Is being real analytic at a point equivalent of matching the Taylor Series around that point?

150 Views Asked by At

Let $U \subset \mathbb{R}$ be an open interval and let $x_0 \in U$. Let $f: U \to \mathbb{R}$ be defined by a power series around $x_0$ with radius of convergence $R > 0$,

$$f(x) = \sum_{n=0}^{\infty} a_n (x - x_0)^n$$

where $(x_0 - R, x_0 + R) \subseteq U$.

Clearly, $f$ is $C^\infty$ on $(x_0 - R, x_0 + R)$, but:

With no other assumptions, is it true that $f$ is analytic at $x_0$?

If yes, how to prove it? If not, what would be a counter-example?


My thoughts:

For a moment this seemed trivially "yes", but then I paused. Let's look at the definition on Wikipedia:

A function $f$ defined on some subset of the real line is said to be real analytic at a point $x_0$ if there is a neighborhood $D$ of $x_0$ on which $f$ is real analytic.

I know that $f$ is equal to its Taylor Series centered at $x_0$, but according to the definition above, the function $f$ must also agree to other Taylor Series, centered at the nearby points as well. But what do I know about the points near $x_0$? Do the taylor series centered at them also agree to the function?