[...]
rwg>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 r>egularized q-trig identities . All we do is replace some occurrences of 2 with 1+q, and some occurrences of 0 with 1-q !-) --rwg
E.g., using the above and some of the identities in http://arxiv.org/abs/1006.5652 <http://arxiv.org/abs/1006.5652%7D>, rcosq[q, (1 + q) x] == 2 rcosq[q^2, q x] rsinq[-q, (-1 + q) x] rsinq[q^2, q x] + rcosq[-q, (-1 + q) x] (1 - 2 rsinq[q^2, q x]^2) rsinq[q, (1 + q) x] == 2 rcosq[-q, (-1 + q) x] rcosq[q^2, q x] rsinq[q^2, q x] + rsinq[-q, (-1 + q) x] (-1 + 2 rsinq[q^2, q x]^2) where we replaced some occurrences of 0 with rsinq[-q, (-1 + q) x]. And some occurrences of 1 with the rcosq. Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless the author means Jesse's or Michael's. --rwg Dear Dentist: If at some point I offer to tell you the hiding place of Ayman al-Zawahiri, it means you are overirrigating.