Re: Attractor set representation

From: Chris Zanek (chris_znk_at_yahoo.com)
Date: 07/07/04 

Date: 7 Jul 2004 07:45:05 -0700

nonlinear5@yahoo.com (Eugene Kononov) wrote in message news:<d4a4fd16.0407061836.57240d23@posting.google.com>...
> chris_znk@yahoo.com (Chris Zanek) wrote in message news:<fc074de2.0407040929.6396ed95@posting.google.com>...
> > I have the system:
> > f2(t + 1) = f2(t) + f6(t)
> > f4(t + 1) = f4(t) + f5(t)
> > f5(t + 1) = f2(t) * f4(t)
> > f6(t + 1) = f5(t) - f6(t)
> > f2(0) = -1.3355323936581297761
> > f4(0) = -0.2712765738091898738
> > f5(0) = -0.37703949692556708642
> > f6(0) = -0.16382178649293054848
> >
> > Is there a "relatively fast" (direct) method to calculate all f6(t)
> > for f5(t) = a (a is real value), with 20 accurate decimal digits?
> >
>
> Your system looks pretty straightforward (from computational
> complexity point of view), so I can't see how it can take longer than
> a minute to generate a million or so iterations, even with the
> precision that you specified.
The problem is that I want to calculate all possible f6(t) values when
f5(t) is a given constant named a. Using the algorithm from my
previous post on this topic I need roughly 10^9 iterations to find
close value f5(t) to a, i.e. |f5(t) - a| < 10^-7. So for example if I
want 16 accurate decimal digits after the period the number of
iterations will become > 10^18 and therefore the program will require
> 1 minute computational time on PC which works with 2-3 GHz processor
(can you check the exact time?).

Chris

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