Bailey lattice

수학노트
http://bomber0.myid.net/ (토론)님의 2010년 10월 9일 (토) 05:42 판
둘러보기로 가기 검색하러 가기
introduction

Let \(\{\alpha_r\}, \{\beta_r\}\) be a Bailey pair relative to a and set

\(\alpha_0'=0\), \(\alpha_n'=(1-a)a^nq^{n^2-n}(\frac{\alpha_n}{1-aq^{2n}}-\frac{aq^{2n-2}\alpha_{n-1}}{1-aq^{2n-2}})\)\(\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}\)

Then \(\{\alpha_r'\}, \{\beta_r'\}\)  is a Bailey pair relative to \(aq^{-1}\)

 

 

comparison with Bailey chain
  • Bailey chain
    \(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
    \(\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r\)
  • This does not change the parameter a of the Bailey pair.
  • lattice construction changes this

 

 

 

 

corollary

Let \(\{\alpha_r\}, \{\beta_r\}\) be the initial Bailey pair relative to a

apply Bailey chain construction k-i-1 times (Bailey chain)

At the (k-i)th step apply Bailey lattice

apply Bailey chain construction i times again.

Then we get a Bailey pair

\(\{\alpha_r'\}, \{\beta_r'\}\)  is a Bailey pair relative to \(aq^{-1}\).

If we use the defining relation of Bailey pair to \(\{\alpha_r'\}, \{\beta_r'\}\),

\(\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}\)

and take the limit L\to\infty

 

 

 

 

 

 

history

 

 

related items

 

 

encyclopedia

 

 

books

 

[[4909919|]]

 

 

articles
  • A Bailey Lattice
    • Jeremy Lovejoy, Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links