Talk on Rogers-Ramanujan identity

introduction

• 틀:수학노트
• 틀:수학노트
• 틀:수학노트

\[L(x)=\operatorname{Li}_2(x)+\frac{1}{2}\log x\log (1-x)=-\frac{1}{2}\int_{0}^{x}\frac{\log(1-y)}{y}+\frac{\log(1-y)}{1-y}dy\] \[L(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}\] \[L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}\]

• \(q=e^{-t}\) 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

\[\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\] \[r(\tau)=q^{\frac{1}{5}} \frac{H(q)}{G(q)} = \cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\] \[r(0)= \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\]

• singular moduli

$$ r(i)=\cfrac{e^{\frac{-2\pi}{5}}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-4\pi}}{1+\cfrac{e^{-6\pi}}{1+\cdots}}}} $$ $$ r(i)={\sqrt{5+\sqrt{5}\over 2}-{\sqrt{5}+1\over 2}} $$

• j-invariant

$$ j(\tau)=-\frac{(r(\tau)^{20}-228r(\tau)^{15}+494r(\tau)^{10}+228r(\tau)^{5}+1)^3}{r(\tau)^{5}(r(\tau)^{10}+11r(\tau)^{5}-1)^5} $$

related items

• asymptotic analysis of basic hypergeometric series