Analyticity of a function Does this definition contain redundancy: "A function $f$ is said to be analytic at $x_0$ if its Taylor series about $x_0$ exists and converges to $f(x)$ for all $x$ in some interval containing $x_0$". Is anything missing in the following definition: "A function $f$ is said to be analytic at $x_0$ if it has derivatives of all orders at $x_0$".
 A: You can think about the definition of an analytic function (at $x_0$) as consisting of three parts:


*

*The derivatives of $f$ of all orders at $x_0$ exist. This is equivalent to saying "The Taylor series around $x_0$ exists" (because to define the taylor series around $x_0$, you need to know the derivatives of all orders at $x = x_0$).

*The Taylor series converges. This means that the Taylor series has a positive radius of convergence. Even if the deriatives of all orders at $x = x_0$ exists, this doesn't mean that the Taylor series will have a positive radius of convergence (see Borel's lemma).

*The Taylor series converges to $f$. Even if the Taylor series has a positive radius of convergence, it might converge to a function different than $f$ as it happens in the examples provided in the other answers.


Your attempt of simplification only captures the first part.
A: The function $\begin{cases}e^{-\frac{1}{x}}&x>0\\0 & x\leq0\end{cases}$ has derivatives of all order at $x_0=0$, but this function is not analytic. If it were analytic, then a power series would converge to identically zero, since $f$ is identically zero for $x\leq0$.
A: $x\mapsto f(x)=\exp(-1/x^2)$ extended to $f(0)=0$ has all its derivatives equal to 0 at 0 so it cannot be equal to its own Taylor series in any interval around 0.
