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 kth 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 kth 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

    \[ P(t) = (t-x)(t-y)(t-z)=t^3+c_1t^2+c_2t+c_3. \]

Then c_1=-(x+y+z)=0 and

    \[ c_2 = xy + yz + zx = \frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=-3ab. \]

Finally, P(x) = P(y) = P(z) = 0 yields

    \[ (x^3 + y^3 + z^3) + c_1(x^2 +y^2 + z^2) + c_2(x + y + z) + 3c_3 = 0, \]

from which we find

    \[ c_3 = -\frac{x^3+y^3+z^3}{3}=-(a^3 + b^3).\]

Thus x, y, z are the roots of the cubic equation

    \[ t^3 - 3abt - (a^3 + b^3) = 0. \]

Observe that t = a + b is one of the roots. Now we can factor to obtain

    \[ \big(t -(a + b)\big)\big(t^2 + (a + b)t + (a^2 - ab + b^2)\big), \]

and so find the complete solution set:

    \[ S = \{a + b, a\omega + b\bar{\omega},a\bar{\omega}+b\omega\}, \]

where \omega and \bar{\omega}=\omega^2 are the two complex cube roots of 1.

Lozansky, Rousseau – Winning Solutions (1996)