题目

e^sinx麦克劳林展开到x^3
答案说e^x=1+1/2x^2+1/6x^3+o(x^3)
sinx=x-1/3x^3+o(x^3)
所以e^sinx=1+sinx=1/2sinx^2+1/6sinx^3+o(sinx^3)
=1+[x-1/6x^3+o(x^3)]+1/2[x+o(x^3)]^2+1/6[x+o(x^3)]^3+o(x^3) (*)
=1+x+1/2x^2+o(x^3)
请问（*）式是如何得来的.

提问时间：2020-11-21

答案

e^u=1+1/2u^2+1/6u^3+o(u^3)
sinx=x-1/3x^3+o(x^3)
e^sinx=1+sinx+1/2sinx^2+1/6sinx^3+o(sinx^3)
=1+[x-1/3x^3+o(x^3)]+1/2[x-1/3x^3+o(x^3)]^2+1/6[x-1/3x^3+o(x^3)]^3+o(x-1/3x^3+o(x^3))
将[x-1/3x^3+o(x^3)]^2,[x-1/3x^3+o(x^3)]^3,展开时,超过x^3的归到o(x^3)
故写成了1/2[x+o(x^3)]^2,1/6[x+o(x^3)]^3+o(x^3)