Multiply both sides by $p(x)^v$. We get
$$r(x)=r_1(x)p(x)^{v-1}+r_2(x)p(x)^{v-2}+\ldots+r_{v-1}(x)p(x)^{1}+r_v(x).$$
Since $\text{deg}(r_v)<\text{deg}(p)$ we get that $r_v$ is the remainder of the division of $r(x)$ by $p(x)$. Notice that
$$\frac{r(x)-r_v(x)}{p(x)^v}=r_1(x)p(x)^{v-2}+r_2(x)p(x)^{v-3}+\ldots+r_{v-1}(x)$$
is the quotient. So, $r_{v-1}$, $r_{v-2}$, ... can be found in a similar way, by computing the remainder by division by $p(x)$ of this quotient.
$\color{brown}{\textbf{Alternative approach.}}$
From the given equation
$$G(s)=\sum\limits_{i=0}^m\dfrac{c_i}{s-p_i} = h\;\dfrac{\prod\limits_{j=1}^n(s-z_j)}
{\prod\limits_{i=1}^m(s-p_i)},\quad n\le m-1,\tag1$$
immediately should
- $$\text{If}\quad h_1= \lim\limits_{\large s\to \infty}sG(s) = \sum\limits_{i=1}^mc_i\not=0,\quad\text{then}\quad \dbinom nh = \dbinom {m-1}{h_1},\tag{2.1}$$
- $$\text{Else if}\quad h_2= \lim\limits_{\large s\to \infty}s^2G(s) \not=0,\quad\text{then}\quad \dbinom nh = \dbinom {m-2}{h_2},\dots\tag{2.2}$$
Also, easily to check the expression for the residue $\;c_k\;$ of $k-$th pole $p_k$ in the form of
$$c_k =\lim\limits_{\large s\to p_k}(s-p_k)G(s) = h\;\dfrac{\prod\limits_{j=1}^n(p_k-z_j)}
{\prod\limits_{i=1}^m(p_k-p_i)^{\large1-\delta_{\large ki}}},\quad (k=1\dots m),\tag3$$
or
$$\prod\limits_{j=1}^n(p_k-z_j) = b_k,\quad (k=1\dots m),\tag4$$
where
$$b_k = \dfrac{c_k}{h}\prod\limits_{i=1}^m(p_k-p_i)^{\large1-\delta_{\large ki}}\tag5$$
are the constants,
and expression with the Kronekker symbol $\delta$ in $(3),(5)$ means the production of non-zero differences.
Solving of the system $(4)$ does not contain calculating the numerator polynomial and looks as the required alternative approach.
If some of the roots are known, then the model order decreases.
$\color{brown}{\textbf{Simple example.}}$
Let us consider the case $\quad\mathbf{ m=3,\; p_1=-1,\; p_2=0,\; p_3=1,\; c_i =1.}\quad$
$$G(s)=\dfrac1{s+1} + \dfrac1s+\dfrac1{s-1}\quad \left(=\dfrac{3s^2-1}{s^3-s}\right),$$
Then $\;h_1=\lim\limits_{s\to\infty} sG(s) =3\;\Rightarrow\;
\mathbf{h=3,\;n=2},$
and from $(1)$ should
$$G(s) = 3\dfrac{(s-z_1)(s-z_2)}{(s+1)s(s-1)},$$
$$\lim\limits_{s \to-1} (s+1)\left(\dfrac{c_1}{s+1} + \dfrac{c_2}s+\dfrac{c_3}{s-1}\right)
= \lim\limits_{s \to-1} 3\;\dfrac{(s-z_1)(s-z_2)}{s(s-1)},$$
$$(-1-z_1)(-1-z_2) = b_1,$$
$$\mathbf{b_1 = \dfrac{(-1)(-1-1)}3 = \dfrac23},$$
$$b_2 = \dfrac{(0+1)(0-1)}3 = -\dfrac13,\quad b_3 = \dfrac{(1+1)(1+0)}3 = \dfrac23,$$
$$\begin{cases}
(-1-z_1)(-1-z_2) = \dfrac23\\
(-z_1)(-z_2) = -\dfrac13\\
(1-z_1)(1-z_2) = \dfrac23,
\end{cases}\;\Rightarrow\;
\begin{cases}
z_1z_2+(z_1+z_2) + 1 = \dfrac23\\
z_1z_2 = -\dfrac13\\
z_1z_2 -(z_1+z_2)+1 = \dfrac23,
\end{cases}\tag6
$$
$$z_1=\dfrac1{\sqrt3},\quad z_2 = -\dfrac1{\sqrt3}.$$
Therefore, proposed approach works.
$\color{brown}{\textbf{Solving.}}$
Solution of the system $(6)$ can be obtained analytically.
Also, this system can be solved as the optimization task in the form of
$$\vec z = \text{ argmin } \{((-1-z_1)(-1-z_2) - \,^2/_3)^2
+ ((-z_1)(-z_2) + \,^1\!/_3)^2
+((1-z_1)(1-z_2) - \,^2/_3)^2 \}.$$
If $n\ge3,$ then attempts of the analytical solving of $(4)$ possibly lead to the same numerator polynomial. So the numeric solutions (such as the gradient descend method) can be recommended. Multivariable model allows to avoid the roots mixing via their resorting after each iteration step.
The alternative approach is the searching of the roots in the intervals between the poles via the dihotomy algorithm (without derivatives using). In this way, algorithm should use sufficient set of the initial points to obtain exactly $n$ roots.
At last, can be used combined algorithm, which uses
- simple dihotomy to obtain some real roots between the poles,
- the simplified model $(4)$ for the rest of roots in the complex presentation.
Good luck!
Best Answer
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?