An error estimate for fixed point computation of b^x - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Computation (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=8) +--- Thread: An error estimate for fixed point computation of b^x (/showthread.php?tid=173) An error estimate for fixed point computation of b^x - bo198214 - 05/31/2008 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 $x_{n+1}<(\epsilon + x_n)b^{-\epsilon}$.