Solution to The February 16, 2006 PoW

Let <x> = d(x,Z), i.e., the distance x is to the nearest integer. It is well known that S0∞ <10nx>/10n represents a continuous, nowhere differentiable function which has a period of 1. Find the area under the curve on the interval [0,1]. And find the maximum of the curve, and all x-values where the maximum occurs on [0,1].

area figure 1 + area figure 2 + area figure 3 ...+ ...

Let ak(x) = <10kx>/10k. Since each ak(x) is a nonnegative continuous function ( made up of 10k triangles),
∫01S0∞ak(x) dx = S0∞∫01ak(x) dx. Therefore the area under the curve is Sj=0∞ 10j(½)(10-j)(½ ×10-j) = 5/18.

We will use rapid induction to find all the places where the maximum occurs. Note that a0(x) + a1(x) is a graph that has a maximum value of ½ for all x in the interval [.45,.55]. And if we add a2(x) + a3(x), we have a maximum of .5 + .005 when x is in any of the ten intervals {[.4545,.4555], [.4645,.4655], [.4745,.4755], ..., [.5445,.5455]}. When we find the maximum of a0 + ... +a5, each of the ten previous intervals will be cut into ten much smaller intervals as before. We now have a maximum of .50505 when x is in any of the intervals {[.454545,.454555], [.454645,.454655], [.454745,.454755], ..., [.455445,.455455], [.464545,.464555], ..., [.465445,.465455], ..., ...,[.545445,.545454]}. And with rapid induction (or: it is easy to see, or: we leave it to the reader) we find that we have a max of
0.50505050... = 50/99 when x is inside the set .454545... + E = 45/99 + E, where
E = {x: x = Sj=1∞ kj/100j, with each kj = 0, 1, 2, ..., 8, or 9}.
Then .454545... = 45/99 is the smallest such x, and .545454... = 54/99 is the largest such x. Notice E is an uncountable nowhere dense closed set.

Perhaps an easier problem would have been to work on S0∞ <2nx>/2n. Instead of getting ten new intervals for each old interval, we get two new intervals for each previous interval (after adding a2n + a2n+1). The maximum is 2/3 whenever x is an element of {x: x = Sj=1∞ kj/4j, with each kj = 1 or 2} = 1/3 + F where F = {t: t = Sj=1∞ kj/4j, with each kj = 0 or 1}. The largest such x is 2/3.