Gibbs Phenomenon

gibbs-phenomenon.md

Gibbs Phenomenon

By Andy Pan (Qbane), 2016/3/10. (See Gist for revision history.)

This is a test document on MathJax support on various platforms, mainly on printed material.

For large $N$, the graph of the $N$th partial sum of the Fourier series of $f$ on $[-L,L]$ overshoots the graph of the function at a jump discontinuity by approx. $9$ of the magnitude of the jump.

(a) $N$th Partial Sum

We write out the $N$th partial sum of the Fourier series of $f(x)$ and expand its coefficients: $$ \begin{align} S_N(x) &= a_0 + \sum_{n=1}^N \left[ a_n \cos \left(\frac{n\pi x}{L}\right) + b_n \sin \left(\frac{n\pi x}{L}\right) \right] \ &= \frac{1}{2L} \int_{-L}^L f(t) dt \ &+ \frac{1}{L} \sum_{n=1}^N \left[ \int_{-L}^L f(t) \cos\left( \frac{n\pi t}{L} \right) dt \right] \cos\left( \frac{n\pi x}{L} \right) \ &+ \frac{1}{L} \sum_{n=1}^N \left[ \int_{-L}^L f(t) \sin\left( \frac{n\pi t}{L} \right) dt \right] \sin\left( \frac{n\pi x}{L} \right) \ &= \frac{1}{L} \int_{-L}^L f(t) \left{ \frac{1}{2} + \sum_{n=1}^N \cos\left[ \frac{n\pi (t-x)}{L} \right] \right} dt. \end{align} $$

(b) Trigonometric Identity

It can be shown that $\sum _{n=1}^N\cos (n\xi )=\frac{\sin [(N+\frac{1}{2})\xi ]}{2\sin \left(\frac{\xi }{2}\right)}-\frac{1}{2}.$

Here is a trick: multiply it by $\sin \left(\frac{\xi }{2}\right)$ and try to lead to an identity. It can also be simplified by transforming $\cos x=\frac{e^{ix}+e^{-ix}}{2}$ and transforming back to $\sin$ terms. (It would be tricky if you didn't know RHS, though.)

(c) Approximating (a)

For small $x$, $\sin x\approx x$, and using (b), (a) can be expressed as $S_N(x)\approx \frac{1}{2L}{\int }_{-L}^{L}f(t)\frac{\sin [\frac{\pi }{L}(N+\frac{1}{2})(t-x)]}{\frac{\pi }{2L}(t-x)}dt.$

(d) Estimating a Discontinuity

To find out the amount, we use the following function with a jump discontinuity $1$ at $x=L$, $$ f(x) = \begin{cases} 0 & \text{if } -L < x \le x_0, \ 1 & \text{if } x_0 < x < L. \end{cases} $$

Inserting it into $S_N$ yields $$ \begin{align} S_N(x) &\approx \frac 1 {2L} \int_{x_0}^L \frac{\sin \left[ \frac \pi L (N + \frac 1 2) (t-x) \right] }{ \frac{\pi}{2L}(t-x) } dt \ &= \frac 1 \pi \int_{(\pi/L)(N + 1/2)(x_0-x)}^{(\pi/L)(N + 1/2)(L-x)} \frac{\sin s}{s} ds. \end{align}$$

(e) Putting $N\to \mathrm{\infty }$

The integral ${\int }_0^z\frac{\sin t}{t}dt$ is called the sine integral, denoted as $\text{Si}(z)$.

Computing $S_N$ when $N\to \mathrm{\infty }$ diverges to three cases depending on $x$, in which we are only interested in the case that $x>x_0$, that is $(\pi /L)(N+1/2)(x_0-x)\to -\mathrm{\infty }$:

Missing or unrecognized delimiter for \left

$$ \begin{align}
\lim_{N \to \infty} S_N(x) &\approx \frac 1 \pi \int_{(\pi/L)(N + 1/2)(x_0-x)}^\infty \frac{\sin s}{s} ds \\
&= \frac 1 \pi \left{ \lim_{z \to \infty} \text{Si}(z) - \text{Si} \left[ \left(\frac \pi L \right) \left(N + \frac 1 2 \right) \left(x_0-x \right) \right] \right},
\end{align} $$

(f) Evaluating the Sine Integral

$\text{Si}(z)$ at infinity can be computed as follows, $$ \begin{align} \lim_{z \to \infty} \text{Si}(z) &= \int_0^\infty \frac{\sin t}{t} dt \ &= \int_0^\infty \int_0^\infty e^{-st} \frac{\sin t}{t}ds,dt \ &= \int_0^\infty \int_0^\infty e^{-st} \frac{\sin t}{t}dt,ds \ &= \int_0^\infty \frac{1}{s^2 + 1}ds \ &= \frac \pi 2. \end{align} $$

$\text{Si}(z)$ is maximized when $\text{Si}'(z^) = 0$, that is $\frac{\sin z^}{z^} = 0$, and the minimal $x^$ satisfying the condition is $z^{\ast }=\pi$.

Thus the overshoot is at most $$ \frac 1 \pi \left[\frac \pi 2 - \text{Si}(\pi) \right] = \frac 1 2 - \frac{\text{Si}(\pi)}{\pi} \approx 0.0895, $$

which is approximately $9$.

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