Fraction decomposition, how to find coefficients?

Question

How to decompose? Can anybody give a hint $\frac{1}{(x^n-1)^2}$.
$\frac{1}{(x^n-1)^2}=\sum_{i=1}^{n}\frac{a_{i}}{x-\zeta_{i}}+\sum_{i=1}^{n}\frac{b_{i}}{(x-\zeta_{i})^2}$.
$1=\sum_{i=1}^{n}\frac{a_{i}(x^n-1)^2}{x-\zeta_{i}}+\sum_{i=1}^{n}\frac{b_{i}(x^n-1)^2}{(x-\zeta_{i})^2}=f(x)$.
Then if $x=\zeta_{j}$
$1=f(\zeta_{j})=b_{j}(\zeta_{j}-\zeta_{k_{1}})^2(\zeta_{j}-\zeta_{k_{2}})^2...(\zeta_{j}-\zeta_{k_{n-1}})^2$, where $\zeta_{j}\neq \zeta_{k_{i}}$, so I can find $b_{j}$.
What about $a_{j}$?

Answer 1

Here we use that the partial fraction decomposition of a rational function is given by the sum of the principal parts of the Laurent series expansions at the poles. More precisely we have the following Theorem:

Partial Fractions Decomposition Theorem: Suppose that $R(z)/D(z)$ is a rational function with degree of $R$ less than the degree of $D$. Denote by $r_j$ the distinct roots of the denominator $D$ and $m_j$ their multiplicities. Let $S_j(z)$ denote the principal part of the Laurent expansion of $R/D$ at the root $r_j$. The singular point $r_j$ is either removable or a pole of order $\leq m_j$ and \begin{align*} \frac{R(z)}{D(z)}=\sum_{j}S_j(z) \end{align*} See for instance this paper.

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Denoting OPs function by $f_n$ we have \begin{align*} f_n(z)&=\frac{1}{\left(z^n-1\right)^2}=\prod_{k=0}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\qquad\qquad \zeta_k^n=1, 0\leq k\leq n-1 \end{align*} where $\zeta_k$ are the $n$-th roots of unity, where $0\leq k\leq n-1$.

We show the following is valid \begin{align*} \color{blue}{f(z)}&\color{blue}{=\frac{1}{\left(z^n-1\right)^2}}\\ &\color{blue}{=\sum_{q=0}^{n-1}\left(\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_k\right)^2}\right) \left(\frac{1}{\left(z-\zeta_q\right)^2} -2\left(\sum_{{l=0}\atop{l\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_l\right)}\right)\frac{1}{z-\zeta_q}\right)}\tag{1} \end{align*}

We start with the calculation of the coefficient of $\frac{1}{z-\zeta_q}$ by calculating the residue of $f_n$ at $\zeta_q$. We obtain \begin{align*} \color{blue}{\mathrm{Res}\left(f,\zeta_q\right)}&=\lim_{z\to \zeta_q}\frac{d}{dz}\left(\left(z-\zeta_q\right)^2\prod_{k=0}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\right)\\ &=\lim_{z\to \zeta_q}\frac{d}{dz}\left(\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\right)\tag{2}\\ &=\lim_{z\to \zeta_q}\sum_{{l=0}\atop{l\ne q}}^{n-1}\left(\frac{d}{dz}\left(\frac{1}{\left(z-\zeta_l\right)^2}\right)\right) \prod_{{k=0}\atop{{k\ne q}\atop{k\ne l}}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\tag{3}\\ &=\lim_{z\to \zeta_q}\sum_{{l=0}\atop{l\ne q}}^{n-1}\frac{-2}{\left(z-\zeta_l\right)^3} \prod_{{k=0}\atop{{k\ne q}\atop{k\ne l}}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\\ &\,\,\color{blue}{=\sum_{{l=0}\atop{l\ne q}}^{n-1}\frac{-2}{\left(\zeta_q-\zeta_l\right)} \prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}}\tag{4}\\ \end{align*} which is the coefficient of $\frac{1}{z-\zeta_q}$ in (1).

Comment:

• In (2) we cancel the factor $\left(z-\zeta_q\right)^2$.

• In (3) we apply a generalized Leibniz product rule. Recall that for instance \begin{align*} \left(fgh\right)^{\prime}=f^{\prime}gh+fg^{\prime}h+fgh^{\prime} \end{align*}

• In (4) we merge $\frac{1}{\left(z-\zeta_q\right)^2}$ into the product.

We now obtain the coefficient of $\frac{1}{\left(z-\zeta_q\right)^2}$ by calculating the residue of $(z-\zeta_q)f_n$ at $\zeta_q$. We get \begin{align*} f_q(z)&=f(z)\left(z-\zeta_q\right)=\frac{1}{z-\zeta_q}\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\\ \color{blue}{\mathrm{Res}\left(f,\zeta_q\right)}&=\lim_{z\to \zeta_q}\left(z-\zeta_q\right)f_q(z) =\lim_{z\to \zeta_q}\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(z-\zeta_k\right)^2}\\ &\,\,\color{blue}{=\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_k\right)^2}}\tag{5} \end{align*} which is the coefficient of $\left(\frac{1}{z-\zeta_q}\right)^2$ in (1).

We obtain from (5) and (4) \begin{align*} \color{blue}{f_n(z)}&=\frac{1}{\left(z^n-1\right)^2}\\ &=\sum_{q=0}^{n-1}\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_k\right)^2}\frac{1}{\left(z-\zeta_q\right)^2}\\ &\qquad+\sum_{q=0}^{n-1}\sum_{{l=0}\atop{l\ne q}}^{n-1}\frac{-2}{\left(z-\zeta_l\right)} \prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_k\right)^2}\frac{1}{\left(z-\zeta_q\right)}\\ &\,\,\color{blue}{=\sum_{q=0}^{n-1}\left(\prod_{{k=0}\atop{k\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_k\right)^2}\right) \left(\frac{1}{\left(z-\zeta_q\right)^2} -2\left(\sum_{{l=0}\atop{l\ne q}}^{n-1}\frac{1}{\left(\zeta_q-\zeta_l\right)}\right)\frac{1}{z-\zeta_q}\right)} \end{align*} and the claim (1) follows.

Example: $n=3$:

Let's check (1) with a small value $n=3$. We consider \begin{align*} f_3(z)&=\frac{1}{\left(z^3-1\right)^2} \end{align*} with the poles \begin{align*} \zeta_0=1, \quad &\zeta_1=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\\ &\zeta_2=\overline{\zeta_1}=-\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{align*} We obtain according to (1) \begin{align*} \color{blue}{f_3(z)}&=\sum_{q=0}^{2}\left(\prod_{{k=0}\atop{k\ne q}}^2\frac{1}{\left(\zeta_q-\zeta_k\right)^2}\right) \left(\frac{1}{\left(z-\zeta_q\right)^2}-2\sum_{{l=0}\atop{l\ne q}}^2\frac{1}{\left(\zeta_q-\zeta_l\right)}\frac{1}{\left(z-\zeta_q\right)}\right)\\ &=\frac{1}{\left(\zeta_0-\zeta_1\right)^2\left(\zeta_0-\zeta_2\right)^2} \left(\frac{1}{\left(z-\zeta_0\right)^2}-2\left(\frac{1}{\zeta_0-\zeta_1}+\frac{1}{\zeta_0-\zeta_2}\right)\frac{1}{z-\zeta_0}\right)\\ &\qquad+\frac{1}{\left(\zeta_1-\zeta_0\right)^2\left(\zeta_1-\zeta_2\right)^2} \left(\frac{1}{\left(z-\zeta_1\right)^2}-2\left(\frac{1}{\zeta_1-\zeta_0}+\frac{1}{\zeta_1-\zeta_2}\right)\frac{1}{z-\zeta_1}\right)\\ &\qquad+\frac{1}{\left(\zeta_2-\zeta_0\right)^2\left(\zeta_2-\zeta_1\right)^2} \left(\frac{1}{\left(z-\zeta_2\right)^2}-2\left(\frac{1}{\zeta_2-\zeta_0}+\frac{1}{\zeta_2-\zeta_1}\right)\frac{1}{z-\zeta_2}\right)\\ &\,\,\color{blue}{=\frac{1}{9}\left(\frac{1}{(z-1)^2}-\frac{2}{z-1}\right)}\\ &\qquad\color{blue}{-\frac{1+i\sqrt{3}}{18}\left(\frac{1}{\left(z+\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right)^2} +\frac{1+i\sqrt{3}}{z+\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}\right)}\\ &\qquad\color{blue}{-\frac{1-i\sqrt{3}}{18}\left(\frac{1}{\left(z+\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right)^2} +\frac{1-i\sqrt{3}}{z+\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}\right)} \end{align*} which was verified with the help of Wolfram Alpha.

Note: The validity of the Partial Fractions Decomposition Theorem is an immediate consequence of Liouvilles Theorem. A special case dealing with simple poles is considered here.