October 2011 - mathfun@mailman.xmission.com

[math-fun] Antimatter belt around Earth discovered by Pamela craft
by Henry Baker 23 Oct '11

23 Oct '11

FYI -- This discovery answers a question asked several times on this list. The fact that this stuff is right here in the solar system is really surprising! I heard this guy interviewed. He said even larger belts would be found around the larger planets -- enough to fuel an "antimatter drive" for a space craft. The antiprotons start life as anti-neutrons when a cosmic ray hits; then the anti-neutron decays into an anti-proton & an anti-electron & is captured by the magnetic field. http://www.bbc.co.uk/news/science-environment-14405122 BBC News Science & Environment 7 August 2011 Last updated at 05:54 ET Antimatter belt around Earth discovered by Pamela craft A thin band of antimatter particles called antiprotons enveloping the Earth has been spotted for the first time. The find, described in Astrophysical Journal Letters, confirms theoretical work that predicted the Earth's magnetic field could trap antimatter. The team says a small number of antiprotons lie between the Van Allen belts of trapped "normal" matter. The researchers say there may be enough to implement a scheme using antimatter to fuel future spacecraft. The antiprotons were spotted by the Pamela satellite (an acronym for Payload for Antimatter Matter Exploration and Light-nuclei Astrophysics) - launched in 2006 to study the nature of high-energy particles from the Sun and from beyond our Solar System - so-called cosmic rays. These cosmic ray particles can slam into molecules that make up the Earth's atmosphere, creating showers of particles. Many of the cosmic ray particles or these "daughter" particles they create are caught in the Van Allen belts, doughnut-shaped regions where the Earth's magnetic field traps them. Among Pamela's goals was to specifically look for small numbers of antimatter particles among the far more abundant normal matter particles such as protons and the nuclei of helium atoms. 'Abundant source' The new analysis, described in an online preprint, shows that when Pamela passes through a region called the South Atlantic Anomaly, it sees thousands of times more antiprotons than are expected to come from normal particle decays, or from elsewhere in the cosmos. http://arxiv.org/abs/1107.4882 The team says that this is evidence that bands of antiprotons, analogous to the Van Allen belts, hold the antiprotons in place - at least until they encounter the normal matter of the atmosphere, when they "annihilate" in a flash of light. Although normal matter particles outweigh the antiprotons by thousands to one, the band is "the most abundant source of antiprotons near the Earth", said Alessandro Bruno of the University of Bari, a co-author of the work. "Trapped antiprotons can be lost in the interactions with atmospheric constituents, especially at low altitudes where the annihilation becomes the main loss mechanism," he told BBC News. "Above altitudes of several hundred kilometres, the loss rate is significantly lower, allowing a large supply of antiprotons to be produced." Dr Bruno said that, aside from confirming theoretical work that had long predicted the existence of these antimatter bands, the particles could also prove to be a novel fuel source for future spacecraft - an idea explored in a report for Nasa's Institute for Advanced Concepts.

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[math-fun] Silly question (?)
by David Wilson 22 Oct '11

22 Oct '11

How do you show that there is an integer greater than any given real number? The real question is, given the axioms for the real numbers (Dedekind-complete ordered field), how do you define the "real integers" within them. Two possibilities spring to mind. Definition A: The "real integers" are the minimal subset of R that contains 1 and is closed over negation and addition. Definition B: 1 is a "real integer", and there is a unique "real integer" in every half-open unit interval of R. Definition A seems more in the spirit of the number theory definition of Z. However, I do not immediately see how it answers my original question. Definition B immediately answers my original question. For any real number r, there is a unique integer on the half-open interval (r, r+1] which must be greater than r. Presumably definitions A and B are equivalent. Is the proof easy?

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[math-fun] FTL
by Cordwell, William R 18 Oct '11

18 Oct '11

In case this wasn't seen, and at the risk of killing off the neutrino jokes: http://www.technologyreview.com/blog/arxiv/27260/

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[math-fun] Funniest physics joke ever, laugh while it lasts
by Hilarie Orman 18 Oct '11

18 Oct '11

The bartender says "We don't serve neutrinos here." A neutrino walks into a bar. (I read this in The Guardian; Hilarie).

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[math-fun] Draw me. Flunk Art class.
by Bill Gosper 17 Oct '11

17 Oct '11

This polyhedral puzzle piece <http://gosper.org/artclass.png> is drawn to scale with correct perspective. The eight quadrilaterals designated on the bitmap with "circles" all have normals exactly 45° from vertical (= short box axis). --rwg

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[math-fun] Pi, 10 trillion digits
by Simon Plouffe 17 Oct '11

17 Oct '11

Hello, Pi has been computed to 10 trillion digits, by the same team as the previous record using a home pc. http://ja0hxv.calico.jp/pai/pietc.html http://ja0hxv.calico.jp/pai/estart.html Details will follow later, Best regards, Simon Plouffe

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[math-fun] ellipsoid girths "triangle inequality"
by Warren Smith 14 Oct '11

14 Oct '11

> In terms of a, b, c, the girths are > {gab -> 4 a EllipticE[1 - b^2/a^2], gac -> 4 a EllipticE[1 - c^2/a^2], gbc > -> 4 b EllipticE[1 - c^2/b^2]} --at least with MAPLE9's definition of EllipticE, this is wrong [I divided by 4, but otherwise same as you]: >g := (a,b,c) -> (a*EllipticE(1 - b^2/a^2), a*EllipticE(1 - c^2/a^2), b*EllipticE(1 - c^2/b^2)); >g(0.1, 5.0, 0.2); 0.000031428499 + 249.8998057 I, 0.02655964076 + 0.2498348128 I, 5.030078415 note nonreal "girths" I conjecture you really meant g := (a,b,c) -> (max(a,b)*EllipticE(1-min(a,b)^2/max(a,b)^2), max(a,c)*EllipticE(1-min(a,c)^2/max(a,c)^2), max(b,c)*EllipticE(1-min(b,c)^2/max(b,c)^2)); which at least returns positive real output for positive real input. Assuming you can correctly define g(a,b,c)... which should not be hard... then a solution to the posed problem would be got by running g on a lot of random numbers, producing a large set of triples, then using linear programming to find constants f1, f2, f3 such that f1*g1+f2*g2+f3*g3 always exceeded 0 for all triples (g1,g2,g3). Any such f1,f2,f3 would define an analogue of the triangle inequality as requested (at least in the limit of huge random data collection). We may wlog demand f2 and f3 have same sign while f1 has opposite sign, and by symmetry we may demand f2=f3. Just by hand considering these triples (1000,1000,0) and permutations (2*pi, 4, 4) and perms (1,1,1) I observe that g1<g2+g3 is obeyed but g1<0.9999*(g2+g3) can fail;; and g1>0*(g2+g3) is obeyed but g1>0.0001*(g2+g3) can fail. So I believe an inequality of the form g1<K*(g2+g3) is obeyed for some constant K>=1 and I do not know (but computer could pretty much tell us) what the minimum possible K is. The obvious guess is K=1. No inequality of the form g1>K*(g2+g3) is obeys for any K>0. > > This is a routine problem for FindRoot, but it would be nice to see how a, > b, c depend on > the girths in the low eccentricity case, say, by reverting the series > expansions of the girths. > Julian, Corey, and I were surprised by the absence of multivariate reversion > from Macsyma > and Mathematica, and set about building our own. After several hours of > cruel hoodwinking > by Mma's Series facility, we emerged having only solved the ellipsoid case. > Surprisingly(?), > it has at least eight (2^3) solutions, which may help to explain the > shortage of multivariate > reverters. To fourth order, > > {a -> (gab/(2 Pi))(1 - eac + eac^2 + ebc + (eac ebc)/2 - (3 ebc^2)/2 + > 1/8 (-8 eac^3 - eac^2 ebc - 9 eac ebc^2 + 18 ebc^3) + > 1/16 (16 eac^4 + 2 eac^3 ebc - 3 eac^2 ebc^2 + 42 eac ebc^3 - 57 > ebc^4)), > > a -> (gac/(2 Pi))(1 + ebc + 3/2 (eac ebc - ebc^2) + > 3/8 (eac^2 ebc - 7 eac ebc^2 + 6 ebc^3) - > 3/16 (7 eac^2 ebc^2 - 26 eac ebc^3 + 19 ebc^4))} > > {b -> (gab/(2 Pi))(1 + eac - (3 eac^2)/2 - ebc + (eac ebc)/2 + ebc^2 + > 1/8 (18 eac^3 - 9 eac^2 ebc - eac ebc^2 - 8 ebc^3) + > 1/16 (-57 eac^4 + 42 eac^3 ebc - 3 eac^2 ebc^2 + 2 eac ebc^3 + > 16 ebc^4)), > > b -> (gac/(2 Pi))(1 + 2 eac - eac^2/2 - ebc - (eac ebc)/2 + ebc^2 + > 1/8 (6 eac^3 - 5 eac^2 ebc + 7 eac ebc^2 - 8 ebc^3) + > 1/16 (-21 eac^4 + 24 eac^3 ebc - 5 eac^2 ebc^2 - 14 eac ebc^3 + > 16 ebc^4))} > > {c -> (gab/(2 Pi))(1 - eac + eac^2 + eac^4 - ebc - (eac ebc)/2 - (eac^3 > ebc)/ > 8 + ebc^2 - (5 eac^2 ebc^2)/16 - (eac ebc^3)/8 + ebc^4 + > 1/8 (-8 eac^3 + eac^2 ebc + eac ebc^2 - 8 ebc^3)), > > c -> (gac/(2 Pi))(1 - ebc - (3 eac ebc)/2 - (3 eac^2 ebc)/8 + > ebc^2 + (9 eac ebc^2)/8 - (3 eac^2 ebc^2)/16 - ebc^3 - ( > 9 eac ebc^3)/8 + ebc^4)} > > where {eac -> gab/gac - 1, ebc -> gab/gbc - 1}, which are small for low > eccentricity. E.g., for {a -> 5., b -> 4., c -> 3.}, the girth formulæ > give > > {gab -> 28.3617, gac -> 25.527, gbc -> 22.1035}. > > Plugging these back into the reversions: > {{a -> 4.97828, a -> 4.98561}, > {b -> 4.00689, b -> 4.00371}, > {c -> 3.0062, c -> 3.00353}}. > --rwg > > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step) and math.temple.edu/~wds/homepage/works.html

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Re: [math-fun] Asymptotic formula for LCM{1,2,3,...,n} ?
by N. J. A. Sloane 13 Oct '11

13 Oct '11

> Does anyone know of an asymptotic formula for LCM{1,2,3,...,n} ? See A003418 !

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Re: [math-fun] Zillions
by Dan Asimov 13 Oct '11

13 Oct '11

Circa 1980 someone at MIT showed me a little pamphlet floating around, created by Gian-Carlo Rota, of all the integers between I think 1 and 1000, listed purely in English, alphabetically. The idea that this could have any conceivable use struck me as so absurd that I could not stop laughing for quite a while. --Dan Hans wrote: << Don Knuth & Allan Miller: A Programming and Problem-Solving Seminar, . . . . . Chapter 1 of the seminar (Alphabetized Integers) provides a good background discussion, contextualizing the problem. >> Sometimes the brain has a mind of its own.

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[math-fun] "Tutoring" Neil
by Bill Gosper 12 Oct '11

12 Oct '11

Impatient with my bumbling attempts to CAD a puzzle design with Mathematica, Neil fired up a copy of MuPad (a gift from Julian, who won two copies in a contest). NB: "Barf! It can't even do sum(cos(9/(PI*n + sqrt(PI^2*n^2 - 9))/n^2, n=0..infinity)" rwg: "Very funny. That's my lifetime abstrusest identity." NB: "Well, Mathematica can do it." rwg: "And I am the Virgin Mary." NB: Sum[Cos[9/(\[Pi]*n + Sqrt[\[Pi]^2*n^2 - 9])]/n^2, {n, \[Infinity]}] Mma: -Pi^2/(12 E^3) rwg (swallowing expletives): "Change something. Change the 9s." (The generalization is Sum[Cos[Sqrt[\[Pi]^2*n^2 + b^2]]*(-1)^n/n^2, {n, \[Infinity]}] == \[Pi]^2/4*(Sin[b]/b - Cos[b]/3) See http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=e…<http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=e…> ) Mma: <FAIL> rwg&NB: "FEH!" Earlier, watching him blurt a particularly brutish code fragment (and inwardly jealous of his fluency) rwg: "If Knuth were dead, he'd be spinning in his grave." NB: "Algorithms are for people who don't know how to buy RAM." Don't believe him. He has an algorithm that reduces to a few hours his 48day 4x4 puzzle search. (http://nbickford.wordpress.com/) --rwg Re: Fastest computer: Whatever happened to GaAs? DanA: Freudian slip? Her name is Angela Saini. No, just geriatric eyesight. But if the antiBayesian (frequentist?) judge left much wiggle room, what are all those professors yelling about?

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