Algebraic and Geometric Topology , Edition: 3rd by James R. Milgram

By James R. Milgram

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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. 8 Identities Involving the Parameter k = R(q)R2 (q 2 ) Recall again that R(q) denotes the Rogers–Ramanujan continued fraction.

Once again, observe that as q → 0+ , u = O(q 1/5 ), v = O(q 3/5 ), R = O(q −1/3 ), and uR = O(q −2/15 ). Thus, as q → 0+ , we find that u → 0+ , v → 0+ , and uR → +∞. Hence if ω = exp(2πi/3), then u6 + 8u + 4uRω is a value in the second √ quadrant as q → 0+ . Therefore, if we consider the principal argument, then u6 + 8u + 4uRω = x + iy is a value in the first quadrant. 11) 53 √ y → +∞. 41) the expression for 4v4 approaches −∞ as q → 0+ . So v4 is not a solution. Therefore v3 is the desired solution.

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. 27, Entry 20] and [63, p. 17, Entry 3]. 11 in Chapter 3 of this book. The only two proofs in the literature are due to Rogers [236] and Yi [299].

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