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Can someone explain in complete detail with the appropriate convergence arguments of the Gibbs Phenomenon for Fourier Series? I know that the overshoot near a jump does not die out as the frequency increases, but approaches a finite limit. We derived the overshoot in an engineering class and there was a lot of hand-waving. I understand it but I would like to see a complete picture explained here for a general function $f(x)$ on a bounded domain.

Further, the Gibbs Phenomena doesn't occur for some other eigen function expansions like Chebyshev expansion etc. The explanation given in the class was again very crude. Again for these cases could someone explain in detail the connection to the Sturm-Liouville equation and which of these eigenfunction expansions lead to overshoot and which ones do not.

Thanks

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    @Matt, @Didier: I posted the question here after looking up wiki. I am expecting something more. For instance, why does he look at $\displaystyle \lim_{N \rightarrow \infty} f(x_0 + \frac{L}{2N})$ and not say $\displaystyle \lim_{N \rightarrow \infty} f(x_0 + \frac{L}{2N^2})$. Also do these things hold for any eigen function expansion in general?. Also I am not sure about the mode of convergence of these eigen functions.2011-02-07

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Chapter 3 of the book The Gibbs phenomenon in Fourier analysis, splines, and wavelet approximations by Abdul J. Jerri seems to cover the case of a general orthogonal expansion along eigenfunctions of a Sturm-Liouville problem.

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The periodic jump function defined by $j(t):=(\pi -t)/2$ on $]0,2\pi[$ has $\sum_{k=1}^\infty \sin(kt)/k$ as its Fourier series. We now compute the value of the partial sum $s_N$ at the point $t_N:=\pi/N$: $s_N(t_N)=\sum_{k=1}^N{\sin(k\pi/N)\over k}= \sum_{k=1}^N{\pi\over N}{\sin(k\pi/N)\over k\pi/N} \doteq \int_0^\pi {\rm sinc}(t)dt\doteq 1.852$ (the last sum can be interpreted as a Riemann sum for the sinc-integral). On the other hand $\lim_{t\to 0+} j(t)=\pi/2\doteq 1.571$ so that we have a guaranteed overshoot of about 17%.

The choice of the point $t_N$ is nearly "optimal", but I won't go into this here.