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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Hi;

I am not getting that, let me check my notes.

The Laurent series are generated from the Taylor series with a substitution thrown in. Maybe we should do one that has a real pole first.

*Last edited by bobbym (2013-03-21 00:09:01)*

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**zetafunc.****Guest**

I looked up a handout, apparently you just take the reciprocal of z^n and sum it...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Do you have the link?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

Google "calculating laurent series" and click on the first link, titled "[PDF] 1 What is a Laurent series? 2 Calculating the Laurent series..."

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

I see your problem there but how is that answer the Laurent series. It has no principal part!

*Last edited by bobbym (2013-03-21 01:41:00)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

Principle part?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

The part of a Laurent series that has the terms involving negative exponents.

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**zetafunc.****Guest**

According W|A there was a 1/(z - a) term...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

On the last page they give,

but it has no negative exponent.

Oh boy! It is staring me right in the face, I am sorry.

*Last edited by bobbym (2013-03-21 01:48:10)*

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**zetafunc.****Guest**

Yes, that one has negative powers of z.

Do you know anything about the contours themselves? I've heard of a few methods, like the "keyhole" method, or drawing squares or rectangles instead of curves...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

I have some examples of how to do contour integrals but my understanding of the theory is not good.

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**zetafunc.****Guest**

Which examples do you have?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

A couple of hopefully not too difficult ones. They were easier back when I did them, now they might be impossible.

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**zetafunc.****Guest**

Hopefully it will not look like Greek to me... (in which case, I can just get adriana to help)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

If you have one to post go right ahead. Or do you want me to post one?

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**zetafunc.****Guest**

I thought you were going to post one -- I would like to see one of your examples, if possible, thanks.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

How about this one?

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**zetafunc.****Guest**

Would it be appropriate to use partial fractions with imaginary co-efficients here, to generate the Laurent series?

My chosen contour would be the upper half of the plane (a semi-circle from negative infinity to positive infinity), since it encloses the real integral...

I think I can generate the partial fractions but it gets messy. Maybe the method of residues is not the way to go.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Hi;

Let's see what we can do first with the residues.

The poles are

Since we are going from 0 to ∞ we only take the second and fourth poles. Are we okay up to here?

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**zetafunc.****Guest**

Using Cauchy's integral formula I got the wrong answer... I said

Let the contour enclose the part of the plane from 0 to infinity, so only sqrt(i) is the pole included in the contour, right?

Then Cauchy's integral formula says:

Letting n = 1, a = sqrt(i) gives me:

evaluated at z = sqrt(i) gives me an answer of pi/2... but the correct answer is

I am stuck.

---

Sorry, just saw your post. Okay that makes sense, we use those two poles since they are in our region of interest (the real integral doesn't exist below the axis, where the other poles are).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Hi;

so only sqrt(i) is the pole included in the contour, right?

Post #9394, there are two poles. The positive ones count.

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**zetafunc.****Guest**

Okay, so that means we are taking two contours?

I just tried with Cauchy's integral formula and I am getting a division by zero...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Maybe there are conditions for the Cauchy integral that are being violated.

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**zetafunc.****Guest**

I tried again and no division by zero, but I'm getting

which is obviously wrong...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,663

Maybe we can work this with this formula here for the residues it looks easy.

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