ルジャンドルの関係式

数学において、ルジャンドルの関係式(Legendre relation)は第一種完全楕円積分と第二種完全楕円積分の間に成立する恒等式である。 K ( k ) E ( 1 − k 2 ) + E ( k ) K ( 1 − k 2 ) − K ( k ) K ( 1 − k 2 ) = π 2 {\displaystyle K(k)E\left({\sqrt {1-k^{2}}}\right)+E(k)K\left({\sqrt {1-k^{2}}}\right)-K(k)K\left({\sqrt {1-k^{2}}}\right)={\frac {\pi }{2}}} 完全楕円積分の導関数 k d d k E ( k ) = k d d k π 2 ( 1 − ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n 2 n − 1 ) = π 2 ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 ( − 1 2 n − 1 − 1 ) k 2 n = E ( k ) − K ( k ) {\displaystyle {\begin{aligned}k{\frac {d}{dk}}E(k)&=k{\frac {d}{dk}}{\frac {\pi }{2}}\left(1-\sum _{n=1}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}{\frac {k^{2n}}{2n-1}}}\right)\\&={\frac {\pi }{2}}\sum _{n=1}^{\infty }\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}\left(-{\frac {1}{2n-1}}-1\right)k^{2n}\\&=E(k)-K(k)\end{aligned}}} k d d k K ( k ) = k d d k π 2 ( 1 + ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n ) = π 2 ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 ( ( 2 n ) 2 ( 2 n − 1 ) 2 ( 2 n − 2 + 1 ) − 1 2 n − 1 − 1 ) k 2 n = k 2 ( k d d k K ( k ) + K ( k ) ) + E ( k ) − K ( k ) {\displaystyle {\begin{aligned}k{\frac {d}{dk}}K(k)&=k{\frac {d}{dk}}{\frac {\pi }{2}}\left(1+\sum _{n=1}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}k^{2n}}\right)\\&={\frac {\pi }{2}}\sum _{n=1}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}\left({\frac {(2n)^{2}}{(2n-1)^{2}}}(2n-2+1)-{\frac {1}{2n-1}}-1\right)k^{2n}}\\&=k^{2}\left(k{\frac {d}{dk}}K(k)+K(k)\right)+E(k)-K(k)\end{aligned}}} k ( 1 − k 2 ) d d k K ( k ) = E ( k ) − ( 1 − k 2 ) K ( k ) {\displaystyle k(1-k^{2}){\frac {d}{dk}}K(k)=E(k)-(1-k^{2})K(k)} から、微分方程式 d d k ( k ( 1 − k 2 ) d d k K ( k ) ) = E ( k ) − K ( k ) k − ( 1 − k 2 ) E ( k ) − ( 1 − k 2 ) K ( k ) k ( 1 − k 2 ) + 2 k K ( k ) = E ( k ) − K ( k ) − E ( k ) + ( 1 − k 2 ) K ( k ) k + 2 k K ( k ) = k K ( k ) {\displaystyle {\begin{aligned}{\frac {d}{dk}}\left(k(1-k^{2}){\frac {d}{dk}}K(k)\right)&={\frac {E(k)-K(k)}{k}}-(1-k^{2}){\frac {E(k)-(1-k^{2})K(k)}{k(1-k^{2})}}+2kK(k)\\&={\frac {E(k)-K(k)-E(k)+(1-k^{2})K(k)}{k}}+2kK(k)\\&=kK(k)\\\end{aligned}}} が得られるが、ここで k ′ = 1 − k 2 {\displaystyle k'={\sqrt {1-k^{2}}}} とすれば d d k ( k ( 1 − k 2 ) d d k K ( k ′ ) ) = d k ′ d k d d k ′ ( k ( 1 − k 2 ) d k ′ d k d d k ′ K ( k ′ ) ) = k 1 − k 2 d d k ′ ( k 2 1 − k 2 d d k ′ K ( k ′ ) ) = k k ′ d d k ( ( 1 − k ′ 2 ) k ′ d d k ′ K ( k ′ ) ) {\displaystyle {\begin{aligned}{\frac {d}{dk}}\left(k(1-k^{2}){\frac {d}{dk}}K(k')\right)&={\frac {dk'}{dk}}{\frac {d}{dk'}}\left(k(1-k^{2}){\frac {dk'}{dk}}{\frac {d}{dk'}}K(k')\right)\\&={\frac {k}{\sqrt {1-k^{2}}}}{\frac {d}{dk'}}\left(k^{2}{\sqrt {1-k^{2}}}{\frac {d}{dk'}}K(k')\right)\\&={\frac {k}{k'}}{\frac {d}{dk}}\left((1-k'^{2})k'{\frac {d}{dk'}}K(k')\right)\\\end{aligned}}} であるから K ′ ( k ) = K ( k ′ ) {\displaystyle K'(k)=K(k')} も同じ微分方程式の解になる。 Y ( k ) = k ( 1 − k 2 ) K ( k ) {\displaystyle Y(k)={\sqrt {k(1-k^{2})}}K(k)} とすれば d 2 d k 2 Y ( k ) = k ( 1 − k 2 ) d 2 d k 2 K ( k ) + ( 1 − 3 k 2 ) k ( 1 − k 2 ) d d k K ( k ) + 3 k 4 − 6 k 2 − 1 4 k ( 1 − k 2 ) k ( 1 − k 2 ) K ( k ) = 1 k ( 1 − k 2 ) ( k ( 1 − k 2 ) d 2 d k 2 K ( k ) + ( 1 − 3 k 2 ) d d k K ( k ) + 3 k 4 − 6 k 2 − 1 4 k ( 1 − k 2 ) K ( k ) ) = 1 k ( 1 − k 2 ) ( d d k ( k ( 1 − k 2 ) d d k K ( k ) ) + 3 k 4 − 6 k 2 − 1 4 k ( 1 − k 2 ) K ( k ) ) = 1 k ( 1 − k 2 ) ( k K ( k ) + 3 k 4 − 6 k 2 − 1 4 k ( 1 − k 2 ) K ( k ) ) = − ( 1 + k 2 ) 2 4 k 2 ( 1 − k 2 ) 2 Y ( k ) {\displaystyle {\begin{aligned}{\frac {d^{2}}{dk^{2}}}Y(k)&={\sqrt {k(1-k^{2})}}{\frac {d^{2}}{dk^{2}}}K(k)+{\frac {(1-3k^{2})}{\sqrt {k(1-k^{2})}}}{\frac {d}{dk}}K(k)+{\frac {3k^{4}-6k^{2}-1}{4{\sqrt {k(1-k^{2})}}\;k(1-k^{2})}}K(k)\\&={\frac {1}{\sqrt {k(1-k^{2})}}}\left(k(1-k^{2}){\frac {d^{2}}{dk^{2}}}K(k)+(1-3k^{2}){\frac {d}{dk}}K(k)+{\frac {3k^{4}-6k^{2}-1}{4k(1-k^{2})}}K(k)\right)\\&={\frac {1}{\sqrt {k(1-k^{2})}}}\left({\frac {d}{dk}}\left(k(1-k^{2}){\frac {d}{dk}}K(k)\right)+{\frac {3k^{4}-6k^{2}-1}{4k(1-k^{2})}}K(k)\right)\\&={\frac {1}{\sqrt {k(1-k^{2})}}}\left(kK(k)+{\frac {3k^{4}-6k^{2}-1}{4k(1-k^{2})}}K(k)\right)\\&=-{\frac {(1+k^{2})^{2}}{4k^{2}(1-k^{2})^{2}}}Y(k)\\\end{aligned}}} となり、 Y ′ ( k ) = k ( 1 − k 2 ) K ′ ( k ) {\displaystyle Y'(k)={\sqrt {k(1-k^{2})}}K'(k)} も同様である。故に d 2 d k 2 Y ( k ) Y ( k ) = − ( 1 + k 2 ) 2 4 k 2 ( 1 − k 2 ) 2 = d 2 d k 2 Y ′ ( k ) Y ′ ( k ) {\displaystyle {\frac {{\frac {d^{2}}{dk^{2}}}Y(k)}{Y(k)}}=-{\frac {(1+k^{2})^{2}}{4k^{2}(1-k^{2})^{2}}}={\frac {{\frac {d^{2}}{dk^{2}}}Y'(k)}{Y'(k)}}} であるから Y ( k ) d 2 d k 2 Y ′ ( k ) − Y ′ ( k ) d 2 d k 2 Y ( k ) = 0 {\displaystyle Y(k){\frac {d^{2}}{dk^{2}}}Y'(k)-Y'(k){\frac {d^{2}}{dk^{2}}}Y(k)=0} k ( 1 − k 2 ) K ( k ) d 2 d k 2 ( k ( 1 − k 2 ) K ′ ( k ) ) − k ( 1 − k 2 ) K ′ ( k ) d 2 d k 2 ( k ( 1 − k 2 ) K ( k ) ) = 0 {\displaystyle {\sqrt {k(1-k^{2})}}K(k){\frac {d^{2}}{dk^{2}}}\left({\sqrt {k(1-k^{2})}}K'(k)\right)-{\sqrt {k(1-k^{2})}}K'(k){\frac {d^{2}}{dk^{2}}}\left({\sqrt {k(1-k^{2})}}K(k)\right)=0} が成立する。積分して整理すると k ( 1 − k 2 ) ( K ( k ) d d k K ′ ( k ) − K ′ ( k ) d d k K ( k ) ) = C {\displaystyle k(1-k^{2})\left(K(k){\frac {d}{dk}}K'(k)-K'(k){\frac {d}{dk}}K(k)\right)=C} となり、これに d d k K ( k ) = E ( k ) − ( 1 − k 2 ) K ( k ) k ( 1 − k 2 ) {\displaystyle {\frac {d}{dk}}K(k)={\frac {E(k)-(1-k^{2})K(k)}{k(1-k^{2})}}} d d k K ′ ( k ) = d d k K ( 1 − k 2 ) = − k 1 − k 2 E ( 1 − k 2 ) − k 2 K ( 1 − k 2 ) 1 − k 2 k 2 = − E ′ ( k ) − k 2 K ′ ( k ) k ( 1 − k 2 ) {\displaystyle {\frac {d}{dk}}K'(k)={\frac {d}{dk}}K\left({\sqrt {1-k^{2}}}\right)={\frac {-k}{\sqrt {1-k^{2}}}}{\frac {E\left({\sqrt {1-k^{2}}}\right)-k^{2}K\left({\sqrt {1-k^{2}}}\right)}{{\sqrt {1-k^{2}}}\;k^{2}}}=-{\frac {E'(k)-k^{2}K'(k)}{k(1-k^{2})}}} を代入すると K ( k ) E ′ ( k ) + E ( k ) K ′ ( k ) − K ( k ) K ′ ( k ) = − C {\displaystyle K(k)E'(k)+E(k)K'(k)-K(k)K'(k)=-C} が得られる。不完全楕円積分の極限を用いて − C = K ( k ) E ′ ( k ) + E ( k ) K ′ ( k ) − K ( k ) K ′ ( k ) = lim x → 1 F ( x , 0 ) E ( x , 1 ) + E ( x , 0 ) F ( x , 1 ) − F ( x , 0 ) F ( x , 1 ) = lim x → 1 sin − 1 ⁡ x ( x + tanh − 1 ⁡ x − tanh − 1 ⁡ x ) = π 2 {\displaystyle {\begin{aligned}-C&=K(k)E'(k)+E(k)K'(k)-K(k)K'(k)\\&=\lim _{x\to 1}F(x,0)E(x,1)+E(x,0)F(x,1)-F(x,0)F(x,1)\\&=\lim _{x\to 1}\sin ^{-1}x\left(x+\tanh ^{-1}x-\tanh ^{-1}x\right)\\&={\frac {\pi }{2}}\\\end{aligned}}} が得られる。 Wolfram MathWorld: Legendre Relation 表示 編集