当n → ∞ n\to\infty n → ∞
∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = { n − 1 n ⋅ n − 3 n − 2 . . . 3 4 ⋅ 1 2 ⋅ π 2 (if n is even) n − 1 n ⋅ n − 3 n − 2 . . . 2 5 ⋅ 2 3 (if n is odd) = { ( n − 1 ) ! ! n ! ! ⋅ π 2 (if n is even) ( n − 1 ) ! ! n ! ! (if n is odd) \begin{align*} \int^{\dfrac{\pi}{2}}_{0}\sin^{n} xdx&=\int^{\dfrac{\pi}{2}}_{0}\cos^{n} x dx\\ & = \begin{cases} \dfrac{n-1}{n}\cdot \dfrac{n-3}{n-2}...\dfrac{3}{4}\cdot \dfrac{1}{2}\cdot \dfrac{\pi}{2} \ &\text{(if n is even)}\\ \dfrac{n-1}{n}\cdot \dfrac{n-3}{n-2}...\dfrac{2}{5}\cdot \dfrac{2}{3}\ \ &\text{(if n is odd)}\\ \end{cases}\\ &= \begin{cases} \dfrac{(n-1)!!}{n!!}\cdot \dfrac{\pi}{2} \ & \text{(if n is even)}\\ \dfrac{(n-1)!!}{n!!} \ &\text{(if n is odd)}\\ \end{cases} \end{align*} ∫ 0 2 π sin n x d x = ∫ 0 2 π cos n x d x = ⎩ ⎨ ⎧ n n − 1 ⋅ n − 2 n − 3 ... 4 3 ⋅ 2 1 ⋅ 2 π n n − 1 ⋅ n − 2 n − 3 ... 5 2 ⋅ 3 2 (if n is even) (if n is odd) = ⎩ ⎨ ⎧ n !! ( n − 1 )!! ⋅ 2 π n !! ( n − 1 )!! (if n is even) (if n is odd)
Wallis 公式是用于求解圆周率 π \pi π
π = 2 ∏ n = 1 ∞ 4 n 2 4 n 2 − 1 \pi = 2 \prod_{n=1}^\infty \frac{4n^2}{4n^2 - 1} π = 2 n = 1 ∏ ∞ 4 n 2 − 1 4 n 2
这个公式可以通过分析 sin x \sin x sin x x = π 2 x = \frac{\pi}{2} x = 2 π
首先, 我们知道 sin x \sin x sin x
sin x = x ∏ n = 1 ∞ ( 1 − x 2 n 2 π 2 ) \sin x = x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2 \pi^2}\right) sin x = x n = 1 ∏ ∞ ( 1 − n 2 π 2 x 2 )
利用 sin x \sin x sin x π \pi π
∫ 0 π / 2 sin 2 n x d x \int_0^{\pi/2} \sin^{2n} x \, dx ∫ 0 π /2 sin 2 n x d x
该积分随 n n n n n n
∫ 0 π / 2 sin 2 n x d x ≈ π 2 ( 2 n − 1 ) ! ! ( 2 n ) ! ! \int_0^{\pi/2} \sin^{2n} x \, dx \approx \frac{\pi}{2} \frac{(2n-1)!!}{(2n)!!} ∫ 0 π /2 sin 2 n x d x ≈ 2 π ( 2 n )!! ( 2 n − 1 )!!
其中 ( 2 n ) ! ! (2n)!! ( 2 n )!! 2 n ⋅ ( 2 n − 2 ) ⋅ … ⋅ 2 2n \cdot (2n-2) \cdot \ldots \cdot 2 2 n ⋅ ( 2 n − 2 ) ⋅ … ⋅ 2 ( 2 n − 1 ) ! ! (2n-1)!! ( 2 n − 1 )!! 1 ⋅ 3 ⋅ … ⋅ ( 2 n − 1 ) 1 \cdot 3 \cdot \ldots \cdot (2n-1) 1 ⋅ 3 ⋅ … ⋅ ( 2 n − 1 )
随着 n → ∞ n \to \infty n → ∞
lim n → ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ n = 1 π \lim_{n \to \infty} \frac{(2n-1)!!}{(2n)!!} \cdot \sqrt{n} = \frac{1}{\sqrt{\pi}} n → ∞ lim ( 2 n )!! ( 2 n − 1 )!! ⋅ n = π 1
这个结果基于 Stirling 的近似.
Wallis 公式可以通过上述积分和 sin x \sin x sin x sin 2 n x \sin^{2n} x sin 2 n x x = π / 2 x = \pi/2 x = π /2 π \pi π π \pi π
∏ n = 1 ∞ 4 n 2 4 n 2 − 1 = π 2 \prod_{n=1}^\infty \frac{4n^2}{4n^2 - 1} = \frac{\pi}{2} n = 1 ∏ ∞ 4 n 2 − 1 4 n 2 = 2 π
因此, Wallis 公式可以表示为:
π = 2 ∏ n = 1 ∞ 4 n 2 4 n 2 − 1 \pi = 2 \prod_{n=1}^\infty \frac{4n^2}{4n^2 - 1} π = 2 n = 1 ∏ ∞ 4 n 2 − 1 4 n 2
这个公式不仅在数学历史上具有重要意义, 而且对理解 π \pi π
