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。 第一種楕円積分 F ( sin ⁡ α , k ) = ∫ t = 0 sin ⁡ α d t 1 − t 2 1 − k 2 t 2 = ∫ ϕ = 0 α d ϕ 1 − k 2 sin 2 ⁡ ϕ {\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。 F ( sin ⁡ α , k ) = 2 1 + k F ( 1 2 ( 1 + k ) 2 sin 2 ⁡ α + ( 1 − k 2 sin 2 ⁡ ϕ − 1 − sin 2 ⁡ ϕ ) 2 , 2 k 1 + k ) {\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)} 同じく、次の恒等式をガウス変換という。 F ( sin ⁡ α , k ) = 1 1 + k F ( ( 1 + k ) sin ⁡ α 1 + k sin 2 ⁡ α , 2 k 1 + k ) {\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}}} の置換により導かれる。 F ( sin ⁡ α , k ) = ∫ ϕ = 0 α d ϕ 1 − k 2 sin 2 ⁡ ϕ = ∫ ϕ = 0 α cos ⁡ ϕ d ϕ 1 − sin 2 ⁡ ϕ 1 − k 2 sin 2 ⁡ ϕ = ∫ θ = 0 β 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 1 − 4 ( 1 + k ) 2 sin 2 ⁡ θ cos 2 ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ 1 − k 2 4 ( 1 + k ) 2 sin 2 ⁡ θ cos 2 ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ d θ = ∫ θ = 0 β 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 1 − 2 1 + k sin 2 ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ 1 − 2 k 1 + k sin 2 ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ d θ = 2 1 + k ∫ θ = 0 β d θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ = 2 1 + k F ( sin ⁡ β , 2 k 1 + k ) {\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}}} sin ⁡ β {\displaystyle \sin \beta } を陽にすると 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 }}}} sin 2 ⁡ α = 4 sin 2 ⁡ β cos 2 ⁡ β ( 1 + k ) 2 − 4 k sin 2 ⁡ β = 1 − cos 2 ⁡ ( 2 β ) 1 + k 2 + 2 k cos ⁡ ( 2 β ) {\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 )}}} cos 2 ⁡ ( 2 β ) + 2 k sin 2 ⁡ α cos ⁡ ( 2 β ) + k 2 sin 2 ⁡ α − 1 + sin 2 ⁡ α = 0 {\displaystyle \cos ^{2}(2\beta )+2k\sin ^{2}\alpha \cos(2\beta )+k^{2}\sin ^{2}\alpha -1+\sin ^{2}\alpha =0} cos ⁡ ( 2 β ) = − k sin 2 ⁡ α + k 2 sin 4 ⁡ α − k 2 sin 2 ⁡ α + 1 − sin 2 ⁡ α {\displaystyle \cos(2\beta )=-k\sin ^{2}\alpha +{\sqrt {k^{2}\sin ^{4}\alpha -k^{2}\sin ^{2}\alpha +1-\sin ^{2}\alpha }}} sin ⁡ β = 1 − cos ⁡ ( 2 β ) 2 = 1 2 ( 1 + k ) 2 sin 2 ⁡ α + ( 1 − k 2 sin 2 ⁡ ϕ − 1 − sin 2 ⁡ ϕ ) 2 {\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。 ガウス変換は sin ⁡ ϕ = 2 1 + k sin ⁡ θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ {\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}} cos ⁡ ϕ d ϕ = 2 1 + k cos ⁡ θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ d θ + 2 1 + k ( 4 k ( 1 + k ) 2 sin 2 ⁡ θ cos ⁡ θ ) 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ) 2 d θ = 2 1 + k cos ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ) d θ {\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。 F ( α , k ) = ∫ ϕ = 0 α d ϕ 1 − k 2 sin 2 ⁡ ϕ = ∫ ϕ = 0 α cos ⁡ ϕ d ϕ 1 − sin 2 ⁡ ϕ 1 − k 2 sin 2 ⁡ ϕ = ∫ θ = 0 β 2 1 + k cos ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ) 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ − 4 1 + k sin 2 ⁡ θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ − 4 k 1 + k sin 2 ⁡ θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ d θ = ∫ θ = 0 β 2 1 + k cos ⁡ θ 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ) 2 1 − sin 2 ⁡ θ 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ ) 2 d θ = 1 1 + k ∫ θ = 0 β 1 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ θ = 1 1 + k F ( β , 2 k 1 + k ) {\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}}} sin ⁡ β {\displaystyle \sin \beta } を陽にすると sin ⁡ α = 2 1 + k sin ⁡ β 1 + 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ β {\displaystyle \sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}}} sin ⁡ α 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ β = 2 1 + k sin ⁡ β − sin ⁡ α {\displaystyle \sin \alpha {\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}={\frac {2}{1+k}}\sin \beta -\sin \alpha } sin 2 ⁡ α ( 1 − 4 k ( 1 + k ) 2 sin 2 ⁡ β ) = 4 ( 1 + k ) 2 sin 2 ⁡ β − 4 1 + k sin ⁡ β sin ⁡ α + sin 2 ⁡ α {\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 } 4 ( 1 + k ) 2 sin 2 ⁡ β + 4 k ( 1 + k ) 2 sin 2 ⁡ α sin 2 ⁡ β − 4 1 + k sin ⁡ β sin ⁡ α = 0 {\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} sin ⁡ β + k sin 2 ⁡ α sin ⁡ β − ( 1 + k ) sin ⁡ α = 0 {\displaystyle \sin \beta +k\sin ^{2}\alpha \sin \beta -(1+k)\sin \alpha =0} sin ⁡ β = ( 1 + k ) sin ⁡ α 1 + k sin 2 ⁡ α {\displaystyle \sin \beta ={\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }}} である。 次の恒等式を楕円関数の上昇ランデン変換という。 sn ⁡ ( u , 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)\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 ) = 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)={\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 ) = 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)={\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)}}} 次の恒等式を楕円関数の下降ランデン変換という。 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 \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)}}} cn ⁡ ( u , k ) = cn ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) dn ⁡ ( 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 \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)}}} dn ⁡ ( u , k ) = 1 − 1 − k 2 1 + 1 − k 2 − ( 1 − dn 2 ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) ) 1 − 1 − k 2 1 + 1 − k 2 + ( 1 − dn 2 ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) ) {\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)}}} 当初の母数が 0 < k < 1 {\displaystyle 0<k<1} であれば、上昇ランデン変換は母数を増加させ、下降ランデン変換は母数を減少させる。上昇ランデン変換を繰り返すことにより、母数が1に収束し、楕円関数は双曲線関数に近似される。下降ランデン変換を繰り返すことにより、母数が0に収束し、楕円関数は三角関数に近似される。この性質により、ランデン変換は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 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)}}} である。楕円積分のガウス変換により sin ⁡ β = ( 1 + k ) sin ⁡ α 1 + k sin 2 ⁡ α {\displaystyle \sin \beta ={\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }}} のときに u = F ( α , k ) = 1 1 + k F ( β , 2 k 1 + k ) {\displaystyle u=F\left(\alpha ,k\right)={\tfrac {1}{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 ) u , 2 k 1 + k ) = sin ⁡ β {\displaystyle \operatorname {sn} \left((1+k)u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)=\sin \beta } であるから sn ⁡ ( ( 1 + k ) u , 2 k 1 + k ) = ( 1 + k ) sn ⁡ α 1 + k sn 2 ⁡ α {\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、 u {\displaystyle u} を u 1 + k {\displaystyle {\tfrac {u}{1+k}}} に改め、 k {\displaystyle k} を 1 − 1 − k 2 1 + 1 − k 2 {\displaystyle {\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}} に改めれば 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 \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)}}} cn ⁡ ( u , k ) = 1 − sn 2 ⁡ ( u , k ) = cn ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) dn ⁡ ( 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 {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}}} dn ⁡ ( u , k ) = 1 − k 2 sn 2 ⁡ ( u , k ) = 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 ) 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 ) = dn 2 ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) − 2 1 − k 2 1 + 1 − k 2 2 1 + 1 − k 2 − dn 2 ⁡ ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\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。 上昇ランデン変換と下降ランデン変換は虚数変換により交替する。 sn ⁡ ( i u , 1 − k 2 ) = i sc ⁡ ( u , k ) = i sn ⁡ ( u , k ) cn ⁡ ( u , k ) {\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)}}} 上昇ランデン変換により i sn ⁡ ( u , k ) cn ⁡ ( u , k ) = 2 i 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 ) 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 ) = 4 k i ( 1 + k ) 2 sn ⁡ ( 1 + k 2 u , 2 k 1 + k ) cn ⁡ ( 1 + k 2 u , 2 k 1 + k ) dn 2 ⁡ ( 1 + k 2 u , 2 k 1 + k ) − 1 − k 1 + k {\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}}} 虚数変換により 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}}} 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。 Wolfram MathWorld: Landen's Transformation Wolfram MathWorld: Gauss's Transformation Abramowitz and Stegun p.573