From: isaacs@hpcc01.HP.COM (Stan Isaacs)
Newsgroups: alt.fractals
Subject: More on the area of the Mandelbrot Set
Date: 25 Feb 91 07:43:14 GMT
Organization: HP Corp Computing & Services
At the Northern California Section of the MAA meeting this past
weekend, John Ewing gave an interesting talk called "Can We See the
Mandlebrot Set". (He avoided using both the "C" word and the "F" word
after the introduction to the talk. It was not about either chaos or
fractals, but about what the M set is, mathematically.) Anyway, he
discussed two results which we've seen separately in this group in the
last month or so. Namely, that by computing the area of the M-set
using lots of terms in a series (Laurent Series?), the upper bound of
the area seems to converge about at 1.72 (the graph gets quite flat,
and seems to have an asymptote there), and by counting pixals more and
more accurately, you seem to get a lower bound of very close to 1.52.
Both these bounds are close to the values the methods would produce in
the limit - that is, it is NOT the case that these numbers would get
closer if a finer grid were used, or more terms were taken in the
series. So, why the difference of 10% or so? No one knows. One
possibility is that the pixal method misses "hairs" around the border.
(I know thats not described very well; it was late, and I wasn't
taking notes.) There was also a vague theory for the higher number
being possibly wrong. But basically, it is not known, at present,
which of these numbers represents the "real" area (although it is
believed that one of them does.) I'm afraid he didn't give references
(I asked afterwards.)
-- Stan Isaacs
From: jasonp@wam.umd.edu (Jason Stratos Papadopoulos)
Newsgroups: sci.math
Subject: Re: centroid of the Mandelbrot's set
Date: 19 Jun 1996 15:52:41 GMT
Organization: University of Maryland College Park
Zdislav V. Kovarik (kovarik@mcmail.cis.McMaster.CA) wrote:
: :
: :Is the Mandelbrot set measurable? I mean, does it *have* an area?
: :If so, what is it?
: A humble start: It is compact and has interior points, so its Lebesgue
: measure is well-defined, positive and finite.
: Good luck, ZVK (Slavek),
I know almost nothing about this, but see an article by Ewing and Schober,
something like "On the Coefficients of the Reciprocal of the Mandelbrot Set"
in "Mathematical Analysis and Applications". It seems the area of the Mandel-
brot set is bounded above by
Infinity
pi*( 1 - Sum n* b(n)^2 )
n=1
where b(n) are the coefficients of the power series mapping the unit circle
to the exterior of the Mandelbrot set. It converges very slowly (the
terms of the series are very irregular, can be found recursively, and
asymptotically behave as O(n^(-5/4) ). I believe that adding a quarter
million terms yields an area approximately of 1.72 (1.75? something like
that); this is interesting because huge scale Monte Carlo runs bound the
area at about 1.53 (or thereabouts). I've tried accelerating the power series
based on the first 20 or so terms, but haven't had any luck.
As for the centroid question...the centroid is the center of area, right?
If so, it'll be on the real axis (Mandelbrot is symmetric), and can be ap-
proximated roughly by finding the "center of mass" of the big cardioid,
big circle, and the two largish (above/below the cardioid) bulbs all
together.
Take all this cautiously. If anyone's interested, let me know and I'll
dig out a more complete reference list and post it here (these were the
highlights).
jasonp
From: Jeff Leader <JeffLeader@worldnet.att.net>
Newsgroups: sci.math
Subject: Re: centroid of the Mandelbrot's set
Date: 20 Jun 1996 01:26:14 GMT
Organization: AT&T WorldNet Services
jasonp@wam.umd.edu (Jason Stratos Papadopoulos) wrote:
>million terms yields an area approximately of 1.72 (1.75? something like
>that); this is interesting because huge scale Monte Carlo runs bound the
>area at about 1.53 (or thereabouts). I've tried accelerating the power series
>based on the first 20 or so terms, but haven't had any luck.
There was a (fairly) recent article on this 1.5/1.75 in the Monthly or
Math. Mag. that was interesting. Noone knows, but it's clearly an object
with a well-defined, finite (it's a subset of a disk of radius two) area.
Don't know about the centroid...
Newsgroups: sci.math,sci.fractals
From: "Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>
Subject: Re: centroid of the Mandelbrot's set
To: JeffLeader@worldnet.att.net,jhill@nosc.mil
Organization: SAIC
Date: Thu, 20 Jun 1996 23:10:07 GMT
Jeff Leader <JeffLeader@worldnet.att.net> wrote:
>jasonp@wam.umd.edu (Jason Stratos Papadopoulos) wrote:
>
>>million terms yields an area approximately of 1.72 (1.75? something like
>>that); this is interesting because huge scale Monte Carlo runs bound the
>>area at about 1.53 (or thereabouts). I've tried accelerating the power series
>>based on the first 20 or so terms, but haven't had any luck.
>
>There was a (fairly) recent article on this 1.5/1.75 in the Monthly or
>Math. Mag. that was interesting. Noone knows, but it's clearly an object
>with a well-defined, finite (it's a subset of a disk of radius two) area.
>Don't know about the centroid...
>
>
Ah, the old MSet area problem. I compute the area is at least 1.505936.
Three years ago a bunch of us (66) who read sci.fractals computed upper
and lower bounds. We found it is more than 1.5031197 and no greater than
1.5613027. See Y. Fisher and J. Hill, Bounding the Area of the Mandelbrot
Set, Numerische Mathematik,. (Submitted for publication). Available via
World Wide Web (in Postscript format)
http://inls.ucsd.edu/y/Complex/area.ps.Z
As for the centroid? Just guessing, but how about -1/3?
Jay "Not to night honey, it's that Mandelbrot project again" Hill
--
int main(){float g,s,f,r,i;char*_="!/-,;<:!lnb/bh`r/ylqbAmmhI/S/x`K\n";int m,u,
e=0;_[32]++;for(;e<3919;){u=(256*(s=(r=.0325*(m=e%80)-2)*r+(i=.047*(e/80)-1.128
)*i)-96)*s+32*r<3?25:16+32*r+16*s<1?31:0;if(u==(s=f=0))do g=s*s-f*f+r;while((f=
2*s*f+i)*f+(s=g)*g<4&&++u<27);putchar(_[++e>3840&&m<25?31-m:m>78?32:u]^1);}}
From: mert0236@sable.ox.ac.uk (Thomas Womack)
Date: 21 Jun 1996 18:56:03 GMT
Newsgroups: sci.math,sci.fractals
Subject: Re: centroid of the Mandelbrot's set
Jay R. Hill (JAY.R.HILL@cpmx.saic.com) wrote:
: Jeff Leader <JeffLeader@worldnet.att.net> wrote:
: >jasonp@wam.umd.edu (Jason Stratos Papadopoulos) wrote:
: >
: >>million terms yields an area approximately of 1.72 (1.75? something like
: >>that); this is interesting because huge scale Monte Carlo runs bound the
: >>area at about 1.53 (or thereabouts). I've tried accelerating the power series
: >>based on the first 20 or so terms, but haven't had any luck.
: >
: >There was a (fairly) recent article on this 1.5/1.75 in the Monthly or
: >Math. Mag. that was interesting. Noone knows, but it's clearly an object
: >with a well-defined, finite (it's a subset of a disk of radius two) area.
: >Don't know about the centroid...
: >
: >
: Ah, the old MSet area problem. I compute the area is at least 1.505936.
: Three years ago a bunch of us (66) who read sci.fractals computed upper
: and lower bounds. We found it is more than 1.5031197 and no greater than
: 1.5613027. See Y. Fisher and J. Hill, Bounding the Area of the Mandelbrot
: Set, Numerische Mathematik,. (Submitted for publication). Available via
: World Wide Web (in Postscript format)
: http://inls.ucsd.edu/y/Complex/area.ps.Z
: As for the centroid? Just guessing, but how about -1/3?
It's somewhere around -0.288, by straight Monte Carlo. The area is about
1.52, by the same argument.
--
Tom
"The first Ariane 5 flight did not result in validation of Europe's new
launcher"
Newsgroups: sci.math,sci.fractals
From: "Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>
Subject: Re: centroid of the Mandelbrot's set
To: JAY.R.HILL@cpmx.saic.com,jhill@nosc.mil,marblanc@mail.mcnet.ch,Dt812n.8J6@undergrad.math.uwaterloo.ca,1C1992A.1FCA@mail.mcnet.ch,lhf@csg.uwaterloo.ca
Organization: SAIC
Date: Fri, 21 Jun 1996 23:54:57 GMT
Sylvestre Blanc <marblanc@mail.mcnet.ch> wrote:
>I know that my question is little odd, but does anybody know where is
>the centroid of the Mandelbrot's set ?
"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com> wrote:
[snippity-snip]
>As for the centroid? Just guessing, but how about -1/3?
The area of the cardioid is 3pi/8 with centroid at -1/6.
The circle has area 1/16 centered at -1. The combined centroid
is -2/7. Pixel counting (I can't believe I wrote that) gets me
the centroid near -0.288, still close enough to -2/7 to wonder....
>Jay "Not to night honey, it's that Mandelbrot project again" Hill
Jay "Not again or is it still..." Hill
--
int main(){float g,s,f,r,i;char*_="!/-,;<:!lnb/bh`r/ylqbAmmhI/S/x`K\n";int m,u,
e=0;_[32]++;for(;e<3919;){u=(256*(s=(r=.0325*(m=e%80)-2)*r+(i=.047*(e/80)-1.128
)*i)-96)*s+32*r<3?25:16+32*r+16*s<1?31:0;if(u==(s=f=0))do g=s*s-f*f+r;while((f=
2*s*f+i)*f+(s=g)*g<4&&++u<27);putchar(_[++e>3840&&m<25?31-m:m>78?32:u]^1);}}
Newsgroups: sci.math,sci.fractals
From: "Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>
Subject: Re: centroid of the Mandelbrot's set
To: jhill,@,nosc.mil,marblanc,@,mail.mcnet.ch,Dt812n.8J6,@,undergrad.math.uwaterloo.ca,,1C1992A.1FCA,@,mail.mcnet.ch,lhf,@,csg.uwaterloo.ca
Organization: SAIC
Date: Mon, 24 Jun 1996 16:21:26 GMT
"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com> wrote:
>Sylvestre Blanc <marblanc@mail.mcnet.ch> wrote:
>>I know that my question is little odd, but does anybody know where is
>>the centroid of the Mandelbrot's set ?
>
>"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com> wrote:
>[snippity-snip]
>>As for the centroid? Just guessing, but how about -1/3?
>
>The area of the cardioid is 3pi/8 with centroid at -1/6.
>The circle has area 1/16 centered at -1. The combined centroid
^
make this pi/16 _|
>is -2/7. Pixel counting (I can't believe I wrote that) gets me
>the centroid near -0.288, still close enough to -2/7 to wonder....
From an over the weekend run: -.28781.
>
>>Jay "Not to night honey, it's that Mandelbrot project again" Hill
>
Jay
--
int main(){float g,s,f,r,i;char*_="!/-,;<:!lnb/bh`r/ylqbAmmhI/S/x`K\n";int m,u,
e=0;_[32]++;for(;e<3919;){u=(256*(s=(r=.0325*(m=e%80)-2)*r+(i=.047*(e/80)-1.128
)*i)-96)*s+32*r<3?25:16+32*r+16*s<1?31:0;if(u==(s=f=0))do g=s*s-f*f+r;while((f=
2*s*f+i)*f+(s=g)*g<4&&++u<27);putchar(_[++e>3840&&m<25?31-m:m>78?32:u]^1);}}
From: fc3a501@GEOMAT.math.uni-hamburg.de (Hauke Reddmann)
Newsgroups: sci.math,sci.fractals
Subject: Re: centroid of the Mandelbrot's set
Date: 25 Jun 1996 09:24:49 GMT
Organization: University of Hamburg -- Germany
Jay R. Hill (JAY.R.HILL@cpmx.saic.com) wrote:
:
: From an over the weekend run: -.28781.
:
Was this a Monte Carlo run?
(Maybe it is better to add up the pieces: the circle, the
cardiodid, the small circle... until the defining formulae
are to awkward to integrate. Anyone try?)
--
Hauke Reddmann <:-EX8
fc3a501@math.uni-hamburg.de PRIVATE EMAIL
fc3a501@rzaixsrv1.rrz.uni-hamburg.de BACKUP
reddmann@chemie.uni-hamburg.de SCIENCE ONLY
From: rice@servo.eng.sun.com (Daniel Rice)
Newsgroups: sci.math,sci.fractals
Subject: Re: centroid of the Mandelbrot's set
Date: 26 Jun 1996 23:50:42 GMT
Organization: Sun Microsystems Computer Corporation
In article <4qob91$8q4@rzsun02.rrz.uni-hamburg.de>,
Hauke Reddmann <fc3a501@GEOMAT.math.uni-hamburg.de> wrote:
>Jay R. Hill (JAY.R.HILL@cpmx.saic.com) wrote:
>:
>: From an over the weekend run: -.28781.
>:
>Was this a Monte Carlo run?
>(Maybe it is better to add up the pieces: the circle, the
>cardiodid, the small circle... until the defining formulae
>are to awkward to integrate. Anyone try?)
Another approach might be to find the centroids (or areas, or whatever)
of the various iterates of z^2 + c. Is there an analytic technique to
determine the centroid of such a figure (defined by |p(c)| == 2 for some
polynomial p)? Is the desired centroid the limit of the resulting sequence?
Dan
Subject: Mandelbrot Set Area, new result
From: Munafo@prepress.pps.com
Date: 1996/12/23
Organization: PrePRESS SOLUTIONS
Newsgroups: sci.fractals
I have continued to compute the area of the Mandelbrot Set using
my method of averaging 20 runs by the pixel-counting method,
with slightly different grid placements and grid spacings. The
latest result (after about 3.7 trillion Mandelbrot iterations)
is 1.50659230 +- 0.0000006.
Here are the results of a set of such averaged runs. For each
grid size, 20 runs were computed and averaged together.
grid dwell average standard
size limit area deviation
---- ------- ------------- -------------
32 8192 1.511 4 0.037 2
64 16384 1.502 6 0.019 4
128 32768 1.504 84 0.005 68
256 65536 1.506 34 0.002 26
512 131072 1.506 88 0.001 37
1024 262144 1.506 783 0.000 493
2048 524288 1.506 674 0.000 152
4096 1048576 1.506 585 0 0.000 074 1
8192 2097152 1.506 593 8 0.000 027 5
16384 4194304 1.506 588 0 0.000 012 6
32768 8388608 1.506 591 54 0.000 004 48
65536 16777216 1.506 592 30 0.000 001 90
The standard deviation measures how much (on average) each
run differs from the average. This means that the average
itself almost certainly varys even less from the true area.
For 20 samples, the standard error of the mean is about 0.23
of the standard deviation. A safe estimate of the error
would be about 1/3 of the standard deiation.
(My old estimate, from April 13th 1993, was 1.506595
+- 0.000002. That estimate was based on an average of 40
runs, and used a dwell limit of 524288. I should have
subtracted 0.000005 to account for the error in using such
a low dwell limit; I also was a little too "liberal" in my
calculation of the expected error.)
- Robert Munafo
From: edgar@math.ohio-state.edu (G. A. Edgar)
Date: Fri, 05 Feb 1999 15:33:25 -0500
Newsgroups: sci.math
Subject: Re: Mandelbrot set
In article <36BA4B97.F6A08E45@erols.com>, John VanSickle
<vansickl@erols.com> wrote:
> Just an idle question, but has the area of the Mandelbrot set been
> calculated (or estimated to any particular degree of precision)?
The area of the Mandelbrot set is between 1.5 and 1.71. Here are
some messages on the topic collected from usenet two years ago.
Those not identified are from me.
From: edgar@math.ohio-state.edu (G. A. Edgar)
The area of the Mandelbrot set is
A = (1 - sum (n=1 to infinity) n (b_n)^2) Pi
or approximately 2.089.
Here the numbers b_n are the coeffients of the Laurent series about
infinity of the conformal map psi of the exterior of the unit disk
onto the exterior of the Mandelbrot set:
psi(w) = w + sum (n=0 to infinity) b_n w^(-n)
= w - (1/2) + (1/8) w^(-1) - (1/4) w^(-2) + (15/128) w^(-3)
+ 0 w^(-4) - (47/1024) w^(-5) + ...
These coefficients can be computed recursively, but a closed form
is not known. The above approximation comes from using the first
72 terms of the series.
From: scott@ferrari.LABS.TEK.COM (Scott Huddleston)
Newsgroups: alt.fractals
Subject: the Mandelbrot area formula
Date: 7 Dec 90 00:51:07 GMT
Organization: Computer Research Laboratory, Tektronix, Inc., Beaverton OR
A formula for computing the area of the Mandelbrot set was discussed on
this newsgroup recently. Gerald Edgar posted an area estimate based on
72 terms. Yuval Fisher pointed out that a 256-term estimate gives a
much lower value, and he described how to compute coefficients in the area
series.
Using Yuval's suggestion, I computed over 100,000 terms of the series.
The estimate converges very slowly -- a summary appears below. An
inportant point is that every truncation of the series is an
*upper bound* on the area of M. Based on the trend I'd say the true
area is probably less than 1.70, but I won't bet my firstborn on that.
# terms: area estimate
72: area < 2.09288
128: area < 2.02781
180: area < 2.01237
256: area < 1.97752
360: area < 1.94961
512: area < 1.92751
720: area < 1.91255
1024: area < 1.89534
1440: area < 1.87172
2048: area < 1.85461
2880: area < 1.84576
4096: area < 1.83452
5760: area < 1.81649
8192: area < 1.80616
11520: area < 1.79642
16384: area < 1.78636
23040: area < 1.7747
32768: area < 1.76683
46080: area < 1.75936
65536: area < 1.75337
92160: area < 1.74321
115232: area < 1.73847
--
Scott Huddleston
scott@crl.labs.tek.com
From: isaacs@hpcc01.HP.COM (Stan Isaacs)
Newsgroups: alt.fractals
Subject: More on the area of the Mandelbrot Set
Date: 25 Feb 91 07:43:14 GMT
Organization: HP Corp Computing & Services
At the Northern California Section of the MAA meeting this past
weekend, John Ewing gave an interesting talk called "Can We See the
Mandlebrot Set". (He avoided using both the "C" word and the "F" word
after the introduction to the talk. It was not about either chaos or
fractals, but about what the M set is, mathematically.) Anyway, he
discussed two results which we've seen separately in this group in the
last month or so. Namely, that by computing the area of the M-set
using lots of terms in a series (Laurent Series?), the upper bound of
the area seems to converge about at 1.72 (the graph gets quite flat,
and seems to have an asymptote there), and by counting pixals more and
more accurately, you seem to get a lower bound of very close to 1.52.
Both these bounds are close to the values the methods would produce in
the limit - that is, it is NOT the case that these numbers would get
closer if a finer grid were used, or more terms were taken in the
series. So, why the difference of 10% or so? No one knows. One
possibility is that the pixal method misses "hairs" around the border.
(I know thats not described very well; it was late, and I wasn't
taking notes.) There was also a vague theory for the higher number
being possibly wrong. But basically, it is not known, at present,
which of these numbers represents the "real" area (although it is
believed that one of them does.) I'm afraid he didn't give references
(I asked afterwards.)
-- Stan Isaacs
From: shallit@graceland.waterloo.edu (Jeffrey Shallit)
Subject: area of the Mandelbrot set
Organization: University of Waterloo
Date: Tue, 10 Mar 1992 14:52:52 GMT
There was some discussion a while back about the area of the
Mandelbrot set. I just noticed the following article:
J. H. Ewing and G. Schober, The area of the Mandelbrot set,
Numer. Math. 61 (1992), 59-72.
It may be of interest.
Jeff Shallit
shallit@graceland.waterloo.edu
From: edgar@math.ohio-state.edu (G. A. Edgar)
The article computes the area estimate using 240,000 terms. The result
is 1.7274... The behovior of the approximations suggests that the limit
is between 1.66 and 1.71. However, the estimates of the area from below,
using pixel counting show that the area is at least 1.52. The large
gap between the lower bound 1.52 and the upper bound 1.71 may possibly
be an indication that the boundary of the Mandelbrot set has
positive area...
The Ewing and Schober paper cited also explains the computation of
the coefficients b_m in the series.
Date: Fri, 18 Sep 92 10:31:51 EST
From: kbriggs@mundoe.maths.mu.oz.au (Keith Briggs rba8 7088)
Subject: Re: Conformal maps
To: edgar@mps.ohio-state.edu
I have done some refined pixel counting and
my best result is
Total Mset area= 1.499936.
On this basis I conjecture that the exact value is 3/2.
However, this number is not universal in the way that the feigenvalues are,
since it varies with reparameterization of the map z^2+c. I have also
done z^d+c (d=3,4,5..), with no clear pattern yet.
Keith.
Newsgroups: sci.fractals
From: hilljr@jupiter.saic.com (Jay R. Hill)
Subject: My Mandelbrot Flower
Keywords: Mandelbrot area
Organization: SAIC
Date: Mon, 15 Mar 1993 17:06:44 GMT
My Mandelbrot Flower
(C) by Jay Hill, 1993
[poem omitted because of the copyright notice]
(-: --- ;^)
So let's get started. We can for each period, p, count around
the Cardioid. Let's name the buds (i,p), the i-th bud with
period p.
n:=0;
for p:=1 to inf do
for i:=1 to p do
n:=n+1;
But we find we already counted some. For example, (2,4) is
the same bud as (1,2). The modified loop is
n:=0; m:=0;
for p:=1 to inf do
for i:=1 to p do
if gcd(i,p)=1 then
n:=n+1
else
m:=m+1;
where m will count the duplicates. To count the buds on buds,
we must make this algorithm recursive. When we are done, we
can also count the island Mandelbroties, l. They will be
l = 2^p - (n+m) - Mandelbrotie buds.
Table I shows the results of this calculation. We can also
keep a running total of the number of buds. The total for
each doubling of the period is shown in Table II. The last
column shows the ratio of the total with each doubling which
approaches 5.56. An approximate formula for the total is
Total buds, T = 5.56^(ln2(p/2)) = 5.56^(-1+1.442695*ln(p))
= exp( 2.46*ln(p)-1.7 )
The density of buds is approximately
dT/dp = (0.45/p)*exp( 2.46*ln(p) )
Now if we use Milnor's bud radius formula r=sin(pi*i/p)/(p*p),
we can estimate the area of the Cardioid plus attached buds.
The formula is not good for buds on buds, however if I use it
anyway, the area approaches 1.507818. I must point out the
formula is quite inaccurate, with error as large as 100%.
Therefore, measured radii for p less than 11 were substituted
in the revised area estimate, Table III, which approaches
1.504106. This is approximate since the bud radii (area) for
periods beyond 2 are still unknown.
The importance of the bud area as a function of p can be
gauged the total area of all buds and cardioids as a function
of p. An estimate based on 'pixel counting' is shown in Table
IV. The values for p=1,2 can be compared to the exact values.
A(1) = 3*pi/8 = 1.178097245 (calculated=1.17809694),
A(2) = pi/16 = 0.196349541 (calculated=0.19634926).
Their error is about 3e-7. The calculation used 8192x4096
samples in region (-2,-1.125),(0.5,1.125) with an iteration
limit of 8388608. Period detection was used for early exit.
When a period was found, an additional set of iterations were
used with a fuzzy period test to distinguish the real period
from a multiple. A simultaneous graphic display showed the
errors in the period selection to be very few. The undecideds
after 8388608 iterations were counted. Only 73 samples out
of 8987185 were unable to decide to exit or find a period.
We can see from Table IV that accurate estimates of bud radii
will be needed up to p=48 if we want 5 or 6 digit accuracy.
That is 2417 buds on the main Mandelbrot, not an impossible
task. 8^) |-0 >^] |-( (Are we having fun yet?) ;^)
Table I
Period Buds Islands Island buds
1 0 0 0
2 1 0 0
3 2 1 0
4 3 3 0
5 4 11 0
6 6 20 1
7 6 57 0
8 9 108 3
9 10 240 2
10 12 472 11
11 10 1013 0
12 22 1959 29
13 12 4083 0
14 18 8052 57
15 24 16315 26
16 27 32496 117
17 16 65519 0
18 38 130464 286
19 18 262125 0
20 44 523209 517
21 36 1048353 120
22 30 2095084 1013
23 22 4194281 0
24 78 8384100 2262
25 36 16777120 44
26 36 33546216 4083
27 50 67108068 490
28 66 134201223 8241
29 28 268435427 0
30 104 536836484 17417
31 30 1073741793 0
32 81 2147417952 32847
33 60 4294964173 2036
34 48 8589803488 65519
35 72 17179868739 294
36 158 34359469848 135274
37 36 68719476699 0
38 54 137438429148 262125
39 72 274877894595 8178
40 156 549754764132 525192
41 40 1099511627735 0
42 156 2199021133728 1064937
43 42 4398046511061 0
44 110 8796088826787 2098153
44 110 8796088826787 2098153
45 152 17592185993904 33724
46 66 35184363700180 4194281
47 46 70368744177617 0
48 270 140737471477920 8455890
49 78 281474976710172 342
50 140 562949919864320 16779140
51 96 1125899906645935 131054
52 132 2251799746572177 33558501
53 52 4503599627370443 0
54 230 9007199120130408 67370674
55 120 18014398509476663 4162
56 234 36028796750519292 134226594
57 108 72057594037141413 524268
58 84 144115187538984904 268435427
59 58 288230376151711685 0
60 456 576460751228083221 537943113
61 60 1152921504606846920 0
62 90 2305843007066210240 1073741793
63 228 4611686018424240100 2098752
64 243 9223372032559775420 2147516493
Table II
Period Buds Total T(n)/T(n/2)
2 1 1
4 3 6 6.00
8 9 31 5.17
16 27 166 5.35
32 81 879 5.30
64 243 4826 5.49
128 729 26537 5.50
256 2187 147542 5.56
512 6561 817561 5.54
1024 19683 4537366 5.550
2048 59049 25234399 5.561
4096 177147 140622586 5.573
Table III
p Area of attached buds
2 1.37444678594553454
3 1.43048719821709103
4 1.45354470458157996
8 1.48460977988076338
16 1.49712977124274433
32 1.50154588264492087
64 1.50316313962822509
128 1.50373319034984287
256 1.50402541594610357
512 1.50407075095490535
1024 1.50409528851486800
2048 1.50410386745418746
4096 1.50410686280880857
Table IV
p Total Area
1 1.17809694
2 0.19634926
3 0.05651246
4 0.02323865
5 0.01310426
6 0.00892924
7 0.00502947
8 0.00446101
9 0.00293483
10 0.00265840
11 0.00137865
12 0.00211441
13 0.00084858
14 0.00113071
15 0.00094799
16 0.00089401
17 0.00038674
18 0.00079376
19 0.00027878
20 0.00060651
21 0.00039814
22 0.00033678
23 0.00015858
24 0.00049604
25 0.00018691
26 0.00021256
27 0.00019345
28 0.00025916
29 0.00007979
30 0.00028716
31 0.00006403
32 0.00017836
33 0.00011114
34 0.00009857
35 0.00010108
36 0.00020602
37 0.00003872
38 0.00007141
39 0.00007040
40 0.00013813
41 0.00002883
42 0.00012220
43 0.00002397
44 0.00007527
45 0.00007761
46 0.00004191
47 0.00001844
48 0.00011047
49 0.00002933
50 0.00005817
51 0.00003218
52 0.00004794
53 0.00001290
54 0.00006118
55 0.00002933
56 0.00005917
57 0.00002447
58 0.00002212
59 0.00001106
60 0.00007845
61 0.00000888
62 0.00001676
63 0.00003386
64 0.00003302
Total 1.50659429
Warmly,
Jay
From: munafo@gcctech.com
Newsgroups: sci.fractals
Subject: Mandelbrot Set center-of-gravity is -((ln(3)-1/3)^Feig)
Date: Wed, 04 Mar 1998 14:36:24 -0600
Organization: Deja News - The Leader in Internet Discussion
Recently I restarted the computing effort I undertook
about a year ago to
compute the area of the Mandelbrot
Set. After reviewing the notes from that
time I saw that there had
been some interest in the past of computing the
Mandelbrot Set's
"center of gravity" as well. I have also done some
optimizations of
the code in the mean time and fixed a few bugs, and have
started
doing a long run.
The method I use involves counting pixels on a
grid. There is error
in the estimate resulting from grid placement and the
fact that the
grid squares on the boundary get counted as completely in or
completely out when in fact only part of their area is in. It is hard
to
estimate the magnitude of this error, so I use statistical methods
to measure
it. I use a bunch of grids of slightly different grid
spacings with slight
vertical and horizontal offsets, and take the
mean and standard deviation of
all the runs.
These are the figures I've compiled so far. Each run is the
average
of 20 grids for which the standard deviations are shown.
grid
iter area and standard iterations
size limit center-of- deviations
time (seconds)
gravity (20 runs) performance (MFLOPs)
256 65536
1.506 38 0.002 31 18,623,118
-0.286 741 0.000 912 9.4 13.918
512
131072 1.506 87 0.001 35 82,659,183
-0.286 881 0.000 417 40.2 14.376
1024 262144 1.506 782 0.000 493 372,533,718
-0.286 802 0.000 173
176 14.839
2048 524288 1.506 674 0.000 151 1,716,452,044
-0.286 811
0 0.000 074 8 788 15.246
4096 1048576 1.506 584 0 0.000 074 4
7,933,339,783
-0.286 765 8 0.000 024 8 3,580 15.502
8192 2097152
1.506 593 6 0.000 027 7 36,987,339,092
-0.286 769 51 0.000 009 73
16,300 15.852
16384 4194304 1.506 588 1 0.000 012 7 171,493,547,507
-0.286 767 27 0.000 003 20 74,400 16.138
32768 8388608 1.506 591 50
0.000 004 51 814,136,316,927
-0.286 768 37 0.000 001 77 347,000 16.442
65536 16777216 1.506 592 31 0.000 001 90 3,844,750,451,387
-0.286 768
423 0.000 000 510 1,630,000 16.529
131072 33554432 1.506 591 734 0.000
000 624 18,173,551,931,685
-0.286 768 317 0.000 000 295 7,590,000
16.754
For each average, the standard deviation gives the expected error
in
each of the 20 individual grids. The expected error of the mean
is 1/sqrt(19)
of the standard deviation (this is called "the standard
error of the mean").
I use 1/3 as a conservative estimate. So, my
best estimate of the
center-of-gravity is -0.28676832 +- 0.00000010.
For the area, we have
1.50659173 +- 0.00000020.
The Inverse Symbolic Calculator is a resource at
the URL
http://www.cecm.sfu.ca/projects/ISC/ISCmain.html
which can be
used to guess what expression produces a given numeric
value. I plugged in
the values 1.50659173 and 0.28676832. I saw
nothing interesting for the area
but for the center of gravity it
gives
-((ln(3)-1/3) ^ Feig1).
Feig1 is
the larger Faigenbaum constant, whose value is
4.6692016091029906718532038,
and shows up in the Mandelbrot Set as
the ratio of the radii of successive
mu-atoms on the real axis.
Plugging in this value we get the following
hypothetical precise
value of the Mandelbrot Set's center of gravity:
-0.2867682633829350268529586
A few other observations:
- My program
currently runs at about 2.39 million Mandelbrot iterations
per second (each
iteration requires 7 floating-point operations).
The latest model Macs are
about 4 times faster. Intel machines get
nowhere close because they don't
have enough floating-point registers
to keep the pipeline full.
- Each
time I double the grid size it takes almost 4.7 times as long.
The current
rate of progress in computer performance is
about 1.56 per year; at that
rate it takes about 3.5 years
to increase computing power by 4.7.
- Each
time the gridsize is doubled the error goes down to about 0.4
of the
previous value (but it varies from 0.3 to 0.55). It would take
about 8 to
10 years for personal computers to improve enough to give
us one more digit
in the area estimate (given the same run time).
- Robert Munafo
Malden
Massachusetts
4 March 1998
rpm%mrob.uucp@spdcc.com