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Full Version: An error estimate for fixed point computation of b^x
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Ok, we want to compute the lower fixed point $a$ of $b^x$ for, as usual, $1.

We do that by iterating:
$x_{n+1} = b^{x_n}$
$\lim_{n\to\infty} x_n = a$

while iterating we can not directly compute the difference $a-x_n$ but we can compute $\mu_{n+1} = x_{n+1}/x_n$ and we ask our self how close $\mu_n$ must come to 1 such that

$a-x_n<\epsilon$

First we translate the difference into a quotient:

\begin{align*}\frac{b^a}{b^{x_n}}& y_{n+1}:=\frac{a}{x_{n+1}}& \end{align*}

$y_n>1$ is decreasing. Now lets compute
$\mu_{n+1} = \frac{x_{n+1}}{x_n}=\frac{b^{x_n}}{x_n}=\frac{b^{a/y_n}}{a/y_n}
=y_n a^{1/y_n-1} = y_n (y_n x_n)^{1/y_n-1}$

To make an estimate we want to know whether $f(x)=x(xc)^{1/x-1}=c^{-1}(xc)^{1/x}$, $c is decreasing for $x>1$. To decide this we differentiate:
$f'(x)=c^{-1}\left( xc \right) ^{1/x} {\frac {1-\ln \left( xc
\right) }{x^2}}$

$f'(x)>0$ for
$0<1-\ln(xc)$
$xc

This is true for $\epsilon<\log_b(e/c)$ because then $x.

So we know that $f$ is strictly increasing for small enough $\epsilon$. So we get
$y_{n+1} iff $\mu_{n+2}
for $\epsilon<\log_b(e/x_{n+1})$, equivalently:

$\frac{a}{x_{n+1}} iff $\frac{x_{n+2}}{x_{n+1}}
$a - x_n <\epsilon$ iff $x_{n+2}<(b^\epsilon x_{n+1})^{b^{-\epsilon}}$
The right side can be further simplified:
$b^{x_{n+1}} < b^{(\epsilon +x_n)b^{-\epsilon}}$
$x_{n+1}<(\epsilon + x_n)b^{-\epsilon}$

The condition $\epsilon<\log_b(e/x_{n+1})$ is satisfied for $\epsilon < \log_b(e/a)$, i.e. on the right side is a constant $>0$. So during the iteration we can decrease $\epsilon$ according to $x_n$ without fear that $\epsilon$ becomes to small and the iteration does not stop.

Proposition. Let $1, let $a$ be the lower fixed point of $b^x$, let $x_0 and $x_{n+1}=b^{x_n}$ then for each $0<\epsilon < \log_b(e/x_n)$:
$a - x_n <\epsilon$ iff $x_{n+1}<(\epsilon + x_n)b^{-\epsilon}$.