르장드르의 배 공식 유도
📂함수 르장드르의 배 공식 유도 공식 Γ ( 2 r ) = 2 2 r − 1 π Γ ( r ) Γ ( 1 2 + r )
\Gamma (2r) = {{2^{ 2r - 1} } \over { \sqrt{ \pi } } } \Gamma \left( r \right) \Gamma \left( {{1} \over {2}} + r \right)
Γ ( 2 r ) = π 2 2 r − 1 Γ ( r ) Γ ( 2 1 + r )
설명 쪼개지는 모양이 그렇게 예쁘지는 않지만 인수를 작게 나눌 수 있다는 것은 분명 유용한 사실이다. 유도 자체는 베타함수에서 파생된 보조정리를 사용하면 별로 어렵지 않다.
유도 B ( p , q ) = Γ ( p ) Γ ( q ) Γ ( p + q ) = ∫ 0 1 t p − 1 ( 1 − t ) q − 1 d t
B(p,q) = {{\Gamma (p) \Gamma (q)} \over {\Gamma (p+q) }} = \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt
B ( p , q ) = Γ ( p + q ) Γ ( p ) Γ ( q ) = ∫ 0 1 t p − 1 ( 1 − t ) q − 1 d t r : = p = q r:= p=q r := p = q Γ ( r ) Γ ( r ) Γ ( 2 r ) = ∫ 0 1 t r − 1 ( 1 − t ) r − 1 d t
{{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} = \int_{0}^{1} t^{r-1} (1-t)^{r-1} dt
Γ ( 2 r ) Γ ( r ) Γ ( r ) = ∫ 0 1 t r − 1 ( 1 − t ) r − 1 d t t = 1 + s 2 \displaystyle t = {{1+s} \over {2}} t = 2 1 + s λ ( s ) : = ( 1 − s 2 ) r − 1 \lambda (s) := \left( 1 - s^2 \right)^{r-1} λ ( s ) := ( 1 − s 2 ) r − 1 우함수 이므로
Γ ( r ) Γ ( r ) Γ ( 2 r ) = 1 2 ∫ − 1 1 ( 1 + s 2 ) r − 1 ( 1 − s 2 ) r − 1 d s = 1 2 1 + 2 ( r − 1 ) ∫ − 1 1 ( 1 − s 2 ) r − 1 d s = 2 1 − 2 r ⋅ 2 ∫ 0 1 ( 1 − s 2 ) r − 1 d s
\begin{align*}
{{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} =& {{1} \over {2}} \int_{-1}^{1} \left( {{1+s} \over {2}} \right)^{r-1} \left( {{1-s} \over {2}} \right)^{r-1} ds
\\ =& {{1} \over {2^{1 + 2(r-1)} }} \int_{-1}^{1} \left( 1 - s^2 \right)^{r-1} ds
\\ =& 2^{1 - 2r} \cdot 2 \int_{0}^{1} \left( 1 - s^2 \right)^{r-1} ds
\end{align*}
Γ ( 2 r ) Γ ( r ) Γ ( r ) = = = 2 1 ∫ − 1 1 ( 2 1 + s ) r − 1 ( 2 1 − s ) r − 1 d s 2 1 + 2 ( r − 1 ) 1 ∫ − 1 1 ( 1 − s 2 ) r − 1 d s 2 1 − 2 r ⋅ 2 ∫ 0 1 ( 1 − s 2 ) r − 1 d s
베타함수의 삼각함수 표현의 따름정리 : B ( x , y ) = 2 ∫ 0 1 t 2 x − 1 ( 1 − t 2 ) y − 1 d t B(x,y) = 2 \int_{0}^{1} t^{2x-1} \left( 1 - t^2 \right)^{y-1} dt B ( x , y ) = 2 ∫ 0 1 t 2 x − 1 ( 1 − t 2 ) y − 1 d t
위 공식에 x = 1 2 \displaystyle x = {{1} \over {2}} x = 2 1 y = r y = r y = r B ( 1 2 , r ) = 2 ∫ 0 1 ( 1 − t 2 ) r − 1 d t
B \left( {{1} \over {2}} , r \right) = 2 \int_{0}^{1} \left( 1 - t^2 \right)^{r-1} dt
B ( 2 1 , r ) = 2 ∫ 0 1 ( 1 − t 2 ) r − 1 d t Γ ( r ) Γ ( r ) Γ ( 2 r ) = 2 1 − 2 r B ( 1 2 , r ) = 2 1 − 2 r Γ ( 1 2 ) Γ ( r ) Γ ( 1 2 + r )
{{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} = 2^{1 - 2r} B \left( {{1} \over {2}} , r \right) = 2^{1 - 2r} {{\Gamma \left( {{1} \over {2}} \right) \Gamma (r)} \over {\Gamma \left( {{1} \over {2}} + r \right) }}
Γ ( 2 r ) Γ ( r ) Γ ( r ) = 2 1 − 2 r B ( 2 1 , r ) = 2 1 − 2 r Γ ( 2 1 + r ) Γ ( 2 1 ) Γ ( r ) 반사 공식에서 Γ ( 1 2 ) = π \displaystyle \Gamma \left( {1 \over 2} \right) = \sqrt{\pi} Γ ( 2 1 ) = π 이므로
Γ ( r ) Γ ( 2 r ) = 2 1 − 2 r π Γ ( 1 2 + r )
{{\Gamma (r)} \over {\Gamma (2r) }} = 2^{1 - 2r} {{\sqrt{\pi} } \over {\Gamma \left( {{1} \over {2}} + r \right) }}
Γ ( 2 r ) Γ ( r ) = 2 1 − 2 r Γ ( 2 1 + r ) π Γ ( 2 r ) \Gamma (2r) Γ ( 2 r ) Γ ( 2 r ) = 2 2 r − 1 π Γ ( r ) Γ ( 1 2 + r )
\Gamma (2r) = {{2^{2r-1} } \over { \sqrt{ \pi } } } \Gamma \left( r \right) \Gamma \left( {{1} \over {2}} + r \right)
Γ ( 2 r ) = π 2 2 r − 1 Γ ( r ) Γ ( 2 1 + r )
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