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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Nope, that is how he solved them in the book. Do you think I invented this? Also, I check the correct answer and correct choice of poles by trying the others in combination. They do not work.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Can you post how he exactly solved it?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

I have! The only thing I am lacking is a coherent idea on which poles to take. That I did not write down because at the time I understood it better and it was self evident to me.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Word to word?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Yes, as I have written them down. Me and zetafunc did a few also. He came up with essentially the same method. The only problem is the ambiguity in explaining the choice of pole. For the contour I believe they always use the unit circle in these examples.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

No, actually, the contour is chosen and the poles inside the contour are the ones you calculate the residue of.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Okay, choose some arbitrary contour and try to get the answer. This method gets the answer to every one of these problems.

For the pole I choose the red point. Answer will be π / 3.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

I did not say that the choice of contour is arbitrary. I said that the contours are chosen, not residues. In fact, sometimes, you may need to calculate the residue of more than one pole.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Yes, we just did a problem where 2 poles were inside the unit circle. Sometimes you do all the poles. But as you can see from the drawing a circle is the simplest shape to go through the 3 points. It happens to be a unit circle.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

You seem to be arbitrarily choosing which poles you are taking and which not.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Nope, I have a rule. I just do not have any solid justification for it.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

What rule is that?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Because we have intervals of 0 to infinity the bottom pole is out. Then I take the poles that have a positive imaginary part. That leaves the one pole I chose.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Hm, okay. Then, how would you integrate log(x)/((1+x^2)^2) from 0 to infinity?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

We have not finished the other one. If we jump around we will get confused. We should calculate the residues for the other problem before moving on to the next one.

But you have not been listening to me, as far as I know this method only works on rational functions. At least that is what some site I was looking at said. He defines a rational function as p(x) / q(x) where both p(x) and q(x) are polynomials and of course q(x) ≠ 0.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

That site is not correct. All sources I have found have it working on other types if functions as well.

Now, please do the integral I am asking you.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

I was going to say that but zetafunc shows a rather simple one where this method falls flat on its face. There must be some conditions when can use it and when you can not.

I will look at the integral.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

bobbym wrote:

I was going to say that but zetafunc shows a rather simple one where this method falls flat on its face. There must be some conditions when can use it and when you can not.

I will look at the integral.

Actually, that was the point of the integral I gave you. As you will see, the method you will try on it will probably fail, but the Wiki article has no problem doing it.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

The residue method does not work on all cases. There are conditions.

The method I am using there will not work but what method works all the time.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Yes, but the contour integration methid works on that one.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

It will fail on others. Either because of violations or impossibility to do.

This method is a not a general method for all integrals.It works on the examples provided.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

This was just to show that the method you have is not contour integration. But I do not know if it works.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Why do you think it is not contour integration?

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Because I did understand some parts of what I redlad earlier, and some of those parts were not included in your method.

By the way, now when I look at this stuff again, I think only the residue part confused me, so I do thank you for making that part clearer. It seems to be a common ground between your method and the contour integration method. I will have to read about it once more to make sure I understand it.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,138

Hi;

If you look here you will see the formula I am using or misusing and even that the contour is circular and encloses the poles.

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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