# What is the domain of $f(x)=x^x$? [duplicate]

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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?

## marked as duplicate by Ian Miller, S.C.B., user147263, John B, heropupMar 7 '16 at 0:30

• What about $(-\frac12)^{-\frac12}$? – GoodDeeds Mar 6 '16 at 13:55
• I think $(-\sqrt{2})^{-\sqrt{2}}$ is imaginary? And that is why it does not appear? – S.C.B. Mar 6 '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 6 '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 6 '16 at 13:57
• I think $x^x$ is defined for all integers. – nsm Mar 6 '16 at 13:59

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: $$\{q\in \mathbb{Q}|q=\frac{m}{2k+1}\quad , \quad m,k \in \mathbb{Z} \}$$

• 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 6 '16 at 16:16

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 $y=x$).