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

2010년 3월 28일 (일) 06:41 판

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

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

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

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]

Prove
\(\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=\frac{2}{\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))=\frac{2}{\sqrt{7}}(Cl(2\pi /7})+Cl(4\pi/7})-Cl(6\pi/7}))\)
this problem is reduced to proving th
N[7 (Cl[2*Pi/7] + Cl[4 *Pi/7] - Cl[6 *Pi/7]) - 2 (3 Cl[2*th] - 3 Cl[4 *th] + Cl[6 *th]), 20]

Clausen function

\(Cl_2(\theta)=-\int_0^{\theta} \ln |2\sin \frac{t}{2}| \,dt=\sum_{n=1}^{\infty}\frac{\sin (n\theta)}{n^2}\)

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

[[4909919|]]

@@ 35번째 줄: / 35번째 줄: @@
 <h5 style="line-height: 2em; margin: 0px;">an open problem</h5>
-<math>\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}))</math>
+*  Prove<br><math>\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}))</math><br>
+*  this problem is reduced to proving th<br> N[7 (Cl[2*Pi/7] + Cl[4 *Pi/7] - Cl[6 *Pi/7]) -   2 (3 Cl[2*th] - 3 Cl[4 *th] + Cl[6 *th]), 20]<br>