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.


As Martín-Blas Pérez Pinilla commented, you are looking for Laurent series.

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.

| cite | improve this answer | |
  • $\begingroup$ This method seems more rigorous, but is what you did really a Laurent series? All the examples I find of Laurent series don't actually expand about the discontinuity, but expand around well behaved points and produce convergent series up until the discontinuity. You can do likewise coming from the other side of the discontinuity. $\endgroup$ – Novice C Feb 18 '16 at 9:17
  • $\begingroup$ @NoviceC. I totally agree with you, for sure. I considered the simple case given by the OP to show that Taylor could be used as usual removing the discontinuity. $\endgroup$ – Claude Leibovici Feb 18 '16 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.