Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
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
[...]
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.
* Bill Gosper <billgosper@gmail.com> [Aug 31. 2011 13:17]:
[...]
Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless the author means Jesse's or Michael's.
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).
--rwg Dear Dentist: If at some point I offer to tell you the hiding place of Ayman al-Zawahiri, it means you are overirrigating. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
* Joerg Arndt <arndt@jjj.de> [Aug 31. 2011 19:09]:
* Bill Gosper <billgosper@gmail.com> [Aug 31. 2011 13:17]:
[...]
Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless the author means Jesse's or Michael's.
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).
And NOW I see: Cieslinsky has ( f(q*z) - f(q) ) / ( q*z - z ) but that should be ( f(q*z) - f(z) ) / ( q*z - z ). ^ ^ ^ | OMG, a typo !!11!!12 Let's call that the Michealala Jesse James's derivative-oid of the fifth kind (possibly useful for quantum chromium dynamics in triply bretzeled(*) space time), OK? (*) that time has to be salted, and served with beer.
On 8/31/2011 11:30 AM, Joerg Arndt wrote:
* Joerg Arndt<arndt@jjj.de> [Aug 31. 2011 19:09]:
* Bill Gosper<billgosper@gmail.com> [Aug 31. 2011 13:17]:
[...]
Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless the author means Jesse's or Michael's.
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).
And NOW I see: Cieslinsky has
( f(q*z) - f(q) ) / ( q*z - z ) but that should be ( f(q*z) - f(z) ) / ( q*z - z ). ^ ^ ^ | OMG, a typo !!11!!12
Let's call that the Michealala Jesse James's derivative-oid of the fifth kind (possibly useful for quantum chromium dynamics in triply bretzeled(*) space time), OK?
Bretzeled or pretzeled(*)?
(*) that time has to be salted, and served with beer.
(*) I quite like salted pretzels.
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* Robert Smith <quadricode@gmail.com> [Sep 01. 2011 09:16]:
[...]
Let's call that the Michealala Jesse James's derivative-oid of the fifth kind (possibly useful for quantum chromium dynamics in triply bretzeled(*) space time), OK?
Bretzeled or pretzeled(*)?
Location dependent: German: http://de.wikipedia.org/wiki/Bretzel English: http://en.wikipedia.org/wiki/Pretzel Top right image of German version shows exactly what I meant (a Laugenbrezel; en(?): lye pretzel).
[...]
participants (3)
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Bill Gosper -
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Robert Smith