How to solve this tetration equation $\;^n 2 = \;^2 n $?

Question

How would one find all real solutions to the following equation:

$\qquad$ $n^n = 2^{2^{2^{2^{\dots^2}}}} $(where the number of $2$s is equal to $n$)

generalizing to $n$ being a real value. In tetration-notation this is

$\qquad $ find a solution to $\displaystyle \;^2 n = \; ^n 2$ for real $n \ne 2$.

I know one solution is $n = 2$, but I wonder if any other solutions exist.

Edit: Could there be any negative number solutions to this equation?

Answer 1

With a simple method, using the eigendecomposition of the $\small 16 \times 16, 24 \times 24,32 \times 32,48 \times 48,64 \times 64$ - sized truncated Carlemanmatrix for $f(x)=2^x$ , I get

 print(n)
 %191 = 3.41417132461  \\ for size 16x16
 %199 = 3.41417539283  \\ for size 24x24
 %206 = 3.41417594201  \\ for size 32x32
 %333 = 3.41417608600  \\ for size 48x48
 %387 = 3.41417609800  \\ for size 64x64  
 print(n-sqrt(2))
 %192 = 1.99995776224  \\ for size 16x16
 %200 = 1.99996183046  \\ for size 24x24
 %207 = 1.99996237964  \\ for size 32x32
 %388 = 1.99996253563  \\ for size 64x64

Possibly this indicates that $n=2+\sqrt {2}$ would be the extrapolated approximation when size of the matrix increases to infinity, but after the tinyness of the change using that increasing sizes, I think that this idea should be rather unlikely (I'll try the Kneser-solution later)
The values for the comparisions are

   n^n           \\ = 66.1807801442
   tet_poly(2,n) \\ = 66.1807801442

So this is one approximation using a very rough method, which I call "polynomial tetration" because using fractional powers by diagonalization of a finite Carleman matrix this involves to find roots of (finite) polynomials.

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Alternatives known to me:

• In the tetration-forum one can find a procedure which extends Hellmuth Kneser's method, which was developed for basis $e\approx 2.718$ to other bases, which would give a very similar result, (search in the tetrationforum for "Sheldon Levenstein" and "fatou.gp" for a Pari/GP-procedure) By some experiments it seemed that the polynomial method converges to that so-generalized Kneser-method when the polynomial method could use matrices of infinite size, so the above result is possibly much meaningful/appropriate.

• One different, fairly well known method, is based on E. Schröder's method, finding the "Schröder-function" $\sigma()$ first and do the fractional part of the tetratrion based on this function. Unfortunately $\sigma()$ has then complex coefficients because the tetration-base $2$ has only complex fixpoints. I didn't try such a solution at the moment.

• Also I could not test the - (self-claimed to be) best- proposal for the tetration to fractional heights developed by D.Kousnetzov & H.Trappmann (the latter is the founder of the "tetration-forum") . I didn't see a peer-reviewed edition so far, maybe meanwhile it exists. At least it can be downloaded from arXiv, but I were unable to (re-) implement the procedure to do own computations. See the link to the Kousnetzov/Trappmann -paper in the reference-section in the wikipedia-entry for tetration. Also see the longer and more detailed entry in the citizendium

Papers on arXiv

• R. Aldrovandi explaining and using Carleman matrices for tetration

• D.Kouznetsov,H.Trappmann explaining their tetration-proposal in the case of two complex fixpoints (which would fit here with the base $b=2$)

A draft paper initially for the discussion in the tetration-forum on my own webspace:

• G.Helms how different a couple of methods look in a graphical comparision (I used for the example base $b=4$ but I think it gives an idea for the caveats if one decides for one of the methods)

Answer 2

I wanted to add some graphs to Gottfried's solution. First definitions; $\text{sexp}_2(z)= \;^z 2$ which is extended to the complex plane by Kneser's solution. I wrote a program to calculate the slog; which is the inverse of sexp and has some nice uniqueness properties. The fatou.gp program works for a wide range of real and complex sexp bases and is written in pari-gp and is available on this site http://math.eretrandre.org/. Instead of graphing $\text{sexp}_2(x)\;$and $\;x^x$, I will take the $\log_2(x)$ of both equations which works when both are positive. So I am graphing $$\text{sexp}_2(x-1)\;\;\text{vs}\;\;\log_2(x^2)=\frac{x\ln(x)}{\ln(2)}$$

Here is a graph for sexp base 2 which shows a solution at x=0, x=2, and at x=3.4141760984020147407016

What about other bases? Here is a graph for sexp base=2.1150455841, which has a parabolic crossing near 2.5360 so there are only two solutions.

For larger bases, there is only one solution. Here is sexp base e which only has a solution at x=0.