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International Journal of Number Theory Vol. 8, No. 8 (2012) 1977–2002 c World Scientific Publishing Company DOI: 10.1142/S1793042112501126
SOME INVERSE RELATIONS AND THETA FUNCTION IDENTITIES
ZHI-GUO LIU Department of Mathematics East China Normal University 500 Dongchuan Road, Shanghai 200241, P. R. China
[email protected] [email protected]
Received 21 November 2011 Accepted 25 May 2012 Published 20 September 2012 Two pairs of inverse relations for elliptic theta functions are established with the method of Fourier series expansion, which allow us to recover many classical results in theta functions. Many nontrivial new theta function identities are discovered. Some curious trigonometric identities are derived. Keywords: Elliptic function; theta function; entire function; Schr¨ oter formula; Fourier series. Mathematics Subject Classification 2010: 11F11, 11F20, 11F27, 11F30, 30D15, 33E05
1. Introduction The main aim of this paper is using the method of Fourier series expansion to develop an inversion technique to theta function identities. Throughout this paper we assume Im τ > 0 and take q = e2πiτ . To carry out our study, we need the definitions of Jacobi theta functions. Definition 1.1. Jacobi theta functions θj for j = 1, 2, 3, 4 are defined as θ1 (z | τ ) = 2
∞
(−1)k q (2k+1)
2
/8
sin(2k + 1)z,
θ3 (z | τ ) = 1 + 2
k=0
θ2 (z | τ ) = 2
∞
∞
qk
2
/2
cos 2kz,
k=1
q
(2k+1)2 /8
cos(2k + 1)z,
θ4 (z | τ ) = 1 + 2
k=0
∞
(−1)k q k
k=1
1977
2
/2
cos 2kz.
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The main results of this paper are the following two theorems below. Theorem 1.1. Suppose that n is a positive integer and f (z | τ ) is an entire function of z satisfying the functional equations f (z | τ ) = f (z + π | τ ) = q n/2 e2inz f (z + πτ | τ ).
(1.1)
Then for any integer m, there exists a constant am independent of z such that n−1 2imkπ kπ τ = nam e2miz θ3 (nz + mπτ | nτ ). e− n f z + (1.2) n k=0
If am are given as above and α is a positive integer coprime to n, then we have f (z | τ ) =
n−1
aαm e2αmiz θ3 (nz + αmπτ | nτ ).
(1.3)
m=0
In particular, when α = 1, we have f (z | τ ) =
n−1
am e2miz θ3 (nz + mπτ | nτ ).
(1.4)
m=0
Theorem 1.2. Suppose that n is a positive integer and f (z | τ ) is an entire function of z satisfying the functional equations f (z | τ ) = (−1)n f (z + π | τ ) = −q n/2 e2inz f (z + πτ | τ ).
(1.5)
Then for any integer m, there exists a constant am independent of z such that n−1 kπ k − 2imkπ n τ = nam e2miz θ1 (nz + mπτ | nτ ). (−1) e f z+ (1.6) n k=0
If am are given as above and α is a positive integer coprime to n, then we have f (z | τ ) =
n−1
aαm e2αmiz θ1 (nz + αmπτ | nτ ).
(1.7)
m=0
In particular, when α = 1, we have f (z | τ ) =
n−1
am e2miz θ1 (nz + mπτ | nτ ).
(1.8)
m=0
Definition 1.2. We call the case m = 0 of (1.2) is the dual identity of (1.4); and the case m = 0 of (1.6) is the dual identity of (1.8). Formulas (1.2) and (1.3) form a pair of inverse relations, and from one of them we can obtain another. Formulas (1.6) and (1.7) also form a pair of inverse relations. Our idea is to choose a suitable f in Theorem 1.1 or 1.2 to get an identity and then use inverse relations to get another identity. These inverse relations provide a new
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insight to theta function identities, and can be used to derive a multitude of both new and old theta function identities. With our method the dual identities of some well-known identities are found for the first time. To discuss the applications of the above two theorems, we need some basic properties of Jacobi theta functions. The following functional equations will be often used in this paper. Proposition 1.1. With respect to the (quasi) periods π and πτ, Jacobi theta functions satisfy the functional equations θ1 (z | τ ) = −θ1 (z + π | τ ) = −q 1/2 e2iz θ1 (z + πτ | τ ), θ2 (z | τ ) = −θ2 (z + π | τ ) = q 1/2 e2iz θ2 (z + πτ | τ ), θ3 (z | τ ) = θ3 (z + π | τ ) = q 1/2 e2iz θ3 (z + πτ | τ ), θ4 (z | τ ) = θ4 (z + π | τ ) = −q 1/2 e2iz θ4 (z + πτ | τ ). For brevity, in the next proposition we use θk (z) to denote θk (z | τ ) for k = 1, 2, 3, 4. For the half periods π/2, (πτ )/2 and (π + πτ )/2, it is easy to verify the following. Proposition 1.2. For mz = q −1/8 e−iz , Jacobi theta functions θ1 , θ2 , θ3 , and θ4 satisfy the relations θ1 (z + π/2) = θ2 (z), θ1 (z + πτ /2) = imz θ4 (z), θ1 (z + (π + πτ )/2) = mz θ3 (z), θ2 (z + π/2) = −θ1 (z), θ2 (z + πτ /2) = mz θ3 (z), θ2 (z + (π + πτ )/2) = −imz θ4 (z), θ3 (z + π/2) = θ4 (z), θ3 (z + πτ /2) = mz θ2 (z), θ3 (z + (π + πτ )/2) = imz θ1 (z), θ4 (z + π/2) = θ3 (z), θ4 (z + πτ /2) = imz θ1 (z), θ4 (z + (π + πτ )/2) = mz θ2 (z). Definition 1.3. The Euler series E(τ ) and the Dedekind eta function are defined as E(τ ) =
∞
(1 − q n ),
η(τ ) = q 1/(24) E(τ ).
n=1
With the Euler series E(τ ), the celebrated Jacobi triple product identity can be stated as follows. Proposition 1.3. For z = 0 and |q| < 1, we have the triple product identity ∞
(−1)n q n(n−1)/2 z n = (1 − z)E(τ )
n=−∞
∞
(1 − q n z)(1 − q n /z).
n=1
Employing the triple product identity, we can easily derive the infinite product representations of the Jacobi theta functions.
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Proposition 1.4 (Infinite product representations for Jacobi theta functions). θ1 (z | τ ) = 2q 1/8 (sin z)E(τ ) θ2 (z | τ ) = 2q 1/8 (cos z)E(τ ) θ3 (z | τ ) = E(τ ) θ4 (z | τ ) = E(τ )
∞ n=1 ∞
∞
(1 − q n e2iz )(1 − q n e−2iz ),
n=1 ∞
(1 + q n e2iz )(1 + q n e−2iz ),
n=1
(1 + q n−1/2 e2iz )(1 + q n−1/2 e−2iz ), (1 − q n−1/2 e2iz )(1 − q n−1/2 e−2iz ).
n=1
If we use θ1 (z | τ ) to denote the partial derivative of θ1 with respect to z and divide both sides of the first equation in Proposition 1.4 by z and then let z → 0, we obtain the following proposition. Proposition 1.5. Let E(τ ) be the Euler series. Then we have θ1 (0 | τ ) = 2q 1/8 E 3 (τ ) = η 3 (τ ). The remainder of this paper is organized as follows. In Sec. 2, we prove Theorems 1.1 and 1.2 using Fourier series. In Sec. 3, we prove the following two remarkable identities. Some applications are also given. Proposition 1.6. For any integer n ≥ 1, we have n−1 2imkπ kπ τ m2 = nq 2n e2imz θ3 (nz + mπτ | nτ ), e − n θ3 z + n n
(1.9)
k=0
n−1 m=0
q
m2 2n
e
2imz
τ θ3 (nz + mπτ | nτ ) = θ3 z . n
(1.10)
When m = 0, (1.9) becomes a result of Boon et al. [4, Eq. (7)], and when n = 2, (1.10) reduces to the identity τ (1.11) = θ3 (2z | 2τ ) + θ2 (2z | 2τ ). θ3 z 2 In Sec. 4, Theorem 1.1 is used to derive the following two identities, and some special cases are discussed. Proposition 1.7. 2n−1 τ kπ τ kπ − x θ3 z + x + z + θ 3 n 2n n 2n k=0 2τ = 2nθ3 2x θ3 (2nz | 2nτ ). n
(1.12)
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θ3 (z + x | τ )θ3 (z − x | τ ) 2n−1 m2 = q 2n e2mi(z+x) θ3 (2nz + mnπτ | 2n2 τ )θ3 (2x + mπτ | 2τ ).
1981
(1.13)
m=0
Definition 1.4. The multiple theta function Gm,n (τ ) is defined by ∞
Gm,n (τ ) =
q
2 2 l2 1 +l2 +···+ln 2
.
l1 ,l2 ,...,ln =−∞ l1 +l2 +···+ln =m
In Sec. 5, the following two nontrivial theta function identities related to the nth power of Jacobi theta function θ3 are derived, and extensions of these two formulas are also given. Proposition 1.8. If Gm,n (τ ) is defined as in Definition 1.4, then we have n−1 2imkπ kπ − n n e θ3 z + (1.14) τ = nGm,n (τ )e2miz θ3 (nz + mπτ | nτ ), n k=0
θ3n (z | τ ) =
n−1
Gm,n (τ )e2miz θ3 (nz + mπτ | nτ ).
(1.15)
m=0
When m = 0, (1.14) reduces to the known identity [8, Eq. (3.1)] n−1 kπ n τ = nG0,n (τ )θ3 (nz | nτ ). θ3 z + n k=0
Definition 1.5. The Ramanujan theta function φ(q) is defined as ∞ 2 φ(q) := qn . n=−∞
Setting z = 0 in (1.14), replacing τ by 2τ and noting that θ3 (0 | 2τ ) = φ(q), we immediately obtain the following curious formula. Proposition 1.9. If φ(q) is defined as in Definition 1.5. Then we have n
φ (q) =
n−1
Gm,n (2τ )θ3 (2mπτ | 2nτ ).
m=0
In Sec. 6, we prove the following two propositions using Theorem 1.1. Proposition 1.10. If n is an odd integer ≥1, then we have the decomposition formula kπ τ θ z + ny + n−1 3 n k − 2imkπ n (−1) e kπ k=0 θ1 z + τ n nθ3 (ny | τ )θ1 (0 | nτ )θ3 (nz + ny + mπτ | nτ ) . = e2miz θ1 (0 | τ )θ3 (ny + mπτ | nτ )θ1 (nz | nτ )
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Taking y = 0 and 2ny = π in the above equation respectively, we immediately obtain kπ τ θ3 z + n−1 2imkπ nθ (0 | τ )θ1 (0 | nτ )θ3 (nz + mπτ | nτ ) n = e2miz 3 , (−1)k e− n kπ θ1 (0 | τ )θ3 (mπτ | nτ )θ1 (nz | nτ ) k=0 τ θ1 z + n kπ τ θ z + n−1 4 nθ (0 | τ )θ1 (0 | nτ )θ4 (nz + mπτ | nτ ) n k − 2imkπ n = e2miz 4 . (−1) e kπ θ1 (0 | τ )θ4 (mπτ | nτ )θ1 (nz | nτ ) k=0 τ θ1 z + n Proposition 1.11. If n is an odd integer ≥1, then we have n−1 θ1 (0 | τ )θ3 (z + ny | τ )θ1 (nz | nτ ) θ3 (nz + ny + mπτ | nτ ) = . e2miz θ3 (ny | τ )θ1 (0 | nτ )θ1 (z | τ ) θ3 (ny + mπτ | nτ ) m=0
Setting y = 0 and 2ny = π in the above equation respectively, we obtain n−1 θ3 (nz + mπτ | nτ ) θ1 (0 | τ )θ3 (z | τ )θ1 (nz | nτ ) = , e2miz θ3 (0 | τ )θ1 (0 | nτ )θ1 (z | τ ) θ3 (mπτ | nτ ) m=0 n−1 θ1 (0 | τ )θ4 (z | τ )θ1 (nz | nτ ) θ4 (nz + mπτ | nτ ) = . e2miz θ4 (0 | τ )θ1 (0 | nτ )θ1 (z | τ ) θ4 (mπτ | nτ ) m=0
Multiplying both sides of the equation in Proposition 1.10 by q 1/8 and then letting q → 0, we obtain the following curious trigonometric identity. Proposition 1.12. If n is an odd integer ≥1 and 2m < n, then we have n−1
(−1)k
k=0
2imkπ
e− n ne2miz = . kπ sin nz sin z + n
(1.16)
In particular, putting m = 0 in the above equation, we find that n−1 k=0
(−1)k n = , kπ sin nz sin z + n
n ∈ {1, 3, 5, . . .}.
(1.17)
In Sec. 6, we also use Theorem 1.2 to derive the following proposition. Proposition 1.13. For any integer n ≥ 1, we have kπ n−1 2imkπ θ2 z + n τ nθ (0 | τ )θ1 (0 | nτ )θ2 (nz + mπτ | nτ ) = e2miz 2 , e− n kπ θ1 (0 | τ )θ2 (mπτ | nτ )θ1 (nz | nτ ) k=0 θ1 z + τ n n−1 θ1 (0 | τ )θ2 (z | τ )θ1 (nz | nτ ) θ2 (nz + mπτ | nτ ) = . e2miz θ2 (0 | τ )θ1 (0 | nτ )θ1 (z | τ ) θ2 (mπτ | nτ ) m=0
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When m = 0, putting q = 0 in the first equation in Proposition 1.13, we can obtain the known trigonometric identity n−1 kπ cot z + = n cot nz. (1.18) n k=0
If 1 < m < n, by setting q = 0 in the first equation in Proposition 1.13, we can find that n−1 2imkπ kπ ne(2m−n)iz − n . (1.19) e cot z + = n sin nz k=0
In particular, when n is even, we can choose 2m = n to derive that n−1 kπ n k . (−1) cot z + = n sin nz
(1.20)
k=0
In Sec. 7, we use Theorem 1.1 to give a completely new derivation of the Schr¨ oter formula. In Sec. 8, we use Theorem 1.2 to derive two general formulas which including the quintuple product identity, Winquist’s identity and septuple product identity as special cases. 2. Proofs of Theorems 1.1 and 1.2 We will use the techniques of Fourier series expansions to prove Theorems 1.1 and 1.2, and we first prove Theorem 1.1. Proof of Theorem 1.1. If we use ω to denote exp( 2πi n ), then it is easy to verify that n if r ≡ 0 (mod n), r 2r r(n−1) 1 + ω + ω + ···+ ω = (2.1) 0 if r ≡ 0 (mod n). Since f (z + π | τ ) = f (z | τ ), the Fourier series expansion of f has the form f (z | τ ) =
∞
ar e2riz .
r=−∞
Substituting the Fourier series expansion of f into the left-hand side of (1.2) and interchanging the order of summation, we find that n−1 ∞ n−1 2imkπ kπ − n 2irz e f z+ a e ω k(r−m) . τ = r n r=−∞ k=0
k=0
Using (2.1) in the right-hand side of the above equation, we conclude that n−1 ∞ 2imkπ kπ − n 2imz e f z+ am+nr e2inrz . τ = ne n r=−∞ k=0
(2.2)
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Next replacing z by z + πτ in the Fourier series expansion of f and then making use of the functional equation f (z + πτ | τ ) = q −n/2 e−2inz f (z | τ ), we deduce that ∞
ar q r e2riz = q −n/2
r=−∞
∞
an+r e2riz .
r=−∞ 2
It follows that an+r = ar q (n+2r)/2 , which easily yields anr+m = am q (nr +2rm)/2 . Substituting this into (2.2) we obtain (1.2). Now we begin to prove (1.3). Replacing m by αm in (1.2) and then summing the resulting equation over 0 ≤ m ≤ n − 1, we have n−1 n−1 n−1 kπ −αmk τ f z+ ω = n aαm e2αmiz θ3 (nz + αmπτ | nτ ). n m=0 m=0 k=0
Using (1.1), we immediately find that the left-hand side of the above equation equals nf (z | τ ). Thus we arrive at (1.3). Conversely, we can show that (1.4) (of course (1.3)) implies (1.2). Replacing z −lk , we obtain by z + kπ n in (1.4) and then multiplying the resulting equation by ω n−1 kπ τ = ω −lk f z + am e2miz ω (m−l)k θ3 (nz + mπτ | nτ ). n m=0 Summing both sides of the above equation over 0 ≤ k ≤ n − 1 and using (1.1), we obtain n−1 kπ −lk ω f z+ τ = nal e2liz θ3 (nz + lπτ | nτ ), n l=0
which is (1.2) after replacing l by m. This completes the proof of Theorem 1.1. Next we prove Theorem 1.2, and the proof is a bit more complicated than the proof of Theorem 1.1. Proof of Theorem 1.2. It is obvious that g(z | τ ) = einz f (z | τ ) satisfies the functional equation g(z + π | τ ) = g(z | τ ), so we can assume that g(z | τ ) = ∞ 2irz . It follows that r=−∞ cr e f (z | τ ) = e−niz
∞
cr e2irz .
r=−∞
Substituting this into the left-hand side of (1.6), interchanging the order of summation, and using (1.1), we find that n−1 ∞ kπ 2imkπ (2m−n)iz τ = ne (−1)k e− n f z + cm+nr e2inrz . (2.3) n r=−∞ k=0
Now if we replace z by z + πτ in the Fourier series expansion of f and then making use of the functional equation f (z + πτ | τ ) = −q −n/2 e−2inz f (z | τ ), we can
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deduce that ∞
cr q r e2riz = −
r=−∞
∞
cn+r e2riz .
r=−∞
r
It follows that cn+r = −cr q , which easily yields cnr+m = (−1)r cm q (nr(r−1)+2rm)/2 . Hence we find the right-hand side of (2.3) equals ∞ −ncm e2miz (−1)r e(2r+1)niz q nr(r+1)+2(r+1)m r=−∞
= −ncm q (4m−n)/8 e2miz θ1 (nz + mπτ | nτ ). Replacing the right-hand side of (2.3) by the right-hand side of the above equation, and then setting am = −ncm q (4m−n)/8 , we obtain (1.6). Using the same method as in the proof of Theorem 1.1, we can show (1.6) and (1.7) are equivalent. Thus we complete the proof of Theorem 1.2. 3. Proof of Proposition 1.6 In this section we prove Proposition 1.6 using Theorem 1.1. Proof of Proposition 1.6. Using Proposition 1.1, it is easily verified that θ3 (z | nτ ) satisfies the functional equations (1.1). So we can take f (z | τ ) = θ3 (z | nτ ) and (1.2) becomes n−1 2mkπi kπ τ − n e θ3 z + = nam e2imz θ3 (nz + mπτ | nτ ). n n k=0
Using the series representations of θ3 in the above equation and equating the coefficients of e2imz , we obtain m2
am = q 2n .
(3.1)
Combining the above two equations, we immediately obtain the first identity in Proposition 1.6. Substituting (3.1) into (1.4), we arrive at the second identity in Proposition 1.6. This completes the proof of Proposition 1.6. Replacing z by z + (π + πτ )/2 in (1.10) and using Proposition 1.2 to simplify the resulting equation, we can obtain the following proposition. These two identities must be standard and easy to verify, but it is difficult to find them in the literature. Proposition 3.1. n−1
m
(−1) q
m=0
m2 2n
e
2miz
θ1 (nz + mπτ | nτ ) = i
n−1
τ θ1 z , n
τ m2 (−1)m q 2n e2miz θ2 (nz + mπτ | nτ ) = in θ4 z , n m=0 n−1
n is odd,
n is even.
(3.2)
(3.3)
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Letting n = 2 in (3.3) and using Proposition 1.2, we easily find that τ θ4 z = θ3 (2z | 2τ ) − θ2 (2z | 2τ ), 2 which can also be obtained by adding two identities in [12, Eq. (16), p. 476]. Proposition 3.2. Suppose that the function a(τ ) is defined as ∞ q 3n+1 q 3n+2 a(τ ) = 1 + 6 − . 1 − q 3n+1 1 − q 3n+2 n=0
(3.4)
Then we have Ramanujan’s identity [3, Entry 1(v), p. 346] a(τ ) =
η 3 (τ /3) + 3η 3 (3τ ) . η(τ )
(3.5)
Proof. Using Proposition 1.1 and a direct computation, we easily find that θ1 (3z + 2πτ | 3τ ) = −q −1/2 e−6iz θ1 (3z − πτ | 3τ ). Taking n = 3 in (3.2) and using the above equation, we deduce that τ 1/6 2iz 1/6 −2iz q e θ1 (3z + πτ | 3τ ) + q e θ1 (3z − πτ | 3τ ) = θ1 z + θ1 (3z | 3τ ). 3 (3.6) Differentiating both sides of the above equation with respect to z and setting z = 0, we arrive at τ θ (πτ | 3τ ) q 1/6 θ1 (πτ | 3τ ) 2i + 3 1 + 3η 3 (3τ ). (3.7) = η3 θ1 (πτ | 3τ ) 3 By doing logarithmic differentiation of the infinite product representation of θ1 with respect to z, we find that ∞ ∞ θ1 (z | τ ) q n e2iz q n e−2iz = −i − 2i + 2i . θ1 (z | τ ) 1 − q n e2iz 1 − q n e−2iz n=0 n=0
(3.8)
Replacing τ by 3τ in the above equation, putting z = πτ and simplifying, we deduce that (see also [21, 25]) 1 θ1 (πτ | 3τ ) = − (2 + a(τ ))i. θ1 (πτ | 3τ ) 3
(3.9)
Using the infinite product representation of θ1 , we can find that θ1 (πτ | 3τ ) = iq −1/8 E(τ ).
(3.10)
Substituting the above two equation into (3.7), we immediately arrive at (3.5). This completes the proof of Proposition 3.2.
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For brevity, we introduce the multiple q-shifted factorials as follows: (a1 , a2 , . . . , am ; q)∞ =
∞
(1 − a1 q n )(1 − a2 q n ) · · · (1 − am q n ).
(3.11)
n=0
Proposition 3.3. 3
5η (5τ ) − η
3
τ 5
∞ q 5n+1 q 5n+4 =q (q, q , q ; q )∞ 3 + 10 − 1 − q 5n+1 1 − q 5n+4 n=0
∞ q 5n+2 q 5n+3 1/40 2 3 5 5 (q , q , q ; q )∞ 1 + 10 − . −q 1 − q 5n+2 1 − q 5n+3 n=0 9/40
4
5
5
This identity may not be new, but we cannot find it in the literature before. Proof of Proposition 3.3. Taking n = 5 in (3.2) and using Proposition 1.1 to simplify the resulting equation to obtain τ
θ1 z − θ1 (5z | 5τ ) = −q 1/10 e2iz θ1 (5z + πτ | 5τ ) + e−2iz θ1 (5z − πτ | 5τ ) 5
+ q 2/5 e4iz θ1 (5z + 2πτ | 5τ ) + e−4iz θ1 (5z − 2πτ | 5τ ) . Differentiating both sides of the above equation with respect to z and setting z = 0 we obtain τ θ (πτ | 5τ ) − 5η 3 (5τ ) = −q 1/10 θ1 (πτ | 5τ ) 2i + 5 1 η3 5 θ1 (πτ | 5τ ) θ (2πτ | 5τ ) + q 2/5 θ1 (2πτ | 5τ ) 4i + 5 1 . (3.12) θ1 (2πτ | 5τ ) Using the infinite product representation of θ1 , we easily find that θ1 (πτ | 5τ ) = iq 1/8 (q, q 4 , q 5 ; q 5 )∞ , θ1 (2πτ | 5τ ) = iq −3/8 (q 2 , q 3 , q 5 ; q 5 )∞ . Using (3.8) and some elementary calculations, we deduce that
∞ q 5n+1 q 5n+4 θ1 (πτ | 5τ ) = −i 3 + 10 − , 2i + 5 θ1 (πτ | 5τ ) 1 − q 5n+1 1 − q 5n+4 n=0
∞ q 5n+2 q 5n+3 θ1 (2πτ | 5τ ) . = −i 1 + 10 − 4i + 5 θ1 (2πτ | 5τ ) 1 − q 5n+2 1 − q 5n+3 n=0 Substituting (3.13) and (3.14) into (3.12), we complete the Proposition 3.3.
(3.13)
(3.14)
proof of
It is obvious that using the same method we can find the identities similar to Proposition 3.3 for n = 7, 9, 11, . . . .
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4. Proof of Proposition 1.7 In this section we use Theorem 1.1 to prove Proposition 1.7, and some interesting special cases are given. Proof of Proposition 1.7. Using Proposition 1.1, it is easy to verify that the function θ3 (z + x | τ /n)θ3 (z − x | τ /n) satisfies the functional equations (1.1) when n is replaced by 2n. Thus (1.2) becomes 2n−1
e
k=0
− imkπ n
τ kπ τ kπ − x θ3 z + x + θ3 z + 2n n 2n n
= 2nam e2miz θ3 (2nz + mπτ | 2nτ ).
(4.1)
Substituting the series representation of θ3 into both sides of the above equation, and equating the coefficients of e2miz , we deduce that m2 mπτ 2τ am = q 2n e2mix θ3 2x + . (4.2) n n Substituting this equation into (4.1) and then setting m = 0, we obtain (1.12). Substituting (4.2) into (1.4), we deduce that θ3 (z + x | τ /n)θ3 (z − x | τ /n) =
m2 mπτ 2τ q 2n e2mi(z+x) θ3 (2nz + mπτ | 2nτ )θ3 2x + . n n m=0
2n−1
Replacing τ by nτ in the above equation, we obtain (1.13). We finish the proof of Proposition 1.7. When n = 1, (1.12) and (1.13) immediately reduce respectively to the following two known identities (see, for example, [24, Eqs. (1.5) and (1.4)]): 2θ3 (2x | 2τ )θ3 (2z | 2τ ) = θ3 (z + x | τ )θ3 (z − x | τ ) + θ4 (z + x | τ )θ4 (z − x | τ ), (4.3) θ3 (z + x | τ )θ3 (z − x | τ ) = θ3 (2z | 2τ )θ3 (2x | 2τ ) + θ2 (2z | 2τ )θ2 (2x | 2τ ). Proposition 4.1.
π τ π τ 2θ3 (2x | τ )θ3 (2z | τ ) + θ3 z + x + θ3 z − x + 4 2 4 2 πτ πτ + θ3 z + x − θ3 z − x − = 4θ3 (2x | τ )θ3 (4z | 4τ ). 4 2 4 2
(4.4)
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1989
Proof. Taking n = 2 in (1.12) and simplifying, we easily deduce that τ τ 4θ3 (2x | τ )θ3 (4z | 4τ ) = θk z + x θk z − x 2 2 k∈{3,4}
+
k∈{1,−1}
kπ τ kπ τ z − x + θ3 z + x + θ . 3 4 2 4 2
Using (4.3), we find that the first summation on the right-hand side of the above equation equals 2θ3 (2x | τ )θ3 (2z | τ ). This completes the proof of Proposition 4.1. Proposition 4.2. φ2 (q) + φ2 (−q 2 ) = 2φ(q)φ(q 4 ). Proof. Putting z = x = 0 in Proposition 4.1, we immediately find that π τ θ32 (0 | τ ) + θ32 = 2θ3 (0 | τ )θ3 (0 | 4τ ). 42 Setting z = π/4 in (1.11), we deduce that θ3 π4 τ2 = θ4 (0 | 2τ ). It follows that θ32 (0 | τ ) + θ42 (0 | 2τ ) = 2θ3 (0 | τ )θ3 (0 | 4τ ). Replacing τ by 2τ and using the obvious fact φ(q) = θ3 (0 | 2τ ) and φ(−q) = θ4 (0 | 2τ ), we complete the proof of Proposition 4.2. 5. Proof of Proposition 1.8 To prove Proposition 1.8, we need Theorem 1.1. Proof of Proposition 1.8. Using Proposition 1.1, we can verify that the function θ3n (z | τ ) satisfies the functional equations (1.1). Thus we can choose f (z | τ ) = θ3n (z | τ ), and (1.2) becomes n−1
2imkπ e− n θ3n
k=0
kπ z+ τ = nam e2miz θ3 (nz + mπτ | nτ ). n
(5.1)
Substituting the series representation of θ3 into the left-hand side of the above equation, we obtain n−1 k=0
e
− 2imkπ n
∞
q
2 2 l2 1 +l2 +···+ln 2
l1 ,l2 ,...,ln =−∞
= nam e2miz θ3 (nz + mπτ | nτ ).
e2iz(l1 +l2 +···+ln ) e
2ik(l1 +l2 +···+ln )π n
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1990
Equating the coefficients of e2miz on both sides of the above equation, we find that ∞
am = Gm,n (τ ) =
q
2 2 l2 1 +l2 +···+ln 2
.
(5.2)
l1 ,l2 ,...,ln =−∞ l1 +l2 +···+ln =m
Substituting this into (5.1), we obtain (1.14) in Proposition 1.8. Using this value of am in (1.4), we obtain (1.15). This completes the proof of Proposition 1.8. Replacing z by z + (π + πτ )/2 in (1.14) and using Proposition 1.2 to simplify the resulting equation, we conclude the following. Proposition 5.1. If n is odd, we have n−1
(−1)k e−
k=0
2imkπ n
kπ θ1n z + τ n
= i1−n (−1)m e2imz Gm,n (τ )θ1 (nz + mπτ | nτ ), and if n is even, we have n−1
(−1)k e−
2imkπ n
k=0
kπ τ = i−n (−1)m e2imz Gm,n (τ )θ2 (nz + mπτ | nτ ). θ1n z + n
Replacing z by z + (π + πτ )/2 in (1.15) and using Proposition 1.2 to simplify the resulting equation, we obtain the following proposition. Proposition 5.2. If n is odd, we have n−1
(−1)m Gm,n (τ )e2imz θ1 (nz + mπτ | nτ ) = in−1 θ1n (z | τ ),
m=0
and if n is even, we have n−1
(−1)m Gm,n (τ )e2imz θ2 (nz + mπτ | nτ ) = in θ1n (z | τ ).
m=0
If we set l1 = l, l2 = n and l3 = −l − n in G0,3 (τ ), we can find that G0,3 (τ ) =
∞
ql
2
+ln+n2
(5.3)
l,n=−∞
which is the same as a(q) in [5]. Using the variable change l1 = −l, l2 = −n and l3 = 1 − l − n, we deduce that G1,3 (τ ) = q 1/2
∞ l,n=−∞
ql
2
+ln+n2 +n+l
,
(5.4)
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which is essentially the same as c(q) in [5]. With the help of the variable change l1 = l + 1, l2 = n + 1 and l3 = −l − n, we obtain ∞
G2,3 (τ ) = q
ql
2
+ln+n2 +n+l
= q 1/2 G1,3 (τ ).
(5.5)
l,n=−∞
Taking n = 3 in (1.15) and in Proposition 5.2 respectively, and simplifying, we obtain the following proposition. Identity in (5.6) can be found in [26, Proposition 3.1; 30, Eq. (4.20)]. Proposition 5.3. If G0,3 (τ ) and G1,3 (τ ) are given by (5.3) and (5.4), then we have θ33 (z | τ ) = G0,3 (τ )θ3 (3z | 3τ ) + G1,3 (τ )(e2iz θ3 (3z + πτ | 3τ ) + e−2iz θ3 (3z − πτ | 3τ )), θ13 (z | τ ) + G0,3 (τ )θ1 (3z | 3τ )
= G1,3 (τ ) e2iz θ1 (3z + πτ | 3τ ) + e−2iz θ1 (3z − πτ | 3τ ) .
(5.6)
(5.7)
Next we introduce a multiple theta function which is an extension of Gm,n (τ ). Definition 5.1. If y1 + y2 + · · · + yn = 0, we define Gm,n (y1 , y2 , . . . , yn | τ ) as ∞
q
2 2 l2 1 +l2 +···+ln 2
e2i(l1 y1 +l2 y2 +···+ln yn ) .
l1 ,l2 ,...,ln =−∞ l1 +l2 +···+ln =m
It is obvious that when y1 = y2 = · · · = yn = 0, Gm,n (y1 , y2 , . . . , yn | τ ) becomes Gm,n (τ ). We end this section with the following two formulas with many parameters, which will reduce to Proposition 1.8 when y1 = y2 = · · · = yn = 0. One may compare these with the main results in [9]. Proposition 5.4. If y1 + y2 + · · · + yn = 0 and Gm,n are defined as in Definition 5.1, then we have n−1 k=0
e
− 2imkπ n
kπ τ θ3 z + y j + n j=1 n
= nGm,n (y1 , y2 , . . . , yn | τ )e2miz θ3 (nz + mπτ | nτ ), n j=1
θ3 (z + yj | τ ) =
n−1
Gm,n (y1 , y2 , . . . , yn | τ )e2miz θ3 (nz + mπτ | nτ ).
m=0
Applying the imaginary transformation to the case m = 0 of the first identity in the above proposition, we obtain the following proposition.
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Proposition 5.5. Suppose that y1 , y2 , . . . , yn are n complex numbers such that y1 + y2 + · · · + yn = 0. Then we have mn−1
q
k2 /2 2kiz
e
n
θ3 (mz + (yj + km)πτ | m2 nτ )
j=1
k=0
= Fm,n (y1 , y2 , . . . , yn )θ3 (z | τ ), where Fm,n is equal to Fm,n =
mn−1
1
q 2m
2
2 n(r12 +r22 +···+rn )−(r1 y1 +r2 y2 +···+rn yn )− 12 k2
.
k=0 m(r1 +r2 +···+rn )=k
There are some misprints in Fm,n in [9], we have here corrected these misprints. See also [28, 30]. 6. Proofs of Propositions 1.10, 1.11 and 1.13 We begin this section by proving Propositions 1.10 and 1.11 using Theorem 1.1. Proof of Propositions 1.10 and 1.11. If n is odd, then for any y, we can easily verify that the function θ1 (nz − ny | nτ )
θ3 (z + (n − 1)y | τ ) θ1 (z − y | τ )
satisfies the functional equations (1.1). Replacing f (z | τ ) by this function in (1.2) and using the relation θ1 (nz − ny + kπ | nτ ) = (−1)k θ1 (nz − ny | nτ ) in the resulting equation, we find that n−1
(−1)k e−
2imkπ n
k=0
θ1 (nz − ny | nτ )
θ3 (z + (n − 1)y + kπ n | τ )) kπ θ1 (z − y + n | τ )
= nam e2miz θ3 (nz + mπτ | nτ ).
(6.1)
Now we begin to determine the value of am . Letting z → y on both sides of the above equation, we immediately find the right-hand side of the above equation becomes nam e2miy θ3 (ny + mπτ | nτ ),
(6.2)
and at the same time we find that the left-hand side of (6.1) reduces to nθ1 (0 | nτ )θ3 (ny | τ ) , θ1 (0 | τ ) by using the visible fact θ1 (0 | τ ) = 0 and the following limit value: lim
z→y
nθ (0 | nτ ) θ1 (nz − ny | nτ ) = 1 . θ1 (z − y | τ ) θ1 (0 | τ )
(6.3)
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Inverse Relations and Theta Function Identities
1993
If we equate (6.2) and (6.3), we immediately find the value of am given by am =
θ1 (0 | nτ )θ3 (ny | τ )e−2miy . θ1 (0 | τ )θ3 (ny + mπτ | nτ )
(6.4)
Substituting this into (6.1) and replacing z − y by z, we complete the proof of Proposition 1.10. If we substitute (6.4) into (1.4), we can obtain the dual identity of (6.1): θ3 (z + (n − 1)y | τ )θ1 (nz − ny | nτ )θ1 (0 | τ ) θ1 (0 | nτ )θ3 (ny | τ )θ1 (z − y | τ ) =
n−1 m=0
e2mi(z−y)
θ3 (nz + mπτ | nτ ) . θ3 (ny + mπτ | nτ )
Replacing z−y by z in the above equation, we complete the proof of Proposition 1.11. Now we use Theorem 1.2 to prove Proposition 1.13. Proof of Proposition 1.13. For any integer n, we can easily verify that the function θ1 (z + (n − 1)y | τ ) θ1 (nz − ny | nτ ) θ1 (z − y | τ ) satisfies the functional equations (1.5). If we replace f (z | τ ) by this function in Theorem 1.2, we find that (1.6) becomes kπ τ θ1 z + (n − 1)y + n−1 2imkπ n e− n θ1 (nz − ny | nτ ) kπ k=0 τ θ1 z − y + n = nam e2miz θ1 (nz + mπτ | nτ ).
(6.5)
Letting z → y in the above equation and using the same method as in the proof of Proposition 1.10, we deduce that am =
θ1 (0 | nτ )θ1 (ny | τ )e−2miy . θ1 (0 | τ )θ1 (ny + mπτ | nτ )
(6.6)
Substituting this equation into (6.5) and writing z − y as z, we conclude that kπ | τ θ z + ny + n−1 2imkπ 1 n e− n kπ k=0 |τ θ1 z + n = e2miz
nθ1 (ny | τ )θ1 (0 | nτ )θ1 (nz + ny + mπτ | nτ ) . θ1 (0 | τ )θ1 (ny + mπτ | nτ )θ1 (nz | nτ )
(6.7)
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We now substitute (6.6) into (1.8) and replace z − y by z, we deduce that θ1 (z + ny | τ )θ1 (nz | nτ )θ1 (0 | τ ) θ1 (0 | nτ )θ1 (ny | τ )θ1 (z | τ ) =
n−1
e2miz
m=0
θ1 (nz + ny + mπτ | nτ ) . θ1 (ny + mπτ | nτ )
(6.8)
Setting 2ny = π in (6.7) and (6.8), we can obtain the two identities in Proposition 1.13. We complete the proof of Proposition 1.13. 7. One Schr¨ oter’s Formula and Its Dual Form There are several Schr¨ oter formulas for theta functions. One of them (see, for example, [12, p. 470]) can be stated in the following proposition. Proposition 7.1. Let α and β be two positive integers with (α, β) = 1, and n = α2 + β 2 . Then we have θ3 (αz + βv | τ )θ3 (βz − αv | τ ) =
n−1
qm
2
/2 2mi(αz+βv)
e
θ3 (nz + mαπτ | nτ )θ3 (nv + mβπτ | nτ ).
m=0
The dual form of the above Schr¨ oter formula was first derived by Liu and Yang [24] using the imaginary transformation for θ3 . Proposition 7.2. Let α and β be two positive integers with (α, β) = 1, and n = α2 + β 2 . Then we have nθ3 (nz | nτ )θ3 (nv | nτ ) =
n−1 k=0
αkπτ βkπτ τ θ τ . θ3 αz + βv + βz − αv + 3 n n
In this section we use Theorem 1.1 to prove the above two propositions. Proof of Propositions 7.1 and 7.2. Since n = α2 + β 2 , with Proposition 1.1, it is easy to verify that the function θ3 (αz + βv | τ )θ3 (βz − αv | τ ) satisfies (1.1). If we replace f (z | τ ) with this function, then (1.2) becomes nam e2imz θ3 (nz + mπτ | nτ ) n−1 2imkπ αkπτ βkπτ − n e θ3 αz + βv + = τ θ3 βz − αv + τ . n n k=0
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Inverse Relations and Theta Function Identities
1995
On replacing m by mα in the above equation, we find that namα e2imαz θ3 (nz + mαπτ | nτ ) n−1 2imαkπ akπ βkπ − n = e θ3 αz + βv + τ θ3 βz − αv + τ . n n
(7.1)
k=0
Substituting the series representation of θ3 into the above equation, we deduce that ∞
namα e2imαz
q
nl2 2
+lmα 2ilnz
e
l=−∞
=
n−1
e−
2imαkπ n
k=0
∞
q
2 l2 1 +l2 2
e2iz(αl1 +βl2 ) e2vi(βl1 −αl2 ) e
2ik(αl1 +βl2 )π n
.
l1 ,l2 =−∞
Now we equate the coefficients of e2imαz on both sides of the above equation to obtain ∞ 2 l2 1 +l2 q 2 e2vi(βl1 −αl2 ) . amα = l1 ,l2 =−∞ αl1 +βl2 =mα
Since (α, β) = 1, we can make the variable change l1 = m + βl and l2 = −αl to obtain ∞ 2 nl2 amα = q m /2 e2mβvi q 2 +mβl e2nlvi l=−∞
= qm
2
/2 2imβv
e
θ3 (nv + mβπτ | nτ ).
(7.2)
Now we substitute (7.2) into (7.1) to obtain the following general formula: nq m
2
/2 2im(αz+βv)
e
θ3 (nz + mαπτ | nτ )θ3 (nv + mβπτ | nτ ) n−1 2imαkπ akπ βkπ τ θ τ . e− n θ3 αz + βv + = βz − αv + 3 n n
(7.3)
k=0
When m = 0, the above identity reduces to Proposition 7.2. Substituting (7.2) into (1.3), we obtain Proposition 7.1. 8. The Quintuple Product Identity, Winquist’s Identity and the Septuple Product Identity We begin this section by introducing two kinds of theta functions. Definition 8.1. For any integers m and n (≥ 1), we define theta functions Hn,m (z | τ ) and Wn,m (z | τ ) as Hn,m (z | τ ) = e2imz θ1 (nz + mπτ | nτ ) − e−2imz θ1 (nz − mπτ | nτ ), Wn,m (z | τ ) = e2imz θ1 (nz + mπτ | nτ ) + e−2imz θ1 (nz − mπτ | nτ ).
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The main aim of this section is proving the following two expansion formulas using the above theta functions. Theorem 8.1. Suppose that n ≥ 1 is an odd integer and f (z | τ ) is even entire function of z satisfying the functional equations f (z | τ ) = −f (z + π | τ ) = −q n/2 e2inz f (z + πτ | τ ).
(8.1)
Then there exists a constant sequence cm independent of z such that n−1
f (z | τ ) =
2
cm Hn,m (z | τ ).
(8.2)
m=1
The dual identity of the above identity states n−1 kπ k τ = 0. (−1) f z + n
(8.3)
k=0
Theorem 8.2. Suppose that n ≥ 1 is an odd integer and f (z | τ ) is odd entire function of z satisfying the functional equations f (z | τ ) = −f (z + π | τ ) = −q n/2 e2inz f (z + πτ | τ ).
(8.4)
Then there exists a constant sequence cm independent of z such that n−1
2
f (z | τ ) = c0 θ1 (nz | nτ ) +
cm Wn,m (z | τ ).
(8.5)
m=1
The dual identity of the above identity states n−1 kπ k τ = nc0 θ1 (nz | nτ ). (−1) f z + n
(8.6)
k=0
We only prove Theorem 8.1. The proof of Theorem 8.2 is similar to the proof of Theorem 8.1, so is omitted. Proof of Theorem 8.1. Let f (z | τ ) be the given function in Theorem 8.1. Since f (z | τ ) satisfies the conditions of Theorem 1.2, there exists a constant sequence am such that f (z | τ ) = a0 θ1 (nz | nτ ) +
n−1
am e2miz θ1 (nz + mπτ | nτ ).
m=1
Replacing z by −z in the above equation and noting that f (−z | τ ) = f (z | τ ), we deduce that f (z | τ ) = −a0 θ1 (nz | nτ ) −
n−1 m=1
am e−2miz θ1 (nz − mπτ | nτ ).
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Adding the above two equations together and using the definition of Hn,m (z | τ ), we find that n−1 2f (z | τ ) = am Hn,m (z | τ ). m=1
We split the summation on the right-hand side into two parts and obtain n−1
2f (z | τ ) =
2
n−1
am Hn,m (z | τ ) +
m=1
am Hn,m (z | τ ).
m= n+1 2
Making the variable change m → n − m in the second summation, we find that n−1
2f (z | τ ) =
2
n−1
am Hn,m (z | τ ) +
m=1
2
an−m Hn−m,m (z | τ ).
m=1
Using the functional equation θ1 (z + πτ | τ ) = −q −1/2 e−2iz θ1 (z | τ ) and a direct computation, we find that Hn−m,m (z | τ ) = −q m−n/2 Hn,m (z | τ ). Combining the above two equations and taking 2cm = am −q m−n/2 an−m , we deduce that n−1
f (z | τ ) =
2
cm Hn,m (z | τ ),
m=1
which is (8.2). Now we turn to prove (8.3). Replacing z by z + kπ/n in (8.2), we obtain n−1 2 2imkπ kπ k τ = (−1) f z + e n e2imz θ1 (nz + mπτ | nτ ) n m=0 n−1
+
2
e
−2imkπ n
e−2imz θ1 (nz − mπτ | nτ ).
m=0
Summing the above equation over 0 ≤ k ≤ n − 1, we find that n−1 n−1 n−1 2 2imkπ kπ τ = (−1)k f z + e n e2imz θ1 (nz + mπτ | nτ ) n m=0 k=0
k=0
n−1 2
+
n−1
e
−2imkπ n
e−2imz θ1 (nz − mπτ | nτ ).
m=0 k=0
Let δm,n be Kronecker delta function. Since 0 ≤ m ≤ n−1 2 , using (2.1), we easily find that n−1 n−1 2imkπ −2imkπ e n = e n = δm,0 . k=0
k=0
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If the above two equations are combined, then we are led to the identity n−1 kπ k τ = nθ1 (nz | nτ ) − nθ1 (nz | nτ ) = 0, (−1) f z + n k=0
which is (8.3), thus we complete the proof of Theorem 8.1. Now we use Theorem 8.1 to derive the quintuple product identity and its dual identity in the following two propositions. Proposition 8.1 (The quintuple product identity). ∞
(−1)n q n(3n+1)/2 cos(2n + 1)z = E(τ )
n=−∞
θ1 (2z | τ ) . θ1 (z | τ )
This beautiful identity has a very rich history and there are many different proofs of this identity in the literature. For example, [1, 2, 7, 13, 16, 20, 22]. One may consult [3, p. 83] and the survey paper [11] for the various proofs of this identity. An important extension of this identity with many applications has been given in [23]. Proof of Proposition 8.1. Using the first equation in Proposition 1.1, we easily verify that the function θ1 (2z | τ ) θ1 (z | τ ) satisfies the case n = 3 of (8.1) and it is an even function of z. Hence in Theorem 8.1 we can replace f by this function and identity (8.2) becomes θ1 (2z | τ ) = c1 H3,1 (z | τ ) = c1 (e2iz θ1 (3z + πτ | 3τ ) − e−2iz θ1 (3z − πτ | 3τ )). θ1 (z | τ ) Putting z = π/3 in the above equation, we find that c1 = θ1−1 (πτ | 3τ ). It follows that θ1 (2z | τ ) θ1 (πτ | 3τ ) = e2iz θ1 (3z + πτ | 3τ ) − e−2iz θ1 (3z − πτ | 3τ ). θ1 (z | τ ) Substituting θ1 (πτ | 3τ ) = iq −1/8 E(τ ) into the left-hand side of the above equation, and the series representation of θ1 into the right-hand side, we obtain Proposition 8.1. Using (8.3), we immediately obtain the following proposition. Proposition 8.2 (The dual identity of the quintuple product identity). 2π 4π τ τ θ θ1 2z + 2z + 1 θ1 (2z | τ ) 3 3 = 0. + − π 2π θ1 (z | τ ) τ θ1 z + τ θ1 z + 3 3
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Inverse Relations and Theta Function Identities
1999
Putting q = 0 in the above equation, we immediately obtain the nontrivial trigonometric identity 2π 4π sin 2z + sin 2z + sin 2z 3 3 = 0. − π + 2π sin z sin z + sin z + 3 3 Proposition 8.3. If Hn,m (z | τ ) is defined as in Definition 8.1, then we have the Hirschhorn septuple product identity θ1 (z | τ )θ1 (2z | τ ) = q 1/2 θ1 (2πτ | 5τ )H5,1 (z | τ ) − q 1/2 θ1 (πτ | 5τ )H5,2 (z | τ ). The identity was derived by Hirschhorn [15] in 1983, which was subsequently rediscovered by Farkas and Kra [14] in 1999. Other proofs can be found in [10, 18]. Proof of Proposition 8.3. Using the first equation in Proposition 1.1, we easily verify that the even function θ1 (z | τ )θ1 (2z | τ ) satisfies the case n = 5 of (8.1). Hence in Theorem 8.1 we can replace f by this function and identity (8.2) becomes θ1 (z | τ )θ1 (2z | τ ) =
2
ck H5,k (z | τ ).
(8.7)
k=1
Setting z = 0 in the above equation, we immediately find that c1 θ1 (πτ | 5τ ) + c2 θ1 (2πτ | 5τ ) = 0.
(8.8)
Appealing to Jacobi’s triple product identity and a direct computation, we easily find that √ √ π 2π τ θ1 τ = − 5q 1/2 θ1 (πτ | 5τ )θ1 (2πτ | 5τ ) = 5q 1/4 E(τ )E(5τ ). θ1 5 5 Taking z =
π 5
in (8.7) and then using the above equation, we deduce that 2π 4π 2c1 cos θ1 (πτ | 5τ ) + 2c2 cos θ1 (2πτ | 5τ ) 5 5 √ = 5q 1/2 θ1 (πτ | 5τ )θ1 (2πτ | 5τ ).
From (8.8) and (8.9), we can find that c1 = q 1/2 θ1 (2πτ | 5τ ),
c2 = −q 1/2 θ1 (πτ | 5τ ).
Substituting these into (8.7), we obtain the septuple product identity.
(8.9)
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Using (8.3), we immediately obtain the following proposition. Proposition 8.4 (The dual identity of the septuple product identity). 4 kπ 2kπ k τ θ1 2z + τ = 0. (−1) θ1 z + 5 5 k=0
Setting q = 0 in the above equation, we conclude that 4 kπ 2kπ (−1)k sin z + sin 2z + = 0. 5 5 k=0
Next we use Theorem 8.2 to derive Winquist’s identity and its dual identity. Proposition 8.5 (Winquist’s identity). If Wn,m (z | τ ) is defined as in Definition 8.1, and E(τ ) is the Euler series defined in Proposition 1.5, then we have q 1/4 E 2 (τ )θ1 (3y | 3τ )W3,1 (z | τ ) − q 1/4 E 2 (τ )θ1 (3z | 3τ )W3,1 (y | τ ) = θ1 (y | τ )θ1 (z | τ )θ1 (z + y | τ )θ1 (z − y | τ ). This identity was first discovered by Winquist [27]. For different proofs, see [6, 10, 17, 18, 21]. If we divide both sides of the above equation by y and then letting y → 0, we can obtain [19] π π 3 3 θ1 z + τ + θ1 z − τ − θ13 (z | τ ) = 3a(τ )θ1 (3z | 3τ ), 3 3 where a(τ ) = 1 + 6
∞ q 3n+1 q 3n+2 − . 1 − q 3n+1 1 − q 3n+2 n=0
Proof of Proposition 8.5. Using the first equation in Proposition 1.1, we easily verify that the odd function θ1 (z | τ )θ1 (z − y | τ )θ1 (z + y | τ ) satisfies the case n = 3 of (8.4). Hence in Theorem 8.2 we can replace f by this function and identity (8.5) becomes θ1 (z | τ )θ1 (z − y | τ )θ1 (z + y | τ ) = c0 θ1 (3z | 3τ ) + c1 W3,1 (z | τ ). Putting z =
π 3
(8.10)
and performing some calculations, we deduce that c1 = q 1/4 E 2 (τ )
θ1 (3y | 3τ ) . θ1 (y | τ )
(8.11)
Substituting this into (8.10) and then letting z = y in the resulting equation, we find that W3,1 (y | τ ) . (8.12) c0 = −q 1/4 E 2 (τ ) θ1 (y | τ )
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2001
If the above two equations are substituted into (8.10), we obtain Winquist’s identity. Combining this with (8.6), we obtain the following proposition. Proposition 8.6 (The dual identity of Winquist’s identity). θ1 (y | τ )θ1 (z | τ )θ1 (z + y | τ )θ1 (z − y | τ ) π π π − θ1 (y | τ )θ1 z + τ θ1 z + y + τ θ1 z − y + τ 3 3 3 2π 2π 2π + θ1 (y | τ )θ1 z + τ θ1 z + y + τ θ1 z − y + τ 3 3 3 = −3q 1/4 E 2 (τ )θ1 (3z | 3τ )W3,1 (y | τ ). Acknowledgments The author would like to thank the referee for reading the manuscript very carefully and giving many invaluable suggestions that greatly helped to improve the paper. This work was supported by the National Science Foundation of China. References [1] G. E. Andrews, Applications of basic hypergeometric functions, SIAM Rev. 16 (1974) 441–484. [2] W. N. Bailey, An expression for ϑ1 (nz)/ϑ1 (z), Proc. Amer. Math. Soc. 4 (1953) 569–572. [3] B. C. Berndt, Ramanujan’s Notebooks, Part III (Springer-Verlag, New York, 1991). [4] M. Boon, M. L. Glasser, J. Zak and J. Zucker, Additive decompositions of θ functions of multiple arguments, J. Phys. A: Math. Gen. 15 (1982) 3439–3440. [5] J. M. Borwein, P. B. Borwein and F. G. Garvan, Some cubic modular identities of Ramanujan, Trans. Amer. Math. Soc. 343 (1994) 35–47. [6] L. Carlitz and M. V. Subbarao, On a combinatorial identity of Winquist and its generalization, Duke Math. J. 39 (1972) 165–172. [7] H.-C. Chan, Another simple proof of the quintuple product identity, Int. J. Math. Math. Sci. 2005 (2005) 2511–2515. [8] H. H. Chan, Z.-G. Liu and S. T. Ng, Circular summation of theta functions in Ramanujan’s Lost Notebook, J. Math. Anal. Appl. 316 (2006) 628–641. [9] S. H. Chan and Z.-G. Liu, On a new circular summation of theta functions, J. Number Theory 130 (2010) 1190–1196. [10] W. Chu and Q. Yan, Verification method for theta function identities via Liouville’s theorem, Arch. Math. (Basel ) 90 (2008) 331–340. [11] S. Cooper, The quintuple product identity, Int. J. Number Theory 2 (2006) 115–161. [12] A. Enneper, Elliptische Functionen: Theorie und Geschichte (Louis Nebert, Halle, 1890). [13] R. J. Evans, Theta function identities, J. Math. Anal. Appl. 147 (1990) 97–121. [14] H. M. Farkas and I. Kra, On the quintuple product identity, Proc. Amer. Math. Soc. 127 (1999) 771–778.
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