Suppose, for an unknown quantity x ∈ ( 0 , 1 ) you observe the signs (positive, negative or zero) of all possible quantities of the form ∑ a n x n where a n ∈ { − 1 , 1 , 0 } for n ∈ { 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?

Question

Suppose, for an unknown quantity   x  ∈  (  0  ,  1  ) you observe the signs (positive, negative or zero) of all possible quantities of the form   ∑      a    n        x    n   where       a    n    ∈  {  −  1  ,  1  ,  0  } for   n  ∈  {  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?

Chandler Hurley · Accepted Answer

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  +  ⋯. More formally, if   0  &amp;lt;  x  ≤  1  /  2, then      x    k    &amp;gt;      x          k      +      1        +      x          k      +      2        +  ⋯  +      x          k      +      m        ≥        |        a          k      +      1            x          k      +      1        +      a          k      +      2            x          k      +      2        +  ⋯  +      a          k      +      m            x          k      +      m              |    .for any finite value of   m. Thus the sign of   ∑      a    n        x    n   is just the sign of the first nonzero value of an (in the list       a    0    ,      a    1    ,      a    2    ,  …), regardless of where   x falls between   0 and   1      /    2, because the first term of the polynomial is sufficiently &quot;heavy&quot; to override all the rest of the terms.If   x  &amp;gt;  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    ,  … such that      lim          i      →      ∞            ∑          n      =      0          i        a    n        x    n    =  0  ,   and      lim          i      →      ∞            ∑          n      =      0          i        a    n        y    n    ≠  0     (if the limit even exists) if   y  ≠  x  .Such a sequence can be generated using the greedy-type algorithm, and the proof is very similar to the proof that   x  &amp;lt;  1  /  2 fails except the other way around. If   x  &amp;gt;  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.

Suppose, for an unknown quantity x ∈ ( 0 , 1 ) you observe

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