How Does Gibbs Phenomenon Affect Fourier Series Convergence?

likearollings
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Homework Statement



http://img684.imageshack.us/img684/4496/fourier.png

Homework Equations



http://en.wikipedia.org/wiki/Small-angle_approximation

The Attempt at a Solution



This is part of a larger question, I have underlined in Red the areas I am struggling with and have cut out bits I have done.

Any ideas?

I have done some research and found out about 'Gibbs Phenonemon', http://www.sosmath.com/fourier/fourier3/gibbs.html

can't really get it all together to answer the question.

any help is appreciated, thanks! :)
 
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I am not sure which small angle formula to use, Taylor series or sine, or how to get the RHS of that equation...
 
Thanks for all the replies...

I have managed to do the first problem. Now I just need help with the wordy question at the end.

Any help will be massively appreciated
 
You have

f_N(\frac{\pi}{2N}) = \frac 2 \pi\int_0^{\frac \pi {2N}}\frac{\sin(2Nt)}{\sin(t)}\, dt

Letting u = 2Nt, du = 2Ndt gives

\frac 2 \pi\int_0^\pi \frac {\sin u}{\sin(\frac 1 {2N}u)}\frac 1 {2N}\, du

Now if N is large, u/(2N) is small and you can use the fact that for small angles, sin(θ) ≈ θ
in the denominator. That will get you your equation.
 
likearollings said:
Thanks for all the replies...

All what replies? I don't see any except for the one I just posted.
 
LCKurtz said:
All what replies? I don't see any except for the one I just posted.

Hey sorry, I posted that right before you replied, was getting really frustrated (sure you know how it is)! but thanks, I am just reading what you have written and it is a great help, thank you so much
 
LCKurtz said:
You have

f_N(\frac{\pi}{2N}) = \frac 2 \pi\int_0^{\frac \pi {2N}}\frac{\sin(2Nt)}{\sin(t)}\, dt

Letting u = 2Nt, du = 2Ndt gives

\frac 2 \pi\int_0^\pi \frac {\sin u}{\sin(\frac 1 {2N}u)}\frac 1 {2N}\, du

Now if N is large, u/(2N) is small and you can use the fact that for small angles, sin(θ) ≈ θ
in the denominator. That will get you your equation.
I thought that it might involve some substitution like that but I didn't think it would cancel out and work that well. Thanks! great help.

The limits of the integral all work out nicely too, thanks!

Any ideas for the last part of the question?

If you have a look at page 5 of this document it has some suggestions.

http://www.scribd.com/doc/20565822/Fourier-Series

These are the conditions:

http://en.wikipedia.org/wiki/Dirichlet_conditions
 
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