December 2012 - mathfun@mailman.xmission.com

[math-fun] challenging limits
by Bill Gosper 23 Dec '12

23 Dec '12

I thought I mentioned here some months ago Steve Finch's strange Integrate[((2 + 4*v + 3*v^2)* Sqrt[-((1 + 2*v + 3*v^2)/(4 + 8*v + 3*v^2))])/(1 + 2*v)^2, {v, -2, -2/3}] that Mathematica can do INdefinitely, but cannot take the endpoint limits. The integral is the surface area of the convex hull of two edge-to-edge orthogonal unit disks. (http://arxiv.org/abs/1211.4514) Working a couple of days, Neil and I numerically conjectured the messy expression on p10, subsequently reduced, again numerically, to (1/(54*(-I + 2*Sqrt[2])))* (-243*I*EllipticE[(1/81)*(17 - 56*I*Sqrt[2])] + 27*Sqrt[-17 + 56*I*Sqrt[2]]* EllipticE[(1/81)*(17 + 56*I*Sqrt[2])] + 2*(81*Sqrt[2]*EllipticK[1/9] - 81*Sqrt[2]* EllipticK[(1/81)*(17 + 56*I*Sqrt[2])] + 3*(8*I - 7*Sqrt[2])* EllipticPi[(1/27)*(7 - 4*I*Sqrt[2]), (1/81)*(17 - 56*I*Sqrt[2])] + (104*I + 71*Sqrt[2])* EllipticPi[(1/3)*(7 - 4*I*Sqrt[2]), (1/81)*(17 - 56*I*Sqrt[2])])) A completely different problem: Limit[(1/( 2 s))(2 y + s (-1 + t - y + z) - t (1 + t - y + z) + 2 ContinuedFractionK[1/2 n^2 (-8 t - 2 t^2) + 1/2 n (-2 t^2 - 4 s y - 2 t (4 + s (-3 + z))) + 1/2 (-2 t + 3 s t - 5 t^2 - 2 s y + 6 t y - 2 y^2 + 3 t^2 z - t (s + 2 y - 2 z) z), s (2 + t) + n s (2 + t) + 2 y - t (1 + t - y + z), {n, 0, ∞}]), t -> 0, Direction -> -1] == -1 - y + z + (E^-y y^z)/Gamma[z, y] where s:=Sqrt[t (4 + t)]. I can't even do the special case z=1, y=2: Limit[(1/( 2 s))((-2 + s - t) (-2 + t) + 2 cfk[-4 + (1 + 2 n) s (-2 + t) - t (-4 + t + n (1 + n) (4 + t)), (2 + s + n s - t) (2 + t), {n, 0, \[Infinity]}]), t -> 0, Direction -> -1] == 0 I can get these as ratios of 2F1s with exploding parameters... It would help if Mma knew that Limit[Hypergeometric2F1[a, b/t, c, t*z], t -> 0] == HypergeometricPFQ[{a}, {c}, b*z] and didn't hang up for literally hours trying to express the ContinuedFractionK in closed form. --rwg

1 1

Re: [math-fun] Sci-fi myths... "Neutronium"
by Henry Baker 23 Dec '12

23 Dec '12

Interesting calculation. What's the situation with small black holes? I understand that the really tiny ones evaporate pretty quickly due to Hawking radiation, but what about a Moon mass black hole? ---- Also, can a black hole ever rotate so fast as to break apart? I.e., you have a really fast rotating black hole that gets additional mass & angular momentum added until it breaks apart into 2 black holes revolving about one another. There's also a movie on the internet somewhere that shows 2 black holes revolve around one another that eventually merge, because _graviational waves_ can carry off some of the angular momentum. At 12:27 PM 12/23/2012, Warren Smith wrote: >In various sci-fi stories, the wonderful substance "neutronium" plays a role. >This is the superdense stuff neutron stars are made out of, except >some amazing advanced aliens find a way to make or get hold of chunks >of it. > >For example in one fine Larry Niven story, there is a chunk of it >about the mass of Earth's moon orbiting some planet, and foolish >trinocs try to get it due to wrongly thinking it is a valuable slaver >stasis box, thereby meeting their doom from its immense surface >gravity. > >So, let's think a bit about the actual properties of neutronium. Do >they live up to the hopes of sci-fi writers? > >They are correct it is very dense. >The estimated density of the nuclei of atoms averages about dens=4*10^17 kg/m^3, >which is 4*10^14 times denser than water. Thus the mass of the Moon >would fit in >a sphere of radius 35 meters, which I thunk is rather larger than >Niven had in mind. > >The problem is, neutrons are unstable. >Neutron has a mean life of 15 minutes for decay into >proton+electron+antineutrino+energy. >The energy release for decay of a free neutron is 782 KeV >(which is about 1/1201 fraction of its rest-mass energy) >but for neutrons in different atomic nuclei it can differ, >sometimes being as large as 1.5MeV (?). > >Anyhow, for a ball of neutronium of radius=r and mass=M, we expect it >should be unstable to neutron decay if the gravitational escape energy >for a neutron-mass >is less than 782 KeV. >That is, > Mcrit = dens * rcrit^3 * (4*Pi/3) >and > mneutron * G * Mcrit / rcrit = (782 KeV)*c^2 = 1.25290199 Ã 10^(-13) joules >Solving, I find > Mcrit = 9.2 * 10^27 kg = about 4.8 jupiter masses > rcrit = 818 meter > >conclusion: A chunk of neutronium smaller than about 5 jupiter masses >or which fits in a ball of radius<=818 meters cannot exist; if you had >one, it would explode. > >Actually, even larger balls should also be unstable since we do not >need the decay products to escape all the way to infinity -- our >argument really only provides very weak lower bounds on Mcrit and >rcrit.

1 0

Re: [math-fun] challenging limits
by Axel Vogt 23 Dec '12

23 Dec '12

I think (using Maple) that the integral equals Integrate[((2 + 4*v + 3*v^2)* Sqrt[-((1 + 2*v + 3*v^2)/(4 + 8*v + 3*v^2))])/(1 + 2*v)^2, {v, -2, -2/3}] = -8/3*EllipticK(1/3)+8/3*EllipticPi(-1/3,1/3)+3*EllipticE(1/3) I normalized it as one fraction as A/(polynomial of degree 4), did a presentation using partial fraction. Then Maple luckily succeeds for each summand and finally arrived after simplification with the above. A = ((v+2)*(3*v+2))^(1/2)*(3*v^2+4*v+2)*(3*v^2+2*v+1)^(1/2)/(v+2)/(3*v+2)/(1+2*v)^2*I

1 0

[math-fun] Sci-fi myths... "Neutronium"
by Warren Smith 23 Dec '12

23 Dec '12

In various sci-fi stories, the wonderful substance "neutronium" plays a role. This is the superdense stuff neutron stars are made out of, except some amazing advanced aliens find a way to make or get hold of chunks of it. For example in one fine Larry Niven story, there is a chunk of it about the mass of Earth's moon orbiting some planet, and foolish trinocs try to get it due to wrongly thinking it is a valuable slaver stasis box, thereby meeting their doom from its immense surface gravity. So, let's think a bit about the actual properties of neutronium. Do they live up to the hopes of sci-fi writers? They are correct it is very dense. The estimated density of the nuclei of atoms averages about dens=4*10^17 kg/m^3, which is 4*10^14 times denser than water. Thus the mass of the Moon would fit in a sphere of radius 35 meters, which I thunk is rather larger than Niven had in mind. The problem is, neutrons are unstable. Neutron has a mean life of 15 minutes for decay into proton+electron+antineutrino+energy. The energy release for decay of a free neutron is 782 KeV (which is about 1/1201 fraction of its rest-mass energy) but for neutrons in different atomic nuclei it can differ, sometimes being as large as 1.5MeV (?). Anyhow, for a ball of neutronium of radius=r and mass=M, we expect it should be unstable to neutron decay if the gravitational escape energy for a neutron-mass is less than 782 KeV. That is, Mcrit = dens * rcrit^3 * (4*Pi/3) and mneutron * G * Mcrit / rcrit = (782 KeV)*c^2 = 1.25290199 × 10^(-13) joules Solving, I find Mcrit = 9.2 * 10^27 kg = about 4.8 jupiter masses rcrit = 818 meter conclusion: A chunk of neutronium smaller than about 5 jupiter masses or which fits in a ball of radius<=818 meters cannot exist; if you had one, it would explode. Actually, even larger balls should also be unstable since we do not need the decay products to escape all the way to infinity -- our argument really only provides very weak lower bounds on Mcrit and rcrit.

4 4

[math-fun] Season's Greetings
by Henry Baker 23 Dec '12

23 Dec '12

The following Maxima code was sent to the Maxima mailing list by Jaime Villate <villate(a)fe.up.pt>. It takes a little while to run (at least on my machine), so give it some time. Enjoy! a1: matrix([0.85,0],[0,0.85])$ a2: matrix([-0.2,0.26],[-0.23,-0.22])$ a3: matrix([0.15,-0.28],[-0.26,-0.24])$ a4: matrix([0,0],[0,0.16])$ q1: [0,1.6]$ q2: [0,0]$ m1: matrix([1/2,sqrt(3)/6],[-sqrt(3)/6,1/2])$ m2: matrix([1/2,-sqrt(3)/6],[sqrt(3)/6,1/2])$ l: [[-0.4,11],[0,12.1],[0.4,11],[-0.4,11]]$ /* Koch's fractal */ for j thru 3 do (l2: [l[1]], for i thru length(l)-1 do ( p1: l[i], p2: l[i+1], l2: endcons(float((2*p1+p2)/3), l2), l2: endcons(float(first(transpose(m1.p1+m2.p2))), l2), l2: endcons(float((p1+2*p2)/3), l2), l2: endcons(p2,l2)), l: l2)$ /* Barnsley's Iterative Function System */ ifs(prob, mat, point, p0, n) := block ([p:p0, s, r, m:last(prob), data:[p0]], for i thru n do (r: random(m) + 1, s: 0, for j while (r-prob[j]) > 0 do s:j, p: float(mat[s+1].p + point[s+1]), data: cons(first(transpose(p)), data)), data)$ l1: ifs([85,92,99,100], [a1,a2,a3,a4], [q1,q1,q1,q2], [0,0], 30000)$ plot2d([[discrete,l],[discrete,l1]],[style,lines,dots],[legend,false],[color,blue,green],[box,false],[x,-8,8]);

1 0

[math-fun] Levy C curve and von Neumann integers
by Stuart Anderson 22 Dec '12

22 Dec '12

I was looking at the von Neumann integer generator here; http://bmanolov.free.fr/von-neumann-integer-gen.php generated a string; {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}},{{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}}} replaced { with L (left turn and stroke) replaced } with R (right turn and stroke) (ignored the commas) got this; L L R L L R R L L R L L R R R L L R L L R R L L R L L R R R R L L R L L R R L L R L L R R R L L R L L R R L L R L L R R R R R R wrote a little postscript program to see what it would like like as a path; %! /inch {72 mul} def /L {90 rotate 30 0 rlineto } def /R {-90 rotate 30 0 rlineto} def .1 setlinewidth newpath -4 inch -4 inch moveto L L R L L R R L L R L L R R R L L R L L R R L L R L L R R R R L L R L L R R L L R L L R R R L L R L L R R L L R L L R R R R R R stroke showpage seemed to be a fractal dragon of some sort. after doing it for higher integers, found it was the Levy C curve. Seems this construction has been known since the early 70's (Gosper). see item 135, Beeler, M., Gosper, R.W., and Schroeppel, (R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972.); "A faster way to approach the same limiting curve is to substitute the curve itself for each of its 2^2^n segments, starting with a 90 degree "<"." This is essentially the same as the von Neumann construction, where the 90 degree "<" represents the bracketing of the empty set "{}". We can convert the von Neumann string into the Levy’s C curve with these substitutions; Replace the left bracket with a left turn and move 1 unit forward, Replace the right bracket with a right turn and move forward 1 unit, Ignore the comma. Wrote another program to draw the Levy C curve using a recursive construction, for integers 0 to 15; %!PS-Adobe-3.0 %%Creator: postscript by Stuart Anderson, stuart.errol.anderson(a)gmail.com %%CreationDate: 23-12-2012 %%Title: Levy's C curve from von Neumann integers % The von Neumann integer N is a finite set with N elements which are the von Neumann integers 0 to N-1. They are defined recursively as follows: % 0 := {} (the empty set) % N+1 := N ∪ {N} % 0 = {} = ∅. % 1 = {0} = {{}} = {∅} % 2 = {0, 1} = { {}, {{}} } = {∅, {∅}} % 3 = (0, 1, 2} = { {}, {{}}, {{}, {{}}} } = {∅, {∅}, {∅, {∅}}} % . % . % . % Levy’s C curve can be obtained from the von Neumann integers. This is not the way it is usually defined. The usual definition is as follows; % D = ψ1 (D) ∪ ψ2 (D), where ψ1 and ψ2 are similar contractions given by ψ1(z) = ( 1/2 + 1/2 i)z, and ψ2(z) = ( 1/2 − 1/2 i)z + ( 1/2 + 1/2 i). % It has been observed that there are other ways to generate the Levy’s C curve, in particular it was noted by Gosper; see item 135, Beeler, M., Gosper, R.W., and Schroeppel, (R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972.); % "A faster way to approach the same limiting curve is to substitute the curve itself for each of its 2^2^n segments, starting with a 90 degree "<"." % This is essentially the same as the von Neumann construction, where the 90 degree "<" represents the bracketing of the empty set "{}". % We can convert the von Neumann string into the Levy’s C curve with these substitutions; % Replace the left bracket with a left turn and move 1 unit forward, % Replace the right bracket with a right turn and move forward 1 unit, % Ignore the comma. %%Pages: 16 %%BeginProlog /pagebegin { gsave /depth 0 def 1 setlinewidth % set line width /down {/depth depth 1 add def} def /up {/depth depth 1 sub def } def /DoLine % print a vert. line { 0 144 rlineto } def /vonNeumann { down DoLine % print line depth integer le % depth<integer? { 90 rotate vonNeumann -90 rotate vonNeumann} if up } def } def % pagebegin /page-end { grestore showpage } def %%EndProlog %%Page: 1 1 pagebegin /integer 0 def .750 .750 scale 400 400 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 2 2 pagebegin /integer 1 def .750 .750 scale 500 400 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 3 3 pagebegin /integer 2 def .50 .50 scale 800 800 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 4 4 pagebegin /integer 3 def .40 .40 scale 1100 1200 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 5 5 pagebegin /integer 4 def .30 .30 scale 1200 1700 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 6 6 pagebegin /integer 5 def .20 .20 scale 1200 2400 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 7 7 pagebegin /integer 6 def .15 .15 scale 1300 2800 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 8 8 pagebegin /integer 7 def .1 .1 scale 2000 3000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 9 9 pagebegin /integer 8 def .08 .08 scale 3000 4000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 10 10 pagebegin /integer 9 def .06 .06 scale 6000 5000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 11 11 pagebegin /integer 10 def .04 .04 scale 12000 10000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 12 12 pagebegin /integer 11 def .03 .03 scale 15000 18000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 13 13 pagebegin /integer 12 def .025 .025 scale 15000 20000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 14 14 pagebegin /integer 13 def .02 .02 scale 10000 30000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 15 15 pagebegin /integer 14 def .012 .012 scale 12000 40000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Page: 16 16 pagebegin /integer 15 def .007 .007 scale 0 setgray 0 0 100000 120000 rectfill 1 setgray 25000 40000 moveto vonNeumann -90 rotate DoLine stroke page-end %%Trailer %%EOF Stuart Anderson

1 0

[math-fun] Fwd: Mathematician/physicist/inventor Richard Crandall dies at 64
by rcs＠xmission.com 22 Dec '12

22 Dec '12

Bob Baillie passed along this note from the experimental math blog of David Bailey and Jonathan Borwein. --Rich > Forwarded message from rjbaillie(a)frii.com > Subject: Mathematician/physicist/inventor Richard Crandall dies at 64 http://experimentalmath.info/blog/2012/12/mathematicianphysicistinventor-ri…

3 2

Re: [math-fun] Richard Crandall
by Guy Haworth 22 Dec '12

22 Dec '12

Very sad to hear of Prof. Crandall's death. My condolences to his family, friends and colleagues. What I knew of his work was absorbing and interesting. Guy Haworth, Reading

1 0

Re: [math-fun] Greedily packing disks in a sector
by Phil Carmody 22 Dec '12

22 Dec '12

James Propp wrote > (Is there software that makes problems like this easy to > play with? If I had to write the code myself it'd take me > days to do, and there's a good chance that my code would > have mistakes that I'd miss.) You might want to get in touch with any of the 'packomania' contributors. Their software (plural) might be overkill, as they're optimising rather than just greedily rule-following. David Cantrell's code is Mathematica-based (I was tempted to say "alas", but I hear some people like that platform). I'm surprised he's not on this list; he deserves to be, he's one of the few worthwhile contributors to sci.math nowadays. Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

3 2

[math-fun] math-fun monthly password reminder enabled
by rcs＠xmission.com 20 Dec '12

20 Dec '12

The Math-Fun mailing list software offers a monthly password reminder as an option. I set this option to "No" when I moved the list to Xmission, since it's nuisance spam. I've had several complaints that the "send me my password" feature doesn't seem to work. I've temporarily turned ON the monthly password reminder. I expect this will send everyone a reminder on New Year's Day, and again on Ground Hog's Eve. I'll turn the reminder off again in February, so this should give everyone two chances to record their password. Rich

1 0