Where is $x^x$ continuous?

Question

The idea of continuity of a function is something I come across quite regularly, but I've never really understood it well. I'm trying to fix that by looking at some interesting functions.

What happens continuity wise for $x<0$ in this function:

$$f(x)=x^x$$

I can't find a quick answer to this. Apologies if this is a duplicate. I'm looking for a detailed explanation.

Answer 1

As mentioned in another answer, writing $x^x$ is the same as writing $e^{x\log x}$. But complex analysis tells us that there are many possible branches of $\log x$ that you could choose. Namely, you can define it on any simply connected region in $\mathbb C$ not containing zero, and you can add $2\pi i$ as many times as you like.

Since you're asking about real negative values of $x$, you'd need a branch of the logarithm that avoids this half line in the complex plane. For example, you could choose the branch $$\log z=\log|z|+i\theta$$ where $-\frac{\pi}{2}<\theta<\frac{3\pi}{2}$. In any such branch, the complex logarithm is analytic and therefore continuous on the negative real half-line.

To conclude, the answer to your question is that $x^x$ is always continuous for $x<0$ provided you've picked a well-defined meaning of the function.

I apologize that this explanation requires that you've seen complex analysis, but this question cannot be adequately answered otherwise.

Answer 2

If $x<0$, then $x^x$ is not defined. The domain of $x^x$ is $\{\ldots,-3,-2,-1\}\cup[0,\infty)$. In general, you cannot raise negative numbers to general real numbers. Only to integer powers.

For an easy example to understand why, what is $(-1/2)^{-1/2}$? I guess it would be $1/\sqrt{-1/2}$, but now you are taking a square root of a negative number.

Going back to your introduction, $x^x$ is continuous on its domain, if you allow one-sided continuity at $0$ and consider a function on $-\mathbb{N}$ to be continuous. (Are you looking for a proof of that?)

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Let $a,b$ be real. $a^b$ means $\exp(b\ln(a))$. When $a$ is positive, this is fine and means what everyone thinks it means. There is a canonical choice for the value of $\ln(a)$ when $a$ is positive, so you use that, then multiply by $b$, then exponentiate base $e$.

When $a$ is negative, there is no canonical choice anymore for what $\ln(a)$ means. For example, $\exp(\ln(2)-\pi i), \exp(\ln(2)+\pi i), \exp(\ln(2)+3\pi i)$, etc. all equal $-2$, so $\ln(-2)$ has just as much right to be $\ln(2)-\pi i$ as $\ln(2)+\pi i$, etc. It's common to take the branch cut where $\ln(-2)$ is $\ln(2)+\pi i$, but that is a convention, not a canonical choice.

So for negative $a$, since $\ln(a)$ can mean lots of things, so can $b\ln(a)$, and so can $\exp(b\ln(a))$. So there is generally no good single value to assign to "$a^b$" when $a$ is negative.

The exception comes when $b$ is an integer. Then and only then, even though there are multiple values for $b\ln(a)$, they all go to the same place when put through $\exp$. This is because the multitude of values for $\ln(a)$ all differ by some integer multiple of $2\pi i$. So multiplying any two by an integer $b$, you still have two complex numbers that differ by an integer multiple of $2\pi i$. And $\exp$ will send all such numbers to the same place.

Conversely if $a$ is negative and $b$ is not an integer, then there exist two values for $\ln(a)$ such that the corresponding two values of $b\ln(a)$ do not differ by an integer multiple of $2\pi$. And therefore $\exp$ will give distinct outputs for these two numbers. And so $\exp(b\ln(a))$ will have no single meaning.

This still doesn't address when $a=0$. The definition of $a^b$ can be extended continuously to when $a=0$ when $b$ is positive ($0^b=0$), and not when $b$ is negative. This leaves $0^0$, which is well-debated elsewhere on this site. I fall into a camp that interprets $0^0=1$, which happens to make $x^x$ defined and right-continuous at $0$.

So the definition $a^b=\exp(b\ln(a))$ has a single meaning exactly for $(a,b)\in\left(\mathbb{R}_{<0}\times\mathbb{Z}\right)\cup\left(\{0\}\times\mathbb{R}_{\geq0}\right)\cup\left(\mathbb{R}_{>0}\times\mathbb{R}\right)$. Applied to $x^x$, this means the domain is $\{\ldots,-3,-2,-1\}\cup[0,\infty)$.

Answer 3

In the real case, $x^x$ is not defined when $x<0$, except for some particular value. For example, $(-1/2)^{-1/2}$ is not defined.

But we can define $z^z=\exp(z\log z)$ for $z\ne 0$, which is a multi-valued function in complex analysis.

Answer 4

The quickest and easiest way to make a statement on this function's continuity is to take a derivative. This requires logarithmic differentiation. The derivative is: $$f'(x) = x^x(\ln(x) + 1)$$

Note that $\ln(x)$ is only valid for positive entries. So, this tells us that $f$ is continuous for positive numbers.