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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fourier-series
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2More 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. – 2012-11-25
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0Which part of your question is not covered by [this](http://en.wikipedia.org/wiki/Gibbs_phenomenon#The_square_wave_example)? – 2012-11-25
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0@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. – 2012-11-25
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0Sorry 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. – 2012-11-25
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0@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? – 2012-11-25
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0What is the first step that you do not understand in there? As written explicitly there, 9% is the relative error made when one replaces the expectation of the function $t\mapsto\sin(t)/t$ on $[0,\pi]$ by $\pi/2$. – 2012-11-25
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0I understand all the steps, but the final limit he states that the limit is that number on the section "Formal mathematical description of the phenomenon" but I don't understand how he evaluated that limit to get to the value. Also, I don't understand why the example of the square wave is valid as a generalization. – 2012-11-25
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0Let us concentrate on the section *The square wave example*. Do you understand the proof that $S_N(f)(\pi/N)$ converges to the limit which is written there? – 2012-11-25
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0Yes, I do. But how does that generalize? – 2012-11-25
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0You are a little difficult to follow, I must say... So now, the computation of the 9% and the whole proof for the square wave function $w$ are OK? Perfect. All you need to complete the proof is a way to deduce the same result for any function $f$ with a jump at point $x_0$, say. Hint: the function $g=f-tw$ is continuous at $x_0$ for some $t$. What is $S_N(g)$? (Or, if you want a fully cooked solution, see [here](http://www.math.chalmers.se/Math/Grundutb/GU/MMG710/H12/gibbs.pdf).) – 2012-11-26
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0I couldn't do it on my own, but I get it now, thank you. If you make this an answer I'll accept it. – 2012-11-26
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0@IvanLerner Please consider accepting the answer below if it resolved your question satisfactorily. – 2013-11-27
1 Answers
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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.