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| − | + | ==introduction== | |
| + | * {{수학노트|url=로저스-라마누잔_항등식}} | ||
| + | * {{수학노트|url=로저스-라마누잔_연분수}} | ||
| + | * {{수학노트|url=로저스_다이로그_함수_(Rogers'_dilogarithm)}} | ||
| + | :<math>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</math> | ||
| + | :<math>L(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}</math> | ||
| + | :<math>L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}</math> | ||
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| − | + | * <math>q=e^{-t}</math> 으로 두면 <math>t\sim 0</math> 일 때, | |
| + | :<math>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)</math> | ||
| + | :<math>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)</math> | ||
| + | * '''[McIntosh1995]''' 참조 | ||
| + | * 이로부터 다음을 알 수 있다<math>t\to 0</math> 일 때, <math>q=e^{-t}\to 1</math> 으로 두면 | ||
| + | :<math>\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots</math> | ||
| + | :<math>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}}}}</math> | ||
| + | :<math>r(0)= \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots</math> | ||
| + | * singular moduli | ||
| + | :<math> | ||
| + | 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}}}} | ||
| + | </math> | ||
| + | :<math> | ||
| + | r(i)={\sqrt{5+\sqrt{5}\over 2}-{\sqrt{5}+1\over 2}} | ||
| + | </math> | ||
| + | * j-invariant | ||
| + | :<math> | ||
| + | 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} | ||
| + | </math> | ||
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| − | + | ==related items== | |
| − | + | * [[asymptotic analysis of basic hypergeometric series]] | |
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| − | + | [[분류:개인노트]] | |
| − | + | [[분류:talks and lecture notes]] | |
| − | + | [[분류:migrate]] | |
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2020년 11월 16일 (월) 10:06 기준 최신판
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
\[\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\] \[r(\tau)=q^{\frac{1}{5}} \frac{H(q)}{G(q)} = \cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\] \[r(0)= \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots\]
- 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
\[ j(\tau)=-\frac{(r(\tau)^{20}-228r(\tau)^{15}+494r(\tau)^{10}+228r(\tau)^{5}+1)^3}{r(\tau)^{5}(r(\tau)^{10}+11r(\tau)^{5}-1)^5} \]