Domain of x^x for real and negative integer values

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prasannapakkiam
I wish to prove that for f(x)=x^x, its domain is: {x E R, x > 0}U{xEZ,x<0}.
I reevaluated to e^(xlnx), obviously that did not help. Is there an algorithm/formula/something that can evaluate irrational powers, so that it can help me with this?
 
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From what I understand of your question, you are following a misleading path. The function x^x, when x < 0, is defined at every x = a/b (reduced fraction) such as that b is not an even number. About irrational numbers, the question is ambiguous. For the negative irrationals, what does the expression x^x even mean? For positive irrationals, x^x can be expressed as an infinite series and can be approximated. A negative irrational could be approximated as x = a/b with the condition that b is and odd number and hence we get an approximation of x^x, but it can also be approximated as a/b in which b is an even number and hence yield an imaginary approximation. I have to put much thinking into it, but from this my very quick assumption is that x^x for negative irrationals is an absurd expression.
 
:P I only just realized that I did not define b can be odd. Anyway I require a technique to evaluate powers so that I can personally show that irrational numbers below zero cannot be part of the domain of x^x.
 
Have you tried re-evaluating it to [itex]e^{x ln(abs(x))}[/itex]? It evaluates to the same thing, except of course that it has a "larger" domain than x^x.

Edit: Actually, I rescind this statement--it's not true. I think it might still be helpful for evaluating [itex]e^{x ln(x)}[/itex], though, so I'll leave it up.
 
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