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Puzzle 8: Solution

**Q**
For positive *x*<1, consider the alternating sum
*S*(*x*) =
*x* −
*x*^{2} +
*x*^{4} −
*x*^{8} +
*x*^{16} −
*x*^{32} + − …
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*) = *x* − *S*(*x*^{2}),
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*) = *x* − *x*^{2}
+ *S*(*x*^{4})
> *S*(*x*^{4}).
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
log_{4}(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!