# FCF (Functional Continued Fraction) cosh_(cf) (x; a)=cosh(x+a/cosh(x+a/cosh(x+...))) How do you prove that y = cosh_(cf) (x; x) is asymptotic to y = cosh x, as x -> 0 or the graphs touch each other, at x = 0?

Aug 16, 2016

The Socratic graphs touch at x = 0 and the point of contact is (0, 1).

#### Explanation:

At x = 0, for y = cosh_(cf)(x;x), y = 1.

For both, $y \ge 1$.

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

Like cosh x, this is also an even function of x.

Now,

$y '$

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

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

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

At (0, 1),

$y ' = \sinh \left(0\right) \left(2\right) = 0$.

For y = cosh x also, when x = 0, y=1 and y' = sinh x = 0, at (0, 1)

Thus, both touch each other at (0, 1), with cosh_(cf) graph bracing

cosh graph, from above.

The equation of the common tangent is y = 1

Graph of y = cosh x:

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

Graph of the FCF y = cosh(x+x/y):

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

Combined Socratic graph for y = cosh x and the FCF y =

cosh(x+x/y)

and the common tangent y = 1:

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