"베일리 사슬(Bailey chain)"의 두 판 사이의 차이

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9번째 줄: 9번째 줄:
 
==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:<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>
−
*  위에서  <math>\rho_1,\rho_2\to \infty</math> 일 경우, 다음을 얻는다<br><math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math><br><math>\beta^\prime_n = \sum_{r=0}^{n}\frac{a^rq^{r^2}}{(q)_{n-r}}\beta_r</math><br>
+
*  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>
−
* [[베일리 쌍(Bailey pair)과 베일리 보조정리|베일리 쌍(Bailey pair)]] 이 만족하는 관계<br><math>\beta^{'}_n=\sum_{r=0}^{n}\frac{\alpha^{'}_r}{(q)_{n-r}(aq)_{n+r}}</math> 로부터, 다음을 얻는다.<br><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>
+
* [[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>
  
 
 
 
 
19번째 줄: 19번째 줄:
 
==사슬의 반복 적용==
 
==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,:<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>
  
−
* <math>n\to\infty</math> 이면<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>
+
* <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>
  
 
 
 
 
29번째 줄: 29번째 줄:
 
==examples==
 
==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:<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>
−
*  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>
+
*  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>
−
*  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>
+
*  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>
−
*  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=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>
−
*  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=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>
 
*  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>
 
*  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>
+
*  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>
  
 
*  if k is bigger than 2, we get some cases of [[앤드류스-고든 항등식(Andrews-Gordon identity)]]<br>
 
*  if k is bigger than 2, we get some cases of [[앤드류스-고든 항등식(Andrews-Gordon identity)]]<br>

2013년 1월 12일 (토) 09:44 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\[\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\]

 

 

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