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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

I am getting:

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

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.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

What formula?

Also, what is a simple pole?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

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.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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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: 84,748

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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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: 84,748

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 84,748

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)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**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: 84,748

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)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 84,748

Hi;

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

Add that for each pole and then times by 2πi

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 84,748

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

We need two:

n = 1

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

How did you evaluate that last line?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,748

What did you get for the first limit?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**