ランデン変換
ランデン変換 (Landen's transformation) は、数学において楕円積分や楕円関数の母数を増減させる恒等式。楕円関数の数値計算に有用である。
につき、次の恒等式をランデン変換という。
同じく、次の恒等式をガウス変換という。
\frac{\frac{2}{1+k}\sin\theta\cos\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta
\begin{align}\cos\phi{d\phi}
&=\frac{\frac{2}{1+k}\left(\cos^2\theta-\sin^2\theta\right)}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}{d\theta}+\frac{\frac{2}{1+k}\left(\frac{4k}{(1+k)^2}\sin^2\theta\cos^2\theta\right)}{\left(\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^3}{d\theta}\\
&=\frac{\frac{2}{1+k}\left(1-\frac{2}{1+k}\sin^2\theta\right)\left(1-\frac{2k}{1+k}\sin^2\theta\right)}{\left(\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^3}{d\theta}
\end{align}}
の置換により導かれる。
\int_{\phi=0}^{\alpha}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi\\
&=\int_{\phi=0}^{\alpha}\frac{\cos\phi{d\phi}{\sqrt{1-\sin^2\phi}\sqrt{1-k^2\sin^2\phi}\\
&=\int_{\theta=0}^{\beta}\frac{\frac{\frac{2}{1+k}\left(1-\frac{2}{1+k}\sin^2\theta\right)\left(1-\frac{2k}{1+k}\sin^2\theta\right)}{\left(\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^3}{\sqrt{1-\frac{\frac{4}{(1+k)^2}\sin^2\theta\cos^2\theta}{1-\frac{4k}{(1+k)^2}\sin^2\theta}\sqrt{1-k^2\frac{\frac{4}{(1+k)^2}\sin^2\theta\cos^2\theta}{1-\frac{4k}{(1+k)^2}\sin^2\theta}}{d\theta}\\
&=\int_{\theta=0}^{\beta}\frac{\frac{\frac{2}{1+k}\left(1-\frac{2}{1+k}\sin^2\theta\right)\left(1-\frac{2k}{1+k}\sin^2\theta\right)}{\left(\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^3}{\frac{1-\frac{2}{1+k}\sin^2\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\;\frac{1-\frac{2k}{1+k}\sin^2\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}}{d\theta}\\
&=\frac{2}{1+k}\int_{\theta=0}^{\beta}\frac{d\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\\
&=\frac{2}{1+k}F\left(\sin\beta,\frac{2\sqrt{k}{1+k}\right)
\end{align}}
\sin\betaを陽にすると
\frac{\frac{2}{1+k}\sin\beta\cos\beta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta
\sin^2\alpha=\frac{4\sin^2\beta\cos^2\beta}{(1+k)^2-4k\sin^2\beta}=\frac{1-\cos^2(2\beta)}{1+k^2+2k\cos(2\beta)}
\cos^2(2\beta)+2k\sin^2\alpha\cos(2\beta)+k^2\sin^2\alpha-1+\sin^2\alpha=0
\cos(2\beta)=-k\sin^2\alpha+\sqrt{k^2\sin^4\alpha-k^2\sin^2\alpha+1-\sin^2\alpha}
\sin\beta=\frac{1-\cos(2\beta)}{2}=\frac{1}{2}\sqrt{\left(1+k\right)^2\sin^2\alpha+\left(\sqrt{1-k^2\sin^2\phi}-\sqrt{1-\sin^2\phi}\right)^2}
}である。
\frac{\frac{2}{1+k}\sin\theta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta
\begin{align}\cos\phi{d\phi}
&=\frac{\frac{2}{1+k}\cos\theta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}{d\theta}+\frac{\frac{2}{1+k}\left(\frac{4k}{(1+k)^2}\sin^2\theta\cos\theta\right)}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\left(1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^2}{d\theta}\\
&=\frac{\frac{2}{1+k}\cos\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\left(1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)}{d\theta}
\end{align}}
の置換により導かれる。
\int_{\phi=0}^{\alpha}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi\\
&=\int_{\phi=0}^{\alpha}\frac{\cos\phi{d\phi}{\sqrt{1-\sin^2\phi}\sqrt{1-k^2\sin^2\phi}\\
&=\int_{\theta=0}^{\beta}\frac{\frac{\frac{2}{1+k}\cos\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\left(1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)}{\frac{\sqrt{2+2\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}-\frac{4}{1+k}\sin^2\theta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\;\frac{\sqrt{2+2\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}-\frac{4k}{1+k}\sin^2\theta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}}{d\theta}\\
&=\int_{\theta=0}^{\beta}\frac{\frac{\frac{2}{1+k}\cos\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\left(1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)}{\frac{2\sqrt{1-\sin^2\theta}\sqrt{2+2\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}-\frac{4k}{(1+k)^2}\sin^2\theta}{\left(1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\right)^2}{d\theta}\\
&=\frac{1}{1+k}\int_{\theta=0}^{\beta}\frac{1}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}\\
&=\frac{1}{1+k}F\left(\beta,\frac{2\sqrt{k}{1+k}\right)
\end{align}}
\sin\betaを陽にすると
\frac{\frac{2}{1+k}\sin\beta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta
\sin\alpha\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta}=\frac{2}{1+k}\sin\beta-\sin\alpha
\sin^2\alpha\left(1-\frac{4k}{(1+k)^2}\sin^2\beta\right)=\frac{4}{(1+k)^2}\sin^2\beta-\frac{4}{1+k}\sin\beta\sin\alpha+\sin^2\alpha
\frac{4}{(1+k)^2}\sin^2\beta+\frac{4k}{(1+k)^2}\sin^2\alpha\sin^2\beta-\frac{4}{1+k}\sin\beta\sin\alpha=0
\sin\beta+k\sin^2\alpha\sin\beta-(1+k)\sin\alpha=0
\sin\beta=\frac{(1+k)\sin\alpha}{1+k\sin^2\alpha}
}である。
次の恒等式を楕円関数の上昇ランデン変換という。
\frac{\tfrac{2}{1+k}\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k{1+k}\right)\operatorname{cn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
\operatorname{cn}\left(u,k\right)=\frac{\tfrac{2}{1+k}\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)-\tfrac{1-k}{1+k}{\tfrac{4k}{(1+k)^2}\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
\operatorname{dn}\left(u,k\right)=\frac{\tfrac{2k}{1+k}\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)+\tfrac{1-k}{1+k}{\tfrac{4k}{(1+k)^2}\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}}
次の恒等式を楕円関数の下降ランデン変換という。
\frac{\tfrac{2}{1+\sqrt{1-k^2\operatorname{sn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\operatorname{cn}\left(u,k\right)=\frac{\operatorname{cn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)\operatorname{dn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\operatorname{dn}\left(u,k\right)=\frac{\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}-\left(1-\operatorname{dn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)\right)}{\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}+\left(1-\operatorname{dn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)\right)}
}
当初の母数が0<k<1であれば、上昇ランデン変換は母数を増加させ、下降ランデン変換は母数を減少させる。上昇ランデン変換を繰り返すことにより、母数が1に収束し、楕円関数は双曲線関数に近似される。下降ランデン変換を繰り返すことにより、母数が0に収束し、楕円関数は三角関数に近似される。この性質により、ランデン変換は楕円関数の数値計算に有用である。
F\left(\alpha,k\right)=\tfrac{2}{1+k}F\left(\beta,\tfrac{2\sqrt{k{1+k}\right)
\operatorname{sn}\left(u,k\right)=\sin\alpha
\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)=\sin\beta}
\frac{\tfrac{2}{1+k}\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k{1+k}\right)\sqrt{1-\operatorname{sn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\sqrt{1-\left(\tfrac{2\sqrt{k}{1+k}\right)^2\operatorname{sn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}=\frac{\tfrac{2}{1+k}\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)\operatorname{cn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
\operatorname{cn}\left(u,k\right)=\sqrt{1-\operatorname{sn}^2\left(u,k\right)}=\frac{1-\tfrac{2}{1+k}\operatorname{sn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}=\frac{\tfrac{2}{1+k}\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)-\tfrac{1-k}{1+k}{\tfrac{4k}{(1+k)^2}\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
\operatorname{dn}\left(u,k\right)=\sqrt{1-k^2\operatorname{sn}^2\left(u,k\right)}=\frac{1-\tfrac{2k}{1+k}\operatorname{sn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}=\frac{\tfrac{2k}{1+k}\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)+\tfrac{1-k}{1+k}{\tfrac{4k}{(1+k)^2}\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}}
である。楕円積分のガウス変換により
F\left(\alpha,k\right)=\tfrac{1}{1+k}F\left(\beta,\tfrac{2\sqrt{k{1+k}\right)
\operatorname{sn}\left(u,k\right)=\sin\alpha
\operatorname{sn}\left((1+k)u,\tfrac{2\sqrt{k}{1+k}\right)=\sin\beta}
であるが、uを\tfrac{u}{1+k}に改め、kを\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}に改めれば
\frac{\tfrac{2}{1+\sqrt{1-k^2\operatorname{sn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\begin{align}\operatorname{cn}\left(u,k\right)&=\sqrt{1-\operatorname{sn}^2\left(u,k\right)}\\
&=\frac{\operatorname{cn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)\operatorname{dn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\end{align}
\begin{align}\operatorname{dn}\left(u,k\right)&=\sqrt{1-k^2\operatorname{sn}^2\left(u,k\right)}\\
&=\frac{1-\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}\\
&=\frac{\operatorname{dn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)-\tfrac{2\sqrt{1-k^2}{1+\sqrt{1-k^2}}{\tfrac{2}{1+\sqrt{1-k^2}-\operatorname{dn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\end{align}
}となる。
上昇ランデン変換と下降ランデン変換は虚数変換により交替する。
\frac
{\frac{\tfrac{2i}{1+k}\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k{1+k}\right)\operatorname{cn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}{\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
{\frac{\tfrac{2}{1+k}\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)-\tfrac{1-k}{1+k}{\tfrac{4k}{(1+k)^2}\operatorname{dn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}\\
&=\frac
{\tfrac{4ki}{(1+k)^2}\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)\operatorname{cn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)}
{\operatorname{dn}^2\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}{1+k}\right)-\tfrac{1-k}{1+k}\\
\end{align}}
\frac{\tfrac{4k}{(1+k)^2}\operatorname{sc}\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)\operatorname{nc}\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}{\operatorname{dc}^2\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)-\tfrac{1-k}{1+k\\
&=\frac{\tfrac{4k}{(1+k)^2}\operatorname{sn}\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}{\operatorname{dn}^2\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)-\tfrac{1-k}{1+k}\operatorname{cn}^2\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}\\
&=\frac{\tfrac{4k}{(1+k)^2}\operatorname{sn}\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}{\tfrac{2k}{1+k}+\tfrac{2k(1-k)}{(1+k)^2}\operatorname{sn}^2\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}\\
&=\frac{\tfrac{2}{1+k}\operatorname{sn}\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}{1+\tfrac{1-k}{1+k}\operatorname{sn}^2\left(\tfrac{1+k}{2}iu,\tfrac{1-k}{1+k}\right)}\\
\end{align}}
iuをuと書き、\sqrt{1-k^2}をkと書けば
\frac{\tfrac{2}{1+\sqrt{1-k^2\operatorname{sn}\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}{1+\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\operatorname{sn}^2\left(\tfrac{1+\sqrt{1-k^2}{2}u,\tfrac{1-\sqrt{1-k^2}{1+\sqrt{1-k^2}\right)}
\end{align}}
となるが、これは下降ランデン変換である。