UNIFORM PROOFS OF q-SERIES-PRODUCT

UNIFORM PROOFS OF q-SERIES-PRODUCT IDENTITIES SARACHAI KONGSIRIWONG AND ZHI-GUO LIU Abstract. In this paper we give a simple proof of the Jacobi triple product identity by using basic properties of cube roots of unity. Then we give a new proof of the quintuple product identity, the Farkas-Kra septuple product identity and Winquist’s identity by using the Jacobi triple product identity and basic properties of cube and fifth roots of unity. Furthermore, we derive some new product identities by this uniform method. Later, we give some generalizations of those identities. Lastly, we derive some modular equations.

1. Introduction In this paper we give two general methods of proving q-series-product identities. The first method uses basic properties of roots of unity. The second method generalizes S. Bhragava’s argument of proving the quintuple product identity [5]. In several sections we focus on using the first method and give some new identities. Then we give four generalizations of product identities proved by the second method. By comparing the results from the two methods, we obtain new identities; from those identities we can derive some modular equations. All proofs in this paper require the Jacobi triple product identity, except our proof of the identity itself. The Jacobi triple product identity is one of the most important identities in the theory of theta functions. Many proofs are known. The first published proof was given by C. G. J. Jacobi [17]. However, C. F. Gauss [13, p. 464] proved it earlier, since the identity was recorded in his posthumous manuscript. The standard proof using the theory of Jacobi’s theta functions can be found in [8, pp. 67–71]. The triple product identity is a special case of S. Ramanujan’s 1 ψ1 summation theorem which can be found Entry 17 of Chapter 16 in Ramanujan’s second notebook [20]. In [1], G. E. Andrews gave a proof using two simple Euler identities of q-series. In [10], J. A. Ewell established this identity using the Euler pentagonalnumber theorem and a theorem of Gauss. A combinatorial proof has been given in [19] by R. P. Lewis. Recently D. Foata and G.-N. Han [12] gave a nice proof and a history of the triple product identity. Recently, H. S. Wilf [23] proved an equivalent proposition to the triple product identity. G. H. Hardy and E. M. Wright [15, pp. 282–283] also give a simple proof using some properties of fourth roots of unity. In this paper we will give a similar but slightly simpler proof, using cube roots of unity. Furthermore, Key words and phrases. Triple product identity, quintuple product identity, septuple product identity, Winquist’s identity. The research of the second author was supported by Academic Research Fund R146000027112 from the National University of Singapore. The second author would like to thank Professor Hirschhorn for his valuable suggestions. The first author thanks his advisor Professor Bruce C. Berndt for his helpful guidance. 1

2

SARACHAI KONGSIRIWONG AND ZHI-GUO LIU

we use this idea and the triple product identity to prove the quintuple product identity. The quintuple product identity was first published in H. A. Schwarz’s book [21, p. 47] in 1893. G. N. Watson [22] gave a proof in 1929, and consequently the identity is sometimes known as Watson’s quintuple product identity. Since Watson’s proof, many proofs of the identity has been given. A simple proof was given by L. Carlitz and M. V. Subbarao [6]. R. J. Evans [9] gave a short proof using complex function theory. For a history on the identity, see B. C. Berndt’s book [3, p. 83]. Our proof is similar to B. Gordon’s [14] and Bhargava’s [5]. In addition, this method can be used to prove the Farkas-Kra septuple product identity. The septuple product identity was first discovered by H. M. Farkas and I. Kra [11] in 1999. Later Foata and Han [12] gave another proof of the identity and called the identity the Farkas-Kra septuple product identity. The word “septuple”was probably given since there are seven different infinite products in the identity as stated in the two papers, [11] and [12]. However, these seven infinite products can be written as five infinite products. In our proof we use some properties of fifth roots of unity and again the Jacobi triple product identity. The applications of roots of unity do not end here; we can use this method to prove Winquist’s identity. In 1969, L. Winquist [24] gave a simple proof of the congruence p(11n + 6) ≡ 0 (mod 11), where p(n) is the number of partitions of the positive integer n. The key identity he used in his proof is now called Winquist’s identity. Later, Carlitz and Subbarao [7] and M. D. Hirschhorn [16] also gave proofs and generalizations of Winquist’s identity. Recently, S.-Y. Kang [18] gave a simple proof using the Jacobi triple product identity, the quintuple product identity and two other identities from Ramanujan’s notebooks. Our proof seems to be simpler since it uses only the Jacobi triple product identity and some properties of cube roots of unity. To prove our product identities, we first recall some simple facts about roots of unity. Let ω=exp(2πi/3). Then we have ½ 3, n ≡ 0 (mod 3), n 2n 1+ω +ω = (1.1) 0, n 6≡ 0 (mod 3), and, for any complex number x, (1 − x)(1 − xω)(1 − xω 2 ) = 1 − x3 .

(1.2)

Similarly, let ζ = exp(2πi/5). Then we have, for any complex number x, (1 − x)(1 − xζ)(1 − xζ 2 )(1 − xζ 3 )(1 − xζ 4 ) = 1 − x5 .

(1.3)

Let a and q be complex numbers with |q| < 1. Set (a; q)∞ =

∞ Y

(1 − aq n ).

(1.4)

n=0

Using (1.2) and (1.4), we obtain (a; q)∞ (aω; q)∞ (aω 2 ; q)∞ = (a3 ; q 3 )∞ .

(1.5)

UNIFORM PROOFS OF q-SERIES-PRODUCT IDENTITIES

3

Using (1.3) and (1.4), we obtain (a; q)∞ (aζ; q)∞ (aζ 2 ; q)∞ (aζ 3 ; q)∞ (aζ 4 ; q)∞ = (a5 ; q 5 )∞ .

(1.6)

Now we are ready to prove the Jacobi triple product identity, the Watson quintuple product identity, the Farkas-Kra septuple product identity, Winquist’s identity, and some other product identities. 2. The triple product identity In this section we prove the Jacobi triple product identity by using some properties of cube roots of unity. Theorem 1. For each pair of complex numbers z and q, with z 6= 0 and |q| < 1, ∞ X

2

q n z n = (q 2 ; q 2 )∞ (−zq; q 2 )∞ (−z −1 q; q 2 )∞ .

(2.1)

n=−∞

Proof. Let f (z) denote the left side of (2.1). It is evident that this function has a Laurent expansion ∞ X f (z) = an (q)z n . (2.2) n=−∞

It can be easily verified that f (z) satisfies the functional equation f (z) = zqf (zq 2 ).

(2.3)

Substituting (2.2) into this equation and equating coefficients of q n , we obtain an (q) = −an−1 (q)q 2n−1 .

(2.4)

From this we immediately have, for each integer n, 2

an (q) = a0 (q)q n .

(2.5)

Combining this with (2.2) we have ∞ X

(q 2 ; q 2 )∞ (−zq; q 2 )∞ (−z −1 q; q 2 )∞ = a0 (q)

2

qn zn.

(2.6)

n=−∞

Next we calculate a0 (q). Setting z = −q, we find that 0 = a0 (q)

∞ X

(−1)n q n

2 +n

.

(2.7)

n=−∞

Putting x = −qω, −qω 2 in (2.6) and using (1.5), we obtain, respectively, 2

6

6

(1 − ω )(q ; q )∞ = a0 (q)

∞ X

(−1)n q n

2 +n

ωn

(2.8)

ω 2n .

(2.9)

n=−∞

and 6

6

(1 − ω)(q ; q )∞ = a0 (q)

∞ X

(−1)n q n

n=−∞

2 +n

4


Adding these three equations together gives ∞ X 2 2 6 6 (2 − ω − ω )(q ; q )∞ = a0 (q) (−1)n q n +n (1 + ω n + ω 2n ).

(2.10)

n=−∞

Making use of (1.1), we arrive at 6

∞ X

6

(q ; q )∞ = a0 (q)

2 +3n

(−1)n q 9n

.

(2.11)

2 +3n

(2.12)

n=−∞

If we replace q by q 9 and set z = −q 3 (2.6), we find that ∞ X

(q 6 ; q 6 )∞ = a0 (q 9 )

(−1)n q 9n

.

n=−∞

Comparing (2.11) with (2.12), we obtain a0 (q) = a0 (q 9 ). By induction, we have a0 (q) = n a0 (q 9 ) and hence a0 (q) = a0 (0). Taking q = 0 on both sides of (2.6) we find that a0 (0) = 1. Thus we complete the proof of the theorem. ¤ 3. The quintuple product identity In this section, we prove the quintuple product identity. Theorem 2. For any complex numbers z and q, with z 6= 0 and |q| < 1, ∞ X

q 3n

2 +n

(z 3n q −3n − z −3n−1 q 3n+1 )

n=−∞

= (q 2 ; q 2 )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (z 2 ; q 4 )∞ (z −2 q 4 ; q 4 )∞ .

(3.1)

Proof. Let f (z) denote the right hand side of (3.1). Since f (z) is analytic on 0 < |z| < ∞, we can write f as a Laurent series ∞ X an z n . (3.2) f (z) = n=−∞

From the definition of f , we find that f (z) = qz 3 f (q 2 z). Thus, from (3.2), ∞ X n=−∞

an z n =

∞ X

an q 2n+1 z n+3 .

n=−∞

Equating coefficients of z n on both sides, we find that, for each n, an = q 2n−5 an−3 . By iteration, we find that, for each integer n, a3n = q 3n a3n+1 = q

2 −2n

3n2

a1 ,

a0 ,


5

and a3n+2 = q 3n

2 +2n

a2 .

From (3.2), we therefore have ∞ X

f (z) = a0

q

3n2 −2n 3n

z

∞ X

+ a1

n=−∞

q

3n2 3n+1

z

∞ X

+ a2

n=−∞

q 3n

2 +2n

z 3n+2 .

(3.3)

n=−∞

From the definition of f , we also find that f (z) = −z 2 f (z −1 ). Thus, from (3.2), ∞ X

n

an z = −

n=−∞

∞ X

a−n z

n+2

=−

n=−∞

∞ X

a−n+2 z n .

n=−∞

It follows that a2 = −a0 and a1 = 0. From (3.3) and the definition of f , we therefore have (q 2 ; q 2 )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (z 2 ; q 4 )∞ (z −2 q 4 ; q 4 )∞ ∞ X 2 2 = a0 (q 3n −2n z 3n − q 3n +2n z 3n+2 ). (3.4) n=−∞

Next, we calculate a0 . Setting z = q in (3.4), we obtain 0 = a0

∞ X

2 +n

(q 3n

− q 3n

2 +5n+2

).

n=−∞ 2

Putting z = qω, qω in (3.4), and using (1.5), we obtain, respectively, ∞ X (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ 2 2 (q 3n +n − q 3n +5n+2 ω 2 ) = a0 (1 − ω ) 2 4 (q ; q )∞ n=−∞ 2

and

∞ X (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ 2 2 = a0 (q 3n +n − q 3n +5n+2 ω). (1 − ω) 2 4 (q ; q )∞ n=−∞

Adding these three equations, we therefore have ∞ X (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ 2 = a q 3n +n . 0 2 4 (q ; q )∞ n=−∞

(3.5)

Employing the Jacobi triple product identity, we find that the right hand side of (3.5) is (−q 6 ; q 6 )∞ a0 (q 6 ; q 6 )∞ (−q 2 ; q 6 )∞ (−q 4 ; q 6 )∞ = a0 (q 6 ; q 6 )∞ (−q 2 ; q 6 )∞ (−q 4 ; q 6 )∞ (−q 6 ; q 6 )∞ (q 6 ; q 6 )∞ (−q 2 ; q 2 )∞ = a0 (−q 6 ; q 6 )∞

6


By (3.5) and Euler’s Identity, (−q; q)∞ = 1/(q; q 2 )∞ , we conclude that a0 = 1. Substituting a0 in (3.4), we get f (z) = = = =

∞ X

(q 3n

n=−∞ ∞ X

(q 3n

n=−∞ ∞ X

(q 3n

n=−∞ ∞ X

2 −2n

z 3n − q 3n

2 −2n

z 3n − q 3(n+1)

2 −2n

z 3n − q 3n

2 +n

q 3n

2 +2n

z 3n+2 )

2 −2n−2

2 +4n+1

z −3n−1 )

z −3n−1 )

(q −3n z 3n − q 3n+1 z −3n−1 ).

n=−∞

Hence we have finished the proof.

¤

4. The septuple product identity In this section we prove the septuple product identity by using some properties of fifth roots of unity. Theorem 3. For any complex numbers z and q, with z 6= 0 and |q| < 1, (q 4 ;q 10 )∞ (q 6 ; q 10 )∞ (q 10 ; q 10 )∞

∞ X

(−1)n (q 5n

2 +3n

z 5n+3 + q 5n

2 −3n

z 5n )

n=−∞

− (q 2 ; q 10 )∞ (q 8 ; q 10 )∞ (q 10 ; q 10 )∞ = (q

2

∞ X

(−1)n (q 5n

2 +n

2 −n

z 5n+2 + q 5n

n=−∞ 2 2 2 −1 2 2 ; q )∞ (z; q )∞ (z q ; q )∞ (z 2 ; q 2 )∞ (z −2 q 2 ; q 2 )∞ .

z 5n+1 ) (4.1)

Proof. Let f (z) denote the right hand side of (4.1). Since f (z) is analytic on 0 < |z| < ∞, we can write f as a Laurent series ∞ X

f (z) =

an z n .

(4.2)

n=−∞

From the definition of f , we find that f (z) = −q 2 z 5 f (q 2 z). Thus, from (4.2), ∞ X n=−∞

n

an z = −

∞ X

an q 2n+2 z n+5 .

n=−∞

n

Equating coefficients of z on both sides, we find that, for each n, an = −q 2n−8 an−5 .

(4.3)


7

By iteration, we find that, for each integer n, a5n = (−1)n q 5n

2 −3n

a0 ,

2 −n

a1 ,

n 5n2 +n

a2 ,

a5n+1 = (−1)n q 5n a5n+2 = (−1) q a5n+3 = (−1)n q

5n2 +3n

a3 ,

2 +5n

a4 .

and a5n+4 = (−1)n q 5n From (4.2), we therefore have f (z) =a0

∞ X

n 5n2 −3n 5n

(−1) q

n=−∞ ∞ X

+ a2 + a4

z

∞ X

+ a1

(−1)n q 5n

2 −n

z 5n+1

n=−∞ n 5n2 +n 5n+2

(−1) q

z

n=−∞ ∞ X

∞ X

+ a3

(−1)n q 5n

2 +3n

z 5n+3

n=−∞

(−1)n q 5n

2 +5n

z 5n+4 .

(4.4)

n=−∞

From the definition of f , we also find that f (z) = z 3 f (z −1 ). Thus, from (4.2), ∞ X

n

an z =

n=−∞

∞ X

an z

−n+3

=

n=−∞

∞ X

a−n+3 z n .

n=−∞

It follows that a4 = a−1 , a3 = a0 , and a2 = a1 . However, from (4.3), we see that a4 = −a−1 . Thus a4 = 0. Substituting a3 = a0 , a2 = a1 , and a4 = 0 in (4.4) and using the definition of f , we therefore have (q 2 ; q 2 )2∞ (z; q 2 )∞ (z −1 q 2 ; q 2 )∞ (z 2 ; q 2 )∞ (z −2 q 2 ; q 2 )∞ ∞ X 2 = a0 (−1)n q 5n (q −3n z 5n + q 3n z 5n+3 ) n=−∞ ∞ X

+ a1

2

(−1)n q 5n (q −n z 5n+1 + q n z 5n+2 ).

(4.5)

n=−∞

Next, we calculate a0 and a1 . Setting z = 1 in (4.5), we obtain 0 = a0

∞ X

n 5n2

(−1) q

n=−∞

(q

−3n

3n

+ q ) + a1

∞ X

2

(−1)n q 5n (q −n + q n ).

n=−∞

(4.6)

8


Putting z = ζ, ζ 2 , ζ 3 , ζ 4 in (4.5) and using (1.3), we obtain, respectively, 2

3

2

2

10

10

(1 − ζ − ζ + ζ )(q ; q )∞ (q ; q )∞ = a0

∞ X

2

(−1)n q 5n (q −3n + q 3n ζ 3 )

n=−∞ ∞ X

2

(−1)n q 5n (q −n ζ + q n ζ 2 ),

+ a1

(4.7)

n=−∞

2

4

2

2

10

∞ X

10

(1 − ζ − ζ + ζ)(q ; q )∞ (q ; q )∞ = a0

2

(−1)n q 5n (q −3n + q 3n ζ)

n=−∞ ∞ X

2

(−1)n q 5n (q −n ζ 2 + q n ζ 4 ),

+ a1

(4.8)

n=−∞

3

4

2

2

10

10

(1 − ζ − ζ + ζ )(q ; q )∞ (q ; q )∞ = a0

∞ X

2

(−1)n q 5n (q −3n + q 3n ζ 4 )

n=−∞ ∞ X

2

(−1)n q 5n (q −n ζ 3 + q n ζ),

+ a1

(4.9)

n=−∞

and 4

3

2

2

2

10

10

(1 − ζ − ζ + ζ )(q ; q )∞ (q ; q )∞ = a0

∞ X

2

(−1)n q 5n (q −3n + q 3n ζ 2 )

n=−∞ ∞ X

2

(−1)n q 5n (q −n ζ 4 + q n ζ 3 ).

+ a1

(4.10)

n=−∞

Adding (4.6), (4.7), (4.8), (4.9), and (4.10), we therefore have 2

2

10

10

5(q ; q )∞ (q ; q )∞ = 5a0

∞ X

(−1)n q 5n

2 −3n

.

(4.11)

n=−∞

Employing the Jacobi triple product Identity, we find that the right hand side of (4.11) is 5a0 (q 10 ; q 10 )∞ (q 2 ; q 10 )∞ (q 8 ; q 10 )∞ . Therefore (q 2 ; q 2 )∞ a0 = (q 2 ; q 10 )∞ (q 8 ; q 10 )∞ (q 2 ; q 10 )∞ (q 4 ; q 10 )∞ (q 6 ; q 10 )∞ (q 8 ; q 10 )∞ (q 10 ; q 10 )∞ = (q 2 ; q 10 )∞ (q 8 ; q 10 )∞ = (q 4 ; q 10 )∞ (q 6 ; q 10 )∞ (q 10 ; q 10 )∞ . Applying the Jacobi triple product identity in (4.6), we obtain 0 = 2a0 (q 10 ; q 10 )∞ (q 2 ; q 10 )∞ (q 8 ; q 10 )∞ + 2a1 (q 10 ; q 10 )∞ (q 4 ; q 10 )∞ (q 6 ; q 10 )∞ .

(4.12)


9

Setting a0 = (q 4 ; q 10 )∞ (q 6 ; q 10 )∞ (q 10 ; q 10 )∞ in (4.12), we therefore find that a1 = −(q 2 ; q 10 )∞ (q 8 ; q 10 )∞ (q 10 ; q 10 )∞ . Putting a0 = (q 4 ; q 10 )∞ (q 6 ; q 10 )∞ (q 10 ; q 10 )∞ and a1 = −(q 2 ; q 10 )∞ (q 8 ; q 10 )∞ (q 10 ; q 10 )∞ in (4.5), we therefore have finished the proof. ¤ 5. Winquist’s identity In this section we prove Winquist’s identity by using some properties of cube roots of unity. Theorem 4. For any complex number q, with |q| < 1, any nonzero complex numbers a and b, ∞ ∞ X X 2 2 (−1)m+n q (3m +3n +3m+n)/2 m=−∞ n=−∞

× (a−3m b−3n − a−3m b3n+1 − a−3n+1 b−3m−1 + a3n+2 b−3m−1 ) = (q; q)2∞ (a; q)∞ (a−1 q; q)∞ (b; q)∞ (b−1 q; q)∞ × (ab; q)∞ (a−1 b−1 q; q)∞ (ab−1 ; q)∞ (a−1 bq; q)∞ .

(5.1)

Proof. Let g(a, b) denote the right hand side of (5.1). Since h(z) := (z; q)∞ (z −1 q; q)∞ is analytic on 0 < |z| < ∞, we can write h as a Laurent series ∞ X h(z) = an z n . n=−∞

Since g(a, b) =

(q; q)2∞ h(a)h(b)h(ab)h(ab−1 ), ∞ X g(a, b) =

we can write g as a double Laurent series ∞ X cm,n am bn . (5.2)

m=−∞ n=−∞

From the definition of g, we find that g(a, b) = −a3 g(aq, b) and g(a, b) = −b3 qg(a, bq). Thus, from (5.2), we have ∞ ∞ X X

m n

cm,n a b = −

m=−∞ n=−∞

and

∞ X

∞ X

∞ X

cm,n q m am+3 bn

m=−∞ n=−∞

m n

cm,n a b = −

m=−∞ n=−∞ m n

∞ X

∞ X

∞ X

cm,n q n+1 am bn+3 .

m=−∞ n=−∞

Equating coefficients of a b on both sides of the last two equations, we find that, for all nonzero integers m and n, cm,n = −q m−3 cm−3,n ,

10


and cm,n = −q n−2 cm,n−3 . By iteration, we find that, for all integers m and n, 2 −3m)/2

c3m,n = (−1)m q (3m c3m+1,n = (−1)m q

(5.3)

c1,n ,

(5.4)

2 +m)/2

c2,n ,

(5.5)

cm,−1 ,

(5.6)

c3m+2,n = (−1)m q (3m cm,3n−1 = (−1)n q

c0,n ,

(3m2 −m)/2

(3n2 −3n)/2

cm,3n = (−1)n q

(3n2 −n)/2

cm,0 ,

(5.7)

cm,3n+1 = (−1)n q

(3n2 +n)/2

cm,1 .

(5.8)

From the definition of g, we also find that g(a, b) = −a3 g(a−1 , b) and g(a, b) = −b g(a, b−1 ). Thus, from (5.2), we have ∞ X

∞ X

cm,n am bn = −

m=−∞ n=−∞

∞ X

∞ X

∞ X

cm,n a−m+3 bn = −

m=−∞ n=−∞

∞ X

c−m+3,n am bn

m=−∞ n=−∞

and ∞ X

∞ X

m=−∞ n=−∞

m n

cm,n a b = −

∞ X

∞ X

m −n+1

cm,n a b

∞ X

=−

m=−∞ n=−∞

∞ X

cm,−n+1 am bn .

m=−∞ n=−∞

Equating coefficients of am bn on both sides of the last two equations, we find that, for all nonzero integers m and n, c2,n = −c1,n and cm,1 = −cm,0 . Using the last two equations in (5.5) and (5.8), respectively, we obtain, 2 +m)/2

c3m+2,n = −(−1)m q (3m

c1,n

(5.9)

cm,0 .

(5.10)

and cm,3n+1 = −(−1)n q (3n

2 +n)/2


11

Using (5.3), (5.4), (5.6), (5.7), (5.9), and (5.10), we obtain 2 −3m+3n2 −3n)/2

c3m,3n−1 = (−1)m+n q (3m

2 −3m+3n2 −n)/2

c3m,3n = (−1)m+n q (3m c3m,3n+1 = −(−1)m+n q c3m+1,3n−1 = (−1)m+n q c3m+1,3n = (−1)m+n q

c0,0 ,

(3m2 −3m+3n2 +n)/2

(3m2 −m+3n2 −3n)/2 (3m2 −m+3n2 −n)/2

c3m+1,3n+1 = −(−1)m+n q

c0,−1 ,

c1,−1 , c1,0 ,

2 +m+3n2 −3n)/2

c3m+2,3n+1 = (−1)m+n q

(3m2 +m+3n2 −n)/2

(3m2 +m+3n2 +n)/2

(5.11)

c1,0 ,

(3m2 −m+3n2 +n)/2

c3m+2,3n−1 = −(−1)m+n q (3m c3m+2,3n = −(−1)m+n q

c0,0 ,

c1,−1 ,

c1,0 ,

c1,0 .

From the definition of g, we find that g(a, b) = −ab−1 g(b, a). Thus, from (5.2), we have ∞ X

∞ X

m n

cm,n a b = −

m=−∞ n=−∞

∞ X

∞ X

n+1 m−1

cm,n a

b

∞ X

=−

m=−∞ n=−∞

∞ X

cn,m am+1 bn−1 .

m=−∞ n=−∞

Equating coefficients of am bn on both sides of the last equality, we find that, for all nonzero integers m and n, cm,n = −cn+1,m−1 . In particular, we have c0,−1 = −c0,−1 , and c1,0 = −c1,0 . It follows that c0,−1 = c1,0 = 0. Using the definition of g, (5.2), and all nine equalities of (5.11), we obtain (q; q)2∞ (a; q)∞ (a−1 q; q)∞ (b; q)∞ (b−1 q; q)∞ (ab; q)∞ (a−1 b−1 q; q)∞ (ab−1 ; q)∞ (a−1 bq; q)∞ ∞ ∞ X X = cm,n am bn =

m=−∞ n=−∞ ∞ ∞ n X X

2 −3m+3n2 −n)/2

(−1)m+n q (3m

(1 − bq n )a3m b3n c0,0

m=−∞ n=−∞ m+n (3m2 −m+3n2 −3n)/2

+ (−1)

q

m

3m+1 3n−1

(1 − aq )a

b

o c1,−1

(5.12)

Next, we calculate c0,0 and c1,−1 . Setting a = b = q 1/2 in (5.12), we obtain 0=

∞ X

∞ n X

2 +3n2 +2n)/2

(−1)m+n q (3m

(1 − q n+1/2 )c0,0

m=−∞ n=−∞ 2 +2m+3n2 )/2

+ (−1)m+n q (3m

o (1 − q m+1/2 )c1,−1 .

(5.13)

12


Putting a = q 1/2 and b = ωq 1/2 in (5.12), and using (1.5), we obtain −3ω 2 (q 3 ; q 3 )2∞ (q 3/2 ; q 3 )∞ (q 1/2 ; q)∞ ∞ ∞ n X X 2 2 = (−1)m+n q (3m +3n +2n)/2 (1 − ωq n+1/2 )c0,0 m=−∞ n=−∞ m+n (3m2 +2m+3n2 )/2

+ (−1)

q

(1 − q

m+1/2

o

2

)ω c1,−1 .

(5.14)

Putting a = q 1/2 and b = ω 2 q 1/2 in (5.12), and using (1.5), we obtain −3ω(q 3 ; q 3 )2∞ (q 3/2 ; q 3 )∞ (q 1/2 ; q)∞ ∞ ∞ n X X 2 2 = (−1)m+n q (3m +3n +2n)/2 (1 − ω 2 q n+1/2 )c0,0 m=−∞ n=−∞ m+n (3m2 +2m+3n2 )/2

+ (−1)

q

(1 − q

m+1/2

o )ωc1,−1 .

(5.15)

Adding (5.13), (5.14), and (5.15), we have ∞ X

3(q 3 ; q 3 )2∞ (q 3/2 ; q 3 )∞ (q 1/2 ; q)∞ = 3

∞ X

2 +3n2 +2n)/2

(−1)m+n q (3m

c0,0 .

(5.16)

m=−∞ n=−∞

Employing the Jacobi triple product identity on the right hand side of (5.16) , we find that the right side of (5.16) is equal to 3c0,0 (q 3 ; q 3 )2∞ (q 3/2 ; q 3 )2∞ (q 5/2 ; q 3 )∞ (q 1/2 ; q 3 )∞ = 3c0,0 (q 3 ; q 3 )2∞ (q 3/2 ; q 3 )∞ (q 1/2 ; q)∞ . Thus, from (5.16), c0,0 = 1. Setting a = b = ωq 1/2 , ω 2 q 1/2 in (5.12), we obtain, respectively, ∞ X

0=

∞ n X 2 2 (−1)m+n q (3m +3n +2n)/2 (1 − ωq n+1/2 )c0,0

m=−∞ n=−∞ 2 +2m+3n2 )/2

+ (−1)m+n q (3m

(1 − ωq m+1/2 )c1,−1

o (5.17)

and 0=

∞ X

∞ n X

2 +3n2 +2n)/2

(−1)m+n q (3m

(1 − ω 2 q n+1/2 )c0,0

m=−∞ n=−∞

+ (−1)

m+n (3m2 +2m+3n2 )/2

q

2 m+1/2

(1 − ω q

o )c1,−1 .

(5.18)

Adding (5.13), (5.17), and (5.18), we have 0=

∞ X

∞ n X

m=−∞ n=−∞

2 +3n2 +2n)/2

(−1)m+n q (3m

2 +2m+3n2 )/2

c0,0 + (−1)m+n q (3m

o c1,−1 .

(5.19)


13

Setting c0,0 = 1 in (5.19), we conclude that c1,−1 = −1. Substituting c0,0 = 1 and c1,−1 = −1 in (5.12), we obtain (q; q)2∞ (a; q)∞ (a−1 q; q)∞ (b; q)∞ (b−1 q; q)∞ (ab; q)∞ (a−1 b−1 q; q)∞ (ab−1 ; q)∞ (a−1 bq; q)∞ ∞ n ∞ X X 2 2 (−1)m+n q (3m −3m+3n −n)/2 (1 − bq n )a3m b3n = m=−∞ n=−∞

− (−1) =

∞ n X

∞ X

m+n (3m2 −m+3n2 −3n)/2

q

2 −3m+3n2 −n)/2

(−1)m+n q (3m

m

(1 − aq )a

3m+1 3n−1

b

(1 − bq n )a3m b3n

m=−∞ n=−∞ 2 −3m+3n2 −n)/2

− (−1)m+n q (3m =

∞ n X

∞ X

2 −3m+3n2 −n)/2

(−1)m+n q (3m

o

(1 − aq n )a3n+1 b3m−1

o

a3m b3n

m=−∞ n=−∞ 2 −3m+3n2 +n)/2

a3m b3n+1

2 −3m+3n2 −n)/2

a3n+1 b3m−1

2 −3m+3n2 +n)/2

a3n+2 b3m−1

− (−1)m+n q (3m

− (−1)m+n q (3m + (−1)m+n q (3m =

∞ n X

∞ X

2 +3m+3n2 +n)/2

(−1)m+n q (3m

o

a−3m b−3n

m=−∞ n=−∞ 2 +3m+3n2 +n)/2

− (−1)m+n q (3m

a−3m b3n+1

2 +3m+3n2 +n)/2

− (−1)m+n q (3m

a−3n+1 b−3m−1 o 2 2 + (−1)m+n q (3m +3m+3n +n)/2 a3n+2 b−3m−1 . We therefore complete the proof.

¤

6. Other q-series-product identities By the same method that we used to prove the quintuple product identity, the septuple product identity, and Winquist’s identity, we can derive further product identities. Theorem 5. For any complex numbers z and q with z 6= 0 and |q| < 1, 6

6

6

12

(q ; q )∞ (q ; q )∞

∞ X

2

q 3n z 3n

n=−∞ 2

12

10

12

12

12

− q(q ; q )∞ (q ; q )∞ (q ; q )∞ 2

2

2

= (q ; q )∞ (zq; q )∞ (z

−1

∞ X

(q 3n

2 +2n

2 −2n

z 3n+1 + q 3n

z 3n−1 )

n=−∞ q; q 2 )∞ (q 4 ; q 4 )∞ (z 2 q 2 ; q 4 )∞ (z −2 q 2 ; q 4 )∞ .

(6.1)

14



f (z) =

an z n .

(6.2)

n=−∞

From the definition of f , we find that f (z) = q 3 z 3 f (q 2 z). Thus, from (6.2), ∞ X

an z n =

n=−∞

∞ X

∞ X

an q 2n+3 z n+3 =

n=−∞

an−3 q 2n−3 z n .

n=−∞

n

Equating coefficients of z on both sides, we find that, for each n, an = q 2n−3 an−3 . By iteration, we find that, for each integer n, 2

a3n = q 3n a0 , 2 +2n

a1 ,

2 −2n

a−1 .

a3n+1 = q 3n and a3n−1 = q 3n From (6.2), we therefore have f (z) = a0

∞ X

q

3n2 3n

z

+ a1

n=−∞

∞ X

q

3n2 +2n 3n+1

z

+ a−1

n=−∞

∞ X

q 3n

2 −2n

z 3n−1 .

(6.3)

n=−∞

From the definition of f , we also find that f (z) = f (z −1 ). Thus, from (6.2), ∞ X

an z n =

n=−∞

∞ X

an z −n =

n=−∞

∞ X

a−n z n .

n=−∞

It follows that a1 = a−1 . From (6.3), we therefore have f (z) = a0

∞ X

q

3n2 3n

z

+ a1

n=−∞

∞ X

(q 3n

2 +2n

z 3n+1 + q 3n

2 −2n

z 3n−1 ).

(6.4)

n=−∞

Next, we calculate a0 . Setting z = q in (6.4), we obtain, from the definition of f , 0 = a0

∞ X n=−∞

q

3n2 +3n

+ a1

∞ X

(q 3n

n=−∞

2 +5n+1

+ q 3n

2 +n−1

).

(6.5)


15

Putting z = qω, qω 2 in (6.4), and using (1.5), we obtain, respectively, 6

6

12

∞ X

12

3(q ; q )∞ (q ; q )∞ = a0

q

3n2 +3n

∞ X

+ a1

n=−∞

(q 3n

2 +5n+1

ω + q 3n

2 +n−1

2 +5n+1

ω 2 + q 3n

ω 2 ),

(6.6)

n=−∞

and ∞ X

3(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ = a0

2 +3n

q 3n

∞ X

+ a1

n=−∞

(q 3n

2 +n−1

ω).

(6.7)

n=−∞

Adding the last three equations, we therefore have 6

6

12

∞ X

12

6(q ; q )∞ (q ; q )∞ = 3a0

2 +3n

q 3n

.

(6.8)

n=−∞

Employing the Jacobi triple product identity, we find that the right hand side of (6.8) is 3a0 (q 6 ; q 6 )∞ (−q 6 ; q 6 )∞ (−1; q 6 )∞ = 3a0 (q 12 ; q 12 )∞ 2(−q 6 ; q 6 )∞ (q 12 ; q 12 )∞ = 6a0 6 12 . (q ; q )∞ Therefore we have a0 = (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ . Multiplying both sides of (6.6) by ω 2 , we obtain 2

6

6

12

∞ X

12

3ω (q ; q )∞ (q ; q )∞ = a0

q

3n2 +3n

2

ω + a1

n=−∞

∞ X

(q 3n

2 +5n+1

+ q 3n

2 +n−1

ω). (6.9)

n=−∞

Multiplying both sides of (6.7) by ω, we obtain 6

6

12

12

3ω(q ; q )∞ (q ; q )∞ = a0

∞ X

q

3n2 +3n

n=−∞

∞ X

ω + a1

(q 3n

2 +5n+1

+ q 3n

2 +n−1

ω 2 ). (6.10)

n=−∞

Adding (6.5), (6.9), and (6.10), we have 6

6

12

12

−3(q ; q )∞ (q ; q )∞ = 3a1

∞ X n=−∞

2 +5n+1

q 3n

.

(6.11)

16


Employing the Jacobi Triple Product Identity, we find that the right hand side of (6.11) is 3a1 q(q 6 ; q 6 )∞ (−q 8 ; q 6 )∞ (−q −2 ; q 6 )∞ (−q 2 ; q 6 )∞ (1 + q −2 )(−q 4 ; q 6 )∞ 1 + q2 (−q 6 ; q 6 )∞ = 3a1 q −1 (q 6 ; q 6 )∞ (−q 2 ; q 6 )∞ (−q 4 ; q 6 )∞ (−q 6 ; q 6 )∞ (q 6 ; q 6 )∞ (−q 2 ; q 2 )∞ = 3a1 q −1 (−q 6 ; q 6 )∞ (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ = 3a1 q −1 (q 2 ; q 4 )∞ (q 6 ; q 6 )∞ (q 6 ; q 12 )∞ = 3a1 q −1 2 12 (q ; q )∞ (q 6 ; q 12 )∞ (q 10 ; q 12 )∞ (q 6 ; q 6 )∞ = 3a1 q −1 2 12 . (q ; q )∞ (q 10 ; q 12 )∞ = 3a1 q(q 6 ; q 6 )∞

Hence we have a1 = −q(q 2 ; q 12 )∞ (q 10 ; q 12 )∞ (q 12 ; q 12 )∞ . Substituting a0 and a1 in (6.4), we therefore have finished the proof. ¤ Using similar methods, we can establish further identities, which we state without proofs. Using fifth roots of unity, we find that ∞ (q; q 2 )5∞ + 4(q 5 ; q 10 )∞ X 2 (−1)n q 5n z 5n 10 10 5 10 2 5(q ; q )∞ (q ; q )∞ n=−∞ ∞ X ¡ 2n ¢ (q; q 2 )5∞ − (q 5 ; q 10 )∞ n 5n2 5n −2n −1 (−1) q z q z + q z + 5(q 10 ; q 10 )∞ (q 3 ; q 10 )∞ (q 7 ; q 10 )∞ n=−∞ ∞ X ¡ 4n 2 ¢ (q; q 2 )5∞ − (q 5 ; q 10 )∞ n 5n2 5n −4n −2 + (−1) q z q z + q z 5(q 10 ; q 10 )∞ (q; q 10 )∞ (q 9 ; q 10 )∞ n=−∞

= (q; q 2 )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (z 2 q; q 2 )∞ (z −2 q; q 2 )∞ .

(6.12)

Similarly, with the use of fourth roots of unity, we find that ∞ (−q 2 ; q 2 )2∞ (−q 6 ; q 6 )2∞ + (−q 4 ; q 4 )∞ (−q 12 ; q 12 )∞ X 4n2 4n q z 2(q 8 ; q 8 )∞ (−q 8 ; q 8 )2∞ n=−∞ ∞ X ¡ ¢ (−q 2 ; q 2 )2∞ (−q 6 ; q 6 )2∞ 2 −q 8 8 q 4n z 4n q 2n z + q −2n z −1 6 8 2 8 (q ; q )∞ (−q ; q )∞ (−q ; q )∞ n=−∞

+q

2 (−q

2

∞ ; q 2 )2∞ (−q 6 ; q 6 )2∞ − (−q 4 ; q 4 )∞ (−q 12 ; q 12 )∞ X 4n2 +4n 4n+2 q z (q 8 ; q 8 )∞ (−q 4 ; q 8 )2∞ n=−∞

= (zq; q 2 )∞ (z −1 q; q 2 )∞ (z 3 q 3 ; q 6 )∞ (z −3 q 3 ; q 6 )∞

(6.13)


17

and ∞ (−q 4 ; q 4 )∞ (−q 6 ; q 12 )∞ + (−q 2 ; q 2 )2∞ (−q 3 ; q 6 )∞ X 4n2 4n −3n q z (q − q 3n z 3 ) 2(q 8 ; q 8 )∞ (−q 5 ; q 8 )∞ (−q 3 ; q 8 )∞ n=−∞ ∞ (−q 4 ; q 4 )∞ (−q 6 ; q 12 )∞ − (−q 2 ; q 2 )2∞ (−q 3 ; q 6 )∞ X 4n2 4n −n q z (q z − q n z 2 ) 2q(q 8 ; q 8 )∞ (−q; q 8 )∞ (−q 7 ; q 8 )∞ n=−∞

= (zq; q 2 )∞ (z −1 q; q 2 )∞ (z 3 ; q 6 )∞ (z −3 q 6 ; q 6 )∞ .

(6.14)

7. Generalized q-series-product identities In this section, we use Bhargava’s method [5] to prove some generalized q-seriesproduct identities. This method is simpler than the first method. At the same time, it gives infinitely many identities. However, we are unable to use this method to prove Winquist’s identity. Let [x] denote the greatest integer less than or equal to x. The next identity is a generalization of the quintuple product identity. Theorem 6. Let z and q be complex numbers with z 6= 0 and |q| < 1. Then, for any positive integer k, [(k−1)/2]

X

j j2

(−1) q

∞ X

(−1)(k−1)m q (k

m=−∞ ∞ X (k+1)n2 (k+1)n

j=0

×

q

z

2 +k)m2 −(2j+1)km

(q (2j−k)n z j − q (k−2j)n z k−j )

n=−∞ 2

2

= (q ; q )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (q 2k ; q 2k )∞ (z k ; q 2k )∞ (z −k q 2k ; q 2k )∞ .

(7.1)

Proof. Let f (z) denote the right hand side of (7.1). Since f (z) is analytic on 0 < |z| < ∞, we can write f as a Laurent series f (z) =

∞ X

an z n .

(7.2)

n=−∞

From the definition of f , we find that f (z) = qz k+1 f (q 2 z). Thus, from (7.2), ∞ X n=−∞

n

an z =

∞ X

an q

2n+1 n+k+1

z

=

n=−∞

∞ X

an−k−1 q 2n−2k−1 z n .

n=−∞

Equating coefficients of z n on both sides, we find that, for each n, an = an−k−1 q 2n−2k−1 . By iteration, we find that, for all integers n and j, with 0 ≤ j ≤ k, a(k+1)n+j = aj q (k+1)n

2 −kn+2jn

.

18


From (7.2), we therefore have f (z) =

k X

∞ X

aj

q (k+1)n

2 +(2j−k)n

z (k+1)n+j .

(7.3)

n=−∞

j=0

From the definition of f , we also find that f (z) = −z k f (z −1 ). Thus, from (7.2), ∞ X

∞ X

n

an z = −

n=−∞

a−n z

n+k

=−

n=−∞

∞ X

a−n+k z n .

n=−∞

It follows that ak−j = −aj for each integer j. In particular, for even k, ak/2 = −ak/2 and hence ak/2 = 0. From (7.3), we therefore have ∞ ³ X

[(k−1)/2]

f (z) =

X

aj

q

(k+1)n2 +(2j−k)n (k+1)n+j

z

−q

(k+1)n2 +(k−2j)n (k+1)n+k−j

z

´ . (7.4)

n=−∞

j=0

Next, we calculate aj . Applying the Jacobi triple product identity to the definition of f , we then obtain ∞ X

f (z) =

m m m2

(−1) z q

m=−∞

∞ X

(−1)n z kn q kn

2 −kn

.

(7.5)

n=−∞

By (7.2) and (7.5), we have ∞ X

an z n =

n=−∞

∞ X

∞ X

2 +km2 −km

(−1)m+n z n+km q n

.

(7.6)

n=−∞ m=−∞

Equating coefficients of z j on both sides of (7.6), we therefore obtain j j2

aj = (−1) q

∞ X

(−1)(k−1)m q (k

2 +k)m2 −(2j+1)km

.

(7.7)

m=−∞

Substituting (7.7) in (7.4), we have [(k−1)/2]

f (z) =

X j=0

j j2

(−1) q

∞ X

(−1)(k−1)m q (k

m=−∞ ∞ X

×

2 +k)m2 −(2j+1)km

¡ ¢ 2 q (k+1)n z (k+1)n q (2j−k)n z j − q (k−2j)n z k−j .

(7.8)

n=−∞

Putting (7.8) together with the definition of f , we therefore have finished the proof. ¤ Note that the quintuple product identity can be derived by putting k = 2 in (7.1). Next we prove a generalization of Theorem 5.


19

Theorem 7. Let z and q be complex numbers with z 6= 0 and |q| < 1. Then, for any positive integer k, ∞ X 1 − (−1)k 2 2 2 (k+1)/2 (k+1)2 /4 (−1) q (−1)(k−1)m q (k +k)m +(k +k)m 2 m=−∞ ∞ X

×

q (k+1)n

2 −(k+1)n

z (k+1)n−(k+1)/2

n=−∞

+

∞ X

(−1)

q

m=−∞

+

2

q (k+1)n z (k+1)n

n=−∞

[k/2]

X

∞ X

(k−1)m (k2 +k)m2

∞ X

j j2

(−1) q

(−1)(k−1)m q (k

2 +k)m2 −2jkm

m=−∞

j=1 ∞ X

×

2

q (k+1)n z (k+1)n (q 2jn z j − q −2jn z −j )

n=−∞ 2

2

= (q ; q )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (q 2k ; q 2k )∞ (z k q k ; q 2k )∞ (z −k q k ; q 2k )∞ .

(7.9)

Proof. Let f (z) denote the right hand side of (7.9). Since f (z) is analytic on 0 < |z| < ∞, we can write f as a Laurent series ∞ X f (z) = an z n . (7.10) n=−∞

From the definition of f , we find that f (z) = q k+1 z k+1 f (q 2 z). Thus, from (7.10), ∞ X

n

an z =

n=−∞

∞ X

an q

2n+k+1 n+k+1

z

∞ X

=

n=−∞

an−k−1 q 2n−k−1 z n .

n=−∞

n

Equating coefficients of z on both sides, we find that, for each n, an = an−k−1 q 2n−k−1 . By iteration, we find that, for all integers n and j with −(k + 1)/2 ≤ j ≤ k/2, a(k+1)n+j = aj q (k+1)n

2 +2jn

.

If k is even, we will consider f as f (z) =

k/2 X

aj

j=−k/2

∞ X

q (k+1)n

2 +2jn

z (k+1)n+j .

(7.11)

n=−∞

Similarly, if k is odd, we will write f as (k−1)/2

f (z) =

X

j=−(k+1)/2

aj

∞ X n=−∞

q (k+1)n

2 +2jn

z (k+1)n+j .

(7.12)

20


From the definition of f , we also find that f (z) = f (z −1 ). Thus, from (7.10), ∞ X

∞ X

n

an z =

n=−∞

a−n z n .

n=−∞

It follows that a−j = aj for each integer j. Next, we calculate aj . Applying the Jacobi triple product identity to the definition of f , we then obtain ∞ X

f (z) =

∞ X

2

(−1)m z m q m

m=−∞

2

(−1)n z kn q kn .

(7.13)

n=−∞

By (7.10) and (7.13), we have ∞ X

∞ X

n

an z =

n=−∞

∞ X

(−1)m+n z n+km q n

2 +km2

.

(7.14)

n=−∞ m=−∞

Equating coefficients of z j on both sides of (7.14), we therefore obtain ∞ X

j j2

aj = (−1) q

(−1)(k−1)m q (k

2 +k)m2 −2jkm

.

(7.15)

m=−∞

Substituing (7.15) in (7.12) and using the fact that aj = a−j for each positive integer j, we have, for even k, (k+1)/2

f (z) =

X

∞ X

j j2

(−1) q

2 +k)m2 −2jkm

m=−∞

j=−(k−1)/2 ∞ X

×

(−1)(k−1)m q (k

q (k+1)n

2 +2jn

z (k+1)n+j

n=−∞

=

∞ X

(k−1)m (k2 +k)m2

(−1)

q

m=−∞

∞ X

2

q (k+1)n z (k+1)n

n=−∞

k/2 ∞ X X 2 2 j j2 + (−1) q (−1)(k−1)m q (k +k)m −2jkm m=−∞

j=0

×

∞ X n=−∞

¡ ¢ 2 q (k+1)n z (k+1)n q 2jn z j + q −2jn z −j .

(7.16)


21

Similarly, if k is odd, we obtain ∞ X

(k+1)/2 (k+1)2 /4

f (z) = (−1)

q

(−1)(k−1)m q (k

2 +k)m2 +(k 2 +k)m

m=−∞ ∞ X

× +

q (k+1)n

n=−∞ ∞ X

2 −(k+1)n

(k−1)m (k2 +k)m2

(−1)

q

m=−∞

+

∞ X

j j2

(−1) q

2

q (k+1)n z (k+1)n

(−1)(k−1)m q (k

2 +k)m2 −2jkm

m=−∞

j=0 ∞ X

×

∞ X n=−∞

(k+1)/2

X

z (k+1)n−(k+1)/2

¡ ¢ 2 q (k+1)n z (k+1)n q 2jn z j + q −2jn z −j .

(7.17)

n=−∞

Putting (7.16) and (7.17) together with the definition of f , we therefore have finished the proof. ¤ Note that if we substitute k = 2 in (7.9), then we obtain Theorem 5. The next identity is a generalization of the septuple product identity. Theorem 8. Let z and q be complex numbers with z 6= 0 and |q| < 1. Then, for any positive integer k, k X

∞ X

j j 2 −j

(−1) q

(−1)m q (4k

2 +k)m2 +(1−4j)km

m=−∞

j=−k+1 ∞ X

×

¡ ¢ 2 (−1)n q (4k+1)n z (4k+1)n q (2j−2k−1)n z j + q (2k+1−2j)n z 2k+1−j

n=−∞ 2

2

= (q ; q )∞ (z; q 2 )∞ (z −1 q 2 ; q 2 )∞ (q 2k ; q 2k )∞ (z 2k ; q 2k )∞ (z −2k q 2k ; q 2k )∞ .

(7.18)

Proof. Let f (z) denote the right hand side of (7.18). Since f (z) is analytic on 0 < |z| < ∞, we can write f as a Laurent series f (z) =

∞ X

an z n .

(7.19)

n=−∞

From the definition of f , we find that f (z) = −q 2k z 4k+1 f (q 2 z). Thus, from (7.19), ∞ X n=−∞

n

an z = −

∞ X n=−∞

an q

2n+2k n+4k+1

z

=−

∞ X n=−∞

an−4k−1 q 2n−6k−2 z n .

22


Equating coefficients of z n on both sides, we find that, for each n, an = −an−4k−1 q 2n−6k−2 .

(7.20)

By iteration, we find that, for all integers n and j, with −k ≤ j ≤ 3k, 2 +(2j−2k+1)n

a(4k+1)n+j = (−1)n aj q (4k+1)n

.


3k X

∞ X

aj

j=−k

(−1)n q (4k+1)n

2 +(2j−2k−1)n

z (4k+1)n+j .

(7.21)

n=−∞

From the definition of f , we also find that f (z) = z 2k+1 f (z −1 ). Thus, from (7.19), ∞ X

∞ X

n

an z =

n=−∞

a−n z

n+2k+1

∞ X

=

n=−∞

a−n+2k+1 z n .

n=−∞

It follows that a2k+1−j = aj for each integer j. In particular, a−k = a3k+1 . However, from (7.20), we get a3k+1 = −a−k and hence a−k = 0. Then we can rewrite (7.21) as k X

f (z) =

aj

j=−k+1

∞ X

2

(−1)n q (4k+1)n z (4k+1)n

n=−∞

¡ ¢ × q (2j−2k−1)n z j + q (2k+1−2j)n z 2k+1−j .

(7.22)

Next, we calculate aj . Applying the Jacobi triple product identity to the definition of f , we then obtain ∞ ∞ X X 2 2 f (z) = (−1)m z m q m −m (−1)n z kn q kn −kn . (7.23) m=−∞

n=−∞

By (7.19) and (7.23), we have ∞ ∞ X X n an z = n=−∞

∞ X

(−1)m+n z n+2km q n

2 −n+km2 −km

.

(7.24)

n=−∞ m=−∞ j

Equating coefficient of z on both sides of (7.24), we therefore obtain ∞ X 2 2 2 aj = (−1)j q j −j (−1)m q (4k +k)m +(1−4j)km .

(7.25)

m=−∞

Substituting (7.25) in (7.22), we have f (z) =

k X

j j 2 −j

(−1) q

(−1)m q (4k

2 +k)m2 +(1−4j)km

m=−∞

j=−k+1

×

∞ X

∞ X

¡ ¢ 2 (−1)n q (4k+1)n z (4k+1)n q (2j−2k−1)n z j + q (2k+1−2j)n z 2k+1−j . (7.26)

n=−∞


23

Putting (7.26) together with the definition of f , we therefore have finished the proof. ¤ Note that the septuple product identity can be derived by substituting k = 1 in (7.18). Theorem 9. Let z and q be complex numbers with z 6= 0 and |q| < 1. Then, for any positive integer k, ∞ X

∞ X

m (4k2 +k)m2

(−1) q

m=−∞

+

2

(−1)n q (4k+1)n z (4k+1)n

n=−∞

2k X

(−1)j q j

2

∞ X

(−1)m q (4k

2 +k)m2 −4jkm

m=−∞

j=1 ∞ X

×

¡ ¢ 2 (−1)n aj q (4k+1)n z (4k+1)n q 2jn z j + q −2jn z −j

n=−∞ 2

= (q ; q 2 )∞ (zq; q 2 )∞ (z −1 q; q 2 )∞ (q 2k ; q 2k )∞ (z 2k q k ; q 2k )∞ (z −2k q k ; q 2k )∞ .

(7.27)


f (z) =

an z n .

(7.28)

n=−∞

From the definition of f , we find that f (z) = −q 4k+1 z 4k+1 f (q 2 z). Thus, from (7.28), ∞ X n=−∞

an z n = −

∞ X

an q 2n+4k+1 z n+4k+1 = −

n=−∞

∞ X

an−4k−1 q 2n−4k−1 z n .

n=−∞

Equating coefficients of z n on both sides, we find that, for each n, an = −an−4k−1 q 2n−4k−1 .

(7.29)

By iteration, we find that, for all integers n and j, with −2k ≤ j ≤ 2k, 2 +2jn

a(4k+1)n+j = (−1)n aj q (4k+1)n

.


2k X j=−2k

aj

∞ X

(−1)n q (4k+1)n

n=−∞

From the definition of f , we also find that f (z) = f (z −1 ).

2 +2jn

z (4k+1)n+j .

(7.30)

24


It follows that a−j = aj for each integer j. Then we can rewrite (7.30) as ∞ X

f (z) = a0

2

(−1)n q (4k+1)n z (4k+1)n

n=−∞ 2k X

+

∞ X

aj

¡ ¢ 2 (−1)n q (4k+1)n z (4k+1)n q 2jn z j + q −2jn z −j .

(7.31)

n=−∞

j=1

Next, we calculate aj ’s. Applying the Jacobi triple product identity to the definition of f , we then obtain ∞ ∞ X X 2 2 f (z) = (−1)m z n q n (−1)n z 2km q km . (7.32) n=−∞

m=−∞

By (7.28) and (7.32), we have ∞ X

∞ X

n

an z =

n=−∞

∞ X

(−1)m+n z n+2km q n

2 +km2

.

(7.33)

n=−∞ m=−∞

Equating coefficients of z j on both sides of (7.33), we therefore obtain aj = (−1)j q j

∞ X

2

(−1)m q (4k

2 +k)m2 −4jkm

.

(7.34)

m=−∞

Substituting (7.34) in (7.31), we have f (z) =

∞ X

(−1) q

m=−∞

+

∞ X

m (4k2 +k)m2

n=−∞

2k X

(−1)j q j

2

∞ X

(−1)m q (4k

2 +k)m2 −4jkm

m=−∞

j=1

×

2

(−1)n q (4k+1)n z (4k+1)n

∞ X

¡ ¢ 2 (−1)n q (4k+1)n z (4k+1)n q 2jn z j + q −2jn z −j .

(7.35)

n=−∞

Putting (7.35) together with the definition of f , we therefore have finished the proof. ¤ Note that if we set k = 1 in (7.27), then we find that ∞ X

m 5m2

(−1) q

m=−∞

−q +q = (q

2

∞ X

2

(−1)n q 5n z 5n

n=−∞ ∞ X

m 5m2 −2m

(−1) q

m=−∞ ∞ X 4

∞ X

¡ ¢ 2 (−1)n q 5n z 5n q 2n z + q −2n z −1

n=−∞ ∞ X m 5m2 −4m

(−1) q

¢ ¡ 2 (−1)n q 5n z 5n q 4n z 2 + q −4n z −2

m=−∞ n=−∞ 2 −1 2 2 ; q )∞ (zq; q )∞ (z q; q 2 )∞ (z 2 q; q 2 )∞ (z −2 q; q 2 )∞ .

(7.36)


25

Comparing (7.36) with (6.12), we find that ∞ X (q; q 2 )5∞ + 4(q 5 ; q 10 )∞ 1 2 = 2 2 2 (−1)m q 5m , 2 10 10 5 10 2 5(q; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ m=−∞ ∞ X (q; q 2 )5∞ − (q 5 ; q 10 )∞ q 2 = − (−1)m q 5m −2m , 2 10 10 3 10 7 10 2 2 2 5(q; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ m=−∞

and ∞ X (q; q 2 )5∞ − (q 5 ; q 10 )∞ q4 2 = 2 2 2 (−1)m q 5m −4m . 2 10 10 10 9 10 5(q; q )∞ (q ; q )∞ (q; q )∞ (q ; q )∞ (q ; q )∞ m=−∞

By using the Jacobi triple product identity, we can write these three identities as Ã ∞ !2 X (q 2 ; q 2 )2∞ 2 {(q; q 2 )5∞ + 4(q 5 ; q 10 )∞ } = (−1)m q 5m , (7.37) 5(q; q 2 )∞ m=−∞ (q 2 ; q 2 )2∞ − {(q; q 2 )5∞ − (q 5 ; q 10 )∞ } = 5q(q; q 2 )∞ and (q 2 ; q 2 )2∞ {(q; q 2 )5∞ − (q 5 ; q 10 )∞ } = 4 2 5q (q; q )∞

Ã

!2

∞ X

(−1)m q

5m2 −2m

,

(7.38)

,

(7.39)

m=−∞

Ã

∞ X

!2 m 5m2 −4m

(−1) q

m=−∞

respectively. 8. Modular equations So far we has shown two methods of proving product identities. At the end of the last section, we have found out that they sometimes give two different kinds of identities. After comparing the two identities derived from the two methods, we therefore obtain new identities. We will show later that some of these identities can give us modular equations. Using Ramanujan’s notation, we set ϕ(q) :=

∞ X

2

q n = (q 2 ; q 2 )∞ (−q; q 2 )2∞ =

n=−∞

and ψ(q) :=

∞ X

(−q; −q)∞ , (q; −q)∞

(8.1)

(q 2 ; q 2 )∞ . (q; q 2 )∞

(8.2)

q n(n+1)/2 = (q; q)∞ (−q; q)2∞ =

n=0

These two product formulas in (8.1) and (8.2) can be deduced by the Jacobi triple product identity. For convenience, we set χ(q) = (−q; q 2 )∞ .

26


Then we can write (7.37) as ϕ2 (−q) + 4ψ 2 (q)χ(−q)χ(−q 5 ) = 5ϕ2 (−q 5 ).

(8.3)

Next we introduce some basic concepts of modular equations. For each nonnegative integer n, let (a)n = a(a + 1)(a + 2) · · · (a + n − 1). The ordinary of Gaussian hypergeometric function 2 F1 is defined by 2 F1 (a, b; c; z)

:=

∞ X (a)n (b)n n=0

(c)n n!

zn,

where a, b, c, and z are complex numbers such that c is not a nonpositive integer and |z| < 1. We now state some results from the theory of elliptic functions. Theorem 10. Let q and x be real number between 0 and 1. If Ã ¡1 1 ¢! F , ; 1; 1 − x 2 1 2 2 ¡1 1 ¢ q = exp −π , 2 F1 2 , 2 ; 1; x then

µ z := 2 F1

1 1 , ; 1; x 2 2

(8.4)

¶ = ϕ2 (q).

(8.5)

The proof of Theorem 10 can be found in [3, pp. 98–101]. Definition 11. Suppose that, for some positive integer n, ¡1 1 ¡1 1 ¢ ¢ 2 F1 2 , 2 ; 1; 1 − β 2 F1 2 , 2 ; 1; 1 − α ¡1 1 ¢ =n ¡1 1 ¢ . 2 F1 2 , 2 ; 1; β 2 F1 2 , 2 ; 1; α

(8.6)

Then a modular equation is a relation between α and β induced by (8.6). We often say that β has degree n over α. The multiplier m (of degree n) for a modular equation of degree n is defined by ¡1 1 ¢ z 2 F1 2 , 2 ; 1; α ¡1 1 ¢ = 1, m= zn 2 F1 2 , 2 ; 1; β where zr = ϕ2 (q r ). From Definition 11, we find that ¡1 1 ¢ ¢ ¡1 1 n 2 F1 2 , 2 ; 1; β 2 F1 2 , 2 ; 1; 1 − β ¡1 1 ¡1 1 ¢= ¢, =n m 2 F1 2 , 2 ; 1; α 2 F1 2 , 2 ; 1; 1 − α

(8.7)

and we obtain the following theorem. Theorem 12 (Method of Reciprocation). If α, β, and m are replaced by 1−β, 1−α, and n/m respectively, where n is the degree of the modular equation, then another modular equation of degree n is obtained. We say that each modular equation is the reciprocal of the other.


27

For further details of the proof of Theorem 12, see Berndt’s book [3, pp. 216–217]. In order to transform theta-function identities to modular equations, we need the following theorem. Theorem 13. We have ϕ(−q) = ϕ(q 2 ) =

z(1 − x)1/4

p √ ¡ ¢1/2 z/2 1 + 1 − x p

ψ(q) = ψ(−q) = ψ(q 4 ) =

√

z/2(x/q)1/8

p

z/2{x(1 − x)/q}1/8

√ 1p z/2{(1 − 1 − x)/q}1/2 2

χ(−q) = 21/6 (1 − x)1/12 (q/x)1/24

(8.8) (8.9) (8.10) (8.11) (8.12) (8.13)

For the proofs, see [3, pp. 122–125]. Now we are ready to transform (8.3) to a modular equation stated in the next theorem. Theorem 14 (Modular Equations of Degree 5). Suppose that β has degree 5 over α. Then µ ¶1/2 µ 5 ¶1/24 µ ¶1/12 1−α α 1−α 5 4/3 +2 = (8.14) 5 1−β β (1 − β) m and µ ¶1/2 µ ¶1/24 µ ¶1/12 β (1 − β)5 β 4/3 +2 = m. (8.15) α 1−α α5 Proof. We can prove (8.14) by substituting (8.8), (8.10), and (8.13) in (8.3). To prove (8.15), we apply Theorem 12 to (8.14). ¤ We can get more modular equations by similar methods. By letting k = 3 in Theorem 7, we find that q

4

∞ X

q

12m2 +12m

m=−∞ ∞ X

+

m=−∞ ∞ X

−q =

2

q 12m

∞ X

q 4n

n=−∞ ∞ X

2 −4n

z 4n−2

2

q 4n z 4n

n=−∞

q

12m2 −6m

∞ X

2

q 4n z 4n (q 2n z − q −2n z −1 )

m=−∞ n=−∞ 2 2 2 (q ; q )∞ (zq; q )∞ (z −1 q; q 2 )∞ (q 6 ; q 6 )∞ (z 3 q 3 ; q 6 )∞ (z −3 q 3 ; q 6 )∞ .

(8.16)

28


Comparing (8.16) with (6.13), we therefore have (−q 2 ; q 2 )2∞ (−q 6 ; q 6 )2∞ + (−q 4 ; q 4 )∞ (−q 12 ; q 12 )∞ (q 8 ; q 8 )∞ (−q 8 ; q 8 )2∞ ∞ X 1 2 = 2 2 q 12m (8.17) 6 6 (q ; q )∞ (q ; q )∞ m=−∞

and 2q 2

(−q 2 ; q 2 )2∞ (−q 6 ; q 6 )2∞ − (−q 4 ; q 4 )∞ (−q 12 ; q 12 )∞ (q 8 ; q 8 )∞ (−q 4 ; q 8 )2∞ ∞ X q4 2 q 12m +12m . (8.18) = 2 2 6 6 (q ; q )∞ (q ; q )∞ m=−∞

Replacing q 2 by q and using (8.1) and (8.2), we can write (8.17) and (8.18), respectively, as ψ(q)ψ(q 3 ) + ψ(−q)ψ(−q 3 ) = 2ψ(q 4 )ϕ(q 6 ) (8.19) and ψ(q)ψ(q 3 ) − ψ(−q)ψ(−q 3 ) = 2qϕ(q 2 )ψ(q 12 ).

(8.20)

Using these two identities, we have the following modular equations of degree 3. Theorem 15 (Modular Equations of Degree 3). Suppose that β has degree 3 over α. Then p √ (αβ)1/8 + {α(1 − α)β(1 − β)}1/8 = {(1 − 1 − α)(1 + 1 − β)}1/2 , (8.21) p √ (αβ)1/8 − {α(1 − α)β(1 − β)}1/8 = {(1 + 1 − α)(1 − 1 − β)}1/2 , (8.22) p √ {(1 − α)(1 − β)}1/8 + {α(1 − α)β(1 − β)}1/8 = {(1 − α)(1 + β)}1/2 , (8.23) p √ {(1 − α)(1 − β)}1/8 − {α(1 − α)β(1 − β)}1/8 = {(1 + α)(1 − β)}1/2 , (8.24) (αβ)1/4 + {(1 − α)(1 − β)}1/4 = 1, ½ ¾1/8 ½ ¾1/8 (1 − β)3 (1 − α)3 1/4 2(αβ) = − , 1−α 1−β and

½ 1/4

2{(1 − α)(1 − β)}

=

α3 β

¾1/8

½ −

β3 α

(8.25) (8.26)

¾1/8 .

(8.27)

Proof. Substituting (8.10), (8.11), (8.12), and (8.9) in (8.19) and (8.20), we obtain (8.21) and (8.22), respectively. Multiplying (8.21) and (8.22), we then obtain (8.25). Squaring both sides of (8.21) and (8.22) and then subtracting the second identity from the first, we then have (8.26). To prove (8.23), (8.24), and (8.27), we apply Theorem 12 to (8.21), (8.22) and (8.26),respectively. ¤


29

Note that we can obtain further theta-function identities by letting k = 3 in Theorem 6 and comparing the identity with (6.14). However, these theta-function formulas do not give us any modular equations. Equality (8.3) can be found in [2, p. 19]. We can find (8.20) in [4, p. 356] and we can make one change in the proof of (8.20) to get (8.19). We also can find (8.25) in [3, p. 230]. However, the proofs in this section are new. References [1] G. E. Andrews, A simple proof of Jacobi’s triple product identity, Proc. Amer. Math. Soc. 16 (1965), 333–334. [2] G. E. Andrews and B. C. Berndt, Ramanujan’s Lost Notebook, Part I, Springer-Verlag, New York, to appear. [3] B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991. [4] B. C. Berndt, Ramanujan’s Notebooks, Part V, Springer-Verlag, New York, 1998. [5] S. Bhargava, A simple proof of the quintuple product identity, J. Indian Math. Soc. (N.S.) 61 (1995) 226–228. [6] L. Carlitz and M. V. Subbarao, A simple proof of the quintuple product identity, Proc. Amer. Math. Soc. 32 (1972), 42–44. [7] L. Carlitz and M. V. Subbarao, On a combinatorial identity of Winquist and its generalization, Duke Math. J. 39 (1972), 165–172. [8] K. Chandrasekharan, Elliptic Functions, Springer-Verlag, Berlin Heideberg, 1985. [9] R. J. Evans, Theta function identities, J. Math. Anal. Appl. 147 (1990), 97–121. [10] J. A. Ewell, An easy proof of the triple-product identity, Amer. Math. Monthly 88 (1981), 270–272. [11] H. M. Farkas and I. Kra, On the quintuple product identity, Proc. Amer. Math. Soc. 88 (1999), 771–778. [12] D. Foata and G.-N. Han, The triple, quintuple and septuple product identities revisited, in The Andrews festschrift : seventeen papers on classical number theory and combinatorics, D. Foata and G.-N. Han, eds., Springer-Verlag, New York, 2001, 323–334. [13] C. F. Gauss, Hundert Theoreme u ¨ber die neuen Transscendenten, Werke, Bd. 3, Königlichen Gesell. Wiss., Göttingen, 1876, 461–469. [14] B. Gordon, Some identities in combinatorial analysis, Quad. J. Math. (Oxford) (2) 12 (1961), 285–290. [15] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Clarendon Press, Oxford, 1983. [16] M. D. Hirschhorn, A generalisation of Winquist’s identity and a conjecture of Ramanujan, J. Indian Math. Soc. 51 (1987), 49–55. [17] C. G. J. Jacobi, Fundamenta Nova Theoriae Functionum Ellipticarum, Fratrum Bornträger Regiomonti, 1829; Gesammelte Werke, Erster Band, G. Reimer, Berlin, 1881. [18] S.-Y. Kang, A new proof of Winquist’s identity, J. Combin. Theory 78 (1997), 313–318. [19] R. P. Lewis, A combinatorial proof of the triple product identity, Amer. Math. Monthly 91 (1984), 420–423. [20] S. Ramanujan, Notebooks (2 volumes), Tata Institute of Fundamental Research, Bombay, 1957. [21] H. A. Schwarz, Formeln und Lehs¨ atze zum Gebrauche der Elliptischen Funcktionen. Nach Vorlesungen und Aufzeichnungen des Herrn Prof. K. Weierstrass, Zweite Ausgabe, Erste Abteilung, Springer, Berlin, 1893. [22] G. N. Watson, Theorems stated by Ramanujan (VII): theorems on contiued fractions, J. London Math. Soc. 4 (1929), 231–237. [23] H. S. Wilf, The number-theoretic content of the Jacobi triple product identity, in The Andrews festschrift : seventeen papers on classical number theory and combinatorics, D. Foata and G.-N. Han, eds., Springer-Verlag, New York, 2001, 227–229.

30


[24] L. Winquist, An elementary proof of p(n) ≡ 0 (mod 11), J. Combin. Theory 6 (1969), 56–59. (Sarachai Kongsiriwong) 1409 W. Green St., Department of Mathematics, University of Illinois, Urbana, IL 61801 E-mail address: [email protected] (Zhi-Guo Liu) Department of Mathematics, East China Normal University, Shanghai 200062, P. R. China E-mail address: [email protected]