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+--- 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, \( 1<b<e^{1/e} \).
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:
\(
\begin{align*}\frac{b^a}{b^{x_n}}&<b^\epsilon\\
y_{n+1}:=\frac{a}{x_{n+1}}&<b^\epsilon
\end{align*} \)
\( y_n>1 \) 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 \( f(x)=x(xc)^{1/x-1}=c^{-1}(xc)^{1/x} \), \( c<a \) 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
\( 0<1-\ln(xc) \)
\( xc<e \)
This is true for \( \epsilon<\log_b(e/c) \) because then \( x<b^\epsilon<\frac{e}{c} \).
So we know that \( f \) is strictly increasing for small enough \( \epsilon \). So we get
\( y_{n+1}<b^\epsilon \) iff \( \mu_{n+2}<b^\epsilon (b^\epsilon x_{n+1})^{b^{-\epsilon}-1} \)
for \( \epsilon<\log_b(e/x_{n+1}) \), equivalently:
\( \frac{a}{x_{n+1}}<b^\epsilon \) iff \( \frac{x_{n+2}}{x_{n+1}}<b^\epsilon (b^\epsilon x_{n+1})^{b^{-\epsilon}-1} \)
\( a - x_n <\epsilon \) 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 \( 1<b<e^{1/e} \), let \( a \) be the lower fixed point of \( b^x \), let \( x_0<a \) and \( x_{n+1}=b^{x_n} \) then for each \( 0<\epsilon < \log_b(e/x_n) \):
\( a - x_n <\epsilon \) iff \( x_{n+1}<(\epsilon + x_n)b^{-\epsilon} \).