The Gibbs Phenomenon

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 $x_1 = \frac{\pi}{2N}$.

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.

Substitute the peak first, before taking any limit: put $x_N = x_1 = \frac{\pi}{2N}$, so $2Nx_N = \pi$ is fixed and $S_N(x_N) = \frac{2}{\pi}\int_0^{\pi}\frac{\sin t}{2N\sin(t/2N)}\,dt$ — the domain is now the fixed interval $(0, \pi]$, independent of $N$. For each fixed $t$, $2N\sin(t/2N) \to t$ (since $\sin v \sim v$), so the integrand converges pointwise to $\frac{\sin t}{t}$.

★ Here — exactly once — we use analysis, applying dominated convergence on the fixed domain $(0, \pi]$: for $t \in (0, \pi]$, $v = \frac{t}{2N} \in [0, \frac{\pi}{2}]$ and $\sin v \ge \frac{2v}{\pi}$ yields $2N\sin(t/2N) \ge \frac{2t}{\pi}$, i.e. $\frac{t}{2N\sin(t/2N)} \le \frac{\pi}{2}$.

Hence $\left\lvert\frac{\sin t}{2N\sin(t/2N)}\right\rvert = \frac{t}{2N\sin(t/2N)}\cdot\left\lvert\frac{\sin t}{t}\right\rvert \le \frac{\pi}{2}\left\lvert\frac{\sin t}{t}\right\rvert \le \frac{\pi}{2}$, and the dominating function $\frac{\pi}{2}\left\lvert\frac{\sin t}{t}\right\rvert$ (with $\left\lvert\frac{\sin t}{t}\right\rvert \le 1$) is integrable on the finite interval $(0, \pi]$. Dominated convergence gives $S_N(x_N) \to \frac{2}{\pi}\int_0^{\pi}\frac{\sin t}{t}\,dt = \frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898$.

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 $\approx 1.85194$.

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.