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Let $n >2$ be an integer, and let $0<s \le \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ be real.

Is there a way to find explicitly the unique solution $0<a<\frac{1}{n}$ to the equation $a(1-a)^{n-1} = s$?

This is the smallest root of the polynomial $p(a)=a(1-a)^{n-1}$.

The condition $s \le \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ comes from the fact that $$\max_{a \in (0,1)}a(1-a)^{n-1}=\frac{1}{n}(1-\frac{1}{n})^{n-1},$$

so if $s > \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ there is no solution in $(0,1)$. Moreover, since the function $f(a)=a(1-a)^{n-1}$ is increasing in $(0,\frac{1}{n}]$, we see that if $0=f(0)<s<f(\frac{1}{n})= \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ then there exist a unique solution $a \in (0,\frac{1}{n}]$ as mentioned above.

Comment: This question is related to this problem.

Of course, there is no general formula for the roots of polynomials of order 5 and higher, but this is a very specific polynomial, and I am asking only for the smallest root.

If an explicit formula is out of reach, what estimates can be done? Is there a known method for estimating the smallest root of a polynomial?

I am interested in estimates from above and from below.

Edit: It seems any formula will be extremely messy, even for $n=3$.

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    $\begingroup$ There is a very likely scenario such that there is no root in the given range. $\endgroup$ – Simply Beautiful Art Sep 17 '17 at 15:36
  • $\begingroup$ When $n\to\infty$? $\endgroup$ – Did Sep 17 '17 at 15:43
  • $\begingroup$ The conditions seem strange. First, $s<a$ always, and for large $n$ it should really be $s\lt\lt a$. Yet these facts are not included in the problem setup. $\endgroup$ – András Salamon Sep 17 '17 at 16:21
  • $\begingroup$ @AndrásSalamon I elaborated on the technical conditions. (Didn't want to scare people away:). $\endgroup$ – Asaf Shachar Sep 17 '17 at 17:01
  • $\begingroup$ @SimplyBeautifulArt I elaborated on the technical conditions. (Didn't want to scare people away:). $\endgroup$ – Asaf Shachar Sep 17 '17 at 17:02

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