(2)f(x)=sinx 解因f"(x)=sin(x+n·)(m=0,1,2,…) 有f(0)=0,f(0)=1f”(0)=0,∫"(0)=-1,…, f2(0)=0,f(2k+)(0)=(-1) 于是∑ f"(0) X=y-—+ +…+(-1 3!5!7! (2n+1)! ∑+的 (2k+1)! 且其收敛区间为(-∞,+∞) 2k+3 而0≤风(x)=Nix+ 2k+3 (2k+3)!(2k+3)10 ( ) ( ) sin( ) ( 0,1,2, ) 2 n f x x n n  解 因 = +  = 有 f f (0) 0, (0) 1 = =  (2 ) (2 1) (0) 0, (0) ( 1) , k k k f f + = = − f f   (0) 0, (0) 1, , = = − ( ) 3 5 7 2 1 0 (0) ( 1) ! 3! 5! 7! (2 1)! n n n n n f x x x x x x n n  + = = − + − + + − +  + 于是 且其收敛区间为(－∞,+ ∞). 2 1 0 ( 1) (2 1)! k k k x k  + = = − +  (2) ( ) sin f x x = 2 3 2 3 2 3 0 ( ) sin[ ] 2 (2 3)! (2 3)! k k k k x x R x x k k   + + +  = +   + + 而