将cosx在x=π/4处展开成幂级数,求详解.
问题描述：
将cosx在x=π/4处展开成幂级数,求详解.
答
解:cosx的各阶导数在x=π/4处的值为:
根2/2 n=0
-根2/2 n=1
-根2/2 n=2
根2/2 n=3
根2/2 n=4
n为求导阶数
根据泰勒级数展开:
cosx=根2/2- 根2/2(x-4)-根2/2(x-4)^2+根2/2(x-4)^3+
根2/2(x-4)^4+...
反正符号的规律就是每4位为1周期
n=4k+1 or 4k+2时 为负
n=4k+3 or 4k时 为正 k取非负整数
答
cosx=cos(π/4＋x-π/4)
=cosπ/4cos(x-π/4)-sinπ/4sin(x-π/4)
=√2/2 [cos(x-π/4)-sin(x-π/4)]
=√2/2× 【1-(x-π/4)^2/2!+(x-π/4)^4/4!-.-[(x-π/4)-(x-π/4)^3/3!+(x-π/4)^5/5!+.]】
x∈R