Limit of a sequence of polygons. - MathOverflow

Limit of a sequence of polygons.

2011-01-03T11:51:19Z

Begin with a polygon $P_0$.

Place two points on every edge of the polygon such that they divide each side equally into three parts. Create a new polygon $P_1$ by connecting all new points with lines.

If we begin with a square and iterate this process, what is the limit as the number of iterations approaches infinity? It is clearly not a circle, but what the correct answer is, I do not know.

Answer by Joseph O'Rourke for Limit of a sequence of polygons.

2011-01-03T14:09:11Z

The left figure shows the first three iterations, the right after 10 iterations, when the polygon has 4096 vertices.

Answer by Bill Thurston for Limit of a sequence of polygons.

2011-01-03T14:11:50Z

Corrected per Thorny's comment: The limit curve does not appear smooth, as is visible in this picture of the first 12 polygons obtained by the process, which become visually indistiguishable at the end:

Despite appearances, and despite my initial assertion, the limit curve actually is smooth. Here is a sequence of closeups of one of the apparent corners, enlarging by a factor of 2 in successive images:

As Thorny pointed out, for the midpoint of any edge, there is an affine transformation fixing the point and taking the limit curve to itself, shrinking the edge by a factor of 3, and shrinking the plane mod the tangent to the edge by a factor of 3, and with only one eigenvector, which is tangent to the edge. Therefore, despite my initial impression, the limit curve is actually tangent to the $k$th stage polygon at the midpont of each of its edges. It looks sharp because the slopes converge slowly.

To understand this better, think of a new process scaled up by a factor of 3. If we describe the polygon by a sequence of vectors ${v_1, \dots, v_k}$ representing the sides, at each stage we interpolate the sum of each pair of successive vectors in circular order. The slopes of the vectors are rational, and the interpolation is by Fary addition (adding numerators and denominators). For instance, if we start with the vectors $(1,0)$ and $(0,1)$, then after 4 subdivisions we get this sequence of slopes:

{0, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1, 4/5, 3/4, 5/7, 2/3, 5/8, 3/5, 4/7, 1/2, 3/7, 2/5, 3/8, 1/3, 2/7, 1/4, 1/5}

All rational slopes are obtained in this way. Since the curves are convex and the slopes are dense in the limit, the limiting curve is $C^1$, but it is not twice differentiable everywhere.

You can't expect this kind of curve to have a name (in most cases).

Answer by lhf for Limit of a sequence of polygons.

2011-01-03T15:34:31Z

You've described de Rham's trisection method (Un peu de mathématiques à propos d'une courbe plane. Elemente der Math. 2, (1947). 73–76, 89–97.) It's the first example of a corner-cutting method. For an analysis of convergence, see Carl de Boor's Cutting corners always works (Comput. Aided Geom. Design 4 (1987), no. 1-2, 125–131). See also Chaikin's corner-cutting method and subdivision surfaces.

Answer by Gerald Edgar for Limit of a sequence of polygons.

2011-01-03T17:11:31Z

As noted, goes back to de Rham. Found also in:

Georges de Rham, "Sur quelques courbes définies par des equations functionnelles". Univ. e Politec. Torino. Rend. Sem. Mat. 16 1956/1957 101–113.

English tranlation in by book CLASSICS ON FRACTALS.