例2将f(x)=sinx展开成的幂级数 解f((x)=sim(x+),f(0)=4兀 ∫(2(0)=0,f(2+)(0)=(-1)",(n=0,1,2,…) 生且r"x)=mx+ n7 ≤1x∈(-0,+0) 2 2n+1 工工 sinx=x-x+x3-…+(-1) 3!5! (2n+1)! x∈(-∞,+o) 王页下例 2 将f (x) = sin x展开成x的幂级数. 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, ) ( ) = ( ) f x 且 n ) 2 sin(  + n x  1 x  (−,+ )  + +  = − + − + − + (2 1)! ( 1) 5!1 3!1 sin 2 1 3 5 nx x x x x n nx(−,+ )