How to do this partial decomposition fraction?

Question

I factorize the equation $(x+m)^n-x^n$ into $\prod_{q=1}^{n-1}(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))$

Then how do I do the partial fraction to $$\frac{1}{\prod_{q=1}^{n-1}(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))}$$

Answer 1

Suppose $f$ is a function which can be written as a product over its roots as $$f(x)=\prod_{f(w)=0}(x-w),$$ where each root $w$ only contributes one term to the product. With the added condition that $f(w)=0\Rightarrow f'(w)\ne 0$, we can show that $$\frac{1}{f(x)}=\sum_{f(w)=0}\frac{1}{f'(w)(x-w)}.$$ This answer will be a proof of that fact.

We start by claiming that $$\frac1{f(x)}=\prod_{f(w)=0}\frac{1}{x-w}=\sum_{f(w)=0}\frac{b(w)}{x-w}$$ for some coefficients $b(w)$. We multiply both sides by $f(x)$, and see that this becomes $$1=\sum_{f(w)=0}b(w)\prod_{f(\phi)=0,\phi\ne w}(x-\phi).$$ Then for any root $z$, we can plug in $x=z$ and get $$1=\sum_{f(w)=0}b(w)\prod_{f(\phi)=0,\phi\ne w}(z-\phi)=b(z)\prod_{f(\phi)=0,\phi\ne z}(z-\phi),$$ because each term in the sum vanishes except for the term $w=z$. Hence $$b(z)=\prod_{f(\phi)=0,\phi\ne z}\frac1{z-\phi}.$$ On the other hand, we have that $$\ln f(x)=\sum_{f(w)=0}\ln(x-w),$$ so that $$\frac{f'(x)}{f(x)}=\sum_{f(w)=0}\frac{1}{x-w}.$$ This is $$f'(x)=\sum_{f(w)=0}\frac{f(x)}{x-w}=\sum_{f(w)=0}\prod_{f(\phi)=0,\phi\ne w}(x-\phi).$$ Again, plug in $x=z$ for any root $z$, and get $$f'(z)=\prod_{f(\phi)=0,\phi\ne z}(z-\phi)=\frac1{b(z)}.$$ Our proof is done.

I will leave it to you to work out the details for applying this formula to your case.

Answer 2

Hint

$$\frac{1}{\prod_{q=1}^{n-1}(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))} = \sum_{q=1}^{n-1}\frac{A_q}{(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))} $$

This gives $$1=\sum_{q=1}^{n-1}A_q\frac{\prod_{q=1}^{n-1}(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))}{(x-\frac{m}{2}(i \cot\frac{\pi q}{n}-1))} $$

Cancel the fractions on the RHS, and then, for $1 \leq j \leq n-1$ plugin $$x=\frac{m}{2}(i \cot\frac{\pi j}{n}-1)$$ With the exception of $q=j$, all the other terms will become $0$.