Trying to understand the function $y = x^x$

Question

I am trying to understand the function $y=x^x$:

1) Why is $0^0$ not defined? Why isn't it defined as $0^0=1$? The limit of the function for $x\to0$ also goes to $1$

2) Why is it only defined for $x>0$ and not for $x\le0$? For example $(-2)^{(-2)}=\frac{1}{4}$ which is defined.

Answer 1

$0^0$ is not defined since the limits of $x^0$ and $0^x$ differ as $x\to 0^+$. Just as with $\frac00$, $0^0$ is not the limit of $x^x$, it is the limit of $x^y$ as $(x,y)\to (0,0)$.

And yes, $x^x$ is defined on the negative integers, but not on any other negative numbers.

Answer 2

The first answer gives a good explanation for why $0^0$ is not defined. Here's a more in-depth look at why you can't define $x^x$ on the negative real numbers:

Firstly, the function is defined on all the negative integers $(-m)^{(-m)}=1/(-m)^m$. But once you start trying to extend the domain of $x^x$ to arbitrary negative fractions, you'll get stuff like $(-1/2)^{(-1/2)}$, which in fact is a complex number.

The problem, of course, is that this depends on the choice of representation of the rational number $x$. For example, $x=-1/3=-2/6$ are both the same number, but when you take $x^x$ the odd denominator of $-1/3$ gives you back a real number, whilst the even denominator of $-2/6$ gives you back a complex number. Thus the function isn't well-defined on the negative rationals (all this depending, I guess, on the "multi-valued" branching of the complex square root/$n$th root function).

Answer 3

Another way of looking at this function is to rewrite it as $$y=e^{\ln x^x}=e^{x\ln x}.$$ The answers to your questions then follow easily:

1) $\lim_{x\to 0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x}$. This limit does not exist since $\lim_{x\to 0^+}\ne\lim_{x\to0^-}$. Note that $\lim_{x\to0^+}x\ln x = 1$, but $\lim_{x\to0^-}x\ln x$ does not exist.

2) Using complex logarithm $(-2)^{-2}=e^{-2\ln(-2)}=e^{-2(\ln|2|+in\pi)}=e^{-2\ln|2|}e^{-2in\pi}=|2|^{-2}\cdot 1=\frac{1}{4}$. Like this you can compute $y=x^x$ for any $x\in\mathbb{C}$ (thus also any negative real numbers).

As mentioned, in the case of negative real numbers (and for that matter any complex numbers) we obtain complex solutions, e.g. $(-\frac{1}{2})^{-1/2}=e^{-1/2\ln(-1/2)}=e^{-1/2(\ln|1/2|+in\pi)}=(\frac{1}{2})^{-1/2}e^{-in\frac{\pi}{2}}=\{\pm\sqrt{2}, \pm\sqrt{2}i\}$, depending on $n\in\mathbb{Z}$.