Does Wolfram Alpha fail for $x^x$?

Question

WolframAlpha generates the following graphic for $f_{(x)} = x^x$:

$f_{(x)} = x^x$

Can anyone explain me why this graphic looks like above? I mean why the real part for negative, non-integer, but rational numbers looks like that?for $0^0$?

To clarify my question:

We know that:

$$a^{m/n} = sqrt(a^m, n)$$

So, for example:

$${-1.1}^{-1.1} = 1 / {sqrt({-1.1}^{11}, 10)}$$ that is a complex number, not a real one

Answer 1

By definition $x^x = \exp(x \log(x))$. Presumably Mathematica is using the principal branch of the logarithm. Then for $x < 0$, $\log(x) = \log(|x|) + i \pi$, and so $x^x = \exp(x \log |x|) (\cos(\pi x) + i \sin(\pi x))$.

Answer 2

The point at $0^0$ does not exist, however the points very near it and the limit $x^x$ as $x\to 0$ do exist. As for the behavior of the graph in the negative real numbers, $x^x=(-a)^{-a}$ for some positive $a$ in the real numbers. By Euler's identity, $e^{\pi i}+1=0$, so we can write $-a$ as $e^{\pi i}a$, and $(e^{\pi i}a)^{-a}=e^{-a\pi i}a^{-a}$. Since $e^{\theta i}=\cos \theta+i\sin \theta$, we have $x^x=(-x)^x(\cos (x\pi)+i\sin(x\pi))$ for negative $x$, which is the graph Wolfram is giving you (note that some of the claims made here cannot simply be assumed, but this is just so you understand why Wolfram is making funny shapes.).