Solving a transcendental equation

Question

How do I go about solving the following equation?

$$x = A + B \log\left( \cosh\left(\frac{x}{C}\right)\right)$$

Answer 1

Set $t=e^{x/C}$. The equation turns to

$$C\log t=A+B\log\left(\frac12\left(t+\frac1t\right)\right),$$ which can be rewritten as $$at^b-t^2-1=0,$$ with $a=2e^{-A/B}$ and $b=(C+1)/B$.

There will be closed formulas (some terrible, solutions of quartics) at least for $b=-6,-4,-2,-1,-\frac23,0,\frac12,\frac23,1,\frac43,\frac32,2,\frac83,3,4,6,8$.

For instance, with $b=4$, $$x=C\ln\sqrt{\pm\frac{1\pm\sqrt{1+4a}}{2a}},$$ and $b=-1$ $$x=C\ln\left(\sqrt[3]{\sqrt{\frac{a^2}4+\frac1{27}}+\frac a2}+\sqrt[3]{-\sqrt{\frac{a^2}4+\frac1{27}}+\frac a2}\right).$$

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For a numerical solution (using Newton for instance), rewrite as $$\log\cosh u=cu+d,$$ where $x=Cu$, and use $$\color{blue}{\log\cosh u}\approx\color{green}{\frac{u^2}2}$$ for small $u$ and $$\color{blue}{\log\cosh u}\approx\color{magenta}{|u|-\log2}$$ for large $u$.

This should give you good starting points by intersecting with the straight line.

Answer 2

If you let $u=e^{x/BC}$, you can rewrite the equation in the form

$$ku^C = u^B+u^{-B}$$

where $k=2e^{-A/B}$. If $B$ and $C$ are commensurable, this can be transformed into a polynomial equation in a variable that converts back to $x$. But that's about as far as it goes, I'd say.

(Yves Daoust and I posted almost simultaenously.)