February 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Just a rock and a string
by N. J. A. Sloane 09 Feb '11

09 Feb '11

Maybe someone already answered this, but if not, this is a famous problem in geometry. See Croft's paper "A net to hold a sphere", available on the web. I saw this mentioned in the collection of problems in geometry edited by (I think) Goodman, Pollack, et al. It might also be mentioned in Croft and Guy's Unsolved Problems in Geometry. Neil

1 0

[math-fun] Earth = mathematical ellipsoid ??
by Henry Baker 09 Feb '11

09 Feb '11

One of the standard GPS models of the Earth is that of an "ellipsoid", where a cross-section through the N-S pole is an ellipse. http://en.wikipedia.org/wiki/Reference_ellipsoid But I seem to recall that a fast-rotating body of liquid produces a funny discus-like form. Given a molten liquid of "constant density" -- e.g., mercury -- held together by Newtonian gravity, what mathematical shape does it assume as it rotates faster & faster? How fast can it rotate before the ellipsoidal shape no longer applies?

2 2

Re: [math-fun] just rubber bands
by James Propp 08 Feb '11

08 Feb '11

Since no one else has jumped in to reply to Jim Propp's post, I'll take a stab at it. Maybe one reason nobody responded to the question is that it isn't really a mathematical question, or even a single question; it's an amorphous stew of several questions, mixing math, physics, and perhaps esthetics. On the subject of esthetics, let me say that I don't find the rubber-band cube in the video at all appealing; it's a far cry from the sort of elegance one finds in modular origami (let me mention here the work of Jeannine Mosely, who takes common objects like business cards or egg cartons and unleashes their latent potential as modules). On the subject of physics, let's notice that the frictional properties of rubber bands play a role in the rubber-band ball. Once one of the outmost rubber-bands start to deviate from a great circle, it will experience even more contractive force, causing it to shrink further, unless there's friction. I'm not sure what specific question (if any) Jim has in mind, but let me mention that, just as one can nearly cover the surface of a cube of side-length L by rubber-bands of length 4L, one can nearly cover the surface of a tetrahedron of side-length L by rubber-bands of length 2L. To see where the rubber-bands go, draw a line connecting the midpoints of two opposite edges of the tetrahedron, and now look at how planes perpendicular to this line cut the tetrahedron. The cross-sections are rectangles of varying shape with common perimeter 2L. (To see this, check that the perimeter varies linearly as one moves the cutting plane, and that it starts and ends with value 2L.) James Propp

5 5

[math-fun] Stumbling across a 2005 4F3 item
by Bill Gosper 07 Feb '11

07 Feb '11

Date: Fri, 16 Dec 2005 13:37:17 -0800 (PST) From: "R. William Gosper" <rwg(a)osots.com> Subject: [math-fun] more hyjynx [...] My POWERSERIES function got a double sum instead of Clausen's 4F3[z] for not knowing the well-poised, 3-balanced (sesquiSaalschutzian?), terminating 4F3[1] in the inner sum. "SesquiPfaffian" is more just, but also more confusing. I got ambitious and derived the whole 4F3[a,b,c,d;e,f,g] (and 3F2[z]) matrix system, and extracted the general, noterminating case: 1 hyper_f ([a, b, c, c - b - a + -], 4, 3 2 1 [c - a + 1, c - b + 1, b + a + -], 1) = 2 1 (b + a - -)! (c - a)! (c - b)! (sqrt(%pi) (c - 2 b - 2 a)! 2 1 1 /((a - -)! (b - -)! (c - 2 a)! (c - 2 b)! (c - b - a)!) 2 2 - hyper_f ([1, c - 2 a + 1, c - 2 b + 1, c - b - a + 1], 4, 3 3 [c + 1, c - 2 b - 2 a + 2, c - b - a + -], 1) 2 c + 1 1 /((a - 1)! (b - 1)! (----- - b - a) c! (c - b - a + -)!)) 2 2 Notice the rhs 4F3 getting divided by infinity exactly when an upper parameter on the left is a nonpositive integer. Also note the weird oo - oo when (c+1)/2 = a+b. As usual, the matrices provide the complete system of identities contiguous to the above. [...] ------------------ I.e., termination of lhs provides a closed form. The lhs remains finite for c=2a+2b-1, so it must be that block([fancy_display:false],hyper_f[3,2]([b,a,((b+a)/2)],[b+a,((b+a+1)/2)],1), print(%%=hypersimp(%%)),0)$ b + a b + a + 1 hyper_f ([b, a, -----], [b + a, ---------], 1) = 3, 2 2 2 2 b + a + 1 %pi gamma (---------) 2 --------------------------- 2 a + 1 2 b + 1 gamma (-----) gamma (-----) 2 2 which I found startling because it's balanced like zeta(3/2) and nonterminating and square. But as you see, Macsyma already knows it, so I simply forgot installing Clausen's theorem. This was a specialization of 4x4 matrices to 3x3s. The old message used "3 balanced" to mean like zeta(3), in violation of the convention that n-balanced means like zeta(n+1), in disregard of that consarned invisible lower parameter "1". We can probably get away with calling zeta(2)-like "Pfaffian", since the completely unrelated standard usage is as a noun. --rwg

1 0

[math-fun] Spherical trig question
by David Wilson 07 Feb '11

07 Feb '11

On the unit sphere, measuring angles in radians. Place an arc with measure A on the equator. Place arcs with measures B and C at each end of A running north along meridians (perpendicular to A). Join the northern endpoints of B and C with a great circle arc. What is the area of the spherical quadrilateral thus formed? ----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1204 / Virus Database: 1435/3426 - Release Date: 02/06/11

2 1

[math-fun] Fwd: A bunch of conjectural series for 1/pi
by rcs＠xmission.com 06 Feb '11

06 Feb '11

Here's another batch of formulas for 1/pi. This is a pretty eclectic subject, so I'm splitting off a separate interest group for this sort of thing. I imagine it will simply be a list of a half dozen email addresses. Drop me a note if you want in. Rich rcs(a)xmission.com ----- Forwarded message from zwsun(a)nju.edu.cn ----- Date: Sun, 6 Feb 2011 13:44:41 -0600 From: Zhi-Wei SUN <zwsun(a)nju.edu.cn> Reply-To: Number Theory List <NMBRTHRY(a)listserv.nodak.edu>, Zhi-Wei SUN <zwsun(a)nju.edu.cn> Subject: A bunch of conjectural series for 1/pi To: NMBRTHRY(a)listserv.nodak.edu Dear number theorists, In 1914 Ramanujan recorded 17 Ramanujan-type series for 1/pi (which are all hyper-geometric). [Actually the identity Sum_{k=0,1,2,...}(4k+1)binom(-1/2,k)^3 = 2/pi on Ramanujan's list was first obtained by G. Bauer in 1859.] This is one of the main achievements of Ramanujan. Recall that for integers b and c we let T_n(b,c) denote the coefficient of x^n in the expansion of (x^2+bx+c)^n. In several messages sent to Number Theory List in 2011, I announced five conjectural series for 1/pi involving T_n(b,c) with bc and b^2-4c nonzero. As Ramanujan found nearly twenty series for 1/pi, I felt pressure in the past days, and now I'm glad that I can add 8 more new conjectural series for 1/pi so that I have a dozen of conjectural identities for 1/pi of the new type. Here is a list of the 8 new identities for 1/pi. CONJECTURE (Zhi-Wei Sun, Feb. 5-6, 2011). We have (1) Sum_{k=0,1,...}(340k+59)binom(2k,k)^2*T_{2k}(62,1)/(-480^2)^k = 120/pi, (2) Sum_{k=0,1,...}(13940k+1559)binom(2k,k)^2*T_{2k}(322,1)/(-5760^2)^k = 4320/pi, (3) Sum_{k=0,1,...}(10k+1)binom(2k,k)^2*T_{2k}(38,1)/240^{2k} = 15*sqrt(6)/(4*pi), (4) Sum_{k=0,1,...}(120k+13)binom(2k,k)^2*T_{2k}(18,1)/320^{2k} = 12*sqrt(15)/pi, (5) Sum_{k=0,1,...}(21k+2)binom(2k,k)^2*T_{2k}(30,1)/896^{2k} = 5*sqrt(7)/(2*pi), (6) Sum_{k=0,1,...}(56k+5)binom(2k,k)^2*T_{2k}(322,1)/48^{4k} = 72*sqrt(7)/(5*pi), (7) Sum_{k=0,1,...}(195k+14)binom(2k,k)^2*T_{2k}(102,1)/10400^{2k} = 85*sqrt(39)/(12*pi), (8) Sum_{k=0,1,...}(14280k+899)binom(2k,k)^2*T_{2k}(198,1)/39200^{2k} = 1155*sqrt(6)/pi. Note that all the series (1)-(8) converge at geometric rates with ratios -64/225, -81/1600, 4/9, 1/16, 1/49, 81/256, 1/625, 1/2401 respectively. In particular, the 8th series converges very fast. If we denote by r(n) the partial sum in (8) with k from 0 to n divided by the right-hand side of the identity, then via Mathematica we see that |r(15)-1|<10^{-50} and |r(30)-1|<10^{-100}. I include all my announced series for 1/pi and their related congruences in my preprint (arXiv:1101.0600) which will be polished later. Also, in my opinion, there should be some (but very few) series for 1/pi of the forms Sum_{k=0,1,...}(a+d*k)binom(6k,3k)binom(3k,k)T_k(b.c)/m^k and Sum_{k=0,1,...}(a+d*k)binom(2k,k)binom(3k,k)T_{3k}(b.c)/m^k, with a,b,c,d,m integers and m*bcd(b^2-4c) nonzero. I have not yet found such series for 1/pi. It seems that all known methods used to establish Ramanujan-type series do not work for my conjectural series which are of a more sophisticated type. In view of the 8+5 conjectural series for 1/pi, I hope that my conjectures could interest some mathematicians and stimulate the birth of a new tool or a new theory. For convenience, here I also collect the 5 announced identities for 1/pi conjectured by me. (9) Sum_{k=0,1,...}(30k+7)binom(2k,k)^2*T_k(1,16)/(-256)^k = 24/pi, (10) Sum_{k=0,1...}(15k+2)binom(2k,k)binom(3k,k)*T_k(18,6)/972^k = 45*sqrt(3)/(4*pi), (11) Sum_{k=0,1,...}(40k+3)binom(4k,2k)binom(2k,k)T_k(98,1)/112^{2k} = 70*sqrt(21)/(9*pi), (12) Sum_{k=0,1,...}(26k+5)binom(2k,k)^2*T_{2k}(7,1)/(-48^2)^k = 48/(5*pi), (13) Sum_{k=0,1,...}(8k+1)binom(2k,k)^2*T_{2k}(10,1)/96^{2k} = 10*sqrt(2)/(3*pi). Also, I'd like to mention that my following open conjectural identities announed in 2010 are still open. (14) Sum_{k>0}(10k-3)8^k/(k^3*binom(2k,k)^2*binom(3k,k)) = pi^2/2, (15) Sum_{k>0}(35k-8)81^k/(k^3*binom(2k,k)^2*binom(4k,2k)) = 12*pi^2, (16) Sum_{k>0}(28k^2-18k+3)(-64)^k/(k^5*binom(2k,k)^4*binom(3k,k)) = -14*zeta(3), (17) Sum_{k>0}(15k-4)(-27)^{k-1}/(k^3*binom(2k,k)^2*binom(3k,k)) = K, where K:=Sum_{k>0}(k/3)/k^2, (18) Sum_{k>0}(5k-1)(-144)^k/(k^3*binom(2k,k)^2*binom(4k,2k)) = -45K/2. Finally, I beg your pardon if you are not interested in my recent messages on series for powers of pi. Any comments are welcome! Zhi-Wei Sun http://math.nju.edu.cn/~zwsun ----- End forwarded message -----

1 0

[math-fun] is a lever arm the same as velocity?
by Gary Antonick 06 Feb '11

06 Feb '11

Hey all, I was wondering about something perhaps obvious to many with a mechanics/vector background. (Apologies if this is too basic compared the usual math-fun stuff.) Am wondering if there's some fundamental principle behind levers and colliding objects.. or if they're somehow describing the same thing. Here's what I mean. Take several masses on a fulcrum. A mass of 1 at distance 2 from the fulcrum plus a mass of 2 at distance 3 balances a mass of 2 at distance 4. 1x2 + 2x3 = 2x4 of 2 + 6 = 8 In other words, M1xD1 + M2xD2 = M3D3 This looks a lot like conservation of momentum (inelastic, in this case) if you replace the distances with velocities. I thought that was interesting. So I wondered... is this a coincidence? Or are these describing the same thing, conceptually? And if they're describing the same thing conceptually, then what's the kinetic energy formula equivalent? In other words, colliding objects conserve 1/2MV^2, or 1/2MVV. Replace one of the V's with a D and you get 1/2MVD. What is 1/2MVD when applied to objects balanced on each side of a fulcrum? Using the example above, the 2x4 could be balanced by 1x2 and infinitely many other mass distances: 2x4 = 1x2 + 2x3 or 2x4 = 1x2 + 1x6 etc This all balance just like many different post-collision velocity combinations will preserve momentum. But what's the kinetic energy equivalent? Am guessing there's only one solution that preserves.... something. Am not sure what it is. Rotational energy? I was about to calculate but am not sure how to start. If the lever broke above the fulcrum and the different weights began to rotate around the fulcrum they'd each have a different velocity. I suppose the bottom-line question is: if you had a mass of 2 a distance 4 from a fulcrum and a mass of 1 a distance 2 on the other side, what additional mass would you add on the side of the smaller mass (the 1) so that it not only balanced, but the x (am not sure what this would be) energy on each side was the same as well? And a related point. Rotation around a fulcrum would create a circle, obviously, which is what's created by the formula for kinetic energy. A coincidence? All the best, Gary

1 0

[math-fun] Kindergarten example of almost-integer mystery
by Dan Asimov 05 Feb '11

05 Feb '11

A crossword puzzle with a 23x23 grid is 529/441 times as large as one with a 21x21 grid. 529/441 is almost 1.2, and the difference D satisfies 1/D = 2204.999999999853571+. Its difference from an integer satisfies |2205 - 1/D| < 1.5 e-10 --Dan Those who sleep faster get more rest.

3 2

Re: [math-fun] Integral House
by Dan Asimov 04 Feb '11

04 Feb '11

Robert Baillie wrote, re James Stewart the calculus book zillionaire: << An 18,000 square foot house? What a pig. >> This is perhaps mitigated by his offering his house's auditorium, seating 150, for benefit performances about eight times a year. (For what he deems "good causes".) It is kind of amazing, though, that he lives in this 5-story house *alone*. --Dan Those who sleep faster get more rest.

1 0

[math-fun] Integral House
by Hans Havermann 04 Feb '11

04 Feb '11

A story about mathematician James Stewart's $24-million "Integral House" in yesterday's Toronto Star: http://www.thestar.com/news/article/933017

4 3