#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
Sep 10, 2016

Answer:

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

Explanation:

y is a hyperbolic cosine value, and so, #abs y >=1# and the FCF

graph is symmetrical with respect to y-axis.

#T_2(x)=2x^2-1#

The FCF is generated by

#y=cosh(T_2(x)(1+1/y))#

A discrete analog for approximating y is the nonlinear difference

equation

#y_n=cosh((2x^2-1)(1+1/y_(n-1)))#.

Here, x = 1.25.

Making 37 iterations, with starter # y_0=cosh (1) = 1.54308..#,

long precision 18-sd y = 18-sd

#y_37 =6.00560689395441650#

with #Deltay_36=y_37-y_36=0#, for this precision.

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.