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Author Topic: Is there anything novel left to do in M-like escape-time fractals in 2d?  (Read 80130 times)
Description: I think there may be. Your opinions are greatly desired.
0 Members and 1 Guest are viewing this topic.
Apophyster
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Posts: 124


« Reply #255 on: August 01, 2012, 09:35:40 AM »

Here are (with any luck) the pix that didn't stick last time, 2,
3, and 4.

Just so you know, I was able to view the images!
Lurker the Fred E
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Love and money are just passengers.
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fracmonk
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Posts: 356


« Reply #256 on: August 01, 2012, 04:52:15 PM »

Apophyster- Yeah, when the bot finally let me put them up...

Element90- Not familiar with your generator, [but that's a bit misleading, since cubing exponents would never produce that pic.]  Also, exponents have to be integer, to be clearer.

element90-  I'm sorry, I was dead wrong about that [in brackets just above]. 
I translated it to fractint formula as soon as I got back:

e90cubinv {;sample test
  z=p1, c=pixel:;z0=-1,1
    t=z*z*z
    z=(t+(1/t))*c
    |z| < p3
}

-and from it, I got a symmetrical ring pic not unlike those
found in post 229, all M2, which I'll post later, if you like, having
already reached today's limit.  It has a 6-sym.  I totally
CRINGE when I do stuff like that.
 
Sorry!

I felt so bad about that that I made a special trip back to
amend this!  Please forgive my intolerable error.



Asdam- I purposely give only a few, so that people DIY using the principles given.  It's the only way they'll move to understand what's going on with this stuff.


Part 2:

1+sqrt(2) (more or less) is the center of the 1st pic, where the
external M2 and a residual antenna extending from the unit
circle last touch before separating when the d value assignment
exceeds 1.828427124746189... on the way to 2.  I stared at this
number for awhile before I saw the pattern in its digits; it's
2*sqrt(2)-1 (more or less...)

This number was refined in precision by looking for a cross in
the second pic (greatly enlarged) while keeping either the
intersection resulting in the cross (d too small), or a gap (d
too large) in the center of the pic, which also has 1+sqrt(2)
at its center (more or less) as a repelling point.  In this
version, the external M is still attached to the unit circle (d
too small, but not by much).  It doesn't lend itself to
exactitude, since both d and the center of these pix are based
on the same trancendental number.

Notice in the unit circle (1st pic left) the similarity of the
contents there with the 1st pic in post 250(?).  You can't miss
it, it's the only one there, unfortunately.

When the d value enlarges, as in the third pic, there was promise
of mapping the cardioid to the unit circle, but probably can't
happen until d reaches infinity (so...No...).  In fact, the set
would then BECOME the unit circle, but there are easier ways to
get one, like the Julia set for f(z)=z^2+0.

Doesn't the second bulb there (last pic) look a little odd, though?
Check where the branching is, nearer the ends of the dendrites than
in standard M.  The behavior is at once similar and different
compared with standard M, once again.

Later.


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« Last Edit: August 01, 2012, 08:20:36 PM by fracmonk, Reason: fix error » Logged
fracmonk
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« Reply #257 on: August 01, 2012, 04:55:43 PM »

While I'm here, let me lay this on you: (preprepared)

The reason why I had returned to this area of my study was inspired
by a theory about information falling into black holes winding up
accumulating on the surface (or event horizon) of them in 2d form. 
Then, there are entwined particle theories, and symmetry-breaking,
to which the "nemesis" effect might relate.  But you must understand
that I have a slightly smaller budget than CERN devoted to physics
fairy stories.  What can you really find out about a bearing ball
by throwing thousands of basketballs at it, only to see them turn
into smaller balls, assuming your elaborate detection equipment is
not lying to you, and telling you what you want to hear, so you can
add a few more canards to an already inelegant theory?  Keep it
alive when it really wants to die, but you've invested too much
reputation and funding into it?  (And dispose of billions of dollars'
and euros' worth of industrial capacity that could have otherwise
improved the lot of millions of people living today?  A better method
for destroying the world?) 
O.K., I'm a little cranky right now...

So I might as well say here that this is just a small sample of the
curiosities available in multiplicative inverses and their likeness
to squaring when observed as fractals in the complex realm.  There's
more to be found, and not enough time in my life to do it all.

The same could be said for Multipowerbrots, which are algebraically
related.  It is like finding new universes; except it is merely a part
of this one, a fragment of its DNA, if you will, except more
picturesque than chemical DNA.  That is, if you go with the theory
that mathematics is the DNA of the design of the universe, then it is
like that. 
Junk DNA of the universe?

The reason why I decided to begin sharing all this stuff when I did
was the effects of age.  I had begun worrying how long I would have
the ability to do it.  It takes intermediate skill, which is a
no-man's land in math, where you are either a beginner or halfway to
Gauss, commanding the arcane notation and nomenclature that preserves
a priesthood from newcomers, who seem to get lost there, as I do. 
Sadly, there doesn't appear to be any good bridge between the two. 
And then, when you do get there, would you write a 15-page proof for
2+2, or try to sell a math that doesn't quite follow existing rules?
At least I know this is not that.

Honestly, I expected more interest, but we're not in the '90's any
more.  While there are more people acquainted with fractals today
than ever before, the interest is more casual than at that time and
lacks the enthusiasm for depth of understanding that existed before. 

Maybe the scientific community overreached in places when it came to
the application of chaos, but in many cases, it found paydirt.  I am
very likely overreaching here myself, but without throwing questions
out there, there is zero probability of answers, as opposed to small
but nonzero chances.  All this stuff could easily have died with me
otherwise, and may anyway.  I was only lucky enough to stumble into
some pretty interesting things.  Why did they appear so, and what do
they mean?  I see no direct line between this stuff and some future
application, and therefore, almighty money, but, fool that I am, I
believe that some things are beautiful in themselves, and worth
appreciating for that alone.

If you do not seek, you will not find.

Later.
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Apophyster
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Posts: 124


« Reply #258 on: August 02, 2012, 11:20:59 AM »

Regarding post #58: imo, well said.
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Alef
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« Reply #259 on: August 05, 2012, 03:23:03 PM »

It wouldb be more ease to look by yourself if formula file would had been present;)
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fracmonk
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Posts: 356


« Reply #260 on: August 06, 2012, 07:40:40 PM »

Asdam-  My formula files are a work in progress; there are too many unresolved questions in them, and untested conditions.  In 3 basic volumes, there are simply connected, multiply connected, and totally experimental kinds.  It would cause too much confusion.  There is adequate info here to see the patterns among them that make them work.  (so stop whining!)

Just kidding.  You can do it.




Thought for the day-   When a value is really large, its multiplicative inverse may go unnoticed...

Later.
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element90
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« Reply #261 on: August 07, 2012, 12:41:05 PM »

In my earlier post I used the the following formula:

z = c(0.5*z^3 + 0.5*z^(-3))

where c is the location in the complex plane and the initial value of z is also the location in the complex plane.

The implementation of the formula allows for non-integer and non-integer complex powers just to add extra flexibility as is the use of separate power parameters instead of using one and using two terms with the same power where one is the inverse of the other.

The result is a ring of 6 Mandelbrots linked together in a similar way to an image already posted by you. The use of the location in the complex plane is unsatisfactory as determining the factors to adjust z^n and z^-n to get well formed Mandelbrots is tricky. This formula only appears to produce Mandelbrot islands and I'm inclined to believe that multibrots won't appear at all.

It is far easier to use the critical values as the start value adjusting the factors to get a a critical value of 1.

So using z = c(z^3 + (z^(-3)) and z0 the resulting picture is as follows:



Comparing this image with that for the previous formula using the initial value of the location in the complex plane the only difference I've noticed is the size.

I then wondered what would happen if I used 2 and -4 as the powers again I got 6 Mandelbrots in a ring but the connections between them was different.

z = c(2z^2 + z^(-4))



So what of z = c(5z + z^(-5)) would there be a ring of 6 Mandelbrot? Well, No:



The limit used for this one is norm(z) > 16000000 instead of the usual norm(z) > 16.

The images of powers 2 & -4 and -2 & 4 are only slightly different, so what of z = c(5z^(-1) + z^5)? The difference is anything but slight:



So far, ignoring signs the powers have all added up to 6 resulting in 6 Mandelbrots, 6 buds or 6 lobes in terms of the largest features. So what happens if the powers have the same sign and allow add up to 6, will the result also have a six of the most prominent features? The answer is no.

For z = c(z^5 - 5z) the answer is 4 major lobes:



For z = c(2z^2 - z^4) the answer is 2 Mandelbrots:



Now for powers that are both negative, for z = c(2z^(-2) - z^(-4)) the answer is two distorted Mandelbrots:



And finally z = c(5z^(-1) - z^(-5)), the answer is a group of four.



The final picture does contain properly formed Mandelbrot islands which can be found easily by zooming in.

After much experimentation there appears to be a rule that the number of the most prominent features increases by 1 as the sum of the powers (ignoring signs) is increased by one. For the initial example of the ring of 6 Mandelbrots, the minimum power, whether positive or negative, is two, so the smallest ring of Mandelbrots is 4. I'm not sure about the number of variations in the construction of the ring, there can only be one version of the 4 Mandelbrot ring, there doesn't seem to be any difference between the two possible variations of the of the 5 Mandelbrot ring but there are at least two different versions of the 6 Mandelbrot ring.
« Last Edit: May 26, 2014, 11:25:04 AM by element90 » Logged

Elelemt90 Fractals blog www.element90.wordpress.com
fracmonk
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Posts: 356


« Reply #262 on: August 07, 2012, 07:16:14 PM »

element90-   ABSOLUTELY SUPERB STUFF!  I'm especially intrigued by the second-to last one.  Will give them a serious look.
You should know, however, that they're not printing right, at least for me where I am.

Later.
« Last Edit: August 07, 2012, 07:20:47 PM by fracmonk » Logged
element90
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« Reply #263 on: August 08, 2012, 12:28:59 PM »

Here are more:

z = c(-3z^(-3) + z^(-9)), z0 = 1, c = location in the complex plane



Now for non-integer powers.

z = c(z^2.5 + z^(-2.5)), z0 = 1, c = location in the complex plane



And the first attempt a throwing imaginary numbers into the mix.

z = c(z^(2i) + z^(-2)), z0 = 0.623832938258 - 0.258400063681i, c = location in the complex plane

Fortunately my calculator has complex numbers so working out the critical value was straight forward.



The picture produced using an imaginary power is similar to those produced using the compasses formula with imaginary powers. Like compasses the image is sensitive to the bailout value, for the above image I used norm(z) > 1.6e8.

I've started to look at using imaginary powers for both terms and they are even more sensitive to the bailout value and zooming out is required before anything can be seen. The width in the complex plane of the picture above is around 0.84 for a fractal with with powers of 2i and -2i and a bailout of norm(z) > 1.6e15 the width required is about 800000. Experimenting I found that size of the image was affected by the size of the bailout value, for a bailout of norm(z) > 1.6e9 the width required is about 2000.

I'll show pictures using imaginary powers for both terms in my next post and in the meantime I'll move onto complex powers.
« Last Edit: May 26, 2014, 11:28:40 AM by element90 » Logged

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fracmonk
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« Reply #264 on: August 10, 2012, 04:26:41 PM »


I had to take a close look at element90's function that makes
those interesting wide loops in the dendrites.  I had run
across similar shapes in my experiments, but never got them
to close like that.  Another thing I hadn't realized was that
functions containing only negative exponents (or, restated,
having z only in denominator expressions) could escape.  It
never occurred to me!  Is it still considered degree 4?  My
Fractint version looks like this:

e90circs(xyaxis) {;z0=+,-sqrt(2)
  z=sqrt(2), c=pixel:;no
    s=z*z           ;periodicity
    t=s*s           ;testing!
    z=(1/s-(1/t))*c
    |z| < p3
}

I tend to favor functions that can be programmed so
efficiently.  This because of faith in Ockham's razor, which
usually suggests that simpler explanations are likelier to be
true, but I suspect that natural phenomena tend to follow
simpler fractal rules as well, perhaps as a related
consequence.  It is why that kind interests me more than more
complicated functions.  Consider the logistic equation if you
need an example.  (Off topic:  As I recall, I think it was
Bob Weir at a Dead concert some 25 years ago or more, while
tuning up, saying: "...  Is it not so?  Is it not written in
the sky?" -and it STILL makes me laugh!)

The critical point was a fortunate early guess on my part, and
so, not too terribly hard to find.  Some detailed insight into what
methodology element90 uses to find usable critical points
would probably enlighten us all.  I know I'm not that good at
finding them myself. 

The first pic below shows the whole set, a "large" one, for
one that fits within computation limits for Fractint without
errors and cutoffs, AFAICT.  Ones that do at all are usually
much smaller.

The second is a closeup of a couple bulbs on the main bulb. 
Large and small loops join at key points on the familiar
branches.  They are infinite in number, but to fractal buffs,
that's nothing new...

One thing I have never seen before is dendrites emanating from
the Seahorse valley cusp.  The third pic shows a closeup of it. 
Notice how the loops rejoin the same bulb to the left, while
they bridge between the bulbs to the right.  There's a lot
going on in this object.  In the last pic, zooming in further
on that dendrite main ring in a different location, a mini
there is connected in the same way that the 2 largest M2 shapes
are, at the main bulb's pinch point.  This is apparently common
there.


More about that...later...


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element90
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« Reply #265 on: August 10, 2012, 06:39:59 PM »

To finding the critical points, I just solved the differential of the fractal function set to zero.

i.e. taking the general form

z(n+1) = f(z(n))

I solve

f'(z) = 0

So with

f(z) = c(alpha*z^beta + gamma*z^delta)
f'(z) = c(beta*alpha^(beta-1) + delta*gamma*z^(delta-1))

Using

alpha = 1
beta = 2
gamma = 1
delta = -3

c(2*1*z + (-3)*(1)z^-4) = 0
c*2*z = c*3*z^-4

c cancels out leaving

2z = 3z^-4

cross multiplying gives

z^5 = 1.5

so z = 1.0844717712 to the accuracy available to my calculator (an HP-42S that I've had for some time ...).

It is easier to select the values for alpha and gamma to give a power of z equal to 1 and hence a critical point of 1, you don't have to of course, you can have it equal a negative number, an imaginary number or indeed a complex number and then determine whatever the critical value happens to be, for example z^5 = -1 would yield a critical point of 0.80916994375 + 0.587785252292i according to the HP-42S. The resulting picture will be identical in structure only the size will change.

Since my calculator can to do complex numbers I can "easily" work out the critical points when using imaginary and complex powers. I've only successfully used this technique when the solution to f'(z) = 0 is straight forward.

Fracmonk, it seems that you're having fun with this variation, I haven't explored the variant the variant that much, yet... No pictures today.
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element90
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« Reply #266 on: August 13, 2012, 01:13:16 PM »

I've been side tracked, so no pictures using complex powers.

First of all, I've tried inverting the complex plane which leads to the teardrop shaped inverse Mandelbrot. For the example I'll show first the fractal in the unaltered complex plane:

Using z = c(alpha*z^beta + gamma*z^delta) where zed on the left hand side of = is z(n+1) and the zeds on the right hand side are z(n)

alpha = -3
beta = -2
gamma = 2
delta = -3

giving a critical point of 1 so z0 is 1.



Saturn has a transforms tab where the complex plane can be transformed so that c = 1/c or c = c^(-1) which modifies the formula so that it becomes:

z = (alpha*z^beta + gamma*z^delta)/c

since c cancels out in the calculation of the critical value 1/c also cancels out so using the same parameters z0 remains at 1.

The inverse version looks like this:



In common with most Saturn fractals there a z = transform(z) line applied before fractal formula if a transform or transforms have been defined (also using the transforms tab).

For

alpha = 1
beta = 6
gamma = 1
delta = -6

a 12 Mandelbrot ring would be produced, however using the simplest transform "translation" with a value of 2 the formula becomes:

z = c*(alpha*(z+2)^beta + gamma*(z+2)^delta)

so

f(z) = c*(alpha*(z+2)^beta + gamma*(z+2)^delta)
f'(z) = c*(beta*alpha*(z+2)^(beta-1) + delta*gamma*(z+2)^(delta-1)

so substituting values we get

f'(z) = c*(6*1*(z+2)^5 - 6*1*(z+2)^(-7))

to find the critical value

c*(6*(z+2)^5 - 6*(z+2)^(-7)) = 0

which is

6*c*(z+2)^5 = 6*c*(z+2)^(-7)

or more concisely

(z+2)^12 = 1

since the 12th root of 1 is 1

z+2 = 1

so the critical value, z and hence z0 is -1.

The 12 Mandelbrot ring has fragmented, is disconnected and there are only 6 of the largest Mandelbrots.




« Last Edit: May 26, 2014, 11:39:58 AM by element90 » Logged

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« Reply #267 on: August 13, 2012, 07:11:06 PM »

The last two are really interesting! More should be done with those...
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fracmonk
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« Reply #268 on: August 14, 2012, 04:33:44 PM »

element90-

I think I've been hijacked!  Very interesting ride...

With post 265, this may be redundant, but I preprepared
this, and it may show how our thinking somewhat overlaps.  
Bear in mind that I haven't been able to have a look here
since I posted last.  This is how cave men do it:

(And I wish I could take your posts back with me printed,
so I could study them, but I can't from here...it'll have to
wait, but you seem to have things well in hand...)

I'd been busy over the weekend, finding myself
mournfully sans Bunny, following my own advice of DIY,
using trial-and-error, fractint testing, a calculator,
and a lot of good ol' human pattern recognition to
"crack" this problem of finding critical points, and
remember, you crawl before you walk in this life.  

I came up with this:

For functions (from element90) with this structure:

f(z)->c((1/(z^a))-(1/(z^b))),

Where a and b are real integers and a>b, and there are
no coefficients, a critical point (for init z) can be
found using the following formula:

              __________
     (a-b)/ a-b
z0=  \  /  ___ + 1
         \/    b

(Copy+Paste the following into Notepad just as it
is if the version above doesn't space right in your
display.  I anticipated that sort of problem here, but
now I notice that the same applies to formulae I've
given.  They'll look right in old fashioned typewriter
spacing:)

          __________
    (a-b)/ a-b
z0=  \  /  ___ + 1
      \/    b

Or, the verbal explanation:

You use the (a-b)th root; if it's 1, there's no radical;
if it's 2, then it's sqrt; if it's 3, then 3rd root, etc.
Take the (a-b)th root of ((a-b)/b)+1 within the radical
to get z0.  Most yield infinite digit strings and require
more precision for better results.

For example, for a=2, b=1, the answer is 2.  All a with
b=1 result in an infinite index set, like pic 1 below (the
origin is in the light circle in the lower right here).  
It is NOT a contortion of Mandelbrot's set, AFAICT.  Check
the progression of dendrite branchings from bulb to bulb
of comparable size.  Each successive large bulb adds 2,
INCREASING with distance from the cusp.

Since element90 appears to like sixes, consider a=8, b=2.  
When you take the 6th root of 4, (same as 3rd root of 2,
BTW), you get 1.2599211... as the critical point for
pic 2 below.  
If b>=2, the index set is of finite size.

While one could use:

e90circsuniv {;z0=? no periodicity
  z=p1, c=pixel:;   testing!
    z=((1/(z^p2))-(1/(z^p3)))*c
    |z| < 1000000
}

-univerally for functions of this structure, it would be
terribly slow.  For the example above, it would be better
to use:

e90circs82(xyaxis) {;z0=6throot(4)=
  z=p1, c=pixel:;    cuberoot(2)=
    r=z*z           ;1.2599211...          
    s=r*r           ;no
    t=s*s           ;periodicity
    z=(1/t-(1/r))*c ;testing!
    |z| < p3
}

The programming for your generator may have different
requirements, of course, but these fractint examples
should be easy enough to follow.  (Why I stick with it.)

The lowest combination of integer exponents resulting in
a finite sized index set is a=3, b=2, giving pic 3.  Pic
4 shows a completely dendritic Julia set for c=20.25,
the precise distance in which the main loop dendrite
crosses the real axis (Pic 3 right).  Compare the famous
f(z)->z^2+i.  
Good stuff for number theorists.


Later.





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« Last Edit: August 14, 2012, 04:42:06 PM by fracmonk » Logged
fracmonk
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« Reply #269 on: August 16, 2012, 04:58:41 PM »

Below is the beginning of an endless progression of simply
connected functions inspired by the formula structure
introduced here by element90.  In this case, the exponents
are then positive ones.  The structure goes as follows:

f(z)->((((z^a)-(z^b))c)-1)^2

in which integer a-b=1, and critical points are z0=b/a, so
that the critical points of the below pix, in order, are
1/2, 2/3, 3/4, and 4/5.  It turns out that a-b=1 appears to
be the only case in which simply connected index sets, in
one piece, as those shown, can be formed.  Counting the
stems from the bulbs, the number of branchings on the
largest dendrite to left in each is 2a-2.  They get more
"furry" as the exponents increase.  (So do up, say, a=98,
b=97, with z0=97/98, if you like.  Go ahead, make my
day...)  Sadly, Fractint has a hard time with these, and
errors increase as exponents increase, and the sets get
larger too.  Actually, some set exponent combination could
act as a benchmark test for the accuracy of both hardware
and software calculation schemes that makers claim can
handle such.  It's a killer.  It seems most computers
these days are geared for tying everyone up with twit'er
and fakebook, etc., and not traditional computing,
ironically, (since they still call them computers) fast as
they may be.  

As an example, the formula for the last one shown here is:

DM54(xaxis) {;deg 10
  c=pixel, z=p1, d=p2:;quasi M2
    q=z*z       ;d=1:z0=4/5=.8
    r=q*q       ;only!
    s=r*z       ;nonstandard
    t=(s-r)*c-d ;dendritic
    z=t*t       ;structure
    |z| < p3
}
  
Note:
If a-b>1 AND b=1, then the sets will look similar to the
a-b=1 case, but will be surrounded by infinite numbers of
satellite islands of varing size, unless I missed something
and have that all wrong (which is always more than
possible!).  For the curious, for b=1, I used sqrt(3)=
.5773503... for a=3, and cuberoot(4)=.6299605... for a=4, to
get coherent shapes, but as a "critical point", in those
cases, it would only indicate the existence of one or more
non-escaping points at or about that location in the island-
bearing Julia sets, not indicating more than partial
connectedness, reflected in the nature of the resulting
index sets themselves.  You've seen that effect before here
in this thread, in post 41, and very recently, 266.
  

Off topic:  I LOST the Olympics!  And I wasn't even trying!
My arms are too short for world-class swimming, and too long
for weightlifting, and worst of all, I gave up letting
coaches abuse me (not sexually, but the usual classical
abuse) long ago.  I found I was just not specialized enough
for the robot junkyard.  And of course, now I'm...old...
...and maybe in the way...
(For Fred)


Later.


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« Last Edit: August 16, 2012, 05:06:00 PM by fracmonk, Reason: lookups » Logged
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