TITLE: Why does substitution work for a maclaurin series but not a taylor series?
QUESTION [6 upvotes]: Say I was trying to calculate the taylor expansion of $\sin(x^2)$ around $x = 0$.
I could assume that $u = x^2$ and solve for taylor expansion around $x=0$ of $\sin(u)$. I would just need to substitute $x^2$ back in for $u$ when I am completed. 
I have been informed that this process of substitution works only for the maclaurin series and not for any taylor series centered about a non-zero point? Why is this the case? Why is substitution even allowed in the first place?

REPLY [2 votes]: What do you mean with not working? 
To avoid convergence issues we assume that the taylor expansion equals the function on all of $\mathbb{R}$
$$f(x)=\sum_{n=0}^\infty a_n(x-a)^n \quad x\in\mathbb{R}.$$
Say you wanna find a taylor expansion for $g(x)=f(x^2)$. Sure, it is legal to enter this equation with $x^2$. 
$$g(x)=f(x^2)=\sum_{n=0}^\infty a_n(x^2-a)^n .$$
But this is no longer a taylor expansion, because of the square in $a_n(x^2-a)^n$.  Generally there is no can-opener method to convert this series expansion to a taylor expansion with terms $b_n(x-a)^n$