May 2009 - mathfun@mailman.xmission.com

Re: [math-fun] x^2=0 but x=1
by Dan Asimov 20 May '09

20 May '09

Are we to assume x,y,z,w are members of some kind of ring? --Dan Emeric wrote: << I am interested in an analytic device to handle the following simple situation. I have a polynomial P in several variables, x,y,z,u, say. It is of degree 2 in each variable. I would like to evaluate the polynomial when x^2 = 0, y^2 = 0, z^2 = 0, u^2 = 0 BUT x=y=z=u=1. Many thanks for any comment, hint, input. >> _____________________________________________________________________ "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] (no subject)
by deutsch 20 May '09

20 May '09

I am interested in an analytic device to handle the following simple situation. I have a polynomial P in several variables, x,y,z,u, say. It is of degree 2 in each variable. I would like to evaluate the polynomial when x^2 = 0, y^2 = 0, z^2 = 0, u^2 = 0 BUT x=y=z=u=1. Many thanks for any comment, hint, input. Emeric

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Re: [math-fun] Now they tell me: Mathematica for $295
by Dan Asimov 20 May '09

20 May '09

I might like to find out why it's called a home "edition", not just a home "deal". --Dan << I wish I had known about this before I shelled out big bucks. But maybe others can benefit from this. For $295, you can get a full-featured 32-bit version of Mathematica for use at home. Commercial uses are prohibited. So is using it for the class you teach at a school. But for true math nerds, like me, who consider Mathematica to be recreational software that we use at home, this sounds like a great bargain! http://www.wolfram.com/products/mathematicahomeedition/ >> _____________________________________________________________________ "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] donate mathematics books to who?
by Andy Latto 18 May '09

18 May '09

Any suggestions? On Mon, May 18, 2009 at 5:39 PM, Varda Haimo <varda.haimo(a)verizon.net>wrote: > > Many years ago, before the dawn of time, I studied Mathematics, and I > have a number of pointy-headed math books to show for it. But now, I > am starting to covet the shelf space where they have sat undisturbed for > oh these many years. Rather than haul them down to Goodwill, or a > general book donation site, I'd like to give them to an organization or > whatever, that could actually make good use of them. Has anyone done > this? What is the best way to donate books so that they reach people > who could really use them? > > Thanks for any ideas on the subject. > > Varda > > --~--~---------~--~----~------------~-------~--~----~ > You received this message because you are subscribed to the Google Groups > "xBBN" group. > To post to this group, send email to xbbn(a)googlegroups.com > To unsubscribe from this group, send email to > xbbn+unsubscribe(a)googlegroups.com <xbbn%2Bunsubscribe(a)googlegroups.com> > For more options, visit this group at > http://groups.google.com/group/xbbn?hl=en > -~----------~----~----~----~------~----~------~--~--- > > -- Andy.Latto(a)pobox.com

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Re: [math-fun] this can't be new
by James Propp 16 May '09

16 May '09

I wrote: >So the theorem says we can make a deterministic choice that makes >the counts #{i: 1 leq i leq k and s_i = s} stay within 1 of their >expected values under random choice, and achieves this simultaneously >for all k geq 1 and all s in S. In fact, the theorem says that we can achieves this result by making choices in a particular (and particularly simple) way. That is, we present an algorithm, not an existence proof. I'll describe the algorithm intuitively, since it may make some bells go off in someone's head: For intuition, think of k as time, and imagine that S is a set of people each of whom has an appetite-level (pi_1 (s) + ... + pi_k (s)) - #{i: 1 leq i leq k and s_i = s} at time k. That is, your appetite goes up by pi_k (s) at time k unless you're the lucky person who gets a biscuit at time k, in which case your appetite goes down by 1 - pi_k (s). Everyone starts out with appetite 0. You die of hunger if your appetite ever rises above 1, and die of overfeeding if your appetite ever falls below -1. The person you should feed next is the person who'll die of hunger soonest if left unfed, not including those people who can't be fed right now (because they'll die of overfeeding if you do). Proving that this works isn't hard (though it also isn't hard to flail around for a while and not succeed in proving it if you don't have the right idea). But the problem and the algorithm and the proof are sufficiently simple that I'm sure this is all old stuff. I also wrote: >If you like, you can take pi_1 = pi_2 = ... = some fixed distribution >pi; this special case is not any easier than the general result >(which is a little tricky but is not in the end very deep). In fact, pi_1 = pi_2 = ... is the case we are most interested in (even though our construction works for the general case). So even if you know of prior work on this special case alone (not the general case), we'd be very interested. Whoa, it's 10 a.m., and thinking about all these people getting fed is making me hungry; gotta go get some breakfast! Maybe some muffins? :-) Jim Propp

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[math-fun] this can't be new
by James Propp 16 May '09

16 May '09

Omer Angel, Ander Holroyd, James Martin and I have proved a result that I'm 90% sure isn't new, but we haven't been able to find a reference: Suppose we have discrete probability distributions pi_1, pi_2, ... on some not necessarily finite set S. Then it is possible to choose a sequence s_1, s_2, ... in S such that for any k geq 1 and any s in S, #{i: 1 leq i leq k and s_i = s} lies within 1 of pi_1 (s) + pi_2 (s) + ... + pi_k (s). Note that pi_1 (s) + pi_2 (s) + ... + pi_k (s) is exactly the expected value of #{i: 1 leq i leq k and s_i = s} if we were to choose s_i randomly in accordance with pi_i for all i leq k. So the theorem says we can make a deterministic choice that makes the counts #{i: 1 leq i leq k and s_i = s} stay within 1 of their expected values under random choice, and achieves this simultaneously for all k geq 1 and all s in S. If you like, you can take pi_1 = pi_2 = ... = some fixed distribution pi; this special case is not any easier than the general result (which is a little tricky but is not in the end very deep). Has anyone seen this before? It seems like the sort of thing operations researchers would have discovered decades ago. Thanks, Jim

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Re: [math-fun] cyclotomic factorization
by Phil Carmody 14 May '09

14 May '09

From: rwg(a)sdf.lonestar.org > > I thought the appearance of 2s (or 3s ...) in the > coefficients of factor(x^n-1) > was equivalent to >= three (or four ...) distinct > primes dividing n, > but it seems they can't all be 3 mod 4 since n=231 fails. What's "failure"? There are cyclotomic polynomials Phi(pqr) with maximum height 1 where p,q,r == 1 (mod 4) and cyclotomic polynomials with huge maximum height where p,q,r == 3 (mod 4), such as pqr=79.131.199 with height 41. Phil

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Re: [math-fun] Slight change in the rules of chess
by Henry Baker 12 May '09

12 May '09

"Terrorist chess" ? If a pawn takes a higher valued piece in this way, perhaps that pawn is sacrificed, too? "Suicide bomber chess" ? At 04:42 PM 5/11/2009, Mark D. Niemiec wrote: >This would make pawns potentially more deadly than any other piece on the board, >because they would threaten not only the squares that pieces landed in, but >also the squares that they pass through.

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Re: [math-fun] Slight change in the rules of chess
by Dan Asimov 12 May '09

12 May '09

I'd describe en passant differently, involving only two moves: If A's pawn P moves two squares from the home rank, and if B has a pawn p immediately to the right or left of P, then -- only on the next move -- p can capture P by moving diagonally, just as if P had moved only one square, not two. --Dan << The "en passant" rule says that if A's pawn moves two squares from the home rank, attacking B's pawn, and on the next play, B moves the attacked pawn one square forward to avoid capture, that A, on the next move, can capture B's pawn in the same rank as if A's pawn had moved only one square on the earlier move. . . . >> _____________________________________________________________________ "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] Slight change in the rules of chess
by Mark D. Niemiec 11 May '09

11 May '09

> On the other hand, I think there should be another slight modification to > the rules of chess that I think is consistent. > > The "en passant" rule says that if A's pawn moves two sqaures from the home > rank, attacking B's pawn, and on the next play, B moves the attacked pawn > one square forward to avoid capture, that A, on the next move, can capture > B's pawn in the same rank as if A's pawn had moved only one square on the > earlier move. But that same logic would apply to any piece that moved next > to A's pawn in the same rank on the move following the A's two-square move. > I argue that it should be possible to capture any piece en passant. This would make pawns potentially more deadly than any other piece on the board, because they would threaten not only the squares that pieces landed in, but also the squares that they pass through. Also, if you grant this power to pawns, why not grant it to every other piece as well, to be consistent? Unfortunately, this would allow rooks and queens to project impassible barriers all the way across the board. What makes the en-passent capture rule special is not the virtue of the capturing pawn, but rather of the captured pawn, because it moves twice instead of once. If one is to extend the en-passent capture rule, it makes more sense to allow any piece to capture a pawn en-passent, rather than the other way around. (This is not to say that your suggestion is not an interesting one, but as I mentioned above, it would make the game VERY different, and would have to be studied carefully.) -- Mark D. Niemiec <mniemiec(a)interserv.com>

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