This is a follow up to If $\sum\limits_{n=1}^{\infty}e_n x^n = 0$ always implies $\sum\limits_{n=1}^{\infty}e_n a_n = 0$, then $(a_n) = (x^n)$?.
For $x \in \mathbb R$ we define the property $P(x)$ as follows: For every sequence $(a_n)$ with
- $a_1 = x$ and
- for every sequence $(e_n) \subset \{-1, 0,1\}$ with $\sum_{n=1}^\infty e_n x^n = 0$ we have $\sum_{n = 1}^\infty e_n a_n = 0$
we get $(a_n) = (x^n)$.
In the linked question it is shown, that for all $x \in [2/(\sqrt{5}+1),1)$ the property $P(x)$ holds.
Can we characterize all $x \in \mathbb{R}$ with $P(x)$?
It is not hard to make the following observations:
- $P(0)$, $P(1/2)$ and $P(1)$ are true
- $P(x)$ implies $P(-x)$
- $P(x)$ is false for $x \in (0,1/2)$ and $x > 2$, since only $(e_n) = 0$ can be used in point 2 above.
What can be said about $x \in (1/2, 2/(\sqrt{5}+1))$ and $x \in (1,2)$?
Thank you.