# 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

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

につき、次の恒等式をランデン変換という。

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

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

### ランデン変換の導出

ランデン変換は

${\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta \cos \theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}$
{\displaystyle {\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。

{\displaystyle {\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}}}

${\displaystyle \sin \beta }$ を陽にすると

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

である。

### ガウス変換の導出

ガウス変換は

${\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}$
{\displaystyle {\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。

{\displaystyle {\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 }}}\\&={\frac {1}{1+k}}F\left(\beta ,{\frac {2{\sqrt {k}}}{1+k}}\right)\end{aligned}}}

${\displaystyle \sin \beta }$ を陽にすると

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

である。

## 楕円関数のランデン変換

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

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

### 導出

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

のときに

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

であるから

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

である。楕円積分のガウス変換により

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

のときに

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

であるから

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

であるが、${\displaystyle u}$ ${\displaystyle {\tfrac {u}{1+k}}}$ に改め、${\displaystyle k}$ ${\displaystyle {\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}}$ に改めれば

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

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

{\displaystyle {\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}}}

{\displaystyle {\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}}}

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

{\displaystyle {\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}}}

となるが、これは下降ランデン変換である。