"베일리 사슬(Bailey chain)"의 두 판 사이의 차이

수학노트
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(같은 사용자의 중간 판 2개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==개요==
 
==개요==
  
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*  기존의 베일리 쌍 relative to <em>a </em> 로부터 새로운 베일리 쌍 relative to <em>a</em> 을 얻는 방법:<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>:<math>\beta^\prime_n = \sum_{r=0}^{n}\frac{(\rho_1;q)_r(\rho_2;q)_r(aq/\rho_1\rho_2;q)_{n-r}(aq/\rho_1\rho_2)^r}{(q;q)_{n-r}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_r</math><br>
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*  기존의 베일리 쌍 relative to <em>a </em> 로부터 새로운 베일리 쌍 relative to <em>a</em> 을 얻는 방법:<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>:<math>\beta^\prime_n = \sum_{r=0}^{n}\frac{(\rho_1;q)_r(\rho_2;q)_r(aq/\rho_1\rho_2;q)_{n-r}(aq/\rho_1\rho_2)^r}{(q;q)_{n-r}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_r</math>
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*  위에서  <math>\rho_1,\rho_2\to \infty</math> 일 경우, 다음을 얻는다:<math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math>:<math>\beta^\prime_n = \sum_{r=0}^{n}\frac{a^rq^{r^2}}{(q)_{n-r}}\beta_r</math><br>
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*  위에서  <math>\rho_1,\rho_2\to \infty</math> 일 경우, 다음을 얻는다:<math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math>:<math>\beta^\prime_n = \sum_{r=0}^{n}\frac{a^rq^{r^2}}{(q)_{n-r}}\beta_r</math>
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* [[베일리 쌍(Bailey pair)과 베일리 보조정리|베일리 쌍(Bailey pair)]] 이 만족하는 관계:<math>\beta^{'}_n=\sum_{r=0}^{n}\frac{\alpha^{'}_r}{(q)_{n-r}(aq)_{n+r}}</math> 로부터, 다음을 얻는다.:<math>\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n'=0}^{n}\frac{a^{n'}q^{n'{^2}}}{(q)_{n-n'}}\beta_{n'}</math><br>
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* [[베일리 쌍(Bailey pair)과 베일리 보조정리|베일리 쌍(Bailey pair)]] 이 만족하는 관계:<math>\beta^{'}_n=\sum_{r=0}^{n}\frac{\alpha^{'}_r}{(q)_{n-r}(aq)_{n+r}}</math> 로부터, 다음을 얻는다.:<math>\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n'=0}^{n}\frac{a^{n'}q^{n'{^2}}}{(q)_{n-n'}}\beta_{n'}</math>
  
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==사슬의 반복 적용==
 
==사슬의 반복 적용==
  
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*  사슬 구성을 여러번 반복하면,:<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>
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*  사슬 구성을 여러번 반복하면,:<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>
  
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* <math>n\to\infty</math> 이면:<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>
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* <math>n\to\infty</math> 이면:<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>
  
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==examples==
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==예==
  
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*  initial Bailey pair:<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>:<math>\beta_{L}=\delta_{L,0}</math><br> For example, if a=1,:<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>
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*  초기 베일리 쌍
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*  result of Bailey chain applied k-times:<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>:<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>
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:<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>:<math>\beta_{L}=\delta_{L,0}</math> For example, if a=1,:<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>
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* obtained q-series identity:<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>
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*  베일리 사슬을 k번 적용하면, 다음을 얻는다
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*  Setting k=1, a=1, we get the Euler pentagonal number theorem:<math>(q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}</math><br>
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:<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>
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*  Setting k=2, a=1, we get one of RR identity:<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>
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:<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>
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*  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>
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* 얻어진 항등식
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* We frequently use Jacobi triple product identity:<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>
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:<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>
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*  <math>k=1, a=1</math>이면, [[오일러의 오각수정리(pentagonal number theorem)]]를 얻는다
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:<math>(q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}</math>
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*  <math>k=2, a=1</math>이면, [[로저스-라마누잔 항등식]]의 하나를 얻는다
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:<math>\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} </math>
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*  <math>k=2, a=q</math>이면, 또다른 [[로저스-라마누잔 항등식]]을 얻는다
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:<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>
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* 결과를 표현하기 위해 [[자코비 삼중곱(Jacobi triple product)]]이 종종 사용된다
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:<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>
  
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*  if k is bigger than 2, we get some cases of [[앤드류스-고든 항등식(Andrews-Gordon identity)]]<br>
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*  <math>k>2</math>인 경우는, [[앤드류스-고든 항등식(Andrews-Gordon identity)]]의 일부를 얻을 수 있다
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*  모든 [[앤드류스-고든 항등식(Andrews-Gordon identity)]]  을 증명하려면, 베일리 격자(Bailey lattice) 가 필요하다<br>
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*  모든 [[앤드류스-고든 항등식(Andrews-Gordon identity)]] 을 얻기 위해서는, [[베일리 격자(Bailey lattice)]]의 방법이 필요하다
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==역사==
 
==역사==
  
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* [[수학사 연표]]
 
* [[수학사 연표]]
  
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==메모==
 
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* Math Overflow http://mathoverflow.net/search?q=
 
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==관련된 항목들==
 
==관련된 항목들==
  
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==수학용어번역==
 
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* 단어사전<br>
 
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** http://translate.google.com/#en|ko|
 
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** http://ko.wiktionary.org/wiki/
 
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* 발음사전 http://www.forvo.com/search/
 
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
 
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
 
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* [http://www.kss.or.kr/pds/sec/dic.aspx 한국통계학회 통계학 용어 온라인 대조표]
 
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
 
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* [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]
 
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==사전 형태의 자료==
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==사전 형태의 자료==
  
 
* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
 
* http://en.wikipedia.org/wiki/
 
* http://en.wikipedia.org/wiki/
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* [http://eom.springer.de/default.htm The Online Encyclopaedia of Mathematics]
 
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* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
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* [http://eqworld.ipmnet.ru/ The World of Mathematical Equations]
 
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==리뷰논문, 에세이, 강의노트==
 
  
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==관련논문==
 
==관련논문==
  
 
* Paule, [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.71.6203 The Concept of Bailey Chains]
 
* Paule, [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.71.6203 The Concept of Bailey Chains]
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* 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>
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* 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.
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* http://www.jstor.org/action/doBasicSearch?Query=
 
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* http://www.ams.org/mathscinet
 
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* http://dx.doi.org/
 
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[[분류:q-급수]]
 
[[분류:q-급수]]

2020년 12월 28일 (월) 02:25 기준 최신판

개요

  • 기존의 베일리 쌍 relative to a 로부터 새로운 베일리 쌍 relative to a 을 얻는 방법\[\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n\]\[\beta^\prime_n = \sum_{r=0}^{n}\frac{(\rho_1;q)_r(\rho_2;q)_r(aq/\rho_1\rho_2;q)_{n-r}(aq/\rho_1\rho_2)^r}{(q;q)_{n-r}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_r\]
  • 위에서 \(\rho_1,\rho_2\to \infty\) 일 경우, 다음을 얻는다\[\alpha^\prime_n= a^nq^{n^2}\alpha_n\]\[\beta^\prime_n = \sum_{r=0}^{n}\frac{a^rq^{r^2}}{(q)_{n-r}}\beta_r\]
  • 베일리 쌍(Bailey pair) 이 만족하는 관계\[\beta^{'}_n=\sum_{r=0}^{n}\frac{\alpha^{'}_r}{(q)_{n-r}(aq)_{n+r}}\] 로부터, 다음을 얻는다.\[\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n'=0}^{n}\frac{a^{n'}q^{n'{^2}}}{(q)_{n-n'}}\beta_{n'}\]



사슬의 반복 적용

  • 사슬 구성을 여러번 반복하면,\[\sum_{r=0}^{n}\frac{a^{kr}q^{kr^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n_1=0}^{n}\sum_{n_2=0}^{n_1}\cdots\sum_{n_k=0}^{n_{k-1}}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n-n_{1}}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\]
  • \(n\to\infty\) 이면\[\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\]



예

  • 초기 베일리 쌍

\[\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\]\[\beta_{L}=\delta_{L,0}\] For example, if a=1,\[\alpha_{L}=(-1)^{L}q^{L(L-1)/2}(1+q^{L})=(-1)^{L}(q^{(3L^2-L)/2}+q^{(3L^2+L)/2})\]

  • 베일리 사슬을 k번 적용하면, 다음을 얻는다

\[\alpha_{L}=(-1)^{L}a^{kL}q^{kL^{2}+L^2/2-L/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\] \[\beta_{L}=\sum_{L\geq n_1\geq\cdots\geq n_{k-1}\geq0}\frac{a^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2}}{(q)_{L-n_1}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\]

  • 얻어진 항등식

\[\frac{1}{(q)_{\infty}}\sum_{r=-\infty}^{\infty}(-1)^{r}q^{r((2k+1)r+1-2jk)/2}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+j(n_1+\cdots+n_{k})}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\]

\[(q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}\]

\[\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} \]

\[\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty}\]

\[\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\]


역사



관련된 항목들

사전 형태의 자료



관련논문

원본 주소 "https://wiki.mathnt.net/index.php?title=베일리_사슬(Bailey_chain)&oldid=48019"