The minimum of $f(x)=1+x+\cdots+x^{2n}$ 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
 A: Yes, it has no zeros, because $f(1)=2n+1\ne0$ and because, if $x\ne1$, $f(x)=\frac{x^{2n+1}-1}{x-1}$, which is never equal to $0$.
However, it is not true that the minimum is attained at some of $[0,1]$. For instance, of $n=2$, the the minimum is attained at about $-0.606$.
A: We can establish the minimum $x_0\in(-1,0)$ without explicitly finding it. This can be done by first re-writing, $$f(x)=\frac{x^{2n+1}-1}{x-1}$$ and then differentiating, $$f'(x)=\frac{2nx^{2n+1}-(2n+1)x^{2n}+1}{(x-1)^2}$$ Descarte's Rule of Signs reveals that $f'(x)$ has only one negative real root, call it $\alpha$. In particular, we see that $f'(-1)=-n$ and $f'(0)=1$, so we know $\alpha\in(-1,0)$, thus we know that $f(x)$ is decreasing (negative derivative) from $(-\infty,\alpha)$ and is increasing (positive derivative) from $(\alpha,0)$. Thus, since $f(x)>1$ for $x>0$ and $f(0)=1$ and $f(-1)=0$, then we are assured that $f(\alpha)<0$ and will be the absolute minimum of the whole function $f(x)$.

Finding $\alpha$ will most likely be done with an approximation (instead of closed form). There are a number of ways to approximate $\alpha$. One particularly straight forward approach, is to half the intervals testing if you are positive or negative at each step using the derivative. Since you know $\alpha\in[-1,0]$ this method will reach a really good approximation very quickly.
A: @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 $x_{2n}$.
