ランデン変換
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)}}}
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}}}
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。