of $\frac{P(x)}{Q(x)}$ is (as long as $a$ is a simple root of $Q$)

$$\lim_{x \to a} \frac{(x - a) P(x)}{Q(x)} = \frac{P(a)}{Q'(a)}.$$

This formula is tremendously useful (for example I recently used it on this math.SE question) but does not seem to be widely known.

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Actually the 2nd formula is pretty well-known if you consider the limit is the residue of P(x)/Q(x) at a. (At least I find it on Wikipedia and a UG textbook on Mathematics in Physics.) – kennytm Sep 16 '10 at 19:41
So it is. But back in my high school days when we were forced to do many partial fraction expansions I don't recall ever having come across it... – Qiaochu Yuan Sep 16 '10 at 19:46
My sympathies on being subjected to "high school math", wherein there is an artificial need for "tasks" for students, which (perversely) argues against making things as easy as they can be. :) – paul garrett Jul 26 '11 at 18:38