December 2018 - mathfun@mailman.xmission.com

[math-fun] Exotic radices
by Keith F. Lynch 19 Dec '18

19 Dec '18

Any real number except 0, 1, and -1 can be used as the radix (base) in a place value system intended to represent all real numbers. But what if we restrict it to radices such that all integers terminate? Any integer radix greater than 1 will work. So will integer roots of those integers, for instance the square root of 2 or the fifth root of 12. Phi will work. No transcendental radix will work. The reciprocals and the negatives of anything that will work will also work. I'm pretty sure that rational numbers, other than integers and their reciprocals, won't work. Is the complete solution set known? Is it dense? Once you've chosen a radix, what digits should be allowed? The standard is 0 through R-1, where R is the ceiling of the radix. (Assuming the radix is greater than 1.) But as balanced ternary demonstrates, that's not the only choice. Will any set of R consecutive digits do, so long as one of them is 0? For instance almost-balanced decimal, where the decimal digits are -4 through +5? Will any other set do? I'll save complex radices for another day. And I'm pretty sure that quaternion radices won't work, i.e. can't represent all quaternions given nothing but real integer digits.

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Re: [math-fun] Egyptian Fractions
by Keith F. Lynch 18 Dec '18

18 Dec '18

"Lucas, Stephen K - lucassk" <lucassk(a)jmu.edu> wrote: > At a higher level, he has a whole bunch more information of Egyptian > fractions at > https://www.ics.uci.edu/~eppstein/numth/egypt/ Interesting. Thanks. It is perhaps unfortunate that there are so many different ways to represent the same number with EFs, many of them quite long. As such, it seems unlikely that advanced aliens will use EFs as their standard way of representing numbers. Similarly with some of the other suggested ways of representing numbers. I was hoping to find a way as succinct and unique as the usual place value notation, but completely different. Is there any reasonable or interesting subset of the reciprocal positive integers for which all EFs of rational numbers still terminate? (It's always possible to skip the first N of them for any N, but not very reasonable or interesting.) The powers of 2 of course give an almost unique representation of every positive number, but they only terminate for dyadic fractions. (Almost unique because each dyadic fraction can end with infinitely many 0s or with infinitely many 1s.) The primes are an especially poor choice. Almost nothing terminates. For instance 18/55 = 1/5 + 1/11 + 1/29 + 1/541 + 1/30829 + .... I did notice that some primes are much likelier than others to show up in a GPEF (greedy prime Egyptian fraction). Highly composite numbers (A002182) are even worse. Not only does almost nothing terminate, but since the sum of their reciprocals converges to about 2.13, nothing larger that can be represented at all.

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[math-fun] Rock music, math & extraterrestrials
by Henry Baker 18 Dec '18

18 Dec '18

>From NPR's "Fresh Air": Terry Gross spoke with Queen lead guitarist *Brian May* in 2010 about recording the many vocals in 'Bohemian Rhapsody,' writing the anthem 'We Will Rock You' and getting a ***PhD in astrophysics***. https://16683.mc.tritondigital.com/NPR_381444908/media-session/dd64f39b-11e… ----- According to Queen guitarist & astrophysicist Brian May, the stomping sound in the original recording of 'We Will Rock You' involved summing a multiplicity of recordings of stomping but with differing delays of *prime numbers*. Presumably, this fact could be confirmed by some sort of matched filtering using FFT's on this recording. ----- This human sound artifact falls into the bizarre category of "better than real", because it would be difficult -- if not impossible -- to organize a real crowd to stand only at prime number distances from the stage. Another example is that of ancient Greek temple columns, which are NOT absolutely straight, but are elegantly curved to *appear* straight in their actual setting. https://en.wikipedia.org/wiki/Entasis "An early view, often articulated and still widespread, espoused by Hero of Alexandria, is that *entasis* corrects the optical illusion of concavity in the columns that the fallible human eye would create if the correction were not made." ----- I think that incorporating something "better than real" into a signal would alert an intelligent receiver that the sender not only understood the reality, but was also capable of "hacking it" until it became real-er than real. This sort of signal is probably more likely to catch the attention of some extraterrestrial than some of the silliness that has been previously proposed. For example, what if someone on Earth computed the hologram which would *correct* for the gravitational lensing in some far-off galaxy. Then the Earth's signal would stand out like a sore thumb in any statistical analysis of the incoming photon data.

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Re: [math-fun] Quotients of Lie groups by discrete subgroups
by Dan Asimov 17 Dec '18

17 Dec '18

I managed to screw up this statement: ----- SO(3) / I = S^3 / (2I). Strikingly, 2I is the *only* subgroup of S^3 for which SO(3) is a homology 3-sphere. ----- which should read instead: ----- SO(3) / I = S^3 / (2I). Strikingly, 2I is the *only* subgroup of S^3 for which S^3 / (2I) is a homology 3-sphere. ----- (And by S^3, I mean the double cover of P^3 = SO(3) both as a manifold and as a group, making S^3 into a Lie group (also known as Spin(3)). —Dan ----- >I don't know much about this, but it's so interesting that I thought >I'd share it with math-fun. > >The rotation group of 3-space R^3, called SO(3), is homeomorphic to >P^3, (real) projective 3-space. Its unique invariant metric, up to >scaling, is that of P^3 induced from > > P^3 = S^3 / (x ~ -x) > >which just means this metric on P^3 is inherited from the 3-sphere S^3, >and in fact is locally isometric to it. > >Now suppose H is a finite subgroup of SO(3). Then H is known to be >isomorphic to one of the following: > >* a cyclic group C_n of order n, > >* a dihedral group of order 2n, > >* the tetrahedral group T of order 12 (also known as A_4), > >* the octahedral group O of order 24, > >* the icosahedral group I of order 60 (also known as A_5). > >Since SO(3) is a Lie group*, any finite >subgroup G of SO(3) gives rise to the left coset space > > SO(3) / H = {g H | g in SO(3)} > >whose elements by this definition are indeed copies of the subgroup K, >translated on the left by some element g of SO(3). Unless H is a normal >subgroup (the only nontrivial one is {-I, I}), this is not a group. > >But suppose H is a discrete subgroup of SO(3) with the property that its >commutator subgroup is the whole group: > > [H, H] = H. > >The word for this is: H is a "perfect" group. > >Then interesting things happen! The coset space SO(3) / H is what's called >a homology sphere: A 3-dimensional manifold with the same homology groups >as the 3-sphere S^3 (H_0(S^3) = H_3(S^3) = Z and H_1(S^3) = H_2(S^3) = 0). > >And if you take H = I, promoting it by a mere one place in the alphabet, >then SO(3) / H = SO(3) / I is a homology sphere whose fundamental group >is the binary icosahedral group 2I, whose order is 120. > >Not surprisingly we can get the same coset space by factoring the double cover >of SO(3) = P^3 out by the double cover of I, namely 2I: > > SO(3) / I = S^3 / (2I). > >Strikingly, 2I is the *only* subgroup of S^3 for which SO(3) is a homology >3-sphere. > >And there is not enough space in this forum to go into detail, but the so-called >"Poincaré dodecahedral space" S^3 / (2I), seems to be uniquely responsible for the >disruption of low-dimensional topology in 4 dimensions. Where as you may know there >are uncountably many smooth structures on R^4, while every other R^n has just one. -----

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[math-fun] Quotients of Lie groups by discrete subgroups
by Dan Asimov 17 Dec '18

17 Dec '18

I don't know much about this, but it's so interesting that I thought I'd share it with math-fun. The rotation group of 3-space R^3, called SO(3), is homeomorphic to P^3, (real) projective 3-space. Its unique invariant metric, up to scaling, is that of P^3 induced from P^3 = S^3 / (x ~ -x) which just means this metric on P^3 is inherited from the 3-sphere S^3, and in fact is locally isometric to it. Now suppose H is a finite subgroup of SO(3). Then H is known to be isomorphic to one of the following: * a cyclic group C_n of order n, * a dihedral group of order 2n, * the tetrahedral group T of order 12 (also known as A_4), * the octahedral group O of order 24, * the icosahedral group I of order 60 (also known as A_5). Since SO(3) is a Lie group*, any finite subgroup G of SO(3) gives rise to the left coset space SO(3) / H = {g H | g in SO(3)} whose elements by this definition are indeed copies of the subgroup K, translated on the left by some element g of SO(3). Unless H is a normal subgroup (the only nontrivial one is {-I, I}), this is not a group. But suppose H is a discrete subgroup of SO(3) with the property that its commutator subgroup is the whole group: [H, H] = H. The word for this is: H is a "perfect" group. Then interesting things happen! The coset space SO(3) / H is what's called a homology sphere: A 3-dimensional manifold with the same homology groups as the 3-sphere S^3 (H_0(S^3) = H_3(S^3) = Z and H_1(S^3) = H_2(S^3) = 0). And if you take H = I, promoting it by a mere one place in the alphabet, then SO(3) / H = SO(3) / I is a homology sphere whose fundamental group is the binary icosahedral group 2I, whose order is 120. Not surprisingly we can get the same coset space by factoring the double cover of SO(3) = P^3 out by the double cover of I, namely 2I: SO(3) / I = S^3 / (2I). Strikingly, 2I is the *only* subgroup of S^3 for which SO(3) is a homology 3-sphere. And there is not enough space in this forum to go into detail, but the so-called "Poincaré dodecahedral space" S^3 / (2I), seems to be uniquely responsible for the disruption of low-dimensional topology in 4 dimensions. Where as you may know there are uncountably many smooth structures on R^4, while every other R^n has just one. —Dan ————— * A Lie (pron. "LEE") group is a manifold M that is also a group having continuous multiplication M x M —> M via (x, y) |—> xy, and also continuous inversion M —> M via x |—> x^(-1).

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[math-fun] coded message
by Cris Moore 17 Dec '18

17 Dec '18

My daughter found a t-shirt with this on it (I hope the color comes through): 1/5 2/5 1/5 1/5 Hint: Dr. Seuss. - Cris

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Re: [math-fun] coded message
by Dan Asimov 17 Dec '18

17 Dec '18

Duh. How about "One point-two, two point-two, red point-two, blue point-two" ? —Dan I wrote: ----- All I can see is ASCII ----- 1/5 2/5 1/5 1/5 ----- But based on the hint that it might be in color, plus Dr. Seuss, I will guess that it's One __, Two __, Red __, Blue __. Other than that, ??? -----

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Re: [math-fun] coded message
by Dan Asimov 17 Dec '18

17 Dec '18

All I can see is ASCII ----- 1/5 2/5 1/5 1/5 ----- But based on the hint that it might be in color, plus Dr. Seuss, I will guess that it's One __, Two __, Red __, Blue __. Other than that, ??? —Dan Cris Moore wrote: ----- My daughter found a t-shirt with this on it (I hope the color comes through): 1/5 2/5 1/5 1/5 Hint: Dr. Seuss. -----

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[math-fun] coded message
by Cris Moore 17 Dec '18

17 Dec '18

Ah, I see the mailing list sticks to plain text. So here’s a link: https://www.teepublic.com/t-shirt/699403-one-fifth-two-fifth-red-fifth-blue… <https://www.teepublic.com/t-shirt/699403-one-fifth-two-fifth-red-fifth-blue…> - Cris _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] sin(x/2)
by Bill Gosper 17 Dec '18

17 Dec '18

Trig books give (c104) HALFANGLES(SIN(X/2)); sqrt(1 - cos(x)) (d104) ---------------- sqrt(2) Mathematica doesn't even bother. Drawbacks: It's in terms of cos instead of sin, and numerically rancid for small x. I like Plot[Sin[x/2] - Sin[x]/(Sqrt[1 - Sin[x]] + Sqrt[1 + Sin[x]]), {x, -1.6, 1.6}], which misleadingly portrays floating point noise until Plot[Sin[x/2] - Sin@x/(√(1 - Sin@x) + √(1 + Sin@x)), {x, -1.6, 1.6},PlotRange -> All] reveals __________________________________________/ / You can actually paste the above Plot commands into https://lab.open.wolframcloud.com/objects/wpl/GetStarted.nb to see them. —rwg

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