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.
How can I prove that the Gibbs phenomenon overshoot for a Fourier Series is approximately 9%?
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0@IvanLerner Please consider accepting the answer below if it resolved your question satisfactorily. – 2013-11-27
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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
- Compute the indicated Fourier series and consider $F_N(x)$, the $N$th partial Fourier sum of $f$.
- 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.
- From there, it's easy to find the critical number of interest; call this $x^*$.
- Evaluating $F_N(x^*)$, the resulting sum can be massaged a bit and then recognized as a Riemann sum.
- The Riemann sum is one corresponding to ${1\over \pi}\int_0^\pi {\sin x\over x}\,dx$.
- Since $N$ is finite, we have $F_N(x^*)\approx {1\over \pi}\int_0^\pi {\sin x\over x}\,dx\approx 0.589\dots$.
- 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.