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.


  • $\log^n(x)$ 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.


  • 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 $a=e$?

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 $0$.

  • $\begingroup$ 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. $\endgroup$ – Լ.Ƭ. Aug 23 '13 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.