ランデン変換 – Wikipedia

Posted on October 30, 2020 by lordneo

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。

Table of Contents

楕円積分のランデン変換とガウス変換[編集]

第一種楕円積分

F(sin⁡α,k)=∫t=0sin⁡αdt1−t21−k2t2=∫ϕ=0αdϕ1−k2sin2⁡ϕ{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 }}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e5887ea658acf46dbb74c4cf592d3a0a0af1c4f2" aria-hidden="true" alt="{displaystyle Fleft(sin alpha ,kright)=int _{t=0}^{sin alpha }{frac {dt}{{sqrt {1-t^{2}}}{sqrt {1-k^{2}t^{2}}}}}=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}}" width="27419.2" height="3663.7">$

につき、次の恒等式をランデン変換という。

F(sin⁡α,k)=21+kF(12(1+k)2sin2⁡α+(1−k2sin2⁡ϕ−1−sin2⁡ϕ)2,2k1+k){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)}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/6e2f04d7a0f0ca7d435e54444f8affec1383d2ed" aria-hidden="true" alt="{displaystyle Fleft(sin alpha ,kright)={frac {2}{1+k}}Fleft({frac {1}{2}}{sqrt {left(1+kright)^{2}sin ^{2}alpha +left({sqrt {1-k^{2}sin ^{2}phi }}-{sqrt {1-sin ^{2}phi }}right)^{2}}},{frac {2{sqrt {k}}}{1+k}}right)}" width="38040.8" height="3376.7">$

同じく、次の恒等式をガウス変換という。

F(sin⁡α,k)=11+kF((1+k)sin⁡α1+ksin2⁡α,2k1+k){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)}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/175affcc26b36b4e01ebf2f41074fc239c2421c7" aria-hidden="true" alt="Fleft(sin alpha ,kright)={frac {1}{1+k}}Fleft({frac {(1+k)sin alpha }{1+ksin ^{2}alpha }},{frac {2{sqrt {k}}}{1+k}}right)" width="19664.1" height="2802.6">$

ランデン変換の導出[編集]

ランデン変換は

sin⁡ϕ=21+ksin⁡θcos⁡θ1−4k(1+k)2sin2⁡θ{displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta cos theta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0ff10ec52593a4ff5f863f01ec11758f734f382b" aria-hidden="true" alt="{displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta cos theta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}" width="11438.9" height="4596.6">$

cos⁡ϕdϕ=21+k(cos2⁡θ−sin2⁡θ)1−4k(1+k)2sin2⁡θdθ+21+k(4k(1+k)2sin2⁡θcos2⁡θ)(1−4k(1+k)2sin2⁡θ)3dθ=21+k(1−21+ksin2⁡θ)(1−2k1+ksin2⁡θ)(1−4k(1+k)2sin2⁡θ)3dθ{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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ac47c3ae4d73d3201129c3ef0412f9e5766acbe8" aria-hidden="true" alt="{displaystyle {begin{aligned}cos phi {dphi }&={frac {{frac {2}{1+k}}left(cos ^{2}theta -sin ^{2}theta right)}{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}{dtheta }+{frac {{frac {2}{1+k}}left({frac {4k}{(1+k)^{2}}}sin ^{2}theta cos ^{2}theta right)}{left({sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{3}}}{dtheta }&={frac {{frac {2}{1+k}}left(1-{frac {2}{1+k}}sin ^{2}theta right)left(1-{frac {2k}{1+k}}sin ^{2}theta right)}{left({sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{3}}}{dtheta }end{aligned}}}" width="28395.4" height="12131.3">$

の置換により導かれる。

F(sin⁡α,k)=∫ϕ=0αdϕ1−k2sin2⁡ϕ=∫ϕ=0αcos⁡ϕdϕ1−sin2⁡ϕ1−k2sin2⁡ϕ=∫θ=0β21+k(1−21+ksin2⁡θ)(1−2k1+ksin2⁡θ)(1−4k(1+k)2sin2⁡θ)31−4(1+k)2sin2⁡θcos2⁡θ1−4k(1+k)2sin2⁡θ1−k24(1+k)2sin2⁡θcos2⁡θ1−4k(1+k)2sin2⁡θdθ=∫θ=0β21+k(1−21+ksin2⁡θ)(1−2k1+ksin2⁡θ)(1−4k(1+k)2sin2⁡θ)31−21+ksin2⁡θ1−4k(1+k)2sin2⁡θ1−2k1+ksin2⁡θ1−4k(1+k)2sin2⁡θdθ=21+k∫θ=0βdθ1−4k(1+k)2sin2⁡θ=21+kF(sin⁡β,2k1+k){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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5d2346d8d8fe68f3d51065bddb1c528131ce5e5f" aria-hidden="true" alt="{displaystyle {begin{aligned}Fleft(sin alpha ,kright)&=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}&=int _{phi =0}^{alpha }{frac {cos phi {dphi }}{{sqrt {1-sin ^{2}phi }}{sqrt {1-k^{2}sin ^{2}phi }}}}&=int _{theta =0}^{beta }{frac {frac {{frac {2}{1+k}}left(1-{frac {2}{1+k}}sin ^{2}theta right)left(1-{frac {2k}{1+k}}sin ^{2}theta right)}{left({sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{3}}}{{sqrt {1-{frac {{frac {4}{(1+k)^{2}}}sin ^{2}theta cos ^{2}theta }{1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}{sqrt {1-k^{2}{frac {{frac {4}{(1+k)^{2}}}sin ^{2}theta cos ^{2}theta }{1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}}{dtheta }&=int _{theta =0}^{beta }{frac {frac {{frac {2}{1+k}}left(1-{frac {2}{1+k}}sin ^{2}theta right)left(1-{frac {2k}{1+k}}sin ^{2}theta right)}{left({sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{3}}}{{frac {1-{frac {2}{1+k}}sin ^{2}theta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}};{frac {1-{frac {2k}{1+k}}sin ^{2}theta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}{dtheta }&={frac {2}{1+k}}int _{theta =0}^{beta }{frac {dtheta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}&={frac {2}{1+k}}Fleft(sin beta ,{frac {2{sqrt {k}}}{1+k}}right)end{aligned}}}" width="29121.6" height="30932.1">$

sin⁡β{displaystyle sin beta }

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/47d2d2720a724f3127503eeb994f24377afc8af2" aria-hidden="true" alt="{displaystyle sin beta }" width="1969.7" height="1080.4">$ を陽にすると

である。

ガウス変換の導出[編集]

ガウス変換は

sin⁡ϕ=21+ksin⁡θ1+1−4k(1+k)2sin2⁡θ{displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta }{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/fa196aba0c09e75273e94412e580338b02b56cc0" aria-hidden="true" alt="{displaystyle sin phi ={frac {{frac {2}{1+k}}sin theta }{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}" width="13162.4" height="4596.6">$

cos⁡ϕdϕ=21+kcos⁡θ1+1−4k(1+k)2sin2⁡θdθ+21+k(4k(1+k)2sin2⁡θcos⁡θ)1−4k(1+k)2sin2⁡θ(1+1−4k(1+k)2sin2⁡θ)2dθ=21+kcos⁡θ1−4k(1+k)2sin2⁡θ(1+1−4k(1+k)2sin2⁡θ)dθ{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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c11355ec29c87b0b7c12ef96c490aa83843e98a6" aria-hidden="true" alt="{displaystyle {begin{aligned}cos phi {dphi }&={frac {{frac {2}{1+k}}cos theta }{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}{dtheta }+{frac {{frac {2}{1+k}}left({frac {4k}{(1+k)^{2}}}sin ^{2}theta cos theta right)}{{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}left(1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{2}}}{dtheta }&={frac {{frac {2}{1+k}}cos theta }{{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}left(1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)}}{dtheta }end{aligned}}}" width="37551.6" height="11557.2">$

の置換により導かれる。

F(α,k)=∫ϕ=0αdϕ1−k2sin2⁡ϕ=∫ϕ=0αcos⁡ϕdϕ1−sin2⁡ϕ1−k2sin2⁡ϕ=∫θ=0β21+kcos⁡θ1−4k(1+k)2sin2⁡θ(1+1−4k(1+k)2sin2⁡θ)2+21−4k(1+k)2sin2⁡θ−41+ksin2⁡θ1+1−4k(1+k)2sin2⁡θ2+21−4k(1+k)2sin2⁡θ−4k1+ksin2⁡θ1+1−4k(1+k)2sin2⁡θdθ=∫θ=0β21+kcos⁡θ1−4k(1+k)2sin2⁡θ(1+1−4k(1+k)2sin2⁡θ)21−sin2⁡θ2+21−4k(1+k)2sin2⁡θ−4k(1+k)2sin2⁡θ(1+1−4k(1+k)2sin2⁡θ)2dθ=11+k∫θ=0β11−4k(1+k)2sin2⁡θ=11+kF(β,2k1+k){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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1f65bc3c7da0a23a995f5f4ead322e562c57b459" aria-hidden="true" alt="{displaystyle {begin{aligned}Fleft(alpha ,kright)&=int _{phi =0}^{alpha }{frac {dphi }{sqrt {1-k^{2}sin ^{2}phi }}}&=int _{phi =0}^{alpha }{frac {cos phi {dphi }}{{sqrt {1-sin ^{2}phi }}{sqrt {1-k^{2}sin ^{2}phi }}}}&=int _{theta =0}^{beta }{frac {frac {{frac {2}{1+k}}cos theta }{{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}left(1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)}}{{frac {sqrt {2+2{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}-{frac {4}{1+k}}sin ^{2}theta }}{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}};{frac {sqrt {2+2{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}-{frac {4k}{1+k}}sin ^{2}theta }}{1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}}}}{dtheta }&=int _{theta =0}^{beta }{frac {frac {{frac {2}{1+k}}cos theta }{{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}left(1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)}}{frac {2{sqrt {1-sin ^{2}theta }}{sqrt {2+2{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}{left(1+{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}right)^{2}}}}{dtheta }&={frac {1}{1+k}}int _{theta =0}^{beta }{frac {1}{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}theta }}}&={frac {1}{1+k}}Fleft(beta ,{frac {2{sqrt {k}}}{1+k}}right)end{aligned}}}" width="32355.7" height="32654.3">$

sin⁡β{displaystyle sin beta }

である。

楕円関数のランデン変換[編集]

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)=21+ksn⁡(1+k2u,2k1+k)cn⁡(1+k2u,2k1+k)dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a52810fd4dc98f489d2041753954cd6184df7114" aria-hidden="true" alt="{displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="19917.4" height="4381.3">$

cn⁡(u,k)=21+kdn2⁡(1+k2u,2k1+k)−1−k1+k4k(1+k)2dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e26ff0984459b30dbc3e2725b720563723036da6" aria-hidden="true" alt="{displaystyle operatorname {cn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="16753.9" height="4596.6">$

dn⁡(u,k)=2k1+kdn2⁡(1+k2u,2k1+k)+1−k1+k4k(1+k)2dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/02e2e3a6368869064f94e273b6c4619e36db2bb1" aria-hidden="true" alt="{displaystyle operatorname {dn} left(u,kright)={frac {{tfrac {2k}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)+{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="16865.9" height="4596.6">$

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)=21+1−k2sn⁡(1+1−k22u,1−1−k21+1−k2)1+1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c567de8bc1ab056412bf1b2aa7bd3cd9fe53f36d" aria-hidden="true" alt="{displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+{sqrt {1-k^{2}}}}}operatorname {sn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}}" width="21670" height="6821.1">$

cn⁡(u,k)=cn⁡(1+1−k22u,1−1−k21+1−k2)dn⁡(1+1−k22u,1−1−k21+1−k2)1+1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/4b6d38d384e868bb94c91510fad1b66dfad480ea" aria-hidden="true" alt="{displaystyle operatorname {cn} left(u,kright)={frac {operatorname {cn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)operatorname {dn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}}" width="26911.3" height="6821.1">$

dn⁡(u,k)=1−1−k21+1−k2−(1−dn2⁡(1+1−k22u,1−1−k21+1−k2))1−1−k21+1−k2+(1−dn2⁡(1+1−k22u,1−1−k21+1−k2)){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1c77c07899a0f027b31bc2c98c8c3b4c191945cc" aria-hidden="true" alt="{displaystyle operatorname {dn} left(u,kright)={frac {{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}-left(1-operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)right)}{{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}+left(1-operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)right)}}}" width="24635.3" height="6821.1">$

当初の母数が

0<k<1{displaystyle 0

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c548b9b4ff336614c8116ed1520e374cba8a2f1b" aria-hidden="true" alt="{displaystyle 0<k<1}" width="4190.6" height="936.9">$ 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[編集]

楕円積分のランデン変換により

sin⁡α=21+ksin⁡βcos⁡β1−4k(1+k)2sin2⁡β{displaystyle sin alpha ={frac {{frac {2}{1+k}}sin beta cos beta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}beta }}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1a42178258e961fda04617dae066222157d1f823" aria-hidden="true" alt="{displaystyle sin alpha ={frac {{frac {2}{1+k}}sin beta cos beta }{sqrt {1-{frac {4k}{(1+k)^{2}}}sin ^{2}beta }}}}" width="11586.9" height="4596.6">$

のときに

u=F(α,k)=21+kF(β,2k1+k){displaystyle u=Fleft(alpha ,kright)={tfrac {2}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d9848e3b4b279f5332221d8b84dea36bf81d2435" aria-hidden="true" alt="{displaystyle u=Fleft(alpha ,kright)={tfrac {2}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}" width="12978" height="2085">$

sn⁡(u,k)=sin⁡α{displaystyle operatorname {sn} left(u,kright)=sin alpha }

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c8219eaef5040a4e46050d6c16e55b4253b18956" aria-hidden="true" alt="{displaystyle operatorname {sn} left(u,kright)=sin alpha }" width="6639.9" height="1223.9">$

sn⁡(1+k2u,2k1+k)=sin⁡β{displaystyle operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/060c3bc8e55ae3945ac64332128947cef5a94c79" aria-hidden="true" alt="{displaystyle operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }" width="9772.6" height="2085">$

であるから

sn⁡(u,k)=21+ksn⁡(1+k2u,2k1+k)1−sn2⁡(1+k2u,2k1+k)1−(2k1+k)2sn2⁡(1+k2u,2k1+k)=21+ksn⁡(1+k2u,2k1+k)cn⁡(1+k2u,2k1+k)dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/17db52ab648aa0fb8dda42533214aacb19f70e52" aria-hidden="true" alt="{displaystyle operatorname {sn} left(u,kright)={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right){sqrt {1-operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}{sqrt {1-left({tfrac {2{sqrt {k}}}{1+k}}right)^{2}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="39526.8" height="5673">$

cn⁡(u,k)=1−sn2⁡(u,k)=1−21+ksn2⁡(1+k2u,2k1+k)dn⁡(1+k2u,2k1+k)=21+kdn2⁡(1+k2u,2k1+k)−1−k1+k4k(1+k)2dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/23133fb3da1499ffcb015520d779dc118d1113ae" aria-hidden="true" alt="{displaystyle operatorname {cn} left(u,kright)={sqrt {1-operatorname {sn} ^{2}left(u,kright)}}={frac {1-{tfrac {2}{1+k}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}={frac {{tfrac {2}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="36675" height="4596.6">$

dn⁡(u,k)=1−k2sn2⁡(u,k)=1−2k1+ksn2⁡(1+k2u,2k1+k)dn⁡(1+k2u,2k1+k)=2k1+kdn2⁡(1+k2u,2k1+k)+1−k1+k4k(1+k)2dn⁡(1+k2u,2k1+k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/822195b8af9ad294d3a737fc42703ae8ae32def6" aria-hidden="true" alt="{displaystyle operatorname {dn} left(u,kright)={sqrt {1-k^{2}operatorname {sn} ^{2}left(u,kright)}}={frac {1-{tfrac {2k}{1+k}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}={frac {{tfrac {2k}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)+{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}" width="37929.1" height="4596.6">$

である。楕円積分のガウス変換により

sin⁡β=(1+k)sin⁡α1+ksin2⁡α{displaystyle sin beta ={frac {(1+k)sin alpha }{1+ksin ^{2}alpha }}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a10d8ae9eb2c35693f3042131fd30ec66a89eaf0" aria-hidden="true" alt="{displaystyle sin beta ={frac {(1+k)sin alpha }{1+ksin ^{2}alpha }}}" width="8891" height="2730.8">$

のときに

u=F(α,k)=11+kF(β,2k1+k){displaystyle u=Fleft(alpha ,kright)={tfrac {1}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/85c9850cf92d4e3e48073112fc4ac071e7236f30" aria-hidden="true" alt="{displaystyle u=Fleft(alpha ,kright)={tfrac {1}{1+k}}Fleft(beta ,{tfrac {2{sqrt {k}}}{1+k}}right)}" width="12978" height="2085">$

sn⁡(u,k)=sin⁡α{displaystyle operatorname {sn} left(u,kright)=sin alpha }

sn⁡((1+k)u,2k1+k)=sin⁡β{displaystyle operatorname {sn} left((1+k)u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f3f45538462836140b9ace5089eeb387b835753f" aria-hidden="true" alt="{displaystyle operatorname {sn} left((1+k)u,{tfrac {2{sqrt {k}}}{1+k}}right)=sin beta }" width="11163.4" height="2085">$

であるから

sn⁡((1+k)u,2k1+k)=(1+k)sn⁡α1+ksn2⁡α{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 }}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/bf764eb47de241d35b16bccff97268b87c1c9053" aria-hidden="true" alt="{displaystyle operatorname {sn} left((1+k)u,{tfrac {2{sqrt {k}}}{1+k}}right)={frac {(1+k)operatorname {sn} alpha }{1+koperatorname {sn} ^{2}alpha }}}" width="14502.5" height="2659.1">$

であるが、

u{displaystyle u}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c3e6bb763d22c20916ed4f0bb6bd49d7470cffd8" aria-hidden="true" alt="u" width="572.5" height="721.6">$ を

u1+k{displaystyle {tfrac {u}{1+k}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/beb2bfd74b8458d862ad3ba986c50eccb683d6b7" aria-hidden="true" alt="{displaystyle {tfrac {u}{1+k}}}" width="1633.1" height="1439.2">$ に改め、

k{displaystyle k}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40" aria-hidden="true" alt="k" width="521.5" height="936.9">$ を

1−1−k21+1−k2{displaystyle {tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/235db7ab833f69fd92adf019bfe34c162fb69326" aria-hidden="true" alt="{displaystyle {tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}}" width="3603" height="3233.2">$ に改めれば

cn⁡(u,k)=1−sn2⁡(u,k)=cn⁡(1+1−k22u,1−1−k21+1−k2)dn⁡(1+1−k22u,1−1−k21+1−k2)1+1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2){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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cf6894729b8dbbfb6d180c817f6f778e2a7049be" aria-hidden="true" alt="{displaystyle {begin{aligned}operatorname {cn} left(u,kright)&={sqrt {1-operatorname {sn} ^{2}left(u,kright)}}&={frac {operatorname {cn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)operatorname {dn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}end{aligned}}}" width="27234.9" height="8973.9">$

dn⁡(u,k)=1−k2sn2⁡(u,k)=1−1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2)1+1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2)=dn2⁡(1+1−k22u,1−1−k21+1−k2)−21−k21+1−k221+1−k2−dn2⁡(1+1−k22u,1−1−k21+1−k2){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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/14fc3c7d7470193dfb4e8bf7d0952329138e552a" aria-hidden="true" alt="{displaystyle {begin{aligned}operatorname {dn} left(u,kright)&={sqrt {1-k^{2}operatorname {sn} ^{2}left(u,kright)}}&={frac {1-{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}&={frac {operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)-{tfrac {2{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}}{{tfrac {2}{1+{sqrt {1-k^{2}}}}}-operatorname {dn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}end{aligned}}}" width="22155.5" height="15862.7">$

となる。

虚数変換[編集]

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⁡(iu,1−k2)=isc⁡(u,k)=isn⁡(u,k)cn⁡(u,k){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)}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/79a420c63ee92e5cdb0cb80ae3c010976730e95f" aria-hidden="true" alt="{displaystyle operatorname {sn} left(iu,{sqrt {1-k^{2}}}right)=ioperatorname {sc} left(u,kright)={frac {ioperatorname {sn} left(u,kright)}{operatorname {cn} left(u,kright)}}}" width="17687.3" height="2802.6">$

上昇ランデン変換により

isn⁡(u,k)cn⁡(u,k)=2i1+ksn⁡(1+k2u,2k1+k)cn⁡(1+k2u,2k1+k)dn⁡(1+k2u,2k1+k)21+kdn2⁡(1+k2u,2k1+k)−1−k1+k4k(1+k)2dn⁡(1+k2u,2k1+k)=4ki(1+k)2sn⁡(1+k2u,2k1+k)cn⁡(1+k2u,2k1+k)dn2⁡(1+k2u,2k1+k)−1−k1+k{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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d3b73d51475b77111ff8f4af37f42ea4a80babf2" aria-hidden="true" alt="{displaystyle {begin{aligned}{frac {ioperatorname {sn} left(u,kright)}{operatorname {cn} left(u,kright)}}&={frac {frac {{tfrac {2i}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}{frac {{tfrac {2}{1+k}}operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}{{tfrac {4k}{(1+k)^{2}}}operatorname {dn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}}}&={frac {{tfrac {4ki}{(1+k)^{2}}}operatorname {sn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)operatorname {cn} left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)}{operatorname {dn} ^{2}left({tfrac {1+k}{2}}u,{tfrac {2{sqrt {k}}}{1+k}}right)-{tfrac {1-k}{1+k}}}}end{aligned}}}" width="22022" height="14714.6">$

虚数変換により

sn⁡(iu,1−k2)=4k(1+k)2sc⁡(1+k2iu,1−k1+k)nc⁡(1+k2iu,1−k1+k)dc2⁡(1+k2iu,1−k1+k)−1−k1+k=4k(1+k)2sn⁡(1+k2iu,1−k1+k)dn2⁡(1+k2iu,1−k1+k)−1−k1+kcn2⁡(1+k2iu,1−k1+k)=4k(1+k)2sn⁡(1+k2iu,1−k1+k)2k1+k+2k(1−k)(1+k)2sn2⁡(1+k2iu,1−k1+k)=21+ksn⁡(1+k2iu,1−k1+k)1+1−k1+ksn2⁡(1+k2iu,1−k1+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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9a9454bad660218a99f230d019bc991d9855c896" aria-hidden="true" alt="{displaystyle {begin{aligned}operatorname {sn} left(iu,{sqrt {1-k^{2}}}right)&={frac {{tfrac {4k}{(1+k)^{2}}}operatorname {sc} left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)operatorname {nc} left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}{operatorname {dc} ^{2}left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)-{tfrac {1-k}{1+k}}}}&={frac {{tfrac {4k}{(1+k)^{2}}}operatorname {sn} left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}{operatorname {dn} ^{2}left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)-{tfrac {1-k}{1+k}}operatorname {cn} ^{2}left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}}&={frac {{tfrac {4k}{(1+k)^{2}}}operatorname {sn} left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}{{tfrac {2k}{1+k}}+{tfrac {2k(1-k)}{(1+k)^{2}}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}}&={frac {{tfrac {2}{1+k}}operatorname {sn} left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}{1+{tfrac {1-k}{1+k}}operatorname {sn} ^{2}left({tfrac {1+k}{2}}iu,{tfrac {1-k}{1+k}}right)}}end{aligned}}}" width="26919.6" height="18302.5">$

iu{displaystyle iu}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/43afd34ebc4535fa25f11835b762b45c44bff73f" aria-hidden="true" alt="{displaystyle iu}" width="918" height="936.9">$ を

u{displaystyle u}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c3e6bb763d22c20916ed4f0bb6bd49d7470cffd8" aria-hidden="true" alt="u" width="572.5" height="721.6">$ と書き、

1−k2{displaystyle {sqrt {1-k^{2}}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cf042a9d40a9be3499955ea8a75cde63081bb31f" aria-hidden="true" alt="{displaystyle {sqrt {1-k^{2}}}}" width="3699.4" height="1510.9">$ を

k{displaystyle k}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40" aria-hidden="true" alt="k" width="521.5" height="936.9">$ と書けば

sn⁡(u,k)=21+1−k2sn⁡(1+1−k22u,1−1−k21+1−k2)1+1−1−k21+1−k2sn2⁡(1+1−k22u,1−1−k21+1−k2){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}}}

$<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ecab43e8f332ef9a92573c00fa83b7d5f78078af" aria-hidden="true" alt="{displaystyle {begin{aligned}operatorname {sn} left(u,kright)&={frac {{tfrac {2}{1+{sqrt {1-k^{2}}}}}operatorname {sn} left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}{1+{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}operatorname {sn} ^{2}left({tfrac {1+{sqrt {1-k^{2}}}}{2}}u,{tfrac {1-{sqrt {1-k^{2}}}}{1+{sqrt {1-k^{2}}}}}right)}}end{aligned}}}" width="21993.5" height="6821.1">$

となるが、これは下降ランデン変換である。

ランデン変換 – Wikipedia

楕円積分のランデン変換とガウス変換[編集]

ランデン変換の導出[編集]

ガウス変換の導出[編集]

楕円関数のランデン変換[編集]

導出[編集]

虚数変換[編集]

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