Decomposition of a rational function

Question

I want to decompose the rational function

$$ \cfrac{P(s)}{Q(s)}=\cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=1}^n (s+b_i)} $$

where $a_i>0$ for every $i=1,\dots,m$, $b_i>0$ for every $i=1,\dots,n$ and $n>m$.

In other words, I'm looking for coefficients $x_j$, $j=1\dots,n$, such that

$$ \cfrac{P(s)}{Q(s)}=\cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n}\quad\Longleftrightarrow\quad \sum_{j=1}^n x_j \prod_{\substack{i=1\\i\neq j}}^n (s+b_i) = \prod_{i=1}^m (s+a_i) $$

Making some attempts with Mathematica for low values of $m$ and $n$ it seems that the $x_j$ are given by

$$ x_j=\cfrac{\prod\limits_{i=1}^m (a_i-b_j)}{\prod\limits_{\substack{i=1\\i\neq j}}^n (b_i-b_j)} $$

So my questions are:

1) Can I say beforehand that there exist unique such $x_j$s?

2) How can I prove that $x_j$ has in general (as it seems) the form above?

Answer 1

These are real (or complex) polynomials? This is known as "partial fractions", perhaps first seen by the typical student in elementary calculus courses.

(1) Suposing that your $P(s)/Q(s)$ in lowest terms, the necessary and sufficient condition that there exists $x_1,\dots,x_n$ such that $$ \cfrac{P(s)}{Q(s)}=\cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} $$ Is: $m

(2) If such $x_j$ exist, they are unique.
Let's show $x_1$ is unique. Suppose $$ \cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} =\cfrac{u_1}{s+b_1}+\dots+\cfrac{u_n}{s+b_n}. $$ Multiply by $s+b_1$, then take the limit as $s \to -b_1$. Result: $x_1=u_1$. Uniqueness for the others is the same.

(3) They have the form you describe.
Let's prove it for $x_1$. We have $$ \cfrac{x_1}{s+b_1}+\dots+\cfrac{x_n}{s+b_n} = \cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=1}^n (s+b_i)} $$ Multiply by $s+b_1$ to get $$ x_1+ (s+b_1)\left(\cfrac{x_2}{s+b_2}+\dots+\cfrac{x_n}{s+b_n}\right) = \cfrac{\prod\limits_{i=1}^m (s+a_i)}{\prod\limits_{i=2}^n (s+b_i)} $$ Take the limit as $s \to -b_1$ [remembering that $b_j \ne b_1$ for all other $b_j$ ] $$ x_1 + 0 = \cfrac{\prod\limits_{i=1}^m (-b_1+a_i)}{\prod\limits_{i=2}^n (-b_1+b_i)} $$ Does that match your claimed formula?