ガウス変換の導出
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2018/09/19 10:13 UTC 版)
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。
※この「ガウス変換の導出」の解説は、「ランデン変換」の解説の一部です。
「ガウス変換の導出」を含む「ランデン変換」の記事については、「ランデン変換」の概要を参照ください。
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