September 2011 - mathfun@mailman.xmission.com

Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
by Bill Gosper 02 Sep '11

02 Sep '11

>rwg> Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless >> the author means Jesse's or Michael's. > joerg> http://en.wikipedia.org/wiki/Q-derivative says > "The q-derivative is also known as the Jackson derivative." > > (I never heard that name before, though). Look closer at the eqn! --rwg

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Re: [math-fun] Connectedness
by Dan Asimov 02 Sep '11

02 Sep '11

I'm just going to hazard a guess, but I suspect that the following discrete subset D of 3-space would require a set of at least (Hausdorff) dimension (dim_H) = 2 to be added to it in order to obtain a connected set. Namely, let discrete set D := {(x,y,z) = (K/N, L/N, 1/N) | (K,L,N) in ZxZxZ+}. As a discrete set, dim_H(D) = 0. Since its limit points constitute R^2 x {0}, and since dim_H(QxQ) = 2 (Q = rationals), I suspect that any countable dense subset of R^2 has Hausdorff dimension = 2, and so intuitively I'd guess that nothing of lower Hausdorff would suffice to add to D so as to form a connected set. --Dan David M. asked: << Myself and another member at http://www.fractalforums.com/ have been having a small discussion about connectedness. Specifically the idea that given any disconnected set of a given dimension d then it's always possible to construct a set with dimension d+1 such that the new set is connected - if considering fractional dimensions then make that floor(d)+1. I have never done any topology beyond the very basics nor any other math relating to general connectedness and am just wondering if the above is correct and if so how goes the proof, when was it proved and who proved it ? >> Sometimes the brain has a mind of its own.

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[math-fun] euler's infinite product, primes of the form 4n+1, 4n+3... ??
by Simon Plouffe 01 Sep '11

01 Sep '11

Hello, I am working on a simple problem which is the following, consider the Euler product with primes, infinity --------' ' | | 1 | | -------- = Zeta(2) | | 1 | | 1 - ---- p = 2 2 p this is well known. Now if we take only the primes of the form 4n+1 and 4n+3 and split (avoiding the number 2), then the product is p1*p3 = Pi^2/8 , p1 is the Euler product with primes of the form 4n+1 and p3 = with the primes of the form 4n+3. Now these 2 values are approximately : p1 = 1.05618212168678504596905 and p3 = 1.16807558536624241169267 The product of p1*p3 being as expected Pi^2/8. the good question is : What is p1 made of actualy ? and p3 ? I tried all the tricks I know and came with NO answer at all. My first guess was a simple rational combination : this is wrong, my second guess was that p1 is a algebraic multiple of Pi : wrong too. And then log(p1) is related to some log(Gamma(a/b)) values ? There are no numerical evidence of that, then I tried all the tables, all the algorithms against that number and my quest was incorrect. Does anybody knows what are these numbers ? Is this a trivial question related to Dirichlet series, I don't get it. The values are correct to about 8 digits. I also tried with the exponent being, 3,4 and 8 : nothing too, a complete mystery. Any clue would help, thanks in advance. My gut feeling is that p1 should be simple (as well as p3 ). Bonne journÃ©e Ã tous. Simon Plouffe

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Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
by Bill Gosper 01 Sep '11

01 Sep '11

> Joerg> One can regularize to get that one >> (via abs( qexp(I*x) ) == 1 for some nice qexp()). >> The paper >> Jan L.\ Cie\'{s}li\'{n}ski: >> {Improved $q$-exponential and $q$-trigonometric functions}, >> arXiv:1006.5652v1 [math.CA], (29-June-2010). >> URL: \url{http://arxiv.org/abs/1006.5652}.} >> does it for the version where limit_{q-->1} gives >> the usual functions (exp(), cos(), sin()). >> rwg> Caution to the casual: In this paper, superscripts of e, E and Script-E > are *not* exponents. In fact, I don't even see a relation of the form > Script-Exp_q1(x)^2 ~ > Script-Exp_q2(2x), so I don't see much motivation, beyond nice > q-difference and > nice Pythagoras. Can we have q, Pythagoras, *and* identities? Ohh, sort of. Using Cie ́slin ́ski's regularized q-exponential rexpq[q_, x_] -> QPochhammer[-1/2 (1 - q) x, q]/QPochhammer[1/2 (1 - q) x, q] and your bisected product observation (below), i.e., QPochhammer[q x, q^2]^2/QPochhammer[-q x, q^2]^2 == ( QPochhammer[-x, -q] QPochhammer[x, q])/(QPochhammer[-x, q] QPochhammer[x, -q]) we have rexpq[q^2, q x]^2 == rexpq[-q, (-1 + q) x] rexpq[q, (1 + q) x] which, as q->1, becomes (e^x)^2 = ?(-1) * e^(2x), where ?(-1) must be some flavor of 1. (Cie ́slin ́ski doesn't explain negative q.) This could take some getting used to. >[...] >Joerg> The equation in question is >> E(+q^2, -q*x)^2 = E(+q, -x) * E(-q, +x) >> >> I stopped investigating, however, after looking >> into Whittaker/Watson chapter.21 and suspecting >> my finding is really a specialization of stuff >> given there. It's just a bisected product but it might be a key to finding rexpq and regularized q-trig identities . All we do is replace some occurrences of 2 with 1+q, and some occurrences of 0 with 1-q !-) --rwg

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