November 2015 - mathfun@mailman.xmission.com

[math-fun] David Wilson's 1-D gun problem, cropping up again and again
by Michael Kleber 28 Nov '15

28 Nov '15

In case anyone is tracing the instances of it cropping up: the "1-dimensional gun" model, which David Wilson first posted about here in June of 2012, has come up again. I think this is among the best open problems ever discussed here. Recall the model: a gun at x=0 fires a bullet along the +x axis at time t=0,1,2,3,... with velocity chosen uniformly at random from [0,1]. If two bullets meet, then they annihilate each other. David's original question was whether the speed of the rightmost bullet tends to 1 over time, but given this overall model there are lots of interesting questions to ask. IBM Research's "Ponder This" from May 2014 asked about the finite case, specifically the probability that if you fire 20 bullets, they all annihilate: https://www.research.ibm.com/haifa/ponderthis/challenges/May2014.html The "solution" was posted there, claiming a formula for all even n, which appears to be empirically correct but was not presented with a correct proof. On Quora, someone asked, maybe in December 2014 (?), for the infinite-bullets case, for "probability that a bullet would reach infinity", without putting any work into defining what that means: https://www.quora.com/A-gun-shoots-bullets-towards-infinity-at-a-constant-r… Then on Reddit earlier this month, someone asked for "the probability that at least one bullet escapes to infinity if you fire n bullets", with also a nod at the n->infinity case: https://www.reddit.com/r/mathriddles/comments/3sa4ci/colliding_bullets/ The comments there include an interesting observation or two in among lots of junk. Two days later it hopped to Math StackExchange http://math.stackexchange.com/questions/1526292/colliding-bullets The top comment there offers the same formula as in the IBM soluiton for the finite-n probability of all bullets annihilating, but again without a solid proof. --Michael -- Forewarned is worth an octopus in the bush.

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[math-fun] Fibonacci polynomial puzzle
by Fred Lunnon 28 Nov '15

28 Nov '15

Define Fibonacci polynomials F_n(Y) for integer n , variable Y via F_0(Y) = 0 , F_1(Y) = F_{-1}(Y) = 1 , F_{n+2}(Y) - Y F_{n+1}(Y) - F_n(Y) = 0 . See https://en.wikipedia.org/wiki/Fibonacci_polynomials http://mathworld.wolfram.com/FibonacciPolynomial.html --- though compare alternative (marginally less tractable in practice) http://oeis.org/wiki/Fibonacci_polynomials . The polynomial sequence shares elementary properties with the Fibonacci number sequence F_n : in particular strong divisibility, established via essentially the same argument --- see (engagingly pedestrian) https://proofwiki.org/wiki/Divisibility_of_Fibonacci_Numbers [Whence this property may be inferred for other sequences --- 14 and counting --- in OEIS. Strong divisibility, that is, not engaging pedestrianism.] I hope to post a more comprehensive account in a few days, explaining the how these matters bear on eccentric gear trains; meanwhile the following modest though positive advance afforded me such satisfaction, that I am unable to resist a little waving it about. **Puzzle** Show that there is a constant t such that for all odd prime p , F_2p(Y) = s Y F_p(Y) F_p(Z) ; where s = (-1)^((p-1)/2) and Z^2 = -(Y^2 + t) . Evaluate t , and deduce that F_p(Y) is irreducible. For example when p = 7 , F_14(Y) = Y (Y^6+5*Y^4+6*Y^2+1) (Y^6+7*Y^4+14*Y^2+7) , irreducibly. Fred Lunnon ____________________________________________________________ n F_n(Y) 0 0 1 1 2 Y 3 Y^2 + 1 4 Y^3 + 2*Y 5 Y^4 + 3*Y^2 + 1 6 Y^5 + 4*Y^3 + 3*Y 7 Y^6 + 5*Y^4 + 6*Y^2 + 1 8 Y^7 + 6*Y^5 + 10*Y^3 + 4*Y 9 Y^8 + 7*Y^6 + 15*Y^4 + 10*Y^2 + 1 10 Y^9 + 8*Y^7 + 21*Y^5 + 20*Y^3 + 5*Y 11 Y^10 + 9*Y^8 + 28*Y^6 + 35*Y^4 + 15*Y^2 + 1 12 Y^11 + 10*Y^9 + 36*Y^7 + 56*Y^5 + 35*Y^3 + 6*Y 13 Y^12 + 11*Y^10 + 45*Y^8 + 84*Y^6 + 70*Y^4 + 21*Y^2 + 1 14 Y^13 + 12*Y^11 + 55*Y^9 + 120*Y^7 + 126*Y^5 + 56*Y^3 + 7*Y 15 Y^14 + 13*Y^12 + 66*Y^10 + 165*Y^8 + 210*Y^6 + 126*Y^4 + 28*Y^2 + 1 16 Y^15 + 14*Y^13 + 78*Y^11 + 220*Y^9 + 330*Y^7 + 252*Y^5 + 84*Y^3 + 8*Y ____________________________________________________________

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[math-fun] E = something K + something/K
by Bill Gosper 28 Nov '15

28 Nov '15

Specifically, EllipticE[m] == EllipticK[m] + (Pi^2 q Inactive[D][Log[EllipticTheta[4, 0, q]], q])/EllipticK[m] or EllipticE[m] == EllipticK[m] - (I Pi (Derivative[1][DedekindEta][-((I Log[q])/(2 Pi))]/ DedekindEta[-((I Log[q])/(2 Pi))] - Derivative[1][DedekindEta][-((I Log[q^2])/(2 Pi))]/ DedekindEta[-((I Log[q^2])/(2 Pi))]))/EllipticK[m] E.g., associated with the 5th singular value, Out[332]= EllipticE[1/2 - Sqrt[-2 + Sqrt[5]]] == EllipticK[1/2 - Sqrt[-2 + Sqrt[5]]] + ( 160 Sqrt[5] \[Pi]^2 - 5^(1/4) (-4 + Sqrt[10 (1 + Sqrt[5])]) Gamma[1/20]^2 Gamma[9/ 20]^2)/(3200 \[Pi] EllipticK[1/2 - Sqrt[-2 + Sqrt[5]]]) The harmless little quartic (actually octic) surd required enhanced interrogation. In[337]:= Log[EllipticNomeQ[1/2 - Sqrt[-2 + Sqrt[5]]]]/\[Pi] // FunctionExpand Out[337]= -(EllipticK[1/2 + Sqrt[-2 + Sqrt[5]]]/ EllipticK[1/2 - Sqrt[-2 + Sqrt[5]]]) In[339]:= N[%^2] Out[339]= 5. Express K as Gammas via etas: In[344]:= EllipticK[(16*DedekindEta[\[Tau]]^8*DedekindEta[4*\[Tau]]^16)/ DedekindEta[2*\[Tau]]^24] == (Pi*DedekindEta[2*\[Tau]]^10)/ (2*DedekindEta[\[Tau]]^4* DedekindEta[4*\[Tau]]^4) /. \[Tau] -> Sqrt[-5]/2 Out[344]= EllipticK[( 16 DedekindEta[(I Sqrt[5])/2]^8 DedekindEta[2 I Sqrt[5]]^16)/ DedekindEta[I Sqrt[5]]^24] == (\[Pi] DedekindEta[I Sqrt[5]]^10)/( 2 DedekindEta[(I Sqrt[5])/2]^4 DedekindEta[2 I Sqrt[5]]^4) Are these the droids we're looking for? In[345]:= RootApproximant[N[%[[1, 1]], 69]] Out[345]= Root[1 - 72 #1 + 88 #1^2 - 32 #1^3 + 16 #1^4 &, 1] In[347]:= ToRadicals[%%] // FullSimplify Out[347]= 1/2 - Sqrt[-2 + Sqrt[5]] Yup. Replace the etas from the table. In[350]:= %344 /. etavals Out[350]= EllipticK[((1 + Sqrt[5])^9 (5 - Sqrt[5] - 2 Sqrt[-110 + 50 Sqrt[5]])^2 (1 + 2/GoldenRatio^(3/2)))/( 20480 GoldenRatio)] == ( 2 2^(3/4) Sqrt[ GoldenRatio/((5 - Sqrt[5] - 2 Sqrt[-110 + 50 Sqrt[5]]) (1 + 2/ GoldenRatio^(3/2)) \[Pi])] Gamma[1/20] Gamma[9/20])/( 5^(3/8) (1 + Sqrt[5])^(15/4)) In[351]:= FullSimplify[Rule @@ %] Out[351]= EllipticK[1/2 - Sqrt[-2 + Sqrt[5]]] -> (Gamma[1/20] Gamma[9/20])/( 4 (5 (85 + 38 Sqrt[5]))^(1/4) Sqrt[(25 - 11 Sqrt[5]) \[Pi]]) Dispel yucky surds. In[387]:= Denominator[%351[[2]]]/Sqrt[\[Pi]] Out[387]= 4 Sqrt[25 - 11 Sqrt[5]] (5 (85 + 38 Sqrt[5]))^(1/4) In[388]:= Log[Prepend[{2, 5, GoldenRatio}, %]] Out[388]= {Log[4 Sqrt[25 - 11 Sqrt[5]] (5 (85 + 38 Sqrt[5]))^(1/4)], Log[2], Log[5], Log[GoldenRatio]} In[389]:= %.FindIntegerNullVector[%] Out[389]= -20 Log[2] - 5 Log[5] + 8 Log[4 Sqrt[25 - 11 Sqrt[5]] (5 (85 + 38 Sqrt[5]))^(1/4)] + 2 Log[GoldenRatio] In[401]:= FullSimplify /@ (%351[[1]] -> %351[[2]]*E^(%389/8) /. GoldenRatio -> \[Phi]) Out[401]= EllipticK[1/2 - Sqrt[-2 + Sqrt[5]]] -> ( 5^(3/8) \[Phi]^(1/4) Gamma[9/20] Gamma[21/20])/Sqrt[2 \[Pi]] This must be my yucky day! Back to E: In[402]:= %332 /. % /. \[Phi] -> GoldenRatio Out[402]= EllipticE[1/2 - Sqrt[-2 + Sqrt[5]]] == ( 160 Sqrt[5] \[Pi]^2 - 5^(1/4) (-4 + Sqrt[10 (1 + Sqrt[5])]) Gamma[1/20]^2 Gamma[9/ 20]^2)/(1600 5^(3/8) GoldenRatio^(1/4) Sqrt[2 \[Pi]] Gamma[9/20] Gamma[21/20]) + ( 5^(3/8) GoldenRatio^(1/4) Gamma[9/20] Gamma[21/20])/Sqrt[2 \[Pi]] In[403]:= FullSimplify /@ % Out[403]= EllipticE[1/2 - Sqrt[-2 + Sqrt[5]]] == ( 5^(1/4) (2 Sqrt[10] + 5 Sqrt[1 + Sqrt[5]]) Gamma[1/20] Gamma[9/ 20]^2 + (20 Sqrt[2] \[Pi]^2)/Gamma[21/20])/( 40 2^(3/4) 5^(7/8) (1 + Sqrt[5])^(1/4) Sqrt[\[Pi]] Gamma[9/20]) which is the best I could do. --rwg

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[math-fun] KIC 8462852
by Warren D Smith 27 Nov '15

27 Nov '15

http://iopscience.iop.org/article/10.1088/2041-8205/814/1/L15 KIC 8462852: THE INFRARED FLUX Massimo Marengo, Alan Hulsebus, and Sarah Willis 2015 November 19 The Astrophysical Journal Letters, Volume 814, Number 1 says it likes the comet-swarm theory. What about alien megastructures theory? Not even mentioned, at least not in the abstract,

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[math-fun] A257007 in three different contexts (?)
by Neil Sloane 26 Nov '15

26 Nov '15

Dear Math Fun: The sequence https://oeis.org/A257007 just arose in two possibly different contexts. The sequence appears to be the same as the function Phi(n) in Theorem 4 of this paper: P. Kleban and A. Özlük. A farey fraction spin chain. Communications in mathematical physics, 203(3):635-647, 1999. Also http://arxiv.org/pdf/cond-mat/9808182.pdf. I discovered this when I was looking at a recent thesis: M. P. Technau, The Calkin-Wilf Tree and a Trace Condition, Master's Thesis, Julius-Maximilian’s University of Würzburg, Faculty for Mathematics and Computer Science, 2015; http://www.mathematik.uni-wuerzburg.de/~steuding/technau.pdf On page 14 there is a triangle of numbers. The first column of the triangle seems to match A257007, and the row sums in the triangle are twice the first column. I created three new sequences, A264597, A264598, A264599 from this paper. It would be nice to know for certain that it is the same sequence (A257007) in all three cases! Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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[math-fun] Mathematica nth root of unity question
by Dan Asimov 25 Nov '15

25 Nov '15

I learned (or was reminded) only recently that the real and imaginary parts of any nth root of unity can each be expressed in terms of (an iterated rational expression in) radicals of integers (using only roots lower than the nth). 1) Is there an easy proof of this? 2) How do I get Mathematica to output these expressions? (Everything I've tried so far gets only the dumbest error messages.) For instance, Wikipedia lists the real part of (exp(2pi*i/7)) as follows: cos(2pi/7) = (-1 + ((7 + 21*sqrt(-3)/2)^(1/3) + ((7 - 21*sqrt(-3)/2)^(1/3))/6 . I'm confident that several of the 2^2 3^2 = 36 choices of roots give the correct answer, but: which ones? This may be easy to guess, but how about cos(6pi/23) ? 3) There ought to be a notation that doesn't leave these choices unspecified. —Dan

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[math-fun] Asimov re roots of unity
by Warren D Smith 25 Nov '15

25 Nov '15

Dan Asimov: I learned (or was reminded) only recently that the real and imaginary parts of any nth root of unity can each be expressed in terms of (an iterated rational expression in) radicals of integers (using only roots lower than the nth). For instance, Wikipedia lists the real part of (exp(2pi*i/7)) as follows: cos(2pi/7) = (-1 + ((7 + 21*sqrt(-3)/2)^(1/3) + ((7 - 21*sqrt(-3)/2)^(1/3))/6 --WDS: Gauss famously showed that the nth roots of unity are constructible iff the odd part of n is a product of distinct Fermat primes. The truth of Asimov's claim follows IF we know that the cyclotomic polynomials always have solvable Galois group. I believe the Galois group for them always arises from (Z mod m), hence is abelian, hence solvable, thus proving Asimov's claim. https://en.wikipedia.org/wiki/Cyclotomic_polynomial http://www.math.niu.edu/~beachy/aaol/galois.html#cyclotomic https://en.wikipedia.org/wiki/Solvable_group -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] laser cool liquids?
by Warren D Smith 25 Nov '15

25 Nov '15

The idea that this method could be of practical interest sounds like bullshit hype. Don't be another sucker for bogus science hype (born every minute). Because you need like 1 blue photon per molecule, and if even a tiny fraction of those photons, like 1%, get wasted/lost, that'd be hugely inefficient. And.. sure enough, looking at the paper they are using megawatt per cm^2 irradiance levels for their "refrigerator." My god. > It's true that lasers are often used as heating implements, but a less > common use is the cooling of gases and crystals. Now for the first time, a > team of researchers has now used an infrared laser to cool liquid by about > 20 degrees Celsius (36 degrees Fahrenheit). > > The study, led by Paden Roder of the University of Washington, has been > published in the Proceedings of the National Academy of Sciences, and can be > viewed on arXiv. > > http://arxiv.org/abs/1503.07265 -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] proportional representation
by Warren D Smith 24 Nov '15

24 Nov '15

V voters score each of C candidates on an 0-to-1 real scale. Those are the ballots. The goal is to use them to elect W winners, 0<W<C. The question is how. I've been working on this for some time and every system I know of seems unsatisfactory in some easily identifiable way or another. Maybe one of you will come up with a new idea that will be enough to break the logjam; or perhaps it just is impossible to satisfy all of my desiderata. Desiderata include * PR: if all voters & candidates are "colored," each voter scores same-color candidates 1 and other-color candidates 0, then the parliament should have same color composition as electorate (up to unavoidable errors caused by integer roundoff, or if too few candidates of some color run) * Stronger PR: if there also are uncolored candidates, who each receive scores from voters that depend only on the candidate, not on the voter, then the colored part of the parliament should have same color-composition as electorate. * Monotonicity: if voters raise a candidate's score, that should help him win. In particular, if parliament A gets the same scores or better than parliament B in all cases, with at least some "better," then B should not be the winner-set. * Simplicity of description of the system. * Single-winner case (W=1) should just be "highest average score wins." * Scaling invariance: if all input scores are multiplied by some C>0, the winning parliament should be unaltered. * I also have a notion of "continuity" that I want, but won't formulate it here. I've got systems that satisfy any of these desiderata, but none satisfying all of them simultaneously. The best I've been able to do so far is satisfy all except "stronger PR." I can prove any system that does all that must solve NP-hard problems. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] "OneSixOne" palindrome approx to phi
by Henry Baker 24 Nov '15

24 Nov '15

At 01:38 PM 11/23/2015, Henry Baker wrote: >But why stop with 161; cf(phi) is also a palindrome. > >Harmony and balance will cost you 2400 euros. > >Smaller amounts of harmony and balance are denoted by "Tau" and cost you only 900 euros. > >So, for which bases is phi a palindrome for the first 3 digits? If phi is truncated to 3 base-b digits, I get palindromes for bases b=4, 10, 26, etc.: (the second number is ln(b), so it appears that the sequence is growing logarithmically) b= 4 1.386294361119891 b= 10 2.302585092994046 b= 26 3.258096538021482 b= 68 4.219507705176107 b= 178 5.181783550292085 b= 466 6.144185634125646 b= 1220 7.106606137727303 b= 3194 8.069029328774958 b= 8362 9.03145291191651 b= 21892 9.993876552263808 b= 57314 10.95630020095731

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