January 2014 - mathfun@mailman.xmission.com

[math-fun] Penrose tilings, pharmaceuticals and crankery
by Warren D Smith 12 Jan '14

12 Jan '14

The result the cranks claimed "We show that when projecting an edge-transitive N-dimensional polytope onto an M-dimensional subspace of R^N, the sums of the squares of the original and projected edges are in the ratio N/M." is in fact true provided we randomly rotate the polytope before projecting it, and take expectation values. Now given that that is true, it should also be true for certain subgroups of the full rotation group SO(N). In particular, the symmetry group of the polytope itself (if it is edge transitive) is a reasonable candidate. Now Goucher's counterexample of a Rhombus is not allowed since fails to have edge transitive sym group (at least, if you demand group consist wholy of rotations). To be allowed, rhombus would need to be square. And in that case the claim is true, in fact equivalent to Pythagoras's theorem. I think claim is true for hypercubes more generally (?). An actual counterexample would be the degenerate "polytope" consisting of a single edge, but I guess they'd whine that that was degenerate. Re the larger claim they are cranks, well, see http://www.quantumgravityresearch.org/ it appears to be a weird little research institute founded by a rich wacko, but it is not yet obvious to me they are cranks. In particular I think the claim above Goucher said was "obviously wrong" is likely in fact correct, and with a correct proof, albeit you might want to be a bit more careful about exactly what is meant by "group" (rotations only) and exactly what polytopes are allowed (nondegenerate only). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Penrose tilings, pharmaceuticals and crankery
by Adam P. Goucher 12 Jan '14

12 Jan '14

Greetings. Today I received a rather unusual and completely unsolicited e-mail from one Klee Irwin, mentioning the glider I found on a Penrose tiling cellular automaton: > ----- Original Message ----- > From: Klee Irwin <Klee(a)QuantumGravityResearch.org> > Sent: 01/12/14 09:07 AM > To: apgoucher(a)gmx.com > > Hi, Adam. Congrats on the Penrose tile glider. Would you have time to do some paid consulting for our group? Approaching this with the deserved amount of suspicion and scepticism, I proceeded to Google `Klee Irwin'. Apparently he owns several businesses, including a nutriceutical company that was sued $2.65 million for selling pills with an illegal concentration of lead. I've read numerous complaints from people who were apparently defrauded by him. Also, there are news stories of him investing six-figure sums in certain companies. His `group' has also published some mathematical papers, including this one: http://arxiv.org/pdf/1210.1446v1.pdf Note that the claimed result is actually false; consider projecting a rhombus onto a diagonal, for instance. Even if it were true, it would surely be a trivial application of Huygens-Steiner or some variant thereof. Judging by this evidence, Irwin appears to be a crank. So, why has this person contacted me ex caeruleo? Sincerely, Adam P. Goucher

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Re: [math-fun] Penrose tilings, pharmaceuticals and crankery
by Adam P. Goucher 12 Jan '14

12 Jan '14

Joerg, Thank you very much for the advice. I've forwarded this to Tim Hutton and Andrew Trevorrow, who also seem to have been similarly targeted. `Ex caeruleo' is Latin for `out of the blue' (quite literally), which is an English colloquialism for `completely unexpectedly'. I see that this could quite easily be lost in translation! Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Joerg Arndt > Sent: 01/12/14 05:54 PM > To: math-fun > Subject: Re: [math-fun] Penrose tilings, pharmaceuticals and crankery > > * Adam P. Goucher <apgoucher(a)gmx.com> [Jan 12. 2014 18:31]: > > Greetings. > > > > Today I received a rather unusual and completely > > unsolicited e-mail from one Klee Irwin, mentioning the > > glider I found on a Penrose tiling cellular automaton: > > > > > ----- Original Message ----- > > > From: Klee Irwin <Klee(a)QuantumGravityResearch.org> > > > Sent: 01/12/14 09:07 AM > > > To: apgoucher(a)gmx.com > > > > > > Hi, Adam. Congrats on the Penrose tile glider. Would you have time to do some paid consulting for our group? > > > > [...] > > > > > > > So, why has this person contacted me > > He wants to "buy" your name/brain to gain (scientific) credibility. > > You will have to check that he won't claim any sort > of cooperation with you. > > Suggest to just give a short "No thanks.", then play dead. > Don't get drawn into any conversions of any kind! > > Do keep copies of all his mails. > > > > ex caeruleo? > > What does "ex caeruleo" mean? (learning English on math-fun: fun!) > > > > > Sincerely, > > > > > > Adam P. Goucher > > > > Best, jj > > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Penrose tilings, pharmaceuticals and crankery
by Warren D Smith 12 Jan '14

12 Jan '14

The result the cranks claimed "We show that when projecting an edge-transitive N-dimensional polytope onto an M-dimensional subspace of R^N, the sums of the squares of the original and projected edges are in the ratio N/M." is in fact true provided we randomly rotate the polytope before projecting it, and take expectation values. Now given that that is true, it should also be true for certain subgroups of the full rotation group SO(N). In particular, the symmetry group of the polytope itself (if it is edge transitive) is a reasonable candidate. Now Goucher's counterexample of a Rhombus is not allowed since fails to have edge transitive sym group (at least, if you demand group consist wholy of rotations). To be allowed, rhombus would need to be square. And in that case the claim is true, in fact equivalent to Pythagoras's theorem. I think claim is true for hypercubes more generally (?). An actual counterexample would be the degenerate "polytope" consisting of a single edge, but I guess they'd whine that that was degenerate. Re the larger claim they are cranks, well, see http://www.quantumgravityresearch.org/ it appears to be a weird little research institute founded by a rich wacko. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Big Splat theory
by Bill Gosper 11 Jan '14

11 Jan '14

Jim>Has anyone ever proposed in the Annals of Improbable Research or a similar venue that, in addition to undergoing periodic reversals of its magnetic field, Earth is also subject to occasional short-lasting reversals of its gravitational field? Such a phenomenon would handily explain selective extinction of megafauna: a mouse could survive an upward fall for five seconds followed by a downward fall for five seconds followed by an extreme deceleration, but a mastodon would not fare as well. :-) <Jim Propp If I suddenly found myself levitating, I'd blame a kickass gravity wave.

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[math-fun] Kazakh Professor Claims Solution of Another Millennium Prize Problem
by Ray Tayek 11 Jan '14

11 Jan '14

http://science.slashdot.org/story/14/01/11/1715227/kazakh-professor-claims-… Posted by <mailto:soulskill@slashdot.org>Soulskill on Saturday January 11, 2014 @12:59PM from the awaiting-peer-review dept. An anonymous reader writes "Kazakh news site BNews.kz reports that Mukhtarbay Otelbaev, Director of the Eurasian Mathematical Institute of the Eurasian National University, is <http://bnews.kz/en/news/post/180213/>claiming to have found the solution to another Millennium Prize Problems. His paper, which is called 'Existence of a strong solution of the Navier-Stokes equations' and is <http://www.math.kz/images/journal/2013-4/Otelbaev_N-S_21_12_2013.pdf>freely available online (PDF in Russian), may present a solution to the fundamental partial differentials equations that <http://www.claymath.org/millenium-problems/navier%E2%80%93stokes-equation>describe the flow of incompressible fluids for which, until now, only a subset of specific solutions have been found. So far, only one of the <http://www.claymath.org/millennium-problems>seven Millennium problems was solved the Poincaré conjecture, <http://science.slashdot.org/story/03/04/15/1337219/poincar-conjecture-may-b…>by Grigori Perelman in 2003. If Otelbaev's solution is confirmed, not only it might be the first time that the $1 million offered by the Clay Millennium Prize will find a home (Perelman <http://science.slashdot.org/story/10/07/02/153208/grigory-perelman-turns-do…>refused the prize in 2010), but also engineering libraries will soon have to update their Fluid Mechanic books." --- co-chair http://ocjug.org/

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[math-fun] Stupid math question: least-squares
by Henry Baker 11 Jan '14

11 Jan '14

Does anything interesting/fun happen when you apply so-called "least(?)-squares" algorithms to finite fields -- e.g., GF(2) ? In particular, the theory of pseudo-inverses, etc. Also, what about least-squares in fields with non-standard orderings (valuations) ? Any references would be appreciated. Thx.

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Re: [math-fun] How I found big dual palindromes
by Keith F. Lynch 11 Jan '14

11 Jan '14

Joerg Arndt <arndt(a)jjj.de> wrote: > What the OEIS sequence number of this one? A060792. Clarification: I said, "on average each solution ought to be e^7.5 (about 1700)." The e^7.5 is also intended to be very approximate. The way I phrased it sounds like I meant it to be exact. Maybe, to get a little more precision, I should rewrite my numerical integration program so it doesn't crap out at 2^1024, which is the largest floating point number my machine will allow. >> [ I'm looking for an IT job in the DC area. So is Alan, and so are >> many other experienced programmers I know. Employers simply aren't >> hiring. Not even the ones who complain to Congress that they can't >> find anyone to hire. ] > The complaint "we cannot find programmers" is really > "we cannot find DIRT CHEAP programmers", that's why. > Same here in Germany. Nobody is offering low salaries. It's very puzzling. It's like playing the lottery: There's a tiny chance that you'll make tons of money, but most likely you'll make nothing. I have over 40 years experience programming in numerous languages, and I'm very good at it, but nobody returns my phone calls or emails. Not even recruiters.

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[math-fun] Magic square variant
by David Wilson 11 Jan '14

11 Jan '14

I was interested in whether there were 3 x 3 squares with row sums equal and column products equal. It turns out these are rather easy to generate. I found the following squares 3 9 12 8 10 6 15 4 5 with row sum 24 and column product 360. I believe 24 is the minimum possible row sum. I also found 1 10 15 12 6 8 20 4 2 With row sum 26 and column product 240, which I believe to be the minimum possible column product. Just in case anyone cares.

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[math-fun] Mile Marker 420 Becomes 419.99 to Thwart Thieves
by Henry Baker 11 Jan '14

11 Jan '14

FYI -- Heck, in Berkeley, the sign would have read "0644" or "0x1A4"; in Princeton, perhaps 7*5*3*2^2... http://abcnews.go.com/Weird/wireStory/mile-marker-420-41999-thwart-thieves-… Mile Marker 420 Becomes 419.99 to Thwart Thieves DENVER January 11, 2014 (AP) Colorado officials think a difference of one-hundredth of a mile will be enough to stop thieves from stealing the mile marker 420 sign along Interstate 70. Amy Ford of the Colorado Department of Transportation says the "MILE 420" sign near Stratton was stolen for the last time sometime in the last year, and officials replaced it with a sign that says "MILE 419.99." Ford says it's the only "420" sign to be replaced in the state that recently legalized recreational marijuana. Most highways aren't long enough to need one. The number "420" has long been associated with marijuana, though its origins as a shorthand for pot are murky. Mile 419.99, about 25 miles from the Kansas border, isn't the only place in Colorado with a fractional mile marker. Cameron Pass in Larimer County has a "MILE 68.5" sign after frequent thefts of the "MILE 69" sign.

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