On Sums of Triangular Numbers - CiteSeerX

An Identity of Ramanujan and The Representation of Integers as Sums of Triangular Numbers All the correspondence should be sent to Prof. Zhi-Guo Liu Department of Mathematics, East China Normal University, Shanghai, 200062, P. P. China Email address: (1) [email protected] (2) [email protected] Running Title:

On Sums of Triangular Numbers

1

An Identity of Ramanujan and The Representation of Integers as Sums of Triangular Numbers Zhi-Guo Liu Department of Mathematics, East China Normal University, Shanghai, 200062, People’s Republic of China In memory of Robert Rankin

Abstract Let k be a positive number and tk (n) denote the number of representations of n as a sum of k triangular numbers. In this paper, we will calculate t2k (n) in the spirit of Ramanujan. We first use the complex theory of elliptic functions to prove a theta-function identity. Then from this identity we derive two Lambert series identities, one of them is a well-known identity of Ramanujan. Using a variant form of Ramanujan’s identity, we study two classes of Lambert series and derive some theta function identities related to these Lambert series . We calculate t12 (n), t16 (n), t20 (n), t24 (n), and t28 (n) using these Lambert series identities. We also rederive a recent result of H. H. Chan and K. S. Chua [6] about t32 (n). In addition, we derive some identities among the Ramanujan function τ (n), the divisor function σ11 (n), and t24 (n). Our methods do not depend upon the theory of modular forms and are somewhat more transparent.

Key words: elliptic functions, theta functions, Lambert series, triangular numbers, Ramanujan τ (n)function, Jacobi’s identity, modular forms. 2000 Mathematics Subject Classification:Primary– 11F11, 11F12, 11F27, 33E05.

1

Introduction

Throughout this paper we will use q to denote exp(πiτ ) with Imτ > 0. If k ≥ 1 is a positive integer, we then let rk (n) denote the number of representations of n as a sum of k squares. We also, let tk (n) denote the number of representations of sum of k triangular numbers. Following Ramanujan (see, for

2

example [4, p. 3, Equation(1.5)]), we use the theta functions φ(q) and ψ(q) defined by φ(q) =

∞ X

2

qn

and

ψ(q) =

n=−∞

∞ X

1

q 2 n(n+1) .

(1.1)

n=0

Consequently, the generating functions for rk (n) and tk (n) are φk (q) =

∞ X

rk (n)q n

and

ψ k (q) =

n=0

∞ X

tk (n)q n .

(1.2)

n=0

The study of rk (n) and tk (n) has a long history and many mathematicians have made contributions to this object. Now it still is one of the fundamental object in number theory . From (1.2) we know that deriving explicit expansions for φk (q) and ψ k (q) are the keys of calculating rk (n) and tk (n). The calculations of φk (q) are closely related to those of ψ k (q). One can obtain the formula for φk (q) using some simple modular transformations to the formula for ψ k (q) ; conversely, we can obtain the formula for ψ k (q) from φk (q). This phenomenon was perhaps first noticed by the author in [17]. In many cases, the calculation of ψ k (q) is simpler than that of φk (q). To evaluate φk (q) and ψ k (q) , one usual uses the theory of elliptic functions and the theory of modular forms. In this paper we will provide a different approach. Our methods do not depend upon the theory of modular forms and are also different from the method of elliptic functions. The main tool of ours is identity (5.1) below. It is surprising that this identity is very useful and powerful in the computations of ψ 2k (q). Our method is simpler than other known methods and one key advantage of our method is that one can follow easily. In Section 2 we introduce some basic facts about the classical theta functions. In Section 3 we prove a theta-function identity by using the complex theory of elliptic functions. In Section 4 we prove one well-known theta-function identity involving Lambert series by using the theta-function identity proved in Section 3, and calculate t2k (n) for 2k = 4, 6, 8. In Section 5 we derive a variant form of an identity of Ramanujan. In Section 6 and 7 we study two classes of Lambert series using the aforementioned identity. In Section 8 we derive two identities for ψ 12 (q). In Section 9 we obtain three identities for ψ 16 (q), ψ 20 (q), and ψ 28 (q), respectively. In Section 10 we establish some identities relating ψ 24 (q) and ψ 32 (q). In the final Section 11 we establish some identities among τ (n), t24 (n), and σ11 (n). Lastly, we introduce two lemmas that will be used in the sequel of this paper. Lemma 1 Let σk (n) denote the sum of the kth powers of the divisors of n , namely, σk (n) =

X

dk .

(1.3)

d|n

Then we have

∞ ∞ X X nk q n = σk (n)q n n 1 − q n=1 n=1

3

(1.4)

and

∞ ∞ X X (2n + 1)k q 2n+1 = σk (2n + 1)q 2n+1 , 2(2n+1) 1 − q n=0 n=0

(1.5)

By convention, we use σ(n) to denote the number of positive divisors of the positive integer n. Proof. Expanding 1/(1 − q n ) as a geometric series and reversing the order of summation, we readily find that (1.4). The proof of (1.5) is similar and so we omit it. Lemma 2 There holds the identity   ∞ ∞ k n X X X nπ 1 n q  sin = dk sin( dπ) q 2n+1 . 2n 1 − q 2 2 n=0 n=1

(1.6)

d|2n+1

The proof is simple and we leave it to the reader. In this paper will also make use of the following standard notation: (z; q)∞ =

∞ Y

(1 − zq n ).

(1.7)

n=0

2

Some basic facts about Jacobi’s theta functions

In this section we will introduce some fundamental facts about classical theta functions. Let q = eπiτ , where Im τ > 0. Then for a complex number z, the Jacobi theta functions [24, pp. 463-464] are defined by θ1 (z|q) θ2 (z|q) θ3 (z|q) θ4 (z|q)

∞ X

1

= −iq 4 1

= q4 = =

1

(−1)n q n(n+1) e(2n+1)iz = 2q 4

n=−∞ ∞ X n(n+1) (2n+1)iz

q

e

n=−∞ ∞ X n2 2niz

(−1)n q n(n+1) sin(2n + 1)z,

(2.1)

n=0 ∞ X

1

= 2q 4

q n(n+1) cos(2n + 1)z,

(2.2)

n=0

q e

n=−∞ ∞ X

∞ X

=1+2

∞ X

2

q n cos 2nz,

(2.3)

n=1 2

(−1)n q n e2niz = 1 + 2

n=−∞

∞ X

2

(−1)n q n cos 2nz.

(2.4)

n=1

Employing the Jacobi triple product identity, one can derive the infinite product expansions for Jacobi’s theta functions θ1 (z|q), θ2 (z|q), θ3 (z|q), and θ4 (z|q), namely, 1

θ1 (z|q) =

2q 4 (sin z)(q 2 ; q 2 )∞ (q 2 e2iz ; q 2 )∞ (q 2 e−2iz ; q 2 )∞ ,

θ2 (z|q) =

2q 4 (cos z)(q 2 ; q 2 )(−q 2 e2iz ; q 2 )∞ (−q 2 e−2iz ; q 2 ),

(2.6)

θ3 (z|q) =

(q 2 ; q 2 )∞ (−qe2iz ; q 2 )∞ (qe−2iz ; q 2 )∞ ,

(2.7)

θ4 (z|q) =

(q 2 ; q 2 )∞ (qe2iz ; q 2 )∞ (qe−2iz ; q 2 )∞

(2.8)

1

4

(2.5)

(see, for example, [24, pp. 469-470]). Using these product expansions for the Jacobi theta functions, by a direct computation, we readily find that θ10 (0|q)θ1 (2z|q) = 2θ1 (z|q)θ2 (z|q)θ3 (z|q)θ4 (z|q).

(2.9)

From this identity we see that with the (quasi) periods π and πτ the zero points of θ1 (2z|q) in the period parallelogram are 0,

π , 2

π + πτ , 2

πτ . 2

(2.10)

When we let z = 0 in (2.1)-(2.8), they yield the following special cases: θ10 (0|q)

1

= 2q 4

∞ X

1

(−1)n (2n + 1)q n(n+1) = 2q 4 (q 2 ; q 2 )3∞ ,

(2.11)

n=0

θ2 (0|q)

1

∞ X

1

= 2q 4 ψ(q 2 ) = 2q 4

1

q n(n+1) = 2q 4 (q 2 ; q 2 )∞ (−q 2 ; q 2 )2∞ ,

(2.12)

n=0

θ3 (0|q) θ4 (0|q)

∞ X

= φ(q) =

2

q n = (q 2 ; q 2 )∞ (−q; q 2 )2∞ ,

n=−∞ ∞ X

= φ(−q) =

2

(−1)n q n = (q 2 ; q 2 )∞ (q; q 2 )2∞ ,

(2.13) (2.14)

n=−∞

where θ10 (z|q) denotes the partial derivative of θ1 (z|q) with respect to z. With respect to the (quasi) periods π and πτ , we have the functional equations θ1 (z + π|q) = −θ1 (z|q)

and

θ1 (z + πτ |q) = −q −1 e−2πiz θ1 (z|q).

(2.15)

The four Jacobi theta functions are obviouly related and we have the following relations: 1 1 1 θ1 (z + π|q) = θ2 (z|q), θ1 (z + (π + πτ )|q) = q − 4 e−πiz θ3 (z|q), 2 2 1 − 14 −πiz θ1 (z + πτ |q) = iq e θ4 (z|q). 2

(2.16)

Taking the logarithmic derivative of both sides of the last identity in (2.16) with respect to z, we find that

πτ ¯¯ ´ θ10 ³ θ0 z+ ¯q = −i + 4 (z|q). θ1 2 θ4

(2.17)

Differentiate the above equation with respect to z, k times, and we find that µ

θ10 θ1

¶(k) ³ z+

πτ ¯¯ ´ ¯q = 2

µ

θ40 θ4

¶(k) (z|q),

k ≥ 1.

(2.18)

Employing the infinite product expansions for θ1 (z|q) and θ4 (z|q) , we can deduce respectively the trigonometric series expansions for the logarithmic derivatives of θ1 (z|τ ) and θ4 (z|τ ), namely, ∞ X θ10 q 2n (z|q) = cot z + 4 sin 2nz θ1 1 − q 2n n=1

5

(2.19)

and

∞ X θ40 qn (z|q) = 4 sin 2nz. θ4 1 − q 2n n=1

(2.20)

It is well-known that the Laurent expansion formula for cot z is cot z =

1 z z3 2z 5 − − − + ···. z 3 45 945

Hence, we substitute (2.21) into (2.19) and reverse the order of summation to get à ! à ! ∞ ∞ X X θ10 nq 2n 1 n3 q 2n 1 1 − 1 − 24 z− 1 + 240 z3 (z|q) = 2n 2n θ1 z 3 1 − q 45 1 − q n=1 n=1 à ! ∞ 5 2n X 2 n q − 1 − 504 z5 + · · · . 2n 945 1 − q n=1

3

(2.21)

(2.22)

An Identity for Theta Function θ1 (z|q)

We begin this section with the following theta-functions identity. It is, perhaps, a new theta function identity. This identity plays a fundamental role in this paper. Theorem 1 Let θ1 (z|q) be defined by (2.1). Then for any complex numbers u, v, and w, we have θ12 (v|q)θ1 (w + u|q)θ1 (w − u|q) − θ12 (u|q)θ1 (w + v|q)θ1 (w − v|q) = θ12 (w|q)θ1 (v + u|q)θ1 (v − u|q).

(3.1)

To prove the identity, we need the following fundamental theorem of elliptic functions (see, for example, [7, p. 22, Theorem2]). This theorem is quite useful in proving theta-function identities . Recently, in [12, 13, 14, 15, 16], we have used it to derive many important theta function identities. Theorem 2 The sum of all the residues of an elliptic function at the poles inside a period-parallelogram is zero. Proof. Let u, v, w be three distinct parameters and v, w are different from 0, 21 π, 12 (π + πτ ), 21 πτ . We consider the function f (z) =

θ1 (2z|q)θ1 (z − u|q)θ1 (z + u|q) , θ12 (z|q)θ1 (z − v|q)θ1 (z + v|q)θ1 (z − w|q)θ1 (z + w|q)

(3.2)

where 0 < v, w < π. Using (2.15), we can readily verify that f (z) is an elliptic function with periods π and πτ . The poles of f (z) in the period parallelogram with vertices located at 0, π, π + πτ, πτ are at z = 0, v, π − v, w, and π − w; and all of them are simple poles. Let res(f ; α) denote the residue of f (z) at α. From Theorem 2, we have res(f ; 0) + res(f ; v) + res(f ; π − v) + res(f ; w) + res(f ; π − w) = 0. 6

(3.3)

We now begin to compute the residues of f (z) at z = 0, v, π − v, w, and π − w, respectively. By L’Hˆopital’s rule, we find that res(f ; 0)

=

lim zf (z)

z→0

θ1 (z − u|q)θ1 (z + u|q) θ1 (z − v|q)θ1 (z + v|q)θ1 (z − w|q)θ1 (z + w|q) zθ1 (2z|q) × lim 2 z→0 θ1 (z|q) 2θ12 (u|q) . = − 0 θ1 (0|q)θ12 (w|q)θ12 (v|q) =

lim

z→0

(3.4)

Employing the same type argument as that of (3.4), we find that res(f ; v) = res(f ; π − v) == −

θ1 (v − u|q)θ1 (v + u|q) θ10 (0|q)θ12 (v|q)θ1 (w − v|q)θ1 (w + v|q)

(3.5)

and res(f ; w) = res(f ; π − w) =

θ1 (w − u|q)θ1 (w + u|q) 0 θ1 (0|q)θ12 (w|q)θ1 (w − v|q)θ1 (w

+ v|q)

.

(3.6)

Substituting (3.4), (3.5) and (3.6) into (3.3), (3.1) follows. By analytic continuation, we know (3.1) holds for all u, v, w. This completes the proof of Thereom 1. Corollary 1 Let θ2 (z|q), θ3 (z|q), and θ4 (z|q) be defined by (2.2), (2.3), and (2.4), respectively. Then θ22 (u|q)θ2 (v + w|q)θ2 (v − w|q) + θ42 (w|q)θ4 (u + v|q)θ4 (u − v|q) = θ32 (v|q)θ3 (u + w|q)θ3 (u − w|q).

(3.7)

Proof. Replacing u by u+ 12 π, v by v + 12 (π +πτ ), and w by w+ 12 πτ in (3.1), respectively, simplifying the resulting equation using (2.16), we obtain (3.7). This complete the proof of Corollary 1. Taking u = v = w = 0 in (3.7), we obtain Jacobi’s quartic identity [24, p. 467] θ34 (0|q) = θ24 (0|q) + θ44 (0|q).

(3.8)

An interesting proof of this identity is given in [9]. In [16], the author has generalized it to a theta function identity with three parameters , namely, θ3 (x|q)θ3 (y|q)θ3 (z|q)θ3 (x + y + z|q) =

θ1 (x|q)θ1 (y|q)θ1 (z|q)θ1 (x + y + z|q) +θ2 (x|q)θ2 (y|q)θ2 (z|q)θ2 (x + y + z|q) +θ4 (x|q)θ4 (y|q)θ4 (z|q)θ4 (x + y + z|q).

From which we can simply derive many modular identities of degree 2 through 7. 7

(3.9)

4

A theta-function identity involving Lambert series and the computations of t4 (n), t6 (n), and t8 (n)

4.1

A theta-function identity involving Lambert series and the compuation of t4 (n)

Theorem 3 Let θ1 (z|q) be defined as in (2.1). Then we have µ 0 ¶0 µ 0 ¶0 θ1 θ1 θ1 (u − v|q)θ1 (u + v|q) (u|q) − (v|q) = θ10 (0|q)2 . θ1 θ1 θ12 (u|q)θ12 (v|q)

(4.1)

This is a well-known theta-function identity (see, for example, [23, p. 325, Equation(1.7)] and [16, p. 143, Equation(7.1)]). Here we will rederive it from (3.1). Proof. Differentiating (3.1) with respect to w, twice, by using the method of logarithmic differentiation, we obtain © ª θ12 (v|q)θ1 (w + u|q)θ1 (w − u|q) φ21 (w) + φ01 (w) © ª −θ12 (u|q)θ1 (w + v|q)θ1 (w − v|q) φ22 (w) + φ02 (w) © ª = 2 θ10 (w|q)2 + θ1 (w|q)θ100 (w|q) θ1 (v + u|q)θ1 (v − u|q),

(4.2)

where φ1 (w) = φ2 (w) =

θ10 θ0 (w + u|q) + 1 (w − u|q), θ1 θ1 θ10 θ10 (w + v|q) + (w − v|q). θ1 θ1

(4.3)

When w = 0, (4.2) reduces to (4.1). This completes the proof of Theorem 3. Replacing u by u + 12 πτ , and v by v + 21 πτ in (4.1), and then using (2.15) , (2.16), and (2.18) in the resulting equation, we obtain Corollary 2 Let θ1 (z|q) and θ4 (z|q) be defined by (2.1) and (2.4), respectively. Then µ 0 ¶0 µ 0 ¶0 θ4 θ4 θ1 (u − v|q)θ1 (u + v|q) (u|q) − (v|q) = −θ10 (0|q)2 . θ4 θ4 θ42 (u|q)θ42 (v|q)

(4.4)

Replacing the left sides of (4.1) and (4.4) by the trigonometric series expansions of the logarithmic derivatives of θ1 (z|q) and θ4 (z|q), respectively, we find that Corollary 3 We have cot2 v − cot2 u + 8 and 8

∞ X nq 2n θ1 (u − v|q)θ1 (u + v|q) (cos 2nu − cos 2nv) = θ10 (0|q)2 2n 1 − q θ12 (u|q)θ12 (v|q) n=1

∞ X

nq n θ1 (u − v|q)θ1 (u + v|q) (cos 2nu − cos 2nv) = −θ10 (0|q)2 . 2n 1 − q θ42 (u|q)θ42 (v|q) n=1 8

(4.5)

(4.6)

Setting u = 12 π and v = 14 π in (4.5), and then writing q 2 as q, we find that θ44 (0|q)

∞ 2n X nq n n nq − 16 (−1) 1 − qn 1 − q 2n n=1 n=1 ½ ¾ ∞ ∞ X X nq n nq n nq n n = 1+8 (−1)n − 8 (−1) − 1 − qn 1 − qn 1 + qn n=1 n=1

= 1+8

= 1+8

∞ X

∞ X

(−1)n

(−1)n

n=1

nq n . 1 + qn

(4.7)

Replacing q by −q in the above equation, after simple reduction, we obtain the following identity due to Jacobi [4, p. 15]. A recent proof of this identity can be found in [1]. Theorem 4 Let φ(q) be defined as in (1.1). Then 4

φ (q) =

θ34 (0|q)

∞ X

=

1+8

nq n 1 + (−q)n n=1

=

1+8

∞ ∞ X X nq n nq 4n − 32 . 1 − qn 1 − q 4n n=1 n=1

Consequently, we have r4 (n) = 8σ1 (n) − 32σ1

³n´ 4

.

(4.8)

(4.9)

By setting u = 0 and v = 12 π in (4.6) , we obtain the following identity. Theorem 5 Let ψ(q) be defined as in (1.2). Then qψ 4 (q 2 ) =

∞ X (2n + 1)q 2n+1 1 4 θ2 (0|q) = , 16 1 − q 2(2n+1) n=0

(4.10)

and thus t4 (n) = σ1 (2n + 1).

(4.11)

Equation (4.11) is [20, Theorem 3]. For an account of (4.12) one may consult [5].

4.2

Lambert series expansions for θ24 (0|q) + θ34 (0|q)

From (4.8) and (4.10), we can obtain the following identity, which plays a pivotal role in the computation of t2k (n). Theorem 6 There holds the identity θ24 (0|q) + θ34 (0|q) = 1 + 24

9

∞ X nq n . 1 + qn n=1

(4.12)

Proof. By (4.8) and (4.10), we have θ24 (0|q) + θ34 (0|q) ∞ ∞ X X nq n (2n + 1)q 2n+1 = 1+8 + 16 1 + (−q)n 1 − q 2(2n+1) n=1 n=0 = 1+8 +8

∞ ∞ X X 2nq 2n (2n + 1)q 2n+1 + 8 2n 1+q 1 − q 2n+1 n=1 n=0 ∞ ∞ X X (2n + 1)q 2n+1 (2n + 1)q 2n+1 + 8 2n+1 1+q 1 − q 2n+1 n=0 n=0

∞ ∞ X X nq n (2n + 1)q 2n+1 + 16 1 + qn 1 − q 2n+1 n=1 n=0 ½ ¾ ∞ ∞ X X nq n nq n 2nq 2n = 1+8 + 16 − 1 + qn 1 − qn 1 − q 2n n=1 n=0

= 1+8

= 1 + 24

∞ X nq n . 1 + qn n=1

(4.13)

This proves (4.12) and so we complete the proof of Theorem 6.

4.3

The computations of t6 (n) and t8 (n)

Dividing both sides of (4.5) and (4.6), respectively, by u − v and then letting v → u gives Theorem 7 With θ1 (z|q) and θ4 (z|q) be defined by (2.1) and (2.4), respectively. Then µ

θ10 θ1

¶00

∞ X θ1 (2u|q) n2 q 2n (z|q) = cot u(1 + cot u) − 16 sin 2nu = θ10 (0|q)3 4 2n 1 − q θ1 (u|q) n=1

and

µ −

θ40 θ4

2

2

(4.14)

(z|q) = 16

∞ X n2 q n θ1 (2u|q) sin 2nu = θ10 (0|q)3 4 . 2n 1 − q θ4 (u|q) n=1

(4.15)

¶00

Now we use (4.15) to calculate t6 (n) and t8 (n). Setting u = 14 π in (4.15), we obtain ∞ ³ nπ ´ X n2 q n 1 2 θ2 (0|q 2 )θ44 (0|q 2 ) = sin 4 1 − q 2n 2 n=1

(4.16)

∞ ³ X ´ X 1 2 1 θ2 (0|q 2 )θ44 (0|q 2 ) = d2 sin( dπ) q 2n+1 . 4 2 n=0

(4.17)

Appealing to (1.6), we have

d|2n+1

Writing q as iq(i =

√

−1), (4.17) becomes ∞ ³ X ´ X 1 2 1 θ2 (0|q 2 )θ34 (0|q 2 ) = (−1)n d2 sin( dπ) q 2n+1 . 4 2 n=0 d|2n+1

10

(4.18)

Subtracting (4.17) from (4.18) and using Jacobi’s identity, (3.8), we obtain ∞ ³ X X

θ26 (0|q 2 ) = −8

n=0

d|4n+3

´ 1 d2 sin( dπ) q 4n+3 . 2

(4.19)

´ 1 d2 sin( dπ) q n . 2

(4.20)

Hence we obtain the following theorem. Theorem 8 Let ψ(q) be defined as in (1.2). Then 8ψ 6 (q) = −

∞ ³ X X n=0

d|4n+3

Consequently, we have t6 (n) = −

1 1 X 2 1 d sin( dπ) = 8 2 8 d|4n+3

X

d2 −

d|4n+3 d≡3 (mod 4)

1 8

X

d2 .

(4.21)

d|4n+3 d≡1 (mod 4)

There is an interesting history about t6 (n). One may consult [18, pp. 78-79]. Equation (4.21) is equivalent to [20, Theorem 4]. Other recent proofs have also been given by Kac and Wakimoto [10] and B. C. Berndt [5]. Dividing both sides of (4.15) by u, and then lettig u → 0, we obtain the following identity: ∞ X n3 q n 1 4 = θ2 (0|q)θ34 (0|q) = qψ 8 (q). 2n 1 − q 16 n=1

(4.22)

Employing (1.4), we find that ∞ X n3 q n 1 − q 2n n=1

= =

∞ ∞ X X n3 q n n3 q 2n − 1 − q n n=1 1 − q 2n n=1 ∞ ³ X n=1

n ´ σ3 (n) − σ3 ( ) q n . 2

(4.23)

Combining (4.22) and (4.23) gives Theorem 9 There holds the identity 8

qψ (q) =

∞ ³ X n=1

n ´ σ3 (n) − σ3 ( ) q n . 2

(4.24)

Consequently, we have n t8 (n) = σ3 (n) − σ3 ( ). 2

(4.25)

Identity (4.22) is due to Legendre [11, p. 133]. It was stated without proof by Ramanujan [21, p. 144]. Ewell [8] proved it by using Jacobi triple product identity and another theta function identity. Equation (4.23) is equivalent to [20, Theorem 5].

11

5

An equivalent form of an identity of Ramanujan

In this section we will prove the following identity. This identity has appeared in [16], but the proof given here is different from that in [16]. Theorem 10 There holds the identity (

)2 ∞ ∞ X X nq 2n nq n 1 − 24 + 24 cos nu 1 − q 2n 1 − q 2n n=1 n=1 = 1 + 240

∞ ∞ X X n3 q 2n n3 q n + 48 cos nu. 2n 1−q 1 − q 2n n=1 n=1

(5.1)

Proof. We multiply both sides of (3.1) by w5 θ1 (2w|q)/θ16 (w|q). Then we obtain θ12 (v|q)f1 (w) − θ12 (u|q)f2 (w) = θ1 (v + u|q)θ1 (v − u|q)f3 (w),

(5.2)

where f1 (w) = f2 (w) = f3 (w) =

θ1 (2w|q) , θ16 (w|q) θ1 (2w|q) w5 θ1 (w + v|q)θ1 (w − v|q) 6 , θ1 (w|q) θ1 (2w|q) w5 4 . θ1 (w|q) w5 θ1 (w + u|q)θ1 (w − u|q)

(5.3)

Differentiating f1 (w) with respect to w, four times, by using the method of logarithmic differentiation, we readily find that (4)

f1 (w) =

where φ(w) =

n f1 (w) φ4 (w) + 6φ2 (w)φ0 (w) + 4φ(w)φ00 (w) o +3φ0 (w)2 + φ000 (w) ,

5 θ0 θ0 θ0 θ0 − 6 1 (w|q) + 2 1 (w|q) + 1 (w − u|q) + 1 (w + u|q). w θ1 θ1 θ1 θ1

(5.4)

(5.5)

Applying (2.22) in (5.5), we find that à ! à ! ∞ ∞ X X 2 nq 2n 2 n3 q 2n φ(w) = 1 − 24 w− 1 + 240 w3 + O(w5 ) 2n 2n 3 1 − q 9 1 − q n=1 n=1 +

θ0 θ10 (w − u|q) + 1 (w + u|q). θ1 θ1

(5.6)

From the above equation, we readily find that φ(0) = 0,

φ00 (0) = 0,

φ0 (0) =

12

2 A(u), 3

4 φ000 (0) = − B(u), 3

(5.7)

where µ 0 ¶0 ∞ X nq 2n θ1 + 3 (u|q) 2n 1−q θ1 n=1 ( ) ∞ 2n X nq 1 1 1 + cot2 u + (1 − cos 2nu) , − 24 12 8 1 − q 2n n=1

A(u) =

1 − 24

=

(5.8)

and B(u)

= =

µ ¶000 ∞ X n3 q 2n 3 θ10 (u|q) − 1 − q 2n 2 θ1 n=1 (µ ) ¶2 ∞ 1 1 1 1 X n3 q 2n 2 + cot u + (5 + cos 2nu) . 576 12 8 12 n=1 1 − q 2n

1 + 240

(5.9)

Employing (5.7)-(5.9) in (5.4) and then setting z = 0, we find that (4)

f1 (0) = −

ª 8θ12 (u|q) © 2 A (u) − B(u) . 3θ10 (0|q)5

(5.10)

ª 8θ12 (v|q) © 2 A (v) − B(v) . 0 5 3θ1 (0|q)

(5.11)

We proceed as in (5.4)-(5.10) to obtain (4)

f2 (0) = −

From (2.22) and (4.14), we can deduce that f3 (w) = 2θ10 (0|q)3 w2 + O(w6 ).

(5.12)

It follows that (4)

f3 (0) = 0.

(5.13)

Differentiating (5.2) with respect to w, five times, and then taking w = 0, we have (4)

(4)

(4)

θ12 (v|q)f1 (0) − θ12 (u|q)f2 (0) = θ1 (v + u|q)θ1 (v − u|q)f3 (0).

(5.14)

Substituting (5.10), (5.11), and (5.13) into (5.14), we obtain, for all u, v not equal integral multiples of π, A2 (u) − B(u) = A2 (v) − B(v). Using (5.8) and (5.9), we find that

=

576 lim (A2 (v) − B(v)) v→0 (∞ ¶) X nq 2n µ 1 1 2 2 2 lim sin nv + cot v sin nv v→0 1 − q 2n 3 2 n=1 (∞ )2 X nq 2n + lim (1 − cos 2nz) v→0 1 − q 2n n=1 13

(5.15)

− =

∞ X 1 n3 q 2n (5 + cos 2nz) lim 12 v→0 n=1 1 − q 2n

∞ ∞ 1 X n3 q 2n 1 X n3 q 2n − 2 n=1 1 − q 2n 2 n=1 1 − q 2n

= 0.

(5.16)

Thus, letting v → 0 in (5.15) and then using (5.16), we obtain A2 (u) − B(u) = 0.

(5.17)

Substituting (5.8) and (5.9) into the above equation, we obtain (

)2 µ 0 ¶0 ∞ X nq 2n θ1 1 − 24 +3 (u|q) 1 − q 2n θ1 n=1 µ ¶000 ∞ X n3 q 2n 3 θ10 = 1 + 240 − (u|q). 1 − q 2n 2 θ1 n=1

(5.18)

Replacing the logarithmic derivative of θ1 (z|q) by its trigonometric series expansion in this equation, we obtain the following identity of Ramanujan [21, p. 139]. Theorem 11 We have (

)2 ∞ X 1 1 nq 2n 2 cot u + + (1 − cos 2nu) 8 12 n=1 1 − q 2n ¾2 ½ ∞ 1 1 X n3 q 2n 1 cot2 u + + (5 + cos 2nu). = 8 12 12 n=1 1 − q 2n

(5.19)

Writing u as u + 12 πτ and then using (2.18), (5.18) becomes (

)2 µ 0 ¶0 ∞ X nq 2n θ4 1 − 24 +3 (u|q) 1 − q 2n θ4 n=1 µ ¶000 ∞ X n3 q 2n 3 θ40 = 1 + 240 − (u|q). 1 − q 2n 2 θ4 n=1

(5.20)

Substituting (2.20) into the above equation, and then writing 2u as u, we obatin (5.1). This completes the proof of Theorem 10.

6

The first class of Lambert seires

In this section we will consider the Lambert seires T2k (q) defined by T2 (q) = 1 + 24

∞ X nq n 1 + qn n=1

and T2k (q) =

14

∞ X n2k−1 q n , 1 − q 2n n=1

k ≥ 2.

(6.1)

We now begin to deduce a recurrence formula for T2k (q) using (5.20). It is well known that the Maclaurin series for cos nu is cos nu =

∞ X

(−1)k

k=0

(nu)2k . (2k)!

(6.2)

Substituting (6.2) into (5.1), inverting the order of summation, we obtain the identity à !2 ∞ 2k X k u T2 (q) + 24 (−1) T2k+2 (q) (2k)! k=1

=

∞ ∞ X X n3 q 2n u2k 1 + 240 T2k+4 (q). + 48T (q) + 48 (−1)k 4 2n 1−q (2k)! n=1

(6.3)

k=1

Comparing the coefficients of u2m on either side, we obtain the following recurence formula among T2k (q), k = 1, 2, 3, · · ·. This formula allows us to express T2k (q) in terms of T2 (q) and T4 (q). Theorem 12 If T2k (q) are defined by (6.1). Then we have T22 (q) = 1 + 240 T2m+4 (q)

∞ X n3 q 2n + 48T4 (q), 1 − q 2n n=1

(6.4)

= T2 (0|q)T2m+2 (q) +12

m−1 X k=1

(2m)! T2k+2 (q)T2m+2−2k (q). (2k)!(2m − 2k)!

(6.5)

In particular, T6 (q)

= T2 (q)T4 (q),

(6.6)

T8 (q)

= T2 (q)T6 (q) + 72T42 (q),

(6.7)

T10 (q)

= T2 (q)T8 (q) + 360T4 (q)T6 (q),

(6.8)

T12 (q)

= T2 (q)T10 (q) + 672T4 (q)T8 (q) + 840T62 (q).

(6.9)

Utilizing (4.10), (4.22), and (6.5) we can write T2k (q) in terms of θ2 (0|q) and θ3 (0|q). Here we shall use (6.6)-(6.9) to express T2k (q) in terms of theta functions θ2 (0|q) and θ3 (0|q) for 6 ≤ 2k ≤ 12. For simplicity, we define x(q) :=

θ24 (0|q) θ34 (0|q)

and

z(q) := θ34 (0|q) = φ4 (q).

(6.10)

Then (4.10) and (4.22) can be written in the following forms: T2 (q) T4 (q)

= z(q)(1 + x(q)), 1 2 = qψ 8 (q) = z (q)x(q). 16

(6.11) (6.12)

Substituting the above two equations into (6.4), we immediately obtain the following identity [4, p. 49, Equation(12. 21)] 1 + 240

∞ X ¡ ¢ n3 q 2n = z 2 (q) 1 − x(q) + x2 (q) . 2n 1 − q n=1

15

(6.13)

Substituting (6.11) and (6.12) into (6.6), we find that T6 (q) =

¢ 1 3 ¡ z (q) x(q) + x2 (q) . 16

(6.14)

Similarly, from (6.7), (6.8), and (6.9), we find respectively that ¢ 1 4 ¡ z (q) 2x(q) + 13x2 (q) + 2x3 (q) , 32 ¢ 1 5 ¡ T10 (q) = z (q) x(q) + 30x2 (q) + 30x3 (q) + x4 (q) , 16 ¢ 1 6 ¡ z (q) 2x(q) + 251x2 (q) + 876x3 (q) + 251x4 (q) + 2x5 (q) . T12 (q) = 32 T8 (q) =

7

(6.15) (6.16) (6.17)

The second class of Lambert series

In this section we will investigate the Lambert series defined as S2k (q) =

∞ X (2n + 1)2k−1 q n+1/2 . 1 − q 2n+1 n=0

(7.1)

And the main result of this section is the following recurrence relation. Using this recurrence , we can express S2k (q) in terms of z(q) and x(q). Theorem 13 Let T2k (q) be defined by (6.1) and S2k (q) by (7.1). Then we have S2m+4 (q) = T2 (q)S2m+2 (q) + 48

m X k=1

(2m)!4k T2k+2 (q)S2m+2−2k (q). (2k)!(2m − 2k)!

(7.2)

Proof. We recall the identity in (5.1), namely, ( )2 ∞ ∞ X X nq 2n nq n 1 − 24 + 24 cos nu 1 − q 2n 1 − q 2n n=1 n=1 ∞ ∞ X X n3 q 2n n3 q n = 1 + 240 + 48 cos nu. 1 − q 2n 1 − q 2n n=1 n=1

(7.3)

Replacing q by −q, this identity then becomes ( )2 ∞ ∞ n X X nq 2n n nq 1 − 24 + 24 (−1) cos nu 1 − q 2n 1 − q 2n n=1 n=1 = 1 + 240

∞ ∞ X X n3 q n n3 q 2n + 48 (−1)n cos nu. 2n 1−q 1 − q 2n n=1 n=1

(7.4)

We subtract (7.4) from (7.3) and then replace q 2 by q to get

=

∞ X (2n + 1)3 q n+1/2 cos(2n + 1)u 1 − q 2n+1 n=0 Ã∞ ! X (2n + 1)q n+1/2 cos(2n + 1)u 1 − q 2n+1 n=0 à ! ∞ ∞ X X nq n nq n × 1 − 24 + 48 cos 2nu . 1 − qn 1 − q 2n n=1 n=1

16

(7.5)

Substituting cos(2n + 1)u = 1 +

∞ X

(−1)k

k=1

(2n + 1)2k 2k u (2k)!

and

cos 2nu = 1 +

∞ X

(−1)k

k=1

(2n)2k 2k u (2k)!

(7.6)

into (7.5), inverting the order of summation, we obtain the identity ∞ X

=

(−1)k

k=0 ̰ X

u2k S2k+4 (q) (2k)!

! Ã ! ∞ 2k X u2k k k u S2k+2 (q) × T2 (q) + 48 T2k+2 (q) . (−1) 4 (−1) (2k)! (2k)! k

k=0

(7.7)

k=1

Equating the coefficients of u2m on both sides of (7.7), we obtain (7.2). This complete the proof of Theorem 13. The first few special cases of (7.2) are S4 (q) =

T2 (q)S2 (q),

(7.8)

S6 (q) =

S4 (q)T2 (q) + 192S2 (q)T4 (q),

(7.9)

S8 (q) =

T2 (q)S6 (q) + 1152T4 (q)S4 (q) + 768S2 (q)T6 (q),

(7.10)

S10 (q) =

T2 (q)S8 (q) + 2880T4 (q)S6 (q) + 11520T6 (q)S4 (q) + 3072T8 (q)S2 (q),

(7.11)

S12 (q) =

T2 (q)S10 (q) + 5376T4 (q)S8 (q) + 53760T6 (q)S6 (q) +86016T8 (q)S4 (q) + 12288T10 (q)S2 (q).

(7.12)

Using the product expansions for θ2 (0|q) in (2.12) and θ3 (0|q) in (2.13), we readily find that 4θ2 (0|q)θ3 (0|q) = θ22 (0|q 1/2 ).

(7.13)

Replacing q 2 by q in (4.10) and then using (7.13), we obtain S2 (q) =

1 2 θ (0|q)θ32 (0|q). 4 2

(7.14)

Applying (6.10) to the right side of this equation, we find that S2 (q) =

p 1 z(q) x(q). 4

(7.15)

Substituting (6.11) and (7.15) into (7.8), we find that S4 (q) =

1 2 p z (q) x(q)(1 + x(q)). 4

(7.16)

In the same way, we also find that S6 (q) =

1 2 p z (q) x(q)(1 + 14x(q) + x2 (q)), 4 17

(7.17)

S8 (q) = S10 (q) = S12 (q) =

¡ ¢ 1 4 p z (q) x(q) 1 + 135x(q) + 135x2 (q) + x3 (q) , 4 ¡ ¢ 1 4 p z (q) x(q) 1 + 1228x(q) + 5478x2 (q) + 1228x3 (q) + x4 (q) , 4 ³ 1 4 p z (q) x(q) 1 + 11069x(q) + 165826x2 (q) + 165826x3 (q) 4 ´ +11069x4 (q) + x5 (q) .

(7.18) (7.19)

(7.20)

Using Jacobi’s quartic identity, (3.8), and the definitions of x(q) and z(q), we find that z(−q) = z(q)(1 − x(q)) and

x(−q) = −

x(q) . 1 − x(q)

(7.21)

By direct computaions, we find that S2 (q 2 ) =

1 (T2 (q) − T2 (−q)) 48

S2k (q 2 ) =

and

1 (T2k (q) − T2k (−q)) 2

, k ≥ 2.

(7.22)

Combining (6.11),(7.21), and the first identity in (7.22), we find that S2 (q 2 ) =

1 z(q)x(q). 16

(7.23)

Combining (6.12),(7.21), and the second identity in (7.22) with k = 2, we find that S4 (q 2 ) =

1 2 z (q)x(q)(2 − x(q)). 32

(7.24)

Similarly, we also have

8

S6 (q 2 )

=

S8 (q 2 )

=

S10 (q 2 )

=

S12 (q 2 )

=

1 3 z (q)x(q)(1 − x(q) + x2 (q)), 16 1 4 z (q)x(q)(2 − x(q))(2 − 2x(q) + 17x2 (q)), 64 1 5 z (q)x(q)(1 − 2x(q) + 78x2 (q) − 77x3 (q) + 31x4 (q)), 16 1 6 z (q)x(q)(2 − x(q))(1 − x(q) + x2 (q))(1 − x(q) + 691x2 (q)). 32

(7.25) (7.26) (7.27) (7.28)

Two identities for ψ 12 (q)

Theorem 14 Let t2k (n) and σ2k (n) be defined by (1.2) and (1.3), respectively. Then ∞ X

σ5 (2n + 1)q 2n+1 = 256

n=0

∞ X

t12 (n)q 2n+3 + q(q 2 ; q 2 )12 ∞.

(8.1)

n=0

Furthermore, if we define a(n) as the coefficient of q n in q(q 2 ; q 2 )12 ∞ =

∞ X

a(n)q 2n+1 ,

(8.2)

n=0

then we have 256t12 (n) = σ5 (2n + 3) − a(n + 1). 18

(8.3)

Equation (8.1) is the same as [20, Theorem 7]. For an account one may consult [18]. Proof. Using Jacobi’s quartic identity, (3.8), and the product expansion formulas for θ2 (0|q), θ3 (0|q), and θ4 (0|q), we readily find that z 3 (q)x(q)(1 − x(q)) = 16q(q 2 ; q 2 )12 ∞

and z(q)x(q) = θ24 (0|q) = 16qψ 4 (q 2 ).

(8.4)

Hence the left side of (7.25) is equal to 1 3 1 12 2 z (q)x(q)(1 − x(q)) + z 3 (q)x3 (q) = q(q 2 ; q 2 )12 ∞ + 256qψ (q ). 16 16 It follows that S6 (q 2 ) = Substituting

∞ X (2n + 1)5 q 2n+1 3 12 2 = q(q 2 ; q 2 )12 ∞ + 256q ψ (q ). 2(2n+1) 1 − q n=0

∞ ∞ X X (2n + 1)5 q 2n+1 = σ5 (2n + 1)q 2n+1 2(2n+1) 1 − q n=0 n=0

and q 3 ψ 12 (q 2 ) =

∞ X

t12 (n)q 2n+3

(8.5)

(8.6)

(8.7)

(8.8)

n=0

into (8.6), equating the coefficients of q 2n+3 , we obtain (8.1), thereby we complete the proof of Theorem 14. Now we state the other formula for ψ 12 (q), which is different from the identity in (8.1). Theorem 15 Let T2k (q) and S2k (q) be defined by (6.1) and (7.1), respectively, then T4 (q 2 )S2 (q 2 ) = q 3 ψ 12 (q 2 ).

(8.9)

Proof. Using the product expansion formulas for θ2 (0|q) and θ3 (0|q), we find that p z(q) x(q) = 4q 1/2 ψ 4 (q).

(8.10)

Multiplying (6.12) and (7.15), we immdediately have T4 (q)S2 (q) =

´3 p 1 ³ z(q) x(q) = q 3/2 ψ 12 (q). 64

(8.11)

Replacing q by q 2 in the above equation, we complete the proof of Theorem 15. Comparing (8.6) and (8.9), we find the following identity, which may be a new identity. Theorem 16 With T2k (q) and S2k (q) defined by (6.1) and (7.1), respectively, 2 2 2 q(q 2 ; q 2 )12 ∞ = S6 (q ) − 256S2 (q )T4 (q ).

19

(8.12)

9 9.1

Identities relating ψ 16 (q), ψ 20 (q), and ψ 28 (q) An identity relating ψ 16 (q)

Theorem 17 Let S2k (q) be the Lambert series defined by (7.1). Then 192

∞ X

t16 (n)q 2n+4 = 192q 4 ψ(q 2 )16 = S2 (q 2 )S6 (q 2 ) − S42 (q 2 ).

(9.1)

n=0

Consequently, we have 192t16 (n − 1) =

n ³ X

´ σ1 (2k + 1)σ5 (2n − 2k + 1) − σ3 (2k + 1)σ3 (2n − 2k + 1) .

(9.2)

k=0

Proof. By (7.16), (7.17), and (7.18) and some elementary evaluations, we find that S2 (q)S6 (q) − S42 (q) 1 4 1 = z (q)x(q)(1 + 14x(q) + x2 (q)) − z 4 (q)x(q)(1 + x(q))2 16 16 p 3 4 = (z(q) x(q)) . 4

(9.3)

Substituting (8.10) into the left side of the above equation and then changing q as q 2 , we obtain (9.1). We compare the coefficients of q n on both sides of (9.1) to obtain (9.2). This complete the proof of Theorem 17.

9.2

An identity relating ψ 20 (q)

Theorem 18 Let T2k (q) and S2k (q) be the Lambert series defined by (6.1) and (7.1), respectively. Then 72

∞ X

t20 (n)q 2n+5 = 120q 5 ψ(q 2 )20 = T8 (q 2 )S2 (q 2 ) − S4 (q 2 )T6 (q 2 ).

(9.4)

n=0

Proof. We recall the two identities in (6.7) and (7.8), namely, T8 (q)

= T2 (q)T6 (q) + 72T42 (q),

(9.5)

S4 (q)

= T2 (q)S2 (q).

(9.6)

Eliminating T2 (q) between the above two equations, we arrive at 72S2 (q)T42 (q) = T8 (q)S2 (q) − S4 (q)T6 (q).

(9.7)

Substituting (6.12) and (7.15) into the left side of the above equation, we find that p 9 (z(q) x(q))5 = T8 (q)S2 (q) − S4 (q)T6 (q). 128

(9.8)

Using (8.10) in the left side of the above equation and then changing q as q 2 , we obtain (9.4). Thus we complete the proof of Theorem 18 . 20

9.3

An identity for t28 (n)

Theorem 19 Let T2k (q) and S2k (q) be defined by (6.1) and (7.1), respectively. Then we have 21T8 (q 2 )S6 (q 2 ) + 2T4 (q 2 )S10 (q 2 ) − 23T12 (q 2 )S2 (q 2 ) = 725760q 7 ψ 28 (q 2 ).

(9.9)

Proof. From (6.12), (6.15), (6.17), (7.14), (7.17), and (7.19), by direct computations, we find that 21T8 (q)S6 (q) + 2T4 (q)S10 (q) − 23T12 (q)S2 (q) =

p 2835 (z(q) x(q))7 . 64

(9.10)

Substituting (8.10) into the left side of the above equation and then replacing q by q 2 , we obtain (9.9). This completes the proof of Theorem 19.

10 10.1

Identities relating ψ 24 (q) and ψ 32 (q) Two formulas for ψ 24 (q)

Theorem 20 We have T24 (q) =

∞ X n11 q n 24 3 24 = 2072q 2 (q 2 ; q 2 )24 ∞ + q(q; q)∞ + 176896q ψ (q). 2n 1 − q n=1

(10.1)

Proof. By elementary computation, we find for any a that 2 + 251a + 876a2 + 251a3 + 2a4 = 1382a2 + 259a(1 − a)2 + 2(1 − a)4 .

(10.2)

Thus from (6.17), we have

= = =

T12 (q) ¡ ¢ 1 6 z (q)x(q) 2 + 251x(q) + 876x2 (q) + 251x3 (q) + 2x4 (q) 32 ¡ ¢ 1 6 z (q)x(q) 1382x(q)2 + 259x(q)(1 − x(q))2 + 2(1 − x(q))4 32 ´2 259 ¡ p ¢2 691 ³ 1 z(q) x(q) + z 3 (q)x(q)(1 − x(q)) + z 6 (q)x(q)(1 − x(q))4 . 16 32 16

(10.3)

Using Jacobi’s quartic identity, (3.8), and the product expansion formulas for θ2 (0|q), θ3 (0|q), and θ4 (0|q), we readily find that z 6 (q)x(q)(1 − x(q))4 = 16q(q; q)24 ∞.

(10.4)

Substituting (8.4) and (10.4) into the left side of (10.1), we find that identity in (10.1) holds. This completes the proof of Theorem 20. We recall the Ramanujan τ (n)-function [22, p. 197, Equation(6.1.13)] defined as the coefficient of q n in q(q; q)24 ∞ =

∞ X n=1

21

τ (n)q n .

(10.5)

Then q 2 (q 2 ; q 2 )24 ∞ =

∞ X

n τ ( )q n , 2 n=1

(10.6)

if we agree that τ (x) means 0 when x ia not an integer. Substituting q 3 ψ 24 (q) =

∞ X

t24 (n − 3)q n ,

(10.7)

n=3

(10.5), and (10.6) into (10.1), we obtain the following theorem [20, Theorem 8]. Theorem 21 We have ∞ ∞ ³ ³ n ´´ X X n11 q n = σ (n) − σ qn 11 11 2n 1 − q 2 n=1 n=1

= 176896

∞ X

t24 (n − 3)q n +

n=1

∞ X

τ (n)q n + 2072

n=1

∞ X

n τ ( )q n . 2 n=1

(10.8)

Consequently, n n 176896t24 (n − 3) = σ9 (n) − σ9 ( ) − τ (n) − 2072τ ( ). 2 2

(10.9)

Here, for convenience, we assume t24 (n) = 0 for n is not a nonnegative integer. Theorem 22 Let T2k (q) be defined by (6.1). Then we have 72q 3 ψ 24 (q) = T4 (q)T8 (q) − T62 (q).

(10.10)

Proof. Multiplying both sides of (6.7) by T4 (q) yields T4 (q)T8 (q) = T2 (q)T4 (q)T6 (q) + 72T43 (q).

(10.11)

Eliminating T2 (q) between this equation and (6.6), we immediately have 72T43 (q) = T4 (q)T8 (q) − T62 (q).

(10.12)

Substituting (6.2) into the left hand side of the above equation yields (10.10). We complete the proof of Theorem 22 This identity can also be found in [6].

10.2

An identity for ψ 32 (q)

Theorem 23 Let T2k (q) be defined by (6.1). Then we have 75600T44 (q) = 75600q 4 ψ 32 (q) 25 21 = T4 (q)T12 (q) + T82 (q) − T6 (q)T10 (q). 4 4 22

(10.13)

This beautiful identity is due to Chan and Chua [6], in which the authors use modular forms to prove the identity. Here we will give a simple proof. Proof. Eliminating T2 (q) between (6.7) and (6.8), we arrive at T42 (q)T8 (q) − 5T4 (q)T62 (q) =

¢ 1 ¡ 2 T8 (q) − T6 (q)T10 (q) . 72

(10.14)

Eliminating T2 (q) between (6.6) and (6.9), we arrive at T4 (q)T12 (q) = T6 (q)T10 (q) + 672T42 (q)T8 (q) + 840T4 (q)T62 (q).

(10.15)

Multiplying both sides of (10.12) by T4 (q) gives 72T44 (q) = T42 (q)T8 (q) − T4 (q)T62 (q).

(10.16)

Eliminating T42 (q)T8 (q) and T4 (q)T62 (q) among (10.14), (10.15), and (10.16), we obtain (10.13). This completes the proof of Theorem 23.

11

Some identities among τ (n), t24 (n), and σ11 (n)

Throughout this section we assume that p is prime. The dissection operator Up [2, p. 161] operating P∞ on f (q) = n=0 a(n)q n is define by Up {f (q)}

∞ X

=

a(np)q n

n=0 p−1

1 1X f (ωpj q p ), p j=0

=

(11.1)

where ωp =exp( 2πi p ). It is obvious that for an integer k ≥ 1, Upk {f (q)} =

∞ X

a(npk )q n .

(11.2)

n=0

Theorem 24 We have ³ n ´ 176896t24 (2k n − 3) = 211k σ11 (n) − σ11 ( ) − τ (2k n) − 2072τ (2k−1 n). 2

(11.3)

When k = 0, the above equation reduces to the following formula for t24 (n) [20, Theorem 8]: ³n´ n . (11.4) 176896t24 (n − 3) = σ11 (n) − σ11 ( ) − τ (n) − 2072τ 2 2 Replacing n by 2n + 1, the above equation reduces to 176896t24 (2n − 2) = σ11 (2n + 1) − τ (2n + 1).

(11.5)

It follows that [19, p. 103, Corollary 6.4.7] τ (2n + 1) ≡ σ11 (2n + 1) 23

(mod 176896).

(11.6)

Proof. Using

∞ ∞ ∞ 11 n X X X n11 q n n11 q 2n n n q 12 + (−1) = 2 , 1 − q 2n n=1 1 − q 2n 1 − q 4n n=1 n=1

we find that

( U2k

∞ X n11 q n 1 − q 2n n=1

) = 211k

∞ X n11 q n . 1 − q 2n n=1

(11.7)

(11.8)

We recall the identity in (10.8), namely, ∞ X n11 q n 1 − q 11n n=1

= 176896

∞ X

t24 (n − 3)q n +

n=1

∞ X

τ (n)q n + 2072

n=1

∞ X

n τ ( )q n . 2 n=1

(11.9)

We operate both sides of this identity by U2k , and we find that 211k

∞ X n11 q n 1 − q 2n n=1

= 176896

∞ X n=1

Substituting

t24 (2k n − 3)q n +

∞ X

τ (2k n)q n + 2072

n=1

∞ X

τ (2k−1 n)q n .

(11.10)

n=1

∞ ∞ ³ X X n11 q n n ´ n = σ (n) − σ ( ) q 11 11 1 − q 2n 2 n=1 n=1

(11.11)

into (11.10), and then comparing the coefficients of q n , we obtain (11.3). We complete the proof of Theorem 24. Theorem 25 We have

µ ¶ n n n ´ 11 k−1 ) σ11 ( ) − σ11 ( ) σ11 (p ) σ11 (n) − σ11 ( ) − p σ11 (p 2 p 2p 1 k k k = 176896t24 (p n − 3) + τ (p n) + 2072τ ( p n), 2 k

³

(11.12)

where p is an odd prime. When n = 1, the above equation reduces to the following identity [20, p. 89, Equation (9)]: 176896t24 (pk − 3) = σ11 (pk ) − τ (pk ),

(11.13)

where p is an odd prime. Proof. If p doesn’t divide n, it is well known that (1 − xy)(1 − xyωpn )(1 − xyωp2n ) · · · (1 − xyωp(p−1)n ) = 1 − xp y p ,

24

(11.14)

where ωp =exp( 2πi p ). Taking logarithmic derivative of the above equation with respect to x, and then setting x = 1, we obtain (p−1)n yωpn yωp2n y yωp py p + . + + · · · + = n 2n (p−1)n 1 − y 1 − yωp 1 − yωp 1 − yp 1 − yωp

Hence, if we take f (q) =

∞ X nr q n , 1 − qn n=1

(11.15)

(11.16)

then using (11.15), we find that p−1 X

f (ωpk q)

k=0

= pr+1

= pr+1

∞ X nr q np + 1 − q np n=1

∞ X n6≡0

(mod p) n=1

nr ωprn q n 1 − q n ωprn

∞ ∞ X X nr q np nr q np + p np 1−q 1 − q np n6≡0 (mod p) n=1 n=1

2 ∞ ∞ X X nr q np nr q np r r+1 = p(1 + p ) −p . 1 − q np 1 − q np2 n=1 n=1

Thus, we have

(

∞ X nr q n 1 − qn n=1

Up It follows that

) = (1 + pr )

∞ ∞ X X nr q n nr q np r − p . n 1−q 1 − q np n=1 n=1

(11.17)

(11.18)

(

) ∞ X nr q n 1 − qn n=1 ) ) ½ (k+1)r ½ kr ¾ (X ¾ (X ∞ ∞ p −1 nr q n p −1 nr q np r = −p pr − 1 1 − qn pr − 1 1 − q np n=1 n=1 Upk

= σr (pk )

∞ ∞ X X nr q n nr q np r k−1 − p σ (p ) . r 1 − qn 1 − q np n=1 n=1

(11.19)

If p is an odd prime, writing q as q 2 , the above equation becomes (∞ ) ∞ ∞ X nr q 2n X X nr q 2n nr q 2np k k r k−1 (p ) (p ) Up = σ − p σ . r r 2n 2n 1−q 1−q 1 − q 2np n=1 n=1 n=1 Subtracting (11.20) from (11.19), we immediately have (∞ ) ∞ ∞ X X X nr q n nr q n nr q np k k r k−1 Up = σ (p ) − p σ (p ) . r r 1 − q 2n 1 − q 2n 1 − q 2np n=1 n=1 n=1

(11.20)

(11.21)

Acting on both sides of (11.9) by Upk , and using (11.2) and (11.21), we find that σ11 (pk )

∞ ∞ X X n11 q n n11 q np 11 k−1 − p σ (p ) 11 2n 1−q 1 − q 2np n=1 n=1

= 176896

∞ X

t24 (pk n − 3)q n +

n=1

∞ X n=1

25

τ (pk n)q n + 2072

∞ X

1 τ ( pk n)q n . 2 n=1

(11.22)

Equating the coefficients of q n on both sides, we obtain (11.12). This completes the proof of Theorem 25. Acknowledgements This work is supported, in part, by Academic Research Fund R146000027112 from the National University of Singapore. The author is grateful for the referee for his many helpful suggestions. He also would like thanks Li-Chien Shen and Yi-Fan Yang for their comments.

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