🚀 go-pugleaf

In article <1993Aug16.185623.23103@ast.saic.com>, hilljr@jupiter.saic.com (Jay R. Hill) writes:
|> In article <1993Aug12.192446.18556@ast.saic.com>, hilljr@jupiter.saic.com 
|>   (Jay R. Hill) writes:
|> |> In article <93222.174612GHGAPAC@cc1.kuleuven.ac.be>, 
|> |>   <GHGAPAC@cc1.kuleuven.ac.be> writes:
|> |> [...]
|> |> |> I ported part ... of fractINT to macintosh. ... I have some help 
|> |> |> ... They told me something about periodicity checking, which is
|> |> |> ...  Can anyone give me some help about this principle ?
|> |> Last February and March, as part of the Mandelbrot area calculation 
|> |> thread, Robert Munafo <mrob@world.std.com> suggested periodicity 
|> |> checking [...]   We used this idea in GIPPP#3.   [...]
|> |> I also discussed how you can test for an orbit being chaotic without
|> |> having to iterate to 'limit'.  Since the most interesting patterns
|> |> are near component edges, these tests can save a lot of time.
 
Dave Uherka <UHERKA%NDSUVM1.BITNET@uga.cc.uga.edu> wrote to the recipients 
of list FRAC-L <FRAC-L%GITVM1.BITNET@uga.cc.uga.edu>
>Jay Hill mentioned a test for chaotic orbits of the Mandelbrot
>function f(z) = z^2 + c.  Theoretically, for which c is the corresponding
>orbit chaotic?  Is the orbit chaotic only for a certain set of c's
>dense in the boundary of the Mandelbrot set?

Darcy Boese <dboese@SPARTAN.AC.BROCKU.CA> replied
>I've been told (by Jay himself) that he has a simple (relatively) method of
>picking out those chaotic orbital starting points.  For example, take the
>2-bud, which is a circle.  The border points which are at a bud converge to
>a cycle.  These points are at a rational multiple of Pi (ie. 0 degrees, 180
>degrees, etc).
>Then to get chaotic points you pick a non-rational multiple.
>In other words, the chaotic orbitals are anywhere along the border of any
>bud where it does not split off onto another bud.

Kerry Mitchell <mitchell@MITCH.LARC.NASA.GOV> added
->Here is an easy way to do this.  For the main cardioid, the boundary is given
->by x = 0.5 * (1-cos(theta)) * cos(theta) + .25, y = 0.5 *
->(1-cos(theta)) * sin(theta).  Theta of course, ranges from 0 to 2pi.
->To get a non-bud point,
->you would need an irrational multiple of pi for theta.  That is, let theta =
->k * 2pi, where k is irrational.  Since pi is itself an irrational
->(transcendental) number, any rational multiple of pi is irrational, so
->it suffices to have the product k*2pi be a rational number, so that k
->is necessarily irrational.

->What I do is pick a rational approximation for pi, say pie=22/7.  (This
->gives a value of theta that is close to a bud-point, but still
->rational.)  Then, I pick a fraction, say k=1/2.  Now, let theta =
->k*2*pie = 22/7 =/= pi.  Typically, I use these x & y values as the
->parameter for Julia sets; the interior (when viewed using Fractint and
->inside=mag) contains much more interesting structure than if I had
->chosen a bud-tangency point.  Also, this internal structure changes
->significantly and non-periodically with iteration number, while the
->boundary settles down after 100 iterations or so.

I agree with the above.  Chaotic points can be found on the edges of
the components, both cardioids and the attached buds.  They are between
the bud attachment points.  There is a direct map from irretionals and
rationals to these poionts though the angle theta.  So there is more
of the boundary chaotic than periodic.  There are also chaotic points
along the filements connecting the island cardioids.  These are harder
to locate off the real line (the only easy filement to find). 

By the way folks, check out Darcy's fine program, MOE, which shows 
these orbits.  It opens with a display of one of the chaotic points I 
posted earlier this year.  You can get it from wuarchive.wustl.edu
/archive/mirrors/msdos/graphics or from garbo.

Jay 'Not tonight honey, I've got a chaotic orbit' Hill
-- 
{ hilljr@jupiter.saic.com } begin writeln(3*ln(640320)/sqrt(163):17:15) end.
void main(){double sqrt(), y=1/sqrt(2.), a=.5, m=1,z; int n=0;
for(;m*=2,z=sqrt(1-y*y),y=(1-z)/(1+z),a=a*(1+y)*(1+y)-m*y,n<4;n++);
printf("%17.15lf\n",1/a);}