Question

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`?

Answer 1

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.

Answer 2

Assuming that y = g(x) ( in my view f(x) ) is the inverse, the distance is 0, everywhere.

Explanation:

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}