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**zetafunc.****Guest**

It's possible that they have run into him already. Maybe that's why H was locked up, nobody believed what she saw.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Well, he seems to have his own girlfriend and he looks like he has a big crush on her.

What did she see?

**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**

Nothing, I was joking.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Anyway, how are the questions coming?

**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 did my 270th today, but I have to stop for a bit to do some other work, annoyingly.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

270! Very good.

**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 wanted to complete all of them but I don't think that's possible now! It's tempting just to jump to solutions...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

That is not a bad idea when you have little time left.

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**

Trouble is, even understanding the solutions can be difficult sometimes, as we saw with the cakes question.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

It is better than not seeing the solution at all. Sometimes you see a solution and it takes a long time before you understand 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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**zetafunc.****Guest**

Yes, I have found that to be true, particularly for STEP.

I am stuck on another question...

Find

giving your answer in terms of the tangent of a single angle.

This is the first part of STEP I 1987 Q7.

I remembered this Fourier series:

If you differentiate that, multiply both sides by -1, you get

I thought plugging in x = 2pi/23 might help, but I just couldn't simplify the RHS to the series they state in the question.

For some reason though, 1/2 tan(pi/23) IS the right answer -- which is what you get when you let x = 2*pi/23.

What can I do here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

The identity is true.

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**

But how do I show that it is?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Working on 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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**zetafunc.****Guest**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Okay got that series too.

The question is a partial sum while you are trying to use an infinite series on the RHS.

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, I remembered it from a post of yours almost 2 years ago.

**zetafunc.****Guest**

I was hoping to somehow show that the infinite sum on the RHS converged to the sum given in the question...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

How are we supposed to change the infinite series we get to the partial sum in the question?

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**

Not sure, I thought I could simplify the infinite sum somehow, such that I'd find lots of the terms cancelled and left the sum given in the question.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I am not seeing any way to do that.

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 couldn't either -- the RHS doesn't converge to that sum.

Will have to think up another way. There is a method on TSR, but it looks long.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I am looking at it now. I do not get it.

I am noticing one thing now, the original question posted is not complete. The series could go one more term and still be true!

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**

It doesn't -- the sum I posted is definitely what appears on the paper. sin(46pi/23) is also zero.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Yes, I was just remarking that the next term also works. This is as you say due to the fact that it is a multiple of 23.

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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