Suppose one follows the Taylor series formula for a function, defining a function as $$ \sum_{n=0}^{ \infty} \frac{f^{n}(x)|_{x=a}}{n!}(x-a)^n.$$
Then, suppose you want this function to instead be defined as a power series since that is a little bit simpler.
You might think to simply take $a=0$, except there are plenty of functions like $\ln(x)$ or $\sqrt{x}$ which don't have a derivative at $x=0$. In such a case, you'd have to pick $a$ to be something else, so how can you convert that function's Taylor series to a power series that results in just $x^n$ instead of $(x-a)^n$?
The simple answer is that you can't. If I would write $$\ln(x)=\sum_{n=0}^\infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$\frac d{dx} \ln x=\frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$\sum_{n=1}^\infty na_n x^{n-1}$$ In the limit $x\to 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $\sum_j^\infty b_j(n) x_0^j$, but there is no guarantee that these series converge.