From:isaacs@hpcc01.HP.COM (Stan Isaacs)Newsgroups:alt.fractalsSubject:More on the area of the Mandelbrot SetDate:25 Feb 91 07:43:14 GMTOrganization: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.mathSubject:Re: centroid of the Mandelbrot's setDate:19 Jun 1996 15:52:41 GMTOrganization: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.mathSubject:Re: centroid of the Mandelbrot's setDate:20 Jun 1996 01:26:14 GMTOrganization: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.fractalsFrom:"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>Subject:Re: centroid of the Mandelbrot's setTo:JeffLeader@worldnet.att.net,jhill@nosc.milOrganization:SAICDate: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 GMTNewsgroups:sci.math,sci.fractalsSubject: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.fractalsFrom:"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>Subject:Re: centroid of the Mandelbrot's setTo: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.caOrganization:SAICDate: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.fractalsFrom:"Jay R. Hill" <JAY.R.HILL@cpmx.saic.com>Subject:Re: centroid of the Mandelbrot's setTo:jhill,@,nosc.mil,marblanc,@,mail.mcnet.ch,Dt812n.8J6,@,undergrad.math.uwaterloo.ca,,1C1992A.1FCA,@,mail.mcnet.ch,lhf,@,csg.uwaterloo.caOrganization:SAICDate: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.fractalsSubject:Re: centroid of the Mandelbrot's setDate:25 Jun 1996 09:24:49 GMTOrganization: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.fractalsSubject:Re: centroid of the Mandelbrot's setDate:26 Jun 1996 23:50:42 GMTOrganization: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 resultFrom:Munafo@prepress.pps.comDate:1996/12/23Organization:PrePRESS SOLUTIONSNewsgroups: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 -0500Newsgroups:sci.mathSubject: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.fractalsSubject:the Mandelbrot area formulaDate:7 Dec 90 00:51:07 GMTOrganization: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.fractalsSubject:More on the area of the Mandelbrot SetDate:25 Feb 91 07:43:14 GMTOrganization: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 setOrganization:University of WaterlooDate: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 ESTFrom:kbriggs@mundoe.maths.mu.oz.au (Keith Briggs rba8 7088)Subject:Re: Conformal mapsTo: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.fractalsFrom:hilljr@jupiter.saic.com (Jay R. Hill)Subject:My Mandelbrot FlowerKeywords:Mandelbrot areaOrganization:SAICDate: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.comNewsgroups:sci.fractalsSubject:Mandelbrot Set center-of-gravity is -((ln(3)-1/3)^Feig)Date:Wed, 04 Mar 1998 14:36:24 -0600Organization: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