Let $f$ be a strictly positive function. Then we can compute the Taylor expansion of $\sqrt{f(x)}$ at zero. Wolfram alpha gives the first terms as
$$\sqrt{f(x)}= \sqrt{f(0)} + \frac{f'(0)x}{2\sqrt{f(0)}}+...$$
Now, I was wondering if there is a closed representation of this as a power series? At first glance this does not look at hard, as we just apply chain and quotient-rule, but I could not come up with a closed equation.
Making the problem more general, if you consider $\sqrt[n]{f(x)}$ $$\sqrt[n]{f(x)}=f(0)^{\frac{1}{n}}+\frac{ f(0)^{\frac{1}{n}-1} f'(0)}{n}x+\frac{ f(0)^{\frac{1}{n}-2} \left(nf(0) f''(0)-n f'(0)^2+f'(0)^2\right)}{2 n^2}x^2+O\left(x^3\right)$$ which is what Wolfram Alpha gave you for $n=2$.
Otherwise, you could consider $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ $$\sqrt[n]{f(x)}=\sqrt[n]{f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)}$$ and use the generalized binomial theorem writing $$\sqrt[n]{f(x)}=\sqrt[n]{f(0)}\left(1+\frac{f'(0)}{f(0)}x+\frac12 \frac{f''(0)}{f(0)}x^2+O\left(x^3\right)\right)^{\frac 1n}$$ using $$(1+z)^{\frac 1n}=1+\frac{1}{n}z+\frac{(1-n) }{2 n^2}z^2+O\left(z^3\right)$$ in which $$z=\frac{f'(0)}{f(0)}x+\frac12 \frac{f''(0)}{f(0)}x^2+O\left(x^3\right)$$ For sure, the same applies with more terms.