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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: 100,319

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.** **Thinking is cheating.**

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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: 100,319

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.** **Thinking is cheating.**

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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: 100,319

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.** **Thinking is cheating.**

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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: 100,319

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.** **Thinking is cheating.**

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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: 100,319

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.** **Thinking is cheating.**

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**zetafunc.****Guest**

Which part?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 100,319

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.** **Thinking is cheating.**

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