"5th order mock theta functions"의 두 판 사이의 차이

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imported>Pythagoras0
imported>Pythagoras0
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==introduction==
 
==introduction==
 
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:<math>f_0(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_{n}}</math>
   
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:<math>f_1(q) = \sum_{n\ge 0} {q^{n^2+n}\over (-q;q)_{n}}</math>
 
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:<math>\phi_0(q) = \sum_{n\ge 0} {q^{n^2}(-q;q^2)_{n}}</math>
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:<math>\phi_1(q) = \sum_{n\ge 0} {q^{(n+1)^2}(-q;q)_{n}}</math>
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:<math>\psi_0(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)/2}(-q;q)_{n}}</math>
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:<math>\psi_1(q) = \sum_{n\ge 0} {q^{n(n+1)/2}(-q;q)_{n}}</math>
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:<math>\chi_0(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n}} = 2F_0(q)-\phi_0(-q)</math>
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:<math>\chi_1(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n+1}} = 2F_1(q)+q^{-1}\phi_1(-q)</math>
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:<math>F_0(q) = \sum_{n\ge 0} {q^{2n^2}\over (q;q^2)_{n}}</math>
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:<math>F_1(q) = \sum_{n\ge 0} {q^{2n^2+2n}\over (q;q^2)_{n+1}}</math>
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:<math>\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})}</math>
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:<math>\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}) }</math>
 
   
 
   
  

2013년 8월 6일 (화) 13:34 판

introduction

\[f_0(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_{n}}\] \[f_1(q) = \sum_{n\ge 0} {q^{n^2+n}\over (-q;q)_{n}}\] \[\phi_0(q) = \sum_{n\ge 0} {q^{n^2}(-q;q^2)_{n}}\] \[\phi_1(q) = \sum_{n\ge 0} {q^{(n+1)^2}(-q;q)_{n}}\] \[\psi_0(q) = \sum_{n\ge 0} {q^{(n+1)(n+2)/2}(-q;q)_{n}}\] \[\psi_1(q) = \sum_{n\ge 0} {q^{n(n+1)/2}(-q;q)_{n}}\] \[\chi_0(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n}} = 2F_0(q)-\phi_0(-q)\] \[\chi_1(q) = \sum_{n\ge 0} {q^{n}\over (q^{n+1};q)_{n+1}} = 2F_1(q)+q^{-1}\phi_1(-q)\] \[F_0(q) = \sum_{n\ge 0} {q^{2n^2}\over (q;q^2)_{n}}\] \[F_1(q) = \sum_{n\ge 0} {q^{2n^2+2n}\over (q;q^2)_{n+1}}\] \[\Psi_0(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q)(1-q^4)(1-q^6)(1-q^9)...(1-q^{5n+1})}\] \[\Psi_1(q) = -1 + \sum_{n \ge 0} { q^{5n^2}\over(1-q^2)(1-q^3)(1-q^7)(1-q^8)...(1-q^{5n+2}) }\]


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