I can decompose
$${1\over(x-a)(x-b)} = {1\over(a-b)}({1\over(x-a)}-{1\over(x-b)}) = {1\over(a-b)}({x-b-x+a\over(x-a)(x-b)}) = {a-b \over (a-b)(x-a)(x-b)}$$
and
$${x\over(x-a)(x-b)} = {1\over(a-b)}({a\over(x-a)}-{b\over(x-b)})$$
however fail to decompose $$x^2\over (x-a)(x-b)$$ because standard procedure (substituting $x\to a$ and $x\to b$) produces
$${1\over(a-b)} ({a^2\over x-a} - {b^2\over x-b}) = {1\over(a-b)} ({a^2x - ba^2 - xb^2 + ab^2 \over (x-a)(x-b)}) = {1\over(a-b)} ({x(a^2 - b^2) - ab(a-b) \over (x-a)(x-b)}) = {x(a+b) - ab \over (x-a)(x-b)}$$
This surprisingly implies that $x^2 = x(a+b)-ab$.
I realized that failure is caused by degree of nominator reaching the degree of denominator. I think that this is not an issue particularly for me since I can factor out the $x^n$ as long as I am finding generating function power series coefficients and x just stands for shift. But I am still curious to know
- why
dont tutorials discuss the root of the failure(though you can answer this also) - is the failure at higher degrees?
- Is it related to why are there more poles than zeroes?
- What is the point of $x^2 = x(a+b)-ab$?
Most simply put if you have $$\frac A{x-a}+\frac B{x-b}$$ and add these together, you get $${(A+B)x-aB-bA\over(x-a)(x-b)}.$$ So such a decomposition can only produce a numerator whose deg is at most 1.