en>EmausBot: Bot: Migrating 1 interwiki links, now provided by Wikidata on d:Q12139807

2013-04-28T13:17:28Z

Bot: Migrating 1 interwiki links, now provided by Wikidata on d:Q12139807

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	~~Hi there~~. ~~Let me begin by introducing~~ the ~~author~~, ~~her name~~ is ~~Sophia~~. ~~Kentucky~~ is ~~where I~~'~~ve usually been living~~. ~~To climb~~ is ~~something I really enjoy doing~~. He is ~~an info officer~~.<br><br>~~my website best psychic readings~~ ([~~http~~://~~jplusfn~~.~~gaplus~~.kr/xe/~~qna~~/~~78647 click~~ for more ~~info~~])		{{Lowercase\|title=''q''-Vandermonde identity}}
			In [[mathematics]], in the field of [[combinatorics]], the '''''q''-Vandermonde identity''' is a [[q-analogue\|''q''-analogue]] of the [[Chu-Vandermonde identity]]. Using standard notation for [[q-binomial coefficient\|''q''-binomial coefficients]], the identity states that

			:<math>
			\binom{m + n}{k}_{\!\!q}
			=
			\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)}.
			</math>
			(The nonzero contributions to this sum come from values of ''j'' such that the ''q''-binomial coefficients on the right side are nonzero, that is, <math>\max(0, k - m) \le j \le \min(n, k)</math>.)

			==Other conventions==
			As is typical for ''q''-analogues, the ''q''-Vandermonde identity can be rewritten in a number of ways. In the conventions common in applications to [[quantum groups]], a different ''q''-binomial coefficient is used. This ''q''-binomial coefficient, which we denote here by <math>B_q(n,k)</math>, is defined by <math> B_q(n, k) = q^{-k(n-k)} \binom{n}{k}_{\!\!q^2}</math>. (In particular, it is the unique shift of the "usual" ''q''-binomial coefficient by a power of ''q'' such that the result is symmetric in ''q'' and <math>q^{-1}</math>.) Using this ''q''-binomial coefficient, the ''q''-Vandermonde identity can be written in the form
			:<math>
			B_q(m + n,k)
			=
			q^{n k}
			\sum_{j}
			q^{-(m+n)j}
			B_q(m,k - j) B_q(n,j). \,
			</math>

			==Proofs of the identity==

			As with the (non-''q'') Chu-Vandermonde identity, there are several possible proofs of the ''q''-Vandermonde identity. We give one proof here, using the [[q-binomial theorem\|''q''-binomial theorem]].

			One standard proof of the Chu-Vandermonde identity is to expand the product <math>(1 + x)^m (1 + x)^n</math> in two different ways. Following Stanley,<ref>[[#EC1\|Stanley (2011)]], Solution to exercise 1.100, p. 188.</ref> we can tweak this proof to prove the ''q''-Vandermonde identity, as well. First, observe that the product
			:<math>(1 + x)(1 + qx) \cdots (1 + q^{m + n - 1}x)</math>
			can be expanded by the ''q''-binomial theorem as
			:<math>(1 + x)(1 + qx) \cdots (1 + q^{m + n - 1}x)</math>
			::<math> = \sum_k q^{k(k-1)/2} \binom{m + n}{k}_{\!\!q} x^k.</math>
			Less obviously, we can write
			:<math>(1 + x)(1 + qx) \cdots (1 + q^{m + n - 1}x)</math>
			::<math> = \left((1 + x)\cdots (1 + q^{m - 1}x)\right) \cdot \left((1 + (q^m x))(1 + q(q^m x)) \cdots (1 + q^{n - 1}(q^m x))\right)</math>
			and we may expand both subproducts separately using the ''q''-binomial theorem. This yields
			:<math>(1 + x)(1 + qx) \cdots (1 + q^{m + n - 1}x)</math>
			::<math> = \left(\sum_i q^{i(i - 1)/2} \binom{m}{i}_{\!\!q} x^i \right) \cdot \left(\sum_i q^{mi + i(i - 1)/2} \binom{n}{i}_{\!\!q} x^i \right).</math>
			Multiplying this latter product out and combining like terms gives
			:<math> \sum_k \sum_j \left(q^{j(m - k + j) + k(k - 1)/2} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q}\right)x^k.</math>
			Finally, equating powers of <math>x</math> between the two expressions yields the desired result.

			This argument may also be phrased in terms of expanding the product <math>(A + B)^m(A + B)^n</math> in two different ways, where ''A'' and ''B'' are [[Operator (mathematics)\|operator]]s (for example, a pair of matrices) that "''q''-commute," that is, that satisfy <math> BA = q AB</math>.

			<!-- On one hand, expanding each factor separately, we have by the [[q-binomial theorem \| ''q''-binomial theorem]] that

			:<math>
			(A + B)^m(A + B)^n \,
			</math>

			::<math>
			= \left(\sum_{i=0}^m
			\binom{m}{i}_{\!\!q}A^{i}B^{m-i}\right)
			\left(\sum_{j=0}^n
			\binom{n}{j}_{\!\!q}A^{j}B^{n-j}\right)
			</math>

			::<math> = \sum_{i,j}
			\binom{m}{i}_{\!\!q}
			\binom{n}{j}_{\!\!q} A^{i}B^{m-i}A^{j}B^{n-j}
			</math>

			::<math> = \sum_{i,j}
			\binom{m}{i}_{\!\!q}
			\binom{n}{j}_{\!\!q} q^{j(m-i)}A^{i+j}B^{m+n-i-j},
			</math>

			where in the last step we make repeated use of the ''q''-commutation relation to switch the positions of ''A'' and ''B'' factors. We can rewrite this sum by collecting like terms, which gives

			: <math> (A + B)^m (A + B)^n =
			\sum_k \left(\sum_{j}
			\binom{m}{k-j}_{\!\!q}
			\binom{n}{j}_{\!\!q} q^{j(m-k+j)}\right)
			A^k B^{m + n - k}.</math>

			On the other hand, we can expand <math>(A + B)^m(A + B)^n</math> more easily by noting that <math>(A + B)^m(A + B)^n = (A + B)^{m + n}</math> and so

			:<math>(A + B)^m (A + B)^n = \sum_k \binom{m + n}{k}_{\!\! q} A^k B^{m + n - k}.</math>

			Equating coefficients of the two expressions gives
			:<math>
			\binom{m + n}{k}_{\!\!q}
			=
			\sum_{j}
			\binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)},
			</math>
			as claimed. -->

			=== Proof by counting subspaces ===

			== Notes ==
			{{Reflist}}

			== References ==
			* {{cite book \|title=Enumerative Combinatorics, Volume 1
			\|ref=EC1
			\|author1=Richard P. Stanley
			\|authorlink=Richard P. Stanley
			\|edition= 2
			\|url=http://www-math.mit.edu/~rstan/ec/ec1.pdf
			\|year=2011
			\|accessdate=August 2, 2011}}

			* Exton, H. (1983), ''q-Hypergeometric Functions and Applications'', New York: Halstead Press, Chichester: Ellis Horwood, 1983, ISBN 0853124914, ISBN 0470274530, ISBN 978-0470274538

			* {{cite journal \|author1=Gaurav Bhatnagar \|title=In Praise of an Elementary Identity of Euler \|year=2011 \|volume=18 \|issue=2 \|journal=Electronic J. Combinatorics, P13, 44pp. \|arxiv=1102.0659}}

			* {{cite journal \|author1=Victor J. W. Guo \|title=Bijective Proofs of Gould's and Rothe's Identities \|doi=10.1016/j.disc.2007.04.020 \|year=2008 \|journal=Discrete Mathematics \|volume=308 \|issue=9 \|pages=1756 \|arxiv=1005.4256}}

			* {{cite arXiv \|eprint=math/0309108 \|author1=Sylvie Corteel \|author2=[[Carla Savage]] \|title=Lecture Hall Theorems, q-series and Truncated Objects \|class=math.CO \|year=2003}}

			{{DEFAULTSORT:Q-Vandermonde Identity}}
			[[Category:Combinatorics]]
			[[Category:Q-analogs\|*]]
			[[Category:Mathematical identities]]

en>JohnBlackburne: /* See also */ disambiguate Rotation group to Rotation group SO(3)

2012-02-05T03:21:52Z

See also: disambiguate Rotation group to Rotation group SO(3)

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Hi there. Let me begin by introducing the author, her name is Sophia. Kentucky is where I've usually been living. To climb is something I really enjoy doing. He is an info officer.<br><br>my website best psychic readings ([http://jplusfn.gaplus.kr/xe/qna/78647 click for more info])

Inversion transformation - Revision history

en>EmausBot: Bot: Migrating 1 interwiki links, now provided by Wikidata on d:Q12139807

en>JohnBlackburne: /* See also */ disambiguate Rotation group to Rotation group SO(3)