$f(x)=x^{x}$ what happens when $x$ is a negative irrational number?

Question

Just looking at negative numbers, $x^{x}$ is defined for all rational numbers (on the real plane) in all instances except whenever $x=\large \frac {2a+1}{2b}$ where $(a, b)$ are integers . However, what happens when $x$ takes a negative irrational value, such as $- \pi$? Is $f(x)$ defined there? I don't even think it can be defined as the limit as $x$ approaches those values since I think the function is discontinuous on the negative $x$ axis because of the $x=\large \frac {2a+1}{2b}$ exception.

Answer 1

$$a^b = e^{b\log a}$$

for all complex and real $a \neq 0,b$.

In general, $x^x$ when $x<0$ becomes:

$$ x^x = \\ e^{x\log(x)} = \\ e^{x(\log|x| + i \pi (2n+1))} =\\ |x|^x e^{i x\pi (2n+1)} =\\ |x|^x\cos(\pi x (2n+1))+i |x|^x\sin(\pi x (2n+1)) $$

where $n \in \mathbb{Z}$ decides which branch of the logarithm to use.

Answer 2

The correct answer is "it's complicated". Or rather, "there's more than one reasonable meaning for $x^x$ when $x <0$".

The function $x \mapsto x^x$ is analytic on the positive real axis $\mathbb R_{>0}$, and so extends uniquely to a function in an open neighborhood of $\mathbb R_{>0}$ in $\mathbb C$. An explicit description that makes it easier to see what's going on is that $x^x = \exp( x \log x)$, where $\log$ is the natural logarithm. The function $\exp$ is well-defined for all complex numbers by $\exp(x) = \sum_{n\geq 0} \frac{x^n}{n!}$. The function $\log x$ is a bit more complicated. When $|x-1|<1$, we can define $\log x = \sum_{n\geq 1} \frac{(1-x)^n}n$. You can give $\log$ a larger domain by using its Taylor expansion around some larger positive real number.

The first problem is that $\log$ does not have a single-valued expansion to all of $\mathbb C$. The problem is the following. Take the function $x \mapsto \sum_{n\geq 0} \frac{x^n}{n!}$, say, with domain $\{x : |x-1|<1\}$, and try to extend it to $\mathbb C \smallsetminus\{0\}$ by walking around the origin in the counterclockwise direction. As you walk, provided you avoid the origin, you can always extend the function — keep re-expressing your function in Taylor series, and then walk a little more, and then compute a new Taylor series. If you go in the counterclockwise direction, then when you get to $-1$, for example, you decide that $\log(-1) = i\pi$. Now keep walking. By the time you get back to $1$, you discover that your function no longer evaluates to $0$ at $1$, but rather to $2\pi i$. Oops! $2\pi i \neq 0$.

The usual fix for this is to choose a branch cut $B$, which is any continuous non-self-intersecting path that starts at $0$ and ends at $\infty$ (aside: you can map the plane $\mathbb C$ injectively into the sphere $S^2$ by stereographic projection, and then "$\infty$" means the north pole; so it's just one point, but it is in all directions). The simplest way to choose a branch cut is to choose an angle $\theta \in (0,2\pi)$, and take the cut to run along the ray $re^{i\theta}$ for $r \in \mathbb R_{>0}$. Provided $B$ does not intersect $\mathbb R_{>0}$, the function $\log$ extends uniquely from $\mathbb R_{>0}$ to $\mathbb C \smallsetminus B$. This extension will have a jump discontinuity along the branch cut $B$: it will jump by $2\pi i$ as you cross in the clockwise direction. Knowing how the extension jumps allows different people to compare their extensions if they choose different branch cuts.

You asked about $x^x$ when $x$ is negative. Well, choose a branch cut then that doesn't intersect either $\mathbb R_{>0}$ or $\mathbb R_{<0}$. There are essentially two ways to make this choice: either the branch cut is in the upper half plane (positive imaginary part), or it is in the lower half. Either case gives a well-defined notion of "$\log x$", and hence "$x \log x$" and "$\exp(x\log x)$".

How do the two values compare? Well, let $\log_+ x$ denote the value of $\log x$ when the branch cut is along $-i\mathbb R_{>0}$, and $\log_- x$ the value using the branch cut $i\mathbb R_{>0}$. Then for $x\in \mathbb R_{<0}$, $\log_\pm x = \log (-x) \pm i\pi$. So the two values of $x \log x$ differ by $2\pi i x$. Therefore the ratio of the two values of "$x^x$" is $\exp(2\pi i x)$.

Note that $\exp(2\pi i x) = 1$ iff $x$ is an integer. Therefore the symbol "$x^x$" is well-defined without extra choices (like a branch cut) only when $x$ is an integer. Otherwise, it has many reasonable meanings.

By the way, how are you defining defining $x^x$ when $x$ is a negative rational number?

Answer 3

It is defined, but the answer is complex (pun intended). If: $$y=(-\pi)^{-\pi} \ \to \ \log y = -\pi \log (-\pi) = -\pi(\log \pi +\pi i)$$ So that: $$y = e^{-\pi(\log \pi +\pi i)}=e^{-\pi\log \pi}e^{-\pi^2 i}=-\pi^{-\pi}\sin(\pi^2)i+\pi^{-\pi}\cos(\pi^2)$$

Answer 4

If $x\in\mathbb{R}^{-}\setminus\mathbb{Q}^{-}$, we have $x\notin\mathbb{R}^{+}\cup\{\dfrac{m}{2k+1}/(m,k)\in\mathbb{Z}^2\}$, and thus $x^x\notin\mathbb{R}$ (at least for the known negative irrational numbers). The value would be of the form $x^x=(-x)^x\cos{((2n+1)\pi{x})}+i\sin{((2n+1)\pi{x})}$