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I'm trying to figure out the domain of the function $y=x^x$. When I graph it, it appears to be defined on $[0, \infty)$, but then when I plug in individual negative numbers, for some of them I get real numbers, such as $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$. However, for other numbers such as $x=2.2$, the result is imaginary.

On what real numbers is $y=x^x$ defined?

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  • $\begingroup$ @ حكيمالفيلسوفالضائع $\endgroup$ – user99680 Jun 18 '14 at 18:39
  • $\begingroup$ Ah, yes, sorry, $\endgroup$ – user99680 Jun 18 '14 at 18:39
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The standard domain for $x^x$, as a Real-valued function of a Real variable is $(0, \infty)$. There are values in the negative Real line where $x^x$ is defined, but there are no open sets in the negative Real line where $x^x$ is Real-valued.

Notice that for negative Real numbers, if $x=p/q$; $p$ a negative integer , q odd, then $x^x=(p/q)^{p/q}=((p/q)^p)^{1/q} $ is the $q-th$ root of a negative number, which exists(as a Real number) iff $q$ is odd. I think this is the entire set where $x^x$ is defined (As a Real number) on the negative Real axis. EDIT: As correctly pointed out by Dave Renfro, this works only for $p,q$ relatively-prime; otherwise , you run into the problem of having many representations for $p/q$, e.g., $1/3,2/6,...$ so that the function is defined for some representations but not for others.

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  • $\begingroup$ What rule is there, if any, for what negative valued numbers are defined for $x^x$? $\endgroup$ – scrblnrd3 Jun 18 '14 at 18:42
  • $\begingroup$ I think it is the set: {m/n: m<0 an integer, n>0 and n is odd}.You want to avoid taking square roots of negative numbers. Then $(m/n)^{m/n}=((m/n)^m)^{1/n}$ is the n-th root of a negative number, which is defined only if $n$ is odd. $\endgroup$ – user99680 Jun 18 '14 at 18:47
  • $\begingroup$ @scrblnrd3: I edited the comment into the answer. $\endgroup$ – user99680 Jun 18 '14 at 18:55
  • $\begingroup$ What you've defined doesn't seem to be a function. If $f$ is a function and $x_1 = x_2$ is in the domain, then $f(x_1) = f(x_2).$ Note, however, that while $\frac{1}{3} = \frac{2}{6},$ what you've described makes $f(1/3)$ defined and $f(2/6)$ undefined. To fix this, you need to say that if $x$ is a negative rational number, we define the value of $f(x)$ by first expressing $x = \frac{p}{q}$ where $p$ and $q$ are relatively prime integers with $p < 0$ and $q > 0,$ or something along these lines. $\endgroup$ – Dave L. Renfro Jun 18 '14 at 19:03
  • $\begingroup$ @DaveL.Renfro: Thanks, you're correct, I edited it out in my answer. $\endgroup$ – user99680 Jun 18 '14 at 19:06
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Observing that $$ x^x=e^{x\log x} $$ you'll notice that $x$ must be strictly positive. Hence the domain is $D=\mathbb R_{>0}=]0,+\infty[$.

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  • $\begingroup$ But I think this definition holds only for x>0. Otherwise, $(-1)^{-1}=-1$ $\endgroup$ – user99680 Jun 18 '14 at 19:27
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    $\begingroup$ This is not a good because consider $f:\mathbb{R} \to \mathbb{R}$ given by $f(x)=x$ clearly $x=e^{\log(x)}$ therefore $x$ must be non-negative... but this is nonsense. $\endgroup$ – Squirtle Jun 18 '14 at 19:37

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