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?

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Answer 1 · 2016-09-08T12:07:29

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

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}