Exploiting Symmetry in Polynomial Coefficients

Exploiting symmetry in polynomial coefficients

(Note: I believe the title refers to one-variable polynomials, since I clicked on that category to get here. If someone wants to post on multi-variable polynomials I guess they can)

There are numerous kinds of symmetry that sometimes occur in polynomials. But firrst let me go on a slightly irrelevant tangent.

Let $P(x) = x^2 - 4x+3, Q(x) = 3x^2-4x+1$. Then the roots of $P$ are $1,3$ and the roots of $Q$ are $1, \dfrac{1}{3}$

In general, let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ for $a_0 \neq 0$. Let $Q(x) = a_0x^n + a_{1}x^{n-1} + ... + a_n$. We claim the roots of $P,Q$ are recipricols of each other.

This is because, by Vieta's Formulas, if $r_1,r_2,..r_n$ are the roots of $P$, and $s_1, s_2, ..s_n$ are the roots of $Q$, then we have some equations:

$r_1r_2...r_n = (-1)^n \dfrac{a_0}{a_n}$ which is the recipricol of $(-1)^n \dfrac{a_n}{a_0} = s_1s_2...s_n$. We can pair up similar equations to get the desired result.

Now back to the question... Suppose we have a symmetric polynomial- that is, $a_0=a_n,a_1=a_{n-1}$, etc.

Let's take a larger symmetric polynomial for a more instructive example:

$x^6 + 2x^4 + 3x^3 + 2x^2 + 1=0$. Dividing both sides by $x^3$ yields

$\left( x^3 + \dfrac{1}{x^3} \right) + 2 \left( x + \dfrac{1}{x} \right) + 3 = 0$

Then we let $x + \frac{1}{x} = t$, and the equation is equivalent with

$t^3 - 3t + 2t + 3 = 0$ or $t^3-t+3=0$. Although $t$ is not a nice number, this method allows one to reduce the degree of polynomials by a lot.

Another sidenote: We can prove that $x^n + \dfrac{1}{x^n}$ is a polynomial in $x + \dfrac{1}{x}$.

Note that if $b_n = x^n + \dfrac{1}{x^n}$ then we have a recursion:

$b_n = b_{n-1}b_1 - b_{n-2}$.

Thus, we may use induction. Our base cases $n=0,1$ are easy. Now for $n \ge 2$, since $b_{n-1}, b_1, b_{n-2}$ are each polynomials in $b_1$, any product or sum of them will be too, yielding that $b_n$ is also a polynomial, which gives us the desired.

Now, I will do the exact reverse process and see what hapens...

Let's take the equation $x^3 - 3x +1 = 0$ and do the reverse: Let $x = w + \dfrac{1}{w}$ for a possibly complex value of $w$. Then our equation becomes

$w^6+w^3+1=0$

This is very easy to solve! It becomes $\dfrac{x^9-1}{x^3-1} = 0$, which I'll leave up to the reader to solve.

Why is this equation easy to solve? It's because saying $x = w + \dfrac{1}{w}$ is really saying $x = 2 \cos \theta$, given that $|w|=1$.

And guess what happens when we plug in $x = 2 \cos \theta$ into the equation? We get

$2(4 \cos^3 \theta - 3 \theta) = -1$, or, by the triple-angle formula, $2 \cos 3\theta = -1$, which is very easy to solve! Thus, whenever you see an equation that looks like a trig identity, the substitution described may help a lot.