Given #f(x) = cosh(x+a/cosh(x+a/cosh( cdots)))# and #g(x)# its inverse, what is the minimum distance between then for #a > 0#?

We seek the minimum distance between the function:

# f(x) = cosh(x+a/cosh(x+a/cosh( x+a/cosh(cdots))) ) #

and its inverse, for #a gt 0#

Let, #y=f(x)#, Then we can write this repeated recursive function definition concisely as:

# y = cosh(x+a/y ) = coshw#, say where #w=x+a/y#

Now, we attempt to find the inverse function, #f^(-1)(x)#. Using the definition of the hyperbolic cosine:

# cosh w = (e^w+e^-w)/2 #

so then:

# y = coshw = (e^w+e^-w)/2 #

# :. 2y = e^w+e^-w #

# :. 2ye^w = e^(2w)+1 #

# :. e^(2w) -2ye^w + 1 = 0 #

# :. (e^(w) -y)^2 - y^2+1= 0 #

# :. (e^(w) -y)^2 = y^2-1 #

# :. e^(w) -y = +-sqrt(y^2-1) #

# :. e^(w) = y +-sqrt(y^2-1) #

# :. w = ln(y +-sqrt(y^2-1)) #

# :. x+a/y = ln(y +-sqrt(y^2-1)) #

# :. x = ln(y +-sqrt(y^2-1)) -a/y #

Therefore although we cannot write an exact function for the original repeated recursive function, #f(x)#, we can write an exact form for its inverse, #f^(-1)(x)#

# :. f^(-1)(x) = ln(x +-sqrt(x^2-1)) -a/x #

In order to minimize the distance between any function and its inverse, which are reflection in the line #y=x#, we can reduce the problem to that of minimizing the distance, #D#, say, between either function and #y=x#, then the distance we seek is #2D#.

Now, the distance from the point #(p,q)# and the line #Ax+By+C=0# is given by:

# D = |Ap+Bq+C|/sqrt(A^2+B^2) #

So, we can find the distance between a generic coordinate #(x,f^(-1)(x))# on #f^(-1)(x)# and #y=x# (or #x-y=0#), using:

# D = |-x+ln(x +-sqrt(x^2-1)) -a/x|/sqrt(1^2+(-1)^2) #

# :. 2D^2 = ln(x +-sqrt(x^2-1)) -a/x - x #

Differentiating wrt #x#, we have:

# 4(dD)/(dx) = ( 1+- x/sqrt(x^2-1) )/(x +-sqrt(x^2-1))+1/x^2-1 #

At a critical point we require that the derivative vanish leading to the two possible outcomes:

# ( 1+ x/sqrt(x^2-1) )/(x +sqrt(x^2-1))+1/x^2-1 = 0 #

# ( 1- x/sqrt(x^2-1) )/(x -sqrt(x^2-1))+1/x^2-1 = 0 #

Solving these equations (which are independent of #a#), we get approximate real solutions:

# x ~~ +- 1.7742 #

From which we can calculate the corresponding value of #D#.

I will return to the solution if time permits.

Disambiguation Note:

Inversion means writing a relation y = f(x) as

x = inverse of y = #f^(-1)(y)#,

for an x-interval, in which the relation is bijective, #1 - 1#.

f is a function operator and #f^(-1)# is its inverse operator. So,

the successive operations # f^(-1) f# and #ff^(-1)# both

produce the multiplication unit operator 1. Also, #f^(-1) f^(-1)# is f.

And so, if y = f(x),

#x = f^(-1)f(x) = f^(-1)(f(x)) = f^(-1)(y)#

I think that is high time to do away with the calling

#y = f^(-1)(x)#,

obtained by the swapping #(x, y) to (y, x)#,

as the inverse.

Importantly, the graphs for both are one and the same.

Swapping x and y rotates this graph and makes it different.

I affirm that the distance is 0, as the two graphs are one and the

same. For the common graph, a = 1. The inverse had to be defined

in two pieces, #x >=0# and #x <= 0#, as

#x = ln(y+(y^2-1)^0.5)# and #x = - ln(y+(y^2-1)^0.5)#,

respectively.

Graph for y = cosh(x+1/y) = the given FCF ( Functional Continued

Fraction ) expansion as f(x):
graph{(x - ln(y+(y^2-1)^0.5))(x + ln(y+(y^2-1)^0.5))=0}