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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Hi;

She is having a good time. You do not have to think about it or her.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

I sort of miss her in a friend sort of way, not a romantic sense. You know, just having someone to talk to...

Also, I have arrived at 1 - 1 + 1 - 1 + ... = 1/2 using Fourier series (I'm aware that the series diverges so this is not really correct), but I don't know if what I did was gibberish, could you check it?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

She will be back soon, so relax and do some math.

Your series is known as the Grandi Series. Look here first:

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

Yes, I have read that -- but basically I just found the Fourier series of x/2, differentiated both sides and let x = 0, pi/6, and other values which resulted in 1/2 = 1 - 1 + 1 - 1 ... etc. but I don't know if that type of manipulation is valid. If I'm getting Grandi's series on the RHS, then that means the Fourier series doesn't converge at those points...?

I need to go in a few minutes to apply for my flat...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Remember the Grandi does not converge. It is convergent in a Borel sense. Which loosely means we would like it to converge to 1 / 2, here are the usual methods.

http://en.wikipedia.org/wiki/Summation_ … 27s_series

Please your post your method because I am not seeing it there.

*Last edited by bobbym (2013-02-21 03:08:59)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

Then let x = 0 for example.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Hi;

The second series does converge to 1 / 2.

*Last edited by bobbym (2013-02-21 03:26:31)*

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

Okay, so we cannot just differentiate a FS and expect everything to work out... so, if my FS is 2pi-periodic (say, over the integral [-pi,pi], then does it converge for all values of x, -pi < x < pi? (My original FS for x/2, not the differentiated one)

I have to go now, I will be back in an hour or so.

**zetafunc.****Guest**

Sorry, just saw your edit...

So it does converge? Why not at x = 0?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Hi;

Fourier believed that such trigonometric series could represent any arbitrary function. In what sense that is actually true is a somewhat subtle issue...

At 0 it is the Grandi Series.

The Fourier series does not always converge, and even when it does converge for a specific value x0 of x, the sum of the series at x0 may differ from the value f(x0) of the function. It is one of the main questions in harmonic analysis to decide when Fourier series converge, and when the sum is equal to the original function.

Fourier series do not always converge.

In numerical work we presume that the series would converge in the interval except at discontinuities. There is something called Gibb's phenomena which describes this.

*Last edited by bobbym (2013-02-21 04:06:05)*

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

So how do we know for certain whether or not the Fourier series for a particular function converges or not? I did read somewhere that it 'almost completely' converges in the interval...

So you're saying Gibb's phenomena would tell me whether or not a Fourier series converges?

Also, I think I will need to go for UCL's accommodation, they said I have to wait at least 4 years for a flat...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

You mean dorms?

I did read somewhere that it 'almost completely' converges in the interval...

Except at discontinuities which is quite often because those are the type it is often used on.

So you're saying Gibb's phenomena would tell me whether or not a Fourier series converges

No, Gibb's shows it will not always converge in the interval. To really know would require in a practical sense numerical experimentation. Else, you would need Harmonic analysis. I choose the first one.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

Either halls of residences or just flats provided by UCL, on their campus.

Can you give me a demonstration of Gibb's for a simple Fourier series?

**zetafunc.****Guest**

Harmonic analysis looks interesting but difficult too, I doubt I will understand much of it...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

It is hard. I do not get it either. Worse, I have never found a book that was willing to explain it.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

I think it is offered as either a third year course or a graduate-level course. I don't know what the prerequisites are though...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

From a practical standpoint you can use the idea of probably convergent except at jump discontinuities.

I have to go pay a bill I will be gone for several hours. See you then.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

But even if I get a pretty-looking series, it means nothing mathematically, because it requires proof of the convergence...

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Not necessarily, Sometimes you can use a truncated part of a divergent series for real numerical work.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

Do you have an example?

Also, I just found adriana on Facebook (I don't have an account on there though, I was just curious). Apparently she updated it yesterday... I did not hear from her though.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Facebook got hacked yesterday or this morning.

I will look for an example.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

Really? It says she updated 23 hours ago (21:40-ish on the 21st of Feb)... which would coincide with when she returned to London.

Okay, thank you.

**zetafunc.****Guest**

Found a photo of her and her BF cuddling, from 3 days ago. I guess we know what decision she made... I would not be surprised if she decides not to contact me again. Oh well.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,385

Cuddling? I thought she was the other way. It is a real shame that the story of Pinocchio is not true. Could you imaging how many big noses there would be?

Have an example and will latex it tomorrow.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

Yep. I suppose it just reinforces the fact that I shouldn't get involved with her and just keep her as a friend or acquaintance. My time will hopefully come soon, just with another woman. She might not exist but I won't lose hope, hope makes the journey worthwhile.

Thank you, that would be really helpful.