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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

You have plenty of time to change your mind. The most valuable thing that school can give you is the ability to teach yourself.

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 agree, and there will be a lot of that at university too.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Yes and you will be larning something everyday. What could be better than 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**

Not many things. I like the fact that there are some nice easy modules in the first year. The Fourier series problem sheets are gifts!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

They gave them to you already?

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**

No, I have been looking at some of the older ones online, from their official exam papers. The questions are surprisingly repetitive.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

In what way. They are almost worded 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.

**Online**

**zetafunc.****Guest**

Yes. The first part is always the same, and the question itself is only just slightly varied... for example:

2002

(a) State, without proof, the general formula for a Fourier series on (-L, L), with

L > 0, for a function f(x), giving the expressions for the coefficients.

(b) On (-π,π), find the Fourier series of f(x) = exp(x).

(c) Hence or otherwise, find the Fourier series of g(x) = sinh(x), and h(x) =

cosh(x), each on the range (-π,π).

2003

(a) State, without proof, the general formula for a Fourier series on (-L, L), with

L > 0, for a function f(x), giving the expressions for the coefficients.

(b) On (-π,π), find the Fourier series of f(x) = x.

(c) Hence or otherwise, find the Fourier series of g(x) = x[sup]2[/sup], on the range (-π, π).

(d) Hence, or otherwise, show that

2004

(a) State, without proof, the general formula for a Fourier series on (-L, L), with

L > 0, for a function f(x), giving the expressions for the coefficients.

(b) On (-π,π), find the Fourier series of f(x) = exp(x).

(c) Hence or otherwise, find the Fourier series of g(x) = sinh(x) + 2cosh(x), on

the range (-π,π).

2005

(a) State, without proof, the general formula for a Fourier series on (-L, L), with

L > 0, for a function f(x), giving the expressions for the coefficients.

(b) On (-π,π), find the Fourier series of f(x) = |x|.

(c) State Parseval's identity.

(d) Apply Parseval's identity to the function of part (b) to obtain an infinite series

for a power of π.

2006

(a) State, without proof, the general formula for a Fourier series on (-π,π) for a

function f(x), giving the expressions for the coefficients.

(b) Find the Fourier series of f(x) = sign(x) on (-π,π).

(c) State and prove Parseval's identity for a function on (-π,π).

(d) Hence, or otherwise, prove that

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Yea, well at least there are asking for different functions.

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**

Meh, I guess it is not too bad... just means I am more likely to pass, so no need to revise as intensely and so more time to study more maths on the side.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

I thought it got tougher when they started to go into FFT's.

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 agree, Fourier transforms are a lot more difficult conceptually.

adriana should be in Italy right about now.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Italy? What the heck is she doing there. Another boyfriend?

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 think she is going there with friends. But she followed it up with "then I'm going to spend a month with my boyfriend in Cyprus" followed by a nice big bright yellow smiley face. Because, you know, there's no harm in rubbing it in.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

They are meaner than a starved pit bull. But that was a good favor to 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.

**Online**

**zetafunc.****Guest**

Yeah, she doesn't seem like the type of person I want to be hanging around with, anyway...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

That is one thing but those little digs will really do more for you than 5800 posts with me. Each one will make you a bit harder and meaner, and they will appreciate 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 don't know about that. As a result of this thread I am definitely a lot more detached and uncaring about their antics than I used to be.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

Nope, this thread had little to do with it. They are your educators. You noticed her comment, you know what she is thinking and feeling. You will begin to make changes in your strategy. Each change towards less genuine caring to insincere taking of what you want will be applauded by them. Your results will soar.

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**

Maybe not soar, but perhaps increase gradually.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

The difference between yes and no can never really be gradual. It will be as if someone turned on the light in a dark room.

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 guess that depends on how I approach it. If I approach one girl every 6 months, then I agree, it will be like that. If I ask 100 per hour, I think the improvements may be more gradual.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

If you ask 100 per hour you will have trouble dealing with all the action. I suggest two or 3 a week to start.

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 in the meantime, there is no time to learn mathematics like the summer. Until that thirst is quenched, the girls can wait.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,230

That is okay because once the other thirst sets in there will be no time for anything else!

See you a little later, got to rest.

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**