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The function $f(x) = x^x$ gives a complex number only if x has an even denominator. I'm not sure about irrational numbers. Why, then, is the best graph I can find of that function that of Wolfram Alpha, which plots adjacent sinusoidal waves for the real and imaginary parts, even at well-defined values such as $x = -\frac{1}{3}$? (And that's the plot that includes imaginary numbers; the real-valued plot shows nothing below zero.) I suspect this might be because the results are computed in a way that requires imaginary numbers.

The number of values for which the function is defined is infinite, but the values for which it is not are also infinite. So how should the graph look below zero?

(Note that in order to avoid another point where the function may be undefined I'm assuming that $0^0 = 1$.)

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Although it doesn't deal directly with $x^x$, here's a WolframAlpha blog post that details how real and complex roots are currently treated by WolframAlpha.

As far as $x^x$ goes, the plot does indeed show the real and imaginary parts of the principal value of $x^x$ on the same axis. Another approach is to plot the complex point itself in a plane that's perpendicular to the $x$-axis at the corresponding point. This leads to a spectacular image called the $x^x$-spindle, which was described in a great paper in Mathematics Magazine back in 1996. This looks like so:

enter image description here

Using the fact that the complex logaritm is multi-valued, this can be generalized to obtain more threads on the spindle:

enter image description here

It sounds like you've seen the claim that $(p/q)^{p/q}$ is defined for $p$ negative and $q$ odd and positive. Thus the graph might look something like so.

enter image description here

From the complex perspective, the dots arise as spots where one of the spiral threads punctures the $x$-$z$ plane.

Note that the Mathematica code for these images is all provided in this answer over on mathematica.SE.

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  • $\begingroup$ What is the minimum value in image #1? Also, $(-1)^{-1}$ is $-1$ and not $1$. $\endgroup$
    – Lee Sleek
    May 17, 2013 at 18:41
  • $\begingroup$ @LeeSleek I don't understand the question. $\endgroup$
    – Mark McClure
    May 17, 2013 at 18:45
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    $\begingroup$ @Lee Sleek: You can find the minimum of $x^x$ for $x > 0$ by calculus, using a method often called logarithmic differentiation. Let $y = x^{x}.$ Taking the logarithm of both sides gives $\ln y = \ln\left(x^x\right) = x\ln x.$ Now find $y'$ by implicit differentiation. You'll get $y' = x^x \left(\ln x + 1 \right).$ Setting this equal to $0$ and solving for $x$ gives $x = e^{-1} = 0.367879 \dots$ See also this WolframAlpha webpage and these related examples. $\endgroup$
    – Dave L. Renfro
    May 20, 2013 at 20:20
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    $\begingroup$ thank you, that explanation is phenomenal. I'm fascinated. It's crazy I did so much math and never really looked at x power x $\endgroup$
    – v.oddou
    May 26, 2020 at 15:51
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    $\begingroup$ @v.oddou Thanks, I'm glad you like it! Here's an interactive 3D version of the spindle: observablehq.com/@mcmcclur/the-xxx-xxx-spindle $\endgroup$
    – Mark McClure
    May 26, 2020 at 15:59

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