라마누잔-셀베르그 연분수

개요

• Ramanujan-Göllnitz-Gordon 연분수
• [Duke2005] (9.1)\[u(\tau)={\sqrt{2}q^{1/8} \over 1+ } {q \over 1+q+} {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\sqrt{2}q^{1/8}\prod_{n=1}^{\infty}(1+q^{n})^{(-1)^{n}}=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}\]

\[v(\tau)={q^{1/2} \over 1+q + } {q \over 1+q^2+} {q^2 \over 1+q^3} \cdots=q^{1/2}\prod_{n=1}^{\infty}(1-q^{n})^{(\frac{8}{n})}=q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\]

• 셀베르그

\[S_1(q)=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}=u(\tau)=\sqrt{2}\frac{\eta(\tau)\eta^{2}(4\tau)}{\eta^{3}(2\tau)}\] \[S_2(q)=q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(q;q^{2})_{\infty}}=q^{1/8}\frac{(-q^{2};q^{2})_{\infty}(q^2;q^{2})_{\infty}}{(q;q^{2})_{\infty}(q^2;q^{2})_{\infty}} =\frac{\eta(4\tau)}{\eta(\tau)}\] S1 , S2은 [Chan2009] 의 표기

• q-series 의 공식 모음

메모

• Math Overflow http://mathoverflow.net/search?q=

관련논문

• From a Ramanujan-Selberg continued fraction to a Jacobian identity
  • Hei-Chi ChanJournal: Proc. Amer. Math. Soc. 137 (2009), 2849-2856.
• Modular relations and explicit values of Ramanujan-Selberg continued fractions
  • Nayandeep Deka Baruah and Nipen Saikia, 2006
• Explicit evaluations of a Ramanujan-Selberg continued fraction
  • Liang-Cheng Zhang, 2002