# New PDF release: Ramanujan's Lost Notebook: Part I

By George E. Andrews

ISBN-10: 038725529X

ISBN-13: 9780387255293

ISBN-10: 038728124X

ISBN-13: 9780387281247

Within the library at Trinity university, Cambridge in 1976, George Andrews of Pennsylvania country college stumbled on a sheaf of pages within the handwriting of Srinivasa Ramanujan. quickly distinctive as "Ramanujan’s misplaced Notebook," it includes substantial fabric on mock theta services and surely dates from the final yr of Ramanujan’s existence. during this ebook, the workstation is gifted with extra fabric and specialist statement.

Similar nonfiction_3 books

Download e-book for iPad: Rock Damage and Fluid Transport, Part I (Pageoph Topical by Georg Dresen, Ove Stephansson, Arno Zang

Mechanical houses and fluid delivery in rocks are in detail associated as deformation of an outstanding rock matrix instantly impacts the pore area and permeability. this can bring about brief or everlasting alterations of pore pressures and potent pressures inflicting rock power to alter in area and time.

Additional resources for Ramanujan's Lost Notebook: Part I

Example text

Replacing q by −q in (i) and subtracting the result from (i), we ﬁnd that (ϕ(q) − ϕ(−q)) + ϕ(q 5 ) − ϕ(−q 5 ) = 2q 4/5 R−1 (q 4 ) f (q, q 9 ) − f (−q, −q 9 ) . With the use of [61, p. 5), the equation above can be rewritten in the form 4qψ(q 8 ) + 4q 5 ψ(q 40 ) = 4q 9/5 R−1 (q 4 ) ∞ q 20n 2 +12n . n=−∞ We now deduce (iii) from the equation above by dividing both sides by 4q and then replacing q by q 1/4 . So it suﬃces to prove (i). 2, ϕ(−q) + ϕ(−q 5 ) = (q; q)∞ (q 5 ; q 5 )∞ + (−q; q)∞ (−q 5 ; q 5 )∞ = (q; q)∞ (−q; q)∞ 1+ (−q; q)∞ (q 5 ; q 5 )∞ (q; q)∞ (−q 5 ; q 5 )∞ = (q; q)∞ (−q; q)∞ 1+ (−q; q 5 )∞ (−q 2 ; q 5 )∞ (−q 3 ; q 5 )∞ (−q 4 ; q 5 )∞ (q; q 5 )∞ (q 2 ; q 5 )∞ (q 3 ; q 5 )∞ (q 4 ; q 5 )∞ = (q; q)∞ (−q; q)∞ 1+ f (q, q 4 )f (q 2 , q 3 ) f (−q, −q 4 )f (−q 2 , −q 3 ) = (q; q)∞ (−q; q)∞ f (−q, −q 4 )f (−q 2 , −q 3 ) + f (q, q 4 )f (q 2 , q 3 ) f (−q, −q 4 )f (−q 2 , −q 3 ) .

7. 1. 9), then (i) (ii) χ(q 5 ) , χ(q) χ(−q) . ψ 2 (q) − 5qψ 2 (q 5 ) = f 2 (−q) χ(−q 5 ) ϕ2 (q) − 5ϕ2 (q 5 ) = −4f 2 (−q 2 ) Proof of (i). 4), ϕ2 (−q) − 5ϕ2 (−q 5 ) = (ϕ2 (−q) − ϕ2 (−q 5 )) ψ 2 (q) . 9), ϕ2 (−q) − 5ϕ2 (−q 5 ) = −4f (−q, −q 9 )f (−q 3 , −q 7 ) = −4 (q 2 ; q 2 )2∞ (q 5 ; q 10 )2∞ (q; q 2 )2∞ (q 10 ; q 10 )2∞ (q; q 2 )∞ (q 2 ; q 2 )2∞ (q 5 ; q 10 )2∞ (q 5 ; q 10 )∞ (q; q 2 )2∞ 28 1 Rogers–Ramanujan Continued Fraction – Modular Properties = −4f 2 (−q 2 ) (q 5 ; q 10 )∞ (q; q 2 )∞ = −4f 2 (−q 2 ) χ(−q 5 ) .

The proof is similar if ω = exp(4πi/3). 3. Let a, b, c, and d be√any real numbers and√assume that we consider only principal arguments. If 3 a + bi = c + di, then 3 a − bi = c − di. √ Proof. Let 3 a + bi = c + di. Then a + bi = (c + di)3 = c3 − 3cd2 + (3c2 d − d3 ) i. Hence a = c3 − 3cd2 and b = 3c2 d − d3 . Therefore a − bi = c3 − 3cd2 − (3c2 d − d3 ) i = (c − di)3 . √ Since we consider only the principal argument, 3 a − bi = c − di, which proves the lemma. 2. 11) 47 which is found on page 321 in Ramanujan’s second notebook [227]; see [39, p.