[math-fun] q-trigs with sin^2 + cos^2 == 1
* Bill Gosper <billgosper@gmail.com> [Aug 29. 2011 16:54]:
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Unfortunately sin^2+cos^2 /=1 in q-land.
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()). (Implementations are given in http://www.jjj.de/pari/qtrig2r.gpi the nonregularized ones are in http://www.jjj.de/pari/qtrig2.gpi ) I recently did essentially the same with the products prod(n>=0, 1 - x*q^n) and prod(n>=0, 1 + x*q^n) (See http://www.jjj.de/pari/qtrig1r.gpi the nonregularized ones are in http://www.jjj.de/pari/qtrig1.gpi ) [the majority of lines in both files are relations given as comments; no need to understand pari syntax] Are the latter ones easily related to your q-trigs? For those in qtrig2r.gpi one can happily rediscover all kinds of modular equations. I was slightly excited to find a two-parameter version of Theta4(q^2)^2 = Theta4(q) * Theta3(q) which, expressed in my q-exp() is E(+q^2, -q^2)^2 = E(+q, -q) * E(-q, +q) 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. Btw. your paper stops at the prime p=5. I could, with some effort, get one modular equation for p=7 (indeed Somos' q56_18_240a) but none for p>7. The _only_ thing I rely on is that my program expressing things as eta-products is working (and it's hard to imagine things going wrong there), the rest comes from (the analogs of) most simple trigonometric identities such as tan() * cot() == 1. (That is why I still much like my q-trigs...) cheers, jj
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Joerg Arndt