# General solution to the polynomial of form $x^a(1-x)^b=c$

Given $x\in(0,1)$, I am interested in finding the solution of polynomials in the form of $x^a(1-x)^b=c,$ where $a,b$ are both positive integers, and $c$ is carefully chosen so that a root in $(0,1)$ exists.

For $a+b<5$ the roots can be found painstakingly with radicals, however the problem won't be straightforward for larger $a,b$ due to the Abel–Ruffini theorem. Generally, I know a root can be found using Jacobi's $\vartheta$ functions, but their complexity makes them very difficult to apply in practice.

So here is my question: can the root to this polynomial in $(0,1)$ be expressed, or approximated, by a "nicer" function? By "nicer", I mean a combination of elementary functions and "common" special functions (e.g. Beta, Bessel, etc.).

• @Nemo This equation can be reduced to trinomial equation How? Note that the $a,b$ here and the $n$ in the trinomial equation are all assumed to be positive integers. – dxiv Oct 22 '16 at 7:04
• @dxiv , I answered your question below. – Nemo Oct 31 '16 at 12:32

The equation $x^a(1-x)^b=c$ can be written as a generalized trinomial equation $$Ax^p+x=1\qquad(p=-\frac{a}{b},~A=c^{{1}/{b}}).$$ Ramanujan showed that the root of the equation $Ax^p+x=1$ is given by the power series $$x=1-A+\sum_{n=2}^\infty\frac{(-A)^n}{n!}\prod_{k=1}^{n-1}(1+pn-k).$$ So, according to these formulas we have for the root of the equation $x^a(1-x)^b=c$ \begin{align} x&=1-c^{1/b}+\sum_{n=2}^\infty\frac{(-c^{1/b})^n}{n!}\prod_{k=1}^{n-1}(1-an/b-k)\\&=1-c^{1/b}-\sum_{n=2}^\infty\frac{c^{n/b}}{n!}\frac{\Gamma(an/b+n-1)}{\Gamma(an/b)}. \end{align} For rational $a/b$ this series can be written in terms of finite sum of hypergeometric functions (see Glasser's paper for details https://arxiv.org/abs/math/9411224). For given $a,b$ Mathematica can easily simplify this expression in terms of generalized hypergeometric functions.

According to dxiv's analysis there are 2 real roots in the interval $(0,1)$. Let's denote them as $x_1$ and $x_2$. The above formula gives one of the roots, say $x_1$. To obtain $1-x_2$ interchange $a$ and $b$ in this formula.

$\it{Example}$. Let $a=2,b=3$, then the real root $x_1$ of the equation $x^2(1-x)^3=c$ is $$x_1=\frac{3}{5}+\frac{2}{5} \, _4F_3\left(-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5};\frac{1}{3},\frac{1}{2},\frac{2}{3};\frac{3125 c}{108}\right)-c^{1/3}{}_4F_3\left(\frac{2}{15},\frac{8}{15},\frac{11}{15},\frac{14}{15};\frac{2}{3},\frac{5}{6},\frac{4}{3};\frac{3125 c}{108}\right)-\frac{2}{3}c^{2/3} {}_4F_3\left(\frac{7}{15},\frac{13}{15},\frac{16}{15},\frac{19}{15};\frac{7}{6},\frac{4}{3},\frac{5}{3};\frac{3125 c}{108}\right).$$ For the other root $x_2$ we have $$x_2=\frac{3}{5}-\frac{3}{5}{}_4F_3\left(-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5};\frac{1}{3},\frac{1}{2},\frac{2}{3};\frac{3125 c}{108}\right)+c^{1/2}{}_4F_3\left(\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10};\frac{5}{6},\frac{7}{6},\frac{3}{2};\frac{3125 c}{108}\right),$$ Numerical check confirms that these formulas are true.

It's always hard to prove that a simple closed form does not exist, but I don't see one here.

That said, for $a,b \gt 0$ (not necessarily integers) $f(x)=x^a(1-x)^b$ is a nice concave function on $[0,1]$, with $f(0)=f(1)=0$ and a maximum at $x=\frac{a}{a+b}$.

It follows that for every $0 \le c \lt f(\frac{a}{a+b})=\frac{a^a b^b}{(a+b)^{a+b}} = M$ the equation $f(x)=c$ will have two solutions, one in each of $[0, M)$ and $(M,1]$ respectively.

Using for example the Newton method the two roots can be found by iterating: $$x_{n+1} = x_n - \frac{x_n^a(1-x_n)^b-c}{x_n^{a-1}(1-x_n)^{b-1}(a - (a+b)x_n)}$$ with starting points $x_0=\frac{a}{2(a+b)}$ and $x_0=\frac{a+ 2b}{2(a+b)}$ respectively.