The parity of the function $f(x)=x^{2^{x^2}}$?

Question

Someone claims that $f(x)=x^{2^{x^2}}$ is an even function which I highly doubt.

I claim that $f(-0.5)=(-0.5)^{\sqrt[4]{2}}$ which is an imaginary number and $f(0.5)$ is a real number.

Can I have some confirmation regarding this?

Answer 1

It depends on the domain we're talking about. If we take it over the integers, then it actually is an even function; this follows from the fact that $2^{x^2}$ is an even number for any integer $x$ other than $0$. Moreover, since $2^{x^2}=2^{(-x)^2}$ and $x^{c}=(-x)^{c}$ for even $c$, we get the desired answer.

However, if we take it over the real numbers then, well, firstly, it's not quite clear that the function is well-defined (we have to be careful about exponentiation of negative numbers), but your objection certainly holds in that case.