4

$$ \lim_{n\to \infty}\left(\sum_{k=0}^n x^{2^k}\right)^n=0 $$ always seems to have two real solutions. One trivial $x_0=0$ and another around $x_1=-0.65862...$ (see W|A @ $n=13$). Where does this value converge to?

Neither Wolfram nor ISC+ give a good hint about that...

draks ...
  • 18,449
  • 1
    In other words, $\sum \limits_{n=0}^\infty x^{2^n}$. – Karolis Juodelė Dec 11 '13 at 23:24
  • @KarolisJuodelė no $\left(\sum_{k=0}^\infty x^{2^k}\right)^\infty=0$ – draks ... Dec 11 '13 at 23:27
  • 2
    Unless I have miscalculated, it seems to me that any number of solutions for $x$ exist such that the given limit goes to $0$... In particular, many values of $x\in\left[-\frac 12,\frac 12\right]$ will have a sum convergence to a value smaller than $1$, which makes the limit converge to $0$. – abiessu Dec 11 '13 at 23:39
  • Was the exponent maybe intended to be $\frac1n$? With the exponent $n$, every $x$ such that $-1 < \sum\limits_{k=0}^\infty x^{2^k} < 1$ would produce a $0$. – Daniel Fischer Dec 11 '13 at 23:44
  • @DanielFischer I thought it will converge to a certain value, but numerical examples might be misleading... – draks ... Dec 11 '13 at 23:51

2 Answers2

4

There is no one "nontrivial" root. For any sufficiently small (in absolute value) $x$ we have $$\sum_{k=0}^n x^{2^k}<\sum_{k=1}^\infty |x|^k=\frac{|x|}{1-|x|}<1$$ and since this bound is uniform in $n$, we see that the limit of the $n$th powers goes to $0$.

Alex Becker
  • 60,569
4

I don't think there is a known closed form for the zero. The function $f(x) = \sum_{k=0}^{\infty} x^{2^k}$ was studied by Mahler in his paper On a special function, Journal of Number Theory, Vol. 12, Issue 1, Feb. 1980, pp. 20–26 (PDF link) and he studied the zeros of its partial sums in the paper On the zeros of a special sequence of polynomials, Mathematics of Computation , Vol. 39, No. 159, Jul. 1982, pp. 207-212 (PDF link).

You can be sure that there are exactly two real zeros since

  • $f''(x) > 0$ (at most two roots),
  • $f(0) = 0$ (found one),
  • $f'(0) > 0$ (changes sign here),
  • $f(x) \to +\infty$ as $x \to -1^-$ (must change sign again).
Hans
  • 9,804
Antonio Vargas
  • 24,993