TITLE: The minimum of $f(x)=1+x+\cdots+x^{2n}$
QUESTION [6 upvotes]: For the function $$
f(x)=\sum_{k=0}^{2n}{x^k}
$$
I think this function has no zeros, and if ${f(x_0)}$ is the minimum point, then ${x_0\in \left[-1,0 \right] }$.
For $1+x+x^2+x^3+x^4$ and $1+x+x^2$, I can get the minimum by calculating $f'(x)$ and then using the root fomula. But this method doesn't work for $n\geqslant 3$ since there is no root fomula for equations higher than the fifth degree.
I have no idea for the situation where  $n\geqslant 3$, so any advice could be helpful

REPLY [10 votes]: @ClaudeLeibovici and I had a lot of fun with this one. The link to the arXiv article below provides a derivation of the minimum of this polynomial for any $n\in\Bbb N$.
Exact and approximate solutions to the minimum of $1+x+\cdots+x^{2n}$

Summary of paper:
To find the minimum of this polynomial, denoted $f_{2n}(x)$, we first derived the relationship
$$
\inf_x f_{2n}(x)=\frac{1+2n}{1+2n(1-x_{2n})},
$$
with $x_{2n}$ being the argument of the minimum. It was then shown that $x_{2n}$ satisfies
$$
x_{2n}^{2n}\left(1-x_{2n}+\tfrac{1}{2n}\right)-\tfrac{1}{2n}=0,
$$
and an exact solution of this equation was subsequently found via Lagrange inversion. For the purposes of numerical computation we also derived a faster converging perturbation series for $x_{2n}$.