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Consider the function $f(x) = x^x$.

Wolfram alpha tells me that this function's domain is $x : x>0$, $x \in \mathbb{R}$. I can't see why it cannot be defined for a number like $(-2)$. I mean $(-2)^{-2}=0.25$, the same Wolfram Alpha told me. I realize that fractional powers for negative numbers may cause problems, but it could be defined for integers. Thanks for any help.

Newb
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Shubham
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2 Answers2

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For most purposes (including the hotly debated question of what to make of $0^0$) you can consider any instances of exponentiation $x^y$ (where $x,y$ can stand for expressions) to stand for on of two quite disparate definitions that happen to coincide on the intersection of their domains:

  1. If $y$ designates an integer, then $x^y$ is defined algebraically; recursively by $x^0=1$ and $x^{n+1}=xx^n$ for the case $y\geq0$, and provided $x$ is invertible by $x^{-n}=(x^{-1})^n$ for $y<0$.

  2. In other cases one must assume that $x$ is real and positive, and $x^y$ stands for $\exp(y\ln x)$ (note that I did not write $e^{y\ln x}$, which would result in a circular definition). Here $\exp$ is a perfectly defined function $\Bbb C\to\Bbb C$, so one can allow $y$ to be any complex number (or one could even amuse oneself by taking square matrices for $y$), but $x$ must be restricted to avoid ambiguity of $\ln x$. One can of course extend the definition by making choices for $\ln x$, but using a bare $x^y$ for such cases would be confusing, and also one must be aware that many properties of exponentiation will start to fail.

Given this, the real function $x\to x^x$ can only be defined using the second variant, which justifies taking the domain to be the (strictly) positive reals. One could extend to domain to contain the non-positive integers as well (using the first definition), but mixing the two definitions of $x^y$ in a single usage is generally not a very fruitful idea.

Marc van Leeuwen
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  • So that means if y is not an integer, x must be real and positive. Got it! – Shubham Nov 26 '13 at 06:20
  • Any justifications, for the transformation x^y=exp(ylnx) ? – Shubham Nov 27 '13 at 09:37
  • and yes, should we be careful in reporting (-2)^(-2)=0.25? – Shubham Nov 27 '13 at 09:43
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    @Shubham: Well for $y$ integer it gives $x^y=\exp(y\ln x)=\exp(\ln x)^y=x^y$ so it coincides with the other definition when both are applicable; also one has for instance $\exp((y_1+y_2)\ln x)=\exp(y_1\ln x)\exp(y_2\ln x)$ giving $x^{y_1+y_2}=x^{y_1}x^{y_2}$, a desirable property of exponentiation. You should view it as the (ultimate) definition of $x^y$ for real $x>0$. – Marc van Leeuwen Nov 27 '13 at 09:45
  • @Shubham: I said clearly: $x$ must be positive real for $x^y=\exp(y\ln x)$ to apply, so for $(-2)^{-2}$ one should use the other (alegbraic) definition, which indeed gives $(-\frac12)^2=\frac14$. – Marc van Leeuwen Nov 27 '13 at 09:48
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Even if you take $x=0$, $f(x)=0^0 $ is an indeterminate form. $f(x)$ is defined for all positive values of $x$

sam_3314
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  • But of course $x^x \to 1$ as $x \to 0^+$ so you could extend the domain from $(0,\infty )$ to $\left[ 0,\infty \right)$ and retain continuity. – Jeppe Stig Nielsen Oct 24 '16 at 16:25