"라마누잔-셀베르그 연분수"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| − | + | ==개요== | |
| − | + | * [[Ramanujan-Göllnitz-Gordon 연분수]] | |
| + | * '''[Duke2005] '''(9.1):<math>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}}</math> | ||
| + | :<math>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}}</math> | ||
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| − | < | + | * 셀베르그 |
| + | :<math>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)}</math> | ||
| + | :<math>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)}</math> S1 , S2은 '''[Chan2009]''' 의 표기 | ||
| + | * [[q-series 의 공식 모음]] | ||
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| − | + | ==메모== | |
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| − | + | * Math Overflow http://mathoverflow.net/search?q= | |
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| − | + | ==관련된 항목들== | |
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| − | + | ==관련논문== | |
| − | * [ | + | * [http://dx.doi.org/0.1090/S0002-9939-09-09835-9 From a Ramanujan-Selberg continued fraction to a Jacobian identity] |
| − | + | ** Hei-Chi ChanJournal: Proc. Amer. Math. Soc. 137 (2009), 2849-2856. | |
| − | * http:// | + | * [http://dx.doi.org/10.1155/IJMMS/2006/54901 Modular relations and explicit values of Ramanujan-Selberg continued fractions] |
| − | * http://www. | + | ** Nayandeep Deka Baruah and Nipen Saikia, 2006 |
| + | * [http://www.ams.org/proc/2002-130-01/S0002-9939-01-06183-4/home.html Explicit evaluations of a Ramanujan-Selberg continued fraction] | ||
| + | ** Liang-Cheng Zhang, 2002 | ||
| − | [[ | + | [[분류:q-급수]] |
| − | + | [[분류:연분수]] | |
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2020년 12월 28일 (월) 02:16 기준 최신판
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
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
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