# A Note on Four-Variable Reciprocity Theorem

Authors: Chandrashekar Adiga; P. S. Guruprasad
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We give new proof of a four-variable reciprocity theorem using Heine's transformation, Watson's transformation, and Ramanujan'sψ11-summation formula. We also obtain a generalization of Jacobi's triple product identity.


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## Body

## 1. Introduction

Throughout the paper, we let|q|<1 and we employ the standard notation:
(1.1)(a)0:=(a;q)0=1,(a)∞:=(a;q)∞=∏n=0∞(1-aqn),(a)n:=(a;q)n=(a;q)∞(aqn;q)∞,-∞<n<∞.
Ramanujan [1] stated several q-series identities in his “lost’’ notebook. One of the beautiful identities is the two-variable reciprocity theorem.Theorem 1.1 (see [2]).
Forab≠0,
(1.2)ρ(a,b)-ρ(b,a)=(1b-1a)(aq/b)∞(bq/a)∞(q)∞(-aq)∞(-bq)∞,
where
(1.3)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 [3] using four-free-variable identity and Jacobi's triple product identity. Further, Andrews [4] applied (1.2) in proving Euler partition identity analogues stated in [1]. Somashekara and Fathima [5] established an equivalent version of (1.2) using Ramanujan's 1ψ1 summation formula [6] and Heine's transformation [7, 8]. Berndt et al. [9] also derived (1.2) using the same above mentioned two transformations. In fact, Berndt et al. [9] in the same paper have given two more proofs of (1.2) one employing the Rogers-Fine identity [10] and the other is purely combinatorial. Using the q-binomial theorem:
(1.4)∑n=0∞(a)n(q)ntn=(at)∞(t)∞,|t|<1,|q|<1,
Kim et al. [11] gave a much different proof of (1.2). Guruprasad and Pradeep [12] also devised a proof of (1.2) using the q-binomial theorem. Adiga and Anitha [13] established (1.2) along the lines of Ismail's proof [14] 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 [2] constructed a proof of (1.2) along the lines of Venkatachaliengar's proof of the Ramanujan 1ψ1 summation formula [6, 15].In [2] Kang proved the following three- and four-variable generalizations of (1.2).For|c|<|a|<1 and |c|<|b|<1,
(1.5)ρ3(a,b,c)-ρ3(b,a,c)=(1b-1a)(c)∞(aq/b)∞(bq/a)∞(q)∞(-c/a)∞(-c/b)∞(-aq)∞(-bq)∞,
where
(1.6)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,(1.7)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
(1.8)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 [2] 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 [16] 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 [19] also established (1.5).Kang [2] 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 [8]:
(1.9)∑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:
(1.10)∑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:
(1.11)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(1.12)∑n=-∞∞qn(n+1)/2zn=(q)∞(-zq)∞(-1z)∞,z≠0,|q|<1.
Andrews [22] gave a proof of (1.12) using Euler identities. Combinatorial proofs of Jacobi's triple product identity were given by Wright [23], Cheema [24], and Sudler [25]. We can also find a proof of (1.12) in [26]. Using (1.12), Hirschhorn [27, 28] established Jacobi's two-square and four-square theorems.Somashekara and Fathima [5] and Kim et al. [11] established
(1.13)∑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].


## 2. Proof of (1.7)—The Four-Variable Reciprocity Theorem

On employingq-binomial theorem, we have
(2.1)∑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)m∑n=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
(2.2)∑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+1∑n=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+1∑n=0m(cq/a)n(q)n(-a)n.
Substituting this in (2.1), we obtain
(2.3)∑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=0∞∑n=0m(b/d)m(dq/a)m(cq/a)n(-a)n(cq/a)m+1(q)n.
Now,(2.4)(1+a-1)(-dq)∞(-bq)∞∑m=0∞∑n=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+1∑n=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
(2.5)∑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.)Changingc to -c/q, d to -d/q in (2.5), we get
(2.6)∑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
(2.7)∑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
(2.8)(d)∞(c)∞(-bq)∞(-aq)∞[1b∑m=0∞(-aq/c)m(-d/b)m+1(-cb)m-1a∑m=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
(2.9)∑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
(2.10)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
(2.11)1b∑n=0∞(-aq/c)n(-d/b)n+1(-cb)n-1a∑n=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(2.12)(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
(2.13)∑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
(2.14)∑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,
(2.15)(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(2.16)(-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).



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