小结 1.周期函数的傅里叶展开 周期为2/的函数(x满足f(x+2)=f(x) 对应方程:有边界条件 f(x)=a+∑{ ak cos k k +b, sin kax cos-,dx=0(k≠0) k f(scolds 1·sn=dx=0. n刀r b=,f(5) kTs COS dx=0(k≠m) kTx sin 2(k=0) sn-ax=0(k≠n), (k≠0) nZA cOS小结 1.周期函数的傅里叶展开 周期为 2l 的函数 f(x) 满足 f (x + 2l) = f (x) 对应方程：有边界条件 ( ) { cos sin }. 1 0 l k x b l k x f x a a k k k     = = + +      − − − − −  =  =   =   =  =  l l l l l l l l l l dx l n x l k x dx k n l n x l k x dx k n l n x l k x dx l k x dx k l k x cos sin 0. sin sin 0 ( ), cos cos 0 ( ), 1 sin 0, 1 cos 0 ( 0),                = =   − − ( )sin . 1 ( ) cos , 1          d l k f l b d l k f l a l l k l l k k     = = 1 ( 0) 2 ( 0) k k  k