Ramanujan-Göllnitz-Gordon 연분수

개요

• Göllnitz\[1+q+{q^{2} \over 1+q^{3} + } {q^{4} \over 1+q^{5}+} {q^{6} \over \cdots}=\frac{(q^{3};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}=\frac{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}\]
• [Gordon1965]

\[1+{q \over 1+q^2 + } {q^3 \over 1+q^4+} {q^5 \over 1+q^6} \cdots=\frac{(q^{2};q^{8})_{\infty}(q^{3};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{5};q^{8})_{\infty}(q^{6};q^{8})_{\infty}}\]

 

라마누잔의 결과

• Berndt, notebook V entry 22 p. 50\[{1 \over 1+} {q+q^2 \over 1+} {q^4 \over 1+} {q^3+q^6 \over 1+}{q^8 \over 1+\cdots} =\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\]

 

 

 

모듈라 함수

• fractional power

\[{q^{1/2} \over 1+q+} {q^2 \over 1+q^3 + } {q^4 \over 1+q^5 + } {q^6 \over 1+q^7+\cdots} =q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\]

• [Duke2005] (9.4)

 

 

메모

 

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

 

 

 

 

리뷰논문, 에세이, 강의노트

• W. Duke [Duke2005]Continued fractions and modular functions, Bull. Amer. Math. Soc. 42 (2005), 137-162

 

관련논문

• [Gordon1965] Basil Gordon Some continued fractions of the Rogers-Ramanujan type,  Duke Math. J. Volume 32, Number 4 (1965), 741-748.