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?

Answer 1

See 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.
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Answer 2

Contractive behaviour.

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.
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Answer 3

[e_{cf}(0.1; 1) \approx 1.880789470]

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Answer 4

To prove that ( e_{cf}(0.1;1) ) is approximately ( 1.880789470 ) when ( a = e ), you would use the formula for the Functional Continued Fraction (F C F) of exponential class:

[ a_{cf}(x;b) = a^{(x+\frac{b}{a^{(x +\frac{b}{a^{(x +...)}))}}}} ]

Substitute ( a = e ), ( x = 0.1 ), and ( b = 1 ) into the formula and compute the value iteratively until convergence. The result will be approximately ( 1.880789470 ).

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