제1종타원적분 K (complete elliptic integral of the first kind)

이 항목의 스프링노트 원문주소

 

 

개요

 

 

 

란덴변환

K(\frac{2\sqrt{k}}{1+k})=(1+k)K(k)

k'=\sqrt{1-k^2}라 두면

2K(\frac{1-k'}{1+k'})=(1+k')K(k)

 

 

초기하함수를 이용한 표현

 

K(k) = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{n!(1)_n}k^{2n} = \frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)

(증명)

K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \int_0^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n}{n!} k^{2n}\sin^{2n}\theta{d\theta}

\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n} (감마함수) 이므로

K(k) = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{n!(1)_n}k^{2n} = \frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)■

 

 

맴돌이군

 

 

 

singular values

 

 

special values of K(k)

K(0) = \frac{\pi}{2}

K(1) = \infty

K(\frac{1}{\sqrt{2}})=\frac{1}{4}B(1/4,1/4)=\frac{\Gamma(\frac{1}{4})^2}{4\sqrt{\pi}}=1.8540746773\cdots

K(2\sqrt{2}-2)

K(\sqrt{2}-1)=\frac{\sqrt{\sqrt{2}+1}}{2^{13/4}}B(\frac{1}{8},\frac{3}{8})=\frac{\sqrt{\sqrt{2}+1}\Gamma(\frac{1}{8})\Gamma(\frac{3}{8})}{2^{13/4}\sqrt{\pi}}

K\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)=\frac{\sqrt[4]{3}\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{4\sqrt{\pi}}=2.768063\cdots

K\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)=\frac{\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{4\sqrt[4]{3}\sqrt{\pi}}=1.5981420\cdots

K\left(3-2\sqrt{2}}\right)=\frac{(2+\sqrt{2})\Gamma(\frac{1}{4})^2}{16\sqrt{\pi}}=1.58255\cdots

 

 

special value의 계산

K(\frac{1}{\sqrt{2}})=\frac{\Gamma(\frac{1}{4})^2}{4\sqrt{\pi}}=1.8540746773\cdots

K(2\sqrt{2}-2)

(증명)

K(\frac{2\sqrt{k}}{1+k})=(1+k)K(k)

 

 

K(\sqrt{2}-1)=\frac{\sqrt{\sqrt{2}+1}\Gamma(\frac{1}{8})\Gamma(\frac{3}{8})}{2^{13/4}\sqrt{\pi}}

(증명)

 

 

 

K\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)=\frac{\sqrt[4]{3}\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{4\sqrt{\pi}}=2.768063\cdots

 

(증명)

\cos \frac{\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}, \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} 이므로 위에서 얻은 결과를 활용하면,

K\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)=\frac{1}{2}\int_{0}^{\infty} \frac{du}{\sqrt{u (u^2 - \sqrt{3}u + 1)}}

여기서 v=\sqrt{3}u-1 으로 치환하면, u(u^2 - \sqrt{3}u+ 1) = 3^{-3/2}(1 + v^3)

\int_{0}^{\infty} \frac{du}{\sqrt{u (u^2 - \sqrt{3}u + 1)}}=\sqrt[4]{3}\int_{-1}^{\infty} \frac{dv}{\sqrt{v^3+1}}=\sqrt[4]{3}(\int_{-1}^{0} \frac{dv}{\sqrt{v^3+1}}+\int_{0}^{\infty} \frac{dv}{\sqrt{v^3+1}})

=\sqrt[4]{3}(\int_{0}^{1} \frac{dv}{\sqrt{1-v^3}}+\int_{0}^{\infty} \frac{dv}{\sqrt{1+v^3}})=\frac{\sqrt[4]{3}\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{2\sqrt{\pi}}=5.536129\cdots

마지막에서 다음을 이용하였음. (이에 대한 증명은 오일러 베타적분 항목 참조)

\int_{0}^{1} \frac{dv}{\sqrt{1-v^3}}=\frac{\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{6\sqrt{\pi}}

\int_{0}^{\infty} \frac{dv}{\sqrt{1+v^3}}=\frac{ \Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{3\sqrt{\pi }} ■

K\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)=\frac{\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{4\sqrt[4]{3}\sqrt{\pi}}=1.5981420\cdots

(증명)

* \frac{K'}{K}\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)= \sqrt{3} 을 이용할 수도 있고, 다음과 같이 직접 증명도 가능  *

\cos \frac{5\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}, \cos \frac{5\pi}{6}=-\frac{\sqrt{3}}{2} 이므로 위에서 얻은 결과를 활용하면,

K\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)=\frac{1}{2}\int_{0}^{\infty} \frac{du}{\sqrt{u (u^2 + \sqrt{3}u + 1)}}

여기서 v=\sqrt{3}u+1 으로 치환하면, u(u^2 + \sqrt{3}u+ 1) = 3^{-3/2}(v^3-1)

\int_{0}^{\infty} \frac{du}{\sqrt{u (u^2+ \sqrt{3}u + 1)}}=\sqrt[4]{3}\int_{1}^{\infty} \frac{dv}{\sqrt{v^3-1}}=\frac{\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{2\sqrt[4]{3}\sqrt{\pi}}=3.1962840\cdots

 

K\left(3-2\sqrt{2}}\right)=\frac{(2+\sqrt{2})\Gamma(\frac{1}{4})^2}{16\sqrt{\pi}}=1.58255\cdots

(증명)

란덴변환을 이용

 K(\frac{2\sqrt{k}}{1+k})=(1+k)K(k)

k=3-2\sqrt{2}라 하면, 

\frac{2\sqrt{k}}{1+k}=\frac{1}{\sqrt{2}}

이로부터 

K(\frac{1}{\sqrt{2}})=(4-2\sqrt{2})K(3-2\sqrt{2})

K(\frac{1}{\sqrt{2}})=\frac{\Gamma(\frac{1}{4})^2}{4\sqrt{\pi}}=1.8540746773\cdots 로부터 

K\left(3-2\sqrt{2}}\right)=\frac{(2+\sqrt{2})\Gamma(\frac{1}{4})^2}{16\sqrt{\pi}}=1.58255\cdots

 

 

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