ランデン変換
楕円関数のランデン変換 > 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。