general form of coefficients in partial fraction decomposition

Question

Given $$ \frac{P(s)}{Q(s)} = \frac{A_{1}}{s-r_{1}} + \dots + \frac{A_n }{s-r_n},$$ where $P(s)$ is a polynomial with degree less than $n$ and $Q(s)$ is a polynomial with degree $n$ and with $r_1 ,\dots , r_n$ distinct zeros.

I'm trying to show that $$ A_k = \frac{P(r_k)}{Q'(r_k)} , \quad k = 1 , \dots , n,$$ with which I'm having trouble.

My attempt is as follows , \begin{gather} \frac{P(s)}{Q(s)} = \frac{A_1 \sum_{j \neq 1}^{n}(s-r_j) + \dots + A_n \sum_{j \neq n}^{n}(s-r_j)}{\prod\limits_{j\le n} (s-r_j)} = \frac{\sum_{i=1}^{n} A_i \left(\sum_{j\neq i}^{n} (s-r_j)\right)}{\prod\limits_{j\le n} (s-r_j)} \\ \frac{P(s)}{Q'(s)} = \frac{\sum_{i=1}^{n} A_i \left(\sum_{j\neq i}^{n} (s-r_j)\right)}{\sum_{j=1}^{n+1} (s-r_j)' \prod\limits_{1 \le i \le n+1 \\ \ \ \ \ \ j \neq i}(s-r_i)} \end{gather} Honestly I'm not even sure what I'm doing at this point, I don't know if I'm taking the right avenue, there must be another elegant way. Any tips and hints would be appreciated.

Answer 1

Multiply both sides by $s-r_k$.

Then it follows that $$A_k=\lim_{s\to r_k}(s-r_k)\frac{P(s)}{Q(s)}=\frac{P(r_k)}{Q'(r_k)}.$$

Can you fill in the details?

Answer 2

Another way, entirely algebraic, is to look at the polynomials

$$L_k(z)=\frac{\prod_{1\leq i\leq n\\i\neq k}(z-r_i)}{\prod_{1\leq i\leq n\\i\neq k}(r_k-r_i)}$$

These are Lagrange interpolating polynomials for the points $r_k$.

Note that $L_k(r_j)=\begin{cases}0,&j\neq k\\1,&j=k\end{cases}$ and they have degree $n-1$. Call these formulas (2).

Therefore, $\sum_{k=1}^nP(r_k)L_k(z)$ is a polynomial of degree $\leq n-1$ and equal to $P$ at the $n$ points $r_1,r_2,...,r_n$. Therefore, it must be equal to $P$.

Remember that $$Q'(z)=\sum_{k=1}^n\prod_{1\leq i\leq n\\i\neq k}(z-r_i)$$ This is just applying the product rule for the derivative.

You see that $$Q'(r_k)=\prod_{1\leq i\leq n\\i\neq k}(r_k-r_i)$$

To see uniqueness, assume that $A_k$ and $A_k'$ are two sets of constants that satisfy your decomposition. Subtracting both equation you get that $$0=\sum_{k=1}^{n}\frac{A_k-A_k'}{(z-r_k)}=\sum_{k=1}^{n}(A_k-A_k')L_k(z)$$

Using (2) and evaluating at $z=r_k$ we get that $0=A_k-A_k'$.