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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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