# How can I prove that the Gibbs phenomenon overshoot for a Fourier Series is approximately 9%?

The question is pretty self explanatory, I'm studying Fourier Series with the book Mathematical Methods for Physicists written by Arfken and it does not explain that.

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More precisely: If $S_N(f)$ is the $N$-the partial sum of the Fourier series for $f$, then $\lim_{N\to\infty}S_N(f)(x_0+(L/2N))=f(x_0^+)+(.09)(f(x_0^+)-f(x_0^-))$ ... and similarly for the left limit. – Matt Nov 25 '12 at 17:42
Which part of your question is not covered by this? – Did Nov 25 '12 at 17:48
@did I had already looked at it, and in the general description of the phenomenon, it just states that the limit gives you a 9% overshoot, but it doesn't actually show a proof. The example of the square wave is a particular case and shows the same thing but without any proof. – Ivan Lerner Nov 25 '12 at 17:54
Sorry but this is not true: the section of the WP page I referred you to does explains a proof of the phenomenon for the square wave. As it happens, to study the proof explained in this case would lead you quite easily to the general case. Your call, I guess. – Did Nov 25 '12 at 17:59
@did Then I think I didn't understood how WP got to the actual number. Could you explain it in another way, or explain whats in there? – Ivan Lerner Nov 25 '12 at 18:05

As mentioned in the comments, study the argument here carefully. I'll add a little play-by-play commentary to guide you.

Basically, the steps for the canonical example $f(x)=\operatorname{sign}(x)$ on $-\pi<x<\pi$ are:

1. Compute the indicated Fourier series and consider $F_N(x)$, the $N$th partial Fourier sum of $f$.
2. We want to find the value of the first positive local max of $F_N(x)$. Do this by standard calculus followed by some trigonometry to sum the resulting cosine series in terms of a single sine function.
3. From there, it's easy to find the critical number of interest; call this $x^*$.
4. Evaluating $F_N(x^*)$, the resulting sum can be massaged a bit and then recognized as a Riemann sum.
5. The Riemann sum is one corresponding to ${1\over \pi}\int_0^\pi {\sin x\over x}\,dx$.
6. Since $N$ is finite, we have $F_N(x^*)\approx {1\over \pi}\int_0^\pi {\sin x\over x}\,dx\approx 0.589\dots$.
7. This last value is the value of that first positive local max and thus the overshoot above the true value at $x^*$ (which is of course 1/2) is about $0.0589-0.5=0.089$, i.e., about 8.9% of the unit jump here.

I hope that sketch is helpful.

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