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The graph of the function $y=x^{\frac{1}{x}}$ for positive $x$ is as shown below:

enter image description here

When I calculated $y$ for negative values of $x$ only some of the values between $0$ and $-1$ and only those for which $x$ is odd(whole number), were given by Excel, and they are:

enter image description here

If the function behaves so nicely for positive $x$, why not so nicely for negative $x$?

If $y=-x^{\frac{1}{-x}}=\frac{1}{-x^{\frac{1}{x}}}$, why don't we take the reciprocal of all the values of $y$ for positive $x$ and negate them to plot the negative X-axis?

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  • $\begingroup$ The function is not defined for $x \le 0$ ... and, once more, Excel does not know it ! $\endgroup$ – Claude Leibovici Sep 13 '14 at 14:39
  • $\begingroup$ It's complex valued for $x\le0$. $\endgroup$ – m0nhawk Sep 13 '14 at 14:40
  • $\begingroup$ @m0nhawk. It seems that you are correct. Complex values for $x<-1$. $\endgroup$ – Claude Leibovici Sep 13 '14 at 14:44
  • $\begingroup$ When $x$ is negative, $x^a$ generally isn't defined, with a couple of exceptions: (1) when $a$ is an integer; (2), according to some conventions, when $a$ is a rational number whose denominator can be taken to be odd. That's because if $a = p/q$, with $q$ odd, it makes sense to write $x^a = (\sqrt[q]{x})^p$ even when $x$ is negative. Excel seems to be following these conventions. $\endgroup$ – Dave Sep 13 '14 at 14:45
  • $\begingroup$ @ClaudeLeibovici I plotted $\Im x^{1/x}$ in Mathematica, it's not zero for $(-1,0)$. $\endgroup$ – m0nhawk Sep 13 '14 at 14:52
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The complex $\log$ map is multivalued and this forces $f$ to be multivalued for $x<0$.

$$\log(k,z)=\ln|z|+(\arg(z)+2k\pi)i,\,\,\,k\in\mathbb{Z}$$

If $z<0$, then $\arg(z)=\pi$

and therefore,

$$[z^{1/z},k]\to\exp\left(\frac{\log(k,z)}{z}\right),\,\,\,k\in\mathbb{Z}$$

consequently for negative $z$ the function is equivalent to the multivalued map:

$$[z^{1/z},k]\to\exp\left(\frac{\ln|z|+(\pi+2k\pi)i}{z}\right),\,\,\,k\in\mathbb{Z}$$

Some Maple code for verification:

 restart;
 f := proc (x) options operator, arrow; x^(1/x) end proc
 fn := proc (k, x) options operator, arrow;
 exp((ln(abs(x))+I*(Pi+2*k*Pi))/x) end proc

and now check:

 f(-2/3); evalc(%)

(3/4*I)*sqrt(2)*sqrt(3)

 fn(0, -2/3); evalc(%) #check using principal branch of log

(3/4*I)*sqrt(6)

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For negative values of $x$, Excel seems to give you an answer for $x^{1/x}$ only when it knows the exponent $1/x$ is a rational number that can be written with an odd denominator.

For example, $(-3)^{-1/3} = 1/\sqrt[3]{-3}$.

If it is doing this, then you probably won't always have a negative answer. For example, for $x = -3/2$, you'd expect to get $1/(\sqrt[3]{-3/2})^2$, which is positive.

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  • $\begingroup$ if $x=-3/2$ then $y=\frac{1}{(\sqrt{-3/2})^3}$ $\endgroup$ – Vikram Sep 13 '14 at 17:20
  • $\begingroup$ That would be $x^x$, not $x^{1/x}$. $\endgroup$ – Dave Sep 13 '14 at 17:22
  • $\begingroup$ yes, you are right $\endgroup$ – Vikram Sep 13 '14 at 17:27

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