Puzzle 8: Solution

Q     For positive x<1, consider the alternating sum
S(x) = xx2 + x4x8 + x16x32 + − …
Does S(x) approach a limit as x approaches 1 from below, and if so what is this limit?

A     Since S satisfies the functional equation

S(x) = xS(x2),
it is clear that if S(x) has a limit as x approaches 1 then that limit must be 1/2. One might guess that S(x) in fact approaches 1/2, and numerical computation supports this guess — at first. But once x increases past 0.9 or so, the approach to 1/2 gets more and more erratic, and eventually we find that
S(0.995) = 0.50088… > 1/2.
Iterating the functional equation, we find
S(x) = xx2 + S(x4) > S(x4).
Therefore the fourth root, 16th root, 64th root, … of 0.995 are all values of x for which
S(x) > S(0.995) > 1/2.
Since these roots approach 1, we conclude that in fact S(x) cannot tend to 1/2 as x approaches 1, and thus has no limit at all!

So what does S(x) do as x approaches 1? It oscillates infinitely many times, each oscillation about 4 times quicker than the previous one; see the graph of S(x) for x in [0,0.9995]. (Apply the “magnifying glass” to the top right corner to see the first few oscillations.) If we change variables from x to log4(log(1/x)), we get in the limit an odd periodic oscillation of period 1 that's almost but not quite sinusoidal, with an amplitude of approximately 0.00275. Remarkably, the Fourier coefficients can be obtained exactly, but only in terms of the Gamma function evaluated at the pure imaginary number π i / ln(2) and its odd multiples!