Context
I started with the following (very common) problem:
Given this polynomial $p(x)$, calculate the sum/the sum of the squares/of the cubes of the roots"
So I wanted to see if I could find the sum of the $n$-th powers:
- If $p(x)=ax+b$, this is trivial: $$\displaystyle x_1^n=\left(-\frac{b}{a}\right)^n$$
- If $p(x)=ax^2+bx+c$, this is easy: $$\displaystyle x_1^n+x_2^n=\left(\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right)^{n}+\left(\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right)^{n}=\frac{(-1)^n}{2^{n-1}a^n}\sum_{k=0}^{\left\lfloor{\frac{n}{2}}\right\rfloor}\binom{n}{2k}b^{n-2k}\Delta^k$$
- If $p(x)=ax^3+bx^2+cx+d$, it's starting to get complicated.
I considered the depressed form: $x^3+cx+d$ to start where the roots are: $$\begin{cases} x_k=2\sqrt{\frac{|c|}{3}}\cos \left(\frac{1}{3}\text{acos}\left(\frac{3d}{2c}\sqrt{\frac{3}{|c|}}\right)-{\frac{2\pi k}{3}}\right)&\left(\frac{c}{3}\right)^3+\left(\frac{d}{2}\right)^2<0,c<0\\ x_1=-2\frac{|d|}{d}\sqrt{\frac{|c|}{3}}\cosh\left(\frac{1}{3}\text{acosh}\left(\frac{3|d|}{2|c|}\sqrt{\frac{3}{|c|}}\right)\right)&\left(\frac{c}{3}\right)^3+\left(\frac{d}{2}\right)^2>0, c<0\\ x_1=-2\sqrt{\frac{c}{3}}\sinh \left(\frac{1}{3}\text{asinh}\left(\frac{3d}{2c}\sqrt{\frac{3}{c}}\right)\right)&c>0\end{cases}$$
And I got this formula:
$$x_1^n+x_2^n+x_3^n=\sum_{k=\left\lceil{\frac{n}{3}}\right\rceil}^{\left\lfloor{\frac{n}{2}}\right\rfloor}(-1)^k\frac{n}{k}\binom{k}{n-2k}\frac{d^{n-2k}}{c^{n-3k}}\qquad n\in\mathbb{N}^{+}$$
Question
I started with this problem, but then I found it more interesting to try to solve another, in fact in the case of the third degree polynomial it was necessary to solve the following sum:
$$\bbox[15px,#FFF8FD,border:5px groove #8312FE]{\sum_{k=1}^{3}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{3}\right)^n=\sum_{k=\left\lceil{\frac{n}{3}}\right\rceil}^{\left\lfloor{\frac{n}{2}}\right\rfloor}\frac{n}{k}\binom{k}{n-2k}\left(\frac{3}{4}\right)^k \left(\frac{x}{3}\right)^{n-2k}\qquad n\in\mathbb{N}^{+}}$$
Which is a fully polynomial expression (and I would be interested in generalizing this formula)
I wanted to find a closed formula for the following sum:
$$\sum_{k=1}^{m}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{m}\right)^n\qquad n\in\mathbb{N}^{+}$$
But seeing the first cases I realized that it is more interesting to find the solutions for odd $m$, in fact:
For $m=1$
$$\sum_{k=1}^{1}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{1}\right)^n=x^n$$ For $m=2$
$$\sum_{k=1}^{2}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{2}\right)^n=\left(\left(-1\right)^{n}+1\right)\left(\frac{x+1}{2}\right)^{\frac{n}{2}}$$ For $m=3$
$$\sum_{k=1}^{3}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{3}\right)^n=\sum_{k=\left\lceil{\frac{n}{3}}\right\rceil}^{\left\lfloor{\frac{n}{2}}\right\rfloor}\frac{n}{k}\binom{k}{n-2k}\left(\frac{3}{4}\right)^k \left(\frac{x}{3}\right)^{n-2k}$$
If $m$ is even and $n$ is odd we always have $0$, if $m$ and $n$ are even we have a nesting of square roots as a solution.
So I think it is more interesting to study the following problem:
$$\bbox[15px,#E0EFFF,border:5px groove #0058B3]{\sum_{k=1}^{2m+1}\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{2m+1}\right)^n\qquad n\in\mathbb{N}^{+}}$$
Reason of interest
It is possible to generalize Cardano's formula to find the roots of particular polynomials of degree higher than the third.
Given the following polynomial:
$$\sum_{k=0}^{n}\frac{2n+1}{2k+1}\binom{n+k}{2k}\alpha^{n-k}x^{2k+1}=2\beta$$
Its roots are:
$$\bbox[15px,#FFF8FD,border:5px groove #8312FE]{\begin{cases} x_k=2\sqrt{|\alpha|}\cos\left(\frac{1}{2n+1}\text{acos}\left(\frac{(-1)^{n}\beta}{\alpha^{n}\sqrt{|\alpha|}}\right)+\frac{2k\pi}{2n+1}\right)&\beta^2+\alpha^{2n+1}<0,\alpha<0\\ x_0=2\frac{|\beta|}{\beta}\sqrt{|\alpha|}\cosh\left(\frac{1}{2n+1}\text{acosh}\left(\frac{|\beta|}{|\alpha^n|\sqrt{|\alpha|}}\right)\right)&\beta^2+\alpha^{2n+1}>0,\alpha<0\\ x_0=2\sqrt{\alpha}\sinh\left(\frac{1}{2n+1}\text{asinh}\left(\frac{\beta}{\alpha^{n}\sqrt{\alpha}}\right)\right)&\alpha>0 \end{cases}}$$
$$x_0=\sqrt[2n+1]{\beta+\sqrt{\beta^2+\alpha^{2n+1}}}+\sqrt[2n+1]{\beta-\sqrt{\beta^2+\alpha^{2n+1}}}$$
$n=1:\quad x^3+3ax=2\beta$
$n=2:\quad x^5+5\alpha x^3+5\alpha^2x=2\beta$
$n=3:\quad x^7+7\alpha x^5+14\alpha^2 x^3+7\alpha^3 x=2\beta$
In fact, $x_k^n$ is of the form (unless multiplicative constants):
$$x_k^n=\cos\left(\frac{2k\pi-\operatorname{cos^{-1}}(x)}{2m+1}\right)^n$$
