Solving this equation for p

Question

I have the equation: $$ t=\sqrt{\frac{m}{gp}} \cosh^{-1}{\left(e^{\dfrac{hp}{m}}\right)} $$ Where $\cosh^{-1}$ is the inverse hyperbolic cosine, and $m$, $g$, $h$ and $t$ are positive, real valued constants, and $p$ is a non-zero positive real number.

I need to solve this equation for $p$, but I cannot successfully separate p. I have thought about making $t=f(p)$, and finding the inverse, but wolfram alpha "cannot find the result in standard mathematical functions"

Is this equation inseparable? If yes, then what method can I use to numerically approximate $p$ for a given $m$, $g$, $h$ and $t$?

Answer 1

$\cosh(x)$ and $\exp(x)$ are almost the same thing if $x$ is large, so if $\frac{hp}{m}$ is large an approximate solution is given by the solution of $$ t=\sqrt{\frac{m}{gp}}\frac{hp}{m} = \sqrt{\frac{h^2 p}{gm}} $$ namely $p=\frac{gmt^2}{h^2}$. If $\frac{hp}{m}$ is not large, we may take such approximation as the starting value of Newton's method and find an accurate approximation of the solution pretty fast. We may also improve such algorithm by considering that $\cosh^{-1}(e^x)\approx x+\log(2)$ is tight for any $x$ moderately large, hence a better starting point is:

$$ p = \frac{g m t^2-2 h m\log(2)+m t \sqrt{g^2 t^2- 4g h\log(2)}}{2 h^2}. $$

Answer 2

Make the following change of variables: $$ x=t\,\sqrt{\frac{g\,p}{m}},\quad a=\frac{h}{t^2\,g}. $$ The equation can then be rewritten as $$ \cosh x=e^{ax^2}. $$ Clearly $x=0$ is a solution. If $a\ge1/2$ then $e^{ax^2}>\cosh x$ for $x>0$, and there are no more solutions. If $0Newton's method to find the solution.