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#?

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Answer 1 · 2016-08-16T03:11:48

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

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}