Download Ramanujan's Lost Notebook: Part I by George E. Andrews PDF

By George E. Andrews

Within the library at Trinity collage, Cambridge in 1976, George Andrews of Pennsylvania nation college chanced on a sheaf of pages within the handwriting of Srinivasa Ramanujan. quickly detailed as "Ramanujan’s misplaced Notebook," it comprises enormous fabric on mock theta services and definitely dates from the final 12 months of Ramanujan’s existence. during this publication, the computing device is gifted with extra fabric and specialist remark.

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Extra resources for Ramanujan's Lost Notebook: Part I

Example text

In his notebooks [227, p. 362], Ramanujan introduced the parameter k := R(q)R2 (q 2 ) and asserted that R5 (q) = k 1−k 1+k 2 and R5 (q 2 ) = k 2 1+k 1−k . 1), see [39, Entry 24] or [63, pp. 12–14, Entry 1(i)]. 2 in Chapter 2. To prove the several identities involving k stated by Ramanujan in his lost notebook, we need the following relations between the Rogers–Ramanujan continued fraction and theta functions. 1 (p. 26). Let µ := µ(q) := R(q)R(q 4 ) and ν := ν(q) := R2 (q 1/2 )R(q)/R(q 2 ). Then (i) (ii) ϕ(q) 1+µ = , 1−µ ϕ(q 5 ) 1+ν ψ(q) = .

First recall that v − u2 . 3) v + u2 This modular equation is found in Ramanujan’s notebooks [227, vol. 2, p. 326]; the first proof was given in [39, p. 31, Entry 24(i)] and later reproduced in [63, Chapter 32, Entry 1, p. 12]. It is also given in a fragment with the publication of his lost notebook [228, p. 365, Entry (10)(a)]. 10 of Chapter 3. 3), we find that uv 2 = vw2 = w − v2 . 6) by u2 and then adding the resulting equalities. Accordingly, v (uw + u2 w2 )v 2 + (u3 w + u2 )v + w(u2 w2 − 1) = v(1 + uw) uwv 2 + u2 v + w(uw − 1) = 0.

6) and Proof. 5). 5) follows easily by noting that α = −(ζ + ζ −1 ); and so ζ 3 − ζ 4 = α(1 − ζ 2 ). 3 with n = 5, a = −ζ, and b = −ζ 4 q 1/5 , and the observations made above, f (−ζ, −ζ 4 q 1/5 ) = f (−q 2 , −q 3 ) − ζf (−q 3 , −q 2 ) + ζ 2 q 1/5 f (−q 4 , −q) − ζ 3 q 3/5 f (−q 5 , −1) + ζ 4 q 6/5 f (−q 6 , −q −1 ) = (1 − ζ) f (−q 2 , −q 3 ) − βq 1/5 f (−q, −q 4 ) , since ζ 2 + ζ 3 = −β. 6). 2. Let n be a positive integer not divisible by 5, and set ζ = e2πi/5 . Then 4 (1 + αζ nj q n/5 + ζ 2nj q 2n/5 ) = (1 − q n )2 .

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