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Consider the function $f(t)=\exp^{(t)}(0)$ where $\exp^{(0)}(0)=0$ and $\exp^{(t+1)}(0)=\exp(\exp^{(t)}(0))$. That is, $$f(t)=\underbrace{\exp(\exp(\dots\exp}_{t}(0)\dots)).$$ Such a function is not short of prior study on this website before, as finding such an analytic interpolation allows for construction of functions like $h(x)$, where $$h(h(x))=\exp(x),$$ by taking $h(x)=f\left(\frac{1}{2}+f^{-1}\left(x\right)\right)$, as described in this answer. There are infinitely many continuous constructions for $f$ over the reals, but AFAIK, an analytic, or even an infinitely smooth example is elusive.

As inspired by this video, my question is, using Operator Calculus, is it possible to solve for a (potentially analytic) solution to $f$?

My idea comes from the fact that, by definition, $$f(x+1)=\exp(f(x))\iff\tau f(x)=\exp(f(x)),$$ where $\tau$ is the Shift Operator. Hence, using Operator Calculus, we find the expression $$e^D f(x) = e^{f(x)}\iff\sum_{n=0}^\infty \frac{D^n f(x)}{n!}=\sum_{n=0}^\infty \frac{(f(x))^n}{n!},$$ where $D=\frac{d}{dx}$. Furthermore, $f$ also has the property that $$\frac{f'(x)}{f(x)}=f'(x-1)\iff \tau\frac{Df(x)}{f(x)}=Df(x).$$ From these facts, is it possible to solve for a solution?

Update: Using the aforementioned method, I was able to deduce the inductive relation: $$D^{n+1}f(1)=\sum_{k=0}^n{n \choose k} (D^{n-k+1}f(0))(D^{k}f(1)).$$ Thus, assuming $D^{m}f(0)=m!c_m$ for $m\in\mathbb{N}$ when $$f(x)=\sum_{n=0}^\infty c_n x^n,$$ we find that $$D^{n+1}f(1)=\sum_{k=0}^n \frac{n!}{k!}(n-k+1)c_{n-k+1}(D^{k}f(1)).$$ With this, given that $$D^{n}f(1)=\sum_{k=n}^{\infty}\frac{k!}{(k-n)!}c_k x^{k-n},$$ I assume we have enough information to solve for the coefficients $(c_k)_{k\in\mathbb{N}}$. However, this seemingly surmounts to solving a countably infinite system of equations, to which I am dumbfounded at where to begin.

Graviton
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  • There certainly does not exist any (even non-continuous) $f\colon\Bbb R\to\Bbb R$ such that $f(t)=\exp(f(t-1))$ holds for all $t\in\Bbb R$. Indeed, $\exp(f(t-1))>0$, hence $f$ is bounded from below. If $s$ is a lower bound for $f$, then so is $e^s$. As $e^s\ge s+1$, there is no greatest lower bound, which is absurd.

    Likewise, consider any $U\subseteq \Bbb R$ such that $U+1\subseteq U$ and $U$ contains some closed subinterval of length $1$. Then one readily shows that if $f\colon U\to\Bbb R$ is continuous such that $f(x+1)=\exp(f(x))$ for all $x\in U$, then $U$ is bounded from below.

     – Hagen von Eitzen Sep 24 '22 at 09:03
  • @HagenvonEitzen My goal of finding $f(t)$ "over $\mathbb{R}$" does not necessarily mean I wish for $f(t)$ to be defined for all $t\in\mathbb{R}$, just at least over some continuum, which is certainly possible. This answer even explicitly constructs a continuously differentiable $f:(-1,\infty)\to\mathbb{R}$. I am looking for something similar, just $C^\infty$. So far I have found a $C^4$ construction, and numerical methods for any $C^n$ construction, but no methods for $C^\infty$. – Graviton Sep 24 '22 at 09:13

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