This started off with me goofing off on Twitter and quickly led to this question to which I didn't know the answer.

Let $T_2(x) = x^x$, $T_3(x) = x^{x{^x}}$, $T_4(x) = x^{x^{x^x}}$, and so forth.

Is there a $C^{\infty}$ elementary function $f(x)$ (it can be piecewise-defined) with $\displaystyle \lim\limits_{x \to \infty} \frac{f(x)}{T_k(x)} = \infty$ for all $k$? If so, what is a (possibly piecewise-defined) formula for such an $f$?

  • $\begingroup$ (I tried $f(x) = T_k(x)$ on [k+(1/4), k+(3/4)] and linearly interpolating between endpoints, but I couldn't find a way to make that both smooth (convoluting with Gaussians would do that, but it wouldn't be elementary then) and elementary (it's already elementary but not smooth.) $\endgroup$ Jan 4 '19 at 9:03
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    $\begingroup$ Possibly related: math.stackexchange.com/questions/1892870/… $\endgroup$
    – Martin R
    Jan 4 '19 at 9:03
  • $\begingroup$ (The Smooth Approximation Theorem from differential topology also does not guarantee an elementary solution.) $\endgroup$ Jan 4 '19 at 9:04
  • $\begingroup$ @Martin R. Mmm, the second answer in that question in close, but doesn't allow piecewise-defined functions; I already found an elementary piecewise-defined function that contradicts the second answer. $\endgroup$ Jan 4 '19 at 9:09
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    $\begingroup$ I am no expert on this topic at all. My guess would be that a similar approach as in this comment might work: Define $f$ on the positive integers as $f(k) = T_k(k)$, and smoothly interpolate between the integers. $\endgroup$
    – Martin R
    Jan 4 '19 at 9:22


There are no such elementary functions, the below constructions are non-elementary but still interesting.

For starters, one can construct a simple $C^k$ function for any $k\in\mathbb N$ simply by integrating as follows:

$$\int_1^x\int_1^{x_1}\dots\int_1^{x_k}T_{\lfloor x_{k+1}\rfloor}(\lfloor x_{k+1}\rfloor)~\mathrm dx_{k+1}~\mathrm dx_k\dots\mathrm dx_2~\mathrm dx_1$$

which is trivially $k$ times continuously differentiable and grows faster than $T_m(x)$ for all $m\in\mathbb N$.

For $C^\infty$ functions, I give special mentions to Wojowu a.k.a. LittlePeng9 for fast growing analytic functions. I will give a slightly simplified construction here.

Let $f:\mathbb C\mapsto\mathbb C$ satisfy the following properties:

  1. $f$ is entire.

  2. $|f(z)|\le z$ for all $|z|<1$.

Consider the following function:

$$g(z):=\sum_{n=1}^\infty f^n\left(\frac z{2^n}\right)$$

where $f^n$ denotes function iteration i.e. $f^2(x)=f(f(x)),f^3(x)=f(f(f(x))),$ etc.

For $|z|<1$ we have


it follows that $|g(z)|\le|z|$ when $|z|<1$ by the geometric series.

Likewise, since $\lim_{n\to\infty}z/2^n=0$ for all $z$, from the Weierstrass M-test, it follows that $\sum_{n=1}^\infty f^n\left(\frac z{2^n}\right)$ converges uniformly everywhere and hence is entire.

Now, on the other hand, if $f$ is increasing on $\mathbb R^+$ (and hence positive on $\mathbb R^+$), then for all $N\in\mathbb N$ and $x\in\mathbb R^+$, we have $g(x)\ge f^N(x/2^N)$.

And since $g$ satisfies all the conditions that $f$ required, this can be repeatedly applied to generate increasingly faster growing analytic functions.

Take, for example $f(z)=\frac12(e^z-1)$, which satisfies all of the requirements. The corresponding $g$ defined above hence grows faster than $f^N(x/2^N)$ for all $N$, and hence faster than your functions, namely since:


For justification of the last limit, one can easily see that $f(x)$ is eventually greater than $2^x$, and that $x<2^x\le x^x\le(2^x)^x=2^{x^2}\le2^{2^x}\le x^{x^x}\le\dots$


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