The classification of and description of the root near $1/2$ of $x^d +x^\left({d-1}\right) + x^\left({d-2}\right) + \dots + x^2 + x - 1=0$

Question

In Arturas Dubickas paper "On the number of reducible polynomials of bounded naive height", manuscripta math. 144, 439–456 (2014) he discusses a bounding polynomial $x^d + x^\left({d-1}\right) + x^\left({d-2}\right) + \dots + x^2 + x - 1$ real root that is near $1/2$ called ${\theta}_{0}$ where $1/2 < {\theta}_{0} \le 1$ of this polynomial of degree $d$. The roots are involved in the papers volume integrals.

My question is this polynomial known with possible references. This is close to the All One Polynomials (AOP).

And the characterization of this root ${\theta}_{0}$. I get from successive approximation $${\theta}_{0}\sim \frac{1}{2} + \frac{1}{2 (2^\left({d-1}\right)-d-2)}$$ I am looking for more detailed description of this root. Also is it valid that when taking the limit as $d \rightarrow \infty$ that ${\theta}_{0} \rightarrow 1/2$. This what I see numerically.

Answer 1

Your polynomial is $$p(x) = x^d + \ldots + x - 1 = \frac{x^{d+1}-1}{x-1} - 2 = \frac{x^{d+1}-2x + 1}{x-1}$$ thus you are looking for a root near $1/2$ of $$ x^{d+1} - 2 x + 1$$

More generally, consider $$Q(x,a) = a x^{d+1} - 2 x + 1$$ Using Lagrange Inversion, the root of this near $1/2$ has a very nice series expansion in powers of $a2^{-d-1}$:

$$\frac{1}{2} + \sum_{j=1}^\infty \frac{(a 2^{-d-1})^j}{2} \prod_{k=2}^j \left(1 + \frac{jd}{k}\right) $$

This should converge to your root at $a=1$.

Answer 2

so we have some polynomial: $$P(x)=-1+\sum_{i=1}^dx^i$$ We can try and use NR: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Now we know that: $1/2<\theta_0\le1$ so we can take $x_0=1/2$ or $x_0=1$, either is valid. If we take $x_0=1$ we get: $$x_1=1-\frac{-1+\sum1^i}{\sum i\times 1^{i-1}}=1-\frac{2(d-1)}{d(d+1)}$$ and then continue this iterative process. I'm not sure how to evaluate the convergence of this sequence but could this be a way?

Answer 3

This is too long for a comment.

Starting from @Robert Israel's elegant answer, for the specific case $a=1$, using Pochhammer symbols, the expansion write $$x_d=\frac 12+\frac 12\sum_{j=1}^\infty \frac{ (d j+1)_j}{j!\, (d j+1)\,2^{(d+1) j}}$$ For discrete values of $d$, this reduces to hypergeometric functions $$\left( \begin{array}{cc} d & 2\, x_d \\ 3 & \, _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{16}{2 7}\right) \\ 4 & \, _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{2}{4},\frac{3}{4} ,\frac{5}{4};\frac{3125}{8192}\right) \\ 5 & \, _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{2}{5} ,\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{729}{3125}\right) \\ 6 & \, _6F_5\left(\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7} ;\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{7}{6};\frac{823543}{5971968 }\right) \\ 7 & \, _7F_6\left(\frac{1}{8},\frac{2}{8},\frac{3}{8},\frac{4}{8},\frac{5}{8},\frac{6}{8} ,\frac{7}{8};\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7},\frac{8}{ 7};\frac{65536}{823543}\right) \\ 8 & \, _8F_7\left(\frac{1}{9},\frac{2}{9},\frac{3}{9},\frac{4}{9},\frac{5}{9},\frac{6}{9} ,\frac{7}{9},\frac{8}{9};\frac{2}{8},\frac{3}{8},\frac{4}{8},\frac{5}{8},\frac{6}{ 8},\frac{7}{8},\frac{9}{8};\frac{387420489}{8589934592}\right) \end{array} \right)$$ where very clear patterns appear for all arguments except the last one I have not been able to identify (I would really be interested to know what it is).

Computed for $50$ significant digits, these are the exact solutions of the equation.

Answer 4

From the posted Lagrange Inversion infinite series solution we get a further infinite series solution based on Pochhammer symbols that yield a generalized hypergeometric function solution for this root near $1/2$. After generalizing from the posted comments we can write $${\theta}_{d} = \frac{1}{2} {}_d{F}_{d - 1} \left({\left\{{\frac{1}{d + 1}, \frac{2}{d + 1}, \dots, \frac{d}{d + 1}}\right\}, \left\{{\frac{2}{d}, \frac{3}{d}, \dots, \frac{d - 1}{d}, \frac{d + 1}{d}}\right\}, \frac{\left({d + 1}\right)^{d + 1}}{{2}^{d + 1}\, {d}^{d}}}\right)$$

I think that this is a known result of solving the trinomial ${x}^{d} + a\, x + b = 0$. When $d = 5$ this is a well known quintic solution (later transformed into the Bring quintic form). I am looking for a reference of this result for degree $d$.