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I've read that in the Gibbs phenomenon, partial Fourier series will over- or underestimate a function's value in neighborhoods of jump discontinuities. Specifically, the maximum error will converge to the height of the jump discontinuity times $ \mathrm{Si}(\pi)/\pi - 1/2 \approx $ 8.949%. But I notice, visually at least, that there are also 2nd, 3rd, 4th etc. artifacts behind the first that also appear to converge to a fixed proportion of the height in the limit. I've highlighted these artifacts using the approximation for the square wave picture from Wikipedia (link below).

artifacts of Gibbs phenomenon

Based on the calculation done at Wikipedia, I surmised these proportions could be explicitly calculated, via

$ \frac{\pi}{4} + (-1)^{n-1} \frac{\pi}{2} e_n = \lim_{N\to\infty}S_N f \left( \frac{2\pi n}{2N} \right) \quad \dots \implies $

$ (-1)^{n-1} e_n = \frac{n}{\pi} \mathrm{Si}\left( \frac{\pi}{n} \right) - \frac{1}{2} . $

However, this can't be correct because it predicts that the error of the artifacts grow larger in the sequence instead of smaller (where the latter is obviously the reality). What gives?

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    @Rahul Narain: Ah, you're correct. I missed an $ n $ in a denominator and ended up with an extra as a result. For some reason I kept overlooking the error over and over again, bah.2011-07-02

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As per protocol, I'm posting my comment here so that the question can be marked as answered. Actually, I'll elaborate what was first merely a comment-worthy hunch into a somewhat justified answer.

If you cannot capture all the frequencies of a given signal $f(x)$, then the best you can do is exactly capture all the frequencies up to a certain maximum. That is, given a maximum frequency $\omega$, the closest signal in a least-squares sense is an ideally low-pass filtered version of the signal, $g(x) := f(x) * (2\omega/\pi)\operatorname{sinc}(2\omega x)$, where I'm using the unnormalized $\operatorname{sinc}(x) = \sin(x)/x$. Now in most(?) piecewise smooth signals with a discontinuity, for large enough $\omega$, the discontinuity "looks" like a scaled, translated version of a step function, so let's restrict our attention to the standard Heaviside step function, $H(x) = \begin{cases}0 & \text{if } x \lt 0, \\ 1 & \text{if } x \ge 0.\end{cases}$ In this case, $H(x)$ is invariant to scaling of $x$, so it doesn't matter what $\omega$ is; let's choose $\omega = 1/2$ for convenience.

When $f(x) = H(x)$ and $\omega = 1/2$, the convolution becomes $\begin{align}g(x) &= H(x)*\operatorname{sinc}(x)/\pi \\ &= 1/\pi \int_{-\infty}^x \operatorname{sinc}(x) dx \\ &= \operatorname{Si}(x)/\pi + 1/2.\end{align}$ The extrema of this function occur at the zeros of $\operatorname{sinc}(x)$, which are $x = n\pi$ for $n = 1, 2, \ldots$ The corresponding maximum deviations from the limit $\lim_{x\to\infty} g(x) = 1$ are $g(n\pi) - 1 = \operatorname{Si}(n\pi)/\pi - 1/2.$