Re: Attractor set representation

From: Jeffrey H (aptsolutions_at_mindspring.com)
Date: 09/21/04

  • Next message: Chris Zanek: "Re: Attractor set representation" 
    Date: Tue, 21 Sep 2004 18:58:09 GMT

    chris_znk@yahoo.com (Chris Zanek) wrote:

    >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

    f4 = 2f2 + f6 + k is consistent because for t+1,
    f4(t) + f5(t) = 2f2(t) + 2f6(t) + f5(t) - f6(t) + k
    This immediately removes one variable, and if we
    remove f6 we have

    f2(t + 1) = -f2(t) + f4(t) - k
    f4(t + 1) = f4(t) + f5(t)
    f5(t + 1) = f2(t) * f4(t)

    This set of 3 equations should be faster and
    easier to work with. It should also be obvious
    how to calculate k from the initial conditions
    and how to calculate f6 at any point.

  • Next message: Chris Zanek: "Re: Attractor set representation"

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