I've been playing with Q-factorials: QF(0) == 1 QF(N+1) == QF(N) * (1 + Q + Q^2 + ... + Q^N). When Q=1, this gives QF(N) = N!. The degree of QF(N) is Triangle(N-1) = N(N-1)/2. Q-binomials seem to be polynomials: QB(A,B) = QF(A+B) / (QF(A) QF(B)). The degree of QB(A,B) is AB. The degree of QB(A,B,C) = QF(A+B+C) / (QF(A)QF(B)QF(C)) is AB+AC+BC. Etc. Perhaps we can use some of our standard bag of tricks for defining QF(N) for non-integer N. Put QF(Z) ~= Q ^ [ Z(Z-1)/2 ] for RealPart(Z) large, and divide by (Q^Z-1)/(Q-1) to go from Q(Z) to Q(Z-1). There's a set of functional equations similar to X! (X+1/2)! = (2X+1)! * stuff. Presumably there's also a reflection equation with QF(Z) QF(-Z) = more-stuff that looks like pi Z / sin(pi Z). Rich
On 6/8/06, Schroeppel, Richard <rschroe@sandia.gov> wrote:
I've been playing with Q-factorials: There's a really nice little expository book by Kac and Cheung, _Quantum Calculus_, that covers all of this stuff. Mathworld also has a bunch of pages on it.
Perhaps we can use some of our standard bag of tricks for defining QF(N) for non-integer N.
The q-Gamma function is \int_0^{\infty} x^{t-1} E_q^{qx} d_q x where E_q = \sum_{j=0}^{\infty} q^{j(j-1)/2} x^j / [j]! [j]! = [j] [j-1] [j-2] ... [1] [j] = (q^j - 1) / (q - 1) \int f(x) d_q x = \sum_{n=0}^{\infty} a_n x^{n+1} / [n+1]! + C and f(x) = \sum_{n=0}^{\infty} a_n x^n / [n]! We can integrate suitably nice functions with the Jackson integral \int f(x) d_q x = (1-q) x \sum_{j=0}^{\infty} q^j f(q^j x)
Rich
-- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
On Thu, 8 Jun 2006, Schroeppel, Richard wrote:
I've been playing with Q-factorials:
QF(0) == 1 QF(N+1) == QF(N) * (1 + Q + Q^2 + ... + Q^N).
There are people who make a living creating "q-analogs" of well-known functions, identities, etc. You can find some leads here: http://mathworld.wolfram.com/q-Analog.html My favorite is the q-binomial(n,k) which as q --> 1 goes to the usual binomial(n,k). Among other things q-binomial(n,k) is the number of k-dimensional subspaces of an n-dimensional vector space over a field with q elements. It is a polynomial in q and the coefficients have a combinatorial interpretation.
Rich wrote:
I've been playing with Q-factorials:
QF(0) == 1 QF(N+1) == QF(N) * (1 + Q + Q^2 + ... + Q^N). [...] Q-binomials seem to be polynomials: QB(A,B) = QF(A+B) / (QF(A) QF(B)).
Other people have offered pointers and commented on q-Gamma, but the self-contained explanation of the Q-binomials is so nice that I can't let this go by. (I use this example as my introduction to q-deformations for grad students interested in the overlap of combinatorics and Lie theory.) The q-binomial is indeed a polynomial. The easy way to see that is to check the initially non-obvious recurrence relation, in your notation, QB(A,B) = QB(A-1,B) + Q^A QB(A,B-1) So what's really going on here? Well, think of the non-Q version, Binomial(a,b) = (a+b)!/a!b!, as counting the number of north/east grid walks from the southwest to northeast corners of an a-by-b rectangle. In that case, the usual recurrence relation B(a,b) = B(a-1,b) + B(a,b-1) reflects the fact that any such path begins with either a step north or east, after which it's a walk in a box with one side-length decreased. The Q version works exactly the same way, except that now, each walk contributes Q^w, where w is the number of unit boxes to the south/east of the walk. Some walks begin with a step east, in which case it's a normal walk in an (a-1)-by-b box. Others begin with a step north, in which case the entire bottom row of boxes are under the line, and the weight of the remaining walk in the a-by-(b-1) box should be increased by a (multiplied by Q^a). I can't stop without pointing out one last thing. In the non-Q version, a step north and a step east "commute", in that a walk containing ...NE... and the same with ...EN... get you to the same place and are otherwise indistinguishable. But in the Q version, these walks differ by a factor of Q, because one box has changed between being over and under the walk. This is a baby example of how the "generating function" Q is the same as the "quantum commutation" Q (where ab = Qba). --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
eclark>There are people who make a living creating "q-analogs" of well-known functions, identities, etc. You can find some leads here: http://mathworld.wolfram.com/q-Analog.html That page is OK, except it mentions the inferior Exton book and not the bible: Gaspar & Rahman's Basic Hypergeometric Series. You could regard q-(basic hypergeometric) series as sums of q-factorials (there is even a "q-Gosper" algorithm, not mine), but the q-pochhammer notation is handier: n-1 (a;q) := (1-a)(1-qa)...(1-q a), n which can be used to define q-factorial at complex z: (q;q) -z oo QF(z) := (1-q) ----------, z+1 (q ;q) oo (using Rich's notation). Unanalogously to ordinary factorials and pochhammers, you'd need a log to define (a;q)_z in terms of q-factorials: a z QF(z + a - 1) (q ; q) = (1 - q) -------------, z QF(a - 1) but all you need is (a; q) oo (a; q) := -----------. z z (a q ; q) oo If this is news to anyone, rereading some of my old emails should make more sense now, e.g., Subject: the q Filbert matrix [long] [...] I tried substituting a couple of candidates for q-Fibonaccis to see if inverting made huge q-polynomials. It is tempting to define qib(0):=0, qib(1):=1, n - 2 qib(n) = qib(n - 1) + q qib(n - 2)), for then n 1 floor(- - -) 2 2 ==== 2 \ k qib(n) = > qbn(n - k - 1, k) q / ==== k = 0 where qbn is the q-binomial coefficient (q; q) n qbn(n, k) = -------------------. (q; q) (q; q) k n - k [...] and Subject: a month in the laboratory ... the probability that an nxn bitmatrix will have (mod 2) rank = k is 2 n - k + 1 k + 1 (n - k) (q ; q) (q ; q) q k n - k P(n,k) := -------------------------------------------, (q; q) n - k if the entries are 1 with probability p = 1-q . Thus the peculiar base q identity Sum P(n,k) = 1. k In particular, the probability of being nonsingular is just (q;q)_n. rcs>There's a set of functional equations similar to X! (X+1/2)! = (2X+1)! * stuff. n n n n z (q ; q ) (1 - q) QF(n z, q) n oo QF(z, q ) = ------------------------------------------------, n - 1 n - 1 n z - ----- /===\ n 2 | | k n (q; q) (1 - q ) | | QF(z - -, q ) oo | | n k = 1 whose q->1 limit is the usual n - 1 n - 1 ----- ----- 2 2 2 %pi (n z)! z! = ------------------------- n - 1 /===\ n z + 1/2 | | i n | | (z - -)! | | n i = 1 rcs>Presumably there's also a reflection equation with QF(Z) QF(-Z) = more-stuff that looks like pi Z / sin(pi Z). A q-factorial reflection formula was the motivation for my q-sin, q-cos definitions in "Experiments and discoveries in q-Trigonometry" in proceedings of 2000 U. Florida (Gainesville) conference: Symbolic Computation, Number Theory, Special Functions, Physics & Combinatorics, (Garvan & Ismail, eds.) pp79-106. The expression (q;q)_oo persists in the reflection formula, hence my excitement at reexpressing it as a simple Theta_2. (The paper gives it as a q-extension of pi.) Unfortunately(?), there are several other plausible q-extensions of sin. --rwg Whale oil is renewable energy.
I gurgled:
The expression (q;q)_oo persists in the reflection formula, hence my excitement at reexpressing it as a simple Theta_2.
DAWK! The equivalent and marginally nicer formula, inf pi 1/6 /===\ theta (--, q ) 1/24 | | n 1 3 q | | (1 - q ) = -----------------, | | sqrt(3) n = 1 has lurked for years as an immediate consequence of two neighboring identities in my Theta demo notebook, www.tweedledum.com/rwg/thet.gif, a scrollable bitmap too tall for IE. So the q-factorial reflection formula can be written 2 2 QF(- z, q ) QF(z - 1, q ) = 2 %pi ------ log(q) 4 %pi 1/3 (z - 1) z 2 theta (0, %e ) theta (---, q ) q (1 - q ) 2 1 3 ---------------------------------------------------------, 2 %pi ------ 2 %pi 1/6 log(q) 3 theta (---, q ) theta (%pi z, %e ) 1 3 1 and numerous clumsy expressions in the thet.gif Formulary (e.g., d45, d47, d102, d84, ...) can be simplified: 1 - n ----- n - 1 n %pi 1/3 2 n /===\ theta (---, q ) 3 theta (n z, q ) | | %pi j 1 3 1 | | theta (z + -----, q) = ----------------------------------------, | | 1 n %pi n/3 j = 0 theta (---, q ) 1 3 n - 1 /===\ | | %pi j | | theta (z + -----, q) = | | 2 n j = 0 1 - n ----- n %pi 1/3 2 %pi n theta (---, q ) 3 theta (n (z + ---), q ) 1 3 1 2 ------------------------------------------------, %pi n/3 theta (---, q ) 1 3 1 - n ----- n - 1 n %pi 1/3 2 n /===\ theta (---, q ) 3 theta (n z, q ) | | %pi j 1 3 4 | | theta (z + -----, q) = ----------------------------------------, | | 4 n %pi n/3 j = 0 theta (---, q ) 1 3 n - 1 /===\ | | %pi j | | theta (z + -----, q) = | | 3 n j = 0 1 - n ----- n %pi 1/3 2 %pi n theta (---, q ) 3 theta (n (z + ---), q ) 1 3 4 2 ------------------------------------------------, %pi n/3 theta (---, q ) 1 3 even ---- 2 /===\ | | %pi j even/2 %pi 1/3 %pi even/3 | | theta (-----, q) = theta (---, q ) theta (---, q ) | | 1 even 1 3 1 3 j = 1 2 even sqrt(-------------------------------------), even/2 + 1 theta (0, q) theta (0, q) 3 3 4 odd - 1 ------- 2 odd - 3 /===\ ------- | | %pi j 2 %pi 1/3 %pi odd/3 | | theta (-----, q) = theta (---, q ) theta (---, q ) | | 1 odd 1 3 1 3 j = 1 1 - odd ------- 4 3 sqrt(odd). --rwg
I always thought n /===\ k (q; q) | | 1 - q n n! := Gamma(n + 1) = | | ------ = --------, q q | | 1 - q n k = 1 (1 - q) (for integer n), based on the notion that "q-number" n := n 1 - q ------. 1 - q But O. Pavlyk points out that quantum theorists seem to prefer -n n q - q "q-number" n := --------, and thus -1 q - q 1 k n -- - q 2 2 /===\ k (q ; q ) | | q n n! := | | ------- = --------------------, q | | 1 (n - 1) n k = 1 - - q --------- q 2 2 n q (1 - q ) (http://arxiv.org/abs/math.QA/0207244, http://arxiv.org/abs/quant-ph/9506036). Does anyone know the motivation for this? It wouldn't be inconceivable that the boring definition n! = Gamma(n+1) break down in q-land. Physicists tend to bend definitions to their narrow purposes rather than the greater good of mathematics, so let's see what happens to the reflection and tuplication formulae. Using qfac for the traditional(?) function and qfaq for the "quantized" one, Reflection: 2 3 (z - 1/2) (q; q) sqrt(q) (1 - q) inf qfac(- z, q) qfac(z - 1, q) = -----------------------------------, 2 2 %pi ------ 2 %pi log(q) sqrt(- ------) theta (z, %e ) log(q) 1 2 2 3 2 (q ; q ) (1 - q ) inf qfaq(- z, q) qfaq(z - 1, q) = ---------------------------------------, 2 %pi ------ 3/4 %pi log(q) q sqrt(- ------) theta (z, %e ) log(q) 1 so qfaq is nicer! m-Tuplication: m m - 1 m m m n /===\ (q ; q ) qfac(m n, q) (1 - q) | | k m inf | | qfac(n - -, q ) = -----------------------------------, | | m m - 1 k = 0 m n - ----- m 2 (q; q) (1 - q ) inf m - 1 /===\ | | k m | | qfaq(n - -, q ) = | | m k = 0 (m - 1) (12 m n - 5 m + 1) m -------------------------- 2 m 2 m 12 2 m n (q ; q ) qfaq(m n, q) q (1 - q ) inf --------------------------------------------------------------------. m - 1 m n - ----- 2 2 2 m 2 (q ; q ) (1 - q ) inf Gack! I guess quantum hackers don't tuplicate. I haven't looked at the respective q-extensions of Sum x^n/n!, e.g., but it's hard to imagine anything that would redeem that tuplication mess. --rwg
Oops! I left a factor of pi out of the reflection theta-ands. Should be 2 3 (z - 1/2) (q; q) sqrt(q) (1 - q) inf qfac(- z, q) qfac(z - 1, q) = -------------------------------------, 2 2 %pi ------ 2 %pi log(q) sqrt(- ------) theta (%pi z, %e ) log(q) 1 2 2 3 2 -3/4 (q ; q ) (1 - q ) q inf qfaq(- z, q) qfaq(z - 1, q) = --------------------------------------. 2 %pi ------ %pi log(q) sqrt(- ------) theta (%pi z, %e ) log(q) 1 Well, that didn't arouse any q-slingers (why isn't PATHETIC the opposite of APATHETIC?), but maybe this will: How would you like your computer algebra system to say -1/24 log(q) q eta(------) instead of qpoch(q,q,oo) ? 2 i pi Rationale: the latter is less simple because it is a special function of three arguments instead of one. And speaking of logs (and 7), I invoke the coffeehouse clause of the math-fun charter to mention one-log fires. A storm on 1 Jan felled a lot of nearby trees (one fatal to a bicyclist), including a 6" thick mystery hardwood that splintered down to broomstraws. I sawed off an adjoining segment, stood it in a fireplace on the frayed end, lit it with no paper or kindling, and it burnt slowly and completely. Is this a hint to commercial firewood cutters? (Forget chainsaws--all you need is a big drill and quarter sticks of dynamite.) Another idea: drill a tight cluster of seven, say, holes down the center of a fat log; stand on end; and light the thin walls between the holes. --rwg Save Hangar One
participants (5)
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Edwin Clark -
Michael Kleber -
Mike Stay -
R. William Gosper -
Schroeppel, Richard