March 2006 - mathfun@mailman.xmission.com

[math-fun] 3D geometry program
by Bernie Cosell 13 Mar '06

13 Mar '06

My wife has taken a fancy to playing around with 3d objects. Things like making a cube with circles inscribed in each face and rotating it around to see what the various projections look like, or fiddling with a menger sponge [and again, rotating it around to see what it looks like], or intersecting tetrahedra... Anyhow, I've been looking around some and I haven't had much/any luck finding a (windows) app that'll do it -- it seems that there are a lot of 2D drawing programs, and 3D modeling/animation programs (that are generally way too complicated (and generally expensive) for the sort of hacking my wife wants to do). Any of you know of an app that might do the job? Thanks!! /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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[math-fun] BBC's "Negative Numbers"
by Henry Baker 11 Mar '06

11 Mar '06

Fascinating audio program on negative & complex (root of negative) numbers -- highly recommended: http://www.bbc.co.uk/radio4/history/inourtime/inourtime.shtml http://downloads.bbc.co.uk/rmhttp/downloadtrial/radio4/inourtime/inourtime_…

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Re: [math-fun] Cardinality puzzle - answer
by dasimov＠earthlink.net 10 Mar '06

10 Mar '06

Franklin wrote: << OK. I'm still refining this problem. The problem now is to describe a maximal uncountable antichain of subsets of Z, where no component set is either finite or counter-finite (i.e., Z-S must not be finite). And yes, I chose the word describe deliberately; the existence of such a set is easy to establish >> Hey, you could've said "nice example" when I solved your last question (:-)>. Anyhow, I've found as cute a little antichain as you could ask for that consists solely of subsets of Z that are each infinite and co-infinite. (I'm not sure how hard this is, but it seemed hard to me.) Without Zorn's Lemma, however, I don't yet see how to extend it to a maximal antichain. --Dan

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Re: [math-fun] Cardinality puzzle - answer
by dasimov＠earthlink.net 10 Mar '06

10 Mar '06

Franklin wrote: << Here's a related problem. Can you describe a *maximal* uncountable antichain? >> << More difficult: describe an (uncountable) maximal antichain where every set is infinite. (I don't know the answer to this second question. It might be: no, no such description is possible.) >> The antichain A = { Z - {n} : n is in Z} is maximal and contains only infinite sets: (Any other proper subset of Z must be missing at least 2 integers m and n, so is a proper subset of Z - {n}, so cannot be added to A -- and of course Z can't either. ) << [Gareth wrote:] >. . . Next: what's the largest a *chain* [of subsets of Z] can be? Obviously aleph-nought. At least one integer must be added at each step. >> There is a subtle flaw in the argument here, which is why this is such a great puzzle. --Dan

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[math-fun] Cardinality puzzle
by dasimov＠earthlink.net 09 Mar '06

09 Mar '06

Let a collection C of subsets of the integers be called an "antichain" if there exists no pair of sets X,Y in C such that X is a subset of Y. Puzzle: What is the largest cardinality that an antichain can have? --Dan

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[math-fun] Cardinality non-puzzle
by dasimov＠earthlink.net 09 Mar '06

09 Mar '06

Okay, now I'm wondering if a set of S any cardinality K > aleph_0 has standard formulas for the maximum cardinality C(K) of a chain of its subsets, and A(K) of an antichain of its subsets. For finite K: * C(K) is clearly K+1. * The non-obvious formula for A(K) is given by Sperner's Theorem. There's a nice discussion and short proof of it in "Proofs from The Book", 2nd ed. (there is a 3rd). But you may enjoy thinking about it before reading about it. --Dan

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[math-fun] Kimberlike sequences (for k=2 to k=9)
by Eric Angelini 09 Mar '06

09 Mar '06

Computed by Gilles Sadowski : http://www.cetteadressecomportecinquantesignes.com/Kimberlike02.htm Looks like, yes, it is always possible to fill the "holes"... [Can't see any pattern, though] Best, É. (and a special thanks to Paul for his interesting private mail)

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RE: [math-fun] sudoku, uniqueness, proofs
by Cordwell, William R 08 Mar '06

08 Mar '06

-----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Gary McGuire Sent: Wednesday, March 08, 2006 5:24 AM To: math-fun Subject: Re: [math-fun] sudoku, uniqueness, proofs Michael Kleber wrote: [deleted part] Dan wrote: >Are you sure that such solutions *are* given full credit in Putnam exams and >Olympiads? I don't know if that's the case. No, I'm not sure. I thought there would be some readers of this group who know. I know this particular problem is easy to solve the other way, but I've seen tricky problems which are made much easier by reasoning "here I am sitting in a Putnam/whatever, therefore this problem has a nice solution, therefore....." [deleted part] Gary McGuire _______________________________________________ I coach a MathCounts (grades 6-8) team, as well as high school students preparing for various contests. On many of the lower-level contests (say, MathCounts or the AMC->12 and the AIME), the exam only asks for the answer. In this case, I think that noticing, say, that a problem doesn't seem to have enough information, and fixing an apparently arbitrary value that will allow one to easily solve the problem (within the logical constraints) falls under a mixture of math and test-taking skills. For the AMC->12, many of the kids are, in ability, close to making the cutoff (a score of 100) for qualifying for the American Invitational (AIME), so I also have them work out the strategy for how many questions they should answer in certain situations. The AMC scores the 25-question exam by giving 6 points for each correct answer (multiple choice), 2.5 points for each unanswered problem, and 0 for each incorrect problem. If one's goal is to attain a score of >= 100, then, there's no advantage in answering 12 questions vs. 11 questions (and a possible disadvantage), but one needs to answer at least 11 questions. Qualifying for the AIME is not a zero-sum game, so I think that it is reasonable to point out this strategy to the kids. By the time that they reach the USAMO, they will be writing proofs for the problems; I've not graded those exams, but I would be surprised if a student got full credit for assuming that a solution was unique as part of his proof, unless he could show it later. Philosophically, I admit to being bothered when, say, a basketball team is in the lead and it intentionally delays, to deny the other team a chance at winning. I understand that this is within the rules, but it bugs me. I don't see the above strategies in that light, as one is not denying his competitors the same chance. However, I would not categorize using the strategies as "being better at math". --Bill Cordwell

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[math-fun] Kimberlike sequences
by Eric Angelini 08 Mar '06

08 Mar '06

Hello Seqfan and Math-fun, please have a look there, if you have a couple of minutes: http://www.cetteadressecomportecinquantesignes.com/Kimberlike01.htm Best, É.

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[math-fun] RE: Kimberlike sequences
by Eric Angelini 08 Mar '06

08 Mar '06

[Paul] : > Perhaps a deeper look at the hidden structure beneath Clark Kimberling's sequences will shed light on yours. ... thank you, Paul, for this interesting post -- and no, I had never heard before about Beatty sequences. I will plunge asap into this matter and try to see if it can be of any help in answering my questions. Thanks again -- and especially for this: > Here is how to determine the parameter r of the Beatty sequences > from the distance factor d. > The r is a root of quadratic equation 1 - d*x - x^2 = 0, or > r=(sqrt(d^2+4)-d)/2. ... Very interesting ! Best, É.

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