Square Orbit
(3 · Σ_(k=1)^(1 + floor(mod(t, 10))) (−1)^(k−1) · sin((2k−1) · u)/(2k−1)², 0, 3 · Σ_(k=1)^(1 + floor(mod(t, 10))) (−1)^(k−1) · sin((2k−1) · (u + π/2))/(2k−1)²)
Open in the app The dials and keys named below are the app's.
What it draws
Both coordinates are the same Fourier partial sum, a quarter period apart. The series Σ (−1)^(k−1) · sin((2k−1)u)/(2k−1)² is the triangle wave, and the second copy takes u + π/2 in place of u. The y expression is 0, so the path stays flat on the ground.
One term each makes them sin(u) and cos(u), which is a circle of radius 3. The clock adds a term a second, and by the tenth the path is a diamond: at every sample |x| + |z| ≈ 3.7, which is the equation of a square standing on its corner.
Why a triangle wave squares it
A square traced at constant speed moves along one edge, turns, and moves along the next. On the right-hand edge one coordinate holds steady while the other runs from one end to the other, so each coordinate rises linearly, holds, falls linearly, holds. That is a triangle wave, and the two coordinates are offset by a quarter of a period because the edges are.
The corners sharpen but never quite arrive. Measured from the centre, a true square reaches its corners at √2 ≈ 1.414 times its edge distance. This path reaches 1.348 at four terms and 1.385 at ten, closing on the value without meeting it.
Try
- Watch one cycle from Rewind. The circle grows corners over ten seconds at Speed 1, then snaps back to a circle.
- Raise Trail to 20. Older outlines stay behind, so every stage of the convergence is on screen at once.
- Drop the alternating sign, using sum(sin((2k−1)u)/(2k−1)², k, 1, 8) for both. The harmonics no longer line up with the corners, and the ratio falls to 1.16, much nearer a circle.
- Change both denominators to (2k−1). That series is the square wave, not the triangle, and its overshoot throws the path far past the corners: the ratio leaps to 3.3 rather than settling near 1.41.