1

A coefficient symmetric polynomial is a polynomial $p(x)=a_nx^n+...+a_0$ of even degree, Such that for every $k$, $a_k=a_{n-k}$ such as $3x^3+2x^2+2x+3$.

In general, to find the roots of such a polynomial: $$a_0x^n+a_1x^{n-1}+...+xa_1+a_0=0$$ Since we know $a_0\ne 0$, we can divide by $x^{\frac{n}{2}}$ $$a_0x^{\frac{n}{2}}+a_1x^{\frac{n}{2}-1}...\frac{a_1}{x^{\frac{n}{2}-1}}+ \frac{a_0}{x^{\frac{n}{2}}}=0$$ $$a_0(x^{\frac{n}{2}}+\frac{1}{x^{\frac{n}{2}}})+ a_1(x^{\frac{n}{2}-1}+\frac{1}{x^{\frac{n}{2}-1}})...$$ Now, we substitute $y=x+\frac{1}{x}$
Here is my question, in general, for any n, $x^n+\frac{1}{x^n}$ is a polynomial in $y$.
Is there a genreral formula for this polynomial?
Examples:
$x+\frac{1}{x}\mapsto y$
$x^2+\frac{1}{x^2}\mapsto y^2-2$
$x^3+\frac{1}{x^3}\mapsto y^3-3y$
I failed to see a consistent pattern.

razivo
  • 2,205
  • Yes, Chebyshev's polynomials give you the expression. The link doesn't go exactly at the formula. It is a few paragraphs below: $T_n\left(\frac{x+x^{-1}}{2}\right)=\frac{x^n+x^{-n}}{2}$. – plop May 19 '21 at 20:45
  • 1
    There is also the easy to derive recurrence $x^{n+1}+\frac{1}{x^{n+1}}=\left(x+\frac{1}{x}\right)\left(x^{n}+\frac{1}{x^{n}}\right)-\left(x^{n-1}+\frac{1}{x^{n-1}}\right)$. – dxiv May 19 '21 at 21:59

0 Answers0