A TRANSFORMATION FORMULA FOR A SPECIAL BILATERAL BASIC HYPERGEOMETRIC 12 ψ12 SERIES ZHIZHENG ZHANG and QIUXIA HU Abstract. In this short note, we shall make use of decomposition of series to derive a transformation formula for a bilateral basic hypergeometric 12 ψ12 series. 1. Introduction Throughout this note, we shall adopt some definitions and notations from [1]. The q-shifted factorial is defined by (a; q)0 = 1, (a; q)n = n−1 Y (1 − aq k ), n = 1, 2, · · · , k=0 and (a; q)∞ = JJ J I II Go back Full Screen Close Quit ∞ Y (1 − aq k ). k=0 In this paper, during the process of the computations we shall also make use of the following notation: n (−q/a)n q ( 2 ) (a; q)−n = , n = 1, 2, . . . . (q/a; q)n Received March 15, 2008; revised March 14, 2009. 2000 Mathematics Subject Classification. Primary 33D15. Key words and phrases. q-series; bilateral basic hypergeometric series; decomposition of series. For products of q-shifted factorials, we use the short notation (a1 , a2 , . . . , qr ; q)n = (a1 ; q)n (a2 ; q)n . . . (ar ; q)n where n is an integer or infinity. Basic and bilateral basic hypergeometric series are defined by X ∞ i1+s−r n (a1 , a2 , . . . , ar ; q)n h a1 , a2 , . . . , ar φ ; q, z = (−1)n q ( 2 ) zn, r s b1 , b2 , . . . , bs (q, b1 , b2 , . . . , bs ; q)n n=0 and r ψs a1 , b1 , ∞ is−r X n (a1 , a2 , . . . , ar ; q)n h a2 , . . . , ar (−1)n q ( 2 ) zn, ; q, z = b2 , . . . , bs (b , b , . . . , b ; q) 1 2 s n n=−∞ respectively. In this short note, we make use of the idea decomposition of series to derive a formula for a bilateral basic hypergeometric 12 ψ12 series. 2. Main results JJ J I II Go back Full Screen Close Quit In the proof of Theorem 1, we use of the following very-well-poised 8 φ7 transformation formula: 1 1 1 1 1 1 a2 q a, qa 2 , −qa 2 , y2, −y 2 , (yq) 2 , −(yq) 2 , x ; q; 2 1 1 1 1 1 1 1 1 8 φ7 y x a 2 , −a 2 , aqy − 2 , −aqy − 2 , aq 2 y − 2 , −aq 2 y − 2 , aq/x (1) 2 2 2 (aq, a q/y ; q)∞ a q y, xy/a = ; q, 2 2 φ1 aq/x (aq/y, a2 q/y; q)∞ y x 2 provided | ya2 xq | < 1, which is equivalent to [1, Equation (3.4.7)] by a substitution of variables. Theorem 1. For |q| < 1 and |q 3 /y 2 x| < 1, we have 5/2 q , −q 5/2 , y 1/2 , qy 1/2 , −y 1/2 , −qy 1/2 , (yq)1/2 , ψ 12 12 1/2 1/2 2 −1/2 3 −1/2 2 −1/2 3 −1/2 3/2 −1/2 q , −q , q y , q y , −q y , −q y , q y , (2) q6 q 3/2 y 1/2 , −(yq)1/2 , −q 3/2 y 1/2 , x, xq ; q2 , 4 2 5/2 −1/2 3/2 −1/2 5/2 −1/2 2 3 q y , −q y , −q y , q /x, q /x y x 3 2 3 2 q (q , q /y ; q)∞ y, xy/q ; q, 2 . = 2 2 φ1 q 2 /x (q /y, q 3 /y; q)∞ y x Proof. We first write out the left-hand side of (1) explicitly: 2 n ∞ X (a, qa1/2 , −qa1/2 , y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)n a q . (3) 2x 1/2 , −a1/2 , aqy −1/2 , −aqy−1/2, aq 1/2 y −1/2 , −aq 1/2 y −1/2 , aq/x; q) y (q, a n n=0 JJ J I II Go back Full Screen Close Quit JJ J I II Letting a = q in (3) and after some elementary manipulations, we get 3 n ∞ X (q 3 ; q 2 )n (y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)n q 2 2 −1/2 2 −1/2 3/2 −1/2 3/2 −1/2 2 (q; q )n (q y , −q y ,q y , −q y , q /x; q)n y 2 x n=0 3 2n ∞ X (q 3 ; q 2 )2n (y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)2n q = 2x 2 ) (q 2 y −1/2 , −q 2 y −1/2 , q 3/2 y −1/2 , −q 3/2 y −1/2 , q 2 /x; q) y (q; q 2n 2n n=0 3 2n+1 ∞ X q (q 3 ; q 2 )2n+1 (y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)2n+1 + 2x 2) 2 y −1/2, −q 2 y −1/2, q 3/2 y −1/2, −q 3/2 y −1/2 , q 2/x; q) y (q; q (q 2n+1 2n+1 n=0 3 2n ∞ X q (q 3 ; q 2 )2n (y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)2n = 2 2 −1/2 2 −1/2 3/2 −1/2 3/2 −1/2 2 (q; q )2n (q y , −q y ,q y , −q y , q /x; q)2n y 2 x n=0 3 2n −1 X (q 3 ; q 2 )2n (y 1/2 , −y 1/2 , (yq)1/2 , −(yq)1/2 , x; q)2n q + . 2x 2 ) (q 2 y −1/2 , −q 2 y −1/2 , q 3/2 y −1/2 , −q 3/2 y −1/2 , q 2 /x; q) y (q; q 2n 2n n=−∞ According to the definition of bilateral basic hypergeometric series and combining the two sums of above, the consequence is just the left-hand side of (2). By (1) the desired result is immediate. Go back Full Screen Close Quit Note that the left-hand side of (2) can be written in the following compact form: 3 2n ∞ X (1 − q 1+4n ) (y; q)4n (x; q)2n q . 3 /y; q) 2 /x; q) 2x (1 − q) (y (q y 4n 2n n=−∞ Acknowledgment. The authors are very grateful to the anonymous referee for comments and valuable suggestions which helped a lot to improve the original version. This research is supported by the National Natural Science Foundation of China (Grant No. 10771093). 1. Gasper G. and Rahman M., Basic Hypergeometric Series, 2nd Edition, Encyclopedia of Mathematics and Its Applications 96, Cambridge University Press, Cambridge, 2004. Zhizheng Zhang, Center of Combinatorics and LPMC, Nankai University, Tianjin 300071, P. R. China, e-mail: zhzhzhang-yang@163.com Qiuxia Hu, Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P. R. China, e-mail: huqiuxia306@163.com JJ J I II Go back Full Screen Close Quit