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[math-fun] Four-star Hanoi: a number-theoretic oddity.
by Fred lunnon 08 Jul '07

08 Jul '07

Sometime in the mid-1990's, Paul Stockmeyer proposed a variation of the Reve's puzzle (Tower of Hanoi with 4 pins) which I propose to christen "four-star": one of the intermediate pins is distinguished, and every move is obliged to involve this pin either as origin or as destination. [ See "Variations on the Four-Post Tower of Hanoi Puzzle" at http://www.cs.wm.edu/~pkstoc/toh.html ] The analysis proceeds much as for the Reve's puzzle. To transfer n discs from origin pin to destination pin: (1) transfer n-k smaller discs from origin to (other) intermediate; (2) transfer k larger discs from origin to destination; (3) transfer n-k smaller discs from intermediate to destination; To choose the optimum value for k, Stockmeyer constructs an integer sequence a_n by arranging the set {2^i 3^j} in ascending order of magnitude, then sets k = k_n = 1 + [log(a_n)/log3]; e.g. (suitably fudged at n = 0) n = 0 1 2 3 4 5 6 7 8 9 10 11 12 a_n = ? 1 2 3 4 6 8 9 12 16 18 24 27 k_n = 0 1 1 2 2 2 2 3 3 3 3 3 4 The number of moves is then optimised at 2 (a_0 + ... + a_n), However, a much simpler algorithm would be simply to set k_n = [sqrt(2 n log2/log3) + 0.1809] ?? giving the same result for n < 1737 [which should satisfy the most enthusiastic apocryphal Brahmin!] It's not hard to see why such a formula might provide a good approximation; what I find surprising is that it fails only after such a protracted run. Fred Lunnon

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[math-fun] Re: <rational>!
by Bill Gosper 04 Jul '07

04 Jul '07

____________________________________________________________________________________ Boardwalk for $500? In 2007? Ha! Play Monopoly Here and Now (it's updated for today's economy) at Yahoo! Games. http://get.games.yahoo.com/proddesc?gamekey=monopolyherenow (Piece of sh*t Yahoo indeed linebroke my attachment. Hopefully, you were able to repair it. And hopefully, this is the last email they'll pollute for me.) The matrix product I gave yesterday can be unspecialized to give the oft-needed Gamma quotient for 2F1[a,b,c]: [ Gamma(c) Gamma(c - b - a) ] [ 0 ------------------------- ] [ Gamma(c - a) Gamma(c - b) ] = [ ] [ 0 1 ] [ (k+c-a) (k+c-b) ] [ --------------- + k ] oo [ c-b-a+1 ] /===\ [ 2 (k + -------) ] | | [ 1 k-a+1 k-b+1 k+c-a k+c-b 2 ] | | [ - ----- ----- ------------- ----------- ------------------- ] | | [ c-b-a c-b-a+1 c-b-a ] k = 0 [ 4 k+1 k+c k + ----- + 1 k + ------- ] [ 2 2 ] [ ] [ 0 1 ] (The matrix input form is [(k-a+1)*(k-b+1)*(k+c-a)*(k+c-b)/(4*(k+1)*(k+c)*(k+(c-b-a)/2+1)*(k+(c-b-a+1)/2)), ((k+c-a)*(k+c-b)/(2*(k+(c-b-a+1)/2))+k)/(c-b-a);0,1]) --rwg

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[math-fun] <rational>!
by Bill Gosper 04 Jul '07

04 Jul '07

(Some of my eavesdroppers' ISPs seem to be blacklisting osots.com, so I'm sending from Yahoo, and setting Rich another chore.) Does anyone know how to widen Yahoo's composition window to a usable width? Meanwhile, I'll try an attachment. If it linebreaks that too, I'll switch to gmail. (Sorry, Rich!) --rwg ____________________________________________________________________________________ Pinpoint customers who are looking for what you sell. http://searchmarketing.yahoo.com/ Rich's old Hakmem (http://www.inwap.com/pdp10/hbaker/hakmem/geometry.html) ####### ITEM 1 (Schroeppel): (1/3)! and (2/3)! are interexpressible. (1/4)! and (3/4)! are interexpressible. Thus these two pairs are of dimensionality one. (1/10)! and (2/10)! are sufficient to express (N/10)! for all N. (1/12)! and (2/12)! are sufficient to express (N/12)! for all N. (1/3)! and (1/4)! are sufficient to express (N/12)! for all N. Thus the three cases above are of dimensionality two. PROBLEM: Find some order to this dimensionality business. The reflection and multiplication formulas: pi Z Z! (-Z)! = --------- sin(pi Z) (N-1)/2 -NZ-1/2 (2 pi) N (NZ)! = Z! (Z-1/N)! (Z-2/N)! ... (Z-(N-1)/N)! ######## can now be rephrased as an applied math problem: Write a simplfier that recognizes when a monomial of fractional factorials is elementary. E.g., 1 2 7/8 (sqrt(6) + sqrt(2)) %pi (--)! (-)! 4 3 sqrt(-----------------------) 12 3 2 ---------- = ------------------------------------. 1 27 (-)! 4 I know of no CAS that can do this. (But haven't tried Mma 6.0). Presumably from its lack of a D.E., we "know" Gamma has no nice series. But the above implies 1/8 %pi 1 1 3 sqrt(-----------------) Beta(-, -) sqrt(6) + sqrt(2) 4 3 sqrt(---------------------------------------) 1 2 (-)! = ---------------------------------------------, 4 2 further evidence that <rational>! = algebraic(hypergeometrics(rationals)). Lastly, I have a 105 page Table of the Gamma Function for Complex Arguments, National Bureau of Standards Applied Mathematics Series : 34 (Issued Agust 6, 1954). >From the preface: The table of the complex gamma function presented in this volume is of fundamental importance in both pure and applied mathematics. The tabulation was prompted by urgent and specific needs in the fields of atomic and nuclear research. [...] Does anyone know of a physics formula with a naked Gamma? I'm guessing that they really needed only Betas and similar Gamma quotients that are merely hypergeometric sums, but involving multiple arguments, making them impractical to tabulate. "Scientist don't need computers-- all we need are a few computers to make them tables." Refutation: oo [ 2 2 2 ] /===\ [ (k + a) (k + b) (2 k + a) (2 k + b) - k ] | | [ ------------------------------------------ ------------------------ ] | | [ 2 b + a b + a + 1 b + a ] | | [ 4 (k + 1) (k + ----- + 1) (k + ---------) ] k = 0 [ 2 2 ] [ ] [ 0 1 ] [ 1 ] [ 0 ---------- ] = [ Beta(a, b) ] [ ] [ 0 1 ] (To hell with hypergeometric notation, but this is essentially a 4F3[1/4] that improves on some F[1/8]/F[1/8]s in Gosper, Strip mining in the abandoned orefields of Nineteenth Century mathematics, in Computers in Mathematics (Lecture Notes in Pure and Applied Mathematics, vol. 125), D. Chudnovsky & R. Jenks, eds. pp 261-283, m. dekker, N. Y.) I.e., a decently convergent series, but of two arguments, so you can't tabulate it for complex a,b. --rwg Video games are a feminist plot to render boys academically noncompetitive.

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[math-fun] Passive walking & hopping
by Henry Baker 27 Jun '07

27 Jun '07

Cool "passive" walking & hopping toys: http://www.hevanet.com/psmfg/web%20site/passive.html

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[math-fun] Rolling & hopping ovals?
by Henry Baker 27 Jun '07

27 Jun '07

In Fig. 4 of the following paper, http://ruina.tam.cornell.edu/research/topics/locomotion_and_robotics/papers… , an "oval" is show to roll and hop along a horizontal line without losing any energy. The center-of-gravity (cog) of the oval is a periodic sequence of upward-facing circular arcs and downward facing parabolic arcs. The speed of travel along the x-coordinate is synchronized with the rotation of the oval, so that the oval touches down in a way entirely symmetrical with the way it lifts off. Furthermore, only one side of the oval ever touches the ground, as the other side is up in the air when the oval revolves before touching down again (I assume that this situation could be generalized to do more than 1/2 rotation in the air). The question: what is the mathematical shape of the oval to achieve this overall lossless rolling & hopping behavior? (I don't know the answer, and the paper didn't give any references that I could download from the Internet.) Additional questions might be: does the oval have to be symmetrical about its short axis? (Maybe not.) Does the cog have to traverse a circular arc during the actual rolling motion? (I think not, so long as the velocity & angular velocity on point of contact and velocity & angular velocity are the "same" (modulo mirror images).) Has anyone else run across this type of oval before? Apparently, a football rolling end-over-end can approximate some of this behavior. I'm surprised that no one has built a children's toy based on this principle -- a hopping hoop would be much more fun than a simple rolling hoop.

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RE : [math-fun] Fungi and scientific points of view
by Eric Angelini 24 Jun '07

24 Jun '07

Neil: >>It's scary in the way that The Hot Zone was: rusts and molds could easily destroy civilisation. >James Brown: > Not only does The Triumph focus on eye-catching but unrepresentative diseases, but much of it is written in a flippant manner which does not help to explain some unfamiliar and occasionally obscure science. (...) http://www.the-ba.net/the-ba/CurrentIssues/ReportsandPublications/ScienceAn… <http://www.the-ba.net/the-ba/CurrentIssues/ReportsandPublications/ScienceAn…> I'll order this bookasap to make my opinion! Best, (thanks for having pointed that book, Neil) E.

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[math-fun] Fungi
by N. J. A. Sloane 23 Jun '07

23 Jun '07

Funsters, I just read a truly great - funny, scholarly and deeply scary - book called "The Triumph of the Fungi - A Rotten History" by Nicholas P. Money. It's the kind of book you come across every ten years that you want to buy copies of for all your friends. It got a great review in Nature. It's scary in the way that The Hot Zone was: rusts and molds could easily destroy civilisation. Neil

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[math-fun] moderator outage for Math-Fun
by Richard Schroeppel 22 Jun '07

22 Jun '07

I'm off to see Dan Asimov's wedding. I'll be out of town for three weeks. Michael Kleber (michael.kleber at gmail.com) is the temporary Math-Fun moderator. Rich

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[math-fun] Still an open problem?
by Christian Boyer 20 Jun '07

20 Jun '07

Euler had published the complete parametric solution in rationals of x^3 + y^3 = z^3 + w^3 (1) but as said by Hardy and Wright in their book: "The problem of finding all integral solutions is more difficult". In the Rocky Mountain Journal of Mathematics, 1998, Ajai Choudhry published a "complete" parametric solution in integers of (1). But it seems that some number theorists don't agree with this solution or any other claimed solution. Do you know if the complete parametric solution in integers of (1) is still an open problem? Christian.

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Re: [math-fun] Continuous derivations on the reals
by Eugene Salamin 19 Jun '07

19 Jun '07

There has been a data type error here. A derivation is a linear operator on the space of functions, not a function. If 1 denotes the constant function, then d(1) = d(1.1) = d(1).1 + 1.d(1) = 2.d(1), so d(1) = 0, and thus constant functions are in the null space of all derivations. On the other hand, I seen no reason why d(x) must be 1. If d(x) = q (i.e. the function q(x)), then for any function f, it looks like d(f) = q.D(f), where D is the standard differentiation operator. Gene ----- Original Message ---- From: "mcintosh(a)servidor.unam.mx" <mcintosh(a)servidor.unam.mx> To: math-fun(a)mailman.xmission.com Cc: mcintosh(a)servidor.unam.mx Sent: Tuesday, June 19, 2007 10:57:40 AM Subject: Re: [math-fun] Continuous derivations on the reals Quoting David Wilson <davidwwilson(a)comcast.net>: > Let [1] d(uv) = d(u) v + u d(v) > > Then [2] d(u^2) = 2u d(u) not when the quantities are noncommutative, as is typically the case when the object of discussion is a Lie Algebra. As for this implying a vanishing Maclaurin series, what does this say about ordinary functions, say polynomials, which do indeed have (non-zero) derivatives? - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ____________________________________________________________________________________ No need to miss a message. Get email on-the-go with Yahoo! Mail for Mobile. Get started. http://mobile.yahoo.com/mail

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