베일리 사슬(Bailey chain)

수학노트
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개요

  • 기존의 베일리 쌍 relative to 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\[\frac{1}{(aq)_{\infty}}\sum_{n=0}^{\infty}a^{kn}q^{kn^{2}}\alpha_{n}=\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}\]

 

 

  • 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

 

 


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