Bounds on the root of $x^{d+1} - x^d - 1$

Question

For integer $d>0$, consider the polynomial $f(x) = x^{d+1} - x^d - 1$. It is easy to see that $f(x)<0$ for $x\in[0,1]$ and $f$ is strictly increasing for $x\geq 1$, so there is one unique positive real root $x_0$. It is also easy to see that $x_0\in (1,2)$.

I would now like to have some precise bounds on $x_0$ in terms of $d$. The asymptotics of $x_0(d)$ for $d\to\infty$ would also be interesting.

The best I could do is this:

• Lower bound: $1 + \frac{1}{d+1}$ (found by applying Newton's method at $x=1$ to the reflected polynomial, easy enough to check as well)
• Upper bound: $2\cdot\frac{d+1+2^{-d}}{d+2}$ (found by applying Newton's method to the original polynomial at $x = 2$)

I wonder if one can do any better than that.

As for the asymptotics, I suspect that it will be something like $x_0(d) \sim 1 + c\cdot\frac{\log d}{d}$, but I have no idea how to get there.

Update: I just remembered I also showed the bound $x_0 \leq \sqrt[d]{d+1}$, which indeed has the asymptotics $1 + \log d/d + O(\log^2 d/d^2)$.

To show this, note that $$(d+2)^d = e^{d \ln(d+2)} < e^{(d+1)\ln(d+1)} = (d+1)^{d+1}$$ because $d/\ln(d+1) < (d+1)/\ln(d+2)$ since $g(x) = x/\ln(x+1)$ is increasing. Thus, we have $$\frac{d+2}{d+1} < \sqrt[d]{d+1}$$ and thereby $$f(\sqrt[d]{d+1}) = \sqrt[d]{d+1} (d+1) - d - 2 > 0$$

Answer 1

Let $y = x-1$. Then $$ x^{d+1} - x^d - 1 = (x-1)x^d - 1 = y(y+1)^d - 1 $$ so it suffices to find asympotics for $y_0>0$ satisfying $y_0(1+y_0)^d = 1$.

The claim is that $$ y_0 = (1+o(1))\frac{\log d}{d} $$ as $d\to\infty$. It suffices to show the following two statements: for an arbitrary (but fixed) $r\in(0,1)$, we have $y_0\ge r\frac{\log d}{d}$ for sufficiently large $d$, and for an arbitrary (but fixed) $s>1$, we have $y_0\le s\frac{\log d}{d}$ for sufficiently large $d$.

To show the former, we use the inequality $y(1+y)^d\le ye^{dy}$ (which follows because $\log(1+y)\le y$ for all $y$). Plugging in $y_- = r\frac{\log d}{d}$ for $0<r<1$ gives $$ y_-e^{dy_-} = r\frac{\log d}{d}e^{r\log d} = r\frac{\log d}{d}d^r = r\frac{\log d}{d^{1-r}}\to 0 $$ as $d\to\infty$. This means that $y_-(1+y_-)^d\le y_-e^{dy_-}<1$ for sufficiently large $d$, and since $y\mapsto y(1+y)^d$ is increasing in $y$ for $y>0$, it follows that $y_-\le y_0$ if $y_0(1+y_0)^d = 1$, i.e. $y_0\ge r\frac{\log d}{d}$ for sufficiently large $d$, given a fixed $r\in(0,1)$.

To show the latter, we note that $$\log(1+y) = (1-o(1))y\text{ as }y\to 0$$ since the derivative of $\log(1+y)$ at $y=0$ is $1$. In particular, for $y_+ = s\frac{\log d}{d}$ for a fixed $s>1$, we have $$ d\log(1+y_+) = d(1-o(1))s\frac{\log d}{d} > \log d $$ for sufficiently large $d$. Thus $$ y_+(1+y_+)^d = s\frac{\log d}{d}e^{d\log(1+y_+)} > s\frac{\log d}{d}e^{\log d} = s\log d > 1 $$ for sufficiently large $d$. This means that $y_0\le y_+$, i.e. $y_0\le s\frac{\log d}{d}$ for sufficiently large $d$, given a fixed $s>1$.

Answer 2

I got a tip from Bruno Salvy that, in the end, led me to a proof that

$$x_0(d) = 1 + \frac{W(d)}{d} + \frac{\log^2 d}{2d^2} + o\left(\frac{\log^2 d}{d^2}\right)\quad\text{as}\ d\to\infty$$ where $W(x)$ is the principal branch of the Lambert $W$ function.

I won't reproduce the computation here in full, but the proof is basically analogous to Joey Zou's answer: Write $f(x) = x(1+x)^d - 1$ and define $y_c(d) := W(d)/d(1 + c\log d/d)$. Then, after some tedious (but routine) asymptotic reasoning, one finds that for any fixed $c\in(0,1)-\{\tfrac{1}{2}\}$:

$$f(y_c(d)) \sim (c-\tfrac{1}{2}) \frac{\log^2 d}{d}\quad\quad\text{as} \ d\to\infty$$

This means that:

• $f(y_c(d)) < 0$ for sufficiently large $d$ if $c < \frac{1}{2}$
• $f(y_c(d)) > 0$ for sufficiently large $d$ if $c > \frac{1}{2}$

and therefore

• $x_0(d) - 1 - W(d)/d < cW(d)\log d/d^2$ for sufficiently large $d$ if $c < \frac{1}{2}$
• $x_0(d) - 1 - W(d)/d > cW(d)\log d/d^2$ for sufficiently large $d$ if $c > \frac{1}{2}$

and thereby $$x_0(d) - 1 - \frac{W(d)}{d} \sim \frac{W(d)\log d}{2d^2} \sim \frac{\log^2d}{2d^2}$$ and that concludes the proof.