August 2015 - mathfun@mailman.xmission.com

[math-fun] black holes could trigger collapse of unive
by Henry Baker 04 Aug '15

04 Aug '15

FYI -- Tiny black holes could trigger collapse of universeÂ-except that they don't http://news.sciencemag.org/europe/2015/08/tiny-black-holes-could-trigger-co… http://arxiv.org/abs/1501.04937

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[math-fun] A New Type of Pentagonal Tiling
by Hans Havermann 04 Aug '15

04 Aug '15

Casey Mann, Jennifer McLoud, and David von Derau have come up with what they say is a new (15th) pentagonal tile that admits a periodic tiling of the plane: https://twitter.com/gelada/status/626810879435083776

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Re: [math-fun] nuclide map
by rcs＠xmission.com 03 Aug '15

03 Aug '15

<from Ed Fredkin, last week. referring to Gosper's request for a nuclide chart --Rich> ---- Date: Tue, 28 Jul 2015 15:46:01 +0000 Subject: Re: nuclide map From: Edward Fredkin <ed(a)fredkin.com> Yes! Look at http://www.nndc.bnl.gov/chart/help/index.jsp Ed

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[math-fun] What does a comet look like?
by Warren D Smith 03 Aug '15

03 Aug '15

http://www.esa.int/Our_Activities/Space_Science/Rosetta/Highlights/Landing_… gives pictures from far away to close up (the Philae lander largely failed, but they were able to obtain these) -- Warren D. Smith http://RangeVoting.org

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Re: [math-fun] Quadrangles
by Walter Trump 03 Aug '15

03 Aug '15

Assume d-c = c-b = b-a = d For Adam's 6 problems the smallest side has following length: y = -2d y = -d x = -4/3d y = 3d x = 0 x = 0 Therefore there is only one family of quadrangles. The one mentioned by Eric. Walter > Date: Sun, 2 Aug 2015 22:53:36 +0200 > From: "Adam P. Goucher" <apgoucher(a)gmx.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] Quadrangles > Message-ID: > > <trinity-6082781c-d98c-470a-8d37-d3953acff651-1438548816659@3c app-mailcom-bs03> > > Content-Type: text/plain; charset=UTF-8 > > Suppose a > b > c > d without loss of generality. If the > right-angles were opposite each other, then we would have w^2 > + x^2 = y^2 + z^2, where {w,x,y,z} = {a,b,c,d} in some order. > It is obvious that the only possibility is a^2 + d^2 = b^2 + > c^2 (lest we violate some of the inequalities). > > Add to this the assumption that the four sides are in > arithmetic progression, we get a + d = b + c, thus a^2 + d^2 > + 2ad = b^2 + c^2 > + 2bc, thus ad = bc. This implies {a, d} and {b, c} are both valid > solution-pairs to some quadratic equation, contradicting the > distinct-ness of {a,b,c,d}. > > So we can assume the two right-angles are adjacent, and that > the four sides are w,x,y,z (with z slanted). Then the > relevant equation > is: > > z^2 = x^2 + (w - y)^2 (assuming wlog w > y) > > We also have z > x, obviously, so there are six possibilities > for the order (which are six algebraically distinct problems): > > z > x > w > y > z > w > x > y > z > w > y > x > w > z > x > y > w > z > y > x > w > y > z > x > > Now, being in an arithmetic progression gives two real > constraints, and the aforementioned one is a third > constraint, suggesting that the family of solutions has > dimension 1. This in turn suggests that up to similarity > there are finitely many solutions for each of the six problems. > > > Sincerely, > > > Adam P. Goucher > > > Sent: Sunday, August 02, 2015 at 8:11 PM > > From: "Eric Angelini" <Eric.Angelini(a)kntv.be> > > To: math-fun <math-fun(a)mailman.xmission.com> > > Subject: [math-fun] Quadrangles > > > > > > Hello MathFunsters, > > Are there quadrangles having: > > - two right angles > > - and their four sides in arithmetic > > progression? > > I know only of one family of such > > quadrangles -- the ones build > > on the pythagorian triplet 3,4,5. > > The simplest example has the > > shape of an U with a tilted roof: > > - the left vertical wall has size 3, > > - the horizontal ground has size 4, > > - the right vertical wall has size 6, > > - the tilted roof has size 5. > > > > Best, > > ?.

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[math-fun] Quadrangles
by Eric Angelini 02 Aug '15

02 Aug '15

Hello MathFunsters, Are there quadrangles having: - two right angles - and their four sides in arithmetic progression? I know only of one family of such quadrangles -- the ones build on the pythagorian triplet 3,4,5. The simplest example has the shape of an U with a tilted roof: - the left vertical wall has size 3, - the horizontal ground has size 4, - the right vertical wall has size 6, - the tilted roof has size 5. Best, É.

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[math-fun] two "trivia"
by Bill Gosper 02 Aug '15

02 Aug '15

A miscellaneous nontriangular matrix product: [ sqrt(%pi) ] [ 0 0 --------- ] [ 1 1 5 1 ] [ 2 3 ] oo [ 0 -- - ----- - - --- ] [ gamma (-) ] /===\ [ 16 2 4 8 j ] [ 4 ] | | [ 64 j ] [ ] | | [ ] = [ 2 3 ] | | [ 1 1 2 ] [ 4 gamma (-) ] j = 1 [ (- - ---) 0 1 ] [ 4 ] [ 2 4 j ] [ 0 0 ----------- ] [ ] [ 3/2 ] [ 0 0 1 ] [ %pi ] [ ] [ 0 0 1 ] (Does anybody know how to make these archaic ASCII displays in Mathematica, for our archaic math-fun server?) A260747, Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > lg(max(A(n),A(p),A(q))/15) 13, 21, 23, 26, 37, 39, 42, 46, 47, 52, 73, 74, 78, 81, 83, 84, 92, 94, 97, 99, Note the gap of 73 - 52 = 21 at position 10. There are apparently no larger gaps, and the next gap of 21 is at position 17651. --rwg

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[math-fun] BarnesG ("superfactorial") tuplication formula
by Bill Gosper 02 Aug '15

02 Aug '15

is not in gosper.org/facfun.pdf because I couldn't relieve Product[BarnesG[-(i/n) + z], {i, 0, -1 + n}] == E^(((-1 + n^2)*Zeta'[-1])/n)* n^(BernoulliB[2, 1 + n*(-2 + z)]/(2*n))* BarnesG[2 + n*(-2 + z)]^(1/n)* (n*(-2 + z))!^(2 - 1/n - z)* Product[(-2 - i/n + z)!^(-1 - i/n + z), {i, 0, -1 + n}] of the explicit Product. And it requires branch diddling for "inexact" z. (If lhs is positive, just Abs the rhs.) But it's still useful. --rwg

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[math-fun] Barn raising
by Bill Gosper 01 Aug '15

01 Aug '15

Mathematica should be able to get Product[(3^(-(1/2) + 3*n)*Gamma[2/21 + n]*Gamma[4/21 + n]*Gamma[5/7 + n])/ (2*Pi*Gamma[3*n]), {n, Infinity}] == (BarnesG[4/3]*BarnesG[5/3])/ (BarnesG[23/21]*BarnesG[25/21]*BarnesG[12/7]) In[885]:= N[Rule @@ %] During evaluation of In[885]:= NIntegrate::slwcon:Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >> During evaluation of In[885]:= NIntegrate::ncvb:NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in n near {n} = {7273.96592}. NIntegrate obtained -0.000408638628 and 8.024253968625528`*^-8 for the integral and error estimates. >> Out[885]= 0.9878943769692339 -> 0.9878939404389084 simply by expressing the finite product using BarnesG, then taking the limit using "Barnes' Stirling's formula". --rwg

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