If $f$ is analytic on $\mathbb{R}$, is it necessary that $f = \sum_{n = 0}^{\infty} a_{n} x^{n}$ converges for all $x \in \mathbb{R}$?

I have two questions regarding analyticity. They are pretty easy, and I think I have them correct, but I just want to make sure.

First, regarding the question in the title, I think that it is necessary. By the definition of analyticity, we must have the power series in some radius of the function. So, if it is analytic on all of $$\mathbb{R}$$, it must be within some radius (and thus convergent) for all of $$\mathbb{R}$$.

Second, if $$f$$ is analytic, is it necessary for $$\text{exp}(f)$$ to be analytic? Pretty sure that this again is necessary since $$e^{x}$$ is analytic, and the composition of analytic functions is analytic.

• This is correct. – Zanzi Nov 16 '18 at 8:13
• Only an accumuled point is required. – Zanzi Nov 16 '18 at 8:14
• Counterexample to question 1: $f(x)=\frac1{1+x^2}$. – Kemono Chen Nov 16 '18 at 8:14
• Yes. This is correct. – Andrej Nov 16 '18 at 8:16
• @KemonoChen Why not an official answer even it is short? – Paul Frost Nov 16 '18 at 8:21

Counterexample to question 1: $$f(x)=\frac1{1+x^2}$$. The Maclaurin series of it does not converge everywhere is because $$f$$ is a meromorphic function in $$\mathbb{C}$$.
If $$f(z)$$ is analytic on $$\mathbb{C}$$, the series you gave is convergent everywhere.