Question

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?

Answer 1

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(x+1/y)`

At x=0,

y=cosh( 1/y).

Using the iteration of the discrete analog

`y_n=cosh(1/y_(n-1)), n=1, 2, 3, ...,`

with starter `y_0=cosh (1)`.

8-sd approximation

y=1.3071725.

Now,

`y'=(cosh(x+1/y))'`

`=sinh(x+1/y)(x+1/y)'`

`=sinh(x+1/y)(1-1/y^2y')`

At x = 0,

`y'=sinh(1/y)(1-1/y^2y')`

Substituting y = 1.3071725 and solving for y',

`y'=sinh(1/1.3071725)/(1+sinh(1/1.3071725)/(1.3071725.^2))`

`=0.56398068`, nearly.

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

`x = ln(y+sqrt(y^2 - 1))-1/y`:

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

Observe that `x >=-1 and y >=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}