The Gibbs Phenomenon

Differentiate $S_N(x) = \frac{4}{\pi}\sum_{k=1}^{N}\frac{\sin((2k-1)x)}{2k-1}$ to prove $S_N'(x) = \frac{2}{\pi}\cdot\frac{\sin(2Nx)}{\sin x}$ . Show the critical points in $(0, \pi)$ are $x_m = \frac{m\pi}{2N}$ , with the first peak at $x_1 = \frac{\pi}{2N}$ . (Complete proof)

\sum_{k=1}^{N}\cos((2k-1)x) = \Re\sum_{k=1}^{N}e^{i(2k-1)x} = \frac{\sin(2Nx)}{2\sin x}

S_N'(x) = \frac{4}{\pi}\cdot\frac{\sin(2Nx)}{2\sin x} = \frac{2}{\pi}\cdot\frac{\sin(2Nx)}{\sin x}

x_m = \frac{m\pi}{2N}\quad (m = 1, \dots, 2N-1)

N = 12

: the first peak sits at

x_1 = \frac{\pi}{24} \approx 0.1309

— the derivative identity's zeros pin down the peak positions.

Show proof

The sum is finite, so termwise differentiation is legitimate: $S_N'(x) = \frac{4}{\pi}\sum_{k=1}^{N}\cos((2k-1)x)$ . View the cosine sum through complex exponentials: $\sum_{k=1}^{N}\cos((2k-1)x) = \Re\sum_{k=1}^{N}e^{i(2k-1)x}$ .
It is a geometric series: $\sum_{k=1}^{N}e^{i(2k-1)x} = e^{ix}\frac{1 - e^{i2Nx}}{1 - e^{i2x}}$ . Factoring, $1 - e^{i2Nx} = -2i\,e^{iNx}\sin(Nx)$ and $1 - e^{i2x} = -2i\,e^{ix}\sin x$ , so the ratio is $\frac{e^{iNx}\sin(Nx)}{e^{ix}\sin x} = e^{i(N-1)x}\frac{\sin(Nx)}{\sin x}$ . Multiplying by the leading $e^{ix}$ gives $e^{iNx}\frac{\sin(Nx)}{\sin x}$ , whose real part is $\frac{\cos(Nx)\sin(Nx)}{\sin x} = \frac{\sin(2Nx)}{2\sin x}$ .
Hence $S_N'(x) = \frac{4}{\pi}\cdot\frac{\sin(2Nx)}{2\sin x} = \frac{2}{\pi}\cdot\frac{\sin(2Nx)}{\sin x}$ — the Dirichlet kernel in closed form for the odd frequencies.
On $(0, \pi)$ we have $\sin x \ne 0$ , so critical points come from $\sin(2Nx) = 0$ : $2Nx = m\pi$ , i.e. $x_m = \frac{m\pi}{2N}$ for $m = 1, \dots, 2N-1$ . On $(0, \frac{\pi}{2N})$ both $\sin(2Nx) > 0$ and $\sin x > 0$ , so $S_N'(x) > 0$ ; starting from $S_N(0) = 0$ the sum climbs monotonically to its first maximum (the peak) at $x_1 = \frac{\pi}{2N}$ .

Show the first peak's height tends to $\frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898$ , marking the step that passes the limit $N \to \infty$ under the integral as the analysis step. Also prove $\mathrm{Si}(t) = \int_0^t \frac{\sin u}{u}\,du$ attains its global maximum at $t = \pi$ . (Complete except for the honestly marked analysis step)

S_N(x) = \frac{2}{\pi}\int_0^{2Nx}\frac{\sin t}{2N\sin(t/2N)}\,dt

\xrightarrow{\text{analysis step}} \frac{2}{\pi}\int_0^{2Nx}\frac{\sin t}{t}\,dt

\mathrm{Si}'(t) = \frac{\sin t}{t} > 0 \quad \text{on } (0, \pi)

N = 12

: measured peak

\approx 1.17956

against the theoretical limit

\frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898

— error under

0.0006

.

Show proof

Since $S_N(0) = 0$ , we have $S_N(x) = \int_0^x S_N'(u)\,du$ . Substituting the previous problem's identity gives $S_N(x) = \frac{2}{\pi}\int_0^x \frac{\sin(2Nu)}{\sin u}\,du$ ; the change of variables $t = 2Nu$ ( $dt = 2N\,du$ ) yields $S_N(x) = \frac{2}{\pi}\int_0^{2Nx}\frac{\sin t}{2N\sin(t/2N)}\,dt$ — pure algebra so far.
For each fixed $t$ , $2N\sin(t/2N) \to t$ (since $\sin v \sim v$ ). ★ Here — exactly once — we use analysis: for $x \le \frac{\pi}{2}$ (in particular at the peak $x_1 = \frac{\pi}{2N}$ ), $t \in (0, 2Nx]$ gives $v = \frac{t}{2N} \in [0, \frac{\pi}{2}]$ , and $\sin v \ge \frac{2v}{\pi}$ yields $2N\sin(t/2N) \ge \frac{2t}{\pi}$ . So the integrand is dominated by $\frac{\pi}{2}\cdot\frac{\lvert\sin t\rvert}{t} \le \frac{\pi}{2}$ . Dominated convergence passes the limit inside: $S_N(x) \to \frac{2}{\pi}\int_0^{2Nx}\frac{\sin t}{t}\,dt = \frac{2}{\pi}\mathrm{Si}(2Nx)$ .
At the first peak $x_1 = \frac{\pi}{2N}$ we have $2Nx_1 = \pi$ , so the peak height tends to $\frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898$ . Numerical cross-check: it agrees with the Fourier scene's observed $\mathrm{GIBBS\_PEAK} = 1.178979744472167$ to within $10^{-9}$ .
Since $\mathrm{Si}'(t) = \frac{\sin t}{t} > 0$ , $\mathrm{Si}$ is strictly increasing on $(0, \pi)$ . For $t > \pi$ , split the tail $\int_{\pi}^{t}\frac{\sin u}{u}\,du$ into lobes $L_k = \int_{k\pi}^{(k+1)\pi}\frac{\sin u}{u}\,du$ ( $k \ge 1$ ): they alternate in sign, shrink because $\lvert L_{k+1}\rvert = \int_{k\pi}^{(k+1)\pi}\frac{\lvert\sin v\rvert}{v+\pi}\,dv < \int_{k\pi}^{(k+1)\pi}\frac{\lvert\sin v\rvert}{v}\,dv = \lvert L_k\rvert$ , and the first lobe $L_1 < 0$ . Every partial sum of this alternating series is negative (even counts are sums of negative pairs, odd counts add one more negative lobe), so $\mathrm{Si}(t) < \mathrm{Si}(\pi)$ for $t > \pi$ . Hence $\mathrm{Si}$ attains its global maximum at $t = \pi$ , with value $\approx 1.85194$ .

Compute that the overshoot is about $8.949\%$ of the jump (size 2), and prove why it never disappears as $N \to \infty$ . (Complete proof)

\frac{\frac{2}{\pi}\mathrm{Si}(\pi) - 1}{2} \approx 0.08948987 \approx 8.949\%

S_N\!\left(\frac{t}{2N}\right) \;\xrightarrow[N\to\infty]{}\; \frac{2}{\pi}\mathrm{Si}(t) \quad \text{(N-independent limit)}

Even at

N = 12

the overshoot is already

8.949\%

of the jump — and growing

N

leaves this fraction untouched.

Show proof

The peak height tends to $\frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898$ while the square wave sits at $1$ , so the overshoot is $1.17898 - 1 = 0.17898$ . The jump runs from $-1$ to $1$ (size $2$ ), hence the overshoot fraction is $\frac{0.17898}{2} = 0.08948987 \approx 8.949\%$ .
Why it never dies: in the rescaled coordinate $t = 2Nx$ the whole curve converges to the $N$ -independent limit function $\frac{2}{\pi}\mathrm{Si}(t)$ (previous problem). The first bump has a fixed shape and height in $t$ -coordinates.
Back in the original $x$ -coordinate the $t$ -interval is squeezed by $1/(2N)$ : the bump's width $\sim \pi/N \to 0$ narrows, but its height tends to the constant $\frac{2}{\pi}\mathrm{Si}(\pi) - 1 \approx 0.179$ . In one sentence: the bump narrows like $1/N$ but never shrinks.
This is why the Fourier scene's numerical observation had to be exactly $\frac{2}{\pi}\mathrm{Si}(\pi)$ — analysis catching up to prove the constant the numerics had already seen.

Examples & Exercises