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。
楕円積分のランデン変換とガウス変換
第一種楕円積分
[math]F\left(\sin\alpha,k\right)=\int_{t=0}^{\sin\alpha}\frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}=\int_{\phi=0}^{\alpha}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}[/math]
につき、次の恒等式をランデン変換という。
[math]F\left(\sin\alpha,k\right)=\frac{2}{1+k}F\left(\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},\frac{2\sqrt{k}}{1+k}\right)[/math]
同じく、次の恒等式をガウス変換という。
[math]F\left(\sin\alpha,k\right)=\frac{1}{1+k}F\left(\frac{(1+k)\sin\alpha}{1+k\sin^2\alpha},\frac{2\sqrt{k}}{1+k}\right)[/math]
ランデン変換の導出
ランデン変換は
[math]\sin\phi=\frac{\frac{2}{1+k}\sin\theta\cos\theta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}}[/math]
[math]\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}[/math]
の置換により導かれる。
[math]\begin{align}F\left(\sin\alpha,k\right) &=\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}[/math]
[math]\sin\beta[/math]を陽にすると
[math]\sin\alpha=\frac{\frac{2}{1+k}\sin\beta\cos\beta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta}}[/math]
[math]\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)}[/math]
[math]\cos^2(2\beta)+2k\sin^2\alpha\cos(2\beta)+k^2\sin^2\alpha-1+\sin^2\alpha=0[/math]
[math]\cos(2\beta)=-k\sin^2\alpha+\sqrt{k^2\sin^4\alpha-k^2\sin^2\alpha+1-\sin^2\alpha}[/math]
[math]\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}[/math]
である。
ガウス変換の導出
ガウス変換は
[math]\sin\phi=\frac{\frac{2}{1+k}\sin\theta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\theta}}[/math]
[math]\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}[/math]
の置換により導かれる。
[math]\begin{align}F\left(\alpha,k\right) &=\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}[/math]
[math]\sin\beta[/math]を陽にすると
[math]\sin\alpha=\frac{\frac{2}{1+k}\sin\beta}{1+\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta}}[/math]
[math]\sin\alpha\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta}=\frac{2}{1+k}\sin\beta-\sin\alpha[/math]
[math]\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[/math]
[math]\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[/math]
[math]\sin\beta+k\sin^2\alpha\sin\beta-(1+k)\sin\alpha=0[/math]
[math]\sin\beta=\frac{(1+k)\sin\alpha}{1+k\sin^2\alpha}[/math]
である。
楕円関数のランデン変換
次の恒等式を楕円関数の上昇ランデン変換という。
[math]\operatorname{sn}\left(u,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)}[/math]
[math]\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)}[/math]
[math]\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)}[/math]
次の恒等式を楕円関数の下降ランデン変換という。
[math]\operatorname{sn}\left(u,k\right)=\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)}[/math]
[math]\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)}[/math]
[math]\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)}[/math]
当初の母数が[math]0\lt k\lt 1[/math]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。
導出
楕円積分のランデン変換により
[math]\sin\alpha=\frac{\frac{2}{1+k}\sin\beta\cos\beta}{\sqrt{1-\frac{4k}{(1+k)^2}\sin^2\beta}}[/math]
のときに
[math]u=F\left(\alpha,k\right)=\tfrac{2}{1+k}F\left(\beta,\tfrac{2\sqrt{k}}{1+k}\right)[/math]
[math]\operatorname{sn}\left(u,k\right)=\sin\alpha[/math]
[math]\operatorname{sn}\left(\tfrac{1+k}{2}u,\tfrac{2\sqrt{k}}{1+k}\right)=\sin\beta[/math]
であるから
[math]\operatorname{sn}\left(u,k\right)=\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)}[/math]
[math]\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)}[/math]
[math]\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)}[/math]
である。楕円積分のガウス変換により
[math]\sin\beta=\frac{(1+k)\sin\alpha}{1+k\sin^2\alpha}[/math]
のときに
[math]u=F\left(\alpha,k\right)=\tfrac{1}{1+k}F\left(\beta,\tfrac{2\sqrt{k}}{1+k}\right)[/math]
[math]\operatorname{sn}\left(u,k\right)=\sin\alpha[/math]
[math]\operatorname{sn}\left((1+k)u,\tfrac{2\sqrt{k}}{1+k}\right)=\sin\beta[/math]
であるから
[math]\operatorname{sn}\left((1+k)u,\tfrac{2\sqrt{k}}{1+k}\right)=\frac{(1+k)\operatorname{sn}\alpha}{1+k\operatorname{sn}^2\alpha}[/math]
であるが、[math]u[/math]を[math]\tfrac{u}{1+k}[/math]に改め、[math]k[/math]を[math]\tfrac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}[/math]に改めれば
[math]\operatorname{sn}\left(u,k\right)=\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)}[/math]
[math]\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}[/math]
[math]\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}[/math]
となる。
虚数変換
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。
[math]\operatorname{sn}\left(iu,\sqrt{1-k^2}\right)=i\operatorname{sc}\left(u,k\right)=\frac{i\operatorname{sn}\left(u,k\right)}{\operatorname{cn}\left(u,k\right)}[/math]
上昇ランデン変換により
[math]\begin{align}\frac{i\operatorname{sn}\left(u,k\right)}{\operatorname{cn}\left(u,k\right)} &=\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}[/math]
虚数変換により
[math]\begin{align}\operatorname{sn}\left(iu,\sqrt{1-k^2}\right) &=\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}[/math]
[math]iu[/math]を[math]u[/math]と書き、[math]\sqrt{1-k^2}[/math]を[math]k[/math]と書けば
[math]\begin{align}\operatorname{sn}\left(u,k\right) &=\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}[/math]
となるが、これは下降ランデン変換である。