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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.

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  • $\begingroup$ A quick remark: if the roots of the polynomial are real, the question is quite easy. $\endgroup$ – Igor Rivin Dec 1 '14 at 15:13
  • $\begingroup$ It suffices to solve the problem for $\sum_k|\alpha_k|^2$, not that it is any easier. $\endgroup$ – Liviu Nicolaescu Dec 1 '14 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 $a_{n-2}/a_n$. $\endgroup$ – MassiveJack Dec 1 '14 at 18:34
  • $\begingroup$ Another not so helpful comment:$\sum_k |\alpha_k|^2$ is a semialgebraic function in the variables $a_j,\bar{a}_k$. $\endgroup$ – Liviu Nicolaescu Dec 1 '14 at 20:21
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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.

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  • $\begingroup$ 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 $M$ ? $\endgroup$ – MassiveJack Dec 4 '14 at 12:11
  • $\begingroup$ @MassiveJack actually that's a mistake, I meant to write $\alpha_i \beta_i$. $\endgroup$ – Will Sawin Dec 4 '14 at 13:20

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