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.
1. Introduction
For convenience, we let |q|<1 throughout the paper. We employ the standard notation (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)]):(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 Section 2, 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 Section 3, 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 number z with z≠0, one has
(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 Theorem 2.1, 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
(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)∞,(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
(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)∞,(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
(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,(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
(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
(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 Theorem 2.1.
Set
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
f(z,q)=∑nan(q)zn,0<|z|<∞.
By (2.10), we can easily verify
f(z,q)=z4f(zq,q).
Combining (2.11) and (2.12) gives
∑mam(q)zm=∑mqm-4am-4(q)zm.
Equate the coefficients of zm on both sides to get
am(q)=qm-4am-4(q).
Using the above relation, we obtain
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
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
f(z,q)=f(qz,q).
This gives
∑mam(q)zm=∑mq-ma-m(q)zm.
Then we have
am(q)=q-ma-m(q).
Set m=1 to get
a1(q)=q-1a-1(q).
By this relation, (2.16) reduces to
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).
Putting z=1 in (2.21) gives
0=a0(q)∑mq2m2-2m+a1(q)∑mq(1/2)(m2-m)+a2(q)∑mq2m2.
Set z=-1 in (2.21) to get
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
(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
a1(q)=-(q3,-q3,-q3;q3)∞.
Add (2.22) and (2.23) to get
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
(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
a0(q)=(q12,-q6,-q6;q12)∞.
Similarly, subtracting (2.24) from (2.26) and, then using (1.2), we have
(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
a2(q)=2q(q12,-q12,-q12;q12)∞,
which completes the proof.
3. Some Applications
In this section, we deduce many modular identities from Theorem 2.1.
Corollary 3.1.
One has
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
(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)∞,(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)∞,(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)∞,(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)∞,(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.
Replace q in (2.1) by q2 and, then, z by -q. Using (1.2) in the resulting identity gives (3.2).
Replace q in (2.1) by q3 and, then, z by q. Using (1.2) in the resulting identity gives (3.3).
Replace q 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
((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
(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).
Replace q in (2.1) by q5 and, then, z by q2. Using (1.2) in the resulting identity gives (3.5).
Replace q 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
(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
limz→z0f(z)1-az=limz→z0g(z)1-az.
Setting z0=a=1 in (3.10) and by L’Hospital’s rule on the right-hand side, we have
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(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(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
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
1+ωn+ω2n={3,n≡0(mod3),0,n≢0(mod3)
in the above identity and, then, replacing q3 by q, we get
(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].
Acknowledgment
This research is supported by the Shanghai Natural Science Foundation (Grant no. 10ZR1409100), the National Science Foundation of China (Grant no. 10771093), and the Natural Science Foundation of Education Department of Henan Province of China (Grant no. 2007110025).
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