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Re: [math-fun] m & m's
by asimovd＠aol.com 06 May '03

06 May '03

Thane writes: << My son complained that his bag of m&m's contained "all browns, and no blues." I don't think blue used to be an M&M color when I was a kid. Brown, green, orange and red, yes. It's hard for me to remember yellows. Maybe just a few. Maybe the company tries to instill a sense of mystery about its identical-tasting candies by making one color "rarer" than another. Then again, maybe blue food coloring is more expensive than the others. --Dan

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[math-fun] Sequences -> "models"; the automated physicist
by Henry Baker 05 May '03

05 May '03

After seeing the success of the integer sequence and the ability of systems to "guess" the function, I would propose a new system that would analogously "guess" one or more "models" of how to produce that sequence, given a "noisy" set of data points. [If this has already been done before, then I'd be interested in links/pointers/references.] Take the following example. A particular human has the capability to run fast for a short while, run further if he/she slows down some, and run even further if he/she slows down a lot. For example, a human might be able to run a 9 second 100 yard dash, or a 3:40 mile, or a 2:10:00 marathon (26 miles). (I apologize for the use of English units -- we're only using units of the "willing", right now!) There might be a variety of physiologic reasons why longer distances require slowing down -- e.g., a variety of energy storage and release mechanisms, where the fastest methods get depleted quickly. One might conceive of a mathematical system that was given various datapoints such as the ones given above, and the system would hypothesize a family of models that "fit" those data points, and might then ask for additional datapoints in order to distinguish the models. Thus, for example, the system might be given one datapoint -- a 9 second 100 yard point. The system would hypothesize that the person runs at a constant speed. You then tell the system that the person runs a 3:40 mile, which is 8 yds/sec versus 11.11 yds/sec for the 100 yard dash. The system then posits some two-level component (smooth) system wherein there are two energy-releasing mechanisms, one fast and one slow, and the fast one "runs out of gas" before the distance of a mile is reached. The system then might ask one or more questions regarding datapoints shorter than 100 yards, intermediate points, or points for distances longer than one mile. If the system is given the marathon datapoint (46,145 yards in 7800 seconds, or 5.92 yards/sec), this point might not introduce more complexity into the system in terms of additional energy systems, but might simply bound the various coefficients more tightly. The idea of this system is not merely to interpolate values, for which there are thousands of different methods, but to come up with one or more "natural" models, which could eventually be used (with much additional research) to start to propose hypotheses about how the human system might actually work, at least in terms of the most important factors for these distances and times. Obvious interpolation schemes would include polynomial and rational functions, sums of exponentials, sums of exponentials times polynomials, etc. I would imagine that such a system ought to be smart enough to "rediscover" a significant number of the simpler "laws of nature" -- e.g., Ohm's law, Colomb's law, etc. Henry

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[math-fun] Re: Consecutive Integers [was Quadratic Question]
by Dean Hickerson 05 May '03

05 May '03

Edwin Clark (eclark(a)math.usf.edu) asked: > Let E=[e_1,e_2,...,e_k] be a list of integers > satisfying 1 <= e_1 <= e_2 <= . . . <= e_k. > > Let S(E) be the set of all positive integers n such that n = Prod{p_i^e_i} > where the p_i's are distinct primes. > > Say that the integers in S(E) have factorization pattern E. I believe this > is a known concept. ... > Is it the case that if 1 is not in E then S(E) contains no consecutive > pair of integers? No. There are at least two counterexamples among the solutions of 23 x^2 + 1 = 8 y^2. The positive solutions of this are given by x(0) = 23; y(0) = 39; x(n) = 24335 x(n-1) + 14352 y(n-1); y(n) = 41262 x(n-1) + 24335 y(n-1). (Note that x(n) is always divisible by 23, so 23 x(n)^2 and 8 y(n)^2 are consecutive powerful numbers.) For n=8, 23 x(n)^2 and 8 y(n)^2 are in S([2,2,2,2,3]); their factorizations are 23^3 1361^2 39157289^2 13475265289^2 43842128059343041^2 and 2^3 3^2 13^2 509^2 61853469184829429986091801293807909^2 For n=9, they're in S([2,2,2,2,2,3]), with factorizations 23^3 151^2 2281^2 23447^2 6461369^2 29366147657538184938924047^2 and 2^3 3^2 13^2 35872892507^2 5281732523824487^2 8087241110442161^2. Dean Hickerson dean(a)math.ucdavis.edu

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[math-fun] Re: Quadratic question
by Jpr2718＠aol.com 04 May '03

04 May '03

David Wilson asked: > For which (a, b) do all solutions of > a x^2 - b y^2 = [1] > have the form (r k, s j) for some fixed r, s > 1? D. T. Walker, ``On the Diophantine Equation $mX^2 - nY^2 = \pm 1$, American Mathematical Monthly, Volume 74, Issue 5, (May 1967), pp. 504-513, has a theorem that answers Professor Wilson's query. Consider (1) $aX^2 - bY^2 = \pm 1$. Theorem 9 in the above is (a and b are positive integers, not squares (and whose product is not a square)): If (1) is solvable for both of a, b > 1 and has r \sqrt{a} + s \sqrt{b} as its smallest solution; and if i is a nonnegative integer, then the formula r_i \sqrt{a} + s_i \sqrt{b} = (r \sqrt{a} + s \sqrt{b} )^{2i + 1} gives all possible solutions of (1). So r will always divide r_i and s will always divide s_i. Any a, b missing from the list in Professor Clark's reply apparently are not there because either the equation has no solution for those values, or because at least one of r, s is 1. John John Robertson 973-331-9978 jpr2718(a)aol.com

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[math-fun] Re: Quadratic question
by Jpr2718＠aol.com 02 May '03

02 May '03

David Wilson asked: > For which (a, b) do all solutions of > a x^2 - b y^2 = [1] > have the form (r k, s j) for some fixed r, s > 1? Consider the binary quadratic form equation Ax^2 + Bxy + Cy^2 = N, with discriminant D = B^2 - 4AC. Any two solutions x, y and x', y' are equivalent if one can be derived from the other by x = ((u-Bv)/2)x' - Cvy', y = Avx' + ((u+Bv)/2)y', where u, v is any solution to X^2 - DY^2 = 4. When N = 1, any two solutions are equivalent. (See the section ``2 The structure of solutions'' in Keith Matthews, The Diophantine equation ax^2 + bxy + cy^2 = N, D = b^2 - 4ac > 0, Journal de Theorie des Nombres de Bordeaux, 14 (2002) 257-270.) Let r, s be the minimal positive solution to ax^2 - by^2 = 1. Then (4ar^2 - 2)^2 - 4ab(2rs)^2 = 4. Assume, for the moment, that 4ar^2-2, 2rs is the minimal positive solution to X^2 - 4abY^2 = 4. If x', y' is a solution to ax^2 - by^2 = 1, and if x' is a multiple of r and y' is a multiple of s (as they are for the minimal solution), then x and y from the above formulas are also multiples of r and s respectively (because v = 2rs). So, by induction, all solutions are multiples of r and s. This says that solutions to ax^2 - by^2 = 1 can fail to have x and y be multiples of r, s only when 4ar^2-2, 2rs is not the minimal positive solution to X^2 - 4abY^2 = 4. I do not know whether this can occur. But, instead of Professor Wilson's question, we have the question as to whether there are any equations ax^2 - by^2 = 1 where x and y for any solution are _not_ multiples of those for the minimal solution. Richard Mollin probably discusses this, and it seems to me there is a paper in the Monthly from the 1950s that may address this issue. John John Robertson 973-331-9978 jpr2718(a)aol.com

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[math-fun] Quadratic question
by David Wilson 02 May '03

02 May '03

When I solved [1] 8 x^2 - 343 y^2 = 1, all solutions (x, y) were of the form (26041 j, 3397 k). An similar situation appears to be true of 22 x^2 - 29 y^2 = 1, where all solutions (x, y) seem to be of the form (31 j, 27 k). For which (a, b) do all solutions of a x^2 - b y^2 = 0 have the form (r k, s j) for some fixed r, s > 1?

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Re: [math-fun] hex on manifolds
by Dan Hoey 01 May '03

01 May '03

Bill Thurston writes: > Does anyone have some thoughts about generalizations of hex to > various manifolds? > ... For CP^2, a good rule seems to be to declare the winner to be > anyone such that H^2(their colors) is non-zero. Forgive me if this is exactly what you said, but I suggested playing hex on the projective plane a few years ago. Bill Taylor pointed out the natural winning configuration: a path that is not contractible to a point. (If that isn't what you said, what's CP^2?) You can play on a "Y" board, which is graph-isomorphic to a hemi-icosahedron with the sides triangulated. Edge moves are made by playing two antipodal stones. On the commercial board (available from Kadon), you play on the vertices of the triangulation. If you prefer ASCII art, I wrote a program that produces drawings where you play in the spaces. Here are three sizes of board. For the smallest, there's only one starting move, only one response, and then two of the four third moves are winners. The next size up has two openers--are they both winners? ____ ____ / \__ ____/ \ / __/ \__ __/ \__ \ \__/ \ \__ __/ __/ \__/ / \ \____/ \ __/ \____/ / \ \ \__/ \ \ __/ / \__ \ \ ____ / / \____/ / __/ \____/ __/ \__/ / / \____ \__/ / \__/ / / \__/ / \ \ / __/ \__ / \__/ __/ __/ __/ __/ \ \ \__/ \__/ \__ \____ \ \__/ / \__/ \__/ \ \____/ / \ / \ \____/ \__ \____/ / / / \__/ \__/ / / / __/ \__ \____ \ \__ \__ / \____/ \ \ \ \__/ \__ \____/ \__ \__/ \__/ \__/ / \ \__/ / / \ \____/ \__ \__ \ \ \__/ / / \ \__/ / __/ \__ \____/ \__ \__ \____/ \__ \__/ / / \ \__/ \__ \____/ \ \__ \__/ \ \__/ \ \ / / / \ \____/ \ \____/ \__ \__ \____/ / / \ \__/ / __/ \ \____/ \ \ \__/ \__ \__/ / / \ \__/ \__ \____/ \ \__ \__ \__ \__/ \ \__/ / / \ \__/ \ \____/ \__/ \__ \____/ / / \ \__/ / / \ \____/ \ \ \ \____/ / / / \ \____/ \____/ \ \__ \__ \__ \ \__/ \__/ \____/ \ \____/ \__/ \__/ \__ / / \ / \ / \ \____/ \ \ \ \ \__/ \__/ \__/ \____/ \__ \__ \__ \__ \__ / \ / \ / \ / \ / \__/ \__/ \__/ \__/ \ \ \__/ \__/ \____/ \__/ \ \ \ \ \ / / \ / \ / \ / \ __/ __/ __/ __/ / \__/ \__/ \__/ \____/ \__/ \__/ \__/ \__/ \__/ / \ / \ / \ / \____/ / / / / \ \__/ \__/ \____/ / \____/ __/ __/ __/ / / \ / \____/ \____/ / \__/ \__/ \__/ \ \ \__/ / \ / \____/ / / / \__/ / \ \ \__/ / \____/ __/ __/ / \ \ \__/ __/ \____/ / \__/ \__/ \ \__/ / \__/ / \____/ / / / / \ \ \____/ / \____/ __/ \ \ \__/ __/ \____/ / \__/ \__/ / \__/ __/ \____/ __/ / \ \ \____/ __/ \__/ \ \__/ __/ \____/ __/ / / \__/ __/ \____/ \ \ \____/ __/ \__/ __/ \____/ / \__/ __/ \ \____/ \____/ By the way, did anyone figure out toral 4x4 tic-tac-toe? Dan Hoey(a)AIC.NRL.Navy.Mil

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[math-fun] math writing
by Richard Schroeppel 01 May '03

01 May '03

Having identified deficiencies in popular math writing, the obvious next step is to do better. If the people doing the writing have better examples to work from, maybe the result will be more accurate. Anyone care to tackle Dan's comment > (such as assuming that n-manifolds are > subsets of n+1 space) and come up with an accurate explanation of manifold, understandable at the level of a college graduate who majored in, say, chemistry? Rich rcs(a)cs.arizona.edu

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[math-fun] Re: nonlinear constraints on hypergeometric parameters--threat or menace?
by Robert S. Maier 01 May '03

01 May '03

A couple of years ago on this list, Bill Gosper asked about hypergeometric identities in which the parameters are constrained nonlinearly, instead of linearly. The example he gave was the summation formula > c + b + a + 4 >hyper_f ([c - a + 1, c + b, c - b + 1, c + a, -------------], > 5, 4 3 > > 3 c + b + a + 1 1 > [b + a, c, c + -, -------------], -) > 2 3 4 > > (b + a - 1)! (2 c + 1)! > = -------------------------------------------, > c (c + b + a + 1) (c + a - 1)! (c + b - 1)! > which holds if c = a*b. I've just written a paper on how Euler's transformation can be extended from the Gauss hypergeometric function to hypergeometric functions of higher order. It's at http://www.arxiv.org/math.CA/0302084 . The natural analogues of Euler's transformation turn out to involve, lo and behold, nonlinear parametric constraints. As a corollary, I've derived a curious nonlinearly parametrized summation formula for 2F1(-1), namely 2F1 (6t^2 - 11t + 6, 4; -6t^2 + 5t + 3; -1) = -t (6t+1) (6t^2 - 5t - 2). This can alternatively be derived from classical contiguous function relations, by uniformizing a genus-0 algebraic curve. But it certainly looks mysterious. In the course of writing the paper, I discovered an even more intriguing identity discovered by Niblett (a student of Chaundy's, I think) in the 1950s. The citation is J. D. Niblett, "Some Hypergeometric Identities", Pacific J. Math. 2 (1953), 219-224. Niblett's identity is a quadratic function transformation, which expresses 5F4(x) in terms of 3F2(-4x/(1-x)^2) whenever the hypergeometric parameters satisfy a certain nonlinear condition. There are three free parameters. Does anyone have any other examples of hypergeometric transformations or summation formulas with nonlinearly constrained parameters? How to verify any such identity algorithmically is an interesting problem. --Robert

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Re: [math-fun] cubic quaternions
by Mike Stay 01 May '03

01 May '03

>What you are doing >wrong is not to notice that if w is the exchange factor to move R >across Q in one direction it is (w inverse} which you need to move >it in the other direction, not w. It doesn't matter whenever >w = (w inverse), e.g. quaternions with w = -1. Ah! Thank you. >regards, > >Harold V. McIntosh -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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