Question

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`?

Answer 1

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

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

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

At (0, 1),

`y'=sinh (0)(2)=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}