When algorithms collide: An infinite sum that isn't (or is it?)

This page went online Dec. 26, 2001

Peano, Hilbert and others devised space-filling curves about a century ago, helping to initiate the field of topology. So, from a topological standpoint, the result below is unremarkable.

Yet, philosophically it seems curious that the same structure can yield two different values.

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It seems quite reasonable that an area under a curve is ever better approximated by summing areas defined by rectangular strips that fit the area ever more closely. At the limit of width w = 0, the area is defined as the infinite sum of vertical heights f(x) between x=a and x=b. This calculus approximation algorithm is standard.

Yet it is possible to devise an algorithm which seems to yield a sum of y heights that clashes with the area algorithm.

We create a sawtooth path between some curve f(x) and a finite x axis interval by this means: Divide the interval by h, keeping h a positive odd integer. Then trace the path a to f(a) to f(a+1/h) to f(a+2/h) to a+2/h to a+3/h to f(a+3/h) ... to f(a+n/h) to a+n/h = b. The length of this path is always longer than the path connecting the discrete points of f(x). [There are more than two points between f(a) and f(a+1/h) for the sawtooth path but only two points for the other path.]

As h approaches infinity, the number of sawteeth approaches infinity. At the limit of h=infinity, the sawteeth have narrowed to the infinitesimal area f(x), where the route is presumably twice f(x). But the infinite sum of f(x) cannot be infinite if the area is finite (fits inside a finite square). Quite often, the infinite sum (the integral) is less than the continuous arc length for f(x).

So this would seem to indicate that the sawtooth function must be considered undefined at h = infinity.

The special case of the unit square makes this plain:

Use the unit square lying on the positive x axis between origin and x=1. Divide 1 by the odd (for convenience) positive integer h. Trace a sawtooth path from 0,0 to 0,1 to 1/h,1 to 1/h,0 to 2/h,0 to 2/h,1 and so on to 1,0. The sawtooth path length is h+2. As h approaches infinity, the sawtooth path length also nears infinity.

It would seem that at the limit, the 'infinitesimal entity' is twice the height f(x). The sum is the integral of f(x), which, by the fundamental theorem of calculus, is the area quantified as 1.

So that at infinity, if the sawtooth function exists, the sawtooth route length collapses to distance 1.

But would a long-lived or very speedy salesman traveling through all the points of a sawtooth curve find that the sawtooth route is optimal at infinity?

The area is rarely directly proportional to some x interval or radius. For example, take any finite radius circle and define its radius as 1. Then the area is less than the circumference. But take exactly the same circle and define its radius as 3, and now the area is greater than the circumference. Now make a sawtooth path through half of the same circle.

The salesman will find, as h increases, that his path length exceeds the area quantity for every h after some h=k. If the sawtooth algorithm is completable (an 'actual infinity' is agreed), then he would appear to find that the optimal route is the sawtooth route.

But, there is an infinity of areas that can be assigned to the semicircle. For every area quantity there exists a k after which h means that the sawtooth path length is longer.

This observation can be generalized.

So we conclude that the sawtooth function is either unbounded or undefined at h's infinite limit.

If we say it is unbounded, however, we would need to explain why we are forbidden to sum the f(x) heights.

As my son Jim, a Cornell topologist, points out, sawtooth functions are well-known for producing curves with infinite perimeters that contain a finite area, the Koch snowflake being a famous example.

Of course the issue here is not that such functions occur but that the logical conclusion of this particular sawtooth algorithm yields an anomolous result.