"함수값의 계산"의 두 판 사이의 차이

수학노트
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1번째 줄: 1번째 줄:
 
* [[삼각함수]]
 
* [[삼각함수]]
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* [[정오각형]], [[황금비]]<br><math>\cos\frac{2\pi}{5}=\frac{\sqrt5 -1}{4}</math><br><math>z^4+z^3+z^2+z^1+1=0</math><br> 복소평면상에서 <math>z</math> 의 <math>x</math> 좌표는 <math>\frac{-1+\sqrt{5}}{4} , \frac{-1-\sqrt{5}}{4}</math> 로 주어짐.<br>
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* [[정오각형]], [[황금비]]:<math>\cos\frac{2\pi}{5}=\frac{\sqrt5 -1}{4}</math>:<math>z^4+z^3+z^2+z^1+1=0</math><br> 복소평면상에서 <math>z</math> 의 <math>x</math> 좌표는 <math>\frac{-1+\sqrt{5}}{4} , \frac{-1-\sqrt{5}}{4}</math> 로 주어짐.<br>
 
* [[가우스와 정17각형의 작도]]
 
* [[가우스와 정17각형의 작도]]
  
9번째 줄: 9번째 줄:
 
* <math>16\cos{2\pi\over17} = -1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ 2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}} </math><br>
 
* <math>16\cos{2\pi\over17} = -1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ 2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}} </math><br>
  
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* [[타원 모듈라 j-함수 (elliptic modular function, j-invariant)|j-invariant]]<br><math> j(\sqrt{-1})=1728</math><br><math>j(\frac {-1+\sqrt{-3}}{2})=0</math><br><math>j(\frac {-1+\sqrt{-43}} {2})=-884736744</math><br><math>j(\frac {-1+\sqrt{-67}} {2})=147197952744</math><br><math> j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744</math><br>
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* [[타원 모듈라 j-함수 (elliptic modular function, j-invariant)|j-invariant]]:<math> j(\sqrt{-1})=1728</math>:<math>j(\frac {-1+\sqrt{-3}}{2})=0</math>:<math>j(\frac {-1+\sqrt{-43}} {2})=-884736744</math>:<math>j(\frac {-1+\sqrt{-67}} {2})=147197952744</math>:<math> j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744</math><br>
  
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* [[정수에서의 리만제타함수의 값]]<br><math>\zeta(2n) =(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}, n \ge 1</math>여기서 <math>B_{2n}</math>은 [[베르누이 수|베르누이수]]. <br><math>\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1</math><br><math>\zeta(0)=-\frac{1}{2}</math><br>
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* [[정수에서의 리만제타함수의 값]]:<math>\zeta(2n) =(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}, n \ge 1</math>여기서 <math>B_{2n}</math>은 [[베르누이 수|베르누이수]]. :<math>\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1</math>:<math>\zeta(0)=-\frac{1}{2}</math><br>
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* [[디리클레 베타함수]]<br><math>\beta(0)= \frac{1}{2},  \beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4},  \beta(3)\;=\;\frac{\pi^3}{32}, \beta(5)\;=\;\frac{5\pi^5}{1536}, \beta(7)\;=\;\frac{61\pi^7}{184320}</math><br>
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* [[디리클레 베타함수]]:<math>\beta(0)= \frac{1}{2},  \beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4},  \beta(3)\;=\;\frac{\pi^3}{32}, \beta(5)\;=\;\frac{5\pi^5}{1536}, \beta(7)\;=\;\frac{61\pi^7}{184320}</math><br>
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* [[감마함수]]<br><math>\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}</math><br>
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* [[감마함수]]:<math>\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}</math><br>
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* [[다이감마 함수(digamma function)|Digamma 함수]]<br><math>\psi(1) = -\gamma\,\!</math><br><math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma</math><br><math>\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma</math><br><math>\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma</math><br><math>\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma</math><br><math>\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma</math><br>
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* [[다이감마 함수(digamma function)|Digamma 함수]]:<math>\psi(1) = -\gamma\,\!</math>:<math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma</math>:<math>\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma</math>:<math>\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma</math>:<math>\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma</math>:<math>\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma</math><br>
 
* [[다이로그 함수(dilogarithm)|Dilogarithm]]
 
* [[다이로그 함수(dilogarithm)|Dilogarithm]]
  

2013년 1월 12일 (토) 10:59 판

\(\cos \frac{2\pi}{17}= \frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ \sqrt{68+12\sqrt{17}-4{\sqrt{170+38\sqrt{17}}}} }{16}\)

  • \(\cos {\frac{2\pi}{3}} = -\frac{1}{2}\)
  • \(\cos {\frac{2\pi}{5}} = \frac{-1+\sqrt{5}}{4}\)
  • \(16\cos{2\pi\over17} = -1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ 2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}} \)
  • j-invariant\[ j(\sqrt{-1})=1728\]\[j(\frac {-1+\sqrt{-3}}{2})=0\]\[j(\frac {-1+\sqrt{-43}} {2})=-884736744\]\[j(\frac {-1+\sqrt{-67}} {2})=147197952744\]\[ j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744\]
  • 정수에서의 리만제타함수의 값\[\zeta(2n) =(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}, n \ge 1\]여기서 \(B_{2n}\)은 베르누이수. \[\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1\]\[\zeta(0)=-\frac{1}{2}\]
  • 디리클레 베타함수\[\beta(0)= \frac{1}{2}, \beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4}, \beta(3)\;=\;\frac{\pi^3}{32}, \beta(5)\;=\;\frac{5\pi^5}{1536}, \beta(7)\;=\;\frac{61\pi^7}{184320}\]
  • 감마함수\[\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\]
  • Digamma 함수\[\psi(1) = -\gamma\,\!\]\[\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma\]\[\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma\]\[\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma\]\[\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma\]\[\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma\]
  • Dilogarithm

\(\mbox{Li}_{2}(0)=0\)

\(\mbox{Li}_{2}(1)=\frac{\pi^2}{6}\)

\(\mbox{Li}_{2}(-1)=-\frac{\pi^2}{12}\)

\(\mbox{Li}_{2}(\frac{1}{2})=\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)\)

\(\mbox{Li}_{2}(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\)

\(\mbox{Li}_{2}(\frac{-1-\sqrt{5}}{2})=-\frac{\pi^2}{10}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\)

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