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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

$F\left(\sin \alpha ;k\right)=\int _{t=0}^{\sin \alpha }{\frac {dt}{{\sqrt {1-t^{2}}}{\sqrt {1-k^{2}t^{2}}}}}=\int _{\phi =0}^{\alpha }{\frac {d\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}$

につき、次の恒等式をランデン変換という。

$F\left(\sin \alpha ;k\right)={\frac {2}{1+k}}F\left({\frac {1}{2}}{\sqrt {\left(1+k\right)^{2}\sin ^{2}\alpha +\left({\sqrt {1-k^{2}\sin ^{2}\alpha }}-{\sqrt {1-\sin ^{2}\alpha }}\right)^{2}}};{\frac {2{\sqrt {k}}}{1+k}}\right)$

同じく、次の恒等式をガウス変換という。

$F\left(\sin \alpha ;k\right)={\frac {1}{1+k}}F\left({\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }};{\frac {2{\sqrt {k}}}{1+k}}\right)$

ランデン変換の導出

ランデン変換は

$\sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta \cos \theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}$
${\begin{aligned}\cos \phi {d\phi }&={\frac {{\frac {2}{1+k}}\left(\cos ^{2}\theta -\sin ^{2}\theta \right)}{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}{d\theta }+{\frac {{\frac {2}{1+k}}\left({\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{d\theta }\\&={\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{d\theta }\end{aligned}}$

の置換により導かれる。

${\begin{aligned}F\left(\sin \alpha ;k\right)&=\int _{\phi =0}^{\alpha }{\frac {d\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\\&=\int _{\phi =0}^{\alpha }{\frac {\cos \phi {d\phi }}{{\sqrt {1-\sin ^{2}\phi }}{\sqrt {1-k^{2}\sin ^{2}\phi }}}}\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{{\sqrt {1-{\frac {{\frac {4}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta }{1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}{\sqrt {1-k^{2}{\frac {{\frac {4}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta }{1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}}{d\theta }\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{{\frac {1-{\frac {2}{1+k}}\sin ^{2}\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}\;{\frac {1-{\frac {2k}{1+k}}\sin ^{2}\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}{d\theta }\\&={\frac {2}{1+k}}\int _{\theta =0}^{\beta }{\frac {d\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}\\&={\frac {2}{1+k}}F\left(\sin \beta ;{\frac {2{\sqrt {k}}}{1+k}}\right)\end{aligned}}$

$\sin \beta$ を陽にすると

$\sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta \cos \beta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}$
$\sin ^{2}\alpha ={\frac {4\sin ^{2}\beta \cos ^{2}\beta }{(1+k)^{2}-4k\sin ^{2}\beta }}={\frac {1-\cos ^{2}(2\beta )}{1+k^{2}+2k\cos(2\beta )}}$
$\cos ^{2}(2\beta )+2k\sin ^{2}\alpha \cos(2\beta )+k^{2}\sin ^{2}\alpha -1+\sin ^{2}\alpha =0$
$\cos(2\beta )=-k\sin ^{2}\alpha +{\sqrt {k^{2}\sin ^{4}\alpha -k^{2}\sin ^{2}\alpha +1-\sin ^{2}\alpha }}$
$\sin \beta ={\sqrt {\frac {1-\cos(2\beta )}{2}}}={\frac {1}{2}}{\sqrt {\left(1+k\right)^{2}\sin ^{2}\alpha +\left({\sqrt {1-k^{2}\sin ^{2}\alpha }}-{\sqrt {1-\sin ^{2}\alpha }}\right)^{2}}}$

である。

ガウス変換の導出

ガウス変換は

$\sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}$
${\begin{aligned}\cos \phi {d\phi }&={\frac {{\frac {2}{1+k}}\cos \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}{d\theta }+{\frac {{\frac {2}{1+k}}\left({\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta \cos \theta \right)}{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{2}}}{d\theta }\\&={\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{d\theta }\end{aligned}}$

の置換により導かれる。

${\begin{aligned}F\left(\alpha ;k\right)&=\int _{\phi =0}^{\alpha }{\frac {d\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\\&=\int _{\phi =0}^{\alpha }{\frac {\cos \phi {d\phi }}{{\sqrt {1-\sin ^{2}\phi }}{\sqrt {1-k^{2}\sin ^{2}\phi }}}}\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{{\frac {\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4}{1+k}}\sin ^{2}\theta }}{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}\;{\frac {\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4k}{1+k}}\sin ^{2}\theta }}{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}}{d\theta }\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{\frac {2{\sqrt {1-\sin ^{2}\theta }}{\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}{\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{2}}}}{d\theta }\\&={\frac {1}{1+k}}\int _{\theta =0}^{\beta }{\frac {1}{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}{d\theta }\\&={\frac {1}{1+k}}F\left(\beta ;{\frac {2{\sqrt {k}}}{1+k}}\right)\end{aligned}}$

$\sin \beta$ を陽にすると

$\sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}}$
$\sin \alpha {\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}={\frac {2}{1+k}}\sin \beta -\sin \alpha$
$\sin ^{2}\alpha \left(1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta \right)={\frac {4}{(1+k)^{2}}}\sin ^{2}\beta -{\frac {4}{1+k}}\sin \beta \sin \alpha +\sin ^{2}\alpha$
${\frac {4}{(1+k)^{2}}}\sin ^{2}\beta +{\frac {4k}{(1+k)^{2}}}\sin ^{2}\alpha \sin ^{2}\beta -{\frac {4}{1+k}}\sin \beta \sin \alpha =0$
$\sin \beta +k\sin ^{2}\alpha \sin \beta -(1+k)\sin \alpha =0$
$\sin \beta ={\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }}$

である。

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。

$\operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$
$\operatorname {cn} \left(u,k\right)={\frac {\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}{{\tfrac {2k}{1+k}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$
$\operatorname {dn} \left(u,k\right)={\frac {\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)+{\tfrac {1-k}{1+k}}}{{\tfrac {2}{1+k}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$

次の恒等式を楕円関数の下降ランデン変換という。

$\operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}$
$\operatorname {cn} \left(u,k\right)={\frac {\operatorname {cn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\operatorname {dn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}$
$\operatorname {dn} \left(u,k\right)={\frac {{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}-\left(1-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\right)}{{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}+\left(1-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\right)}}$

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。

導出

楕円積分のランデン変換により

$\sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta \cos \beta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}$

のときに

$u=F\left(\alpha ,k\right)={\tfrac {2}{1+k}}F\left(\beta ,{\tfrac {2{\sqrt {k}}}{1+k}}\right)$
$\operatorname {sn} \left(u,k\right)=\sin \alpha$
$\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)=\sin \beta$

であるから

$\operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right){\sqrt {1-\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}}{\sqrt {1-\left({\tfrac {2{\sqrt {k}}}{1+k}}\right)^{2}\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}}={\frac {{\tfrac {2}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$
$\operatorname {cn} \left(u,k\right)={\sqrt {1-\operatorname {sn} ^{2}\left(u,k\right)}}={\frac {1-{\tfrac {2}{1+k}}\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}={\frac {{\tfrac {2}{1+k}}\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}{{\tfrac {4k}{(1+k)^{2}}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$
$\operatorname {dn} \left(u,k\right)={\sqrt {1-k^{2}\operatorname {sn} ^{2}\left(u,k\right)}}={\frac {1-{\tfrac {2k}{1+k}}\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}={\frac {{\tfrac {2k}{1+k}}\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)+{\tfrac {1-k}{1+k}}}{{\tfrac {4k}{(1+k)^{2}}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}$

である。楕円積分のガウス変換により

$\sin \beta ={\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }}$

のときに

$u=F\left(\alpha ,k\right)={\tfrac {1}{1+k}}F\left(\beta ,{\tfrac {2{\sqrt {k}}}{1+k}}\right)$
$\operatorname {sn} \left(u,k\right)=\sin \alpha$
$\operatorname {sn} \left((1+k)u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)=\sin \beta$

であるから

$\operatorname {sn} \left((1+k)u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)={\frac {(1+k)\operatorname {sn} \alpha }{1+k\operatorname {sn} ^{2}\alpha }}$

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

$\operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}$
${\begin{aligned}\operatorname {cn} \left(u,k\right)&={\sqrt {1-\operatorname {sn} ^{2}\left(u,k\right)}}\\&={\frac {\operatorname {cn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\operatorname {dn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\end{aligned}}$
${\begin{aligned}\operatorname {dn} \left(u,k\right)&={\sqrt {1-k^{2}\operatorname {sn} ^{2}\left(u,k\right)}}\\&={\frac {1-{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\\&={\frac {\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)-{\tfrac {2{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}}{{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\end{aligned}}$

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。

$\operatorname {sn} \left(iu,{\sqrt {1-k^{2}}}\right)=i\operatorname {sc} \left(u,k\right)={\frac {i\operatorname {sn} \left(u,k\right)}{\operatorname {cn} \left(u,k\right)}}$

上昇ランデン変換により

${\begin{aligned}{\frac {i\operatorname {sn} \left(u,k\right)}{\operatorname {cn} \left(u,k\right)}}&={\frac {\frac {{\tfrac {2i}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}{\frac {{\tfrac {2}{1+k}}\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}{{\tfrac {4k}{(1+k)^{2}}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}}\\&={\frac {{\tfrac {4ki}{(1+k)^{2}}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}}\\\end{aligned}}$

虚数変換により

${\begin{aligned}\operatorname {sn} \left(iu,{\sqrt {1-k^{2}}}\right)&={\frac {{\tfrac {4k}{(1+k)^{2}}}\operatorname {sc} \left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)\operatorname {nc} \left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}{\operatorname {dc} ^{2}\left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)-{\tfrac {1-k}{1+k}}}}\\&={\frac {{\tfrac {4k}{(1+k)^{2}}}\operatorname {sn} \left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}{\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)-{\tfrac {1-k}{1+k}}\operatorname {cn} ^{2}\left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}}\\&={\frac {{\tfrac {4k}{(1+k)^{2}}}\operatorname {sn} \left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}{{\tfrac {2k}{1+k}}+{\tfrac {2k(1-k)}{(1+k)^{2}}}\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}}\\&={\frac {{\tfrac {2}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}{1+{\tfrac {1-k}{1+k}}\operatorname {sn} ^{2}\left({\tfrac {1+k}{2}}iu,{\tfrac {1-k}{1+k}}\right)}}\\\end{aligned}}$

$iu$ を $u$ と書き、 ${\sqrt {1-k^{2}}}$ を $k$ と書けば

${\begin{aligned}\operatorname {sn} \left(u,k\right)&={\frac {{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\end{aligned}}$