f(x)cosnxdx, bn=I f(sin ndx 丌 丌 计算函数f(x)= 1-丌<x<0 的傅里叶系数: 10<x<丌 在[-r,上f(x) cos nx是奇函数,所以积分为0,即 an=0(n=0,1,2,……) sinni dx=l(Isinnxdx+I[mI-sinnxdx Ir cosn+ 6=--[-coSnT-coSn+l 丌n 1丌 2 4 1-(-1)”]=nz n=1,3,5, 1丌 0n=2,4,6,… 上页 下页上页 返回 下页  − =    a f x nxdx n ( )cos 1   − =    b f x nxdx n ( )sin 1  计算函数      − −   =   x x f x 1 0 1 0 ( ) 的傅里叶系数 在[− ]上f(x)cos nx是奇函数所以积分为0 即 an =0 (n =0, 1, 2,   );    = = − +  − −        0 0 1 sin 1 ( 1)sin 1 ( )sin 1 b f x nxdx nxdx nxdx n [1 cos cos 1] 1 ] cos [ 1 ] cos [ 1 0 = 0 − + − = −  −  +      n n n n nx n nx n 2 = [1−(−1)n ]     =  =  = 0 2, 4, 6, 1, 3, 5, 4 n n n     = = − +  − −        0 0 1 sin 1 ( 1)sin 1 ( )sin 1 b f x nxdx nxdx nxdx n [1 cos cos 1] 1 ] cos [ 1 ] cos [ 1 0 = 0 − + − = −  −  +      n n n n nx n nx n 2 = [1−(−1) n ]     =  =  = 0 2, 4, 6, 1, 3, 5, 4 n n n  返回