Understanding the Gibbs Phenomenon in Fourier Series: A Step-by-Step Guide

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In summary, the given formula involves a limit as N goes to infinity, and shows the relationship between a sum and an integral involving sine functions. The 't' that is mentioned may be a substitution and the problem has been solved by correcting a variable.
  • #1
joriarty
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Show that [PLAIN]http://img829.imageshack.us/img829/3411/screenshot20101011at115.png [Broken]

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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  • #2
[tex]\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[/tex]

:wink:
 
  • #3
Borek said:
[tex]\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[/tex]

: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
 
  • #4
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.
 

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