q-SERIES IDENTITIES AND VALUES OF CERTAIN L-FUNCTIONS
George E. Andrews, Jorge Jiménez-Urroz and Ken Ono
Appearing in the Duke Mathematical Journal.
1. Introduction and Statement of Results.
As usual, define Dedekind’s eta function η(z) by the infinite product
(1.1)
η(z) := q
1/24
∞
Y
(1 − q n ).
(q := e2πiz throughout)
n=1
In a recent paper, Zagier [Z; Theorem 2] proved that (note. empty products equal 1 throughout)
(1.2)
∞
X
η(24z) − q(1 − q 24 )(1 − q 48 ) · · · (1 − q 24n ) = η(24z)D(q) + E(q)
n=0
where the series D(q) and E(q) are defined by
∞
∞
1 X q 24n
1 X
1
=− +
d(n)q 24n = − + q 24 + 2q 48 + 2q 72 + 3q 96 + . . . ,
D(q) = − +
24n
2 n=1 1 − q
2 n=1
2
∞
2
1 X 12
1
5
7
11
E(q) =
nq n = q − q 25 − q 49 + q 121 + . . . .
2 n=1 n
2
2
2
2
Here d(n) denotes the number of positive divisors of n. This identity plays an important
role in Zagier’s work on Vassiliev invariants in knot theory [Z].
Two other similar identities were known, and they were noticed by the first author in
connection with one of Ramanujan’s mock theta functions. In [A2], the first author proved
that
∞ X
η(48z)
η(48z)
M1 (q)
24
48
24n
(1.3)
− q(1 + q )(1 + q ) · · · (1 + q ) =
D(q) +
,
η(24z)
η(24z)
2
n=0
The first and third authors thank the National Science Foundation for its generous research support. The
second author thanks PB98-0067 for their support. The third author thanks the Alfred P. Sloan Foundation
and the David and Lucile Packard Foundation for their support.
Typeset by AMS-TEX
1
2
(1.4)
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
∞ X
η(48z)
n=0
q
−
24
72
η(24z) (1 − q )(1 − q ) · · · (1 − q 24(2n+1) )
=
η(48z)
M1 (q)
D(q 2 ) +
η(24z)
2
where M1 (q) is the mock theta function given by
(1.5)
∞
X
2
q 12n +12n+1
M1 (q) = q +
= q + q 25 − q 49 + 2q 73 − · · · .
24 )(1 + q 48 ) · · · (1 + q 24n )
(1
+
q
n=1
The q-series of the function M1 (q) was the focus of two extensive studies [A-D-H, C]. Although M1 (q) is not the Fourier expansion of a modular form, these works show
√ that the
coefficients of M1 (q) are given by a Hecke character for the quadratic field Q( 6). In particular, M1 (q) enjoys nice properties that one expects for certain weight 1 cusp forms. For
these reasons, we shall refer to M1 (q) and M2 (q) (defined in (1.8)) as mock theta functions
although they do not exactly fit Ramanujan’s original definition [A3; p. 291].
In view of identities (1.2-4), it is natural to investigate the general behavior of q-series
which are obtained by summing the iterated differences between an infinite product and its
truncated products. Here we establish two general theorems which yield infinitely many
such identities, and we illustrate how such identities are useful in determining the values at
negative integers for certain L-functions.
We shall employ the standard notation
∞
Y
(1 − Aq j )
(A; q)n =
,
(1 − Aq n+j )
j=0
(1.6)
and throughout we assume that |q| < 1 and that the other parameters are restricted to
domains that do not contain any singularities of the series or products under consideration.
Theorem 1.
∞ X
(t; q)∞
(t; q)n
−
(a; q)∞
(a; q)n
n=0
∞
X
(q/a; q)n (a/t)n
=
(q/t; q)n
n=1
(t; q)∞
+
(a; q)∞
∞
X
∞
∞
∞
X q n t−1
X tq n
X aq n t−1
qn
+
−
−
1 − q n n=1 1 − q n t−1 n=0 1 − tq n n=0 1 − aq n t−1
n=1
!
Theorem 2.
∞ X
(a; q)∞ (b; q)∞
(a; q)n (b; q)n
−
(q; q)∞ (c; q)∞
(q; q)n (c; q)n
n=0
(b; q)∞ (a; q)∞
=
(c; q)∞ (q; q)∞
∞
X
∞
∞
X aq n
X (c/b; q)n bn
qn
−
−
1 − q n n=0 1 − aq n n=1 (a; q)n (1 − q n )
n=1
!
.
.
q-SERIES IDENTITIES
3
Many interesting specializations of these two theorems yield identities for modular forms
that are eta-products (including
√ identities (1.2-4)). Here we highlight ten of these identities.
First we fix notation. We let Θ be the operator defined by
!
∞
∞
X
X
√
√
n
(1.7)
Θ
na(n)q n .
a(n)q
=
n=0
n=0
It is easy to see that the series E(q) in (1.2) is given by
√
E(q) = Θ (η(24z)) /2.
In addition to the mock theta function M1 (q), we shall require the mock theta function
M2 (q) defined by
∞
X
2
(−1)n q 24n −1
M2 (q) =
= −q 23 − q 47 − · · · .
24 )(1 − q 72 ) · · · (1 − q 24(2n−1) )
(1
−
q
n=1
(1.8)
See [A-D-H] for a detailed study of this function.
The ten eta-products F1 (z), F2 (z), . . . , F10 (z) we consider are of the form
Fi (z) = η ai (δi z)η bi (2δi z)
with ai 6= 0. Obviously, each Fi (z) is a modular form of weight (ai + bi )/2. For each Fi (z)
we define quantities ci and fi (j), which are not necessarily unique, for which
(1.9)
Fi (z) = ci
∞
Y
fi (j).
j=1
These are listed in the table below.
Table 1.
i
1
2
3
4
5
6
7
8
9
10
Fi (z)
1/η(24z)
η(2z)/η 2 (z)
η(8z)/η 2 (16z)
η(48z)/η(24z)
η(48z)/η(24z)
η(24z)/η(48z)
η(24z)/η(48z)
η(24z)
2
η (z)/η(2z)
η 2 (16z)/η(8z)
δi
24
1
8
24
24
24
24
24
1
8
ci
q −1
1
q −1
q
q/(1 − q 24 )
q −1
q −1
q
1
q/(1 − q 8 )
fi (j)
1/(1 − q 24j )
(1 + q j )/(1 − q j )
(1 − q 16j−8 )/(1 − q 16j )
1 + q 24j
1/(1 − q 24(2j+1) )
1/(1 + q 24j )
1 − q 24(2j−1)
1 − q 24j
(1 − q j )/(1 + q j )
(1 − q 16j )/(1 − q 16j+8 )
4
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
If δ ∈ {1, 8, 24}, then let dδ (n) be the divisor function defined by
P

d(n) = d|n 1
if δ = 24,

P
d
if δ = 8,
(1.10)
dδ (n) =
d|n (−1)

P

if δ = 1.
d|n odd 1
Also, for each i define αi by
(1.11)
αi =
− 12
if (ai + 2bi )δi = 24,
0
otherwise.
Notice that αi = −1/2 if and only if the order of vanishing of Fi (z) at ∞ is 1. The last
quantity we require is γi which is defined by
2
if i = 5, 7,
(1.12)
γi =
1
otherwise.
Theorem 3. If 1 ≤ i ≤ 10, then


∞
n
X
Y
Fi (z) − ci
fi (j) = (1 + [1/δi ]) Fi (z)Di (q) + Gi (q)
n=0
j=1
where [•] denotes the greatest integer function,
Di (q) = αi +
∞
X
dδi (n)q δi γi n
n=1
and

0



 M (q)/2
1
Gi (q) =

2M2 (q)/γi


√

(−αi + [2/δi ]) Θ (Fi (z))
if i = 1, 2, 3,
if i = 4, 5,
if i = 6, 7,
if i = 8, 9, 10.
The three forms F1 (z), F2 (z) and F3 (z) have weight -1/2 and the four forms F4 (z), F5 (z),
F6 (z) and F7 (z) have weight 0. The remaining three forms have weight 1/2. The series
G4 (z), G5 (z), G6 (z) and G7 (z) are mock theta functions, whereas G8 (q), G9 (q) and G10 (q)
are the half-derivatives of F8 (z), F9 (z) and F10 (z). In other words, the “error series” Gi (q)
in Theorem 3 satisfy

if Fi (z) has weight -1/2,

0
if Fi (z) has weight 0,
Fi (z) −→ Gi (q) ∼ Mock Theta function

√
Θ(Fi (z))
if Fi (z) has weight 1/2.
q-SERIES IDENTITIES
5
Although these identities are elegant in their own right, they are also often useful in
calculating the values of L-functions at negative integers. In particular, they lead to analogs
of the classical result
∞
X
t
tn
n+1
,
=
1
+
(−1)
ζ(1
−
n)
·
et − 1
(n
−
1)!
n=1
where ζ(s) is the Riemann zeta-function. In this direction, Zagier used (1.2) to show that
−t/24
(1.13) −e
∞
X
−t
−2t
(1 − e )(1 − e
−nt
) · · · (1 − e
n=0
∞
1X
tn
)=
(−1/24)n · L(χ12 , −2n − 1) · ,
2 n=0
n!
where χ12 is the Dirichlet character with modulus 12 defined by


1
−1
χ12 (n) :=


0
if n ≡ 1, 11 (mod 12),
if n ≡ 5, 7 (mod 12),
otherwise.
Here we illustrate the generality of this phenomenon by proving the following theorems.
Theorem 4. As a power series in t, we have
∞
∞
1 X (1 − e−t )(1 − e−2t ) · · · (1 − e−nt ) X
tn
n n+1
− ·
=
(−1)
(4
−
1)
·
ζ(−2n
−
1)
·
.
4 n=0 (1 + e−t )(1 + e−2t ) · · · (1 + e−nt ) n=0
n!
In addition to ζ(s), we shall consider the Dirichlet L-function
(1.14)
L(χ2 , s) :=
∞
X
χ2 (n)
,
s
n
n=1
where


1
χ2 (n) :=
−1


0
(1.15)
if n ≡ 1, 7 (mod 8),
if n ≡ 3, 5 (mod 8),
otherwise.
Theorem 5. As a power series in t, we have
−t/8
−2e
∞
∞
X
X
tn
(1 − e−2t )(1 − e−4t ) · · · (1 − e−2nt )
n
=
(−1/8)
·
L(χ
,
−2n
−
1)
·
.
2
n!
(1 + e−t )(1 + e−3t ) · · · (1 + e−(2n+1)t ) n=0
n=0
6
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
We shall also consider the Hecke L-function
(1.16)
L(ρ, s) =
∞
X
a(n)
:=
s
n
n=1
X
χ(a)N a−s
√
a⊆Z[ 6]
√
√
where χ is the order 2 character of conductor 4(3 + 6) on ideals in Z[ 6] defined by
yx 12
i
if y is even,
x
(1.17)
χ(a) :=
iyx+1 12
if y is odd,
x
√
when a = (x + y 6). If 1 ≤ r < 48 is an integer, then let Lr (ρ, s) be the partial L-function
defined by
(1.18)
X
Lr (ρ, s) :=
n≡r (mod 48)
a(n)
.
ns
By the orthogonality of the Dirichlet characters modulo 48 and the analytic continuation of
the associated twists of L(ρ, s), each Lr (ρ, s) has an analytic continuation to C.
Theorem 6. As a power series in t, we have
t/24
−2e
∞
X
−t
−3t
(1−e )(1−e
−(2n−1)t
) · · · (1−e
)=
n=0
∞
X
(−1/24)n ·(L23 (ρ, −n) + L47 (ρ, −n))·
n=0
tn
.
n!
Theorem 7. As a power series in t, we have
−t/24
−2e
∞
X
(1 − e−t )(1 + e−2t ) · · · (1 + (−1)n e−nt )
n=0
=
∞
X
n=0
(−1/24)n · (L1 (ρ, −n) − L25 (ρ, −n)) ·
tn
.
n!
In §2 we recall certain facts about q-series and basic hypergeometric series, and we prove
Theorems 1 and 2. In §3 we prove Theorem 3 and in §4 we prove Theorems 4, 5, 6, and 7.
In §5 we examine the partition theoretic consequences of the identities for F1 (z) and F8 (z).
In §6 we give a few more identities which are related to eta-products. The most interesting
of these is
∞ ∞
X
X
1
1
1
−
=
(d(n) + m(n)) q 24n ,
2 (24z)
2 (1 − q 24 )2 (1 − q 48 )2 · · · (1 − q 24n )2
2 (24z)
η
q
η
n=0
n=1
where m(n) denotespthe number
of middle divisors of n. A divisor is a middle divisor if it
√
lies in the interval [ n/2, 2n).
q-SERIES IDENTITIES
7
Acknoweledgements
The authors thank Scott Ahlgren, Andrew Granville, Don Zagier and the referee for a variety
of useful comments and suggestions.
2. Preliminaries and important facts.
The ten identities in Theorem 3 rely on Theorems 1 and 2. Here we prove Theorems 1
and 2, and we begin with the following observation which follows from Abel’s Theorem.
P∞
n
Proposition 2.1. Suppose that f (z) =
is analytic for |z| < 1. If α is a
n=0 α(n)z
complex number for which
P∞
(1)
n=0 (α − α(n)) < +∞,
(2) limn→+∞ n(α − α(n)) = 0,
then
∞
X
d
(1 − z)f (z) =
(α − α(n)).
lim
z→1− dz
n=0
(2.1)
In all of our applications, f (z) will have a pole of order 1 at z = 1, so the limz→1−
can be replaced by a simple evaluation. We shall employ the standard notation of basic
hypergeometric series [G-R; pp. 3-4, 125]:
(2.2)
r+1 φr
(2.3)
r ψr
a0 , a1 , . . . , ar ;
q, z
b1 , . . . , b r
a1 , a2 , . . . , ar ;
q, z
b1 , b 2 , . . . , b r
=
∞
X
(a0 ; q)j (a1 ; q)j · · · (ar ; q)j z j
j=0
(q; q)j (b1 ; q)j · · · (br ; q)j
,
∞
X
(a1 ; q)j (a2 ; q)j · · · (ar ; q)j z j
.
=
(b
;
q)
(b
;
q)
·
·
·
(b
;
q)
1
j
2
j
r
j
j=−∞
We shall also require Heine’s transformation [G-R; p. 9]:
(b; q)∞ (az; q)∞
a, b;
c/b, z;
(2.4)
q, z =
× 2 φ1
q, b ;
2 φ1
c
az
(c; q)∞ (z; q)∞
Ramanujan’s summation [G-R; p. 126]:
(q; q)∞ (b/a, q)∞ (az; q)∞ (q/az; q)∞
a;
(2.5)
q, z =
,
1 ψ1
b
(b; q)∞ (q/a; q)∞ (z; q)∞ (b/az; q)∞
and the Rogers-Fine identity [F; p. 15]:
(2.6)
2
∞
∞
X
X
(a; q)n tn
(a; q)n (atq/b; q)n (1 − atq 2n )bn tn q n −n
(1 − t)
=
.
(b; q)n
(b; q)n (tq; q)n
n=0
n=0
8
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
Throughout we assume that |q| < 1 and that the other parameters are restricted to
domains that do not contain any singularities of the series or products under consideration.
For succinctness of notation, we define the differential operator by
f (z) = f 0 (1).
(2.7)
Proof of Theorem 1. By Proposition 2.1, we have that
∞ X
(t; q)∞
(a; q)∞
n=0
(t; q)n
−
(a; q)n
= (1 − z)
∞
X
(t; q)n z n
.
(a;
q)
n
n=0
By (2.2-6), this equals
= (1 − z)
=
1 ψ1
t;
q, z
a
−1
X
(t; q)n z n
−
(a; q)n
n=−∞
!
∞
X
(q/a; q)n (a/(tz))n
(q; q)∞ (a/t; q)∞ (tz; q)∞ (q/tz; q)∞
− (1 − z)
(a; q)∞ (q/t; q)∞ (zq; q)∞ (a/tz; q)∞
(q/t; q)n
n=1
!
.
Differentiating this last expression with respect to z and then setting z = 1 yields the result.
Q.E.D.
Proof of Theorem 2. By Proposition 2.1, (2.2) and (2.4), we have that
∞ X
(a; q)∞ (b; q)∞
n=0
(a; q)n (b; q)n
−
(q; q)∞ (c; q)∞
(q; q)n (c; q)n
a, b;
= (1 − z) 2 φ1
q, z
c
=
(b; q)∞ (az; q)∞
(c; q)∞ (zq; q)∞
∞
X
(c/b; q)n (z; q)n bn
1+
(q; q)n (az; q)n
n=1
!!
.
Noting that (z; q)n = −(q; q)n−1 when n > 0, we differentiate this last expression with
respect to z and then set z = 1. This yields the result.
Q.E.D.
3. Proof of Theorem 3.
In this section we prove each of the ten identities using the facts in §2.
Case of F1 (z): This appears implicitly in [F; p. 14]. It is the instance of Theorem 2 where
a = 0 and b = c.
q-SERIES IDENTITIES
9
Case of F2 (z): This is the instance of a = −q and b = c in Theorem 2.
Case of F3 (z): In Theorem 2, replace q by q 2 , then set b = c and a = q.
Case of F4 (z): This result was proved in [A2; eq. (1.4)]. It follows from Theorem 1 with
t = −q with a → 0.
Case of F5 (z): This result was proved in [A2; eq. (1.5)]. In Theorem 2 replace q by q 2 ,
then set a = q 2 , b = 0 and c = q 3 . This yields, after multiplication by (1 − q)−1 ,
∞ X
1
1
−
2)
(q;
q
(q; q 2 )n+1
∞
n=0
2
∞
X
(−1)n−1 q n +2n
1
=
(q; q 2 )∞ n=1 (q 2 ; q 2 )n (1 − q 2n)
2
∞
X
1
(−1)n−1 q n +2n ((1 − q −n ) + q −n )
=
(q; q 2 )∞ n=1
(q 2 ; q 2 )n (1 − q 2n )
∞
X
∞
2
2
X (−1)n−1 q n +n
(−1)n q n +n
+
(q 2 ; q 2 )n (1 + q n ) n=1 (q 2 ; q 2 )n (1 − q 2n )
n=1
1
=
(q; q 2 )∞
!
.
By [F; p. 14, eq. (12.42)], we find that this equals
1
=
lim 3 φ2
2(q; q 2 )∞ τ →0
q/τ, q/τ, −1;
q, −τ 2
−q, −q
1
+
(q; q 2 )∞
∞
1 X qn
− +
2 n=1 1 − q n
!
.
By [G-R; p. 241, eq. (III.9)] with a = b = q/τ , c = −1 and d = e = −q, this equals
1
=
lim 3 φ2
2(q; q 2 )∞ (−q; q)∞ τ →0
2
∞
1
1 X q n +n
=
+
2 n=0 (−q; q)n
(q; q 2 )
q/τ, −τ, q;
q, −τ
−q, −qτ
!
∞
1 X qn
− +
.
2 n=1 1 − q n
1
+
(q; q 2 )∞
∞
1 X qn
− +
2 n=1 1 − q n
This is the identity.
Case of F6 (z): In Theorem 2 let a = q, b = 0 and c = −q. This yields
∞ X
1
1
−
(−q; q)∞
(−q; q)n
n=0
2
∞
X
1
q (n +n)/2
=
(−q; q)∞ n=1 (q; q)n (1 − q n )
2
∞
X
1
q (n +n)/2 (1 − (−1)n−1 + (−1)n−1 )
=
.
(−q; q)∞ n=1
(q; q)n (1 − q n )
!
10
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
By [F; p. 14, eq. (12.42)], this equals
2
∞
∞
X
X
1
2
q 2n +3n+1
qn
+
=
(−q; q)∞ n=0 (q; q)2n+1 (1 − q 2n+1 ) (−q; q)∞ n=1 1 − q n
∞
X
2q
1
qn
q/τ, q/τ, q; 2 2 3
=
φ
q
,
τ
q
+
lim
3
2
3
3
q ,q
(−q; q)∞ (1 − q)2 τ →0
(−q; q)∞ n=1 1 − q n
By [G-R; p. 241, eq. (III.10)] with q replaced by q 2 followed by letting a = c = −q/τ, b = q
and d = e = q 3 , this equals
!
∞
∞
n
X
X
2q(q; q 2 )∞
1
1
q
(q 2 ; q 2 )m q m +
− +
.
=
(−q; q)∞ (q 3 ; q 2 )2∞ (1 − q)2 m=0
(q; q 2 )∞
2 n=1 1 − q n
!
∞
∞
n
X
X
1
q
1
=2
(q 2 ; q 2 )m q m +
− +
2)
(q;
q
2 n=1 1 − q n
∞
m=0
!
∞
∞
n−1 n2
n
X
X
(−1)
q
q
1
1
=2
.
+
− +
2
2
(q; q )n
(q; q )∞
2 n=1 1 − q n
n=1
This is the identity for F6 (z).
Case of F7 (z) : In Theorem 1 replace q by q 2 , and then set a = 0 and t = q.
Case of F8 (z) : This is identity (1.2), and it is Theorem 2 in [Z]. We include a proof for
completeness. In Proposition 2.1 set α = (q; q)∞ and let α(n) := (q; q)n . This yields
∞
X
((q; q)∞ − (q; q)n ) = (1 − z)
n=0
∞
X
(q; q)n z n
n=0
= −
=−
(3.1)
∞
X
(q; q)n−1 q n z n
n=1
∞
X
(q; q)n−1 nq n .
n=1
Now it is immediate (because the partial sums equal the partial products) that
1−
(3.2,)
∞
X
zq n (zq; q)n−1 = (zq; q)∞
n=1
and applying to (3.2) we find that
(3.3)
−
∞
X
n=1

q n ((zq; q)n−1 ) = −1 − (q; q)∞ −1 +
∞
X
j=1
j

q 
.
1 − qj
q-SERIES IDENTITIES
11
Now set b = 0, a = z 2 q and t = z 2 in (2.6), and after simplification we obtain upon
multiplication by z
(3.4) z − z
3
∞
X
2
2
n
(z q; q)n−1 (z q) = z +
n=1
∞
X
n
(−1)
q
(3n2 −3n)/2 6n−1
z
+q
(3n2 +n)/2 6n+1
z
.
n=1
Noting that zf (z 2 ) = f (1) + 2f (z) we apply to (3.4) and find that
1+
∞
X
(−1)n (6n − 1)q (3n
2
−n)/2
n=1
+
∞
X
(−1)n (6n + 1)q (3n
2
+n)/2
n=1
∞
X
= (q; q)∞ + 2 1 − z
!
n n
(zq; q)n−1 z q
n=1
= (q; q)∞ − 2 ((q; q)∞ − 1) − 2
∞
X
n
(q; q)n−1 nq − 2
n=1
∞
X
q n ((zq; q)n−1 )
n=1
(3.5)
∞
X
= 2 − (q; q)∞ + 2


((q; q)∞ − (q; q)n ) + 2 −1 − (q; q)∞ −1 +
n=0
∞
X
j=1
j

q 
,
1 − qj
by (3.1) and (3.3). This is the identity.
Case of F9 (z): In Theorem 2, let a = b = q and c = −q. This yields
(3.6)
∞ ∞
X
(q; q)n
(q; q)∞ X (−q; q)n−1 q n
(q; q)∞
−
.
= −2
n)
(−q;
q)
(−q;
q)
(−q;
q)
(q;
q)
(1
−
q
∞
n
∞
n
n=0
n=1
Now in (2.6) set t = z, a = zq, b = −zq (cf. [F; p. 15, eq. (14.31)]). This yields after
simplification
(3.7)
∞
∞
X
X
2
(zq; q)n z n
=1+2
(−z 2 )n q n .
(1 − z)
(−zq; q)n
n=0
n=1
Applying to (3.7) yields
4
∞
X
2
(−1)n nq n =
n=1
= (1 − z)2 φ1
q, zq;
q, z .
−zq
12
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
By (2.4) with a = q, b = zq and c = −zq, and also (3.6), we find that this equals
(zq; q)∞
−1, z;
=
(3.8)
q, zq .
2 φ1
zq
(−zq; q)∞


∞
∞
∞
j
j
X
X
−q
q 
(q; q)∞ X (−q; q)n−1 q n
(q; q)∞ 
−
−2
=
(−q; q)∞ j=1 1 − q j j=1 1 + q j
(−q; q)∞ n=1 (q; q)n (1 − q n )
∞
∞ X
(q; q)∞
(q; q)∞ X q j
(q; q)n
+
−
.
= −2
(−q; q)∞ j=1 1 − q 2j n=0 (−q; q)∞
(−q; q)n
Case of F10 (z): In Theorem 2 replace q by q 2 , and then set a = b = q 2 and c = q 3 . This
yields, after multiplication by (1 − q)−1
(3.9)
∞ 2 2
X
(q ; q )∞
n=0
(q; q 2 )∞
(q 2 ; q 2 )n
−
(q; q 2 )n+1
∞
(q; q 2 )n q 2n
(q 2 ; q 2 )∞ X
.
=−
(q; q 2 )∞ n=1 (q 2 ; q 2 )n (1 − q 2n )
Now in (2.6) replace q by q 2 , then set a = zq and b = zq 2 (cf. [F; p. 16, eq. (14.4)]). This
yields, after simplification,
(3.10)
∞
∞
X
X
2
(zq; q 2 )n z n q n
z n q (n +n)/2 .
=
2
2
(zq ; q )n
n=0
n=0
Applying to (3.10), we find that
∞
X
nq
(n2 +n)/2
n=1
= 2 φ1
zq, q 2 ; 2
q , zq .
zq 2
This equals, by (2.4) with q replaced by q 2 and z replaced by zq with a = zq, b = q 2 and
c = zq 2 ,
(q 2 ; q 2 )∞ (z 2 q 2 ; q 2 )∞
z, zq; 2 2
(3.11)
=
q ,q
2 φ1
z2 q2
(zq 2 ; q 2 )∞ (zq; q 2 )∞
z, zq; 2 2
2 2
= (q ; q )∞ (−zq; q)∞ 2 φ1
q ,q
z2 q2
∞
∞
(q 2 ; q 2 )∞ X
(q; q 2 )n q 2n
(q 2 ; q 2 )∞ X q j
−
=
(q; q 2 )∞ j=1 1 + q j
(q; q 2 )∞ n=1 (q 2 ; q 2 )n (1 − q 2n )
∞
∞ (q 2 ; q 2 )∞ X (−1)j q j X (q 2 ; q 2 )∞
(q 2 ; q 2 )n
=
+
−
.
(q; q 2 )∞ j=1 1 − q j
(q; q 2 )∞
(q; q 2 )n+1
n=0
We use (3.9) in the last step above.
q-SERIES IDENTITIES
13
4. Proof of Theorems 4, 5, 6, and 7.
In this section we prove Theorems 4, 5, 6, and 7. The proofs are similar to the proof of
[Th. 3, Z], and so we give a brief proof of Theorem 4 and we give sketches of the remaining
cases. In each case, it is well known that the relevant L-function has an analytic continuation
to C (with the exception of a simple pole at s = 1 for ζ(s)) and a functional equation via a
Mellin transformation.
Proof of Theorems 4, 5, 6, and 7.
Case of Theorem 4: The identity for F9 (z) in Theorem 3 is
(4.1)
∞ X
n=0
(1 − q)(1 − q 2 ) · · · (1 − q n )
F9 (z) −
(1 + q)(1 + q 2 ) · · · (1 + q n )
= 2F9 (z)
∞
X
!
d1 (n)q n
√
+ 2 Θ (F9 (z)) .
n=1
It is well known that
(4.2)
F9 (z) = 1 + 2
∞
X
2
(−1)n q n .
n=1
Notice that F9 (z) vanishes to infinite order as q → 1. Therefore, we replace q by e−t with
t & 0. Now define coefficients c9 (n) and b9 (n) by the asymptotic expansions
∞
∞
X
(1 − e−t )(1 − e−2t ) · · · (1 − e−nt ) X
−
=
c9 (n)tn
−t )(1 + e−2t ) · · · (1 + e−nt )
(1
+
e
n=0
n=0
(4.3)
and
−t
H9 e
=4
∞
X
n
−n2 t
(−1) ne
∼
n=1
∞
X
b9 (n)tn
as t & 0.
n=0
Now we observe, by (4.1) and (4.2), that c9 (n) = b9 (n) for all n. On the other hand
Z
∞
−t
H9 e
t
s−1
dt = 4
0
∞
X
n
Z
∞
2
e−n t ts−1 dt.
(−1) n
0
n=1
By replacing T by n2 t, we find that
Z
∞
(4.4)
−t
H9 e
0
t
s−1
∞
X
(−1)n
dt = 4Γ(s)
= 4Γ(s)(41−s − 1)ζ(2s − 1).
2s−1
n
n=1
To compute the coefficients b9 (n), notice that
Z
(4.5)
∞
H9
0
e−t ts−1 dt =
Z
0
∞
N
−1
X
n=0
!
n
N
b9 (n)t + O(t ) t
s−1
dt =
N
−1
X
n=0
c9 (n)
+ F9 (s),
s+n
14
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
where F9 (s) is analytic for Re(s) > −N . The residue at s = −n of (4.4) is
b9 (n) =
(−1)n
· 4(41+n − 1) · ζ(−2n − 1).
n!
This completes the proof of Theorem 4.
Case of Theorem 5: It is well known that
(4.6)
F10 (z) =
∞
X
2
q (2n+1) .
n=0
It is easy to see that the identity in Theorem 3 for F10 (z) is equivalent to
∞ X
n=0
q 1/8
(1 − q 2 )(1 − q 4 ) · · · (1 − q 2n )
F10 (z/8) −
×
1 − q (1 − q 3 )(1 − q 5 ) · · · (1 − q 2n+1 )
(4.7)
∞
1 X
= F10 (z/8) − +
d8 (n)q n
2 n=1
!
+
∞
2
1X
(2n + 1)q (2n+1) /8 .
2 n=0
Observe that F10 (z/8) vanishes to infinite order as q → −1. Define the coefficients c10 (n)
and b10 (n) by the asymptotic expansions
−t/8
−e
∞
∞
X
X
(1 − e−2t )(1 − e−4t ) · · · (1 − e−2nt )
=
c10 (n)tn
−t
−3t
−(2n+1)t
(1 + e )(1 + e ) · · · (1 + e
) n=0
n=0
and
∞
∞
X
1X
2
H10 e−t =
(2n + 1)(−1)n(n+1)/2 e−t(2n+1) /8 ∼
b10 (n)tn
2 n=0
n=0
as t & 0.
Therefore, by replacing q by −e−t with t & 0 and ζ8 := eπi/8 , (4.6) and (4.7) imply that
c10 (n) = b10 (n) for all n and, on the other hand
Z
∞
−t
H10 e
0
(4.8)
t
s−1
Z ∞
∞
2
1X
n(n+1)/2
dt =
(2n + 1)(−1)
e−(2n+1) t/8 ts−1 dt
2 n=0
0
Z
∞
∞
2
1X
=
χ2 (n)n
e−n t/8 ts−1 dt
2 n=1
0
=
1 s
8 Γ(s)L(χ2 , 2s − 1).
2
The rest of the proof is identical to the proof of Theorem 4.
q-SERIES IDENTITIES
15
Case of Theorem 6: By [C], it is known that the coefficients a(n) defining L(ρ, s) in (1.16)
are defined by
∞
X
(4.9)
a(n)q n = M1 (q) + 2M2 (q).
n=1
In this case we use the identity for F7 (z) in Theorem 3
(4.10)
!
∞ ∞
X
X
F7 (z) − q −1 (1 − q 24 )(1 − q 72 ) · · · (1 − q 24(2n−1) ) = F7 (z)
d(n)q 24n + M2 (q).
n=0
n=1
Moreover, notice in (1.8) that the non-zero coefficients of M2 (q) are supported on those
exponents n ≡ 23 (mod 24). Now observe that F7 (z/24) vanishes to infinite order as q → 1.
The rest of the argument is virtually identical to those above.
Case of Theorem 7: In this case we consider the identity for F4 (z) in Theorem 3
(4.11)
!
∞
∞
X
X
1
1
d(n)q 24n + M1 (q).
F4 (z) − q(1 + q 24 )(1 + q 48 ) · · · (1 + q 24n ) = F4 (z) − +
2 n=1
2
n=0
Notice that F4 (z/24) vanishes to infinite order as q → −1. Arguing as before, we consider
the asymptotic t-series expansion of
−t/24
−ζ24 e
∞
X
(1 − e−t )(1 + e−2t ) · · · (1 + (−1)n e−nt )
n=0
where ζ24 = eπi/24 . The rest of the proof is a routine exercise using ζ24 , the Mellin transform,
and the fact that the non-zero coefficients of M1 (q) are supported on those exponents n ≡ 1
(mod 24).
Q.E.D.
5. Partition theoretic consequences.
Recall that a partition of a non-negative integer N is any nonincreasing sequence of
positive integers whose sum is N . If p(N ) denotes the number of partitions of N , then we
have
∞
∞
X
Y
1
1
(5.1)
=
p(N )q 24N −1 = q −1
= q −1 + q 23 + 2q 47 + 3q 71 + · · · .
24n
η(24z)
1
−
q
n=1
N =0
If pe (N ) (resp. po (N )) denote the number of partitions of N into an even (resp. odd)
number of distinct parts, then Euler’s Pentagonal Number Theorem asserts that the Fourier
expansion of η(24z) ∈ S1/2 (Γ0 (576), χ) is
(5.2)
η(24z) =
∞
X
N =0
(pe (N ) − po (N )) q
24N +1
=q−q
25
−q
49
+ ··· =
∞
X
n=1
χ(n)q n
2
16
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
where
(5.3)


1
χ(n) :=
−1


0
if n ≡ ±1 (mod 12),
if n ≡ ±5 (mod 12),
otherwise.
As a consequence for our identity for F1 (z) = 1/η(24z), we obtain the following partition
theoretic result which is equivalent to the observation made by Erdös [E] in the beginning
of his study of the asymptotics of p(N ) by elementary means.
Theorem 5.1. If n is a positive integer, then let an (N ) denote the number of partitions of
N into parts not exceeding n. For every positive integer N we have
(N + 1)p(N ) =
N
X
p(N − n)d(n) +
n=1
N
X
an (N ).
n=1
Proof. Here we use the identity for F1 (z) = 1/η(24z) in the form
∞ X
1
(5.4)
P (q) −
= P (q)D(q),
2 ) · · · (1 − q n )
(1
−
q)(1
−
q
n=0
where
P (q) =
D(q) =
∞
X
n=0
∞
X
n
p(n)q =
∞
Y
1
,
n
1
−
q
n=1
d(n)q n .
n=1
If m is a non-negative integer, then the series
(5.5)
∞
X
1
m+1
P (q) −
=q
+
bm (n)q n
2
m
(1 − q)(1 − q ) · · · (1 − q )
n=m+2
for some sequence of integers bm (n).
Now let N be a positive integer, then it is easy to see that the coefficient of N on the
right hand side of (5.4) is
N
X
p(N − n)d(n).
n=1
Therefore, to prove the result it suffices to show that the coefficient of q N on the left hand
side of of (5.4) is
(5.6)
(N + 1)p(N ) −
N
X
n=1
an (N ).
q-SERIES IDENTITIES
17
By letting m = N as in (5.5), it is easy to see that the coefficient of q N on the left hand
side of (5.4) is the coefficient of q N in
N X
n=0
1
P (q) −
(1 − q)(1 − q 2 ) · · · (1 − q n )
.
Claim (5.6) follows from the obvious fact that
∞
X
an (N )q n =
N =0
1
.
(1 − q)(1 − q 2 ) · · · (1 − q n )
This completes the proof.
Q.E.D.
As with Theorem 5.1, Zagier’s identity for F8 (z) = η(24z) has an interesting partition
theoretic consequence which is similar in flavor to Euler’s Pentagonal Number Theorem
(5.2). We require some notation. If π is a partition, then let mπ denote its largest part.
Moreover, if N is an integer, then let Se (N ) (resp. So (N )) denote the set of partitions of
N into an even (resp. odd) number of distinct parts. Using this notation, define the two
partition functions Ae (N ) and Ao (N ) by
X
(5.7)
Ae (N ) :=
mπ ,
π∈Se (N )
(5.8)
Ao (N ) :=
X
mπ
π∈So (N )
Theorem 5.2. If N is a positive integer, then
Ae (N ) − Ao (N ) =

k

 (−1) (3k)
X
=
(−1)k d N − (3k 2 + k)/2 + (−1)k (3k − 1)


k∈Z
0
if N =
if N =
3k2 +k
2
3k2 −k
2
with k ≥ 0,
with k ≥ 1,
otherwise.
Proof. If E(q) is the q-series defined by
E(q) =
∞
Y
(1 − q n ) = 1 − q − q 2 + · · · ,
n=1
then we claim that
(5.9)
∞
X
n=0
∞
X
E(q) − (1 − q)(1 − q ) · · · (1 − q ) = −
nq n (1 − q)(1 − q 2 ) · · · (1 − q n−1 ).
2
n
n=1
18
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
As described above, we have that
E(q) =
∞
Y
n
(1 − q ) =
n=1
∞
X
(pe (n) − po (n)) q n .
n=0
Since E(q) is the generating function for the number of partitions into an even number of
distinct parts minus the number into an odd number of distinct parts, it is easy to see that
E(q) − (1 − q)(1 − q 2 ) · · · (1 − q n )
is the generating function whose coefficient of q N is the number of partitions of N into
an even number of distinct parts where at least one part exceeds n minus the number of
partitions of N into an odd number of distinct parts where at least one part exceeds n.
Consequently, each partition π into distinct parts is counted with multiplicity (−1)nπ mπ
where nπ is the number of parts in π. This implies (5.9).
It is an easy exercise to deduce Theorem 5.2 from the identity in Theorem 1 for F8 (z) =
η(24z).
Q.E.D.
6. Related Results
The uniform nature of the ten results proved in Theorem 3 makes it natural to group
those identities together. There are, however, a variety of results that can be deduced from
Theorems 1 and 2. Here we record a number of further identities.
p
√
Theorem 6.1. Let m(n) denote the number of divisors of n in the interval [ n/2, 2n).
Moreover, let de (n) (resp. do (n)) denote the number of even (resp. odd) divisors of n. The
following identities are true:
∞ X
(6.1)
n=0
1
1
− 2 24 24 2
2
η (24z) q (q ; q )n
∞ 2
X
η (2z)
(6.2)
n=0
(6.3)
(6.4)
(−q; q)2n
−
η 4 (z)
(q; q)2n
∞
X
1
= 2
(d(n) + m(n)) q 24n ,
η (24z) n=1
∞
4η 2 (2z) X
= 4
do (2m + 1)q 2m+1 ,
η (z) m=0
∞ 2
∞
X
η (8z)
(q 8 ; q 16 )2n
η 2 (8z) X
−
=
(de (n) − 2do (n)) q 8n ,
4 (16z)
2 (q 16 ; q 16 )2
4 (16z)
η
q
η
n
n=0
n=1
∞ X
n=0
1
1
−
24
η(24z) q(q ; q 24 )2n
1
=
η(24z)
∞
X
2
∞
X
(−1)n−1 q 24n
48n
d(n)q
+
1 − q 48n
n=1
n=1
!
,
q-SERIES IDENTITIES
(6.5)
∞ X
η(2z)
n=0
(−q; q)2n
−
2
η (z)
(q; q)2n
(6.6)
∞ X
η(8z)
(q 8 ; q 16 )2n
η(8z)
= 2
−
2
16
16
η (16z) q(q ; q )2n
4η (16z)
n=0
19
∞
2η(2z) X q 2n − (−q)n
,
= 2
η (z) n=1 1 − q 4n
1−
∞
X
2
q 8n + 2
n=−∞
∞
X
!
(de (n) − do (n)) q 8n
.
n=1
Proof. Here we prove these identities.
Case of (6.1): Let a = b = 0 and c = q in Theorem 2. This yields
!
2
∞
∞
∞ X
X
X
1
qn
(−1)n−1 q (n +n)/2
1
1
−
+
.
=
(6.7)
(q; q)2∞
(q; q)2n
(q; q)2∞ n=1 1 − q n n=1
1 − qn
n=0
In order to complete the proof of (6.1), we must establish that
2
∞
∞
X
X
(−1)n−1 q (n +n)/2
n
(6.8)
m(n)q =
.
n
1
−
q
n=1
n=1
To see (6.8),pwe first note
√ that m(n) equals d(n) minus the number of divisors in the two
intervals [1, n/2) and [ 2n, n]. Hence, we have that
∞
∞
∞
X
X
X
qn
n
(n)(2n)
n
2n
3n
m(n)q =
−
q
1
+
2q
+
2q
+
2q
+
·
·
·
1 − q n n=1
n=1
n=1
(6.9)
∞
X
∞
2
X (1 + q n )q 2n
qn
=
−
.
1 − q n n=1
1 − qn
n=1
Next in [G-R, eq. (III.17), p. 242] set a = e = z and let b, c, d and f → +∞. This yields
!
2
∞
∞
2n 2n2 n
2
X
X
(z; q)n (1 − zq )q z
(q; q)∞
(z; q)n (−1)n q (n +n)/2
1+
.
=
(zq; q)∞
(q; q)2n (1 − z)
(q; q)n
n=1
n=0
Hence we get
2
∞
X
(−1)n−1 q (n +n)/2
=
1 − qn
n=1
2
∞
X
(z; q)n (−1)n q (n +n)/2
=
(q; q)n
n=0
(q; q)∞
=
(zq; q)∞
2
∞
X
(z; q)2n (1 − zq 2n )q 2n z n
1+
(q; q)2n (1 − z)
n=1
∞
X
∞
X (1 + q n )q 2n
qn
=
−
1 − q n n=1
1 − qn
n=1
(6.10)
=
∞
X
n=1
m(n)q n ,
2
!
20
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
which proves (6.8) and therefore (6.1).
Case of (6.2): To prove this identity let a = b = −q and c = q in Theorem 2. The result
now follows easily by combining the resulting Lambert series.
Case of (6.3): To prove this identity replace q by q 2 and set a = b = q and c = q 2 in
Theorem 2. The result now follows easily by combining the resulting Lambert series.
Case of (6.4): To prove this identity replace q by q 2 and set a = b = 0 and c = q 2 in
Theorem 2.
Case of (6.5): This identity follows from Theorem 2 by replacing q by q 2 and the setting
a = −q, b = −q 2 and c = q.
Case of (6.6): Surprisingly, (6.6) is more intricate. Here we use Proposition 2.1 with
α(n) = (q; q 2 )2n /(q 2 ; q 2 )2n .
Consequently (noting that f (z) = 21 f (z 2 )), we find that
∞ X
(q; q 2 )∞
(q; q 2 )2n
− 2 2
(q 2 ; q 2 )∞
(q ; q )2n
n=0
∞
X
(q; q 2 )2n z n
= (1 − z)
(q 2 ; q 2 )2n
n=0
∞
X
z n (q; q 2 )n (1 + (−1)n )
1
2
= (1 − z )
2
2(q 2 ; q 2 )n
n=0
(qz; q 2 )∞
(−zq; q 2 )∞
1
+ (1 − z) ·
= (1 + z)
4
(zq 2 ; q 2 )∞
(−zq 2 ; q 2 )∞
(
)
∞ 1 (q; q 2 )∞
(−q; q 2 )∞
2(q; q 2 )∞ X
q 2n
q 2n−1
=
−
+ 2 2
−
.
4 (q 2 ; q 2 )∞
(−q 2 ; q 2 )∞
(q ; q )∞ n=1 1 − q 2n
1 − q 2n−1
Identity (6.6) now follows by recalling [A1; p. 21, eq. (2.2.10)], with z = 1, that
∞
X
(−q; q 2 )∞ (q 2 ; q 2 )∞
n2
=
q
.
(q; q 2 )∞ (−q 2 ; q 2 )∞
n=−∞
Q.E.D.
Finally, we recall the following attractive and very elementary result on partitions into
consecutive integers (cf. [L, p. 85, Problem 4]).
q-SERIES IDENTITIES
21
Proposition 6.2. The number of partitions of n into consecutive integers equals the number
of odd divisors of n.
Proof. This result is easily deduced from the formula for the sum of an arithmetic progression. It is also directly deduced from the generating function identity [M, p. 28]
2
∞
X
q (m +m)/2
q 2m−1
=
.
2m−1
m
1
−
q
1
−
q
m=1
m=1
∞
X
Q.E.D.
From our proof of (6.1), we may easily deduce the following result for ce (n) (resp. co (n))
the number of partitions of n into an even (resp. odd) number of consecutive integers.
Theorem 6.3. For every positive integer n we have
co (n) − ce (n) = m(n).
Proof. This is immediate from (6.10):
2
∞
∞
X
X
(−1)n−1 q (n +n)/2
m(n)q =
(co (n) − ce (n)) q n .
=
n
1
−
q
n=1
n=1
n=1
∞
X
n
Q.E.D.
References
[A1]
[A2]
[A3]
[A-D-H]
[C]
[E]
[F]
[G-R]
[L]
[M]
[Z]
G. E. Andrews, The theory of partitions, Cambridge University Press, 1984.
G. E. Andrews, Ramanujan’s “Lost” Notebook V: Euler’s partition identity, Advances in Mathematics 61 (1986), 156-164.
G. E. Andrews, Mock theta functions, Proc. Symp. Pure Math. 49 (1989), 283-298.
G. E. Andrews, F. Dyson and D. Hickerson, Partitions and indefinite quadratic forms, Inventiones Mathematicae 91 (1988), 391-407.
H. Cohen, q-identities for Maass waveforms, Inventiones Mathematicae 91 (1988), 409-422.
P. Erdös, On an elementary proof of some asymptotic formulas in the theory of partitions,
Annals of Mathematics 43 (1942), 437-450.
N. J. Fine, Basic hypergeometric series and applications, Math. Sruveys and Monographs, Vol.
27, Amer. Math.Soc., Providence, 1988.
G. Gasper and M. Rahman, Basic hypergeometric series, Ency. of Math. and its Appl., Vol. 2,
Addison-Wesley, Reading, 1976.
W. J. Leveque, Topics in Number Theory, Vol. 1, Addison-Wesley, Reading, 1956.
P. A. MacMahon, Combinatory Analysis, Vol. 2, Cambridge Univ. Press, Cambridge, 1916
(reissued by Chelsea, New York, 1960).
D. Zagier, Vassiliev invariants and a strange identity related to the Dedekind eta-function, to
appear, Topology.
Department of Mathematics, Penn State University, University Park, Pa. 16802
22
GEORGE E. ANDREWS, JORGE JIMÉNEZ-URROZ AND KEN ONO
E-mail address: andrews@math.psu.edu
Dept. Matemáticas, Facultad de Ciencias, Universidad Autónoma de Madrid, 28049, Madrid,
España.
E-mail address: jorge.jimenez@uam.es
Department of Mathematics, Penn State University, University Park, Pa. 16802
E-mail address: ono@math.psu.edu
Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706
E-mail address: ono@math.wisc.edu