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**zetafunc.****Guest**

Find Laurent expansions for:

valid in the annuli;

(a) 0 ≤ |z| < 1,

(b) 1 < |z| < 3,

(c) 3 < |z|.

---

I've found a Laurent expansion but I'm not sure what to do about the different annulus ranges.

which forms the geometric series

or

which I think simplifies to

but I don't know how to address the problem of the annulus ranges given for (a), (b) and (c). I think mine is valid for the range in (a), but I don't know what to do about the others.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

Hi;

The poles of that are -1 and 3.

Trouble is, when I expand around either one I do not get your principal part.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I know those are the poles (from the quadratic), but how can I use that here? I thought the principal part was the part of the series involving negative powers -- which part is my principal part here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

The principle part is the part involving negative powers. To find the Laurent you expand around one of the poles. Which pole did you use?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I'm not sure what you mean -- I just got f(z) into a form which I could generate the series of. How do I expand around a pole?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

Hi;

I just do not see any negative powers in your series. Why is it a Laurent series?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I don't know. The lecture notes I am looking at say "find Laurent expansions for the function f(z)" and gives the annulus ranges.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

I can get two Laurent series for that but I do not remember how to handle the annulus stuff. I did not write it down either so I will have to rediscover the method.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Go here:

tinyurl. com /bqwts5a

(remove spaces)

and go to Part 6. In there is the exercise I am working on. Unfortunately they have not provided solutions to Parts 5-7, only 1-4.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

Hi;

They do mention expanding around a singularity.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Which part?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

Hi;

p57 and on when they start expanding around z0. This is the pole, the roots of the denominator. You can use the integral or algebraic tricks to get the Laurent series.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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