华中师范大学：《数学分析》课程PPT教学课件（讲稿）第十一章（11.7.3）傅里叶系数续

一22(xk3x0的傅里叶系数 计算f()=n x<丌 C(x)+=x; 0 f(x)cosnxdx=(x)cosnxdx+-xcosnxdx 丌yr 丌 2 n=1.3.5 n2x(cosnT-1) n=7T 0n=2.4.6 在[-x上f(x) )sin nx是奇函数,其积分为0,即 bn=0(n=1,2,…) 上页 下页

上页 返回 下页         = = − + =    − − 0 0 0 1 ( ) 1 ( ) 1 a f x dx x dx xdx     = = − + − −        0 0 cos 1 ( )cos 1 ( )cos 1 a f x nxdx x nxdx x nxdx n     =  − =  = − = 0 2, 4, 6, 1, 3, 5, 4 (cos 1) 2 2 2 n n n n n     在[−,]上 f(x)sin nx是奇函数, 其积分为0, 即 bn =0 (n=1, 2,   ).         = = − + =    − − 0 0 0 1 ( ) 1 ( ) 1 a f x dx x dx xdx          = = − + =    − − 0 0 0 1 ( ) 1 ( ) 1 a f x dx x dx xdx     = = − + − −        0 0 cos 1 ( )cos 1 ( )cos 1 a f x nxdx x nxdx x nxdx n     =  − =  = − = 0 2, 4, 6, 1, 3, 5, 4 (cos 1) 2 2 2 n n n n n     计算      − −   =   x x x x f x 0 0 ( ) 的傅里叶系数 返回