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

2010년 3월 31일 (수) 04:41 판

Dedekind zeta funciton evaluated at 2 gives a number related to volume of 3-manifold
Bloch-Wigner dilogarithm is involved

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

[[4909919|]]

@@ 12번째 줄: / 12번째 줄: @@
 #  KnotData[]<br> KnotData["FigureEight", "HyperbolicVolume"]<br> N[%, 20]<br>
+*  Dedekind zeta funciton evaluated at 2 gives a number related to volume of 3-manifold<br>
+* [http://pythagoras0.springnote.com/pages/4633853 Bloch-Wigner dilogarithm] is involved<br>
-* [http://pythagoras0.springnote.com/pages/4633853 Bloch-Wigner dilogarithm]<br>
-복소이차수체의 [http://pythagoras0.springnote.com/pages/4533335 데데킨트 제타함수]
-<math>\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|})</math>
-<math>\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})</math>
-<math>\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}))</math>
-#  L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]<br> N[Sum[JacobiSymbol[a, 7]*L[Exp[2 I*Pi*a/7]], {a, 1, 6}], 20]<br> N[L[Exp[2 I*Pi/7]] + L[Exp[4 I*Pi/7]] - L[Exp[6 I*Pi/7]], 20]<br>