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I was investiging the series $\sum_{k=0}^\infty{x^{(2^k)}}$ where $x\in\mathbb{R}$, and came up with this question:

Is there a closed form of the negative root of $\sum_{k=0}^\infty{x^{(2^k)}}=0$ ?

On Desmos, taking the first $5$ terms, and taking the first $1001$ terms, both give the negative root as $-0.6586$ to $4$ decimal places.

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Here are my thoughts. According to this, "there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients". So if we take the first five terms of the series, I would not expect a closed form of the negative root. But, based on my experience with power series, if we take the limit as the number of terms approaches $\infty$, I would be more inclined to believe that a closed form might exist.

Dan
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    "there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients" -- In general, this is true. However, it makes assumptions on the kinds of functions/operations permissible, and this is a general statement about a formula working for all possible quintic polynomials (of certain coefficients). Closed forms may exist for special cases, the obvious one being $f(x) = x^n - 1$ having a root at $x=1$ always. Consequently, I wouldn't take the insolvability of the quintic as a reason to believe a closed form does not exist for this. – PrincessEev Jul 24 '22 at 03:41
  • @EeveeTrainer Agreed. To clarify: If we take the first five terms of the series, I would not expect that a closed form must exist. – Dan Jul 24 '22 at 03:44
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    The tenth term evaluated at the root is already pretty small, $(-0.6586)^{512} = 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000136\ldots$, so taking $1001$ terms won't make much of a difference. – Théophile Jul 24 '22 at 03:46
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    The paper linked in this answer on MSO lists the negative root $−0.6586...$ numerically, not mentioning a closed form or other references. – dxiv Jul 24 '22 at 03:49
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    I will be surprised if it admits a closed-form expression. Indeed, $f(z)=\sum_{n=0}^{\infty} z^{2^n}$ is a quintessential example of lacunary functions. Since most, if not all, well-studied special functions do not show lacunary behavior, I believe that $f$ is not linked to any such functions. It seems to me that this strongly suggests that the negative zero of $f$ is not linked to such special functions. – Sangchul Lee Jul 24 '22 at 04:17
  • as @EveeTrainer mentions polynomials can in general only be sure to be solved in finite number of steps of */-+ and roots if they are degree 4 or less, but if other things are allowed such as transcendental functions, trigonometric polynomials and others, closed forms could exist for larger degrees as well. – mathreadler Jul 24 '22 at 04:58
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    $A339253$ in $OEIS$ gives $86$ digits – Claude Leibovici Jul 24 '22 at 06:47
  • @ClaudeLeibovici Thanks - good idea to check OEIS (I had checked Wolfram, but hadn't thought about checking OEIS). Of course, appearing in OEIS does not mean there is no closed form (e.g. $\pi$). Incidentally, the coordinates of the minimum point on the curve $(-0.3828964..., -0.2143301...)$, do not appear in OEIS. – Dan Jul 24 '22 at 07:47

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