$N$ roots of "geometric" polynomial

Question

For arbitrary $N$, find all roots (or some recursive formula for them) of the "geometric" polynomial equation

$$ \sum_{k=1}^{N-1} x^k = 1 $$

I tried to sum up the LHS which gives, by the geometric series,

$$ \frac{x^{N}-x}{x-1} = 1 \qquad \text{or} \qquad \frac{x^{N}-1}{x-1} = 2 $$ or, for $x\ne1$, $x^{N}-2x +1 = 0$ but that doesn't lead me further.

Answer 1

$x=1$ is a root for all $N$. It is easy to see (using Rolles's theorem) that if $N$ is even then there is another real root in $(0,1)$, and that if $N$ is odd then there are another two real roots, one in $(0,1)$ and the other in $(-2,-1)$. The rest of the roots are complex. As to finding an explicit formula for them, I doubt there is one. The following are the roots in $(0,1)$ for $n=4,5$ (computed with Mathematica.)
$$ \frac{1}{3} \left(-1-\frac{2}{\sqrt[3]{17+3 \sqrt{33}}}+\sqrt[3]{17+3 \sqrt{33}}\right) $$

$$ -\frac{1}{4}-\frac{1}{4 \sqrt{\frac{3}{-5-56 \sqrt[3]{\frac{2}{65+3 \sqrt{1689}}}+2\ 2^{2/3} \sqrt[3]{65+3 \sqrt{1689}}}}}+\frac{1}{2} \sqrt{-\frac{5}{6}+\frac{14}{3} \sqrt[3]{\frac{2}{65+3 \sqrt{1689}}}-\frac{1}{3} \sqrt[3]{\frac{1}{2} \left(65+3 \sqrt{1689}\right)}+\frac{5}{2} \sqrt{\frac{3}{-5-56 \sqrt[3]{\frac{2}{65+3 \sqrt{1689}}}+2\ 2^{2/3} \sqrt[3]{65+3 \sqrt{1689}}}}} $$

Answer 2

Towards a solution : a graphical, heuristic approach, which doesn't pretend to be rigorous.

Let us recall the two forms :

$$\sum_{k=1}^{N-1} x^k = 1 \ \ \tag{1}$$

equivalent, when $x \neq 1$ to

$$x^N = 2x - 1 \ \ \tag{2}$$

About real roots : considering (2) as the equation governing the abscissas of the intersection points of the curves with equation $y=2x-1$ and $y=x^N$, we "see" that

• if $N$ even, there is a single real root ($x=1$ is a "false root").

• whereas, if $N$ is odd, there are two real roots, one positive, the other negative.

Now, by using a CAS (Computer Algebra System) giving the roots, I saw that they are rather close to the $N$th roots of unity (except for the positive root). Moreover, this proximity increases with $N$ (with the exception of the real positive root, which tends to $\tfrac12$).

Here is such a graphical representation :

Fig. 1 : (reference to form (1)) Case $N-1$ odd : $N=10$ here. The circles are the roots of the considered polynomial. The stars are the $9$th roots of unity.

Fig. 2 : Case $N-1$ even : $N=11$ here. The circles are the roots of the considered polynomial. The stars are the $10$th roots of unity.

On these figures, the roots look to be (except, once again, the real positive root) approximately on a circle shifted by $-\frac{1}{2(N-1)}$, providing a rather satisfactory approximate general expression for the roots under the form :

$$z_k=\left(1+\frac{1}{2(N-1)}\right)e^{2i\pi k/N}-\frac{1}{2(N-1)}$$

$$\text{for} \ k=1,2,..., (N-1)$$

The error (distance between exact and approximate values) has an order of magnitude $10^{-2}$ for $n=14$ for example.

Once again, the real positive root tends to $1/2$.