When I do a search on Wolfram|Alpha, I get that $x≥0$, except I thought that this isn't exactly the case. Because it is defined for all real numbers greater than zero, but there are infinitely many negative numbers for which the function is also defined. So why can't we extend the domain to numbers less than zero? It isn't continuous, I understand, but I suppose this is linked to the domain of the Dirichlet function?
What exactly is the domain of $x^x$?
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functions
continuity
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0Well, it's clearly devined for negative integers... For non-integers, however, you get into multivalue problems. For instance, is $(-1/2)^{-1/2}$ equal to $0.707i$, or $-0.707i$? – 2017-01-03
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0@Arthur how about some values with real solutions that are non-integers? Does the domain contain these values? – 2017-01-03
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0Well, $(-2.6)^{-13/5}$ has five different possible values in the complex plane. Which one do you choose? Well, here there is one that is actually a negative number because $5$ is odd, but try $-2.75$ instead, for instance. – 2017-01-03
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0@Arthur, my apologies, that was a mistake. I mean to extend this to any non-integer negative value in which the has a real output. – 2017-01-03
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0Well, _could_ give a real value to any rational number that has an odd denominator, but then you get into philosophical issues with $\frac12 = \frac24$. – 2017-01-03