## Symmetric polynomials

**Definition**

*A polynomial is symmetric, if * (the equality is understood as the polynomial identity).

**Examples**

Some of them can be expressed through others.

**Problem**

Let . Prove that the sum of powers can be expressed through

.

**Solution**

The first case can be seen outright: .

To check the general case, multiply by : we get . This identity can be rewritten as follows:

One can resolve this identity and obtain the recursive (inductive) formula . With the initial conditions , this allows to compute all one by one.

**Theorem**

*Any symmetric polynomial in two variables can be expressed as a polynomial in .*

**Proof**

A polynomial is symmetric if and only if each its homogeneous component (collection of terms of the same degree) is also symmetric. Thus it is sufficient to prove the theorem only for homogeneous symmetric polynomials.

A homogeneous polynomial in two variables is symmetric if and only if its string of coefficients is palindromic: etc. In this case the terms can be rearranged as follows:

Since each sum of powers is already expressed as a polynomial in , we obtain the required expression for the homogeneous polynomial .

**Uniqueness of the representation**

Can a symmetric polynomial admit two different representations via ? It is sufficient to study only the trivial case of identically zero polynomial (why?).

**Theorem**

*If a polynomial is identically zero as polynomial in , then it is identically zero also as a polynomial in . *

The idea of the proof is to introduce the notion of a “leading term” for polynomials in two variables and for polynomials in the variables such a way that the leading terms can be easily compared for and its expansion.

**Example (bad)**

Declare for a polynomial the leading term be the monomial of the highest degree (with nonzero coefficient) and, in case there are several such terms, the one whose degree in is the highest.

How to find the leading term of a polynomial ? One has to find the monomial with the highest value of , but what next? in the case there are several such monomials with the nonzero coefficients, all of them will contribute to the leading term of the expansion, so theoretically can cancel each other and as a result we cannot guarantee that the outcome will be nonzero.

One has to be more smart in defining what is the leading term.

**Example (good)**

Declare the leading term of a polynomial the monomial of highest degree in with a nonzero coefficient. In case there are several such monomials, choose the one which has largest degree also in . (Such order is called “lexicographical”: it is used to arrange words in the dictionaries. The order is alphabetic in the initial letter, and all words starting with the same letter, are arranged alphabetically in the second letter. If necessary, the process can continue further).

How to determine the leading term of the composite polynomial ? Clearly, one has to choose the terms of the highest total degree (which will give the degree in ). In case there are several nonzero terms with the same value , clearly, the largest degree in will contribute the term with the largest value of (why?). Such term is only one: the corresponding coefficient will give the coefficient before the leading term in the expansion of .

**Proof of the uniqueness theorem**

Assume that the polynomial is nonzero, i.e., not all are zeros. Choose the coefficient with the largest value of , and in case there are several such, take the one with the largest . As follows from the above (good) Example, the lexicographically leading term of the expansion will have the same nonzero coefficient, thus after expansion we get nonzero polynomial, contrary to our assumption.

**Problems for homework/class**

Solve the following systems of equations, using their symmetry.

**Problem**

Compose a quadratic equation, whose roots are cubes of the equation .

**Solution**

Denoting the roots , we know that . It is required to find an equation whose roots will be . By the same Vieta formulas, , . Expressing these symmetric functions of via , we obtain the answer without solving explicitly any of the equations.

Generalizing this trick, we come to the following conclusion: *any rational symmetric function of the roots of a quadratic equation can be expressed as a rational function of (without radicals).*

**Further observations**

- Since the degree of is 2, it enters in the expression of symmetric polynomials of degree in the degree not exceeding . This in many cases simplifies factorization of symmetric polynomials into symmetric factors.
- Palindromic polynomials in one variable, having the form with for all can be reduced to a symmetric form.If the degree is even, then after division by we obtain a rational function which is in fact a polynomial of degree in the variable (prove it!).
If is odd, then is divisible by , i.e., , and the quotient is a palindromic polynomial of an even degree. Indeed, the divisibility follows from the fact that is an alternating sum of the coefficients , which cancel each other because of the parity together with palindromic condition, thus .

The second assumption follows from the following obvious characterization: is palindromic, if and only if . Applied to the representation , we see that , which implies that is palindromic.

This observation allows to reduce degree of palindromic polynomial equations by half.

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