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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

$$ \frac1{x^n - 1} = \frac{a_0}{x - 1} + \frac{a_1}{x - \zeta} + \frac{a_2}{x - \zeta^2} + \ldots + \frac{a_{n-1}}{x - \zeta^{n-1}}. $$

Then for $0 \leq i \leq n -1$, $$ a_i = \lim_{x \to \zeta^i} \frac{x - \zeta^i}{x^n - 1} = \frac1{(\zeta^{i} - 1) \cdots (\zeta^i - \zeta^{i-1}) (\zeta^i - \zeta^{i + 1}) \cdots (\zeta^i - \zeta^{n-1})}. $$ 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.2010-09-17
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    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.]2010-09-17

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