# 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?

May 8, 2018

We seek the minimum distance between the function:

$f \left(x\right) = \cosh \left(x + \frac{a}{\cosh} \left(x + \frac{a}{\cosh} \left(x + \frac{a}{\cosh} \left(\cdots\right)\right)\right)\right)$

and its inverse, for $a > 0$

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

$y = \cosh \left(x + \frac{a}{y}\right) = \cosh w$, say where $w = x + \frac{a}{y}$

Now, we attempt to find the inverse function, ${f}^{- 1} \left(x\right)$. Using the definition of the hyperbolic cosine:

$\cosh w = \frac{{e}^{w} + {e}^{-} w}{2}$

so then:

$y = \cosh w = \frac{{e}^{w} + {e}^{-} w}{2}$

$\therefore 2 y = {e}^{w} + {e}^{-} w$

$\therefore 2 y {e}^{w} = {e}^{2 w} + 1$

$\therefore {e}^{2 w} - 2 y {e}^{w} + 1 = 0$

$\therefore {\left({e}^{w} - y\right)}^{2} - {y}^{2} + 1 = 0$

$\therefore {\left({e}^{w} - y\right)}^{2} = {y}^{2} - 1$

$\therefore {e}^{w} - y = \pm \sqrt{{y}^{2} - 1}$

$\therefore {e}^{w} = y \pm \sqrt{{y}^{2} - 1}$

$\therefore w = \ln \left(y \pm \sqrt{{y}^{2} - 1}\right)$

$\therefore x + \frac{a}{y} = \ln \left(y \pm \sqrt{{y}^{2} - 1}\right)$

$\therefore x = \ln \left(y \pm \sqrt{{y}^{2} - 1}\right) - \frac{a}{y}$

Therefore although we cannot write an exact function for the original repeated recursive function, $f \left(x\right)$, we can write an exact form for its inverse, ${f}^{- 1} \left(x\right)$

$\therefore {f}^{- 1} \left(x\right) = \ln \left(x \pm \sqrt{{x}^{2} - 1}\right) - \frac{a}{x}$

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 $2 D$.

Now, the distance from the point $\left(p , q\right)$ and the line $A x + B y + C = 0$ is given by:

$D = | A p + B q + C \frac{|}{\sqrt{{A}^{2} + {B}^{2}}}$

So, we can find the distance between a generic coordinate $\left(x , {f}^{- 1} \left(x\right)\right)$ on ${f}^{- 1} \left(x\right)$ and $y = x$ (or $x - y = 0$), using:

$D = | - x + \ln \left(x \pm \sqrt{{x}^{2} - 1}\right) - \frac{a}{x} \frac{|}{\sqrt{{1}^{2} + {\left(- 1\right)}^{2}}}$

$\therefore 2 {D}^{2} = \ln \left(x \pm \sqrt{{x}^{2} - 1}\right) - \frac{a}{x} - x$

Differentiating wrt $x$, we have:

$4 \frac{\mathrm{dD}}{\mathrm{dx}} = \frac{1 \pm \frac{x}{\sqrt{{x}^{2} - 1}}}{x \pm \sqrt{{x}^{2} - 1}} + \frac{1}{x} ^ 2 - 1$

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

$\frac{1 + \frac{x}{\sqrt{{x}^{2} - 1}}}{x + \sqrt{{x}^{2} - 1}} + \frac{1}{x} ^ 2 - 1 = 0$

$\frac{1 - \frac{x}{\sqrt{{x}^{2} - 1}}}{x - \sqrt{{x}^{2} - 1}} + \frac{1}{x} ^ 2 - 1 = 0$

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

$x \approx \pm 1.7742$

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

Jun 20, 2018

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 = ${f}^{- 1} \left(y\right)$,

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 $f {f}^{- 1}$ both

produce the multiplication unit operator 1. Also, ${f}^{- 1} {f}^{- 1}$ is f.

And so, if y = f(x),

$x = {f}^{- 1} f \left(x\right) = {f}^{- 1} \left(f \left(x\right)\right) = {f}^{- 1} \left(y\right)$

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

$y = {f}^{- 1} \left(x\right)$,

obtained by the swapping $\left(x , y\right) \to \left(y , x\right)$,

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, $x \ge 0$ and $x \le 0$, as

$x = \ln \left(y + {\left({y}^{2} - 1\right)}^{0.5}\right)$ and $x = - \ln \left(y + {\left({y}^{2} - 1\right)}^{0.5}\right)$,

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}