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Mathematical Problems in Engineering
VolumeΒ 2011Β (2011), Article IDΒ 462507, 11 pages
http://dx.doi.org/10.1155/2011/462507
Research Article

A Sextuple Product Identity with Applications

1Department of Mathematics, East China Normal University, Shanghai 200241, China
2Department of Mathematics, Luoyang Normal University, Luoyang, Henan 471022, China

Received 21 February 2011; Revised 20 March 2011; Accepted 24 March 2011

Academic Editor: MingΒ Li

Copyright Β© 2011 Jun-Ming Zhu. This 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.

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 (π‘ž;π‘ž)4∞, are obtained from this sextuple product identity.

1. Introduction

For convenience, we let |π‘ž|<1 throughout the paper. We employ the standard notation (π‘Ž;π‘ž)∞=βˆžξ‘π‘›=0(1βˆ’π‘Žπ‘žπ‘›),(π‘Ž,𝑏,…,𝑐;π‘ž)∞=(π‘Ž;π‘ž)∞(𝑏;π‘ž)βˆžβ‹―(𝑐;π‘ž)∞.(1.1) 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)]):ξ‚€π‘žπ‘ž,𝑧,𝑧;π‘žβˆž=βˆžξ“π‘›=βˆ’βˆž(βˆ’1)π‘›π‘ž(1/2)𝑛(π‘›βˆ’1)𝑧𝑛,𝑧≠0.(1.2)

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 βˆ‘π‘› for βˆ‘βˆžπ‘›=βˆ’βˆž 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 ξ‚€π‘žπ‘ž,𝑧,𝑧;π‘žβˆžξ‚΅π‘ž3,𝑧3,π‘ž3𝑧3;π‘ž3ξ‚Άβˆž=ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›π‘ž2𝑛2βˆ’2𝑛𝑧4π‘›ξ€·π‘ž+212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›π‘ž2𝑛2+1𝑧4𝑛+2βˆ’ξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›π‘ž(1/2)(𝑛2βˆ’π‘›)𝑧2𝑛+1.(2.1)

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 ξ€·π‘ž8,π‘ž3,π‘ž5;π‘ž8ξ€Έβˆžξ€·π‘ž24,π‘ž9,π‘ž15;π‘ž24ξ€Έβˆž+π‘ž2ξ€·π‘ž8,π‘ž,π‘ž7;π‘ž8ξ€Έβˆžξ€·π‘ž24,π‘ž3,π‘ž21;π‘ž24ξ€Έβˆž=ξ€·π‘ž2,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž2ξ€Έβˆžξ€·π‘ž6,π‘ž3,π‘ž3;π‘ž6ξ€Έβˆž,ξ€·π‘ž(2.2)8,π‘ž,π‘ž7;π‘ž8ξ€Έβˆžξ€·π‘ž24,π‘ž9,π‘ž15;π‘ž24ξ€Έβˆžξ€·π‘žβˆ’π‘ž8,π‘ž3,π‘ž5;π‘ž8ξ€Έβˆžξ€·π‘ž24,π‘ž3,π‘ž21;π‘ž24ξ€Έβˆž=ξ€·π‘ž2,π‘ž,π‘ž;π‘ž2ξ€Έβˆžξ€·π‘ž6,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž6ξ€Έβˆž.(2.3)

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 (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž+(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆžξ€·π‘ž=24,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž4ξ€Έβˆžξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆž,(2.4)(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžβˆ’(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆžξ€·π‘ž=2π‘ž4,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž4ξ€Έβˆžξ€·π‘ž12,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆž.(2.5)

Proof. By (1.2), we have (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž=14(π‘ž,βˆ’1,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’1,βˆ’π‘ž3;π‘ž3ξ€Έβˆž=14ξ“π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)ξ“π‘›π‘ž(3/2)(𝑛2βˆ’π‘›)=ξ“π‘šπ‘ž2π‘š2+π‘šξ“π‘›π‘ž6𝑛2+3𝑛=ξ“π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+6𝑛+π‘ž4ξ“π‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+18𝑛+π‘ž3ξ“π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+18𝑛+π‘žπ‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+6𝑛,(2.6)(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆž=12(π‘ž,𝑖,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–,π‘–π‘ž3;π‘ž3ξ€Έβˆž=12ξ“π‘š(βˆ’1)π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)π‘–π‘šξ“π‘›(βˆ’1)π‘›π‘ž(3/2)(𝑛2βˆ’π‘›)𝑖3𝑛=ξ“π‘š(βˆ’1)π‘šπ‘ž2π‘š2+π‘šξ“π‘›(βˆ’1)π‘›π‘ž6𝑛2+3𝑛=ξ“π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+6𝑛+π‘ž4ξ“π‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+18π‘›βˆ’π‘ž3ξ“π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+18π‘›ξ“βˆ’π‘žπ‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+6𝑛.(2.7) Adding (2.6) and (2.7), we have (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž+(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆžξ“=2π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+6𝑛+2π‘ž4ξ“π‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+18π‘›ξ€·π‘ž=216,βˆ’π‘ž6,βˆ’π‘ž10;π‘ž16ξ€Έβˆžξ€·π‘ž48,βˆ’π‘ž18,βˆ’π‘ž30;π‘ž48ξ€Έβˆž+2π‘ž4ξ€·π‘ž16,βˆ’π‘ž2,βˆ’π‘ž14;π‘ž16ξ€Έβˆžξ€·π‘ž48,βˆ’π‘ž6,βˆ’π‘ž42;π‘ž48ξ€Έβˆž.(2.8) By (2.2), we have (2.4).
Subtracting (2.7) from (2.6), we obtain (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžβˆ’(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆž=2π‘ž3ξ“π‘šπ‘ž8π‘š2+2π‘šξ“π‘›π‘ž24𝑛2+18𝑛+2π‘žπ‘šπ‘ž8π‘š2+6π‘šξ“π‘›π‘ž24𝑛2+6𝑛=2π‘ž3ξ€·π‘ž16,βˆ’π‘ž6,βˆ’π‘ž10;π‘ž16ξ€Έβˆžξ€·π‘ž48,βˆ’π‘ž6,βˆ’π‘ž42;π‘ž48ξ€Έβˆžξ€·π‘ž+2π‘ž16,βˆ’π‘ž2,βˆ’π‘ž14;π‘ž16ξ€Έβˆžξ€·π‘ž48,βˆ’π‘ž18,βˆ’π‘ž32;π‘ž48ξ€Έβˆž.(2.9) Replacing π‘ž in (2.3) by βˆ’π‘ž2 and, then, applying the resulting identity to the above equation, we get (2.5). This completes the proof.

Proof of Theorem 2.1. Set ξ‚€π‘žπ‘“(𝑧,π‘ž)=π‘ž,𝑧,𝑧;π‘žβˆžξ‚΅π‘ž3,𝑧3,π‘ž3𝑧3;π‘ž3ξ‚Άβˆž.(2.10) Then 𝑓 is an analytic function of 𝑧 in the annulus 0<|𝑧|<∞. Put 𝑓(𝑧,π‘ž)=π‘›π‘Žπ‘›(π‘ž)𝑧𝑛,0<|𝑧|<∞.(2.11) By (2.10), we can easily verify 𝑓(𝑧,π‘ž)=𝑧4𝑓(π‘§π‘ž,π‘ž).(2.12) Combining (2.11) and (2.12) gives ξ“π‘šπ‘Žπ‘š(π‘ž)π‘§π‘š=ξ“π‘šπ‘žπ‘šβˆ’4π‘Žπ‘šβˆ’4(π‘ž)π‘§π‘š.(2.13) Equate the coefficients of π‘§π‘š on both sides to get π‘Žπ‘š(π‘ž)=π‘žπ‘šβˆ’4π‘Žπ‘šβˆ’4(π‘ž).(2.14) Using the above relation, we obtain π‘Ž4π‘šβˆ’1(π‘ž)=π‘ž2π‘š2βˆ’3π‘šπ‘Žβˆ’1(π‘ž),π‘Ž4π‘š(π‘ž)=π‘ž2π‘š2βˆ’2π‘šπ‘Ž0π‘Ž(π‘ž),4π‘š+1(π‘ž)=π‘ž2π‘š2βˆ’π‘šπ‘Ž1(π‘ž),π‘Ž4π‘š+2(π‘ž)=π‘ž2π‘š2π‘Ž2(π‘ž).(2.15) Substituting the above four identities into (2.11), we have 𝑓(𝑧,π‘ž)=π‘Žβˆ’1(π‘ž)π‘šπ‘ž2π‘š2βˆ’3π‘šπ‘§4π‘šβˆ’1+π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘šπ‘§4π‘š+π‘Ž1(π‘ž)π‘šπ‘ž2π‘š2βˆ’π‘šπ‘§4π‘š+1+π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2𝑧4π‘š+2.(2.16) By (2.10), we also have ξ‚€π‘žπ‘“(𝑧,π‘ž)=𝑓𝑧,π‘ž.(2.17) This gives ξ“π‘šπ‘Žπ‘š(π‘ž)π‘§π‘š=ξ“π‘šπ‘žβˆ’π‘šπ‘Žβˆ’π‘š(π‘ž)π‘§π‘š.(2.18) Then we have π‘Žπ‘š(π‘ž)=π‘žβˆ’π‘šπ‘Žβˆ’π‘š(π‘ž).(2.19) Set π‘š=1 to get π‘Ž1(π‘ž)=π‘žβˆ’1π‘Žβˆ’1(π‘ž).(2.20) By this relation, (2.16) reduces to 𝑓(𝑧,π‘ž)=π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘šπ‘§4π‘š+π‘Ž1(π‘ž)π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)𝑧2π‘š+1+π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2𝑧4π‘š+2.(2.21) Now, it remains to determine π‘Ž0(π‘ž), π‘Ž1(π‘ž), and π‘Ž2(π‘ž).
Putting 𝑧=1 in (2.21) gives 0=π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘š+π‘Ž1(π‘ž)π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)+π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2.(2.22) Set 𝑧=βˆ’1 in (2.21) to get 4(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž=π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘šβˆ’π‘Ž1(π‘ž)π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)+π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2.(2.23) Taking 𝑧=𝑖 in (2.21) and noting that βˆ‘π‘š(βˆ’1)π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)=0, we have (π‘ž,𝑖,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–,π‘–π‘ž3;π‘ž3ξ€Έβˆž=π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘šβˆ’π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2.(2.24) Subtracting (2.23) from (2.22) and noting that βˆ‘π‘šπ‘ž(1/2)(π‘š2βˆ’π‘š)=2(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)∞, we obtain π‘Ž1ξ€·π‘ž(π‘ž)=βˆ’3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž.(2.25) Add (2.22) and (2.23) to get 2(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž=π‘Ž0(π‘ž)π‘šπ‘ž2π‘š2βˆ’2π‘š+π‘Ž2(π‘ž)π‘šπ‘ž2π‘š2.(2.26) Adding (2.24) and (2.26) and, then, using (1.2) in the resulting equation, we obtain (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆž+(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆž=2π‘Ž0ξ€·π‘ž(π‘ž)4,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž4ξ€Έβˆž.(2.27) By (2.4), we have π‘Ž0ξ€·π‘ž(π‘ž)=12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆž.(2.28) Similarly, subtracting (2.24) from (2.26) and, then using (1.2), we have (π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžβˆ’(π‘ž,π‘–π‘ž,βˆ’π‘–π‘ž;π‘ž)βˆžξ€·π‘ž3,βˆ’π‘–π‘ž3,π‘–π‘ž3;π‘ž3ξ€Έβˆž=π‘Ž2ξ€·π‘ž(π‘ž)4,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž4ξ€Έβˆž.(2.29) Applying (2.5) to this equation gives π‘Ž2ξ€·π‘ž(π‘ž)=2π‘ž12,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆž,(2.30) 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(π‘ž;π‘ž)3βˆžξ€·π‘ž3;π‘ž3ξ€Έ3∞=ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›2𝑛(4π‘›βˆ’1)π‘ž2𝑛2βˆ’2π‘›ξ€·π‘ž+212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›(2𝑛+1)(4𝑛+1)π‘ž2𝑛2+1βˆ’ξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›π‘›(2𝑛+1)π‘ž1/2(𝑛2βˆ’π‘›).(3.1)

Proof. Dividing both sides of (2.1) by (1βˆ’π‘§)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 ξ€·π‘ž24,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž24ξ€Έβˆžξ€·π‘ž8,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž8ξ€Έβˆž+4π‘ž4ξ€·π‘ž24,βˆ’π‘ž24,βˆ’π‘ž24;π‘ž24ξ€Έβˆžξ€·π‘ž8,βˆ’π‘ž8,βˆ’π‘ž8;π‘ž8ξ€Έβˆžξ€·π‘ž+2π‘ž6,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž6ξ€Έβˆžξ€·π‘ž2,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž2ξ€Έβˆž=ξ€·π‘ž2,βˆ’π‘ž,βˆ’π‘ž;π‘ž2ξ€Έβˆžξ€·π‘ž6,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž6ξ€Έβˆž,ξ€·π‘ž(3.2)36,βˆ’π‘ž18,βˆ’π‘ž18;π‘ž36ξ€Έβˆžξ€·π‘ž12,βˆ’π‘ž4,βˆ’π‘ž8;π‘ž12ξ€Έβˆž+2π‘ž5ξ€·π‘ž36,βˆ’π‘ž36,βˆ’π‘ž36;π‘ž36ξ€Έβˆžξ€·π‘ž12,βˆ’π‘ž2,βˆ’π‘ž10;π‘ž12ξ€Έβˆžξ€·π‘žβˆ’π‘ž9,βˆ’π‘ž9,βˆ’π‘ž9;π‘ž9ξ€Έβˆžξ€·π‘ž3,βˆ’π‘ž,βˆ’π‘ž2;π‘ž3ξ€Έβˆž=(π‘ž;π‘ž)βˆžξ€·π‘ž3;π‘ž3ξ€Έβˆž,ξ€·π‘ž(3.3)12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ€·π‘ž4,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž4ξ€Έβˆžξ€·π‘žβˆ’π‘ž12,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ€·π‘ž4,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž4ξ€Έβˆž=ξ€·π‘ž2;π‘ž2ξ€Έβˆžξ€·βˆ’π‘ž;π‘ž2ξ€Έβˆžξ€·π‘ž6;π‘ž6ξ€Έβˆžξ€·βˆ’π‘ž3;π‘ž6ξ€Έβˆž,ξ€·π‘ž(3.4)60,βˆ’π‘ž30,βˆ’π‘ž30;π‘ž60ξ€Έβˆžξ€·π‘ž20,βˆ’π‘ž8,βˆ’π‘ž12;π‘ž20ξ€Έβˆž+2π‘ž9ξ€·π‘ž60,βˆ’π‘ž60,βˆ’π‘ž60;π‘ž60ξ€Έβˆžξ€·π‘ž20,βˆ’π‘ž2,βˆ’π‘ž18;π‘ž20ξ€Έβˆžβˆ’π‘ž2ξ€·π‘ž15,βˆ’π‘ž15,βˆ’π‘ž15;π‘ž15ξ€Έβˆžξ€·q5,βˆ’π‘ž,βˆ’π‘ž4;π‘ž5ξ€Έβˆž=(π‘ž;π‘ž)βˆžξ€·π‘ž,π‘ž4;π‘ž5ξ€Έβˆžξ€·π‘ž5;π‘ž5ξ€Έβˆžξ€·π‘ž3,π‘ž12;π‘ž15ξ€Έβˆž,ξ€·π‘ž(3.5)60,βˆ’π‘ž30,βˆ’π‘ž30;π‘ž60ξ€Έβˆžξ€·π‘ž20,βˆ’π‘ž4,βˆ’π‘ž16;π‘ž20ξ€Έβˆž+2π‘ž7ξ€·π‘ž60,βˆ’π‘ž60,βˆ’π‘ž60;π‘ž60ξ€Έβˆžξ€·π‘ž20,βˆ’π‘ž6,βˆ’π‘ž14;π‘ž20ξ€Έβˆžξ€·π‘žβˆ’π‘ž15,βˆ’π‘ž15,βˆ’π‘ž15;π‘ž15ξ€Έβˆžξ€·π‘ž5,βˆ’π‘ž2,βˆ’π‘ž3;π‘ž5ξ€Έβˆž=(π‘ž;π‘ž)βˆžξ€·π‘ž2,π‘ž3;π‘ž5ξ€Έβˆžξ€·π‘ž5;π‘ž5ξ€Έβˆžξ€·π‘ž6,π‘ž9;π‘ž15ξ€Έβˆž.(3.6)

Proof. Replace π‘ž in (2.1) by π‘ž2 and, then, 𝑧 by βˆ’π‘ž. Using (1.2) in the resulting identity gives (3.2).
Replace π‘ž in (2.1) by π‘ž3 and, then, 𝑧 by π‘ž. Using (1.2) in the resulting identity gives (3.3).
Replace π‘ž in (2.1) by π‘ž4 and, then, 𝑧 by π‘ž. Using (1.2) and the fact that (π‘ž4,βˆ’π‘ž,βˆ’π‘ž3;π‘ž4)∞=(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)∞ in the resulting identity, we obtain π‘žξ€·ξ€·48,βˆ’π‘ž24,βˆ’π‘ž24;π‘ž48ξ€Έβˆž+2π‘ž6ξ€·π‘ž48,βˆ’π‘ž48,βˆ’π‘ž48;π‘ž48ξ€Έβˆžπ‘žξ€Έξ€·4,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž4ξ€Έβˆžξ€·π‘žβˆ’π‘ž12,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ€·π‘ž4,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž4ξ€Έβˆž=ξ€·π‘ž2;π‘ž2ξ€Έβˆžξ€·βˆ’π‘ž;π‘ž2ξ€Έβˆžξ€·π‘ž6;π‘ž6ξ€Έβˆžξ€·βˆ’π‘ž3;π‘ž6ξ€Έβˆž.(3.7) By (1.2), we have ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆž=ξ“π‘›π‘ž6𝑛2=ξ“π‘›π‘ž6(2𝑛)2+ξ“π‘›π‘ž6(2𝑛+1)2=ξ€·π‘ž48,βˆ’π‘ž24,βˆ’π‘ž24;π‘ž48ξ€Έβˆž+2π‘ž6ξ€·π‘ž48,βˆ’π‘ž48,βˆ’π‘ž48;π‘ž48ξ€Έβˆž.(3.8) Combining (3.7) and (3.8) gives (3.4).
Replace π‘ž in (2.1) by π‘ž5 and, then, 𝑧 by π‘ž2. Using (1.2) in the resulting identity gives (3.5).
Replace π‘ž in (2.1) by π‘ž5 and, then, 𝑧 by π‘ž. 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 (π‘ž;π‘ž)4∞ from (2.1).

Corollary 3.3. One has (π‘ž;π‘ž)4βˆžξ“=2π‘šπ‘ž2π‘š2𝑛2π‘›π‘ž6𝑛2+2𝑛+2π‘žπ‘šπ‘ž2π‘š2+2π‘šξ“π‘›(2π‘›βˆ’1)π‘ž6𝑛2βˆ’4𝑛+ξ“π‘šπ‘ž2π‘š2+π‘šξ“π‘›(2𝑛+1)π‘ž(1/2)(3𝑛2+𝑛).(3.9)

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 π‘Žπ‘§0=1, we have lim𝑧→𝑧0𝑓(𝑧)1βˆ’π‘Žπ‘§=lim𝑧→𝑧0𝑔(𝑧)1βˆ’π‘Žπ‘§.(3.10) Setting 𝑧0=π‘Ž=1 in (3.10) and by L’Hospital’s rule on the right-hand side, we have ξ€·π‘ž0=3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›(2𝑛+1)π‘ž(1/2)(𝑛2βˆ’π‘›)ξ€·π‘žβˆ’212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›(4𝑛+2)π‘ž2𝑛2+1βˆ’ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›4π‘›π‘ž2𝑛2βˆ’2𝑛.(3.11) Let πœ”=𝑒(2/3)πœ‹π‘–. Putting 𝑧0=πœ” and π‘Ž=πœ”2 in (3.10) and noting πœ”3𝑛=1 for any integer 𝑛, we have ξ€·π‘ž3(1βˆ’πœ”)3;π‘ž3ξ€Έ4∞=ξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›(2𝑛+1)π‘ž(1/2)(𝑛2βˆ’π‘›)πœ”2(π‘›βˆ’1)ξ€·π‘žβˆ’212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›(4𝑛+2)π‘ž2𝑛2+1πœ”π‘›βˆ’1βˆ’ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›4π‘›π‘ž2𝑛2βˆ’2π‘›πœ”π‘›.(3.12) Taking 𝑧0=πœ”2 and π‘Ž=πœ” in (3.10), we obtain 3ξ€·1βˆ’πœ”2π‘žξ€Έξ€·3;π‘ž3ξ€Έ4∞=ξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›(2𝑛+1)π‘ž(1/2)(𝑛2βˆ’π‘›)πœ”π‘›βˆ’1ξ€·π‘žβˆ’212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›(4𝑛+2)π‘ž2𝑛2+1πœ”2(π‘›βˆ’1)βˆ’ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›4π‘›π‘ž2𝑛2βˆ’2π‘›πœ”2𝑛.(3.13) Adding the above three identities together gives 9ξ€·π‘ž3;π‘ž3ξ€Έ4∞=ξ€·π‘ž3,βˆ’π‘ž3,βˆ’π‘ž3;π‘ž3ξ€Έβˆžξ“π‘›(2𝑛+1)π‘ž(1/2)(𝑛2βˆ’π‘›)ξ€·1+πœ”π‘›βˆ’1+πœ”2(π‘›βˆ’1)ξ€Έξ€·π‘žβˆ’212,βˆ’π‘ž12,βˆ’π‘ž12;π‘ž12ξ€Έβˆžξ“π‘›(4𝑛+2)π‘ž2𝑛2+1ξ€·1+πœ”π‘›βˆ’1+πœ”2(π‘›βˆ’1)ξ€Έβˆ’ξ€·π‘ž12,βˆ’π‘ž6,βˆ’π‘ž6;π‘ž12ξ€Έβˆžξ“π‘›4π‘›π‘ž2𝑛2βˆ’2𝑛1+πœ”π‘›+πœ”2𝑛.(3.14) Using the fact 1+πœ”π‘›+πœ”2𝑛=ξƒ―3,𝑛≑0(mod3),0,𝑛≒0(mod3)(3.15) in the above identity and, then, replacing π‘ž3 by π‘ž, we get (π‘ž;π‘ž)4∞=(π‘ž,βˆ’π‘ž,βˆ’π‘ž;π‘ž)βˆžξ“π‘›(2𝑛+1)π‘ž(1/2)(3𝑛2+𝑛)ξ€·π‘žβˆ’4π‘ž4,βˆ’π‘ž4,βˆ’π‘ž4;π‘ž4ξ€Έβˆžξ“π‘›(2𝑛+1)π‘ž6𝑛2+4π‘›ξ€·π‘žβˆ’24,βˆ’π‘ž2,βˆ’π‘ž2;π‘ž4ξ€Έβˆžξ“π‘›2π‘›π‘ž6𝑛2βˆ’2𝑛.(3.16) Replacing 𝑛 in the last two sums on the right-hand side of the above identity by βˆ’π‘› 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).

References

  1. B. C. Berndt, Ramanujan's Notebooks. Part III, Springer, New York, NY, USA, 1991.
  2. S. Kongsiriwong and Z.-G. Liu, β€œUniform proofs of q-series-product identities,” Results in Mathematics, vol. 44, no. 3-4, pp. 312–339, 2003.
  3. W. N. Bailey, β€œOn the simplification of some identities of the Rogers-Ramanujan type,” Proceedings of the London Mathematical Society. Third Series, vol. 1, no. 3, pp. 217–221, 1951. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  4. S. Cooper, β€œA new proof of the Macdonald identities for Anβˆ’1,” Journal of the Australian Mathematical Society. Series A, vol. 62, no. 3, pp. 345–360, 1997. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  5. S. Cooper, β€œThe Macdonald identities for G2 and some extensions,” New Zealand Journal of Mathematics, vol. 26, no. 2, pp. 161–182, 1997. View at Zentralblatt MATH
  6. S.-L. Chen and S.-S. Huang, β€œNew modular relations for the GΓΆllnitz-Gordon functions,” Journal of Number Theory, vol. 93, no. 1, pp. 58–75, 2002. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  7. E. G. Bakhoum and C. Toma, β€œMathematical transform of traveling-wave equations and phase aspects of quantum interaction,” Mathematical Problems in Engineering, vol. 2010, Article ID 695208, 15 pages, 2010. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  8. C. Cattani, β€œShannon wavelets for the solution of integrodifferential equations,” Mathematical Problems in Engineering, vol. 2010, Article ID 408418, 22 pages, 2010. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  9. M. Li, S. C. Lim, and S. Chen, β€œExact solution of impulse response to a class of fractional oscillators and its stability,” Mathematical Problems in Engineering, vol. 2011, Article ID 657839, 9 pages, 2011. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH