沃利斯乘积

沃利斯乘積，又称沃利斯公式，由數學家約翰·沃利斯在1655年时发现。 ∏ n = 1 ∞ 2 n 2 n − 1 ⋅ 2 n 2 n + 1 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ = π 2 . {\displaystyle \prod _{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}.} 今日多數的微積分教科書透過比較 ∫ 0 π sin n ⁡ x d x {\displaystyle \int _{0}^{\pi }\sin ^{n}xdx} 在n是奇數或是偶數，甚至是接近無窮大的情況下，發現即使將n增加一就會發生不一樣的情形。在那時，微積分尚未存在，而且有關數學收斂的分析工具也還未俱全，所以完成這證明較現今有相當的難度。從現在來看，從欧拉公式中的正弦展開式得到此乘積是必然的結果。 sin ⁡ ( x ) x = ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ⋯ = ∏ n = 1 ∞ ( 1 − x 2 n 2 π 2 ) , {\displaystyle {\frac {\sin(x)}{x}}=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right),} 在x = π/2時 2 π = ∏ n = 1 ∞ ( 1 − 1 4 n 2 ) = ( 1 − 1 2 2 ) ( 1 − 1 2 2 ⋅ 4 ) ( 1 − 1 2 2 ⋅ 9 ) ⋯ {\displaystyle {\frac {2}{\pi }}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)=\left(1-{\frac {1}{2^{2}}}\right)\left(1-{\frac {1}{2^{2}\cdot 4}}\right)\left(1-{\frac {1}{2^{2}\cdot 9}}\right)\cdots } π 2 = ∏ n = 1 ∞ ( 4 n 2 4 n 2 − 1 ) = ∏ n = 1 ∞ ( 2 n ) ( 2 n ) ( 2 n − 1 ) ( 2 n + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ {\displaystyle {\begin{aligned}{\frac {\pi }{2}}&{}=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\&{}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots \end{aligned}}} 先考慮不定積分 ∫ sin n ⁡ x d x {\displaystyle \int \sin ^{n}xdx} 有 ∫ sin n ⁡ x d x {\displaystyle \int \sin ^{n}xdx} = − ∫ sin n − 1 ⁡ x d cos ⁡ x {\displaystyle =-\int \sin ^{n-1}xd\cos x} = − cos ⁡ x sin n − 1 ⁡ x + ∫ cos ⁡ x d sin n − 1 ⁡ x {\displaystyle =-\cos x\sin ^{n-1}x+\int \cos xd\sin ^{n-1}x} = − cos ⁡ x sin n − 1 ⁡ x + ∫ ( n − 1 ) sin n − 2 ⁡ x cos 2 ⁡ x d x {\displaystyle =-\cos x\sin ^{n-1}x+\int (n-1)\sin ^{n-2}x\cos ^{2}xdx} = − cos ⁡ x sin n − 1 ⁡ x + ( n − 1 ) ∫ sin n − 2 ⁡ x ( 1 − sin 2 ⁡ x ) d x {\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}x(1-\sin ^{2}x)dx} = − cos ⁡ x sin n − 1 ⁡ x + ( n − 1 ) ∫ sin n − 2 ⁡ x d x − ( n − 1 ) ∫ sin n ⁡ x d x {\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}xdx-(n-1)\int \sin ^{n}xdx} 故 ∫ sin n ⁡ x d x = − 1 n cos ⁡ x sin n − 1 ⁡ x + n − 1 n ∫ sin n − 2 ⁡ x d x {\displaystyle \int \sin ^{n}xdx=-{\frac {1}{n}}\cos x\sin ^{n-1}x+{\frac {n-1}{n}}\int \sin ^{n-2}xdx} ∫ 0 π 2 sin n ⁡ x d x = n − 1 n ∫ 0 π 2 sin n − 2 ⁡ x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}xdx={\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}xdx} 對整數m ∫ 0 π 2 sin 2 m ⁡ x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx} = 2 m − 1 2 m ∫ 0 π 2 sin 2 m − 2 ⁡ x d x {\displaystyle ={\frac {2m-1}{2m}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-2}xdx} = 2 m − 1 2 m 2 m − 3 2 m − 2 ∫ 0 π 2 sin 2 m − 4 ⁡ x d x {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-4}xdx} = . . . {\displaystyle =...} = 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 ∫ 0 π 2 sin 0 ⁡ x d x {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\sin ^{0}xdx} = 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 π 2 {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}} 另一方面 ∫ 0 π 2 sin 2 m + 1 ⁡ x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx} = 2 m 2 m + 1 ∫ 0 π 2 sin 2 m − 1 ⁡ x d x {\displaystyle ={\frac {2m}{2m+1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx} = 2 m 2 m + 1 2 m − 2 2 m − 1 ∫ 0 π 2 sin 2 m − 3 ⁡ x d x {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-3}xdx} = . . . {\displaystyle =...} = 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 ∫ 0 π 2 sin ⁡ x d x {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}\int _{0}^{\frac {\pi }{2}}\sin xdx} = 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}} 兩式相除得 ∫ 0 π 2 sin 2 m ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x = 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 π 2 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 {\displaystyle {\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {{\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}}{{\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}}}} 故 π 2 = 2 1 2 3 4 3 4 5 . . . 2 m 2 m − 1 2 m 2 m + 1 ∫ 0 π 2 sin 2 m ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x = ∫ 0 π 2 sin 2 m ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x ∏ n = 1 m 2 n 2 n − 1 ⋅ 2 n 2 n + 1 {\displaystyle {\frac {\pi }{2}}={\frac {2}{1}}{\frac {2}{3}}{\frac {4}{3}}{\frac {4}{5}}...{\frac {2m}{2m-1}}{\frac {2m}{2m+1}}{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}\prod _{n=1}^{m}{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}} 又因為 1 = ∫ 0 π 2 sin 2 m + 1 ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x < ∫ 0 π 2 sin 2 m ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x < ∫ 0 π 2 sin 2 m − 1 ⁡ x d x ∫ 0 π 2 sin 2 m + 1 ⁡ x d x = 2 m + 1 2 m {\displaystyle 1={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {2m+1}{2m}}} 由夾擠定理知 lim m → ∞ 1 = lim m → ∞ 2 m + 1 2 m = 1 {\displaystyle \lim _{m\to \infty }1=\lim _{m\to \infty }{\frac {2m+1}{2m}}=1} 故 π 2 = ∏ n = 1 ∞ 2 n 2 n − 1 ⋅ 2 n 2 n + 1 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ {\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots } 我們可將上述的正弦乘積式化為泰勒级数： x ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ⋯ = x − 1 3 ! x 3 + 1 5 ! x 5 − ⋯ {\displaystyle x\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =x-{\frac {1}{3!}}x^{3}+{\frac {1}{5!}}x^{5}-\cdots }