# What is the inverse function of f(x) = cosh(x+a/cosh(x+a/cosh(x+cdots))) with domain and range?

Aug 9, 2016

cosh^(-1)(x+a/cosh_(cf)(x; a))

#### Explanation:

I confine myself to FCF-naming of the function. For me, the strain is

inevitable.

${\cosh}_{c f} \left(x , a\right) = f \left(x\right)$

For this FCF,

cosh_(cf)(x;a)=cosh(x+a/cosh_(cf)(x; a)).

The operand is clear in the cosh function, in contrast to either f(x)

or cosh_(cf)(x; a)

Inversely,

the inverse of cosh_(cf)(x, a))

= the inverse of the equivalent cosh(x+a/(cosh_(cf)(x; a))).

Now, the inverse is cosh^(-1)(x+a/cosh_(cf)(x; a))

For a = 1, I use here the inverse for the FCF y = cosh(x+1/y) as

$x = \ln \left(y + \sqrt{{y}^{2} - 1}\right) - \frac{1}{y}$ ( see the answer by Cesareo), for making

graph that reveals domain and range.

The graphs for y = f(x) and its inverse $x = {f}^{- 1} y$ are one and

the same.

As any cosh value $\ge 1$ for any x,

The domain/range: $x \in \left(- \infty , \infty\right)$ and $y \ge 1$.

Graph of y = cosh(x + 1/y):

Note that there is no axis of symmetry.

The lowest point (-1, 1) is plotted in the graph.

graph{(x-ln(y+(y^2-1)^0.5)+1/y)(x+ln(y+(y^2-1)^0.5)+1/y)((x+1)^2+(y-1)^2-.004)=0 [-5 5 -1 4]}

Combined graph below, for this and y = cosh x reveals the

patterns.

This graph is my present to those (Caserio, George et al) who had

shown keen interest in FCF.

graph{(x-ln(y+(y^2-1)^0.5)+1/y)(x+ln(y+(y^2-1)^0.5)+1/y)(x-ln(y+(y^2-1)^0.5))(x+ln(y+(y^2-1)^0.5))=0[-5 5 0 10]}

.

Aug 9, 2016

$g \left(x\right) = {\log}_{e} \left(x \pm \sqrt{{x}^{2} - 1}\right) - \frac{a}{x}$

#### Explanation:

From

$y = \cosh \left(x + \frac{a}{y}\right)$ calling $z = x + \frac{a}{y}$ we have

$y = \frac{{e}^{z} + {e}^{- z}}{2}$ giving

${e}^{2 z} - 2 y {e}^{z} + 1 = 0$. Solving for ${e}^{z}$ gives

${e}^{z} = y \pm \sqrt{{y}^{2} - 1}$ but

$z = {\log}_{e} \left(y \pm \sqrt{{y}^{2} - 1}\right) = x + \frac{a}{y}$

Finally

$x = {\log}_{e} \left(y \pm \sqrt{{y}^{2} - 1}\right) - \frac{a}{y}$

then

$g \left(x\right) = {\log}_{e} \left(x \pm \sqrt{{x}^{2} - 1}\right) - \frac{a}{x}$ is the inverse. This inverse is not bijective. Attached a plot with $a = 1$, showing in red $f \left(x\right)$ and in blue and green the two leafs of $g \left(x\right)$