# Show that the iterated $\ln^{[n]}$ of tetration(x,n) is nowhere analytic

$$f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n$$

The conjecture is that $f(x)$ is monotonic and infinitely differentiable at the real axis, but nowhere analytic; because at each point on the real axis, the Taylor series has a zero radius of convergence. The function is well defined at the real axis, but not as well behaved in the complex plane. $f(x)$ is a useful function for comparing tetration for different bases, to see how much faster one base grows than another, where x can be thought of as the tetration base. See my post, Comparison between two tetrations showing the evaulation of $f(\pi)$.

How difficult would it be to show that $f(x)$ is infinitely differentiable, and nowhere analytic? Below, is a graph of $f(x)$, showing the "logarithmic singularity" at the real axis near 1.805; for $x >\approx1.805$, $f(x)$ is real valued and converges nicely at the real axis.

One simplification I found, also shows that $f(x)$ behaves somewhat like $\ln(x)+ \ln(\ln(x))$ as x increases. The $x \uparrow \uparrow n$ term grows very rapidly, so that the $\ln(\ln(x))$ term becomes insignificant in the equation below, that can be used to evaluate $f(x)$. At the same time as the $x \uparrow \uparrow n$ term in the denominator becomes insignificant at the real axis, it becomes less and less well behaved in the complex plane.

$$f_n(x) = \ln^{[n]} x \uparrow\uparrow n$$ $$f_{n+1}(x) = \ln^{[n]} ((x \uparrow\uparrow n)\ln(x) )$$ $$f_{n+2}(x) = \ln^{[n]} ((x \uparrow\uparrow n)\ln(x) + \ln(\ln(x)))$$ $$f_{n+2}(x) = \ln^{[n]} ((x \uparrow\uparrow n)\ln(x) \times (1+ \frac{\ln(\ln(x))}{(x \uparrow\uparrow n)\ln(x)})$$ $$f_{n+2}(x) = \ln^{[n]} ( \exp^{on}(f_{n+1}) \times (1+ \frac{\ln(\ln(x))}{ \exp^{on}(f_{n+1}) })$$

After this, the algebra gets messy.... but here's the next step I took. Eventually, you get to an equation of $f_{n+2}(x) \approx f_{n+1}(x) +$ reciprical of a superexponential product.

$$f_{n+2}(x) \approx f_{n+1} + \frac{\ln(\ln(x))} { \exp^{on}(f_{n+1}) \times \exp^{on-1}(f_{n+1}) \times \exp^{on-2}(f_{n+1}) \times ... \exp(f_{n+1}) } + O\frac{1}{(\exp^{on}(f_{n+1}) )^2}$$

For this function, "Nowhere analytic" means the function has a well defined Taylor series at the real axis, but the Taylor series has a zero radius of convergence, so that that the function is not equal to its Taylor series. This is because the Taylor series terms eventually grow faster than any exponential series, so that for any value of r you pick, for n large enough, $|a_n|>r^n$, or equivalently, $\ln(|a_n|)>n\ln(r)$. I think the key to showing this is to look at the function $f_n(x)-f_{n-1}(x)$, as n increases.

In the complex plane, $f_6(x)$ has a singularity near x=1.96219034541054 + 0.254713677298596i. The $f_6$ singularities occur where $\exp^{4}(f_{5}) =-\ln(\ln(x))$. As x gets larger, the singularities get closer to the real axis, and also, as n gets larger, the singularities get closer to the real axis.

• At the same time, one must show that the series of functions converges. So that the taylor series coefficients $a_n$ for $f_m - f_{m-1}$, for smaller enough values of $a_n$ also get arbitrarily small, $|a_n|<r^n$, for arbitrarily small values of r. Equivalently, $\ln(|a_n|)<n\ln(r)$, for arbitrarily small r. Paradoxically, at the same time, for large enough values of n, because the function is nowher analytic, we need to show that $\ln(|a_n|)>n\ln(r)$ for arbitrarily large r! – Sheldon L Feb 26 '14 at 14:35
• I think the way to go about this is to leverage the chaos-theoretic properties of the tetration. In particular, can one use those properties to show that $\bigcup_{n=0}^{\infty} \{ z \uparrow \uparrow n = 0 : z \in \mathbb{C}\}$ has a point in every neighborhood of every point of the positive real axis? – The_Sympathizer Jun 21 '14 at 11:08
• @mike4ty4, that might work -- $(z+\delta)\uparrow\uparrow n =0$, as n goes to $\infty$, $\delta$ gets arbitrarily small. But, isn't there the possibility that the singularity for n at $\delta$ is no longer relevant for n+1? The only way I can think of to formalize it is to show the Taylor coefficients for n+1 and th Taylor coefficients for n are very very close to each other, and that all of the Taylor series coefficients are converging, Coo, still thinking... Someday, I'll present a brute force solution. – Sheldon L Jun 24 '14 at 12:48
• I'm not sure what you mean by the "singularity is no longer relevant". The singularity won't just disappear, I don't believe, since $\log^n$ always has a singularity at $z = 0$, so any point on the holomorphic $z \uparrow \uparrow n$ where it is $0$ will be a singularity when $\log^n$ is applied. Of course, it could be possible that the "strength" of the singularity diminishes to nothing as $n \rightarrow \infty$, as in $\frac{1}{n} \log(z)$. – The_Sympathizer Jun 25 '14 at 1:58
• (and therefore, the limit has no (non-removable) singularities) – The_Sympathizer Jun 25 '14 at 4:47