By Robert J. Walker

This advent to algebraic geometry examines how the newer summary suggestions relate to conventional analytical and geometrical difficulties. The presentation is saved as ordinary as attainable, because the textual content can be utilized both for a starting path or for self-study.

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4. 1 is complete. 2 (p. 5). If |b| < 1 and a is an arbitrary complex number, then ∞ ∞ (−1)n (−q; q)n (−aq/b; q)n bn (−1)n (−aq/b; q)n bn q n(n+1)/2 = . 2 (aq; q )n+1 (−b; q)n+1 n=0 n=0 Proof. 1), set h = 2 and t = q 2 , and replace b, c, and a by −b, aq, and aq, respectively. 1) with h = 1, q replaced by q 2 , t = q 2 , and a, b, and c replaced by −b, −bq, and aq 2 , respectively. 1) with q replaced by q 2 , α = −aq/b, β = −bq 2 , and τ = −bq to deduce that, after multiplying both sides by 1/(1 + b), ∞ (−aq/b; q 2 )n (−bq)n 2) (−b; q n+1 n=0 ∞ = (−aq/b; q 2 )n (−aq 2 /b; q 2 )n (−bq 2 )n (−bq)n q 2n (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 ∞ 2 −2n (1 − aq 4n+2 ) 2 (−aq/b; q 2 )n (−aq 2 /b; q 2 )n b2n q 2n +n (1 − aq 4n+2 ) = (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 = = ∞ 2 ∞ 2 (−aq/b; q)2n b2n q 2n +n (1 − aq 4n+2 ) (−b; q)2n+2 n=0 (−aq/b; q)2n b2n q 2n (−b; q)2n+1 n=0 ∞ (−aq/b; q)2n b2n q 2n = (−b; q)2n+1 n=0 +n 1+ 2 +n −bq 2n+1 − aq 4n+2 1 + bq 2n+1 ∞ − (−aq/b; q)2n+1 b2n+1 q (n+1)(2n+1) (−b; q)2n+2 n=0 ∞ = (−1)n (−aq/b; q)n bn q n(n+1)/2 , (−b; q)n+1 n=0 which is the desired result.

11). For any complex number a, ∞ (q 2 ; q 4 )∞ ∞ (aq 2 ; q 2 )n q n(n+1)/2 (aq 2 ; q 4 )n q 4n = (aq 4 ; q 4 )∞ (q; q)n (q 2 ; q 2 )2n n=0 n=0 2 ∞ + (aq 2 ; q 4 )∞ 2 (aq 4 ; q 4 )n q 4n +4n+1 . (q 2 ; q 2 )2n+1 n=0 Proof. 13, replace q by q 2 and set b = −1/q. This yields ∞ 1 an q 2n = 4 ; q 4 ) (−q; q 2 ) 2 ; q 4 ) (−q; q 2 ) (q (aq n n ∞ ∞ n=0 + ∞ (aq 2 ; q 4 )n q 4n (q 2 ; q 2 )2n n=0 1 4 4 (aq ; q )∞ (−q; q 2 )∞ ∞ 2 2 (aq 4 ; q 4 )n q 4n +4n+1 . 1). More precisely, let h = 2, c = −q, and a = 0, and let b tend to 0.

17) which is implicit in the work of Ramanujan in his lost notebook [244]. 2]. 10 involving two additional parameters. 10 (p. 10). ∞ 1 qn = 2 (q)n (q)2∞ n=0 ∞ (−1)n q n(n+1)/2 . 18) n=0 Proof. 1), set h = 1, t = c = q, and a = 0, and then let b → 0. 10 follows immediately. 11 (p. 10). ∞ ∞ 1 q 2n = 2 2 (q) (q) n ∞ n=0 (−1)n q n(n+1)/2 1+2 . 19) n=1 Proof. 1), set h = 1, a = 0, c = q, and t = q 2 . Now let b → 0 to deduce that ∞ ∞ n+1 1 q 2n n1 − q q n(n+1)/2 = (1 − q) (−1) 2 2 (q) (q) 1 − q n ∞ n=0 n=0 1 = (q)2∞ = 1 (q)2∞ ∞ ∞ n n(n+1)/2 (−1) q n=0 (−1)n+1 q (n+1)(n+2)/2 + n=0 ∞ (−1)n q n(n+1)/2 1+2 .

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