"베일리 사슬(Bailey chain)"의 두 판 사이의 차이

2011년 11월 12일 (토) 06:12 판

이 항목의 수학노트 원문주소

베일리 사슬(Bailey chain)

개요

Bailey chain construction

we derive a new Bailey pair relative to a from a known Bailey pair relative to a
\(\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n\)
\(\beta^\prime_n = \sum_{j=0}^{n}\frac{(\rho_1;q)_j(\rho_2;q)_j(aq/\rho_1\rho_2;q)_{n-j}(aq/\rho_1\rho_2)^j}{(q;q)_{n-j}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_j\)
this is called the Bailey chain construction

special case of Bailey chain

by taking \(\rho_1,\rho_2\to \infty\) in the original Bailey chain , we get
\(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
\(\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r\)
applying the Bailey pair equation
\(\beta^{'}_L=\sum_{r=0}^{L}\frac{\alpha^{'}_r}{(q)_{L-r}(aq)_{L+r}}\)
to the new Bailey pair, we get
\(\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n_1=0}^{n}\frac{a^{n_1}q^{n_1^2}\beta_{n_{1}}}{(q)_{n-n_{1}}}\)

repetition of chain construction

repeating this construction several times
\(\sum_{r=0}^{n}\frac{a^{kr}q^{kr^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n_1=0}^{n}\sum_{n_2=0}^{n_1}\cdots\sum_{n_k=0}^{n_{k-1}}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n-n_{1}}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\)

Taking \(n\to\infty\), we get
\(\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\)

examples

initial Bailey pair
\(\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
\(\beta_{L}=\delta_{L,0}\)
For example, if a=1,
\(\alpha_{L}=(-1)^{L}q^{L(L-1)/2}(1+q^{L})=(-1)^{L}(q^{(3L^2-L)/2}+q^{(3L^2+L)/2})\)
result of Bailey chain applied k-times
\(\alpha_{L}=(-1)^{L}a^{kL}q^{kL^{2}+L^2/2-L/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
\(\beta_{L}=\sum_{L\geq n_1\geq\cdots\geq n_{k-1}\geq0}\frac{a^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2}}{(q)_{L-n_1}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
obtained q-series identity
\(\frac{1}{(q)_{\infty}}\sum_{r=-\infty}^{\infty}(-1)^{r}q^{r((2k+1)r+1-2jk)/2}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+j(n_1+\cdots+n_{k})}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
Setting k=1, a=1, we get the Euler pentagonal number theorem
\((q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}\)
Setting k=2, a=1, we get one of RR identity
\(\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} \)
Setting k=2, a=q, we get one of RR identity
\(\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty}\)
We frequently use Jacobi triple product identity
\(\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\)

if k is bigger than 2, we get some cases of 앤드류스-고든 항등식(Andrews-Gordon identity)
모든 앤드류스-고든 항등식(Andrews-Gordon identity) 을 증명하려면, 베일리 격자(Bailey lattice) 가 필요하다

@@ 1번째 줄: / 1번째 줄: @@
+<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 수학노트 원문주소</h5>
+* [[베일리 사슬(Bailey chain)]]
+<h5>개요</h5>
+<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey chain construction</h5>
+*  we derive a new Bailey pair relative to <em>a</em> from a known Bailey pair relative to <em>a</em><br><math>\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n</math><br><math>\beta^\prime_n = \sum_{j=0}^{n}\frac{(\rho_1;q)_j(\rho_2;q)_j(aq/\rho_1\rho_2;q)_{n-j}(aq/\rho_1\rho_2)^j}{(q;q)_{n-j}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_j</math><br>
+*  this is called the Bailey chain construction<br>
+<h5 style="line-height: 2em; margin: 0px;">special case of Bailey chain</h5>
+*  by taking  <math>\rho_1,\rho_2\to \infty</math> in the original Bailey chain , we get <br><math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math><br><math>\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r</math><br>
+*  applying the Bailey pair equation<br><math>\beta^{'}_L=\sum_{r=0}^{L}\frac{\alpha^{'}_r}{(q)_{L-r}(aq)_{L+r}}</math><br> to the new Bailey pair, we get<br><math>\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n_1=0}^{n}\frac{a^{n_1}q^{n_1^2}\beta_{n_{1}}}{(q)_{n-n_{1}}}</math><br>
+<h5 style="line-height: 2em; margin: 0px;">repetition of chain construction</h5>
+*  repeating this construction several times<br><math>\sum_{r=0}^{n}\frac{a^{kr}q^{kr^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n_1=0}^{n}\sum_{n_2=0}^{n_1}\cdots\sum_{n_k=0}^{n_{k-1}}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n-n_{1}}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}</math><br>
+*  Taking <math>n\to\infty</math>, we get<br><math>\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}</math><br>
+<h5 style="line-height: 2em; margin: 0px;">examples</h5>
+*  initial Bailey pair<br><math>\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}</math><br><math>\beta_{L}=\delta_{L,0}</math><br> For example, if a=1,<br><math>\alpha_{L}=(-1)^{L}q^{L(L-1)/2}(1+q^{L})=(-1)^{L}(q^{(3L^2-L)/2}+q^{(3L^2+L)/2})</math><br>
+*  result of Bailey chain applied k-times<br><math>\alpha_{L}=(-1)^{L}a^{kL}q^{kL^{2}+L^2/2-L/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}</math><br><math>\beta_{L}=\sum_{L\geq n_1\geq\cdots\geq n_{k-1}\geq0}\frac{a^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2}}{(q)_{L-n_1}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br>
+*  obtained q-series identity<br><math>\frac{1}{(q)_{\infty}}\sum_{r=-\infty}^{\infty}(-1)^{r}q^{r((2k+1)r+1-2jk)/2}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+j(n_1+\cdots+n_{k})}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br>
+*  Setting k=1, a=1, we get the Euler pentagonal number theorem<br><math>(q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}</math><br>
+*  Setting k=2, a=1, we get one of RR identity<br><math>\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} </math><br>
+*  Setting k=2, a=q, we get one of RR identity<br>  <math>\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty}</math><br>
+*  We frequently use Jacobi triple product identity<br><math>\sum_{n=-\infty}^\infty  z^{n}q^{n^2}= \prod_{m=1}^\infty  \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)</math><br>
+*  if k is bigger than 2, we get some cases of [[앤드류스-고든 항등식(Andrews-Gordon identity)]]<br>
+*  모든 [[앤드류스-고든 항등식(Andrews-Gordon identity)]]  을 증명하려면, 베일리 격자(Bailey lattice) 가 필요하다<br>
+<h5>역사</h5>
+* http://www.google.com/search?hl=en&tbs=tl:1&q=
+* [[수학사연표 (역사)|수학사연표]]
+<h5>메모</h5>
+* Math Overflow http://mathoverflow.net/search?q=
+<h5>관련된 항목들</h5>
+<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">수학용어번역</h5>
+*  단어사전<br>
+** http://translate.google.com/#en|ko|
+** http://ko.wiktionary.org/wiki/
+* 발음사전 http://www.forvo.com/search/
+* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
+** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
+* [http://www.kss.or.kr/pds/sec/dic.aspx 한국통계학회 통계학 용어 온라인 대조표]
+* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
+* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
+<h5>사전 형태의 자료</h5>
+* http://ko.wikipedia.org/wiki/
+* http://en.wikipedia.org/wiki/
+* [http://eom.springer.de/default.htm The Online Encyclopaedia of Mathematics]
+* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
+* [http://eqworld.ipmnet.ru/ The World of Mathematical Equations]
+<h5>리뷰논문, 에세이, 강의노트</h5>
+<h5>관련논문</h5>
+* Paule, [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.71.6203 The Concept of Bailey Chains]
+* George E. Andrews[http://projecteuclid.org/euclid.pjm/1102708707 Multiple series Rogers-Ramanujan type identities.], Pacific J. Math.  Volume 114, Number 2 (1984), 267-283.<br>
+* http://www.jstor.org/action/doBasicSearch?Query=
+* http://www.ams.org/mathscinet
+* http://dx.doi.org/
+<h5>관련도서</h5>
+*  도서내검색<br>
+** http://books.google.com/books?q=
+** http://book.daum.net/search/contentSearch.do?query=

"베일리 사슬(Bailey chain)"의 두 판 사이의 차이

2011년 11월 12일 (토) 06:12 판

목차

이 항목의 수학노트 원문주소

개요

Bailey chain construction

special case of Bailey chain

repetition of chain construction

examples

역사

메모

관련된 항목들

수학용어번역

사전 형태의 자료

리뷰논문, 에세이, 강의노트

관련논문

관련도서

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