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.

  • $\begingroup$ A quick remark: if the roots of the polynomial are real, the question is quite easy. $\endgroup$
    – Igor Rivin
    Dec 1, 2014 at 15:13
  • $\begingroup$ It suffices to solve the problem for $\sum_k|\alpha_k|^2$, not that it is any easier. $\endgroup$ Dec 1, 2014 at 18:08
  • $\begingroup$ Yes, this is equivalent, but I choose to put emphasis on the real part of the sum of the $\alpha_i\bar{alpha_j}$'s because, as noticed by Igor Rivin, it gives the solution when all the roots are real since that sum is then equal to $a_{n-2}/a_n$. $\endgroup$ Dec 1, 2014 at 18:34
  • $\begingroup$ Another not so helpful comment:$\sum_k |\alpha_k|^2$ is a semialgebraic function in the variables $a_j,\bar{a}_k$. $\endgroup$ Dec 1, 2014 at 20:21

1 Answer 1


You may notice that your formula is very similar to the formula for the roots of the polynomial. That is not an accident - they are basically equally difficult. So in particular there is no formula in radicals for $n\geq 5$, and you don't want to see the formula in radicals for $n=4$.

proof: Take any formula in terms of arithmetic operations, radicals and conjugation. Because the other operations commute with conjugation, we can assume that conjugation comes first - i.e. the function is an algebraic function, expressible using radicals, of $a_0,\dots, a_n, \overline{a_0}, \dots, \overline{a_n}$.

We can express $a_0,\dots, a_n, \overline{a_0}, \dots, \overline{a_n}$ as algebraic functions of $\alpha_1\dots, \alpha_n, \overline{\alpha_1},\dots,\overline{\alpha_n}$. The mean is also an algebraic function of these numbers. So we are seeking an identity of two algebraic functions. Here is the key point: The subset of $\mathbb C^{2n}$ with coordinates $\alpha_1,\dots,\alpha_n,\beta_1,\dots,\beta_n$ defined by the relations $\beta_k=\overline{\alpha_k}$ is Zariski dense, so any algebraic equation that holds there holds for all $\alpha_i,\beta_i$.

Now we're essentially done, because we have a question of pure Galois theory. We have three fields:

$K = \mathbb C( \alpha_1,\dots, \alpha_n,\beta_1,\dots,\beta_n)$

$L= \mathbb C( a_1,\dots, a_n, b_1,\dots b_n) $ (where $b_i$ are the symmetric polynomials of the $\beta_i$.)

$M = \mathbb C( a_1,\dots, a_n, b_1,\dots b_n, \sum_{i=1}^n \alpha_i \beta_i) $

We want to know whether $M$ is a radical extension of $L$. This happens if and only if its normal closure is solvable. It is contained in $K$, which has Galois group $S_n \times S_n$ over $M$. The subgroup that fixes $\sum_{i=1}^n a_i b_i$ is clearly the diagonal $S_n$. For $n>2$, the diagonal subgroup contains no normal subgroup, and so the normal closure of $M$ over $L$ is $K$. For $n>4$, this is not solvable.

  • $\begingroup$ Thank you for your answer! I'm a probabilist and, unfortunately, my last contact with Galois theory dates back 10 years ago, as I was a student. So, I apologize if my question is sutpid, but why are you considering $\sum a_i b_i$ instead of $\sum \alpha_i\beta_i$ in the definition of $M$ ? $\endgroup$ Dec 4, 2014 at 12:11
  • $\begingroup$ @MassiveJack actually that's a mistake, I meant to write $\alpha_i \beta_i$. $\endgroup$
    – Will Sawin
    Dec 4, 2014 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.