December 2010 - mathfun@mailman.xmission.com

[math-fun] Baserunner's Optimal Path _is_ optimal, after all!
by Henry Baker 27 Dec '10

27 Dec '10

Ok, after rotating the n-gon, I finally got the catenary curves orbiting the n-gon to behave as advertised. Here are the results for the n-gon inscribed in the unit circle "orbit" problem: The acceleration vector a, |a|=1: a = [-1/(1+(Bt)^2), -Bt/(1+(Bt)^2)] (B>0) We focus on the path segment from z=[z0x,0] to z=[cos(pi/n),sin(pi/n)], where t goes from t=0 to t=tau. The velocity vector starts vertically v=[0,v0y] when t=0, and becomes parallel to [-sin(pi/n),cos(pi/n)] at t=tau. Let D = 1+cos(pi/n)*asinh(tan(pi/n))/tan(pi/n). B^2 = D/2/cos(pi/n)^2 B = sqrt(D/2)/cos(pi/n) v0y = sqrt(2)(sqrt(D)-cos(pi/n)/sqrt(D)) z0x = 2*(cos(pi/n)^3-1)/cos(pi/n)/D+(2-cos(pi/n)^2)/cos(pi/n) tau = sin(pi/n)*sqrt(2/D) Time = 2*n*tau Percent2pi = Time/2/pi n=2,B=1.1548311487678426E+16, v0y=1.414213562373095, z0x=4.595339800345407E-15, Time=5.656854249492381, Percent2pi=0.90031631615711 n=3,B=1.661428901969911, v0y=1.059537362214989, z0x=0.96408602060014, Time=6.255040365008328, Percent2pi=0.99552059333041 n=4,B=1.274058570137272, v0y=1.016897637016049, z0x=0.99490419407429, Time=6.279146177038037, Percent2pi=0.99935715247218 n=5,B=1.156505513430619, v0y=1.006591443623277, z0x=0.99878961735182, Time=6.282222778602842, Percent2pi=0.99984680881914 n=6,B=1.102711571990029, v0y=1.003097017726408, z0x=0.99961516043849, Time=6.282878865388524, Percent2pi=0.99995122827418 n=7,B=1.073049873498318, v0y=1.00164592643468, z0x=0.9998520082811, Time=6.28306738560551, Percent2pi=0.99998123219859 n=8,B=1.054794889128636, v0y=1.000955201356099, z0x=0.99993488083036, Time=6.283133399939409, Percent2pi=0.99999173870614 n=16,B=1.013052493113118, v0y=1.000058267649409, z0x=0.99999902987078, Time=6.283184533299711, Percent2pi=0.99999987683319 n=32,B=1.003225198474834, v0y=1.000003619879664, z0x=0.99999998501889, Time=6.283185295227248, Percent2pi=0.99999999809773 n=64,B=1.000803965385648, v0y=1.000000225903149, z0x=0.9999999997666, Time=6.283185306993373, Percent2pi=0.99999999997036 Note that the main vertex does not lie on the X axis, so z0x /= 1, but slightly smaller, as the curve bows in slightly inward from a circular path. It is this slight bowing that produces the advantage over the circular orbit. So the catenaries always beat the circle (just barely). It's amazing how close the circle is to being optimal even at n=3. Note that the formula even "works" for n=2! (Sorta. The 2-gon line segment goes from [0,-1] to [0,+1]. B = oo, z0x=0.) This formula may be the most obscure formula for pi ever invented! I apologize for bothering everyone with my bad math. I thoroughly enjoyed the exercise, tho. It was fun getting Macsyma to produce closed form solns for this problem.

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Re: [math-fun] Baserunner's Optimal Path **isn't optimal**
by Henry Baker 26 Dec '10

26 Dec '10

I've now analyzed the square and concluded that the catenary paths as described in the "Baserunner" paper are _not_ optimal for an "orbit" ! [Carozza, et al. Baserunner's Optimal Path. Math. Intelligencer (October, 2009).] The time for a circular path about a square is 99.8% of the time for the catenary path. Very, very close, but not optimal; the circle appears to be optimal. Here's the math. We place a square with vertices at (+-1,0), (0,+-1). (x0,y0)=(1,0), (v0x,v0y)=(0,v0y), (a0x,a0y)=(-1,0). Acceleration. ax=-1/(1+(Bt)^2) ay=-Bt/(1+(Bt)^2) for some B>0 to be determined. At t=tau (tau to be determined), ax=ay. (We assume that acceleration is smooth (no jerk) for the entire orbit.) Solving for tau, we get tau=1/B. Velocity. vx=-asinh(Bt)/B vy=-(sqrt(1+(Bt)^2)-1)/B+v0y At t=tau, vx=-vy. Solving for v0y, we get v0y=(asinh(1)+sqrt(2)-1)/B ~ 1.295587149392638/B. Position. x=(B^2-Bt*asinh(Bt)+sqrt(1+(Bt)^2)-1)/B^2 y=-(asinh(Bt)+Bt(sqrt(1+(Bt)^2)-2*asinh(1)-sqrt(8)))/(2*B^2) At t=tau, x=y. Solving for B, we get B=sqrt((3*asinh(1)+2-sqrt(2))/2) ~ 1.270808246488339. Backsubstituting, we get v0y ~ 1.019498538015292, tau ~ 0.78690077969145. The total time to "orbit" the square is 8*tau ~ 6.295206237531611 ~ 1.001913190486088*2pi. So, the "Baserunner" catenary orbit around a square is approx .2% _longer_ than a circular orbit. --- If we do the same calculations for an equilateral triangle, we get B = sqrt(7*asinh(sqrt(3))/(2*sqrt(3))) ~ 1.631321852686744. v0y = sqrt(2)*(asinh(sqrt(3))+sqrt(3))/sqrt(7*sqrt(3)*asinh(sqrt(3))) ~ 1.079091776648307. tau = sqrt(3)/B=sqrt(6*sqrt(3)/(7*asinh(sqrt(3)))) ~ 1.061746831084396. The total time to "orbit" the triangle is 6*tau ~ 6.370480986506378 ~ 1.013893538875424*2pi. So, the "Baserunner" catenary orbit around a triangle is approx 1.4% _longer_ than a circular orbit. --- For n=5, time = 1.000452457288549*2pi. For n=6, time = 1.000143694223457*2pi. For n=7, time = 1.000055234883237*2pi. For n=8, time = 1.000024299612241*2pi. For n=16, time = 1.000000361898094*2pi. For n=32, time = 1.00000000558828*2pi. For n=64, time = 1.000000000087061*2pi. For n=128, time = 1.000000000001358*2pi. --- So, it appears that the _only_ time the catenary orbit wins is for n=2. My only assumption that may not be warranted is the assumption of "no jerk". Perhaps a very small jerk half-way between vertices could help, but I'm not convinced.

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Re: [math-fun] Adam's pi machine
by Adam P. Goucher 25 Dec '10

25 Dec '10

>> I have cross-posted this message to the Life List, in case anyone >> there has a copy of this file at 10^14 generations or later. > > A list about the Game of Life is interesting, but the crossposted > address didn't show up on the math-fun list and googling has only > shown postings about Life on other lists. Yes, it's in a non-Google-searchable area of webspace. > Where is the Life List? The Life List is members-only and has a strict policy of not disclosing its location, considering what happened to the public Life forum, comp.theory.cell-automata. Anyway, if you are interested in joining, send an e-mail to one of the seven moderators: * David Bell * Matthew Cook (owner) * Dave Greene * Alan Hensel * Dean Hickerson * Stephen Silver * Jason Summers Dean Hickerson is on math-fun, so he'll be able to invite you. Sincerely, Adam P. Goucher

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Re: [math-fun] e^(pi rt 163) =
by Bill Gosper 24 Dec '10

24 Dec '10

mrob> Your sequence appears in another series sum (but with non-alternating signs) in a 2003 paper (He and Jejjala, "Modular Matrix Models", arXiv:hep-th/0307293v2 , page 20 ) Possibly also related is the discussion at: mathoverflow.net/questions/4775/why-are-powers-of-exppisqrt163-almost-integ… the end: David Speyer, Kevin Buzzard, and "FC") which appears to use the same power series. ----------------- Yow, I hope this merely reflects your deft googlemanship rather than your idea of fun reading. Weird! Their series is q + 744 + 196884/q + 167975456/q^2 + 180592706130/q^3 + 217940004309744/q^4 + 282054965806724344/q^5 + ... which they apparently get from 1/InverseSeries[1728*KleinInvariantJ[Log[q]/2/Pi/I]+O[q]^9] Plugging in q->-640320^3, both series seem to work!? Maybe just for this q? Maybe because they actually diverge? Maybe because this Mma kernel has (certifiably) lost its mind? Intriguingly, the paper gives a more direct formula as a modular transformation of a ratio of two 2F1s, but I can't get it to work. (Authors seem to underutilize the to-TeX features of their algebra systems.) Anyway, the inverse reciprocal J series is clearly what I was hoping for. My greedily calculated one was running out of convergence. Reciprocating now to q->E^(-Pi*Sqrt[163]), we have singular value #163 as K(k_163) = K((Theta2(q)/Theta3(q))^4) = K(1/4 (2 - 209 Sqrt[163]) + (50 (889 Sqrt[3] + 209 Sqrt[163]))/(1 + 557403 Sqrt[489])^(1/3) + (5 (889 Sqrt[3] - 209 Sqrt[163]) (1 + 557403 Sqrt[489])^(1/3))/5336) = Pi/2*EllipticTheta[3, 0, q]^2 So everything is made out of thetas, and the thetas are all made out of Eta[q] and Eta[q^2]. Recall the "Beegner" formula Eta[E^(-2*Sqrt[ 163]*Pi)]^8== Product[Gamma[k/ 163]^(2*JacobiSymbol[k, 163]),{k,166}]/(32*a* 163*Pi^2) where a is the real root of 3*a^2* 163*KleinInvariantJ[((Sqrt[163]*I + 1)/2)]^(1/3)*(-1)^(2/3) + 2*a^3== 163^3 i.e. a -> 163/ 2 (53360 + 1/3 (4102147072512054 - 30099762 Sqrt[489])^(1/3) + (2 (75965686528001 + 557403 Sqrt[489]))^(1/3)) (probably simplifiable) and Eta[E^(-Sqrt[163]*Pi)]/Eta[E^(-2*Sqrt[163]*Pi)] is algebraic: ((Eta[E^(-Sqrt[163]*Pi)]/Eta[E^(-2*Sqrt[163]*Pi)])^-8= 4270934400 - 334525400 Sqrt[163] + 72963522596556133540/(2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^( 1/3) - (5714944154168505080 Sqrt[ 163])/(2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^(1/3) + 1/2 (2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^(1/3), probably simplifiable.) Just for yucks, the "closed" form is EllipticEta[E^(-2 Sqrt[163]*Pi)]^4-> -(4096 2^(129/163) (-20 - (200 2^(2/3))/(40027 + 1881 Sqrt[489])^( 1/3) + (40027 + 1881 Sqrt[489])^(1/3)/2^(2/3)) Pi^17 G[1/ 326] G[9/326] G[21/326] G[25/326] G[33/326] G[41/326] G[49/ 326] G[53/326] G[57/326] G[61/326] G[65/326] G[69/326] G[77/ 326] G[81/326] G[43/163] G[46/163] G[47/163] G[51/163] G[54/ 163] G[55/163] G[56/163] G[58/163] G[60/163] G[167/326] G[179/ 326] G[199/326] G[227/326])/(3 G[-(1/163)] G[3/326] G[5/326] G[ 7/326] G[11/326] G[13/326] G[17/326] G[19/326] G[23/326] G[27/ 326] G[29/326] G[31/326] G[37/326] G[45/326] G[59/326] G[63/ 326] G[67/326] G[73/326] G[75/326] G[79/326] G[42/163] G[89/ 326] G[99/326] G[50/163] G[101/326] G[103/326] G[105/326] G[107/ 326] G[109/326] G[117/326] G[123/326] G[125/326] G[127/326] G[ 129/326] G[66/163] G[137/326] G[139/326] G[141/326] G[72/163] G[ 147/326] G[149/326] G[153/326] G[157/326] G[159/326] G[175/ 326] G[183/326] G[191/326] G[193/326] G[207/326] G[211/326] G[ 215/326] G[231/326] G[233/326] G[239/326] G[241/326] G[243/ 326] G[128/163] G[130/163] G[138/163] G[287/326] G[305/326] G[ 154/163] G[311/326])/.G->Gamma --rwg

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[math-fun] Life-like systems that are continuous and real-valued
by Robert Munafo 24 Dec '10

24 Dec '10

Fans of Conway's Game of Life may appreciate this system, which has similar characteristics but has a continuously variable state at each point in a continuous domain (vs. cellular automata, which are discrete in both aspects). Here are some patterns: mrob.com/pub/comp/xmorphia/catalog.html and relevant videos: youtube.com/watch?v=IBrPnp3025I youtube.com/watch?v=-k98XOu7pC8 youtube.com/watch?v=ypYFUGiR51c The system includes still-lifes, spaceships, rotaters (the nearest equivalent to oscillators), chaotic patterns and a bunch of other stuff. It is a common reaction-diffusion model but with very specific parameter values. More here: mrob.com/pub/comp/xmorphia/uskate-world.html -- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] e^(pi rt 163) =
by N. J. A. Sloane 23 Dec '10

23 Dec '10

Concerning Bill Gosper's conjectured (but apparently incorrect beyond a certain point, according to a subsequent posting by Robert Munafo) series, I added it to the OEIS as A178448. I also added the sequence of coefficients of 1/j_revert: A178451, where j_revert is A091406, the reversion of the j-function A000521. Robert, could you add a further sequence, the correct version of Bill's, if you found it? Thanks! Neil

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Re: [math-fun] e^(pi rt 163) =
by N. J. A. Sloane 22 Dec '10

22 Dec '10

aaah! typo! Concerning Bill Gosper's conjectured (but apparently incorrect beyond a certain point, according to a subsequent posting by Robert Munafo) series, I added it to the OEIS as A178449 [I had mistyped A178448, sorry] I also added the sequence of coefficients of 1/j_revert: A178451, (which is really close to Bill's, but different) where j_revert is A091406, the reversion of the j-function A000521. Robert, could you add a further sequence, the correct version of Bill's, if you found it? Thanks! Neil

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[math-fun] e^(pi rt 163) =
by Bill Gosper 22 Dec '10

22 Dec '10

s^3 + 744 - 196884/s^3 + 167975456/s^6 - 180592706130/s^9 + 217940004309743/s^12 - 19517553165954887/s^15 + 74085136650518742/s^18 - ... where s = 640320. Only the 1st three terms are in EIS. Are the rest well defined? --rwg

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[math-fun] More definitive reference (was Re: e^(pi rt 163) =)
by Robert Munafo 22 Dec '10

22 Dec '10

(I made a mistake, 217940004309743 != 217940004309744 . . . My laziness, I apologize! /-: ) But I think I found a better answer to your question. See members.bex.net/jtcullen515/math7.htm (Jim Cullen, "An approximation of pi from Monster Group symmetries"). He starts with s^3 + 744, then proposes successively better approximations to e^(pi sqrt(163)) : s^3 + 744 - 196884/s^3 s^3 + 744 - 196884/s^3 + 167975456/s^6 s^3 + 744 - 196884/s^3 + 167975456/s^6 - 180592706130/s^9 then states that the next term is "barely identifiable as the integer 217940004309743" and suggests that further terms do not add to the elegance (of extending the near-exact approximation). He then goes on to that other series with the 217940004309744 term. So I guess no, those coefficients are probably not definitive, but still if they're in a paper somewhere they might as well also be in OEIS. - Robert On Wed, Dec 22, 2010 at 00:09, Robert Munafo <mrob27(a)gmail.com> wrote: > [...] (He and Jejjala, "Modular Matrix Models", arXiv:hep-th/0307293v2 , > page 20 ) > [...] > mathoverflow.net/questions/4775/why-are-powers-of-exppisqrt163-almost-integ… > > [...] > On Tue, Dec 21, 2010 at 21:38, Bill Gosper <billgosper(a)gmail.com> wrote: > >> s^3 + 744 - 196884/s^3 + 167975456/s^6 - 180592706130/s^9 >> + 217940004309743/s^12 - 19517553165954887/s^15 + 74085136650518742/s^18 - >> ... >> >> where s = 640320. Only the 1st three terms are in EIS. Are the rest >> well defined? >> --rwg > > -- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com

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[math-fun] Orbiting n-gons
by Henry Baker 21 Dec '10

21 Dec '10

I've computed a generalization of the orbit around the regular n-gon inscribed in the unit circle, n=2,3,... This version simply increments angle of the constant absolute value (=1) acceleration vector discretely, allowing the orbiting body to hop over the next vertex, just grazing it with its parabolic path. Thus, a 12-gon would be like a clock, with a vertex at each number, and the acceleration vector would change angle discretely half-way between each pair of adjacent numbers. Note that n=2 simply moves along the line segment from (1,0) to (-1,0) and back again, but constantly accelerating either towards (-1,0) or (1,0). The cool thing about the parabolic segments is that one can compute their length _in closed form_. n = number of vertices in n-gon. tau = time to go from vertex to next acceleration shift; T = 2*n*tau = total time for the orbit around the n-gon. T = 4*n*sin(pi/n)/sqrt(3+cos(2pi/n)) L = total length of orbit. L = 4*n*(sin(pi/n)+cos(pi/n)^2*asinh(tan(pi/n)))/(3+cos(2pi/n)) r = radius of orbit at acceleration shift (perigee?). r = 4*cos(pi/n)/(3+cos(2pi/n)) R = radius of orbit at vertex (apogee?). R = 1. v0 = velocity at vertex/apogee. |v0| = 2*cos(pi/n)/sqrt(3+cos(2pi/n)) v1 = velocity at perigee. |v1| = 4/(3+cos(2pi/n)) The total time T has the following series approximation: T(n) ~ 2*pi*[1+(1/12)*(pi/n)^2-(11/480)*(pi/n)^4-(491/40320)*(pi/n)^6-(11159/5806080)*(pi/n)^8...] T(1)=0 (I don't know what this means!). T(2)=sqrt(2^5) ~ 90% of 2pi T(3)=sqrt(6^3/5) ~105% of 2pi T(4)=sqrt(2^7/3) ~104% of 2pi T(5) messy ~103% of 2pi T(6) ~102% of 2pi T(7) ~101.6% of 2pi T(8) ~101.2% of 2pi T(16) ~100.3% of 2pi T(32) ~100.1% of 2pi So T(n)>2*pi, except for n=1,2. I.e., this discrete "clock" form of acceleration shifting only betters a circular orbit for the line segment case. Total length L has the approximation: L(n) ~ 2*pi*[1-(7/120)*(pi/n)^4-(41/2520)(pi/n)^6+(83/120960)*(pi/n)^8+...] (Note missing (pi/n)^2 term.) L(n)<2*pi. L(1)=0 L(2)=4 ~ 64% of 2pi L(3)=(6/5)(asinh(sqrt(3))+2*sqrt(3)) ~ 91% of 2pi L(4)=(8*asinh(1)+sqrt(2^7))/3 ~ 97% of 2pi L(5) ~ 99% of 2pi L(6) ~ 99.5% of 2pi L(7) ~ 99.7% of 2pi L(8) ~ 99.85% of 2pi L(16) ~ 99.99% of 2pi L(32) ~ 99.9995% of 2pi L(n)/T(n) is the mean linear speed of the orbiting object over a complete orbit, compared with a circular orbit. L(n)/T(n)=(1+cos(pi/n)*asinh(tan(pi/n))/tan(pi/n))/(3+cos(2pi/n)) L(1)/T(1)=? L(2)/T(2)=1/sqrt(2) ~ 70.7% L(3)/T(3)=asinh(sqrt(3))/sqrt(30)+sqrt(2/5) ~ 87.3% L(4)/T(4)=asinh(1)/sqrt(6)+1/sqrt(3) ~ 93.7% L(5)/T(5) ~ 96.2% L(6)/T(6) ~ 97.5% L(7)/T(7) ~ 98.2% L(8)/T(8) ~ 98.6% L(16)/T(16) ~ 99.7% L(32)/T(32) ~ 99.9%

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