Question

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

Answer 1

`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_(cf)(x, a) = f(x)`

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(y+sqrt(y^2-1))-1/y` ( see the answer by Cesareo), for making

graph that reveals domain and range.

Indeed, x admits negative values.

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

the same.

As any cosh value `>=1` for any x,

The domain/range: `x in (-oo, oo)` and `y>=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]}

.

Answer 2

`g(x) = log_e(x pm sqrt(x^2-1))-a/x`

Explanation:

From

`y = cosh(x+a/y)` calling `z = x+a/y` we have

`y = (e^z+e^{-z})/2` giving

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

`e^z = y pm sqrt(y^2-1)` but

`z = log_e(y pm sqrt(y^2-1)) = x + a/y`

Finally

`x = log_e(y pm sqrt(y^2-1)) -a/y`

then

`g(x) = log_e(x pm sqrt(x^2-1))-a/x` is the inverse. This inverse is not bijective. Attached a plot with `a = 1`, showing in red `f(x)` and in blue and green the two leafs of `g(x)`