Extending Tetration to The Reals
Idea
How do we define
tetration, ${}^hb$, for $h\in\mathbb{R}$?
Backround
According to Wikipedia,
"At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex [heights]."
There is, however, a significant amount of research going into this question. See, for instance, the
tetrationforum.
Tetrations of e
Consider a function $f(x)$ with the properties that $f(0)=0$ and $f(x+1)=\exp(f(x))$. Notice that such a function could serve as a definition for $\exp^{x}0$ (where $g^{n}=\underbrace{g\circ g \circ \cdots \circ g}_{n}$ represents
iteration). By doing so, this function allows us to do several novel things.
Motivation
For instance, we may then define
$$\exp^x(a)=f(x+f^{-1}(a)).$$
This would, furthermore, allow us to define a continuum of
hyperoperations. Define, inductively,
$$a\bullet_0 b=a+b\text{ and }a\bullet_{n+1}b=\exp(\ln(a)\bullet_n\ln(b)).$$
Thus, for $n=0$ and $n=1$ we recover addition and multiplication. However for $n=2$, we get not our usual exponentiation, but the
"product-log" which is an alternative generalization of multiplication that is both
commutative and
assosiative. In fact, all hyperoperations $\bullet_n$ have these properties, by definition, via the fact that $\exp$ serves as a
homomorphism for them.
Hence, the existence of such an $f(x)$ would allow for defining
$$a \bullet_x b = \exp^x(\exp^{-x}a+\exp^{-x}b)$$
for any $x\in\mathbb{R}$. One motivation of this is for the generalizing of
asymptotics. That is to say, consider two functions $\alpha(t)$ and $\beta(t)$, we say $$\alpha\sim_{x}\beta\iff\lim_{t\to\infty}\alpha(t) \bullet_x-_x\beta(t)=e_x,$$
where $-_x (\cdot)$ and $e_x$ are the inverse and identity element of $\bullet_x$, respectively (I.e; $b\bullet_x-_x b = e_x$). This is inherently a measurement of how closely two functions approximate eachother. For instance, $\sqrt{1+t^2}\sim_0 t$ means that their differences converge to $e_0=0$, since
$$\lim_{t\to\infty}\sqrt{1+t^2}\circ_0-_0 t=\sqrt{1+t^2}-t=0=e_0.$$
Meanwhile,
Striling's approximation is of order $\sim_1$, meaning their ratios converge to $e_1=1$. We may then ask about the minimal $x$ s.t. $\alpha\sim_x\beta$. For example, it's clearly established that the minimal $x$ s.t.
$$n!\sim_x\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
lies within $x\in(0,1]$, but what is the value exactly?
Non-uniqueness problem
Actually, it is quite easy to define such an $f(x)$. Simply take any $\phi(x)$ s.t. $\phi(0)=0$, and define
$$f(x)=\begin{cases}
\ln(f(x+1)) &: x \lt 0,\\
\phi(x) &: 0 \le x \lt 1,\\
\exp(f(x-1)) &: x \ge 1.
\end{cases}$$
Even with the restriction that $f(x)$ is
continuous, there are infinite solutions. Ideally, we'd like an
analytic solution, which is equivalent to asserting that
$$\frac{d^n}{dx^n}\phi(x)|_{x=1}=\frac{d^n}{dx^n}\exp\phi(x)|_{x=1}$$
for all $n\in\mathbb{N}$.
We have found an explicit quartic solution for $\phi(x)$ where the above holds up to $n=3$.
$$\phi(x)=(6-2\sqrt{6})x+\frac{1}{5}\left((-36+14\sqrt{6})x^2+(-28+12\sqrt{6})x^3+(39-16\sqrt{6})x^4\right),$$
which, interestingly, has coefficients which are all elements of
$\mathbb{Q}(\sqrt{6})$.
$$\text{Graph 1. Interactive graph of $f(x)$ given our $n=3$ approximation for $\phi(x)$.}$$
Numerical methods have also provided us with an approximate solution up to order $n=18$.
$$\text{Graph 2. Interactive graph of $f(x)$ given our $n=18$ approximation for $\phi(x)$.}$$
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