A symmetric polynomial is a polynomial where if you switch any pair of variables, it remains the same. For example, is a symmetric polynomial, since switching any pair, say and , the resulting polynomial is the same as the initial polynomial. Symmetric polynomials can often be found in Vieta's formula and Newton's identities.
The polynomial is symmetric if for any permutation of ,
Informally speaking, this means that all the variables play the same role in the polynomial and replacing one by another would have no effect. If is a symmetric polynomial, then
Perhaps the most simple example is
Or, a slightly more complex example in three variables is
As atoms build up matter, elementary symmetric polynomials are the building blocks for all symmetric polynomial. The elementary symmetric polynomials for a polynomial consists of variables, , and are defined as
In general, for
Elementary symmetric polynomials in variables, , can be generated using the polynomial
where the elementary symmetric polynomial, , will be the coefficient of .
List all elementary symmetric polynomials that consist of variables .
Fundamental Theorem of Symmetric Polynomials
Every symmetric polynomial can be rewritten in terms of the elementary symmetric polynomials .
Express only in terms of the elementary symmetric polynomials.
When factoring symmetric polynomials, it's useful to make use of the fundamental theorem of symmetric polynomials and rewrite the original symmetric polynomial completely in terms of the elementary symmetric polynomials, because then you can factor more easily. When you are dealing with power sums, Newton's identities are helpful in that they allow you to express the power sums in terms of elementary symmetric polynomials.
The above polynomial is a symmetric polynomial because you can switch any two variables and still get the same polynomial.
Thus, we can express it in terms of the elementary symmetric polynomials. In this case, we can use Newton's identities for the power sum portion which can be rewritten as , where is the degree elementary symmetric polynomial and is the degree power sum.
This gives us
Since , this is equal to
Let be positive real numbers. Define as the arithmetic mean, and as the geometric mean. Then, the AM-GM inequality states that .
We have symmetric polynomials on both sides, so we can use elementary symmetric polynomials to restate it as .
Now let be the harmonic mean. Then, the AM-HM inequality states that . Again, both sides are symmetric polynomials, but it's not very obvious how to express using elementary symmetric polynomials. Let's try to add the fractions:
Now we recognize both the numerator and denominator as elementary symmetric polynomials:
Thus, the inequality becomes or .
What would we get if we start from the GM-HM inequality? Try it!We have
It would be nicer if we could derive more inequalities with every elementary symmetric polynomial. Again, consider the set of positive real numbers and all their elementary symmetric polynomials :
We define the elementary symmetric mean as , where is the elementary symmetric polynomial of . Do you see whyNote that gives us the number of distinct terms of the elementary symmetric polynomial , which is equivalent to seeing in how many ways we can choose a product consisting of variables from the set . This is just the binomial coefficient .
Let's start with two important theorems:
Newton's inequality is
with equality if and only if . (Link to the proof)
Raise both sides to the power:
If we take the product from to of both sides, we obtain another theorem:
Almost everything cancels, and we end up with
So we introduce the second theorem:
Maclaurin's inequality is
with equality if and only if .
That allow us to obtain the following chain of inequalities:
Note that if we take we obtain the AM-GM inequality!
Prove that for positive reals where , the following inequality holds:
By Maclaurin's inequality, we have
We also know that , so
Notice that when we obtain the famous inequality .
Problem Solving Principle: If you can switch variables, do so.
An antisymmetric polynomial is one such that
Show that any antisymmetric polynomial can be expressed as where is symmetric.
The first step is certainly to show that divides . This is straightforward from the remainder factor theorem. We need to show that . To do so, notice that by definition of antisymmetry Hence, Now, let us use our problem solving principle here: Thus, is symmetric.
Problem Solving Principle: Express your symmetric polynomials using the elementary polynomials.
Let Assuming the system has a solution, what is the relationship between and
Let us use the principle we just stated: The rest of the problem is trivial now. Substitute in the second equation to get Furthermore, substitute that into the last equation:
Problem Solving Principle: Replace uncomfortable terms with variables in the hope that symmetric polynomials could be formed.
Clearly, we are uncomfortable with the term under But hey! The nice thing is that which is a pretty simple equation.
Let us replace these terms with and
Now, let us use the previous problem solving principle: By substitution, it is quite simple to find the value of . And now that we know and , we could use the quadratic formula to obtain them separately. (That is, if they exist, of course.)
Then we report the value of as the answer.