为什么函数可以展为麦克劳林级数RT

网友回答

The basic idea is that the polynomial functions P = {x^n,n= 0,1,2,...} are linearly independent,so P should expand to an (infinite-dimensional) subspace.Maclaurin series of a function F can be seen as a projection of F onto that subspace P.
======以下答案可供参考======
供参考答案1:
在关于原点对称的定义域(-R,R)上有各阶导数即可~~当然它是不是收敛还要看余项是否趋于0~~