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I started to think about the function $f(x) =x^{1/x} $ and how it behaves. How would you show for which negative real values of x that $f(x)$ gives real values?

I tried by rewritting, set $y=-x$ and we get $\displaystyle-\frac{(-1)^{((y-1)/y)} }{y^{(1/y)} } $. The nominator will be complex if $y>1$. Is there any other way to show this? By using euler's formula maybe? Since it is basically just the finding the roots for all negative x.

Quanto
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Plebbut
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  • See the related https://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero for some further insights. – Michael Hoppe Nov 11 '19 at 18:07
  • Another post on the subject https://math.stackexchange.com/questions/3387603/how-can-i-find-the-domain-of-fx-x1-x-on-the-negative-numbers/3388547#3388547 – zwim Nov 11 '19 at 18:50

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This is real when $x$ is negative only when $x$ is also an odd integer or a rational with odd numerator, for if $x$ were an even integer, we have complex values. If $x$ is rational, then it should have the form $$\frac{\text{odd integer}}{\text{nonzero integer}},$$ for the same reason as in the integer case. If $x$ is irrational, then we might want to write the expression as $$e^{\log x^{1/x}}=e^{\frac{\log x}{x}},$$ which is not then real.

Thus, in summary, if $x<0,$ then it should either be an odd integer, or a rational with odd numerator as well.

Allawonder
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The expression $x^{1/x}$ would always yield a real value for any $x<0$. To see so, let $x=-a$, with $a>0$,

$$x^{1/x}=(-a)^{-1/a} =(-a)^{2(-\frac1{2a})}=(a^2)^{-\frac1{2a}} =\frac1{(a^2)^{\frac1{2a}}}=\frac1{a^{\frac1{a}}}$$

However, the real value obtained above may not be unique. For example, if $a$ is an odd natural number, that is, $a=2k+1$, the expression $x^{1/x}$ could also lead to,

$$x^{1/x}=[-(2k+1)]^{-1/(2k+1)} = \frac1{[-(2k+1)]^{1/(2k+1)} } =-\frac1{(2k+1)^{1/(2k+1)} } = -\frac1{a^{\frac1{a}}} $$

The reason is that $f(x) = x^{1/x}$ is undefined for $x<0$.

Quanto
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    This is incorrect; $(-2)^{-\frac{1}{2}}$ takes only two values, namely $\pm i \frac{\sqrt 2}{2}$, while no irrational takes real values either; you need a rational with odd numerator (so odd/odd or odd/even); $x^{\frac{1}{x}}$ is in general an infinite set of complex numbers for any $x \ne 0$ though it is finite for rational $x$, while it has a principal positive value for $x >0$ – Conrad Nov 11 '19 at 19:15
  • Which part is incorrect? – Quanto Nov 11 '19 at 19:19
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    $(-a)^{-1/a} =(-a)^{2(-\frac1{2a})}$ is incorrect as complex powers do not work that way unless they are integral – Conrad Nov 11 '19 at 19:19
  • It is in real space, which is the focus here – Quanto Nov 11 '19 at 19:21
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    Put $a=1$ and your equality would give $-1=1$; once you take non integral powers of negative numbers, you get into complex numbers whether you like it or not – Conrad Nov 11 '19 at 20:02
  • For odd a, $x^x$ could assumes values of either -1 and 1 in real space, which is the point made. The disconnect is the answer is restricted to real space discussion while the comments assume otherwise. – Quanto Nov 11 '19 at 20:11
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    $(-1)^{(-1)}$ is never $1$, not sure where you learned your math – Conrad Nov 11 '19 at 20:12
  • Yes that works because it is of the form mentioned in my comment (rational odd/odd) while the real number is negative not positive ( there are 3 values here with one real) – Conrad Nov 11 '19 at 20:25
  • The point is $x^x$ as a function for $x<0$ is undefined in real space, which was made via an example. No need to get smart about some math, or being too complex. – Quanto Nov 11 '19 at 20:30
  • @Quanto the main point here is that $a^b$ is in general complex when $a<0.$ In this case, the manipulations you performed are not generally valid. One has to be careful with indices, even in cases where the base is positive -- for example when the index is a non reduced rational. – Allawonder Nov 12 '19 at 04:48
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    @Allawonder - thanks for the note – Quanto Nov 12 '19 at 05:04