June 2016 - mathfun@mailman.xmission.com

[math-fun] The sign of the Minkowski metric --- a smoking gun at last?
by Fred Lunnon 21 Jun '16

21 Jun '16

For as long as I have known not very much about physics , experts have disagreed over the sign of the Minkowski metric: apparently particle physicists prefer +++- , relativists ---+ [Wikipedia]. An attempt to resolve the issue via geometric algebra, combined with an exotic experimental investigation into neutrinoless double beta decay ultimately foundered [Ber01]; passim. [ To conflate a metric with a Clifford algebra is in any case dubious, since mathematicians seem unable to agree on whether to associate +++- with Cl(3,1,0) or Cl(1,3,0) --- leading to confusion elsewhere in this area. ] My own less adventurous investigations encountered similar controversy over the signature for Clifford algebra tailored to 3-space Euclidean geometry: I unhesitatingly selected Cl(3,0,1) in order to retain positive magnitudes for all versors, whereas roboticists [Sel05] follow a preference of Ian Porteous for Cl(0,3,1) . [ This puzzled me until I realised that practical robotics can presumably find little application for an odd-grade versor. ] However, while contemplating the the computation of a versor X transforming one vector frame F into another G , I noticed what might just constitute a decisive practical distinction, at any rate between Cl(p,q) and Cl(q,p) --- when p (alone) is even, the sign of G = 1/X F X is switched by substituting for X its dual X~ !! [ Trivial example when p+q = 1 --- F = G = [e] , X = 1 , X~ = e ; in Cl(0, 1) we have 1/X F X = G , 1/X~ F X~ = -G ; whereas in Cl(1, 0) no versor transforms F to -G . ] At the moment any physical and geometrical significance of this is unclear to me. In particular, in Cl(p,q,r) when r > 0 duality is a much more delicate matter: indeed when all p,q,r > 0 , a consistent definition of the concept seems impossible to formulate. [ In this fashion I extricate myself from having to admit that perhaps Jon Selig was right after all. ] Can any physicists out there cast further light? Fred Lunnon https://en.wikipedia.org/wiki/Minkowski_space @book{[Sel05], author = {Jon.~M.~Selig}, title = {Geometric Fundamentals of Robotics}, publisher = {Springer}, year = 2005, comment = {ISBN 0-387-20874-7 ; includes Euclidean subspaces} } @article{[Ber01], author = {Marcus Berg and C\'ecile DeWitt-Morette and Shangjr Gwo and Eric Kramer}, title = {The Pin Groups in Physics --- C, P, and T}, journal = {Reviews in Mathematical Physics}, volume = {13}, year = 2001, pages = {953--1034}, note = {August, 2001; \url{arXiv:math-ph/0012006v1}},

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[math-fun] Data storage in buckminsterfullerene
by Keith F. Lynch 20 Jun '16

20 Jun '16

The recent discussion about data storage in diamonds made me wonder about data storage in another stable form of carbon: C60, buckyballs. How many bits can be stored in one buckyball, i.e. fullerene molecule? It's a truncated icosahedron with 32 faces (of which 12 are regular pentagons and 20 are regular hexagons), 90 edges, and 60 vertices. The 60 vertices are carbon atoms. Data can be stored by making some of them C12 and some of them C13. But how much data? The base-2 log of the number of distinct arrangements of C12 and C13 atoms. But how many arrangements are there? If the molecules were not free to rotate, there would be 2^60 arrangements, hence 60 bits. But of course the molecules are free to rotate. The rotation group is order 60. So there are at least 2^60/60 arrangements, hence at least 54 bits. Can anyone think of a practical way to count the arrangements? The obvious way would be to try all 2^60 arrangements and eliminate the ones that are rotations of arrangements that have already been found. But 2^60 is more than 10^18, so that's just barely computationally feasible. Is there a better way? Has someone counted them already? Thanks. This number might seem to be uselessly small, but it isn't, since data could be stored in a large set of buckyballs. Each molecule could contain a 50-bit sequence number and several bits of data, giving several hundred terabytes of total storage. There would be many duplicates to ensure that at least one molecule with each sequence number is found. Several hundred terabytes may not seem like much by today's standards, but it can store the whole of Wikipedia, plus more books than anyone can read in a lifetime, plus more movies than anyone is likely to view in a lifetime. It's a lot larger than the genome of any organism, so it's a lot more practical than living DNA data storage. And 2^50 buckyballs times 600 duplicates of each of them would mass less than one milligram.

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[math-fun] OEIS differences (was Re: And I missed this Bernoulli bizarreness)
by Keith F. Lynch 20 Jun '16

20 Jun '16

Bill Gosper <billgosper(a)gmail.com> wrote: > Michael Somos <http://oeis.org/wiki/User:Michael_Somos> (Feb 01 2004) > discovered the remarkable fact that A001067 <http://oeis.org/A001067> > is different from A046968 <http://oeis.org/A046968>, even though > they agree for the first 573 terms. > Hey, Bernoulli(n+1)/n/(n+1) agrees with Bernoulli(n+1)/(n+1) for > 1147 terms! What's the most number of terms two OEIS sequences match for? (Not counting sequences which aren't known for sure to ever differ.)

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[math-fun] Maximal 3D unit distance graphs.
by Ed Pegg Jr 20 Jun '16

20 Jun '16

In 3D space, with n distinct points, what is the maximal number of unit distance lengths? I'm trying to build up a sequence for OEIS, but I'm only positive I've got the right values on two or three of the entries. Here's what I have so far. V -- E -- figure 4 -- 6 -- tetrahedron 5 -- 9 -- triangular bipyramid 6 -- 12 -- octahedron 7 -- 15 -- pentagonal bipyramid 8 -- 18 -- snub disphenoid or Raiskii spindle 9 -- 21 -- triaugmented triangular prism 10 -- 25 -- Nechustan spindle 11 -- 28 -- Augmented Nechustan spindle 12 -- 31 -- Double Pacman spindle 13 -- 36 -- Cuboctahedron + center 14 -- 40 -- Cuboctahedron + center + pyramid 15 -- 45 -- Icosahedron + 3 internal points 16 -- 50 -- Icosahedron + 4 internal points http://math.stackexchange.com/questions/1830194/maximal-unit-lengths-in-3d-… has a bit more info. Ed Pegg Jr

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Re: [math-fun] Maximal 3D unit distance graphs.
by apgoucher＠gmx.com 20 Jun '16

20 Jun '16

I think I can increase 16/50 to 16/51, but I'll need to verify that all points are distinct: Operate on a sphere S of unit radius, and let ABC be a fixed equilateral triangle on S, all of whose edges have length pi/3 (i.e. OABC is a regular tetrahedron where O is the centre of S). Now we want points A', B' such that AA'B'B is a spherical 'rhombus' (all edges are unit length). There is a set of such solutions parametrised by S^1. Let C' be the third point such that A'B'C' is an equilateral triangle oriented in the *opposite* way to ABC. By the intermediate value theorem, there is some choice of parameter which makes CC' = 1. Then OABC and OA'B'C' are regular tetrahedra sharing a vertex (the origin) with AA' = BB' = CC' = 1. Take the Minkowski sum of these two tetrahedra; this yields 16 points with 51 edges of unit length. -- APG. -----Original message----- Sent: Monday, 20 June 2016 at 21:44:03 From: "Tom Rokicki" <rokicki(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: Re: [math-fun] Maximal 3D unit distance graphs. I can bump your 12/31 value to 12/32. Start with your 6/12 octohedron; place it flat on a surface with two of the faces parallel to the surface. Copy the octohedron exactly one unit off the surface and add edges between the two pairs; we are now at 12 nodes and 24+6 or 30 edges, and the top octohedron has two degrees of freedom (fixing the bottom octohedron). Move the top octohedron so that one of its vertices forms the apex of a tetrahedron from the base triangle of the bottom octohedron; this lets you add two more edges from that vertex, and also two more edges from the corresponding upper plane, for a total of 32 edges. If this isn't clear I can draw a diagram. On Mon, Jun 20, 2016 at 7:52 AM, Ed Pegg Jr <ed(a)mathpuzzle.com> wrote: > In 3D space, with n distinct points, what is the maximal number of unit > distance lengths? > > I'm trying to build up a sequence for OEIS, but I'm only positive I've got > the right values on two or three of the entries. Here's what I have so > far. > > V -- E -- figure > 4 -- 6 -- tetrahedron > 5 -- 9 -- triangular bipyramid > 6 -- 12 -- octahedron > 7 -- 15 -- pentagonal bipyramid > 8 -- 18 -- snub disphenoid or Raiskii spindle > 9 -- 21 -- triaugmented triangular prism > 10 -- 25 -- Nechustan spindle > 11 -- 28 -- Augmented Nechustan spindle > 12 -- 31 -- Double Pacman spindle > 13 -- 36 -- Cuboctahedron + center > 14 -- 40 -- Cuboctahedron + center + pyramid > 15 -- 45 -- Icosahedron + 3 internal points > 16 -- 50 -- Icosahedron + 4 internal points > > > http://math.stackexchange.com/questions/1830194/maximal-unit-lengths-in-3d-… > has a bit more info. > > Ed Pegg Jr > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > -- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] -- _______________________________________________ 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] convergent Stirling's formula
by Bill Gosper 20 Jun '16

20 Jun '16

On 2016-06-18 23:26, rcs(a)xmission.com wrote: > I'm hoping for something like a recurrence with a bounded number of terms. > The goal is a tolerable formula for arbitrary precision gamma. > The wikipedia page gives/linksto several convergent formulas, but they > all have disabling features that spoil their use in long summations. > > Rich For stuff like sum 1/(n+a)! you can just use a matrix product. So you're thinking more arcane. Something like sum 1/f(n)! ? --rwg > > ----------- > Quoting rwg <rwg(a)sdf.org>: >> On 2016-06-18 17:50, rcs(a)xmission.com wrote: >>> Do you have a formula, or a recurrence, for the Poch form? >> >> Ouch! I had this in the clipboard but forgot to paste it: >> https://en.wikipedia.org/wiki/Stirling%27s_approximation#A_convergent_versi… >> --rwg >>> >>> Converting the denominators to factorials/2, starting at 12, >>> I find the numerators are >>> 1, 5, 59, 1218, 38376, ... >>> When these are divided by 12, 60, 360, 2520, 20160, >>> we get the fractions below. >>> >>> I tried OEIS and Superseeker, no joy. >>> >>> It would be nice to have a better formula for X! >>> >>> Rich >>> >>> >>> --------- >>> Quoting Bill Gosper <billgosper(a)gmail.com>: >>>> ... But with "rising factorials" (Pochhammers) >>>> instead of x^n, and weighted row sums of Stirling's Triangle >>>> instead of Bernoullis, >>>> Out[839]= >>>> 1 1 >>>> ---------- + ------------------ + >>>> 12 (1 + x) 12 (1 + x) (2 + x) >>>> >>>> 59 >>>> --------------------------- + >>>> 360 (1 + x) (2 + x) (3 + x) >>>> >>>> 29 >>>> ---------------------------------- + >>>> 60 (1 + x) (2 + x) (3 + x) (4 + x) >>>> >>>> 533 >>>> ------------------------------------------- + >>>> 280 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) >>>> >>>> 1577 >>>> --------------------------------------------------- >>>> 168 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) >>>> >>>> >>>> >>>> + 280361 / >>>> >>>> (5040 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) (7 + x)) >>>> >>>> Both of these series get added to the usual Stirling approximation >>>> ln(x^x/e^x ?(2?x)). The coefficients of this convergent series >>>> actually outgrow the Bernoullis, but the Pochhammers grow faster >>>> still. ("Pochhammers" is so appealing compared to "rising factorial >>>> powers". Too bad that, as usual, Pochhammer didn't invent them.) >>> >>>

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[math-fun] Two things I should've known about Stirling's formula
by Bill Gosper 19 Jun '16

19 Jun '16

https://en.wikipedia.org/wiki/Stirling%27s_approximation#A_convergent_versi… 1) DeMoivre had it first, but didn't recognize the √2π. 2) There's a convergent series (whose?) for ln x!, corresponding to the usual divergent In[837]:= BernoulliB[n + 1] ----------------- n ----------------- n + 1 Table[-----------------, {n, 11}] n x 1 1 1 1 {----, 0, -(------), 0, -------, 0, -(-------), 0, 12 x 3 5 7 360 x 1260 x 1680 x 1 691 -------, 0, -(----------)} 9 11 1188 x 360360 x (Divergent because Bernoulli_n, like n!, outgrows x^n for any x.) But with "rising factorials" (Pochhammers) instead of x^n, and weighted row sums of Stirling's Triangle instead of Bernoullis, Out[839]= 1 1 ---------- + ------------------ + 12 (1 + x) 12 (1 + x) (2 + x) 59 --------------------------- + 360 (1 + x) (2 + x) (3 + x) 29 ---------------------------------- + 60 (1 + x) (2 + x) (3 + x) (4 + x) 533 ------------------------------------------- + 280 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) 1577 --------------------------------------------------- 168 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) + 280361 / (5040 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) (7 + x)) Both of these series get added to the usual Stirling approximation ln(x^x/e^x √(2πx)). The coefficients of this convergent series actually outgrow the Bernoullis, but the Pochhammers grow faster still. ("Pochhammers" is so appealing compared to "rising factorial powers". Too bad that, as usual, Pochhammer didn't invent them.) Expanding the convergent series termwise about ∞, In[841]:= Series[Out[839], {x, Infinity, 9}] Out[841]= SeriesData[x, Infinity, 1 1 1 1 69311 {--, 0, -(---), 0, ----, 0, -(----), -(-----), 12 360 1260 1680 180 3892619 -------}, 1, 10, 1] 360 I.e., (with enough terms) we "recover" the divergent series from the convergent one! --rwg

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[math-fun] sumplosion
by Thane Plambeck 18 Jun '16

18 Jun '16

Here's a web-playable version of a solitaire number-adding game we (counterwave) just put up on the web for fun. it's similar to, or perhaps identical to, something discussed on math-fun I think....i'd welcome pointers to that game if anyone recalls it. http://sumplosion.com -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] And I missed this Bernoulli bizarreness
by Bill Gosper 18 Jun '16

18 Jun '16

A090495 Numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-1))). 12 574, 1185, 1240, 1269, 1376, 1906, 1910, 2572, 2689, 2980, 3238, 3384, 3801, 3904, 4121, Michael Somos <http://oeis.org/wiki/User:Michael_Somos> (Feb 01 2004) discovered the remarkable fact that A001067 <http://oeis.org/A001067> is different from A046968 <http://oeis.org/A046968>, even though they agree for the first 573 terms. Hey, Bernoulli(n+1)/n/(n+1) agrees with Bernoulli(n+1)/(n+1) for 1147 terms! Do[If[Numerator[BernoulliB[n + 1]/n/(n + 1)] != Numerator[BernoulliB[n + 1]/(n + 1)], Print[n.GCD[n, Numerator[BernoulliB[n + 1]/(n + 1)]]]], {n, 3511}] // tim 1147.37 2369.103 2479.37 2537.59 2751.131 8.898235 secs (Don't laugh. You should usually think in terms of Bernoulli *polynomials* .) As http://mathworld.wolfram.com/StirlingsSeries.html points out, Bernoulli(2*n)/(2*n*(2*n-1))) are the coefficients of the (divergent) Stirling series I mentioned yesterday. --rwg Mathematica now has anagrams: "deeeiimnorsstt" -> {"endometritises", "densitometries"}, ... "adegiiinnorstt" -> {"disorientating", "disintegration"}, ... "aeegiinorrsttv" -> {"tergiversation", "interrogatives"}, ... "eiinoopprssstu" -> {"superpositions", "propitiousness"}, ... Except that endometritises should be endometritides, which it doesn't even know: In[903]:= DictionaryWordQ@"endometritides" Out[903]= False

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[math-fun] Problem for John's son
by Zak Seidov 18 Jun '16

18 Jun '16

Just a year ago, John was older than his son by 6 times, and now only by 5. How old is John now? -- Zak Seidov

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