Need help solving equation involving $\cosh$

Question

I am trying to solve this equation for $a$

$$R= (a)\cosh\left(\frac{l}{a}\right)$$

where $R$ and $l$ are real positive constants.

I tried breaking $\cosh$ into exponentials but this didnt seem to help.

Answer 1

You might look at the Lambert W function It is basically the only tool we have for things that combine polynomials and exponentials. Whether you consider a solution in terms of that acceptable is a matter of taste. Many people don't accept it. I didn't look whether I could solve this in terms of W. We could define a function $M(R)$ that gives the answer you want, but that wouldn't help much. A numerical solution is probably what you are in for.

Answer 2

Note that there is a bit of scaling that can be done to remove one parameter: if $t = a/\ell$ and $r = R/\ell$, the equation becomes $r = t \cosh(1/t)$. Now, besides numerical methods, you might try a series solution (for one of the two branches of the solution): for large $r$, $$ t = r- \dfrac{1}{2! r}-\dfrac{7}{4! r^3}-\dfrac{241}{6! r^5}-\dfrac{19279}{8! r^7} + \ldots $$

Answer 3

$\large\mbox{Hints}:$

1. $\displaystyle{% \mbox{Define}\ \mu \equiv {R \over \ell}\,, \quad x \equiv {\ell\ {\rm sgn}\left(\mu\right) \over a}.\ \mbox{Then,}\quad \left\vert\mu\right\vert\, x = \cosh\left(x\right) }$.
2. $\displaystyle{% \mbox{The solutions satisfy}\ x > {1 \over \left\vert\mu\right\vert} }$.
3. There is $\it one$ solution $\left(~x_{t}~\right)$ which is valid for a particular valor $\mu_{t}$ of $\mu$. $x_{t}$ and $\mu_{t}$ are determined by the pair of equations $$ \left.% \begin{array}{rcl} \cosh\left(x_{t}\right) & = & \left\vert\mu_{t}\right\vert\, x_{t} \\[1mm] \sinh\left(x_{t}\right) & = & \left\vert\mu_{t}\right\vert \end{array}\right\} \quad\mbox{which yield}\quad \left\{% \begin{array}{rcl} \left\vert\mu_{t}\right\vert & = & \sinh\left(\sqrt{1 + {1 \over \mu_{t}^{2}}\,}\,\right) \\[1mm] x_{t} & = & \sqrt{1 + {1 \over \mu_{t}^{2}}\,} \end{array}\right. $$
4. When $\left\vert\mu\right\vert < \left\vert\mu_{t}\right\vert$, there is not any real solution.
5. When $\left\vert\mu\right\vert > \left\vert\mu_{t}\right\vert$, we have $\it two$ solutions $\left(~x_{< \atop >}~\right)$: $$ {1 \over \left\vert\mu\right\vert} <\ x_{<}\ <\ x_{t}\,, \qquad\qquad x_{>}\ >\ x_{t} $$
6. When $\left\vert\mu\right\vert \gg 1,\quad$ $\displaystyle{% x_{<} \approx {1 \over \left\vert\mu\right\vert} + {1 \over \left\vert\mu\right\vert^{3}} }$

At least, the reader can take from here toward a $\it\mbox{numerical calculation}$.

Answer 4

Almost ten years too late !

Continuing the expansion given in @Robert Israel and transforming it as a Padé approximant of low degrees $$t=r+\frac{60 \left(11303-9954 r^2\right) r}{60 \left(19908 r^2-34219\right) r^2+398027}+O\left(\frac{1}{r^9}\right)$$

Concerning the left branch of the curve, as @Ross Millikan wrote, we can take advantage of Lambert function since,if $t$ is small, then $$r(t)=t \cosh \left(\frac{1}{t}\right) \sim \frac{1}{2} te^{\frac{1}{t}}$$ from which $$r(t)=\frac{1}{2} te^{\frac{1}{t}}\implies t_0=-\frac{1}{W_{-1}\left(-\frac{1}{2 r(t)}\right)}$$ Now, using one single iteration of Halley method gives extremely good results

$$\left( \begin{array}{ccc} r & \text{estimate} & \text{solution} \\ 10 & \color{red}{0.2222421500725}861711913621 & 0.2222421500725973096761854 \\ 20 & \color{red}{0.1862331343297978}293505787 & 0.1862331343297978675828843 \\ 30 & \color{red}{0.1705612198514985}297213084 & 0.1705612198514985313258937 \\ 40 & \color{red}{0.161086762368383517}3629720 & 0.1610867623683835175403648 \\ 50 & \color{red}{0.1544933072614891940}642392 & 0.1544933072614891940971162 \\ 60 & \color{red}{0.1495266555608954227}465585 & 0.1495266555608954227549607 \\ 70 & \color{red}{0.14559003584384098636}28814 & 0.1455900358438409863655546 \\ 80 &\color{red}{ 0.14235706580617812806}55875 & 0.1423570658061781280665845 \\ 90 &\color{red}{ 0.1396315693952796847}498242 & 0.1396315693952796847502436 \\ 100 & \color{red}{0.137287269123803503467}7868 & 0.1372872691238035034679807 \end{array} \right)$$

Using Householder method would significantly improve.