Taylor series at two different points If I have a function $y(x)$ for which there is a Taylor series about $x=1$ that has an infinite radius of convergence, and I also have a Taylor series for $y(x)$ about $x=0$ (with unknown radius of convergence), is it necessary that the radius of convergence of the second Taylor series must be infinity?
If yes, what is the proof and if no, can someone post a counter-example?
 A: If the function is real-analytic, then the answer is yes. Then the Taylor series about $1$ gives you an entire holomorphic function
$$f(z) = \sum_{n=0}^\infty \frac{y^{(n)}(1)}{n!}(z-1)^n$$
by extending the series representation to complex arguments, and $y$ is the restricition of $f$ to real arguments. The Taylor series of a holomorphic function $h\colon U\to \mathbb{C}$ about $z_0\in U$ converges in every disk with centre $z_0$ that is contained in $U$. For $U = \mathbb{C}$, that is every disk, hence the radius of convergence of the Taylor series of an entire function about any centre is infinite.
If the function is merely smooth, the radius of convergence of the Taylor series about $0$ can be finite (it can even be $0$). For an example, consider
$$y(x) = \begin{cases} \quad 0 &, \lvert x\rvert \geqslant \frac{1}{2} \\ \exp \Bigl( \frac{1}{4x^2-1}\Bigr) &, \lvert x\rvert < \frac{1}{2}. \end{cases}$$
Then $y$ vanishes identically in a neighbourhood of $1$, so its Taylor series about $1$ is $0$, which has infinite radius of convergence (and the Taylor series even converges to $y$ in a neighbourhood of $1$), but the Taylor series about $0$ has radius of convergence $\frac{1}{2}$.
