함수값의 계산

• 삼각함수의 값

\[\cos {\frac{2\pi}{3}} = -\frac{1}{2}\]

• 정오각형, 황금비

\[\cos\frac{2\pi}{5}=\frac{\sqrt5 -1}{4}\]\[z^4+z^3+z^2+z^1+1=0\] 복소평면상에서 \(z\) 의 \(x\) 좌표는 \(\frac{-1+\sqrt{5}}{4} , \frac{-1-\sqrt{5}}{4}\) 로 주어짐.

• 가우스와 정17각형의 작도

\[\cos \frac{2\pi}{17}= \frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ \sqrt{68+12\sqrt{17}-4{\sqrt{170+38\sqrt{17}}}} }{16}\] \[16\cos{2\pi\over17} = -1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+ 2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}} \]

• j-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 베르누이 수.  \[\zeta(-n)=-\frac{B_{n+1}}{n+1}, n \ge 1\]\[\zeta(0)=-\frac{1}{2}\]

• 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

\[\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\]

• 다이감마 함수(digamma function)

\[\psi(1) = -\gamma\,\!\] \[\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma\] \[\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma\]\[\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma\]\[\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma\] \[\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma\]

• 다이로그 함수의 special value 계산

\[\mbox{Li}_{2}(0)=0\] \[\mbox{Li}_{2}(1)=\frac{\pi^2}{6}\] \[\mbox{Li}_{2}(-1)=-\frac{\pi^2}{12}\] \[\mbox{Li}_{2}(\frac{1}{2})=\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)\] \[\mbox{Li}_{2}(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\] \[\mbox{Li}_{2}(\frac{-1-\sqrt{5}}{2})=-\frac{\pi^2}{10}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})\]

• 르장드르 카이 함수
• 일종타원적분 K (complete elliptic integral of the first kind)