Possible Tetration extension for a specific interval (part 1) My friend and I have been developing an extension of tetration for non integer values.
We managed to get definitions of extensions for : ${^r}x$.

*

*$x$>0.

*$r$ not equal to any whole number below -1.
Here, I will only present the extention for values of x belonging to the interval $I$ = $]1,e^{1/e}[$ and $I' = [e^{-e}, 1[$
So, we know that ${^\infty}x$ here converges to a value we will call $\tau$ (for example, ${^\infty}({\sqrt2}) = 2$). Thus, we can notice that for any value of x belonging in the interval $I$,
$$\lim_{r\rightarrow{+}\infty}(\frac{{^{r+1}}x-\tau}{{^r}x-\tau})=ln(\tau)$$
$$\text{Let's call "}\ln(\tau)\text{"} \lambda$$
I haven't demonstrated that yet, but I think it wouldn't be too hard to do.
$$({^{23}}\sqrt2-2)/({^{22}}\sqrt2-2)=0.6931203376$$
(which is pretty close to ln(2) which is 0.6931471806)
This was just an example, and you can try for different values of x, it always works.
So, we can drive the formula
$$\lim_{r\rightarrow{+}\infty}({^{r+1}}x)=(({^r}x-\tau)*\lambda) +\tau$$
So we can also deduce :
$$\lim_{r\rightarrow{+}\infty}({^{r+k}}x)=(({^r}x-\tau)*\lambda^k) +\tau$$
And since tetration can be expressed recursively (meaning taking n times the $\log_x({^r}x)={^{r-n}}x$, and that we defined ${^{r+k}}x$ for any value of k (including non integer values), we can drive our final formula :
$${^r}x =(\lim_{n\rightarrow{+}\infty}(\log_{x} ^{n-k}((({^n}x-\tau)*\lambda^p) +\tau))$$
Where $k+p=r$, $k$ being an integer and $|p|<1$, and $\log_x^n$ means taking n times the log (or iterating n times the log).
With this formula, we can define ${^r}x$ for values of r lower than -2, BUT NOT whole negatives below or equal to minus 2. For example, $r=-1$ is ok, same for $r=-3.64, r=-15.9$, but $r=-2,r=-3,r=-4$, ... is forbidden.
The formula looks hard but is really easy to understand, for example, to calculate ${^{0.5}}(\sqrt2)$, we take a big value of ${^n}(\sqrt2)$ (like $n=23$), we substract $2$, multiply by $\ln(2)^{0.5}$, we add $2$, and take the $\log_{\sqrt2}$ $n$ times ($23$ times here). And we get a good approximation.
With this formula, I have the following results :
${^{-0.5}}({\sqrt2})≈0.6290566121$.
${^{1.3}}({\sqrt2})≈1.49334127$.
${^{-{\sqrt2}}}({\sqrt2})≈-1$.
${^{-{2.5}}}({\sqrt2})≈0.8390270267+9.064720284i$
etc...
Here is the graph of ${^{x}}({\sqrt2})$ from -2 to 4.

(It seems like this function is symmetrical, meaning ${^x}(\sqrt2)=a\leftrightarrow{^{-a}}(\sqrt2)=-x$)
ASLO, this formula can be applied for numbers x belonging to the internal $I' = [e^{-e}, 1[$. The main difference here, is that $\lambda$ is negative, implying that real non integer tetration gives complex results.
For example, here are some values of ${^x}0.5$ :
${^{0.5}}0.5 ≈ 0.6297281585+0.2178921319i$.
${^{-1.9}}0.5 ≈ 2.487456488+1.718168223i$.
etc...
Here is the graph of ${^x}0.5~$ from -2 to 5. (red is the real part, blue is the imaginary part)

For ${^r}1$, I think it makes sense to define it as being equal to $1$ for at least all r belonging to $]-1,+\infty[$.
Anyway, tell me what you think of this. Do you think this extension is correct? Plausible? Or bad?
EDIT : apparently my conjecture about ${^r}(\sqrt2)$ being symmetrical is false.
 A: Good idea! As far as I see, this might be seen as a "light" version of the method which has been invented by Ernst Schröder.
To make this a bit better visible let me simplify your notation and then compare your with Schröder's formula.
The value $\,^\infty x$ is a fixpoint; let's denote it as $\tau$, then let us denote its logarithm with $\lambda$.
Then your formula $\lim_{r\rightarrow{+}\infty}({^{r+k}}x)=(({^r}x-V_r)*\ln(V_r)^k) +V_r$ can be written as
$$ \lim_{r\rightarrow{+}\infty}{^{r+k}}x = (({^r}x-\tau) \cdot \lambda^k + \tau \tag 1
$$
Schröder has a version of this with some function
$$\sigma(z)=1 \cdot z+b_2 z^2 + b_3 z^3 +... \tag {2a}$$ and the inverse function $$\sigma^{\circ-1}(z)=1 \cdot z+c_2 z^2 + c_3 z^3 +\cdots \qquad \text{ . }\tag {2b}$$
First he computes $$ w=\sigma(x-\tau) \tag {3a}$$ -which might roughly be seen as your initial expression $ {^r}x-\tau$  and inserts this into $\sigma^{\circ -1}()$  as
$$ v = \sigma^{\circ -1}(w \cdot \lambda^k) \tag {3b}$$
which is more explicite
$$ v= (w \lambda^k) + c_2 (w \lambda^k)^2 + c_3 (w \lambda^k)^3 + ... \tag {3c}$$
and gets then
$$ {^k}x = v+\tau \tag 4 $$
Because the powerseries $\sigma()$ and $\sigma^{\circ -1}()$ have a small range of convergence it is better, like you do it, to use $r$'th iterates towards the fixpoint
$$ w_r=\sigma( {^r}x-\tau) \\ 
  v_r=\sigma^{\circ -1}( w_r \cdot \lambda^k ) \tag {5a} $$
and then
$$
 {^k}x= {^{-r}} \left( v_r  + \tau \right) \tag {5b} 
$$
It becomes visible now, how your idea gives an approximation to the Schröder-mechanism, because by (eq 3c) one can see that your version just uses the linear term. This can be defended: because if we use $w_r$ with large $r$ then $w_r$ is small, and the second and all further terms vanish when $w_r$ vanishes.

There is one important note to be appended with the use of fixpoints and shifting to fixpoints: for instance base $\sqrt 2$ has two fixpoints, not only $\tau_1=2$ but as well $\tau_2=4$ (and moreover infinitely many complex fixpoints). Which fixpoint do you use? Schröder requires you use the "attracting" fixpoint (if there is one), but the "repelling" one can be operated if we do not use the exponentiation $ {^k}x$ but the logarithmizing
$ {^{-k}}x$ and insert $\tau_2 (=4)$ in the formulae.
This looks totally compatible - but it isn't! We get different values for the fractional iterates.
I think it is important to be aware of this incompatibility over fixpoints.

One more general remark: I think it is always meaningful to recall that any found (or let it as well be "any intended") method plays in a garden of various methods and there is not yet consensus, which is "the best". To get a feeling for this I recommend to look at a small treatize where I compare 5 methods visually, and show the differences in the complex plane. The method you found here is very close to that which I invented there, only that this simple method was based on linearization of the log-polar expression of the iterates ${^r}x-\tau$ near the fixpoint: its there with the label "poor man's Schröder implementation". Have fun!
