Can we find a unique solution to this system of inequalities This is a question I chanced upon while thinking about my research. Suppose, for an unknown quantity $x\in(0,1)$ you observe the signs (positive, negative or zero) of all possible quantities of the form $\sum a_nx^n$ where $a_n\in\{-1,1,0\}$ for $n\in \{0,1,2,3...\}$. 
Of course, if you encounter something like $x=0$ or $x^2+x=0$ among the system of (in)equalities, you can find a unique solution for $x$.
But, can you always find a unique solution for $x$ from there given the infinite set of (in)equalities at your disposal? Or is there a counter-example?
I am more than happy to get hints or possible literature recommendations which might help me find the answer myself. 
 A: You cannot distinguish between values less than 1/2 (including 1/2 if all your polynomials are finite).  The key idea is that $1/2 = 1/4 + 1/8 + 1/16 + \cdots$.  More formally, if $0 < x \leq 1/2$, then $$x^k > x^{k+1}+x^{k+2}+\cdots+x^{k+m} \geq \Big|a_{k+1}x^{k+1}+a_{k+2}x^{k+2}+\cdots+a_{k+m}x^{k+m}\Big|.$$ for any finite value of $m$.  Thus the sign of $\sum a_nx^n$ is just the sign of the first nonzero value of $a_n$ (in the list $a_0, a_1, a_2, \ldots$), regardless of where $x$ falls between 0 and 1/2, because the first term of the polynomial is sufficiently "heavy" to override all the rest of the terms.
If $x > 1/2$, then I think you should be able to uniquely determine $x$ from the signs of the polynomials.  This follows from the following claim: for each $x$ in $(1/2, 1)$, there exists an infinite sequence $a_0, a_1, a_2, \ldots$ such that $$\lim_{i\to\infty}\sum_{n=0}^{i} a_nx^n = 0,\text{ and}$$$$\lim_{i\to\infty}\sum_{n=0}^{i} a_ny^n \neq 0\ \text{(if the limit even exists) if }y \neq x.$$  Such a sequence can be generated using the greedy-type algorithm, and the proof is very similar to the proof that $x < 1/2$ fails except the other way around.  If $x > 1/2$ then each individual term can be outweighed by finitely many future terms in the opposite direction.  So a greedy algorithm (by growing the list one term at a time by picking whatever sign brings you closer to 0) will bring the partial sums closer and closer to 0 over time.
I skipped some steps/rigor so if you want me to elaborate anywhere just make a comment!
