例1将f(x)=e展开成幂级数 解f(x)=c,f(0)=1.(n=0.1,2, ex<)1+x+-x2+…+-x"+ 2! WM>0,在-M,M上/(x)=e2≤e (n=0,1,2,…) ∴ex=1+x+-x2+…+-xm+ 2! 由于M的任意性,即得 ex=1+x+x2+…+x"+…x∈(-00+0 2!
例1 解 将 ( ) 展开成幂级数. x f x = e ( ) , (n) x f x = e (0) 1. ( 0,1,2, ) f (n) = n = x + + ++ x n + n e x x ! 1 2! 1 1 2 M 0, 在[−M, M]上 n x f (x) = e ( ) M e (n = 0,1,2, ) x = + + ++ x n + n e x x ! 1 2! 1 1 2 由于M的任意性, 即得 ( , ) ! 1 2! 1 1 2 = + + + + x + x − + n e x x x n
例2将f(x)=sinx展开成x的幂级数 解f(x)=si(x+)/(0)=m元 2 f(0)=0,f(2m+1(0)=(-1)2,(n=0,1,2,) 且 n7 f(m(x)=sin(x+)≤1x∈(-∞+∞) 2 2n+1 sinx=x-x3+=x3-…+(-1) 3 5 2n+ x∈(=000
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(−,+ )