December 2013 - mathfun@mailman.xmission.com

[math-fun] There must be something wrong with Mathematica's Round function--
by Bill Gosper 06 Dec '13

06 Dec '13

It has no effect on this 4th Grade homework problem: "Estimate: Workers in Italy receive an average of 42 vacation days per year. Workers in the United States receive an average of 13 vacation days per year. Explain how to use rounding to estimate how many more vacation days workers in Italy receive each year." --rwg Another nice one: "Write 47,000 in expanded form." I suggested forty-seven Ms. This is a private school. Everybody says parTISSiple.

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[math-fun] quantum entanglement = "wormholes"??
by Warren D Smith 06 Dec '13

06 Dec '13

http://web.mit.edu/newsoffice/2013/you-cant-get-entangled-without-a-wormhol… This is the craziest thing I've seen for a while. Could something like this be true? The old "Einstein Podolsky Rosen paradoxes" about quantum-entangled far-distant particles and measurement apparently "transmitting information faster than light" would be resolved in an insane way by proposing the two particles are NOT way far-distant preventing "information transmission"... since they are actually nearby in the true metric because linked by a "wormhole." Some papers underlying this: Kristan Jensen and Andreas Karch: Holographic Dual of an Einstein-Podolsky-Rosen Pair has a Wormhole, Phys. Rev. Lett. 111 (Nov 2013) 211602 http://arxiv.org/abs/1307.1132 Julian Sonner: Holographic Schwinger Effect and the Geometry of Entanglement, Phys. Rev. Lett. 111 (Nov 2013) 211603 http://arxiv.org/abs/1307.6850 Juan Maldacena, Leonard Susskind: Cool horizons for entangled black holes, http://arxiv.org/abs/1306.0533 H. Nikolic: Can a wormhole be interpreted as an EPR pair? http://arxiv.org/abs/1307.1604 Hrant Gharibyan, Robert F. Penna: Are entangled particles connected by wormholes? Support for the ER=EPR conjecture from entropy inequalities, http://arxiv.org/abs/1308.0289 Mariano Chernicoff, Alberto Guijosa, Juan F. Pedraza: Holographic EPR Pairs, Wormholes and Radiation, J.High Energy Phys. 1310 (2013) 211 http://arxiv.org/abs/1308.3695

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Re: [math-fun] Fermat abundance
by Bill Gosper 06 Dec '13

06 Dec '13

rcs> Your solution comes from (3^64 + i)/(1+i). > Whenever X = 2N+1 is odd, (X+i)/(1+i) generates the > decomposition (X2 + 1)/2 = 2N2 + 2N + 1 = (N+1)^2 + N^2. > > The 128 solutions: Reduce is obviously counting all orders > and all sign combinations as different. In this scheme, > 5 = 4+1 = 2^2+1^2 has 8 solutions as a2+b2. Right. To just count them: In[16]:= SquaresR[2, (3^128 + 1)/2] Out[16]= 128 > Normally, we'd > say that 5 splits as the sum of two squares in 1 way. > Right. To get just those, In[17]:= Length[PowersRepresentations[(3^128 + 1)/2, 2, 2]] Out[17]= 16 --rwg (As always, docfiend NeilB is showing me these things.) > Now 3^128+1 = > 2 257 275201 138424618868737 3913786281514524929 153849834853910661121 > > Ignoring the 2, each of these primes is 256K+1, and hence 4J+1, and > splits (uniquely) into two squares. E.g. 257 = 16^2 + 1^2. > Each additional 4J+1 prime doubles the number of solutions, as we > may choose either sign for the imaginary terms in the product > (16 +- i) (524 +- 25i) (10663711 +- 4970904i) > (1503205840 +- 1286140927i) (11066340060 +- 5602316711i). > > --- > Slightly related: I once went looking for Fermatish bases N, > so that N+1, N2+1, N4+1, N8+1, &c are all prime. > For N=2: 2^2^K + 1 is prime for K=0-4, but composite when K=5. > For N=6: 7, 37, and 1297 are prime, but 6^8+1 = 17*98801. > I found a few N which worked for K=0-5, but no K=0-6 primes. > N = 7072833120, 9736020616, and 12852419340 work for K=0...5, > but generate composites when K=6. > Standard conjectures assert that any finite set of Ks should have > an infinite set of prime-generating bases N. Gee, and you can't even find six in a row? Tsk. > > Rich > > ------ > Quoting Bill Gosper <billgosper(a)gmail.com>: > >> Reduce[(3^128 + 1)/2 == a^2 + b^2, {b, a}, Integers] >> finds 128 solutions, including >> (b == 1716841910146256242328924544640 && >> a == 1716841910146256242328924544641) >> --rwg

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[math-fun] Fermat abundance
by Bill Gosper 06 Dec '13

06 Dec '13

Reduce[(3^128 + 1)/2 == a^2 + b^2, {b, a}, Integers] finds 128 solutions, including (b == 1716841910146256242328924544640 && a == 1716841910146256242328924544641) --rwg

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[math-fun] hypocycloids
by Bill Gosper 06 Dec '13

06 Dec '13

Jon Ziegler points out some nice stuff @ http://johncarlosbaez.wordpress.com/2013/12/03/rolling-hypocycloids ,<http://johncarlosbaez.wordpress.com/2013/12/03/rolling-hypocycloids> to which I add Julian's http://gosper.org/sidereal.gif (the numerals record full rotations) and http://gosper.org/tri-penta.gif , showing how the quadricuspid hypocycloid is a trisectrix and the epicycloid is a pentasectrix. The former construction produces all three angular solutions, plus five other angles for which we've never found a purpose. The epicycloid produces all five solutions plus three "superfluons". Both animations are supplanted by Manipulators in http://gosper.org/tri-pentasect.cdf , which includes the source code and some explanatory text. It might be instructive to roll hypocycloids on epicycloids, and vice versa. --rwg <http://johncarlosbaez.wordpress.com/2013/12/03/rolling-hypocycloids>

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[math-fun] Quaternion GCD and canonical associates
by Fred Lunnon 02 Dec '13

02 Dec '13

A earlier comment observed that GCD is defined only up to unit association, which led me to reconsider the problem of defining a canonical representative among the (left or right) unit associates of a quaternion. This is not the sort of question which would normally concern a mathematician, who is liable to snort that one should simply deal with the class of associates in its own right. Nor is it likely to worry a computer programmer, for whom an adequate solution comprises sorting the set and extracting its first member. [It might though come within the ambit of a maverick like Don Knuth; but as far as I am aware, he has not concerned himself with quaternions to date.] Dealing with classes rather than individual quaternions is certainly feasible in mainstream computer algebra systems, though somewhat clumsier to program and slower to execute. Sorting is dependent on the convention adopted to order quadruplets, leading to incompatible results between multiple systems. For Q = w + x i + y j + z k a Lipschitz integer (say), a representative may be designated based on the fact that there is in general a unique member with 0 < w > |x|,|y|,|z| ; or alternatively with x,y,z > 0 . For both definitions, further arrangements must be made to deliver a unique result in subcases where the inequalities degenerate to equality. While such straightforward algorithms are adequate for integers, they are less satisfactory for real components, on the grounds of discontinuity in the presence of small rounding errors when component values are approximately equal. As a result, it becomes complicated to decide whether two such classes are approximately equal. I attempt to resolve this difficulty as follows. Define the "separation" of component to be the minimum difference between its absolute value and that of any other component. Then the canonical associate has real part equal to the largest component having maximum separation. This associate is unique unless (a) that component vanishes, in which case select the associate for which the largest component with submaximum separation also is positive; or (b) all components have equal size and separation zero, in which case select (1 + i + j + k)|w| or its conjugate. Both subcases deliver a unique result unless the quaternion vanishes, when there is anyway only 1 associate instead of the usual 8. Examples (canonical right-associates): Q = 3 i - 7 j - 6 k , separation 3, 3, 1, 1 , Q (-i) = 3 + 6 j - 7 k general case; Q = 3 i - 4 j - 6 k , separation 3, 1, 1, 2 , Q (-1) = -3 i + 4 j + 6 k subcase (a); Q = 1 - i + j - k , separation 0, 0, 0, 0 , Q k = 1 + i + j + k subcase (b). Subcase (b) remains the focus of a weakness with respect to continuity, since eg. Q = (1 + i + j + k) + E where ||E|| is small will inevitably be re-equipped with spurious random "canonical" signs depending on the errors E . There seems no simple corrective available, short of specifying the error bound as a parameter of the separation function; in which case one might as well adopt a less elaborate scheme as for integers. Fred Lunnon

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[math-fun] GCD[1,π]
by Bill Gosper 02 Dec '13

02 Dec '13

OK, I give up. Especially since anyone who agrees with me can just In[895]:= Unprotect[GCD]; GCD[a_, b_] /; a/b ∉ Rationals = 0; In[896]:= GCD[1/6, 1/9] Out[896]= 1/18 In[897]:= {GCD[π, 1], GCD[1, π]} Out[897]= {0, 0} In[908]:= GCD[E,π] Out[908]= GCD[E, π] In[909]:= GCD[Sqrt[2], π] Out[909]= 0 Mma is surprisingly accommodating this way. E.g., Unprotect[Numerator, Denominator]; Numerator[ComplexInfinity] = 1; Denominator[ComplexInfinity] = 0; Off[Power::infy] Then if In[903]:= Clear[mediant]; mediant[a__] := Divide @@ (Plus @@ # & /@ {Numerator[{a}], Denominator[{a}]}) E.g., In[904]:= mediant[a/b, c/d, e/f] Out[904]= (a + c + e)/(b + d + f) you can In[907]:= mediant[0, 1/0] Out[907]= 1 which is acutally useful for Farey interpolation. --rwg

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Re: [math-fun] Eccentricity of hyperbolic path of deflected light
by Adam P. Goucher 01 Dec '13

01 Dec '13

>From the first sesquipage of these revision notes: http://cp4space.files.wordpress.com/2013/04/kepler.pdf you can derive the orbit equation, namely: r = (L^2/GMm^2) / (1 + epsilon cos(theta)) Now consider the perihelion and aphelion to get simultaneous equations for epsilon. At these two points, the velocity is entirely tangential so is given by: v = L/mr and therefore the kinetic energy is: T = L^2/2mr^2 and therefore: E = L^2/2mr^2 - GMm/r Those are sufficiently many equations to derive your result by routine algebraic manipulation. Sincerely, Adam P. Goucher http://cp4space.wordpress.com > ----- Original Message ----- > From: Mike Stay > Sent: 11/30/13 11:45 PM > To: math-fun > Subject: [math-fun] Eccentricity of hyperbolic path of deflected light > > I'm trying to follow the derivation in > http://arxiv.org/pdf/physics/0508030v4.pdf, but I have no idea where > equation 5 comes from: > ------ > For a light particle with mass m, total energy E and angular momentum > L (with respect to the Sun's center) and where the Sun has mass M, the > eccentricity epsilon is > > epsilon = sqrt(1 + (2 E L^2)/(G^2 m^3 M^2)) (5) > ------ > > I could probably get some of that by dimensional analysis, but > certainly not all of it. How would one derive this? > -- > Mike Stay - metaweta(a)gmail.com > http://www.cs.auckland.ac.nz/~mike > http://reperiendi.wordpress.com > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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