# 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?

Aug 4, 2016

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:

${F}_{a , b , x} \left(t\right) = {a}^{x + \frac{b}{t}}$

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

For example, ${2016}^{\frac{1}{2016}}$ is a fixed point of $x \to {x}^{x}$, but is not a solution of ${x}^{{x}^{{x}^{{x}^{\ldots}}}} = 2016$ (There is no solution).

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

Then:

${F}_{a , b , x} \left(t\right) = {e}^{0.1 + \frac{1}{1.880789470}}$

$\approx {e}^{0.1 + 0.5316916199}$

$= {e}^{0.6316916199}$

$\approx 1.880789471 \approx t$

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 $t$.

$\frac{d}{\mathrm{ds}} {F}_{e , 1 , 0.1} \left(s\right) = \frac{d}{\mathrm{ds}} {e}^{0.1 + \frac{1}{s}} = - \frac{1}{s} ^ 2 {e}^{0.1 + \frac{1}{s}}$

So we find:

$F {'}_{e , 1 , 0.1} \left(t\right) = - \frac{1}{t} ^ 2 {e}^{0.1 + \frac{1}{t}} = - \frac{1}{t} ^ 2 \cdot t = - \frac{1}{t} \approx - 0.5316916199$

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:

$F {'}_{e , 1 , 0.1} \left(s\right) = - \frac{1}{s} ^ 2 {e}^{0.1 + \frac{1}{s}} < 0$

That is ${F}_{e , 1 , 0.1} \left(s\right)$ is strictly monotonically decreasing.

Hence $t$ is the unique stable fixed point.

Aug 5, 2016

Contractive behaviour.

#### Explanation:

With $a = e$ and $x = {x}_{0}$ the iteration follows as

${y}_{k + 1} = {e}^{{x}_{0} + \frac{b}{y} _ k}$ and also
${y}_{k} = {e}^{{x}_{0} + \frac{b}{y} _ \left\{k - 1\right\}}$

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

Substracting both sides

${y}_{k + 1} - {y}_{k} = {e}^{{x}_{0}} \left({e}^{\frac{b}{y} _ k} - {e}^{\frac{b}{y} _ \left\{k - 1\right\}}\right)$

but in first approximation

${e}^{\frac{b}{y} _ k} = {e}^{\frac{b}{y} _ \left\{k - 1\right\}} + \frac{d}{{\mathrm{dy}}_{k - 1}} \left({e}^{\frac{b}{y} _ \left\{k - 1\right\}}\right) \left({y}_{k} - {y}_{k - 1}\right) + O \left({\left({y}_{k - 1}\right)}^{2}\right)$

or

${e}^{\frac{b}{y} _ k} - {e}^{\frac{b}{y} _ \left\{k - 1\right\}} \approx - b \frac{{e}^{\frac{b}{y} _ \left\{k - 1\right\}}}{{y}_{k - 1}} ^ 2 \left({y}_{k} - {y}_{k - 1}\right)$

To have a contraction we need

$\left\mid {y}_{k + 1} - {y}_{k} \right\mid < \left\mid {y}_{k} - {y}_{k - 1} \right\mid$

This is attained if

$\left\mid {e}^{{x}_{0}} b \frac{{e}^{\frac{b}{y} _ \left\{k - 1\right\}}}{{y}_{k - 1}} ^ 2 \right\mid < 1$. Supposing $b > 0$ and $k = 1$ we have.

${x}_{0} + \frac{b}{y} _ 0 < 2 {\log}_{e} \left({y}_{0} / b\right)$

So given ${x}_{0}$ and $b$ this relationship allow us to find the initial iteration under contractive behaviour.