"Talk on Rogers-Ramanujan identity"의 두 판 사이의 차이

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

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

related items

asymptotic analysis of basic hypergeometric series

@@ 1번째 줄: / 1번째 줄: @@
 ==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>
-* [http://pythagoras0.springnote.com/pages/3004578 로저스-라마누잔 연분수와 항등식]<br>
-* [http://pythagoras0.springnote.com/pages/4855791 로저스 dilogarithm]<br><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><br><math>L(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}</math><br><math>L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}</math><br>
+* <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>
-* <math>q=e^{-t}</math> 으로 두면 <math>t\sim 0</math> 일 때,<br><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><br><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><br>
-* '''[McIntosh1995]''' 참조<br>
-*  이로부터 다음을 알 수 있다<br><math>t\to 0</math> 일 때, <math>q=e^{-t}\to 1</math> 으로 두면<br><math>\frac{H(1)}{G(1)} = \sqrt{\frac{5-\sqrt{5}}{5+\sqrt{5}}}=\varphi-1=0.618\cdots</math><br>
-<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>
-==history==
-* http://www.google.com/search?hl=en&tbs=tl:1&q=
 ==related items==
-* [[asymptotic analysis of basic hypergeometric series]]<br>
+* [[asymptotic analysis of basic hypergeometric series]]
-==encyclopedia==
-* http://en.wikipedia.org/wiki/
-* http://www.scholarpedia.org/
-* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
-==books==
-* [[2010년 books and articles]]<br>
-* http://gigapedia.info/1/
-* http://gigapedia.info/1/
-* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
-[[4909919|4909919]]
-==articles==
-* http://www.ams.org/mathscinet
-* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
-* [http://arxiv.org/ ]http://arxiv.org/
-* http://pythagoras0.springnote.com/
-* http://math.berkeley.edu/~reb/papers/index.html
-* http://dx.doi.org/
-==question and answers(Math Overflow)==
-* http://mathoverflow.net/search?q=
-* http://mathoverflow.net/search?q=
-==blogs==
-*  구글 블로그 검색<br>
-** http://blogsearch.google.com/blogsearch?q=
-** http://blogsearch.google.com/blogsearch?q=
-==experts on the field==
-* http://arxiv.org/
-==links==
-* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
-* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
-* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
-* http://functions.wolfram.com/
-*
 [[분류:개인노트]]
-[[분류:talks]]
 [[분류:talks and lecture notes]]
+[[분류:migrate]]

"Talk on Rogers-Ramanujan identity"의 두 판 사이의 차이

2020년 11월 16일 (월) 10:06 기준 최신판

introduction

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