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.

  • $\begingroup$ 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 $\endgroup$ – 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 $0<y\leq 1$ 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.

  • $\begingroup$ ...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. $\endgroup$ – Milo Brandt Dec 5 '18 at 3:02
  • $\begingroup$ 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. $\endgroup$ – user3412516 Dec 10 '18 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.