September 2010 - mathfun@mailman.xmission.com

[math-fun] Attention Bay Area funsters
by Thane Plambeck 22 Sep '10

22 Sep '10

1) I'm hosting a Martin Gardner party at a loft in San Francisco (SOMA neighborhood, and walkable from the Caltrain terminus at 6th & King) the evening of Thu Oct 21. Please send me email if you're possibly interested in coming. See also this site, which describes the locations of other (worldwide!) Martin Gardner parties to be held at the same time http://www.g4g-com.org/ 2) Terence Tao is giving interesting-sounding lecture(s) at Stanford in two weeks http://math.stanford.edu ---------- Forwarded message ---------- From: <spk(a)math.stanford.edu> To: <announce(a)math.stanford.edu> Date: Tue, 21 Sep 2010 14:26:22 -0700 Subject: Special Math talk at Stanford: Professor Terence Tao, UCLA Dear friends and colleagues, I am writing to invite you to a public lecture at Stanford University by Professor Terence Tao entitled “The Cosmic Distance Ladder.” The lecture will be given at 7:30 pm on Wednesday October 6th, 2010 in Cubberley Auditorium on the Stanford campus. Details about the lecture can be found in the attachment to this email. Terence Tao, professor of mathematics at the UCLA, was awarded the Fields Medal in 2006, the mathematical equivalent of the Nobel Prize. A native of Australia, Tao is the recipient of numerous awards, including the Salem Prize, the Bôcher Prize, the SASTRA Ramanujan Prize, the Clay Research Award, the Nemmers Prize, and, in 2010, the Pólya Prize. In the citation for Tao's Fields Medal, the selection committee wrote: “Terence Tao is a supreme problem-solver whose spectacular work has had an impact across several mathematical areas. He combines sheer technical power, an other-worldly ingenuity for hitting upon new ideas, and a startlingly natural point of view...” Professor Tao’s lecture is part of a series of events sponsored by the Department of Mathematics. It is free and open to the public. We hope you will join us. Parking at Stanford is unrestricted and plentiful after 4:00pm. The following searchable campus map may be of use in planning your visit. http://campus-map.stanford.edu/index.cfm?=03-300 Professor Tao will also be giving the Bergman Lectures in the Mathematics Department on October 5th and 7th. These lectures are intended for a general mathematics audience. Details can be found on the Department’s webpage at http://math.stanford.edu/ If you have any questions, please feel free to contact us at cahill(a)math.stanford.edu or (650) 723-2602. Sincerely yours, Steve Kerckhoff, Chair Department of Mathematics Stanford University -- Thane Plambeck tplambeck(a)gmail.com http://thaneplambeck.typepad.com/

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Re: [math-fun] sums of seven cubes
by Dan Asimov 22 Sep '10

22 Sep '10

Ah, yes, heard of that. According to Wikipedia: The only difference is that Waring asked about the smallest number g(p) of summed pth powers that can represent *all* integers K, not just sufficiently large ones. Hardy and Littlewood then defined G(p) as the smallest number of pth powers adequate to represent all *sufficiently large* integers. In case anyone's interested, here's what Wikipedia says: << G(3) is at least four (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 109, 1290740 is the last to require six cubes, and the number of numbers between N and 2N requiring five cubes drops off with increasing N at sufficient speed to have people believe G(3)=4; the largest number now known not to be a sum of four cubes is 7373170279850, and the authors give reasonable arguments there that this may be the largest possible. 13792 is the largest number to require seventeen fourth powers (Deshouillers, Hennecart and Landreau showed in 2000 that every number between 13793 and 10245 required at most sixteen, and Kawada, Wooley and Deshouillers extended Davenport's 1939 result to show that every number above 10220 required no more than sixteen). Sixteen fourth powers are always needed to write a number of the form 31·16n. 617597724 is the last number less than 1.3 × 109 which requires ten fifth powers, and 51033617 the last number less than 1.3 × 109 which requires eleven. Using his improved Hardy-Littlewood method, I. M. Vinogradov has shown that G(k) <= p*(3*log(p) + 11) Using his p-adic form of the Hardy-Littlewood-Ramanujan-Vinogradov method A.A. Karatsuba proved the following new bound for: G(p) < 2p*log(p) + 2p*loglog(p) + 12p . >> Also: << 4 ≤ G(2) ≤ 4 4 ≤ G(3) ≤ 7 16 ≤ G(4) ≤ 16 6 ≤ G(5) ≤ 17 9 ≤ G(6) ≤ 21 8 ≤ G(7) ≤ 33 32 ≤ G(8) ≤ 42 13 ≤ G(9) ≤ 50 12 ≤ G(10) ≤ 59 12 ≤ G(11) ≤ 67 16 ≤ G(12) ≤ 76 14 ≤ G(13) ≤ 84 15 ≤ G(14) ≤ 92 16 ≤ G(15) ≤ 100 64 ≤ G(16) ≤ 109 18 ≤ G(17) ≤ 117 27 ≤ G(18) ≤ 125 20 ≤ G(19) ≤ 134 25 ≤ G(20) ≤ 142 >> More at < http://en.wikipedia.org/wiki/Waring's_problem >. --Dan Allan wrote: << Affirmative. Your conjecture is Waring's Problem, 1770. Proved by Hilbert, 1905. It's finding M and N for a given p that is really hard. I wrote: << Does this imply the possibility that: For every integer p > 0 there exists a number N(p) of pth powers, and a minimum M(p) > 0, such that every integer K > M(p) is the sum of N(p) nonnegative pth powers. ? >> >> _____________________________________________________________________ "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] domino networks reference sought
by Michael Beeler 22 Sep '10

22 Sep '10

A month or two ago, someone (Ed Pegg Jr. ?) posed a problem I'm calling "domino network". Surely this isn't new, but I didn't find any mention via some web searches. Anyone know of a reference? Problem: Label the nodes of a 4x7 rectangular graph with 0 through 9 (repeats allowed), such that each of the 45 edges corresponds to a different domino in the set of 45 dominoes with 0 (blank) through 9 pips, without doubles. The above problem is impossible (sketch below). I'm looking into related problems: varying the pip limit, whether doubles are included or not, and using various simple shapes on a squre, triangular, or hexagonal grid. Thanks for any comments! -- Mike Beeler Proof by contradiction. Suppose there is a solution. Consider the set of nodes labeled "0". Together, they must connect to 9 other nodes: once to a "1", once to a "2", etc. This is true of the nodes with any given label. So, the solution has 10 sets of nodes, each set connecting to exactly 9 other nodes. The given graph has 4 nodes (the corners) with valence 2; 14 nodes (sides) of valence 3; and 10 of valence 4. There is no way to partition these 28 nodes into 10 sets with the valence sum in each set being 9. I'm calling these "domino networks" because "domino graph" seems to already mean other things. (A particular 6-node graph, per MathWorld; or, a graph with a domino at each node, and links between dominos that have no number of pips in common, per the maa.org page by Ed. There is IBM Lotus Domino, a server product, so "domino network" is unfortunately a bit ambiguous too.)

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Re: [math-fun] sums of seven cubes
by Dan Asimov 22 Sep '10

22 Sep '10

Interesting. Does this imply the possibility that: For every integer p > 0 there exists a number N(p) of pth powers, and a minimum M(p) > 0, such that every integer K > M(p) is the sum of N(p) nonnegative pth powers. ? What is known about this? --Dan Rich quoted: << arXiv:1009.3983 Date: Tue, 21 Sep 2010 02:24:50 GMT (10kb) Title: Every even number greater than 454 is the sum of seven cubes Authors: Noam D. Elkies Categories: math.NT Comments: 9 pages MSC-class: Primary 11P05 \\ It is conjectured that every integer N>454 is the sum of seven nonnegative cubes. We prove the conjecture when N is congruent to 2 mod 4. This result, together with a recent proof for 4|N, shows that the conjecture is true for all even N. \\ ( http://arxiv.org/abs/1009.3983 , 10kb) >> _____________________________________________________________________ "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] sums of seven cubes
by rcs＠xmission.com 22 Sep '10

22 Sep '10

arXiv:1009.3983 Date: Tue, 21 Sep 2010 02:24:50 GMT (10kb) Title: Every even number greater than 454 is the sum of seven cubes Authors: Noam D. Elkies Categories: math.NT Comments: 9 pages MSC-class: Primary 11P05 \\ It is conjectured that every integer N>454 is the sum of seven nonnegative cubes. We prove the conjecture when N is congruent to 2 mod 4. This result, together with a recent proof for 4|N, shows that the conjecture is true for all even N. \\ ( http://arxiv.org/abs/1009.3983 , 10kb)

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Re: [math-fun] Corn on the cob
by Dan Asimov 22 Sep '10

22 Sep '10

In grad school I lived at International House for 1 1/2 years, where the math students would usually try to sit at the same table. When there were sufficiently many of us at one table as compared with students in other fields, the conversation inevitably turned to our common field of interest; when this happened we termed it "critical math". We eventually classified all math into two categories: gooey and crunchy. This was a question of continuous vs. discrete math, but probably analysts are gooey and algebraists are gooey. (Not sure about people who study Lie groups.) We tried to figure out correlations between gooey/crunchy and other behaviors, but nothing ever seemed backed up by evidence. Finally this corn-eating thing accomplishes this long-sought goal! Fascinating. --Dan << Are you an analyst or an algebraist? If you're presented with a delicious butter-dripping corn cob to eat, how do you eat it? According to http://bentilly.blogspot.com/2010/08/analysis-vs-algebra-predicts-eating.ht… your answers to these two questions are strongly correlated. For what it's worth, the alleged correlation works in my case; what about other munsters? >> _____________________________________________________________________ "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] McMullen's iterative cubic solver
by Henry Baker 22 Sep '10

22 Sep '10

McMullen mentions that Newton's method applied to q(x)=p(x)/r(x)=(x^3+a*x+b)/(3*a*x^2+9*b*x-a^2) is a more reliable iteration for solving the cubic p(x)=x^3+a*x+b than Newton's method applied to p(x)=x^3+a*x+b itself. "Braiding of the attractor and the failure of iterative algorithms", Invent. math. 91, 259-272 (1988). "Families of rational maps and iterative root-finding algorithms", Annals of Math., 125 (1987), 467-493. (Both papers are available for free online.) However, McMullen doesn't discuss where his algorithm came from, except the following quote in [Families...]: "The success of this algorithm can be understood by checking that the _points of inflection_ of q(x) coincide with the zeros of p(x). Thus Newton's method certainly converges if we take one of the points of inflection of q as our initial guess." I assume that the rationale for providing the denominator r(x) in q(x) is to avoid the well-known cycling discussed in good high school algebra classes. But McMullen doesn't provide any additional detail. I noticed that McMullen's approach wasn't discussed in the Wikipedia entry for cubics. Does anyone here have any thoughts on how to make McMullen's method more obvious?

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[math-fun] Did Jupiter and Saturn play pinball with Uranus?
by Henry Baker 20 Sep '10

20 Sep '10

FYI -- I couldn't find out if these simulations utilized general relativity. I would guess not, although I don't think that it would affect the outcome. http://www.newscientist.com/article/mg20727785.300-did-jupiter-and-saturn-p… Did Jupiter and Saturn play pinball with Uranus? * 17 September 2010 URANUS may have been batted back and forth between Jupiter and Saturn before being flung out to its present location, new simulations suggest. Previous modelling has shown that Jupiter and Saturn moved out of their initial orbits in the early solar system, scattering nearby objects. In some simulations, this led to Uranus crossing the path of Saturn, which could then have flung it towards Jupiter, which lobbed it back to Saturn. The process might have happened three times before Uranus was finally ejected beyond Saturn, to where it now resides. Hurling Uranus would have caused Jupiter and Saturn to recoil, further shifting their orbits. New simulations led by Alessandro Morbidelli of the CÃ´te d'Azur Observatory in France suggest this pinball game, which would have lasted just 100,000 years, fits with observations. In an alternate scenario, Jupiter and Saturn moved to their orbits over 5 million years by simply flinging away space rocks, but this would have visibly scarred the asteroid belt (Astronomical Journal, in press). "The evolution of the giant planets has been more violent than we thought," Morbidelli says.

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[math-fun] Cubic eqn with 3 real zeros, again
by Henry Baker 20 Sep '10

20 Sep '10

Has anyone seen the following treatment of the real cubic? A cubic equation with 3 real zeros can be reduced to either (your choice): 4*c^3-3*c=c3 or 4*s^3-3*s=-s3 where c,c3 reminds us of cos(a),cos(3*a), and s,s3 reminds us of sin(b),sin(3*b), respectively. Now Newton's method _can_ be used to compute one of the zeros, but an unfortunate choice of starting point will attempt to converge on a double (or almost double) root, thereby seriously slowing the convergence. For example, Newton on the cosine version gives c' = (c3+8*c^3)/[3*(2*c-1)*(2*c+1)] which leads to very unfortunate behavior for c near +- 1/2. But if we recognize that we are attempting to solve the 1-dimensional projection of a 2-dimensional problem, then we can avoid this convergence problem. Specifically, if we are given 4*c^3-3*c=c3, then we first compute s3=sqrt(1-c3^2), and then instead solve the problem z^3 = z3 = = c3+i*s3 = (c+i*s)^3 = c3+i*s3 = c^3 - 3*c*s^2 - i*(s^3 - 3*s*c^2) = c^3 - 3*c*(1-c^2) - i*(s^3 - 3*s*(1-s^2)) = c^3 - 3*c +3*c^3 - i*(s^3 - 3*s + 3*s^3) = 4*c^3 - 3*c - i*(4*s^3 - 3*s) = c3 - i*(-s3) = c3 + i*s3 But Newton's method on the complex problem z^3=z3 is much more robust, so long as we stay away from z=0: z' = (z3+2*z^3)/(3*z^2) (Throw away s=imagpart(z) & keep c=realpart(z) for the answer. But before doing so, we can trivially produce the other two roots with complex multiplications: w*z and w^2*z, where w is cube root of unity: w^3=1.) But we know that |z|^2 must be 1, so we can easily avoid the region near z=0. On modern hardware with multiple floating point units, complex arithmetic may have no more latency than real arithmetic, so this method may be no slower, and (due to faster convergence) may actually be faster.

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[math-fun] radiculus?
by Bill Gosper 20 Sep '10

20 Sep '10

I seem to recall at least one old book used it to mean sqrt(1-x^2), but I can't find it on the Web. --rwg Another piece of false memory syndrome: An expansion due to Newton of a function in terms of both derivatives and differences, which I thought was improbably buried in A&S somewhere.

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