$\sum(\log^2(root))$ of an nth order polynomial.

Question

Is there an approach to find the $\sum_{over all roots}(\log^2(root))$ of an nth order polynomial of the form

$a_nx^n+a_{n-1}x^{n-1}+.......+a_0$; in terms of the coefficients $a_n$, $a_{n-1}$, $a_{n-2}$,.....,$a_{0}$ (i.e., without explicitly computing the roots and then taking Log and squaring and summing up) ?

Answer 1

I don't think there's a stantard way of doing that. However, if you replace $\log^2(root)$ by $\log(root)$ you can do the following $$\sum_{r,P(r)=0}log(r)=log(\mbox{product of roots})= \log a_0,$$

That is assuming all logs are defined. An approach to your problem would be to find a functional property verified by $\log^2$ and see if it helps, but I'm a bit skeptic.

UPDATE: see the other clever answer.

Answer 2

You already asked a similar question and you are going to receive a similar answer.
Assuming that all the roots of the given polynomial lie in the ball $\|z-1\|<1$, $$ \log^2(1+z) = \sum_{n\geq 2}\frac{2H_{n-1}}{n}(-1)^n z^n $$ leads to $$ \sum_{\zeta}\log^2(\zeta) = \sum_{n\geq 2}\frac{2H_{n-1}}{n}(-1)^n\!\!\!\underbrace{\sum_{\zeta}(\zeta-1)^n}_{\substack{\text{only depends on}\\\text{the coefficients of p.}}}\!\!\!. $$