华里士公式的推导及其推广
基础知识
华里士公式
I n = ∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = { n − 1 n n − 3 n − 2 ⋯ 2 3 n i s o d d , n − 1 n n − 3 n − 2 ⋯ 1 2 π 2 n i s e v e n \Large \begin{aligned} I_n = \int_{0}^{\frac{\pi}{2}} \sin^n{x} \mathrm{d}x = \int_{0}^{\frac{\pi}{2}} \cos^n{x} \mathrm{d}x = \begin{cases} \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{2}{3} &&n\ is\ odd,\\ \\ \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{1}{2} \frac{\pi}{2} &&n\ is\ even \end{cases} \end{aligned} In=∫02πsinnxdx=∫02πcosnxdx=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧nn−1n−2n−3⋯32nn−1n−2n−3⋯212πn is odd,n is even
基础公式的推导
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\large \begin{aligned} I_n &= -\int_{0}^{\frac{\pi}{2}} \sin^{n-1}{x} \mathrm{d}{\cos{x}} \\[2ex] &= -\left. \sin^{n-1}x\cdot\cos{x} \right|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n - 1)\cdot\cos^2{x} \cdot \sin^{n-2}{x} \mathrm{d}{x} \\[2ex] &= (n-1) \int_{0}^{\frac{\pi}{2}} (1-\sin^2{x}) \cdot \sin^{n-2}{x} \mathrm{d}x \\[2ex] &= (n-1) I_{n-2} - (n-1) I_n \end{aligned}
In=−∫02πsinn−1xdcosx=−sinn−1x⋅cosx∣∣∣02π+∫02π(n−1)⋅cos2x⋅sinn−2xdx=(n−1)∫02π(1−sin2x)⋅sinn−2xdx=(n−1)In−2−(n−1)In
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In=nn−1In−2
分情况讨论,求出通项公式,即可得原式成立。
华里士公式的推广
推广公式
∫ 0 m 2 π sin n x d x = { m I n n i s e v e n , I n n i s o d d , m = 2 k + 1 , 2 I n n i s o d d , m = 4 k + 2 , 0 n i s o d d , m = 4 k \Large \begin{aligned} &\int_{0}^{\frac{m}{2}{\pi}} \sin^n{x} \mathrm{d}x = \begin{cases} m I_n &n\ is\ even,\\ \\ I_n &n\ is\ odd,\ \ m=2k+1,\\ \\ 2I_n &n\ is\ odd,\ \ m=4k+2,\\ \\ 0 &n\ is\ odd,\ \ m=4k \end{cases} \end{aligned} ∫02mπsinnxdx=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧mInIn2In0n is even,n is odd, m=2k+1,n is odd, m=4k+2,n is odd, m=4k
∫ 0 m 2 π cos n x d x = { m I n n i s e v e n , I n n i s o d d , m = 4 k + 1 , − I n n i s o d d , m = 4 k + 3 , 0 n i s o d d , m = 2 k \Large \begin{aligned} &\int_{0}^{\frac{m}{2}{\pi}} \cos^n{x} \mathrm{d}x = \begin{cases} m I_n &n\ is\ even,\\ \\ I_n &n\ is\ odd,\ \ m=4k+1,\\ \\ -I_n &n\ is\ odd,\ \ m=4k+3,\\ \\ 0 &n\ is\ odd, \ \ m=2k \end{cases} \end{aligned} ∫02mπcosnxdx=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧mInIn−In0n is even,n is odd, m=4k+1,n is odd, m=4k+3,n is odd, m=2k
当 m = 1 m=1 m=1 时,该公式退化为原华里士公式。
推广公式的图像理解
这是 f ( x ) = sin 3 x f(x)=\sin^3x f(x)=sin3x 的图像(对应n为奇数时的情况):
由图可知, f ( x ) f(x) f(x) 是以 2 π 2\pi 2π 为周期的周期函数,且函数在 [ 0 , π 2 ] [0, \frac{\pi}{2}] [0,2π] 与 [ π 2 , π ] [\frac{\pi}{2}, \pi] [2π,π] 的部分各自与x轴围成的面积相等,函数在 [ 0 , π ] [0, \pi] [0,π] 与 [ π , 2 π ] [\pi, 2\pi] [π,2π] 的部分各自与x轴围成的面积也相等。
这是 f ( x ) = sin 4 x f(x)=\sin^4x f(x)=sin4x 的图像(对应n为偶数时的情况):
由图可知, f ( x ) f(x) f(x) 是以 π \pi π 为周期的周期函数,且函数在 [ 0 , π 2 ] [0, \frac{\pi}{2}] [0,2π] 与 [ π 2 , π ] [\frac{\pi}{2}, \pi] [2π,π] 的部分各自与x轴围成的面积相等。
故无论m的值为多少,积分值都是 [ 0 , π 2 ] [0, \frac{\pi}{2}] [0,2π] 的 m m m 倍。
推广公式的特例
m=2
∫ 0 π sin n x d x = 2 I n ∫ 0 π cos n x d x = { 2 I n n i s e v e n 0 n i s o d d \Large \begin{aligned} &\int_{0}^{\pi} \sin^n{x} \mathrm{d}x = 2I_n \\[4ex] \Large &\int_{0}^{\pi} \cos^n{x} \mathrm{d}x = \begin{cases} 2I_n \quad &n\ is\ even \\[2ex] 0 \quad &n\ is\ odd \end{cases} \end{aligned} ∫0πsinnxdx=2In∫0πcosnxdx=⎩⎪⎪⎪⎨⎪⎪⎪⎧2In0n is evenn is odd文章来源:https://www.toymoban.com/news/detail-414201.html
m=4
∫ 0 2 π sin n x d x = ∫ 0 2 π cos n x d x = { 4 I n n i s e v e n 0 n i s o d d \Large \begin{aligned} \int_{0}^{2\pi} \sin^n{x} \mathrm{d}x = \int_{0}^{2\pi} \cos^n{x} \mathrm{d}x = \begin{cases} 4I_n \quad &n\ is\ even \\[2ex] 0 \quad &n\ is\ odd \end{cases} \end{aligned} ∫02πsinnxdx=∫02πcosnxdx=⎩⎪⎪⎪⎨⎪⎪⎪⎧4In0n is evenn is odd文章来源地址https://www.toymoban.com/news/detail-414201.html
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