By Andrews G.E., Berndt B.P. (eds.)

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3) and 2 ψ2 aq (q/c, q/d, aq/e, aq/f ; q)∞ e, f ; q, = aq/c, aq/d ef (aq, q/a, aq/(cd), aq/(ef ); q)∞ √ √ ∞ a3 q (q a, −q a, c, d, e, f ; q)n √ √ × ( a, − a, aq/c, aq/d, aq/e, aq/f ; q)n cdef n=−∞ n 2 qn . 4). However, one can also give a proof in the spirit of Ismail’s aforementioned proof [184]. 9). 2) in full, because the identity holds on a convergent sequence within the domain of analyticity |γ| < 1. 2) to itself. 1) of Part I [31, p. 262]. To see this, replace c by aq/γ. 1) from Part I. 4) in general, since the identity holds on a convergent sequence within the domain of analyticity |γ| < 1.

The desired result now follows by analytic continuation in a. 5. Our second proof is taken from a paper by Berndt, Kim, and Yee [73]. 3) can be written in the equivalent form ∞ ∞ am q m(m+1) (−aq)n (−aq 2n+2 ; q 2 )∞ . 2 and Auxiliary Results 27 Note that (−aq 2n+2 ; q 2 )∞ generates partitions into distinct even parts, each greater than or equal to 2n + 2, with the exponent of a denoting the number of parts. Let m be the number of parts generated by a partition arising from (−aq 2n+2 ; q 2 )∞ .

M. Somos has observed that if we set ∞ F (a, b; q) := (−bq; q 2 )∞ 2 an q n , 2 2 (q ; q )n (−bq; q 2 )n n=0 then F (a, b; q) = F (b, a; q). 1). 12, set n = 1, and replace q by q 2 , a by a/q, and b by b/q. 1). 2 (p. 42). If a is any complex number, then ∞ ∞ 2 2 a2n q 4n an q n = (aq; q 2 )∞ . 4 4 2 2 (q ; q )n (q ; q )n (aq; q 2 )n n=0 n=0 Proof. 6). Then set b = −aq/t and let t → 0. 3 (p. 26). ∞ (q; q 2 )∞ 2 ∞ q 2n +n (−1)n q n(n+1)/2 = . (q 2 ; q 2 )n (q 2 ; q 2 )n n=0 n=0 √ √ Proof. 2, replace q by q, and then set a = − q.

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