October 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Petersen Graph, or Hemi-dodecahedron
by Dan Asimov 08 Oct '08

08 Oct '08

Ed wrote: << I made a demo on the 57-cell. I seem to be the first to notice that the skeleton is the Perkel graph. http://demonstrations.wolfram.com/The57Cell/ I tried to build the 57-cell in 4D. I thought I would start by finding a really nice hemi-dodecahedron, or Petersen graph. . . . . . . >> That's a cool demo! As an abstract regular polytope, the 57-cell has as its group of symmetries L(2,19), aka PSL(2,19). The lowest dimension of a permutation representation of this group is 20. So to embed the 2-skeleton of the 57-cell (171 dodecahedra suitably identified) in R^n so that its symmetries are all elements of O(n), we'd need n >= 20. --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] Chestnut (fwd)
by Richard Guy 08 Oct '08

08 Oct '08

Was any progress ever made on Max's problem?? Thanks in anticipation of a response. R. ---------- Forwarded message ---------- Date: Thu, 23 Nov 2006 17:04:09 -0800 From: Max A. <maxale(a)gmail.com> To: David Wilson <davidwwilson(a)comcast.net> Cc: Sequence Fans <seqfan(a)ext.jussieu.fr> Subject: Re: Chestnut On 10/21/06, David Wilson <davidwwilson(a)comcast.net> wrote: > > This is a chestnut that I am convinced it is true, but I cannot prove it. I > have posed it on math-fun occasionally, and never got an answer. I would > like to see it posed to some better minds that some of you might know, e.g, > JHC or other number theorist. > > Let S and T be sets of real numbers. Call T a divider of S if some element > of T lies strictly between any two elements of S. > > For integer n >= 1, let Fence(n) be the set of all rationals with > denominator n, that is, { k/n : k in Z }. > > For real set S, let f(S) be the least n such that Fence(n) is a divider of > S, if such an n exists. > > Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator > <= n, that is, { 1/b : 1 <= b <= n } > > Let Farey(n) be the set of all rationals on [0,1] with denominator <= n, > that is, { a/b : 0 <= a <= b, 1 <= b <= n } > > Is f(Farey(n)) = f(Recip(n)) for every n? > > The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the > OEIS, I cannot find it in the short time I have.

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Re: [math-fun] Financial Crisis = Mathematical Malpractise ?
by Steve Witham 08 Oct '08

08 Oct '08

>From: Henry Baker <hbaker1(a)pipeline.com> >Subject: [math-fun] Financial Crisis = Mathematical Malpractise ? > >FYI -- According to Nassim Nicholas Taleb, [...] >most of the blame for the current financial crisis >can be laid at the feet of the "quants" [...]. There is a crisis this month, and everybody who had their head in the sand last month wants a theory to explain why they were not informed. http://blogs.cgdev.org/globaldevelopment/2008/09/crisis_not_if_we_take_a_lo… The term "black swan" is a favorite of mine because of the Black Swan River in Australia. Australians are sometimes puzzled by the name (I have heard) since "Black" is redundant for Australian swans. >From: "Fred lunnon" <fred.lunnon(a)gmail.com> > >The remark about globalisation raises an issue which I have never seen >explored technically: the network aspects of financial markets. >Even as crude a model as a resonant electrical circuit is sufficient to To me, "resonant circuit" implies a simple circuit with a sharp peak in frequency response, while "globalization" suggests lots of connections with different time constants, blurring out the response. >suggest that, left to its own devices, such a network is certain >--- under (in)appropriate conditions --- to become unstable. But these conditions could be as rare as the ones that cause all the air molecules to rush to the corner of the room. >An engineer equipped with a suitable model may able to design the >network to avoid this development. In the situation to hand, there is >(as far as I am aware) no engineer in charge. While a market is loosely similar to an electrical circuit, that doesn't mean an engineer could design anything so complex. But we do know that good, highly-complex designs exist without intelligent designers. >Another feature of networks [from percolation theory rather than >basic physics] >is that they go critical: as traffic approaches the maximum capacity, a small >increase traffic precipitates sudden gridlock over a large region. >As a result, >a crisis arising from other causes is intensified by the inability of the >participants to communicate at all. > >It seems a shame that we cannot make more use of the mathematics and physics >we already have! Not sure of the percolation model's applicability to financial crises, but complexity theorists (including Mandelbrot himself, as Henry mentions) are swarming all over economics. --Steve

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Re: [math-fun] Financial Crisis = Mathematical Malpractise ?
by Dan Asimov 07 Oct '08

07 Oct '08

Fred L. writes (among other things): << Another feature of networks [from percolation theory ...] is that they go critical: as traffic approaches the maximum capacity, a small increase traffic precipitates sudden gridlock over a large region. As a result, a crisis arising from other causes is intensified by the inability of the participants to communicate at all. >> Percolation theory is really incredibly interesting. For those who don't know about it: One example is, consider the hexagonal tessellation of the plane. Let 0 < p < 1. Suppose each tile is colored White or Black independently with respective probability p or 1-p. Then each tile belongs to a maximum connected piece of the same color. What is the probability I(p) there exists a connected piece of infinitely many White tiles? Turns out there is a number p_crit in (0,1) such that I(p) = 0 for p < p_crit, and I(p) = 1 for p > p_crit for this lattice (and other "lattices" as well). For the triangular lattice, p_crit = 1/2. The proofs of such things tend to be incredibly long and complicated. (For the triangular lattice, I believe I(p) = 1 for p = p_crit = 1/2 as well.) Last I checked there were very few results for this kind of "site percolation" for "lattices" naturally occurring in dimensions >= 3. (The original motivation for studying this stuff was as a model for how a fractal-ish rock decides whether to let water flow through it or not.) ----------------------------------------- << . . . Mr Taleb is credited with making himself 40M$, presumably on the back of short-selling. [I'm not going to criticise him on thos grounds --- if people are permitted to profit by backing stocks to rise, then it makes no sense to forbid them to profit by backing them to fall --- at any rate, provided it's their own money they're risking.] >> I wouldn't blame Mr Taleb for the way he made his money, but there are times it may make sense to ban betting that a stock price will go down -- as the U.S. Securities and Exchange Commission did recently. If enough people are betting that stocks will go down, they may use their influence to try to make that happen, and that would add to the probability the stock markets will crash -- or crash worse than it already has. ------------------------------------------ By the way, when I worked for Bank of America Global Trading ('87-'91), I learned (as many of you probably already know) that the Glass-Steagall Act, passed during the First Great Depression, forbade banks from dealing in many kinds of securities, especially stocks. The intent was to add stability to financial markets by decoupling major market sectors. But the Glass-Steagall Act was repealed in 1999. I'm puzzled that I haven't encountered in the media anyone's mentioning that repeal as a possible factor in what's happening now. Anyone have (non-partisan) comments on this? --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] Simultaneous solutions of 2 eqns., 2 unknowns
by Eugene Salamin 03 Oct '08

03 Oct '08

----- Original Message ---- From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Friday, September 26, 2008 1:55:02 PM Subject: [math-fun] Simultaneous solutions of 2 eqns., 2 unknowns I have 2 equations in x & y, each of form trig(x) + P(x,y) = 0 where trig is sin or cos, and P is a real polynomial of degree 3 or 4. ________________________________ In Maple 12: > fsolve({cos(y) = x^4+y^4, sin(x) = x^3+y^3}, {x, y}); {x = 0.9667490675, y = -0.4317536212} Gene

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Re: [math-fun] Beastly number problem
by Michael Reid 03 Oct '08

03 Oct '08

i blathered: > > A beastly number (A051003) is a number with substring 666. > > > > It is easy enough to prove that for each positive integer n, there is an > > integer k such that kn is beastly. Let f(n) be the smallest such k. > > > > Show that f(n) is bounded over all integers n. What is the largest possible > > value of f(n)? [ ... ] > to get the largest value of f(n) that occurs, i found it easier to > consider final digits and split it into numerous cases; perhaps there > is an easier calculation using initial digits. or maybe not. let g(n) denote the smallest positive integer k , such that kn begins with the digits 666 . what is the largest value of the function g(n) ? mike

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[math-fun] Number Theory Puzzle
by rcs＠xmission.com 02 Oct '08

02 Oct '08

Puzzle: While futzing around with "prime formulas", I noticed that 8^8 - 7^7 = 3 * 19 * 19 * 14731, and then 14^14 - 13^13 = 3 * 61 * 61 * 968299894201. This clue leads to the following guess: When N = 6K+1, (N^2+N+1)^2 | 3 * [(N+1)^(N+1) - N^N]; and When N = 6K+4, (N^2+N+1)^2 | 3 * [(N+1)^(N+1) + N^N]. For example, N=16: 273^2 | 3 * (17^17 + 16^16). True or False? Rich

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Re: [math-fun] Beastly number problem
by Michael Reid 02 Oct '08

02 Oct '08

> A beastly number (A051003) is a number with substring 666. > > It is easy enough to prove that for each positive integer n, there is an > integer k such that kn is beastly. Let f(n) be the smallest such k. > > Show that f(n) is bounded over all integers n. What is the largest possible > value of f(n)? the largest possible value is 4263 , which occurs as f(15625) , and occurs infinitely often as f(15625 * (10^t + 1)) for sufficiently large t . as allan's post suggests, it is easy to get a universal upper bound on f(n) by considering initial digits. indeed, let m = ceiling(log_10(n)) so that 10^(m-1) < n <= 10^m , and let k = ceiling(666 * 10^m / n) . then k <= ceiling(666 * 10^m / (10^(m-1) + 1)) <= 6660 , and kn begins with the digits 666 , so that f(n) <= 6660 , independently of n . to get the largest value of f(n) that occurs, i found it easier to consider final digits and split it into numerous cases; perhaps there is an easier calculation using initial digits. note that f(n) = f(10n) , so we need only consider those n not divisible by 10 . if gcd(n, 10) = 1 , then kn = 666 mod 1000 for some 0 < k < 1000 , so f(n) < 1000 . if n is even, consider the largest power of 2 that divides n . as noted, we may assume that 5 does not divide n . if the largest power of 2 that divides n is 2^1 , then kn = 666 mod 1000 for some 0 < k < 500 , whence f(n) < 500 . if the largest power of 2 that divides n is 2^2 , then kn = 6664 mod 10000 for some 0 < k < 2500 , whence f(n) < 2500 . if the largest power of 2 that divides n is 2^3 , then kn = 6664 mod 10000 for some 0 < k < 1250 , whence f(n) < 1250 . if the largest power of 2 that divides n is 2^4 , then kn = 66608 mod 100000 for some 0 < k < 6250 , whence f(n) < 6250 . if n is divisibly by 2^5 , then kn = 66624 mod 100000 for some 0 < k < 3125 , whence f(n) < 3125 . now consider the case when n is divisible by 5 , but not by 2 . if the largest power of 5 that divides n is 5^1 , then kn = 6665 mod 10000 for some 0 < k < 2000 , whence f(n) < 2000 . if the largest power of 5 that divides n is 5^2 , then kn = 66625 mod 100000 for some 0 < k < 4000 , whence f(n) < 4000 . if the largest power of 5 that divides n is 5^3 , then kn = 66625 mod 100000 for some 0 < k < 800 , whence f(n) < 800 . if the largest power of 5 that divides n is 5^4 , then kn = 666875 mod 1000000 for some 0 < k < 1600 , whence f(n) < 1600 . if the largest power of 5 that divides n is 5^5 , then kn = 6665625 mod 10000000 for some 0 < k < 3200 , whence f(n) < 3200 . if the largest power of 5 that divides n is 5^6 , then kn = 66609375 mod 100000000 for some 0 < k < 6400 , whence f(n) < 6400 . if the largest power of 5 that divides n is 5^7 , then kn = 66640625 mod 100000000 for some 0 < k < 1280 , whence f(n) < 1280 . if the largest power of 5 that divides n is 5^8 , then kn = 666015625 mod 1000000000 for some 0 < k < 2560 , whence f(n) < 2560 . if n is divisible by 5^9 , then kn = 666015625 mod 1000000000 for some 0 < k < 512 , whence f(n) < 512 . these give the slightly better upper bound f(n) <= 6249 . however, we can consider each of these cases in more detail, as i had my computer do. for example, suppose n is divisible by 5^8 , but not by 5^9 , nor by 2 . there are 1024 residue classes of such n modulo 10^9 , and for each, we verify that there is 0 < k <= 1705 such that the digit string 666 occurs within the last 9 digits of kn . (this only requires that kn is one of the residues: 666015625, 666406250 or 666796875 .) this shows that f(n) <= 1705 for all n in that case above. i checked all the cases above, and found that f(n) <= 4263 in all cases, with smaller upper bounds in all cases except for the case when 5^6 and 2^0 are the largest powers of 5 and 2 dividing n . (it took much longer to code it than it took to execute (1.1 secs), and there is some possibility of error.) moreover, 4263 indeed occurs as f(5^6) , so that is the largest value. i did not give it much thought, but i suppose it is conceivable that the upper bound from examining the cases in detail could not be obtained. if that happens, perhaps it would suffice to extend the cases to consider higher powers of 2 and/or 5 . mike

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Re: [math-fun] number anagrams
by Dan Asimov 02 Oct '08

02 Oct '08

This isn't exactly what you asked about, but close, with simultaneous arithmetic and permutational equivalence: ELEVEN PLUS TWO = TWELVE PLUS ONE --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] number anagrams (french)
by Eric Angelini 02 Oct '08

02 Oct '08

Hello Erich, with this : « CINQUANTE ET UN + ONZE - SIX = CENT UN + QUINZE - SOIXANTE » (found there : http://www.cetteadressecomportecinquantesignes.com/Eleven.htm ) ... and using your building method, I get (in french) : 11,651 = 101,175 ... or : ONZE mille SIX cent CINQUANTE ET UN = CENT UN mille cent SOIXANTE-QUINZE Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Erich Friedman Envoyé : jeudi 2 octobre 2008 15:26 À : math-fun Objet : [math-fun] number anagrams Some English names for numbers can be anagrammed to other numbers. There are some trivial cases like 67 = 76 (SIXTY SEVEN = SEVENTY SIX) that just involve splitting and rearranging compound words. Then there are slightly more interesting cases like 112 = 211 (ONE HUNDRED TWELVE = TWO HUNDRED ELEVEN) but this still corresponds to anagramming the digits of the numbers. What is the smallest pair of English numbers that are anagrams without the underlying numbers being anagrams? The best I could do is 1,008,020,068 = 10,010,082,086 (ONE BILLION EIGHT MILLION TWENTY THOUSAND SIXTY EIGHT = TEN BILLION TEN MILLION EIGHTY TWO THOUSAND EIGHTY SIX) but I suspect there is a smaller solution. I'm not a polyglot, so what are the smallest solutions in other languages? Erich Friedman _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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