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。
sn  ( u , k ) = 2 1 + k sn  ( 1 + k 2 u , 2 k 1 + k ) 1 − sn 2  ( 1 + k 2 u , 2 k 1 + k ) 1 − ( 2 k 1 + k ) 2 sn 2  ( 1 + k 2 u , 2 k 1 + k ) = 2 1 + k sn  ( 1 + k 2 u , 2 k 1 + k ) cn  ( 1 + k 2 u , 2 k 1 + k ) dn  ( 1 + k 2 u , 2 k 1 + k ) {\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)}}} cn  ( u , k ) = 1 − sn 2  ( u , k ) = 1 − 2 1 + k sn 2  ( 1 + k 2 u , 2 k 1 + k ) dn  ( 1 + k 2 u , 2 k 1 + k ) = 2 1 + k dn 2  ( 1 + k 2 u , 2 k 1 + k ) − 1 − k 1 + k 4 k ( 1 + k ) 2 dn  ( 1 + k 2 u , 2 k 1 + k ) {\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)}}} dn  ( u , k ) = 1 − k 2 sn 2  ( u , k ) = 1 − 2 k 1 + k sn 2  ( 1 + k 2 u , 2 k 1 + k ) dn  ( 1 + k 2 u , 2 k 1 + k ) = 2 k 1 + k dn 2  ( 1 + k 2 u , 2 k 1 + k ) + 1 − k 1 + k 4 k ( 1 + k ) 2 dn  ( 1 + k 2 u , 2 k 1 + k ) {\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)}}}
楕円積分のガウス変換により
sin  β {\displaystyle \sin \beta } を陽にすると
楕円積分のランデン変換により
sin  α = 2 1 + k sin  β cos  β 1 − 4 k ( 1 + k ) 2 sin 2  β {\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
u = F ( α , k ) = 2 1 + k F ( β , 2 k 1 + k ) {\displaystyle u=F\left(\alpha ,k\right)={\tfrac {2}{1+k}}F\left(\beta ,{\tfrac {2{\sqrt {k}}}{1+k}}\right)} sn  ( u , k ) = sin  α {\displaystyle \operatorname {sn} \left(u,k\right)=\sin \alpha } sn  ( 1 + k 2 u , 2 k 1 + k ) = sin  β {\displaystyle \operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)=\sin \beta }
であるから