"함수값의 계산"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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| (같은 사용자의 중간 판 2개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
* [[삼각함수의 값]] | * [[삼각함수의 값]] | ||
:<math>\cos {\frac{2\pi}{3}} = -\frac{1}{2}</math> | :<math>\cos {\frac{2\pi}{3}} = -\frac{1}{2}</math> | ||
| − | * [[정오각형]], | + | * [[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> | :<math>\cos\frac{2\pi}{5}=\frac{\sqrt5 -1}{4}</math>:<math>z^4+z^3+z^2+z^1+1=0</math> | ||
| − | + | 복소평면상에서 <math>z</math> 의 <math>x</math> 좌표는 <math>\frac{-1+\sqrt{5}}{4} , \frac{-1-\sqrt{5}}{4}</math> 로 주어짐. | |
* [[가우스와 정17각형의 작도]] | * [[가우스와 정17각형의 작도]] | ||
:<math>\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}</math> | :<math>\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}</math> | ||
| − | :<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 | + | :<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> |
* [[타원 모듈라 j-함수 (elliptic modular function, j-invariant)|j-invariant]] | * [[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(\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{-43}} {2})=-884736744</math> | ||
| − | :<math>j(\frac {-1+\sqrt{-67}} {2})=147197952744</math>:<math> j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744</math | + | :<math>j(\frac {-1+\sqrt{-67}} {2})=147197952744</math>:<math> j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744</math> |
* [[정수에서의 리만제타함수의 값]] | * [[정수에서의 리만제타함수의 값]] | ||
| − | :<math>\zeta(2n) =(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}, n \ge 1</math> | + | :<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]]. |
| − | :<math>\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1</math>:<math>\zeta(0)=-\frac{1}{2}</math | + | :<math>\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1</math>:<math>\zeta(0)=-\frac{1}{2}</math> |
* [[디리클레 베타함수]] | * [[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 | + | :<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> |
* [[감마함수]] | * [[감마함수]] | ||
| − | :<math>\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}</math | + | :<math>\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}</math> |
* [[다이감마 함수(digamma function)]] | * [[다이감마 함수(digamma function)]] | ||
:<math>\psi(1) = -\gamma\,\!</math> | :<math>\psi(1) = -\gamma\,\!</math> | ||
:<math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \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}{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 | + | :<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> |
* [[다이로그 함수의 special value 계산]] | * [[다이로그 함수의 special value 계산]] | ||
:<math>\mbox{Li}_{2}(0)=0</math> | :<math>\mbox{Li}_{2}(0)=0</math> | ||
2020년 12월 28일 (월) 03:13 기준 최신판
\[\cos {\frac{2\pi}{3}} = -\frac{1}{2}\]
\[\cos\frac{2\pi}{5}=\frac{\sqrt5 -1}{4}\]\[z^4+z^3+z^2+z^1+1=0\] 복소평면상에서 \(z\) 의 \(x\) 좌표는 \(\frac{-1+\sqrt{5}}{4} , \frac{-1-\sqrt{5}}{4}\) 로 주어짐.
\[\cos \frac{2\pi}{17}= \frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ \sqrt{68+12\sqrt{17}-4{\sqrt{170+38\sqrt{17}}}} }{16}\] \[16\cos{2\pi\over17} = -1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ 2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}} \]
\[ j(\sqrt{-1})=1728\]\[j(\frac {-1+\sqrt{-3}}{2})=0\] \[j(\frac {-1+\sqrt{-43}} {2})=-884736744\] \[j(\frac {-1+\sqrt{-67}} {2})=147197952744\]\[ j(\frac {-1+\sqrt{-163}} {2})=-262537412640768744\]
\[\zeta(2n) =(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}, n \ge 1\]여기서 \(B_{2n}\)은 베르누이 수. \[\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1\]\[\zeta(0)=-\frac{1}{2}\]
\[\beta(0)= \frac{1}{2}, \beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4}, \beta(3)\;=\;\frac{\pi^3}{32}, \beta(5)\;=\;\frac{5\pi^5}{1536}, \beta(7)\;=\;\frac{61\pi^7}{184320}\]
\[\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\]
\[\psi(1) = -\gamma\,\!\] \[\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma\] \[\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma\]\[\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma\]\[\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma\] \[\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma\]
\[\mbox{Li}_{2}(0)=0\] \[\mbox{Li}_{2}(1)=\frac{\pi^2}{6}\] \[\mbox{Li}_{2}(-1)=-\frac{\pi^2}{12}\] \[\mbox{Li}_{2}(\frac{1}{2})=\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)\] \[\mbox{Li}_{2}(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{-1-\sqrt{5}}{2})=-\frac{\pi^2}{10}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\]