Understanding the Gibbs Phenomenon in Fourier Series

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Show that [PLAIN]http://img829.imageshack.us/img829/3411/screenshot20101011at115.png

I really don't know where to start with this. It is the very last question of an assignment on Fourier series and the Gibbs Phenomenon, if that is relevant I can give more details but I don't think it is. It's just algebra from here.

I have no idea where the 't' has come from (not mentioned at any earlier stage in the assignment), but I think this might be some sort of substitution? Perhaps it is some sort of standard integral I'm not familiar with?

Could I please have a hint if you can see how this could be done? Thanks :)
 
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\lim _{N \rightarrow \infty} \frac 4 \pi \sum ^{N-1} _{n=0} \frac {\sin(\frac {(2n+1)\pi} {2n})} {2n+1} = \frac 2 \pi \int _0 ^\pi \frac {\sin (t)} {t} dt

:wink:
 
Borek said:
\lim _{N \rightarrow \infty} \frac 4 \pi \sum ^{N-1} _{n=0} \frac {\sin(\frac {(2n+1)\pi} {2n})} {2n+1} = \frac 2 \pi \int _0 ^\pi \frac {\sin (t)} {t} dt

:wink:

I'm not sure what you're trying to say there, beyond restating the initial problem. I just managed to figure it out anyway, I had an 'n' in there that should have been an 'N'. Problem solved, assignment done :D
 
I have seen you struggling with the LaTeX and finally deciding to post an image - as you see, forum LaTeX is perfectly capable of displaying the formula.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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