Res. Lett. Inf. Math. Sci., 2003, Vol.5, pp 23-59
Available online at http://iims.massey.ac.nz/research/letters/
23
Cubic elliptic functions
Shaun Cooper
Institute of Information & Mathematical Sciences
Massey University at Albany, Auckland, New Zealand
s.cooper@massey.ac.nz
The function:
?(?; q) = ? + 3
??
k=1
sin(2k?)qk
k(1 + qk + q2k)
occurs in one of Ramanujan?s inversion formulas for elliptic integrals. In this article, a common
generalization of the cubic elliptic functions
g1(?; q) =
1
6
+
??
k=1
qk
1 + qk + q2k
cos k?,
g2(?; q) =
1
2
sin ?2
sin 3?2
+
??
k=1
?3(k)qk
1? qk
cos k?,
is given. The function g1 is the derivative of Ramanujan?s function ? (after rescaling), and
?3(n) = 0, 1 or ?1 according as n ? 0, 1 or 2 (mod 3), respectively, and |q| < 1. Many
properties of the common generalization, as well as the functions g1 and g2, are proved.
1 Introduction
Suppose Re t > 0 and let q = e?2pit. The function
?(?; q) =
1
4
cot
?
2
+
??
n=1
qn
1? qn
sinn? (1.1)
plays an important role in Ramanujan?s paper (16). For example, Ramanujan (16, eq. (17)) proved
that
?(?; q)2 =
(
1
4
cot
?
2
)2
+
??
n=1
qn
(1? qn)2
cosn? +
1
2
??
n=1
nqn
1? qn
(1? cosn?), (1.2)
and he used this to prove many identities for elliptic functions.
Venkatachaliengar (19, p. 42) generalized Ramanujan?s formula (1.2). Let
F (x, y; q) =
??
n=1
(1? xyqn?1)(1? x?1y?1qn)(1? qn)2
(1? xqn?1)(1? x?1qn)(1? yqn?1)(1? y?1qn)
24 S. Cooper
and
?(z; q) =
1
2
+
??
n=??
n6=0
zn
1? qn
, for |q| < |z| < 1, (1.3)
=
1 + z
2(1? z)
+
??
n=1
qn
1? qn
(zn ? z?n), for |q| < |z| < |q|?1. (1.4)
The function ? is related to the function ? in (1.1) by
?(ei?) = 2i?(?).
Venkatachaliengar?s generalization of Ramanujan?s identity (1.2) is
F (x, y; q)F (x, z; q) = x
?
?x
F (x, yz; q) + F (x, yz; q)(?(y; q) + ?(z; q)). (1.5)
See (9, p. 66), (10, Thm. 2.2), or (19, p. 37, eqs. (3.2), (3.3)). Letting y ? 1/z and then setting
x = ei?, z = ei?, we get
F (ei?, ei?; q)F (ei?, e?i?; q) = 2 (??(?; q)? ??(?; q)) . (1.6)
See (9, p. 90, eq. (3.23)), (10, (2.41)), (19, p. 112, eq. (6.50)) for the details. This is equivalent
to (20, p. 451, Ex. 1):
?(?)? ?(?) = ?
?(? + ?)?(? ? ?)
?2(?)?2(?)
,
where ? is the Weierstrass elliptic function with periods 2pi and 2piit, and ? is the corresponding
Weierstrass sigma function. Ramanujan?s formula (1.2) may be obtained by expanding (1.6) in
powers of ? and extracting coefficients of ?0. See (19, pp. 42?45, eq. (3.40)).
The aim of this article is to give analagous results for the functions
g1(?; q) =
1
6
+
??
n=1
qn
1 + qn + q2n
cosn?,
g2(?; q) =
1
2
sin ?2
sin 3?2
+
??
n=1
?3(n)qn
1? qn
cosn?. (1.7)
The antiderivative of the function g1 occurs in one of Ramanujan?s inversion formulas for elliptic
integrals. Several properties of g1 were established by Berndt, Bhargava and Garvan (4). They
used the notation v(z, q), where
g1(?; q) =
1
6
v(ei?, q).
We shall begin by observing that g1 and g2 have a common generalization. A number of basic
properties of the generalization, and the functions g1 and g2 are given in Section 2. Power series
expansions for g1 and g2 in terms of the corresponding Eisenstein series are given in Section 3. In
Section 4, we prove the transformation formulas
g1(?; e
?2pit) =
1
t
?
3
g2
(
i?
3t
; e?
2pi
3t
)
,
g2(?; e
?2pit) =
1
t
?
3
g1
(
i?
t
; e?
2pi
3t
)
.
Expanding in powers of ?, we obtain transformation formulas for various Eisenstein series. A
detailed analysis of these Eisenstein series is given in Section 5.
Cubic elliptic functions 25
Connections between the functions g1, g2 and the Hirschhorn-Garvan-J. Borwein cubic theta func-
tions are given in Section 6 and connections with the Weierstrass ? function are given in Section
7. We indicate how to obtain addition formulas for g1 and g2 in Section 7. Various formulas for
g1, g2 g?1 and g
?
2, involving infinite products, are given in Section 8.
An analogue of Venkatachaliengar?s formula (1.5) for the function G is given in Section 9, and an
analogue of (1.6) is given in Section 10. Fourier series for g21 and g
2
2 are given as a consequence.
These are analogues of Ramanujan?s formula (1.2).
In Section 11, we introduce the cubic transcendentals Z andX. We express Ramanujan?s Eisenstein
series P (q), Q(q), R(q), as well as P (q3), Q(q3) and R(q3), in terms of Z and X. In Section 12,
the differential equations satisfied by g1 and g2 are given. Lastly, in Section 13 we prove some
recurrence relations for cubic Eisenstein series.
2 Definitions and basic properties
2.1 The functions F and G
Let
F (x, y; q) =
(xy, qx?1y?1, q, q; q)?
(x, qx?1, y, qy?1; q)?
. (2.1)
Here we are using the standard notation
(x; q)? =
??
n=1
(1? xqn?1),
(x1, x2, ? ? ? , xn; q)? = (x1; q)?(x2; q)? ? ? ? (xn; q)?.
It is straightforward to check that
F (x, y; q) = F (y, x; q), (2.2)
F (x, y; q) = ?F (x?1, y?1; q), (2.3)
F (x, y; q) = xF (x, qy; q) = yF (qx, y; q). (2.4)
By Ramanujan?s 1?1 summation formula (1, p. 502), (2, Ch. 16, Entry 17), (17, Ch. 16, Entry
17),
F (x, y; q) =
??
n=??
xn
1? yqn
, for |q| < |x| < 1. (2.5)
Since F is symmetric in x and y,
F (x, y; q) =
??
n=??
yn
1? xqn
, for |q| < |y| < 1. (2.6)
Let
G(x, y; q) =
1
i
?
3
(
F (x, ?y; q)? F (x, ?2y; q)
)
. (2.7)
Observe that G is not symmetric in x and y.
26 S. Cooper
Expanding in powers of x using (2.5) gives
G(x, y; q) =
1
i
?
3
??
n=??
(
xn
1? ?yqn
?
xn
1? ?2yqn
)
=
??
n=??
xnqny
1 + yqn + y2q2n
=
y
1 + y + y2
+
??
n=1
xnqny
1 + yqn + y2q2n
+
??
n=1
x?nqny?1
1 + y?1qn + y?2q2n
.
(2.8)
This series converges for |q| < |x| < |q|?1.
Expanding in powers of y using (2.6) gives
G(x, y; q) =
1
i
?
3
??
n=??
(
?nyn
1? xqn
?
?2nyn
1? xqn
)
=
??
n=??
?3(n)yn
1? xqn
. (2.9)
This converges for |q| < |y| < 1. Another form may be obtained from this as follows:
G(x, y; q) =
??
n=1
?3(n)yn
1? xqn
+
??
n=1
?3(?n)y?n
1? xq?n
=
??
n=1
?3(n)yn(1? xqn + xqn)
1? xqn
+
??
n=1
?3(n)x?1y?nqn
1? x?1qn
=
??
n=1
?3(n)y
n +
??
n=1
?3(n)
(
xynqn
1? xqn
+
x?1y?nqn
1? x?1qn
)
=
y
1 + y + y2
+
??
n=1
?3(n)
(
xynqn
1? xqn
+
x?1y?nqn
1? x?1qn
)
. (2.10)
This converges for |q| < |y| < |q|?1.
Also from (2.9) we have
G(x, y; q) =
??
n=??
y3n+1
1? xq3n+1
?
??
n=??
y3n?1
1? xq3n?1
= yF (qx, y3; q3)? y?1F (q?1x, y3; q3). (2.11)
2.2 The functions g1 and g2
Let
g1(?; q) =
1
2
G(ei?, 1; q), (2.12)
g2(?; q) =
1
2
G(1, ei?; q). (2.13)
Cubic elliptic functions 27
Equations (2.8) and (2.10) immediately give
g1(?) =
1
6
+
??
n=1
qn
1 + qn + q2n
cosn? (2.14)
=
1
6
+
1
2
??
n=1
?3(n)
(
qnei?
1? qnei?
+
qne?i?
1? qne?i?
)
, (2.15)
g2(?) =
1
2
sin ?2
sin 3?2
+
??
n=1
?3(n)qn
1? qn
cosn? (2.16)
=
1
2
sin ?2
sin 3?2
+
??
n=1
(
ei?qn
1 + ei?qn + e2i?q2n
+
e?i?qn
1 + e?i?qn + e?2i?q2n
)
.
(2.17)
Equations (2.14) and (2.16) are Fourier series for g1 and g2. They converge for |q| < |ei?| < |q|?1, or
equivalently, ?2piRe t < Im ? < 2piRe t. Equations (2.15) and (2.17) are the analytic continuations
of g1 and g2, and are valid for all complex values of ?.
Equation (2.11) gives
g1(?; q) =
1
2
(
F (qei?, 1; q3)? F (q?1ei?, 1; q3)
)
, (2.18)
g2(?; q) =
1
2
(
ei?F (q, e3i?; q3)? e?i?F (q?1, e3i?; q3)
)
. (2.19)
These expressions are valid for all complex values of ?.
The locations of the poles and periodicity properties are readily determined from (2.15) and (2.17).
Theorem 2.20. Let q = e?2pit, where Re t > 0. Then
1. g1(? + 2pi; q) = g1(?; q)
g1(? + 6piit; q) = g1(?; q).
2. g1(?; q) is meromorphic on C, with simple poles at ? = 2pim + 2piint, m, n ? Z, n 6? 0
(mod 3), and no other singularities. The residue at each pole is 12i?3(n).
3. g2(? + 2pi; q) = g2(?; q)
g2(? + 2piit; q) = g2(?; q).
4. g2(?; q) is meromorphic on C, with simple poles at ? = 2pin/3 + 2piimt, m, n ? Z, n 6? 0
(mod 3), and no other singularities. The residue at each pole is ?1
2
?
3
?3(n).
Proof
Let z = ei?. The Fourier series (2.14) shows g1(? + 2pi; q) = g1(?; q).
Next, from (2.15) we have
g1(? + 6piit; q)? g1(?; q)
=
1
2
??
n=1
(
q3n+1z
1? q3n+1z
?
q3n+2z
1? q3n+2z
+
q3n?5z?1
1? q3n?5z?1
?
q3n?4z?1
1? q3n?4z?1
)
?
1
2
??
n=1
(
q3n?2z
1? q3n?2z
?
q3n?1z
1? q3n?1z
+
q3n?2z?1
1? q3n?2z?1
?
q3n?1z?1
1? q3n?1z?1
)
=
1
2
(
?qz
1? qz
+
q2z
1? q2z
+
q?2z?1
1? q?2z?1
?
q?1z?1
1? q?1z?1
)
= 0.
28 S. Cooper
Equation (2.15) implies that g1(?, q) is meromorphic and has simple poles at ? = 2pim + 2piint,
n 6? 0 (mod 3), and no other singularities. The residue at ? = 2piit may be calculated as follows.
Res (g1(?; q), ? = 2piit) = lim
??2piit
(? ? 2piit)g1(?; q)
=
1
2
lim
??2piit
(? ? 2piit)
qe?i?
1? qe?i?
=
1
2
lim
??2piit
qe?i?
d
d? (1? qe
?i?)
=
1
2i
.
Similarly, Res (g1(?; q), ? = 4piit) = ? 12i . The residue at the singularity ? = 2pim+ 2piint is there-
fore 12i?3(n) by the periodicity properties.
The corresponding properties for the function g2 may be proved in the same way, using (2.17).
2.3 The functions ?, h1 and h2
The function ? is defined by (1.1). Its analytic continuation, quasi-periodicity properties and the
location of its singularities are given by
Theorem 2.21. Let q = e?2pit, where Re t > 0. Then
1.
?(?; q) =
1
4
cot
?
2
+
1
2i
??
n=1
(
qnei?
1? qnei?
?
qne?i?
1? qne?i?
)
, (2.22)
valid for all complex values of ?.
2.
?(? + 2pi; q) = ?(?; q)
?(? + 2piit; q) = ?(?; q)?
i
2
.
3. The function ?(?; q) is meromorphic on C, with simple poles at ? = 2pin + 2piimt, and no
other singularities. The residue at each pole is 12 .
Proof
Starting with (1.1) and writing z = ei?, we have
?(?; q) =
1
4
cot
?
2
+
??
n=1
qn
1? qn
sinn?
=
1
4
cot
?
2
+
1
2i
??
n=1
qn
1? qn
(
zn ? z?n
)
=
1
4
cot
?
2
+
1
2i
??
n=1
??
m=1
(
qmnzn ? qmnz?n
)
=
1
4
cot
?
2
+
1
2i
??
m=1
(
qmz
1? qmz
?
qmz?1
1? qmz?1
)
(2.23)
=
1
4
cot
?
2
+
1
2i
??
m=1
(
qmei?
1? qmei?
?
qme?i?
1? qme?i?
)
.
Cubic elliptic functions 29
This proves the first part of the theorem.
The quasi-periodicity properties, location of the singularities and values of the residues can be
determined from (2.22) using the same procedure as in the proof of Theorem 2.20.
For future reference, we define the functions h1 and h2 by
h1(?; q) = ?
?(?; q)? ??(?; q3)
=
??
n=1
nqn
1? qn
cosn? ?
??
n=1
nq3n
1? q3n
cosn?, (2.24)
h2(?; q) = ?
?(?; q)? 9??(3?; q3)
=
9
8
csc2
3?
2
?
1
8
csc2
?
2
+
??
n=1
nqn
1? qn
cosn? ? 9
??
n=1
nq3n
1? q3n
cos 3n?.
(2.25)
The analytic continuations, periodicity properties and location of poles follow right away from the
definition of these functions and Theorem 2.21.
2.4 The cubic theta functions
The cubic theta functions are defined by
a?(q, z) =
??
m=??
??
n=??
qm
2+mn+n2zn, (2.26)
a(q, z) =
??
m=??
??
n=??
qm
2+mn+n2zm?n, (2.27)
b(q, z) =
??
m=??
??
n=??
qm
2+mn+n2?m?nzn, (2.28)
c(q, z) =
??
m=??
??
n=??
q(m+
1
3 )
2+(m+ 13 )(n+
1
3 )+(n+
1
3 )
2
zm?n, (2.29)
where ? = exp(2pii/3) and |q| < 1. When z = 1, we will denote the functions a?(q, 1) = a(q, 1)
simply by a(q). Similarly, we will abbreviate b(q, 1) and c(q, 1) to b(q) and c(q), respectively. The
functions a?(q, z), a(q, z), b(q, z) and c(q, z) were introduced1 by Hirschhorn et. al. (12). They
showed (12, (1.22), (1.23)) that
b(q, z) = (q; q)?(q
3; q3)?
(qz; q)?(qz?1; q)?
(q3z; q3)?(q3z?1; q3)?
(2.30)
c(q, z) = q
1
3 (q; q)?(q
3; q3)?(1 + z + z
?1)
(q3z3; q3)?(q3z?3; q3)?
(qz; q)?(qz?1; q)?
.
(2.31)
1The function c(q, z) in (12) differs from the one defined here by a factor of q
1
3 .
30 S. Cooper
We also record, for future reference, the properties
a(q) =
??
m=??
??
n=??
qm
2+mn+n2 (2.32)
= 1 + 6
??
n=1
(
q3n?2
1? q3n?2
?
q3n?1
1? q3n?1
)
(2.33)
= 1 + 6
??
n=1
qn
1 + qn + q2n
, (2.34)
b(q) =
??
m=??
??
n=??
qm
2+mn+n2?m?n (2.35)
=
(q; q)3?
(q3; q3)?
, (2.36)
c(q) =
??
m=??
??
n=??
q(m+
1
3 )
2+(m+ 13 )(n+
1
3 )+(n+
1
3 )
2
(2.37)
= 3q
1
3
(q3; q3)3?
(q; q)?
, (2.38)
a(q)3 = b(q)3 + c(q)3. (2.39)
Equations (2.32), (2.35) and (2.37) follow from (2.26) ? (2.29), by definition. Equations (2.33) and
(2.34) are proved in (6, (2.21), (2.25)), and (2.36) and (2.38) follow from (2.30) and (2.31). Proofs
of equation (2.39) are given in (5), (6) and (15).
3 Laurent series expansions
Theorem 3.1. Define the cubic Bernoulli numbers sn by
1
2
sinh ?2
sinh 3?2
=
??
n=0
sn
n!
?n.
Let q = e?2pit, where Re t > 0. For n = 1, 2, 3, ? ? ? , let
E2n(q) = ?
B2n
4n
+
??
k=1
k2n?1qk
1? qk
, n = 1, 2, 3, ? ? ? , (3.2)
S0(q) =
1
6
+
??
k=1
qk
1 + qk + q2k
, (3.3)
S2n(q) =
??
k=1
k2nqk
1 + qk + q2k
, n = 1, 2, 3, ? ? ? , (3.4)
E2n(?3; q) = s2n +
??
k=1
k2n?3(k)qk
1? qk
, n = 0, 1, 2, ? ? ? , (3.5)
E(1)2n (q) = E2n(q)? E2n(q
3), n = 1, 2, 3, ? ? ? , (3.6)
E(2)2n (q) = E2n(q)? 3
2nE2n(q
3), n = 1, 2, 3, ? ? ? . (3.7)
Then
S0(q) = E0(?3; q) =
a(q)
6
,
Cubic elliptic functions 31
and
?(?; e?2pit) =
1
2?
+
??
n=1
(?1)n?1
(2n? 1)!
E2n(q)?
2n?1,
g1(?; q) =
??
n=0
(?1)n
(2n)!
S2n(q)?
2n,
g2(?; q) =
??
n=0
(?1)n
(2n)!
E2n(?3; q)?
2n,
h1(?; e
?2pit) =
??
n=0
(?1)n
(2n)!
E(1)2n+2(q)?
2n,
h2(?; e
?2pit) =
??
n=0
(?1)n
(2n)!
E(2)2n+2(q)?
2n.
The series expansion for ? is valid for 0 < |?| < min{2pi, |2piit+ 2pik|, k ? Z}, and the others for
|?| < min{2pi, |2piit+ 2pik|, k ? Z}.
Proof
The first result is a restatement of (2.33) and (2.34). The expansion for ? follows by expanding
(1.1) in powers of ?, with the help of the expansion
1
2
cot
?
2
=
1
?
+
??
n=1
B2n(?1)n
(2n)!
?2n?1.
The expansions for g1 and g2 follow by expanding the Fourier series (2.14) and (2.16) in powers
of ?. The expansions for h1 and h2 follow from the expansion for ? by using the definitions (2.24)
and (2.25).
Remark 3.8. We shall call the numbers s0, s1, s2, ? ? ? , the cubic Bernoulli numbers. These num-
bers were introduced by Liu (15, eqs. (1.11), (1.13)), who expressed them in terms of derivatives
of the cotangent function evaluated at pi/3. From the definition, it is clear that s2n+1 = 0. The
first few values of s2n are:
s0 = 16 , s2 = ?
1
9 , s4 =
1
3
s6 = ? 73 , s8 =
809
27 , s10 = ?
1847
3 ,
s12 = 556013 , s14 = ?
6921461
3 , s16 =
126235201
3 .
In Section 5, we will show that (?1)ns2n > 0. Observe also that
1
2
sin ?2
sin 3?2
=
??
n=0
(?1)ns2n
(2n)!
?2n.
For future reference, we make the definitions
P (q) = ?24E2(q) = 1? 24
??
n=1
nqn
1? qn
, (3.9)
Q(q) = 240E2(q) = 1 + 240
??
n=1
n3qn
1? qn
, (3.10)
R(q) = ?504E2(q) = 1? 504
??
n=1
n5qn
1? qn
. (3.11)
32 S. Cooper
4 The modular transformation
4.1 The modular transformation for cubic theta functions
Theorem 4.1. Suppose Re t > 0. Then
a?(e?2pit, ei?) =
1
t
?
3
exp
(
??2
6pit
)
a
(
e?
2pi
3t , e
?
3t
)
,
a(e?2pit, ei?) =
1
t
?
3
exp
(
??2
2pit
)
a?
(
e?
2pi
3t , e
?
t
)
,
b(e?2pit, ei?) =
1
t
?
3
exp
(
??2
6pit
)
c
(
e?
2pi
3t , e
?
3t
)
,
c(e?2pit, ei?) =
1
t
?
3
exp
(
??2
2pit
)
b
(
e?
2pi
3t , e
?
t
)
.
Proof
See (11, Theorem 5.12).
Corollary 4.2. Suppose Re t > 0. Then
a(e?2pit) =
1
t
?
3
a
(
e?
2pi
3t
)
,
b(e?2pit) =
1
t
?
3
c
(
e?
2pi
3t
)
,
c(e?2pit) =
1
t
?
3
b
(
e?
2pi
3t
)
.
Proof
Let ? = 0 in Theorem 4.1. Also see (5, (2.2)) or (11, Corollary 5.19).
4.2 The modular transformation for g1 and g2
Theorem 4.3.
g1(?; e
?2pit) =
1
t
?
3
g2
(
i?
3t
; e?
2pi
3t
)
g2(?; e
?2pit) =
1
t
?
3
g1
(
i?
t
; e?
2pi
3t
)
.
Proof
Observe that by Theorem 2.20, the functions g1(?; e?2pit) and
1
t
?
3
g2
(
i?
3t
; e?
2pi
3t
)
both have simple
poles at ? = 2pim+2piint, m, n ? Z, n 6? 0 (mod 3), and no other singularities. Furthermore, the
residue of each function at each pole is 12i?3(n). Therefore the difference
g1(?; e
?2pit)?
1
t
?
3
g2
(
i?
3t
; e?
2pi
3t
)
is entire. Again by Theorem 2.20, the difference is doubly periodic, and therefore by Liouville?s
theorem, it is a constant. The value of the constant can be found by plugging in a value for ?, for
Cubic elliptic functions 33
example ? = 0:
g1(?; e
?2pit)?
1
t
?
3
g2
(
i?
3t
; e?
2pi
3t
)
= g1(0; e
?2pit)?
1
t
?
3
g2
(
0; e?
2pi
3t
)
=
1
6
(
a(e?2pit)?
1
t
?
3
a
(
e?
2pi
3t
))
= 0, (4.4)
by Lemma (4.2). This proves the first part of the theorem.
The second part follows from the first part by replacing t with 1/3t, then replacing ? with i?/t,
rearranging and using the fact that g1 and g2 are even functions of ?.
4.3 The modular transformation for ?, h1 and h2
The transformation properties for the functions h1 and h2 are obtained from the corresponding
transformation formula for ?.
Lemma 4.5. Let Re t > 0. Then
?(?; e?2pit) =
i
t
?
(
i?
t
; e?
2pi
t
)
?
?
4pit
.
Proof
See (11, (4.7)) or (19, pp. 32?35).
Theorem 4.6. Suppose Re t > 0. Then
h1(?; e
?2pit) =
1
9t2
h2
(
i?
3t
; e?
2pi
3t
)
?
1
6pit
,
h2(?; e
?2pit) =
1
t2
h1
(
i?
t
; e?
2pi
3t
)
+
1
2pit
.
Proof
Differentiating the result in Lemma 4.5 with respect to ? gives
??(?; e?2pit) = ?
1
t2
??
(
i?
t
; e?
2pi
t
)
?
1
4pit
.
Replacing t with 3t in Lemma 4.5 and then differentiating with respect to ? gives
??(?; e?6pit) = ?
1
9t2
??
(
i?
3t
; e?
2pi
3t
)
?
1
12pit
.
Therefore, on writing q = e?2pit and p = e?2pi/3t, we have
h1(?; q) = ?
?(?; q)? ??(?; q3)
= ?
1
t2
??
(
i?
t
; p3
)
?
1
4pit
+
1
9t2
??
(
i?
3t
; p
)
+
1
12pit
=
1
9t2
(
??
(
i?
3t
; p
)
? 9??
(
i?
t
; p3
))
?
1
6pit
=
1
9t2
h2
(
i?
3t
; p
)
?
1
6pit
.
This proves the first part of the theorem. The second part follows from the first by replacing t
with 1/3t, and then replacing ? with ?/it.
34 S. Cooper
4.4 Transformation of Eisenstein series
Corollary 4.7. Suppose Re t > 0 and let q = e?2pit, p = e?
2pi
3t . Then, for m ? 0,
S2m(q) =
(?1)m
?
3
(3t)2m+1
E2m(?3; p),
E2m(?3; q) =
(?1)m
t2m+1
?
3
S2m(p).
Proof
These follow by expanding the equations in Theorem 4.3 in powers of ? using Theorem 3.1, and
equating coefficients of ?2m.
Remark 4.8. A different proof of Corollary 4.7 was given by Chan and Liu (7). Another proof of
this result (for m ? 1) will be given in the next section.
Corollary 4.9. Suppose Re t > 0 and let q = e?2pit, p = e?2pi/3t. Then
E2(q) = ?
1
t2
E2(p
3)?
1
4t
, (4.10)
E(1)2 (q) =
1
9t2
E(2)2 (p)?
1
6pit
, (4.11)
E(2)2 (q) =
1
t2
E(1)2 (p) +
1
2pit
, (4.12)
and for n = 2, 3, 4, ? ? ? , we have
E2m(q) =
(?1)m
t2m
E2m(p
3), (4.13)
E(1)2m(q) =
(?1)(m?1)
32mt2m
E(2)2m(p), (4.14)
E(2)2m(q) =
(?1)(m?1)
t2m
E(1)2m(p). (4.15)
Proof
Substitute the series expansions from Lemma 3.1 into Lemma 4.5 and Theorem 4.6.
5 More on Eisenstein series
The Bernoulli numbers {Bn} are defined by
x
ex ? 1
=
??
n=0
Bn
xn
n!
,
and it is well known (for example, see (1, p. 12)), that for each positive integer n,
??
k=1
1
k2n
=
(?1)n+122n?1
(2n)!
B2npi
2n.
The analogous result for the cubic Bernoulli numbers is
Cubic elliptic functions 35
Theorem 5.1. Let the cubic Bernoulli numbers {sn} be defined by
1
2
sinh x2
sinh 3x2
=
??
n=0
sn
xn
n!
,
or equivalently by
1
2
sin x2
sin 3x2
=
??
n=0
(?1)nsn
xn
n!
.
Then
1. For all complex numbers ? ,
??
k=??
?3(k)
? + k
=
2pi
?
3
sin pi?3
sinpi?
. (5.2)
The series on the left converges for all complex values of ? (except for integer values of the
form 3k?1, where there are simple poles), but not absolutely. An absolutely convergent series
is given by
??
k=??
?3(k)
(
1
? + k
?
1
k
)
=
2pi
?
3
sin pi?3
sinpi?
. (5.3)
2. For each non-negative integer n,
??
k=1
?3(k)
k2n+1
=
(?1)n22n+1
32n+
1
2 (2n)!
s2npi
2n+1.
3. (?1)ns2n > 0.
Proof
Starting with the partial fractions expansion of the cotangent (1, p. 11)
pi cotpix =
1
x
+
??
n=1
(
1
x+ n
+
1
x? n
)
,
we obtain
pi
3
(
cot
pi
3
(? + 1)? cot
pi
3
(? ? 1)
)
=
1
? + 1
+
(
1
? + 4
+
1
? ? 2
)
+
(
1
? + 7
+
1
? ? 5
)
+ ? ? ?
?
1
? ? 1
?
(
1
? + 2
+
1
? ? 4
)
?
(
1
? + 5
+
1
? ? 7
)
+ ? ? ?
=
??
k=??
?3(k)
? + k
. (5.4)
Next, using
cot(x+ y)? cot(x? y) =
2 sin 2y
cos 2x? cos 2y
and
1
1 + 2 cos 2x
=
sinx
sin 3x
,
36 S. Cooper
we get
pi
3
(
cot
pi
3
(? + 1)? cot
pi
3
(? ? 1)
)
=
2pi/
?
3
1 + 2 cos 2pi?3
=
2pi
?
3
sin pi?3
sinpi?
. (5.5)
Equating (5.4) and (5.5) proves (5.2).
The series on the left hand side of (5.2) can be seen to be convergent by considering the real
and imaginary parts of the terms. It is clear that the series is not absolutely convergent, since
?
?
?
?3(k)
?+k
?
?
? = O
(
1
k
)
. The series (5.2) and (5.3) converge to the same value because
?3(k)
k
= ?
?3(?k)
?k
,
and the series in (5.3) converges absolutely because
?
?
??3(k)
(
1
?+k ?
1
k
)?
?
? = O
(
1
k2
)
. This completes
the proof of the first part of the Theorem.
Expanding the left hand side of (5.2) in powers of ? gives
??
k=??
?3(k)
? + k
= 2
??
k=1
k?3(k)
k2 ? ?2
= 2
??
k=1
?3(k)
k
1
(
1? ?
2
k2
)
= 2
??
k=1
?3(k)
k
??
n=0
?2n
k2n
= 2
??
n=0
(
??
k=1
?3(k)
k2n+1
)
?2n, (5.6)
valid for |? | < 1. Expanding the right hand side of (5.2) in powers of ? gives
2pi
?
3
sin pi?3
sinpi?
=
4pi
?
3
??
n=0
(?1)ns2n
(2n)!
(
2pi?
3
)2n
. (5.7)
Equating coefficients of ?2n in (5.6) and (5.7) completes the proof of the second part of the Theorem.
The third part of the Theorem follows immediately from the second part, since the sum of the
series is positive.
Lemma 5.8. Suppose Im ? > 0. Then
??
k=??
1
(? + k)n+1
=
(?2pii)n+1
n!
??
k=1
kne2pii?k, if n ? 1, (5.9)
??
k=??
?3(k)
(? + k)n+1
=
2pi
n!
?
3
(
?
2pii
3
)n ??
k=1
?3(k)k
ne2pii?k/3, if n ? 0.
(5.10)
Proof
The first of these is a standard result, for example, see (13, p. 226, eq. (8.9)) or (14, p. 65, Th.
Cubic elliptic functions 37
4). We shall prove the second part. By Theorem 5.1 we have, on writing u = e2pii?/3,
??
k=??
?3(k)
? + k
=
2pi
?
3
sin pi?3
sinpi?
=
2pi
?
3
(u1/2 ? u?1/2)
(u3/2 ? u?3/2)
=
2pi
?
3
u(1? u)
1? u3
=
2pi
?
3
??
k=1
?3(k)u
k
=
2pi
?
3
??
k=1
?3(k)e
2pii?k/3.
This proves the k = 0 case of the Lemma. The general case follows from this by first rewriting the
left hand side as an absolutely convergent series using (5.3):
??
k=??
?3(k)
(
1
? + k
?
1
k
)
=
2pi
?
3
??
k=1
?3(k)e
2pii?/3,
and then differentiating n times with respect to ? .
Theorem 5.11. Suppose Re t > 0, let ? = it, q = e?2pit = e2pii? . Then
? ?
(m,n) 6=(0,0)
1
(m+ n?)2j
= 2
(2pii)2j
(2j ? 1)!
E2j(q) (5.12)
??
m=??
?
n 6?0 (mod 3)
1
(m+ n?)2j
= 2
(2pii)2j
(2j ? 1)!
E(1)2j (q) (5.13)
?
m6?0 (mod 3)
??
n=??
1
(m+ n?)2j
=
?2
(2j ? 1)!
(
2pii
3
)2j
E(2)2j (q
1
3 ) (5.14)
??
m=??
??
n=??
?3(n)
(m+ n?)2j+1
= ?2
(2pii)2j+1
(2j)!
S2j(q) (5.15)
??
m=??
??
n=??
?3(m)
(m+ n?)2j+1
=
?2i
?
3
(2j)!
(
2pii
3
)2j+1
E2j(?3; q
1
3 ).
(5.16)
Equations (5.12)?(5.14) hold for j ? 2, and (5.15)?(5.16) hold for j ? 1.
Proof
The first of these is a standard result, for example, see (13, p. 226, eq. (8.10)). The second and
38 S. Cooper
third results follow from the first. For example
??
m=??
?
n 6?0 (mod 3)
1
(m+ n?)2j
=
? ?
(m,n) 6=(0,0)
1
(m+ n?)2j
?
? ?
(m,n) 6=(0,0)
1
(m+ 3n?)2j
= 2
(2pii)2j
(2j ? 1)!
E2j(q)? 2
(2pii)2j
(2j ? 1)!
E2j(q
3)
= 2
(2pii)2j
(2j ? 1)!
E(1)2j (q),
and this is (5.13). Equation (5.14) is obtained similarly.
Next, by Lemma 5.8,
??
m=??
??
n=??
?3(n)
(m+ n?)2j+1
= 2
??
n=1
?3(n)
??
m=??
1
(m+ n?)2j+1
= 2
??
n=1
?3(n)
??
k=1
(?2pii)2j+1
(2j)!
k2jqnk
= ?2
(2pii)2j+1
(2j)!
??
k=1
k2j
??
n=1
?3(n)q
nk
= ?2
(2pii)2j+1
(2j)!
??
k=1
k2j
qk ? q2k
1? q3k
= ?2
(2pii)2j+1
(2j)!
S2j(q).
This proves (5.15). Finally, by Theorem 5.1 and Lemma 5.8,
??
m=??
??
n=??
?3(m)
(m+ n?)2j+1
=
?
m6=0
?3(m)
m2j+1
+
??
m=??
?
n 6=0
?3(m)
(m+ n?)2j+1
= 2
??
m=1
?3(m)
m2j+1
+ 2
??
n=1
(
??
m=??
?3(m)
(m+ n?)2j+1
)
= 2
??
m=1
?3(m)
m2j+1
+
4pi
(2j)!
?
3
(
?
2pii
3
)2j ??
n=1
??
m=1
?3(m)m
2je2piimn?/3
=
(?1)j22j+2pi2j+1
32j+
1
2 (2j)!
s2j +
(?1)j22j+2pi2j+1
32j+
1
2 (2j)!
??
m=1
?3(m)m2jq
m
3
1? q
m
3
=
(?1)j22j+2pi2j+1
32j+
1
2 (2j)!
E2j(?3; q
1
3 ),
which proves (5.16).
Remark 5.17. Corollary 4.7 in the case m ? 1, and Corollary 4.9 in the case m ? 2 follow
immediately from Theorem 5.11. The cases m = 0 of Corollary 4.7 and m = 1 of Corollary 4.9
Cubic elliptic functions 39
do not follow from Theorem 5.11, because the double series on the left hand sides do not converge
absolutely.
Corollary 5.18. Let ?mn = 2pim+ 2piint and ??mn =
2pim
3 + 2piint. Then
g1(?; q) = g1(0; q) +
1
2i
??
m=??
??
n=??
?3(n)
[
1
? ? ?mn
+
?
(?mn)2
+
1
?mn
]
,
g2(?; q) = g2(0; q)?
1
2
?
3
??
m=??
??
n=??
?3(m)
[
1
? ? ??mn
+
?
(??mn)2
+
1
??mn
]
.
Proof
The functions g??1 (?; q) and
1
i
??
m=??
??
n=??
?3(n)
(? ? ?mn)3
are both doubly periodic with periods 2pi and
6piit. Furthermore, they both have poles at ? = ?mn, n 6? 0 (mod 3), and no other singularities,
and the singular parts of both functions at each pole are identical. Consequently their difference
is an entire, doubly periodic function, which is therefore constant, i.e.,
g??1 (?; q) =
1
i
??
m=??
??
n=??
?3(n)
(? ? ?mn)3
+ c.
The value of c may be found by plugging in a value for ?, for example ? = 0. Using Theorems 3.1
and 5.11, we obtain c = 0.
Applying
? ?
0 d? to both sides and using the fact that g
?
1(0; q) = 0, we get
g?1(?; q) = ?
1
2i
??
m=??
??
n=??
?3(n)
[
1
(? ? ?mn)2
?
1
(?mn)2
]
.
Applying
? ?
0 d? again we complete the proof of the first part of the theorem.
The second part may be proved similarly, or obtained from the first part using the modular
transformation.
6 Connection between cubic elliptic functions and cubic theta functions
The cubic elliptic functions g1(?; q), g2(?; q) are related to the cubic theta functions b(q, z), c(q, z)
as follows.
Theorem 6.1.
g1(?; q) =
1
4
(q; q)2?(q
3; q3)2?
(q2; q2)?(q6; q6)?
b(q,?ei?)
b(q, ei?)
?
1
12
b(q)2
b(q2)
(6.2)
g2(?; q) =
1
2
q
1
2
(q; q)2?(q
3; q3)2?
(q
1
2 ; q
1
2 )?(q
3
2 ; q
3
2 )?
ei?
c(q, q
1
2 ei?)
c(q, ei?)
?
1
6
c(q)2
c(q
1
2 )
. (6.3)
Proof
As usual, let q = e?2pit, Re t > 0. Let us put
B(?; q) =
b(q,?ei?)
b(q, ei?)
.
40 S. Cooper
Clearly B(?; q) is periodic with period 2pi. Next, from (2.28) or (2.30) we find (12, (1.17)) that
b(q, z) = z2q3b(q, zq3),
and therefore B(?; q) is periodic with period 6piit.
From (2.30), we see that B(?; q) has simple poles at ? = 2pim+ 2piint, m, n ? Z, n 6? 0 (mod 3),
and no other singularities. We calculate the residue at 2piit:
Res(B(?; q); ? = 2piit)
= lim
??2piit
(? ? 2piit)
b(e?2pit,?ei?)
b(e?2pit, ei?)
= lim
??2piit
(? ? 2piit)
(1? e?2pit?i?)
(?qei?; q)?(?qe?i?; q)?
(?q3ei?; q3)?(?q3e?i?; q3)?
(q3ei?; q3)?(q3e?i?; q3)?
(qei?; q)?(q2e?i?; q)?
= ?i
(?q2; q)?(?1; q)?
(?q4; q3)?(?q2; q3)?
(q4; q3)?(q2; q3)?
(q2; q)?(q; q)?
= ?2i
(q2; q2)?(q6; q6)?
(q; q)2?(q3; q3)2?
,
after simplification. Similarly,
Res(B(?; q); ? = 4piit) = 2i
(q2; q2)?(q6; q6)?
(q; q)2?(q3; q3)2?
,
and by the periodicity properties of B, we obtain
Res(B(?; q); ? = 2pim+ 2piint) =
2
i
?3(n)
(q2; q2)?(q6; q6)?
(q; q)2?(q3; q3)2?
.
By Theorem 2.20, it follows that
g1(?; q)?
1
4
(q; q)2?(q
3; q3)2?
(q2; q2)?(q6; q6)?
b(q,?ei?)
b(q, ei?)
is doubly periodic and entire. Therefore by Liouville?s theorem it is constant. Letting ? = 0 we
find that the value of the constant is given by
g1(0; q)?
1
4
(q; q)2?(q
3; q3)2?
(q2; q2)?(q6; q6)?
b(q,?1)
b(q, 1)
=
1
6
a(q)?
1
4
(q; q)2?(q
3; q3)2?
(q2; q2)?(q6; q6)?
(?q; q)2?
(?q3; q3)2?
(q3; q3)2?
(q; q)2?
=
1
6
a(q)?
1
4
(q3; q3)6?
(q; q)2?
(q2; q2)?
(q6; q6)3?
=
1
6
a(q)?
1
12
c(q)2
c(q2)
= ?
1
12
b(q)2
b(q2)
.
The last step follows by (12, (1.29)). This completes the proof of the first part of the Theorem.
The second part may be proved similarly. Alternatively, it can be deduced from the first part by
applying the modular transformation and using Theorems 4.1 and 4.3.
Remark 6.4. Equation (6.2) was proved by Berndt et. al. (4, Lemma 8.2). The proof we have
given here is simpler.
Cubic elliptic functions 41
7 Connection between cubic elliptic functions and the Weierstrass ?
function
In this section, we establish connections between the cubic elliptic functions g1, g2 and the Weier-
strass ? function. The results in this section will be used in Section 12 to obtain differential
equations for g1 and g2, and also in Section 13 to prove recurrence relations for the Eisenstein
series S2n(q) and E2n(?3; q).
Theorem 7.1.
(g1(?; q)? g1(0; q))
(
??(?; q3)? ??(2piit; q3)
)
=
S2(q)
4
,
(g2(?; q)? g2(0; q))
(
??(?; q)? ??
(
2pi
3
; q
))
=
E2(?3; q)
4
.
Proof
Let
?(?) = g1(?; q)? g1(0; q),
?(?) = ??(?; q3)? ??(2piit; q3).
By Theorems 2.20 and 3.1,
? ? is periodic with periods 2pi and 6piit;
? ? has simple poles at ? = 2pim+ 2piint, n 6? 0 (mod 3), and no other singularities;
? ? has zeros of order 2 at ? = 2pim+ 6piint, and no other zeros.
Similarly, by Theorem 2.21, and the fact that ?? is an even function, we have that
? ? is periodic with periods 2pi and 6piit;
? ? has simple zeros at ? = 2pim+ 2piint, n 6? 0 (mod 3), and no other zeros;
? ? has poles of order 2 at ? = 2pim+ 6piint, and no other singularities.
It follows that the product ?(?)?(?) is a doubly periodic function with no zeros or poles, and
therefore is a constant. Letting ? ? 0 and using the expansions in Theorem 3.1, we find that the
value of the constant is S2(q)/4. This proves the first part of the theorem.
The second part may either be proved similarly, or obtained from the first part using the transfor-
mation t? 1/3t.
The Weierstrass ? function with periods ?1 and ?2 is defined by
?(?;?1, ?2) =
1
?2
+
? ?
(m,n) 6=(0,0)
(
1
(? ?m?1 ? n?2)2
?
1
(m?1 + n?2)2
)
.
It can be shown that (9), (10)
?(?; 2pi, 2piit) = ?2??(?; q)?
P (q)
12
, (7.2)
where q = e?2pit. Using this in Theorem 7.1, we obtain
42 S. Cooper
Theorem 7.3.
g1(?; q)? g1(0; q) =
S2(q)
2 (?(2piit; 2pi, 6piit)? ?(?; 2pi, 6piit))
,
g2(?; q)? g2(0; q) =
E2(?3; q)
2
(
?( 2pi3 ; 2pi, 2piit)? ?(?; 2pi, 2piit)
) .
A slightly different formulation of Theorem 7.3 will be given in Section 12.
We conclude this Section by observing that addition formulas for g1 and g2 which express g1(?+?)
(resp. g2(?+ ?)) in terms of g1(?), g1(?), g?1(?) and g
?
1(?), (resp. g2(?), g2(?), g
?
2(?) and g
?
2(?))
can be obtained from Theorem 7.3 and the addition formula for the Weierstrass ? function:
?(?+ ?) + ?(?) + ?(?) =
1
4
(
??(?)? ??(?)
?(?)? ?(?)
)2
.
8 Infinite product representations
The functions g1 and g2 may be expressed as differences of infinite products.
Theorem 8.1.
g1(?; q) =
(q; q)3?
(q3; q3)?
(
sin( ?2 +
pi
3 )
sin ?2
??
n=1
1? 2qn cos(? + 2pi3 ) + q
2n
1? 2qn cos ? + q2n
?
sin( ?2 ?
pi
3 )
sin ?2
??
n=1
1? 2qn cos(? ? 2pi3 ) + q
2n
1? 2qn cos ? + q2n
)
(8.2)
g2(?; q) =
3(q3; q3)3?
(q; q)?
(
e?i?
(q2e3i?, qe?3i?; q3)?
(q3e3i?, e?3i?; q3)?
?ei?
(qe3i?, q2e?3i?; q3)?
(e3i?, q3e?3i?; q3)?
)
. (8.3)
Proof
Equation (8.2) follows from (2.1), (2.7) and (2.12), and some simplification. Equation (8.3) is
obtained from (2.1) and (2.19). Alternatively, (8.3) can be proved by applying the modular trans-
formation t? 13t to (8.2).
The derivatives g?1 and g
?
2 can both be written as single infinite products.
Theorem 8.4.
d
d?
g1(?; q) = ?q(q; q)?(q
3; q3)3? sin ?
(q3e2i?, q3e?2i?; q3)?(q3ei?, q3e?i?; q3)2?
(qei?, qe?i?; q)2?
d
d?
g2(?; q) = (q; q)
3
?(q
3; q3)? sin ?(qe
2i?, qe?2i?; q)?
?
sin2 ?2
sin2 3?2
(qei?, qe?i?; q)2?
(q3e3i?, q3e?3i?; q3)2?
.
Cubic elliptic functions 43
Proof
dg1
d?
= ?
??
n=1
nqn sinn?
1 + qn + q2n
= ?
1
2i
??
n=1
n(qn ? q2n)(ein? ? e?in?)
1? q3n
= ?
1
2i
??
n=1
(
nqnein?
1? q3n
?
nqne?in?
1? q3n
?
nq2nein?
1? q3n
+
nq2ne?in?
1? q3n
)
. (8.5)
Observe that
??
n=1
nqnein?
1? q3n
=
??
n=1
n(qn ? q4n + q4n)ein?
1? q3n
=
??
n=1
nqnein? +
??
n=1
nq4nein?
1? q3n
=
qei?
(1? qei?)2
+
??
n=1
nq4nein?
1? q3n
= ?
1
4
csc2
? + 2piit
2
+
??
n=1
nq4nein?
1? q3n
,
and similarly
??
n=1
nq2nein?
1? q3n
= ?
1
4
csc2
? + 4piit
2
+
??
n=1
nq5nein?
1? q3n
.
Substituting these into (8.5) and using (1.1) we get
dg1
d?
= ?
1
2i
(
?
1
4
csc2
? + 2piit
2
+ 2
??
n=1
nq3n
1? q3n
cosn(? + 2piit)
)
+
1
2i
(
?
1
4
csc2
? + 4piit
2
+ 2
??
n=1
nq3n
1? q3n
cosn(? + 4piit)
)
= i
(
??(? + 2piit; e?6pit)? ??(? + 4piit; e?6pit)
)
.
Finally, using (1.6) and (2.1), this becomes
dg1
d?
=
i
2
F (ei??4pit, ei??2pit; e?6pit)F (e?i?+4pit, ei??2pit; e?6pit)
=
i
2
(q3e2i?, e?2i?, q?1, q4, q3, q3, q3, q3; q3)?
(q2ei?, qe?i?, qei?, q2e?i?, q?2e?i?, q5ei?, qei?, q2e?i?; q3)?
=
i
2
(1? e?2i?)
(1? q?1)
(1? q)
(1? q2ei?)
(1? q?2e?i?)
?
(q3e2i?, q3e?2i?; q3)?(q; q)?(q3; q3)3?
(qei?, q2ei?, qe?i?, q2e?i?; q3)2?
= ?q(q; q)?(q
3; q3)3? sin ?
(q3e2i?, q3e?2i?; q3)?(q3ei?, q3e?i?; q3)2?
(qei?, qe?i?; q)2?
.
This proves the first part of the theorem.
The second part may be proved similarly, or by using the modular transformation t? 1/3t.
44 S. Cooper
Remark 8.6. The first part of Theorem 8.4 was proved by Berndt et. al. (4, Lemma 8.5), using
the 6?6 summation formula. The proof we have given here is simpler.
9 A multiplicative identity for G
In this section, we give a multiplicative identity for G, which is analogous to Venkatachaliengar?s
identity (1.5) for F . We begin with some preparatory lemmas.
Lemma 9.1.
F (x, y; q) + F (x, ?y; q) + F (x, ?2y; q) = 3F (x, y3; q3).
?(x; q) + ?(?x; q) + ?(?2x; q) = 3?(x3; q3),
Proof
The first of these follows immediately from the series representation (2.5). The second follows from
(1.3).
Lemma 9.2. Let
f(x) =
??
n=??
cnx
n,
in some annulus r1 < |x| < r2. Let
sift(f(x);x,m, k) =
??
n=??
cmn+kx
mn+k.
Then
sift(?(x; q);x, 3, 0) = ?(x3; q3),
sift(?(x; q);x, 3, 1) = xF (x3, q; q3),
sift(?(x; q);x, 3, 2) = x2F (x3, q2; q3),
sift(F (x, y; q); y, 3, 0) = F (x, y3; q3),
sift(F (x, y; q); y, 3, 1) = yF (qx, y3; q3),
sift(F (x, y; q); y, 3, 2) = y2F (q2x, y3; q3).
Proof
These all follow directly from the series representations (2.5) and (1.3).
Lemma 9.3. Suppose the series
??
n=?? anx
n and
??
n=?? bnx
n both converge in the annulus
r1 < |x| < r2. Then in this annulus,
??
n=??
anx
n
??
n=??
bnx
n +
?
n
an?
2nxn
?
n
?nbnx
n +
?
n
an?
nxn
?
n
?2nbnx
n
= 3
[
?
n
a3nx
3n
?
n
b3nx
3n +
?
n
a3n+1x
3n+1
?
n
b3n+1x
3n+1
+
?
n
a3n+2x
3n+2
?
n
b3n+2x
3n+2
]
.
Cubic elliptic functions 45
Proof
Write
?
anx
n =
?
a3nx
3n +
?
a3n+1x
3n+1 +
?
a3n+2x
3n+2
and sift each of the other series
?
an?nxn,
?
an?2nxn,
?
bnxn,
?
bn?nxn and
?
bn?2nxn sim-
ilarly. The result follows by expanding and simplifying.
Theorem 9.4.
G(x, y; q)G(x, z; q)
= x
?
?x
(
F (x, yz; q)? F (x, y3z3; q3)
)
+
(
F (x, yz; q)? F (x, y3z3; q3)
) (
?(y3; q3) + ?(z3; q3)
)
?yzF (qx, y3z3; q3)
(
yF (q, y3; q3) + zF (q, z3; q3)
)
?y2z2F (q2x, y3z3; q3)
(
y2F (q2, y3; q3) + z2F (q2, z3; q3)
)
.
Proof
Using the definition (2.7) and the multiplicative identity (1.5), we obtain
G(x, y; q)G(x, z; q)
= ?
1
3
(
F (x, ?y; q)? F (x, ?2y; q)
) (
F (x, ?z; q)? F (x, ?2z; q)
)
= ?
1
3
(
F (x, ?y; q)F (x, ?z; q) + F (x, ?2y; q)F (x, ?2z; q)
?F (x, ?2y; q)F (x, ?z; q)? F (x, ?y; q)F (x, ?2z; q)
)
= ?
1
3
x
?
?x
(
F (x, ?yz; q) + F (x, ?2yz; q)? 2F (x, yz; q)
)
?
1
3
F (x, ?2yz; q)(?(?y; q) + ?(?z; q))
?
1
3
F (x, ?yz; q)(?(?2y; q) + ?(?2z; q))
?
1
3
F (x, yz; q)(?(?y; q) + ?(?2y; q) + ?(?z; q) + ?(?2z; q)).
Applying Lemmas 9.1, 9.2 and 9.3 to this gives
G(x, y; q)G(x, z; q)
= x
?
?x
(
F (x, yz; q)? F (x, y3z3; q3)
)
+
1
3
F (x, yz; q) (?(y; q) + ?(z; q))
?F (x, y3z3; q3)
(
?(y3; q3) + ?(z3; y3)
)
?yzF (qx, y3z3; q3)
(
yF (q, y3; q3) + zF (q, z3; q3)
)
?y2z2F (q2x, y3z3; q3)
(
y2F (q2, y3; q3) + z2F (q2, z3; q3)
)
+
1
3
F (x, yz; q)
(
3?(y3; q3)? ?(y; q) + 3?(z3; q3)? ?(z; q)
)
.
Simplifying, we complete the proof.
46 S. Cooper
10 Squares
An important special case of Theorem 9.4 is the limiting case y ? 1/z. We begin by proving a
number of preliminary lemmas which will be useful in computing this limit. Let
?(x; q) = x
d
dx
?(x; q)
=
??
n=??
n6=0
nxn
1? qn
, if |q| < |x| < 1,
=
x
(1? x)2
+
??
n=1
nqn(xn + x?n)
1? qn
, if |q| < |x| < |q|?1, (10.1)
?(x; q) = x
d
dx
F (q, x; q3)
=
??
n=??
n6=0
nxn
1? q3n+1
, if |q3| < |x| < 1,
=
x
(1? x)2
+
??
n=1
(
nq3n+1xn
1? q3n+1
+
nq3n?1x?n
1? q3n?1
)
, if |q|3 < |x| < |q|?3.
(10.2)
Observe that, from (1.1), (2.24) and (2.25), we have
?(ei?; q) = 2??(?; q), (10.3)
?(ei?; q)? ?(ei?; q3) = 2h1(?; q), (10.4)
?(ei?; q)? 9?(e3i?; q3) = 2h2(?; q). (10.5)
Lemma 10.6.
lim
y?1/z
x
?
?x
F (x, yz; q) = ?(x; q),
lim
y?1/z
x
?
?x
F (x, y3z3; q3) = ?(x; q3).
Proof
The first of these follows by expanding F in powers of x using (2.5), computing the partial derivative
and then evaluating the limit. The second part follows from the first part, by replacing y, z and q
by their cubes.
Lemma 10.7.
lim
t?1
(1? t)F (x, t; q) = 1,
lim
y?1/z
(1? y3z3)F (x, yz; q) = 3,
lim
y?1/z
(1? y3z3)F (x, y3z3; q) = 1.
Proof
The first part follows from the definition (2.1). The second and third parts follow using 1?y3z3 =
(1? yz)(1 + yz + y2z2) and simple changes of variable.
Cubic elliptic functions 47
Lemma 10.8.
lim
y?1/z
?(y3; q3) + ?(z3; q3)
1? y3z3
= ??(z3; q3),
lim
y?1/z
F (q, z3; q3)? F (q, y?3; q3)
1? y3z3
= ??(z3; q).
Proof
From (1.4), we obtain ?(t; q) = ??(t?1; q). Therefore, putting w = y?1 we get
lim
y?1/z
?(y3; q3) + ?(z3; q3)
1? y3z3
= ? lim
w?z
w3
?(z3; q3)? ?(w3; q3)
z3 ? w3
= ?z3
d
dt
?(t; q3)
?
?
?
?
t=z3
= ??(z3; q3).
Similarly,
lim
y?1/z
F (q, z3; q3)? F (q, y?3; q3)
1? y3z3
= lim
w?z
?w3
F (q, z3; q3)? F (q, w3; q3)
z3 ? w3
= ?z3
d
dt
F (q, t; q3)
?
?
?
?
t=z3
= ??(z3; q).
Lemma 10.9.
lim
y?1/z
yz2F (qx, y3z3; q3)F (q, z3; q3) + y4z2F (q2x, y3z3; q3)F (q2, y3; q3)
= ?z?(z3; q) + zF (q, z3; q3)
(
G(x, 1; q)?
1
3
)
,
lim
y?1/z
y2zF (qx, y3z3; q3)F (q, y3; q3) + y2z4F (q2x, y3z3; q3)F (q2, z3; q3)
= ?z?1?(z?3; q) + z?1F (q, z?3; q3)
(
G(x, 1; q)?
1
3
)
.
Proof
First, using (2.3) and (2.4), we have
F (q2, y3; q3) = ?y?3F (q, y?3; q3).
48 S. Cooper
Using this, together with Lemma 10.8, we obtain
lim
y?1/z
yz2F (qx, y3z3; q3)F (q, z3; q3) + y4z2F (q2x, y3z3; q3)F (q2, y3; q3)
= lim
y?1/z
yz2
(
F (qx, y3z3; q3)F (q, z3; q3)? F (q2x, y3z3; q3)F (q, y?3; q3)
)
= z lim
y?1/z
F (qx, y3z3; q3)(1? y3z3)
(
F (q, z3; q3)? F (q, y?3; q3)
1? y3z3
)
+z lim
y?1/z
F (q, y?3; q3)
(
F (qx, y3z3; q3)? F (q2x, y3z3; q3)
)
= ?z?(z3; q) + zF (q, z3; q3)
?
n 6=0
(
qnxn
1? q3n
?
q2nxn
1? q3n
)
= ?z?(z3; q) + zF (q, z3; q3)
?
n 6=0
xnqn
1 + qn + q2n
= ?z?(z3; q) + zF (q, z3; q3)
(
G(x, 1; q)?
1
3
)
.
This proves the first part. The second part follows from this by change of variable.
Lemma 10.10.
zF (q, z3; q3) + z?1F (q, z?3; q3) = G(1, z; q),
z?(z3; q) + z?1?(z?3; q) =
1
3
?(z; q)? ?(z3; q3)?
1
3
G(1, z; q).
Proof
The first of these follows from (2.11), using (2.2). The second is proved by series manipulations.
From (10.1) and (10.2), we have
1
3
?(z; q)? ?(z3; q3)? z?(z3; q)? z?1?(z?3; q)
=
1
3
z
(1? z)2
+
1
3
??
n=1
nqn
1? qn
(zn + z?n)
?
z3
(1? z3)2
?
??
n=1
nq3n
1? q3n
(z3n + z?3n)
?
z4
(1? z3)2
?
??
n=1
(
nq3n+1z3n+1
1? q3n+1
+
nq3n?1z?3n+1
1? q3n?1
)
?
z?4
(1? z?3)2
?
??
n=1
(
nq3n+1z?3n?1
1? q3n+1
+
nq3n?1z3n?1
1? q3n?1
)
=
1
3
z
(1? z)2
?
z3
(1? z3)2
?
z4
(1? z3)2
?
z?4
(1? z?3)2
+
1
3
??
n=1
(3n? 2)q3n?2
1? q3n?2
(z3n?2 + z?(3n?2))
+
1
3
??
n=1
(3n? 1)q3n?1
1? q3n?1
(z3n?1 + z?(3n?1))
?
??
n=0
nq3n+1
1? q3n+1
(z3n+1 + z?(3n+1))?
??
n=1
nq3n?1
1? q3n?1
(z3n?1 + z?(3n?1)).
Cubic elliptic functions 49
Simplifying, we obtain
1
3
?(z; q)? ?(z3; q3)? z?(z3; q)? z?1?(z?3; q)
=
z
3(1 + z + z2)
+
1
3
??
n=1
q3n?2
1? q3n?2
(z3n?2 + z?(3n?2))
?
1
3
??
n=1
q3n?1
1? q3n?1
(z3n?1 + z?(3n?1))
=
z
3(1 + z + z2)
+
1
3
??
n=1
?3(n)qn
1? qn
(zn + z?n)
=
1
3
G(1, z; q),
by (2.10).
Theorem 10.11.
G(ei?, ei?; q)G(ei?, e?i?; q) + 4g1(?; q)g2(?; q) = 2h1(?; q) +
2
3
h2(?; q).
Proof
First rewrite Theorem 9.4 in the form
G(x, y; q)G(x, z; q)
= x
?
?x
(
F (x, yz; q)? F (x, y3z3; q3)
)
+
(
F (x, yz; q)? F (x, y3z3; q3)
) (
?(y3; q3) + ?(z3; q3)
)
?
(
y2zF (qx, y3z3; q3)F (q, y3; q3) + y2z4F (q2x, y3z3; q3)F (q2, z3; q3)
)
?
(
yz2F (qx, y3z3; q3)F (q, z3; q3) + y4z2F (q2x, y3z3; q3)F (q2, y3; q3)
)
.
Now take the limit as y ? 1/z, using Lemmas 10.6 ? 10.9, to get
G(x, z; q)G(x, z?1; q)
= ?(x; q)? ?(x; q3)? 2?(z3; q3)
+z?(z3; q)? zF (q, z3; q3)
(
G(x, 1; q)?
1
3
)
+z?1?(z?3; q)? z?1F (q, z?3; q3)
(
G(x, 1; q)?
1
3
)
.
Apply Lemma 10.10 to this, and simplify to get
G(x, z; q)G(x, z?1; q)
= ?(x; q)? ?(x; q3) +
1
3
?(z; q)? 3?(z3; q3)?G(x, 1; q)G(1, z; q).
Setting x = ei?, z = ei?, and using the definitions (2.12), (2.13), (10.4) and (10.5), we complete
the proof.
Theorem 10.11 immediately implies the following results of Liu (15, Theorems 5 and 7):
50 S. Cooper
Corollary 10.12.
(
g1(?; q) +
a(q)
12
)2
=
h1(?)
2
+
9P (q3)? P (q)
144
+
a(q)2
144
,
(
g2(?; q) +
a(q)
12
)2
=
h2(?)
6
+
P (q3)? P (q)
48
+
a(q)2
144
.
Proof
Take ? = 0 and ? = 0 in Theorem 10.11, respectively, use (2.24) and (2.25) and complete the
square.
Corollary 10.13.
a(q)2 =
3
2
P (q3)?
1
2
P (q),
E(1)4 (q) = 6S0(q)S2(q),
E(2)4 (q) = 18E0(?3; q)E2(?3; q).
Proof
Letting ? = 0 in either part of Corollary 10.12, and using (2.24) (or (2.25)), we get
a(q)2
16
=
9P (q3)? P (q)
144
+
P (q3)? P (q)
48
+
a(q)2
144
.
Simplifying, we get the first part.
The second and third parts follow by equating coefficients of ?2 in Corollary 10.12.
Remark 10.14. Corollary 10.13 was given by Ramanujan (16, eq. (19)). He obtained it by
putting ? = 2pi/3 in his identity (1.2). A more general result than Corollary 10.13 can be obtained
by equating coefficients of ?2n. This will be given in Section 13.
Corollary 10.15.
G(ei?, ei?; e?2pit)G(ei?, e?i?; e?2pit) =
1
3t2
G(e
?
t , e
?
3t ; e?
2pi
3t )G(e
?
t , e?
?
3t ; e?
2pi
3t ).
Proof
By Theorem 10.11, followed by Theorems 4.3 and 4.6, and then Theorem 10.11 again, we obtain
G(ei?, ei?; e?2pit)G(ei?, e?i?; e?2pit)
= 2h1(?; e
?2pit) +
2
3
h2(?; e
?2pit)? 4g1(?; e
?2pit)g2(?; e
?2pit)
= 2
(
1
9t2
h2
(
i?
3t
; e?
2pi
3t
)
?
1
6pit
)
+
2
3
(
1
t2
h1
(
i?
t
; e?
2pi
3t
)
+
1
2pit
)
?
4
3t2
g2
(
i?
3t
; e?
2pi
3t
)
g1
(
i?
t
; e?
2pi
3t
)
=
1
3t2
(
2h1
(
i?
t
; e?
2pi
3t
)
+
2
3
h2
(
i?
3t
; e?
2pi
3t
)
? 4g1
(
i?
t
; e?
2pi
3t
)
g2
(
i?
3t
; e?
2pi
3t
))
=
1
3t2
G(e
?
t , e
?
3t ; e?
2pi
3t )G(e
?
t , e?
?
3t ; e?
2pi
3t ).
Cubic elliptic functions 51
11 The transcendentals Z and X
Definition 11.1. Let
Z = Z(q) = a(q),
X = X(q) =
c(q)3
a(q)3
.
The main result of the section is Theorem 11.11, which expresses various Eisenstein series in terms
of Z and X. We begin with some lemmas.
Lemma 11.2.
1?X(q) =
b(q)3
a(q)3
,
Z(e?2pit) =
1
t
?
3
Z
(
e?
2pi
3t
)
,
X(e?2pit) = 1?X
(
e?
2pi
3t
)
.
Proof
The first part follows from equation (2.39), and the other two parts follow from Lemma 4.2.
Lemma 11.3.
(
1
8
cot2
?
2
+
1
12
+
??
n=1
nqn
1? qn
(1? cosn?)
)2
=
(
1
8
cot2
?
2
+
1
12
)2
+
1
12
??
n=1
n3qn
1? qn
(5 + cosn?).
Proof
This was given by Ramanujan (18, eq. (18)). It is equivalent to the differential equation satisfied
by the Weierstrass ? function:
???(z) = 6?2(z)? g2/2.
Lemma 11.4.
a
(a2, qa?2; q)?(q; q)6?
(a, qa?1; q)4?
=
a(1 + a)
(1? a)3
+
??
m=1
m2qm
1? qm
(am ? a?m).
Proof
Multiply both sides of (1.6) by ei?/(ei? ? ei?), take the limit as ? ? ?, and finally put a = ei?.
This is equivalent to (20, p. 459, ex. 24).
Lemma 11.5.
b(q)3 = ?9E2(?3; q), (11.6)
c(q)3 = 27S2(q), (11.7)
a(q)4 =
1
10
(
Q(q) + 9Q(q3)
)
. (11.8)
52 S. Cooper
Proof
Let a = e2pii/3 in Lemma 11.4 and simplify to obtain (11.6). Replace q with q3 in Lemma 11.4, let
a = q and simplify to obtain (11.7). Take ? = 2pi/3 in Lemma 11.3 and simplify to get
(
3
2
P (q3)?
1
2
P (q)
)2
=
1
10
(Q(q) + 9Q(q3)).
Now apply Corollary 10.13 on the left to complete the proof of (11.8).
Remark 11.9. Equation (11.7) was given by Ramanujan; see (4, Theorem 8.7). Equation (11.8)
was given by Berndt et. al. (4, Corollary 4.6). Equations (11.6) and (11.7) were given without
proof by J. M. and P. B. Borwein (5, Remark 2.4 (iii)). All of (11.6)?(11.8) were given by Liu
(15, eqs. (1.15), (1.16) and (1.19)). The proof of Lemma 11.5 that we have outlined above is
substantially the same as Liu?s.
Theorem 11.10.
q(q; q)24? =
1
27
Z12X(1?X)3, (11.11)
q3(q3; q3)24? =
1
39
Z12X3(1?X), (11.12)
q
dX
dq
= Z2X(1?X), (11.13)
3
2
P (q3)?
1
2
P (q) = Z2, (11.14)
P (q) = Z2(1? 4X) + 12ZX(1?X)
dZ
dX
, (11.15)
P (q3) = Z2(1?
4
3
X) + 4ZX(1?X)
dZ
dX
, (11.16)
Q(q) = Z4(1 + 8X), (11.17)
Q(q3) = Z4(1?
8
9
X), (11.18)
R(q) = Z6(1? 20X ? 8X2), (11.19)
R(q3) = Z6(1?
4
3
X +
8
27
X2). (11.20)
Proof
Equations (11.11) and (11.12) follow from Definition 11.1, Lemma 11.2 and the infinite product
formulas (2.36) and (2.38).
Taking the logarithm of (11.11) and applying q
d
dq
gives
1? 24
??
n=1
nqn
1? qn
= q
d
dq
log
(
1
27
Z12X(1?X)3
)
,
which is equivalent to
P (q) = q
dX
dq
(
12
Z
dZ
dX
+
1
X
?
3
1?X
)
. (11.21)
Applying the same procedure to (11.12) leads to
3P (q3) = q
dX
dq
(
12
Z
dZ
dX
+
3
X
?
1
1?X
)
. (11.22)
Cubic elliptic functions 53
Subtracting (11.21) from (11.22) and dividing by 2 gives
1
2
(
3P (q3)? P (q)
)
= q
dX
dq
(
1
X
+
1
1?X
)
.
Simplifying using Corollary 10.13 we obtain
q
dX
dq
= Z2X(1?X).
This proves (11.13).
Equation (11.14) is just a restatement of the first part of Corollary 10.13. Equations (11.15) and
(11.16) are obtained by substituting (11.13) into (11.21) and (11.22).
By (3.3), (3.5), Corollary 10.13 and Lemma 11.5, we have
E(1)4 (q) =
1
27
a(q)c(q)3,
E(2)4 (q) = ?
1
3
a(a)b(q)3.
Expressing E(1)4 and E
(2)
4 in terms of Q(q) and Q(q
3) using (3.6), (3.7) and (3.10), and expressing
a(q), b(q), c(q) in terms of X and Z using Definition 11.1 and Lemma 11.2, we get
Q(q)?Q(q3) =
80
9
Z4X,
81Q(q3)?Q(q) = 80Z4(1?X).
Solving for Q(q) and Q(q3) we obtain (11.17) and (11.18).
Next, using Jacobi?s discriminant
Q(q)3 ?R(q)2 = 1728q(q; q)24?,
(see (18, p. 144) for a simple proof), and making use of (11.11) and (11.17), we obtain
R(q)2 = Q(q)3 ? 1728q(q; q)24?
= Z12(1 + 8X)3 ?
1728
27
Z12X(1?X)3
= Z12(1? 20X ? 8X2)2.
Taking square roots and comparing the coefficients of q0 to determine the sign, we obtain (11.19).
Equation (11.20) is obtained in the same way, using (11.12) and (11.18).
12 Differential equations
Lemma 12.1. Let ?(?) = ?(?; 2pi, 2piit) and q = e?2pit. Then
??(?)2 = 4?(?)? g?2?(?)? g?3,
where2
g?2 = 60
?
(m,n) 6=(0,0)
1
(2pin+ 2piimt)4
=
Q(q)
12
,
g?3 = 140
?
(m,n) 6=(0,0)
1
(2pin+ 2piimt)6
=
R(q)
216
.
2The Weierstrassian parameters g?2 and g?3 should not be confused with our functions g1 and g2.
54 S. Cooper
Proof
This is a statement of the differential equation satisfied by the Weierstrass ? function. See, for
example, (8, Ch. 3 and 6) or (19).
Theorem 12.2. Writing g1 = g1(?; q) and g2 = g2(?; q), we have
(
dg1
d?
)2
= ?
2
27
(
g1 ?
Z
6
)
(
54g31 + 27Zg
2
1 ? Z
3(1?X)
)
,
(
dg2
d?
)2
= ?
2
9
(
g2 ?
Z
6
)
(
54g32 + 27Zg
2
2 ? Z
3X
)
.
Proof
Using Lemma 13.5 and the table after Theorem 13.11, observe that Theorem 7.3 may be written
in the form
g1 ?
Z
6
=
?Z3X
54(?+ Z2/12)
. (12.3)
Differentiating and squaring we get
(g?1)
2 =
Z6X2
542
(??)2
(?+ Z2/12)4
.
Using Lemma 12.1 and (12.3), this becomes
(g?1)
2 =
542
Z6X2
(
g1 ?
Z
6
)4(
4?3 ?
Q(q3)
12
??
R(q3)
216
)
.
Using (11.18) and (11.20) and rearranging, we get
(g?1)
2 =
542
Z6X2
(
g1 ?
Z
6
)4(
4?3 ?
Z4(9? 8X)
108
??
Z6(1? 43X +
8
27X
2)
216
)
=
542
Z6X2
(
g1 ?
Z
6
)4
(
4
(
?+
Z2
12
)3
? Z2
(
?+
Z2
12
)2
+
2XZ4
27
(
?+
Z2
12
)
?
X2Z6
272
)
.
Using (12.3) again, this simplifies to
(g?1)
2 = ?4
(
g1 ?
Z
6
)(
g31 +
Z
2
g21 +
Z3(X ? 1)
54
)
.
Rearranging, we complete the proof of the first part of the Theorem.
The second part follows from the first by the modular transformation, using Theorem 4.3 and
Lemma 11.2.
Remark 12.4. The first part of Theorem 12.2 was first proved by Berndt et. al. (4, p. 4209, eq.
(8.32)). The proof we have given here has also been found independently by Chan and Liu (7). It
would be useful to have proof of Theorem 12.2 in the style of Venkatachaliengar (19, pp. 11?13).
Cubic elliptic functions 55
13 Recurrences for the Eisenstein series S2n(q) and E2n(?3; q)
Expanding the results in Corollary 10.12 in powers of ? using the series expansions in Theorem
3.1, we obtain the following results of Liu (15, Theorems 6 and 8):
Theorem 13.1.
1
3
E2(q)? E2(q
3) =
1
4
E(1)2 (q) +
1
12
E(2)2 (q) = S0(q)
2 = E0(?3; q)
2,
and, for n ? 1,
1
2
E(1)2n+2(q) = 3S0(q)S2n(q) +
n?1?
k=1
(
2n
2k
)
S2k(q)S2n?2k(q), (13.2)
1
6
E(2)2n+2(q) = 3E0(?3; q)E2n(?3; q) +
n?1?
k=1
(
2n
2k
)
E2k(?3; q)E2n?2k(?3; q). (13.3)
Remark 13.4. Equation (13.3) can be deduced from (13.2) (and vice versa) by the modular trans-
formation, using Corollaries 4.7 and 4.9.
Lemma 13.5. Let q = e?2pit. Then
??(2piit; q3) =
1
2
(
E2(q)? E2(q
3)
)
,
?(2piit; 2pi, 6piit) = ?
Z2
12
,
??(?; q3)? ??(2piit; q3) = ?
1
2?2
?
3
2
S0(q)
2 +
??
n=1
(?1)n
(2n)!
E2n+2(q
3)?2n.
Proof
From (1.1),
??(?; q) = ?
1
8
csc2
?
2
+
??
n=1
nqn
1? qn
cosn?.
Therefore
??(2piit; q3) =
q
2(1? q)2
+
1
2
??
n=1
nq3n(qn + q?n)
1? q3n
=
q
2(1? q)2
?
1
2
??
n=1
nqn +
1
2
??
n=1
n(qn + q2n)
1? q3n
=
1
2
??
n=1
n(qn + q2n + q3n)
1? q3n
?
1
2
??
n=1
nq3n
1? q3n
=
1
2
??
n=1
nqn
1? qn
?
1
2
??
n=1
nq3n
1? q3n
=
1
2
(E2(q)? E2(q
3)).
This proves the first part.
The second part follows from the first part, using (7.2) and (11.14).
56 S. Cooper
Next, using Theorem 3.1 and the first part of the lemma, we get
??(?; q3)? ?(2piit; q3)
= ?
1
2?2
+
??
n=1
(?1)n?1
(2n? 2)!
E2n(q
3)?2n?2 ?
1
2
(E2(q)? E2(q
3))
= ?
1
2?2
+
(
3
2
E2(q
3)?
1
2
E2(q)
)
+
??
n=1
(?1)n
(2n)!
E2n+2(q
3)?2n.
Applying Theorem 13.1 we complete the proof.
Theorem 13.6. For n = 1, 2, 3, ? ? ? ,
S2n+2(q) = 3(2n+ 1)(2n+ 2)S0(q)
2S2n(q)
?2(2n+ 1)(2n+ 2)
n?1?
j=1
(
2n
2j
)
S2j(q)E2n+2?2j(q
3),
E2n+2(?3; q) = ?9(2n+ 1)(2n+ 2)E0(?3; q)
2E2n(?3; q)
?2(2n+ 1)(2n+ 2)
n?1?
j=1
(
2n
2j
)
E2j(?3; q)E2n+2?2j(q).
Proof
Expand both sides of Theorem (7.1) in powers of ? using Theorem 3.1 and Lemma 13.5, and equate
coefficients of ?2n. This proves the first part. The second part follows from the first, using the
transformation t? 1/3t.
Remark 13.7. Chan and Liu (7) have obtained a formula for S2n purely in terms of S2k, with
k < n, by differentiating the first result in Theorem 12.2.
Lemma 13.8. For n = 2, 3, 4, ? ? ? ,
E2n(q) =
?
2j+3k=n
Kj,kQ(q)
jR(q)k, (13.9)
E2n(q
3) = Z2npn(X), (13.10)
where Kj,k are rational numbers, and pn(X) is a polynomial in X with rational coefficients and
degree b2n/3c.
Proof
A proof of (13.9) has been given by Ramanujan (18, p. 141). Equation (13.10) follows from (13.9)
by induction and making use of (11.18) and (11.20).
Theorem 13.11. S0(q) = E0(?3; q) = Z6 , and for n = 1, 2, 3, ? ? ? ,
S2n(q) = Z
2n+1Pn(X), (13.12)
E2n(?3; q) = 3
nZ2n+1Pn(1?X), (13.13)
where Pn is a polynomial with rational coefficients and degree ? b(2n+ 1)/3c.
Cubic elliptic functions 57
Proof
Equation (13.12) follows by induction from Theorem 13.6, using Lemma 13.8. The bound on the
degree of Pn follows because
b
2j
3
c+ b
2n+ 1? 2j
3
c ? b
2n+ 1
3
c.
Equation (13.13) follows from (13.12) using Corollary 4.7 and Lemma 11.2.
The first few instances of Theorem 13.11 are as follows:
S0 =
1
6
Z
S2 =
1
27
Z3X
S4 =
1
27
Z5X
S6 =
1
27
Z7X
(
1 +
4
3
X
)
S8 =
1
27
Z9X
(
1 + 8X +
80
81
X2
)
S10 =
1
27
Z11X
(
1 + 36X +
848
27
X2
)
S12 =
1
27
Z13X
(
1 +
448
3
X +
12448
27
X2 +
6080
81
X3
)
S14 =
1
27
Z15X
(
1 + 604X +
422432
81
X2 +
289792
81
X3 +
70400
729
X4
)
E0(?3, q) =
1
6
Z
E2(?3, q) =
1
9
Z3(1?X)
E4(?3, q) =
1
3
Z5(1?X)
E6(?3, q) = Z
7(1?X)
(
1 +
4
3
(1?X)
)
E8(?3, q) = 3Z
9(1?X)
(
1 + 8(1?X) +
80
81
(1?X)2
)
E10(?3, q) = 9Z
11(1?X)
(
1 + 36(1?X) +
848
27
(1?X)2
)
E12(?3, q) = 27Z
13(1?X)
(
1 +
448
3
(1?X) +
12448
27
(1?X)2
+
6080
81
(1?X)3
)
E14(?3, q) = 81Z
15(1?X)
(
1 + 604(1?X) +
422432
81
(1?X)2
+
289792
81
(1?X)3 +
70400
729
(1?X)4
)
.
58 S. Cooper
References
[1] G. E. Andrews, R. Askey and R. Roy, Special functions, Encyclopedia of Mathematics and its
Applications, 71. Cambridge University Press, Cambridge, 1999.
[2] B. C. Berndt, Ramanujan?s Notebooks, Part III, Springer-Verlag, 1991.
[3] B. C. Berndt, Ramanujan?s Notebooks, Part V, Springer-Verlag, 1998.
[4] B. C. Berndt, S. Bhargava and F. G. Garvan, Ramanujan?s theories of elliptic functions to
alternative bases, Trans. Amer. Math. Soc., 347 (1995), 4163?4244. Reprinted in (3, Chapter
33, pp. 89?181)
[5] J. M. Borwein and P. B. Borwein, A cubic counterpart of Jacobi?s identity and the AGM,
Trans. Amer. Math. Soc. 323 (1991), no. 2, 691?701.
[6] J. M. Borwein, P. B. Borwein and F. G. Garvan, Some cubic modular identities of Ramanujan,
Trans. Amer. Math. Soc. 343 (1994), no. 1, 35?47.
[7] H. H. Chan and Z.-G. Liu, Analogues of Jacobi?s inversion formula for the incomplete elliptic
integral of the first kind, Advances in Mathematics 174 (2003), 69?88.
[8] K. Chandrasekharan, Elliptic functions, Springer-Verlag, Berlin, 1985.
[9] S. Cooper, The development of elliptic functions according to Ramanujan and Venkatachalien-
gar, Proceedings of The International Conference on the Works of Srinivasa Ramanujan (C.
Adiga and D. D. Somashekara, eds.), University of Mysore, Manasagangotri, Mysore - 570
006, India (2001) 81?99.
Available electronically at http://iims.massey.ac.nz/research/letters/ (see vol. 1, 2000)
[10] S. Cooper, On sums of an even number of squares, and an even number of triangular num-
bers: an elementary approach based on Ramanujan?s 1?1 summation formula, q-Series with
Applications to Combinatorics, Number Theory and Physics (B. C. Berndt and K. Ono, eds.),
Contemporary Mathematics, No. 291, American Mathematical Society, Providence, RI (2001)
115?137.
[11] S. Cooper, Cubic theta functions, Journal of Computational and Applied Mathematics, to
appear.
[12] M. Hirschhorn, F. Garvan and J. Borwein, Cubic analogues of the Jacobian theta functions
?(z, q), Canadian J. Math. 45 (4), 1993, 673?694.
[13] A. W. Knapp, Elliptic curves, Mathematical Notes, 40. Princeton University Press, Princeton,
NJ, 1992
[14] M. I. Knopp, Modular functions in analytic number theory, Markham Publishing Co., Chicago,
Ill. 1970.
[15] Zhi-Guo Liu, Some Eisenstein series identities associated with the Borwein functions, Sym-
bolic computation, number theory, special functions, physics and combinatorics (Gainesville,
FL, 1999), 147?169, Dev. Math., 4, Kluwer Acad. Publ., Dordrecht, 2001.
[16] S. Ramanujan, On certain arithmetical functions, Trans. Camb. Phil. Soc., 22 (9), 1916,
159?184. Reprinted in (18, 136?162).
[17] S. Ramanujan, Notebooks, (2 volumes), Tata Institute of Fundamental Research, Bombay,
1957.
Cubic elliptic functions 59
[18] S. Ramanujan, Collected papers, AMS Chelsea Publishing, Providence, Rhode Island, 2000.
[19] K. Venkatachaliengar, Development of Elliptic Functions according to Ramanujan, Depart-
ment of Mathematics, Madurai Kamaraj University, Technical Report 2, 1988.
[20] E. T. Whittaker and G. N. Watson, A course of modern analysis, Reprint of the fourth (1927)
edition. Cambridge University Press, Cambridge, 1996.