2012-11-17 00:20:59 8 Comments

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

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## 6 comments

## @Mathematically Encrypted 2020-03-31 03:38:05

We want to find a solution for $z$ within the following equation: $$ _{}^yx=z $$ Taking the super logarithm (slog) of base $x$ on both sides gives: $$ y=\text{slog} _x(z) $$ Wikipedia shows a quadratic approximation the super logarithm as follows: $$ \text{slog}_a(b)\approx\begin{cases}\text{slog}_a(a^b)-1\text{, if }b\leq 0\\ -1+\frac{2\ln a}{1+\ln a}b+\frac{1-\ln a}{1+ln a}b^2\text{, if }0<b\leq 1\\ \text{slog}_a(\log _a(b))\text{, if }b>1\end{cases} $$ Assuming that the condition $0<z\leq 1$ is true, our equation then becomes: $$ y\approx -1+\frac{2\ln x}{1+\ln x}z+\frac{1-\ln x}{1+ln x}z^2 $$ Rearranging gives: $$ \frac{1-\ln x}{1+ln x}z^2+\frac{2\ln x}{1+\ln x}z-y-1\approx 0 $$ Using the quadratic formula and simplifying gives the final result of: $$ z\approx1+\frac{1\mp\sqrt{1+y\left(1-\ln ^2x\right)} }{\ln x -1} $$ Again, assuming that the following condition is true: $$ 0<z\leq 1\text{.}\\ \text{(I would greatly appreciate if someone found boundaries for $x$ and $y$ to satisfy this condition)} $$

## @Simply Beautiful Art 2020-03-26 15:24:01

Disclaimer: My proofs of the following facts may be faulty, but numerical computations suggest otherwise.

The factorial can be uniquely extended by enforcing the condition that it grows at a certain asymptotic rate, namely that $(n+x)!\sim n^x\cdot n!$ as $n\to\infty$ by inducting on the fact that $(n+1)!\sim n\cdot n!$.

Similarly, tetration can be uniquely extended by enforcing a condition on its asymptotic rate of growth when it converges monotonically. With the help of some calculus, one can show that for $1<a<e^{1/e}$ we have

$$^{n+1}a-{}^\infty a\sim({}^na-{}^\infty a)\ln({}^\infty a)$$

and hence the natural condition is

$${}^{n+x}a-{}^\infty a\sim({}^na-{}^\infty a)\ln({}^\infty a)^x$$

which, when rearranged in terms of ${}^xa$ gives us the definition:

$$^xa=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a+({}^na-{}^\infty a)\ln({}^\infty a)^x)\tag{$\star$}$$

where $\log^{\circ n}$ is the $n$-fold logarithm e.g. $\log^{\circ2}(x)=\log(\log(x))$.

A chain of posts discusses this definition, including proofs that this does in fact give a tetration, as it satisfies the basic properties:

$^0a=1$

$^{x+1}a=a^{({}^xa)}$

as well as having the additional advantages that it is unique, satisfies nice asymptotics, and is analytic. Due to the last property, it is possible to analytically continue this definition outside of the initial interval, though the computation is largely difficult for two main reasons:

The definition itself requires large amounts of precision, since $^na-{}^\infty a$ involves large amounts of cancellation and loss of significant figures. This can be somewhat avoided, however, with the use of acceleration techniques such as Aitken's delta squared method.

Analytic continuation is a fairly difficult task to perform numerically, and I've yet to find an alternative representation of this tetration that converges further out. This can be somewhat circumvented using modern techniques, such as conformal mappings (e.g. expanding ${}^x(a+t/(1-t))$ about $t=0$), to allow for convergence even on the entire interval $[1,\infty)$.

## @Mark Hunter 2019-09-03 06:53:03

Some attempts at defining $e$ to itself $s$ times when $s$ is a real number besides just a whole number involve adding a sort of socket variable to turn the problem into finding continuous iterates of $exp(x)$.

We want to find a unique function $exp_s (x)$ of a single variable $x$ where $s$ is a parameter for the number of iterations.

... $exp_1$(x) = $exp(x)$

and for $s$ and $t$ real

... $exp_s (exp_t (x)) = exp_{s + t} (x)$.

Then $exp_s (1)$ will answer your question.

It’s hard to find a continuou iterate of a function that doesn't have a fixed point, like $exp(x)$ on the real line. H. Kneser’s method involves going into the complex plane to find a fixed point. See the links other people here have provided.

The trouble is that there is more than one fixed point in the complex plane, which leads to singularities and non-uniqueness. There are ways of dealing with this but they don’t convince everybody.

George Szekeres (1911 - 2005) tackled the problem solely in the real domain. His method is explained, and some serious gaps in his argument patched up, in the article:

“The Fourth Operation” http://ariwatch.com/VS/Algorithms/TheFourthOperation.htm

## @Douglas Shamlin 2017-09-18 23:18:17

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).

## @Simply Beautiful Art 2020-03-26 14:45:58

If I am interpreting this right, then it strongly fails to give reasonable tetration, as it implies things such as ${}^n2\to1$. There is also no motivation for this definition provided here.

## @Gottfried Helms 2014-02-11 17:10:05

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.

## @The_Sympathizer 2012-11-17 01:27:27

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

inverseof 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.

## @Lucas 2012-11-17 01:50:51

In that article I like the graph that goes outside of it's grid. Seems appropriate for hyperexponentiation.

## @Sheldon L 2012-11-25 17:14:30

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.