# Laurent Series

1. Nov 8, 2009

### nk735

1. The problem statement, all variables and given/known data

Hello, i'm having massive troubles with Laurent series'. I'm pretty shocking with series', so i'm probably making some fundamental mistake that you'll want to slap me for, but everytime I try one of these questions i'm wrong ever so slightly. I've attached a couple of examples of how I suck at this, any feedback would be greatly appreciated.

Note; I don't just need solutions to these problems, I need to understand the entire process. If anyone can be bothered, once I know how exactly i'm wrong, a several step walkthough on these problems would be infinitely helpful.

2. Relevant equations

Taylor Series
f(z) = sum((1/n!).((f^n)(z_0)).(z-z_0)^n)

Standard Taylor Series Expansions
1/(1-z) = sum(z^n)

exp(z) = sum((z^n)/n!)

3. The attempt at a solution

Attached

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2. Nov 9, 2009

### clamtrox

You have to do the series expansions around the point that you are told, not around zero. For example, e^z around z=1 is $$e^z = e + e (z-1) + e/2 (z-1)^2 +...$$

One thing that you can do to check if your answer is even plausible is to check how it behaves around the pole you're expanding around of. For example again, the result that you got for 1/[z^2(z-1)] is obviously wrong, since the leading order term is proportional to 1/(z-1)^2 instead of 1/(z-1).

3. Nov 9, 2009

### nk735

I'm sorry, I still don't quite understand. So my problem's going from the problem statement to the Taylor series? Could you perhaps show me exactly where in my workings I am wrong and the correct alternative?

Thankyou

4. Nov 9, 2009

### clamtrox

Okay, let $$f(z) = \frac{1}{z^2(z-1)}.$$

First, note that there's a first order pole at z=1. This is why you're doing a Laurent series, not a Taylor series (Taylor coefficients would be infinite).

Now, to solve it, you can expand the part without the pole; g(z) = 1/z^2. To expand it around z0=1, just write it out as an ordinary Taylor series,

$$g(z) = \sum_{n=0}^{\infty} \frac{g^{(n)}(z_0)}{n!} (z-z_0)^n$$

This is easy to do and the answer is

$$g(z) = \sum_{n=0}^{\infty} (-1)^n (n+1) (z-1)^n.$$

Now, just write f(z) with the help of this (notice how it's now convenient to multiply the pole inside the sum):

$$f(z) = g(z) \frac{1}{z-1} = \sum_{n=0}^{\infty} (-1)^n (n+1) (z-1)^{n-1}.$$

5. Nov 9, 2009

### nk735

Ahh, I get it now. Just did a few more problems and managed to (finally) get them right.

Thankyou very much for you help and patience.