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.).

  • 1
    $\begingroup$ @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. $\endgroup$ – dxiv Oct 22 '16 at 7:04
  • 2
    $\begingroup$ @dxiv , I answered your question below. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.