例2将∫(x)=sinx展开成x幂级数 解f(x)=sin(x+),∫(0)=+兀 f2(0)=0,f(+1(0)=(-1)",(n=0,1,2,…) n70 且f(x)=sin(x+)≤1x∈(-∞+) 2 131 2n+1 .sinx=x-x3+x3-…+(-1) 十 3 5 (2n+1)! ∈ +oo 上一页下一页返回解 ), 2 ( ) sin( ( )  = + n f x x n , 2 (0) sin ( )  = n f n (0) 0, (2 )  = n f (0) ( 1) , (2n 1) n f = − + (n = 0,1,2, )  + +  = − + − + − + (2 1)! ( 1) 5! 1 3! 1 sin 2 1 3 5 n x x x x x n n x (− ,+ ) ( ) = ( ) f x n ) 2 sin(  + n 且 x  1 x(−,+) 例2 将 f (x) = sin x 展开成 x 幂级数