# 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

See explanation...

#### Explanation:

Let

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,

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

Note that by itself,

For example,

However, let us consider

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

To prove that it is stable, consider the derivative near

#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

Also note that for any non-zero Real value of

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

That is

Hence

Contractive behaviour.

#### Explanation:

With

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

Substracting both sides

but in first approximation

or

To have a contraction we need

This is attained if

So given