ランデン変換
sn  ( i u , 1 − k 2 ) = 4 k ( 1 + k ) 2 sc  ( 1 + k 2 i u , 1 − k 1 + k ) nc  ( 1 + k 2 i u , 1 − k 1 + k ) dc 2  ( 1 + k 2 i u , 1 − k 1 + k ) − 1 − k 1 + k = 4 k ( 1 + k ) 2 sn  ( 1 + k 2 i u , 1 − k 1 + k ) dn 2  ( 1 + k 2 i u , 1 − k 1 + k ) − 1 − k 1 + k cn 2  ( 1 + k 2 i u , 1 − k 1 + k ) = 4 k ( 1 + k ) 2 sn  ( 1 + k 2 i u , 1 − k 1 + k ) 2 k 1 + k + 2 k ( 1 − k ) ( 1 + k ) 2 sn 2  ( 1 + k 2 i u , 1 − k 1 + k ) = 2 1 + k sn  ( 1 + k 2 i u , 1 − k 1 + k ) 1 + 1 − k 1 + k sn 2  ( 1 + k 2 i u , 1 − k 1 + k ) {\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}}}