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Let $f(x) : \mathbb{R} \to \mathbb{R}$ be a positive real analytic function of $x$. My question is, how can I prove or disprove that $g(x) := \sqrt{f(x)}$ is also real analytic? I am new to this stuff, and my definition of real analytic functions is something that has a Taylor series expansion. However, I am having trouble implementing this definition on $g(x)$ directly. Is there any other way to check real analyticity? Thanks!

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Okay, I think I have an idea: from Andres Caicedo's answer on this MSE post, we know that the composition of real analytic functions is real analytic. So, we can write $$\sqrt{f} = e^{\text{log}(\sqrt{f})} = e^{\frac{1}{2}\text{log }f}.$$ I think this solves the problem?

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  • $\begingroup$ To be a completely correct solution, you should add something like: The function $\sqrt x$ is analytic on $(0,\infty)$. $\endgroup$ – Julián Aguirre Nov 23 '15 at 17:20

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