Question

The FCF (Functional Continued Fraction) `cosh_(cf) (x;a)=cosh (x+a/cosh(x+a/cosh(x+...)))`. How do you prove that this FCF is an even function with respect to both x and a, together?and `cosh_(cf) (x; a) and cosh_(cf) (-x;a)` are different?

Answer 1

`cosh_(cf)(x;a)=cosh_(cf)(-x;a) and cosh_(cf)(x;-a)=cosh_(cf)(-x;-a)`.

Explanation:

As cosh values are `>=1`, any y here `>=1`

Let us show that y = cosh(x+1/y) = cosh(-x+1/y)

Graphs are made assigning `a = +-1`. The corresponding two
structures of FCF are different.

Graph for y = cosh(x+1/y). Observe that a = 1, x >=-1

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

Graph for y = cosh(-x+1/y). Observe that a = 1, x <=1

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

Combined graph for y = cosh(x+1/y) and y = cosh(-x+1/y)

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

Likewise, it is shown that y = cosh(-x-1/y) = cosh(-x-1/y).

Graph for y = cosh(x-1/y). Observe that a = -1, x >=1

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

Graph for y = cosh(-x-1/y). Observe that a = -1, x <=-1

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

Combined graph for y = cosh(x-1/y) and y = cosh(-x-1/y)

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