I was working on finding a series expression for a function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(x^y) = f(x)^{f(y)}$ along the way for construction of such a function I came across a limit that I don't seem to have any tools to evaluate:


If we use the usual notation for tetration as $^{n}x$ then I am interested in evaluating


$$ \lim\_{\epsilon \rightarrow 1} \frac{^{\epsilon}2-2}{\epsilon - 1}$$


How can such a limit be "made sense of" and then evaluated? (I realize a closed form is probably out of question but hopefully a series representation might still be possible).


I did some digging around and it seems like [Hellmuth Kneser](https://en.wikipedia.org/wiki/Hellmuth_Kneser)'s tetration is the commonly accepted answer for tetration with fractional real values. At least as per wikipedia.


It also seems there are some formulas for tetration such as the one given [here](https://mathoverflow.net/questions/259278/an-explicit-series-representation-for-the-analytic-tetration-with-complex-height) by what appears to be James D Nixon.


I could try to evaluate the limit in terms of one of those formulae and see if anything comes out but before I went through that I wanted to see if some hyperoperator experts might be able to suggest something better.



Finding closed forms/related constants to a limit involving tetration
There is an implementation around (Pari/GP; in the [tetration-forum](https://math.eretrandre.org/tetrationforum/showthread.php?tid=1017)) which claims to have a Kneser-implementation. It is a bit difficult to handle, so I'll show here a simpler version (essentially polynomial) of a tetration-function which seems to approximate that Kneser function when the polynomial's order is increased.  

With a polynomial of order 32 I got for your problem a value of about $a=1.2329216$


***Added***: *I got the mentioned Kneser-implementation work; it gives a taylorseries in $h$ (where the spurious imaginary parts of the coefficients are deleted) as*



```
 f(h)=  2.0 
    + 1.2329216 21568706952691849*h
    + 0.3920521 821375754199047770*h^2 
    + 0.2175339 103974922699983933*h^3 
    + 0.087772 79346256013703882103*h^4
    + 0.0409164 3821897543029382908*h^5 + O(h^6)

```

*where the coefficient at $h^1$ is the first derivative with respect to $h$ which I've approximated by my method.  

To compare, the method described below gives (when the same parameters are inserted)*



```
q&d(h) = 2 
    + 1.2329216 07*h
    + 0.3920521 614*h^2
    + 0.2175339 098*h^3
    + 0.087772 80375*h^4
    + 0.0409164 4709*h^5 + O(h^6)

```

*[end addition]*




---


This is how I got my approximate solution:


* get `pc` as the vector of first 32 coefficients of the power series for $f(x)=2^x$
* From pc generate the (truncated!) Carlemanmatrix `F` for the function $f(x)$.  

The vector `pc` and thus `F` should have infinite size, but of course we must cut it to something we can handle and which gives not too much crap, at least in a certain range of data.
* By the principle of (infinite) Carlemanmatrixes, powers of `F` give iterates of $f(x)$. Fractional powers give fractional iterates- of course, practically this is always only approximated.
* Fractional powers of `F` can be achieved, when we diagonalize `F = M * D * W`(where I write `W`for `M^-1` for simplicity) and compute $F^h = M\*D^h\*W$


The approximation runs now



```
 z_0=1;
   m_z0 = vector(32,j, sum(k=0,31, z_0^k * M[1+k,j] )   )     
 h=1.1
   z_h=sum(k=0,31, mz_0[1+k]  *  D[1+k]^h  *  W[1+k,2])  

```

$z\_h$ is now (approximately) the $h$'th iterate from $z\_0$ to $z\_h$.


Letting $h$ go downwards to $h=1$ it seems that we approximate the value I've given above. And because I've tested earlier this method against the (claimed) Kneser's method and have empirically found improving convergence to that results when increasing the size of the matrix (which means increased order of the underlying polynomial) I think, that we are near the true Kneser's solution here.


*Remark: the matrices `M` and `W` look ugly in the view of giving coefficients for a polynomials (ideally we should have power series with infinitely many coefficients) and should surely be evaluated only for $z\_0$ and $h$ near zero. But tests with $z\_0=1$ and various heights $h$ gave meaningful results such that $z\_{h\_1+h\_2} = (z\_{h1})\_{h\_2}$ .*


***{update***  

A much better way to compute the derivative of *d/dh F^h* is to use the derivative of the diagonalmatrix $D^h$ with repect to *h*. The code



```
 z_0=1;
    m_z0=vector(32,j,   sum(k=0,31, z_0^k * M[1+k,j] )  )     
 h=1 
    z_h=sum(k=0,31, m_z0[1+k] * log(D[1+k]) * D[1+k]^h * W[1+k,2])  

```

$\hspace{120 px}$ represents the derivative of `F` with respect to $h$ and gives immediately the first derivative with respect to $h$ at $z\_0=1$ and $h=1$ ***}***


This Q&D-method can easily be reconstructed using Pari/GP. Note that the diagonalization needs large value for the internal decimal precision, I just used *prec=800* (dec digits) for the matrixsize *32x32*.


Short protocol ("a" means your sought value $a={z\_h-2 \over h-1}$):



```
  h                   z_h          a
1.0001,          2.0001233, 1.2329608
1.000001,        2.0000012, 1.2329220
1.000000001,     2.0000000, 1.2329216
1.0000000000001, 2.0000000, 1.2329216

```



---

 
I did not find some closed form for this via google, W|A and inverse symbolic calculator, so possibly the same coefficient, just for the base $e$ instead, might be better known: 
The computation for the base $b=e$ instead of $b=2$ gives the following approximation:



```
1.1000000, 3.0407711, 3.2248931
1.0100000, 2.7481969, 2.9915037
1.0010000, 2.7212519, 2.9700850
1.0001000, 2.7185786, 2.9679609
1.0000100, 2.7183115, 2.9677487
1.0000010, 2.7182848, 2.9677274
1.0000001, 2.7182821, 2.9677253
1.0000000, 2.7182819, 2.9677251
1.0000000, 2.7182818, 2.9677251
1.0000000, 2.7182818, 2.9677251

```

*But no success for finding a closed form representation so far.*




---


***Added:*** With that idea of a derivative with respect to the height-parameter the idea of constructing a powerseries from the sequence of higher derivatives comes to mind. With the adapting of my derivative-formula (see par. "update") I construct a powerseries in $h$ which gives the $h$'th iterates starting from $z\_0=0$. This are the first *32* coefficients:
$$ g(h)=
\small \begin{array} {rl}
 1 & \\ 
 +0.8893649182531790 & \*h \\ 
 +0.008676688896772925 & \*h^{2} \\ 
 +0.09523868310780171 & \*h^{3} \\ 
 -0.005752348511600163 & \*h^{4} \\ 
 +0.01296662374986303 & \*h^{5} \\ 
 -0.002196108907293045 & \*h^{6} \\ 
 +0.001996803031014021 & \*h^{7} \\ 
 -0.0005633994435825211 & \*h^{8} \\ 
 +0.0003482750168520301 & \*h^{9} \\ 
 -0.0001285546189885988 & \*h^{10} \\ 
 +0.00006709638266032520 & \*h^{11} \\ 
 -0.00002830779288402342 & \*h^{12} \\ 
 +0.00001380563705669089 & \*h^{13} \\ 
 -0.000006205175034368587 & \*h^{14} \\ 
 +0.000002957461339961192 & \*h^{15} \\ 
 -0.000001369764406300663 & \*h^{16} \\ 
 +0.0000006496675894767698 & \*h^{17} \\ 
 -0.0000003055056195438239 & \*h^{18} \\ 
 +0.0000001451337771274319 & \*h^{19} \\ 
 -0.00000006884715898369447 & \*h^{20} \\ 
 +0.00000003282131706893498 & \*h^{21} \\ 
 -0.00000001565967274219413 & \*h^{22} \\ 
 +0.000000007493152321891950 & \*h^{23} \\ 
 -0.000000003590500482888162 & \*h^{24} \\ 
 +0.000000001723905030908289 & \*h^{25} \\ 
 -0.0000000008288804738313791 & \*h^{26} \\ 
 +0.0000000003991564395933366 & \*h^{27} \\ 
 -1.924678487785590E-10 & \*h^{28} \\ 
 +9.292387572960278E-11 & \*h^{29} \\ 
 -4.491503965074198E-11 & \*h^{30} \\ 
 +2.173332915755159E-11 & \*h^{31}
 \end{array} $$


This is then good at least for $h= 0 \ldots 1$ and should reproduce the Kneser-solution by $z\_h=\exp\_2^{\circ h}(0)=g(h-1)$ .  

*(This part is just current experimenting and might have errors)*


