三角函数的正交性及其公式推导 |
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三角函数的正交性
废话不多说,直接上公式: ∫ − π π ( sin n x ) ( sin m x ) d x = 0 , 其 中 n ≠ m , n , m = 0 , 1 , 2 , ⋯ ∫ − π π ( sin n x ) ( cos m x ) d x = 0 , 其 中 n ≠ m , n , m = 0 , 1 , 2 , ⋯ ∫ − π π ( cos n x ) ( cos m x ) d x = 0 , 其 中 n ≠ m , n , m = 0 , 1 , 2 , ⋯ \begin{aligned} & \int_{-\pi}^{\pi} (\sin nx) (\sin mx)~ dx = 0 ,其中 ~ n \neq m,n,m=0,1,2,\cdots \\\\ & \int_{-\pi}^{\pi} (\sin nx) (\cos mx)~ dx = 0 ,其中 ~ n \neq m,n,m=0,1,2,\cdots \\\\ & \int_{-\pi}^{\pi} (\cos nx) (\cos mx)~ dx = 0 ,其中 ~ n \neq m,n,m=0,1,2,\cdots \\\\ \end{aligned} ∫−ππ(sinnx)(sinmx) dx=0,其中 n=m,n,m=0,1,2,⋯∫−ππ(sinnx)(cosmx) dx=0,其中 n=m,n,m=0,1,2,⋯∫−ππ(cosnx)(cosmx) dx=0,其中 n=m,n,m=0,1,2,⋯ 其他公式: ∫ − π π ( cos m x ) ( cos m x ) d x = π , 其 中 m = 0 , 1 , 2 , ⋯ \begin{aligned} & \int_{-\pi}^{\pi} (\cos mx) (\cos mx) ~dx = \pi,其中~m=0,1,2,\cdots \\\\ \end{aligned} ∫−ππ(cosmx)(cosmx) dx=π,其中 m=0,1,2,⋯ 公式推导∫ − π π ( sin n x ) ( sin m x ) d x = − 1 2 [ ∫ − π π cos ( n + m ) x d x − ∫ − π π cos ( n − m ) x d x ] ( 积 化 和 差 ) = − 1 2 [ 1 n + m sin ( n + m ) x ∣ − π π − 1 n − m sin ( n − m ) x ∣ − π π ] = 0 + 0 = 0 \begin{aligned} & \int_{-\pi}^{\pi} (\sin nx) (\sin mx)~ dx \\\\ = ~~& -\frac{1}{2} [ \int_{-\pi}^{\pi} \cos(n+m)x ~dx - \int_{-\pi}^{\pi} \cos(n-m) x~dx ] ~~~~(积化和差)\\\\ = ~~ &-\frac{1}{2} [ \frac{1}{n+m} \sin(n+m)x \mid_{-\pi}^{\pi} - \frac{1}{n-m} \sin(n-m)x \mid_{-\pi}^{\pi}] \\\\ = ~~ & 0+0 \\\\ = ~~ & 0 \end{aligned} = = = = ∫−ππ(sinnx)(sinmx) dx−21[∫−ππcos(n+m)x dx−∫−ππcos(n−m)x dx] (积化和差)−21[n+m1sin(n+m)x∣−ππ−n−m1sin(n−m)x∣−ππ]0+00 ∫ − π π ( cos n x ) ( cos m x ) d x = 1 2 [ ∫ − π π cos ( n − m ) x d x + ∫ − π π cos ( n + m ) x d x ] ( 积 化 和 差 ) = 1 2 [ 1 n − m sin ( n − m ) x ∣ − π π + 1 n + m sin ( n + m ) x ∣ − π π ] = 0 + 0 = 0 \begin{aligned} & \int_{-\pi}^{\pi} (\cos nx) (\cos mx)~ dx \\\\ = ~~& \frac{1}{2} [ \int_{-\pi}^{\pi} \cos(n-m)x ~dx + \int_{-\pi}^{\pi} \cos(n+m) x~dx ] ~~~~(积化和差)\\\\ = ~~ &\frac{1}{2} [ \frac{1}{n-m} \sin(n-m)x \mid_{-\pi}^{\pi} + \frac{1}{n+m} \sin(n+m)x \mid_{-\pi}^{\pi}] \\\\ = ~~ & 0+0 \\\\ = ~~ & 0 \end{aligned} = = = = ∫−ππ(cosnx)(cosmx) dx21[∫−ππcos(n−m)x dx+∫−ππcos(n+m)x dx] (积化和差)21[n−m1sin(n−m)x∣−ππ+n+m1sin(n+m)x∣−ππ]0+00 ∫ − π π ( cos m x ) ( cos m x ) d x = ∫ − π π 1 2 [ 1 + cos 2 m x ] d x = 1 2 [ ∫ − π π 1 d x + ∫ − π π cos 0 x cos 2 m x d x ] = 1 2 ∫ − π π 1 d x = π \begin{aligned} & \int_{-\pi}^{\pi} (\cos mx) (\cos mx)~ dx \\\\ = ~~ & \int_{-\pi}^{\pi} \frac{1}{2} [1+\cos 2mx] dx\\\\ = ~~& \frac{1}{2} [\int_{-\pi}^{\pi} 1 ~dx + \int_{-\pi}^{\pi} \cos 0x \cos 2mx ~dx] \\\\ = ~~ & \frac{1}{2} \int_{-\pi}^{\pi} 1 ~dx \\\\ = ~~ & \pi \end{aligned} = = = = ∫−ππ(cosmx)(cosmx) dx∫−ππ21[1+cos2mx]dx21[∫−ππ1 dx+∫−ππcos0xcos2mx dx]21∫−ππ1 dxπ 其他公式同理 参考资料纯干货数学推导_傅里叶级数与傅里叶变换_Part1_三角函数的正交性:https://www.bilibili.com/video/BV1Et411R78v 考研必备数学公式大全:https://blog.csdn.net/zhaohongfei_358/article/details/106039576 |
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