By George E. Andrews
In the spring of 1976, George Andrews of Pennsylvania country college visited the library at Trinity university, Cambridge, to check the papers of the past due G.N. Watson. between those papers, Andrews came upon a sheaf of 138 pages within the handwriting of Srinivasa Ramanujan. This manuscript used to be quickly distinct, "Ramanujan's misplaced notebook." Its discovery has usually been deemed the mathematical similar of discovering Beethoven's 10th symphony.
The "lost computing device" includes substantial fabric on mock theta features and so certainly emanates from the final 12 months of Ramanujan's existence. it's going to be emphasised that the cloth on mock theta features may be Ramanujan's private paintings. Mathematicians are most likely a number of a long time clear of a whole knowing of these capabilities. greater than 1/2 the cloth within the ebook is on q-series, together with mock theta services; the remainder half bargains with theta functionality identities, modular equations, incomplete elliptic integrals of the 1st type and different integrals of theta services, Eisenstein sequence, specific values of theta services, the Rogers-Ramanujan persisted fraction, different q-continued fractions, different integrals, and components of Hecke's thought of modular forms.
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Additional info for Ramanujan's Lost Notebook
11 in Chapter 3 of this book. The only two proofs in the literature are due to Rogers  and Yi . 12) is quartic in each of u and v. We thus use Ferrari’s method [277, pp. 94–96] to solve for each of u and v. 13) uv − u v − 3u v + v − u = 0. 13) as a quartic equation in u, we rewrite it in the form u4 + 1 3 3 2 1 u + u − vu − 2 = 0. 15) First, we brieﬂy explain Ferrari’s method. 16) we ﬁrst determine a, b, and k such that p x4 + px3 + qx2 + rx + s + (ax + b)2 = x2 + x + k 2 2 . 17). 18) 2ab + r = kp, ⎪ ⎪ ⎩ 2 b +s = k2 .
4), we deduce that √ ψ(q 2 ) + q 5ψ(q 10 ) = = (−ζ 3 q 2 ; q 2 )∞ (−ζ 2 q 2 ; q 2 )∞ (q 2 ; q 2 )∞ (ζq; q 2 )∞ (−ζ 2 q; q)∞ (−ζ 3 q; q)∞ (ζ 4 q; q 2 )∞ (q 2 ; q 2 )∞ (ζq; q 2 )∞ (−ζ 2 q; q 2 )∞ (−ζ 3 q; q 2 )∞ (ζ 4 q; q 2 )∞ f (−q 2 ) = n odd which proves (iii). 8 Identities Involving the Parameter k = R(q)R2 (q 2 ) 33 Proof of (iv). The proof of (iv) is similar to that of (ii). 4) to ﬁnd that √ (q 2 ; q 4 )∞ ψ(q 2 ) − q 5ψ(q 10 ) = f 2 (−q 2 ) 10 20 (q ; q )∞ = f (−q 2 ) = 1 √ ψ(q ) + q 5ψ(q 10 ) 2 (ζq; q 2 )∞ (−ζ 2 q; q 2 )∞ (−ζ 3 q; q 2 )∞ (ζ 4 q; q 2 )∞ (ζq 2 ; q 4 )∞ (ζ 2 q 2 ; q 4 )∞ (ζ 3 q 2 ; q 4 )∞ (ζ 4 q 2 ; q 4 )∞ f (−q 2 ) (−ζq; q 2 )∞ (ζ 2 q; q 2 )∞ (ζ 3 q; q 2 )∞ (−ζ 4 q; q 2 )∞ f (−q 2 ) = (1 − αq n + q 2n ) n odd (1 + βq n + q 2n ) , n odd as desired.
12) are congruent to 3 5 (mod 1). 12), we conclude that −q 3/5 f 3 (−q 5 ) 3 + R3 (q) R2 (q) ∞ = −q 1/5 (−1)n (10n + 3)q (5n+3)n/2 n=−∞ and q 3/5 f 3 (−q 5 ) 1 − 3R2 (q) R3 (q) ∞ (−1)n (10n + 1)q n(5n+1)/2 . 2) now follow, respectively, from the last two equalities. 11) It had been thought that Ramanathan  published the ﬁrst proof of the factorization theorems below. However, possibly due to an attempt to be brief, the argument for a key step is absent. This important step, an application of an addition theorem for theta functions due to Ramanujan and found in Ramanujan’s notebooks , is perhaps the most diﬃcult part of the proof.