Domain of the n composed logarithms on x.

Question

As we know, the real logarithm has the domain

$$ D_1 = \{x : x \in \mathbb{R}, x > 0\} $$

What is the logarithmic domain of "higher order" logarithms, at index n? For example, it seems that

$$ D_2 = \{x : x \in \mathbb{R}, x > 1\} $$

which would be the domain of

$$ f(x) = \log(\log(x)) $$

$ D_3 $ is a little hard to imagine. I think my real question deals with formalization, and the extended case of $ D_n $ however.

There may be connections to the iterated logarithm, which says how many iterations are necessary before a value breaks in a certain base. See the wikipedia article.

Answer 1

If you denote by $\ln^{(n)}(x)$ the iterated logarithm and by ${^n}e=e^{e^{\ldots^{e}}}$ (height $n$) iterated exponentiation of the base $e$ (as per the comment), we have by definition:

$$\ln^{(n)}({^n}e)=1\Rightarrow$$ $$\ln^{(n+1)}({^n}e)=\ln(1)=0$$

Apply for $n=1$ and we get:

$$\ln(\ln(e))=0$$

So $D_2$ should be $D_2=\{x\in\mathbb{R}\colon x\ge e\}$. This however breaks the pattern, because the range of $\ln(\ln(x))$ can be negative for this case, if we extend the domain of $x$ to be $x\gt 1$.

You can't do this for higher iterates however, because negative ranges are not allowed in the domain of $\ln$. Therefore:

$$D_1=\{x\in\mathbb{R}\colon x\gt 0\}$$ $$D_2=\{x\in\mathbb{R}\colon x\gt 1\}$$ $$D_3=\{x\in\mathbb{R}\colon x\ge {^2}e\}$$ $$D_4=\{x\in\mathbb{R}\colon x\ge {^3}e\}$$

and in general ($n\ge 3$):

$$D_n=\{x\in\mathbb{R}\colon x\ge {^{n-1}}e\}$$

Answer 2

Actually, now that I've looked into it, the table on the page seems to suggest that the iterated logarithm values are the way you compute the bounds on the $ D_n $ domains, and it switches at the tetration integer values.

The tetration operation on the base defines the bounds of the domains $ D_n $.

Let b be the base of the logarithm in question, so that $ f_1(x) = \log_b(x) $

We can say

$$ D_1 = \{x : x \in \mathbb{R}, x > 0\} $$ $$ D_2 = \{x : x \in \mathbb{R}, x > b\} $$

And for n > 2, we apply the iterated logarithm to say that

$$ D_3 = \{x : x \in \mathbb{R}, x > b^b\} $$ $$ D_4 = \{x : x \in \mathbb{R}, x > (b^b)^b\} $$ $$ ... $$ $$ D_n = \{x : x \in \mathbb{R}, x > tetra(b,n - 1)\} $$