ランデン変換
ランデン変換は
sin  φ = 2 1 + k sin  θ cos  θ 1 − 4 k ( 1 + k ) 2 sin 2  θ {\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta \cos \theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}} cos  φ d φ = 2 1 + k ( cos 2  θ − sin 2  θ ) 1 − 4 k ( 1 + k ) 2 sin 2  θ d θ + 2 1 + k ( 4 k ( 1 + k ) 2 sin 2  θ cos 2  θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 3 d θ = 2 1 + k ( 1 − 2 1 + k sin 2  θ ) ( 1 − 2 k 1 + k sin 2  θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 3 d θ {\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}}}
i u {\displaystyle iu} を u {\displaystyle u} と書き、 1 − k 2 {\displaystyle {\sqrt {1-k^{2}}}} を k {\displaystyle k} と書けば
sn  ( u , k ) = 2 1 + 1 − k 2 sn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\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。