How can i find the domain of $f(x)$= $x^{1/x}$ on the negative numbers?

Question

I have been thinking of which negative values of $x$ yield a real number when plugged on $f(x)$ = $x^{1/x}$,

It is clear to see that it does not work for negative even numbers $(-2)^{-1/2}$ = $(\frac{1}{\sqrt{-2}})$ is not real,

but for negative odd numbers it holds, $(-3)^{-1/3}$ = $\frac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{-3}}$ = $\frac{1}{-\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{3}}$which is real.

For some fractions it also holds, for example $(-\frac{3}{2})^{-2/3}$ is real, but for others it does not.

My question is if there is any easy way of finding all values for which $f(x)$ is real, and if there is any irrational number that yields a real value.

Answer 1

Non-integral power of negative real number is always non-real for instance $(-2)^{2.1}$ because $-1=e^{i\pi}$ so this nuber equal $2^{2.1} e^{2.1i}=2^{2.1}[\cos(2.1)\pi+i\sin(2.1 \pi)]$. But the integral powers of a negative number can be a real number, so $f(x)=x^{1/x}$ for isolated points(s) such as $x=-1$ is real, and this is just one point with no-neighborhood namely $f(x)$ is non-real on the left and the right of $x=-1$ as $f(-.99)$ and $f(-1.01)$ are non-real.

The function $g(x)=x^x$ for negative $x$ is real only at isolated points $x\in I$ but in their neighborhood f(x) is non real. That is why we say its domain is only $(0,\infty)$.

Similarly, the domain of $f(x)=x^{1/x}$ where it is real is $(0,\infty)$, Only at isolated points $x=-1,-3,-5,..$ is real and these isolated points are not included in the domain.

Answer 2

I have already discussed some related problem here: Is $(-1)^{2.16}$ a real number?

Basically for negative reals $x$, the value of $x^y$ can be extended via rational exponentiation to all $y\in\mathbb Q_{odd}$, the rational numbers represented by an irreducible fraction whose denominator is an odd natural.

This is made possible by the fact that $x\mapsto x^q$ is an odd function so $\sqrt[q]{x}=x^\frac 1q$ is well defined for odd $q$.

Let set $f(x)=x^{1/x}$

According to the previous paragraph $f(x)$ would be defined for $x<0$ when $\frac 1x\in\mathbb Q_{odd}\iff \frac 1x=\frac pq\iff x=\frac qp$ with $q$ odd.

Also since $\dfrac{\ln(x)}x\to -\infty$ when $x\to 0^+$ we can extend $f$ by continuity in zero with the value $f(0)=0$.

All these considerations result in the domain $$\left\{-\frac pq\mid (p,q)\in\mathbb N^2,\ \gcd(p,q)=1,\ p\text{ odd}\right\}\cup[0,+\infty)$$

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Now we could wonder if the domain of $f$ may be extended even more while considering complex calculation.

For $x<0$ we have $\quad\ln(x)=\ln(-|x|)=\ln|x|+\ln(-1)=\ln|x|+(2k+1)i\pi\quad$ for $k\in\mathbb Z$.

$x^{\frac 1x}=\exp\left(\frac 1x(\ln|x|+(2k+1)i\pi)\right)=\underbrace{|x|^\frac 1x}_{\in\mathbb R}\times \underbrace{\exp\left(-i\frac{(2k+1)\pi}{|x|}\right)}_{\Large z_k}$

As you suggested we could extend the domain of $f$ to any $x$ for which $\{z_k\in\mathbb R\mid k\in\mathbb Z\}$ is a non-empty singleton.

$z_k\in\mathbb R\iff \exists n\in\mathbb Z\mid -\frac{(2k+1)\pi}{|x|}=n\pi\iff \exists n\in\mathbb Z\mid x=\frac{2k+1}{n}$

And we notice that it means $\frac 1x\in\mathbb Q_{odd}$ as seen in first paragraph, in particular we do not have to bother about unicity (the singleton property) since we already have a definition for such calculation via rational exponentiation.

So we conclude that we cannot extend more the domain of $f$ than what we already had without considering complex calculation.