Suppose the polynomial $p(x) = a_n x^n + ... + a_1 x + a_0 $ has only positive real roots given by $(\alpha_1, \alpha_2, ..., \alpha_{n})$. It is already known that if someone has the power sum of roots of the polynomial then it is possible to find the individual roots by using the Newton's identities method.
Now, let's define a new $\tilde{p}(x)$ that is version of $p(x)$ with the $a_n$ terms somehow slightly corrupted. Is it possible to find real and positive roots close to the ones of the $p(x)$ having only the power sum of roots of the polynomial $\tilde{p}(x)$?
i) the roots of the polynomial $p$ are continuous functions of the $(a_k)$.
ii) Assume that $p$ has several $>0$ roots of the order $10^{-10}$ and several $>0$ roots of the order $10^{10}$ and that you consider a polynomial $q=\sum_i b_ix^i$ close to $p$. We assume also that the roots of $p$ are simple (otherwise they may become non-real for $q$).
There are $2$ problems. Firstly, the roots of $q$ are close to the roots of $p$; then the $|b_k-a_k|$ must be small enough for the roots of $p$ to remain positive.
Secondly, if you explicitly know $q$, then, with some standard software, you can obtain good approximations of the roots of $q$.
Unfortunately, you know only the $S_k((r_i)_i)$ here the $(r_i)_i$ are the roots of $q$; then the system to be solved is very unstable. For example, with Maple, fsolve is unable to recover the $(r_i)_i$.