# #T_n(x)# is the Chebyshev polynomial of degree n. The FCF #cosh_(cf )(T_n (x); T_n (x))=cosh(T_n(x)+(T_n(x))/cosh(T_n (x) + ...)), x >= 1#. How do you prove that the 18-sd value of this FCF for #n=2, x =1.25# is #6.00560689395441650?

##### 1 Answer

#### Answer:

See the explanation and the super Socratic graphs, for this complicated FCF

#### Explanation:

y is a hyperbolic cosine value, and so,

graph is symmetrical with respect to y-axis.

The FCF is generated by

A discrete analog for approximating y is the nonlinear difference

equation

Here, x = 1.25.

Making 37 iterations, with starter

long precision 18-sd y = 18-sd

with

graph{(2x^2-1-(y/(1+y))ln(y+(y^2-1)^0.5))(x-1.25)((x-1.25)^2+(y-6)^2-.001) =0[-2 2 0 10)]}

Graph for 6-sd in y(1.25) = 6.00561:

graph{(2x^2-1-(y/(1+y))ln(y+(y^2-1)^0.5))((x-1.25)^2+(y-6)^2-.001) =0[1.2499998 1.2500001 6.0056 6.00561]}

I expect applications of this type of FCF, in computer

approximations.

Observe that, despite being an even function, in the middle, the

graph is absent, and this is discontinuity.