"Bailey pair and lemma"의 두 판 사이의 차이

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* [http://arxiv.org/abs/0910.2062v2 50 Years of Bailey's lemma]<br>
 
* [http://arxiv.org/abs/0910.2062v2 50 Years of Bailey's lemma]<br>
 
**  S. Ole Warnaar, 2009<br>
 
**  S. Ole Warnaar, 2009<br>
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* [http://dx.doi.org/10.1142/S0217751X97001110 Virasoro character identities from the Andrews–Bailey construction]<br>
 
* [http://dx.doi.org/10.1142/S0217751X97001110 Virasoro character identities from the Andrews–Bailey construction]<br>
 
**  Foda, O., Quano, Y.-H, Int. J. Mod. Phys. A 12, 1651–1675 (1997)<br>
 
**  Foda, O., Quano, Y.-H, Int. J. Mod. Phys. A 12, 1651–1675 (1997)<br>
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* [http://dx.doi.org/10.1112%2Fjlms%2Fs1-37.1.504 Wilfrid Norman Bailey]<br>
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**  Slater, L. J. (1962), Journal of the London Mathematical Society. Second Series 37: 504–512<br>
 
* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>
 
* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>
 
**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167<br>
 
**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167<br>

2010년 6월 11일 (금) 09:31 판

introduction
  •  q-Pfaff-Sallschutz sum

 

 

Bailey lemma

 

 

 

Bailey pair
  • the sequence \(\{\alpha_r\}, \{\beta_r\}\) satisfying the following is called a Bailey pair
    \(\beta_L=\sum_{r=0}^{L}\frac{\alpha_r}{(q)_{L-r}(aq)_{L+r}}\)
  • conjugate Bailey pair  \(\{\delta_r\}, \{\gamma_r\}\)
    \(\gamma_L=\sum_{r=L}^{\infty}\frac{\delta_r}{(q)_{r-L}(aq)_{r+L}}\)

 

 

Bailey chain
  • we derive a new Bailey chain from a known Bailey pair
    \(\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n\)
    \(\beta^\prime_n = \sum_{j\ge0}\frac{(\rho_1;q)_j(\rho_2;q)_j(aq/\rho_1\rho_2;q)_{n-j}(aq/\rho_1\rho_2)^j}{(q;q)_{n-j}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_j\)
  • corollary. by taking  \(\rho_1,\rho_2\to \infty\) , we get 
     
    \(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
    \(\beta^\prime_n = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_j\)

 

 

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