By George E. Andrews, Bruce C. Berndt

Within the spring of 1976, George Andrews of Pennsylvania kingdom college visited the library at Trinity university, Cambridge, to check the papers of the overdue G.N. Watson. between those papers, Andrews found a sheaf of 138 pages within the handwriting of Srinivasa Ramanujan. This manuscript used to be quickly certain "Ramanujan's misplaced notebook." The "lost computing device" includes enormous fabric on mock theta capabilities and so certainly emanates from the final yr of Ramanujan's lifestyles. it's going to be emphasised that the cloth on mock theta capabilities could be Ramanujan's inner most paintings.

**Read Online or Download Ramanujan's Lost Notebook: Part II PDF**

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**Extra info for Ramanujan's Lost Notebook: Part II**

**Sample text**

106–107], [75, p. 182]. 19 (p. 26). Deﬁne the coeﬃcients cn , n ≥ 0, by ∞ cn λn := n=0 (−aλ)∞ . (bλ)∞ (cλ)∞ Then ∞ ∞ cn q n(n+1)/2 = (−cq)∞ n=0 (−a/b)n bn q n(n+1)/2 . 23) Proof. 3), ∞ cn λn = n=0 (−aλ)∞ 1 (bλ)∞ (cλ)∞ ∞ = (−a/b)m (bλ)m (q)m m=0 ∞ λn = n=0 m+k=n Hence, for n ≥ 0, cn = m+k=n ∞ k=0 (cλ)k (q)k (−a/b)m bm ck . (q)m (q)k (−a/b)m bm ck . 4), ∞ ∞ cn q n(n+1)/2 = n=0 n=0 m+k=n ∞ = (−a/b)m bm ck q (m+k)(m+k+1)/2 (q)m (q)k (−a/b)m bm q m(m+1)/2 (q)m m=0 ∞ = m m(m+1)/2 (−a/b)m b q (q)m m=0 ∞ k=0 (cq m+1 )k q k(k−1)/2 (q)k (−cq m+1 )∞ ∞ = (−cq)∞ (−a/b)m bm q m(m+1)/2 , (q)m (−cq)m m=0 which is what we wanted to prove.

2 and Auxiliary Results 37 The latter entry and the next entry are the analytic versions of the two famous G¨ ollnitz–Gordon identities [157]. They can also be found in Slater’s list [262, equations (36), (34)], but with q replaced by −q. The G¨ollnitz– Gordon identities have played a seminal role in the subsequent development of the theory of partition identities. They were ﬁrst studied in this regard by H. G¨ ollnitz [156] and by B. Gordon [158], [159]. A generalization by Andrews [10] led to a number of further discoveries culminating in [16].

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 .