# A Sextuple Product Identity with Applications

Authors: Jun-Ming Zhu
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

We get a new proof of a sextuple product identity depending on the Laurent expansion of an analytic function in an annulus. Many identities, including an identity for(q;q)∞4, are obtained from this sextuple product identity.


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## Body

## 1. Introduction

For convenience, we let|q|<1 throughout the paper. We employ the standard notation (1.1)(a;q)∞=∏n=0∞(1-aqn),(a,b,…,c;q)∞=(a;q)∞(b;q)∞⋯(c;q)∞.
Series product has been an interesting topic. The Jacobi triple product is one of the most famous series-product identity. We announce it in the following (see, e.g., [1, page 35, Entry 19] or [2, Equation (2.1)]):(1.2)(q,z,qz;q)∞=∑n=-∞∞(-1)nq(1/2)n(n-1)zn,z≠0.It is well known that an analytic function has a unique Laurent expansion in an annulus. Bailey [3] used this property to prove the quintuple product identity. By this approach, Cooper [4, 5] and Kongsiriwong and Liu [2] proved many types of the Macdonald identities and some other series-product identities. In this paper, we use this method to deal with a sextuple product identity.In Section2, we present the sextuple product identity ((2.1) below) and its proof. Our identity is equivalent to [2, Equation (8.16)] by Kongsiriwong and Liu, which is the simplification of [2, Equation (6.13)]. Kongsiriwong and Liu got [2, Equation (8.16)] from a more general identity. In this section, we give it a direct proof.In Section3, we get many identities from this sextuple product identity.To simplify notation, we often write∑n for ∑n=-∞∞ in the following when no confusion occurs.


## 2. A New Proof of the Sextuple Product Identity

The starting point of our investigation in this section is the identity in the following theorem.Theorem 2.1.
For any complex numberz with z≠0, one has
(2.1)(q,z,qz;q)∞(q3,z3,q3z3;q3)∞=(q12,-q6,-q6;q12)∞∑nq2n2-2nz4n+2(q12,-q12,-q12;q12)∞∑nq2n2+1z4n+2-(q3,-q3,-q3;q3)∞∑nq(1/2)(n2-n)z2n+1.Before the proof of Theorem2.1, we need some preparations. The two identities in the following lemma are from [6]. We write them in this version.Lemma 2.2.
One has(2.2)(q8,q3,q5;q8)∞(q24,q9,q15;q24)∞+q2(q8,q,q7;q8)∞(q24,q3,q21;q24)∞=(q2,-q2,-q2;q2)∞(q6,q3,q3;q6)∞,(2.3)(q8,q,q7;q8)∞(q24,q9,q15;q24)∞-q(q8,q3,q5;q8)∞(q24,q3,q21;q24)∞=(q2,q,q;q2)∞(q6,-q6,-q6;q6)∞.Proof.
For (2.2), see [6, Equation (3.18)]. Equation (2.3) is from [6, Equation (3.21)]. Its proof is similar to that of [6, Equation (3.18)].The lemma above is used to prove the following two identities.Lemma 2.3.
One has(2.4)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞+(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=2(q4,-q4,-q4;q4)∞(q12,-q6,-q6;q12)∞,(2.5)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞-(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=2q(q4,-q2,-q2;q4)∞(q12,-q12,-q12;q12)∞.Proof.
By (1.2), we have
(2.6)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞=14(q,-1,-q;q)∞(q3,-1,-q3;q3)∞=14∑mq(1/2)(m2-m)∑nq(3/2)(n2-n)=∑mq2m2+m∑nq6n2+3n=∑mq8m2+2m∑nq24n2+6n+q4∑mq8m2+6m∑nq24n2+18n+q3∑mq8m2+2m∑nq24n2+18n+q∑mq8m2+6m∑nq24n2+6n,(2.7)(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=12(q,i,-iq;q)∞(q3,-i,iq3;q3)∞=12∑m(-1)mq(1/2)(m2-m)im∑n(-1)nq(3/2)(n2-n)i3n=∑m(-1)mq2m2+m∑n(-1)nq6n2+3n=∑mq8m2+2m∑nq24n2+6n+q4∑mq8m2+6m∑nq24n2+18n-q3∑mq8m2+2m∑nq24n2+18n-q∑mq8m2+6m∑nq24n2+6n.
Adding (2.6) and (2.7), we have
(2.8)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞+(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=2∑mq8m2+2m∑nq24n2+6n+2q4∑mq8m2+6m∑nq24n2+18n=2(q16,-q6,-q10;q16)∞(q48,-q18,-q30;q48)∞+2q4(q16,-q2,-q14;q16)∞(q48,-q6,-q42;q48)∞.
By (2.2), we have (2.4).
Subtracting (2.7) from (2.6), we obtain
(2.9)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞-(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=2q3∑mq8m2+2m∑nq24n2+18n+2q∑mq8m2+6m∑nq24n2+6n=2q3(q16,-q6,-q10;q16)∞(q48,-q6,-q42;q48)∞+2q(q16,-q2,-q14;q16)∞(q48,-q18,-q32;q48)∞.
Replacing q in (2.3) by -q2 and, then, applying the resulting identity to the above equation, we get (2.5). This completes the proof.Proof of Theorem2.1.
Set(2.10)f(z,q)=(q,z,qz;q)∞(q3,z3,q3z3;q3)∞.
Then f is an analytic function of z in the annulus 0<|z|<∞. Put
(2.11)f(z,q)=∑nan(q)zn,0<|z|<∞.
By (2.10), we can easily verify
(2.12)f(z,q)=z4f(zq,q).
Combining (2.11) and (2.12) gives
(2.13)∑mam(q)zm=∑mqm-4am-4(q)zm.
Equate the coefficients of zm on both sides to get
(2.14)am(q)=qm-4am-4(q).
Using the above relation, we obtain
(2.15)a4m-1(q)=q2m2-3ma-1(q),a4m(q)=q2m2-2ma0(q),a4m+1(q)=q2m2-ma1(q),a4m+2(q)=q2m2a2(q).
Substituting the above four identities into (2.11), we have
(2.16)f(z,q)=a-1(q)∑mq2m2-3mz4m-1+a0(q)∑mq2m2-2mz4m+a1(q)∑mq2m2-mz4m+1+a2(q)∑mq2m2z4m+2.
By (2.10), we also have
(2.17)f(z,q)=f(qz,q).
This gives
(2.18)∑mam(q)zm=∑mq-ma-m(q)zm.
Then we have
(2.19)am(q)=q-ma-m(q).
Set m=1 to get
(2.20)a1(q)=q-1a-1(q).
By this relation, (2.16) reduces to
(2.21)f(z,q)=a0(q)∑mq2m2-2mz4m+a1(q)∑mq(1/2)(m2-m)z2m+1+a2(q)∑mq2m2z4m+2.
Now, it remains to determine a0(q), a1(q), and a2(q).
Puttingz=1 in (2.21) gives
(2.22)0=a0(q)∑mq2m2-2m+a1(q)∑mq(1/2)(m2-m)+a2(q)∑mq2m2.
Set z=-1 in (2.21) to get
(2.23)4(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞=a0(q)∑mq2m2-2m-a1(q)∑mq(1/2)(m2-m)+a2(q)∑mq2m2.
Taking z=i in (2.21) and noting that ∑m(-1)mq(1/2)(m2-m)=0, we have
(2.24)(q,i,-iq;q)∞(q3,-i,iq3;q3)∞=a0(q)∑mq2m2-2m-a2(q)∑mq2m2.
Subtracting (2.23) from (2.22) and noting that ∑mq(1/2)(m2-m)=2(q,-q,-q;q)∞, we obtain
(2.25)a1(q)=-(q3,-q3,-q3;q3)∞.
Add (2.22) and (2.23) to get
(2.26)2(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞=a0(q)∑mq2m2-2m+a2(q)∑mq2m2.
Adding (2.24) and (2.26) and, then, using (1.2) in the resulting equation, we obtain
(2.27)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞+(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=2a0(q)(q4,-q4,-q4;q4)∞.
By (2.4), we have
(2.28)a0(q)=(q12,-q6,-q6;q12)∞.
Similarly, subtracting (2.24) from (2.26) and, then using (1.2), we have
(2.29)(q,-q,-q;q)∞(q3,-q3,-q3;q3)∞-(q,iq,-iq;q)∞(q3,-iq3,iq3;q3)∞=a2(q)(q4,-q2,-q2;q4)∞.
Applying (2.5) to this equation gives
(2.30)a2(q)=2q(q12,-q12,-q12;q12)∞,
which completes the proof.


## 3. Some Applications

In this section, we deduce many modular identities from Theorem2.1.Corollary 3.1.
One has(3.1)3(q;q)∞3(q3;q3)∞3=(q12,-q6,-q6;q12)∞∑n2n(4n-1)q2n2-2n+2(q12,-q12,-q12;q12)∞∑n(2n+1)(4n+1)q2n2+1-(q3,-q3,-q3;q3)∞∑nn(2n+1)q1/2(n2-n).Proof.
Dividing both sides of (2.1) by (1-z)2, letting z→1, and then using L’Hospital’s rule twice on the right-hand side gives (3.1).Corollary 3.2.
One has(3.2)(q24,-q12,-q12;q24)∞(q8,-q4,-q4;q8)∞+4q4(q24,-q24,-q24;q24)∞(q8,-q8,-q8;q8)∞+2q(q6,-q6,-q6;q6)∞(q2,-q2,-q2;q2)∞=(q2,-q,-q;q2)∞(q6,-q3,-q3;q6)∞,(3.3)(q36,-q18,-q18;q36)∞(q12,-q4,-q8;q12)∞+2q5(q36,-q36,-q36;q36)∞(q12,-q2,-q10;q12)∞-q(q9,-q9,-q9;q9)∞(q3,-q,-q2;q3)∞=(q;q)∞(q3;q3)∞,(3.4)(q12,-q6,-q6;q12)∞(q4,-q4,-q4;q4)∞-q(q12,-q12,-q12;q12)∞(q4,-q2,-q2;q4)∞=(q2;q2)∞(-q;q2)∞(q6;q6)∞(-q3;q6)∞,(3.5)(q60,-q30,-q30;q60)∞(q20,-q8,-q12;q20)∞+2q9(q60,-q60,-q60;q60)∞(q20,-q2,-q18;q20)∞-q2(q15,-q15,-q15;q15)∞(q5,-q,-q4;q5)∞=(q;q)∞(q,q4;q5)∞(q5;q5)∞(q3,q12;q15)∞,(3.6)(q60,-q30,-q30;q60)∞(q20,-q4,-q16;q20)∞+2q7(q60,-q60,-q60;q60)∞(q20,-q6,-q14;q20)∞-q(q15,-q15,-q15;q15)∞(q5,-q2,-q3;q5)∞=(q;q)∞(q2,q3;q5)∞(q5;q5)∞(q6,q9;q15)∞.Proof.
Replaceq in (2.1) by q2 and, then, z by -q. Using (1.2) in the resulting identity gives (3.2).
Replaceq in (2.1) by q3 and, then, z by q. Using (1.2) in the resulting identity gives (3.3).
Replaceq in (2.1) by q4 and, then, z by q. Using (1.2) and the fact that (q4,-q,-q3;q4)∞=(q,-q,-q;q)∞ in the resulting identity, we obtain
(3.7)((q48,-q24,-q24;q48)∞+2q6(q48,-q48,-q48;q48)∞)(q4,-q4,-q4;q4)∞-q(q12,-q12,-q12;q12)∞(q4,-q2,-q2;q4)∞=(q2;q2)∞(-q;q2)∞(q6;q6)∞(-q3;q6)∞.
By (1.2), we have
(3.8)(q12,-q6,-q6;q12)∞=∑nq6n2=∑nq6(2n)2+∑nq6(2n+1)2=(q48,-q24,-q24;q48)∞+2q6(q48,-q48,-q48;q48)∞.
Combining (3.7) and (3.8) gives (3.4).
Replaceq in (2.1) by q5 and, then, z by q2. Using (1.2) in the resulting identity gives (3.5).
Replaceq in (2.1) by q5 and, then, z by q. Using (1.2) in the resulting identity gives (3.6).Obviously, using the same method above, we can get more identities from (2.1).Now, we deduce a formula for(q;q)∞4 from (2.1).Corollary 3.3.
One has(3.9)(q;q)∞4=2∑mq2m2∑n2nq6n2+2n+2q∑mq2m2+2m∑n(2n-1)q6n2-4n+∑mq2m2+m∑n(2n+1)q(1/2)(3n2+n).Proof.
Denote the left-hand side of (2.1) by f(z) and the right-hand side of (2.1) by g(z). Let z0 be a zero point of f(z). Because (2.1) holds in 0<|z|<∞, z0 is also a zero point of g(z). If az0=1, we have
(3.10)lim⁡z→z0f(z)1-az=lim⁡z→z0g(z)1-az.
Setting z0=a=1 in (3.10) and by L’Hospital’s rule on the right-hand side, we have
(3.11)0=(q3,-q3,-q3;q3)∞∑n(2n+1)q(1/2)(n2-n)-2(q12,-q12,-q12;q12)∞∑n(4n+2)q2n2+1-(q12,-q6,-q6;q12)∞∑n4nq2n2-2n.
Let ω=e(2/3)πi. Putting z0=ω and a=ω2 in (3.10) and noting ω3n=1 for any integer n, we have
(3.12)3(1-ω)(q3;q3)∞4=(q3,-q3,-q3;q3)∞∑n(2n+1)q(1/2)(n2-n)ω2(n-1)-2(q12,-q12,-q12;q12)∞∑n(4n+2)q2n2+1ωn-1-(q12,-q6,-q6;q12)∞∑n4nq2n2-2nωn.
Taking z0=ω2 and a=ω in (3.10), we obtain
(3.13)3(1-ω2)(q3;q3)∞4=(q3,-q3,-q3;q3)∞∑n(2n+1)q(1/2)(n2-n)ωn-1-2(q12,-q12,-q12;q12)∞∑n(4n+2)q2n2+1ω2(n-1)-(q12,-q6,-q6;q12)∞∑n4nq2n2-2nω2n.
Adding the above three identities together gives
(3.14)9(q3;q3)∞4=(q3,-q3,-q3;q3)∞∑n(2n+1)q(1/2)(n2-n)(1+ωn-1+ω2(n-1))-2(q12,-q12,-q12;q12)∞∑n(4n+2)q2n2+1(1+ωn-1+ω2(n-1))-(q12,-q6,-q6;q12)∞∑n4nq2n2-2n(1+ωn+ω2n).
Using the fact
(3.15)1+ωn+ω2n={3,n≡0(mod⁡3),0,n≢0(mod⁡3)
in the above identity and, then, replacing q3 by q, we get
(3.16)(q;q)∞4=(q,-q,-q;q)∞∑n(2n+1)q(1/2)(3n2+n)-4q(q4,-q4,-q4;q4)∞∑n(2n+1)q6n2+4n-2(q4,-q2,-q2;q4)∞∑n2nq6n2-2n.
Replacing n in the last two sums on the right-hand side of the above identity by -n and, then, applying (1.2) to the resulting equation, we get Corollary 3.3.


## 4. Conclusion

Besides the Jacobi triple product (1.2), well-known series-product identities are known as the quintuple product identity, the Winquist identity, and so forth. The formula (2.1) is also such an identity. Recently, we also obtain some other identities of this kind, including the simplifications of the formulae [2, Equations (6.12) and (6.14)], with a different method. These identities are widely used in number theory, combinatorics, and many other fields. literature on this topic abounds. In (2.1), if we replace z by e2iz, then the right-hand side of (2.1) turns into fourier series. For recent papers on the applications of fourier analysis, we refer the readers to [7–9].



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