THE RAMANUJAN JOURNAL 4, 43–50, 2000 c 2000 Kluwer Academic Publishers. Manufactured in The Netherlands. °
The Borweins’ Cubic Theta Function Identity and Some Cubic Modular Identities of Ramanujan ZHI-GUO LIU Xinxing Education College, Xinxiang, Henan, P.R. China
[email protected]
Dedicated to George E. Andrews on the occasion of his 60th Birthday Received July 1, 1998; Accepted November 16, 1998
Abstract. In this paper the author will give new proofs of the Borweins’ cubic theta function identity and a related identity relying on the properties of elliptic functions and the technique of comparing constant terms. Key words: theta functions, elliptic functions, modular equations, modular forms 1991 Mathematics Subject Classification: Primary–11F11, 11F27, 33E05
1.
Introduction
In [4] Borwein, Borwein and Garvan introduce three functions, a(q) = b(q) = c(q) =
∞ X n,m=−∞ ∞ X n,m=−∞ ∞ X
qm
2
+mn+n 2
ωm−n q m
2
,
+mn+n 2
(ω := e2πi/3 ),
q (m+ 3 ) +(m+ 3 )(n+ 3 )+(n+ 3 ) . 1 2
1
1
1 2
n,m=−∞
They prove the following important result by using the Euler q-binomial theorem and the theta transformation formulas and the properties of modular equations. (q; q)3∞ , (q 3 ; q 3 )∞ (q 3 ; q 3 )3∞ , c(q) = 3q 1/3 (q; q)∞ a 3 (q) = b3 (q) + c3 (q), a(q)a(q 2 ) = b(q)b(q 2 ) + c(q)c(q 2 ). b(q) =
(1.1) (1.2) (1.3) (1.4)
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(1.3) was first established by Borwein and Borwein in [3] by using modular function techniques. Later Hirschhorn, Garvan and Borwein in [5] using only the Jacobi triple product identity prove the above equation again. From (1.3) and (1.4) we can easily obtain the following modular identities of Ramanujan [4, 5]. Ã µ ¶ ¶ !3 µ η(q 9 ) 3 η(q 9 ) 12 = 1+9 , 1 + 27 η(q) η(q) Ã ¶ !Ã ¶! ¶ µ µ µ η(q 18 ) 3 η(q 3 )η(q 6 ) 4 η(q 9 ) 3 , 1+9 = 1 + 9 1+9 η(q) η(q 2 ) η(q)η(q 2 )
(1.5) (1.6)
where, here and throughout the paper; η(q) := q 1/24 (q; q)∞ , q = eiπ τ , I mτ > 0 and ∞ Y
(a; q)∞ =
(1 − aq n ).
n=0
Both (1.5) and (1.6) appear in Ramanujan’s second notebook [6]. (1.3) is Entry 1(iv) of Chapter 20 [1, p. 354] and (1.6) is equivalent to the first equation on page 259. In this paper we will use the residue theorem of elliptic functions and the method of equating the constant terms of Fourier expansions to give new proofs of (1.3) and (1.4). More importantly we will obtain the Lambert series expansions of a(q)a(q 2 ), b(q)b(q 2 ) and c(q)c(q 2 ) in the course of our investigations. Note that the residue theorem of elliptic functions states that the sum of the residues of an elliptic function vanishes in the period parallelogram. 2.
Proofs
The Jacobi theta functions θ1 (z | q), and θ3 (z | q) are usually defined by [7, pp. 463–464] θ1 (z | q) := −i
∞ X
(−1)n q (n+ 2 ) e(2n+1)i z 1 2
n=−∞
= 2
∞ X
(−1)n q (n+ 2 ) sin(2n + 1)z, 1 2
n=0
θ3 (z | q) :=
∞ X
2
q n e2ni z
n=−∞
= 1+2
∞ X
2
q n cos 2nz
n=0
It is easy to show that [7, p. 464] ¶ µ π + πτ θ1 z + = q −1/4 e−iπ z θ3 (z | q). 2
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CUBIC THETA FUNCTION IDENTITY
The Jacobi theta functions θ1 (z | q), and θ3 (z | q) have the following product representations [7, pp. 469–470] θ1 (z | q) = 2q 1/4 sin z(q 2 ; q 2 )∞ (q 2 e2i z ; q 2 )∞ (q 2 e−2i z ; q 2 )∞ , θ3 (z | q) = (q 2 ; q 2 )∞ (−qe2i z ; q 2 )∞ (−qe−2i z ; q 2 )∞ . In this paper we will need the following special values of θ1 (z | q) θ 0 (0 | q) := θ10 (q) = 2q 1/4 (q 2 ; q 2 )3∞ , µ1 ¯ ¶ √ π ¯¯ q = 3q 1/4 (q 6 ; q 6 )∞ , θ1 ¯ 3 ¯ ¶ µ πτ ¯¯ q = iq −1/12 (q 2/3 ; q 2/3 )∞ . θ1 3 ¯
(2.1)
It is well-known that [7, p. 489] ∞ X q 2n sin 2nz θ10 (z | q) = cot z + 4 . θ 1 − q 2n n=1
Therefore
à µ ¯ ¶ ¶! ∞ µ X q 3n+2 1 q 3n+1 θ10 π ¯¯ 1/2 − q = √ 1+6 . θ 3¯ 1 − q 3n+1 1 − q 3n+2 3 n=0
From the product representations of the θ3 (z | q) it is easy to show ¯ ¶ µ µ (q 6 ; q 6 )∞ π¯ π θ3 (3z | q 3 ) = 2 2 3 θ3 (z | q)θ3 z + ¯¯ q θ3 z − (q ; q )∞ 3 3
(2.2)
¯ ¶ ¯ ¯q . ¯
By equating the constant terms on both sides of the above equation we obtain X (q 2 ; q 2 )3∞ 2 2 2 = ωn−m q n +m + p 6 6 (q ; q )∞ m+n+ p=0 =
∞ X
ωn−m q 2m
2
+2mn+2n 2
,
m,n=−∞
By writing q 2 for q in the above equation we have [4, 5] b(q) =
∞ X
ωn−m q m
2
n,m=−∞
+mn+n 2
=
(q; q)3∞ . (q 3 ; q 3 )∞
We construct the following elliptic function f (z) =
θ1 (2z | q)θ1 (z | q 1/3 ) , θ12 (z | q)θ1 (3z | q 3 )θ1 (z − x | q)θ1 (z + x | q)
(2.3)
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where 0 < x < π . It is easy to check that f (z) is an elliptic function with period π and π τ . Its poles are at 0, π/3, 2π/3, x, π − x, and all poles are simple. It is easy to show that: (q 2/3 ; q 2/3 )3 1 , Res( f ; 0) = − q −11/12 2 2 3 6 6 3 ∞ 2 3 (q ; q )∞ (q ; q )∞ θ1 (x | q) 1 θ1 (x | q 1/3 ) Res( f ; x) = Re( f ; π − x) = q −1/4 2 2 3 2 , 2 (q ; q )∞ θ1 (x | q)θ1 (3x | q 3 ) µ ¶ µ ¶ π 2π 1 θ1 (x | q) Res f ; = Res f ; = − q −11/12 2 2 2 6 6 3 , 3 3 6 (q ; q )∞ (q ; q )∞ θ1 (3x | q 3 ) where Res( f ; α) means the residue of f (z) at α. From the residue theorem of elliptic functions we have θ13 (x | q) = 3q 2/3 π+πτ 2
Replacing x by x +
(q 6 ; q 6 )3∞ θ1 (x | q 1/3 ) − b(q 2/3 )θ1 (3x | q 3 ). (q 2 ; q 2 )∞
(2.4)
in the above equation we get
θ33 (x | q) = 3q 2/3
(q 6 ; q 6 )3∞ θ3 (x | q 1/3 ) + b(q 2/3 )θ3 (3x | q 3 ). (q 2 ; q 2 )∞
(2.5)
By equating the coefficients of e2πi x on both sides of the above equation we have X
c(q 2 ) = q −2/3
qm
2
+n 2 + p 2
m+n+ p=1 ∞ X
=
q 2(m+ 3 ) +2(m+ 3 )(n+ 3 )+2(n+ 3 ) = 3q 2/3 1 2
1
1
1 2
n,m=−∞
(q 6 ; q 6 )3∞ . (q 2 ; q 2 )∞
By writing q 2 for q we have [4, 5] c(q) =
∞ X
q (m+ 3 ) +(m+ 3 )(n+ 3 )+(n+ 3 ) = 3q 1/3 1 2
1
1
1 2
n,m=−∞
(q 3 ; q 3 )3∞ . (q; q)∞
(2.6)
By comparing the constant terms on both sides of (2.5) and writing q 2 for q we have a(q 3 ) = c(q 3 ) + b(q). From the above equation one can easily obtain (1.3) (see [4, 5] for details). We construct the following elliptic function: f (z) =
θ1 (2z | q)θ1 (6z | q 6 ) . θ12 (z | q)θ1 (2z | q 2 )θ12 (3z | q 3 )
(2.7)
CUBIC THETA FUNCTION IDENTITY
47
f (z) is an elliptic function with periods π and π τ . π/3, 2π/3 are its simple poles and 0 is its pole of order 2. It is easy to know that: µ ¶ µ ¶ π 2π 1 (q 12 ; q 12 )2 Res f ; = Res f ; = q −3/4 6 6 7∞ . 3 3 9 (q ; q )∞ Let F(z) = z 3 f (z). It is easy to show that 3 θ0 θ0 θ0 θ0 θ0 F0 (z) = − 2 1 (z | q) + 2 1 (2z | q) − 2 1 (2z | q 2 ) − 6 1 (3z | q 3 ) + 6 1 (6z | q 6 ) F z θ1 θ1 θ1 θ1 θ1 ∞ ∞ 2n 4n X X q q 16 (sin 4nz − sin 2nz) − 8 sin 4nz = − z + O(z 3 ) + 8 3 1 − q 2n 1 − q 4n n=1 n=1 − 24
∞ X n=1
∞ X q 6n q 12n sin 6nz + 24 sin 12nz. 6n 1−q 1 − q 12n n=1
So we have µ
à ! ¶ ∞ ∞ ∞ ∞ X X X X 16 nq 2n nq 4n nq 6n nq 12n F0 0 (0) = − +6 + 27 − 54 1−3 F 3 1 − q 2n 1 − q 4n 1 − q 6n 1 − q 12n n=1 n=1 n=1 n=1
Hence 2 (q 12 ; q 12 )3 Res( f ; 0) = − q −3/4 6 6 6 4 4 3∞ 2 2 3 9 (q ; q )∞ (q ; q )∞ (q ; q )∞ ! Ã ∞ ∞ ∞ ∞ X X X X nq 2n nq 4n nq 6n nq 12n × 1−3 +6 + 27 − 54 1 − q 2n 1 − q 4n 1 − q 6n 1 − q 12n n=1 n=1 n=1 n=1 By the residue theorem of elliptic functions we know 1−3
∞ ∞ ∞ ∞ X X X X nq 2n nq 4n nq 6n nq 12n + 6 + 27 − 54 = b(q 2 )b(q 4 ). 2n 4n 6n 12n 1 − q 1 − q 1 − q 1 − q n=1 n=1 n=1 n=1
After writing q 2 for q we obtain the following interesting result: 1−3
∞ ∞ ∞ ∞ X X X X nq n nq 2n nq 3n nq 6n +6 + 27 − 54 = b(q)b(q 2 ). n 2n 3n 6n 1 − q 1 − q 1 − q 1 − q n=1 n=1 n=1 n=1 (2.8)
By applying the residue theorem to the following elliptic function f (z) =
θ1 (2z | q)θ1 (z | q 1/6 ) , θ12 (z | q)θ1 (z | q 1/2 )θ12 (z | q 1/3 )
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we obtain à ! ∞ ∞ ∞ ∞ X X X X nq n nq 2n nq 3n nq 6n 9 −2 − +2 = c(q)c(q 2 ). 1 − qn 1 − q 2n n=1 1 − q 3n 1 − q 6n n=1 n=1 n=1 Let
(2.9)
¯ ¢ ¡ θ12 2z + π3 ¯ q ¯ ¢ ¡ ¯ ¢. f (z) = ¡ θ1 z − π3 ¯ q 1/2 θ13 z ¯ q 1/2
f (z) is an elliptic function with only one pole of order 3 at 0. Let F(z) := z 3 f (z). Then we have ¯ ¯ ¶ µ ¶ µ ∞ X F0 θ10 π ¯¯ 1/2 θ10 π ¯¯ qn 3 (z) = z + O(z ) − 12 sin 2nz − z − ¯q +4 2z + ¯ q . F 1 − qn θ 3 θ 3 1
n=1
1
Noting that [3] ¶ ∞ µ X q 3n+2 q 3n+1 − a(q) = 1 + 6 , 1 − q 3n+1 1 − q 3n+2 n=0 we have
µ ¯ ¶ µ ¯ ¶ θ10 π ¯¯ 1/2 1 θ10 π ¯¯ F0 (0) = q q = √ (a(q) + 4a(q 2 )), +4 F θ1 3 ¯ θ1 3 ¯ 3 µ 0 ¶0 µ 0 ¶0 µ ¯ ¶ µ 0 ¶0 µ ¯ ¶ ∞ X F θ1 π ¯¯ 1/2 θ π ¯¯ nq n q q (0) = 1 − 24 − +8 1 ¯ n F 1−q θ1 3 θ1 3¯ n=1 (2.10) ∞ ∞ X X nq n nq 2n 25 − 32 = − − 20 3 1 − qn 1 − q 2n n=1 n=1 − 36
∞ ∞ X X nq 3n nq 6n + 288 . 1 − q 3n 1 − q 6n n=1 n=1
In [1, p. 460, Entry3(i)], it is shown that 1 + 12
∞ ∞ X X nq n nq 3n − 36 = a 2 (q). 1 − qn 1 − q 3n n=1 n=1
Hence µ
F0 (0) F
¶2 =
∞ ∞ X X 8 17 nq n nq 3n − 12 a(q)a(q 2 ) + +4 n 3 3 1−q 1 − q 3n n=1 n=1
+ 64
∞ ∞ X X nq 2n nq 6n − 192 . 2n 1−q 1 − q 6n n=1 n=1
(2.11)
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CUBIC THETA FUNCTION IDENTITY
From the residue theorem we know õ Res( f ; 0) = F(0)
F0 F
¶0
µ (0) +
¶2 ! F0 (0) = 0. F
F(0) 6= 0, so substituting (2.10) and (2.11) into the above equation we get the following interesting result 1+6
∞ ∞ ∞ ∞ X X X X nq n nq 2n nq 3n nq 6n − 12 + 18 − 36 = a(q)a(q 2 ). n 2n 3n 6n 1 − q 1 − q 1 − q 1 − q n=1 n=1 n=1 n=1 (2.12)
It is easy to see that (2.8), (2.9) and (2.12) trivially imply (1.4). It should be pointed out that by using the method of this paper we can derive many interesting results about a(q), b(q) and c(q). For examples: By considering f (z) =
θ16 (z
θ13 (2z | q 1/2 ) , | q 1/2 )θ12 (3z | q 3/2 )
we obtain [2] 1 + 240 1 + 240
∞ X n3q n = 9a 4 (q) − 8a(q)b3 (q), 1 − qn n=1
∞ X 1 n 3 q 3n = (a 4 (q) + 8a(q)b3 (q)). 3n 1 − q 9 n=1
(2.13) (2.14)
By considering f (z) =
θ19 (z
θ13 (2z | q 1/2 ) , | q 1/2 )θ1 (3z | q 3/2 )
we obtain [2] 1 − 504 1 − 504
∞ X n5q n = −27a 6 (q) − 8b6 (q) + 36a 3 (q)b3 (q), n 1 − q n=1
∞ X 1 n 5 q 3n (−a 6 (q) + 8b6 (q) + 20a 3 (q)b3 (q)). = 3n 1 − q 27 n=1
By considering f 1 (z) =
θ1 (2z | q 1/2 )θ1 (5z | q 5/2 ) , θ13 (3z | q 3/2 )
(2.15) (2.16)
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and f 2 (z) =
θ1 (5z | q 15/2 )θ1 (2z | q 3/2 ) , θ12 (z | q 1/2 )
we are led to the following results respectively 1+6
∞ ∞ X X 1 nq 3n nq 15n − 30 = (a(q)a(q 5 ) + 2b(q)b(q 5 )), 3n 15n 1 − q 1 − q 3 n=1 n=1
1+6
∞ ∞ X X nq n nq 5n − 30 = a(q)a(q 5 ) + 2c(q)c(q 5 ). n 5n 1 − q 1 − q n=1 n=1
(2.17) (2.18)
Acknowledgments This work was supported in part by the K.C. Wong Education Foundation at Hong Kong. The author would like to thank the SMS at Sussex University for their hospitality. He would especially like to thank R.P. Lewis. He would also like to thank the referees and M.D. Hirschhorn for their valuable comments. References 1. B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, N.Y., 1991. 2. B.C. Berndt, S. Bhargava, and F.G. Garvan, “Ramanujan’s theories of elliptic functions to alternative bases,” Trans. Amer. Math. Soc. 347 (1995) 4136–4244. 3. J.M. Borwein and P.B. Borwein, “A cubic counterpart of Jacobi’s identity and AGM,” Trans. Amer. Math. Soc. 323 (1991) 691–701. 4. J.M. Borwein, P.B. Borwein, and F.G. Garvan, “Some cubic modular identities of Ramanujan,” Trans. Amer. Math. Soc. 343 (1994) 35–47. 5. M. Hirschhorn, F. Garvan, and J.M. Borwein, “Cubic analogues of the Jacobian theta function θ (z, q),” Canad. J. Math. 45 (1993) 673–694. 6. S. Ramanjan, Notebooks, 2 vols., Tata Institute of Fundamental Research, Bombay, 1957. 7. E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 4th edn., Cambridge Univ. Press, Cambridge, 1966.
