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?

1 Answer
Sep 8, 2016

Answer:

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 + pi /y ) = pi^( pi( 1 + 1/y ) )#.

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

#Deltay_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]}