"Bailey lattice"의 두 판 사이의 차이

2010년 10월 9일 (토) 04:28 판

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 \(aq^{-1}\)

corollary

apply

history

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related items

encyclopedia

http://en.wikipedia.org/wiki/
http://www.scholarpedia.org/
Princeton companion to mathematics(Companion_to_Mathematics.pdf)

books

[[4909919|]]

articles

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

The Bailey lattice
- David Bressoud, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.
The Bailey Lattice
- A. Agarwal, G.E. Andrews, and D. Bressoud, J. Indian Math. Soc. 51 (1987), 57-73.
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@@ 1번째 줄: / 1번째 줄: @@
 <h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">introduction</h5>
-\alpha_n'=(1-a)a^nq^{n^2-n}{
+Let <math>\{\alpha_r\}, \{\beta_r\}</math> be a Bailey pair relative to a and set
+<math>\alpha_0'=0</math>, <math>\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}})</math><math>\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}</math>
+Then <math>\{\alpha_r'\}, \{\beta_r'\}</math>  is a Bailey pair relative to <math>aq^{-1}</math>
+<h5 style="line-height: 2em; margin: 0px;">corollary</h5>
+apply