Solution of $(1-x)^p= x$ in $(0,1/n)$

Question

Let $n,p$ be positive integers. The equation $$ (1-x)^p = x $$ has a unique solution $x_p$ in the interval $(0,1)$. This follows by the monotonicity properties of $(1-x)^p$ and $x$.

My question is: given $n$ is it possible to determine $p$ such that $x_p$ satisfies $$ 0< x_p \le \frac 1n? $$

Bernoullis inequality $$ (1-x)^p \ge 1 - px, \quad x\in (0,1), $$ only yields a lower bound $$ x_p \ge \frac1{p+1}. $$

Answer 1

Given $n$, it is the set of $p$ so that $$\left(1-\frac{1}{n}\right)^p\leq\frac 1 n$$

Taking the log of both sides, this means:

$$p(\log (n-1)-\log n) \leq -\log n$$ Since we are dealing with negative numbers, that means:

$$p\geq \frac{\log n}{\log n-\log(n-1)}$$

Answer 2

First of all, since $0\lt x\lt 1$,$$(1-x)^p=x\iff 1-x=x^{1/p}\iff x+x^{1/p}-1=0.$$ Then, letting $f(x)$ be the LHS, we have for $x\gt 0$ $$f'(x)=1+\frac{1}{p}x^{1/p-1}\gt 0.$$ Hence, we know $f(x)$ is strictly increasing.

Hence, with $f(0)=-1\lt 0$ and $f(1)=1\gt 0$, we know $$0\lt x_p\le \frac 1n\iff f\left(\frac 1n\right)\ge 0\iff \frac 1n+\left(\frac 1n\right)^{1/p}-1\ge 0.$$

Solving this for $p$ will give you the same answer as the one Thomas Andrews gives.