Is square root of Taylor series of $f(x)$ equivalent to the Taylor series of square root of $f(x)$ Mathematica treats two expressions as they are equivalent:
Sqrt[Series[y[x], {x, x0, 1}]]

Series[Sqrt[y[x]], {x, x0, 1}]

Is that mathematically justified? Is the square root of the Taylor expansion of $f(x)$ equal to the expansion of the $\sqrt{f(x)}$ at some point $a$?
$$\sqrt{\sum_{i=0}^\infty f^{(i)}(a)(x-a)^i}=\sum_{i=0}^\infty\ g^{(i)}(a)(x-a)^i$$
where $g(x)=\sqrt{f(x)}$
 A: I presume $f(x)$ is analytic in a neighbourhood of $x=a$.  
If $a$ is a zero of $f$ with odd multiplicity, $f(x)$ does not have a square root that is analytic at $a$, so neither expression exists.  If $f(a) \ne 0$ or $a$ is a root of even multiplicity, then it does have a square root that is analytic at $a$.
If that square root is $g(x) = g_0 + g_1 (x-a) + g_2 (x-a)^2 + \ldots$, 
then $f(x) = g(x)^2 = g_0^2 + 2 g_0 g_1 (x-a) + (g_1^2 + 2 g_0 g_2) (x-a)^2 + \ldots$, i.e. the square of the Taylor series for $g$ at $a$ is a series in powers of $x-a$ that converges in some neighbourhood of $a$ to $f(x)$, and therefore must be the Taylor series for $f(x)$.
Thus the Taylor series for $\sqrt{f(x)}$ is a square root of the Taylor series for $f(x)$.
Notice that I'm saying "a square root", rather than "the square root".  There are in general two, and you have to choose one.  In the 
case of the series, if $f(a) > 0$ I expect Mathematica will choose
the branch of the square root that is positive at $a$.  For the square root of the function, especially if $a \ne 0$ I'm not sure Mathematica would choose that same branch.
