On the 3 ψ 3 Basic. Bilateral Hypergeometric Series Summation Formulas

Transcription

1 International JMath Combin Vol4 (2009), On the 3 ψ 3 Basic Bilateral Hypergeometric Series Summation Formulas K RVasuki and GSharath (Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysore , INDIA) vasuki kr@hotmailcom, sharath gns@rediffmailcom Abstract: HExton has recorded two basic bilateral hypergeometric series summation formula without proof on page 305 of his book entitled q- hypergeometric functions and applications In this paper, we give a proof of them Key Words: Basic hypergeometric series, basic bilateral hypergeometric series, contiguous functions AMS(2000): 33D15 1 Introduction We follow the standard notation of q-series [4] and we always assume that q < 1 The q-shifted factorials (a; q) n and (a; q) are defined as 1, if n 0, (a; q) n (a) n : (1 a)(1 aq)(1 aq 2 )(1 aq n ), if n 1 and (a; q) (a) : (1 a)(1 aq)(1 aq 2 ) The basic hypergeometric series r+1 ϕ r is defined by r+1ϕ r a 1,a 2,,a r+1 ; z (a 1 ) n (a 2 ) n (a r+1 ) n : z n, z < 1 b 1,b 2,,b r (q) n (b 1 ) n (b 2 ) n (b r ) n n0 One of the most classical identities in q-series is the q-binomial theorem, due to Cauchy: 1 Received Oct3, 2009 Accepted Nov 8, 2009

2 42 K RVasuki and GSharath 1ϕ 0 a ; z (az), z < 1, (11) (z) Another classical q-series identity in q-series is Heine s q-analogue of the Gauss 2 F 1 summation formula: 2ϕ 1 a, b c ; c (c/a) (c/b), ab (c) (c/ab) c < 1 (12) ab Heine deduced (12) as a particular case of his transformation formula [5] 2ϕ 1 a, b c ; z (b) (az) 2ϕ 1 (c) (z) c/b, z az ; b, z < 1, b < 1 (13) Another interesting transformation formula due to Sear s [7] is 3ϕ 2 ; de (e/a) (de/bc) a, d/b, d/c 3ϕ 2 d, e abc (e) (de/abc) d, de/bc ; e/a, (14) de/abc < 1, e/a < 1 The basic bilateral hypergeometric series r ψ r is defined by rψ r a 1,a 2,,a r ; z (a 1 ) n (a 2 ) n (a r ) n : z n, b 1,b 2,,b r (b 1 ) n (b 2 ) n (b r ) n b1b2 br a 1a 2 a r < z < 1 There are many generalizations of q-binomial theorem (11) of which, one of the interesting is the following Ramanujan s 1 ψ 1 summation [1] [6]: 1ψ 1 a b ; z (az) (b/a) (q/az) (q), b/a < z < 1 (15) (z) (q/a) (b/az) (b) A variety of proofs have been given of (15) For more details of (15), one may refer [1], [4] H Exton [3, p 305] has given following two 3 ψ 3 basic bilateral series summation formula without proof : q d, bq, c ; 1 a (1 (b/c))(d/b) (bq/a) (q) 2 (1 (1/c))(q/b) (q/a) (bq) (d), (16) d < 1, 1/a < 1and q d, bq, c ; q (1 (c/b))(d/b) (bq/a) (q) 2, (17) a (1 c)(q/b) (q/a) (bq) (d)

3 On the Basic Bilateral Hypergeometric Series Summation Formulas 43 d/q < 1, q/a < 1 Exton [3, p 305] has incorrectly given (q/c) instead of (q/b) in the denominator of (17) W Chu [2], deduced (16) and (17) as a special cases of his integralsummation formula In this paper, we give a proof of (16) and (17) on the lines of G E Andrews and R Askey [1] proof of (15) 2 Proof of (16) and (17) Lemma 21 We have a d (1 d) d, e, f ; z +(1 (a/d)) 3 ψ 3 dq,e,f ; z (1 d)(1 (e/q))(1 (f/q)) a, b/q, c/q z(1 (b/q))(1 (c/q)) d, e/q, f/q ; z, (21) ((d/q) b) 1 b d, e, f ; z (d a) a/q, bq, c (q a) 3 ψ 3 ; z d, e, f z((a/q) b)(1 c) a, bq, cq (1 e)(1 f) d, eq, fq ; z, (22) b(d 1)(d a) d(d b) d, e, f ; z (1 (a/d)) 3 ψ 3 dq,e,f ; z (d 1)(d a)(1 (e/q))(1 (f/q)) + z((d/q) (b/q))(1 (c/q))(q a) a/q, b, c/q d, e/q, f/q ; z, (23) [ ] (d a) a/q, bq, c f (a/q) ; z ((d/q) f)(f (a/q)) (q a) d, e, f (1 f) a/q, bq, c ((d/q) f) a, bq, c ; zq d, e, fq (1 f) d, e, fq ; z (24) Proof of (21) It is easy to see that a 3 ψ 3 dq,e,f ; zq + a(1 d) d d, e, f ; z a d [ (a) n (b) n (c) n z n dq n ] (dq) n 1 (e) n (f) n (1 dq n ) +1

4 44 K RVasuki and GSharath Hence, a d (a) n (b) n (c) n (dq) n (e) n (f) n z n a 3 ψ 3 dq,e,f ; zq + a(1 d) d d, e, f ; z a d 3 ψ 3 dq,e,f ; z (25) Also, we have d, e, f ; z a 3 ψ 3 d, e, f ; zq (a) n+1 (b) n (c) n z n (d) n (e) n (f) n Thus, (1 (d/q))(1 (e/q))(1 (f/q)) z(1 (b/q))(1 (c/q)) d, e, f ; z a 3 ψ 3 d, e, f ; zq (1 (d/q))(1 (e/q))(1 (f/q)) z(1 (b/q))(1 (c/q)) (a) n (b/q) n (c/q) n (d/q) n (e/q) n (f/q) n z n a, b/q, c/q d/q, e/q, f/q ; z (26) Changing d to dq in (26) and then adding resulting identity with (25), we obtain (21) Proof of (22) We have ((d/q) b) (1 b) (a) n (b) n (c) n z n (d a) (d) n (e) n (f) n (q a) (q/a) n (bq) n (c) n z n (d) n (e) n (f) n (a) n 1 (bq) n 1 (c) n (d) n (e) n (f) n z n [ ((d/q) b)(1 aq n 1 ) ((d/q) (a/q))(1 bq n ) ] (a) n 1 (bq) n 1 (c) n (d) n (e) n (f) n z n ((a/q) b)(1 dq n 1 ) z((a/q) b)(1 c) (1 e)(1 f) (a) n (bq) n (cq) n (d) n (eq) n (fq) n z n This proves (22) Proofof(23) Changing b to b/q, c to c/q, e to e/q and f to f/q in (22), and multiplying throughout by q(1 d)(1 (e/q))(1 (f/q)) z(1 (c/q))(d b) and adding the resulting identity with (21), we find (23)

5 On the Basic Bilateral Hypergeometric Series Summation Formulas 45 Proof of (24) From [8], we have (1 f) 3 ψ 3 d, e, f ; z ((d/q) f) 3 ψ 3 and z(1 a)(1 b)(1 c) (1 fq)(1 e) d, e, fq ; zq aq, bq, cq, d, eq, fq 2 ; z ((d/q) a) (1 a) d, e, f ; z ((d/q) f) (1 f) aq,b,c d, e, fq ; z z(f a)(1 b)(1 c) aq, bq, cq (1 f)(1 fq)(1 e) 3 ψ 3 d, eq, fq 2 ; z aq, bq, cq Eliminating 3 ψ 3 d, eq, fq ; z between above two identities and then replacing a by a/q and b by bq, we obtain(24) Proof of (16) Setting c cq, e bq, f c and z 1/a in (24), we deduce that [ ] (d a) c (a/q) (q a) 2ψ 2 a/q, cq d, c ;1/a ((d/q) c)(c (a/q)) (1 c) 1ψ 1 a/q a ; q/a ((d/q) c) (1 c) 1ψ 1 a d ;1/a Employing (15) in the right side of the above, we obtain a/q, cq 2ψ 2 d, c ;1/a 0 (27) Let f(d) 3 ψ 3 q d, bq, c ;1/a

6 46 K RVasuki and GSharath As a function of d, f(d) is clearly analytic for d < 1and a > 1 Setting c cq, e bq, f c and z 1/a in (23) and then employing (27), we find that f(d) (1 (d/b)) f(dq) (28) (1 d) Iterating (28) n 1 times, we find that f(d) (d/b) n (d) n f(dq n ) Since f(d) isanalyticfor d < 1, a > 1, by letting n,weobtain f(d) (d/b) f(0) (d) Setting c cq, d c, e bq in (14), we deduce that q 3ϕ 2 ;1/a bq, c (bq/a) (q) (bq) (1/a) n0 (a) n (c/b) n (1/q) n (bq/a) n (q) n (c) n (q) n 1 (bq/a) (q) (1 a)(1 (c/b))(1 (1/q))(bq/a) (bq) (1/a) (1 q)(1 c) n0 (aq) n (cq/b) n (1) n (q) n (cq) n (q 2 (bq/a) n ) n b(1 (c/b))(bq/a) (q) (1 c)(bq) (q/a) Thus, f(q) (1 (b/c))(bq/a) (q) (1 (1/c))(bq) (q/a) Setting d q in (29), and using the above, we find that f(0) (1 (b/c))(bq/a) (q) 2 (1 (1/c))(bq) (q/a) (q/b) Using this in (29), we deduce that f(d) (1 (b/c))(bq/a) (d/b) (q) 2 (1 (1/c))(bq) (q/a) (q/b) (d)

7 On the Basic Bilateral Hypergeometric Series Summation Formulas 47 This completes the proof of (16) Proof of (17) Setting c cq, e bq, f c and z q/a in (24) and then employing (15), we find that 2ψ 2 a/q, cq d, c ; q/a 0 (29) Let f(d) : 3 ψ 3 q d, bq, c ; q/a As a function of d, f(d) is clearly analytic for d < 1, when q/a < 1 Setting c cq, e bq, f c and z q/a in (23) and then employing (210), we find that f(d) (1 (d/b)) f(dq) (1 d) Iterating the above n 1 times, we get f(d) (d/b) n (d) n f(dq n ) Since f(d) isanalyticfor d < 1, q/a < 1, by letting n, we obtain f(d) (d/b) f(0) (d) Setting c bq, z q 2 /a in (13) and employing (11), we obtain 2ϕ 1 a, b bq ; q2 /a (bq/a) (q) b(bq) (q/a) Also by (12), we deduce that n1 (a) n (b) n (q) n (bq) n (q/a) n (bq/a) (q) (bq) (q/a)

8 48 K RVasuki and GSharath Thus, q f(q) 3 ϕ 2 ; q/a bq, c 1 2 ϕ 1 a, b (1 c) bq ; q/a c 2 ϕ 1 a, b bq ; q2 /a (1 (c/b))(bq/a) (q) (1 c)(q/a) (bq) Now setting in d q in (211) and employing above, we find that f(0) (1 (c/b))(bq/a) (q) 2 (1 c)(q/a) (bq) (q/b) Using this in (211), we deduce (17) Acknowledgement Authors are thankful to DST, New Delhi for awarding research project [No SR/S4/MS:517/08] under which this work has been done References [1] G E Andrews, and R Askey, A simple proof of Ramanujan s summation of the 1 ψ 1, Aequationes Math, 18 (1978), [2] W Chu, Partial fractions and bilateral summations, J Math Phys, 35 (1994), [3] H Exton, q-hypergeometric Functions and Appplications, Ellis Horwood Series in Mathematics and its Application, Chichester, 1983 [4] G Gasper and M Rahman, On Basic hypergeometric Series, Second Ed, Encycl Math Applics, Vol 35, Cambridge University Press, Cambridge, 2004 [5] E Heine, Untersuchungen über die Reihe 1+ (1 qα )(1 q β ) (1 q)(1 q γ ) x + (1 qα )(1 q α+1 )(1 q β )(1 q β+1 ) (1 q)(1 q 2 )(1 q γ )(1 q γ+1 x 2 +, ) JReine Angew Math, 34 (1847), [6] S Ramanujan, Notebooks (2 Volumes), Tata Institute of Fundamenal Research, Bombay, 1957 [7] D B Sears, Transformations of basic hypergeometric functions of special type, Proc London Math Soc, 52 (1951 ), [8] KRVasuki, Abdulrawf A A Kahtan and GSharath, On certain continued fractions related to 3 ψ 3 basic bilateral hypergeometric functions, Preprint