June 2011 - mathfun@mailman.xmission.com

[math-fun] 300 anti-hydrogen atoms stored for 16 minutes
by Henry Baker 06 Jun '11

06 Jun '11

FYI -- I'd still be interested in the details of the matter/antimatter annihilation; e.g., does the electron have to find the anti-electron & the proton find the anti-proton? Does this annihilation violate the "conservation of information" principle ? (I assume that no anti-neutrons were harmed in this experiment.) http://www.telegraph.co.uk/science/science-news/8558603/Scientists-trap-and… Scientists 'trap' and study elusive anti-matter Scientists have trapped and stored antihydrogen atoms for a record 16 minutes, a stunning technical feat that promises deeper insights into the mysteries of anti-matter. In the study, published in the journal Nature Physics, researchers report trapping some 300 antiatoms 7:00AM BST 06 Jun 2011 Particles and anti-particles annihilate each other in a small flash of energy when they collide. At the moment of the big bang, nearly 14 billion years ago, matter and anti-matter are thought to have existed in equal quantities. If that balance had persisted, the observable Universe we inhabit would never have come into being. For unknown reasons -- and fortunately for us -- Nature seemed to have a slight preference for matter, and today anti-matter is rare. This asymmetry remains one of the greatest riddles in particle physics. But ongoing low-energy experiments with hydrogen atoms could be a key step toward solving it. "We can keep the antihydrogen atoms trapped for 1,000 seconds. This is long enough to begin to study them -- even with the small number that we can catch so far," said Jeffrey Hangst, spokesman for the ALPHA team conducting the tests at the European Organisation for Nuclear Research (CERN) in Geneva. In the study, published in the journal Nature Physics, researchers report trapping some 300 antiatoms. Scientists used CERN's high-energy accelerator to create the antihydrogen atoms, and then chilled them to near-zero temperatures. The aim is to use laser and microwave spectroscopy to compare the immobilised particles to their hydrogen counterparts. The same team succeeded last year in trapping dozens of anti-matter atoms and holding them in place for a fraction of a second, a world first at the time. But that was not long enough for the excitable particles to settle into the stable "ground" state needed for precise measurements. The new benchmark extended this storage time 5,000 fold, making it possible to carry out crucial experiments. Scientists will now look for "violations" or discrepancies in something called the charge-parity-time reversal (CPT) symmetry. Measurements of trapped antihydrogen are due to get underway shortly, and could yield results before the end of the year.

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[math-fun] Paradox
by Adam P. Goucher 06 Jun '11

06 Jun '11

Select two random integers, a and b. Clearly, the probability of b being larger than a is 100%, as the expected value of b is infinite, yet the value of a is finite. Similarly, the probability of a being larger than b is 100%, by symmetry. What have I done wrong? Sincerely, Adam P. Goucher

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Re: [math-fun] Escape-time fractals
by Dan Asimov 05 Jun '11

05 Jun '11

Thanks a lot, David. I appreciate your taking the time to explain terms. --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] Colourings of the plane
by Bill Gosper 05 Jun '11

05 Jun '11

Corey Ziegler Hunts wants to know what's wrong with the 3-coloration: (rational, rational) -> red ; (irrational, irrational) -> indigo; <discordant> -> blood green. He claims the validity of this solution is equivalent to the nonexistence of a continuous function that complements rationality on some interval. --rwg > Dear all, > > Can the plane be n-coloured, such that there are no mono- > chromatic continuous curves of nonzero length? > > > > For n = 4, there is the 'Millar colouring': > > Colour a point (x,y): > > * red if x and y are both rational; > * green if x is irrational, y is rational; > * yellow if x is rational, y is irrational; > * blue if x and y are both irrational. > > It is obvious that no monochromatic continuous curves can exist, > as both the ordinate and abcissa are fixed (as if they varied by a > nonzero amount, they would cross an infinity of rational and > irrational points). > > > > With a little effort, it is possible to define a 3-colouring with > similar properties: > > * red if x and y are both irrational; > * red if x is a dyadic rational and y is irrational; > * green if x is a non-dyadic rational and y is irrational; > * blue if x is irrational and y is rational; > * blue if x is a non-dyadic rational and y is rational; > * green if x is a dyadic rational and y is rational; > > (A dyadic rational has a terminating binary expansion; a non-dyadic > rational has a recurring binary expansion; an irrational has an > aperiodic binary expansion.) > > It is clear that no red and blue curves exist, as they can only be > horizontal lines, and are terminated by the infinity of green points > on the lines. Also, no green lines exist, as dyadic and non-dyadic > rationals are separated by an infinity of irrational numbers. > > > > For 2-colourings, this is not so easy. The 'reduced Millar colouring', > where a point is red if x and y are both either rational or irrational, > and blue otherwise, does not satisfy the property. For example, > the curve y = x is exclusively red. > > A variant, the 'Elliott colouring' was proposed, with the following > rules: > > * red if x and y are both irrational; > * blue if x is irrational, y is rational; > * blue if x is a dyadic rational, y is irrational; > * red if x is a dyadic rational, y is rational. > > However, Appendix I shows how the reals can be continuously > deformed to map the ordinary rationals to dyadic rationals, > thereby rendering the Elliott colouring equivalent to the reduced > Millar colouring. > > > So, I leave the n = 2 case as an 'exercise for the reader'. > > > > Sincerely, > > > Adam P. Goucher >

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[math-fun] Why math(s) matters, part 239...
by Henry Baker 05 Jun '11

05 Jun '11

>From comp.risks: even lawyers need to know math while doing God's work... Date: Wed, 01 Jun 2011 08:41:08 +0100 From: Martyn Thomas <martyn(a)thomas-associates.co.uk> Subject: Another role for provers? The complexity of an esoteric Hong Kong financial instrument has come back to haunt Goldman Sachs after an simple typographical slip threatened to cost it HK$350m (27m UK pounds). The error appeared in the small print of a phone book-sized prospectus accompanying the issue in February of four so-called "exchange-traded warrants" which offered exposure to Japan's Nikkei index of leading shares. In a formula to calculate the value of the warrants a multiplication symbol appeared where their should have been a division. The potentially costly error appeared in the bank's paperwork despite it having been scrutinised and approved by the Hong Kong stock exchange. Such warrants are hugely popular in Hong Kong, with 14,400 similar products said to have been issued last year by large investment banks. It was not until the end of March -- almost seven weeks after the warrants had been issued -- that a lawyer from Goldman reported the mistake to the stock exchange. For almost two hours the price of warrants began to soar until trading was suspended at the bank's request. Goldman has offered to buy back warrants at a 10% premium, an offer accepted by 75% of holders. However, a hard core of large investors believe they are contractually entitled to considerably more. One told the Economist magazine the bank's offer was worth HK$10m, whereas a strict application of the formula suggested the warrants could be worth $350m. http://www.guardian.co.uk/business/2011/may/31/goldman-sachs-libya-investme… (final paragraphs).

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[math-fun] Favourite prime numbers
by Dan Asimov 04 Jun '11

04 Jun '11

((( My favorite number is 24 hands down, because of its amazing properties like showing up in Dedekind eta, 1^2 + ... + 24^2 = 70^2 being the only such relation, the 24-dimensional Leech lattice, and 24 being the largest number N such that all smaller numbers K with GCD(K,N) = 1 satisfy K^2 == 1 (mod N). At least some of these are interconnected. ))) But how about prime numbers? After the first few, how do you distinguish, say, 101, 103, 107, 109 ? Are there standard measures of some sort that distinguish among prime numbers? --Dan Sometimes the brain has a mind of its own.

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[math-fun] Really, really large catenaries
by Henry Baker 04 Jun '11

04 Jun '11

The usual catenary is formed by a hanging chain in a gravity field which is uniform -- i.e., the case for chains which are small relative to the size of the Earth. Is there a closed form solution to "large" catenaries which have a size as large as, or larger than the Earth? In particular, consider a chain whose links are _repulsed_ by a spherical (non-rotating) Earth (e.g., proportional to -1/r^2). What are the shapes of these curves? A large enough chain would go completely around the Earth. I guess there would be solutions where such a closed chain would touch the Earth at just one point. I don't know if there would be stable solutions where the chain would not touch the Earth at all. If rotation is introduced, things probably get a lot weirder.

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[math-fun] Colourings of the plane
by Adam P. Goucher 04 Jun '11

04 Jun '11

Dear all, Can the plane be n-coloured, such that there are no mono- chromatic continuous curves of nonzero length? For n = 4, there is the 'Millar colouring': Colour a point (x,y): * red if x and y are both rational; * green if x is irrational, y is rational; * yellow if x is rational, y is irrational; * blue if x and y are both irrational. It is obvious that no monochromatic continuous curves can exist, as both the ordinate and abcissa are fixed (as if they varied by a nonzero amount, they would cross an infinity of rational and irrational points). With a little effort, it is possible to define a 3-colouring with similar properties: * red if x and y are both irrational; * red if x is a dyadic rational and y is irrational; * green if x is a non-dyadic rational and y is irrational; * blue if x is irrational and y is rational; * blue if x is a non-dyadic rational and y is rational; * green if x is a dyadic rational and y is rational; (A dyadic rational has a terminating binary expansion; a non-dyadic rational has a recurring binary expansion; an irrational has an aperiodic binary expansion.) It is clear that no red and blue curves exist, as they can only be horizontal lines, and are terminated by the infinity of green points on the lines. Also, no green lines exist, as dyadic and non-dyadic rationals are separated by an infinity of irrational numbers. For 2-colourings, this is not so easy. The 'reduced Millar colouring', where a point is red if x and y are both either rational or irrational, and blue otherwise, does not satisfy the property. For example, the curve y = x is exclusively red. A variant, the 'Elliott colouring' was proposed, with the following rules: * red if x and y are both irrational; * blue if x is irrational, y is rational; * blue if x is a dyadic rational, y is irrational; * red if x is a dyadic rational, y is rational. However, Appendix I shows how the reals can be continuously deformed to map the ordinary rationals to dyadic rationals, thereby rendering the Elliott colouring equivalent to the reduced Millar colouring. So, I leave the n = 2 case as an 'exercise for the reader'. Sincerely, Adam P. Goucher ***** Appendix I ***** Consider the following sequences, which can be derived from the Stern-Brocot tree: [0] [0,1/2] [0,1/3,1/2,2/3] [0,1/4,1/3,2/5,1/2,3/5,2/3,1/4] ... For each rational in the interval [0,1), there is a constant k such that for all n > k, the rational appears in the nth sequence. A similar sequence of sequences can be created for the dyadic rationals in the interval [0,1): [0] [0,1/2] [0,1/4,1/2,3/4] [0,1/8,1/4,3/8,1/2,5/8,3/4,7/8] ... This gives a natural bijection between the rationals in [0,1) to the dyadic rationals in [0,1), which can be extended to a bijection between all rationals and all dyadic rationals. It also extends to a mapping from the reals to the reals, which operates as follows: 1. Express the real number, x, as its integer part followed by binary expansion: x = n . a_1 a_2 a_3 a_4 a_5 ... 2. Let L_0 = n, H_0 = n+1. 3. Iterate for all natural numbers i: (a) If a_i = 0, let L_i = L_(i-1), H_i = mediant(L_(i-1),H_(i-1)) (b) If a_i = 1, let H_i = H_(i-1), L_i = mediant(L_(i-1),H_(i-1)) 4. The sequences L_i and H_i converge to a real number, y, from below and above, respectively. ***** End of appendix I *****

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Re: [math-fun] Favourite prime numbers
by Bill Gosper 04 Jun '11

04 Jun '11

rcs>I think you need a sign bit for radix 1+i. Knuth likes i-1, which doesn't need the sign bit. Rich --- Quoting Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>>: > Mike Stay>My favorite prime is 1+i.>> One good reason: It's the base of a binary number system,>> with bits 0 and 1, that can represent all complex numbers>> (without needing a sign bit). The pure fractions represent>> a simply connected region coincident with the twindragon>> fractal (File:Dragon_tiling5.svg> <http://en.wikipedia.org/wiki/File:Dragon_tiling5.svg>) .>> --rwg>> Start with 0. Then iterate>> Union[%, 1 + %]/(1 + I); a few times. Then>> ListPlot[{Re[#], Im[#]} & /@ %, PlotStyle -> PointSize[Large]]>------------ Ouch, Mike Farmwald, Robert Maas, and I made that picture in the later editions of Vol2! But I misrecalled it as base 1-i, and figured it's always safe to conjugate. i+1 draws a twindragon all right, but with the origin at the tip, so it does need a sign bit. -(1+i) does the trick. And it's prime, technically. Does it share other attributes that endear 1+i to Mike S? --rwg

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Re: [math-fun] math-fun Digest, Vol 100, Issue 3; 1^2 + ... + 24^2 = 70^2
by Stuart Anderson 04 Jun '11

04 Jun '11

While 1^2 + ... + 24^2 = 70^2 cannot be packed into a square of size 70, 2 (1^2 + ... + 24^2) = 70x140 can be packed, Gunter Sterten found this; http://www.squaring.net/140x70-2x24squares.png. He said it was at the edge of what could be packed with his program 4(1^2 + ... + 24^2) = 140^2 can also be packed, Eric Friedman found this; http://www.squaring.net/140x140-4x24squares.jpg by hand Stuart Anderson > > On Fri, Jun 3, 2011 at 8:49 AM, Dan Asimov <dasimov(a)earthlink.net> wrote: > >> ((( >> My favorite number is 24 hands down, because >> of its amazing properties like showing up >> in Dedekind eta, 1^2 + ... + 24^2 = 70^2 >> being the only such relation, the 24-dimensional >> Leech lattice, and 24 being the largest number N >> such that all smaller numbers K with GCD(K,N) = 1 >> satisfy K^2 == 1 (mod N). At least some of these >> are interconnected. >> ))) >>

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