"3-manifolds and their invariants"의 두 판 사이의 차이
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| 6번째 줄: | 6번째 줄: | ||
# L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]<br> f[x_, y_] :=<br> L[x] + L[1 - x*y] + L[y] + L[(1 - y)/(1 - x*y)] + L[(1 - x)/(1 - x*y)]<br> Print["five term relation"]<br> Table[f[i, j], {i, 0.1, 0.9, 0.1}, {j, 0.1, 0.9, 0.1}] // TableForm<br> N[3 L[Exp[2 I*Pi/3]], 20]<br> N[2 L[Exp[I*Pi/3]], 20]<br> N[3 (L[Exp[2 I*Pi/3]] - L[Exp[4 I*Pi/3]])/2, 20]<br> N[Pi^2*L[Exp[2 I*Pi/3]]/(3 Sqrt[3]), 20]<br> | # L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]<br> f[x_, y_] :=<br> L[x] + L[1 - x*y] + L[y] + L[(1 - y)/(1 - x*y)] + L[(1 - x)/(1 - x*y)]<br> Print["five term relation"]<br> Table[f[i, j], {i, 0.1, 0.9, 0.1}, {j, 0.1, 0.9, 0.1}] // TableForm<br> N[3 L[Exp[2 I*Pi/3]], 20]<br> N[2 L[Exp[I*Pi/3]], 20]<br> N[3 (L[Exp[2 I*Pi/3]] - L[Exp[4 I*Pi/3]])/2, 20]<br> N[Pi^2*L[Exp[2 I*Pi/3]]/(3 Sqrt[3]), 20]<br> | ||
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| 11번째 줄: | 13번째 줄: | ||
<h5 style="line-height: 2em; margin: 0px;">Volume of knot complement</h5> | <h5 style="line-height: 2em; margin: 0px;">Volume of knot complement</h5> | ||
| − | KnotData[]<br> KnotData["FigureEight", "HyperbolicVolume"]<br> N[%, 20] | + | # KnotData[]<br> KnotData["FigureEight", "HyperbolicVolume"]<br> N[%, 20]<br> |
| 25번째 줄: | 27번째 줄: | ||
<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_{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 | + | <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> | <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> | ||
2010년 3월 25일 (목) 17:56 판
introduction
- volume of figure 8 knot complement
- 2.02988321281930725
\(V=\frac{9\sqrt{3}}{\pi^2}\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=3D(e^{\frac{2i\pi}{3}})=2D(e^{\frac{i\pi}{3}})=2.029883212819\cdots\)
- 2.02988321281930725
- what is \(\zeta_{\mathbb{Q}(\sqrt{-3})}(2)\)? numrically 1.285190955484149
- L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]
f[x_, y_] :=
L[x] + L[1 - x*y] + L[y] + L[(1 - y)/(1 - x*y)] + L[(1 - x)/(1 - x*y)]
Print["five term relation"]
Table[f[i, j], {i, 0.1, 0.9, 0.1}, {j, 0.1, 0.9, 0.1}] // TableForm
N[3 L[Exp[2 I*Pi/3]], 20]
N[2 L[Exp[I*Pi/3]], 20]
N[3 (L[Exp[2 I*Pi/3]] - L[Exp[4 I*Pi/3]])/2, 20]
N[Pi^2*L[Exp[2 I*Pi/3]]/(3 Sqrt[3]), 20]
Volume of knot complement
- KnotData[]
KnotData["FigureEight", "HyperbolicVolume"]
N[%, 20]
복소이차수체의 데데킨트 제타함수
\(\zeta_{K}(2)=\frac{\pi^2}{6\sqrt{|d_K|}}\sum_{(a,d_k)=1} (\frac{d_K}{a})D(e^{2\pi ia/|d_k|})\)
\(\zeta_{\mathbb{Q}\sqrt{-3}}(2)=\frac{\pi^2}{6\sqrt{3}}(D(e^{2\pi i/3})-D(e^{4\pi i/3}))=\frac{\pi^2}{3\sqrt{3}}D(e^{2\pi i/3})\)
\(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))\)
- 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]
history
encyclopedia
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/Figure-eight_knot_(mathematics)
- http://en.wikipedia.org/wiki/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
[[4909919|]]
articles
- 논문정리
- http://www.ams.org/mathscinet
- http://www.zentralblatt-math.org/zmath/en/
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html[1]
- http://front.math.ucdavis.edu/search?a=&t=&c=&n=40&s=Listings&q=
- http://www.ams.org/mathscinet/search/publications.html?pg4=AUCN&s4=&co4=AND&pg5=TI&s5=&co5=AND&pg6=PC&s6=&co6=AND&pg7=ALLF&co7=AND&Submit=Search&dr=all&yrop=eq&arg3=&yearRangeFirst=&yearRangeSecond=&pg8=ET&s8=All&s7=
- http://dx.doi.org/
question and answers(Math Overflow)
blogs
experts on the field