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Suppose a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is analytic at 0, i.e., it is given by its Taylor series on a neighborhood $U$ of 0. Then is it necessarily true that $f$ is analytic at every point in $U$ ?

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Yes. You can find a proof of this fact in Chapter 8 of Baby Rudin ("Principles of Mathematical Analysis"). –  Jesse Madnick Apr 2 '11 at 4:06
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Yes it is true. The way I'm used to thinking of it is by extending the function to the complex plane, where to show that a function is analytic it suffices to show it is differentiable. If the power series of $f$ at $0$ has radius of convergence $r$, then this power series can be used to extend $f$ to the open disk of radius $r$ centered at $0$. Because power series are differentiable and complex differentiable functions are analytic, $f$ will be complex analytic on that disk, and therefore its restriction to the real line will be real analytic.

If you want to stick to real variables, see Chapter 1 of A primer of real analytic functions by Krantz and Parks (Corollary 1.2.4).

Edit: Thanks to Jesse Madnick for pointing out in a comment that it's also Theorem 8.4 in Principles of mathematical analysis, 3rd Ed. by Rudin.

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