# Attempt on fractional tetration

To start off, I was looking at the following ingeniously made form of the Gamma function:

$$\Gamma(z+1)=\lim_{n\to\infty}\frac{n!(n+1)^z}{(1+z)(2+z)\cdots(n+z)}$$

which lies on the back of

$$1=\lim_{n\to\infty}\frac{n!(n+1)^z}{(n+z)!}$$

for all integer $z$. One them multiplies through by $z!$ and use the recursive formula for the factorial to reach the above formula.

In the same light, I was wondering if a limit definition of tetration was possible. Consider the following:

$$a^{a^{a^{\dots}}}=\underbrace{a\uparrow a\uparrow a\uparrow \dots\uparrow}_nb=a\uparrow_nb$$

And then consider the following:

$$a\uparrow_nf(n)$$

Particularly, I was wondering if there was a continuous function $f:\mathbb R\to\mathbb C$ such that

$$\lim_{n\to\infty}a\uparrow_nf(n)=c$$

for some constant $c$. From there, one could imagine something like...

\begin{align}a\uparrow_{1/2}c&=a\uparrow_{1/2}\lim_{n\to\infty}a\uparrow_nf(n)\\&=\lim_{n\to\infty}a\uparrow_{n+1/2}f(n)\\&=\lim_{n\to\infty}a\uparrow_nf(n-\frac12)\end{align}

Does this seem reasonable? Does anyone know much about if this is a good path towards defining fractional ordered tetration?

## 1 Answer

It could work. I haven't seen anyone try this method though. It's probably because of the fact that the function f(n) is hard to define in general. For example if a=2 (tetration base 2), then, the limit of 𝑎↑𝑛𝑏 diverges quickly even when you extend b to complex numbers,however, there are fixed points but I think they are all repelling fixed points. Therefore, for most bases, the function has to converge quickly enough for the limit to be a constant. The difficult question to answer is how quickly? Also how do you calculate the fixed points if their repelling points, as you can't just iterate the function a^x.

Edit: The function would have to either quickly approach one of the fixed points so that the limit is the fixed point or approach one of the fixed point slower so that the limit can be a different constant.