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For some positive integer $k$ and a vector $s = (s_0,s_1,\dots,s_k) \in \left\{ -1,1 \right\}^{k+1}$ of signs, let

$$ p_{k,s} = s_0 1 + s_1 x + s_2 x^2 + \dots + s_k x^k $$

be a "signed" geometric progression, viewed as a polynomial in $x$. Let $U$ be the set of real zeros of all such polynomials, i.e. $x \in U$ iff there exists some $k \in \mathbb{N}$ and $s \in \left\{ -1,1 \right\}^{k+1}$ such that $p_{k,s}(x) = 0$.

In this post, which motivated my question, it was shown that all elements of $U$ must have absolute value less than two. It was also shown there (see the comment of @JeppeStigNielsen) that $-2$ and $2$ are accumulation points of $U$.

Question: Is $U$ dense in $(-2,2)$?

n_flanders
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    The same proof shows that all elements of $U$ must hae absolute value at least $\frac12$ (note that $u$ is a root of such a polynomial if and only if $\frac1u$ is). Computations certainly suggest that $U$ is dense in $[-2,-\frac12]\cup[\frac12,2]$. – Greg Martin Apr 07 '23 at 19:52
  • @GregMartin Thanks for the very insightful comment. Unfortunately, your claim regarding the zeros is only true if the signs are assigned symmetrically. However, the zeros of $x + x^2 + \dots + x^k - 1$ converge to $1/2$ and an elementary argument shows that they can't have smaller absolute value, so maybe I misunderstood you. In any case a very helpful observation! – n_flanders Apr 10 '23 at 08:03
  • I'm not saying that the roots of a single $p_{k,s}$ are closed under taking reciprocals; I'm saying that the roots of the collection of all possible $p_{k,s}$ are closed under taking reciprocals. – Greg Martin Apr 10 '23 at 15:44
  • Ah, understood, thanks for the clarification! – n_flanders Apr 10 '23 at 19:41

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