Dears,
I am interesting to know how many (real) roots can have the polynomial $p(x):=1+a_{1}x+\ldots+a_{n}x^{n}$ in the interval $(-1,0)$, where $a_{i}\in\{-1,1\}$ for all $i=1,\ldots,n$. I think that the Descartes' rule of signs (or even the Sturm's Th.) can not be applied here because we don't know the $a_{i}$ coefficients, only that $a_{i}\in\{-1,1\}$ and not all are equal, i.e. there is at least one change of sign.
From computer experiences, I "suspect" that p(x) can have only one root in the interval (-1,0). What do you think?
Thank for your time!
There can be as many roots in this interval as you like.
Namely, start with the polynomial $f_1(x) = -x^2-x+1$ which has one root $\phi^{-1}$ strictly between $0$ and $1$.
Next we can consider $$f_2(x) = f_1(x)f_1(x^3) = x^8+x^7-x^6+x^5+x^4-x^3-x^2-x+1$$ which has all its coefficients $\pm 1$ and whose roots include $\phi^{-1}$ and $\phi^{-1/3}$. And in general $$ f_n(x) = \prod_{j=0}^{n-1} f_1(x^{3^j}) = \sum_{i=0}^{3^n-1} (-1)^{(\text{number of nonzero digits in base-3 rep of }i)}\, x^i$$ is a polynomial of degree $3^n$ with coefficients $\pm 1$ and $n$ roots strictly between $0$ and $1$.
In order to meet your precise specification, negate all of the odd-degree coefficients to move all the $n$ roots to $(-1,0)$.