Series expantion of a function around an undefined point

Given a function which is undefined in at least one point, such as

$$f(x) = \frac{x^2}{(x-a)(x-b)},$$

how do you find the series expansion about that undefined point? The problem with Taylor expanding about these points is that $f(a)$, or $f(b)$, is obviously undefined. Moreover, for this function at least, the problem doesn't stop there since all higher order terms will also be undefined at either $a$ or $b$.

In practice, I am interested in the form of the equation near either $a$ or $b$. I've thought of expanding the reciprocal function around those points and then inverting whatever approximation I want to make, but I'm unsure if this is correct. For example take the expansion around $x=a$, the reciprocal is given as

$$(f(x))^{-1} = g(x) = \frac{(x-a)(x-b)}{x^2}.$$

Then Taylor expanding around $a$,

$$g(x) = 0 + \frac{a-b}{a^2}(x-a) + \mathcal{O}(x-a)^2.$$

Therefore, really close to $a$ we can ignore higher order terms, and we say that

$$f(x) \approx \frac{a^2}{(x-a)(a-b)}.$$

My issue is if it is legitimate? I got rid of the problem of expanding $f(x)$ by having the undefined term be $0$ in the inverse and trivially allows me to consider the next term.

• – Martín-Blas Pérez Pinilla Feb 18 '16 at 7:46
• Beware : you are mixing $f^{-1}(x)$ and $1/f (x)$. – Jean Marie Feb 18 '16 at 8:18

Let us take the case of $$f(x)= \frac{x^2}{(x-a)(x-b)}$$ that you want to expand around the dicontinuity $x=a$. Consider the function $$g(x)=(x-a)f(x)= \frac{x^2}{x-b}$$ and, now, use Taylor for $g(x)$. You will get $$g(x)=\frac{a^2}{a-b}+\frac{\left(a^2-2 a b\right) }{(a-b)^2}(x-a)+\frac{b^2 }{(a-b)^3}(x-a)^2+O\left((x-a)^3\right)$$ Back to $f(x)$, divide each term by $(x-a)$ and you are done.