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We usually only see the graph $y=x^x$ for $x>0$, because $x^x$ is a complex number for most negative values of $x$. Yet here is a full graph of $y=x^x$ on the real line:enter image description here

This graph may seem like it's not even a function, failing the vertical line test, but what's actually going on is that the graph contains infinitely many holes. It's true that $x^x$ is not a real number for almost all values of $x<0$, but it is real in the rare situation when $x$ is a rational number which has an odd denominator when written in simplest form. In that case, $x^x$ is a positive real number when $x$ can be written as an even number divided by an odd number, and a negative real number when $x$ can be written as an odd number divided by an odd number. So just those rational points are being graphed, but since the rational numbers with odd denominators are dense in the real numbers it looks like we have continuous curves.

My question is, what are the continuous curves that seem to be there? Any continuous curve defined on a dense subset of the reals can be uniquely extended to a continuous function on all the reals. So what continuous function passes through all the points $(x,x^x)$ where $x<0$ and $x$ can be written as an even number divided by an odd number? (I'd ask the analogous question about the second curve, but it seems to be a mirror image of the first.)

Any help would be greatly appreciated.

Thank You in Advance.

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    $\begingroup$ Perhaps, my answer to this question helps? $\endgroup$ Mar 2 '14 at 1:47
  • $\begingroup$ @MarkMcClure Yes it does, thanks. $\endgroup$ Mar 2 '14 at 2:22
  • $\begingroup$ fantastic question $\endgroup$
    – v.oddou
    May 26 '20 at 15:38
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The continuous curves that pass through the points satisfying $y = x^x$ on $x \in (-\infty, 0)$ are simply $$g(x) = \pm \frac{1}{|x|^{-x}}.$$

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  • $\begingroup$ Thanks, I guess I missed the obvious. $\endgroup$ Mar 2 '14 at 2:21
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    $\begingroup$ Why not write $g(x) = \pm |x|^x$? $\endgroup$ Sep 6 '15 at 12:09
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    $\begingroup$ Hmm, maybe it does make sense to write $\pm\frac{1}{|x|^{|x|}}$ or $\pm\frac{1}{(-x)^{-x}}$ to make the relation with $x^x$ more clear. $\endgroup$ Sep 6 '15 at 12:36
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I would like to mention that your claim about extending a continuous function on a dense set, is not quite correct. For example consider $f(x)={1\over x}$ defined on $\mathbb{R}-\{0\}$. This is continuous on a dense set, but $f(0)$ cannot be defined to maintain continuity. You need the fact that the limit exists at every real number.

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  • $\begingroup$ nice point. now it's only missing a proof :) $\endgroup$
    – v.oddou
    May 26 '20 at 15:41
  • $\begingroup$ proof is a trivial exercise. $\endgroup$ May 26 '20 at 23:21

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