0.m≠n sin mx sin ndx T. n =n cos mx cos ndx=(0,m≠ .=/ sInner cos nx dk=0.(其中m,n=1,2,…)
, , 0, sin sin = = − m n m n mx nxdx , , 0, cos cos = = − m n m n mx nxdx sin cos = 0. − mx nxdx (其中m,n = 1,2, )
函数展开成傅里叶级数 问题:1.若能展开,41,b是什么? 2展开的条件是什么? 1.傅里叶系数 若有f(x)=0+∑(ac0skx+ b sinka) 2 (1)求a1 ∫nf(x)dx=Jmax+∑( a, cos kar+ b sin kor)lx 2 k=1
三、函数展开成傅里叶级数 问题: 1.若能展开, ai ,bi 是什么? 2.展开的条件是什么? 1.傅里叶系数 = + + =1 0 ( cos sin ) 2 ( ) k ak kx bk kx a 若有 f x (1) . 求a0 dx a kx b kx dx a f x dx k k k [ ( cos sin )] 2 ( ) 1 0 = + + − = − −
∫2a+2 a cos kodex+ b sin kdx 2 o.2 ao= f(r) 2 (2)求an ∫nf(x) cos nxd= 2 z cos neda k=lk cos kx cos nx a-yo /m sin kx cos nxdxI +∑
2 , 2 0 = a = − a f (x)dx 1 0 dx a kxdx b kxdx a k k k k cos sin 2 1 1 0 − = − = − = + + (2) . 求an = − − nxdx a f x nxdx cos 2 ( )cos 0 [ cos cos sin cos ] 1 − − = + a kx nxdx + bk kx nxdx k k
cos ndx n J-T f(x)cos ndx (n=1,2,3,…) (3)求bn f(x)sin ndx SOrT sinned 2 +Elas cos kx sin ndx +b. sin kc sinnxdxI=b,t, b= f()sin ndx (n=1,2,3, T
= − an nxdx 2 cos = , an = − an f (x)cos nxdx 1 (n = 1,2,3, ) (3) . 求bn = − bn f (x)sinnxdx 1 (n = 1,2,3, ) − − = nxdx a f x nxdx sin 2 ( )sin 0 [ cos sin sin sin ] 1 − − = + a kx nxdx + bk kx nxdx k k = , n b
傅里叶系数 f()cos ndx, (n=0, 1, 2,". f(xsin ndt, (n=1, 2. T f()cos nxd, (n=0, 1, 2, .) 或 兀0 b f(x) sinned,(n=1,2,…
= = = = − − ( )sin , ( 1,2, ) 1 ( )cos , ( 0,1,2, ) 1 b f x nxdx n a f x nxdx n n n = = = = 2 0 2 0 ( )sin , ( 1,2, ) 1 ( )cos , ( 0,1,2, ) 1 b f x nxdx n a f x nxdx n n n 或 傅里叶系数