# On a New Summation Formula for2ψ2 Basic Bilateral Hypergeometric Series  and Its Applications

Authors: D. D. Somashekara; K. Narasimha Murthy; S. L. Shalini
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We have obtained a new summation formula for2ψ2 bilateral basic hypergeometric series by the method of parameter augmentation and demonstrated its various uses leading to some development of etafunctions, q-gamma, and q-beta function identities.


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## Body

## 1. Introduction

The summation formulae for hypergeometric series form a very interesting and useful component of the theory of (basic) hypergeometric series. Theq-binomial theorem of Cauchy [1] is perhaps the first identity in the class of the summation formulae, which can be stated as (1.1)∑k=0∞(a)k(q)kzk=(az)∞(z)∞,|z|<1,|q|<1,
where(1.2)(a)∞:=(a;q)∞:=∏k=0∞(1-aqk),(a)k:=(a;q)k:=(a)∞(aqk)∞,kisaninteger.
For more details about the q-binomial theorem and about the identities which fall in this sequel, one may refer to the book by Gasper and Rahman [2]. Another famous identity in the sequel is the Ramanujan's 1ψ1 summation formula [3](1.3)∑k=-∞∞(a)k(b)kzk=(az)∞(q)∞(q/az)∞(b/a)∞(z)∞(b)∞(b/az)∞(q/a)∞,|ba|<|z|<1,|q|<1.
There are a number of proofs of the1ψ1 summation formula (1.3) in the literature. For more details, one refers to the book by Berndt [4] and a recent paper of Johnson [5].In this paper, we derive a new summation formula for2ψ2 basic bilateral hypergeometric series using the1ψ1 summation formula (1.3) by the method of parameter augmentation. We then use the formula to derive the q-analogue of Gauss summation formula and to obtain a number of etafunction, q-gamma, and q-beta function identities, which complement the works of Bhargava and Somashekara [6], Bhargava et al. [7], Somashekara and Mamta [8], Srivastava [9], and Bhargava and Adiga [10].First, we recall thatq-difference operator and the q-shift operator are defined by (1.4)Dqf(a)=f(a)-f(aq)a,ζ(f(a))=f(aq),
respectively. In [11], Chen and Liu have constructed an operator θ as (1.5)θ=ζ-1Dq,
and thereby they defined the operator E(bθ) as(1.6)E(bθ)=∑k=0∞(bθ)kqk(k-1)/2(q;q)k.
Then, we have the following operator identities [12, Theorem 1]:(1.7)E(bθ){(at;q)∞}=(at,bt;q)∞,E(bθ){(as,at;q)∞}=(as,at,bs,bt;q)∞(abst/q;q)∞,|abstq|<1.Further, the Dedekind etafunction is defined by(1.8)η(τ):=eπiτ/12∏k=1∞(1-e2πikτ):=q1/24(q;q)∞,
where q=e2πiτ, and Im (τ)>0.Jackson [13] defined the q-analogue of the gamma function by(1.9)Γq(x)=(q;q)∞(qx;q)∞(1-q)1-x,0<q<1.
In his paper on the q-gamma and q-beta function, Askey [14] has obtained q-analogues of several classical results about the gamma function. Further, he has given the definition for q-beta function as(1.10)Bq(x,y)=(1-q)∑n=0∞qnx(qn+1)∞(qx+y)∞.
In fact, he has shown that(1.11)Bq(x,y)=Γq(x)Γq(y)Γq(x+y).In Section2, we prove our main result. In Section 3, we deduce the well-known q-analogue of the Gauss summation formula and some etafunction, q-gamma, and q-beta function identities.


## 2. Main Result

Theorem 2.1.
If0<|z|<1,|q|<1, then
(2.1)∑k=-∞∞(a)k(bc/azq)k(b)k(c)kzk=(az)∞(q)∞(q/az)∞(b/a)∞(c/a)∞(bc/azq)∞(z)∞(b)∞(c)∞(b/az)∞(c/az)∞(q/a)∞.Proof.
Ramanujan's1ψ1 summation formula (1.3) can be written as
(2.2)∑k=0∞(a)k(b)kzk+∑k=1∞(q/b)k(q/a)k(baz)k=(az)∞(q)∞(q/az)∞(b/a)∞(z)∞(b)∞(b/az)∞(q/a)∞.
This is the same as
(2.3)∑k=0∞(a)kzk{(bqk)∞(baz)∞}+∑k=1∞(-1)kqk(k+1)/2(q/a)k(1az)k{(bq-k)∞(baz)∞}=(az)∞(q)∞(q/az)∞(z)∞(q/a)∞{(ba)∞}.
On applying E(cθ) to both sides with respect to b, we obtain
(2.4)∑k=0∞(a)kzk{(bqk)∞(b/az)∞(cqk)∞(c/az)∞(bcqk/azq)∞}+∑k=1∞(-1)kqk(k+1)/2(q/a)k(1az)k{(bq-k)∞(b/az)∞(cq-k)∞(c/az)∞(bcq-k/azq)∞}=(az)∞(q)∞(q/az)∞(z)∞(q/a)∞{(ba)∞(ca)∞}.
Multiplying (2.4) throughout by {(bc/azq)∞/(b)∞(c)∞}, we obtain
(2.5)∑k=0∞(a)k(bc/azq)k(b)k(c)kzk+∑k=1∞(a)-k(bc/azq)-k(b)-k(c)-kz-k=(az)∞(q)∞(q/az)∞(b/a)∞(c/a)∞(bc/azq)∞(z)∞(b)∞(c)∞(b/az)∞(c/az)∞(q/a)∞,
which yields (2.1).


## 3. Some Applications of the Main Identity

The following identity is the well-knownq-analogue of the Gauss summation formula.Corollary 3.1 (see [15]).
If|q|<1, |γ/αβ|<1, then
(3.1)∑k=0∞(α)k(β)k(q)k(γ)k(γαβ)k=(γ/α)∞(γ/β)∞(γ)∞(γ/αβ)∞.Proof.
Puttinga=α, b=q, c=γ, and z=γ/αβ in (2.1), we obtain (3.1).Corollary 3.2.
If|q|<1, then
(3.2)∑k=-∞∞(q2;q4)k(1-q4k+1)(q3;q4)kqk=q-1/8η7(2τ)η3(τ)η2(4τ),(3.3)∑k=-∞∞(-q2;q2)k(1+q2k+1)2(-q;q2)kqk=2q-3/4η6(4τ)η3(2τ),(3.4)∑k=-∞∞(-q;q2)k(-q2;q2)k(1-q2k+1)2(q;q2)n2qk=2q-3/4η8(4τ)η7(2τ),(3.5)∑k=-∞∞(1+q)(-q;q2)k(-q2;q2)k(1-q2k+2)(q2;q2)k2qk=η(2τ)η2(τ),(3.6)∑k=-∞∞(q2;q6)k(1-q6k+1)(q5;q6)kq3k=q-1/4η4(2τ)η2(τ),(3.7)∑k=-∞∞(1+q2)2(-q2;q2)k-1(-q2;q2)k+1(1-q2k+2)2(q2;q2)k2qk=η2(4τ)η4(2τ).Proof.
Puttinga=z=q1/4, b=q3/4, c=q5/4,  and then changing q to q4 in (2.1), we obtain
(3.8)∑k=-∞∞(q;q4)k(q2;q4)k(q;q4)k+1(q3;q4)kqk=(q2;q4)∞4(q4;q4)∞2(q;q4)∞3(q3;q4)∞3.
Simplifying the right hand side and then using (1.8), we obtain (3.2).
Similarly, puttinga=-q1/2, z=q1/2, b=-q3/2, c=-q3/2,  and then changing q to q2, we obtain (3.3). Putting  a=-q1/2,    z=q1/2,  b=c=q3/2, and then changing q to q2, we obtain (3.4). Putting a=-q1/2,    z=q1/2,  b=q, c=q2, and then changing q to q2, we obtain (3.5). Putting a=q1/6,    z=q1/2, b=q7/6,  c=q5/6, and then changing q to q6, we obtain (3.6). Finally, putting  a=-1, z=q,  b=c=q2,  and then changing q to q2, we obtain (3.7).Corollary 3.3.
If0<q<1,  0<x,  y<1, and 0<x+y<1, then
(3.9)Bq3(x,y)=Γq(1-x+y)Γq(x-y)(1-q)2(1-qy)2∑k=-∞∞(q1-x)k(qx+y)k(q1+y)k2qky.Proof.
Puttinga=q1-x,  z=qy, and b=c=q1+y in (2.1), we get
(3.10)∑k=-∞∞(q1-x)k(qx+y)k(q1+y)k2qky=(q1-x+y)∞(q)∞(qx-y)∞(qy)∞(q1+y)∞(q1+y)∞((qx+y)∞(qx)∞)3.
On using (1.9), (1.10), and (1.11), we obtain (3.9).Corollary 3.4.
If0<q<1, 0<x, y<1, and 1<x+y<2, then
(3.11)Bq2(x,y)=Γq(y-x)Γq(x-y+1)Γq(x+y-1)Γq(y)Γq(1+x)∑k=-∞∞(q-x)k(qx+y-1)k(qy)k2qky.Proof.
Puttinga=q-x,  z=b=c=qy,  in (2.1), and then using (1.9), (1.10), and (1.11), we obtain (3.11).Corollary 3.5.
If0<x, y<1, and 0<x+y<1, then
(3.12)B3(x,y)=Γ(1-x+y)Γ(x-y)y2[∑k=0∞(1-x)k(x+y)k(1+y)k2+∑k=1∞(-y)k2(x)k(1-x-y)k].Proof.
Lettingq→1 in (3.9), we obtain (3.12).Corollary 3.6.
If0<x,  y<1, and 1<x+y<2, then
(3.13)B2(x,y)=Γ(y-x)Γ(x-y+1)Γ(x+y-1)Γ(y)Γ(1+x)×[∑k=0∞(-x)k(x+y-1)k(y)k2+∑k=1∞(1-y)k2(1+x)k(2-x-y)k].Proof.
Lettingq→ 1 in (3.11), we obtain (3.13).Corollary 3.7.
If0<q<1,  0<x, y<1, and 0<x+y<1, then
(3.14)Bq(x,y)=Γq(y)Γq(1-y)∑k=0∞(q1-x-y)k(qy)k(q)k2qkx.Proof.
Puttinga=q1-x-y, z=qx, and b=c=q in (2.1), we obtain (3.14).



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*Source: 132081-2011-03-20.xml*