"3rd order mock theta functions"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
imported>Pythagoras0
11번째 줄: 11번째 줄:
 
 
 
 
  
==shadow==
 
 
* <math>\Theta(24z)=q-5q^{25}+7q^{49}-11q^{121}+13q^{169}-\cdots</math>
 
* <math>M_f(z)=q^{-1}f(q^{24})+\frac{i}{\sqrt{3}}\int_{}^{}\frac{\Theta(24z)}{}dz</math>
 
 
 
 
 
 
<math>f(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_n^2} =1+\sum_{n\ge 1} \frac{q^{n^2}}{(1+q)^2(1+q^2)^2\cdots{(1+q^{n})^2}}={2\over \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^nq^{3n^2/2+n/2}\over 1+q^n}</math>
 
  
 
==harmonic weak Maass==
 
==harmonic weak Maass==
28번째 줄: 19번째 줄:
  
 
 
 
 
 +
 +
==shadow==
 +
 +
* <math>\Theta(24z)=q-5q^{25}+7q^{49}-11q^{121}+13q^{169}-\cdots</math>
 +
* <math>M_f(z)=q^{-1}f(q^{24})+\frac{i}{\sqrt{3}}\int_{}^{}\frac{\Theta(24z)}{}dz</math>
 +
 +
<math>f(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_n^2} =1+\sum_{n\ge 1} \frac{q^{n^2}}{(1+q)^2(1+q^2)^2\cdots{(1+q^{n})^2}}={2\over \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^nq^{3n^2/2+n/2}\over 1+q^n}</math>
  
 
 
 
 

2012년 11월 2일 (금) 05:56 판

introduction

 

 

 

 


harmonic weak Maass

We have a weight k=1/2, harmonic weak Maass form $h_3$ under \(\Gamma(2)\) defined by \[h_3(\tau)=q^{-1/24}f(q)+R_3(q)\] where \(R_3(\tau)=\sum_{n\equiv 1\pmod 6}\operatorname{sgn}(n)\beta_{1/2}(n^2y/6)q^{-n^2/24}\) where \(\displaystyle \beta(t) = \int_t^\infty u^{-1/2} e^{-\pi u} \,du=2\int_{\sqrt{x}}^{\infty} e^{-\pi t^2}\,dt\) Note that this can be rewritten as \[R_3(\tau)=(i/2)^{k-1} \int_{-\overline\tau}^{i\infty} (z+\tau)^{-k}\overline{g(-\overline z)}\,dz\]

  • shadow = weight 3/2 theta function \[g_3(z)=\sum_{n\equiv 1\pmod 6}nq^{n^2/24}\]

 

shadow

  • \(\Theta(24z)=q-5q^{25}+7q^{49}-11q^{121}+13q^{169}-\cdots\)
  • \(M_f(z)=q^{-1}f(q^{24})+\frac{i}{\sqrt{3}}\int_{}^{}\frac{\Theta(24z)}{}dz\)

\(f(q) = \sum_{n\ge 0} {q^{n^2}\over (-q;q)_n^2} =1+\sum_{n\ge 1} \frac{q^{n^2}}{(1+q)^2(1+q^2)^2\cdots{(1+q^{n})^2}}={2\over \prod_{n>0}(1-q^n)}\sum_{n\in Z}{(-1)^nq^{3n^2/2+n/2}\over 1+q^n}\)

 

articles

 

 

history

 

 

related items

 

 

encyclopedia


 

 

books

 


 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links