Talk on Rogers-Ramanujan identity
introduction
\[L(x)=\operatorname{Li}_2(x)+\frac{1}{2}\log x\log (1-x)=-\frac{1}{2}\int_{0}^{x}\frac{\log(1-y)}{y}+\frac{\log(1-y)}{1-y}dy\] \[L(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}\] \[L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}\]
- \(q=e^{-t}\) 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
\[\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\]
\[r(\tau)=q^{\frac{1}{5}} \frac{H(q)}{G(q)} = \cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\]
\[r(0)= \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\]