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I am investigating the family of functions $$\log_{(n)}(x):=\log\circ \cdots \circ \log(x)$$

Is there a known smooth interpolation function $H(\alpha, x)$ such that $H(n,x)=\log_{(n)}(x)$ for $n\in\mathbb{N}$?

Gottfried Helms
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pre-kidney
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    The iterated logarithm is usually denoted $\log^n(x)$ or $\log^{(n)}(x)$; it is easy to confuse $\log_n(x)$ with log base $n$ (which uses the same notation). – Mario Carneiro Jun 09 '13 at 02:29
  • What is smooth interpolation function? – Paul Jun 09 '13 at 02:29
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    @Paul I think he wants a continuous function $H(\alpha,x)$ such that $H(0,x)=x$ and $H(\alpha+1,x)=\log H(\alpha,x)$ for all $\alpha\in{\Bbb R}$. – Mario Carneiro Jun 09 '13 at 02:31
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    Please revert the notation change - in analytic number theory (which is where my application lies), $\log^n(x)$ is used to denote $(\log(x))^n$ and the subscript notation is somewhat standard... – pre-kidney Jun 09 '13 at 02:31
  • @pre-kidney Is $\log_{(n)}(x)$ better? I think it would not be wise to use conflicting notation here, for a general mathematical audience. – Mario Carneiro Jun 09 '13 at 02:34
  • It's sort of awkward, but I guess it's fine if you think it makes things more readable :) – pre-kidney Jun 09 '13 at 02:34
  • @pre-kidney The criterion that $H(n,x)=\log_{(n)}(x)$ is not a very strong one: I can just take $H(\alpha,x)=H(\lfloor\alpha\rfloor,x)+(\alpha-\lfloor\alpha\rfloor)(H(1+\lfloor \alpha\rfloor,x)-H(\lfloor\alpha\rfloor,x))$ to make a continuous function given the specified special cases for $\alpha\in{\Bbb Z}$. – Mario Carneiro Jun 09 '13 at 02:38
  • How smooth do you need it to be ? Are you looking for a formula or just existence ? – justt Jun 09 '13 at 02:39
  • Your function is not smooth! In particular, $\frac{\partial H}{\partial \alpha}$ does not exist for integer $\alpha$! – pre-kidney Jun 09 '13 at 02:40
  • (see https://en.wikipedia.org/wiki/Smooth_function if you are still confused) – pre-kidney Jun 09 '13 at 02:43
  • @pre-kidney So by "smooth" you mean $C^\infty$? Well, then replace $(\alpha-\lfloor\alpha\rfloor)$ by $f(\alpha-\lfloor\alpha\rfloor)$, where $f(x)=\int_0^x g(x),dx/\int_0^1 g(x),dx$ and $g(x)=e^{1/(x(x-1))}$ on $x\in(0,1)$ and $g(x)=0$ elsewhere. The function $g(x)$ is a $C^\infty$ function that has a "bump" on the interval $(0,1)$, and $f(x)$ is a $C^\infty$ interpolation function that goes from $0$ to $1$. – Mario Carneiro Jun 09 '13 at 02:53
  • Sure, but I was looking for a less artificial example. Perhaps consider the analogy with the Gamma function, where there is an additional log-convexity requirement. – pre-kidney Jun 09 '13 at 02:55
  • @pre-kidney That's why I pointed out that the essential feature here is the functional relationship $H(\alpha+1,x)=\log H(\alpha,x)$ for all $\alpha\in{\Bbb R}$, just as the gamma function satisfies $\Gamma(x+1)=x\Gamma(x)$ for all $x\in{\Bbb R}$. This constraint is very nontrivial, and if you additionally specify that it be analytic, you get a unique function $H$ satisfying the requirements. You may want to check out http://en.wikipedia.org/wiki/Tetration, which deals with a similar situation (extending the domain of a function defined by recursion). – Mario Carneiro Jun 09 '13 at 03:02
  • Yes, your functional equation is the correct version of what I was trying to convey in the original post - thank you for clarifying. – pre-kidney Jun 09 '13 at 03:20
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    Look at "tetration" which is interpolation of the inverse function, $\exp(x)$, so $\exp(\exp(...\exp(x)))) = \exp^{\circ h}(x) = \exp_{(h)}(x) =_{x=1} \ ^h e $ and the interpolation of the iterated log is then the same as taking h negative. See https://en.wikipedia.org/wiki/Tetration – Gottfried Helms Jun 09 '13 at 03:23
  • This seems to be the correct approach, thanks for the suggestion! – pre-kidney Jun 09 '13 at 03:30

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Your function $H(\alpha,x)$ is the same as $\exp^{-\alpha}(x)$ in the notation of wikipedia, and as such it has been widely studied. Two good sources are the Citizendium page and also this great paper by Dmitri Kouznetsov, whose results I have managed to reproduce.

D. Kouznetsov (July 2009). "Solution of $F(z+1)=\exp(F(z))$ in complex z-plane". Mathematics of Computation 78 (267): 1647–1670.

The function he studies is $F(-\alpha)=H(\alpha,1)$ in your notation. Moreover, $H(\alpha,x)$ for different $x$ are just shifted versions of each other, in the sense that $H(\alpha,y)=H(\beta+\alpha,1)$ for some $\beta$ dependent on $y$. This is because if there is some $\beta$ such that $H(0,y)=H(\beta,1)$, then if I define $H'(\alpha)=H(\beta+\alpha,1)$, we have $H'(0)=H(0,y)$ and $H'(\alpha+1)=\log H'(\alpha)$, so $H'$ satisfies the same functional equations as $H$ and thus is the same function (assuming the additional translation-invariant regularity properties, which means analyticity and an asymptotic form as $\alpha\to i\infty$, see Kouznetsov's paper for details).

Over the reals, the fact that $H(0,y)=H(\beta,1)$ has a solution for any $y$ follows from the fact that the range of $H(\beta,1)$ for $\beta\in(-2,\infty)$ is all of ${\Bbb R}$, since $\lim_{\beta\to-2}H(\beta,1)=-\infty$ and $\lim_{\beta\to\infty}H(\beta,1)=\infty$. Over the complexes, it follows from Picard's little theorem, since $H$ is non-constant.

Mario Carneiro
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