Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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**zetafunc.****Guest**

darn, I guess Fourier Series cannot provide me with an elegant solution.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

The funny thing is,

**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 what I was trying to show -- but I can't do it...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Are you allowed to use a table of sums?

**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**

Nope, we get a formula booklet with just some standard Taylor series, trig formulae, stats tables, etc.

I'm also a bit worried about using tricks like Fourier series because to score a full 20 marks on one question, your argument has to be flawless -- for instance, I probably won't use contour integration in STEP, since I'd need to justify a lot of stuff. As a result, I tend to just look for the solution they want me to find.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

That is like asking a runner to use only one leg.

**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**

Rules are rules unfortunately.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Without one, summing the partial series by series methods will be tough.

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've been able to reduce the problem to

with multiple use of the sum to product trig formulae. But I'm not sure if that helps.

**zetafunc.****Guest**

W|A is saying I'm wrong, will have to try another approach.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Hi;

It is not going anywhere from there that I can see. You have an idea?

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**

TSR's solution makes sense to me now, seems pretty simple now that I think about it...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Very good then. Glad you figured it out.

I need a little break be back soon.

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**

The hardest part is the recognition of things like sin(22pi/23) = sin(pi/23). Other than that, it is just a case of equating imaginary parts with De Moivre's. I'll remember this trick for similar series!

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Hi;

Very good work.

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**

You mean, very good work to whoever made that solution... I simply read it.

Nothing from adriana this weekend. Apparently she, her sister and her brother have all deleted their Facebook accounts...?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

That means she is going undercover.

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**

Undercover? Her Google+ profile is still active though...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Yes but getting off of facebook is sort of sacrilegious.

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 it is strange to see adriana, her sister and her brother all come off of it at once...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Sounds like they are all preparing to be abducted by aliens...

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 just assumed it was because the exam season is approaching, although she never did this during January I don't think... she's also not been active on chat as far I have been able to see. Maybe she is genuinely busy.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Maybe, but not as busy as you.

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 would think she should be busier than me because when I last talked to adriana she had only done 10 STEP questions.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

That suggests she is not busy at all. Perhaps between parties she is working on one.

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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