How equation can be solved

Question

I have equation $x^{t}(1-x)^{1-t}=a$. Is there any ideas how to solve this equation for x? It's looks like convolution, but I have no idea how to solve this. I realy need a help and ideas. Thank you!

Answer 1

Let us consider that we look for the zero's of function $$f(x)=x^{t}(1-x)^{1-t}-a\tag 1$$ for $0 < t <1$ and $0 < x <1$.

The cases of $t=0$ and $t=1$ are trivial, as is the case of $t=\frac 12$ which gives $x_{\pm}=\frac{1}{2} \left(1\pm\sqrt{1-4 a^2}\right)$.

Its derivatives are $$f'(x)=(1-x)^{-t} (t-x) x^{t-1}$$ $$f''(x)=(t-1) t (1-x)^{-t-1} x^{t-2}$$

The first derivative cancels when $x=t$ and $$x=t\implies f(t)=(1-t)^{1-t} t^t-a\implies f''(t)=-(1-t)^{-t} t^{t-1}$$ $f''(t) <0 \, \, \forall t$ makes $f(t)$ to correspond to a maximum and then, since $f(0)=f(1)=-a$ , $(1)$ has two solutions if $a < (1-t)^{1-t} t^t$.

The solutions will be $$0 <x_1 < t \qquad \text{and}\qquad t < x_2 < 1$$ To get estimates, the simplest we can do is to expand $f(x)$ as a Taylor series built at $x=t$ $$f(x)=\left((1-t)^{1-t} t^t-a\right)-\frac{1}{2} \left((1-t)^{-t} t^{t-1}\right) (x-t)^2+O\left((x-t)^3\right)$$ and, ignoring the higher order terms, this lead to the estimates $$x_{\pm}^{est}=t\pm\sqrt{2} t^{-t} \sqrt{-t^{t+1} \left(a (1-t)^t+(t-1) t^t\right)}$$

Just to prevent the estimates to be outside the range $0 x < 1$, use $$x_-^{est}=\max\left(\epsilon,t-\sqrt{2} t^{-t} \sqrt{-t^{t+1} \left(a (1-t)^t+(t-1) t^t\right)} \right)$$ $$x_+^{est}=\min\left(1-\epsilon,t+\sqrt{2} t^{-t} \sqrt{-t^{t+1} \left(a (1-t)^t+(t-1) t^t\right)} \right)$$ ($\epsilon \approx 0.01$ for example) and use Newton method starting from these guesses. We could also systematically start using $x_-=\frac t2$ and $x_+=\frac{1-t}2$ if prefered

For illustration purposes, let us use $a=\frac 13$ and a few values of $t$.

• for $t=\frac 14$, we have $x_-=0.01$ and $x_+\approx 0.644531$. Starting from these values, Newton iterates would be $$\left( \begin{array}{ccc} n & x_-^{(n)} & x_+^{(n)} \\ 0 & 0.010000000 & 0.64453052 \\ 1 & 0.012560308 & 0.75596917 \\ 2 & 0.012831112 & 0.74517623 \\ 3 & 0.012833458 & 0.74505785 \\ 4 & & 0.74505784 \end{array} \right)$$ which are the solutions for eight significant figures.

• for $t=\frac 23$, we have $x_-\approx 0.261128$ and $x_+= 0.99$. Starting from these values, Newton iterates would be $$\left( \begin{array}{ccc} n & x_-^{(n)} & x_+^{(n)} \\ 0 & 0.26112753 & 0.99000000 \\ 1 & 0.21474730 & 0.97292716 \\ 2 & 0.21755633 & 0.96169382 \\ 3 & 0.21756788 & 0.95983099 \\ 4 & & 0.95979509 \\ 5 & & 0.95979508 \end{array} \right)$$ which are the solutions for eight significant figures.

  • for $t=\frac {11}{24}$, we have $x_-\approx 0.050096$ and $x_+\approx 0.866571$. Starting from these values, Newton iterates would be $$\left( \begin{array}{ccc} n & x_-^{(n)} & x_+^{(n)} \\ 0 & 0.05009570 & 0.86657097 \\ 1 & 0.09109635 & 0.84964479 \\ 2 & 0.10294484 & 0.84886276 \\ 3 & 0.10353862 & 0.84886124 \\ 4 & 0.10353996 & \end{array} \right)$$ which are the solutions for eight significant figures.

For sure, we could speed up the convergence replacing Newton by Halley of Householder methods but hte principle will remain the same.

Edit

It seems that the roots can be at least approximated by $$x_-^{est}=a^{\frac{1}{t}} \qquad \text{and} \qquad x_+^{est}=1-a^{\frac{1}{1-t}}$$ For the worked examples, this would give $\{0.0123,0.7689\}$, $\{0.1925,0.9630\}$ and $\{0.0910,0.8684\}$ which seem to be more than acceptable.