IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation37039010.1155/2009/370390370390Research ArticleA Note on Four-Variable Reciprocity TheoremAdigaChandrashekarGuruprasadP. S.BulboacăTeodorDepartment of Studies in MathematicsUniversity of MysoreManasagangotriMysore 570 006Indiauni-mysore.ac.in200912012010200912092009071220092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We give new proof of a four-variable reciprocity theorem using Heine's transformation, Watson's transformation, and Ramanujan's ψ11-summation formula. We also obtain a generalization of Jacobi's triple product identity.

1. Introduction

Throughout the paper, we let |q|<1 and we employ the standard notation: (a)0:=(a;q)0=1,(a):=(a;q)=n=0(1-aqn),(a)n:=(a;q)n=(a;q)(aqn;q),-<n<. Ramanujan  stated several q-series identities in his “lost’’ notebook. One of the beautiful identities is the two-variable reciprocity theorem.

Theorem 1.1 (see [<xref ref-type="bibr" rid="B19">2</xref>]).

For ab0, ρ(a,b)-ρ(b,a)=(1b-1a)(aq/b)(bq/a)(q)(-aq)(-bq), where ρ(a,b):=(1+1b)n=0(-1)nqn(n+1)/2anb-n(-aq)n.

In the recent past many new proofs of (1.2) have been found. The first proof of (1.2) was given by Andrews  using four-free-variable identity and Jacobi's triple product identity. Further, Andrews  applied (1.2) in proving Euler partition identity analogues stated in . Somashekara and Fathima  established an equivalent version of (1.2) using Ramanujan's 1ψ1 summation formula  and Heine's transformation [7, 8]. Berndt et al.  also derived (1.2) using the same above mentioned two transformations. In fact, Berndt et al.  in the same paper have given two more proofs of (1.2) one employing the Rogers-Fine identity  and the other is purely combinatorial. Using the q-binomial theorem: n=0(a)n(q)ntn=(at)(t),|t|<1,|q|<1, Kim et al.  gave a much different proof of (1.2). Guruprasad and Pradeep  also devised a proof of (1.2) using the q-binomial theorem. Adiga and Anitha  established (1.2) along the lines of Ismail's proof  of Ramanujan's 1ψ1 summation formula. Further, they showed that the reciprocity theorem (1.2) leads to a q-integral extension of the classical gamma function. Kang  constructed a proof of (1.2) along the lines of Venkatachaliengar's proof of the Ramanujan 1ψ1 summation formula [6, 15].

In  Kang proved the following three- and four-variable generalizations of (1.2).

For |c|<|a|<1 and |c|<|b|<1, ρ3(a,b,c)-ρ3(b,a,c)=(1b-1a)(c)(aq/b)(bq/a)(q)(-c/a)(-c/b)(-aq)(-bq), where ρ3(a,b,c):=(1+1b)n=0(c)n(-1)nqn(n+1)/2anb-n(-aq)n(-c/b)n+1,a,cb-q-n, and for |c|,|d|<|a|, |b|<1,

ρ4(a,b,c,d)-ρ4(b,a,c,d)=(1b-1a)(d)(c)(cd/ab)(aq/b)(bq/a)(q)(-d/a)(-d/b)(-c/a)(-c/b)(-aq)(-bq), where ρ4(a,b,c,d):=(1+1b)n=0(d)n(c)n(cd/ab)n(1+cdq2n/b)(-1)nqn(n+1)/2anb-n(-aq)n(-c/b)n+1(-d/b)n+1,a,cb,db-q-n. Kang  established (1.5) on employing Ramanujan's 1ψ1 summation formula and Jackson's transformation of 2ϕ1 and 2ϕ2-series. Recently (1.5) was derived by Adiga and Guruprasad  using q-binomial theorem and Gauss summation formula. Somashekara and Mamta [17, 18] obtained (1.5) using the two-variable reciprocity theorem (1.2), Jackson's transformation, and again two-variable reciprocity theorem by parameter augmentation. Zhang  also established (1.5).

Kang  established (1.7) on employing Andrews's generalization of 1ψ1 summation formula, Sears's transformation of 3ϕ2-series, and a limiting case of Watson's transformation for a terminating very well-poised 8ϕ7-series : n=0(α)n(β)n(γ)n(δ)n(ϵ)n(1-αq2n)qn(n+3)/2(αq/β)n(αq/γ)n(αq/δ)n(αq/ϵ)n(q)n(1-α)(-α2βγδϵ)n=(αq)(αq/δϵ)(αq/δ)(αq/ϵ)n=0(δ)n(ϵ)n(αq/βγ)n(αq/β)n(αq/γ)n(q)n(αqδϵ)n. Recently Ma [20, 21] proved a six-variable generalization and a five-variable generalization of (1.2). The main purpose of this paper is to provide a new proof of (1.7) using (1.9), Heine's transformation: n=0(α)n(β)n(q)n(γ)nzn=(γ/β)(βz)(γ)(z)n=0(αβz/γ)n(β)n(βz)n(q)n(γβ)n,|q|<1,|z|<1,|γ|<|β|<1 and Ramanujan's 1ψ1 summation formula:   1ψ1(a;b;z):=n=-(a)n(b)nzn=(b/a)(az)(q/az)(q)(q/a)(b/az)(b)(z),|q|<1,|ba|<|z|<1.

Jacobi's triple product identity states that n=-qn(n+1)/2zn=(q)(-zq)(-1z),z0,|q|<1. Andrews  gave a proof of (1.12) using Euler identities. Combinatorial proofs of Jacobi's triple product identity were given by Wright , Cheema , and Sudler . We can also find a proof of (1.12) in . Using (1.12), Hirschhorn [27, 28] established Jacobi's two-square and four-square theorems.

Somashekara and Fathima  and Kim et al.  established n=0(-1)nanb-nqn(n+1)/2(-a)n+1-n=0(-1)na-(n+1)bn+1qn(n+1)/2(-b)n+1=(aq/b)(b/a)(q)(-a)(-b). Note that (1.13) which is equivalent to (1.2) may be considered as a two-variable generalization of (1.12). Corteel and Lovejoy [29, equation ( 1.5)] have given a bijective proof of (1.13) using representations of over partitions. All the reciprocity theorems (1.2), (1.5), and (1.7) are generalizations of Jacobi's triple product identity (1.12).

We also obtain a generalization of Jacobi's triple product identity (1.12) which is due to Kang .

2. Proof of (<xref ref-type="disp-formula" rid="EEq1.3">1.7</xref>)—The Four-Variable Reciprocity Theorem

On employing q-binomial theorem, we have n=0(-cq)n(-dq)n(-aq)n(-bq)nqn=(-dq)(-bq)n=0(-cq)n(-bqn+1)(-aq)n(-dqn+1)qn=(-dq)(-bq)m=0(b/d)m(q)m(-dq)mn=0(-cq)n(-aq)n(qm+1)n. On using Heine's transformation (1.10) with α=-cq, β=q, γ=-aq, z=qm+1, we have n=0(-cq)n(-aq)n(qm+1)n=(qm+2)(-a)(qm+1)(-aq)n=0(cqm+2/a)n(qm+2)n(-a)n=(q)m(1+a)(-a)-m-1(cq/a)m+1n=0(cq/a)n+m+1(q)n+m+1(-a)n+m+1=(q)m(1+a)(-a)-m-1(cq/a)m+1[n=0(cq/a)n(q)n(-a)n-n=0m(cq/a)n(q)n(-a)n]=(q)m(-a)-m-1(-cq)(cq/a)m+1(-aq)-(q)m(1+a)(-a)-m-1(cq/a)m+1n=0m(cq/a)n(q)n(-a)n. Substituting this in (2.1), we obtain n=0(-cq)n(-dq)n(-aq)n(-bq)nqn=(-dq)(-cq)(-a)(-bq)(-aq)m=0(b/d)m(cq/a)m+1(dqa)m+(1+a-1)(-dq)(-bq)m=0n=0m(b/d)m(dq/a)m(cq/a)n(-a)n(cq/a)m+1(q)n. Now,

(1+a-1)(-dq)(-bq)m=0n=0m(b/d)m(dq/a)m(cq/a)n(-a)n(cq/a)m+1(q)n=(1+a-1)(-dq)(-bq)n=0(b/d)n(-dq)n(q)n(1-cqn+1/a)m=0(bqn/d)m(dq/a)m(cqn+2/a)m=(1+a-1)(-dq)(-bq)n=0(b/d)n(-dq)n(q)n(1-cqn+1/a)·(1-cqn+1/a)(1-dq/a)m=0(b/c)m(cqn+1/a)m(dq2/a)m,(onusing(1.10)withα=bqnd,β=q,γ=cqn+2a,z=dqa)=(1+a-1)(-dq)(-bq)m=0(b/c)m(cq/a)m(dq/a)m+1n=0(b/d)n(-dqm+1)n(q)n=(1+a-1)(-dq)(-bq)m=0(b/c)m(cq/a)m(dq/a)m+1·(-bqm+1)(-dqm+1)=(1+a-1)m=0(b/c)m(-dq)m(-bq)m(dq/a)m+1(cqa)m. Substituting (2.4) in (2.3), we obtain n=0(-cq)n(-dq)n(-aq)n(-bq)nqn=(-dq)(-cq)(-a)(-bq)(-aq)m=0(b/d)m(cq/a)m+1(dqa)m+(1+a-1)m=0(b/c)m(-dq)m(-bq)m(dq/a)m+1(cqa)m=(-dq)(-cq)(-a)(-bq)(-aq)m=0(b/c)m(dq/a)m+1(cqa)m+(1+a-1)m=0(b/c)m(-dq)m(-bq)m(dq/a)m+1(cqa)m. (Here, we used (1.10) with α=b/d, β=q, γ=cq2/a, z=dq/a.)

Changing c to -c/q, d to -d/q in (2.5), we get n=0(c)n(d)n(-aq)n(-bq)nqn=(d)(c)(-a)(-bq)(-aq)m=0(-bq/c)m(-d/a)m+1(-ca)m+(1+a-1)m=0(-bq/c)m(d)m(-bq)m(-d/a)m+1(-ca)m. Interchanging a and b in (2.6), we have n=0(c)n(d)n(-aq)n(-bq)nqn=(d)(c)(-b)(-bq)(-aq)m=0(-aq/c)m(-d/b)m+1(-cb)m+(1+b-1)m=0(-aq/c)m(d)m(-aq)m(-d/b)m+1(-cb)m. Subtracting (2.6) from (2.7), we deduce that (d)(c)(-bq)(-aq)[1bm=0(-aq/c)m(-d/b)m+1(-cb)m-1am=0(-bq/c)m(-d/a)m+1(-ca)m]=(1+b-1)m=0(-aq/c)m(d)m(-aq)m(-d/b)m+1(-cb)m-(1+a-1)m=0(-bq/c)m(d)m(-bq)m(-d/a)m+1(-ca)m. Now change a to -b/d, b to -c/a, and z to -d/a in (1.11) to obtain n=1(-b/d)n(-c/a)n(-da)n+n=0(-aq/c)n(-dq/b)n(-cb)n=(cd/ab)(b/a)(aq/b)(q)(-c/a)(-c/b)(-d/a)(-dq/b). Changing n to n+1 in the first summation of the above identity and then multiplying both sides by (1+d/b)-1, we find that 1(1+d/b)n=0(-b/d)n+1(-c/a)n+1(-da)n+1+n=0(-aq/c)n(-d/b)n+1(-cb)n=(1-ba)(cd/ab)(bq/a)(aq/b)(q)(-c/a)(-c/b)(-d/a)(-d/b). Using (1.10) with α=-bq/c, β=q, γ=-dq/a, and z=-c/a in the first summation of the above identity and then multiplying both sides by 1/b, we get 1bn=0(-aq/c)n(-d/b)n+1(-cb)n-1an=0(-bq/c)n(-d/a)n+1(-ca)n=(1b-1a)(cd/ab)(bq/a)(aq/b)(q)(-c/a)(-c/b)(-d/a)(-d/b). Substituting (2.11) in (2.8), we see that

(1b-1a)(cd/ab)(c)(d)(bq/a)(aq/b)(q)(-aq)(-bq)(-c/a)(-c/b)(-d/a)(-d/b)=(1+b-1)m=0(-aq/c)m(d)m(-aq)m(-d/b)m+1(-cb)m-(1+a-1)m=0(-bq/c)m(d)m(-bq)m(-d/a)m+1(-ca)m. Now setting α=-cd/b, β=cd/ab, γ=c, δ=q, and ϵ=d in (1.9) and then multiplying both sides by 1/(1+d/b)(1+c/b), we obtain n=0(cd/ab)n(c)n(d)n(1+cdq2n/b)qn(n+1)/2(-1)nanb-n(-aq)n(-c/b)n+1(-d/b)n+1=n=0(-aq/c)n(d)n(-aq)n(-d/b)n+1(-cb)n. Interchanging a and b in (2.13), we have n=0(cd/ab)n(c)n(d)n(1+cdq2n/a)qn(n+1)/2(-1)nbna-n(-bq)n(-c/a)n+1(-d/a)n+1=n=0(-bq/c)n(d)n(-bq)n(-d/a)n+1(-ca)n. Substituting (2.13) and (2.14) in (2.12), we deduce (1.7).

Theorem 2.1 (A four-variable generalization of Jacobi's triple product identity).

For |c|,|d|<|a|, |b|<1, (cd/ab)(c)(d)(b/a)(aq/b)(q)(-a)(-b)(-c/a)(-c/b)(-d/a)(-d/b)=m=0(d)m(-cq-m/a)m(-1)mamb-mqm(m+1)/2(-a)m+1(-d/b)m+1-m=0(d)m(-cq-m/b)m(-1)ma-(m+1)bm+1qm(m+1)/2(-b)m+1(-d/a)m+1.

Proof.

Employing (-aqc)m=(ac)mqm(m+1)/2(-cq-ma)m,(-bqc)m=(bc)mqm(m+1)/2(-cq-mb)m in the right side of (2.12) and then multiplying both sides by b/(1+a)(1+b), we obtain (2.15).

Acknowledgment

The authors thank the anonymous referee for several helpful comments.

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