"복소 이차 수체의 데데킨트 제타함수 special values"의 두 판 사이의 차이

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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 수학노트 원문주소</h5>
 
<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 수학노트 원문주소</h5>
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* [[복소이차수체의 데데킨트 제테함수]]
 
* [[복소이차수체의 데데킨트 제테함수]]
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32번째 줄: 26번째 줄:
 
<h5 style="margin: 0px; line-height: 2em;"><math>s=2</math> 에서의 값</h5>
 
<h5 style="margin: 0px; line-height: 2em;"><math>s=2</math> 에서의 값</h5>
  
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*  복소이차수체의 경우<br><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><br><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><br> 여기서 <math>D(z)</math>는 [[블로흐-비그너 다이로그(Bloch-Wigner dilogarithm)|Bloch-Wigner dilogarithm]]<br>
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*  복소이차수체의 경우<br><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><br><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><br> 여기서 <math>D(z)</math>는 [[블로흐-비그너 다이로그(Bloch-Wigner dilogarithm)]]<br>
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*  예<br><math>\zeta_{\mathbb{Q}\sqrt{-1}}(2)=1.50670301</math><br><math>\zeta_{\mathbb{Q}\sqrt{-2}}(2)=1.75141751\cdots</math><br><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})=1.285190955484149\cdots</math><br><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}))=1.89484145</math><br><math>\zeta_{\mathbb{Q}\sqrt{-11}}(2)=1.49613186</math><br>
  
 
 
 
 
  
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">introduction</h5>
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* http://books.google.co.kr/books?id=yrmT56mpw3kC&pg=PA367&dq=smallest+norms+of+prime+ideals&hl=ko&ei=IMRTTIaRGoqWvAP88MUZ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDgQ6AEwAw#v=onepage&q=smallest%20norms%20of%20prime%20ideals&f=false<br>
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*  복소이차수체의 [[데데킨트 제타함수]]<br><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><br>
 
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*  Note that<br>
 
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**  the Clausen function and the Bloch-Wigner dilogarithms are same if <math>z=e^{i\theta}</math><br><math>\operatorname{Cl}_2(\theta)=-\int_0^{\theta} \ln |2\sin \frac{t}{2}| \,dt=\sum_{n=1}^{\infty}\frac{\sin (n\theta)}{n^2}</math><br><math>D(z)=\text{Im}(\operatorname{Li}_2(z))+\log|z|\arg(1-z)</math><br>
 
  
 
 
 
 
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<h5 style="line-height: 2em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">a few examples</h5>
 
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* http://books.google.co.kr/books?id=yrmT56mpw3kC&pg=PA367&dq=smallest+norms+of+prime+ideals&hl=ko&ei=IMRTTIaRGoqWvAP88MUZ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDgQ6AEwAw#v=onepage&q=smallest%20norms%20of%20prime%20ideals&f=false<br>
 
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<math>\zeta_{\mathbb{Q}\sqrt{-1}}(2)=1.50670301</math>
 
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<math>\zeta_{\mathbb{Q}\sqrt{-2}}(2)=1.75141751\cdots</math>
 
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<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})=1.285190955484149\cdots</math>
 
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<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}))=1.89484145</math>
 
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<math>\zeta_{\mathbb{Q}\sqrt{-11}}(2)=1.49613186</math>
 
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#  Cl[x_] := Im[PolyLog[2, Exp[I*x]]]<br> disc[n_] := NumberFieldDiscriminant[Sqrt[-n]]<br> L2[n_] :=<br>  1/Sqrt[Abs[disc[n]]]*<br>   Sum[JacobiSymbol[disc[n], k] Cl[2 Pi*k/Abs[disc[n]]], {k, 1,<br>     Abs[disc[n]] - 1}]<br> Zeta2[n_] := L2[n]*Pi^2/6<br> Zeta2[1]<br>
 
  
 
 
 
 
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<math>L_{-3}(2)=\frac{2}{\sqrt{3}}D(e^{\frac{2\pi  i}{3}})</math>
 
<math>L_{-3}(2)=\frac{2}{\sqrt{3}}D(e^{\frac{2\pi  i}{3}})</math>
  
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*  2.02988321281930725<br><math>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</math><br> where D is [[블로흐-비그너 다이로그(Bloch-Wigner dilogarithm)|Bloch-Wigner dilogarithm]].<br>
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*   <br> 2.02988321281930725<br><math>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</math><br>
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*  what is <math>\zeta_{\mathbb{Q}(\sqrt{-3})}(2)</math>? numerically 1.285190955484149<br>
 
  
 
 
 
 

2012년 6월 1일 (금) 10:03 판

이 항목의 수학노트 원문주소

 

 

개요

 

 

 

\(s=1\) 에서의 값
  • 이차 수체에 대한 디리클레 class number 공식
  • 복소이차수체의 경우
    \(K=\mathbb{Q}(\sqrt{-q})\), \(q \geq 7\) , \(q \equiv 3 \pmod{4}\) 인 경우
    \(d_K=-q\)
    \(\chi(a)=\left(\frac{a}{q}\right)\)
    \(\chi(-1)=-1\), \(\tau(\chi)=i\sqrt{q}\)
    \(L(1,\chi)= \frac{- \pi\sqrt{q}}{q^2}\sum_{a=1}^{q-1}\left(\frac{a}{q}\right) a=\frac{\pi h_K}{\sqrt{q}}\)
    \(h_K=-\sum_{a=1}^{q-1}\left(\frac{a}{q}\right)\frac{a}{q}\)
     
    \(K=\mathbb{Q}(\sqrt{-q})\)  , \(q \geq 5\) ,  \(q \equiv 1 \pmod{4}\) 인 경우
    \(d_K=-4q\)
    \(\chi(-1)=-1\), \(\tau(\chi)=2i\sqrt{q}\)
    \(L(1,\chi)= -\frac{ \pi\sqrt{q}}{8q^2}{\sum_{(a,4q)=1}\chi(a) a=\frac{\pi h_K}{2\sqrt{q}}\)
    \(h_K=-\frac{1}{4}\sum_{(a,4q)=1}\left(\frac{a}{q}\right)\frac{a}{q}\)

 

 

\(s=2\) 에서의 값
  • 복소이차수체의 경우
    \(\zeta_{K}(2)=\frac{\pi^2}{6\sqrt{|d_K|}}\sum_{(a,d_k)=1} (\frac{d_K}{a})D(e^{2\pi ia/|d_k|})\)
    \(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))\)
    여기서 \(D(z)\)는 블로흐-비그너 다이로그(Bloch-Wigner dilogarithm)
  • 예
    \(\zeta_{\mathbb{Q}\sqrt{-1}}(2)=1.50670301\)
    \(\zeta_{\mathbb{Q}\sqrt{-2}}(2)=1.75141751\cdots\)
    \(\zeta_{\mathbb{Q}\sqrt{-3}}(2)=\frac{\pi^2}{6\sqrt{3}}(D(e^{2\pi i/3})-D(e^{4\pi i/3}))=\frac{\pi^2}{3\sqrt{3}}D(e^{2\pi i/3})=1.285190955484149\cdots\)
    \(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))=1.89484145\)
    \(\zeta_{\mathbb{Q}\sqrt{-11}}(2)=1.49613186\)

 

 

 

 

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
    \(V(4_{1})=\frac{9\sqrt{3}}{\pi^2}\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=3D(e^{\frac{2i\pi}{3}})=2D(e^{\frac{i\pi}{3}})=2.029883212819\cdots\)

 

 

역사

 

 

 

메모
  • \(s=1\) 에서의 \(L_{d_K}'(1)\)의 값
    \(L_{d_K}'(1)=\frac{2\pi h_K(\gamma+\ln 2\pi)}{w_K \cdot \sqrt{|d_K|}}-\frac{\pi}{\sqrt{|d_K|}}\sum_{(a,d_K)=1}\chi(a)\log\Gamma (\frac{a}{|d_K|})\)
  • L-함수의 미분 항목 참조

 

 

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