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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
@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 $x_{2n}$.
It seems possible to generate an estimate of the value of $x$ where the function $$f(x)=\sum_{k=0}^{2n}{x^k}=\frac{x^{2 n+1}-1}{x-1}$$ is minimum. As shown in other answers, the problem is to approximate the zero of function $$g(x)=2nx^{2n+1}-(2n+1)x^{2n}+1$$ This have been done using the first iteration of an high order iterative method starting with $x=-1$ as an initial guess. The result is given by $$x_0=-\frac{256 n^6+10240 n^5+75296 n^4+212560 n^3+273764 n^2+159400 n+33199}{2 (2 n+3) \left(64 n^5+2688 n^4+20224 n^3+53604 n^2+56320 n+20001\right)}$$
Now, one single interation of Newton method is supposed to sufficiently polish the solution (it would not be any problem to use Halley or Householder methods in place of Newton).
Below are reported some results ($x_1$ by Newton). $$\left( \begin{array}{cccc} n & x_0 & x_1 & \text{solution} \\ 1 & -0.500137 & -0.500000 & -0.500000 \\ 2 & -0.605614 & -0.605830 & -0.605830 \\ 3 & -0.669716 & -0.670334 & -0.670332 \\ 4 & -0.713566 & -0.714543 & -0.714538 \\ 5 & -0.745773 & -0.747064 & -0.747054 \\ 6 & -0.770592 & -0.772160 & -0.772142 \\ 7 & -0.790396 & -0.792205 & -0.792178 \\ 8 & -0.806624 & -0.808643 & -0.808605 \\ 9 & -0.820203 & -0.822403 & -0.822353 \\ 10 & -0.831760 & -0.834116 & -0.834053 \\ 11 & -0.841736 & -0.844223 & -0.844148 \\ 12 & -0.850449 & -0.853047 & -0.852958 \\ 13 & -0.858136 & -0.860826 & -0.860724 \\ 14 & -0.864977 & -0.867742 & -0.867627 \\ 15 & -0.871111 & -0.873936 & -0.873809 \\ 16 & -0.876648 & -0.879520 & -0.879381 \\ 17 & -0.881675 & -0.884583 & -0.884433 \\ 18 & -0.886264 & -0.889197 & -0.889037 \\ 19 & -0.890473 & -0.893421 & -0.893252 \\ 20 & -0.894349 & -0.897304 & -0.897127 \\ 21 & -0.897932 & -0.900888 & -0.900703 \\ 22 & -0.901257 & -0.904206 & -0.904015 \\ 23 & -0.904351 & -0.907289 & -0.907091 \\ 24 & -0.907239 & -0.910161 & -0.909958 \\ 25 & -0.909942 & -0.912843 & -0.912636 \end{array} \right)$$
It is possible to do better for $x_0$ at the price of an higher order method.
Edit
For the fun of it, I used the same procedure with two more orders. The result write, just as before, $$x_0=-\frac{\sum_{k=0}^8 a_k\,n^k}{\sum_{k=0}^8 b_k\,n^k}$$ and the coefficients are $$\left( \begin{array}{ccc} k & a_k & b_k \\ 0 & 2300106 & 8278851 \\ 1 & 13284828 & 37098288 \\ 2 & 30007592 & 67149620 \\ 3 & 34417264 & 63318640 \\ 4 & 21546112 & 33284272 \\ 5 & 7272832 & 9611392 \\ 6 & 1199488 & 1382080 \\ 7 & 73216 & 75520 \\ 8 & 512 & 512 \end{array} \right)$$
Just to give an idea, for $n=25$ this produces for $x_0$ a value of $-0.912088$.
Update
Up to now, we just focused on the value of $x$ for which $f(x)$ is minimum. What is interesting if that when $f'(x_0)=0$ by elimination we have the simple $$f(x_0)=\frac{2 n+1}{1+2 n(1-x_0)}$$
Using the last set of coefficients given in the edit, we have $$\left( \begin{array}{ccccc} n & f_{\text{min}}^{\text{est}} & f_{\text{min}}^{\text{calc}} & x_0^{\text{calc}}& x_0^{\text{est}}\\ 1 & 0.749996 & 0.750000& -0.500000 & -0.500011 \\ 2 & 0.673570 & 0.673553& -0.605830 & -0.605782 \\ 3 & 0.635115 & 0.635094& -0.670332 & -0.670271 \\ 4 & 0.611584 & 0.611567& -0.714538 & -0.714486 \\ 5 & 0.595555 & 0.595543& -0.747054 & -0.747017 \\ 6 & 0.583866 & 0.583858& -0.772142 & -0.772117 \\ 7 & 0.574928 & 0.574922& -0.792178 & -0.792159 \\ 8 & 0.567852 & 0.567846& -0.808605 & -0.808586 \\ 9 & 0.562099 & 0.562091& -0.822353 & -0.822327 \\ 10 & 0.557321 & 0.557309& -0.834053 & -0.834013 \\ 11 & 0.553284 & 0.553267& -0.844148 & -0.844089 \\ 12 & 0.549825 & 0.549801& -0.852958 & -0.852876 \\ 13 & 0.546825 & 0.546793& -0.860724 & -0.860614 \\ 14 & 0.544196 & 0.544156& -0.867627 & -0.867487 \\ 15 & 0.541872 & 0.541823& -0.873809 & -0.873636 \\ 16 & 0.539802 & 0.539743& -0.879381 & -0.879173 \\ 17 & 0.537945 & 0.537876& -0.884433 & -0.884189 \\ 18 & 0.536269 & 0.536190& -0.889037 & -0.888755 \\ 19 & 0.534749 & 0.534660& -0.893252 & -0.892932 \\ 20 & 0.533363 & 0.533263& -0.897127 & -0.896769 \\ 21 & 0.532094 & 0.531984& -0.900703 & -0.900306 \\ 22 & 0.530926 & 0.530807& -0.904015 & -0.903580 \\ 23 & 0.529850 & 0.529719& -0.907091 & -0.906618 \\ 24 & 0.528852 & 0.528712& -0.909958 & -0.909447 \\ 25 & 0.527926 & 0.527777& -0.912636 & -0.912088 \end{array} \right)$$
Update
After this question (which looks quite similar), I thought that it could be interesting to look for the zero of function $$h(x)=\log \left((2 n+1) x^{2 n}-2 n x^{2 n+1}\right)$$ Expanding it as a series around $x=-1$ and using series reversion, we end with $$x=-1+z+\sum_{k=2}^\infty (-1)^{k+1}\frac {P_k(n) } { 2^{k-1}\, k! \,(4 n+1)^{k-1} }\,z^k$$ with $$z=\frac{(4 n+1) \log (4 n+1)}{4 n (2 n+1)}$$ The first $P_k(n)$ are $$\left( \begin{array}{cc} 2 & 8 n+1 \\ 3 & 64 n^2+8 n-1 \\ 4 & 512 n^3+32 n^2-24 n-1 \\ 5 & 4096 n^4+256 n^3+32 n^2+104 n+13 \end{array} \right)$$
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 $-0.606$.
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.
