Zeta value at 2
http://bomber0.myid.net/ (토론)님의 2012년 6월 1일 (금) 09:30 판
introduction
- 복소이차수체의 데데킨트 제타함수
\(\zeta_{K}(2)=\frac{\pi^2}{6\sqrt{|d_K|}}\sum_{(a,d_k)=1} (\frac{d_K}{a})D(e^{2\pi ia/|d_k|})\) - Note that
- the Clausen function and the Bloch-Wigner dilogarithms are same if \(z=e^{i\theta}\)
\(\operatorname{Cl}_2(\theta)=-\int_0^{\theta} \ln |2\sin \frac{t}{2}| \,dt=\sum_{n=1}^{\infty}\frac{\sin (n\theta)}{n^2}\)
\(D(z)=\text{Im}(\operatorname{Li}_2(z))+\log|z|\arg(1-z)\)
- the Clausen function and the Bloch-Wigner dilogarithms are same if \(z=e^{i\theta}\)
a few examples
\(\zeta_{\mathbb{Q}\sqrt{-1}}(2)=1.50670301\)
\(\zeta_{\mathbb{Q}\sqrt{-2}}(2)=1.75141751\cdots\)
\(\zeta_{\mathbb{Q}\sqrt{-3}}(2)=\frac{\pi^2}{6\sqrt{3}}(D(e^{2\pi i/3})-D(e^{4\pi i/3}))=\frac{\pi^2}{3\sqrt{3}}D(e^{2\pi i/3})=1.285190955484149\cdots\)
\(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))=1.89484145\)
\(\zeta_{\mathbb{Q}\sqrt{-11}}(2)=1.49613186\)
- Cl[x_] := Im[PolyLog[2, Exp[I*x]]]
disc[n_] := NumberFieldDiscriminant[Sqrt[-n]]
L2[n_] :=
1/Sqrt[Abs[disc[n]]]*
Sum[JacobiSymbol[disc[n], k] Cl[2 Pi*k/Abs[disc[n]]], {k, 1,
Abs[disc[n]] - 1}]
Zeta2[n_] := L2[n]*Pi^2/6
Zeta2[1]
figure eight knot complement
\(V=\frac{9\sqrt{3}}{\pi^2}\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=3D(e^{\frac{2i\pi}{3}})=2D(e^{\frac{i\pi}{3}})=2.029883212819\cdots\)
\(\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=\frac{\pi^2}{3\sqrt{3}}D(e^{\frac{2\pi i}{3}})\)
\(L_{-3}(2)=\frac{2}{\sqrt{3}}D(e^{\frac{2\pi i}{3}})\)
- 2.02988321281930725
\(V(4_{1})=\frac{9\sqrt{3}}{\pi^2}\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=3D(e^{\frac{2i\pi}{3}})=2D(e^{\frac{i\pi}{3}})=2.029883212819\cdots\)
where D is Bloch-Wigner dilogarithm. - what is \(\zeta_{\mathbb{Q}(\sqrt{-3})}(2)\)? numerically 1.285190955484149
history