"Ramanujan-Göllnitz-Gordon 연분수"의 두 판 사이의 차이

2012년 11월 2일 (금) 07:28 판

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

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

fractional power
\(q^{1/2}({1 \over {1+q+}} {q^2 \over 1+q^3 + } {q^4 \over 1+q^5 + {}} {q^6 \over 1+q^7} } \cdots)=q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\)
[Duke2005] (9.4)

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

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