June 2012 - mathfun@mailman.xmission.com

[math-fun] Lie theory, ODEs, and doubt re "decision procedures"
by Warren Smith 02 Jun '12

02 Jun '12

I think the Lie techniques have been implemented in MAPLE ODE-solver and probably elsewhere. My experience is MAPLE's ODE solver is very powerful, way stronger than I am. Is it "a decision procedure"? This I do not know. There is a decision procedure for indefinite integration by Risch, and various mathematica etc claim to have implemented it, but I claim they are lying. They pretended to implement it, but not really -- it's too hard and too slow so they had to cheat and lie. I.e. if MAPLE or mathematica fails to integrate something, then I deny that's a proof it cannot be done. So given this disbelief, it is logically forced that I must also disbelieve that these softwares can solve any ODE or by failing prove unsolvable. It would be nice, by the way, if Mathematica were to print out "I have really fully completed the Risch procedure, no kidding, and proven this integral is un-doable" but I have never seen any such system print such a thing out. Anyhow, why don't you just feed in the air-drag ODE system to such packages and see what comes out? [As has now been pointed out, "Ray's" constant of the motion was known in 1860, and just having that does not show me how to solve the whole ball of wax except via leaving it in the form of an integral or by resorting to infinite series of some kind -- I doubt Ray fully solved the system to find position as function of time in closed form, and doubt it is possible to do so. I was initially impressed that this constant of the motion had escaped notice for 300+ years and he just found it, but it appears not and it appears the press got hold of this and overhyped it as "solving 300-year old problem!" while actually it was "nice, but not a full solve which is probably impossible" and much of what he found was already published albeit largely forgotten.]

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[math-fun] nonrecursive doubling of continued fractions
by dupre＠math.rutgers.edu 02 Jun '12

02 Jun '12

I have been reading Bill Gosper's and Adolph Hurwitz' recursive presentation of finding the half or double(and more) of a continued fraction. I have been working on unrolling the recursion in the case of multiplying and dividing a continued fraction by 2 and 3. Does anyone know(or care) about such a nonrecursive (and general)expression or algorithm? I find that this unrolling for a general continued fraction has some very interesting properties. Art DuPre'

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[math-fun] Lie theory
by Dan Asimov 02 Jun '12

02 Jun '12

Could someone please say what they mean when they use the term "Lie theory" ? I mean, I know a little about Lie groups and stuff but I don't know how this would apply to finding a formula for the path of a projectile with quadratic air resistance. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Checkers "solved"? Think so.
by Warren Smith 02 Jun '12

02 Jun '12

Perhaps this has some amusement value. Can these examples be made even more dramatic than they are? ---- Here is a checkers position I created to annoy Jonathan Schaefer, leader of the "chinook" project which solved the game of checkers (forced draw). View in constant-width font. .W.W.W.W W.W.W.w. .w.w.w.. ........ ...b.... ........ ...W.... ........ W=white king, w=white checker, white is moving North. White to move. In this position, white has 8 kings plus 4 checkers, while black has only a single checker. Nevertheless, black soon wins the game. Here's basically the same thing backed up a little to make it a bit more amusing, black to move and win: .W.W.W.W W.W.W.w. .w.w.... ..b...B. .......B ........ ...W.w.. ........ In view of this, should the Chinook team still contend they solved checkers? I took a look at their paper in Science and it looks to me like they really did solve checkers. Their strategy was to use chinook search + heuristic evaluator + perfect endgame tables. They assumed (perhaps falsely) that a heuristic eval > X was a "win" and a heuristic eval < -X was a "loss." They then pseudo-solved the game. Then they increased the value of X and did it again. They only claimed a full proof once X had increased all the way to a genuine win. So this kind of counterexample shows you need to go a pretty long way and cannot cheat by using too-small X... but the chinook team claims not to have cheated (although I think their initial efforts had cheated in this way). Warren D. Smith http://RangeVoting.org

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[math-fun] Teenager's new ODE solution; where's Lie Theory?
by Henry Baker 01 Jun '12

01 Jun '12

After thinking about this for a couple of days, I seem to recall that there was already a decision procedure for ODE's based on Lie Theory. (Olver, et al). Obviously, I must have been wrong. Does anyone here understand why Lie Theory wasn't able to solve the system of equations for trajectories with quadratic air resistance?

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[math-fun] Kimberling's polynomial problem
by Warren Smith 01 Jun '12

01 Jun '12

>It appears that a denominator may be a polynomial whose roots are the >reciprocal-products 1/(r(i)*s(j)), where r(i) ranges through all the >roots of q(x) and s(j) ranges through all the roots of v(x). >Clark Kimberling --This polynomial A(x) is expressible in terms of the original polynomials q(x) and v(x) by the use of "resultant" -- you can evaluate using determinants without computing any roots, just do rational operations. Specifically, the polynomial B(x) whose roots are the products r(i)*s(j) should be B(x) = Resultant_y [q(y), y^degree(v) * v(x/y)] and then A(x) is the "reverse" of B(x) to reciprocate its roots, A(x) = B(1/x) * x^degree(B). The "numerator" polynomial Kimberling also wanted could then be obtained by simply solving for it by given the known values of the first degree # of outputs, i.e. series coefficients, of the generating function. This is just a polynomial multiplication. -- Warren D. Smith http://RangeVoting.org

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[math-fun] projectile with quadratic air drag solution
by Warren Smith 01 Jun '12

01 Jun '12

Somebody pointed out that S.Ray's constant of the motion was actually stated in different notation about p.209 in Traite de balistique by Isidore Didion (French 2nd edition published 1860 is http://books.google.com/books?id=2C9LAAAAMAAJ) For long range battleship type gunnery, the Ray solution would be useless because the atmosphere gets exponentially thinner at higher altitude which should be a major effect that he can't handle. However, a perturbational analysis where the drag coefficient is regarded as small, start with the trivial parabolic no-drag solution then perturb it via series expansion... that still should be doable.

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Re: [math-fun] projectile with quadratic air drag solution
by Henry Baker 01 Jun '12

01 Jun '12

"Purchase this article online: US Dollar price: $33.00. There are no additional delivery charges." At 09:54 AM 6/1/2012, Warren Smith wrote: >Somebody also pointed out this paper (which I have not seen) > >K. Yabushita, M. Yamashita, K. Tsuboi: An Analytic Solution of >Projectile Motion with the Quadratic Resistance Law Using the Homotopy >Analysis Method, Journal of Physics A: Mathematical and Theoretical 40 >(2007) 8403-8416.

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Re: [math-fun] projectile with quadratic air drag solution
by Henry Baker 01 Jun '12

01 Jun '12

I'm actually more interested in the relativistic raytracer for computer graphics developed by one of the other contestants: http://jugend-forscht-sachsen.de/2012/teilnehmer/fachgebiet/projekt/3/id/5 Google translation: Members: Julius Kunze School: Johannes-Kepler-Gymnasium, Chemnitz Development of relativistic raytracer The project includes: A derivation of the optical effects occurring in relativistic constant (high) rates on the basis of special relativity. The programming of a simulation program using a proprietary ray tracer and a special interactive user interface for creating images and videos. An illustration of the effects occurring on the basis of realistic examples of images and videos. At 09:54 AM 6/1/2012, Warren Smith wrote: >The kid who solved the projectile problem was Shouryya Ray, born India >now living Germany.

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Re: [math-fun] generating function
by Michael Reid 01 Jun '12

01 Jun '12

> Suppose that a(0), a(1), a(2), ... is a linear recurrence sequence with generating function p(x)/q(x), and that > b(0), b(1), b(2), ... is a linear recurrence sequence with generating function u(x)/v(x). > > Does someone know a generating function for the product sequence a(0)*b(0), a(1)*b(1), a(2)*b*2), ... ? > > It appears that a denominator may be a polynomial whose roots are the reciprocal-products 1/(r(i)*s(j)), where > r(I) ranges through all the roots of q(x) and s(j) ranges through all the roots of v(x). That's right for the denominator. I do not know if is easy to give a formula for the numerator. See Prop. 3.9 on page 9 (and the ensuing remark) of my paper: http://ajc.maths.uq.edu.au/pdf/40/ajc_v40_p115.pdf A standard reference for this sort of thing is Stanley's "Enumerative Combinatorics"; however, when I looked there, he only had the statement about the rationality of the generating function of the pointwise product. I needed the degree of the recurrence (and a little bit more) in my paper. Michael Reid

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