December 2008 - mathfun@mailman.xmission.com

[math-fun] Special Holiday Post: lazy projective space question
by Dan Asimov 22 Dec '08

22 Dec '08

("Lazy" because monkey wonders, monkey posts without thinking about question.) As you may know, the division algebras 1D reals R, 2D complexes C, 4D quaternions H, and 8D octonions O, each have their projective spaces: RP^n, CP^n, HP^n, and OP^n, respectively. (The standard construction is to let A* be the nonzero elements of an associative division algebra A, and to use the equivalence relation (a_1,...,a_(n+1)) ~ lambda*(a_1,...,a_(n+1)), where lambda and all a_j are in A, with at least one a_j being non-zero.) This is an equivalence relation when A is associative, and the set of equivalence classes is defined as RA^n.) It is often stated, however, that OP^n exists solely for n <= 2, since the relation (o_1,...,,o_n,o_(n+1)) ~ lambda*((o_1,...,,o_n,o_(n+1)) is said to be an equivalence relation only when multiplication is associative -- which is not true for arbitrary octonions. There is, however, a careful construction that yields what is generally agreed upon to be OP^2. QUESTION: What exactly goes wrong when one tries to construct OP^n, n >= 3 ??? In particular, suppose one defines a relation among all (n+1)-tuples (o_1,...,o+(n+1)) of octonions by the usual means, as above: (*) (o_1,...,,o_n,o_(n+1)) ~ lambda*((o_1,...,,o_n,o_(n+1)) where lambda, and the o_j's are octonions ... and lambda -- and at least one o_j -- is nonzero. How does this relation fail to be an equivalence relation? Clearly it satisfies reflexivity and symmetry, so the problem is presumably with transitivity. One can always take a relation and discuss the least equivalence relation it defines. SO: What is the least equivalence relation defined by (*) above? Does it make all elements of O^(n+1)* (all (n+1)-tuples of octonions that are not all 0) equivalent? Or what? Be merry, --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] [seqfan] You wouldn't believe (fwd)
by Eric Angelini 22 Dec '08

22 Dec '08

> The next year also could be presented > using just 7's and ones > 7*7*(7*(7-1) -1) ... what about using only 200's and 9's ? Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Alexander Povolotsky Envoyé : samedi 20 décembre 2008 19:06 À : Sequence Fanatics Discussion list Cc : Sequence Fans; Math Fun Objet : Re: [math-fun] [seqfan] You wouldn't believe (fwd) The next year also could be presented using just 7's and ones 7*7*(7*(7-1) -1) _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] The sausage conjecture
by rwg＠sdf.lonestar.org 21 Dec '08

21 Dec '08

> Warut wrote: > > << > Ian Stewart's NewScientist article on the sausage conjecture: > > http://www.newscientist.com/article/mg20026873.800-the-scientific-guide-to-… >>> > > Thanks for posting this, Warut -- it's totally fascinating! > > (I recall hearing about the situation through 4 dimensions > from Neil Sloane some years back.) > > What is most fascinating about this is that in dimensions > up through 4 (well, OK, just 2, 3, and 4), the least-volume > convex set containing n unit balls is a sausage up through > some value of n, and then it becomes a more ball-shaped mass. > > But in sufficiently high dimensions (currently known for > dim >= 42), the sausage (straight-line) configuration of > balls is optimals no matter how many balls are involved. > > Maybe it's just me, but I was astonished to learn even that > for 4 unit balls in 3-space, the tetrahedral configuration > is not the one whose convex hull has the least volume; > it's all 4 balls in a straight line. > > --Dan What I find hardest to believe is the claim that there's only one cutover at 57 spheres in the 3D case. I'd expect it to jitter for a while for 58, 59, ... . --rwg ALGORISMIC MICROGLIAS

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[math-fun] You wouldn't believe (fwd)
by Richard Guy 20 Dec '08

20 Dec '08

Funsters & fansters might be amused by this somewhat unlikely coincidence. I had calculated a couple of hundred terms of a fourth order recurring sequence [it's A005178 in OEIS, if you want details] and was looking for the ranks of apparition of various primes. A 19-digit prime factor of the 53rd term turned up as a substring of the 103rd term! Have a prime time in 7^2 x 41. R. ---------- Forwarded message ---------- Date: Wed, 17 Dec 2008 09:24:06 -0700 (MST) From: Richard Guy <rkg(a)cpsc.ucalgary.ca> To: Hugh Cowie Williams <williams(a)math.ucalgary.ca> Subject: You wouldn't believe Hugh, before sending the final (??) version of the quadric file, I decided I'd do a search for larger primes just to see if there were any double ranks. While searching with 3140540902719737029 which is a factor of a(53), I discovered that it's a substring of a(103): 93118232931779128686097911301602920314054090271973702981658870701 ^^^^^^^^^^^^^^^^^^^ R.

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Re: [math-fun] You wouldn't believe (fwd)
by Dan Asimov 20 Dec '08

20 Dec '08

Richard wrote: << Funsters & fansters might be amused by this somewhat unlikely coincidence. I had calculated a couple of hundred terms of a fourth order recurring sequence [it's A005178 in OEIS, if you want details] and was looking for the ranks of apparition of various primes. A 19-digit prime factor of the 53rd term turned up as a substring of the 103rd term! Have a prime time in 7^2 x 41. R. >> Hmmm, this is just the kind of thing that led to Monstrous Moonshine (and at least one Fields medal). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] The sausage conjecture
by Dan Asimov 20 Dec '08

20 Dec '08

Warut wrote: << Ian Stewart's NewScientist article on the sausage conjecture: http://www.newscientist.com/article/mg20026873.800-the-scientific-guide-to-… >> Thanks for posting this, Warut -- it's totally fascinating! (I recall hearing about the situation through 4 dimensions from Neil Sloane some years back.) What is most fascinating about this is that in dimensions up through 4 (well, OK, just 2, 3, and 4), the least-volume convex set containing n unit balls is a sausage up through some value of n, and then it becomes a more ball-shaped mass. But in sufficiently high dimensions (currently known for dim >= 42), the sausage (straight-line) configuration of balls is optimals no matter how many balls are involved. Maybe it's just me, but I was astonished to learn even that for 4 unit balls in 3-space, the tetrahedral configuration is not the one whose convex hull has the least volume; it's all 4 balls in a straight line. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] The sausage conjecture
by Warut Roonguthai 19 Dec '08

19 Dec '08

Ian Stewart's NewScientist article on the sausage conjecture: http://www.newscientist.com/article/mg20026873.800-the-scientific-guide-to-…

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Re: [math-fun] Fast factoring ??
by Michael Reid 19 Dec '08

19 Dec '08

> > > Thereby reenacting one of my favorite denestings: > > > > > > 1/5 3/5 3/5 2/5 1/5 2/5 > > > 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 > > > (3 - 2 ) = ------------------------------------- , > > > 2/3 > > > 5 > [ ... ] > The ^1/3 has already been highlighted. So far, I've only seen the numeric verifications, but surely a symbolic one would be more appropriate? > verification doesn't require computer algebra. it took me less than half a page to calculate and simplify (3^(1/5) + 4^(1/5) - 54^(1/5) + 72^(1/5))^3 with pen and paper. contrast this to FINDING the identity! mike

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Re: [math-fun] Fast factoring ??
by Phil Carmody 18 Dec '08

18 Dec '08

From: "Fred lunnon" <fred.lunnon(a)gmail.com> > On 12/17/08, rwg(a)sdf.lonestar.org > <rwg(a)sdf.lonestar.org> wrote: > > ... > > Thereby reenacting one of my favorite denestings: > > > > 1/5 3/5 3/5 2/5 1/5 2/5 > > 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 > > (3 - 2 ) = ------------------------------------- , > > 2/3 > > 5 Ug, I hope I've stitched that back together and yahoo mail doesn't murder it on egress. > > a major simplification that looks anything but. > > Indeed! > > u := 3^(1/5); v := 2^(1/5); w := 5^(1/3); > l := (u^3 - v)^1/3; > r := (u + v^2 - v*u^3 + v^3*u^2)/w^2; > evalf(l - r); # -.6607822845 ?! The ^1/3 has already been highlighted. So far, I've only seen the numeric verifications, but surely a symbolic one would be more appropriate? ? f2=Mod(a,a^5-2) Mod(a, a^5 - 2) ? f3=Mod(b,b^5-3) Mod(b, b^5 - 3) I cubed both sides, which is fair as we're entirely within the reals. ? lift(lift((-f2*f3^3 + f2^3*f3^2 + f3 + f2^2)^3)) -25*a + 25*b^3 ? lift(lift((f3^3-f2)*25)) -25*a + 25*b^3 Phil

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[math-fun] Antikythera Device, a working model
by Simon Plouffe 18 Dec '08

18 Dec '08

http://www.networkworld.com/news/2008/121708-antikythera.html?hpg1=bn Hello, that guy made a working model of that calculator, it is surprinsingly complex. it uses gears and a lot of ingenuity to construct this device, it is 2000 years old. simon plouffe

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