This seems to be some elementary point and I have red several questions about it, but I could not have a clarifying answer, so I directly ask myself. I say that a real-valued function $f$ on a real variable (defined on some interval in $\mathbb{R}$) is analytic at $x_0$ if it is $C^\infty$ and such that its Taylor series at $x_0$ converges on some open neighborhood $I_{x_0}$ of $x_0$, and converge to $f$ there.
Now, if we take $y_0$ in such $I_{x_0}$, it seems well known that $f$ is also analytic at $y_0$.
Maybe there is some confusion on the terminology, but my understanding this should mean that there exists $I_{y_0}$ open neighborhood of $y_0$ (presumably $I_{x_0}$ itself) on which the Taylor expansion at $y_0$ (which is a totally different power series from the Taylor expansion at $x_0$) converges again pointwise to $f$. Maybe I am missing something elementary, but I do not see any obvious proof of this fact. If someone can help me to clarify, thank you very much.
The radius of convergency around any point $z_0$ in the complex plane is the distance to the nearest singular point.
For polynomials there is one singular point at $\infty$ only. Polynomials can be re-expanded around an finite point. Differently from real algebra, $\infty$ is a point by the inversion of neighbourhoods of $(0,\infty)$ at the unit circle $z\to 1/z.$
For meromorphic functions (rationals of polynomials) it is the distance to the nearest pole.
For algebraic functions (defined by solutions of algebraic equations e.eg $ z\to w(z): z=w^2$ , it is the distance to the nearest branch point or pole.
In general, for series with an inifinite number of nonzero coefficients, it is the distance to nearest point of divergency from the point of expansion.
For any point $z_1$ inside this circle of convergency of the series in $(z-z_0)^n$ the series can be reorganized in $(z-z_1)^n$, either algebraically or by Taylors formula, convergent in an open neighborhood of however small radius.
It's different radius of convergency will change, maybe randomly, much like the radius of the maximal circle of free view changes during a flight through valleys of the alps at a level of let's say 3000 m.
Classical example:
$$\tan^{-1}(z) = \int\frac{1}{1+z^2} dz = \sum_n \frac{(-1)^n}{2n+1}z^{2n+1} $$
Its evident by an elementary convergency check at the beginning of a course in complex theory of functions, that all term by term integrals and derivatives of a power series have the same radius of convergency.
So in this case, there are two poles at $z=\pm i$ and the radius of convergency grows by $\sqrt{1+x_0^2}$ as the distance from a real point of expansion $x_0$