Proof of Heaviside cover-up method

Question

Can someone please provide me with a proof of the heaviside cover-up method for computing coefficients of the partial fraction expansion of a rational function. I understand that it's related to the residue, but for higher order poles I fail to see the link between the residue and the coefficient for say $c_{n}$ for: $$ \frac{c_n}{(z-z_{0})^n} $$

A link to an existing proof would also be much appreciated (I can't seem to find one online).

EDIT: if you downvote please explain, I do believe that having a proof for this benefits not only me but the larger community in general. Given that it does not seem to be available online.

Answer 1

An example will illustrate it:

$$\frac{1}{x(x-1)(x-2)}=\frac Ax + \frac{B}{x-1} +\frac{C}{x-2}$$

For A we can multiply by $x$ on each side and take the limit as $x$ goes to $0$:

$$\lim_{x \to 0}\frac{1}{(x-1)(x-2)}=\lim_{x \to 0}(A+\frac{B}{x-1}x+\frac{C}{x-2}x)$$

See that every term in the right will go to $0$, except the one that had x in the denominator. Same goes for calculating B and C, but replacing $1$ and $2$ respectively. See that this method only works for calculating the coefficient of the greatest power of each term. For example:

$$\frac{1}{x^3(x-1)}=\frac Ax + \frac{B}{x^2} +\frac{C}{x^3}+\frac{D}{x-1}$$ We can proceed as before to calculate C, multiplying each side by $x^3$ and taking the limit, but see that this is not possible for $x^2$ and $x$. Hope this helps.

Answer 2

Suppose this https://en.wikipedia.org/wiki/Partial_fraction_decomposition is what you were/are looking for.

I don't expect this is a formal proof, but it might help to understand.

Let $f(z)=\frac{c_n}{(z-z_{0})^n}$ and $g(z)=(z-z_{0})^{n-1}f(z)=\frac{c_n}{z-z_{0}}$ so the higher-order pole for $f(z)$ is converted into a simple pole for $g(z)$. $c_n=\operatorname{Res}(f,z_0)=\frac{1}{(n-n)!}\lim_{z\to z_0}\frac{d^{n-n}}{dz^{n-n}} \left[(z-z_0)^n f(z) \right]=\operatorname{Res}(g,z_0)=\lim_{z\to z_0}(z-z_0)g(z)$