I am trying to find an analytic solution for $\beta$ in the following expression. However, Mathematica was not able to manage to get the solution.

Solve[2 Cosh[$\beta \nu$] == Exp[$\beta$], $\beta$]

On the other hand, solution for $\nu$ is as follows:

$\nu=\dfrac{-\cosh ^{-1}\left(\frac{e^{\beta }}{2}\right)+2 i \pi}{\beta }$

However, I can not Inverse this function. $\beta$ and $\nu$ are both positive real numbers.

I would be grateful if you can show me the way to do this.

  • $\begingroup$ One usually does not expect a transcendental equation like yours to have a neat closed-form solution. You can derive a series in $\nu$, however. $\endgroup$ Oct 8, 2016 at 21:01

1 Answer 1


As J.M. pointed out it's a transcendental equation, but you can use InverseFunction to compute numerical values:

nu[\[Beta]_, n_] := (-ArcCosh[Exp[\[Beta]]/2] + 2*\[Pi]*n)/\[Beta];
nu0[\[Beta]_] := nu[\[Beta], 0];
beta = InverseFunction[nu0];

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