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

2 Answers
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_(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]}

.

Aug 9, 2016

#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)#

enter image source here