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Dec 29, 2020 at 18:12 comment added Gottfried Helms @Raffaele - well the question of the domain ... I've often seen that problem, and have put it into discussion, but I myself don't have a reliable answer in many cases. So it must be answered case-by-case; usually my analyses are only taken the most simple assumptions for the domain; but at the moment I'm absorbed elsewhere...
Dec 29, 2020 at 18:09 comment added Raffaele If $f(x)=x^2$ can we say that $f^{(1/2)}=x^{\sqrt 2}$? I am asking because the domain of $x^{\alpha}$ for real exponent is $x>0$ while the domain of $f$ is $\mathbb{R}$ so applying twice $x^{\sqrt 2}$ I get $x^2;\;x>0$ which is not $f(x)$.
Aug 10, 2013 at 9:13 history edited Gottfried Helms CC BY-SA 3.0
added 769 characters in body
Jan 4, 2013 at 11:32 history edited Mario Carneiro CC BY-SA 3.0
fixed LaTeX
Sep 4, 2010 at 15:11 history edited Gottfried Helms CC BY-SA 2.5
just noticed the better link to mathoverflow
Sep 4, 2010 at 14:13 history edited Gottfried Helms CC BY-SA 2.5
extended the explanations
Sep 4, 2010 at 13:42 history answered Gottfried Helms CC BY-SA 2.5