Multiple roots of polynomials with coefficients $\pm 1$ Question P. Can a polynomial $P(x)=\sum_{n=0}^ma_nx^n$ with coefficients $a_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?
Also I am interested in a similar question for analytic functions.
Question A. Let $f(x)=\sum_{n=0}^\infty a_nx^n$ a series with coefficients $a_n\in\{-1,1\}$ such that $\sup_{m\in\mathbb N}|\sum_{n=0}^ma_i|<\infty$. Can the analytic function $f$ have a multiple root in the interval $(0,1)$?
 A: Not an answer, only some sort of evidence for A, hence community wiki.
There seems to be an algorithm that produces a sequence of polynomials $(P_n)$ like in P, with $P_0=1$, $P_{n+1}=P_n\pm x^{n+1}$, $n\geqslant0$, and with $P_n$ having a local minimum at $m_n$ such that the sequence of $m_n$ converges to $\approx0.7257794$ while $P_n(m_n)$ tends to 0.
You do this: let $P_6=1-x-x^2-x^3+x^4+x^5+x^6$; it has a nice local minimum at $m_6\approx0.719842$. Now iterate the following procedure. Find minima of $P_n+x^{n+1}$, $P_n-x^{n+1}$ in the vicinity of $m_n$. Compare absolute values of these two polynomials at these minima. Choose for $P_{n+1}$ the one with smaller absolute value (and for $m_{n+1}$ where that value is attained).
Experimentally, these values become quite small quickly. For example, $P_{500}(m_{500})$ is about $1.644734\times 10^{-70}$.
Of course there is no control of the sequence $P_n(1)$ which is required to remain bounded. In fact $P_3(1)=-2$, $P_7(1)=2$, while $P_n(1)$ is either $-1$, $0$ or $1$ for all other $n$ up to $500$.
