What is the domain of $f(x)=x^x$?

Question

What is the domain of $f(x)=x^x$ ?

I used Wolfram alpha where it is said that the domain is all positive real numbers. Isn't $(-1)^{(-1)} = -1$ ? Why does the domain not include negative real numbers as well?

I also checked graph and its visible for only $x>0$ . Can someone help me clarify this?

I think $(-\sqrt{2})^{-\sqrt{2}}$ is imaginary? And that is why it does not appear? — S.C.B., Mar 06 '16 at 13:56
I think $(\text{negative number})^{(\text{negative number})}$ is often imaginary, so that is why it does not appear. — S.C.B., Mar 06 '16 at 13:57
There is an unambiguous definition for $x^x$ in the following cases: a) $x>0$, where we can let $x^x=\exp(x\ln x)$. -- b) $x=0$ where $0^0=1$ (yes, it is!!) -- c) $x$ a negative integer, where we can obtain $x^x=\frac1{x^{|x|}}$ with $x^{|x|}$ obtained by "repeated multiplication". -- For other $x$ there be dragons — Hagen von Eitzen, Mar 06 '16 at 13:57
@HagenvonEitzen What does there be dragons mean? I thought that was the name of some bad movie. — S.C.B., Mar 06 '16 at 13:59
@MXYMXY I wanted to refer do dangerous and uncharted territory — Hagen von Eitzen, Mar 06 '16 at 16:13
See http://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero — Michael Hoppe, Mar 06 '16 at 18:14

Emilio Novati · Accepted Answer · 2016-03-06T14:33:42.783

6

Write:

$$y=x^x=e^{x\log x}$$

If we want $y \in \mathbb{R}$ we must have $\log x \in \mathbb{R}$ and this is done only if $x> 0$

This is the usual definition for the function $y=f(x)=x^x$ for $x \in \mathbb{R}$, that gives $(0,+\infty)$ as the domain.

If we want $x\in \mathbb{Q}$ than we can define the function as: $$ y=f(x)=x^x=\left( \frac{m}{n}\right)^{\frac{m}{n}} \iff y=\sqrt[n]{x^m} \iff y^n=\left(\frac{m}{n}\right)^m $$

If we define $0^0=1$, this is a real number if $n=2k+1 \quad \forall k\in \mathbb{Z}$ so the domain of the function can be: $$ \{q\in \mathbb{Q}|q=\frac{m}{2k+1}\quad , \quad m,k \in \mathbb{Z} \} $$

edited Mar 06 '16 at 14:33

answered Mar 06 '16 at 14:00

Emilio Novati

62,675

Re "If we define $0^0=1$" I'd like to interject that the definition $0^0=1$ (e.g., as cardinality of the set of maps from the empty set to the empty set) comes in handy before we can even write down a nice definition of the exponential via its series $\sum_{k=0}^\infty\frac1{k!}x^k$, which even uses $0^0=1$ right in our face – Hagen von Eitzen Mar 06 '16 at 16:16

score 0 · Answer 2 · answered Mar 06 '16 at 14:59

The expression $x^y$ can be assigned a reasonable meaning for all real $x$ and all rational numbers of the form $y=m/n$, where $m$ is even and $n$ is odd and positive. Thus $x^y=(x^m)^{1/n}$, interpreted as the unique real $n$th root of $x^m$ (define $0^0$ to be $1$). Since every real number can be arbitrarily well approximated by such "even/odd" rationals, by continuity, a synthetic definition of $x^y$ can be obtained for all real $x$ and $y$. For example, using this definition, a graph can be plotted for the relation $y^y=x^x$, which runs smoothly as a loop through all four quadrants (along with the obvious line $y=x$).

What is the domain of $f(x)=x^x$?

2 Answers2