How to solve this functional equation involving hyperbolic functions?

Question

I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)

$$f(K_1,0)=-\frac{1}{2}\ln\{2\sqrt{\cosh(2K_1)}\}+\frac{1}{2}f(\ln\sqrt{\cosh(2K_1)},0).\qquad(1)$$

They give the following solution (book eq. 14.2.15)

$$f(K_1,0)=-\ln(2\cosh K_1).\qquad(2)$$

I tried to check the solution was correct simply plugging (2) into (1) but end up with a $\ln(\cosh(\ln(\cosh(K_1)))$ thing.

Questions:

1. How do I obtain the solution (2) from (1)?
2. Is there an easy way to check the solution is correct?

Answer 1

Note that \begin{align} \cosh x = \frac{e^x+e^{-x}}{2} \end{align} then you see that \begin{align} 2 \cosh( \ln( \sqrt{\cosh(2K_1)}) =&\ 2\frac{\exp\left(\ln \sqrt{\cosh(2K_1)}\right)+\exp\left(-\ln \sqrt{\cosh(2K_1)} \right)}{2}\\ =&\ 2\frac{\sqrt{\cosh 2K_1}+ (\sqrt{\cosh 2K_1})^{-1}}{2} = \frac{\cosh 2K_1 +1}{\sqrt{\cosh 2K_1}}. \end{align} Then it follows \begin{align} -\frac{1}{2}\ln\left(2\sqrt{\cosh 2K_1} \right)-\frac{1}{2}\ln\left(\frac{\cosh 2K_1 +1}{\sqrt{\cosh 2K_1}} \right)=&\ -\frac{1}{2}\ln\left( 2\cosh(2K_1)+2\right)\\ =&\ -\frac{1}{2}\ln(4\cosh^2 K_1). \end{align}