1
$\begingroup$

I've seen many questions already on Laurent Expansions and how to find examples like my question, however, most of them turn to geometric series and partial fractions, which I understand can be faster but when under time pressure, sometimes I just don't see the trick they want me to use and I would like to know that I always have a backup in a definition that should always work.

However, using this definition I keep getting stuck and most of the time my answer is just not correct. This particular problem is about $f(z)=\frac{1}{z(z-1)}$ where we want to express $f$ as a Laurent series around $0$ for $|z|>1$.

This is how I'd go about this problem:

We know that $f(z)=\frac{1}{z(z-1)}=\sum_{-\infty}^\infty a_nz^n$ for some coefficients $a_n$, where we use the Laurent series around $0$. Both $0$ and $1$ are simple poles, and for any $r>1$, both poles lie in the interior of de circle around $0$ of radius $r$.

We look at the open annulus of $1<r<|z|<R$ where $f$ is holomorphic. Then we know that the $a_n$ from above are given by:

$$a_n=\frac{1}{2\pi i}\int_{|z|=r_0}\frac{f(z)}{z^{n+1}}dz= \frac{1}{2\pi i}\int_{|z|=r_0}\frac{1}{z^{n+2}(z-1)}=: \frac{1}{2\pi i}\int_{|z|=r_0}g(z),$$ where $|z|=r_0$ is any circle in the open annulus ($r<r_0<R)$, on which we integrate in a positive direction.

Both poles $0$ and $1$ are inside this circle of radius $r_0$, and $g$ is holomorphic on the interior of this circle except at these poles. Therefore, we can use the residue sum theorem for $g$ to calculate the above integral, where $0$ is a pole of order $n+2$ and $1$ is a simple pole:

$$Res(g,0)=\lim_{z\to 0}\frac{d^{n+1}}{dz^{n+1}}\Big(\frac{z^{n+2}}{z^{n+2}(z-1)(n+1)!}\Big)= \frac{1}{(n+1)!}\lim_{z\to 0}\Big(\frac{(-1)^{n+1}(n+1)!}{(z-1)^{n+2}}\Big)=-1$$

$$Res(g,1) = \lim_{z\to 1}(z-1)\frac{1}{z^{n+2}(z-1)}=1$$

By the Residue Sum Theorem, the integral is equal to $2\pi i*(Res(g,0) +Res(g,1)) = 0$.

This however would lead to the $a_n$ being all 0, which is not the case for $n<-1$.

I saw the answer model which said that with a trick using geometric series you will get the answer $\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}...$. So clearly the $a_n$ are in fact $0$ for all $n>-2$, but this method does not provide me with $a_n$ for $n<-1$.

The first part of this question was about $|z|<1$ where you could use the geometric series directly; I did use this method as well when solving the first question. In that case, only the pole $0$ was inside the region of integration, and the residue, and thus the $a_n$, was equal to $-1$. This answer was correct, but now for all $n>-2$, and wrong for all $n<-1$ (as it's about a simple pole, the $a_n$ should logically be $0$ anyway for $n<-1$).

Somehow I seem to calculate the $a_n$ only for one half of the series, whereas the coefficients are incorrect for the other half. Where am I doing anything wrong in the steps above?

Edit: Finally I have maybe found what could be the problem; the calculation I used for residues uses $n+1$'th derivatives. As soon as $n<-1$, this calculation is no longer valid and the formula used for a general derivative of $1/z$ is no longer what I say it is. How to proceed now?

$\endgroup$
4
  • $\begingroup$ What's wrong with just writing down the geometric series? One has the expansion $\sum_{k=2}^\infty z^{-k}$, so $a_n=1$ for $n\le-2$ and $a_n=0$ for $n\ge-1$ $\endgroup$ Jun 21 '20 at 17:44
  • $\begingroup$ @AnginaSeng Please read my first paragraph; in this case I could figure it out, I just want the other described method to always work as it's not using any 'tricks'. I'm using this simple example so I can understand the concept, on an exam they might ask something a bit harder in which case I might not see how I can find an already know power series, and I would like to have a backup plan. $\endgroup$
    – Marc
    Jun 21 '20 at 17:47
  • 1
    $\begingroup$ You are belittling the geometric series by dismissing it as a "trick" (your quotation marks). I find that when I have a simple method to solve a problem, it is best to use it, rather than try a more complex method which is both time-consuming and error-prone. Here the simple method is just to write down the series from one's knowledge of the geometric series. The more complex method is to use contour integration; it is easy to fall into error thereby. $\endgroup$ Jun 21 '20 at 18:01
  • $\begingroup$ @AnginaSeng I might have phrased it wrongly; by no means am I trying to say the the method I'm attempting is better than using the geometric series. I promise, if I see how to use it in an exam I will definitely take the shorter, easier and cleaner route. However, as I said, it's happened to me multiple times that I just don't see the creative or clever way in which a problem is asking me to think, and the last resort is using definition and build it from the ground up to see where you get. $\endgroup$
    – Marc
    Jun 21 '20 at 18:15
2
$\begingroup$

When $n\le -2$, $-n\ge2$, $$\int_C \frac{f(z)}{z^{n+1}}\,dz=\int_C\frac{z^{-n-2}\,dz}{z-1}$$ and the integrand has no pole at $z=0$, so only the pole at $z=1$ counts, and there the residue is $1$.

$\endgroup$

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.