Interchanging Taylor series with Maclaurin series Is it always valid to say that the Taylor series of some function $f(x)$ about a point $x = a$ equivalent to the the Maclaurin series of another function $h(x) = f(x+a)$?
For example, is it correct to state the Taylor series of $\arctan(x)$ around $x=1$ is equivalent to the Taylor series of $\arctan(x+1)$ around $x=0$, even though their forms are different (i.e. the latter has terms of powers of $(x-a)$ rather than just $x$)?
Is this true in general for all functions or are there convergence issues?
 A: No.  If we consider an infinitely differentiable function $f(x)$, then we can write down it's Taylor series about the point $x = a$ as
$$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.$$  Now, if we consider the translation $h(x)=f(x+a)$ and compute its Maclaurin series, we get $$\sum_{n=0}^{\infty}\frac{h^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n.$$  Note that $$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \ne \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n$$ because they clearly disagree at $x = a$.  What we can say, however, is that the Taylor series for $f(x)$ is the Maclaurin series for $g(x) = f(x+a)$ shifted to the right by $a$.
A: Let $ f $ be a function defined arround $ x= a $, let us say at $ (a-\eta,a+\eta)$.
then, the function $ g $ given  by $ g(x)=f(x+a) $ will be well defined at $(-\eta,\eta) $.
If $$ g(x)=P_n(x)+x^n\epsilon_1(x)$$
with $\lim_{x\to 0}\epsilon_1(x)=0$
then
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^n\epsilon_2(x)$$
with $ \lim_{x\to a}\epsilon_2(x)=0$.
