Solving this system for Partial Fraction Decomposition

Question

I have a rational fraction $\frac{P(x)}{Q(x)}$ and would transform it into a sum of separate fractions. I know that $\{a_n\}$ is the set of the roots of $Q(x)$ which is of grade $t$, so it has exactly $t$ roots. On the other side, $P(x)=x^{t-1}$, so we should be able to decompose the fraction into a sum of $t-1$ fractions $\frac{C_n}{1-a_n^{-1}x}$ where $C_n$ are constants. Formally:

$$ \frac{x^{t-1}}{\prod_{a\in\{a_n\}}(1-a^{-1}x)}=\sum_{n=0}^{t-1}\frac{C_n}{1-a_n^{-1}x} $$

Now my question is, how to calculate the coefficients $C_n$?

This is what I've made so far: Setting $x=0$ we have that $\sum C_n=1$. Setting $x=m\in\mathbb{N}$ we have $\sum_{n=0}^{t-1} \frac{1}{1-a_n^{-1}m}C_n=\frac{m^{t-1}}{\prod_{a\in\{a_n\}}(1-a^{-1}m)}$. We can thus set up a system of $t-1$ linear equations:

$$ \begin{cases} &C_1&+&C_2&+&\ldots&+&C_{t-1}&=0&\\ \frac{1}{1-a_0^{-1}}&C_1&+\frac{1}{1-a_1^{-1}}&C_2&+&\ldots&+\frac{1}{1-a_{t-1}^{-1}}&C_{t-1}&=\frac{1}{\prod1-a^{-1}}&\\ \frac{1}{1-2a_0^{-1}}&C_1&+\frac{1}{1-2a_1^{-1}}&C_2&+&\ldots&+\frac{1}{1-2a_{t-1}^{-1}}&C_{t-1}&=\frac{1}{\prod1-2a^{-1}}&\\ &\vdots&&\vdots&&\vdots&&\vdots&&\vdots\\ \frac{1}{1-(t-1)a_0^{-1}}&C_1&+\frac{1}{1-(t-1)a_1^{-1}}&C_2&+&\ldots&+\frac{1}{1-(t-1)a_{t-1}^{-1}}&C_{t-1}&=\frac{1}{\prod1-(t-1)a^{-1}}& \end{cases} $$

I made some computer calculations which suggest that $C_n=\prod_{0\leq k\leq t-1\land k\neq n}a_n\frac{1}{a_n-a_k}$, but I don't know how to prove it formally.

Thanks.

Answer 1

Let me use an example with $t=6$.

$$\frac {x^5}{\prod_{n=1}^5 \left(1-\frac{x}{a_n}\right)}=\sum_{n=1}^5 \frac {c_n}{1-\frac{x}{a_n}}$$

Cross multiply to have $$x^5=\Bigg[\sum_{n=1}^5 \frac{c_n}{1-\frac{x}{a_n}}\Bigg]\,\times\, \Bigg[\prod_{n=1}^5\frac{1}{1-\frac{x}{a_n}}\Bigg]$$

Now, take the $\color{red}{\text{limit}}$ of the rhs when $x\to a_1$. You will have $$a_1^5=\frac{\prod_{n=2}^5 (a_i-a_1)}{\prod_{n=2}^5 a_i } c_1\qquad\implies\qquad c_1=a_1^5 \,\prod_{n=2}^5 \frac {a_i}{a_i-a_1}$$

Now, make it general (this is simple - you just need to exclude one term).

Hoping that this helps.