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?
Fix $u>0$. Maclaurin expansion $\sin(u)$ around $u$ is:
$$\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}$$
Now $u$ is fixed, so you can express $u$ as $u=x^2$ for some $x$ and rewrite series:
$$\sin(x^2)\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}$$
Note that it is Taylor series for $\sin(x^2)$ If you have Taylor series for some function $f(x)$ around $x_0\neq 0$, you also can substitute $u=x^2$, but you don't get Taylor series as result:
$$f(u)=\sum_{n=0}^{\infty}a_n(u-x_0)^n$$
If you substitute $u=x^2$ in this case you get:
$$f(x^2)=\sum_{n=0}^{\infty}a_n(x^2-x_0)^n$$
It's something, but it is not Taylor series in form:
$$f(x^2)=\sum_{n=0}^{\infty}b_n(x-x_0)^n$$
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 $b_n(x-a)^n$
