"3-manifolds and their invariants"의 두 판 사이의 차이

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Clausen function
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<math>\mathfrak{I}(\operatorname{Li}_2(e^{i\theta}))=\sum_{n=1}^\infty \frac{\sin n\theta}{n^2}=Cl_2(\theta)</math>
  
 
 
 
 

2010년 3월 28일 (일) 06:13 판

introduction
  • volume of knot complements

 

 

Volume of knot complement
  1. KnotData[]
    KnotData["FigureEight", "HyperbolicVolume"]
    N[%, 20]

 

 

 

복소이차수체의 데데킨트 제타함수

\(\zeta_{K}(2)=\frac{\pi^2}{6\sqrt{|d_K|}}\sum_{(a,d_k)=1} (\frac{d_K}{a})D(e^{2\pi ia/|d_k|})\)

\(\zeta_{\mathbb{Q}\sqrt{-3}}(2)=\frac{\pi^2}{6\sqrt{3}}(D(e^{2\pi i/3})-D(e^{4\pi i/3}))=\frac{\pi^2}{3\sqrt{3}}D(e^{2\pi i/3})\)

\(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))\)

  1. L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]
    N[Sum[JacobiSymbol[a, 7]*L[Exp[2 I*Pi*a/7]], {a, 1, 6}], 20]
    N[L[Exp[2 I*Pi/7]] + L[Exp[4 I*Pi/7]] - L[Exp[6 I*Pi/7]], 20]

 

 

an open problem

\(\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=\frac{2}{\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))=\frac{2}{\sqrt{7}}(Cl(2\pi /7})+Cl(4\pi/7})-Cl(6\pi/7}))\)

 

Clausen function

\(\mathfrak{I}(\operatorname{Li}_2(e^{i\theta}))=\sum_{n=1}^\infty \frac{\sin n\theta}{n^2}=Cl_2(\theta)\)

 

 

 

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[[4909919|]]

 

 

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원본 주소 "https://wiki.mathnt.net/index.php?title=3-manifolds_and_their_invariants&oldid=43535"