The domain of $x^x$?

Question

This one looks simple, but apparently there is something more to it. $$f{(x)=x^x}$$ I read somewhere that the domain is $\Bbb R_+$, a friend said that $x\lt-1, x\gt0$...

I'm really confused, because i don't understand why the domain isn't just all the real numbers. According to any grapher online the domain is $\Bbb R_+$. Any Thoughts on the matter?

Can someone explain what am I missing?

And I'm not sure how your friend would make sense of $f(-3/2)$ either — Ben Grossmann, Dec 02 '15 at 19:48
Write it as $x^x = e^{x\ln(x)}$ not it is easy to determine the domain. — m_gnacik, Dec 02 '15 at 19:48
@Eleven-Eleven, I think if $n$ were odd, you would be fine. Domain should not include $x=\frac{-1}{2n}$, yes? So I suppose there is a class of negative rational numbers you could use for the domain, but certainly not all $x<-1$, as in the comment from Omnomnomnom. — John Molokach, Dec 02 '15 at 19:50
differntiate $x^x$ with respect to $x$ in the case of real $x$ must be $x>0$ — Dr. Sonnhard Graubner, Dec 02 '15 at 19:50
Yes, @JohnMolokach, you are correct...i should have read the part in the OP's comment where $x<-1$... — Eleven-Eleven, Dec 02 '15 at 19:52
@Dr.SonnhardGraubner, why should the derivative define the domain of a function? — John Molokach, Dec 02 '15 at 19:52
@m_gnacik but your equation is only true for $x>0$ while the LHS is defined from some negative values. See https://www.desmos.com/calculator/cftrtgyi75. — John Molokach, Dec 02 '15 at 19:57
Even though one might be able to play around with odd roots versus even roots and so on, given that the tag is real-analysis, the comment of @m_gnacik seems to give the correct answer. — Lee Mosher, Dec 02 '15 at 19:59
@LeeMosher, I agree with you, which is why I elaborated on this in my answer below. — John Molokach, Dec 02 '15 at 20:03
This answer provides a plot of the graph in the real case and indicates how it arises as a slice of a collection of complex graphs clearly indicating the domain from kamil09875's answer. — Mark McClure, Dec 02 '15 at 21:04

Kamil Jarosz · Accepted Answer · 2015-12-03T15:26:56.950

Split it into cases:

When $x=p/q$ where $p\in \mathbb Z,q\in\mathbb N_{>1},p\ne0,\gcd(p,q)=1$, then: $$x^x=\left(\frac{p}{q}\right)^\frac{p}{q}=\sqrt[q]{\left(\frac{p}{q}\right)^p}$$
- when $p<0$ then $$x^x=\sqrt[q]{\left(-\frac{q}{|p|}\right)^{|p|}}$$ if $p$ is even, then $\left(-\frac{q}{|p|}\right)^{|p|}$ is positive, otherwise it's negative and the root doesn't exist for even $q$.
- when $p>0$ then $$x^x=\sqrt[q]{\left(\frac{|p|}{q}\right)^{|p|}}$$ and $\left(\frac{|p|}{q}\right)^{|p|}$ is always positive.
When $x\in\mathbb Z$ the value $x^x$ always exist except $x=0$.
When $x$ is irrational then the only way to define $x^x$ is $$x^x=\exp(x\ln x)$$ and for real numbers we have $x>0$.

Summarizing, $x^x$ exist for all

$x\in\mathbb R_+$
$x\in\mathbb Z_-$
$x\in\left\{ -\frac{p}{q}\in \mathbb Q\colon p,q\in\mathbb N_+ \land \gcd(p,q)=1\land q\text{ is odd}\right\}$

Why we don't see the negative part of the plot

Technical reason: $x^x$ in programs is usually defined as exp(x*log(x)) and the function log(x) is not defined for negative x.
Mathematical reason: set of negative $x$ which $x^x$ exists for is countable. Countable many points is not enough to form a curve.

This function may be plotted with points for negative $x$.

score 0 · Answer 2 · answered Dec 02 '15 at 20:02

I am not sure what your mathematics background is, but the function $f(x)=x^x$ is defined for $\mathbb{R}_+$ as well as a countably infinite set of rational values in $\mathbb{Q}_-$. For example, we can find $f(-\frac n3)$ for all $n\in\mathbb{N}$. In fact, I cannot with confidence write down the entire set of negative values in the domain but any $x=\frac n {1-2n}$ will work for starters.

Some textbooks won't include the negative values, because they form a countable set of numbers that has Lebesgue measure zero.

…and because the resulting function is extremely discontinuous on that countable set. — Akiva Weinberger, Dec 02 '15 at 20:26

The domain of $x^x$?

2 Answers2