# 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?

Sep 10, 2016

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

#### Explanation:

y is a hyperbolic cosine value, and so, $\left\mid y \right\mid \ge 1$ and the FCF

graph is symmetrical with respect to y-axis.

${T}_{2} \left(x\right) = 2 {x}^{2} - 1$

The FCF is generated by

$y = \cosh \left({T}_{2} \left(x\right) \left(1 + \frac{1}{y}\right)\right)$

A discrete analog for approximating y is the nonlinear difference

equation

${y}_{n} = \cosh \left(\left(2 {x}^{2} - 1\right) \left(1 + \frac{1}{y} _ \left(n - 1\right)\right)\right)$.

Here, x = 1.25.

Making 37 iterations, with starter ${y}_{0} = \cosh \left(1\right) = 1.54308 . .$,

long precision 18-sd y = 18-sd

${y}_{37} = 6.00560689395441650$

with $\Delta {y}_{36} = {y}_{37} - {y}_{36} = 0$, 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.