symmetric functions

A symmetric function of $x_1,x_2,\dots,x_n$ is one whose value is unchanged if $x_1,x_2,\dots,x_n$ are permuted arbitrarily. For example, each of the following is a symmetric function of three variables:

$\begin{align*} P(x_1,x_2,x_3) &= x_1x_2+x_2x_3+x_3x_1, \\ Q(x_1,x_2,x_3) &= x^3_1+x^3_2+x^3_3, \\ R(x_1,x_2,x_3) &= \frac{x_2+x_3}{x_1}+\frac{x_3+x_1}{x_2}+\frac{x_1+x_2}{x_3}. \end{align*}$

Certain symmetric functions serve as building blocks for all the rest. Let

$\sigma_k=\sum x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}},$

where the sum is taken over all $\binom{n}{k}$ choices of the indices $i_1,i_2,\dots,i_k$ from $\{1, 2,\dots, n\}$ .
Then $\sigma_k$ is called the $k$ th elementary symmetric function of $x_1,x_2,\dots,x_n$ .

The Symmetric Function Theorem

Every symmetric polynomial function of $x_1,x_2,\dots,x_n$ is a polynomial function of $\sigma_1,\sigma_2,\dots,\sigma_n$ . (And every rational function of $x_1,x_2,\dots,x_n$ is a rational function of $\sigma_1,\sigma_2,\dots,\sigma_n$ .)

E.g. For $n = 3$ the elementary symmetric functions are

$\begin{align*} \sigma_1 &= x_1+x_2+x_3, \\ \sigma_2 &= x_1x_2+x_2x_3+x_3x_1, \\ \sigma_3 &= x_1x_2x_3, \end{align*}$

and the above examples expressed in terms of these are:

$\begin{align*} x_1x_2+x_2x_3+x_3x_1 &= \sigma_2, \\ x^3_1+x^3_2+x^3_3 &= \sigma^3_1-3\sigma_1\sigma_2+3\sigma_3, \\ \frac{x_2+x_3}{x_1}+\frac{x_3+x_1}{x_2}+\frac{x_1+x_2}{x_3} &= \frac{\sigma_1\sigma_2-3\sigma_3}{\sigma_3}. \end{align*}$

Theorem Let $x_1,x_2,\dots,x_n$ be the roots of the polynomial equation

$x^n+c_1x^{n-1}+\cdots+c_n=0,$

and let $\sigma_k$ be the $k$ th elementary symmetric function of the $x_i$ . Then

$\sigma_k=(-1)^k c_k, \qquad \qquad k = 1,2,\dots, n.$

Explanation. On the LHS of the equation

$x^n+c_1x^{n-1}+\cdots+c_n=(x-x_1)(x-x_2)\cdots(x-x_n)$

the coefficient of $x^{n-k}$ is $c_k$ ; on the RHS, the coefficient of $x^{n-k}$ is $(-1)^k$ times the sum of all $\binom{n}{k}$ products of $k$ of the $x_i$ . Thus $c_k=(-1)^k\sigma_k$ .

Problem. Find all solutions of the system of equations

$\begin{align*} x + y + z &= 0, \\ x^2 + y^2 + z^2 &= 6ab, \\ x^3+y^3+z^3 &= 3(a^3+b^3). \end{align*}$

Solution. The LHS of each equation is a symmetric function of $x$ , $y$ , $z$ . This suggests that we can use the information given to construct a polynomial equation whose roots are $x$ , $y$ , $z$ . Let