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+--- Thread: Uniqueness summary and idea (/showthread.php?tid=19)
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RE: Uniqueness summary and idea - bo198214 - 08/16/2007
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 \( f^{\circ -n} \)?
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.
RE: Uniqueness summary and idea - jaydfox - 08/16/2007
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.
RE: Uniqueness summary and idea - bo198214 - 08/16/2007
jaydfox Wrote:You shouldn't lose much precision.
Yes, thats right because you can compute \( f^{\circ n} \) and \( f^{\circ -n} \) to arbitrary exactness.
So did you already determine the formula for the optimal truncation and its error function? (The answer however should go into the thread computing the iterated exp(x)-1).
RE: Uniqueness summary and idea - jaydfox - 08/16/2007
Not yet, I've been busy with my job most of today, just taking opportunities here and there to try to make a few posts. Also, I'm working on trying to figure out if there's a relationship between the graph of \( \mu_e^{\small -1}(x) \) and \( {}^{x} \check \eta \). I think I'm getting close, conceptually, but I'm still not there. That's where most of my attention is focussed at the moment. If I can derive a formula for \( \mu_e^{\small -1}(x) \), then I'll be quite satisfied that all the hard pieces are out of the way, and it'll just be a matter of formalizing everything, including providing proofs with all the gory details.