Is there a way to solve $x\left(\frac{e^x+1}{e^x-1}\right)=4$ for x besides just plugging numbers in?

Question

This comes into play in the equation for the shift in Cosmic Microwave Background (CMB) photon frequency due to inverse Compton scattering:

$\frac{\Delta T}{T_{CMB}} = y \left( x\left(\frac{e^x+1}{e^x-1}\right)-4 \right)$

Where $y$ is essentially the integral of electron pressure, and $x$ is a scaled frequency. To find the null frequency, where the CMB temperature doesn't change, you need to solve for $\Delta T = 0$, and therefore you get what's in the title:

$x\left(\frac{e^x+1}{e^x-1}\right)=4$

I know by plotting the left hand side and then just plugging in numbers that the answer is $x = 3.830016097$ but I was wondering if there was any other way to solve for x.

Answer 1

$$x\left(\frac{e^x+1}{e^x-1}\right)=4$$ $$x\coth\frac{x}{2}=4$$ $$x=4\tanh\frac{x}{2}$$ it is easy to get the solution by iteration $$x_{i+1}=4\tanh\frac{x_i}{2}$$

Answer 2

As said in comments, you can rewrite the equation as $$e^{-x}=-\frac{x-4}{x+4}$$ the solution of which being given in terms of the generalized Lambert function.

From a formal point of view, this is nice but not very practical and numerical method such as Newton will be extremely powerful.

Notice that, since you use Mathematica, you could perform a series expansion to get $$x\left(\frac{e^x+1}{e^x-1}\right)=2+\frac{x^2}{6}-\frac{x^4}{360}+\frac{x^6}{15120}-\frac{x^8}{604800}+\frac{x^{10}}{ 23950080}+O\left(x^{12}\right)$$ Then, use series reversion to get $$x=t+\frac{t^3}{120}+\frac{t^5}{22400}-\frac{t^7}{2688000}-\frac{163 t^9}{19869696000}+O\left(t^{11}\right)$$ where $t=\sqrt{6(y-2)}$. Making $y=4$ would give $$x=\frac{9534829 \sqrt{3}}{4312000}\approx 3.82996 $$ while the exact solution is $3.83002$.