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Truncated Jacobi triple product series
Journal of Combinatorial Theory, Series A 166 (2019) 382–392
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Journal of Combinatorial Theory, Series A www.elsevier.com/locate/jcta
Truncated Jacobi triple product series Chun Wang a , Ae Ja Yee b a
Department of Mathematics, East China Normal University, 500 Dongchuan Road, Shanghai 200241, PR China b Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, USA
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 May 2018 Available online xxxx
In their study of truncated theta series, Andrews-Merca and Guo-Zeng made a conjecture on Jacobi’s triple product identity, which is settled by Mao and Yee independently. In this paper, we reprove this ex-conjecture by providing an explicit series form with nonnegative coefficients, which is reminiscent of the results of Andrews-Merca and Guo-Zeng. We also provide a companion theorem to this ex-conjecture. © 2019 Elsevier Inc. All rights reserved.
MSC: primary 11P81 secondary 05A17 Keywords: Theta series Truncated series Jacobi’s triple product identity
1. Introduction A partition of a positive integer n is a weakly decreasing sequence of positive integers whose sum equals n. Let p(n) be the number of partitions of n for n ≥ 1 with p(0) = 1. Then, the generating function for p(n) is ∞
p(n)q n =
n=0
∞
1 . 1 − qn n=1
E-mail addresses: [email protected] (C. Wang), [email protected] (A.J. Yee). https://doi.org/10.1016/j.jcta.2019.03.003 0097-3165/© 2019 Elsevier Inc. All rights reserved.
C. Wang, A.J. Yee / Journal of Combinatorial Theory, Series A 166 (2019) 382–392
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The reciprocal of the generating function for p(n) has the following series expansion, which is known as Euler’s pentagonal number theorem: ∞
(−1)n q n(3n+1)/2 (1 − q 2n+1 ) =
n=0
∞
(1 − q n ).
n=1
In [6], Andrews and Merca proved the following truncated theorem on Euler’s identity: for m ≥ 1, m ∞ m−1 1 q 2 +(m+1)n n − 1 . (1) (−1)n q n(3n+1)/2 (1 − q 2n+1 ) = 1 + (−1)m−1 m−1 (q; q)∞ n=0 (q; q)n n=m
Here and throughout the paper, we adopt the following standard q-series notation:
(a; q)∞ :=
∞
(1 − aq n ) for |q| < 1,
n=0
(a; q)n :=
(a; q)∞ for any integer n. (aq n ; q)∞
The q-binomial coefficient is defined as ⎧ (q; q)n ⎨ , n n = := (q; q)k (q; q)n−k k k q ⎩ 0,
if 0 ≤ k ≤ n, otherwise.
We sometimes use the following compressed notation: (a1 , a2 , . . . , ar ; q)∞ := (a1 ; q)∞ (a2 ; q)∞ · · · (ar ; q)∞ , (a1 , a2 , . . . , ar ; q)n := (a1 ; q)n (a2 ; q)n · · · (ar ; q)n . The work of Andrews and Merca opened up a new study on truncated series as one can find a series of papers on the subject [7–12,14–16]. In particular, in [10], Guo and Zeng showed truncated theorems on the following Gauss’s theta series identities: ∞
2
(−1)n q n (1 − q 2n+1 ) =
n=0 ∞ n=0
Namely, for m ≥ 1,
(−1)n q n(2n+1) (1 − q 2n+1 ) =
(q; q)∞ , (−q; q)∞ (q 2 ; q 2 )∞ . (−q; q 2 )∞
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m ∞ (−q; q)∞ (−q; q)m (−1; q)n−m q (m+1)n n − 1 n n2 m , (−1) q = 1 + (−1) m (q; q)∞ n=−m (q; q)n n=m+1 (2) and m−1 (−q; q 2 )∞ (−1)n q n(2n+1) (1 − q 2n+1 ) (q 2 ; q 2 )∞ n=0 m−1
= 1 + (−1)
∞ (−q; q 2 )m (−q; q 2 )n−m q 2(m+1)n−m n − 1 . m − 1 q2 (q 2 ; q 2 )n n=m
(3)
At the end of their papers, Andrews and Merca [6, Question (2)], and Guo and Zeng [10, Conjecture 6.1] posed the following conjecture: for 1 ≤ S ≤ R/2, and m ≥ 1, m−1
1
(q R , q S , q R−S ; q R )∞ n=0
(−1)n q
n+1 2
R−nS
1 − q (2n+1)S
(4)
has nonnegative coefficients if m is odd and nonpositive coefficients if m is even. The truncated theorems in (1) and (3) confirm this conjecture for R = 3, S = 1, and R = 4, S = 1, respectively. It is now settled by Mao [14] and Yee [16]. The proof of Mao uses q-series manipulations while the proof of Yee is based on combinatorial argument. However, both did not give an explicit form that is analogous to the right hand sides of (1) and (3). The main purpose of this paper is to reprove this ex-conjecture by providing an explicit series form with nonnegative coefficients, which is reminiscent of the results of Andrews and Merca in (1) and Guo and Zeng in (3). Theorem 1.1. For 1 ≤ S ≤ R/2 and m ≥ 1, m−1
1
(q R , q S , q R−S ; q R )∞ n=0 = 1 + (−1)m−1 q
m 2
R
(−1)n q
∞
n+1
R−nS
2
n=m i+j+h+k=n i,j,h,k≥0
1 − q (2n+1)S
q (mj+hk)R+(h−k)S+nR n−1 . (q R ; q R )i (q R ; q R )j (q R ; q R )h (q R ; q R )k m − 1 qR
As a companion of Theorem 1.1, we obtain the following theorem. Theorem 1.2. For 1 ≤ S ≤ R/2 and m ≥ 0, 1
m
(−1)n q
(q R , q S , q R−S ; q R )∞ n=−m
n 2
R+nS
C. Wang, A.J. Yee / Journal of Combinatorial Theory, Series A 166 (2019) 382–392
m
= 1 + (−1) q
m+1 2
R
∞
n=m+1 i+j+h+k=n i,j,h,k≥0
385
q (mj+h(k−1))R+(h−k)S+nR n−1 . m qR (q R ; q R )i (q R ; q R )j (q R ; q R )h (q R ; q R )k
When R = 2, S = 1, the left hand side in Theorem 1.2 yields that of (2). We note that this theorem yields a weaker result than Theorem 1.1 as pointed out by Guo and Zeng [10, Remark 6.2] for the case when R = 2, S = 1. By taking the following Jacobi’s identity ∞
(−1)n (2n + 1)q
n+1 2
= (q; q)3∞ ,
n=0
Guo and Zeng [10, Conjecture 6.4] conjectured that for m ≥ 1, m n+1 1 n 2 (−1) (2n + 1)q (q; q)3∞ n=0
(5)
has nonnegative coefficients if m is even and nonpositive coefficients if m is odd. This conjecture was first proved by Mao [14] and then proved combinatorially by He, Ji, and Zang [11]. As a corollary of Theorem 1.1, we can also obtain an explicit series form for the truncated series in (5), which reconfirms the conjecture. In the next section, we will prove all the results stated in Introduction. 2. Proofs of Theorems 1.1 and 1.2 A pair of sequences (αn , βn )n≥0 is called a Bailey pair relative to (a, q) if αn and βn are related by the following identity [3, Sect. 3.4]: for n ≥ 0, βn =
n r=0
αr . (q; q)n−r (aq; q)n+r
We need a truncated Bailey transform. Lemma 2.1. If (αn , βn ) is a Bailey pair relative to (a, q) and m is a nonnegative integer, then we have m m (aq; q)m (−1)m q m(m+1)/2 −m (q , aq 1+m ; q)n βn (a, q) = αn (a, q). (q; q)m n=0 n=0
Proof. We recall the following q-series expansion formula of Liu [13, Theorem 9.2]: m (aq, abc/q; q)m (q −m , q/b, q/c; q)n q n An (ab, ac; q)m n=0 (q 2 /abcq m ; q)n
(6)
386
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=
n m (1 − aq 2n )(q −m , a, q/b, q/c; q)n (abcq m−1 )n −n (q , aq n ; q)k q k Ak , m+1 , ab, ac; q) (1 − a)(q, aq n n=0
(7)
k=0
where m is a nonnegative integer and {An } is an arbitrary complex sequence. Replacing b by a−1 q −m in (7) and then letting c → 0, we get m (aq; q)m (−1)m q m(m+1)/2 −m (q , aq 1+m ; q)n An (q; q)m n=0
=
=
n m (1 − aq 2n )(a; q)n (−1)n q n(n−1)/2 −n (q , aq n ; q)k q k Ak (1 − a)(q; q) n n=0 m
n 2n
1 − aq 1−a n=0
k=0
(a; q)n+k (−1)n−k q (q; q)n−k
k=0 n−k 2
Ak .
Since (αn , βn ) is a Bailey pair relative to (a, q) [2, Lemma 3] or [3, Eq. (3.40)], n 1 − aq 2n (a; q)n+j (−1)n−j q αn (a, q) = 1 − a j=0 (q; q)n−j
n−j 2
βj (a, q).
Thus, upon taking An = βn (a, q), we see that Lemma 2.1 follows immediately.
2
Remark 1. The formula of Liu in (7) can be found in [2, Eq. (3.3)]. 2.1. Proof of Theorem 1.1 Recall the following Bailey pair (αn , βn ) relative to (q, q) from [5, p. 98-99]: ⎧ ⎨αn ⎩βn
= (z −n − z n+1 )(−1)n q n(n+1)/2 , (z; q)n+1 (q/z; q)n = . (q 2 ; q)2n
(8)
Lemma 2.2. For m ≥ 0, (−1)m q m(m+1)/2
=
m
m (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 n=0
(−1)n q n(n+1)/2 (z −n − z n+1 ).
n=0
Proof. Apply Lemma 2.1 to the Bailey pair in (8). Then this lemma easily follows and we omit the details. 2
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Remark 2. Note that (−1)m q m(m+1)/2
=
m (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 n=0
m (−1)m−n (q m−n+1 ; q)n q n=0
=
m (−1)n (q n+1 ; q)m−n q n=0
=
m (−1)n q
n+1 2
n=0
m−n+1
n+1 2
(q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 2
(q 1+m ; q)m−n+1 (z; q)m−n+1 (q/z; q)m−n (q; q)2m−2n+1
(q 2m−2n+2 ; q)n (z; q)m−n+1 (q/z; q)m−n . (q; q)n
Thus, if m → ∞, we have
n+1 ∞ (−1)n q 2 (z, q/z; q)∞ = (z, q/z, q; q)∞ . (q; q)n n=0
Therefore, (q, z, q/z; q)∞ =
∞
(z −n − z n+1 )(−1)n q
n+1 2
.
n=0
Now we consider the truncated sum in the following theorem. Theorem 2.3. For m ≥ 0, m n+1 1 (z −n − z n+1 )(−1)n q 2 (q, z, q/z; q)∞ n=0 ⎛ m
= 1 + (−1) q
m+1
∞
2
n=m+1
⎜ ⎜ ⎝
i+j+h+k=n i,j,h,k≥0
⎞
⎟ n n−1 q (m+1)j+hk z h−k ⎟q . (9) m (q; q)i (q; q)j (q; q)h (q; q)k ⎠
Proof. By Lemma 2.2, we have m 1 (z −n − z n+1 )(−1)n q n(n+1)/2 (q, z, q/z; q)∞ n=0
=
m 1 (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (−1)m q m(m+1)/2 (q, z, q/z; q)∞ (q; q)2n+1 n=0
m (−1)m q m(m+1)/2 (q −m ; q)n (q; q)m+n+1 (z; q)n+1 (q/z; q)n = (q, z, q/z; q)∞ (q; q)m n=0 (q; q)2n+1
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388
=
m (−1)m q m(m+1)/2 (q −m ; q)n (q/z; q)n (q 2n+2 ; q)∞ (q/z; q)∞ (q; q)m n=0 (q; q)n (q n+1 ; q)∞ (q m+n+2 ; q)∞ (zq n+1 ; q)∞
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n (q/z; q)n q (n+1)i q (m+n+2)j (q/z; q)∞ (q; q)m n=0 (q; q)n (q; q)i j=0 (q; q)j i=0
×
∞ (z −1 q n+1 ; q)h z h q (n+1)h (q; q)h
h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+2)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ z h q (n+1)h 1 −1 n+h+1 (q; q)h (z q ; q)∞
h=0
=
m ∞ ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+2)j z h q (n+1)h (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j (q; q)h n=0 h=0
×
∞ k=0
=
z
−k (n+h+1)k
q (q; q)k
(−1)m q m(m+1)/2 (q; q)m
∞ i,j,h,k=0 ∞
m m(m+1)/2
=
(−1) q (q; q)m
i,j,h,k=0
m m(m+1)/2
=
(−1) q (q; q)m +
∞ i+j+h+k=m+1 i,j,h,k≥0
m q i+(m+2)j+h+(h+1)k z h−k (q −m ; q)n q (i+j+h+k)n (q; q)i (q; q)j (q; q)h (q; q)k n=0 (q; q)n
q i+(m+2)j+h+(h+1)k z h−k i+j+h+k−m (q ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
(q −m ; q)m
q i+(m+2)j+h+(h+1)k z h−k i+j+k+h−m (q ; q)m (q; q)i (q; q)j (q; q)h (q; q)k ∞
= 1 + (−1)m q m(m+1)/2
i+j+h+k=m+1 i,j,h,k≥0
q i+(m+2)j+h+(h+1)k z h−k i + j + h + k − 1 m (q; q)i (q; q)j (q; q)h (q; q)k
⎛
m m(m+1)/2
= 1 + (−1) q
∞ n=m+1
⎜ ⎜ ⎝
i+j+h+k=n i,j,h,k≥0
⎞
⎟ n n−1 q (m+1)j+hk z h−k ⎟q m (q; q)i (q; q)j (q; q)h (q; q)k ⎠
as desired. Here the fourth equality follows from [1, p. 17, Eq. (2.2.1)] and the third last equality follows from (q k−m ; q)m = 0 if 0 < k ≤ m. 2 With q and z replaced by q R , q S in Theorem 2.3, respectively, Theorem 1.1 follows easily.
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Corollary 2.4. For m ≥ 0, m 1 (2n + 1)(−1)n q n(n+1)/2 (q; q)3∞ n=0
∞
= 1 + (−1)m q m(m+1)/2
n=m+1
i+j+h+k=n i,j,h,k≥0
Proof. Let z → 1 in (9). We omit the details.
q (m+1)j+hk n−1 qn . m (q; q)i (q; q)j (q; q)h (q; q)k 2
Let t(n) be the number of partition triples (λ, μ, ν) such that the total sum of parts of λ, μ and ν is n. Then, its generating functions is ∞ n=0
t(n)q n =
1 . (q; q)3∞
As a consequence of Corollary 2.4, we have the following partition function inequalities, which was the conjecture of Guo and Zeng noted in Introduction. Corollary 2.5. For n, m ≥ 0, m
(−1)
m
j(j + 1) ≥ 0. (−1)j (2j + 1)t n − 2 j=0
2.2. Proof of Theorem 1.2 Recall the following Bailey pair relative to (1, q) [4, Lemma 3]: (α0 , β0 ) = (1, 1) and for n ≥ 1 ⎧ ⎨ ⎩
n
αn = (−1)n (z n q 2 + z −n q (z; q)n (q/z; q)n . βn = (q; q)2n
n+1 2
), (10)
Lemma 2.6. For m ≥ 0, (−1)m q m(m+1)/2
m m n (q −m , q 1+m ; q)n (z; q)n (q/z; q)n = (−1)n z n q 2 . (q; q)2n n=−m n=0
Proof. Apply Lemma 2.1 to the Bailey pair in (10). Then, the proof follows. We omit the details. 2
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Following Remark 2, we have (−1)m q m(m+1)/2
=
m (q −m , q 1+m ; q)n (z; q)n (q/z; q)n (q; q)2n n=0
m (−1)m−n (q m−n+1 ; q)n q
m (−1)n (q n+1 ; q)m−n q n=0
=
2
(q 1+m ; q)n (z; q)n (q/z; q)n
(q; q)2n
n=0
=
m−n+1
m (−1)n q
n+1
n=0
2
n+1 2
(q 1+m ; q)m−n (z; q)m−n (q/z; q)m−n (q; q)2m−2n
(q 1+2m−2n ; q)n (z; q)m−n (q/z; q)m−n . (q; q)n
Thus, if m → ∞, we have
n+1 ∞ (−1)n q 2 (z, q/z; q)∞ = (z, q/z, q; q)∞ . (q; q)n n=0
Therefore,
(q, z, q/z; q)∞ =
∞
(−1)n z n q
n 2
.
n=−∞
Let us consider the truncated sum in the following theorem. Theorem 2.7. For m ≥ 0, m n 1 (−1)n z n q 2 (q, z, q/z; q)∞ n=−m
= 1 + (−1)m q
m+1
∞
2
n=m+1
i+j+h+k=n i,j,h,k≥0
q mj+h(k−1) z h−k n−1 qn . m (q; q)i (q; q)j (q; q)h (q; q)k (11)
Proof. By Lemma 2.6, we have m n 1 (−1)n z n q 2 (q, z, q/z; q)∞ n=−m m 1 (q −m , q 1+m ; q)n (z; q)n (q/z; q)n m m(m+1)/2 = (−1) q (q, z, q/z; q)∞ (q; q)2n n=0
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=
m (−1)m q m(m+1)/2 (q −m ; q)n (q; q)m+n (z; q)n (q/z; q)n (q, z, q/z; q)∞ (q; q)m n=0 (q; q)2n
=
m (−1)m q m(m+1)/2 (q −m ; q)n (z −1 q; q)n (q 2n+1 ; q)∞ (q; q)m (q/z; q)∞ n=0 (q; q)n (q n+1 ; q)∞ (q m+n+1 ; q)∞ (zq n ; q)∞
=
m ∞ (−1)m q m(m+1)/2 (q −m ; q)n (z −1 q; q)n q (n+1)i (q; q)m (q/z; q)∞ n=0 (q; q)n (q; q)i i=0
×
391
∞ ∞ q (m+n+1)j (z −1 q n+1 ; q)h z h q nh (q; q)j (q; q)h j=0 h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+1)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ z h q nh 1 −1 n+h+1 (q; q)h (z q ; q)∞
h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+1)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ ∞ z h q nh q (n+h+1)k z −k (q; q)h (q; q)k
h=0
=
k=0
(−1)m q m(m+1)/2 (q; q)m
∞ i,j,h,k=0 ∞
m m(m+1)/2
=
(−1) q (q; q)m
i,j,h,k=0
m(m+1)/2
=
(−1)m q (q; q)m +
∞ i+j+h+k=m+1 i,j,h,k≥0
m q i+(m+1)j+(h+1)k z h−k (q −m ; q)n q (i+j+h+k)n (q; q)i (q; q)j (q; q)h (q; q)k n=0 (q; q)n
q i+(m+1)j+(h+1)k z h−k (q i+j+h+k−m ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
(q −m ; q)m
q i+(m+1)j+(h+1)k z h−k (q i+j+h+k−m ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
m m(m+1)/2
= 1 + (−1) q
∞ i+j+h+k=m+1 i,j,h,k≥0
m m(m+1)/2
= 1 + (−1) q
∞ n=m+1
q i+(m+1)j+(h+1)k z h−k i + j + h + k − 1 m (q; q)i (q; q)j (q; q)h (q; q)k
i+j+h+k=n i,j,h,k≥0
q mj+h(k−1) z h−k n n−1 q m (q; q)i (q; q)j (q; q)h (q; q)k
as desired. Here the fourth equality follows from [1, p. 17, Eq. (2.2.1)] and the third last equality follows from (q k−m ; q)m = 0 if 0 < k ≤ m. 2
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With q and z replaced by q R , q S in Theorem 2.7, respectively, Theorem 1.2 follows immediately. Remark 3. We note that one can get a truncated theorem on the left hand side of Theorem 2.7 using Theorem 2.3. Acknowledgments The authors are very grateful to the anonymous referee for his/her valuable comments and suggestions. The first author was partially supported by the outstanding doctoral dissertation cultivation plan of action (No. YB2016028). The second author was partially supported by a grant (#280903) from the Simons Foundation. References [1] G.E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976, reprinted, Cambridge University Press, 1998. [2] G.E. Andrews, Connection coefficient problems and partitions, Proc. Sympos. Pure Math. 34 (1979) 1–24. [3] G.E. Andrews, q-Series: Their Development and Application in Analysis, Number Theory, Combinatorics, Physics and Computer Algebra, C.B.M.S. Regional Conference Series in Math., vol. 66, American Math. Soc., Providence, 1986. [4] G.E. Andrews, Bailey chains and generalized Lambert series: I. Four identities of Ramanujan, Illinois J. Math. 36 (1992) 251–274. [5] G.E. Andrews, B.C. Berndt, Ramanujan’s Lost Notebook, Part II, Springer, 2009. [6] G.E. Andrews, M. Merca, The truncated pentagonal number theorem, J. Combin. Theory Ser. A 119 (2012) 1639–1643. [7] G.E. Andrews, M. Merca, Truncated theta series and a problem of Guo and Zeng, J. Combin. Theory Ser. A 154 (2018) 610–619. [8] M. Burnette, L. Kolitsch, Interpreting the truncated pentagonal number theorem using partition pairs, Electron. J. Combin. 22 (2015) 2.55. [9] S.H. Chan, T.P.N. Ho, R. Mao, Truncated series from the quintuple product identity, J. Number Theory 169 (2016) 420–438. [10] V.W.J. Guo, J. Zeng, Two truncated identities of Gauss, J. Combin. Theory Ser. A 120 (2013) 700–707. [11] T.Y. He, K.Q. Ji, W.J.T. Zang, Bilateral truncated Jacobi’s identity, European J. Combin. 51 (2016) 255–267. [12] L. Kolitsch, Another approach to the truncated pentagonal number theorem, Int. J. Number Theory 11 (2015) 1563–1569. [13] Z.-G. Liu, On the q-partial differential equations and q-series, in: The Legacy of Srinivasa Ramanujan, in: Ramanujan Math. Soc. Lect. Notes Ser., vol. 20, Ramanujan Math. Soc., Mysore, 2013, pp. 213–250. [14] R. Mao, Proofs of two conjectures on truncated series, J. Combin. Theory Ser. A 130 (2015) 15–25. [15] C. Wang, A.J. Yee, Truncated Hecke-Rogers type series, preprint. [16] A.J. Yee, A truncated Jacobi triple product theorem, J. Combin. Theory Ser. A 130 (2015) 1–14.
Contents lists available at ScienceDirect
Journal of Combinatorial Theory, Series A www.elsevier.com/locate/jcta
Truncated Jacobi triple product series Chun Wang a , Ae Ja Yee b a
Department of Mathematics, East China Normal University, 500 Dongchuan Road, Shanghai 200241, PR China b Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, USA
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 May 2018 Available online xxxx
In their study of truncated theta series, Andrews-Merca and Guo-Zeng made a conjecture on Jacobi’s triple product identity, which is settled by Mao and Yee independently. In this paper, we reprove this ex-conjecture by providing an explicit series form with nonnegative coefficients, which is reminiscent of the results of Andrews-Merca and Guo-Zeng. We also provide a companion theorem to this ex-conjecture. © 2019 Elsevier Inc. All rights reserved.
MSC: primary 11P81 secondary 05A17 Keywords: Theta series Truncated series Jacobi’s triple product identity
1. Introduction A partition of a positive integer n is a weakly decreasing sequence of positive integers whose sum equals n. Let p(n) be the number of partitions of n for n ≥ 1 with p(0) = 1. Then, the generating function for p(n) is ∞
p(n)q n =
n=0
∞
1 . 1 − qn n=1
E-mail addresses: [email protected] (C. Wang), [email protected] (A.J. Yee). https://doi.org/10.1016/j.jcta.2019.03.003 0097-3165/© 2019 Elsevier Inc. All rights reserved.
C. Wang, A.J. Yee / Journal of Combinatorial Theory, Series A 166 (2019) 382–392
383
The reciprocal of the generating function for p(n) has the following series expansion, which is known as Euler’s pentagonal number theorem: ∞
(−1)n q n(3n+1)/2 (1 − q 2n+1 ) =
n=0
∞
(1 − q n ).
n=1
In [6], Andrews and Merca proved the following truncated theorem on Euler’s identity: for m ≥ 1, m ∞ m−1 1 q 2 +(m+1)n n − 1 . (1) (−1)n q n(3n+1)/2 (1 − q 2n+1 ) = 1 + (−1)m−1 m−1 (q; q)∞ n=0 (q; q)n n=m
Here and throughout the paper, we adopt the following standard q-series notation:
(a; q)∞ :=
∞
(1 − aq n ) for |q| < 1,
n=0
(a; q)n :=
(a; q)∞ for any integer n. (aq n ; q)∞
The q-binomial coefficient is defined as ⎧ (q; q)n ⎨ , n n = := (q; q)k (q; q)n−k k k q ⎩ 0,
if 0 ≤ k ≤ n, otherwise.
We sometimes use the following compressed notation: (a1 , a2 , . . . , ar ; q)∞ := (a1 ; q)∞ (a2 ; q)∞ · · · (ar ; q)∞ , (a1 , a2 , . . . , ar ; q)n := (a1 ; q)n (a2 ; q)n · · · (ar ; q)n . The work of Andrews and Merca opened up a new study on truncated series as one can find a series of papers on the subject [7–12,14–16]. In particular, in [10], Guo and Zeng showed truncated theorems on the following Gauss’s theta series identities: ∞
2
(−1)n q n (1 − q 2n+1 ) =
n=0 ∞ n=0
Namely, for m ≥ 1,
(−1)n q n(2n+1) (1 − q 2n+1 ) =
(q; q)∞ , (−q; q)∞ (q 2 ; q 2 )∞ . (−q; q 2 )∞
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m ∞ (−q; q)∞ (−q; q)m (−1; q)n−m q (m+1)n n − 1 n n2 m , (−1) q = 1 + (−1) m (q; q)∞ n=−m (q; q)n n=m+1 (2) and m−1 (−q; q 2 )∞ (−1)n q n(2n+1) (1 − q 2n+1 ) (q 2 ; q 2 )∞ n=0 m−1
= 1 + (−1)
∞ (−q; q 2 )m (−q; q 2 )n−m q 2(m+1)n−m n − 1 . m − 1 q2 (q 2 ; q 2 )n n=m
(3)
At the end of their papers, Andrews and Merca [6, Question (2)], and Guo and Zeng [10, Conjecture 6.1] posed the following conjecture: for 1 ≤ S ≤ R/2, and m ≥ 1, m−1
1
(q R , q S , q R−S ; q R )∞ n=0
(−1)n q
n+1 2
R−nS
1 − q (2n+1)S
(4)
has nonnegative coefficients if m is odd and nonpositive coefficients if m is even. The truncated theorems in (1) and (3) confirm this conjecture for R = 3, S = 1, and R = 4, S = 1, respectively. It is now settled by Mao [14] and Yee [16]. The proof of Mao uses q-series manipulations while the proof of Yee is based on combinatorial argument. However, both did not give an explicit form that is analogous to the right hand sides of (1) and (3). The main purpose of this paper is to reprove this ex-conjecture by providing an explicit series form with nonnegative coefficients, which is reminiscent of the results of Andrews and Merca in (1) and Guo and Zeng in (3). Theorem 1.1. For 1 ≤ S ≤ R/2 and m ≥ 1, m−1
1
(q R , q S , q R−S ; q R )∞ n=0 = 1 + (−1)m−1 q
m 2
R
(−1)n q
∞
n+1
R−nS
2
n=m i+j+h+k=n i,j,h,k≥0
1 − q (2n+1)S
q (mj+hk)R+(h−k)S+nR n−1 . (q R ; q R )i (q R ; q R )j (q R ; q R )h (q R ; q R )k m − 1 qR
As a companion of Theorem 1.1, we obtain the following theorem. Theorem 1.2. For 1 ≤ S ≤ R/2 and m ≥ 0, 1
m
(−1)n q
(q R , q S , q R−S ; q R )∞ n=−m
n 2
R+nS
C. Wang, A.J. Yee / Journal of Combinatorial Theory, Series A 166 (2019) 382–392
m
= 1 + (−1) q
m+1 2
R
∞
n=m+1 i+j+h+k=n i,j,h,k≥0
385
q (mj+h(k−1))R+(h−k)S+nR n−1 . m qR (q R ; q R )i (q R ; q R )j (q R ; q R )h (q R ; q R )k
When R = 2, S = 1, the left hand side in Theorem 1.2 yields that of (2). We note that this theorem yields a weaker result than Theorem 1.1 as pointed out by Guo and Zeng [10, Remark 6.2] for the case when R = 2, S = 1. By taking the following Jacobi’s identity ∞
(−1)n (2n + 1)q
n+1 2
= (q; q)3∞ ,
n=0
Guo and Zeng [10, Conjecture 6.4] conjectured that for m ≥ 1, m n+1 1 n 2 (−1) (2n + 1)q (q; q)3∞ n=0
(5)
has nonnegative coefficients if m is even and nonpositive coefficients if m is odd. This conjecture was first proved by Mao [14] and then proved combinatorially by He, Ji, and Zang [11]. As a corollary of Theorem 1.1, we can also obtain an explicit series form for the truncated series in (5), which reconfirms the conjecture. In the next section, we will prove all the results stated in Introduction. 2. Proofs of Theorems 1.1 and 1.2 A pair of sequences (αn , βn )n≥0 is called a Bailey pair relative to (a, q) if αn and βn are related by the following identity [3, Sect. 3.4]: for n ≥ 0, βn =
n r=0
αr . (q; q)n−r (aq; q)n+r
We need a truncated Bailey transform. Lemma 2.1. If (αn , βn ) is a Bailey pair relative to (a, q) and m is a nonnegative integer, then we have m m (aq; q)m (−1)m q m(m+1)/2 −m (q , aq 1+m ; q)n βn (a, q) = αn (a, q). (q; q)m n=0 n=0
Proof. We recall the following q-series expansion formula of Liu [13, Theorem 9.2]: m (aq, abc/q; q)m (q −m , q/b, q/c; q)n q n An (ab, ac; q)m n=0 (q 2 /abcq m ; q)n
(6)
386
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=
n m (1 − aq 2n )(q −m , a, q/b, q/c; q)n (abcq m−1 )n −n (q , aq n ; q)k q k Ak , m+1 , ab, ac; q) (1 − a)(q, aq n n=0
(7)
k=0
where m is a nonnegative integer and {An } is an arbitrary complex sequence. Replacing b by a−1 q −m in (7) and then letting c → 0, we get m (aq; q)m (−1)m q m(m+1)/2 −m (q , aq 1+m ; q)n An (q; q)m n=0
=
=
n m (1 − aq 2n )(a; q)n (−1)n q n(n−1)/2 −n (q , aq n ; q)k q k Ak (1 − a)(q; q) n n=0 m
n 2n
1 − aq 1−a n=0
k=0
(a; q)n+k (−1)n−k q (q; q)n−k
k=0 n−k 2
Ak .
Since (αn , βn ) is a Bailey pair relative to (a, q) [2, Lemma 3] or [3, Eq. (3.40)], n 1 − aq 2n (a; q)n+j (−1)n−j q αn (a, q) = 1 − a j=0 (q; q)n−j
n−j 2
βj (a, q).
Thus, upon taking An = βn (a, q), we see that Lemma 2.1 follows immediately.
2
Remark 1. The formula of Liu in (7) can be found in [2, Eq. (3.3)]. 2.1. Proof of Theorem 1.1 Recall the following Bailey pair (αn , βn ) relative to (q, q) from [5, p. 98-99]: ⎧ ⎨αn ⎩βn
= (z −n − z n+1 )(−1)n q n(n+1)/2 , (z; q)n+1 (q/z; q)n = . (q 2 ; q)2n
(8)
Lemma 2.2. For m ≥ 0, (−1)m q m(m+1)/2
=
m
m (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 n=0
(−1)n q n(n+1)/2 (z −n − z n+1 ).
n=0
Proof. Apply Lemma 2.1 to the Bailey pair in (8). Then this lemma easily follows and we omit the details. 2
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Remark 2. Note that (−1)m q m(m+1)/2
=
m (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 n=0
m (−1)m−n (q m−n+1 ; q)n q n=0
=
m (−1)n (q n+1 ; q)m−n q n=0
=
m (−1)n q
n+1 2
n=0
m−n+1
n+1 2
(q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (q; q)2n+1 2
(q 1+m ; q)m−n+1 (z; q)m−n+1 (q/z; q)m−n (q; q)2m−2n+1
(q 2m−2n+2 ; q)n (z; q)m−n+1 (q/z; q)m−n . (q; q)n
Thus, if m → ∞, we have
n+1 ∞ (−1)n q 2 (z, q/z; q)∞ = (z, q/z, q; q)∞ . (q; q)n n=0
Therefore, (q, z, q/z; q)∞ =
∞
(z −n − z n+1 )(−1)n q
n+1 2
.
n=0
Now we consider the truncated sum in the following theorem. Theorem 2.3. For m ≥ 0, m n+1 1 (z −n − z n+1 )(−1)n q 2 (q, z, q/z; q)∞ n=0 ⎛ m
= 1 + (−1) q
m+1
∞
2
n=m+1
⎜ ⎜ ⎝
i+j+h+k=n i,j,h,k≥0
⎞
⎟ n n−1 q (m+1)j+hk z h−k ⎟q . (9) m (q; q)i (q; q)j (q; q)h (q; q)k ⎠
Proof. By Lemma 2.2, we have m 1 (z −n − z n+1 )(−1)n q n(n+1)/2 (q, z, q/z; q)∞ n=0
=
m 1 (q −m ; q)n (q 1+m ; q)n+1 (z; q)n+1 (q/z; q)n (−1)m q m(m+1)/2 (q, z, q/z; q)∞ (q; q)2n+1 n=0
m (−1)m q m(m+1)/2 (q −m ; q)n (q; q)m+n+1 (z; q)n+1 (q/z; q)n = (q, z, q/z; q)∞ (q; q)m n=0 (q; q)2n+1
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388
=
m (−1)m q m(m+1)/2 (q −m ; q)n (q/z; q)n (q 2n+2 ; q)∞ (q/z; q)∞ (q; q)m n=0 (q; q)n (q n+1 ; q)∞ (q m+n+2 ; q)∞ (zq n+1 ; q)∞
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n (q/z; q)n q (n+1)i q (m+n+2)j (q/z; q)∞ (q; q)m n=0 (q; q)n (q; q)i j=0 (q; q)j i=0
×
∞ (z −1 q n+1 ; q)h z h q (n+1)h (q; q)h
h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+2)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ z h q (n+1)h 1 −1 n+h+1 (q; q)h (z q ; q)∞
h=0
=
m ∞ ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+2)j z h q (n+1)h (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j (q; q)h n=0 h=0
×
∞ k=0
=
z
−k (n+h+1)k
q (q; q)k
(−1)m q m(m+1)/2 (q; q)m
∞ i,j,h,k=0 ∞
m m(m+1)/2
=
(−1) q (q; q)m
i,j,h,k=0
m m(m+1)/2
=
(−1) q (q; q)m +
∞ i+j+h+k=m+1 i,j,h,k≥0
m q i+(m+2)j+h+(h+1)k z h−k (q −m ; q)n q (i+j+h+k)n (q; q)i (q; q)j (q; q)h (q; q)k n=0 (q; q)n
q i+(m+2)j+h+(h+1)k z h−k i+j+h+k−m (q ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
(q −m ; q)m
q i+(m+2)j+h+(h+1)k z h−k i+j+k+h−m (q ; q)m (q; q)i (q; q)j (q; q)h (q; q)k ∞
= 1 + (−1)m q m(m+1)/2
i+j+h+k=m+1 i,j,h,k≥0
q i+(m+2)j+h+(h+1)k z h−k i + j + h + k − 1 m (q; q)i (q; q)j (q; q)h (q; q)k
⎛
m m(m+1)/2
= 1 + (−1) q
∞ n=m+1
⎜ ⎜ ⎝
i+j+h+k=n i,j,h,k≥0
⎞
⎟ n n−1 q (m+1)j+hk z h−k ⎟q m (q; q)i (q; q)j (q; q)h (q; q)k ⎠
as desired. Here the fourth equality follows from [1, p. 17, Eq. (2.2.1)] and the third last equality follows from (q k−m ; q)m = 0 if 0 < k ≤ m. 2 With q and z replaced by q R , q S in Theorem 2.3, respectively, Theorem 1.1 follows easily.
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389
Corollary 2.4. For m ≥ 0, m 1 (2n + 1)(−1)n q n(n+1)/2 (q; q)3∞ n=0
∞
= 1 + (−1)m q m(m+1)/2
n=m+1
i+j+h+k=n i,j,h,k≥0
Proof. Let z → 1 in (9). We omit the details.
q (m+1)j+hk n−1 qn . m (q; q)i (q; q)j (q; q)h (q; q)k 2
Let t(n) be the number of partition triples (λ, μ, ν) such that the total sum of parts of λ, μ and ν is n. Then, its generating functions is ∞ n=0
t(n)q n =
1 . (q; q)3∞
As a consequence of Corollary 2.4, we have the following partition function inequalities, which was the conjecture of Guo and Zeng noted in Introduction. Corollary 2.5. For n, m ≥ 0, m
(−1)
m
j(j + 1) ≥ 0. (−1)j (2j + 1)t n − 2 j=0
2.2. Proof of Theorem 1.2 Recall the following Bailey pair relative to (1, q) [4, Lemma 3]: (α0 , β0 ) = (1, 1) and for n ≥ 1 ⎧ ⎨ ⎩
n
αn = (−1)n (z n q 2 + z −n q (z; q)n (q/z; q)n . βn = (q; q)2n
n+1 2
), (10)
Lemma 2.6. For m ≥ 0, (−1)m q m(m+1)/2
m m n (q −m , q 1+m ; q)n (z; q)n (q/z; q)n = (−1)n z n q 2 . (q; q)2n n=−m n=0
Proof. Apply Lemma 2.1 to the Bailey pair in (10). Then, the proof follows. We omit the details. 2
390
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Following Remark 2, we have (−1)m q m(m+1)/2
=
m (q −m , q 1+m ; q)n (z; q)n (q/z; q)n (q; q)2n n=0
m (−1)m−n (q m−n+1 ; q)n q
m (−1)n (q n+1 ; q)m−n q n=0
=
2
(q 1+m ; q)n (z; q)n (q/z; q)n
(q; q)2n
n=0
=
m−n+1
m (−1)n q
n+1
n=0
2
n+1 2
(q 1+m ; q)m−n (z; q)m−n (q/z; q)m−n (q; q)2m−2n
(q 1+2m−2n ; q)n (z; q)m−n (q/z; q)m−n . (q; q)n
Thus, if m → ∞, we have
n+1 ∞ (−1)n q 2 (z, q/z; q)∞ = (z, q/z, q; q)∞ . (q; q)n n=0
Therefore,
(q, z, q/z; q)∞ =
∞
(−1)n z n q
n 2
.
n=−∞
Let us consider the truncated sum in the following theorem. Theorem 2.7. For m ≥ 0, m n 1 (−1)n z n q 2 (q, z, q/z; q)∞ n=−m
= 1 + (−1)m q
m+1
∞
2
n=m+1
i+j+h+k=n i,j,h,k≥0
q mj+h(k−1) z h−k n−1 qn . m (q; q)i (q; q)j (q; q)h (q; q)k (11)
Proof. By Lemma 2.6, we have m n 1 (−1)n z n q 2 (q, z, q/z; q)∞ n=−m m 1 (q −m , q 1+m ; q)n (z; q)n (q/z; q)n m m(m+1)/2 = (−1) q (q, z, q/z; q)∞ (q; q)2n n=0
C. Wang, A.J. Yee / Journal of Combinatorial Theory, Series A 166 (2019) 382–392
=
m (−1)m q m(m+1)/2 (q −m ; q)n (q; q)m+n (z; q)n (q/z; q)n (q, z, q/z; q)∞ (q; q)m n=0 (q; q)2n
=
m (−1)m q m(m+1)/2 (q −m ; q)n (z −1 q; q)n (q 2n+1 ; q)∞ (q; q)m (q/z; q)∞ n=0 (q; q)n (q n+1 ; q)∞ (q m+n+1 ; q)∞ (zq n ; q)∞
=
m ∞ (−1)m q m(m+1)/2 (q −m ; q)n (z −1 q; q)n q (n+1)i (q; q)m (q/z; q)∞ n=0 (q; q)n (q; q)i i=0
×
391
∞ ∞ q (m+n+1)j (z −1 q n+1 ; q)h z h q nh (q; q)j (q; q)h j=0 h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+1)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ z h q nh 1 −1 n+h+1 (q; q)h (z q ; q)∞
h=0
=
m ∞ ∞ (−1)m q m(m+1)/2 (q −m ; q)n q (n+1)i q (m+n+1)j (q; q)m (q; q)n i=0 (q; q)i j=0 (q; q)j n=0
×
∞ ∞ z h q nh q (n+h+1)k z −k (q; q)h (q; q)k
h=0
=
k=0
(−1)m q m(m+1)/2 (q; q)m
∞ i,j,h,k=0 ∞
m m(m+1)/2
=
(−1) q (q; q)m
i,j,h,k=0
m(m+1)/2
=
(−1)m q (q; q)m +
∞ i+j+h+k=m+1 i,j,h,k≥0
m q i+(m+1)j+(h+1)k z h−k (q −m ; q)n q (i+j+h+k)n (q; q)i (q; q)j (q; q)h (q; q)k n=0 (q; q)n
q i+(m+1)j+(h+1)k z h−k (q i+j+h+k−m ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
(q −m ; q)m
q i+(m+1)j+(h+1)k z h−k (q i+j+h+k−m ; q)m (q; q)i (q; q)j (q; q)h (q; q)k
m m(m+1)/2
= 1 + (−1) q
∞ i+j+h+k=m+1 i,j,h,k≥0
m m(m+1)/2
= 1 + (−1) q
∞ n=m+1
q i+(m+1)j+(h+1)k z h−k i + j + h + k − 1 m (q; q)i (q; q)j (q; q)h (q; q)k
i+j+h+k=n i,j,h,k≥0
q mj+h(k−1) z h−k n n−1 q m (q; q)i (q; q)j (q; q)h (q; q)k
as desired. Here the fourth equality follows from [1, p. 17, Eq. (2.2.1)] and the third last equality follows from (q k−m ; q)m = 0 if 0 < k ≤ m. 2
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With q and z replaced by q R , q S in Theorem 2.7, respectively, Theorem 1.2 follows immediately. Remark 3. We note that one can get a truncated theorem on the left hand side of Theorem 2.7 using Theorem 2.3. Acknowledgments The authors are very grateful to the anonymous referee for his/her valuable comments and suggestions. The first author was partially supported by the outstanding doctoral dissertation cultivation plan of action (No. YB2016028). The second author was partially supported by a grant (#280903) from the Simons Foundation. References [1] G.E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976, reprinted, Cambridge University Press, 1998. [2] G.E. Andrews, Connection coefficient problems and partitions, Proc. Sympos. Pure Math. 34 (1979) 1–24. [3] G.E. Andrews, q-Series: Their Development and Application in Analysis, Number Theory, Combinatorics, Physics and Computer Algebra, C.B.M.S. Regional Conference Series in Math., vol. 66, American Math. Soc., Providence, 1986. [4] G.E. Andrews, Bailey chains and generalized Lambert series: I. Four identities of Ramanujan, Illinois J. Math. 36 (1992) 251–274. [5] G.E. Andrews, B.C. Berndt, Ramanujan’s Lost Notebook, Part II, Springer, 2009. [6] G.E. Andrews, M. Merca, The truncated pentagonal number theorem, J. Combin. Theory Ser. A 119 (2012) 1639–1643. [7] G.E. Andrews, M. Merca, Truncated theta series and a problem of Guo and Zeng, J. Combin. Theory Ser. A 154 (2018) 610–619. [8] M. Burnette, L. Kolitsch, Interpreting the truncated pentagonal number theorem using partition pairs, Electron. J. Combin. 22 (2015) 2.55. [9] S.H. Chan, T.P.N. Ho, R. Mao, Truncated series from the quintuple product identity, J. Number Theory 169 (2016) 420–438. [10] V.W.J. Guo, J. Zeng, Two truncated identities of Gauss, J. Combin. Theory Ser. A 120 (2013) 700–707. [11] T.Y. He, K.Q. Ji, W.J.T. Zang, Bilateral truncated Jacobi’s identity, European J. Combin. 51 (2016) 255–267. [12] L. Kolitsch, Another approach to the truncated pentagonal number theorem, Int. J. Number Theory 11 (2015) 1563–1569. [13] Z.-G. Liu, On the q-partial differential equations and q-series, in: The Legacy of Srinivasa Ramanujan, in: Ramanujan Math. Soc. Lect. Notes Ser., vol. 20, Ramanujan Math. Soc., Mysore, 2013, pp. 213–250. [14] R. Mao, Proofs of two conjectures on truncated series, J. Combin. Theory Ser. A 130 (2015) 15–25. [15] C. Wang, A.J. Yee, Truncated Hecke-Rogers type series, preprint. [16] A.J. Yee, A truncated Jacobi triple product theorem, J. Combin. Theory Ser. A 130 (2015) 1–14.