bo198214 Wrote:You mean if you have an attracting fixed point then you simply get an arbitrary small x by applying \( f^{\circ n} \) so that the precision becomes arbitrary big for the subsequent \( f^{\circ t} \) and then you transform it back to the original value by applying \( f^{\circ -n} \)?Exactly, except for f(z)=e^z-1, it's a repelling fixed point, so -n and n would be used instead of n and -n.
Quote:Hm, I dont know whether you loose the achieved precision by transforming it back. This also would only work for fractional iterations, while I was talking about general series.You shouldn't lose much precision. The absolute value of the difference between \( f^{\circ \small -100}(1) \) and \( f^{\circ \small -101}(1) \) is about 2/(100^2). For
\( f^{\circ \small -1000}(1) \) and \( f^{\circ \small -1001}(1) \), it's about 2/(1000^2).
In other words, the precision you lose is on the order of 1/n^2. For n=1000, you only need an extra six decimal places of precision in your iterated step out function. Of course, each iteration will introduce additional errors, so really you need about 6+log_10(1000) = 9 to 10 extra digits of precision. Still, the iterating functions converge quite nicely, so the extra expense of using additional precision for the integer iterations is acceptable.
~ Jay Daniel Fox