# Analysis of the function $y=x^{\frac{1}{x}}$

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

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:

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?

• The function is not defined for $x \le 0$ ... and, once more, Excel does not know it ! – Claude Leibovici Sep 13 '14 at 14:39
• It's complex valued for $x\le0$. – m0nhawk Sep 13 '14 at 14:40
• @m0nhawk. It seems that you are correct. Complex values for $x<-1$. – Claude Leibovici Sep 13 '14 at 14:44
• 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. – Dave Sep 13 '14 at 14:45
• @ClaudeLeibovici I plotted $\Im x^{1/x}$ in Mathematica, it's not zero for $(-1,0)$. – m0nhawk Sep 13 '14 at 14:52

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)

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.

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