February 2013 - mathfun@mailman.xmission.com

[math-fun] new mersenne prime
by Simon Plouffe 06 Feb '13

06 Feb '13

Hello, they have found a 48'th Mersenne prime _http://www.isthe.com/chongo/tech/math/digit/m57885161/prime-c.html_ 2^57885161 - 1 which has 17 000 000 digits, I think it was confirmed today. Best regards, Simon Plouffe

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[math-fun] A family of lattices including both FCC, BCC, and a new(?) lattice
by Warren Smith 05 Feb '13

05 Feb '13

The BCC lattice is, you take the points (x,y,z) with integer coordinates, and then also adjoin a copy of that translated by (1/2, 1/2, 1/2). That is, the vertices of the usual grid of unit cubes, plus points at all the cube centers. Now suppose we change all the "cubes" to aXaXc "bricks" where a*a*c=1 (to keep the point-density at 2). This is a 1-parameter family of "tetragonal" lattices. Some interesting particular cases of it are: BCC: arises when a=c=1. Sphere packing volume-density = pi*sqrt(3)/8 = 0.6801747619 #nearest neighbors = 8 FCC: arises when a=0.8908987180, c=1.259921049, c/a=sqrt(2)=1.414213562. Sphere packing volume-density = pi/sqrt(18) = 0.7404804898 #nearest neighbors = 12 New(?) Lattice: arises when a=1.069913194, c=0.8735804647, c/a=sqrt(3/2)=1.224744871. Sphere packing volume-density = 2*pi/9 = 0.6981317011 #nearest neighbors = 10 -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] lattice neighbors
by Warren Smith 05 Feb '13

05 Feb '13

Maybe Asimov's concept better described as "Voronoi face count" or "Delaunay triangulation graph adjacencies". BCC has 8 nearest and 6 second-nearest neighbors [at sqrt(4/3) times the distance], all 14 Voronoi [truncated octahedron]. FCC has 12 nearest and 6 second-nearest [at sqrt(2) times the distance], but voronoi has only 12 faces [rhombic dodecahedron]. NewLatt has 10 nearest and 4 second-nearest neighbors [at sqrt(3/2) times the distance; sorry, this square is not an integer, but if multiply square by 2, then integer; I should have said, when defining "integral" lattice, that all squared distances are integer multiples of SOMETHING, but not nec. of the shortest squared distance]. I think NewLatt's Voronoi has 14 faces, kind of a distorted version of the BCC Voronoi region. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Anti-aliasing photo detector arrays
by Henry Baker 04 Feb '13

04 Feb '13

Fujifilm has a square 6x6 RGB detector array pattern which is claimed to be better than the standard "Bayer" RGB pattern in eliminating Moire patterns: http://upload.wikimedia.org/wikipedia/en/thumb/f/ff/Xtranscolourfilter.svg/… http://en.wikipedia.org/wiki/Fujifilm_X-Pro1 One additional interesting feature of this array is the slightly different % of RGB: 20 G's, 8 R's, 8 B's (56%, 22%, 22%) The more traditional "Bayer" pattern has 2 G's, 1 R, 1 B (50%, 25%, 25%) http://en.wikipedia.org/wiki/Bayer_filter This Fujifilm pattern would probably be better for large electronic displays than for sensors, because the pixels are so obvious. Both Bayer & Fujifilm patterns can be extended with 2 different Greens: G1 & G2, which gives better coverage of the color spectrum. Question: wouldn't some of the aperiodic/quasicrystal tiling patterns work even better for these purposes?

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Re: [math-fun] important contribution to Indian mathematics
by Bill Gosper 03 Feb '13

03 Feb '13

ARGH! In planes parallel to two edges, we see ellipses of all possible eccentricities! Nice one. --rwg (I tried pasting in both rich and plain mode, and somebody destroys the formatting either way.) DanA>Proof? Yes. (Stare at (*).) --Dan On 2013-02-02, at 5:48 PM, Bill Gosper wrote: DanA> Actually RWG's samosa, and the locus of (*) below in [-1,1]^3, should be identical, both corresponding to all triples of cosines of 3 angles that add up to 2pi. But it seems that peculiarities of the way acos() is interpreted might mess things up. This is the case in the Mac Grapher utility, where only one face of the pillow is displayed for the equation (**) acos(x) + acos(y) + acos(z) = 2pi. PUZZLE: Determine whether any planar geometric circle lies on the locus of (*). --Dan I'll bet not. The surface contains four equilateral triangles. Resting it on a "face" and considering horizontal plane sections, we see this triangle deforming continuously toward infinitesimal circles at the rounded bottom and pointy top. I think other sections would lack symmetry. But we shouldn't forget Villarceau. You say "puzzle". Have you proof? On 2013-02-01, at 2:25 AM, Dan Asimov wrote: (since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that (*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0 whose locus fits RWG's samosa description to a T. Like I said, "This can't be new." In fact, both definitions lead to the same (intractable?) area integral, Integrate[ Sqrt[Sin[t]^2*Sin[u]^2 + (Sin[t]^2 + Sin[u]^2)*Sin[t + u]^2], {t, 0, Pi}, {u, 0, Pi}]== Integrate[ Sqrt[1 + ((-2 + x^2 + y^2)*(-y^2 - 2*x*y*Sqrt[(-1 + x^2)*(-1 + y^2)] + x^2*(-1 + 2*y^2)))/((-1 + x^2)*(-1 + y^2))], {x, -1, 1}, {y, -1, 1}] In[135]:= 4*N[List@@%] Out[135]= {31.4159, 31.4159} so 5 pi /2 remains unproven. Will this be another "Coxeter's Challenge Integral"? In[136]:= 2*Pi^2/15 == Integrate[ArcSec[t]*(2 + 1/Sqrt[t + 3])/Sqrt[t + 1]/(t + 2), {t, 1, 6}] In[137]:= N[%] Out[137]= True Except Coxeter had a geometric proof. We have only a numerical tease. Are there lots of these conjectural integral identities? --rwg

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Re: [math-fun] important contribution to Indian mathematics
by Bill Gosper 03 Feb '13

03 Feb '13

DanA> Actually RWG's samosa, and the locus of (*) below in [-1,1]^3, should be identical, both corresponding to all triples of cosines of 3 angles that add up to 2pi. But it seems that peculiarities of the way acos() is interpreted might mess things up. This is the case in the Mac Grapher utility, where only one face of the pillow is displayed for the equation (**) acos(x) + acos(y) + acos(z) = 2pi. PUZZLE: Determine whether any planar geometric circle lies on the locus of (*). --Dan I'll bet not. The surface contains four equilateral triangles. Resting it on a "face" and considering horizontal plane sections, we see this triangle deforming continuously toward infinitesimal circles at the rounded bottom and pointy top. I think other sections would lack symmetry. But we shouldn't forget Villarceau. You say "puzzle". Have you proof? On 2013-02-01, at 2:25 AM, Dan Asimov wrote: (since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that (*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0 whose locus fits RWG's samosa description to a T. Like I said, "This can't be new." In fact, both definitions lead to the same (intractable?) area integral, Integrate[ Sqrt[Sin[t]^2*Sin[u]^2 + (Sin[t]^2 + Sin[u]^2)*Sin[t + u]^2], {t, 0, Pi}, {u, 0, Pi}]== Integrate[ Sqrt[1 + ((-2 + x^2 + y^2)*(-y^2 - 2*x*y*Sqrt[(-1 + x^2)*(-1 + y^2)] + x^2*(-1 + 2*y^2)))/((-1 + x^2)*(-1 + y^2))], {x, -1, 1}, {y, -1, 1}] In[135]:= 4*N[List@@%] Out[135]= {31.4159, 31.4159} so 5 pi /2 remains unproven. Will this be another "Coxeter's Challenge Integral"? In[136]:= 2*Pi^2/15 == Integrate[ArcSec[t]*(2 + 1/Sqrt[t + 3])/Sqrt[t + 1]/(t + 2), {t, 1, 6}] In[137]:= N[%] Out[137]= True Except Coxeter had a geometric proof. We have only a numerical tease. Are there lots of these conjectural integral identities? --rwg

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[math-fun] Repeating decimal expansion
by Eric Angelini 02 Feb '13

02 Feb '13

Hello Math-Fun, à propos the table "Repeating decimal expansion", here: http://thestarman.pcministry.com/math/rec/RepeatDec.htm ... I read that: 1/2 = 0.5 1/4 = 0.25 1/5 = 0.2 ... but is this equivalent to: 1/2 = 0.499999999999999999999999999999... 1/4 = 0.249999999999999999999999999999... 1/5 = 0.199999999999999999999999999999... And what about: 1/3 = 0.6666666666666666666666666666666... ... re-written as: 1/3 = 0.6599999999999999999999999999999... 1/3 = 0.6666659999999999999999999999999... 1/3 = 0.6666666666666666666666599999999... I'm confused. Best, É.

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[math-fun] TJX markup/text formatter?
by Bill Gosper 02 Feb '13

02 Feb '13

quad> Hello math-fun, I guess Abelson sent a friend of mine a paper in TJX format. Or maybe the text formatter was called TJX? Does anyone know what this is? Any interesting history? Apparently it's an archaic format(ter) before my time. I can't find anything on Google about it. Thanks, Robert Smith This might be the PDP10 follow-on to TJ6: people.csail.mit.edu/bkph/AIM/AIM-164A-OCR-OPT.pdf --rwg (Experiencing a flicker of brain activity from Tastebuds' Zucchini Fish.)

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[math-fun] important contribution to Indian mathematics
by Bill Gosper 01 Feb '13

01 Feb '13

The equation of the samosa! ParametricPlot3D[{Cos[t], Cos[u], Cos[u + t]}, {t, 0, 2 π}, {u, 0, π}] (Rounded regular tetrahedron, four vertices, no edges, but contains the line segments joining the vertices.) Implicit equation: acos(z)=acos(x)+acos(y) Volume pi^2/2 Area (empirical) 5 pi, but I can't get Mma to do the integral. This can't be new. Does this surface have a legitimate name? Can anyone prove the area? --rwg

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[math-fun] What to conclude from physics experiment / voting
by Warren Smith 01 Feb '13

01 Feb '13

(True story, but abstracted.) There were three candidate scientific hypotheses named B, F, and H. It was possible to perform a pairwise comparison experiment to decide which hypothesis worked better to explain that experiment, BUT the situation was (somewhat strangely, but this was the situation) that you only got information about TWO among the candidates, and it decided which two, not you. Anyhow, after laboriously performing 19+26+9+8+22+10=94 experiments, the results were BH=19, HB=26, BF=9, FB=8, FH=22, HF=10 meaning B won 19 of the 45 H-versus-B comparisons, etc. So... which hypothesis is the right one, do you think, and what odds would you take in a bet?

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