# The FCF (Functional Continued Fraction) cosh_(cf) (x;a) = cosh(x+a/cosh(x+a/cosh(x+...))). How do you prove that cosh_(cf) (0;1) = 1.3071725, nearly and the derivative (cosh_(cf) (x;1))'=0.56398085, at x = 0?

Sep 8, 2016

See the explanation and the Socratic graph for y = cosh(x+1/y).

#### Explanation:

Let y = cosh_(cf)(x;1)=cosh(x+1/cosh(x+1/cosh(x+...))).

This FCF is generated by

$y = \cosh \left(x + \frac{1}{y}\right)$

At x=0,

y=cosh( 1/y).

Using the iteration of the discrete analog

${y}_{n} = \cosh \left(\frac{1}{y} _ \left(n - 1\right)\right) , n = 1 , 2 , 3 , \ldots ,$

with starter ${y}_{0} = \cosh \left(1\right)$.

8-sd approximation

y=1.3071725.

Now,

$y ' = \left(\cosh \left(x + \frac{1}{y}\right)\right) '$

$= \sinh \left(x + \frac{1}{y}\right) \left(x + \frac{1}{y}\right) '$

$= \sinh \left(x + \frac{1}{y}\right) \left(1 - \frac{1}{y} ^ 2 y '\right)$

At x = 0,

$y ' = \sinh \left(\frac{1}{y}\right) \left(1 - \frac{1}{y} ^ 2 y '\right)$

Substituting y = 1.3071725 and solving for y',

$y ' = \sinh \frac{\frac{1}{1.3071725}}{1 + \sinh \frac{\frac{1}{1.3071725}}{1.3071725 {.}^{2}}}$

$= 0.56398068$, nearly.

Graph for y = cosh(x+1/y), using the inversion

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

graph{x=ln(y+(y^2-1)^0.5)-1/y}

Observe that $x \ge - 1 \mathmr{and} y \ge 1$

The second graph includes the tangent at x = 0.

graph{(x-ln(y+(y^2-1)^0.5)+1/y)(y-1.307-0.564x)=0}