3
$\begingroup$

I am looking at ways which you can write a function as a series. I am aware that one can use the Taylor series. I am currently trying to understand the Laurent series. I understand there are cases where the Taylor series will not work as all the terms in it cannot have a negative degree. The Laurent series is there to aid this kind of problem.

Upon examining texts, including the wikipedia page I see that for a function $f:C\subset \mathbb{C} \rightarrow \mathbb{C}$, the Laurent series at a point $z_0 \in C$ is given by

\begin{equation} f(z) = \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n \end{equation}

where the $a_n$ is a somewhat a generalized version of the Cauchy integral formula. I have the following questions with regards to this:

  • Why do you need to have an annulus around $z_0$
  • When calculating the $a_n$'s, can you pick any closed curves surrounding $z_0$ within the annulus and integrate along their boundaries?
$\endgroup$
3
$\begingroup$

If $f$ is not analytic at $z_0$ (and the singularity is isolated and not removable), then the Laurent series at $z_0$ will converge in an annulus around $z_0$. Why an annulus? What follows assumes you accept why Taylor series gives you a radius/disk of convergence.

If you ignore negative powers, then it is like a regular Taylor series and you already know Taylor series have a radius of convergence (outer circle) $|z-z_0| < R$.

Now if you focus on the convergence of the negative powers, think of it as a Taylor series of a different change of variable. $\sum_{n > 0} a_{-n} (z-z_0)^{-n}$ becomes $\sum_{n>0} a_{-n} y^n$ for $y = (z-z_0)^{-1}$. Note that there is a radius of convergence for this series, $|y| < r$, which translates to convergence when $|z-z_0| > r^{-1}$.

Conclude convergence in annulus $r^{-1} < |z-z_0| < R$. This is the largest region for which the Laurent series converges (since this is the case for the individual Taylor series).

As for the second part, the answer is yes. In fact the curve is allowed to even leave the annulus as long as you avoid other singularities of the function by consequences of Cauchy's theorem (tells you that you can deform the contour while preserving the integral so long as you don't cross any other singularities)

$\endgroup$

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.