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}