The Ringing That Never Dies

Layer sine waves onto a square wave. The ringing beside each jump never dies out, no matter how large

N

gets.

Examples & Exercises

Compute the sine-series coefficients $b_{2k-1} = \frac{4}{\pi(2k-1)}$ of the square wave by direct integration.

b_n = \frac{2}{\pi}\int_0^{\pi} \sin(nx)\,dx = \frac{2(1-\cos n\pi)}{n\pi}

N = 5

: the first 5 odd harmonics mimic the step.

Show proof

The square wave is odd on $(-\pi, \pi)$ , so cosine coefficients vanish and $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = \frac{2}{\pi}\int_0^{\pi}\sin(nx)\,dx$ .
Integrating gives $\frac{2}{n\pi}(1 - \cos n\pi)$ : $0$ for even $n$ , $\frac{4}{n\pi}$ for odd $n$ .
So $b_{2k-1} = \frac{4}{\pi(2k-1)}$ and $b_{2k} = 0$ .

Verify numerically that the peak near the jump tends to $\approx 1.179$ , i.e. about 9% of the jump. (Numerical check)

\max S_N \to \frac{2\,\mathrm{Si}(\pi)}{\pi} \approx 1.17898

N = 50

: peak

\approx 1.179

— overshooting 1 by

0.179

Show proof

Measuring the peak just right of the jump at $N = 50$ gives $\approx 1.179$ (theory: $\frac{2\,\mathrm{Si}(\pi)}{\pi} \approx 1.17898$ ).
The jump size is $2$ , so the excess $0.179$ above $1$ is about $8.95\%$ of the jump.
This fraction does not go to 0 as $N$ grows — that is the Gibbs phenomenon. (Verified numerically; the rigorous proof is omitted.)

Show algebraically that $S_N$ at $x = \pi/2$ becomes the Leibniz series.

S_N\!\left(\tfrac{\pi}{2}\right) = \frac{4}{\pi}\sum_{k=1}^{N}\frac{(-1)^{k+1}}{2k-1} \to 1

x = \pi/2

the partial sum is

\frac{4}{\pi}

times the Leibniz partial sum.

Show proof

Since $\sin\left((2k-1)\frac{\pi}{2}\right) = (-1)^{k+1}$ , $S_N(\pi/2) = \frac{4}{\pi}\sum_{k=1}^{N}\frac{(-1)^{k+1}}{2k-1}$ .
As $N \to \infty$ , the bracket tends to the Leibniz series $\frac{\pi}{4}$ (series chapter).
Hence $S_N(\pi/2) \to \frac{4}{\pi}\cdot\frac{\pi}{4} = 1$ , matching the square wave's value.