월리스 공식

월리스 공식(Wallis product)은 원주율을 구하는 간단한 공식으로, 존 월리스에 의해 만들어졌다. 정수 n {\displaystyle n} 에 대하여 다음과 같이 놓자. I ( n ) = ∫ 0 π sin n ⁡ x d x {\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx} 그러면 부분 적분에 의하여 다음을 얻는다. I ( n ) = ∫ 0 π sin n ⁡ x d x = − sin n − 1 ⁡ x cos ⁡ x | 0 π − ∫ 0 π ( − cos ⁡ x ) ( n − 1 ) sin n − 2 ⁡ x cos ⁡ x d x = 0 + ( n − 1 ) ∫ 0 π cos 2 ⁡ x sin n − 2 ⁡ x d x , n > 1 = ( n − 1 ) ∫ 0 π ( 1 − sin 2 ⁡ x ) sin n − 2 ⁡ x d x = ( n − 1 ) ∫ 0 π sin n − 2 ⁡ x d x − ( n − 1 ) ∫ 0 π sin n ⁡ x d x = ( n − 1 ) I ( n − 2 ) − ( n − 1 ) I ( n ) = n − 1 n I ( n − 2 ) ⇒ I ( n ) I ( n − 2 ) = n − 1 n {\displaystyle {\begin{aligned}I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}} 이제 2 n {\displaystyle 2n} 과 2 n + 1 {\displaystyle 2n+1} 에 대하여 다음 점화식이 성립한다. I ( 2 n ) = 2 n − 1 2 n I ( 2 n − 2 ) , {\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2),} I ( 2 n + 1 ) = 2 n 2 n + 1 I ( 2 n − 1 ) . {\displaystyle I(2n+1)={\frac {2n}{2n+1}}I(2n-1).} 이때 I ( 0 ) {\displaystyle I(0)} 과 I ( 1 ) {\displaystyle I(1)} 은 다음과 같다. I ( 0 ) = ∫ 0 π d x = x | 0 π = π , I ( 1 ) = ∫ 0 π sin ⁡ x d x = − cos ⁡ x | 0 π = ( − cos ⁡ π ) − ( − cos ⁡ 0 ) = − ( − 1 ) − ( − 1 ) = 2. {\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi ,\\I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2.\end{aligned}}} I ( 2 n ) {\displaystyle I(2n)} 에 대하여 점화식을 활용하면 I ( 2 n ) = ∫ 0 π sin 2 n ⁡ x d x = 2 n − 1 2 n I ( 2 n − 2 ) = 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 I ( 2 n − 4 ) {\displaystyle I(2n)=\int _{0}^{\pi }\sin ^{2n}x\,dx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)} 을 얻고, 마찬가지로 하면 I ( 2 n + 1 ) {\displaystyle I(2n+1)} 에 대하여 = 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 ⋅ 2 n − 5 2 n − 4 ⋅ ⋯ ⋅ 5 6 ⋅ 3 4 ⋅ 1 2 I ( 0 ) = π ∏ k = 1 n 2 k − 1 2 k {\displaystyle ={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdot \cdots \cdot {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I(0)=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}} I ( 2 n + 1 ) = ∫ 0 π sin 2 n + 1 ⁡ x d x = 2 n 2 n + 1 I ( 2 n − 1 ) = 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 I ( 2 n − 3 ) {\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}x\,dx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)} = 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 ⋅ 2 n − 4 2 n − 3 ⋅ ⋯ ⋅ 6 7 ⋅ 4 5 ⋅ 2 3 I ( 1 ) = 2 ∏ k = 1 n 2 k 2 k + 1 {\displaystyle ={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdot \cdots \cdot {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I(1)=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}} 을 얻는다. 한편 sin ⁡ x ≤ 1 {\displaystyle \sin {x}\leq 1} 에서 sin 2 n + 1 ⁡ x ≤ sin 2 n ⁡ x ≤ sin 2 n − 1 ⁡ x , 0 ≤ x ≤ π {\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi } ⇒ I ( 2 n + 1 ) ≤ I ( 2 n ) ≤ I ( 2 n − 1 ) {\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)} 이고, 이것을 I ( 2 n + 1 ) {\displaystyle I(2n+1)} 으로 나누면 ⇒ 1 ≤ I ( 2 n ) I ( 2 n + 1 ) ≤ I ( 2 n − 1 ) I ( 2 n + 1 ) = 2 n + 1 2 n {\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}}\leq {\frac {I(2n-1)}{I(2n+1)}}={\frac {2n+1}{2n}}} 을 얻는다. 샌드위치 정리를 활용하면 ⇒ lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = 1 {\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}=1} lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = π 2 lim n → ∞ ∏ k = 1 n ( 2 k − 1 2 k ⋅ 2 k + 1 2 k ) = 1 {\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}={\frac {\pi }{2}}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=1} ⇒ π 2 = ∏ k = 1 ∞ ( 2 k 2 k − 1 ⋅ 2 k 2 k + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ ⋯ {\displaystyle \Rightarrow {\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \cdots } 을 얻는다. “Integrating Powers and Product of Sines and Cosines: Challenging Problems”.  비에트의 공식 라이프니츠의 공식