Let $P\in\mathbb{C}[X]$ be a complex polynomial of degree $n\geq 2$ with complex roots $\alpha_1, \alpha_2,\ldots, \alpha_n$. My question is about the existence of a formula for the variance of the roots of $P$ in terms of the coefficients of $P$. First, let me fix some notations.

I will denote by $m=m(P)$ the arithmetic mean of the roots $$m=\frac{\alpha_1+\alpha_2+\cdots+\alpha_n}{n},$$ and by $v=v(P)$, the variance $$v=\frac{1}{n}\sum_{i=1}^n\vert \alpha_i-m\vert^2,$$ that is, the arithmetic mean of the squared distance to the mean, which is a quantity of great interest in probability theory. Note that $v=0$ if and only if $P$ has one root of multiplicity $n$.

Le us write $$P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_0=a_n\prod_{i=1}^n(X-\alpha_i).$$

Clearly, the arithmetic mean can be expressed in terms of the coefficients as
$$m=-\frac{a_{n-1}}{na_n}.$$
Consider now the problem of expressing the variance $v$ in terms of the coefficients by a formula involving only *arithmetic operations, radicals and conjugation*
(in what follows, a formula means such a formula).
By a standard computation, the variance can be written as
$$v=(n-1)\vert m\vert^2-\frac{2}{n}Re(\sum_{i<j}\alpha_i\bar{\alpha}_j).$$
So my question becomes :

**Is it possible to find a formula for
$$Re(\sum_{i<j}\alpha_i\bar{\alpha}_j)$$
in terms of the coefficients of $P$ for any given degree $n$, or is there some theoritical obstruction as for expressing the roots themselves ?**

I've worked out the cases $n=2$ and $n=3$ for which the formula looks like that for the roots themselves; so my basic intuition is that it will not be possible in general.

Here are those formulas :

Without loss of generality, we can focus on monic polynomial ($a_n=1$).

The case $n=2$ is particularly simple : a direct calculation shows that $$v=\frac{\vert \alpha_1-\alpha_2\vert^2}{4}=\frac{\vert (\alpha_1+\alpha_2)^2-4\alpha_1\alpha_2\vert}{4}=\frac{\vert a_1^2-4a_0\vert}{4}.$$ In other words, $v=\vert \Delta \vert/4$, where $\Delta$ stands for the discrimant of $P$.

Before considering the case $n=3$, let me make a few observations. For any complex number $t$, the roots of $Q(X)=P(X+t)$ are $\alpha_i-t$, $i=1\ldots n$, and the variance is invariant by translation, hence $v(Q)=v(P)$. On the other hand, the coefficients of $Q$ can be expressed in terms of the coefficients of $P$ by means of arithmetic operations. So, w.l.o.g. one can concentrate on the problem of expressing $v(Q)$ in terms of the coefficients of $Q$ for some well-chosen $t$. A natural choice is to set $t=m$ since, by doing this, we obtain a polynomial $Q$ with $m(Q)=0$.

These remarks show that it is sufficient to consider the case where $P$ is monic, with mean zero. In this case, $$nv(P)=-2Re(\sum_{i<j}\alpha_i\bar{\alpha}_j).$$

Starting directly from Cardan's formulas for the roots of $$P(X)=X^3-pX-q,$$ I found the following expression : $$v=\sqrt[3]{\left\vert\frac{q+\sqrt{\Delta}}{2}\right\vert^2}+\sqrt[3]{\left\vert\frac{q-\sqrt{\Delta}}{2}\right\vert^2},$$ where $\Delta=q^2-4p^3/27$ is the discriminant of the associated quadratic equation.

real, the question is quite easy. $\endgroup$ – Igor Rivin Dec 1 '14 at 15:13