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 Uniqueness summary and idea bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/16/2007, 09:29 PM 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. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/16/2007, 10:03 PM (This post was last modified: 08/16/2007, 10:05 PM by jaydfox.) 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 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/16/2007, 10:13 PM (This post was last modified: 08/16/2007, 10:17 PM by bo198214.) 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). jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/16/2007, 10:30 PM 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. ~ Jay Daniel Fox « Next Oldest | Next Newest »

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