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**zetafunc.****Guest**

But it looks so tedious to find the residues...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

I am getting:

**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**

That is correct... did you use the residues?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Yes, there is a simple formula for computing them when the poles are simple.

**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**

What formula?

Also, what is a simple pole?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

I am the worst at definitions so please check this. I think a simple pole is a pole with multiplicity of one.

**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 by the pole having a multiplicity of one...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Multiplicity of one means there is only one pole like that. Of two, would mean two poles the same.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes but I don't understand how two poles can be the same... how do you mean?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

If the roots of the denominator are all different then the poles are all simple ones because they are of order 1.

Since all 4 roots are distinct they are all simple.

If the roots were 2,2,2,2 then 2 would be a pole of order 4 and not simple.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Ohh I see, so it is sort of like having repeated roots of a polynomial?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Yes, something like that. It may be more complicated and I do not remember. But for our purposes that integral has 4 simple poles. We are interested in only two. The formula becomes very, very easy when the pole is simple.

*Last edited by bobbym (2013-03-22 01:55:32)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

**zetafunc.****Guest**

Are you using this formula?

I cannot get the formula to display properly, it is the one with the limit of z -> z[sub]0[/sub], and the derivative and the factorial.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Hi;

I can not see your latex. I could show you what I am doing, it does look like that, but I am not sure.

*Last edited by bobbym (2013-03-22 02:05:24)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Is there some sort of bug with the latex today, I have tried to type 3 formulae and none of them worked apart from a few...?

I'll take out the limit, maybe this works:

Let z[sub]0[/sub] be a pole of f, of order n. Then,

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Hi;

n is the order of the pole. a is the pole.

Add that for each pole and then times by 2πi

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes, that is the formula I am using. But how do you use it?

For instance, consider

How would you then use your formula to find the residue?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

Let's continue with the other one first. I will post the answer step by step.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

We need two:

n = 1

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

How did you evaluate that last line?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

The limits? L'Hopitals rule and Alpha for the plugging in.

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 am not getting your answer... I am ending up with

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,492

What did you get for the first limit?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**