# A Functional Continued Fraction ( FCF ) is exp_(cf)(a;a;a)=a^(a+a/a^(a+a/a^(a+...))), a > 1. Choosing a=pi, how do you prove that the 17-sd value of the FCF is 39.90130307286401?

Sep 8, 2016

See details in explanation for the derivation. Some Socratic graphs are now included for graphical verification.

#### Explanation:

Let y = exp_(cf)(pi;pi;pi)=pi^(pi+pi/(pi^(pi+pi/(pi^(pi+...))). Then, The

implicit form for this FCF value y is

$y = {\pi}^{\pi + \frac{\pi}{y}} = {\pi}^{\pi \left(1 + \frac{1}{y}\right)}$.

A discrete anolog for approximating y is the nonlinear difference

equation

y_n = pi^( pi ( 1 + 1/y_(n-1))

Adopting this for iteration, with starter value ${y}_{0} = {\pi}^{\pi}$, and

making 15 iterations in long precision arithmetic,.

$y = {y}_{15} = 39.901130307206401$, nearly,

with the forward difference

$\Delta {y}_{14} = {y}_{15} - {y}_{14} = 0$, for this 17-sd precision.

Here, 0 means smallness of order ${10}^{- 18}$.

Scaled local graphs, for cross check:

Use y = exp_(cf)(x;x;x)=x^(x(1+1/y)) for the graph

x-range encloses $\pi$ for each of the two graphs.

y-ranges are appropriate, for precision levels.

The first is for higher precision.

Read y against x = $\pi$
graph{y-x^(x(1+1/y))=0 [3.141592 3.141593 39.9011 39.90115]}

graph{y-x^(x(1+1/y))=0 [1.6 4 0 60]}