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I'm a little bit confused! what is the domain of this function:

$$ \ln(x) ^{ \ln(x) } $$

this function, in fact, is:

$$ \exp(\ln(\ln(x))\cdot\ln(x)) $$ so the domain would be: $$ x>1 $$

But: $x$ can "also" take on the value (for example) :$$\sqrt[3] {e^{-1}}$$

then there would be some other numbers, not included in $x>1$, in the domain of the function.

gebruiker
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Seeen Jim
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  • @Shailesh: we are asking if we can raise a number in the range of $\log$ to the power of itself. – Ross Millikan Dec 19 '15 at 15:35
  • Yes, you are correct. The domain includes values of $x$ for which $x=e^{p/(2q+1)}$ for positive integers $p$ and $q$ – Mark Viola Dec 19 '15 at 15:44
  • This is essentially same as asking the domain of functions like $x^x$. This is a well defined real function for $x>0$, but for $x<0$ it has some nasty behaviours. For instance $(-0.5)^{-0.5}$ is not real anymore. But it gets worse when we ask value of $x^x$, for instance, for negative irrational $x$ - are they real or complex, or possibly undefined etc. But for instance $x=-1$ then $(-1)^{-1}$ is well defined. I've never seen the domain given explicly, and I doubt if there is a simple answer to this – user160738 Dec 19 '15 at 15:47
  • @user160738 See my comment. The domain for $x^x$ includes $x=p/(2q+1)$ for integers $p$ and $q$. – Mark Viola Dec 19 '15 at 15:52
  • @Dr.MV excuse me but why did you exclude negative integers, because i think it works: like $$\sqrt[3] {e^{-1}}$$ ? what you mentioned, is included in x>1, but my question is about numbers less than 1... – Seeen Jim Dec 19 '15 at 15:57
  • We need to exclude negative values of $x$ if we are restricted to real analysis since the logarithm is not defined for negative $x$. And remember $1/e$ is positive. – Mark Viola Dec 19 '15 at 16:01
  • Just nitpicking: the definition of a function is incomplete if it doesn't contain the domain. – Gyro Gearloose Dec 19 '15 at 16:13

3 Answers3

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Analysts like continuous functions. Thus for this problem they would either use domain $x>1$ or else use complex numbers, taking a "principal value" for the logarithms.

Recreational mathematicians like to do things like $\sqrt[3]{e^{-1}}$ for this. But there is no known use of that type of calculation in mathematics itself.

GEdgar
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We can avoid the logs for a start and ask about the domain of $y^y$, then set $y=\ln x$ as $\ln y$ is surjective on the reals. $y^y$ is defined whenever $y \gt 0$ or $y$ is a negative integer, so its domain is $k(k \lt 0, k \in \Bbb Z) \cup k(k \in (0, \infty), k \in \Bbb R)$ That would give the domain of $\ln x^{\ln x}$ as $e^{k}(k \lt 0, k \in \Bbb Z) \cup e^k(k \in (0, \infty), k \in \Bbb R)$ or $e^{k}(k \lt 0, k \in \Bbb Z) \cup m(m \in (1, \infty), m \in \Bbb R)$

More generally, we write $a^b$ with domain $a \gt 0, b\in \Bbb R \cup a \lt 0, b \in \Bbb Z $ (except $0^0$) but it makes some sense to think of the two pieces of the domain as separate functions. The first is $e^{b \ln a}$ and has a nice contiguous domain. The second is $(-1)^be^{b \ln(-a)}$ and because of the $(-1)^b$ piece we can only do it by repeated multiplication, which forces $b$ to be integral. One obstacle to seeing this is that $a \gt 0, b \in \Bbb Z$ belongs to both pieces of the domain. Generally in a problem you will only use one piece of the domain or the other. It will be clear from context which you are using and you can pretend that the other piece of the domain does not exist for that problem.

Ross Millikan
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If you are more interested in continuity than in algebraic properties then you might want to extend $\ln x$ analytically so that it is equal to $\ln|x|+\pi i$ for negative real numbers.

Justpassingby
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