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이 항목의 수학노트 원문주소==
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| 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 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== |
* [[베일리 사슬(Bailey chain)]] | * [[베일리 사슬(Bailey chain)]] | ||
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| − | ==개요 | + | ==개요== |
* 기존의 베일리 쌍 relative to <em>a </em> 로부터 새로운 베일리 쌍 relative to <em>a</em> 을 얻는 방법<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_{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> | * 기존의 베일리 쌍 relative to <em>a </em> 로부터 새로운 베일리 쌍 relative to <em>a</em> 을 얻는 방법<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_{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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| − | <h5 style="line-height: 2em; margin: 0px;">사슬의 반복 적용 | + | <h5 style="line-height: 2em; margin: 0px;">사슬의 반복 적용== |
* 사슬 구성을 여러번 반복하면,<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> | * 사슬 구성을 여러번 반복하면,<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> | ||
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| − | <h5 style="line-height: 2em; margin: 0px;">examples | + | <h5 style="line-height: 2em; margin: 0px;">examples== |
* 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> | * 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> | ||
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| − | ==역사 | + | ==역사== |
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| − | ==메모 | + | ==메모== |
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| − | ==관련된 항목들 | + | ==관련된 항목들== |
| 71번째 줄: | 71번째 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%;">수학용어번역 | + | <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== |
* 단어사전<br> | * 단어사전<br> | ||
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| − | ==사전 형태의 자료 | + | ==사전 형태의 자료== |
* http://ko.wikipedia.org/wiki/ | * http://ko.wikipedia.org/wiki/ | ||
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| − | ==리뷰논문, 에세이, 강의노트 | + | ==리뷰논문, 에세이, 강의노트== |
| 109번째 줄: | 109번째 줄: | ||
| − | ==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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| − | ==관련도서 | + | ==관련도서== |
* 도서내검색<br> | * 도서내검색<br> | ||
** http://books.google.com/books?q= | ** http://books.google.com/books?q= | ||
** http://book.daum.net/search/contentSearch.do?query= | ** http://book.daum.net/search/contentSearch.do?query= | ||
2012년 11월 1일 (목) 12:49 판
이 항목의 수학노트 원문주소==
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
\(\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\)
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
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
수학용어번역==
- 단어사전
- 발음사전 http://www.forvo.com/search/
- 대한수학회 수학 학술 용어집
- 한국통계학회 통계학 용어 온라인 대조표
- 남·북한수학용어비교
- 대한수학회 수학용어한글화 게시판
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/
- The Online Encyclopaedia of Mathematics
- NIST Digital Library of Mathematical Functions
- The World of Mathematical Equations
리뷰논문, 에세이, 강의노트
관련논문
- Paule, The Concept of Bailey Chains
- George E. AndrewsMultiple series Rogers-Ramanujan type identities., Pacific J. Math. Volume 114, Number 2 (1984), 267-283.
관련도서
\(\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n\)
\(\beta^\prime_n = \sum_{r=0}^{n}\frac{(\rho_1;q)_r(\rho_2;q)_r(aq/\rho_1\rho_2;q)_{n-r}(aq/\rho_1\rho_2)^r}{(q;q)_{n-r}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_r\)
\(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
\(\beta^\prime_n = \sum_{r=0}^{n}\frac{a^rq^{r^2}}{(q)_{n-r}}\beta_r\)
\(\beta^{'}_n=\sum_{r=0}^{n}\frac{\alpha^{'}_r}{(q)_{n-r}(aq)_{n+r}}\) 로부터, 다음을 얻는다.
\(\sum_{r=0}^{n}\frac{a^{r}q^{r^2}\alpha_r}{(q)_{n-r}(aq)_{n+r}}=\sum_{n'=0}^{n}\frac{a^{n'}q^{n'{^2}}}{(q)_{n-n'}}\beta_{n'}\)
- 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
\(\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\)
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
\(\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
\(\beta_{L}=\delta_{L,0}\)
For example, if a=1,
\(\alpha_{L}=(-1)^{L}q^{L(L-1)/2}(1+q^{L})=(-1)^{L}(q^{(3L^2-L)/2}+q^{(3L^2+L)/2})\)
\(\alpha_{L}=(-1)^{L}a^{kL}q^{kL^{2}+L^2/2-L/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
\(\beta_{L}=\sum_{L\geq n_1\geq\cdots\geq n_{k-1}\geq0}\frac{a^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2}}{(q)_{L-n_1}(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
\(\frac{1}{(q)_{\infty}}\sum_{r=-\infty}^{\infty}(-1)^{r}q^{r((2k+1)r+1-2jk)/2}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+j(n_1+\cdots+n_{k})}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
\((q)_{\infty}=\sum_{k=-\infty}^\infty(-1)^kq^{k(3k-1)/2}\)
\(\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} \)
\(\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty}\)
\(\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\)
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들