# How to evaluate fractional tetrations?

Recently I've come across 'tetration' in my studies of math, and I've become intrigued how they can be evaluated when the "tetration number" is not whole. For those who do not know, tetrations are the next in sequence of iteration functions. (The first three being addition, multiplication, and exponentiation, while the proceeding iteration function is pentation)

As an example, 2 with a tetration number of 2 is equal to $$2^2$$ 3 with a tetration number of 3 is equal to $$3^{3^3}$$ and so forth.

My question is simply, or maybe not so simply, what is the value of a number "raised" to a fractional tetration number. What would the value of 3 with a tetration number of 4/3 be?

Thanks for anyone's insight

Ah yes, a fave topic of mine. Basically, there is no universally-agreed on way to do this. The problem is, that, in general, there isn't a unique way to interpolate the values of tetration at integer "height" (which is what the "number of exponents in the 'tower'" may be called). So in theory, you could define it to be anything.

In the case of exponentiation, one has the useful identity $a^{n + m} = a^n a^m$, which enables for a "natural" extension to non-integer values of the exponent. Namely, you can see, for example, that $a^1 = a^{1/2 + 1/2} = (a^{1/2})^2$, from which we can say that we need to define $a^{1/2} = \sqrt{a}$ if we want that identity to hold in the extended exponentiation. No such identities exist for tetration.

You may also want to look at Qiaochu Yuan's answer here, where he explores some of this from a viewpoint of higher math:

https://math.stackexchange.com/a/56710/11172

One could, perhaps, compare this problem to the question of the interpolation of factorial $n!$ to non-integer values of $n$. There is, in general, no simple identity that provides a natural extension for this, either. But, when an extension is desired, the usual choice is to use what is called the "Gamma function", defined by

$$\Gamma(x) = \int_{0}^{\infty} e^{-t} t^{x-1} dt$$.

Then, you can extend $n!$ to non-integer $x$ by $x! = \Gamma(x+1)$. However, usually one does not use $x!$ for non-integer factorials, but rather the Gamma function notation.

One can give a uniqueness theorem involving soem simple analytical conditions; it is called the Bohr-Mullerup theorem. In addition, the gamma function has various nice number-theoretic and analytic properties, and turns up in a number of different areas of math.

But in the case of tetration, there are no nice integral representations known. Henryk Trappmann and some others recently proved a theorem that gives a simple uniqueness criterion for the inverse of tetration (with respect to the "height") here, presuming extension not just to the real, but the complex numbers:

http://www.ils.uec.ac.jp/~dima/PAPERS/2009uniabel.pdf

The solution that satisfies the condition is one that was developed by Hellmuth Kneser in the 1940s. I call it "Kneser's tetrational function" or simply "Kneser's function". It defies simple description.

On this site:

http://math.eretrandre.org/tetrationforum/index.php

an algorithm was posted to compute the Kneser solution (though I'm not sure if it's been proven) for various bases of tetration. Using this solution, the answer to your question would be

$$^{4/3} 3_\mathrm{Kneser} = 4.834730793026332...$$

Other interpolations for tetration have been proposed, some of which give different results. But this is the only one that seems to satisfy "nice" properties like analyticity and has a simple uniqueness theorem via its inverse. Yet as I said in the beginning, I don't believe that it's universally agreed by the general mathematical community that this is "the" answer.

• In that article I like the graph that goes outside of it's grid. Seems appropriate for hyperexponentiation. – Lucas Nov 17 '12 at 1:50
• The kneser.gp pari-gp code isn't proven to converge, but empirically, it converges. If it converges, it can be proven that the solution is the same as Kneser's. I wrote the code :). The interesting thing about Kneser's solution, is that as $\Im(z)$ increases, sexp(z) converges towards the $\psi^{-1}(\lambda^z)$ abel function developed from the fixed point via the inverse Schroder series for $\text{base}^z$. Here $\lambda=\log(\text{base})\times\text{fixedpoint}$. As $\Im(z)$ decreases, it converges towards the complex conjugate fixed point $\psi^{-1}(\lambda^z)$ function. – Sheldon L Nov 25 '12 at 17:14

Here is a q&d - implementation in Pari/GP to get some intuition about what is going on at all. The "Kneser-Method" is much more involved, but it seems there is a good possibility, that the simple method below (I call it the "polynomial method") is asymptotic to/approximates the Kneser-method when the size of the matrices gets increased without bounds.

n=32                            \\ Size for matrix and power series. If n=48 ...
default(realprecision,800)      \\  ... choose realprecision at least 2000!
\\  For n=64 Pari/GP needs much more digits
\\       and time
default(format,"g0.12")         \\ display only 12 significant digits

[b =3, bl=log(b)]               \\ we choose exponentiation/tetration to base bb=3

Bb = matrix(n,n,r,c,(bl*(c-1))^(r-1)/(r-1)!) ;   \\ create the Carleman-matrix
\\ for iterable z1 = 3^z0

tmpM=mateigen(Bb);                               \\ invoke diagonalization to
tmpW = tmpM^-1;                                  \\  allow fractional powers of
tmpD=diag(tmpW*Bb*tmpM);                         \\  the matrix Bb

\\ ==============================================================================
h=4/3                                 \\ the tetration-"height" can be fractional;
\\ and is best in the interval 0..1

coeffs=tmpM * vectorv(n,r, tmpW[r,2]*tmpD[r]^h)
\\ coeffs of the new power series for h=4/3

z0 = 1.0                                     \\ default starting value with
\\    tetration is usually z0=1
z1 = sum(k=0, n-1, z0^k * coeffs[1+k])   \\ = z0 tetrated to height ...
\\      ...    4/3 with base 3

\\ results:
\\   4.8347111352647465948 \\ n=32  use matrix-size n=32
\\   4.8347252436478228906 \\ n=48  when run with matrixsize n=48
\\                         \\ n=64 : expected to approximate Kneser-value
\\                         \\        if matrix size is increased
\\   4.834730793026332...  \\ reference by kneser-method as shown by @Mike4ty4


This may be a possible answer.

Let $f(x)=\log_{10}(x+1)$ and $g(x)=10^x-1$ (inverse function of $f$).
Then let $f^n(x) = f(f(\cdots(f(x))\cdots))$ with n $f$'s, similarly for $g$.

$$^{x+2}10\approx\lim_{n\to \infty} g^n(f^n(10^{10})\cdot(\ln10)^x)$$

The values behave fairly well for positive x-values. With $x=0.5$ and $n=40$, I got $^{2.5}10\approx4.106483157\times10^{294}$. With $x=1$ and $n=40$, I got $^310\approx9.881444237\times10^{9,999,999,999}$ (not exact because I only did 40 iterations).