Analytical solution for $\frac{e^{(2x + a)} - 1}{e^{(2x + a)}+1} = x$?

Question

do you know if there is an analytical solution for $\frac{e^{(2x + a)} - 1}{e^{(2x + a)}+1} = x$?

Thank you!

Answer 1

Up to now, there is no analytical solution and numerical methods would be required.

Starting from egreg's approach, it would be quite easy using Newton method $$t_{n+1}=\frac{1}{2} a \coth ^2(t_n)+\coth (t_n)-t_n\, \text{csch}^2(t_n)$$

With regard to the starting value $t_0$, if $a <2$ you can use from Taylor expansion $t_0=\left(\frac {3a}2 \right)^{1/3}$ and if, $a>2$, $t_0=1+\frac a 2$.

For illustration purposes, let us try for $a=1$. The iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 1.144714243 \\ 1 & 1.401951708 \\ 2 & 1.381330937 \\ 3 & 1.381225364 \\ 4 & 1.381225361 \end{array} \right)$$

For $a=4$ $$\left( \begin{array}{cc} n & t_n \\ 0 & 3.000000000 \\ 1 & 2.995005478 \\ 2 & 2.995005229 \end{array} \right)$$

Answer 2

write your equation in the form $$2x+a-\ln\left(\frac{x+1}{1-x}\right)=0$$ and use calculus

Answer 3

If you set $2x+a=2t$, the equation becomes $$ \frac{e^t-e^{-t}}{e^t+e^{-t}}=t-\frac{a}{2} $$ that is, $$ \tanh t=t-\frac{a}{2}\qquad\text{or}\qquad t-\tanh t=\frac{a}{2} $$ Consider the function $f(t)=t-\tanh t$. Since $-1<\tanh t<1$, we see that $\lim_{t\to-\infty}f(t)=-\infty$ and $\lim_{t\to\infty}f(t)=\infty$.

Moreover $$ f'(t)=1-\frac{1}{\cosh^2t}=\tanh^2t $$ Thus the function is increasing and the equation $f(t)=a/2$ has unique solution for every $a$, by the intermediate value theorem.

You can use numeric methods for approximating the solution, given a value for $a$.

If you want an “explicit” solution, call $g$ the inverse function of $f$, so the solution is $t=g(a/2)$ and therefore $$ x=g\biggl(\frac{a}{2}\biggr)-\frac{a}{2} $$ However, there is no expression of $g$ in terms of elementary functions.