April 2003 - mathfun@mailman.xmission.com

[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

But if Q^3 is a scalar q, you have QRQQ = w RQQQ = w q R QRQQ = ww QQQR = w^2 q R so w must equal w^2 -> w can only be 0,1 I was looking at the same thing you described while trying to get an extension of geometric algebra (the even grades of 3-d geometric algebra are the quaternions) where instead of the product rule e_i^2 = (e_i . e_i) + (e_i /\ e_i) = 1+0 = 1 we have e_i^3 = 1 but I spotted the problem above and couldn't think of a solution for it (the commutative case being boring...) It reminded me of anyons, which only work in 2-D, so I think it might be possible to get an arbitrary phase on the exchange operator if you restrict yourself to powers of e_1 and e_2. Rich wrote: >I've been wondering about doing a cubic analog of the quaternions. > >The idea is to have two generators Q and R with Q = cbrt(q) and >R = cbrt(r), and the non-commutative multiplication rule R*Q = w Q*R >(where w = cbrt(1) = (-1 + i sqrt3)/2 = e^(2 pi i/3)). -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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[math-fun] cubic quaternions - zero divisors
by Richard Schroeppel 30 Apr '03

30 Apr '03

Thanks for the suggestions! In the special case Q = cbrt 2, R = cbrt 3, I spotted a zero divisor. (Q-R)^3 = Q^3 - R^3 since the cross terms cancel out. = 2 - 3 = -1 in the special case. So 1 + (Q-R)^3 = 0, and it's the product of 1 + Q - R with 1 - (Q-R) + (Q-R)^2. This construction seems to work on a lot of cases. Rich rcs(a)cs.arizona.edu

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Re: [math-fun] cubic quaternions
by reid＠math.arizona.edu 30 Apr '03

30 Apr '03

andy latto points out, in private e-mail, that my definition of central simple division algebra would also include the complex numbers as a csda over R (which would be incorrect). the condition that i inadvertently omitted is that the center of D is precisely k . (it obviously contains k , but is not allowed to be larger.) with this condition, C / R is excluded. sorry! hopefully there are no further omissions! mike

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RE: [math-fun] cubic quaternions
by Andy Latto 30 Apr '03

30 Apr '03

> -----Original Message----- > From: Richard Schroeppel [mailto:rcs@CS.Arizona.EDU] > Sent: Wednesday, April 30, 2003 1:25 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] cubic quaternions > > > I've been wondering about doing a cubic analog of the quaternions. > I haven't proved "no zero divisors", which is an important theorem > in the quaternion case. Or worked out the Norm formula, which is > needed to compute reciprocals. I think that it's guaranteed that you will have zero divisors. I think there's a theorem that the only finite-dimensional division algebras over the reals are the complex numbers, the quaternions, and the octonions. But I might be confused. Or there might be some property that you need to assume before you get limited to these three, that you don't have and don't need for your "cubic quaterions". How do you check for zero divisors? You can express elements of your structure as 9-vectors, and multiplication as a 9x9 matrix. A nonzero determinant is necessary, but not sufficient, for no zero divisors. If the multiplication was commutative, then the matrix would be symmetric, and you would have no zero divisors just if the matrix is positive definite. But what is the corresponding criterion for assymmetric matrices? Andy Latto andy.latto(a)pobox.com

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[math-fun] cubic quaternions
by Richard Schroeppel 30 Apr '03

30 Apr '03

I've been wondering about doing a cubic analog of the quaternions. The idea is to have two generators Q and R with Q = cbrt(q) and R = cbrt(r), and the non-commutative multiplication rule R*Q = w Q*R (where w = cbrt(1) = (-1 + i sqrt3)/2 = e^(2 pi i/3)). For definiteness, I'm imagining q=2 and r=3. The result seems to be a nine-dimensional space, linear combinations of 1, Q, Q^2, R, R^2, Q R, Q^2 R, Q R^2, Q^2 R^2, with coefficients of the shape a + b w. The type of construction seems to guarantee associativity, needing only to check things like R Q^3 = Q^3 R -- required, since Q^3 is in the ground field which should commute with R, and true, since we pick up three factors of w when we move the R through the Qs while sorting the factors. There are some nice properties like (Q+R)^3 = Q^3 + R^3, which happens because the cross terms like QRR come in three orders and when reordered and collected have coefficient 1+w+w^2 = 0. I haven't proved "no zero divisors", which is an important theorem in the quaternion case. Or worked out the Norm formula, which is needed to compute reciprocals. One proof of the Four Square Theorem (every number is the sum of four squares) uses quternion arithmetic. Perhaps there's a proof of the Nine Cube Theorem lurking somewhere (with q=r=1 ?). Has anyone seen this stuff before? Rich rcs(a)cs.arizona.edu

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Re: [math-fun] cubic quaternions
by reid＠math.arizona.edu 30 Apr '03

30 Apr '03

there is something called a "central simple division algebra", of which your construction is probably a special case. if k is a field, then such an object, D , is a finite-dimensional vector space over k , which has an associative multiplication (which is k-linear in each factor), that is also a division ring. such objects are classified by elements of the brauer group of k , which is the continuous cohomology group H^2(Gal(kbar/k) , kbar^*) (where kbar is the separable closure of k ). the dimension of D over k is the square of the order of the corresponding element of Br(k) . for k = Q , its brauer group is huge, so there are many central simple division algebras. for k = R , its brauer group has order 2. the corresponding csda's are R itself, and the quaternion algebra. i've also seen the term "azumaya algebra" used, but this might also include matrix rings over central simple division algebras. i'm not sure of a good reference off the top of my head, but i think cassels and frohlich "algebraic number theory" has some about this. (maybe also jacobson's "lectures in abstract algebra".) mike

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Re: [math-fun] cubic quaternions
by Henry Baker 30 Apr '03

30 Apr '03

Rich: If this ring is associative, then it has a matrix representation. Have you attempted to find the mapping? Henry ----- At 10:25 AM 4/30/03 -0700, Richard Schroeppel wrote: >I've been wondering about doing a cubic analog of the quaternions. > >The idea is to have two generators Q and R with Q = cbrt(q) and >R = cbrt(r), and the non-commutative multiplication rule R*Q = w Q*R >(where w = cbrt(1) = (-1 + i sqrt3)/2 = e^(2 pi i/3)). >For definiteness, I'm imagining q=2 and r=3. > >The result seems to be a nine-dimensional space, linear combinations >of 1, Q, Q^2, R, R^2, Q R, Q^2 R, Q R^2, Q^2 R^2, with coefficients >of the shape a + b w. > >The type of construction seems to guarantee associativity, needing >only to check things like R Q^3 = Q^3 R -- required, since Q^3 is >in the ground field which should commute with R, and true, since >we pick up three factors of w when we move the R through the Qs >while sorting the factors. > >There are some nice properties like (Q+R)^3 = Q^3 + R^3, which >happens because the cross terms like QRR come in three orders and >when reordered and collected have coefficient 1+w+w^2 = 0. > >I haven't proved "no zero divisors", which is an important theorem >in the quaternion case. Or worked out the Norm formula, which is >needed to compute reciprocals. > >One proof of the Four Square Theorem (every number is the sum of >four squares) uses quternion arithmetic. Perhaps there's a proof >of the Nine Cube Theorem lurking somewhere (with q=r=1 ?). > >Has anyone seen this stuff before? > >Rich rcs(a)cs.arizona.edu

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[math-fun] Probabilistic harmonic series
by asimovd＠aol.com 30 Apr '03

30 Apr '03

I think this is an old problem, almost certainly solved, but I don't think I've ever seen the solution: Given the harmonic series but with an independently chosen random sign in front of each term: ±1 ± 1/2 ± 1/3 ± ..., then a. what is the probability of convergence? b. when it converges, what is the probability distribution of the sum? Similarly, one could ask the same questions a. and b. about the closely related series: u_1 + u_2 / 2 + u_3 / 3 +... where each u_n is a complex number independently chosen at random from the unit circle. Anyone know the answers, or a reference? Remarks: 1. It's a bit tricky to even define rigorously what the probability in question a. means. 2. Intutitively it may seem obvious that the series, at least in the first case of ±1 ± 1/2 ± 1/3 ± ..., converges with probability 1. But it's not quite that obvious, since the "arcsine law" says that if S_k is the sum of k independent choices of ±1 at random, then as k -> oo, the most probable fraction(s) of the time that S_k is positive are 0 and 1 -- rather than the 1/2 that most people would guess. (Ditto, of course, for the most probable fraction(s) of the time S_k is negative.) So it's not clear that the random harmonic series isn't closer to the actual harmonic series than to the series with alternating signs. 3. These can be considered the 1- and 2-dimensional cases of a question that can be easily stated n dimensions: Let {v_k} be vectors chosen independently at random from the unit sphere in R^n, and consider the series v_1 + v_2 / 2 + v_3 / 3 + .... --Dan

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