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.
 A: 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.
A: 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. 
