October 2010 - mathfun@mailman.xmission.com

[math-fun] meaning of life ... today: 101010base2 = 42base10
by Ray Tayek 10 Oct '10

10 Oct '10

--- co-chair http://ocjug.org/

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[math-fun] Hey, that's MY trick.
by Bill Gosper 10 Oct '10

10 Oct '10

Or so I thought. Rich showed me http://arxiv.org/abs/1010.1420 which sez (p 15) "Taking the even part of the continued fraction (corresponding to the 2n-th convergents) on the right-hand side of (37) gives ..." I never heard of "the even part of a continued fraction", but what they mean is to pairwise coalesce terms, which does not obviously produce another C.F. without a slightly tricky maneuver (http://gosper.org/stanfordn2.pdf) with which you can convert to a C.F. any composition of homographic transformations. E.g., pairwise grouping the "period 2" C.F. for e^z in DLMF 4.9.3b (A&S 4.2.40(b)), 2 2 (2 n + 3) z 2 (- z + 4 (n + 1) - 1) exp z = 1 + z + K (------------ // ------------------------), n>=0 2 n + 1 2 n + 1 k terms of which give the Pade(k+1,k). --rwg

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Re: [math-fun] Jim's question for Conway
by James Propp 08 Oct '10

08 Oct '10

Thanks, David! I'll send Thurston an email to see what else I can learn about where the problems originated and how they spread. Jim

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Re: [math-fun] Martin Gardner archive
by James Propp 01 Oct '10

01 Oct '10

Okay, here's another clarification, which I hope will clear up the confusion I created with my original post: 1) I wrote >tiling a triangle of dots of side n with tiles consisting of either >3 dots in a row or 3 dots forming a triangle. but what I should have written was >tiling a triangle of dots of side n with tiles, where EITHER all tiles >consist of 3 dots in a row OR all tiles consist of 3 dots in a triangle. As David Moulton points out, if one intermixes the two kinds of tiles (as the wording of my original post would permit), then such tilings exist for all values of n. But if intermixing is not allowed, things are more interesting. In the case where all tiles consist of 3 dots in a row, no such tiling exists for any positive value of n. In the case where all tiles consist of 3 dots forming a triangle, such a tiling exists precisely when n is congruent to 0, 2, 9, or 11 (mod 12). Jim Propp

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Re: [math-fun] Martin Gardner archive
by James Propp 01 Oct '10

01 Oct '10

I wrote: >Just to be clear: my question for John Conway is whether he learned >about the problem from Gardner, and if not, where he learned about >the problem from. but I wasn't clear about what problem I was talking about! Here's the quote from Gardner's letter, sent to David Klarner on April 16, 1990: You have a triangular array of points, the number of points a multiple of 3. You wish to tile the lattice with lines that cover three points, like so: o / o o / \ o o o \ o---o---o o o o o---o---o It is almost certain that such tiling is never possible, but how to prove it? All my efforts to find a coloring proof have failed. Proving similar impossibility for triangles of three points seems equally hard; perhaps a proof for one would apply to the other. o / \ o---o o o---o / \ \ / o---o o o o o---o o o / \ \ / / \ o---o o o---o o Jim Propp

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[math-fun] National math program of Singapore
by Dan Asimov 01 Oct '10

01 Oct '10

>From the NY Times (more at < http://www.nytimes.com/2010/10/01/education/01math.html >: << The slower pace is a cornerstone of the district’s new approach to teaching math, which is based on the national math system of Singapore and aims to emulate that country’s success by promoting a deeper understanding of numbers and math concepts. Students in Singapore have repeatedly ranked at or near the top on international math exams since the mid-1990s. >> Didn't Richard Guy mention designing the curriculum for Singapore's math program? Richard? --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] Martin Gardner archive
by James Propp 01 Oct '10

01 Oct '10

Just to be clear: my question for John Conway is whether he learned about the problem from Gardner, and if not, where he learned about the problem from. Jim

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[math-fun] Location of the Oracle Death Ray
by Thane Plambeck 01 Oct '10

01 Oct '10

http://maps.google.com/maps?f=q&source=s_q&hl=en&geocode=&q=Redwood+Shores,… Tippy-tapped on my iPad

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[math-fun] Las Vegas Vdara Hotel = Solar furnace
by Henry Baker 01 Oct '10

01 Oct '10

FYI -- You've heard of buildings -- e.g., MIT's main corridor -- that become aligned with the suns rays one day per year. Now we have buildings which focus the sun's rays (inadvertently or vertently) on areas on the ground (e.g., the hotel's pool area). This recalls A.C.Clarke's 1950's short story about the daytime soccer game where all the fans were given programs made of silvered material. After a "bad" call, one of the referees was vaporized... http://www.engadget.com/2010/09/29/vdara-hotel-death-ray-claiming-victims-i… Vdara hotel 'death ray' claiming victims in the Las Vegas Strip By Joseph L. Flatley posted Sep 29th 2010 9:09PM "Death ray" is probably one of the most tantalizing phrases in the blogger's vocabulary, so our ears perked up when we heard talk of one out there in the Las Vegas Strip. A totally unforeseen and unpredictable consequence of building a 57-foot curved mirror in the desert, the Vdara Death Ray is the affectionate nickname given to a phenomenon that's been plaguing poolside loungers staying at the hotel / spa. Among the victims is one Bill Pintas, whose tale of woe was recently printed in the Las Vegas Review-Journal: [A]fter a brief dip in the hotel pool, he was sunning on a recliner. He was on his stomach, relaxed, eyes closed. But suddenly, the lawyer became so uncomfortably hot that he leaped up to move. He tried to put on his flip-flop sandals but, inexplicably, they were too hot to touch. So he ran barefoot to the shade. "I was effectively being cooked," Pintas said. "I started running as fast as I could without looking like a lunatic." Then he smelled an odor, and realized it was coming from his head, where a bit of hair had been scorched. A spokesman for MGM Resorts International (which owns Vdara) claims that it is working to fix the problem. In the meantime, if you're going to be in town for any random tech shows that might pop up, we suggest that you either get a trailer or barricade yourself into your suite at Circus Circus. Both have worked for us in the past.

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[math-fun] eta triples again
by Bill Gosper 01 Oct '10

01 Oct '10

I was surprised to get Jacobi's (least abstruse of all) aequatio satis abstrusa, a polynomial relating eta(q), eta(q^2), eta(q^4), abbreviated {1,2,4}, from the abstruser {1,2,3} := 0= -\[Eta][q]^60 \[Eta][q^2]^24+\[Eta][q]^72 \[Eta][q^3]^12+12 \[Eta][q]^48 \[Eta][q^2]^24 \[Eta][q^3]^12+196608 \[Eta][q]^24 \[Eta][q^2]^48 \[Eta][q^3]^12+16777216 \[Eta][q^2]^72 \[Eta][q^3]^12-196830 \[Eta][q]^36 \[Eta][q^2]^24 \[Eta][q^3]^24-19131876 \[Eta][q]^24 \[Eta][q^2]^24 \[Eta][q^3]^36-387420489 \[Eta][q]^12 \[Eta][q^2]^24 \[Eta][q^3]^48 (spurning Mma's DedekindEta, which expects period ratio tau instead of nome q.) {1,2,3} -> 6/{3,2,1} = {2,3,6} by Jacobi's imaginary transformation {1,2,3} -> {2,4,6} by q -> q^2 resultant({2,3,6},{2,4,6},eta(q^6)] = {2,3,4} (Warning: huge!) resultant({2,3,4},{1,2,3},eta(q^3)) = {1,2,4} quod erat desideratum. With the imaginary reciprocator and resultant you can apparently get any distinct triple of the form {2^a 3^b, 2^c 3^d, 2^e 3^f}, {a,b,c,d,e,f} nonnegative integers. Similarly for p and q, relatively prime, instead of 2 and 3, if you start with {1,p,q}. Strangely, with *three* relative primes, e.g., {2,3,5}, you seem to be limited to the 23 identities {{2,3,5},{4,6,9},{4,6,15},{4,6,25},{4,9,10},{4,9,15},{4,10,15},{4,10,25},{4,15,25},{6,9,10},{6,9,25}, {6,10,15},{6,10,25},{6,15,25},{9,10,15},{9,10,25},{9,15,25},{12,20,45},{12,20,75},{12,50,75}, {18,20,45},{18,45,50},{18,50,75}} The resultants are formidable. It might be easier to use undetermined coefficients on the series expansions. But not much. --rwg

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