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.

  • Yes, there is a connection. Which base did you have in mind? – hardmath Nov 14 '15 at 23:08
  • base e? I was kind of asking abstractly, if you have an answer in an arbitrary base that would be impressive and helpful. – theREALyumdub Nov 14 '15 at 23:08
  • Hint: If $\ln^{(n)}(x)$ denotes the iterated logarithm and ${^n}e=e^{e^{\ldots{e}}}$ ($n$-times) denotes tetration, then the following identity holds for all $n\in\mathbb{N}$: $\ln^{(n)}({^n}e)=1$. – Yiannis Galidakis Nov 14 '15 at 23:34
  • @YiannisGalidakis, do you mean the super logarithm (inverse of tetration)? – theREALyumdub Nov 14 '15 at 23:37
  • I think so, yes, although I am not very familiar with the actual definition Andrew Robbins uses. It's bullet 3 on that page :As the number of times a logarithm must be iterated to get to 1 (the Iterated logarithm) @theREALyumdub – Yiannis Galidakis Nov 14 '15 at 23:42
up vote 1 down vote accepted

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 ($n\ge 3$):

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

  • Much more formal than my answer; I was so busy trying to edit mine to be correct that I missed this. – theREALyumdub Nov 15 '15 at 1:35

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

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

  • Yes, that looks ok. – Yiannis Galidakis Nov 15 '15 at 0:03
  • The exponent should probably be $n-1$. Your $D_2$ is wrong, since $\ln(\ln(1))$ has problems. It should be $x>e$ for base $e$. Then $D_n=\{x:x\in\mathbb{R},x>tetra(e,n-1)\}$ for base $e$. – Yiannis Galidakis Nov 15 '15 at 0:14
  • Does that look better? – theREALyumdub Nov 15 '15 at 1:32
  • Your $D_2$ needs correction. It is $x>1$ (it is the only one that breaks the pattern, see my answer. My comment above has a typo) and when you notate tetration it's prettier to stick to one of the two forms you have: $tetra(b,3)=b^{b^{b}}$, but parentheses go from top to bottom. So it'd be $b^{(b^b)}$, $b^{(b^{(b^b)})}=tetra(b,4)$, etc. – Yiannis Galidakis Nov 15 '15 at 1:55

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