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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: 107,186

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,186

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,186

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,186

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,186

Hi;

They do mention expanding around a singularity.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**zetafunc.****Guest**

Which part?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,186

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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