An error estimate for fixed point computation of b^x bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/31/2008, 04:11 PM (This post was last modified: 05/31/2008, 08:51 PM by bo198214.) Ok, we want to compute the lower fixed point $a$ of $b^x$ for, as usual, $11$ 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 $f(x)=x(xc)^{1/x-1}=c^{-1}(xc)^{1/x}$, $c1$. 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 $0<1-\ln(xc)$ $xc0$. 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

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