Throughout this paper, a , x and q are complex number with | q | < 1 . Here and in what follows, we adopt the standard q-series notation [1] . For any positive integer n,

( a ; q ) 0 : = 1 ,   ( a ; q ) n : = ∏ k = 0 n − 1 ( 1 − a q k ) ,   ( a ; q ) ∞ : = ∏ k = 0 ∞ ( 1 − a q k ) ,

( a 1 , a 2 , a 3 , ⋯ , a m ; q ) n : = ( a 1 ; q ) n ( a 2 ; q ) n ( a 3 ; q ) n ⋯ ( a m ; q ) n ,

( a 1 , a 2 , a 3 , ⋯ , a m ; q ) ∞ : = ( a 1 ; q ) ∞ ( a 2 ; q ) ∞ ( a 3 ; q ) ∞ ⋯ ( a m ; q ) ∞ .

For convenience, we use ( a ) n to denote ( a ; q ) n . We will often use basic properties without reference, such as

( a ; q ) n + k = ( a ; q ) n ( a q n ; q ) k ,   ( a ; q ) ∞ = ( a ; q ) n ( a q n ; q ) ∞ .

The q-binomial coefficient is given for any nonnegative integers M and N by

[ N M ] q : = [ N M ] = ( ( q ; q ) N ( q ; q ) M ( q ; q ) N − M , if   N ≥ M , 0, otherwise .

Following Gasper and Rahman [1] , the bilateral basic hypergeometric series is defined by

r ψ s [ a 1 , a 2 , ⋯ , a r b 1 , b 2 , ⋯ , b s ; q , z ] = ∑ n = − ∞ ∞ ( a 1 , a 2 , ⋯ , a r ; q ) n ( b 1 , b 2 , ⋯ , b s ; q ) n { ( − 1 ) n q ( n 2 ) } s − r z n ,

and the unilateral basic hypergeometric series is defined by

r ϕ s [ a 1 , a 2 , ⋯ , a r b 1 , b 2 , ⋯ , b s ; q , z ] = ∑ n = 0 ∞ ( a 1 , a 2 , ⋯ , a r ; q ) n ( q , b 1 , b 2 , ⋯ , b s ; q ) n { ( − 1 ) n q ( n 2 ) } 1 + s − r z n .

The Ramanujan’s 1 ψ 1 sum ( [1] , Appendix (II. 29)), and the sum of a 1 ϕ 1 series, ( [1] , Appendix (II.5)) are stated as follows.

1 ψ 1 [ a b ; q , z ] = ( q , b / a , a z , q / a z ; q ) ∞ ( b , q / a , z , b / a z ; q ) ∞ , (1.1)

1 ϕ 1 [ a c ; q , c a ] = ( c / a ; q ) ∞ ( c ; q ) ∞ . (1.2)

Among the other formulas needed for this paper, we have separately documented the q-binomial formula and its consequence ( [1] (1.3.2)) in the sequel,

∑ n ≥ 0 ( a ; q ) n ( q ; q ) n x n = ( a x ; q ) ∞ ( x ; q ) ∞ . (1.3)

Setting q → q 2 , a → ∞ , x → x q / a in (1.3), we have

∑ n ≥ 0 ( − 1 ) n q n 2 x n ( q 2 ; q 2 ) n = ( x q ; q 2 ) ∞ . (1.4)

Furthermore, Euler’s formulas (cf. [1] , Corollary 2.2)

∑ n ≥ 0 q ( n 2 ) x n ( q ; q ) n = ( − x , q ) ∞   and   ∑ n ≥ 0 x n ( q ; q ) n = 1 ( x , q ) ∞ . (1.5)

Cauchy’s identity (cf. [2] , Theorem 3.3)

( x ; q ) n = ∑ k = 0 n ( − 1 ) k q ( k 2 ) [ n k ] x k , (1.6)

and the formula ( [1] , Exercise 1.16) which is a special case of the Bailey-Daum sum ( [1] , (1.8.1))

∑ n ≥ 0 ( x ; q ) n q ( n + 1 2 ) ( q ; q ) n = ( − q ; q ) ∞ ( x q ; q 2 ) ∞ . (1.7)

Besides, if a ≠ 0,1 − b q n ≠ 0 , the following identity was given by Ramanujan [3] ,

∑ n ≥ 0 ( − b / a ; q ) n a n q n ( n + 1 ) / 2 ( q ; q ) n ( b q ; q ) n = ( − a q ; q ) ∞ ( b q ; q ) ∞ . (1.8)

A pair of sequences ( α n , β n ) is called a Bailey pair relative to a if

β n = ∑ r = 0 n α r ( q ; q ) n − r ( a q ; q ) n + r . (1.9)

And a conjugate Bailey pair relative to a is a pair of sequences ( δ n , γ n ) satisfying

γ n = ∑ k = n ∞ δ k ( q ; q ) k − n ( a q ; q ) k + n . (1.10)

In fact, the Bailey pair and the conjugate Bailey pair are the special cases of the following Bailey transform.

Lemma 1.1 ( [4] ). Let n be a nonnegative integer and { A n } n = 0 ∞ , { B n } n = 0 ∞ , { C n } n = 0 ∞ , { D n } n = 0 ∞ be sequences of complex numbers. Assuming convergence of the series, if

B n = ∑ j = 0 n   A j U n − j V n + j   and   C n = ∑ j = n ∞   D j U j − n V j + n , (1.11)

then

∑ n = 0 ∞   A n C n = ∑ n = 0 ∞   B n D n .

This terminology was first proposed by Slater [5] . Bailey transform is widely used in mathematics for a long time, especially, in the area of basic hypergeometric series. For example, Andrews [6] , Kim and Lovejoy [7] , and Lovejoy [8] established multiple sums Rogers-Ramanujan type identities and partial theta identities. Andrews and Warnaar [9] applied the Bailey transform to give another proof of false theta functions. Bailey [4] [10] , Bressoud [2] , and Slater [11] [12] used this transform to derive a number of identities of Rogers-Ramanujan type identities. Ji and Zhao [13] established the Hecke-Rogers identities for the universal mock theta functions by means of the Bailey transform.

Notice that if we take

U n = 1 ( q ; q ) n , V n = 1 ( a q ; q ) n

in lemma 1.1, the pair of sequences ( A n , B n ) is a Bailey pair relative to a, and the pair of sequences ( C n , D n ) is a conjugate Bailey pair relative to a.

The main motivation for this work came from some Bailey’s transform of M. E. Bachraoui which appear in [14] , specifically the ( V n ) is the constant sequences with value 1. Applying the ( [14] , Theorem 3, 4, 6) and choosing appropriate D n , we establish some new transformation formulas for q-series.

Theorem 2.1. We have

( a q 1 2 ; q ) ∞ ∑ n ≥ 0 ( a q 1 2 ) n ( 1 − q n ) ( x ; q ) n ( q ; q ) n 2 = ∑ n ≥ 0 ( − 1 ) n ( a x ) n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 . (2.1)

Corollary 2.2. There holds

x n ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n = ∑ k = 0 n ( − 1 ) k q ( k + 1 2 ) − n k ( 1 − q k ) ( x ; q ) k ( q ; q ) k [ n k ] , (2.2)

( q ; q ) ∞ ∑ n ≥ 1 q n ( − q ; q ) n − 1 ( q ; q ) n ( q ; q ) n − 1 = ∑ n ≥ 1 q ( n + 1 2 ) ( − q ; q ) n − 1 ( q ; q ) n ( q ; q ) n − 1 , (2.3)

( − 1 ; q ) ∞ ∑ n ≥ 1 ( − 1 ) n ( − q ; q ) n − 1 ( q ; q ) n ( q ; q ) n − 1 = ∑ n ≥ 1 ( − 1 ) n q ( n 2 ) ( − q ; q ) n − 1 ( q ; q ) n ( q ; q ) n − 1 . (2.4)

Theorem 2.3. We have

( c / a ; q ) ∞ ( c ; q ) ∞ ∑ n ≥ 0 ( c / a ) n ( 1 − q n ) ( x , a ; q ) n ( q ; q ) n 2 = ∑ n ≥ 0 ( − 1 ) n ( c x / a ) n q ( n 2 ) ( 1 − q n ) ( x − 1 , a ; q ) n ( q ; q ) n 2 ( c ; q ) n . (2.5)

Corollary 2.4. There holds

1 ( c ; q ) ∞ ∑ n ≥ 0 ( − 1 ) n c n q ( n 2 ) ( 1 − q n ) ( x ; q ) n ( q ; q ) n 2 = ∑ n ≥ 0 ( c x ) n q n ( n − 1 ) ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 ( c ; q ) n , (2.6)

∑ n ≥ 0 q ( n 2 ) ( 1 − q n ) ( q ; q ) n 2 ( − q n ; q ) ∞ = ∑ n ≥ 0 q n ( n − 1 ) ( 1 − q n ) ( q ; q ) n 2 . (2.7)

Theorem 2.5. We have

( a q 1 2 ; q ) ∞ ∑ n ≥ 0 ( a q 1 2 ) n ( x ; q ) n ( q ; q ) n = ∑ n ≥ 0 ( − 1 ) n ( a x ) n q n 2 2 ( q ; q ) n . (2.8)

Theorem 2.6. There holds

( c / a ; q ) ∞ ( c ; q ) ∞ ∑ n ≥ 0 ( c / a ) n ( x , a ; q ) n ( q ; q ) n = ∑ n ≥ 0 ( − 1 ) n ( c x / a ) n q ( n 2 ) ( a ; q ) n ( q , c ; q ) n . (2.9)

Corollary 2.7. We have

∑ n ≥ 0 ( c / a ) n ( a q k ; q ) n ( q ; q ) n = ( c q k ; q ) ∞ ( c / a ; q ) ∞ , (2.10)

1 ( c ; q ) ∞ ∑ n ≥ 0 ( − 1 ) n c n q ( n 2 ) ( x ; q ) n ( q ; q ) n = ∑ n ≥ 0 ( c x ) n q n ( n − 1 ) ( c , q ; q ) n . (2.11)

Remark. Setting c = − q , x = q in (2.11), we derive

1 ( − q ; q ) ∞ ∑ n = 0 ∞ q ( n + 1 2 ) = ∑ n ≥ 0 ( − 1 ) n q n 2 + n ( q 2 ; q 2 ) n .

Furthermore, we have

∑ n = 0 ∞   q ( n + 1 2 ) = ( q 2 ; q 2 ) ∞ ( − q ; q ) ∞ .

Combining the above two identities, we arrive at

∑ n ≥ 0 ( − 1 ) n q n 2 + n ( q 2 ; q 2 ) n = ( q 2 ; q 2 ) ∞ .

The above identity is a companion to number (23) on Slater’s list [12] as follows.

∑ n ≥ 0 ( − 1 ) n q n 2 ( q 2 ; q 2 ) n = ( q ; q 2 ) ∞ .

Theorem 2.8. We have

( − q 2 ; q 2 ) ∞ ∑ n ≥ 0 q n 2 ( x ; q 2 ) n ( x q 2 n + 2 ; q 4 ) ∞ ( q 2 ; q 2 ) n = ∑ n ≥ 0 q ( n + 1 2 ) ( x ; q 2 ) n ( q ; q ) n . (2.12)

Corollary 2.9. There holds

( − q 2 ; q 2 ) ∞ ∑ n ≥ 0   q n 2 ( q 2 n + 4 ; q 4 ) ∞ = ∑ n ≥ 0   q ( n + 1 2 ) ( − q ; q ) n , (2.13)

( − q 2 ; q 2 ) ∞ ∑ n ≥ 0 q n 2 ( q ; q 2 ) n ( q 2 n + 3 ; q 4 ) ∞ ( q 2 ; q 2 ) n = ∑ n ≥ 0 q ( n + 1 2 ) ( q ; q 2 ) n ( q ; q ) n . (2.14)

Theorem 2.10. We have

( − a q 3 ; q 2 ) ∞ ( b q 2 ; q 2 ) ∞ ∑ n ≥ 0 a n q n 2 + n ( − b a q ; q 2 ) n ( q 2 ; q 2 ) n = ∑ n ≥ 0 a n q ( n + 2 2 ) − 1 ( − b a q ; q 2 ) n ( q ; q ) n ( b q 2 ; q 2 ) n . (2.15)

Proof of Theorem 2.1. Setting

A n = ( − 1 ) n q n 2 ( 1 − q n ) ( x ; q ) n ( q ; q ) n 2 ,   U n = q n 2 2 ( q ; q ) n , B n = x n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 ,   D n = ( − 1 ) n a n

and V n = 1 in Lemma 1.1, we obtain

C n = ∑ k = n ∞   D k U k − n V k + n = ∑ k ≥ 0   D n + k U k = ∑ k ≥ 0 ( − a ) n + k q k 2 2 ( q ; q ) k = ( − a ) n ∑ k ≥ 0 ( − 1 ) k a k q k 2 2 ( q ; q ) k = ( − 1 ) n a n ( a q 1 2 ; q ) ∞ , (3.1)

where the last step follows by (1.4).

Thus,

∑ n ≥ 0   A n C n = ( a q 1 2 ; q ) ∞ ∑ n ≥ 0 ( a q 1 2 ) n ( 1 − q n ) ( x ; q ) n ( q ; q ) n 2 = ∑ n ≥ 0   B n D n = ∑ n ≥ 0 ( − 1 ) n ( a x ) n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 .

This completes the proof.

Proof of Corollary 2.2. Based on (1.4), we can rewrite (2.1) as follows

∑ m ≥ 0 ( − 1 ) m q m 2 2 a m ( q ; q ) m ∑ k ≥ 0 q k 2 ( 1 − q k ) ( x ; q ) k a k ( q ; q ) k 2 = ∑ n ≥ 0 ( − 1 ) n x n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n a n ( q ; q ) n 2 .

Equating terms of the corresponding powers of a n , we achieve

∑ k = 0 n ( − 1 ) n − k q ( n − k ) 2 2 ( q ; q ) n − k ⋅ q k 2 ( 1 − q k ) ( x ; q ) k ( q ; q ) k 2 = ( − 1 ) n x n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 ,

which is (2.2) by some basic simplifications. (2.3) (2.4) follow from (2.1) upon letting a = q 1 2 , x = − 1 ( a = − q − 1 2 , x = − 1 ), respectively.

Proof of Theorem 2.3. We now apply Lemma 1.1 with

A n = ( − 1 ) n q n 2 ( 1 − q n ) ( x ; q ) n ( q ; q ) n 2 , U n = q n 2 2 ( q ; q ) n , B n = x n q n 2 2 ( 1 − q n ) ( x − 1 ; q ) n ( q ; q ) n 2 , D n = ( − 1 ) n ( c / a ) n q − n 2 ( a ; q ) n ( c ; q ) n ,

and V n = 1 . We compute

C n = ∑ k ≥ 0 ( a ; q ) n + k ( c ; q ) n + k ( − 1 ) n + k ( c / a ) n + k q − n + k 2 ⋅ q k 2 2 ( q ; q ) k = ( − 1 ) n ( c / a ) n q − n 2 ( a ; q ) n ( c ; q ) n ∑ k ≥ 0 ( a q n ; q ) k ( c q n , q ; q ) k ( − 1 ) k q ( k 2 ) ( c / a ) k = ( − 1 ) n ( c / a ) n q − n 2 ( a ; q ) n ⋅ ( c / a ; q ) ∞ ( c ; q ) ∞ , (4.1)

where the last step follows by (1.2). Thus, we arrive at

∑ n ≥ 0   A n C n = ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ n ≥ 0 ( c / a ) n ( 1 − q n ) ( x , a ; q ) n ( q ; q ) n 2 = ∑ n ≥ 0   B n D n = ∑ n ≥ 0 ( − 1 ) n ( c x / a ) n q ( n 2 ) ( 1 − q n ) ( x − 1 , a ; q ) n ( q ; q ) n 2 ( c ; q ) n ,

which completes the proof.

Proof of Corollary 2.4. (2.6) follows from (2.5) upon letting a → ∞ and (2.7) follows from (2.6) by letting x = c = − 1 , which proves the desired formula.

Proof of Theorem 2.5. Setting

A n = ( − 1 ) n q n 2 ( x ; q ) n ( q ; q ) n ,   U n = q n 2 2 ( q ; q ) n ,   B n = x n q n 2 2 ( q ; q ) n ,   D n = ( − 1 ) n a n ,

and V n is the constant sequences with value 1 in Lemma 1.1. Due to (3.1), we have

C n = ( − 1 ) n a n ( a q 1 2 ; q ) ∞ .

Thus,

∑ n ≥ 0   A n C n = ( a q 1 2 ; q ) ∞ ∑ n ≥ 0 ( a q 1 2 ) n ( x ; q ) n ( q ; q ) n = ∑ n ≥ 0   B n D n = ∑ n ≥ 0 ( − 1 ) n ( a x ) n q n 2 2 ( q ; q ) n ,

which is (2.8).

Notice that we use (1.4) to express the left-hand side of (2.8)

∑ k ≥ 0 ( − 1 ) k q k 2 2 a k ( q ; q ) k ∑ m ≥ 0 q m 2 ( x ; q ) m a m ( q ; q ) m = ∑ n ≥ 0 ( − 1 ) n x n q n 2 2 a n ( q ; q ) n . (5.1)

Now we equate the terms corresponding to a n in (5.1) to obtain

∑ k = 0 n ( − 1 ) n − k q ( n − k ) 2 2 ( q ; q ) n − k ⋅ q k 2 ( x ; q ) k ( q ; q ) k = ( − 1 ) n x n q n 2 2 ( q ; q ) n ,

which gives the following identity after straightforward simplifications.

x n = ∑ k = 0 n ( − 1 ) k q ( k + 1 2 ) − n k ( x ; q ) k [ n k ] .

The above identity appears in ( [4] , p. 157).

Proof of Theorem 2.6. We apply Lemma 1.1 with

A n = ( − 1 ) n q n 2 ( x ; q ) n ( q ; q ) n ,   U n = q n 2 2 ( q ; q ) n ,   B n = x n q n 2 2 ( q ; q ) n , D n = ( − 1 ) n ( c / a ) n q − n 2 ( a ; q ) n ( c ; q ) n ,

and V n is the constant sequences with value 1. Then due to (4.1), we get

C n = ( − 1 ) n ( c / a ) n q − n 2 ( a ; q ) n ⋅ ( c / a ; q ) ∞ ( c ; q ) ∞ .

Thus,

∑ n ≥ 0   A n C n = ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ n ≥ 0 ( c / a ) n ( a , x ; q ) n ( q ; q ) n = ∑ n ≥ 0   B n D n = ( − 1 ) n ( c x / a ) n q ( n 2 ) ( a ; q ) n ( q , c ; q ) n ,

which yields the desired formula.

Proof of Corollary 2.7. To prove (2.10), we first use (1.6) to express the left-hand side of (2.9) as powers series in x. Then

L. H. S. of (2.9) = ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ n ≥ 0 ( c / a ) n ( a ; q ) n ( q ; q ) n ∑ k = 0 n ( − 1 ) k q ( k 2 ) [ n k ] x k

= ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ k ≥ 0 ( − 1 ) k q ( k 2 ) x k ( q ; q ) k ∑ n = k ∞ ( c / a ) n ( a ; q ) n ( q ; q ) n − k = ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ k ≥ 0 ( − 1 ) k q ( k 2 ) x k ( q ; q ) k ∑ n = 0 ∞ ( c / a ) n + k ( a ; q ) n + k ( q ; q ) n = ( c / a ; q ) ∞ ( c ; q ) ∞ ∑ k ≥ 0 ( ∑ n ≥ 0 ( − 1 ) k q ( k 2 ) ( c / a ) n + k ( a ; q ) n + k ( q ; q ) k ( q ; q ) n ) x k .

Now equate the terms corresponding to x k in (2.9) to obtain

∑ n ≥ 0 ( − 1 ) k q ( k 2 ) ( c / a ) n + k ( a ; q ) n + k ( q ; q ) k ( q ; q ) n = ( c ; q ) ∞ ( c / a ; q ) ∞ ⋅ ( − 1 ) k ( c / a ) k q ( k 2 ) ( a ; q ) k ( c , q ; q ) k ,

which by some basic calculations down to (2.10). (2.11) follows easily from (2.9) upon letting a → ∞ .

Proof of Theorem 2.8. Let us use Lemma 1.1 with a Bailey transform as follows:

A n = q n 2 + n ( q 2 ; q 2 ) n ,     U n = q n 2 + 2 n ( q 2 ; q 2 ) n ,     B n = q ( n + 2 2 ) − 1 ( q ; q ) n ,     D n = q − n ( x ; q 2 ) n ,     V n = 1.

We compute

C n = ∑ k ≥ 0   q − n − k ( x ; q 2 ) n + k ⋅ q k 2 + 2 k ( q 2 ; q 2 ) k = q − n ( x ; q 2 ) n ∑ k ≥ 0 q k ( k + 1 ) ( x q 2 n ; q 2 ) k ( q 2 ; q 2 ) k = q − n ( x ; q 2 ) n ( − q 2 ; q 2 ) ∞ ( x q 2 n + 2 ; q 4 ) ∞ ,

where in the last step, we used the (1.7) with q → q 2 , x → x q 2 n . Then by virtue of Lemma 1.1, we get

∑ n ≥ 0   A n C n = ( − q 2 ; q 2 ) ∞ ∑ n ≥ 0 q n 2 ( x ; q 2 ) n ( x q 2 n + 2 ; q 4 ) ∞ ( q 2 ; q 2 ) n = ∑ n ≥ 0   B n D n = ∑ n ≥ 0 q ( n + 1 2 ) ( x ; q 2 ) n ( q ; q ) n ,

which completes the proof.

Proof of Corollary 2.9. (2.13) and (2.14) follow from (2.12) by letting, respectively, x = q 2 and x = q .

Proof of Theorem 2.10. Let us use Lemma 1.1 with a Bailey transform as follows:

A n = q n 2 + n ( q 2 ; q 2 ) n ,     U n = q n 2 + 2 n ( q 2 ; q 2 ) n ,     B n = q ( n + 2 2 ) − 1 ( q ; q ) n ,     D n = a n ( − b a q ; q 2 ) n ( b q 2 ; q 2 ) n ,     V n = 1.

We compute

C n = ∑ k ≥ 0 a n + k ( − b a q ; q 2 ) n + k ( b q 2 ; q 2 ) n + k ⋅ q k 2 + 2 k ( q 2 ; q 2 ) k = a n ( − b a q ; q 2 ) n ( b q 2 ; q 2 ) n ∑ k ≥ 0 a k q k 2 + 2 k ( − b q 2 n + 1 / a ; q 2 ) k ( q 2 ; q 2 ) k ( b q 2 n + 2 ; q 2 ) k = a n ( − b a q ; q 2 ) n ( − a q 3 ; q 2 ) ∞ ( b q 2 ; q 2 ) ∞ ,

where in the last step, we used (1.8) by letting a = a q , b = b q 2 n . Then by virtue of Lemma 1.1, we get

∑ n ≥ 0   A n C n = ( − a q 3 ; q 2 ) ∞ ( b q 2 ; q 2 ) ∞ ∑ n ≥ 0 a n q n 2 + n ( − b a q ; q 2 ) n ( q 2 ; q 2 ) n = ∑ n ≥ 0   B n D n = ∑ n ≥ 0 a n q ( n + 2 2 ) − 1 ( − b a q ; q 2 ) n ( q ; q ) n ( b q 2 ; q 2 ) n ,

which completes the proof.

By choosing some sequences, we can derive many identities from the Bailey transform. Furthermore, we should study the generalized Bailey transform [14] deeply to establish the multiple parameterized identities. On the other hand, we can also study the mock theta functions or the Rogers-Ramanujan identities through the Bailey transform.

The authors declare no conflicts of interest regarding the publication of this paper.

Hao, L.J. and Xu, L.L. (2023) Some New Transformation Formulas for q-Series through the Bailey Transform. Advances in Pure Mathematics, 13, 651-661. https://doi.org/10.4236/apm.2023.1310045