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Elliptic Functions, Theta Functions
and Identities
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
5.1
The theta-function identity . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2
A conjecture on the relation between Dixon’s doubly-periodic functions and
the theta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
43
43
47
Chapter 1
Introduction
The first function related to theta functions is perhaps the function
∞
(1 − xn z)−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
∞
(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 + y 3 − 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 coordinates 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 periodparallelograms 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
.
f (z)
is infinite, we consider the reciprocal function
Since f (z) is elliptic, so is
1
,
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
2πi
f (z) dz
C
1
2πi
1
=
2πi
1
=
2πi
t+2ω1
t+2ω1 +2ω2
f (z) dz +
=
t
t+2ω1
t+2ω2
t+2ω1
f (z) dz +
t
t+2ω1
t+2ω2
t
f (z) dz +
f (z) dz +
t+2ω1 +2ω2
t
f (z + 2ω1 ) dz +
t
{f (z) − f (z + 2ω2 )} dz −
t
t
f (z + 2ω2 ) dz +
t+2ω1
1
2πi
f (z) dz
t+2ω2
f (z) dz
t+2ω2
t+2ω2
{f (z) − f (z + 2ω1 )} dz
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. ✷
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
℘(z) =
1
+
z2
m,n
1
1
−
(z − 2mω1 − 2nω2 )2 (2mω1 + 2nω2 )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
℘(z) = z −2 +
(z − Ωm,n )−2 − Ω−2
m,n .
m,n
When m and n are such that |Ωm,n | is large, the general term of the series defining ℘(z)
is O(|Ωm,n |−3 ), 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;
℘ (z) = −2
m,n
1
.
(z − Ωm,n )3
It is clear that ℘ (z) is an odd function, whereas ℘(z) is an even function. Also we have
℘ (z + 2ω1 ) = ℘ (z),
℘ (z + 2ω2 ) = ℘ (z),
since the sets of points (Ωm,n − 2ω1 ) and (Ωm,n − 2ω2 ) are both the same as the set Ωm,n
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
∞
1
2
(−1)n q 2 n e2niz ,
ϑ4 (z, q) =
(3.1)
n=−∞
1
1
ϑ1 (z, q) = −ieiz+ 4 πiτ ϑ4 (z + πτ, q)
2
∞
1
1 2
(3.2)
(−1)n q 2 (n+ 2 ) e(2n+1)iz ,
= −i
n=−∞
1
ϑ2 (z, q) = ϑ1 (z + π, q)
2
∞
1
1 2
(3.3)
q 2 (n+ 2 ) e(2n+1)iz
=
n=−∞
and
1
ϑ3 (z, q) = ϑ4 (z + π, q)
2
∞
1
2
q 2 n e2niz .
=
n=−∞
8
(3.4)
Let A be any positive constant. For any complex number z such that |z| ≤ A, we have
1
1
2
2
| q 2 n e±2niz | ≤ | q| 2 n e2nA ,
where n is a positive integer. Thus we have
∞
∞
n
| (−1) q
1 2
n
2
2niz
e
∞
|=
|q
1 2
n
2
2niz
e
n=−∞
n=−∞
1
2
| q| 2 n e2nA .
|≤
n=−∞
∞
n=−∞
Now the d’Alembert’s ratio [16, p.g 22] for the series
∞
n=−∞
which tends to zero as n → ∞. Therefore we have
1
2
1
| q| 2 n e2nA is | q|n+ 2 e2A ,
1
2
| q| 2 n e2nA being absolutely
convergent, which implies that the series
∞
1
2
(−1)n q 2 n e2niz
ϑ4 (z, q) =
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
∞
1
2
(−1)n q 2 n cos 2nz,
ϑ4 (z, q) = 1 + 2
n=1
∞
1
1 2
(−1)n q 2 (n+ 2 ) sin(2n + 1)z,
ϑ1 (z, q) = 2
n=0
∞
ϑ2 (z, q) = 2
(3.5)
q
1
(n+ 12 )2
2
cos(2n + 1)z,
n=0
∞
1
2
q 2 n cos 2nz.
and ϑ3 (z, q) = 1 + 2
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
∞
1
2
1
2
1
2
(−1)n q 2 n e2ni(z+π)
ϑ4 (z + π, q) =
n=−∞
∞
(−1)n q 2 n e2niz e2niπ
=
(3.6)
n=−∞
∞
(−1)n q 2 n e2niz = ϑ4 (z, q),
=
n=−∞
∞
1
2
1
2
(−1)n q 2 n e2ni(z+πτ )
and ϑ4 (z + πτ, q) =
n=−∞
∞
(−1)n q 2 n q n e2niz
=
(3.7)
n=−∞
∞
1
1
= −q − 2 e−2iz
2
(−1)n+1 q 2 (n+1) e2(n+1)iz
n=−∞
1
= −q − 2 e−2iz ϑ4 (z, q).
For ϑ1 (z, q), we have
∞
1
1 2
1
1 2
(−1)n q 2 (n+ 2 ) e(2n+1)i(z+π)
ϑ1 (z + π, q) = −i
n=−∞
∞
(−1)n q 2 (n+ 2 ) e(2n+1)iz e(2n+1)iπ
= −i
(3.8)
n=−∞
∞
1
1 2
(−1)n q 2 (n+ 2 ) e(2n+1)iz = −ϑ1 (z, q),
=i
n=−∞
∞
1
1 2
1
1 2
(−1)n q 2 (n+ 2 ) e(2n+1)i(z+πτ )
and ϑ1 (z + πτ, q) = −i
n=−∞
∞
1
(−1)n q 2 (n+ 2 ) q n+ 2 e(2n+1)iz
= −i
(3.9)
n=−∞
∞
1
1
= iq − 2 e−2iz
n=−∞
= −q
− 12
1 2
(−1)n+1 q 2 ((n+1)+ 2 ) e(2(n+1)+1)iz
e−2iz ϑ1 (z, q).
10
For ϑ2 (z, q), we have
∞
1
1 2
1
1 2
q 2 (n+ 2 ) e(2n+1)i(z+π)
ϑ2 (z + π, q) =
n=−∞
∞
q 2 (n+ 2 ) e(2n+1)iz e(2n+1)iπ
=
(3.10)
n=−∞
∞
1 2
1
q 2 (n+ 2 ) e(2n+1)iz = −ϑ2 (z, q),
=−
n=−∞
∞
1
1 2
1
1 2
q 2 (n+ 2 ) e(2n+1)i(z+πτ )
and ϑ2 (z + πτ, q) =
n=−∞
∞
1
q 2 (n+ 2 ) q n+ 2 e(2n+1)iz
=
(3.11)
n=−∞
∞
1
1
= q − 2 e−2iz
1 2
q 2 ((n+1)+ 2 ) e(2(n+1)+1)iz
n=−∞
=q
− 12
e−2iz ϑ2 (z, q).
For ϑ3 (z, q), we have
∞
1
2
1
2
1
2
q 2 n e2ni(z+π)
ϑ3 (z + π, q) =
n=−∞
∞
q 2 n e2niz e2niπ
=
(3.12)
n=−∞
∞
q 2 n e2niz = ϑ3 (z, q),
=
n=−∞
∞
1
2
1
2
q 2 n e2ni(z+πτ )
and ϑ3 (z + πτ, q) =
n=−∞
∞
q 2 n q n e2niz
=
(3.13)
n=−∞
∞
1
1
= q − 2 e−2iz
2
q 2 (n+1) e2(n+1)iz
n=−∞
=q
− 12
e−2iz 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 (z, q)
be its derivative with respect to z. Then
ϑ (z + π, q)
ϑ (z, q)
=
,
ϑ(z + π, q)
ϑ(z, q)
ϑ (z + πτ, q)
ϑ (z, q)
= −2i +
.
ϑ(z + πτ, q)
ϑ(z, q)
11
(3.14)
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
ϑ4 (z + π, q) = ϑ4 (z, q), and thus we have
ϑ4 (z + π, q)
ϑ4 (z, q)
=
.
ϑ4 (z + π, q)
ϑ4 (z, q)
1
From equation (3.7), we have ϑ4 (z + πτ ) = −q − 2 e−2iz ϑ4 (z, q). This implies that
1
1
ϑ4 (z + πτ ) = −q − 2 e−2iz ϑ4 (z, q) + q − 2 (2i)e−2iz ϑ4 (z, q),
and thus
ϑ4 (z + πτ, q)
ϑ4 (z, q)
= −2i +
ϑ4 (z + πτ, q)
ϑ4 (z, q)
✷
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
ϑ4 (z) by 1 or −q − 2 e−2iz respectively. In the case of ϑ1 (z), the effect of increasing z by π
1
or πτ is the same as multiplying ϑ1 (z) by −1 or −q − 2 e−2iz respectively. ϑ2 (z) and ϑ3 (z)
also exhibit similar properties. Because of this, the theta functions are known as quasi
1
doubly-periodic functions of z, and ±1 and −q − 2 e−2iz 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
2πi
C
ϑ (z)
dz.
ϑ(z)
Using the results from (3.14), we have
1
2πi
C
ϑ (z)
1
dz =
ϑ(z)
2πi
1
=
2πi
t+π
t
ϑ (z) ϑ (z + πτ )
−
ϑ(z)
ϑ(z + πτ )
dz −
1
2πi
t+πτ
t
ϑ (z) ϑ (z + π)
−
ϑ(z)
ϑ(z + π)
t+π
2i dz
t
=1
Therefore ϑ(z) has one and only one simple zero inside C. ✷
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
conclude that the zeros of ϑ2 (z, q), ϑ3 (z, q), ϑ4 (z, q) are of the form
1
π
2
1
π
2
+ mπ + nπτ ,
+ 12 πτ + mπ + nπτ , 12 πτ + mπ + nπτ respectively, m and n being integers.
13
dz
3.3
Infinite products expressions for the theta functions
Theorem 3.3.1 For |q| < 1, we have
∞
∞
(1 − q
ϑ4 (z) = G
n− 12 2iz
1
(1 − q n− 2 e−2iz )
e )
n=1
∞
n=1
1
(1 − 2q n− 2 cos 2z + q 2n−1 ),
=G
n=1
∞
∞
1
8
(1 − q n e−2iz )
n 2iz
(1 − q e )
ϑ1 (z) = 2Gq sin z
n=1
∞
1
n=1
(1 − 2q n cos 2z + q 2n ),
= 2Gq 8 sin z
n=1
∞
∞
1
(1 + q n e−2iz )
(1 + q n e2iz )
ϑ2 (z) = 2Gq 8 cos z
n=1
∞
1
n=1
(1 + 2q n cos 2z + q 2n ),
= 2Gq 8 cos z
n=1
∞
∞
1
1
(1 + q n− 2 e−2iz )
(1 + q n− 2 e2iz )
and ϑ3 (z) = G
n=1
∞
n=1
1
(1 + 2q n− 2 cos 2z + q 2n−1 ),
=G
n=1
where G is a constant independent of z.
Proof
Let f (z) be a function defined as follows:
∞
∞
1
1
(1 − q n− 2 e−2iz ).
(1 − q n− 2 e2iz )
f (z) =
n=1
n=1
1
1
First we see that | − q n− 2 e±2iz | ≤ | q n− 2 |eA in any bounded domain of values of z and
for n >> 0, and the series
∞
n=1
1
q n− 2 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
∞
∞
(1 − q
f (z + π) =
n− 21 2i(z+π)
e
n=1
∞
n=1
∞
1
1
(1 − q n− 2 e−2iz )
(1 − q n− 2 e2iz )
=
1
(1 − q n− 2 e−2i(z+π) )
)
n=1
n=1
= f (z),
and
∞
∞
(1 − q
f (z + πτ ) =
(1 − q n− 2 e−2iz )
e )
n=1
=
3
n+ 12 2iz
n=1
1−q
− 12 −2iz
e
1
2
1 − q e2iz
f (z)
1
= −q 2 e−2iz f (z).
This shows that f (z) is quasi doubly-periodic with periods π and πτ .
1
Now if we differentiate the equations f (z + π) = f (z) and f (z + πτ ) = −q 2 e−2iz f (z) with
respect to z, we get
f (z + π) = f (z)
and
1
1
f (z + πτ ) = −q − 2 e−2iz f (z) + q − 2 (2i)e−2iz f (z),
which imply that
f (z + π)
f (z)
=
,
f (z + π)
f (z)
(3.15)
f (z + πτ )
f (z)
= −2i +
.
f (z + πτ )
f (z)
(3.16)
and
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
2πi
C
f (z)
dz,
f (z)
15
and using the results from (3.15) and (3.16), we have
1
2πi
C
f (z)
1
dz =
f (z)
2πi
1
=
2πi
t+π
t
f (z) f (z + πτ )
−
f (z)
f (z + πτ )
dz −
1
2πi
t+πτ
t
f (z) f (z + π)
−
f (z)
f (z + π)
t+π
2i dz
t
=1
This shows that f (z) has one and only one simple zero inside C.
We now consider the zeros of f (z). It is easy to verify that 12 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 + m1 πτ + m2 π,
2
where m1 and m2 are integers. Therefore f (z) and ϑ4 (z) have the same zeros; and consequently the quotient
ϑ4 (z)
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
respect to periods π and πτ . This means that
ϑ4 (z)
f (z)
is a doubly-periodic, integral function
with no zeros or poles, and so by Theorem 2.2.1 (iv),
ϑ4 (z)
f (z)
is a constant independent of z,
denoted as G, i.e.
ϑ4 (z) = Gf (z)
∞
1
(1 − 2q n− 2 cos 2z + q 2n−1 ).
=G
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
dz
1
1
ϑ1 (z) = −ieiz+ 4 πiτ ϑ4 (z + πτ )
2
1
1
= −iq 8 eiz ϑ4 (z + πτ )
2
∞
1
8
iz
∞
(1 − q n−1 e−2iz )
n 2iz
= −iq e G
(1 − q e )
n=1
∞
1
n=1
∞
(1 − q n e−2iz )
(1 − q n e2iz )
= 2Gq 8 sin z
n=1
∞
1
n=1
(1 − 2q n cos 2z + q 2n ),
= 2Gq 8 sin z
n=1
1
ϑ2 (z) = ϑ1 (z + π)
2
∞
1
8
∞
(1 + q n e−2iz )
n 2iz
= 2Gq cos z
(1 + q e )
n=1
∞
1
n=1
(1 + 2q n cos 2z + q 2n ),
= 2Gq 8 cos z
n=1
1
and ϑ3 (z) = ϑ4 (z + π)
2
∞
=G
∞
(1 + q
n− 12 2iz
1
(1 + q n− 2 e−2iz )
e )
n=1
∞
n=1
1
(1 + 2q n− 2 cos 2z + q 2n−1 )
=G
n=1
✷
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:
ϑ1 (0) = ϑ2 (0)ϑ3 (0)ϑ4 (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)
2
= −4
n2 en πiτ +2niz
2
∂z
−∞
=−
4 ∂ϑ3 (z, q)
.
πi
∂τ
Therefore ϑ3 (z, q) satisfies the partial differential equation
1 ∂ 2 y ∂y
πi
+
= 0.
4 ∂z 2 ∂τ
(3.17)
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 −2 e−8iz f (z).
Therefore the function
f1 (z)
2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
=
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’ Hˆ
opital’s rule [10], we see that
lim
z→0
ϑ1 (z)
ϑ (0)
= 1
.
ϑ1 (2z)
2ϑ1 (0)
18
Therefore letting z → 0, we see that the constant c must be 1.
Therefore we have
ϑ1 (2z)
ϑ2 (z)ϑ3 (z)ϑ4 (z)
=
,
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
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)
+ 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
2ϑ2 (0)ϑ3 (0)ϑ4 (0)ϑ1 (2z)
= ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z) ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z) ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
= ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z)
+ 2ϑ1 (z)ϑ2 (z)ϑ3 (z)ϑ4 (z).
19
(3.20)
Now we divide (3.20) throughout by ϑ1 (z) to get
2ϑ2 (0)ϑ3 (0)ϑ4 (0)
=
ϑ1 (z)
ϑ1 (z)
+ 2ϑ1 (z)
ϑ1 (2z)
ϑ1 (z)
ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ2 (z)ϑ3 (z)ϑ4 (z) + ϑ2 (z)ϑ3 (z)ϑ4 (z)
ϑ2 (z)
ϑ1 (z)
ϑ3 (z)ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)
ϑ3 (z)
ϑ1 (z)
ϑ4 (z) + 2ϑ1 (z)ϑ2 (z)ϑ3 (z)
ϑ4 (z)
ϑ1 (z)
+ 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
ϑ1 (0) = ϑ2 (0) = ϑ3 (0) = ϑ4 (0) = 0.
So taking limit with z → 0 on both sides of (3.21), and using L’ Hˆ
opital’s rule [10], we get
4ϑ2 (0)ϑ3 (0)ϑ4 (0)
=
ϑ1 (0)
ϑ1 (0)
+ 2ϑ1 (0)
ϑ1 (0)
ϑ1 (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)
ϑ2 (0)
ϑ1 (0)
ϑ3 (0)ϑ4 (0) + 2ϑ1 (0)ϑ2 (0)
ϑ3 (0)
ϑ1 (0)
ϑ4 (0) + 2ϑ1 (0)ϑ2 (0)ϑ3 (0)
ϑ4 (0)
ϑ1 (0)
+ 2ϑ2 (0)ϑ3 (0)ϑ4 (0) + 2ϑ2 (0)ϑ3 (0)ϑ4 (0) + 2ϑ2 (0)ϑ3 (0)ϑ4 (0)
which implies
3ϑ2 (0)ϑ3 (0)ϑ4 (0)
ϑ1 (0)
ϑ1 (0)
(3.22)
= 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
ϑ1 (0)
ϑ (0) ϑ3 (0) ϑ4 (0)
= 2
+
+
.
ϑ1 (0)
ϑ2 (0) ϑ3 (0) ϑ4 (0)
Utilizing the differential equation given in (3.17), we have
4 dϑ1 (z, q)
,
πi
dτ
4 dϑ2 (z, q)
ϑ2 (z) = −
,
πi
dτ
4 dϑ3 (z, q)
,
ϑ3 (z) = −
πi
dτ
4 dϑ4 (z, q)
and ϑ4 (z) = −
,
πi
dτ
ϑ1 (z) = −
20
(3.23)
and so (3.23) can be written as
1
dϑ1 (0, q)
ϑ1 (0, q) dτ
1
dϑ2 (0, q)
1
dϑ3 (0, q)
1
dϑ4 (0, q)
=
+
+
.
ϑ2 (0, q) dτ
ϑ3 (0, q) dτ
ϑ4 (0, q) dτ
Integrating with respect to τ , we get
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
8
1
1 2
(−1)n q 2 (n+ 2 ) (2n + 1),
ϑ1 (0, q) = 2q + 2
n=1
∞
1
1
1 2
q 2 (n+ 2 ) ,
ϑ2 (0, q) = 2q 8 + 2
n=1
∞
1
2
q 2n
ϑ3 (0, q) = 1 + 2
n=1
∞
1
2
(−1)n q 2 n .
and ϑ4 (0, q) = 1 + 2
n=1
Therefore by letting q → 0, we will have
1
1
lim q − 8 ϑ1 (0) = 2, lim q − 8 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
ϑ1 (0) = ϑ2 (0)ϑ3 (0)ϑ4 (0).
✷
Theorem 3.3.3 Let G be as defined in theorem 3.3.1. We have
∞
(1 − q n ).
G=
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)),
∞
n=1 (1
where φ(z) = 2Gq 1/8
− 2q n cos 2z + q 2n ). We then have
∞
1
8
(1 − q n )2 ,
ϑ1 (0) = φ(0) = 2q G
n=1
∞
1
(1 + q n )2 ,
ϑ2 (0) = 2q 8 G
n=1
∞
1
(1 + q n− 2 )2 ,
ϑ3 (0) = G
n=1
∞
1
(1 − q n− 2 )2 .
ϑ4 (0) = G
n=1
By lemma 3.3.2, we then have
∞
∞
n 2
(1 − q ) = G
2
n=1
∞
∞
n 2
(1 + q )
n=1
(1 + q
n− 12 2
1
(1 − q n− 2 )2 .
)
n=1
n=1
Since
∞
∞
(1 − q
n− 21
∞
n
(1 − q )
)
n=1
∞
(1 + q
n=1
∞
1
n=1
∞
n=1
(1 + q n )
)
n=1
1
(1 − q 2 n )
=
∞
n− 21
(1 + q 2 n )
n=1
(1 − q n ),
=
n=1
we have
∞
2
(1 − q n )2
G =
n=1
which implies that G = ±
∞
n=1 (1
− q n ).
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
∞
(1 − q n )
G=
n=1
is established. ✷
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
∞
(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 = 0. We define
∞
(1 − zq n−1 ).
(z; q)∞ =
n=1
Then we have the following identity
∞
q
(3n2 +n)
2
(z 3n − z −3n−1 ) = (q; q)∞ (zq; q)∞ (1/z; q)∞ (z 2 q; q 2 )∞ (q/z 2 ; q 2 )∞
(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:
(q; q)∞
ϑ1 (2u, q)
πτ 3
πτ 3
= eiu ϑ4 3u +
, q + e−iu ϑ4 3u −
,q ,
ϑ1 (u, q)
2
2
(4.2)
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 = 0, we
can replace z by −1/x, x = ∞, to get
∞
q
(3n2 +n)
2
∞
3n
((−1/x)
−3n−1
− (−1/x)
)=
n=−∞
q
(3n2 +n)
2
q
(3n2 −n)
2
n=−∞
∞
=
∞
(3n2 +n)
(−1)n
2
+
q
(−1)n x3n+1
x3n
n=−∞
∞
n 3n
(−1) x
+
n=−∞
q
(3n2 +n)
2
(−1)n x3n+1 ,
n=−∞
the second equality follows since by replacing n by −n, we get
∞
q
(3n2 +n)
2
n=−∞
∞
(3n2 −n)
(−1)n
2
=
q
(−1)n x3n .
x3n
n=−∞
Before we look at the right hand side of (4.1), first we see that
∞
2
2
(1 − x2 q 2n−1 )
(x q; q )∞ =
n=1
has odd positive integral powers of q, and
∞
2
2
(1 − x2 q 2n−2 )
(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
(1 − x2 q 2n−1 )(1 − x2 q 2n−2 )
(x q; q )∞ (x ; q )∞ =
n=1
∞
(4.3)
2 n−1
(1 − x q
=
2
) = (x ; q)∞ .
n=1
Using similar argument we also have
∞
2
2
2
2
2
(1 − q 2n−1 /x2 )(1 − q 2n /x2 )
(q/x ; q )∞ (q /x ; q )∞ =
n=1
∞
(4.4)
n
2
2
(1 − q /x ) = (q/x ; q)∞ .
=
n=1
Also we have
∞
(1 − xq n−1 )(1 + xq n−1 )
(x; q)∞ (−x; q)∞ =
n=1
∞
(4.5)
2 2n−2
(1 − x q
=
n=1
25
2
2
) = (x ; q )∞ ,
and
∞
(1 − q n /x)(1 + q n /x)
(q/x; q)∞ (−q/x; q)∞ =
n=1
∞
(4.6)
(1 − q 2n /x2 ) = (q 2 /x2 ; q 2 )∞ .
=
n=1
So we now look at the right hand side of (4.1). Again we can replace z by −1/x to get
(q; q)∞ (zq; q)∞ (1/z; q)∞ (z 2 q; q 2 )∞ (q/z 2 ; q 2 )∞
= (q; q)∞ (x2 q; q 2 )∞ (q/x2 ; q 2 )∞ (−x; q)∞ (−q/x; q)∞
(q; q)∞ (x2 ; q)∞ (q/x2 ; q)∞ (−x; q)∞ (−q/x; q)∞
(using (4.3) and (4.4))
(x2 ; q 2 )∞ (q 2 /x2 ; q 2 )∞
(q; q)∞ (x2 ; q)∞ (q/x2 ; q)∞
=
(using (4.5) and (4.6))
(x; q)∞ (q/x; q)∞
n−1 2
n −2
x(1 − x)(1 − x−2 )(q; q)∞ ∞
x) ∞
n=1 (1 − q
n=1 (1 − q x )
=
∞
n−1 x)
n −1
(1 − x2 )(1 − x−1 ) ∞
n=1 (1 − q
n=1 (1 − q x )
∞
∞
x(q; q)∞ n=1 (1 − q n x2 ) n=1 (1 − q n−1 x−2 )
=
∞
∞
n
n−1 x−1 )
n=1 (1 − q x)
n=1 (1 − q
=
Then the quintuple product identity given in (4.1) become
∞
q
(3n2 −n)
2
∞
n 3n
(−1) x
n=−∞
+
q
(3n2 +n)
2
(−1)n x3n+1
n=−∞
=
n−1 −2
− q n x2 ) ∞
x )
n=1 (1 − q
,
∞
∞
n
n−1 x−1 )
n=1 (1 − q x)
n=1 (1 − q
x(q; q)∞
∞
n=1 (1
and letting x = e2iu and using the definition of the theta functions ϑ1 (z) and ϑ4 (z) given
in (3.1) and (3.2), we finally have
∞
q
(3n2 −n)
2
∞
(−1)n e2ni(3u) +
n=−∞
q
(3n2 +n)
2
(−1)n e2ni(3u) e2iu
n=−∞
=
n−1 −2i(2u)
− q n e2i(2u) ) ∞
e
)
n=1 (1 − q
,
∞
∞
n
2iu
n−1
−2iu
) n=1 (1 − q e
)
n=1 (1 − q e
e2iu (q; q)∞
∞
n=1 (1
which implies that
e−iu ϑ4 (3u −
πτ 3
πτ 3
, q ) + eiu ϑ4 (3u +
,q )
2
2
1
n 2i(2u)
n−1 −2i(2u)
(q; q)∞ (−i)q 8 e2iu ∞
) ∞
e
)
n=1 (1 − q e
n=1 (1 − q
,
=
1
∞
∞
iu
n
2iu
n−1
−2iu
(−i)q 8 e
) n=1 (1 − q e
)
n=1 (1 − q e
26
and thus we have
e−iu ϑ4 (3u −
πτ 3
πτ 3
ϑ1 (2u, q)
, q ) + eiu ϑ4 (3u +
, q ) = (q; q)∞
2
2
ϑ1 (u, q)
which is the equivalent form we desired. ✷
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)
, eiu ϑ4 (3u +
ϑ1 (u, q)
πτ
, q 3 ),
2
and e−iu ϑ4 (3u −
πτ 3
,q )
2
satisfy the transformation formula
g(u + π) = g(u).
The three functions also satisfy
3
g(u + πτ ) = −q − 2 e−6iu g(u),
since
ϑ1 ((2u + πτ ) + πτ, q)
ϑ1 (2(u + πτ ), q)
=
ϑ1 (u + πτ, q)
ϑ1 (u + πτ, q)
1
−q − 2 e−2i(2u+πτ ) ϑ1 (2u + πτ, q)
=
ϑ1 (u + πτ, q)
−1 −2i(2u+πτ ) −4iu
q e
e
ϑ1 (2u, q)
=
− 12 −2iu
−q e
ϑ1 (u, q)
3
ϑ1 (2u, q)
= −q − 2 e−6iu
,
ϑ1 (u, q)
ei(u+πτ ) ϑ4 (3(u + πτ ) +
1
πτ
πτ 3
, q ) = q 2 eiu ϑ4 ((3u +
) + 3πτ, q 3 )
2
2
3
1
= q 2 eiu (−q − 2 )e−2i(3u+
3
πτ
2
= −q − 2 e−6iu eiu ϑ4 (3u +
27
)
ϑ4 ((3u +
πτ 3
, q ),
2
πτ
), q 3 )
2
and
e−i(u+πτ ) ϑ4 (3(u + πτ ) −
1
πτ 3
πτ
, q ) = q − 2 e−iu ϑ4 ((3u −
) + 3πτ, q 3 )
2
2
1
3
πτ
2
= q − 2 e−iu (−q − 2 )e−2i(3u−
3
= −q − 2 e−6iu eiu ϑ4 (3u −
)
ϑ4 ((3u −
πτ
), q 3 )
2
πτ 3
, q ).
2
Using the above transformation formulas we see that the function
g1 (u) =
πτ 3
ϑ1 (u, q) iu
e ϑ4 (3u +
,q )
ϑ1 (2u, q)
2
g2 (u) =
ϑ1 (u, q) −iu
πτ 3
e ϑ4 (3u −
,q )
ϑ1 (2u, q)
2
and the function
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
u = 0 in cell C2 with corners t, t + π2 , t +
π+πτ
2
and t +
function ϑ1 (2u, q) has zeros congruent to u = 0, π2 , πτ
and
2
πτ
.
2
Therefore we see that the
π+πτ
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
only 3 possible simple poles congruent to u = π2 , πτ
and
2
π+πτ
2
in C1 .
Now the function
∞
eiu ϑ4 3u +
3n2 +n
πτ 3
πτ 3
, q + e−iu ϑ4 3u −
,q = 2
(−1)n q 2 cos((6n + 1)u)
2
2
n=−∞
28
vanishes at the point u = π2 . At the point u =
e
iπτ
2
3πτ
πτ 3
+
,q
2
2
ϑ4
+e
−iπτ
2
πτ
,
2
3πτ
πτ 3
−
,q
2
2
ϑ4
∞
=q
∞
1
4
n
(−1) q
3 2
n
2
4inπτ
e
+q
n=−∞
∞
2 +2n+ 1
4
2
3
2 −m− 1
4
(−1)m q 2 m
+
m=−∞
∞
n=−∞
∞
3
2 +2n+ 1
4
(−1)n q 2 n
2 −(m+1)− 1
4
3
(−1)m+1 q 2 (m+1)
+
m=−∞
∞
n=−∞
∞
3
2 +2n+ 1
4
(−1)n q 2 n
=
3
(−1)n q 2 m e2imπτ
m=−∞
3
=
− 14
∞
(−1)n q 2 n
=
we have
3
n=−∞
2 +2m+ 1
4
(−1)m+1 q 2 m
+
m=−∞
= 0,
and so the function also vanishes at u =
poles at u =
π
2
and u =
πτ
2
πτ
.
2
Therefore the function g1 (u) + g2 (u) has no
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
A(q) = 2g3−1 (0)ϑ4 (
πτ 3
, q ),
2
where
g3 (u) =
ϑ1 (2u, q)
.
ϑ1 (u, q)
Since
g3 (0) = 2,
∞
∞
πτ
ϑ4 ( , q 3 ) =
(1 − q 3n )
(1 − q 3(n−1/2) eiπτ )(1 − q 3(n−1/2) e−iπτ )
2
n=1
n=1
∞
∞
(1 − q 3n )
=
n=1
∞
∞
(1 − q 3n−1 )
n=1
(1 − q 3n−2 )
n=1
(1 − q n ) = (q ; q)∞ ,
=
n=1
we have A = (q ; q)∞ and the desired result is obtained. ✷
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 = 0. Defining
(z1 , z2 , z3 , . . . , zn ; q)∞ = (z1 ; q)∞ (z2 ; q)∞ (z3 ; q)∞ . . . (zn ; q)∞
we have the following identity
(q, q, x, q/x; q)∞ (qx2 , q/x2 , x2 , q 2 /x2 ; q 2 )∞
∞
(−1)m q
=
5m2 +m
2
m=−∞
(−1)k q
5k2 +3k
2
∞
x5k+3 +
k=−∞
∞
m
−
∞
(−1) q
5m2 +3m
2
m=−∞
(−1)k q
5k2 −3k
2
x5k
k=−∞
∞
k
(−1) q
k=−∞
5k2 +k
2
∞
x
5k+2
+
(−1)k q
5k2 −k
2
x5k+1
k=−∞
(4.7)
where x = 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
−q −1/4 ϑ1 (u, q)ϑ1 (2u, q)
∞
(−1)m q
=
m=−∞
∞
−
(−1)m q
5m2 +m
2
5m2 +3m
2
e−3iu ϑ4 (5u − 3πτ /2, q 5 ) + e3iu ϑ4 (5u + 3πτ /2, q 5 )
eiu ϑ4 (5u + πτ /2, q 5 ) + e−iu ϑ4 (5u − πτ /2, q 5 ) ,
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
(q, q, e2iu , qe−2iu ; q)∞ (qe4iu , qe−4iu , e4iu , q 2 e−4iu ; q 2 )∞
∞
(1 − q n )(1 − q n )(1 − e2iu q n−1 )(1 − e−2iu q n )(1 − e4iu q 2n−1 )
=
n=1
(1 − e−4iu q 2n−1 )(1 − e4iu q 2(n−1) )(1 − e−4iu q 2n )
∞
(1 − q n )(1 − q n )(1 − e2iu q n−1 )(1 − e−2iu q n )(1 − e4iu q n−1 )(1 − e−4iu q n )
=
n=1
∞
= (−q
− 41
1
8
3iu
−iu
(1 − q n )(1 − e2iu q n−1 )(1 − e−2iu q n )
−iq e
)e
n=1
∞
1
−iq 8 e−2iu
(1 − q n )(1 − e4iu q n−1 )(1 − e−4iu q n )
n=1
= (−q
− 41
)e3iu ϑ1 (−u, q)ϑ1 (−2u, q)
1
= (−q − 4 )e3iu ϑ1 (u, q)ϑ1 (2u, q),
and the right hand side of (4.7) can be re-written as
∞
m
(−1) q
m=−∞
m
(−1) q
(−1) q
5k2 +3k
2
∞
2ik(5u) 6iu
e
e
(−1)k q
+
k
(−1) q
5k2 +k
2
∞
2ik(5u) 4iu
e
e
+
k=−∞
∞
= e3iu
(−1)m q
m=−∞
∞
(−1)m q
5m2 +m
2
5m2 +3m
2
5k2 −3k
2
e2ik(5u)
k=−∞
∞
5m2 +3m
2
m=−∞
− e3iu
k
k=−∞
∞
−
∞
5m2 +m
2
(−1)k q
5k2 −k
2
e2ik(5u) e2iu
k=−∞
e3iu ϑ4 (5u + 3πτ /2, q 5 ) + e−3iu ϑ4 (5u − 3πτ /2, q 5 )
eiu ϑ4 (5u + πτ /2, q 5 ) + e−iu ϑ4 (5u − πτ /2, q 5 )
m=−∞
Combining the above computations, we deduce (4.8). ✷
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),
h2 (u) = e−3iu ϑ4 (5u − 3πτ /2, q 5 ),
h3 (u) = e3iu ϑ4 (5u + 3πτ /2, q 5 ),
h4 (u) = eiu ϑ4 (5u + πτ /2, q 5 ),
h5 (u) = e−iu ϑ4 (5u − πτ /2, q 5 )
satisfy the transformation formula
hj (u + π) = −hj (u), 1 ≤ j ≤ 5,
and the five functions also satisfy the transformation formula
5
hj (u + πτ ) = −q − 2 e−10iu hj (u), 1 ≤ j ≤ 5,
since
h1 (u + πτ ) = ϑ1 (u + πτ, q)ϑ1 (2(u + πτ ), q)
1
1
= −q − 2 e−2iu ϑ1 (u, q)(−q − 2 )e−2i(2u+πτ ) ϑ1 (2u + πτ, q)
1
= −q − 2 e−2iu ϑ1 (u, q)q −2 e−8iu ϑ1 (2u, q)
5
5
= −q − 2 e−10iu ϑ1 (u, q)ϑ1 (2u, q) = −q − 2 e−10iu h1 (u),
h2 (u + πτ ) = e−3i(u+πτ ) ϑ4 (5(u + πτ ) − 3πτ /2, q 5 )
3
5
= q − 2 e−3iu (−q − 2 )e−2i(5u−3πτ /2) ϑ4 (5u − 3πτ /2, q 5 )
5
5
= −q − 2 e−10iu e−3iu ϑ4 (5u − 3πτ /2, q 5 ) = −q − 2 e−10iu h2 (u),
h3 (u + πτ ) = e3i(u+πτ ) ϑ4 (5(u + πτ ) + 3πτ /2, q 5 )
3
5
= q 2 e3iu (−q − 2 )e−2i(5u+3πτ /2) ϑ4 (5u + 3πτ /2, q 5 )
5
5
= −q − 2 e−10iu e3iu ϑ4 (5u + 3πτ /2, q 5 ) = −q − 2 e−10iu h3 (u),
h4 (u + πτ ) = ei(u+πτ ) ϑ4 (5(u + πτ ) + πτ /2, q 5 )
1
5
= q 2 eiu (−q − 2 )e−2i(5u+πτ /2) ϑ4 (5u + πτ /2, q 5 )
5
5
= −q − 2 e−10iu eiu ϑ4 (5u + πτ /2, q 5 ) = −q − 2 e−10iu h4 (u),
32
and
h5 (u + πτ ) = e−i(u+πτ ) ϑ4 (5(u + πτ ) − πτ /2, q 5 )
1
5
= q − 2 e−iu (−q − 2 )e−2i(5u−πτ /2) ϑ4 (5u − πτ /2, q 5 )
5
5
= −q − 2 e−10iu e−iu ϑ4 (5u − πτ /2, q 5 ) = −q − 2 e−10iu h5 (u).
We now consider the functions
∞
(−1)k q
H1 (u) = h2 (u) + h3 (u) = 2
5k2 +3k
2
cos(10k + 3)u
(4.9)
cos(10k + 1)u.
(4.10)
k=−∞
and
∞
(−1)k q
H2 (u) = h4 (u) + h5 (u) = 2
5k2 +k
2
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
H1 (πτ /2) = e3i(πτ /2) ϑ4 (5(πτ /2) + 3πτ /2, q 5 ) + e−3i(πτ /2) ϑ4 (−5(πτ /2) + 3πτ /2, q 5 )
∞
∞
5
5
2
(−1)n q 2 n e2in(4πτ ) + e−3i(πτ /2)
= e3i(πτ /2)
n=−∞
2
(−1)n q 2 n e2in(−πτ )
n=−∞
∞
∞
5
2 +4n+ 3
4
(−1)n q 2 n
=
n=−∞
∞
5
2 +4n+ 3
4
5
2 −(n+1)− 3
4
(−1)n+1 q 2 (n+1)
+
n=−∞
∞
n=−∞
∞
5
2 +4n+ 3
4
(−1)n q 2 n
=
2 −n− 3
4
n=−∞
∞
(−1)n q 2 n
=
5
(−1)n q 2 n
+
5
n=−∞
2 +4n+ 3
4
(−1)n+1 q 2 n
+
n=−∞
= 0,
H2 (πτ /2) = ei(πτ /2) ϑ4 (5(πτ /2) + πτ /2, q 5 ) + e−i(πτ /2) ϑ4 (−5(πτ /2) + πτ /2, q 5 )
∞
∞
= ei(πτ /2)
5
n=−∞
∞
5
2 +3n+ 1
4
(−1)n q 2 n
n=−∞
∞
5
2 +3n+ 1
4
2 −2(n+1)− 1
4
n=−∞
∞
5
2 +3n+ 1
4
(−1)n q 2 n
n=−∞
5
(−1)n+1 q 2 (n+1)
+
n=−∞
∞
=
2 −2n− 1
4
n=−∞
∞
(−1)n q 2 n
=
5
(−1)n q 2 n
+
5
2 +3n+ 1
4
(−1)n+1 q 2 n
+
n=−∞
= 0,
33
2
(−1)n q 2 n e2in(−2πτ )
n=−∞
∞
=
5
2
(−1)n q 2 n e2in(3πτ ) + e−i(πτ /2)
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
∞
k
α = H2 (0) =
(−1) q
5k2 +k
2
∞
,
β = −H1 (0) = −
k=−∞
(−1)k q
5k2 +3k
2
.
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)
∞
(−1)m q
=
m=−∞
∞
−
(−1)m q
5m2 +m
2
5m2 +3m
2
e−3iu ϑ4 (5u − 3πτ /2, q 5 ) + e3iu ϑ4 (5u + 3πτ /2, q 5 )
(4.11)
eiu ϑ4 (5u + πτ /2, q 5 ) + e−iu ϑ4 (5u − πτ /2, q 5 ) .
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
5m2 +m+5k2 +3k
2
(−1)m+k q
2(cos (3π/5) − cos (π/5))
m,k∈Z
= −4 sin (2π/5) sin (π/5)ϑ4 (πτ /2, q 5 )ϑ4 (3πτ /2, q 5 )
∞
∞
(1 − q 5n )
= −4 sin (2π/5) sin (π/5)
n=1
∞
(1 − q 5n−2 )(1 − q 5n−3 )
n=1
(4.12)
∞
5n
×
(1 − q )
n=1
(1 − q
5n−1
)(1 − q
5n−4
)
n=1
∞
∞
5n
= −4 sin (2π/5) sin (π/5)
(1 − q n )
(1 − q )
n=1
n=1
= −4 sin (2π/5) sin (π/5)(q 5 ; q 5 )∞ (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/5 x)(1 − e−2πi/5 x)(1 − e4πi/5 x)(1 − e−4πi/5 x) = (1 − x5 ),
we see that
ϑ1 (π/5, q)ϑ1 (2π/5, q)
∞
= 4q
1/4
(1 − q n )2
sin (π/5) sin (2π/5)
n=1
∞
(1 − e2πi/5 q n )(1 − e−2πi/5 q n )(1 − e4πi/5 q n )(1 − e−4πi/5 q n )
×
(4.13)
n=1
∞
∞
= 4q
1/4
(1 − q 5n )
n
(1 − q )
sin (π/5) sin (2π/5)
n=1
n=1
= 4q 1/4 sin (π/5) sin (2π/5)(q; q)∞ (q 5 ; q 5 )∞
Combining (4.11), (4.12), (4.13), we then have
A(q) = −q −1/4
and thus obtain the desired result. ✷
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 = 0. Then we have the following identity
∞
∞
(−1)m+n q
3m2 +3n2 +3m+n
2
m=−∞ n=−∞
× (a−3m b−3n − a−3m b3n+1 − a−3n+1 b−3m−1 + a3n+2 b−3m−1 )
(4.14)
= (q ; q)2∞ (a, a−1 q, b, b−1 q, ab, a−1 b−1 q, ab−1 , a−1 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)∞
∞
∞
= (q ; q)2∞
(−1)n q
3n(n+1)
2
y −3n
n=−∞
m=−∞
∞
n
∞
−1
− yx (q
; q)2∞
(−1)n q
n(n−1)
6
(−1)m q
y
n=−∞
Let
(−1)m q
xm
3m(m+1)
6
(4.15)
x−3m .
m=−∞
∞
T1 = (q
m(m−1)
6
∞
; q)2∞
n
(−1) q
3n(n+1)
2
y
−3n
n=−∞
m(m−1)
6
xm ,
m=−∞
∞
T2 = yx−1 (q ; q)2∞
(−1)m q
∞
(−1)n q
n(n−1)
6
n=−∞
yn
(−1)m q
3m(m+1)
6
x−3m .
m=−∞
First we observe that
∞
(−1)m q
m(m−1)
6
um
m=−∞
∞
(−1)3k q
=
k(3k+1)
2
(u−3k − u3k+1 ) + (−1)3k+2 q
k=−∞
36
(3k+2)(3k+1)
6
u3k+2 .
So we will have
∞
∞
T1 = (q ; q)2∞
(−1)n q
3n(n+1)
2
y −3n
n=−∞
∞
= (q ; q)2∞
(−1)m q
m(m−1)
6
xm
m=−∞
(−1)n q
3n(n+1)
2
y −3n
n=−∞
∞
(−1)3k q
×
k(3k+1)
2
(x−3k − x3k+1 ) + (−1)3k+2 q
(3k+2)(3k+1)
6
x3k+2
k=−∞
(−1)n+k q
= (q ; q)2∞
3n2 +3n+3k2 +k
2
(y −3n x−3k − y −3n x3k+1 )
n,k∈Z
+ (q ; q)2∞
(−1)n+k q (
(3k+2)(3k+1)
3n2 +3n
+
)
2
6
y −3n x3k+2
n,k∈Z
and
∞
∞
T2 = yx−1 (q ; q)2∞
(−1)n q
n(n−1)
6
yn
n=−∞
∞
= yx−1 (q ; q)2∞
(−1)m q
3m(m+1)
6
x−3m
m=−∞
(−1)3k q
k(3k+1)
2
(y −3k − y 3k+1 ) + (−1)3k+2 q
(3k+2)(3k+1)
6
y 3k+2
k=−∞
∞
(−1)n q
×
3n(n+1)
2
x−3n
n=−∞
= (q ; q)2∞
(−1)n+k q
3n2 +3n+3k2 +k
2
(y −3k+1 x−3n−1 − y −3k+2 x−3n−1 )
n,k∈Z
(−1)n+k q (
+ (q ; q)2∞
(3k+2)(3k+1)
3n2 +3n
+
)
2
6
y 3k+3 x−3n−1
.
n,k∈Z
Doing a change of variables n = −k − 1, we note that
(q ; q)2∞
(−1)n+k q (
(3k+2)(3k+1)
3n2 +3n
+
)
2
6
y 3k+3 x−3n−1
n,k∈Z
= (q ; q)2∞
(−1)(−k−1)+(−n−1) q (
3(−k−1)2 +3(−k−1)
(3(−n−1)+2)(3(−n−1)+1)
+
)
2
6
n,k∈Z
= (q ; q)2∞
(−1)n+k q (
(3k+2)(3k+1)
3n2 +3n
+
)
2
6
y −3n x3k+2
n,k∈Z
37
,
y 3(−n−1)+3 x−3(−k−1)−1
therefore we have
(x, q/x, y, q/y, y/x, qx/y, xy, q/xy, q, q, q, q ; q)∞
= T1 − T2
= (q ; q)2∞
(−1)n+k q
3n2 +3n+3k2 +k
2
(y −3n x−3k − y −3n x3k+1 − y −3k+1 x−3n−1 + y −3k+2 x−3n−1 )
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
q 12 (q ; q)2∞ {ϑ1 (u, q 3 )ϑ1 (3v, q 3 ) − ϑ1 (v, q 3 )ϑ1 (3u, q 3 )}
(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
∞
T1 = (q
∞
; q)2∞
n
(−1) q
3n(n+1)
2
n=−∞
∞
= (q
y
−3n
(−1)m q
xm
m=−∞
∞
2
n ( 3n2 + 3n
) 2n(3iu)
2
; q)2∞
m(m−1)
6
(−1) q
(−1)m q (
e
n=−∞
n2
+n
)
6
6
e2miv
m=−∞
∞
3
5
= −q − 12 e−3iu e−iv (q ; q)2∞ −i
n=−∞
∞
×
−i
1
1 2
(−1)m q 6 (m+ 2 ) e(2m+1)(iv)
m=−∞
= −q
5
− 12
1 2
(−1)n q 2 (n+ 2 ) e(2n+1)(3iu)
1
e−3iu e−iv (q ; q)2∞ ϑ1 (3u, q 3 )ϑ1 (v, q 3 ),
38
∞
T2 = yx−1 (q ; q)2∞
∞
n(n+1)
6
(−1)n q
y −n
n=−∞
=e
e
(q
3m(m+1)
2
x−3m
m=−∞
∞
−2iu 2iv
(−1)m q
; q)2∞
n
(−1) q
2
( n6 + n
)
6
∞
2n(iu)
(−1)m q (
e
n=−∞
3n2
+ 3n
)
2
2
e2m(3iv)
m=−∞
∞
5
1
= −q − 12 e−3iu e−iv (q ; q)2∞ −i
1 2
(−1)n q 6 (n+ 2 ) e(2n+1)(iu)
n=−∞
∞
×
3
1 2
(−1)m q 2 (m+ 2 ) e(2m+1)(3iv)
−i
m=−∞
= −q
5
− 12
1
e−3iu e−iv (q ; q)2∞ ϑ1 (u, q 3 )ϑ1 (3v, q 3 ),
(x, q/x, y, q/y, y/x, qx/y, xy, q/xy, q, q, q, q ; q)∞
= (q, e−2iu , e2iu q ; q)∞ (q, e−2iv , e2iv q ; q)∞ (q, e−2iu e2iv , e2iu e−2iv q ; q)∞ (q, e−2iu e−2iv , e2iu e2iv q ; q)∞
1
1
1
= q − 2 e−3iu e−iv (−iq 8 eiu (q, e−2iu , e2iu q ; q)∞ )(−iq 8 eiv (q, e−2iv , e2iv q ; q)∞ )
1
1
× (−iq 8 ei(u−v) (q, e−2iu e2iv , e2iu e−2iv q ; q)∞ )(−iq 8 ei(u+v) (q, e−2iu e−2iv , e2iu e2iv q ; q)∞ )
1
= q − 2 e−3iu e−iv ϑ1 (u, q)ϑ1 (v, q)ϑ1 (u − v, q)ϑ1 (u + v, q).
Therefore equation (4.15) becomes
1
q − 2 e−3iu e−iv ϑ1 (u, q)ϑ1 (v, q)ϑ1 (u − v, q)ϑ1 (u + v, q)
5
1
1
= q − 12 e−3iu e−iv (q ; q)2∞ {ϑ1 (u, q 3 )ϑ1 (3v, q 3 ) − ϑ1 (3u, q 3 )ϑ1 (v, q 3 )}
and multiply throughout by q 1/2 e3iu eiv , we get the result. ✷
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
ϑ1 (u, q 3 ),
ϑ1 (3u, q 3 ),
ϑ1 (u, q)ϑ1 (u − v, q)ϑ1 (u + v, q)
satisfy the transformation formulas
f (z + π) = −f (z),
f (z + πτ ) = −q −3/2 e−6iz f (z).
39
Therefore the functions
F1 (u) =
F2 (u) =
ϑ1 (u, q)ϑ1 (u − v, q)ϑ1 (u + v, q)
1
ϑ1 (u, q 3 )
ϑ1 (3u, q 3 )
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
ϑ1 (3u, q 3 ) 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
C with corners t, t + π, t + π + 31 πτ and t + 13 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
C3 (v, q)ϑ1 (u, q 3 ) + C2 (v, q)ϑ1 (3u, q 3 ) = 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
C3 (v, q)ϑ1 (v, q 3 ) + C2 (v, q)ϑ1 (3v, q 3 ) = 0.
Hence we have
C3 (v, q) = −C2 (v, q)
ϑ1 (3v, q 3 )
1
ϑ1 (v, q 3 )
,
and we can rewrite (4.17) as
1
1
−C2 (v, q)ϑ1 (3v, q 3 )ϑ1 (u, q 3 ) + C2 (v, q)ϑ1 (v, q 3 )ϑ1 (3u, q 3 )
1
(4.18)
= 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
1
1
−C2 (u, q)ϑ1 (3u, q 3 )ϑ1 (v, q 3 ) + C2 (u, q)ϑ1 (u, q 3 )ϑ1 (3v, q 3 )
1
= C1 (u, q)ϑ1 (u, q 3 )ϑ1 (v, q)ϑ1 (v + u, q)ϑ1 (v − u, q).
40
(4.19)
Dividing (4.18) by (4.19), we see that
C2 (v, q)ϑ1 (v, q)
1
3
C1 (v, q)ϑ1 (v, q )
C2 (u, q)ϑ1 (u, q)
=
1
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
A(q) =
C2 (v, q)ϑ1 (v, q)
1
C1 (v, q)ϑ1 (v, q 3 )
we have
,
1
C1 (v, q)ϑ1 (v, q 3 )
A(q)
C2 (v, q) =
ϑ1 (v, q)
and by substituting it into (4.18) we get
1
1
A(q) −ϑ1 (3v, q 3 )ϑ1 (u, q 3 ) + ϑ1 (v, q 3 )ϑ1 (3u, q 3 )
= ϑ1 (u, q)ϑ1 (v, q)ϑ1 (u + v, q)ϑ1 (u − v, q).
1
1
Let G1 (v) = −ϑ1 (3v, q 3 )ϑ1 (u, q 3 ) + ϑ1 (v, q 3 )ϑ1 (3u, q 3 ), 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
1
1
A(q) −3ϑ1 (u, q 3 )ϑ1 (0, q 3 ) + ϑ1 (0, q 3 )ϑ1 (3u, q 3 )
3
= ϑ31 (u, q)ϑ1 (0, q).
Now from lemma(3.3.2), we have
ϑ1 (0, q) = ϑ2 (0, q)ϑ3 (0, q)ϑ4 (0, q)
∞
= 2q
1
8
∞
n
n 2
(1 − q )(1 + q )
n=1
∞
1
∞
n
(1 − q )(1 + q
n− 21 2
n=1
1
(1 − q n )(1 − q n− 2 )2
)
n=1
(1 − q n )(1 − q n )2 (1 + q n )2 (1 − q 2n−1 )2
= 2q 8
n=1
∞
1
(1 − q n )(1 − q 2n )2 (1 − q 2n−1 )2
= 2q 8
n=1
∞
1
(1 − q n )(1 − q n )2
= 2q 8
n=1
∞
1
1
(1 − q n )3 = 2q 8 (q ; q)3∞ .
= 2q 8
n=1
41
(4.20)
Also we have
∞
1
π
(1 + q n + q 2n )
ϑ1 ( , q) = 2(q ; q)∞ q 8 sin(π/3)
3
n=1
=
√
∞
3q
1
8
(1 − q 3n ) =
√
(4.21)
1
8
3q (q 3 ; q 3 )∞ .
n=1
Therefore by letting u = π3 , and using the identities (4.20), (4.21) and ϑ1 (π, q 3 ) = 0, we
get
1
A(q) = −q 12 (q ; q)2∞
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
ϑ31 (z, q) − ϑ31 (z +
π
π
, q) − ϑ31 (z − , q)
3
3
(q 3 ; q 3 )∞
π
π
= 3a(q)
ϑ1 (z, q)ϑ1 (z + , q)ϑ1 (z − , q),
3
(q; q)∞
3
3
where
∞
(1 − zq n−1 )
(z; q)∞ =
n=1
and
∞
a(q) = 1 + 6
n=0
q 3n+1
q 3n+2
−
1 − q 3n+1 1 − q 3n+2
43
.
Proof
We first consider the functions
π
π
, q) − ϑ31 (z − , q),
3
3
π
π
F2 (z) = ϑ1 (z, q)ϑ1 (z + , q)ϑ1 (z − , q).
3
3
F1 (z) = ϑ31 (z, q) − ϑ31 (z +
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
Fi (z + πτ ) = −q − 2 e−6iz 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
F1 (0) = ϑ31 (0) − ϑ31 ( ) − ϑ31 (− )
3
3
3 π
3 π
= −ϑ1 ( ) + ϑ1 ( ) = 0,
3
3
and for z = −π/3, we have
π
π
2π
F1 (− ) = ϑ31 (− ) − ϑ31 (0) − ϑ31 (− )
3
3
3
π
2π
= ϑ31 (− ) + ϑ31 (−
+ π)
3
3
π
π
= −ϑ31 ( ) + ϑ31 ( ) = 0.
3
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).
44
(5.1)
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
3ϑ21 (0)ϑ1 (0) − 3ϑ21 ( π3 )ϑ1 ( π3 ) − 3ϑ21 ( π3 )ϑ1 ( π3 )
−6ϑ21 ( π3 )ϑ1 ( π3 )
=
ϑ1 (0){ϑ1 ( π3 )ϑ1 ( π3 ) − ϑ1 ( π3 )ϑ1 ( π3 )} − ϑ21 ( π3 )ϑ1 (0)
−ϑ21 ( π3 )ϑ1 (0)
6ϑ1 ( π3 )
=
,
ϑ1 (0)
and by comparing coefficients, we get
A(q) =
6ϑ1 ( π3 )
.
ϑ1 (0)
Now using the product identity of ϑ1 (z) as given in theorem 3.3.1, applying logarithmic
differentiation, we get
∞
ϑ1 (z)
q n sin 2z
= cot z + 4
.
ϑ1 (z)
1 − 2q n cos 2z + q 2n
n=1
Now by the product identity of ϑ( z) given in Theorem 3.3.1, we have
1
π
π
ϑ1 ( ) = 2(q ; q)∞ q 8 sin
3
3
=
=
√
√
∞
(1 − 2q n cos
n=1
∞
1
(1 − q n )(1 + q n + q 2n )
3q 8
n=1
1
8
3q (q 3 ; q 3 )∞ ,
45
2π
+ q 2n )
3
and so we get
∞
=
√
∞
√
q n 23
1
√ +4
1 + q n + q 2n
3
n=1
π
ϑ1 ( ) =
3
1+6
n=1
qn
1 + q n + q 2n
∞
1
= q 8 (q 3 ; q 3 )∞ 1 + 6
n=1
∞
1
= q 8 (q 3 ; q 3 )∞ 1 + 6
1
3q 8 (q 3 ; q 3 )∞
1
q 8 (q 3 ; q 3 )∞
q n − q 2n
1 − q 3n
∞
(q n − q 2n )
n=1
∞ ∞
1
= q 8 (q 3 ; q 3 )∞ 1 + 6
q 3nk
k=0
q (3k+1)n − q (3k+2)n
n=1 k=0
∞ ∞
1
q (3k+1)(n+1) − q (3k+2)(n+1)
= q 8 (q 3 ; q 3 )∞ 1 + 6
1
= q 8 (q 3 ; q 3 )∞ 1 + 6
k=0 n=0
∞
∞
3k+1
q
k=0
∞
1
= q 8 (q 3 ; q 3 )∞ 1 + 6
k=0
∞
q (3k+1)n − q 3k+2
n=0
3k+1
q (3k+2)n
n=0
3k+2
q
q
−
1 − q 3k+1 1 − q 3k+2
1
8
= q (q 3 ; q 3 )∞ a(q).
Using the identity
1
ϑ1 (0) = 2q 8 (q ; q)3∞ ,
we will have
A(q) = 3a(q)
(q 3 ; q 3 )∞
,
(q ; q)3∞
and from (5.1), we conclude that
ϑ31 (z, q) − ϑ31 (z +
π
π
, q) − ϑ31 (z − , q)
3
3
= 3a(q)
(q 3 ; q 3 )∞
π
π
ϑ1 (z, q)ϑ1 (z + , q)ϑ1 (z − , q).
3
(q; q)∞
3
3
✷
Note that the constant
(q 3 ; q 3 )∞
(q; q)3∞
46
is
1
,
b(q)
where b (q) is defined as
∞
2 +mn+n2
ω m−n q m
b(q) :=
,
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 doublyperiodic 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 = cm2 u − αsm u
du
where cm u is another function of u, given in terms of sm u by the equation
cm3 u + sm3 u − 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
cm3 u + sm3 u − 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
ϑ1 (z, q)
ϑ1 (z + π3 , q)
3
ϑ1 (z − π3 , q) 3
ϑ1 (z + π3 , q)
(q 3 ; q 3 )∞
= 3a(q) −
(q; q)3∞
−1+ −
ϑ1 (z, q)
ϑ1 (z + π3 , q)
ϑ1 (z − π3 , q)
−
ϑ1 (z + π3 , q)
,
which implies that
ϑ1 (z, q)
ϑ1 (z + π3 , q)
3
ϑ1 (z − π3 , q) 3
ϑ1 (z + π3 , q)
(q 3 ; q 3 )∞
− 3a(q) −
(q; q)3∞
+ −
ϑ1 (z, q)
ϑ1 (z + π3 , q)
ϑ1 (z − π3 , q)
−
ϑ1 (z + π3 , q)
(5.2)
= 1.
From the definition of the Dixon’s functions smu and cmu, we have
cm3 u + sm3 u − 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 similar to ϑ1 (z, q)/(ϑ1 (z + π3 , q)), with the constant term in the z-expansion of ϑ1 (z, q)/(ϑ1 (z +
π
, q))
3
being 0, and the algebraic expression of cmu is similar to −ϑ1 (z − π3 , q)/(ϑ1 (z + π3 , q)),
with the constant term in the z-expansion of −ϑ1 (z − π3 , q)/(ϑ1 (z + π3 , q)) being 1.
Now in [5], the u-expansion of smu and cmu is given as
1
1
cm u = 1 + αu + α2 u2 + (α3 − 2)u3 + . . .
2
6
1 2 1 2 3
1
sm u = u + αu + α u + (5α3 − 4)u4 + . . . .
2
2
24
(5.4)
It turns out that if we set
z=
ϑ1 (π/3)
u,
ϑ1 (0)
and let
ϑ1 (z, q)
ϑ1 (z + π3 , q)
ϑ1 (z − π3 , q)
and c(u) = −
,
ϑ1 (z + π3 , q)
s(u) =
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
(q 3 ; q 3 )∞
α = a(q) −
(q; q)3∞
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
ϑ1 (z, q)
ϑ1 (z − π3 , q)
3
ϑ1 (z + π3 , q) 3
−1
ϑ1 (z − π3 , q)
(q 3 ; q 3 )∞
ϑ1 (z, q)
= 3a(q) −
3
(q; q)∞
ϑ1 (z − π3 , q)
+ −
ϑ1 (z + π3 , q)
−
ϑ1 (z − π3 , q)
,
we will relate smu with ϑ1 (z, q)/(ϑ1 (z − π3 , q)), the constant term of the z-expansion of
ϑ1 (z, q)/(ϑ1 (z − π3 , q)) being 0, and relate cmu with ϑ1 (z + π3 , q)/(ϑ1 (z − π3 , q)), the constant
term of the z-expansion of ϑ1 (z + π3 , q)/(ϑ1 (z − π3 , q)) being 1. If we now set
z=−
ϑ1 (π/3)
u,
ϑ1 (0)
and let
ϑ1 (z, q)
ϑ1 (z − π3 , q)
ϑ1 (z + π3 , q)
and c1 (u) = −
,
ϑ1 (z − π3 , q)
s1 (u) =
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
α = a(q) −
(q 3 ; q 3 )∞
(q; q)3∞
.
We conjecture that s1 (u) and c1 1(u) are also possible algebraic expressions for smu and
cmu respectively.
49
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50
[12] C.G.J. Jacobi, Ges. Werke, 1838.
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56-59.u
51
[...]... 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... by −1 or −q − 2 e−2iz respectively ϑ2 (z) and ϑ3 (z) also exhibit similar properties Because of this, the theta functions are known as quasi 1 doubly-periodic functions of z, and ±1 and −q − 2 e−2iz 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),... = −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. .. 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... + 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... 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... 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. .. 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 u = 0 in cell C2 with corners t, t + π2 , t + π+πτ 2 and. .. 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... 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 ... of elliptic functions and the theta functions 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. .. 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... 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