Elliptic Functions, Theta Functions and Identities

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Ng Say Tiong

Department of Mathematics National University of Singapore 2004 Elliptic Functions, Theta Functions and Identities

Ng Say Tiong

A THESIS SUBMITTED FOR THE DEGREE OF MASTER OF SCIENCE.

Supervisor: A/P. Chan Heng Huat

Department of Mathematics National University of Singapore 2004 Acknowledgements

I would like to thank my family members and friends for giving me support in the completion of this thesis. Also I would like to thank all the lecturers that have given me guidance in my courses in NUS.

Last but not least, I would like to thank my supervisor, Associate Professor Chan Heng Huat for his patience and guidance throughout the duration of this thesis. Without his support, the completion of this thesis in such a short amount of time would not be possible.

ii Statement of Author’s Contributions

1. The proof to Lemma 3.3.2, which is given as an exercise in [16], is given in detail by the author.

2. Together with Associate Professor Chan Heng Huat, and Professor Liu Zhi Guo, the author provides original proofs for the quintuple product identity in Theorem 4.1.1, the septuple product identity in Theorem 4.2.1 and the Winquist identity in Theorem 4.3.1, based on the theories of elliptic functions and the theta functions.

3. The author provides original proof for the theta function identity in Theorem 5.1.1 based on the theories of elliptic functions and the theta functions.

iii Contents

1 Introduction 1

2 Elliptic Functions 3

2.1 Deﬁnition of elliptic functions ...... 3

2.2 Properties of elliptic functions ...... 4

3 The Theta Functions 8

3.1 Deﬁnition of the theta functions ...... 8

3.2 The zeros of the theta functions ...... 12

3.3 Inﬁnite products expressions for the theta functions ...... 14

4 The Quintuple, Septuple Product Identities and the Winquist Identity 23

4.1 The quintuple product identity ...... 24

4.2 The septuple product identity ...... 30

4.3 The Winquist identity ...... 35

5 A Theta Function Identity and Its Application 43

5.1 The theta-function identity ...... 43

5.2 A conjecture on the relation between Dixon’s doubly-periodic functions and the theta functions ...... 47

iv Chapter 1

Introduction

The ﬁrst function related to theta functions is perhaps the function

∞ Y (1 − xnz)−1, n=1 deﬁned by L. Euler in his studies of the properties of partitions [6]. However, it was C.G.J. Jacobi who deﬁned the more general theta functions in [11], and he developed the theory of theta functions from the theory of elliptic functions. His results were later reprinted in [12].

The theta functions are great tools for solving q-series identities. In latter chapters we will use the theta functions to systematically proved the quintuple product identity [3, p.g 80], the septuple product identity [7] and the Winquist identity [17]. The quintuple product identity has a long history, and is best summarized in [3, p.g 83]. The septuple product identity was ﬁrst discovered by M.D. Hirschhorn [9] in 1983, and later rediscovered by H.M.Farkas and I.Kra in 1999 [7]. In 1999, D.Foata and G.-N. Han showed a connection between the quintuple product identity and the septuple product identity [8], using the Jacobi triple product identity [2, p.g 91] and the quintuple product identity to prove the septuple product identity. The Winquist identity was discovered by L.Winquist in [17], and was used to derive a double series identity for the product series

∞ Y (1 − xn)10, n=1

1 and a proof for the partition function congruence

p (11n + 6) ≡ 0 (mod 11), where p (n) denotes the number of unrestricted partitions of n [1, p.g 1]. Later in 1997, S.-Y. Kang gave a new proof of the Winquist identity based on elementary methods [13].

In [5], A.C.Dixon deﬁned two doubly-periodic functions smu and cmu using the curve x3 + y3 − 3αxy = 1, where α is a parameter of the cubic curve. Dixon showed that these two functions shared many similar properties to the Jacobian elliptic functions snu and cnu [16, p.g 491]. Using a theta function identity proved by Z.-G. Liu in [14], we conjecture possible algebraic expressions for smu and cmu based on the theta functions.

2 Chapter 2

Elliptic Functions

We will begin our discussion with elliptic functions. The properties of elliptic functions that are discussed in this chapter will play an important part in our proofs of theorems in latter chapters.

2.1 Deﬁnition of elliptic functions

Deﬁnition 2.1.1 Let ω1, ω2 be any two numbers (real or complex) whose ratio is complex. A function which satisﬁes the equations

f(z + 2ω1) = f(z), f(z + 2ω2) = f(z),

for all values of z for which f(z) exists, is called a doubly-periodic function of z, with

periods 2ω1, 2ω2. A doubly-periodic function which has no singularities other than poles in the ﬁnite part of the complex plane is called an elliptic function.

To study elliptic functions, we need the geometrical representation of the complex plane aﬀorded by the Argand diagram [16, p.g 9]. Suppose that in the complex plane, we mark

out the points 0, 2ω1, 2ω2, 2ω1 + 2ω2, and generally, all the points whose complex coordi- nates are of the form 2mω1 + 2nω2, where m and n are integers. By joining in succession consecutive points of the set 0, 2ω1, 2ω1 + 2ω2, 2ω2, 0, we will obtain a parallelogram. If

3 there is no point ω inside or on the boundary of this parallelogram (the vertices excepted) such that f(z + ω) = f(z) for all values of z, this parallelogram is called a fundamental period-parallelogram [16, p.g

430] for an elliptic function with periods 2ω1, 2ω2.

It is clear that the complex plane can be covered with a network of parallelograms equal to the fundamental period-parallelogram and similarly situated, each of the points

2ω1 + 2ω2 being a vertex of four parallelograms. These parallelograms are called period- parallelograms or meshes [16, p.g 430], and for all values of z, the points z, z + 2ω1, ... z + 2mω1 + 2nω2, ... manifestly occupy corresponding positions in the meshes. Any of such pair of points are said to be congruent to one another [16, p.g 430]. Due to the fundamental periodic properties of elliptic functions, it follows that an elliptic function takes the same value at every one of a set of congruent points, and so its values in any mesh are just a mere repetition of its values in any other mesh.

However, for integration purposes, it is not convenient to deal with the actual meshes if they have singularities of the integrand on their boundaries. Because of the periodic properties of elliptic functions, nothing is lost by taking a contour, which is not an actual mesh, but a parallelogram obtained by translating a mesh without rotation in such a way that none of the poles of the integrands considered are on the boundaries of the parallelogram, and all the poles considered are inside the parallelogram. Such a parallelogram is called a cell [16, p.g 430]. Again the values assumed by an elliptic function in a cell are just a mere repetition of its value in any mesh.

We will now prove some important facts about elliptic functions.

2.2 Properties of elliptic functions

Theorem 2.2.1 i) The number of poles of an elliptic function in any cell is ﬁnite. ii) The number of zeros of an elliptic function in any cell is ﬁnite.

4 iii) The sum of the residues of an elliptic function, f(z), at its poles in any cell is zero. iv) An elliptic function, f(z), with no pole in a cell is merely a constant.

Proof

i) Suppose there exists an elliptic function such that its number of poles in a cell is inﬁnite, then the poles would have a limit point in the cell, which is an essential singularity of the function. However this contradicts the deﬁnition of an elliptic function.

ii) Suppose there exists an elliptic function f(z) such that its number of zeros in a cell

1 1 is inﬁnite, we consider the reciprocal function f(z) . Since f(z) is elliptic, so is f(z) , and it has an inﬁnite number of poles in a cell by the property of f(z). This is a contradiction, by part(i).

iii) Let C be a contour formed by edges of a cell, whose corners are t, t+2ω1, t+2ω1 +2ω2

and t + 2ω2, t being an arbitrary complex number. The sum of the residues of f(z) at its poles inside C is 1 Z f(z) dz 2πi C 1 Z t+2ω1 Z t+2ω1+2ω2 Z t+2ω2 Z t = f(z) dz + f(z) dz + f(z) dz + f(z) dz 2πi t t+2ω1 t+2ω1+2ω2 t+2ω2 1 Z t+2ω1 Z t+2ω2 Z t Z t = f(z) dz + f(z + 2ω1) dz + f(z + 2ω2) dz + f(z) dz 2πi t t t+2ω1 t+2ω2 1 Z t+2ω1 1 Z t+2ω2 = {f(z) − f(z + 2ω2)} dz − {f(z) − f(z + 2ω1)} dz 2πi t 2πi t = 0,

and the theorem is established.

iv) Let f(z) be an elliptic function such that it has no pole in a cell. Then it is analytic, and consequently bounded, inside and on the boundary of the cell. This implies that there exists a number K such that | f(z)| < K for all z inside or on the boundary of the cell. Since f(z) is doubly periodic, it follows that f(z) is analytic and | f(z)| < K for all values of z except the point ∞. By Liouville’s theorem [16, p.g 105], f(z) is a constant. 2

5 Because of Theorem 2.2.1 part (iii), if an elliptic function f(z) contains poles, then it must have at least 2 poles in a cell, because if it has only one pole in a cell, then the residue of f(z) is non-zero in the cell, contradicting to the result of Theorem 2.2.1 part (iii).

Lastly, we will give an example of elliptic functions to conclude this chapter.

Example 2.2.2 (The Weierstrass ℘ function)

Let ω1, ω2 be two numbers (real or complex) whose ratio is complex. The Weierstrass ℘ function is deﬁned as 1 X0 1 1 ℘(z) = + − , z2 (z − 2mω − 2nω )2 (2mω + 2nω )2 m,n 1 2 1 2 where the summation extends over all integer values m and n, except when m and n are simultaneously zero. We will now show that ℘(z) is indeed an elliptic function.

For convenience, we write Ωm,n in place of 2mω1 + 2nω2, so that

0 −2 X −2 −2 ℘(z) = z + (z − Ωm,n) − Ωm,n . m,n

When m and n are such that |Ωm,n| is large, the general term of the series deﬁning ℘(z) −3 is O(|Ωm,n| ), and thus the series converges absolutely and uniformly with respect to

z, except near its poles, which are the points Ωm,n. This implies that ℘(z) is analytic

throughout the whole complex plane except at the points Ωm,n, where it has double poles. So to show ℘(z) is elliptic, we are left only to show that it is a doubly-periodic function.

Since ℘(z) is uniformly convergent, we can obtain the diﬀerential of ℘(z) with respect to z; X 1 ℘0(z) = −2 . (z − Ω )3 m,n m,n It is clear that ℘0(z) is an odd function, whereas ℘(z) is an even function. Also we have

0 0 ℘ (z + 2ω1) = ℘ (z),

0 0 ℘ (z + 2ω2) = ℘ (z),

since the sets of points (Ωm,n − 2ω1) and (Ωm,n − 2ω2) are both the same as the set Ωm,n 0 0 as m, n, run through all the integers. Integrating the equation ℘ (z + 2ω1) = ℘ (z), we get

℘(z + 2ω1) = ℘(z) + A,

6 where A is a constant. Putting z = −ω1 and using the fact that ℘(z) is an even function, we have A = 0, and so we get

℘(z + 2ω1) = ℘(z).

Using similar method we also get

℘(z + 2ω2) = ℘(z),

and this shows that ℘(z) is doubly-periodic. Hence we conclude that ℘(z) is an elliptic function.

7 Chapter 3

The Theta Functions

In this chapter we will discuss the four theta functions as deﬁned by Jacobi. Jacobi’s analysis on the theta functions was so complete that all the results discussed in this chapter can be found in his works [12].

3.1 Deﬁnition of the theta functions

Deﬁnition 3.1.1 Let τ be a complex number whose imaginary part is positive; and write q = e2πiτ , such that | q| < 1. We deﬁne ∞ X n 1 n2 2niz ϑ4(z, q) = (−1) q 2 e , (3.1) n=−∞

iz+ 1 πiτ 1 ϑ (z, q) = −ie 4 ϑ (z + πτ, q) 1 4 2 ∞ (3.2) X n 1 (n+ 1 )2 (2n+1)iz = −i (−1) q 2 2 e , n=−∞ 1 ϑ (z, q) = ϑ (z + π, q) 2 1 2 ∞ (3.3) X 1 (n+ 1 )2 (2n+1)iz = q 2 2 e n=−∞ and 1 ϑ (z, q) = ϑ (z + π, q) 3 4 2 ∞ (3.4) X 1 n2 2niz = q 2 e . n=−∞

8 Let A be any positive constant. For any complex number z such that |z| ≤ A, we have

1 n2 ±2niz 1 n2 2nA | q 2 e | ≤ | q| 2 e , where n is a positive integer. Thus we have

∞ ∞ ∞ X n 1 n2 2niz X 1 n2 2niz X 1 n2 2nA | (−1) q 2 e | = | q 2 e | ≤ | q| 2 e . n=−∞ n=−∞ n=−∞

∞ 1 2 1 P 2 n 2nA n+ 2 2A Now the d’Alembert’s ratio [16, p.g 22] for the series n=−∞ | q| e is | q| e , ∞ 1 2 P 2 n 2nA which tends to zero as n → ∞. Therefore we have n=−∞ | q| e being absolutely convergent, which implies that the series

∞ X n 1 n2 2niz ϑ4(z, q) = (−1) q 2 e n=−∞

is uniformly convergent in any bounded domain of values of z, since A is any arbitrary constant. Furthermore, since ϑ4(z, q) is a series of analytic functions in any bounded domain of values of z, it is also an integral function, i.e. a function that is analytic everywhere in the complex plane except at ∞. From the deﬁnition of ϑ1(z, q), ϑ2(z, q) and

ϑ3(z, q), it is clear that these three theta functions are also uniformly convergent in any bounded domain of values of z and are also integral functions.

The above four theta functions can also be expressed as

∞ X n 1 n2 ϑ4(z, q) = 1 + 2 (−1) q 2 cos 2nz, n=1 ∞ X n 1 (n+ 1 )2 ϑ1(z, q) = 2 (−1) q 2 2 sin(2n + 1)z, n=0 ∞ (3.5) X 1 (n+ 1 )2 ϑ2(z, q) = 2 q 2 2 cos(2n + 1)z, n=0 ∞ X 1 n2 and ϑ3(z, q) = 1 + 2 q 2 cos 2nz. n=1

It is obvious that ϑ1(z, q) is an odd function of z, whereas ϑ2(z, q), ϑ3(z, q) and ϑ4(z, q) are even functions of z.

9 We will now show the eﬀects of increasing z by periods of π and πτ respectively for the four theta functions. For ϑ4(z, q), we have

∞ X n 1 n2 2ni(z+π) ϑ4(z + π, q) = (−1) q 2 e n=−∞ ∞ X n 1 n2 2niz 2niπ = (−1) q 2 e e (3.6) n=−∞ ∞ X n 1 n2 2niz = (−1) q 2 e = ϑ4(z, q), n=−∞

∞ X n 1 n2 2ni(z+πτ) and ϑ4(z + πτ, q) = (−1) q 2 e n=−∞ ∞ X n 1 n2 n 2niz = (−1) q 2 q e n=−∞ (3.7) ∞ − 1 −2iz X n+1 1 (n+1)2 2(n+1)iz = −q 2 e (−1) q 2 e n=−∞

− 1 −2iz = −q 2 e ϑ4(z, q).

For ϑ1(z, q), we have

∞ X n 1 (n+ 1 )2 (2n+1)i(z+π) ϑ1(z + π, q) = −i (−1) q 2 2 e n=−∞ ∞ X n 1 (n+ 1 )2 (2n+1)iz (2n+1)iπ = −i (−1) q 2 2 e e (3.8) n=−∞ ∞ X n 1 (n+ 1 )2 (2n+1)iz = i (−1) q 2 2 e = −ϑ1(z, q), n=−∞

∞ X n 1 (n+ 1 )2 (2n+1)i(z+πτ) and ϑ1(z + πτ, q) = −i (−1) q 2 2 e n=−∞ ∞ X n 1 (n+ 1 )2 n+ 1 (2n+1)iz = −i (−1) q 2 2 q 2 e n=−∞ (3.9) ∞ − 1 −2iz X n+1 1 ((n+1)+ 1 )2 (2(n+1)+1)iz = iq 2 e (−1) q 2 2 e n=−∞

− 1 −2iz = −q 2 e ϑ1(z, q).

10 For ϑ2(z, q), we have ∞ X 1 (n+ 1 )2 (2n+1)i(z+π) ϑ2(z + π, q) = q 2 2 e n=−∞ ∞ X 1 (n+ 1 )2 (2n+1)iz (2n+1)iπ = q 2 2 e e (3.10) n=−∞ ∞ X 1 (n+ 1 )2 (2n+1)iz = − q 2 2 e = −ϑ2(z, q), n=−∞ ∞ X 1 (n+ 1 )2 (2n+1)i(z+πτ) and ϑ2(z + πτ, q) = q 2 2 e n=−∞ ∞ X 1 (n+ 1 )2 n+ 1 (2n+1)iz = q 2 2 q 2 e n=−∞ (3.11) ∞ − 1 −2iz X 1 ((n+1)+ 1 )2 (2(n+1)+1)iz = q 2 e q 2 2 e n=−∞

− 1 −2iz = q 2 e ϑ2(z, q).

For ϑ3(z, q), we have ∞ X 1 n2 2ni(z+π) ϑ3(z + π, q) = q 2 e n=−∞ ∞ X 1 n2 2niz 2niπ = q 2 e e (3.12) n=−∞ ∞ X 1 n2 2niz = q 2 e = ϑ3(z, q), n=−∞ ∞ X 1 n2 2ni(z+πτ) and ϑ3(z + πτ, q) = q 2 e n=−∞ ∞ X 1 n2 n 2niz = q 2 q e n=−∞ (3.13) ∞ − 1 −2iz X 1 (n+1)2 2(n+1)iz = q 2 e q 2 e n=−∞

− 1 −2iz = q 2 e ϑ3(z, q). From the above relations we are able to show the following lemma:

Lemma 3.1.2 Let ϑ(z, q) to be any one of the above four theta functions and let ϑ 0(z, q) be its derivative with respect to z. Then ϑ 0(z + π, q) ϑ 0(z, q) ϑ 0(z + πτ, q) ϑ 0(z, q) = , = −2i + . (3.14) ϑ(z + π, q) ϑ(z, q) ϑ(z + πτ, q) ϑ(z, q)

11 Proof

We will only show the case for ϑ4(z, q), since the proof is similar for the other three

theta functions. From equation (3.6), we have ϑ4(z + π, q) = ϑ4(z, q). This implies that 0 0 ϑ4 (z + π, q) = ϑ4 (z, q), and thus we have ϑ 0(z + π, q) ϑ 0(z, q) 4 = 4 . ϑ4(z + π, q) ϑ4(z, q)

− 1 −2iz From equation (3.7), we have ϑ4(z + πτ) = −q 2 e ϑ4(z, q). This implies that

0 − 1 −2iz 0 − 1 −2iz ϑ4 (z + πτ) = −q 2 e ϑ4 (z, q) + q 2 (2i)e ϑ4(z, q),

and thus ϑ 0(z + πτ, q) ϑ 0(z, q) 4 = −2i + 4 ϑ4(z + πτ, q) ϑ4(z, q) 2

If there is no ambiguity on the value of q, we will denote ϑi(z, q) as ϑi(z), i = 1, 2, 3, 4.

3.2 The zeros of the theta functions

The effect of increasing z by π or πτ for ϑ4(z) is the same as the effect of multiplying − 1 −2iz ϑ4(z) by 1 or −q 2 e respectively. In the case of ϑ1(z), the effect of increasing z by π − 1 −2iz or πτ is the same as multiplying ϑ1(z) by −1 or −q 2 e respectively. ϑ2(z) and ϑ3(z) also exhibit similar properties. Because of this, the theta functions are known as quasi

− 1 −2iz doubly-periodic functions of z, and ±1 and −q 2 e are called the periodic factors [16, p.g 463].

Because of the quasi-periodic properties of the theta functions, it is obvious that, denoting

any one of the theta functions as ϑ(z), if z0 is a zero of ϑ(z), then

z0 + mπ + nπτ

is also a zero of ϑ(z), for all integral values of m and n. Thus we only consider the zeros of the theta functions within a cell with corners t, t + π, t + π + πτ and t + πτ, where t is an arbitrary complex number.

12 Theorem 3.2.1 Let ϑ(z) be any one of the theta functions, and C be a cell with corners t, t + π, t + π + πτ and t + πτ, where t is a complex number arbitrary chosen such that no zeroes of ϑ(z) lies on the boundaries of C. Then ϑ(z) has one and only one simple zero in the interior of C.

Proof

First note that, since ϑ(z) is an integral function, it is analytic everywhere except at ∞. Therefore it has no pole throughout the ﬁnite part of the complex plane, in particular on the boundaries of C and in the interior of C. Thus the number of zeros of ϑ(z) in the interior of C is given by the following integral:

1 Z ϑ 0(z) dz. 2πi C ϑ(z)

Using the results from (3.14), we have

1 Z ϑ 0(z) 1 Z t+π ϑ 0(z) ϑ 0(z + πτ) 1 Z t+πτ ϑ 0(z) ϑ 0(z + π) dz = − dz − − dz 2πi C ϑ(z) 2πi t ϑ(z) ϑ(z + πτ) 2πi t ϑ(z) ϑ(z + π) 1 Z t+π = 2i dz 2πi t = 1

Therefore ϑ(z) has one and only one simple zero inside C. 2

Note that because the value of q and the value of τ are related, with q = e2πiτ , the cell C in the above theorem varies with the value of q that deﬁnes the theta functions involved.

Also, it is clear that one zero of ϑ1(z) is z = 0 in a cell containing 0, and thus it follows

that the zeros of ϑ1(z) is of the form mπ + nπτ, m and n being integers. Similarly we 1 conclude that the zeros of ϑ2(z, q), ϑ3(z, q), ϑ4(z, q) are of the form 2 π + mπ + nπτ, 1 1 1 2 π + 2 πτ + mπ + nπτ , 2 πτ + mπ + nπτ respectively, m and n being integers.

13 3.3 Inﬁnite products expressions for the theta functions

Theorem 3.3.1 For |q| < 1, we have

∞ ∞ Y n− 1 2iz Y n− 1 −2iz ϑ4(z) = G (1 − q 2 e ) (1 − q 2 e ) n=1 n=1 ∞ Y n− 1 2n−1 = G (1 − 2q 2 cos 2z + q ), n=1

∞ ∞ 1 Y n 2iz Y n −2iz ϑ1(z) = 2Gq 8 sin z (1 − q e ) (1 − q e ) n=1 n=1 ∞ 1 Y n 2n = 2Gq 8 sin z (1 − 2q cos 2z + q ), n=1 ∞ ∞ 1 Y n 2iz Y n −2iz ϑ2(z) = 2Gq 8 cos z (1 + q e ) (1 + q e ) n=1 n=1 ∞ 1 Y n 2n = 2Gq 8 cos z (1 + 2q cos 2z + q ), n=1 ∞ ∞ Y n− 1 2iz Y n− 1 −2iz and ϑ3(z) = G (1 + q 2 e ) (1 + q 2 e ) n=1 n=1 ∞ Y n− 1 2n−1 = G (1 + 2q 2 cos 2z + q ), n=1 where G is a constant independent of z.

Proof

Let f(z) be a function deﬁned as follows:

∞ ∞ Y n− 1 2iz Y n− 1 −2iz f(z) = (1 − q 2 e ) (1 − q 2 e ). n=1 n=1

n− 1 ±2iz n− 1 A First we see that | − q 2 e | ≤ | q 2 |e in any bounded domain of values of z and ∞ 1 P n− 2 for n >> 0, and the series n=1 q is absolutely convergent. Therefore each of the two products in the deﬁnition of f(z) converges uniformly in any bounded domain of values of z. Hence f(z) is analytic throughout the ﬁnite part of the complex plane, and so it is an

14 integral function. Next we note that

∞ ∞ Y n− 1 2i(z+π) Y n− 1 −2i(z+π) f(z + π) = (1 − q 2 e ) (1 − q 2 e ) n=1 n=1 ∞ ∞ Y n− 1 2iz Y n− 1 −2iz = (1 − q 2 e ) (1 − q 2 e ) n=1 n=1 = f(z), and

∞ ∞ Y n+ 1 2iz Y n− 3 −2iz f(z + πτ) = (1 − q 2 e ) (1 − q 2 e ) n=1 n=1 − 1 −2iz ! 1 − q 2 e = 1 f(z) 1 − q 2 e2iz

1 −2iz = −q 2 e f(z).

This shows that f(z) is quasi doubly-periodic with periods π and πτ.

1 −2iz Now if we diﬀerentiate the equations f(z + π) = f(z) and f(z + πτ) = −q 2 e f(z) with respect to z, we get f 0(z + π) = f 0(z)

and

0 − 1 −2iz 0 − 1 −2iz f (z + πτ) = −q 2 e f (z) + q 2 (2i)e f(z),

which imply that f 0(z + π) f 0(z) = , (3.15) f(z + π) f(z) and f 0(z + πτ) f 0(z) = −2i + . (3.16) f(z + πτ) f(z)

Now let C be a cell with corners t, t + π, t + π + πτ and t + πτ. Since f(z) is an integral function, it is analytic everywhere except at ∞. Therefore f(z) has no poles throughout the ﬁnite part of the complex plane, in particular on and within C. Thus the number of zeros of f(z) inside C is given by the integral

1 Z f 0(z) dz, 2πi C f(z)

15 and using the results from (3.15) and (3.16), we have

1 Z f 0(z) 1 Z t+π f 0(z) f 0(z + πτ) 1 Z t+πτ f 0(z) f 0(z + π) dz = − dz − − dz 2πi C f(z) 2πi t f(z) f(z + πτ) 2πi t f(z) f(z + π) 1 Z t+π = 2i dz 2πi t = 1

This shows that f(z) has one and only one simple zero inside C.

1 We now consider the zeros of f(z). It is easy to verify that 2 πτ is a zero of f(z). Thus by the quasi doubly-periodic property of f(z), and that f(z) has one and only one simple zero in the cell C, the zeros of f(z) are just simple zeros at points

1 z = πτ + m πτ + m π, 2 1 2

where m1 and m2 are integers. Therefore f(z) and ϑ4(z) have the same zeros; and con-

ϑ4(z) sequently the quotient f(z) have neither zeros nor poles in the ﬁnite part of the complex plane.

Now by (3.15) and (3.16), we see that f(z) and ϑ4(z) have the same periodicity factors with

ϑ4(z) respect to periods π and πτ. This means that f(z) is a doubly-periodic, integral function

ϑ4(z) with no zeros or poles, and so by Theorem 2.2.1 (iv), f(z) is a constant independent of z, denoted as G, i.e.

ϑ4(z) = Gf(z) ∞ Y n− 1 2n−1 = G (1 − 2q 2 cos 2z + q ). n=1

With the product identity of ϑ4(z) established, the product identities of ϑ1(z), ϑ2(z) and

ϑ3(z) immediately follow since

16 iz+ 1 πiτ 1 ϑ (z) = −ie 4 ϑ (z + πτ) 1 4 2 1 iz 1 = −iq 8 e ϑ (z + πτ) 4 2 ∞ ∞ 1 iz Y n 2iz Y n−1 −2iz = −iq 8 e G (1 − q e ) (1 − q e ) n=1 n=1 ∞ ∞ 1 Y n 2iz Y n −2iz = 2Gq 8 sin z (1 − q e ) (1 − q e ) n=1 n=1 ∞ 1 Y n 2n = 2Gq 8 sin z (1 − 2q cos 2z + q ), n=1 1 ϑ (z) = ϑ (z + π) 2 1 2 ∞ ∞ 1 Y n 2iz Y n −2iz = 2Gq 8 cos z (1 + q e ) (1 + q e ) n=1 n=1 ∞ 1 Y n 2n = 2Gq 8 cos z (1 + 2q cos 2z + q ), n=1 1 and ϑ (z) = ϑ (z + π) 3 4 2 ∞ ∞ Y n− 1 2iz Y n− 1 −2iz = G (1 + q 2 e ) (1 + q 2 e ) n=1 n=1 ∞ Y n− 1 2n−1 = G (1 + 2q 2 cos 2z + q ) n=1 2

We will now determine the explicit value of the constant G. We will need the following lemma:

Lemma 3.3.2 The following identity holds:

0 ϑ1(0) = ϑ2(0)ϑ3(0)ϑ4(0),

0 where ϑ1(z) denotes the derivative of ϑ1(z) with respect to z.

Proof

We first show a differential equation satisfied by the theta functions. Consider ϑ3(z, q); it can be regarded as a function of two independent variables z and τ, since q = e2πiτ . Since

17 ϑ3(z) is uniformly convergent, we can diﬀerentiate it with respect to z or τ any number of times when Imz > 0. In particular, we have

2 ∞ ∂ ϑ3(z, q) X 2 = −4 n2en πiτ+2niz ∂z2 −∞ 4 ∂ϑ (z, q) = − 3 . πi ∂τ

Therefore ϑ3(z, q) satisfies the partial differential equation 1 ∂2y ∂y πi + = 0. (3.17) 4 ∂z2 ∂τ Using similar arguments we can also show that the other three theta functions also satisfy the above partial differential equation.

Now we consider the functions

f1(z) = 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z),

f2(z) = ϑ1(2z)ϑ2(0)ϑ3(0)ϑ4(0).

Using the transformation formulas of the theta functions in (3.6), (3.7), (3.8), (3.9), (3.10),

(3.11), (3.12), (3.13), we see that both the functions f1(z) and f2(z) satisfy the transformation formulas

f(z + π) = f(z),

f(z + πτ) = q−2e−8izf(z).

Therefore the function f (z) 2ϑ (z)ϑ (z)ϑ (z)ϑ (z) 1 = 1 2 3 4 f2(z) ϑ1(2z)ϑ2(0)ϑ3(0)ϑ4(0) is an elliptic function with periods π and πτ. Let C be a cell with corners t, t+π, t+π+πτ and t + πτ, the complex constant t chosen such that the sides of the cell do not contain

any of the possible zeros and poles of the functions f1(z)/f2(z). It is clear that f2(z) has

simple zeros at z = 0, z = π/2, z = πτ/2 and z = π/2 + πτ/2 in C, whereas f1(z) has

simple zeros at z = 0, z = π/2, z = πτ/2 and z = π/2 + πτ/2 in C. Therefore f1(z)/f2(z) is an elliptic function with no poles in C, and by Theorem 2.2.1 (iii), it must be a constant,

i.e. f1(z)/f2(z) = c for some constant c. Using L’ Hˆopital’s rule [10], we see that 0 ϑ1(z) ϑ1(0) lim = 0 . z→0 ϑ1(2z) 2ϑ1(0)

18 Therefore letting z → 0, we see that the constant c must be 1.

Therefore we have ϑ (2z) ϑ (z)ϑ (z)ϑ (z) 1 = 2 3 4 , 2ϑ1(z) ϑ2(0)ϑ3(0)ϑ4(0) and doing some rearrangements, we get

ϑ2(0)ϑ3(0)ϑ4(0)ϑ1(2z) = 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z). (3.18)

Diﬀerentiating (3.18) with respect to z, we have

0 0 0 ϑ2(0)ϑ3(0)ϑ4(0)ϑ1(2z) = ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) (3.19) 0 0 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z), and diﬀerentiate (3.19) with respect to z again, we get

00 2ϑ2(0)ϑ3(0)ϑ4(0)ϑ1(2z)

00 0 0 0 0 = ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 0 0 00 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z) ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 0 0 0 0 + ϑ1(z)ϑ2(z)ϑ3(z) ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 00 0 0 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 0 0 0 0 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) (3.20)

00 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

00 00 00 = ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

00 0 0 0 0 + ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 0 0 0 0 + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z)

0 0 + 2ϑ1(z)ϑ2(z)ϑ3(z)ϑ4(z).

19 Now we divide (3.20) throughout by ϑ1(z) to get 00 ϑ1(2z) 2ϑ2(0)ϑ3(0)ϑ4(0) ϑ1(z) 00 ϑ1(z) 00 00 00 = ϑ2(z)ϑ3(z)ϑ4(z) + ϑ2(z)ϑ3(z)ϑ4(z) + ϑ2(z)ϑ3(z)ϑ4(z) + ϑ2(z)ϑ3(z)ϑ4(z) ϑ1(z) 0 0 0 0 ϑ2(z) 0 ϑ3(z) 0 ϑ4(z) + 2ϑ1(z) ϑ3(z)ϑ4(z) + 2ϑ1(z)ϑ2(z) ϑ4(z) + 2ϑ1(z)ϑ2(z)ϑ3(z) ϑ1(z) ϑ1(z) ϑ1(z) 0 0 0 0 0 0 + 2ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ2(z)ϑ3(z)ϑ4(z) + 2ϑ2(z)ϑ3(z)ϑ4(z) (3.21)

Now we have ϑ1(0) = 0, and from (3.5), it is easy to see that

00 0 0 0 ϑ1(0) = ϑ2(0) = ϑ3(0) = ϑ4(0) = 0.

So taking limit with z → 0 on both sides of (3.21), and using L’ Hˆopital’s rule [10], we get 000 ϑ1 (0) 4ϑ2(0)ϑ3(0)ϑ4(0) 0 ϑ1(0) 000 ϑ1 (0) 00 00 00 = 0 ϑ2(0)ϑ3(0)ϑ4(0) + ϑ2(0)ϑ3(0)ϑ4(0) + ϑ2(0)ϑ3(0)ϑ4(0) + ϑ2(0)ϑ3(0)ϑ4(0) ϑ1(0) 00 00 00 0 ϑ2(0) 0 ϑ3(0) 0 ϑ4(0) + 2ϑ1(0) 0 ϑ3(0)ϑ4(0) + 2ϑ1(0)ϑ2(0) 0 ϑ4(0) + 2ϑ1(0)ϑ2(0)ϑ3(0) 0 ϑ1(0) ϑ1(0) ϑ1(0) 0 0 0 0 0 0 + 2ϑ2(0)ϑ3(0)ϑ4(0) + 2ϑ2(0)ϑ3(0)ϑ4(0) + 2ϑ2(0)ϑ3(0)ϑ4(0) which implies 000 ϑ1 (0) 3ϑ2(0)ϑ3(0)ϑ4(0) 0 ϑ1(0) (3.22) 00 00 00 = 3ϑ2(0)ϑ3(0)ϑ4(0) + 3ϑ2(0)ϑ3(0)ϑ4(0) + 3ϑ2(0)ϑ3(0)ϑ4(0).

Dividing (3.22) throughout by 3ϑ2(0)ϑ3(0)ϑ4(0), we have 000 00 00 00 ϑ1 (0) ϑ2(0) ϑ3(0) ϑ4(0) 0 = + + . (3.23) ϑ1(0) ϑ2(0) ϑ3(0) ϑ4(0) Utilizing the diﬀerential equation given in (3.17), we have 4 dϑ0 (z, q) ϑ000(z) = − 1 , 1 πi dτ 4 dϑ (z, q) ϑ00(z) = − 2 , 2 πi dτ 4 dϑ (z, q) ϑ00(z) = − 3 , 3 πi dτ 4 dϑ (z, q) and ϑ00(z) = − 4 , 4 πi dτ

20 and so (3.23) can be written as

0 1 dϑ1(0, q) 0 ϑ1(0, q) dτ 1 dϑ (0, q) 1 dϑ (0, q) 1 dϑ (0, q) = 2 + 3 + 4 . ϑ2(0, q) dτ ϑ3(0, q) dτ ϑ4(0, q) dτ

Integrating with respect to τ, we get

0 ϑ1(0, q) = Cϑ2(0, q)ϑ3(0, q)ϑ4(0, q), where C is a constant independent of q.

Now from the expressions given in (3.5), we can derive the following:

∞ 1 X 1 1 2 0 8 n 2 (n+ 2 ) ϑ1(0, q) = 2q + 2 (−1) q (2n + 1), n=1 ∞ 1 X 1 (n+ 1 )2 ϑ2(0, q) = 2q 8 + 2 q 2 2 , n=1 ∞ X 1 n2 ϑ3(0, q) = 1 + 2 q 2 n=1 ∞ X n 1 n2 and ϑ4(0, q) = 1 + 2 (−1) q 2 . n=1

Therefore by letting q → 0, we will have

1 1 − 8 0 − 8 lim q ϑ1(0) = 2, lim q ϑ2(0) = 2, lim ϑ3(0) = 1, lim ϑ4(0) = 1, q→0 q→0 q→0 q→0 and so we get C = 1. Thus we obtain the result

0 ϑ1(0) = ϑ2(0)ϑ3(0)ϑ4(0).

Theorem 3.3.3 Let G be as deﬁned in theorem 3.3.1. We have

∞ Y G = (1 − qn). n=1

21 Proof

By the product identity given in Theorem 3.3.1, we can write ϑ1(z) as

ϑ1(z) = sin z(φ(z)),

1/8 Q∞ n 2n where φ(z) = 2Gq n=1(1 − 2q cos 2z + q ). We then have ∞ 1 Y 0 8 n 2 ϑ1(0) = φ(0) = 2q G (1 − q ) , n=1 ∞ 1 Y n 2 ϑ2(0) = 2q 8 G (1 + q ) , n=1 ∞ Y n− 1 2 ϑ3(0) = G (1 + q 2 ) , n=1 ∞ Y n− 1 2 ϑ4(0) = G (1 − q 2 ) . n=1 By lemma 3.3.2, we then have ∞ ∞ ∞ ∞ Y n 2 2 Y n 2 Y n− 1 2 Y n− 1 2 (1 − q ) = G (1 + q ) (1 + q 2 ) (1 − q 2 ) . n=1 n=1 n=1 n=1 Since ∞ ∞ ! ∞ ∞ ! Y n− 1 Y n Y n− 1 Y n (1 − q 2 ) (1 − q ) (1 + q 2 ) (1 + q ) n=1 n=1 n=1 n=1 ∞ ∞ Y 1 n Y 1 n = (1 − q 2 ) (1 + q 2 ) n=1 n=1 ∞ Y = (1 − qn), n=1 we have ∞ Y G2 = (1 − qn)2 n=1 Q∞ n which implies that G = ± n=1(1 − q ). To determine the ambiguity of the sign, we observe that G is an analytic function of q (and consequently one-valued) throughout the domain |q| < 1. From the product identity of ϑ3(z) in theorem 3.3.1, we see that G → 1 as q → 0. Hence the plus sign must always be taken, and thus the result ∞ Y G = (1 − qn) n=1 is established. 2

22 Chapter 4

The Quintuple, Septuple Product Identities and the Winquist Identity

In this chapter, we will derive new proofs of the quintuple product identity, the septuple product identity and the Winquist identity, using properties of the theta functions and the theory of elliptic functions. The quintuple product identity had a very long history, and is best summarized in [3, p.g 83]. The septuple product identity was ﬁrst discovered by M.D. Hirschhorn [9] in 1983, and later rediscovered by H.M.Farkas and I.Kra in 1999 [7]. The Winquist identity was ﬁrst discovered by L.Winquist in [17], and was used to derive a double series identity for the product series

∞ Y (1 − xn)10, n=1 and a proof for the partition function congruence

p (11n + 6) ≡ 0 (mod 11), where p (n) denotes the number of unrestricted partitions of n [1, p.g 1].

Most of the previous proofs given to the three identities involved known q-series identities and direct algebraic manipulation. For the new proofs given in this chapter, however, it involves a systematic way of ﬁrst determining equivalent forms of the identities in terms of the theta functions, and then proved the equivalent theta functions identities using

23 theories of elliptic functions. The advantage of this new method is that it provides us with a systematic approach and direction of proving such identities.

4.1 The quintuple product identity

Theorem 4.1.1 (The quintuple product identity c.f.[15]) Let q and z be complex variables such that | q| < 1, z 6= 0. We deﬁne

∞ Y n−1 (z; q)∞ = (1 − zq ). n=1 Then we have the following identity

∞ 2 X (3n +n) 3n −3n−1 2 2 2 2 q 2 (z − z ) = (q; q)∞(zq; q)∞(1/z; q)∞(z q; q )∞(q/z ; q )∞ (4.1) n=−∞

The quintuple product identity has many applications, one of which is used to prove the Winquist identity [13]. However in this chapter, we will give separate proofs for the two identities, both using properties of the theta functions and the theory of elliptic functions.

Before we prove the quintuple product identity, we need to show an equivalent form of the identity shown in (4.1) in terms of theta functions.

Lemma 4.1.2 Let τ be a complex number whose imaginary part is positive, and q = e2πiτ , such that | q| ≤ 1. Then the quintuple product identity given in (4.1) has the following equivalent form:

ϑ1(2u, q) iu πτ 3 −iu πτ 3 (q; q)∞ = e ϑ4 3u + , q + e ϑ4 3u − , q , (4.2) ϑ1(u, q) 2 2

where ϑ4(z, q) and ϑ1(z, q) are the theta functions given in (3.1) and (3.2), u being a complex variable.

Proof

24 We ﬁrst look at the left hand side of the identity given in (4.1). Since we let z 6= 0, we can replace z by −1/x, x 6= ∞, to get

∞ ∞ ∞ 2 2 n 2 X (3n +n) 3n −3n−1 X (3n +n) (−1) X (3n +n) n 3n+1 q 2 ((−1/x) − (−1/x) ) = q 2 + q 2 (−1) x x3n n=−∞ n=−∞ n=−∞ ∞ ∞ 2 2 X (3n −n) n 3n X (3n +n) n 3n+1 = q 2 (−1) x + q 2 (−1) x , n=−∞ n=−∞

the second equality follows since by replacing n by −n, we get

∞ ∞ 2 n 2 X (3n +n) (−1) X (3n −n) n 3n q 2 = q 2 (−1) x . x3n n=−∞ n=−∞

Before we look at the right hand side of (4.1), ﬁrst we see that

∞ 2 2 Y 2 2n−1 (x q; q )∞ = (1 − x q ) n=1 has odd positive integral powers of q, and

∞ 2 2 Y 2 2n−2 (x ; q )∞ = (1 − x q ) n=1 has even positive integral powers of q (inclusive of the zero power). Therefore multiplying both series together, we get

∞ 2 2 2 2 Y 2 2n−1 2 2n−2 (x q; q )∞(x ; q )∞ = (1 − x q )(1 − x q ) n=1 ∞ (4.3) Y 2 n−1 2 = (1 − x q ) = (x ; q)∞. n=1 Using similar argument we also have

∞ 2 2 2 2 2 Y 2n−1 2 2n 2 (q/x ; q )∞(q /x ; q )∞ = (1 − q /x )(1 − q /x ) n=1 ∞ (4.4) Y n 2 2 = (1 − q /x ) = (q/x ; q)∞. n=1

Also we have ∞ Y n−1 n−1 (x; q)∞(−x; q)∞ = (1 − xq )(1 + xq ) n=1 ∞ (4.5) Y 2 2n−2 2 2 = (1 − x q ) = (x ; q )∞, n=1

25 and

∞ Y n n (q/x; q)∞(−q/x; q)∞ = (1 − q /x)(1 + q /x) n=1 ∞ (4.6) Y 2n 2 2 2 2 = (1 − q /x ) = (q /x ; q )∞. n=1

So we now look at the right hand side of (4.1). Again we can replace z by −1/x to get

2 2 2 2 (q; q)∞(zq; q)∞(1/z; q)∞(z q; q )∞(q/z ; q )∞

2 2 2 2 = (q; q)∞(x q; q )∞(q/x ; q )∞(−x; q)∞(−q/x; q)∞ 2 2 (q; q)∞(x ; q)∞(q/x ; q)∞(−x; q)∞(−q/x; q)∞ = 2 2 2 2 2 (using (4.3) and (4.4)) (x ; q )∞(q /x ; q )∞ (q; q) (x2; q) (q/x2; q) = ∞ ∞ ∞ (using (4.5) and (4.6)) (x; q)∞(q/x; q)∞ x(1 − x)(1 − x−2)(q; q) Q∞ (1 − qn−1x2) Q∞ (1 − qnx−2) = ∞ n=1 n=1 2 −1 Q∞ n−1 Q∞ n −1 (1 − x )(1 − x ) n=1(1 − q x) n=1(1 − q x ) x(q; q) Q∞ (1 − qnx2) Q∞ (1 − qn−1x−2) = ∞ n=1 n=1 Q∞ n Q∞ n−1 −1 n=1(1 − q x) n=1(1 − q x )

Then the quintuple product identity given in (4.1) become

∞ ∞ 2 2 X (3n −n) n 3n X (3n +n) n 3n+1 q 2 (−1) x + q 2 (−1) x n=−∞ n=−∞ x(q; q) Q∞ (1 − qnx2) Q∞ (1 − qn−1x−2) = ∞ n=1 n=1 , Q∞ n Q∞ n−1 −1 n=1(1 − q x) n=1(1 − q x )

2iu and letting x = e and using the deﬁnition of the theta functions ϑ1(z) and ϑ4(z) given in (3.1) and (3.2), we ﬁnally have

∞ ∞ 2 2 X (3n −n) n 2ni(3u) X (3n +n) n 2ni(3u) 2iu q 2 (−1) e + q 2 (−1) e e n=−∞ n=−∞ e2iu(q; q) Q∞ (1 − qne2i(2u)) Q∞ (1 − qn−1e−2i(2u)) = ∞ n=1 n=1 , Q∞ n 2iu Q∞ n−1 −2iu n=1(1 − q e ) n=1(1 − q e ) which implies that πτ πτ e−iuϑ (3u − , q3) + eiuϑ (3u + , q3) 4 2 4 2 1 ∞ ∞ 8 2iu Q n 2i(2u) Q n−1 −2i(2u) (q; q)∞(−i)q e n=1(1 − q e ) n=1(1 − q e ) = 1 ∞ ∞ , 8 iu Q n 2iu Q n−1 −2iu (−i)q e n=1(1 − q e ) n=1(1 − q e )

26 and thus we have

−iu πτ 3 iu πτ 3 ϑ1(2u, q) e ϑ4(3u − , q ) + e ϑ4(3u + , q ) = (q; q)∞ 2 2 ϑ1(u, q) which is the equivalent form we desired. 2

We now prove the quintuple product identity stated in Theorem 4.1.1 using the equivalent form (4.2).

Proof of Theorem 4.1.1

Using the transformation formulas of the theta functions stated in (3.6),(3.7),(3.8) and (3.9), it is easy to see that the following functions

ϑ1(2u, q) iu πτ 3 −iu πτ 3 , e ϑ4(3u + 2 , q ), and e ϑ4(3u − , q ) ϑ1(u, q) 2 satisfy the transformation formula

g(u + π) = g(u).

The three functions also satisfy

− 3 −6iu g(u + πτ) = −q 2 e g(u), since ϑ (2(u + πτ), q) ϑ ((2u + πτ) + πτ, q) 1 = 1 ϑ1(u + πτ, q) ϑ1(u + πτ, q) − 1 −2i(2u+πτ) −q 2 e ϑ (2u + πτ, q) = 1 ϑ1(u + πτ, q) q−1e−2i(2u+πτ)e−4iuϑ (2u, q) = 1 − 1 −2iu −q 2 e ϑ1(u, q)

− 3 −6iu ϑ1(2u, q) = −q 2 e , ϑ1(u, q)

i(u+πτ) πτ 3 1 iu πτ 3 e ϑ (3(u + πτ) + , q ) = q 2 e ϑ ((3u + ) + 3πτ, q ) 4 2 4 2 1 iu − 3 −2i(3u+ πτ ) πτ 3 = q 2 e (−q 2 )e 2 ϑ ((3u + ), q ) 4 2 − 3 −6iu iu πτ 3 = −q 2 e e ϑ (3u + , q ), 4 2

27 and

−i(u+πτ) πτ 3 − 1 −iu πτ 3 e ϑ (3(u + πτ) − , q ) = q 2 e ϑ ((3u − ) + 3πτ, q ) 4 2 4 2 − 1 −iu − 3 −2i(3u− πτ ) πτ 3 = q 2 e (−q 2 )e 2 ϑ ((3u − ), q ) 4 2 − 3 −6iu iu πτ 3 = −q 2 e e ϑ (3u − , q ). 4 2

Using the above transformation formulas we see that the function

ϑ1(u, q) iu πτ 3 g1(u) = e ϑ4(3u + , q ) ϑ1(2u, q) 2

and the function

ϑ1(u, q) −iu πτ 3 g2(u) = e ϑ4(3u − , q ) ϑ1(2u, q) 2 are both elliptic functions with period π and πτ. So we now consider the cell C1 with corners t, t + π, t + π + πτ and t + πτ, the complex constant t chosen such that the sides of the cell do not contain any of the possible zeros and poles of the functions g1(u) and

g2(u). By Theorem 3.2.1, we see that the function ϑ1(2u, q) has one zero congruent to π π+πτ πτ u = 0 in cell C2 with corners t, t + 2 , t + 2 and t + 2 . Therefore we see that the π πτ π+πτ function ϑ1(2u, q) has zeros congruent to u = 0, 2 , 2 and 2 in C1. Now since ϑ1(u, q) has only one zero congruent to u = 0 in C1, we see that the functions g1(u) and g2(u) have π πτ π+πτ only 3 possible simple poles congruent to u = 2 , 2 and 2 in C1. Now the function

∞ 2 iu πτ 3 −iu πτ 3 X n 3n +n e ϑ 3u + , q + e ϑ 3u − , q = 2 (−1) q 2 cos((6n + 1)u) 4 2 4 2 n=−∞

28 π πτ vanishes at the point u = 2 . At the point u = 2 , we have iπτ 3πτ πτ 3 −iπτ 3πτ πτ 3 e 2 ϑ + , q + e 2 ϑ − , q 4 2 2 4 2 2 ∞ ∞ 1 X n 3 n2 4inπτ − 1 X n 3 m2 2imπτ = q 4 (−1) q 2 e + q 4 (−1) q 2 e n=−∞ m=−∞ ∞ ∞ X n 3 n2+2n+ 1 X m 3 m2−m− 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ m=−∞ ∞ ∞ X n 3 n2+2n+ 1 X m+1 3 (m+1)2−(m+1)− 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ m=−∞ ∞ ∞ X n 3 n2+2n+ 1 X m+1 3 m2+2m+ 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ m=−∞ = 0,

πτ and so the function also vanishes at u = 2 . Therefore the function g1(u) + g2(u) has no π πτ poles at u = 2 and u = 2 in C1. This means that g1(u)+g2(u) is an elliptic function with period π and πτ with at most one pole in C1, and since any non-trivial elliptic function must have at least 2 poles by theorem 2.2.1(iii), g1(u) + g2(u) must be a constant, i.e.

ϑ1(u, q) iu πτ 3 ϑ1(u, q) −iu πτ 3 e ϑ4(3u + , q ) + e ϑ4(3u − , q ) = A(q), ϑ1(2u, q) 2 ϑ1(2u, q) 2 where A(q) is a constant in terms of q. Letting u = 0, we see that πτ A(q) = 2g−1(0)ϑ ( , q3), 3 4 2 where ϑ1(2u, q) g3(u) = . ϑ1(u, q) Since

g3(0) = 2,

∞ ∞ πτ Y Y ϑ ( , q3) = (1 − q3n) (1 − q3(n−1/2)eiπτ )(1 − q3(n−1/2)e−iπτ ) 4 2 n=1 n=1 ∞ ∞ ∞ Y Y Y = (1 − q3n) (1 − q3n−1) (1 − q3n−2) n=1 n=1 n=1 ∞ Y n = (1 − q ) = (q ; q)∞, n=1 we have A = (q ; q)∞ and the desired result is obtained. 2

29 4.2 The septuple product identity

In this section, we will discuss a new proof for the septuple product identity based on theta-functions and theories of elliptic functions.

Theorem 4.2.1 (The septuple product identity c.f.[9]) Let q and z be complex variables such that | q| < 1, z 6= 0. Deﬁning

(z1, z2, z3, . . . , zn; q)∞ = (z1; q)∞(z2; q)∞(z3; q)∞ ... (zn; q)∞ we have the following identity

2 2 2 2 2 2 (q, q, x, q/x; q)∞(qx , q/x , x , q /x ; q )∞ ∞ ∞ ∞ ! 2 2 2 X m 5m +m X k 5k +3k 5k+3 X k 5k −3k 5k = (−1) q 2 (−1) q 2 x + (−1) q 2 x m=−∞ k=−∞ k=−∞ ∞ ∞ ∞ ! 2 2 2 X m 5m +3m X k 5k +k 5k+2 X k 5k −k 5k+1 − (−1) q 2 (−1) q 2 x + (−1) q 2 x , m=−∞ k=−∞ k=−∞ (4.7) where x 6= 0.

Again to prove Theorem 4.2.1, we show an equivalent form for (4.7) in terms of theta functions.

Lemma 4.2.2 The equation stated in (4.7) has the following equivalent form

−1/4 −q ϑ1(u, q)ϑ1(2u, q) ∞ 2 X m 5m +m −3iu 5 3iu 5 = (−1) q 2 e ϑ4(5u − 3πτ/2, q ) + e ϑ4(5u + 3πτ/2, q ) m=−∞ ∞ 2 X m 5m +3m iu 5 −iu 5 − (−1) q 2 e ϑ4(5u + πτ/2, q ) + e ϑ4(5u − πτ/2, q ) , m=−∞ (4.8) where Imu > 0.

30 Proof

For the equation stated in (4.7), let x = e2iu. Then the left hand side of (4.7) can be re-written as

2iu −2iu 4iu −4iu 4iu 2 −4iu 2 (q, q, e , qe ; q)∞(qe , qe , e , q e ; q )∞ ∞ Y = (1 − qn)(1 − qn)(1 − e2iuqn−1)(1 − e−2iuqn)(1 − e4iuq2n−1) n=1 (1 − e−4iuq2n−1)(1 − e4iuq2(n−1))(1 − e−4iuq2n) ∞ Y = (1 − qn)(1 − qn)(1 − e2iuqn−1)(1 − e−2iuqn)(1 − e4iuqn−1)(1 − e−4iuqn) n=1 ∞ ! − 1 3iu 1 −iu Y n 2iu n−1 −2iu n = (−q 4 )e −iq 8 e (1 − q )(1 − e q )(1 − e q ) n=1 ∞ ! 1 −2iu Y n 4iu n−1 −4iu n −iq 8 e (1 − q )(1 − e q )(1 − e q ) n=1 − 1 3iu = (−q 4 )e ϑ1(−u, q)ϑ1(−2u, q)

− 1 3iu = (−q 4 )e ϑ1(u, q)ϑ1(2u, q), and the right hand side of (4.7) can be re-written as

∞ ∞ ∞ ! 2 2 2 X m 5m +m X k 5k +3k 2ik(5u) 6iu X k 5k −3k 2ik(5u) (−1) q 2 (−1) q 2 e e + (−1) q 2 e m=−∞ k=−∞ k=−∞ ∞ ∞ ∞ ! 2 2 2 X m 5m +3m X k 5k +k 2ik(5u) 4iu X k 5k −k 2ik(5u) 2iu − (−1) q 2 (−1) q 2 e e + (−1) q 2 e e m=−∞ k=−∞ k=−∞ ∞ 2 3iu X m 5m +m 3iu 5 −3iu 5 = e (−1) q 2 e ϑ4(5u + 3πτ/2, q ) + e ϑ4(5u − 3πτ/2, q ) m=−∞ ∞ 2 3iu X m 5m +3m iu 5 −iu 5 − e (−1) q 2 e ϑ4(5u + πτ/2, q ) + e ϑ4(5u − πτ/2, q ) m=−∞

Combining the above computations, we deduce (4.8). 2

Proof of theorem 4.2.1

We will prove the septuple product identity (4.7) by proving the equivalent identity (4.8). Using the transformation formulas of the theta functions stated in (3.6), (3.7),(3.8) and

31 (3.9), we observe that the following ﬁve functions

h1(u) = ϑ1(u, q)ϑ1(2u, q),

−3iu 5 h2(u) = e ϑ4(5u − 3πτ/2, q ),

3iu 5 h3(u) = e ϑ4(5u + 3πτ/2, q ),

iu 5 h4(u) = e ϑ4(5u + πτ/2, q ),

−iu 5 h5(u) = e ϑ4(5u − πτ/2, q )

satisfy the transformation formula

hj(u + π) = −hj(u), 1 ≤ j ≤ 5,

and the ﬁve functions also satisfy the transformation formula

− 5 −10iu hj(u + πτ) = −q 2 e hj(u), 1 ≤ j ≤ 5, since

h1(u + πτ) = ϑ1(u + πτ, q)ϑ1(2(u + πτ), q)

− 1 −2iu − 1 −2i(2u+πτ) = −q 2 e ϑ1(u, q)(−q 2 )e ϑ1(2u + πτ, q)

− 1 −2iu −2 −8iu = −q 2 e ϑ1(u, q)q e ϑ1(2u, q)

− 5 −10iu − 5 −10iu = −q 2 e ϑ1(u, q)ϑ1(2u, q) = −q 2 e h1(u),

−3i(u+πτ) 5 h2(u + πτ) = e ϑ4(5(u + πτ) − 3πτ/2, q )

− 3 −3iu − 5 −2i(5u−3πτ/2) 5 = q 2 e (−q 2 )e ϑ4(5u − 3πτ/2, q )

− 5 −10iu −3iu 5 − 5 −10iu = −q 2 e e ϑ4(5u − 3πτ/2, q ) = −q 2 e h2(u),

3i(u+πτ) 5 h3(u + πτ) = e ϑ4(5(u + πτ) + 3πτ/2, q )

3 3iu − 5 −2i(5u+3πτ/2) 5 = q 2 e (−q 2 )e ϑ4(5u + 3πτ/2, q )

− 5 −10iu 3iu 5 − 5 −10iu = −q 2 e e ϑ4(5u + 3πτ/2, q ) = −q 2 e h3(u),

i(u+πτ) 5 h4(u + πτ) = e ϑ4(5(u + πτ) + πτ/2, q )

1 iu − 5 −2i(5u+πτ/2) 5 = q 2 e (−q 2 )e ϑ4(5u + πτ/2, q )

− 5 −10iu iu 5 − 5 −10iu = −q 2 e e ϑ4(5u + πτ/2, q ) = −q 2 e h4(u),

32 and

−i(u+πτ) 5 h5(u + πτ) = e ϑ4(5(u + πτ) − πτ/2, q )

− 1 −iu − 5 −2i(5u−πτ/2) 5 = q 2 e (−q 2 )e ϑ4(5u − πτ/2, q )

− 5 −10iu −iu 5 − 5 −10iu = −q 2 e e ϑ4(5u − πτ/2, q ) = −q 2 e h5(u).

We now consider the functions ∞ 2 X k 5k +3k H1(u) = h2(u) + h3(u) = 2 (−1) q 2 cos(10k + 3)u (4.9) k=−∞ and ∞ 2 X k 5k +k H2(u) = h4(u) + h5(u) = 2 (−1) q 2 cos(10k + 1)u. (4.10) k=−∞

Since cos(π/2) = 0, we see that both H1(u) and H2(u) vanish at u = π/2. At u = πτ/2, we have

3i(πτ/2) 5 −3i(πτ/2) 5 H1(πτ/2) = e ϑ4(5(πτ/2) + 3πτ/2, q ) + e ϑ4(−5(πτ/2) + 3πτ/2, q ) ∞ ∞ 3i(πτ/2) X n 5 n2 2in(4πτ) −3i(πτ/2) X n 5 n2 2in(−πτ) = e (−1) q 2 e + e (−1) q 2 e n=−∞ n=−∞ ∞ ∞ X n 5 n2+4n+ 3 X n 5 n2−n− 3 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ ∞ ∞ X n 5 n2+4n+ 3 X n+1 5 (n+1)2−(n+1)− 3 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ ∞ ∞ X n 5 n2+4n+ 3 X n+1 5 n2+4n+ 3 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ = 0,

i(πτ/2) 5 −i(πτ/2) 5 H2(πτ/2) = e ϑ4(5(πτ/2) + πτ/2, q ) + e ϑ4(−5(πτ/2) + πτ/2, q ) ∞ ∞ i(πτ/2) X n 5 n2 2in(3πτ) −i(πτ/2) X n 5 n2 2in(−2πτ) = e (−1) q 2 e + e (−1) q 2 e n=−∞ n=−∞ ∞ ∞ X n 5 n2+3n+ 1 X n 5 n2−2n− 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ ∞ ∞ X n 5 n2+3n+ 1 X n+1 5 (n+1)2−2(n+1)− 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ ∞ ∞ X n 5 n2+3n+ 1 X n+1 5 n2+3n+ 1 = (−1) q 2 4 + (−1) q 2 4 n=−∞ n=−∞ = 0,

33 and therefore both H1(u) and H2(u) also vanish at u = πτ/2. We now consider the equation

H(u) = αH1(u) + βH2(u),

where

∞ ∞ 2 2 X k 5k +k X k 5k +3k α = H2(0) = (−1) q 2 , β = −H1(0) = − (−1) q 2 . k=−∞ k=−∞

It is easy to see that at u = 0, H(0) = 0. Also since H1(u) and H2(u) are even functions,

we know that the zero of H(u) at u = 0 is at least of order 2. Since both H1(u) and H2(u) are both equal to zero at u = π/2 and u = πτ/2, H(u) has zero of order at least 1 at u = π/2 and u = πτ/2. Now consider the cell C with corners t, t + π, t + π + πτ and t + πτ, the complex constant t chosen such that the sides of the cell do not contain any of the zeros of the functions

H(u) and h1(u). Since the only zeros of the function h1(u) in C are a double zero at u = 0 and simple zeros at u = π/2 , u = πτ/2 and u = (π + πτ)/2, we conclude that the

function H(u)/h1(u) has at most one pole in C, and since H(u) and h1(u) satisfy the same

transformation formulas under the period π and πτ, H(u)/h1(u) is therefore an elliptic

function with at most one pole in C. By Theorem 2.2.1 (iii), We know that H(u)/h1(u) is a constant, i.e.

H(u)/h1(u) = A(q),

where A(q) is independent of u. Hence we have

A(q)ϑ1(u, q)ϑ1(2u, q) ∞ 2 X m 5m +m −3iu 5 3iu 5 = (−1) q 2 e ϑ (5u − 3πτ/2, q ) + e ϑ (5u + 3πτ/2, q ) 4 4 (4.11) m=−∞ ∞ 2 X m 5m +3m iu 5 −iu 5 − (−1) q 2 e ϑ4(5u + πτ/2, q ) + e ϑ4(5u − πτ/2, q ) . m=−∞

To determine the constant A(q), we ﬁrst substitute u = π/5 into (4.9) and (4.10), and

using the product identity of ϑ4(z, q) in theorem 3.3.1, we see that the right hand side of

34 (4.11) is

2 2 X m+k 5m +m+5k +3k 2(cos (3π/5) − cos (π/5)) (−1) q 2 m,k∈Z 5 5 = −4 sin (2π/5) sin (π/5)ϑ4(πτ/2, q )ϑ4(3πτ/2, q ) ∞ ∞ Y Y = −4 sin (2π/5) sin (π/5) (1 − q5n) (1 − q5n−2)(1 − q5n−3) n=1 n=1 ∞ ∞ (4.12) Y Y × (1 − q5n) (1 − q5n−1)(1 − q5n−4) n=1 n=1 ∞ ∞ Y Y = −4 sin (2π/5) sin (π/5) (1 − q5n) (1 − qn) n=1 n=1 5 5 = −4 sin (2π/5) sin (π/5)(q ; q )∞(q; q)∞.

On the other hand, using the product identity of ϑ1(z, q) in theorem 3.3.1 and the identity

(1 − x)(1 − e2πi/5x)(1 − e−2πi/5x)(1 − e4πi/5x)(1 − e−4πi/5x) = (1 − x5), we see that

ϑ1(π/5, q)ϑ1(2π/5, q) ∞ Y = 4q1/4 sin (π/5) sin (2π/5) (1 − qn)2 n=1 ∞ Y × (1 − e2πi/5qn)(1 − e−2πi/5qn)(1 − e4πi/5qn)(1 − e−4πi/5qn) (4.13) n=1 ∞ ∞ Y Y = 4q1/4 sin (π/5) sin (2π/5) (1 − qn) (1 − q5n) n=1 n=1 1/4 5 5 = 4q sin (π/5) sin (2π/5)(q; q)∞(q ; q )∞

Combining (4.11), (4.12), (4.13), we then have

A(q) = −q−1/4 and thus obtain the desired result. 2

4.3 The Winquist identity

In this section, we will discuss a new proof for the Winquist identity, which will again based on properties of the theta functions and elliptic functions.

35 Theorem 4.3.1 (Winquist identity c.f.[17]) Let q, a and b be complex variables such that | q| < 1, a , b 6= 0. Then we have the following identity

∞ ∞ 2 2 X X m+n 3m +3n +3m+n (−1) q 2 m=−∞ n=−∞ (4.14) × (a−3mb−3n − a−3mb3n+1 − a−3n+1b−3m−1 + a3n+2b−3m−1)

2 −1 −1 −1 −1 −1 −1 = (q ; q)∞(a, a q, b, b q, ab, a b q, ab , a bq ; q)∞.

Now to prove theorem 4.3.1, we ﬁrst show that equation (4.14) follows from the following equation:

(x, q/x,y, q/y, y/x, qx/y, xy, q/xy, q, q, q, q ; q)∞ ∞ ∞ 2 X n 3n(n+1) −3n X m m(m−1) m = (q ; q) (−1) q 2 y (−1) q 6 x ∞ (4.15) n=−∞ m=−∞ ∞ ∞ X n(n−1) X 3m(m+1) −1 2 n 6 n m 6 −3m − yx (q ; q)∞ (−1) q y (−1) q x . n=−∞ m=−∞

Let ∞ ∞ X 3n(n+1) X m(m−1) 2 n 2 −3n m 6 m T1 = (q ; q)∞ (−1) q y (−1) q x , n=−∞ m=−∞ ∞ ∞ X n(n−1) X 3m(m+1) −1 2 n 6 n m 6 −3m T2 = yx (q ; q)∞ (−1) q y (−1) q x . n=−∞ m=−∞ First we observe that

∞ X m m(m−1) m (−1) q 6 u m=−∞ ∞ X 3k k(3k+1) −3k 3k+1 3k+2 (3k+2)(3k+1) 3k+2 = (−1) q 2 (u − u ) + (−1) q 6 u . k=−∞

36 So we will have

∞ ∞ X 3n(n+1) X m(m−1) 2 n 2 −3n m 6 m T1 = (q ; q)∞ (−1) q y (−1) q x n=−∞ m=−∞ ∞ X 3n(n+1) 2 n 2 −3n = (q ; q)∞ (−1) q y n=−∞ ∞ X 3k k(3k+1) −3k 3k+1 3k+2 (3k+2)(3k+1) 3k+2 × (−1) q 2 (x − x ) + (−1) q 6 x k=−∞ ! X 3n2+3n+3k2+k 2 n+k 2 −3n −3k −3n 3k+1 = (q ; q)∞ (−1) q (y x − y x ) n,k∈Z ! X 3n2+3n (3k+2)(3k+1) 2 n+k ( 2 + 6 ) −3n 3k+2 + (q ; q)∞ (−1) q y x n,k∈Z

and

∞ ∞ X n(n−1) X 3m(m+1) −1 2 n 6 n m 6 −3m T2 = yx (q ; q)∞ (−1) q y (−1) q x n=−∞ m=−∞ ∞ X k(3k+1) (3k+2)(3k+1) −1 2 3k 2 −3k 3k+1 3k+2 6 3k+2 = yx (q ; q)∞ (−1) q (y − y ) + (−1) q y k=−∞ ∞ X n 3n(n+1) −3n × (−1) q 2 x n=−∞ ! X 3n2+3n+3k2+k 2 n+k 2 −3k+1 −3n−1 −3k+2 −3n−1 = (q ; q)∞ (−1) q (y x − y x ) n,k∈Z ! X 3n2+3n (3k+2)(3k+1) 2 n+k ( 2 + 6 ) 3k+3 −3n−1 + (q ; q)∞ (−1) q y x . n,k∈Z

Doing a change of variables n = −k − 1, we note that ! X 3n2+3n (3k+2)(3k+1) 2 n+k ( 2 + 6 ) 3k+3 −3n−1 (q ; q)∞ (−1) q y x n,k∈Z ! X 3(−k−1)2+3(−k−1) (3(−n−1)+2)(3(−n−1)+1) 2 (−k−1)+(−n−1) ( 2 + 6 ) 3(−n−1)+3 −3(−k−1)−1 = (q ; q)∞ (−1) q y x n,k∈Z ! X 3n2+3n (3k+2)(3k+1) 2 n+k ( 2 + 6 ) −3n 3k+2 = (q ; q)∞ (−1) q y x , n,k∈Z

37 therefore we have

(x, q/x, y, q/y, y/x, qx/y, xy, q/xy, q, q, q, q ; q)∞

= T1 − T2 ! X 3n2+3n+3k2+k 2 n+k 2 −3n −3k −3n 3k+1 −3k+1 −3n−1 −3k+2 −3n−1 = (q ; q)∞ (−1) q (y x − y x − y x + y x ) n,k∈Z and letting x = b and y = a, we obtain equation (4.14).

So we will use the alternative form of the Winquist identity (4.15) for our proof of the Winquist identity. We will now show an equivalent form in terms of theta-functions for equation (4.15).

Lemma 4.3.2 The equation (4.15) has the following equivalent form

1 1 1 12 2 3 3 3 3 q (q ; q)∞{ϑ1(u, q )ϑ1(3v, q ) − ϑ1(v, q )ϑ1(3u, q )} (4.16) = ϑ1(u, q)ϑ1(v, q)ϑ1(u − v, q)ϑ1(u + v, q)

Proof

Using (3.2) and Theorem 3.3.1, and letting x = e−2iv, y = e−2iu, we have

∞ ∞ X 3n(n+1) X m(m−1) 2 n 2 −3n m 6 m T1 = (q ; q)∞ (−1) q y (−1) q x n=−∞ m=−∞ ∞ ∞ X 3n2 3n X n2 n 2 n ( 2 + 2 ) 2n(3iu) m ( 6 + 6 ) 2miv = (q ; q)∞ (−1) q e (−1) q e n=−∞ m=−∞ ∞ ! 5 X 3 1 2 − 12 −3iu −iv 2 n 2 (n+ 2 ) (2n+1)(3iu) = −q e e (q ; q)∞ −i (−1) q e n=−∞ ∞ ! X m 1 (m+ 1 )2 (2m+1)(iv) × −i (−1) q 6 2 e m=−∞

5 1 − 12 −3iu −iv 2 3 3 = −q e e (q ; q)∞ϑ1(3u, q )ϑ1(v, q ),

38 ∞ ∞ X n(n+1) X 3m(m+1) −1 2 n 6 −n m 2 −3m T2 = yx (q ; q)∞ (−1) q y (−1) q x n=−∞ m=−∞ ∞ ∞ X n2 n X 3n2 3n −2iu 2iv 2 n ( 6 + 6 ) 2n(iu) m ( 2 + 2 ) 2m(3iv) = e e (q ; q)∞ (−1) q e (−1) q e n=−∞ m=−∞ ∞ ! 5 X 1 1 2 − 12 −3iu −iv 2 n 6 (n+ 2 ) (2n+1)(iu) = −q e e (q ; q)∞ −i (−1) q e n=−∞ ∞ ! X m 3 (m+ 1 )2 (2m+1)(3iv) × −i (−1) q 2 2 e m=−∞

5 1 − 12 −3iu −iv 2 3 3 = −q e e (q ; q)∞ϑ1(u, q )ϑ1(3v, q ),

(x, q/x, y, q/y, y/x, qx/y, xy, q/xy, q, q, q, q ; q)∞

−2iu 2iu −2iv 2iv −2iu 2iv 2iu −2iv −2iu −2iv 2iu 2iv = (q, e , e q ; q)∞(q, e , e q ; q)∞(q, e e , e e q ; q)∞(q, e e , e e q ; q)∞

− 1 −3iu −iv 1 iu −2iu 2iu 1 iv −2iv 2iv = q 2 e e (−iq 8 e (q, e , e q ; q)∞)(−iq 8 e (q, e , e q ; q)∞)

1 i(u−v) −2iu 2iv 2iu −2iv 1 i(u+v) −2iu −2iv 2iu 2iv × (−iq 8 e (q, e e , e e q ; q)∞)(−iq 8 e (q, e e , e e q ; q)∞)

− 1 −3iu −iv = q 2 e e ϑ1(u, q)ϑ1(v, q)ϑ1(u − v, q)ϑ1(u + v, q).

Therefore equation (4.15) becomes

− 1 −3iu −iv q 2 e e ϑ1(u, q)ϑ1(v, q)ϑ1(u − v, q)ϑ1(u + v, q)

5 1 1 − 12 −3iu −iv 2 3 3 3 3 = q e e (q ; q)∞{ϑ1(u, q )ϑ1(3v, q ) − ϑ1(3u, q )ϑ1(v, q )} and multiply throughout by q1/2e3iueiv, we get the result. 2

So to complete the proof of the Winquist identity, we just need to prove the identity (4.16).

Proof of identity (4.16)

Using (3.8) and (3.9), we see that the following functions (view as functions of u),

1 3 ϑ1(u, q 3 ), ϑ1(3u, q ), ϑ1(u, q)ϑ1(u − v, q)ϑ1(u + v, q)

satisfy the transformation formulas

f(z + π) = −f(z), f(z + πτ) = −q−3/2e−6izf(z).

39 Therefore the functions

ϑ1(u, q)ϑ1(u − v, q)ϑ1(u + v, q) F1(u) = 1 ϑ1(u, q 3 ) 3 ϑ1(3u, q ) F2(u) = 1 ϑ1(u, q 3 ) are elliptic functions with period π and πτ. So we consider the cell C with corners t, t+π, t + π + πτ and t + πτ, the complex constant t chosen such that the sides of the cell do not

contain any of the poles of the functions F1(u) and F2(u). In C, the functions ϑ1(u, q) and 3 ϑ1(3u, q ) have one and only one simple zero at u = 0 respectively, whereas the function 1 ϑ1(u, q 3 ) has three simple zeroes in C, since it has one and only one simple zero in the cell 0 1 1 C with corners t, t + π, t + π + 3 πτ and t + 3 πτ. Therefore the functions F1(u) and F2(u)

has two poles in C respectively. Furthermore, since F1(u)/F2(u) is dependent of u, there exists C1(v, q) and C2(v, q) such that the elliptic function C1(v, q)F1(u) + C2(v, q)F2(u) is not identically zero, and has at most one pole in C, and so by theorem 2.2.1(iii) it must

be an expression independent of u, say C3(v, q). Hence we have

1 3 C3(v, q)ϑ1(u, q 3 ) + C2(v, q)ϑ1(3u, q ) = C1(v, q)ϑ1(u, q)ϑ1(u − v, q)ϑ1(u + v, q). (4.17)

Let u = v in (4.17), we get

1 3 C3(v, q)ϑ1(v, q 3 ) + C2(v, q)ϑ1(3v, q ) = 0.

Hence we have 3 ϑ1(3v, q ) C3(v, q) = −C2(v, q) 1 , ϑ1(v, q 3 )

and we can rewrite (4.17) as

3 1 1 3 −C2(v, q)ϑ1(3v, q )ϑ1(u, q 3 ) + C2(v, q)ϑ1(v, q 3 )ϑ1(3u, q ) (4.18) 1 = C1(v, q)ϑ1(v, q 3 )ϑ1(u, q)ϑ1(u + v, q)ϑ1(u − v, q).

Since (4.18) holds for all u and v, we interchange u and v in (4.18) and deduce that

3 1 1 3 −C2(u, q)ϑ1(3u, q )ϑ1(v, q 3 ) + C2(u, q)ϑ1(u, q 3 )ϑ1(3v, q ) (4.19) 1 = C1(u, q)ϑ1(u, q 3 )ϑ1(v, q)ϑ1(v + u, q)ϑ1(v − u, q).

40 Dividing (4.18) by (4.19), we see that

C2(v, q)ϑ1(v, q) C2(u, q)ϑ1(u, q) 1 = 1 . C1(v, q)ϑ1(v, q 3 ) C1(u, q)ϑ1(u, q 3 )

This means that the expression C2(x, q)ϑ1(x, q) 1 C1(x, q)ϑ1(x, q 3 ) must be independent of x and thus it is an expression in terms of q. Let

C2(v, q)ϑ1(v, q) A(q) = 1 , C1(v, q)ϑ1(v, q 3 ) we have 1 C1(v, q)ϑ1(v, q 3 ) C2(v, q) = A(q) ϑ1(v, q) and by substituting it into (4.18) we get

3 1 1 3 A(q) −ϑ1(3v, q )ϑ1(u, q 3 ) + ϑ1(v, q 3 )ϑ1(3u, q )

= ϑ1(u, q)ϑ1(v, q)ϑ1(u + v, q)ϑ1(u − v, q).

3 1 1 3 Let G1(v) = −ϑ1(3v, q )ϑ1(u, q 3 ) + ϑ1(v, q 3 )ϑ1(3u, q ), G2(v) = ϑ1(u, q)ϑ1(v, q)ϑ1(u +

v, q)ϑ1(u − v, q). To get the value of A(q), we consider the coeﬃcients of v of G1(v) and

G2(v) respectively, and deduce that 1 0 3 1 0 1 3 3 0 A(q) −3ϑ (u, q 3 )ϑ (0, q ) + ϑ (0, q 3 )ϑ (3u, q ) = ϑ (u, q)ϑ (0, q). 1 1 3 1 1 1 1

Now from lemma(3.3.2), we have

0 ϑ1(0, q) = ϑ2(0, q)ϑ3(0, q)ϑ4(0, q) ∞ ∞ ∞ 1 Y n n 2 Y n n− 1 2 Y n n− 1 2 = 2q 8 (1 − q )(1 + q ) (1 − q )(1 + q 2 ) (1 − q )(1 − q 2 ) n=1 n=1 n=1 ∞ 1 Y n n 2 n 2 2n−1 2 = 2q 8 (1 − q )(1 − q ) (1 + q ) (1 − q ) n=1 ∞ (4.20) 1 Y n 2n 2 2n−1 2 = 2q 8 (1 − q )(1 − q ) (1 − q ) n=1 ∞ 1 Y n n 2 = 2q 8 (1 − q )(1 − q ) n=1 ∞ 1 Y 1 8 n 3 8 3 = 2q (1 − q ) = 2q (q ; q)∞. n=1

41 Also we have

∞ π 1 Y n 2n ϑ ( , q) = 2(q ; q) q 8 sin(π/3) (1 + q + q ) 1 3 ∞ n=1 (4.21) √ ∞ √ 1 Y 3n 1 3 3 = 3q 8 (1 − q ) = 3q 8 (q ; q )∞. n=1

π 3 Therefore by letting u = 3 , and using the identities (4.20), (4.21) and ϑ1(π, q ) = 0, we get

1 12 2 A(q) = −q (q ; q)∞

and hence proved identity (4.16).

42 Chapter 5

A Theta Function Identity and Its Application

In this chapter, we will discuss a theta-function identity due to Liu [14], and use it to conjecture a relation between the doubly-periodic functions smu and cmu deﬁned by Dixon in [5], and the theta functions. The proof of the theta-function identity given in this chapter is in the spirit of the proofs in the previous chapter, and is diﬀerent to that of Liu in [14].

5.1 The theta-function identity

Theorem 5.1.1 (c.f.[14]) Let ϑ1(z, q) be the theta function as deﬁned in (3.2). Then we have π π ϑ3(z, q) − ϑ3(z + , q) − ϑ3(z − , q) 1 1 3 1 3 3 3 (q ; q )∞ π π = 3a(q) 3 ϑ1(z, q)ϑ1(z + , q)ϑ1(z − , q), (q; q)∞ 3 3 where ∞ Y n−1 (z; q)∞ = (1 − zq ) n=1 and ∞ X q3n+1 q3n+2 a(q) = 1 + 6 − . 1 − q3n+1 1 − q3n+2 n=0

43 Proof

We ﬁrst consider the functions

π π F (z) = ϑ3(z, q) − ϑ3(z + , q) − ϑ3(z − , q), 1 1 1 3 1 3 π π F (z) = ϑ (z, q)ϑ (z + , q)ϑ (z − , q). 2 1 1 3 1 3

Using the transformation formulas of ϑ1(z, q) in (3.8) and (3.9), we see that both F1(z) and F2(z) satisfy the transformation formulas

Fi(z + π) = −Fi(z),

− 3 −6iz Fi(z + πτ) = −q 2 e Fi(z), i = 1, 2.

Therefore the function F1(z)/F2(z) is an elliptic function with periods π and πτ. Thus we consider the cell C with corners t, t + π, t + π + πτ and t + πτ, the complex constant t chosen such that the sides of the cell do not contain any of the possible zeros and poles of the function F1(z)/F2(z).

From theorem 3.2.1, we know that F2(z) has only three simple zeros in C, namely z = 0, z = π/3 and z = −π/3. Moreover, for z = 0, we have π π F (0) = ϑ3(0) − ϑ3( ) − ϑ3(− ) 1 1 1 3 1 3 π π = −ϑ3( ) + ϑ3( ) = 0, 1 3 1 3 and for z = −π/3, we have π π 2π F (− ) = ϑ3(− ) − ϑ3(0) − ϑ3(− ) 1 3 1 3 1 1 3 π 2π = ϑ3(− ) + ϑ3(− + π) 1 3 1 3 π π = −ϑ3( ) + ϑ3( ) = 0. 1 3 1 3

This implies the function F1(z) vanishes at z = 0 and z = −π/3, and therefore the function

F1(z)/F2(z) has at most one pole in C, and by theorem 2.2.1(iii), it must be a constant independent of z. Denoting this constant as A(q), we thus have

F1(z) = A(q)F2(z). (5.1)

44 To ﬁnd A(q), we just need to consider the constant term in the z-expansion of F1(z)/F2(z). Using Maclaurin’s theorem [16, p.g 127], the constant term of the expansion is given as

2 0 2 π 0 π 2 π 0 π 2 π 0 π 3ϑ1(0)ϑ1(0) − 3ϑ1( 3 )ϑ1( 3 ) − 3ϑ1( 3 )ϑ1( 3 ) −6ϑ1( 3 )ϑ1( 3 ) π 0 π π 0 π 2 π 0 = 2 π 0 ϑ1(0){ϑ1( 3 )ϑ1( 3 ) − ϑ1( 3 )ϑ1( 3 )} − ϑ1( 3 )ϑ1(0) −ϑ1( 3 )ϑ1(0) 0 π 6ϑ1( 3 ) = 0 , ϑ1(0)

and by comparing coeﬃcients, we get

0 π 6ϑ1( 3 ) A(q) = 0 . ϑ1(0)

Now using the product identity of ϑ1(z) as given in theorem 3.3.1, applying logarithmic diﬀerentiation, we get

∞ ϑ0 (z) X qn sin 2z 1 = cot z + 4 . ϑ (z) 1 − 2qn cos 2z + q2n 1 n=1

Now by the product identity of ϑ(z) given in Theorem 3.3.1, we have

∞ π 1 π Y n 2π 2n ϑ ( ) = 2(q ; q) q 8 sin (1 − 2q cos + q ) 1 3 ∞ 3 3 n=1 √ ∞ 1 Y n n 2n = 3q 8 (1 − q )(1 + q + q ) n=1 √ 1 3 3 = 3q 8 (q ; q )∞,

45 and so we get √ ∞ n 3 ! √ 0 π 1 X q 2 1 3 3 ϑ ( ) = √ + 4 3q 8 (q ; q ) 1 3 1 + qn + q2n ∞ 3 n=1 ∞ n ! X q 1 3 3 = 1 + 6 q 8 (q ; q ) 1 + qn + q2n ∞ n=1 ∞ n 2n ! 1 3 3 X q − q = q 8 (q ; q ) 1 + 6 ∞ 1 − q3n n=1 ∞ ∞ ! 1 3 3 X n 2n X 3nk = q 8 (q ; q )∞ 1 + 6 (q − q ) q n=1 k=0 ∞ ∞ ! 1 3 3 X X (3k+1)n (3k+2)n = q 8 (q ; q )∞ 1 + 6 q − q n=1 k=0 ∞ ∞ ! 1 3 3 X X (3k+1)(n+1) (3k+2)(n+1) = q 8 (q ; q )∞ 1 + 6 q − q k=0 n=0 ∞ ∞ ∞ !! 1 3 3 X 3k+1 X (3k+1)n 3k+2 X (3k+2)n = q 8 (q ; q )∞ 1 + 6 q q − q q k=0 n=0 n=0 ∞ 3k+1 3k+2 ! 1 3 3 X q q = q 8 (q ; q ) 1 + 6 − ∞ 1 − q3k+1 1 − q3k+2 k=0 1 3 3 = q 8 (q ; q )∞a(q).

Using the identity

1 0 8 3 ϑ1(0) = 2q (q ; q)∞, we will have 3 3 (q ; q )∞ A(q) = 3a(q) 3 , (q ; q)∞

and from (5.1), we conclude that π π ϑ3(z, q) − ϑ3(z + , q) − ϑ3(z − , q) 1 1 3 1 3 3 3 (q ; q )∞ π π = 3a(q) 3 ϑ1(z, q)ϑ1(z + , q)ϑ1(z − , q). (q; q)∞ 3 3 2

Note that the constant 3 3 (q ; q )∞ 3 (q; q)∞

46 is 1 , b(q) where b (q) is deﬁned as

∞ X 2 2 b(q) := ωm−nqm +mn+n , ω = exp(2πi/3). m,n=−∞

It is the constant that appeared in the Borwein’s functions [4, p.g 93].

5.2 A conjecture on the relation between Dixon’s doubly- periodic functions and the theta functions

We will now use the identity shown in Theorem 5.1.1 to conjecture possible algebraic expressions for the Dixon’s functions smu and cmu.

Definition 5.2.1 (c.f [5]) smu is a function of u, where u is a complex number, defined by the differential equation d smu = cm2u − αsm u du where cm u is another function of u, given in terms of sm u by the equation

cm3u + sm3u − 3α sm u cm u = 1, where α is a constant parameter. Also when u = 0, the value of smu and cmu is 0 and 1 respectively.

In [5], Dixon showed that the two functions smu and cmu shared many similar properties to the Jacobian elliptic functions snu and cnu [16, p.g 491]. Because the Dixon’s functions smu and cmu are deﬁned to satisfy the equation

cm3u + sm3u − 3α sm u cm u = 1, it is possible to use the identity given in theorem 5.1.1 to conjecture possible algebraic expressions for the two functions.

47 First we rewrite the identity in theorem 5.1.1 as 3 π 3 ϑ1(z, q) ϑ1(z − 3 , q) π − 1 + − π ϑ1(z + 3 , q) ϑ1(z + 3 , q) 3 3 π (q ; q )∞ ϑ1(z, q) ϑ1(z − 3 , q) = 3a(q) − 3 π − π , (q; q)∞ ϑ1(z + 3 , q) ϑ1(z + 3 , q) which implies that ϑ (z, q) 3 ϑ (z − π , q)3 1 + − 1 3 ϑ (z + π , q) ϑ (z + π , q) 1 3 1 3 (5.2) 3 3 π (q ; q )∞ ϑ1(z, q) ϑ1(z − 3 , q) − 3a(q) − 3 π − π = 1. (q; q)∞ ϑ1(z + 3 , q) ϑ1(z + 3 , q)

From the deﬁnition of the Dixon’s functions smu and cmu, we have

cm3u + sm3u − 3α cm u sm u = 1, (5.3) and that sm(0) = 0, cm(0) = 1. So comparing the two equations (5.2) and (5.3), and the fact that sm(0) = 0, cm(0) = 1, we conjecture that the algebraic expression of smu is simi-

π lar to ϑ1(z, q)/(ϑ1(z + 3 , q)), with the constant term in the z-expansion of ϑ1(z, q)/(ϑ1(z + π π π 3 , q)) being 0, and the algebraic expression of cmu is similar to −ϑ1(z− 3 , q)/(ϑ1(z+ 3 , q)), π π with the constant term in the z-expansion of −ϑ1(z − 3 , q)/(ϑ1(z + 3 , q)) being 1.

Now in [5], the u-expansion of smu and cmu is given as 1 1 cm u = 1 + αu + α2u2 + (α3 − 2)u3 + ... 2 6 (5.4) 1 1 1 sm u = u + αu2 + α2u3 + (5α3 − 4)u4 + .... 2 2 24 It turns out that if we set ϑ1(π/3) z = 0 u, ϑ1(0) and let

ϑ1(z, q) s(u) = π ϑ1(z + 3 , q) π ϑ1(z − 3 , q) and c(u) = − π , ϑ1(z + 3 , q) then s(u) and c(u) have the same expansions as that of smu and cmu in (5.4) respectively, with the parameter α given as 3 3 (q ; q )∞ α = a(q) − 3 . (q; q)∞

48 We conjecture that s(u) and c(u) are possible algebraic expressions for smu and cmu respectively.

Note that there is another possible conjecture for the algebraic expressions of smu and cmu. If we write the identity in theorem 5.1.1 as

3 π 3 ϑ1(z, q) ϑ1(z + 3 , q) π + − π − 1 ϑ1(z − 3 , q) ϑ1(z − 3 , q) 3 3 π (q ; q )∞ ϑ1(z, q) ϑ1(z + 3 , q) = 3a(q) − 3 π − π , (q; q)∞ ϑ1(z − 3 , q) ϑ1(z − 3 , q)

π we will relate smu with ϑ1(z, q)/(ϑ1(z − 3 , q)), the constant term of the z-expansion of π π π ϑ1(z, q)/(ϑ1(z − 3 , q)) being 0, and relate cmu with ϑ1(z + 3 , q)/(ϑ1(z − 3 , q)), the constant π π term of the z-expansion of ϑ1(z + 3 , q)/(ϑ1(z − 3 , q)) being 1. If we now set

ϑ1(π/3) z = − 0 u, ϑ1(0) and let

ϑ1(z, q) s1(u) = π ϑ1(z − 3 , q) π ϑ1(z + 3 , q) and c1(u) = − π , ϑ1(z − 3 , q) it turns out that s1(u) and c1(u) also have the same expansions as that of smu and cmu in (5.4) respectively, with the parameter α given as

3 3 (q ; q )∞ α = a(q) − 3 . (q; q)∞

We conjecture that s1(u) and c11(u) are also possible algebraic expressions for smu and cmu respectively.

49 Bibliography

[1] G.E. Andrews, The Theory of Partitions, Addison-Wesley Pub. Com., 1976.

[2] T.M. Apostol, Modular Functions and Dirichlet Series in Number Theory, 2nd ed., Springer-Verlag, 1990.

[3] B.C. Berndt, Ramanujan’s Notebooks Part III, Springer-Verlag, 1991.

[4] B.C. Berndt, Ramanujan’s Notebooks Part V, Springer-Verlag, 1998.

[5] A.C. Dixon, On the Doubly Periodic Functions Arising out of the Curve x3 + y3 − 3αxy = 1, The Quarterly J. of Pure and Applied Math. 24 (1890), 167-223.

[6] L. Euler, Introductio in Analysin Inﬁnitorum, Lausanne, 1748.

[7] H.M. Farka, and I. Kra, On the Quintuple Product Identity, Proc. Amer. Math. Soc. 27 (1999), 771-778.

[8] D. Foata, and G.-N. Han, The Triple, Quintuple and Septuple Product Identities Re- visited, Seminaire Lotharingien de Combinatoire, vol 42 (1999), 1-12.

[9] M.D. Hirschhorn, A Simple Proof of an Identity of Ramanujan, J. Austral. Math. Soc. Ser. A 34 (1983), 31-35.

[10] R.W. Howell, and J.H. Mathews, Complex Analysis: for Mathematics and Engineer- ing, 4th ed., Jones and Bartlett Pub. Inc., 2001.

[11] C.G.J. Jacobi, Fundamenta Nova Theoriae Functionum Ellipticarum, Konigsberg, 1829.

50 [12] C.G.J. Jacobi, Ges. Werke, 1838.

[13] S.-Y. Kang, A New Proof of Winquist Identity, J. Comb. Theory, Ser. A 78 (1997), 313-318.

[14] Z.-G. Liu, A Theta Function Identity and its Implications, to appear.

[15] H.A. Schwarz, Formeln und Lehrsatze zum Gebrauche der Elliptischen Funkionen. Nach Vorlesungen und Aufzeichnungen des Herrn Prof. K. Weierstrass, Zweite Aus- gabe, Erste Abteilung, Springer, Berlin, 1893.

[16] E.T. Whittaker, and G.N. Watson, A Course of Modern Analysis, 4th ed., Cambridge University Press, Cambridge, 1966.

[17] L.Winquist, An Elementary Proof of p(n) ≡ 0(mod11), J.Comb. Theory 6 (1969), 56-59.u