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 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 $0<y\leq 1$ 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.

Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.

  • ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable. – Milo Brandt Dec 5 at 3:02
  • I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers. – user3412516 Dec 10 at 19:32

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.

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