sigma09-050. 259KB Jun 04 2011 12:10:27 AM

Symmetry, Integrability and Geometry: Methods and Applications

SIGMA 5 (2009), 050, 19 pages

Partial Sums of Two Quartic q-Series
Wenchang CHU

†

and Chenying WANG

‡

†

Dipartimento di Matematica, Universit`
a degli Studi di Salento,
Lecce-Arnesano P. O. Box 193, Lecce 73100, Italy
E-mail: chu.wenchang@unile.it

‡

College of Mathematics and Physics, Nanjing University of Information Science
and Technology, Nanjing 210044, P. R. China
E-mail: wang.chenying@163.com

Received January 20, 2009, in final form April 17, 2009; Published online April 22, 2009
doi:10.3842/SIGMA.2009.050
Abstract. The partial sums of two quartic basic hypergeometric series are investigated
by means of the modified Abel lemma on summation by parts. Several summation and
transformation formulae are consequently established.
Key words: basic hypergeometric series (q-series); well-poised q-series; quadratic q-series;
cubic q-series; quartic q-series; the modified Abel lemma on summation by parts
2000 Mathematics Subject Classification: 33D15; 05A15

1

Introduction and motivation

For two indeterminate x and q, the shifted-factorial of x with base q is defined by

(x; q)0 = 1
and
(x; q)n = (1 − x)(1 − xq) · · · 1 − xq n−1
for n ∈ N.
When |q| < 1, we have two well-defined infinite products
(x; q)∞ =

∞
Y

k=0



1 − qk x

and

(x; q)n = (x; q)∞ / xq n ; q

∞

.

The product and fraction of shifted factorials are abbreviated respectively to
[α, β, . . . , γ; q]n = (α; q)n (β; q)n · · · (γ; q)n ,

 
(α; q)n (β; q)n · · · (γ; q)n
α, β, . . . , γ 
.
q =
A, B, . . . , C n (A; q)n (B; q)n · · · (C; q)n

Following Gasper–Rahman [12], the basic hypergeometric series is defined by

∞ n
  X
os−r  a , a , . . . , a  

a0 , a1 , . . . , ar 
0 1
r
n (n
)
2
(−1) q
q z n ,
q; z =
1+r φs
q, b1 , . . . , bs
b1 , . . . , bs
n
n=0

where the base q will be restricted to |q| < 1 for nonterminating q-series. For its connections to
special functions and orthogonal polynomials, the reader can refer, for example, to the monograph written by Andrews–Askey–Roy [2] and the paper by Koornwinder [15].
In the theory of basic hypergeometric series, there are several important classes, for example,
well-poised [3], quadratic [13, 17], cubic [16] and quartic [10, 11] series. To our knowledge, there
are four quartic series which can be displayed as follows:



n−1
 
 
X


[qa/bd; q]3k
b, d 
b2 d2/q 2  4
5k
Fn (a, b, d) :=
1−q a
qk ;
q
q
q 3 a/b2 d2 k [bd, bd/q, qbd; q 2 ]k q 4 a/b, q 4 a/d
k
k=0

2

W. Chu and C. Wang
n−1
X




 

c2 e2/q 2 a3 
[qa2/ce, q 2 a2/ce, q 3 a2/ce; q 2 ]k
c, e  4
Gn (a, c, e) :=
1−q a
q
q q k ;
qa/c, qa/e k
q 6 a4/c2 e2
(ce/a; q)3k

k
k=0




n−1


X

(q 2 a/bd; q)k
b, d  2 (q 3 a2 /bd; q 6 )k b2 d2 /q 3 a  3 (k2)
Un (a, b, d) :=
1−q 5k a
q
q q ;
5
(bd; q 2 )2k q a2 /b2 d2
(−bd/q 3 a)k q 3 a/b, q 3 a/d
k

k
k=0




n−1

X

(a2 /ce; q 2 )2k qc2 e2 /a2  2 (−a/ce)k
k
qc, qe  3
Vn (a, c, e) :=
1 − q 5k a
q q −( 2 ) .
q
2
3
2
2

5
6
q a /c e
(qce/a; q)k qa/c, qa/e
k
k (q ce; q )k
5k

k=0

By means of the series rearrangement, Gasper and Rahman [10, 11, 12] discovered several
summation and transformation formulae for the nonterminating special cases of Fn (a, b, d) and
Un (a, b, d). The terminating series identities for the last four sums have been established by
Chu [4] and Chu–Wang [8], respectively, through inversion techniques and Abel’s lemma on
summation by parts. For the partial sums Fn (a, b, d) and Gn (a, c, e), the present authors [7]
recently derived several useful reciprocal relations and transformation formulae in terms of wellpoised series.
The purpose of this paper is to investigate the remaining two partial sums Un (a, b, d) and
Vn (a, c, e). By utilizing the modified Abel lemma on summation by parts, we shall show six
unusual transformation formulae with two between them and other four expressing Un (a, b, d)
and Vn (a, c, e) as partial sums of quadratic and cubic series. Several new and known terminating

as well as nonterminating series identities are consequently obtained as particular instances.
In order to make the paper self-contained, we reproduce Abel’s lemma on summation by
parts (cf. [5, 6, 7, 8]) as follows. For an arbitrary complex sequence {τk }, define the backward
and forward difference operators ▽ and ∆, respectively, by
▽τk = τk − τk−1

and

∆τk = τk+1 − τk .

Then Abel’s lemma on summation by parts can be modified as follows:
n−1
X





Bk ▽ Ak = An−1 Bn − A−1 B0 −

k=0

n−1
X

Ak ∆Bk .

k=0

This can be considered as the discrete counterpart for the integral formula
Z

b

a



f (x)g ′ (x)dx = f (b)g(b) − f (a)g(a) −

Z

b

f ′ (x)g(x)dx.

a

In fact, it is almost trivial to check the following expression
n−1
X
k=0

Bk ▽ Ak =

n−1
X
k=0





Bk Ak − Ak−1 =

n−1
X

Ak Bk −

k=0

n−1
X

Ak−1 Bk .

k=0

Replacing k by k + 1 for the last sum, we can reformulate the equation as follows
n−1
X

Bk ▽ Ak = An−1 Bn − A−1 B0 +

k=0

= An−1 Bn − A−1 B0 −

n−1
X

k=0
n−1
X



Ak Bk − Bk+1
Ak ∆Bk ,

k=0

which is exactly the equality stated in the modified Abel lemma.

Partial Sums of Two Quartic q-Series

3

Throughout the paper, if Wn is used to denote the partial sum of some q-series, then the
corresponding letter W without subscript will stand for the limit of Wn (if it exists of course)
when n → ∞.
When applying the modified Abel lemma on summation by parts to deal with hypergeometric
series, the crucial step lies in finding shifted factorial fractions {Ak , Bk } so that their differences
are expressible as ratios of linear factors. This has not been an easy task, even though it is
indeed routine to factorize {Ak , Bk } once they are figured out. Specifically for Un (a, b, d) and
Vn (a, b, d), we shall devise three well-poised difference pairs for each partial sum. This is based
on numerous attempts to detect Ak and Bk sequences such that not only their differences turn
out to be factorizable, but also their combinations match exactly the summands displayed in
Un (a, b, d) and Vn (a, b, d).
The contents of the paper will be organized as follows. In the second section, Un (a, b, d) will
be reformulated through the modified Abel lemma on summation by parts, which lead to three
transformations of Un (a, b, d) into partial sums of quadratic, cubic and quartic series. Then the
third section will be devoted to the transformation formulae of Vn (a, b, d) in terms of partial
sums of quadratic, cubic and quartic series. These transformations on Un (a, b, d) and Vn (a, b, d)
will recover several known identities appeared in Chu–Wang [4, 8] and Gasper–Rahman [12],
and yield a few additional new summation formulae.

2

Transformation and summation formulae for Un (a, b, d)

In this section, we shall investigate the partial sum of quartic q-series Un (a, b, d). Three transformation formulae will be established from Un (a, b, d) to quadratic, cubic and another quartic series, unlike those for Fn (a, b, d) and Gn (a, c, e) shown in [7], where reciprocal relations
and transformations into well-poised partial sums have been derived. As particular cases of
Un (a, b, d), four nonterminating series will be evaluated, including two that appeared in Gasper–
Rahman’s book [12].
In order for the reader to gain an immediate insight into the well-poised structure, we reformulate the series Un (a, b, d) in the following manner:
Un (a, b, d) =

n−1
X
k=0

×

2.1



1 − q 5k a

(b2 d2 /q 3 a; q 3 )k k
[b, d; q 2 ]k
q
[q 3 a/b, q 3 a/d; q 3 ]k (q 5 a2/b2 d2 ; q 2 )k

[q 2 a/bd, q 2 a/bd; q]k (q 3 a2 /bd; q 6 )k
.
[bd, q 2 bd; q 4 ]k
(bd/q 2 a; q −1 )k

Quartic series to quadratic series

Let Ak and Bk be defined by
[q 3 a/bd, q 5 a/bd; q]k (b3 d3 /q 7 a2 ; q 2 )k (q 9 a2 /bd; q 6 )k
,
[bd, q 2 bd; q 4 ]k (q 12 a3 /b3 d3 ; q 3 )k (bd/q 4 a; q −1 )k
[b, d; q 2 ]k
[b2 d2 /q 6 a, q 12 a3 /b3 d3 ; q 3 ]k
Bk = 3
.
[q a/b, q 3 a/d; q 3 ]k [q 9 a2 /b2 d2 , b3 d3 /q 9 a2 ; q 2 ]k
Ak =

We can easily show the relations
a(1 − q 2 /bd)(1 − q 4 /bd)(1 − q 3 a/bd)(1 − q 9 a3 /b3 d3 )
,
(1 − q 2 a/bd)(1 − q 4 a/bd)(1 − q 3 a2 /bd)(1 − q 9 a2 /b3 d3 )

 
An−1 Bn
1 − q 9+3n a3 /b3 d3
b, d  2
R :=
=
q
q 9 a2 /b2 d2
A−1 B0
1 − q 9 a3 /b3 d3
n
̟ := A−1 B0 =

4

W. Chu and C. Wang
 2 2 6  
[q 2 a/bd, q 4 a/bd; q]n
b d /q a  3 (q 3 a2 /bd; q 6 )n
;
×
q
3
−1
q 3 a/b, q 3 a/d
(bd/q 4 ; q 2 )2n
n (bd/q a; q )n

and calculate the finite differences

(1 − q 5k a)(1 − q 6−3k a/b2 d2 )(1 − q 2k+5 a2 /b2 d2 )(1 − q 2k+7 a2 /b2 d2 )
(1 − q 2 a/bd)(1 − q 4 a/bd)(1 − q 3 a2 /bd)(1 − q 9 a2 /b3 d3 )
[q 2 a/bd, q 4 a/bd; q]k (b3 d3 /q 9 a2 ; q 2 )k (q 3 a2 /bd; q 6 )k 2k
q ,
×
[bd, q 2 bd; q 4 ]k (q 12 a3 /b3 d3 ; q 3 )k (bd/q 4 a; q −1 )k
(1 − q 3+5k a)(1 − q 3+k a/bd)(1 − q 9 a2 /b2 d3 )(1 − q 9 a2 /b3 d2 )
∆Bk = −
(1 − q 3 a/b)(1 − q 3 a/d)(1 − q 9 a2 /b2 d2 )(1 − q 9 a2 /b3 d3 )

 
   2 2 6
b d /q a, q 12 a3 /b3 d3  3
b,
d
 2
× 11 2 2 2 3 3 7 2 q
q 2k .
q
6 a/b,
6 a/d
q
q
q a /b d , b d /q a
k
k

▽Ak =

According to the modified Abel lemma on summation by parts, the finite U -sum can be reformulated as follows:
Un (a, b, d)

(1 − q 5 a2 /b2 d2 )(1 − q 7 a2 /b2 d2 )(1 − q 6 a/b2 d2 )
(1 − q 2 a/bd)(1 − q 4 a/bd)(1 − q 3 a2 /bd)(1 − q 9 a2 /b3 d3 )
n−1
X
X

n−1
=
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

k=0

Observing that there holds the equality for the last partial sum
−

n−1
X
k=0

Ak ∆Bk =

Un (q 3 a, b, d) (1 − q 3 a/bd)(1 − q 9 a2 /b2 d3 )(1 − q 9 a2 /b3 d2 )
,
(1 − q 3 a/b)(1 − q 3 a/d)(1 − q 9 a2 /b2 d2 )(1 − q 9 a2 /b3 d3 )

we derive after some simplification the recurrence relation
(q 2 a/bd; q)3 (1 − q 3 a2/bd)(1 − q 9 a2/b2 d3 )(1 − q 9 a2/b3 d2 )
(q 5 a2/b2 d2 ; q 2 )3 (1 − q 3 a/b)(1 − q 3 a/d)(1 − q 6 a/b2 d2 )

(1 − q 2 /bd)(1 − q 4 /bd)(1 − q 3 a/bd)(1 − q 9 a3 /b3 d3 )
.
− a 1 − R(a, b, d)
(1 − q 5 a2 /b2 d2 )(1 − q 7 a2 /b2 d2 )(1 − q 6 a/b2 d2 )

Un (a, b, d) = Un (q 3 a, b, d)

Iterating this equation m-times, we get the following expression

(q 2 a/bd; q)3m [q 3 a2/bd, q 9 a2/b2 d3 , q 9 a2/b3 d2 ; q 6 ]m
(q 5 a2/b2 d2 ; q 2 )3m [q 3 a/b, q 3 a/d, q 6 a/b2 d2 ; q 3 ]m
a (1 − q 2 /bd)(1 − q 4 /bd)(1 − q 3 a/bd)
−
(1 − q 5 a2 /b2 d2 )(1 − q 7 a2 /b2 d2 )(1 − q 6 a/b2 d2 )
 2
m−1
 
X
q a/bd, q 4 a/bd, q 6 a/bd  3
9+9k 3 3 3
×
(1 − q
a /b d ) 3
q 3k
q
q a/b, q 3 a/d, q 9 a/b2 d2
k
k=0
 3 2

9
2
2
3
9
2
3

q a /bd, q a /b d , q a /b d2  6
3k
× 1 − R(q a, b, d)
.
q
q 9 a2 /b2 d2 , q 11 a2 /b2 d2 , q 13 a2 /b2 d2
k

Un (a, b, d) = Un (q 3m a, b, d)

Writing explicitly the R-function by separating k and n factorials

 
1 − q 9+3n+9k a3 /b3 d3
b, d
 2
3k
R(q a, b, d) =
q
q 9+6k a2 /b2 d2
1 − q 9+9k a3 /b3 d3
n

  (q 3+6k a2 /bd; q 6 )
2+3k
4+3k
−6−3k
[q
a/bd, q
a/bd; q]n
q
b2 d2 /a  3
n
×
q
−3−3k bd/a; q −1 )
q 3+3k a/b, q 3+3k a/d
(bd/q 4 ; q 2 )2n
(q
n
n

Partial Sums of Two Quartic q-Series

5


   2 2 6   (q 3 a2 /bd; q 6 )
[q 2 a/bd, q 4 a/bd; q]n
b, d  2
b d /q a  3
n
=
q
9
2
2
2
3
3 a/d q
4
2
3 a; q −1 )
q a /b d
q
a/b,
q
(bd/q ; q )2n
(bd/q
n
n
n
 3


9+3n+9k
3
3
3
3+6n
2
6
3
9
2
2
1−q
a /b d q a/b, q a/d, q a/b d  3 (q
a /bd; q )k
×
q
2
4
6
3
2
6
9+9k
3
3
3
q a/bd, q a/bd, q a/bd
1−q
a /b d
k (q a /bd; q )k



(q 9 a2 /b2 d2 ; q 2 )3k
q 2+n a/bd, q 4+n a/bd, q 6+n a/bd  3
× 9+2n 2 2 2 2
,
q
(q
a /b d ; q )3k q 3+3n a/b, q 3+3n a/d, q 9−3n a/b2 d2
k

and then defining the partial sum of quadratic series in base q 3 by
⋄
Um
(a, b, d)

m−1
X

 
q 2 a/bd, q 4 a/bd, q 6 a/bd  3
=
(1 − q
a /b d ) 3
q
q a/b, q 3 a/d, q 9 a/b2 d2
k
k=0
 3 2


q a /bd, q 9 a2 /b2 d3 , q 9 a2 /b3 d2 
× 9 2 2 2 11 2 2 2 13 2 2 2 q 6 q 3k ,
q a /b d , q a /b d , q a /b d
k
9+9k 3

3 3



we derive the following transformation formula.

Theorem 1 (Transformation between quartic and quadratic series).
(q 2 a/bd; q)3m [q 3 a2/bd, q 9 a2/b2 d3 , q 9 a2/b3 d2 ; q 6 ]m
(q 5 a2/b2 d2 ; q 2 )3m [q 3 a/b, q 3 a/d, q 6 a/b2 d2 ; q 3 ]m
(1 − q 3 a/bd)(1 − bd/q 2 )(1 − bd/q 4 )
=
(1 − q 5 a2/b2 d2 )(1 − q 7 a2/b2 d2 )(1 − b2 d2 /q 6 a)
(
(q 3 a2 /bd; q 6 )n
⋄
⋄
× Um
(a, b, d) − Um
(q 5n a, q 2n b, q 2n d)
(bd/q 3 a; q −1 )n
 2 2 6   )

 
b d /q a  3
b, d  2 [q 2 a/bd, q 4 a/bd; q]n
.
× 9 2 2 2 q
3 a/b, q 3 a/d q
4; q2)
q
q a /b d
(bd/q
2n
n
n

Un (a, b, d) − Un (q 3m a, b, d)

By means of the Weierstrass M -test on uniformly convergent series (cf. Stromberg [19,
p. 141]), we can compute the following limit
lim Un (q

m,n→∞

3m

a, b, d) =

∞
X
k=0

(bd)k

[b, d; q 2 ]k 4(k)
q 2 .
(bd; q 2 )2k

Letting m, n → ∞ in Theorem 1, we obtain the transformation formula.
Proposition 2 (Nonterminating series transformation).
U (a, b, d) =

(1 − q 3 a/bd)(1 − bd/q 2 )(1 − bd/q 4 )
U ⋄ (a, b, d)
(1 − q 5 a2/b2 d2 )(1 − q 7 a2/b2 d2 )(1 − b2 d2 /q 6 a)
∞
2
(q 2 a/bd; q)∞ [q 3 a2/bd, q 9 a2/b2 d3 , q 9 a2/b3 d2 ; q 6 ]∞ X
k [b, d; q ]k 4(k2)
+
.
q
(bd)
(q 5 a2/b2 d2 ; q 2 )∞ [q 3 a/b, q 3 a/d, q 6 a/b2 d2 ; q 3 ]∞
(bd; q 2 )2k
k=0

When bd = q 2+2δ with δ = 0, 1, this proposition results in

   3−6δ 2 5−4δ 2  
a, q 1−2δ a  3
q
a b, q
a /b  6
2+2δ
U (a, b, q
/b) = 3
q
q
1−2δ
3−6δ
2
5−4δ
q a/b, q
ab
q
a ,q
a2
∞
∞
∞
2+2δ
2
X
k
[b, q
/b; q ]k 4( )+(2+2δ)k
q 2
.
×
(q 2+2δ ; q 2 )2k
k=0

6

W. Chu and C. Wang

Taking a = q 1+4δ in this equation and noting that the initial condition
U (q 1+4δ , b, q 2+2δ /b) = 1 − q 1+4δ ,
we recover the following formula due to Andrews [1, equation (4.6)] and Ismail–Stanton [14,
Proposition 6]
 2+2δ 4+4δ  
∞
X
[b, q 2+2δ /b; q 2 ]k 4(k)+(2+2δ)k
q
b, q
/b  6
=
q 2
,
q
q 2+2δ , q 4+4δ
(q 2+2δ ; q 2 )2k
∞
k=0

which leads, under the replacement a → q 2δ a, to the nonterminating series formula.
Corollary 3 (Gasper–Rahman [12, Exercise 3.29(ii),(iii)]).


∞
X
1−q 5k+2δ a b, q 2+2δ/b  2 (a; q)k (q 1+2δ/a; q 3 )k (q 1+2δ a2 ; q 6 )k (k+1)
q 2 (−a)k
q
2+2δ ; q 2 ) [qab, q 3+2δ a/b; q 3 ]
qa2
1 − q 2δ a
(q
2k
k
k
k=0

 
   5 2
3+2δ
qa, q
a  3
q a /b, q 3−2δ a2 b, q 2+2δ b, q 4+4δ/b  6
.
=
q
q
q 5 a2 , q 3−2δ a2 , q 2+2δ , q 4+4δ
qab, q 3+2δ a/b
∞
∞

2.2

Quartic series to cubic series

Define two sequences by
[q 3 a/bd, bd2 /q 4 a; q]k (q 2 b; q 2 )k (q 9 a2 /bd; q 6 )k
,
[q 2 bd, q 9 a2 /bd2 ; q 4 ]k (q 3 a/b; q 3 )k (bd/q 4 a; q −1 )k
(q 4 a/bd; q)k (d/q 2 ; q 2 )k (b2 d2 /q 3 a; q 3 )k (q 9 a2 /bd2 ; q 4 )k
Bk =
.
(bd; q 4 )k (q 6 a/d; q 3 )k (q 7 a2 /b2 d2 ; q 2 )k (bd2 /q 5 a; q)k

Ak =

It is not hard to check the relations
(1 − bd/q 2 )(1 − q 5 a2 /bd2 )(1 − a/b)(1 − bd/q 3 a)
,
(1 − q 2 a/bd)(1 − bd2 /q 5 a)(1 − b)(1 − q 3 a2 /bd)


An−1 Bn
1 − q 5+4n a2 /bd2 b, d/q 2  2
=
R :=
q
q 7 a2 /b2 d2
A−1 B0
1 − q 5 a2 /bd2
n


[q 2 a/bd, q 4 a/bd; q]n b2 d2 /q 3 a  3 (q 3 a2 /bd; q 6 )n
;
×
q
3
−1
a/b, q 6 a/d
(bd/q 2 ; q 2 )2n
n (bd/q a; q )n

̟ := A−1 B0 =

and compute the finite differences

(1 − q 5k a)(1 − q 2−2k /d)(1 − q 5+2k a2 /b2 d2 )(1 − q 3+3k a/d)
(1 − q 2 a/bd)(1 − q 5 a/bd2 )(1 − b)(1 − q 3 a2 /bd)
[q 2 a/bd, bd2 /q 5 a; q]k (b; q 2 )k (q 3 a2 /bd; q 6 )k
qk ,
× 2
[q bd, q 9 a2 /bd2 ; q 4 ]k (q 3 a/b; q 3 )k (bd/q 4 a; q −1 )k
(1 − q 4+5k a)(1 − q 3+k a/bd)(1 − q 2+2k b)(1 − q 9 a2 /b2 d3 )
∆Bk = −
(1 − q 5 a/bd2 )(1 − q 7 a2 /b2 d2 )(1 − q 6 a/d)(1 − bd)
 4
  
   2 2 3    9 2 2 
q a/bd 
d/q 2  2
b d /q a  3
q a /bd  4
×
q
qk .

q
9 a/d q
4 bd q
bd2 /q 4 a k q 9 a2 /b2 d2
q
q
k
k
k

▽Ak =

Then applying the modified Abel lemma on summation by parts, the U -sum can alternatively
be reformulated as follows:
Un (a, b, d)

(1 − q 2 /d)(1 − q 3 a/d)(1 − q 5 a2 /b2 d2 )
(1 − q 2 a/bd)(1 − q 5 a/bd2 )(1 − b)(1 − q 3 a2 /bd)

Partial Sums of Two Quartic q-Series
=

n−1
X
k=0

7

X

n−1
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

Observing that the last partial sum results in
−

n−1
X

Ak ∆Bk =

k=0

Un (q 4 a, q 4 b, d/q 2 ) (1 − q 2 b)(1 − q 3 a/bd)(1 − q 9 a2 /b2 d3 )
,
(1 − bd)(1 − q 6 a/d)(1 − q 5 a/bd2 )(1 − q 7 a2 /b2 d2 )

we derive after some simplification the relation
(b; q 2 )2 (q 2 a/bd; q)2 (1 − q 3 a2 /bd)(1 − q 9 a2 /b2 d3 )
(q 5 a2 /b2 d2 ; q 2 )2 (q 3 a/d; q 3 )2 (1 − q 2 /d)(1 − bd)

(1 − bd/q 2 )(1 − q 5 a2 /bd2 )(1 − a/b)(1 − q 3 a/bd)
.
+ 1 − R(a, b, d)
(1 − d/q 2 )(1 − q 3 a/d)(1 − q 5 a2 /b2 d2 )

Un (a, b, d) = Un (q 4 a, q 4 b, d/q 2 )

Iterating it m-times, we get the following expression

(b; q 2 )2m (q 2 a/bd; q)2m [q 3 a2/bd, q 9 a2/b2 d3 ; q 6 ]m
(q 5 a2 /b2 d2 ; q 2 )2m (q 3 a/d; q 3 )2m [q 2 /d, bd; q 2 ]m
m−1
(1 − bd/q 2 )(1 − q 3 a/bd)(1 − a/b) X (1 − q 5+8k a2 /bd2 )(b; q 2 )2k

Un (a, b, d) = Un (q 4m a, q 4m b, q −2m d)
+

(1 − d/q 2 )(1 − q 3 a/d)(1 − q 5 a2 /b2 d2 )
(q 7 a2 /b2 d2 ; q 2 )2k
k=0




q 2 a/bd, q 5 a/bd  2 q 3 a2/bd, q 9 a2/b2d3  6 2k
4k
4k
−2k
× 1−R(q a, q b, q d)
q q .
q
q 6 a/d, q 9 a/d
q 4 /d, bd/q 2
k
k

Rewriting the R-function explicitly as
4k

4k

R(q a, q b, q

−2k



1 − q 5+4n+8k a2 /bd2 q 4k b, q −2−2k d  2
d) =
q
q 7+4k a2 /b2 d2
1 − q 5+8k a2 /bd2
n

  (q 3+6k a2 /bd; q 6 )
2+2k
4+2k
2
2
3
[q
a/bd, q
a/bd; q]n
b d /q a  3
n
×
q
−3−2k bd/a; q −1 )
a/b, q 6+6k a/d
(q 2k−2 bd; q 2 )2n
(q
n
  2 2 3 n 


2
4
3
2
6
2
[q a/bd, q a/bd; q]n b, d/q  2
b d /q a  3 (q a /bd; q )n
=
q
q
3
−1
a/b,
q 6 a/d
q 7 a2 /b2 d2
(bd/q 2 ; q 2 )2n
n
n (bd/q a; q )n



1 − q 5+4n+8k a2 /bd2 q 2+n a/bd, q 5+n a/bd, q 4 /d, q −2 bd  2
×
q
q 2 a/bd, q 5 a/bd, q 4−2n /d, q 4n−2 bd
1 − q 5+8k a2 /bd2
k
 2n 7 2 2 2  
6
3
3+6n
2
6
(q a/d; q )2k
(q
a /bd; q )k
q b, q a /b d  2
,
× 6+3n
q
3 a2 /bd; q 6 )
(q
a/d; q 3 )2k b, q 7+2n a2 /b2 d2
(q
k
2k

and defining further the finite cubic sum in base q 2 by
m−1
X

 
q 3 a2 /bd, q 9 a2 /b2 d3  6
Um (a, b, d) =
(1 − q
a /bd )
q
q 6 a/d, q 9 a/d
k
k=0



(b; q 2 )2k
q 2 a/bd, q 5 a/bd  2
× 7 2 2 2 2
q 2k ,
q
q 4 /d, bd/q 2
(q a /b d ; q )2k
k
△

5+8k 2

2



we find the following transformation formula.

Theorem 4 (Transformation between quartic and cubic series).
Un (a, b, d) − Un (q 4m a, q 4m b, q −2m d)

(b; q 2 )2m (q 2 a/bd; q)2m [q 3 a2/bd, q 9 a2/b2 d3 ; q 6 ]m
(q 5 a2/b2 d2 ; q 2 )2m (q 3 a/d; q 3 )2m [q 2/d, bd; q 2 ]m

8

W. Chu and C. Wang

(1 − bd/q 2 )(1 − q 3 a/bd)(1 − a/b)
△
△
(a, b, d) − Um
(q 5n a, q 2n b, q 2n d)
Um
=
(1−d/q 2 )(1−q 3 a/d)(1−q 5 a2/b2 d2 )
  2 2 3  


[q 2 a/bd, q 4 a/bd; q]n b, d/q 2  2
b d /q a  3 (q 3 a2 /bd; q 6 )n
×
.
q
q
3
−1
a/b, q 6 a/d
q 7 a2 /b2 d2
(bd/q 2 ; q 2 )2n
n
n (bd/q a; q )n

By means of the Weierstrass M -test, we can compute the limit
lim Un (q 4m a, q 4m b, q −2m d) =

m,n→∞

∞  
X
a k (b2 d2 /q 3 a; q 3 )k
k=0

b

(q 3 a/b; q 3 )k

k+1
2

q 3(

).

The limiting case m, n → ∞ of Theorem 4 leads to the transformation formula.
Proposition 5 (Nonterminating series transformation).
U (a, b, d) =
+

(1 − bd/q 2 )(1 − q 3 a/bd)(1 − a/b)
U △ (a, b, d)
(1 − d/q 2 )(1 − q 3 a/d)(1 − q 5 a2 /b2 d2 )
∞
(q 2 a/bd; q)∞ (b; q 2 )∞ [q 3 a2/bd, q 9 a2/b2 d3 ; q 6 ]∞ X  q 3 a k (b2 d2/q 3 a; q 3 )k
(q 3 a/d; q 3 )∞ [q 2 /d, bd, q 5 a2 /b2 d2 ; q 2 ]∞

k=0

b

(q 3 a/b; q 3 )k

When b = a, this proposition reduces to the following relation
∞
(q 2 /d; q)∞ (a; q 2 )∞ [q 3 a/d, q 9 /d3 ; q 6 ]∞ X (ad2/q 3 ; q 3 )k 3(k)+3k
.
q 2
U (a, a, d) =
(q 3 a/d; q 3 )∞ [q 2 /d, ad, q 5 /d2 ; q 2 ]∞
(q 3 ; q 3 )k
k=0

Taking d = 1 in the last equation and noting U (a, b, 1) = 1 − a, we get
∞
X
(a/q 3 ; q 3 )k
k=0

(q 3 ; q 3 )k

k

q 3(2)+3k =

(a; q 6 )∞
,
(q 3 ; q 6 )∞

which results also from a limiting case of the q-Bailey–Daum formula (cf. [12, II-9]).
Combining the last two equations leads us to the nonterminating series identity.
Corollary 6 (Gasper–Rahman [12, Exercise 3.29(i)]).

2.3

 

∞
 
X

k
1 − q 5k a a, d  2
ad2/q 3  3 (q 2/d; q)k (q 3 a/d; q 6 )k (k2)
− q 3/d
q
q
q
3
3
5
2
2
q , q a/d
q /d
1−a
(ad; q )2k
k
k
k=0



 2


ad2 , q 9/d3  6
q a, q 3/d  2
=
.
q
q
5
2
q 3 , q 6 a/d
ad, q /d
∞
∞

Quartic series to quartic series

Finally, for the two sequences given by
(q 3 a/bd; q)k (q 2 b; q 2 )k (b2 d2 /a; q 3 )k (q 5 a2 /b2 d; q 4 )k
,
(q 2 bd; q 4 )k (q 3 a/b; q 3 )k (q 5 a2 /b2 d2 ; q 2 )k (b2 d/a; q)k
[qa/bd, b2 d/a; q]k (d/q 2 ; q 2 )k (q 3 a2 /bd; q 6 )k
;
Bk =
[bd, qa2 /b2 d; q 4 ]k (q 3 a/d; q 3 )k (bd/q 2 a; q −1 )k
Ak =

we have no difficulty to check the relations
̟ := A−1 B0 =

(1 − b2 d/qa)(1 − b2 d2 /q 3 a2 )(1 − b/a)(1 − bd/q 2 )
,
(1 − bd/q 2 a)(1 − b)(1 − b2 d2 /q 3 a)(1 − b2 d/qa2 )

k

q 3( 2 ) .

Partial Sums of Two Quartic q-Series

9



1 − q n−1 b2 d/a b, d/q 2  2
An−1 Bn
=
R :=
q
A−1 B0
1 − q −1 b2 d/a q 3 a2 /b2 d2
n


[qa/bd, q 2 a/bd; q]n b2 d2 /q 3 a  3 (q 3 a2 /bd; q 6 )n
×
;
q
2
−1
a/b, q 3 a/d
(bd/q 2 ; q 2 )2n
n (bd/q a; q )n

and compute the finite differences
▽Ak =

(1 − q 5k a)(1 − q 2−2k /d)(1 − q 1+k a/bd)(1 − q 3 a2 /b3 d2 )
(1 − b)(1 − q 2 a/bd)(1 − qa2 /b2 d)(1 − q 3 a/b2 d2 )
 2
  
   2 2 3    2 2  
q a/bd 
b d /q a  3
b
qa /b d  4
 2
× 2
q 3k ,
q
q
q
2 bd q
3
5
2
2
2
q
q
a/b
b d/a
q
a
/b
d
k
k
k
k

(1 − q 1+5k a)(1 − q 3k a/b)(1 − q 2+2k b)(1 − q 3+2k a2 /b2 d2 )
(1 − bd)(1 − qa2 /b2 d)(1 − q 3 a/d)(1 − q 2 a/bd)
[qa/bd, b2 d/a; q]k (d/q 2 ; q 2 )k (q 3 a2 /bd; q 6 )k
q −k .
× 4
[q bd, q 5 a2 /b2 d; q 4 ]k (q 6 a/d; q 3 )k (bd/q 3 a; q −1 )k

∆Bk = −

Then by means of the modified Abel lemma on summation by parts, the U -sum can be reformulated as follows:
Un (a, b, d)

(1 − q 2 /d)(1 − qa/bd)(1 − q 3 a2 /b3 d2 )
(1 − b)(1 − q 2 a/bd)(1 − qa2 /b2 d)(1 − q 3 a/b2 d2 )
n−1
X
X

n−1
=
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

k=0

Writing the last partial sum in terms of U -sum as
−

n−1
X

Ak ∆Bk =

k=0

Un (qa, q 4 b, d/q 2 ) (1 − b/a)(1 − q 2 b)(1 − b2 d2 /q 3 a2 )
,
(1 − bd)(1 − b2 d/qa2 )(1 − q 3 a/d)(1 − bd/q 2 a)

we derive after some simplification the following relation

(1 − a/b)(1 − bd/q 2 )(1 − b2 d/qa)(1 − b2 d2 /q 3 a2 )
Un (a, b, d) = 1 − R(a, b, d)
(1 − d/q 2 )(1 − bd/qa)(1 − b3 d2 /q 3 a2 )
(q 2 a/bd) (b; q 2 )2 (1 − b2 d2 /q 3 a)(1 − b2 d2 /q 3 a2 )(1 − b/a)
.
− Un (qa, q 4 b, d/q 2 )
(1 − bd)(1 − q 2/d)(1 − bd/qa)(1 − q 3 a/d)(1 − b3 d2/q 3 a2 )
Iterating it m-times, we get the following expression
 2 2 3 2 
b d /q a  2
m
4m
−2m
Un (a, b, d) = Un (q a, q b, q
d)
q
q 2 /d, bd
m

  (q 2 a/bd; q −1 )
(b; q 2 )2m
b/a, b2 d2/q 3 a  3
m
×
q
3 a/d
2
3
2
3
2
6
q
[bd/qa, bd/q a; q]m
m (b d /q a ; q )m
+

 2 2 2  
m−1
(1 − a/b)(1 − bd/q 2 )(1 − b2 d2 /q 3 a2 ) X
b d /qa  2
5k−1 2
(1
−
q
b
d/a)
q
q 4 /d, bd/q 2
(1 − d/q 2 )(1 − bd/qa)(1 − b3 d2 /q 3 a2 )
k
k=0


k


k

(b; q 2 )2k q 3 b/a, b2 d2/q 3 a  3
qa/bd q −(2)
q
.
× 1−R(q k a, q 4k b, q −2k d)

3 3 2 2 6
q 3 a/d
(bd/a; q)k
k (q b d /a ; q )k

Separating k and n factorials in the R-function


1 − q n+5k−1 b2 d/a q 4k b, q −2−2k d  2
k
4k
−2k
R(q a, q b, q d) =
q
1 − q 5k−1 b2 d/a q 3−2k a2 /b2 d2
n

10

W. Chu and C. Wang

  (q 3 a2 /bd; q 6 )
[q 1−k a/bd, q 2−k a/bd; q]n
q 3k−3 b2 d2 /a  3
n
×
q
k−2 bd/a; q −1 )
q −3k a/b, q 3+3k a/d
(q 2k−2 bd; q 2 )2n
(q
n
  2 2 3  n


3
2
6
2
2
[qa/bd, q a/bd; q]n b, d/q  2
b d /q a  3 (q a /bd; q )n
=
q
q
3
2
2
2
2
2
2
−1
a/b, q 3 a/d
q a /b d
(bd/q ; q )2n
n (bd/q a; q )n
n


1 − q n+5k−1 b2 d/a q 3n−3 b2 d2 /a, q 3−3n b/a, q 3 a/d  3
×
q nk
q
b2 d2 /q 3 a, q 3 b/a, q 3+3n a/d
1 − q 5k−1 b2 d/a
k


(bd/a; q)k (q 2n b; q 2 )2k q 4 /d, bd/q 2 , q −1−2n b2 d2 /a2  2
× −n
,
q
(q bd/a; q)k (b; q 2 )2k q 4−2n /d, q 4n−2 bd, b2 d2 /qa2
k

and then defining the partial sum of quartic series
⋆
Um
(a, b, d)

m−1
X

  (b; q 2 )
b2 d2 /qa2  2
2k
(1 − q
b d/a) 4
=
q
q /d, bd/q 2
(bd/a;
q)
k
k
k=0


k
 3
  − qa/bd k q −(2)
q b/a, b2 d2 /q 3 a  3
,
×
q
3 3 2 2 6
q 3 a/d
k (q b d /a ; q )k


5k−1 2

we establish the following transformation formula.

Theorem 7 (Transformation between two quartic series).
 2 2 3 2 
b d /q a  2
Un (a, b, d) − Un (q m a, q 4m b, q−2m d)
q
bd, q 2 /d
m
 2 2 3


2
(b; q )2m
(q 2 a/bd; q −1 )m
b d /q a, b/a  3
×
q

3
2
3 2 3 2 6
q a/d
[bd/qa, bd/q a; q]m
m (b d /q a ; q )m

(1 − a/b)(1−bd/q 2 )(1−b2 d2 /q 3 a2 )
⋆
⋆
Um
(a, b, d) − Um
(q 5n a, q 2n b, q 2n d)
=
(1−d/q 2 )(1−bd/qa)(1−b3 d2 /q 3 a2 )


  2 2 3  
[qa/bd, q 2 a/bd; q]n b, d/q 2  2
b d /q a  3 (q 3 a2 /bd; q 6 )n
.
×
q
q
2
−1
q 3 a2 /b2 d2
a/b, q 3 a/d
(bd/q 2 ; q 2 )2n
n (bd/q a; q )n
n
In particular for m = n, we have the reduced transformation.

Proposition 8 (Transformation between two quartic series: q 3 a2 = b2 d2 ).
(a; q 3 )n (bd; q 6 )n
(q 3 a/d; q 3 )n (q 3 a/b; q 3 )n−1


(qa/bd; q)n (q 3 a/bd; q)n−1 q 2 b, d  2
×
.
q
q2
(qa/bd; q −1 )n (bd; q 2 )2n−1
n−1

Un (a, b, d) = Un⋆ (q 5n a, q 2n b, q 2n d)

In order to examine the limiting case n → ∞ of the last equation, we write Un⋆ (q 5n a, q 2n b,
explicitly as follows:

q 3/2+2n a/b)

Un⋆ (q 5n a, q 2n b, q 3/2+2n a/b) =

n−1
X

 
q 2−2n
 2
q
q 5/2−2n b/a, q 4n−1/2 a
k
k=0


 3−3n

k

k
q
b/a, q 3n a  3 (q 2n b; q 2 )2k − q n−1/2
×
q −( 2 ) .
q
3/2+3n
3/2−n
6
6
q
b
; q)k (q b; q )k
k (q
(1 − q n+5k+1/2 b)



Inverting the summation index k → n − 1 − k and then applying the relation
(q 2−2n ; q 2 )n−1−k (q 3−3n b/a; q 3 )n−1−k
(q 5/2−2n b/a; q 2 )n−1−k (q 3/2−n ; q)n−1−k

Partial Sums of Two Quartic q-Series
=

11

(q 2 ; q 2 )n−1 (q 3 a/b; q 3 )n−1 1−n2 (q 1/2 ; q)k (q 3/2 a/b; q 2 )k k2 +2k
q
,
q
(q 2 ; q 2 )k (q 3 a/b; q 3 )k
(q 3/2 a/b; q 2 )n−1 (q 1/2 ; q)n−1

we can reformulate the finite sum Un⋆ (q 5n a, q 2n b, q 3/2+2n a/b) as
n−1

(−1)

×

n−1
1/2 ; q) (q 3/2 a/b; q 2 )
(q 2 ; q 2 )n−1 (q 3 a/b; q 3 )n−1 1−n2 X
k
k k(k+2)
k (q
2
q 2
q
(−1)
(q 2 ; q 2 )k (q 3 a/b; q 3 )k
(q 3/2 a/b; q 2 )n−1 (q 1/2 ; q)n−1
k=0

(1 − q 6n−5k−9/2 b)(q 2n b; q 2 )2n−2−2k (q 3n a; q 3 )n−1−k
.
(q 4n−1/2 a; q 2 )n−1−k (q 3/2+3n b; q 3 )n−1−k (q 6 b; q 6 )n−1−k

Substituting this expression into Proposition 8 and then letting n → ∞, we derive the following
transformation formula
U (a, b, q 3/2 a/b) =

(b; q 2 )∞ (a; q 3 )∞ (q 3/2 a; q 6 )∞
(q 3/2 a; q 2 )∞ (q 3/2 b; q 3 )∞ (b; q 6 )∞
∞ 
k (q 1/2 ; q) (q 3/2 a/b; q 2 ) k
X
k
k ( )
q 2 .
×
−q 3/2
(q 2 ; q 2 )k (q 3 a/b; q 3 )k

(1)

k=0

From this transformation, we can derive two new interesting summation formulae. First,
taking b = 1 in this equation and keeping in mind that U (a, 1, d) = 1, we obtain the following
remarkable summation formula.
Corollary 9 (Nonterminating series identity).
∞ 
k (q 1/2 ; q) (q 3/2 a; q 2 ) k
X
(q 3/2 a; q 2 )∞ (q 3/2 ; q 3 )∞ (q 6 ; q 6 )∞
k
k ( )
2 =
− q 3/2
q
.
(q 2 ; q 2 )k (q 3 a; q 3 )k
(q 2 ; q 2 )∞ (q 3 a; q 3 )∞ (q 3/2 a; q 6 )∞
k=0

The special case a = 0 of this corollary recovers an identity of Rogers–Ramanujan type due
to Stanton [18, p. 61]:
∞
X
(−q; q 2 )k
k=0

(q 4 ; q 4 )k

q k(k+2) =

(−q; q 2 )∞ 6
[q , q, q 5 ; q 6 ]∞ .
(q 2 ; q 2 )∞

Combining (1) with Corollary 9 yields another formula for quartic series.
Corollary 10 (Nonterminating series identity).
 3/2
∞
  (q 1/2 ; q) (q 3/2 a; q 6 ) k2 +2k
  
5k
X
a
a/b  2
 3
k
k
k 1 − q a b, q
(−1)
q 2
q
3 a/b, q 3/2 b q
3/2 a; q 2 )
q2
q
1−a
(q
2k
k
k
k=0
 3/2
   6 3/2  
   3
q a, q 3/2  3
b, q a/b  2
q ,q a  6
=
.
q
q
q
3
3/2
2
3/2
b, q 3/2 a/b
q a/b, q b
q ,q a
∞
∞
∞

3

Transformation and summation formulae for Vn (a, b, d)

The quartic series Vn (a, c, e) may be considered as dual one to the Un (a, b, d) in the last section
in the sense that the numerator factorials and denominator factorials are inverted. This section
will be devoted analogously to investigation of summation and transformation formulae for
Vn (a, c, e). As the series Un (a, c, e), the following expression for Vn (a, c, e) makes its well-poised
structure more transparent
Vn (a, c, e) =

n−1
X
k=0

×

(1 − q 5k a)

[qc, qe; q 3 ]k
(qc2 e2 /a2 ; q 2 )k
qk
(q 2 a3 /c2 e2 ; q 3 )k [qa/c, qa/e; q 2 ]k

[a2 /ce, q 2 a2 /ce; q 4 ]k (q −1 a/ce; q −1 )k
.
[qce/a, qce/a; q]k
(q 5 ce; q 6 )k

12

3.1

W. Chu and C. Wang

Quartic series to quadratic series

Let Ak and Bk be defined by
[q 3 c2 e2 /a2 , a4 /qc3 e3 ; q 2 ]k [q 4 c, q 4 e; q 3 ]k
,
[q 2 a3 /c2 e2 , q 6 c3 e3 /a3 ; q 3 ]k [qa/c, qa/e; q 2 ]k
(q 6 c3 e3 /a3 ; q 3 )k (a2 /ce; q 2 )2k (a/q 2 ce; q −1 )k
Bk = 4 3 3 3 2
.
(a /q c e ; q )k [qce/a, q 3 ce/a; q]k (q 5 ce; q 6 )k

Ak =

We can easily show the following relations
(1 − a/qc)(1 − a/qe)(1 − a3 /qc2 e2 )(1 − q 3 c3 e3 /a3 )
,
(1 − qc)(1 − qe)(1 − qc2 e2 /a2 )(1 − a4 /q 3 c3 e3 )


An−1 Bn
1 − q 3+3n c3 e3 /a3 qc2 e2 /a2  2
R :=
=
q
a/qc, a/qe
A−1 B0
1 − q 3 c3 e3 /a3
n



2
2
2
(a /ce; q )2n
qc, qe  3 (a/q ce; q −1 )n
×
;
q
3
[qce/a, q 3 ce/a; q]n a /qc2 e2
(q 5 ce; q 6 )n
n

̟ := A−1 B0 =

and calculate the finite differences
▽Ak =

(1 − q 5k a)(1 − q 2+k ce/a)(1 − q 2 c2 e3 /a3 )(1 − q 2 c3 e2 /a3 )
(1 − qc)(1 − qe)(1 − qc2 e2 /a2 )(1 − q 3 c3 e3 /a4 )
 2 2 2 4 3 3 3  
 
qc,
qe
qc e /a , a /q c e  2
 3
q 2k ,
×
q
2 a3 /c2 e2 , q 6 c3 e3 /a3 q
q
qa/c,
qa/e
k
k

(1 − q 3+5k a)(1 − q 1−3k c2 e2 /a3 )(1 − q 3+2k c2 e2 /a2 )(1 − q 5+2k c2 e2 /a2 )
(1 − q 3 c3 e3 /a4 )(1 − qce/a)(1 − q 3 ce/a)(1 − q 5 ce)
(q 6 c3 e3 /a3 ; q 3 )k (a2 /ce; q 2 )2k (a/q 2 ce; q −1 )k 2k
q .
× 4 3 3 2
(a /qc e ; q )k [q 2 ce/a, q 4 ce/a; q]k (q 11 ce; q 6 )k

∆Bk = −

Applying the modified Abel lemma on summation by parts, we can manipulate the finite V -sum
as follows:
Vn (a, c, e)

(1 − q 2 ce/a)(1 − q 2 c2 e3 /a3 )(1 − q 2 c3 e2 /a3 )
(1 − qc)(1 − qe)(1 − qc2 e2 /a2 )(1 − q 3 c3 e3 /a4 )
n−1
X
X

n−1
=
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

k=0

Noting that the last partial sum results in
−

n−1
X

Ak ∆Bk = Vn (q 3 a, q 3 c, q 3 e)

k=0

(1 − qc2 e2 /a3 )(1 − q 3 c2 e2 /a2 )(1 − q 5 c2 e2 /a2 )
,
(1 − q 3 c3 e3 /a4 )(1 − qce/a)(1 − q 3 ce/a)(1 − q 5 ce)

we derive after some simplification the recurrence relation
(1 − qc)(1 − qe)(1 − qc2 e2 /a3 )(qc2 e2 /a2 ; q 2 )3
(1 − q 5 ce)(1 − q 2 c2 e3 /a3 )(1 − q 2 c3 e2 /a3 )(qce/a; q)3

a(1 − qc/a)(1 − qe/a)(1 − qc2 e2 /a3 )(1 − q 3 c3 e3 /a3 )
.
− 1 − R(a, c, e)
(1 − q 2 ce/a)(1 − q 2 c3 e2 /a3 )(1 − q 2 c2 e3 /a3 )

Vn (a, c, e) = Vn (q 3 a, q 3 c, q 3 e)

Iterating it m-times, we get the following expression
Vn (a, c, e) = Vn (q 3m a, q 3m c, q 3m e)

[qc, qe, qc2 e2 /a3 ; q 3 ]m (qc2 e2 /a2 ; q 2 )3m
[q 5 ce, q 2 c2 e3 /a3 , q 2 c3 e2 /a3 ; q 6 ]m (qce/a; q)3m

Partial Sums of Two Quartic q-Series

13

a (1 − qc/a)(1 − qe/a)(1 − qc2 e2 /a3 )
(1 − q 2 ce/a)(1 − q 2 c2 e3 /a3 )(1 − q 2 c3 e2 /a3 )

m−1
 
X
qc, qe, q 4 c2 e2 /a3  3
3+9k 3 3 3
×
(1 − q
c e /a )
q 3k
q
qce/a, q 3 ce/a, q 5 ce/a
k
k=0
 2 2 2 3 2 2 2 5 2 2 2 

qc e /a , q c e /a , q c e /a  6
× 1 − R(q 3k a, q 3k c, q 3k e)
.
q
q 5 ce, q 8 c2 e3 /a3 , q 8 c3 e2 /a3
k
−

Writing explicitly the R-function by separating k and n factorials


1 − q 3+3n+9k c3 e3 /a3 q 1+6k c2 e2 /a2  2
3k
3k
3k
R(q a, q c, q e) =
q
a/qc, a/qe
1 − q 3+9k c3 e3 /a3
n
 1+3k 1+3k  
(a2 /ce; q 2 )2n
q
c, q
e  3 (q −2−3k a/ce; q −1 )n
× 1+3k
q
−1−3k
3
2
a /c e2
[q
ce/a, q 3+3k ce/a; q]n q
(q 5+6k ce; q 6 )n
n






(a2 /ce; q 2 )2n
qc2 e2 /a2  2
qc, qe  3 (a/q 2 ce; q −1 )n
=
q

q
3
2
2
a /qc e
[qce/a, q 3 ce/a; q]n a/qc, a/qe
(q 5 ce; q 6 )n
n
n
1 − q 3+3n+9k c3 e3/a3 (q 1+2n c2 e2 /a2 ; q 2 )3k (q 5 ce; q 6 )k
1 − q 3+9k c3 e3 /a3 (qc2 e2 /a2 ; q 2 )3k (q 5+6n ce; q 6 )k
 1+3n 1+3n
 
q
c, q
e, qce/a, q 3 ce/a, q 5 ce/a, q 4−3n c2 e2 /a3  3
×
,
q
qc, qe, q 1+n ce/a, q 3+n ce/a, q 5+n ce/a, q 4 c2 e2 /a3
k
×

and then defining the finite quadratic sum in base q 3 by
Vm⋄ (a, c, e)

m−1
X

 
qc, qe, q 4 c2 e2 /a3  3
=
(1 − q
c e /a )
q
qce/a, q 3 ce/a, q 5 ce/a
k
k=0
 2 2 2 3 2 2 2 5 2 2 2 
qc e /a , q c e /a , q c e /a  6
×
q 3k ,
q
q 5 ce, q 8 c2 e3 /a3 , q 8 c3 e2 /a3
k
3+9k 3 3

3



we obtain the following transformation formula.

Theorem 11 (Transformation between quartic and quadratic series).
[qc, qe, qc2 e2 /a3 ; q 3 ]m (qc2 e2 /a2 ; q 2 )3m
[q 5 ce, q 2 c2 e3 /a3 , q 2 c3 e2 /a3 ; q 6 ]m (qce/a; q)3m
(
(1 − a/qc)(1 − a/qe)(1 − qc2 e2 /a3 )
=
V ⋄ (a, c, e) − Vm⋄ (q 5n a, q 3n c, q 3n e)
(1−a/q 2 ce)(1−q 2 c3 e2/a3 )(1−q 2 c2 e3/a3 ) m
)
 2 2 2   
 
(a2 /ce; q 2 )2n
qc, qe  3 (a/q 2 ce; q −1 )n
qc e /a  2
.
×
q
q
a3 /qc2 e2
[qce/a, q 3 ce/a; q]n a/qc, a/qe
(q 5 ce; q 6 )n
n
n

Vn (a, c, e) − Vn (q 3m a, q 3m c, q 3m e)

Letting n → 1 + m, c → a/q and e → q −1−3m in this theorem, we derive the summation
formula, which does not seem to have explicitly appeared previously.
Corollary 12 (Terminating series identity).
k
 −3−6m   
m
 
X
1 − q 5k a
a, q −3m  3 (q 2+3m a; q 2 )2k (−1)k q (2+3m)k−(2)
q
 2
q
q
−1−3m ; q)
q 6+6m a
q 2 , q 2+3m a
1−a
(q 3−3m a; q 6 )k
k
k (q
k
k=0



 6+3m


q5, q7
q
a, q −3m /a  3
 6
q
.
=

9+3m a, q 3−3m /a q
q
q2, q4
m
m

14

3.2

W. Chu and C. Wang

Quartic series to cubic series

Define two sequences by
(a2 /qc2 e; q)k (q 3 c2 e2 /a2 ; q 2 )k (q 4 c; q 3 )k (q 4 a2 /ce; q 4 )k
,
(q 6 c2 e/a; q 4 )k (q 2 a3 /c2 e2 ; q 3 )k (qa/c; q 2 )k (qce/a; q)k
[q 2 a2 /ce, q 6 c2 e/a; q 4 ]k (qe; q 3 )k (a/qce; q −1 )k
Bk = 2
.
[q ce/a, a2 /q 2 c2 e; q]k (q 3 a/e; q 2 )k (q 5 ce; q 6 )k
Ak =

It is not hard to check the relations
(1 − ce/a)(1 − a/qc)(1 − a3 /qc2 e2 )(1 − q 2 c2 e/a)
,
(1 − a2 /q 2 c2 e)(1 − qc2 e2 /a2 )(1 − qc)(1 − a2 /ce)


An−1 Bn
1 − q 2+4n c2 e/a qc2 e2 /a2  2
R :=
=
q
a/qc, q 3 a/e
A−1 B0
1 − q 2 c2 e/a
n

 
(a2 /ce; q 2 )2n
qc, qe  3 (a/qce; q −1 )n
×
;
q
5
6
[ce/a, q 2 ce/a; q]n a3 /qc2 e2
n (q ce; q )n

̟ := A−1 B0 =

and compute the finite differences
▽Ak =

(1 − q 5k a)(1 − q 1+k ce/a)(1 − q 1+2k a/e)(1 − q 2 c3 e2 /a3 )
(1 − q 2 c2 e/a2 )(1 − qc2 e2 /a2 )(1 − qc)(1 − a2 /ce)
 2 2 2    2 2 2  
   2
 
qc
a /ce  4
qc e /a  2
a /q c e 
 3
qk ,
×
q
q
2 a3 /c2 e2 q
6 c2 e/a q
q
q
qa/c
qce/a
k
k
k
k

(1 − q 4+5k a)(1 − q 1−2k c/a)(1 − q 3+2k c2 e2 /a2 )(1 − q 4+3k c)
(1 − q 2 ce/a)(1 − q 2 c2 e/a2 )(1 − q 3 a/e)(1 − q 5 ce)
[q 2 a2 /ce, q 6 c2 e/a; q 4 ]k (qe; q 3 )k (a/qce; q −1 )k k
q .
× 3
[q ce/a, a2 /qc2 e; q]k (q 5 a/e; q 2 )k (q 11 ce; q 6 )k

∆Bk = −

Then by means of the modified Abel lemma on summation by parts, the V -sum can be reformulated as follows:
Vn (a, c, e)

(1 − qce/a)(1 − qa/e)(1 − q 2 c3 e2 /a3 )
(1 − q 2 c2 e/a2 )(1 − qc2 e2 /a2 )(1 − qc)(1 − a2 /ce)
n−1
X
X

n−1
=
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

k=0

By expressing the last partial sum in terms of V -sum as
−

n−1
X
k=0

Ak ∆Bk =

Vn (q 4 a, q 6 c, e) (1 − qc/a)(1 − q 3 c2 e2 /a2 )(1 − q 4 c)
,
(1 − q 2 ce/a)(1 − q 2 c2 e/a2 )(1 − q 3 a/e)(1 − q 5 ce)

we derive after some simplification the following relation
(qc; q 3 )2 (qc2 e2 /a2 ; q 2 )2 (1 − qc/a)(1 − a2 /ce)
(qce/a; q)2 (qa/e; q 2 )2 (1 − q 5 ce)(1 − q 2 c3 e2 /a3 )

(1 − ce/a)(1 − qc/a)(1 − a3 /qc2 e2 )(1 − q 2 c2 e/a)
.
+ 1 − R(a, c, e)
(1 − a/qce)(1 − qa/e)(1 − q 2 c3 e2 /a3 )

Vn (a, c, e) = Vn (q 4 a, q 6 c, e)

Iterating it m-times, we get the following expression
Vn (a, c, e) = Vn (q 4m a, q 6m c, e)

(qc; q 3 )2m (qc2 e2 /a2 ; q 2 )2m [qc/a, a2 /ce; q 2 ]m
(qce/a; q)2m (qa/e; q 2 )2m [q 5 ce, q 2 c3 e2 /a3 ; q 6 ]m

Partial Sums of Two Quartic q-Series

15
m−1

(1 − ce/a)(1 − qc/a)(1 − a3/qc2 e2 ) X (1−q 2+8k c2 e/a)(qc; q 3 )2k
(1 − a/qce)(1 − qa/e)(1 − q 2 c3 e2/a3 )
[q 5 ce, q 8 c3 e2/a3 ; q 6 ]k
k=0



q 3 c/a, a2 /ce  2 (qc2 e2/a2 ; q 2 )2k 2k
4k
6k
q .
× 1 − R(q a, q c, e)
q
3
2
ce/a, q 3 ce/a
k (q a/e; q )2k

+

Rewriting the R-function explicitly as


 
1 − q 2+4n+8k c2 e/a
q 1+4k c2 e2 /a2
 2
q
q −1−2k a/c, q 3+4k a/e
1 − q 2+8k c2 e/a
n

 1+6k

2k
2
2
(q a /ce; q )2n
q
c, qe  3 (q −1−2k a/ce; q −1 )n
× 2k
q
3
[q ce/a, q 2+2k ce/a; q]n a /qc2 e2
(q 5+6k ce; q 6 )n
n



 

(a2 /ce; q 2 )2n
qc, qe  3 (a/qce; q −1 )n
qc2 e2 /a2  2
q
=
q

5
6
a3 /qc2 e2
[ce/a, q 2 ce/a; q]n a/qc, q 3 a/e
n (q ce; q )n
n



1 − q 2+4n+8k c2 e/a ce/a, q 3 ce/a, q 4n a2 /ce, q 3−2n c/a  2
×
q
q n ce/a, q n+3 ce/a, a2 /ce, q 3 c/a
1 − q 2+8k c2 e/a
k

 1+2n 2 2 2 3

1+3n
3
5
6
(q
c; q )2k (q ce; q )k q
c e /a , q a/e  2
×
,
q
(qc; q 3 )2k (q 5+6n ce; q 6 )k qc2 e2 /a2 , q 3+2n a/e
2k

R(q 4k a, q 6k c, e) =

and defining further the finite cubic sum in base q 2 by
Vm△ (a, c, e) =

m−1
X
k=0

 3
 
(qc; q 3 )2k (qc2 e2 /a2 ; q 2 )2k
q c/a, a2 /ce  2
q 2k ,
(1 − q 2+8k c2 e/a)
q
3
3
2 ) [q 5 ce, q 8 c3 e2 /a3 ; q 6 ]
ce/a, q ce/a
(q
a/e;
q
2k
k
k

we establish the following transformation formula.

Theorem 13 (Transformation between quartic and cubic series).
(qc; q 3 )2m (qc2 e2 /a2 ; q 2 )2m [qc/a, a2 /ce; q 2 ]m
(qce/a; q)2m (qa/e; q 2 )2m [q 5 ce, q 2 c3 e2 /a3 ; q 6 ]m

(1 − ce/a)(1 − qc/a)(1 − a3/qc2 e2 )
V △ (a, c, e) − Vm△ (q 5n a, q 3n c, q 3n e)
=
(1 − a/qce)(1 − qa/e)(1 − q 2 c3 e2/a3 ) m


  
 
(a2 /ce; q 2 )2n
qc2 e2 /a2  2
qc, qe  3 (a/qce; q −1 )n
.
×
q
q
5
6
a3 /qc2 e2
[ce/a, q 2 ce/a; q]n a/qc, q 3 a/e
n
n (q ce; q )n

Vn (a, c, e) − Vn (q 4m a, q 6m c, e)

When n → 1 + δ + 2m, e → q −1/2 a3/2 /c and c = q −1−3δ−6m with δ = 0, 1, the last theorem
recovers the following summation formula.
Corollary 14 (Chu [4, Equation (4.8d)]).

n
X
1 − q 5k a
k=0


a
 2
1
1 q
q 2+3n a, q 2 −3n /a 2

1−a


0,

"
#
2a 
=
q

2


 3/2 3/2 q
q a

3m

"



k


 3
1 1
1
1
3
k
(q 2 a 2 ; q 2 )2k (−q 2 a− 2 )k q 2 +3n a 2 , q −3n  3
q −( 2 )
q
1 1
9 3
3
q
(q 2 a 2 ; q)k (q 2 a 2 ; q 6 )k
k

q 3  6
q
q 9/2 a3/2

n − odd;

#

,

n = 2m.

m

Instead, letting n → 1 + m, e → q −1/2 a3/2 /c and a = q −1−4m in Theorem 13, we recover
another terminating series identity.

16

W. Chu and C. Wang

Corollary 15 (Chu and Wang [8, Corollary 40]).

m
 
  (q −2m ; q 2 ) (−q 1+2m )k 
X
k
1 − q 5k−1−4m
qc, q −1−6m  3
q −1−4m
 2
2k
q −( 2 )
q
q
3
−4m
2+2m
−1−4m
−2m
3−6m
6
q
q
/c, q
c
1−q
; q)k (q
; q )k
k
k (q
k=0
 4       2  
q c 2
q c 6
q  2
.
=
q
q
q
3
2
q
q
q c
m
2m
m

3.3

Quartic series to quartic series

Finally, for the two sequences given by
[q 4 a2 /ce, q 2 ce2 /a; q 4 ]k (q 4 c; q 3 )k (a/ce; q −1 )k
,
[qce/a, q 3 a2 /ce2 ; q]k (qa/c; q 2 )k (q 5 ce; q 6 )k
(q 3 a2 /ce2 ; q)k (c2 e2 /qa2 ; q 2 )k (e/q 2 ; q 3 )k (q 2 a2 /ce; q 4 )k
;
Bk =
(ce2 /q 2 a; q 4 )k (q 2 a3 /c2 e2 ; q 3 )k (q 3 a/e; q 2 )k (ce/qa; q)k
Ak =

we have no difficulty to check the relations
(1 − ce/a)(1 − q 2 a2 /ce2 )(1 − a/qc)(1 − ce/q)
,
(1 − a2 /ce)(1 − ce2 /q 2 a)(1 − qc)(1 − qa/ce)


1 − q 2+n a2 /ce2 c2 e2 /qa2  2
An−1 Bn
=
R :=
q
A−1 B0
1 − q 2 a2 /ce2 a/qc, q 3 a/e
n



2
2
2
(a /ce; q )2n
qc, e/q  3 (qa/ce; q −1 )n
;
×
q
6
[ce/qa, ce/a; q]n q 2 a3 /c2 e2
n (ce/q; q )n

̟ := A−1 B0 =

and compute the finite differences

(1 − q 5k a)(1 − q 3k−2 e)(1 − q 1+2k a/e)(1 − q 2k−1 c2 e2 /a2 )
(1 − a2 /ce)(1 − ce2 /q 2 a)(1 − qc)(1 − ce/qa)
2
[a /ce, ce2 /q 2 a; q 4 ]k (qc; q 3 )k (qa/ce; q −1 )k −k
×
q ,
[qce/a, q 3 a2 /ce2 ; q]k (qa/c; q 2 )k (q 5 ce; q 6 )k
(1 − q 1+5k a)(1 − q k ce/a)(1 − q 2k−1 a/c)(1 − q 4 a3 /c2 e3 )
∆Bk = −
(1 − qa/ce)(1 − q 3 a/e)(1 − q 2 a3 /c2 e2 )(1 − ce2 /q 2 a)
 3 2 2   2 2
  
   2 2
 
q a /ce 
c e /qa2  2
e/q 2  3
q a /ce  4
×
qk .
q
5 a/e q
5 a3 /c2 e2 q
2 ce2 /a q
ce/a
q
q
q
k
k
k
k

▽Ak =

Then according to the modified Abel lemma on summation by parts, the V -sum can be reformulated as follows:
Vn (a, c, e)

(1 − e/q 2 )(1 − qa/e)(1 − c2 e2 /qa2 )
(1 − a2 /ce)(1 − ce2 /q 2 a)(1 − qc)(1 − ce/qa)
n−1
X
X

n−1
=
B k ▽ Ak = ̟ R − 1 −
Ak ∆Bk .
k=0

k=0

Observing that the last partial sum results in
−

n−1
X

Ak ∆Bk =

k=0

Vn (qa, q 3 c, e/q 3 ) (1 − ce/a)(1 − a/qc)(1 − q 4 a3 /c2 e3 )
,
(1 − qa/ce)(1 − q 3 a/e)(1 − q 2 a3 /c2 e2 )(1 − ce2 /q 2 a)

we derive after some simplification the following relation
Vn (a, c, e) = a


(1 − a/ce)(1 − q 2 a2 /ce2 )(1 − qc/a)(1 − q/ce) 
R(a, c, e) − 1
2
2
2
2
(1 − qa /c e )(1 − q /e)(1 − qa/e)

Partial Sums of Two Quartic q-Series
− Vn (qa, q 3 c, e/q 3 )

17

(1 − qc)(1 − qc/a)(1 − a/ce)(1 − a2 /ce)(1 − q 4 a3 /c2 e3 )
.
(ce/qa)(1 − q 2 a3 /c2 e2 )(1 − qa2 /c2 e2 )(1 − q 2 /e)(qa/e; q 2 )2

Iterating it m-times, we get the following expression

 
qc/a, a2 /ce  2
m
3m
−3m
Vn (a, c, e) = Vn (q a, q c, q
e)
q
qa2 /c2 e2
m

  (q 4 a3 /c2 e3 ; q 6 )
[a/ce, qa/ce; q]m
qc
 3
m
×
2 a3/c2 e2 , q 2/e q
2
−1
q
(qa/e; q )2m
n (ce/qa; q )m
(1 − a/ce)(1 − qc/a)(1 − q/ce)
−a
(1 − qa2 /c2 e2 )(1 − q 2 /e)(1 − qa/e)
 3
m−1
  (qa/ce; q)
X
k
q c/a, a2/ce  2
k
2+5k 2
2
×
q (2)
(1 − q
a /ce ) 3 2 2 2 q
3
2
q a /c e
k (q a/e; q )2k
k=0

(qc; q 3 )k (q 4 a3 /c2 e3 ; q 6 )k  q 2 a k
.
−
× 1 − R(q k a, q 3k c, q−3k e)
[q 2 a3 /c2 e2 , q 5 /e; q 3 ]k
ce
Writing explicitly the R-function as
k

3k

R(q a, q c, q

−3k


 
1 − q 2+n+5k a2 /ce2
q −1−2k c2 e2 /a2  2
e) =
q
1 − q 2+5k a2 /ce2 q −1−2k a/c, q 3+4k a/e
n
 1+3k −2−3k  
2k
2
2
(q a /ce; q )2n
q
c, q
e  3 (q 1+k a/ce; q −1 )n
× −1−k
q
2+3k
3
2
2
a /c e
(ce/q; q 6 )n
[q
ce/a, q −k ce/a; q]n q
n






(a2 /ce; q 2 )2n
c2 e2 /qa2  2
qc, e/q 2  3 (qa/ce; q −1 )n
=
q
q
3
6
q 2 a3 /c2 e2
[ce/qa, ce/a; q]n a/qc, q a/e
n
n (ce/q; q )n



1 − q 2+n+5k a2 /ce2 q 3 a2 /c2 e2 , q 4n a2 /ce, q 3−2n c/a  2
×
q −nk
q
q 3−2n a2 /c2 e2 , a2 /ce, q 3 c/a
1 − q 5k+2 a2 /ce2
k

 1+3n 5

1−n
3
2
2
3
2
2
(q
a/ce; q)k (q a/e; q )2k q
c, q /e, q a /c e  3
×
,
q
(qa/ce; q)k (q 3+2n a/e; q 2 )2k qc, q 5−3n /e, q 2+3n a3 /c2 e2
k

and then defining the finite sum of quartic series
Vm⋆ (a, c, e) =

m−1
X

  (qa/ce; q)
q 3 c/a, a2 /ce  2
k
(k2)
q
q
3
2
2
2
3
2
q a /c e
k (q a/e; q )2k
k=0



 q 2 a k
qc
 3
× −
(q 4 a3 /c2 e3 ; q 6 )k ,
q
q 2 a3 /c2 e2 , q 5 /e
ce
k
(1 − q 2+5k a2/ce2 )



we find the following transformation formula.

Theorem 16 (Transformation between two quartic series).

 
qc/a, a2 /ce  2
m
3m
−3m
Vn (a, c, e) − Vn (q a, q c, q
e)
q
qa2 /c2 e2
m

  (q 4 a3 /c2 e3 ; q 6 )
[a/ce, qa/ce; q]m
qc
 3
m
×
2 a3/c2 e2 , q 2/e q
2
−1
q
(qa/e; q )2m
(ce/qa; q )m
 n
(1 − a/qc)(1 − a/ce)(1 − ce/q)
=
V ⋆ (a, c, e) − Vm⋆ (q 5n a, q 3n c, q 3n e)
(1 − qa2/c2 e2 )(1 − e/q 2 )(1 − qa/e) m
 2 2

  
 
(a2 /ce; q 2 )2n
c e /qa2  2
qc, e/q 2  3 (qa/ce; q −1 )n
.
×
q
q
6
q 2 a3 /c2 e2
[ce/qa, ce/a; q]n a/qc, q 3 a/e
n (ce/q; q )n
n

18

W. Chu and C. Wang

Letting n → 1+m, c = q −1−3m and e → q 2+3m , we obtain from the last theorem the following
terminating series identity.
Corollary 17 (Terminating series identity).

m
X
1 − q 5k a
k=0

1−a

=

4

   3+3m −3m   (a2 /q; q 2 ) (−a)k
1+k
q
,q
q 3 /a2
 3
 2
2k
q −( 2 )
q
q
3
2+3m
−1−3m
2
6
6
a
q
a, q
a
k
k (q /a; q)k (q ; q )k

[a/q, qa; q]m (q −3m /a; q 2 )m (q −3m a3 ; q 6 )m
.
(q −1−3m a; q 2 )2m (a3 ; q 3 )m (1/a; q −1 )m

Concluding remarks

Recently, hypergeometric series has been found to have elliptic analogue after the pioneering work
of Frenkel–Turaev [9]. Warnaar [20] derived several terminating elliptic series identities. Further
summation formulae have been established by Chu–Jia [6] through Abel’s lemma on summation
by parts. It is plausible that the same approach works also for the elliptic analogue of the quartic
series. All what we have gotten are two terminating elliptic analogues for Corollary 6 plus one
for Corollary 17. Following the notations of [6], they are produced below for reader’s reference.
Theorem 18 (Terminating elliptic series identity).


m
X
θ(q 5k a; p) a, q 2+m  2
[q −m ; q, p]k [q 1−m a; q 6 , p]k (1+k)−mk
q
,
p
q 2

q 1−2m
θ(a; p)
[q 2+m a; q 2 , p]2k
k
k=0

 1+2m 
χ(m = 2n)[q 2 a; q 2 , p]n [q 3 ; q 6 , p]n
q
a  3
k
× (−1)
.
q , p =
3
1−m
q ,q
a
[q 1+2n ; q 2 , p]n [q 4−2n a; q 6 , p]n
k

Theorem 19 (Terminating elliptic series identity).



m
X
[q 2+2m ; q, p]k [q 3+2m a; q 6 , p]k (k)+(3+2m)k
θ(q 5k a; p) a, q −2m  2
q 2
q , p
5+4m
−2m a; q 2 , p]
q
θ(a; p)
[q
2k
k
k=0
 −3−4m 

[q 2 a; q 2 , p]m [q 5 ; q 2 , p]2m [q −4m a; q 6 , p]m
q
a 
× (−1)k 3 3+2m q 3 , p =
.
q ,q
a
[q 3 ; q 2 , p]m [q −2m a; q 2 , p]2m [q 9 ; q 6 , p]m
k

Theorem 20 (Terminating elliptic series identity).

m
X
θ(q 5k a; p)
k=0



[a2 /q; q 2 , p]2k (−a)k −(1+k)
q 3 /a2
 2
q 2
q , p
2+3m
−1−3m
2
6 6
q
a, q
a
θ(a; p)
k [q /a; q, p]k [q ; q , p]k

 3+3m −3m 
[a/q, qa; q, p]m [q −3m /a; q 2 , p]m [q −3m a3 ; q 6 , p]m
q
,q
 3
q
,
p
×
=
.

a3
[q −1−3m a; q 2 , p]2m [a3 ; q 3 , p]m [1/a; q −1 , p]m
k

Acknowledgments

The work of the second author was partially supported by National Science Foundation of China
(Youth grant 10801026).

Partial Sums of Two Quartic q-Series

19

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