Talk on Rogers-Ramanujan identity
imported>Pythagoras0님의 2012년 10월 28일 (일) 17:01 판 (찾아 바꾸기 – “<h5 (.*)">” 문자열을 “==” 문자열로)
introduction
- 로저스-라마누잔 연분수와 항등식
- 로저스 dilogarithm
\(L(x)=\operatorname{Li}_2(x)+\frac{1}{2}\log x\log (1-x)=-\frac{1}{2}\int_{0}^{x}\frac{\log(1-y)}{y}+\frac{\log(1-y)}{1-y}dy\)
\(L(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}\)
\(L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}\)
- \(q=e^{-t}\) 으로 두면 \(t\sim 0\) 일 때,
\(H(q)=\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\)
\(G(q)=\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\) - [McIntosh1995] 참조
- 이로부터 다음을 알 수 있다
\(t\to 0\) 일 때, \(q=e^{-t}\to 1\) 으로 두면
\(\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\)
\(r(\tau)=q^{\frac{1}{5}} \frac{H(q)}{G(q)} = \cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\)
\(r(0)= \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\)
history
encyclopedia
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
articles
- http://www.ams.org/mathscinet
- [1]http://www.zentralblatt-math.org/zmath/en/
- [2]http://arxiv.org/
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html
- http://dx.doi.org/
question and answers(Math Overflow)
blogs
experts on the field