# Tetration of non-integers: is there something wrong with this approach?

I'm trying to figure out a formula for tetration that will work for non-integer heights.

I know the usual recurrence relation for tetration ($$x \in \mathbb{R}, \text{ }n \in \mathbb{N})$$:

$${^{n}x} = \begin{cases} 1 &\text{if }n=0 \\ \\ x^{\left(^{(n-1)}x\right)} &\text{if }n>0 \end{cases}$$

I also know that $$x^y=e^{y \ln x}$$ for positive $$x$$.

I combined these two and formed this recurrence: $${^y}x = f(x,y) = \begin{cases} e^{y \ln x} & \text{if }0 \lt y \le 1 \\ \\ e^{f(x,\text{ }y-1) \ln x} & \text{if }1 \lt y \end{cases}$$

Playing around with this in Maxima, I got correct answers for integer $$y$$, and reasonable-looking answers for non-integers. Yet I have read numerous sources stating that a general formula for tetration is very difficult.

So, my question: have I a correct solution for a limited domain, or am I off in the weeds and it just happens to work for integers?

Thank you.

• it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence – Masacroso Dec 5 '18 at 2:51

The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $$^yx$$ for $$y\in [0,1)$$ and extend it to work for all positive $$y$$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $$^yx$$ to be anything I want in that interval and there's no clear reason to take $$^yx=x^y$$ for $$0 as you do - it makes the function continuous, but I could just as easily take $$^yx=y(x-1)+1$$ to get a continuous answer that will make results that look nice for non-integers.
Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $$n$$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $$x$$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.