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I was wondering if there was a function that extends the domain of the following function to non-negative real numbers. For non-negative integer $n$ and real $y$, $y = f(x,n)$ is given by:

$$f(x,n) = \ln\underset{n-2}{\cdots}\ln x$$

e.g. $f(x,3) = \ln \ln \ln x, f(x,0) = x$.

Do any well known functions, defined instead for $n\in\mathbb{R}_0^+$, have this property?

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This is not a full answer, but some remarks that are related to your problem. So let us start with an abstract setup. Let $f(x)$ be an invertible real valued function. Define a new function $F(n,x)$ defined by $F(0,x)=x$ and $F(n+1,x)=f(F(n),x)$. Since $f$ is invertible, we may extend this to negative numbers by setting $F(-1,x)=f^{-1}(x)$, and iterating backwards. Let us denote $F(r,x)=f^r(x)$ The question is whether nor not we are able to extend this all real numbers (or in your case non-negative reals). But we do not want any old extension, we could simply take a piecewise linear extension (or many other types of extensions in fact). A sensible property to ask for is to ask that $f(r+s)=f^r\circ f^s$ for all real numbers. We might also ask that our function is continuous. To solve this problem in general, it suffices to find a function to Schröder's equation. What this is a pair $(\phi,s)$ where $\phi$ is an invertible real valued function and $s$ is a non-zero (positivity would be nice) such that $$\phi(f(x))=s\phi(x).$$ From this it is easy to see that $f^r(x)=\phi^{-1}(s^r\phi(x))$ solves our problem (although whether or not the solution is unique or not is another issue). A nice condition that makes the existence possible is the existence of a fixed point. Unfortunately, the natural log does not have a fixed point. See Here.

This also seems to be of interest.

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  • $\begingroup$ Cool, seems the problem is much deeper than I thought when I wrote the question. $\endgroup$ – Lucas Jul 4 '13 at 13:15

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