What can be said about the roots of the polynomial $x^{n+1} - (1 - x)^n ( n + x )$?

Question

I have encountered the polynomial equation

$$x^{n+1} = (1 - x)^n ( n + x )$$

where $n \geq 0$, and I am interested in its real roots. The number $n$ can be an integer or, more generally, any positive real number.

The main reason I am posting this very specific question here is that this could be one of those classical equations in some subarea of analysis or algebra, so perhaps somebody recognizes it by name. Of course, somebody might happen to have a good hint. It seems that this polynomial has only one single positive root.

Answer 1

The positive root is indeed unique and single for any real $n>0$. Indeed,

1. for $x>1$, we have $$|(1-x)^n(n+x)|=(x-1)^n(n+x)<\left(\frac{n\cdot (x-1)+(n+x)}{n+1}\right)^{n+1}=x^{n+1}$$ by the (weighted) Arithemtic-Geometric Means inequality.

2. for $0\leqslant x\leqslant 1$ there exists a root in $(0,1)$ since the function $f(x):=(1-x)^n(n+x)-x^{n+1}$ is positive at $x=0$ and negative at $x=1$. To prove that it is unique and simple, note that $$ (1-x)^{-n}x^{-1}f(x)=\frac n{x}+1-\left(\frac{x}{1-x}\right)^n $$ is a strictly decreasing (and with strictly negative derivative) function on $(0,1)$.

Answer 2

Fedor Petrov showed that, for natural $n$, the equation \begin{equation*} x^{n+1}=(1-x)^n(n+x) \tag{10}\label{10} \end{equation*} has exactly one nonnegative root $x=x_n$, and $x_n\in(0,1)$.

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For $x\in(0,1)$, rewriting \eqref{10} as \begin{equation*} \Big(\frac x{1-x}\Big)^n=1+\frac nx, \tag{20}\label{20} \end{equation*} we see that, for each real $h>0$, as $n\to\infty$,

(i) uniformly in $x\in(1/2+h,1)$, the left-hand side of \eqref{20} goes exponentially fast to $\infty$ whereas the right-hand side of \eqref{20} is $O(n)$.

(ii) uniformly in $x\in(0,1/2-h)$, the left-hand side of \eqref{20} goes exponentially fast to $0$ whereas the right-hand side of \eqref{20} is $>1$.

So, for the only nonnegative root $x=x_n$ of \eqref{10} we have \begin{equation*} x_n\to1/2 \end{equation*} (as $n\to\infty$).

(As Emil Jeřábek noted, we also have $x_n>1/2$ for all $n$.)

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Let us consider the possible negative roots $x$ of \eqref{10}. Here we have to distinguish two cases, whether $n$ is even or odd.

Consider first the case when $n$ is even. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} g_n(u):=n\ln\frac u{u+1}-\ln(1-n/u) \tag{30}\label{30} \end{equation*} for $u:=-x>n$. We have \begin{equation*} g'_n(u)=-\frac{n(n+1)}{u(u+1)(u-n)}<0 \end{equation*} and $g_n(u)\to0$ as $u\to\infty$. So, $g_n(u)>0$ for all $u>n$, and hence \eqref{10} has no negative roots $x$ if $n$ is even.

Consider finally the case when $n$ is odd. Then \eqref{10} (or, rather, \eqref{20}), can be rewritten as \begin{equation*} h_n(c):=(c-1)(1+c/n)^n=1\tag{40}\label{40} \end{equation*} for $c:=n/u>1$ and $u:=-x\in(0,n)$. Note that $h_n(c)$ is continuously and strictly increasing in $c>1$ from $0$ to $\infty$. So, equation \eqref{40} has a unique root $c=c_n>1$. Since $(1+c/n)^n\to e^c$, we see that \begin{equation} c_n\to c_\infty:=W(1/e)+1=1.278\ldots, \end{equation} where $c_\infty$ is the unique root $c>1$ of the equation $(c-1)e^c=1$ and $W$ is the Lambert function. So, when $n$ is odd, equation \eqref{10} has exactly one negative root $x=y_n$, and $y_n\sim -n/c_\infty$.