# Power towers: to infinity and all the way back

In the following, let $n$ be a positive integer, all other variables be real (furthermore, $a>1$), all functions be real-valued, and logarithms of negative arguments be undefined.

Let

• $\log^n(x)$ denote the iterated natural logarithm (base $e$), with $x$ in the innermost position,
• $\operatorname{pow}_a^n(x)$ denote the iterated exponentiation (base $a$), with $x$ in the innermost position,

where the superscript ${}^n$ to the right of a function name denotes the number of iterations of the function (not raising its result to a power).

More precisely,

$\hspace{.2in}\begin{cases} \log^1(x) = \ln x \\ \log^{n+1}(x) = \log^n(\ln x) \end{cases}$

$\hspace{.2in}\begin{cases} \operatorname{pow}_a^1(x) = a^x \\ \operatorname{pow}_a^{n+1}(x) = \operatorname{pow}_a^n(a^x) \end{cases}$

For example, $\log^3(x) = \ln \ln \ln x$, and $\operatorname{pow}_a^2(x) = a^{a^x}$.

Now define

$$\boxed{\phantom{\Bigg|}\hspace{0.2in} f_a(x) = \lim\limits_{n\to\infty} \log^n(\operatorname{pow}_a^n(x)) \hspace{0.25in}}$$ In other words, $f_a(x)$ is the limit of the sequence $\{\ln a^x,\ \ln \ln a^{a^x},\ \ln \ln \ln a^{a^{a^x}},\ \dots\}$. Note that the first several elements of the sequence can be simplified, but next ones will end up with a repeated logarithm of a sum with the rest of the power tower sitting inside: $\{x \ln a,\ x \ln a+\ln \ln a,\ \ln\left(a^x \ln a+\ln \ln a\right),\ \ln \ln\left(a^{a^x}\ln a+\ln \ln a\right),\ \dots\}$.

Obviously, $f_e(x)=x$. The behavior of the function for other values of $a$ is more interesting.

Questions:

• Can any non-trivial ($a \ne e$) value of $f_a(x)$ with closed-form arguments be expressed in a closed form in terms of elementary functions, any known special functions, and any known mathematical constants?
• What is the domain of $f_a(1)$? Is $f_a(1)$ an analytic function within its domain?
• What is the domain of $f_2(x)$? Is $f_2(x)$ an analytic function within its domain?
• What it the range of $f_3(x)$?
• What is the value of $\lim\limits_{x \to \infty} \frac{f_2(x)}{x}$, if it exists? What is the asymptotic behavior of $f_2(x)$ as $x \to \infty$?
• What is the value of $\lim\limits_{x \to -\infty} f_3(x)$, if it exists? What is the asymptotic behavior of $f_3(x)$ as $x \to -\infty$?
• What is the Taylor expansion of $f_a(1)$ near $a=e$?
• Why do you say $n \in \Bbb N^+$ when all natural numbers are positive? – Ali Caglayan Aug 16 '13 at 11:25
• @Alizter: Almost all. – Rahul Aug 16 '13 at 12:16
• @RahulNarain It's all within definition so both our statements are incorrect. – Ali Caglayan Aug 16 '13 at 14:33
• – X.C. Aug 16 '13 at 16:14

I fix some $a > e$ and use subscripts for $n$.

Define $f_n(x) = \log^n(\operatorname{pow}_a^n(x))$, so that $f_0(x) = x$, and $f_{n+1}(x) = \log(f_n(a^x))$.

Since $\log$ and $pow_a$ are increasing, it's easy to check that the operator $T : f \mapsto \log \circ f \circ pow_a$ is "increasing" in several ways : it takes increasing functions to increasing functions, and if $f \ge g$ on $\Bbb R^+$, then $T(f) \ge T(g)$ on $\Bbb R$. Hence, since $f_0(x) = x$ is increasing, and $f_1(x) = \log(a^x) \ge \log(e^x) = f_0(x)$ for $x \in \Bbb R^+$, all the $f_n$ are increasing functions of $x$, and $(f_n)$ is an increasing sequence of functions : $f_{n+1}(x) \ge f_n(x)$ except for $x < 0$ and $n = 0$.

Hence, $\forall x \in \Bbb R, f_n(x) \ge f_1(x) = (\log a) x$.

Next, by looking at what $T$ does on affine functions ($L_{A,B}(x) = Ax+B$), we can find an affine upper bound for all the $f_n$ for large enough $x$ : $T(L_{A,B})(x) = \log(Aa^x+B)$, and $\log(Aa^x) = \log A + x \log a = L_{\log a,\log A}(x)$. Finally, a simple comparisons shows that $L_{\log a, \log A}(x) \le T(L_{A,B})(x) \le L_{\log a, \log A}(x) + B/(Aa^x)$

Let $A = \log a > 1$, and let $B = (\log A)/(1-1/A)$.
Then for $x \ge 0$, $L_{\log a, \log A}(x) + B/(Aa^x) \ge L_{\log a, \log A}(x) + B/A = L_{A, B}(x)$.

Now we can show by induction that from $f_1 = L_{A,0}$, we get $L_{A,\log A} \le f_n \le L_{A,B}$ forall $x \ge 0$ and $n \ge 2$.
As for $x \le 0$, we have$L_{A,\log A} \le f_n$ for $n \ge 2$, and $f_n(x) \le L_{A,\log A} + B/(Aa^x)$ for $n \ge 3$.

This gives a uniform bound on all the $f_n$, which proves that $f(x)$ exists forall $x \in \Bbb R$. We get that $f$ is increasing, $f(x) \ge Ax$, $\lim_{x \to +\infty} f(x)/x = A$, and $f$ satisfies the functional equation $f = T(f)$, or also $e^{f(x)} = f(a^x)$.

As for negative $x$, $\lim_{x \to - \infty} f(x) = \lim_{x \to - \infty} \log(f(a^x)) = \lim_{x \to 0} \log(f(0)) \ge \log(f(0))$, with equality if $f$ is continuous at $0$.

• Thanks, mercio! I am awarding the bounty to your answer as the best as of today. But I do not mark it as accepted yet, because I still hope to get more detailed answers to questions I listed above. – Լ.Ƭ. Aug 23 '13 at 19:53