华中师范大学：《数学分析》课程PPT教学课件（讲稿）第十一章（11.7.2）傅里叶系数

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   )