라마누잔-셀베르그 연분수

수학노트
http://bomber0.myid.net/ (토론)님의 2011년 5월 6일 (금) 08:14 판
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introduction
  • [Duke2005] (9.1)
    \(u(\tau)={\sqrt{2}q^{1/8} \over 1+ } {q \over 1+q+} {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\sqrt{2}q^{1/8}\prod_{n=1}^{\infty}(1+q^{n})^{(-1)^{n}}=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}\)
    \(v(\tau)={q^{1/2} \over 1+q + } {q \over 1+q^2+} {q^2 \over 1+q^3} } \cdots=q^{1/2}\prod_{n=1}^{\infty}(1-q^{n})^{(\frac{8}{n})}=q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\)

 

  • Selberg continued fractions [Duke2005] (9.13, 155p)
    \(S_1(q)=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}=u(\tau)=\sqrt{2}\frac{\eta(\tau)\eta^{2}(4\tau)}{\eta^{3}(2\tau)}\)
    \(S_2(q)=q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(q;q^{2})_{\infty}}=q^{1/8}\frac{(-q^{2};q^{2})_{\infty}(q^2;q^{2})_{\infty}}{(q;q^{2})_{\infty}(q^2;q^{2})_{\infty}} =\frac{\eta(4\tau)}{\eta(\tau)}\)
    S1 and S2 are notations from [Chan2009]
  • useful techniques in q-series

 

 

 

relation with other modular functions

 

 

 

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