# ランデン変換 – Wikipedia

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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。

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## 楕円積分のランデン変換とガウス変換

${displaystyle Fleft(sin alpha ,kright)=int _{t=0}^{sin alpha }{frac {dt}{{sqrt {1-t^{2}}}{sqrt {1-k^{2}t^{2}}}}}=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}}$

につき、次の恒等式をランデン変換という。

${displaystyle Fleft(sin alpha ,kright)={frac {2}{1+k}}Fleft({frac {1}{2}}{sqrt {left(1+kright)^{2}sin ^{2}alpha +left({sqrt {1-k^{2}sin ^{2}phi }}-{sqrt {1-sin ^{2}phi }}right)^{2}}},{frac {2{sqrt {k}}}{1+k}}right)}$

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${displaystyle Fleft(sin alpha ,kright)={frac {1}{1+k}}Fleft({frac {(1+k)sin alpha }{1+ksin ^{2}alpha }},{frac {2{sqrt {k}}}{1+k}}right)}$

### ランデン変換の導出

ランデン変換は

${displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta cos theta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}$

{displaystyle {begin{aligned}cos phi {dphi }&={frac {{frac {2}{1+k}}left(cos ^{2}theta -sin ^{2}theta right)}{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}{dtheta }+{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}}}{dtheta }&={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}}}{dtheta }end{aligned}}}

の置換により導かれる。

{displaystyle {begin{aligned}Fleft(sin alpha ,kright)&=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}&=int _{phi =0}^{alpha }{frac {cos phi {dphi }}{{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 }}}}}}{dtheta }&=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 }}}}}{dtheta }&={frac {2}{1+k}}int _{theta =0}^{beta }{frac {dtheta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}&={frac {2}{1+k}}Fleft(sin beta ,{frac {2{sqrt {k}}}{1+k}}right)end{aligned}}}

${displaystyle sin beta }$

を陽にすると

である。

### ガウス変換の導出

ガウス変換は

${displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta }{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}$

{displaystyle {begin{aligned}cos phi {dphi }&={frac {{frac {2}{1+k}}cos theta }{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}{dtheta }+{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}}}{dtheta }&={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)}}{dtheta }end{aligned}}}

の置換により導かれる。

{displaystyle {begin{aligned}Fleft(alpha ,kright)&=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}&=int _{phi =0}^{alpha }{frac {cos phi {dphi }}{{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 }}}}}}{dtheta }&=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}}}}{dtheta }&={frac {1}{1+k}}int _{theta =0}^{beta }{frac {1}{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}&={frac {1}{1+k}}Fleft(beta ,{frac {2{sqrt {k}}}{1+k}}right)end{aligned}}}

${displaystyle sin beta }$

を陽にすると

である。

## 楕円関数のランデン変換

${displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

${displaystyle operatorname {cn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

${displaystyle operatorname {dn} left(u,kright)={frac {{tfrac {2k}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)+{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

${displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+{sqrt {1-k^{2}}}}}operatorname {sn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}}$

${displaystyle operatorname {cn} left(u,kright)={frac {operatorname {cn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)operatorname {dn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}}$

${displaystyle operatorname {dn} left(u,kright)={frac {{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}-left(1-operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)right)}{{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}+left(1-operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)right)}}}$

${displaystyle 0$

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。

### 導出

${displaystyle sin alpha ={frac {{frac {2}{1+k}}sin beta cos beta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}beta }}}}$

のときに

${displaystyle u=Fleft(alpha ,kright)={tfrac {2}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}$

${displaystyle operatorname {sn} left(u,kright)=sin alpha }$

${displaystyle operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }$

であるから

${displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right){sqrt {1-operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}{sqrt {1-left({tfrac {2{sqrt {k}}}{1+k}}right)^{2}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

${displaystyle operatorname {cn} left(u,kright)={sqrt {1-operatorname {sn} ^{2}left(u,kright)}}={frac {1-{tfrac {2}{1+k}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}={frac {{tfrac {2}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

${displaystyle operatorname {dn} left(u,kright)={sqrt {1-k^{2}operatorname {sn} ^{2}left(u,kright)}}={frac {1-{tfrac {2k}{1+k}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}={frac {{tfrac {2k}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)+{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}$

である。楕円積分のガウス変換により

${displaystyle sin beta ={frac {(1+k)sin alpha }{1+ksin ^{2}alpha }}}$

のときに

${displaystyle u=Fleft(alpha ,kright)={tfrac {1}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}$

${displaystyle operatorname {sn} left(u,kright)=sin alpha }$

${displaystyle operatorname {sn} left((1+k)u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }$

であるから

${displaystyle operatorname {sn} left((1+k)u,{tfrac {2{sqrt {k}}}{1+k}}right)={frac {(1+k)operatorname {sn} alpha }{1+koperatorname {sn} ^{2}alpha }}}$

であるが、

${displaystyle u}$

${displaystyle {tfrac {u}{1+k}}}$

に改め、

${displaystyle k}$

${displaystyle {tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}}$

に改めれば

${displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+{sqrt {1-k^{2}}}}}operatorname {sn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}}$

{displaystyle {begin{aligned}operatorname {cn} left(u,kright)&={sqrt {1-operatorname {sn} ^{2}left(u,kright)}}&={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}}}

{displaystyle {begin{aligned}operatorname {dn} left(u,kright)&={sqrt {1-k^{2}operatorname {sn} ^{2}left(u,kright)}}&={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。

### 虚数変換

${displaystyle operatorname {sn} left(iu,{sqrt {1-k^{2}}}right)=ioperatorname {sc} left(u,kright)={frac {ioperatorname {sn} left(u,kright)}{operatorname {cn} left(u,kright)}}}$

{displaystyle {begin{aligned}{frac {ioperatorname {sn} left(u,kright)}{operatorname {cn} left(u,kright)}}&={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}}}

{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}}}

${displaystyle iu}$

${displaystyle u}$

と書き、

${displaystyle {sqrt {1-k^{2}}}}$

${displaystyle k}$

と書けば

{displaystyle {begin{aligned}operatorname {sn} left(u,kright)&={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。

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