"5th order mock theta functions"의 두 판 사이의 차이
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2020년 11월 16일 (월) 09:49 기준 최신판
introduction
\[f_0(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_{n}}\] \[f_1(q) = \sum_{n\ge 0} {q^{n^2+n}\over (-q;q)_{n}}\] \[\phi_0(q) = \sum_{n\ge 0} {q^{n^2}(-q;q^2)_{n}}\] \[\phi_1(q) = \sum_{n\ge 0} {q^{(n+1)^2}(-q;q)_{n}}\] \[\psi_0(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)/2}(-q;q)_{n}}\] \[\psi_1(q) = \sum_{n\ge 0} {q^{n(n+1)/2}(-q;q)_{n}}\] \[\chi_0(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n}} = 2F_0(q)-\phi_0(-q)\] \[\chi_1(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n+1}} = 2F_1(q)+q^{-1}\phi_1(-q)\] \[F_0(q) = \sum_{n\ge 0} {q^{2n^2}\over (q;q^2)_{n}}\] \[F_1(q) = \sum_{n\ge 0} {q^{2n^2+2n}\over (q;q^2)_{n+1}}\] \[\Psi_0(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q)(1-q^4)(1-q^6)(1-q^9)...(1-q^{5n+1})}\] \[\Psi_1(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q^2)(1-q^3)(1-q^7)(1-q^8)...(1-q^{5n+2}) }\]
articles
- Nickolas Andersen, Vector-valued modular forms and the Mock Theta Conjectures, arXiv:1604.05294 [math.NT], April 18 2016, http://arxiv.org/abs/1604.05294
- G.E. Andrews, The fifth and seventh order mock theta functions. Trans. Amer. Math. Soc. 293 (1986), pp. 113–134
- Andrews, George E. (1988), "Ramanujan's fifth order mock theta functions as constant terms", Ramanujan revisited (Urbana-Champaign, Ill., 1987),
- Basil Gordon and Richard J. Mcintosh Modular Transformations of Ramanujan's Fifth and Seventh Order Mock Theta Functions, 2003