算 ∫x-x≤x<0 的傅里叶系数 00<x<丌 f(x)dx xdx 丌z f()cosnxdx x xsinnx cosnx XcosnX 2 (1-cosn) =neT =1,3,5 n2元 0n=2.4.6 丌 f(x)sinnxdx xcosnx sinx 2 xsin ndx=-[ COSn (1)n+l (n=1,2,…) 上页 下页
上页 返回 下页 计算 − = x x x f x 0 0 0 ( ) 的傅里叶系数 2 1 ( ) 1 0 0 = = =− − − a f x dx xdx 2 1 ( ) 1 0 0 = = =− − − a f x dx xdx 2 1 ( ) 1 0 0 = = =− − − a f x dx xdx 0 2 0 ] sin cos [ 1 cos 1 ( )cos 1 − − − = = = + n nx n x nx a f x nxdx x nxdx n = = = − = 0 2, 4, 6, 1, 3, 5, 2 (1 cos ) 1 2 2 n n n n n 0 2 0 ] cos sin [ 1 sin 1 ( )sin 1 − − − = = = − + n nx n x nx b f x nxdx x nxdx n n n n n 1 cos ( 1) − + =− = (n =1 2 ) 0 2 0 ] sin cos [ 1 cos 1 ( )cos 1 − − − = = = + n nx n x nx a f x nxdx x nxdx n 0 2 0 ] sin cos [ 1 cos 1 ( )cos 1 − − − = = = + n nx n x nx a f x nxdx x nxdx n = = = − = 0 2, 4, 6, 1, 3, 5, 2 (1 cos ) 1 2 2 n n n n n 0 2 0 ] cos sin [ 1 sin 1 ( )sin 1 − − − = = = − + n nx n x nx b f x nxdx x nxdx n 0 2 0 ] cos sin [ 1 sin 1 ( )sin 1 − − − = = = − + n nx n x nx b f x nxdx x nxdx n n n n n 1 cos ( 1) − + =− = (n =1 2 )