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Mathematical Problems in Engineering
VolumeΒ 2011Β (2011), Article IDΒ 462507, 11 pages
doi: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 ) ∞ = ∞  𝑛 = 0 ( 1 βˆ’ π‘Ž π‘ž 𝑛 ) , ( π‘Ž , 𝑏 , … , 𝑐 ; π‘ž ) ∞ = ( π‘Ž ; π‘ž ) ∞ ( 𝑏 ; π‘ž ) ∞ β‹― ( 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‘ž π‘ž , 𝑧 , 𝑧  ; π‘ž ∞ = ∞  𝑛 = βˆ’ ∞ ( βˆ’ 1 ) 𝑛 π‘ž ( 1 / 2 ) 𝑛 ( 𝑛 βˆ’ 1 ) 𝑧 𝑛 , 𝑧 β‰  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‘ 𝑛 for βˆ‘ ∞ 𝑛 = βˆ’ ∞ 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 β‰  0 , one has ξ‚€ π‘ž π‘ž , 𝑧 , 𝑧  ; π‘ž ∞ ξ‚΅ π‘ž 3 , 𝑧 3 , π‘ž 3 𝑧 3 ; π‘ž 3 ξ‚Ά ∞ = ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 π‘ž 2 𝑛 2 βˆ’ 2 𝑛 𝑧 4 𝑛 ξ€· π‘ž + 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 π‘ž 2 𝑛 2 + 1 𝑧 4 𝑛 + 2 βˆ’ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 π‘ž ( 1 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) 𝑧 2 𝑛 + 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‘ž 8 , π‘ž 3 , π‘ž 5 ; π‘ž 8 ξ€Έ ∞ ξ€· π‘ž 2 4 , π‘ž 9 , π‘ž 1 5 ; π‘ž 2 4 ξ€Έ ∞ + π‘ž 2 ξ€· π‘ž 8 , π‘ž , π‘ž 7 ; π‘ž 8 ξ€Έ ∞ ξ€· π‘ž 2 4 , π‘ž 3 , π‘ž 2 1 ; π‘ž 2 4 ξ€Έ ∞ = ξ€· π‘ž 2 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 2 ξ€Έ ∞ ξ€· π‘ž 6 , π‘ž 3 , π‘ž 3 ; π‘ž 6 ξ€Έ ∞ , ξ€· π‘ž ( 2 . 2 ) 8 , π‘ž , π‘ž 7 ; π‘ž 8 ξ€Έ ∞ ξ€· π‘ž 2 4 , π‘ž 9 , π‘ž 1 5 ; π‘ž 2 4 ξ€Έ ∞ ξ€· π‘ž βˆ’ π‘ž 8 , π‘ž 3 , π‘ž 5 ; π‘ž 8 ξ€Έ ∞ ξ€· π‘ž 2 4 , π‘ž 3 , π‘ž 2 1 ; π‘ž 2 4 ξ€Έ ∞ = ξ€· π‘ž 2 , π‘ž , π‘ž ; π‘ž 2 ξ€Έ ∞ ξ€· π‘ž 6 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 6 ξ€Έ ∞ . ( 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 ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ + ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ ξ€· π‘ž = 2 4 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 4 ξ€Έ ∞ ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞ , ( 2 . 4 ) ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ βˆ’ ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ ξ€· π‘ž = 2 π‘ž 4 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 4 ξ€Έ ∞ ξ€· π‘ž 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞ . ( 2 . 5 )

Proof. By (1.2), we have ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 1 4 ( π‘ž , βˆ’ 1 , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 1 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 1 4  π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š )  𝑛 π‘ž ( 3 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) =  π‘š π‘ž 2 π‘š 2 + π‘š  𝑛 π‘ž 6 𝑛 2 + 3 𝑛 =  π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 + π‘ž 4  π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛 + π‘ž 3  π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛  + π‘ž π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 , ( 2 . 6 ) ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 1 2 ( π‘ž , 𝑖 , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 1 2  π‘š ( βˆ’ 1 ) π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) 𝑖 π‘š  𝑛 ( βˆ’ 1 ) 𝑛 π‘ž ( 3 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) 𝑖 3 𝑛 =  π‘š ( βˆ’ 1 ) π‘š π‘ž 2 π‘š 2 + π‘š  𝑛 ( βˆ’ 1 ) 𝑛 π‘ž 6 𝑛 2 + 3 𝑛 =  π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 + π‘ž 4  π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛 βˆ’ π‘ž 3  π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛  βˆ’ π‘ž π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 . ( 2 . 7 ) Adding (2.6) and (2.7), we have ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ + ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞  = 2 π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 + 2 π‘ž 4  π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛 ξ€· π‘ž = 2 1 6 , βˆ’ π‘ž 6 , βˆ’ π‘ž 1 0 ; π‘ž 1 6 ξ€Έ ∞ ξ€· π‘ž 4 8 , βˆ’ π‘ž 1 8 , βˆ’ π‘ž 3 0 ; π‘ž 4 8 ξ€Έ ∞ + 2 π‘ž 4 ξ€· π‘ž 1 6 , βˆ’ π‘ž 2 , βˆ’ π‘ž 1 4 ; π‘ž 1 6 ξ€Έ ∞ ξ€· π‘ž 4 8 , βˆ’ π‘ž 6 , βˆ’ π‘ž 4 2 ; π‘ž 4 8 ξ€Έ ∞ . ( 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 ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ βˆ’ ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 2 π‘ž 3  π‘š π‘ž 8 π‘š 2 + 2 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 1 8 𝑛  + 2 π‘ž π‘š π‘ž 8 π‘š 2 + 6 π‘š  𝑛 π‘ž 2 4 𝑛 2 + 6 𝑛 = 2 π‘ž 3 ξ€· π‘ž 1 6 , βˆ’ π‘ž 6 , βˆ’ π‘ž 1 0 ; π‘ž 1 6 ξ€Έ ∞ ξ€· π‘ž 4 8 , βˆ’ π‘ž 6 , βˆ’ π‘ž 4 2 ; π‘ž 4 8 ξ€Έ ∞ ξ€· π‘ž + 2 π‘ž 1 6 , βˆ’ π‘ž 2 , βˆ’ π‘ž 1 4 ; π‘ž 1 6 ξ€Έ ∞ ξ€· π‘ž 4 8 , βˆ’ π‘ž 1 8 , βˆ’ π‘ž 3 2 ; π‘ž 4 8 ξ€Έ ∞ . ( 2 . 9 ) Replacing π‘ž in (2.3) by βˆ’ π‘ž 2 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‘ž 𝑓 ( 𝑧 , π‘ž ) = π‘ž , 𝑧 , 𝑧  ; π‘ž ∞ ξ‚΅ π‘ž 3 , 𝑧 3 , π‘ž 3 𝑧 3 ; π‘ž 3 ξ‚Ά ∞ . ( 2 . 1 0 ) Then 𝑓 is an analytic function of 𝑧 in the annulus 0 < | 𝑧 | < ∞ . Put  𝑓 ( 𝑧 , π‘ž ) = 𝑛 π‘Ž 𝑛 ( π‘ž ) 𝑧 𝑛 , 0 < | 𝑧 | < ∞ . ( 2 . 1 1 ) By (2.10), we can easily verify 𝑓 ( 𝑧 , π‘ž ) = 𝑧 4 𝑓 ( 𝑧 π‘ž , π‘ž ) . ( 2 . 1 2 ) Combining (2.11) and (2.12) gives  π‘š π‘Ž π‘š ( π‘ž ) 𝑧 π‘š =  π‘š π‘ž π‘š βˆ’ 4 π‘Ž π‘š βˆ’ 4 ( π‘ž ) 𝑧 π‘š . ( 2 . 1 3 ) Equate the coefficients of 𝑧 π‘š on both sides to get π‘Ž π‘š ( π‘ž ) = π‘ž π‘š βˆ’ 4 π‘Ž π‘š βˆ’ 4 ( π‘ž ) . ( 2 . 1 4 ) Using the above relation, we obtain π‘Ž 4 π‘š βˆ’ 1 ( π‘ž ) = π‘ž 2 π‘š 2 βˆ’ 3 π‘š π‘Ž βˆ’ 1 ( π‘ž ) , π‘Ž 4 π‘š ( π‘ž ) = π‘ž 2 π‘š 2 βˆ’ 2 π‘š π‘Ž 0 π‘Ž ( π‘ž ) , 4 π‘š + 1 ( π‘ž ) = π‘ž 2 π‘š 2 βˆ’ π‘š π‘Ž 1 ( π‘ž ) , π‘Ž 4 π‘š + 2 ( π‘ž ) = π‘ž 2 π‘š 2 π‘Ž 2 ( π‘ž ) . ( 2 . 1 5 ) Substituting the above four identities into (2.11), we have 𝑓 ( 𝑧 , π‘ž ) = π‘Ž βˆ’ 1  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 3 π‘š 𝑧 4 π‘š βˆ’ 1 + π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š 𝑧 4 π‘š + π‘Ž 1  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ π‘š 𝑧 4 π‘š + 1 + π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 𝑧 4 π‘š + 2 . ( 2 . 1 6 ) By (2.10), we also have ξ‚€ π‘ž 𝑓 ( 𝑧 , π‘ž ) = 𝑓 𝑧  , π‘ž . ( 2 . 1 7 ) This gives  π‘š π‘Ž π‘š ( π‘ž ) 𝑧 π‘š =  π‘š π‘ž βˆ’ π‘š π‘Ž βˆ’ π‘š ( π‘ž ) 𝑧 π‘š . ( 2 . 1 8 ) Then we have π‘Ž π‘š ( π‘ž ) = π‘ž βˆ’ π‘š π‘Ž βˆ’ π‘š ( π‘ž ) . ( 2 . 1 9 ) Set π‘š = 1 to get π‘Ž 1 ( π‘ž ) = π‘ž βˆ’ 1 π‘Ž βˆ’ 1 ( π‘ž ) . ( 2 . 2 0 ) By this relation, (2.16) reduces to 𝑓 ( 𝑧 , π‘ž ) = π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š 𝑧 4 π‘š + π‘Ž 1  ( π‘ž ) π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) 𝑧 2 π‘š + 1 + π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 𝑧 4 π‘š + 2 . ( 2 . 2 1 ) Now, it remains to determine π‘Ž 0 ( π‘ž ) , π‘Ž 1 ( π‘ž ) , and π‘Ž 2 ( π‘ž ) .
Putting 𝑧 = 1 in (2.21) gives 0 = π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š + π‘Ž 1  ( π‘ž ) π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) + π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 . ( 2 . 2 2 ) Set 𝑧 = βˆ’ 1 in (2.21) to get 4 ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š βˆ’ π‘Ž 1  ( π‘ž ) π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) + π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 . ( 2 . 2 3 ) Taking 𝑧 = 𝑖 in (2.21) and noting that βˆ‘ π‘š ( βˆ’ 1 ) π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) = 0 , we have ( π‘ž , 𝑖 , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š βˆ’ π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 . ( 2 . 2 4 ) Subtracting (2.23) from (2.22) and noting that βˆ‘ π‘š π‘ž ( 1 / 2 ) ( π‘š 2 βˆ’ π‘š ) = 2 ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ , we obtain π‘Ž 1 ξ€· π‘ž ( π‘ž ) = βˆ’ 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ . ( 2 . 2 5 ) Add (2.22) and (2.23) to get 2 ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = π‘Ž 0  ( π‘ž ) π‘š π‘ž 2 π‘š 2 βˆ’ 2 π‘š + π‘Ž 2  ( π‘ž ) π‘š π‘ž 2 π‘š 2 . ( 2 . 2 6 ) 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 ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ + ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = 2 π‘Ž 0 ξ€· π‘ž ( π‘ž ) 4 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 4 ξ€Έ ∞ . ( 2 . 2 7 ) By (2.4), we have π‘Ž 0 ξ€· π‘ž ( π‘ž ) = 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞ . ( 2 . 2 8 ) Similarly, subtracting (2.24) from (2.26) and, then using (1.2), we have ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞ βˆ’ ( π‘ž , 𝑖 π‘ž , βˆ’ 𝑖 π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 , βˆ’ 𝑖 π‘ž 3 , 𝑖 π‘ž 3 ; π‘ž 3 ξ€Έ ∞ = π‘Ž 2 ξ€· π‘ž ( π‘ž ) 4 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 4 ξ€Έ ∞ . ( 2 . 2 9 ) Applying (2.5) to this equation gives π‘Ž 2 ξ€· π‘ž ( π‘ž ) = 2 π‘ž 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞ , ( 2 . 3 0 ) 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 ) 3 ∞ ξ€· π‘ž 3 ; π‘ž 3 ξ€Έ 3 ∞ = ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 2 𝑛 ( 4 𝑛 βˆ’ 1 ) π‘ž 2 𝑛 2 βˆ’ 2 𝑛 ξ€· π‘ž + 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) ( 4 𝑛 + 1 ) π‘ž 2 𝑛 2 + 1 βˆ’ ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 𝑛 ( 2 𝑛 + 1 ) π‘ž 1 / 2 ( 𝑛 2 βˆ’ 𝑛 ) . ( 3 . 1 )

Proof. Dividing both sides of (2.1) by ( 1 βˆ’ 𝑧 ) 2 , letting 𝑧 β†’ 1 , and then using L’Hospital’s rule twice on the right-hand side gives (3.1).

Corollary 3.2. One has ξ€· π‘ž 2 4 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 2 4 ξ€Έ ∞ ξ€· π‘ž 8 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 8 ξ€Έ ∞ + 4 π‘ž 4 ξ€· π‘ž 2 4 , βˆ’ π‘ž 2 4 , βˆ’ π‘ž 2 4 ; π‘ž 2 4 ξ€Έ ∞ ξ€· π‘ž 8 , βˆ’ π‘ž 8 , βˆ’ π‘ž 8 ; π‘ž 8 ξ€Έ ∞ ξ€· π‘ž + 2 π‘ž 6 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 6 ξ€Έ ∞ ξ€· π‘ž 2 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 2 ξ€Έ ∞ = ξ€· π‘ž 2 , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž 2 ξ€Έ ∞ ξ€· π‘ž 6 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 6 ξ€Έ ∞ , ξ€· π‘ž ( 3 . 2 ) 3 6 , βˆ’ π‘ž 1 8 , βˆ’ π‘ž 1 8 ; π‘ž 3 6 ξ€Έ ∞ ξ€· π‘ž 1 2 , βˆ’ π‘ž 4 , βˆ’ π‘ž 8 ; π‘ž 1 2 ξ€Έ ∞ + 2 π‘ž 5 ξ€· π‘ž 3 6 , βˆ’ π‘ž 3 6 , βˆ’ π‘ž 3 6 ; π‘ž 3 6 ξ€Έ ∞ ξ€· π‘ž 1 2 , βˆ’ π‘ž 2 , βˆ’ π‘ž 1 0 ; π‘ž 1 2 ξ€Έ ∞ ξ€· π‘ž βˆ’ π‘ž 9 , βˆ’ π‘ž 9 , βˆ’ π‘ž 9 ; π‘ž 9 ξ€Έ ∞ ξ€· π‘ž 3 , βˆ’ π‘ž , βˆ’ π‘ž 2 ; π‘ž 3 ξ€Έ ∞ = ( π‘ž ; π‘ž ) ∞ ξ€· π‘ž 3 ; π‘ž 3 ξ€Έ ∞ , ξ€· π‘ž ( 3 . 3 ) 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞ ξ€· π‘ž 4 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 4 ξ€Έ ∞ ξ€· π‘ž βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞ ξ€· π‘ž 4 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 4 ξ€Έ ∞ = ξ€· π‘ž 2 ; π‘ž 2 ξ€Έ ∞ ξ€· βˆ’ π‘ž ; π‘ž 2 ξ€Έ ∞ ξ€· π‘ž 6 ; π‘ž 6 ξ€Έ ∞ ξ€· βˆ’ π‘ž 3 ; π‘ž 6 ξ€Έ ∞ , ξ€· π‘ž ( 3 . 4 ) 6 0 , βˆ’ π‘ž 3 0 , βˆ’ π‘ž 3 0 ; π‘ž 6 0 ξ€Έ ∞ ξ€· π‘ž 2 0 , βˆ’ π‘ž 8 , βˆ’ π‘ž 1 2 ; π‘ž 2 0 ξ€Έ ∞ + 2 π‘ž 9 ξ€· π‘ž 6 0 , βˆ’ π‘ž 6 0 , βˆ’ π‘ž 6 0 ; π‘ž 6 0 ξ€Έ ∞ ξ€· π‘ž 2 0 , βˆ’ π‘ž 2 , βˆ’ π‘ž 1 8 ; π‘ž 2 0 ξ€Έ ∞ βˆ’ π‘ž 2 ξ€· π‘ž 1 5 , βˆ’ π‘ž 1 5 , βˆ’ π‘ž 1 5 ; π‘ž 1 5 ξ€Έ ∞ ξ€· q 5 , βˆ’ π‘ž , βˆ’ π‘ž 4 ; π‘ž 5 ξ€Έ ∞ = ( π‘ž ; π‘ž ) ∞ ξ€· π‘ž , π‘ž 4 ; π‘ž 5 ξ€Έ ∞ ξ€· π‘ž 5 ; π‘ž 5 ξ€Έ ∞ ξ€· π‘ž 3 , π‘ž 1 2 ; π‘ž 1 5 ξ€Έ ∞ , ξ€· π‘ž ( 3 . 5 ) 6 0 , βˆ’ π‘ž 3 0 , βˆ’ π‘ž 3 0 ; π‘ž 6 0 ξ€Έ ∞ ξ€· π‘ž 2 0 , βˆ’ π‘ž 4 , βˆ’ π‘ž 1 6 ; π‘ž 2 0 ξ€Έ ∞ + 2 π‘ž 7 ξ€· π‘ž 6 0 , βˆ’ π‘ž 6 0 , βˆ’ π‘ž 6 0 ; π‘ž 6 0 ξ€Έ ∞ ξ€· π‘ž 2 0 , βˆ’ π‘ž 6 , βˆ’ π‘ž 1 4 ; π‘ž 2 0 ξ€Έ ∞ ξ€· π‘ž βˆ’ π‘ž 1 5 , βˆ’ π‘ž 1 5 , βˆ’ π‘ž 1 5 ; π‘ž 1 5 ξ€Έ ∞ ξ€· π‘ž 5 , βˆ’ π‘ž 2 , βˆ’ π‘ž 3 ; π‘ž 5 ξ€Έ ∞ = ( π‘ž ; π‘ž ) ∞ ξ€· π‘ž 2 , π‘ž 3 ; π‘ž 5 ξ€Έ ∞ ξ€· π‘ž 5 ; π‘ž 5 ξ€Έ ∞ ξ€· π‘ž 6 , π‘ž 9 ; π‘ž 1 5 ξ€Έ ∞ . ( 3 . 6 )

Proof. Replace π‘ž in (2.1) by π‘ž 2 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‘ž 3 and, then, 𝑧 by π‘ž . Using (1.2) in the resulting identity gives (3.3).
Replace π‘ž in (2.1) by π‘ž 4 and, then, 𝑧 by π‘ž . Using (1.2) and the fact that ( π‘ž 4 , βˆ’ π‘ž , βˆ’ π‘ž 3 ; π‘ž 4 ) ∞ = ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞ in the resulting identity, we obtain π‘ž ξ€· ξ€· 4 8 , βˆ’ π‘ž 2 4 , βˆ’ π‘ž 2 4 ; π‘ž 4 8 ξ€Έ ∞ + 2 π‘ž 6 ξ€· π‘ž 4 8 , βˆ’ π‘ž 4 8 , βˆ’ π‘ž 4 8 ; π‘ž 4 8 ξ€Έ ∞ π‘ž ξ€Έ ξ€· 4 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 4 ξ€Έ ∞ ξ€· π‘ž βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞ ξ€· π‘ž 4 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 4 ξ€Έ ∞ = ξ€· π‘ž 2 ; π‘ž 2 ξ€Έ ∞ ξ€· βˆ’ π‘ž ; π‘ž 2 ξ€Έ ∞ ξ€· π‘ž 6 ; π‘ž 6 ξ€Έ ∞ ξ€· βˆ’ π‘ž 3 ; π‘ž 6 ξ€Έ ∞ . ( 3 . 7 ) By (1.2), we have ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞ =  𝑛 π‘ž 6 𝑛 2 =  𝑛 π‘ž 6 ( 2 𝑛 ) 2 +  𝑛 π‘ž 6 ( 2 𝑛 + 1 ) 2 = ξ€· π‘ž 4 8 , βˆ’ π‘ž 2 4 , βˆ’ π‘ž 2 4 ; π‘ž 4 8 ξ€Έ ∞ + 2 π‘ž 6 ξ€· π‘ž 4 8 , βˆ’ π‘ž 4 8 , βˆ’ π‘ž 4 8 ; π‘ž 4 8 ξ€Έ ∞ . ( 3 . 8 ) Combining (3.7) and (3.8) gives (3.4).
Replace π‘ž in (2.1) by π‘ž 5 and, then, 𝑧 by π‘ž 2 . Using (1.2) in the resulting identity gives (3.5).
Replace π‘ž in (2.1) by π‘ž 5 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 ) 4 ∞  = 2 π‘š π‘ž 2 π‘š 2  𝑛 2 𝑛 π‘ž 6 𝑛 2 + 2 𝑛  + 2 π‘ž π‘š π‘ž 2 π‘š 2 + 2 π‘š  𝑛 ( 2 𝑛 βˆ’ 1 ) π‘ž 6 𝑛 2 βˆ’ 4 𝑛 +  π‘š π‘ž 2 π‘š 2 + π‘š  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 3 𝑛 2 + 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 𝑧 0 be a zero point of 𝑓 ( 𝑧 ) . Because (2.1) holds in 0 < | 𝑧 | < ∞ , 𝑧 0 is also a zero point of 𝑔 ( 𝑧 ) . If π‘Ž 𝑧 0 = 1 , we have l i m 𝑧 β†’ 𝑧 0 𝑓 ( 𝑧 ) 1 βˆ’ π‘Ž 𝑧 = l i m 𝑧 β†’ 𝑧 0 𝑔 ( 𝑧 ) 1 βˆ’ π‘Ž 𝑧 . ( 3 . 1 0 ) Setting 𝑧 0 = π‘Ž = 1 in (3.10) and by L’Hospital’s rule on the right-hand side, we have ξ€· π‘ž 0 = 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) ξ€· π‘ž βˆ’ 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 ( 4 𝑛 + 2 ) π‘ž 2 𝑛 2 + 1 βˆ’ ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 4 𝑛 π‘ž 2 𝑛 2 βˆ’ 2 𝑛 . ( 3 . 1 1 ) Let πœ” = 𝑒 ( 2 / 3 ) πœ‹ 𝑖 . Putting 𝑧 0 = πœ” and π‘Ž = πœ” 2 in (3.10) and noting πœ” 3 𝑛 = 1 for any integer 𝑛 , we have ξ€· π‘ž 3 ( 1 βˆ’ πœ” ) 3 ; π‘ž 3 ξ€Έ 4 ∞ = ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) πœ” 2 ( 𝑛 βˆ’ 1 ) ξ€· π‘ž βˆ’ 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 ( 4 𝑛 + 2 ) π‘ž 2 𝑛 2 + 1 πœ” 𝑛 βˆ’ 1 βˆ’ ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 4 𝑛 π‘ž 2 𝑛 2 βˆ’ 2 𝑛 πœ” 𝑛 . ( 3 . 1 2 ) Taking 𝑧 0 = πœ” 2 and π‘Ž = πœ” in (3.10), we obtain 3 ξ€· 1 βˆ’ πœ” 2 π‘ž ξ€Έ ξ€· 3 ; π‘ž 3 ξ€Έ 4 ∞ = ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) πœ” 𝑛 βˆ’ 1 ξ€· π‘ž βˆ’ 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 ( 4 𝑛 + 2 ) π‘ž 2 𝑛 2 + 1 πœ” 2 ( 𝑛 βˆ’ 1 ) βˆ’ ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 4 𝑛 π‘ž 2 𝑛 2 βˆ’ 2 𝑛 πœ” 2 𝑛 . ( 3 . 1 3 ) Adding the above three identities together gives 9 ξ€· π‘ž 3 ; π‘ž 3 ξ€Έ 4 ∞ = ξ€· π‘ž 3 , βˆ’ π‘ž 3 , βˆ’ π‘ž 3 ; π‘ž 3 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 𝑛 2 βˆ’ 𝑛 ) ξ€· 1 + πœ” 𝑛 βˆ’ 1 + πœ” 2 ( 𝑛 βˆ’ 1 ) ξ€Έ ξ€· π‘ž βˆ’ 2 1 2 , βˆ’ π‘ž 1 2 , βˆ’ π‘ž 1 2 ; π‘ž 1 2 ξ€Έ ∞  𝑛 ( 4 𝑛 + 2 ) π‘ž 2 𝑛 2 + 1 ξ€· 1 + πœ” 𝑛 βˆ’ 1 + πœ” 2 ( 𝑛 βˆ’ 1 ) ξ€Έ βˆ’ ξ€· π‘ž 1 2 , βˆ’ π‘ž 6 , βˆ’ π‘ž 6 ; π‘ž 1 2 ξ€Έ ∞  𝑛 4 𝑛 π‘ž 2 𝑛 2 βˆ’ 2 𝑛 ξ€· 1 + πœ” 𝑛 + πœ” 2 𝑛 ξ€Έ . ( 3 . 1 4 ) Using the fact 1 + πœ” 𝑛 + πœ” 2 𝑛 = ξƒ― 3 , 𝑛 ≑ 0 ( m o d 3 ) , 0 , 𝑛 β‰’ 0 ( m o d 3 ) ( 3 . 1 5 ) in the above identity and, then, replacing π‘ž 3 by π‘ž , we get ( π‘ž ; π‘ž ) 4 ∞ = ( π‘ž , βˆ’ π‘ž , βˆ’ π‘ž ; π‘ž ) ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž ( 1 / 2 ) ( 3 𝑛 2 + 𝑛 ) ξ€· π‘ž βˆ’ 4 π‘ž 4 , βˆ’ π‘ž 4 , βˆ’ π‘ž 4 ; π‘ž 4 ξ€Έ ∞  𝑛 ( 2 𝑛 + 1 ) π‘ž 6 𝑛 2 + 4 𝑛 ξ€· π‘ž βˆ’ 2 4 , βˆ’ π‘ž 2 , βˆ’ π‘ž 2 ; π‘ž 4 ξ€Έ ∞  𝑛 2 𝑛 π‘ž 6 𝑛 2 βˆ’ 2 𝑛 . ( 3 . 1 6 ) Replacing 𝑛 in the last two sums on the right-hand side of the above identity by βˆ’ 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 𝑒 2 𝑖 𝑧 , 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 [79].

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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