The Functional Continued Fraction (F C F) of exponential class is defined by #a_(cf) (x;b) = a^(x+b/(a^(x+b/a^(x +...)))), a > 0#. Upon setting a = e = 2.718281828.., how do you prove that #e_(cf) ( 0.1; 1 ) = 1.880789470#, nearly?

2 Answers
Aug 4, 2016

Answer:

See explanation...

Explanation:

Let #t = a_(cf)(x;b)#

Then:

#t = a_(cf)(x;b) = a^(x+b/a^(x+b/a^(x+b/a^(x+...)))) = a^(x+b/(a_(cf)(x;b))) = a^(x+b/t)#

In other words, #t# is a fixed point of the mapping:

#F_(a,b,x)(t) = a^(x+b/t)#

Note that by itself, #t# being a fixed point of #F(t)# is not sufficient to prove that #t = a_(cf)(x;b)#. There might be unstable and stable fixed points.

For example, #2016^(1/2016)# is a fixed point of #x -> x^x#, but is not a solution of #x^(x^(x^(x^...))) = 2016# (There is no solution).

However, let us consider #a = e#, #x = 0.1#, #b = 1.0# and #t = 1.880789470#

Then:

#F_(a,b,x)(t) = e^(0.1+1/1.880789470)#

#~~e^(0.1+0.5316916199)#

#=e^0.6316916199#

#~~ 1.880789471 ~~ t#

So this value of #t# is very close to a fixed point of #F_(a,b,x)#

To prove that it is stable, consider the derivative near #t#.

#d/(ds) F_(e,1,0.1) (s) = d/(ds) e^(0.1+1/s) = -1/s^2 e^(0.1+1/s) #

So we find:

#F'_(e,1,0.1) (t) = -1/t^2 e^(0.1+1/t) = -1/t^2*t = -1/t ~~ -0.5316916199#

Since this is negative and of absolute value less than #1#, the fixed point at #t# is stable.

Also note that for any non-zero Real value of #s# we have:

#F'_(e,1,0.1) (s) = -1/s^2 e^(0.1+1/s) < 0#

That is #F_(e,1,0.1)(s)# is strictly monotonically decreasing.

Hence #t# is the unique stable fixed point.

Aug 5, 2016

Answer:

Contractive behaviour.

Explanation:

With #a = e# and #x = x_0# the iteration follows as

#y_{k+1} = e^{x_0+b/y_k}# and also
#y_k = e^{x_0+b/y_{k-1}}#

Let us investigate the conditions for a contraction in the iteration operator.

Substracting both sides

#y_{k+1}-y_k = e^{x_0}(e^{b/y_k}-e^{b/y_{k-1}})#

but in first approximation

#e^{b/y_k} = e^{b/y_{k-1}} + d/(dy_{k-1})(e^(b/y_{k-1}))(y_k-y_{k-1}) + O((y_{k-1})^2)#

or

#e^{b/y_k} - e^{b/y_{k-1}} approx -b(e^{b/y_{k-1}})/(y_{k-1})^2(y_k-y_{k-1})#

To have a contraction we need

#abs(y_{k+1}-y_k) < abs(y_k-y_{k-1})#

This is attained if

#abs(e^{x_0}b(e^{b/y_{k-1}})/(y_{k-1})^2)< 1#. Supposing #b > 0# and #k = 1# we have.

#x_0 + b/y_0 < 2 log_e(y_0/b)#

So given #x_0# and #b# this relationship allow us to find the initial iteration under contractive behaviour.