# [WSC17] Extending power towers to complex realms

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# Introduction

I'm a 15 year old student that this year has had the privilege to attend the Wolfram Summer Camp and created, helped by my mentor Greg "Chip" Hurst, a visualization of the results of extending tetration (aka power towers) to different realms. In the following post I will go over the basics of tetration and its properties and then how we expanded it to different number sets.

# The basics of tetration

Tetration is, long story short, repeated powers. You have a base "a" and a height "h" and the result of "a tetrate t" is simply $a^{a^{a^{a..}}}$(h times). To understand it better we can compare it to other arithmetic operations:

Multiplication is repeated addition: a*b = a+a+a+a+...+a (b times)
Exponentiation is repeated multiplication: a^b = a*a*a*a*...*a (b times)
Tetration is repeated powers: a^a^(a...)(b times)


Although there are many different ways of representing tetration the most common one is $$^{z} a$$. One of the most important properties of tetration that should always stay true is the fact that $$^{n} a = a^{\left({^{n-1} a}\right)}$$.

Implementing basic tetration into Mathematica should be easy:

Tetration[a_, k_Integer] := Power @@ Table[a, {k}]
Tetration[2, 3]

16


We can also plot the results:

ListLogPlot[{Table[Tetration[2, x], {x, 0, 4}],
Table[Tetration[x, 2], {x, 1, 5}]}, PlotRange -> All,
Joined -> True, PlotLegends -> {"Tetration[2,x]", "Tetration[x,2]"}]


# The Challenge: Extending tetration to other realms

## - Complex base, ∞ height

Let h(z) be $z^{z^{z^{z..}}}$(∞ times). Now take logarithms both sides resulting in $z^{z^{z^{z..}}} = h(z) \log z = \log h(z)$. Solving for $h(z)$ results in the limit of the function which, as proven by Euler, converges for $\exp(-e) < x < \exp(1/e)$: $h(z) = - W(-\log z)/\log z$ where $W(x)$ is the principal branch of the W Lambert Function which describes the product of logarithms. In Mathematica, such function is called ProductLog[x]. We can plot $h(a)$ for different complex $a$ or for its convergence interval:

h[z_] = -(ProductLog[-Log[z]]/Log[z]);
DensityPlot[Abs[h[x + (I*y)]], {x, -10, 10}, {y, -10, 10},
PlotRange -> All, ColorFunction -> Hue, PlotPoints -> 150]


Plot[h[x], {x, Exp[-E], Exp[1/E]}]


## - Complex base, real height

Although we couldn't find a graphic approximation to real-height tetration, a linear approximation we determined would be the best approach to visualize the problem. This linear approximation follows this piecewise function

Tetration[z_, x_Real] :=
Piecewise[{{Log[z, tetrf[z, x + 1]],
x <= -1}, {1 + x, -1 < x <= 0}, {z^tetrf[z, x - 1], x > 0}}];


Fixing the height to (x = π) and plotting on the height, attending to the argument of the base, results in:

## - Complex base, complex height

According to the following source (https://mathoverflow.net/questions/259278/an-explicit-series-representation-for-the-analytic-tetration-with-complex-height/259371), there is a series analytical representation for complex height tetration. It uses the q-Pochhammer symbol and q Binomials.

The $q$-Pochhammer symbol: $$(p;\,q)_\infty = \prod_{k=0}^\infty(1-p\,q^k$$ $$(p;\,q)_z = \frac{(p;\,q)_\infty}{(p\,q^z;\,q)_\infty}, \quad z\in\mathbb C$$ $$(p;\,q)_n = \prod_{k=0}^{n-1}(1-p\,q^k), \quad n\in\mathbb N$$

The $q$-binomial coefficients: $${r \brack s}_q = \frac{(q;\,q)_r}{(q;\,q)_s \, (q;\,q)_{r-s}}.$$

Series: $$t(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\binom {n-k} 2} {z \brack n}_q {n \brack k}_q ({^k a}).$$

Technically, this series only converges and can therefore have a meaningful value in the interval $-2 < \Re(z)$. However, we created a conjecture by which were able to - meaningfully - extend this interval to $\Re(z) <= -2$ by recursion (until it enters the "convergent" zone). Following the main property of tetration $$^{z} a = a^{\left({^{z-1} a}\right)}$$ we can also deduct $$^{z+1} a = a^{\left({^z a}\right)}$$ . Taking logarithms both sides we end up with $$log_a (^{z+1} a)= ^{z} a$$. After $\lceil -\Re(z) \rceil - 2+1$ iterations, it will jump to the original series.

Tetration[a_, z_ /; Re[z] <= -2] := Log[a, tet[a, z + 1]];


So, the handle-all-cases function would end up being:

ClearAll[Tetration, qBinomial];
Options[Tetration] = {"Terms" -> 4, WorkingPrecision -> 30};
qBinomial[args__] := qBinomial[args] = QBinomial[args];
Tetration[_, -1, ___] = 0;
Tetration[a_, n_Integer /; n >= 0, o___] :=
Tetration[a, n, o] =
Block[{$MaxExtraPrecision = 10000,$IterationLimit = 10000},
Power @@ Table[a, {n}]];
Tetration[a_(*/;1<a<E^(1/E)*), z_ /; Re[z] > -2, OptionsPattern[]] :=
Block[{$MaxExtraPrecision = 10000,$IterationLimit = 10000},
With[{q = -ProductLog[-Log[a]]},
With[{limit =(*NumericalMathNSequenceLimit@*)
N[Total[Table[
Sum[(-1)^(n + k) q^Binomial[n - k, 2] qBinomial[z, n,
q] qBinomial[n, k, q] Tetration[a, k], {k, 0, n}], {n, 0,
OptionValue["Terms"]}]], OptionValue[WorkingPrecision]]},
SetPrecision[limit, 3/4 Precision[limit]]]]];
Tetration[a_, z_, o___] /; Re[z] <= -2 :=
Log[a, Tetration[a, z + 1, o]];


This code is an adapted version of the code posted here, posted by the author of the MSO question linked above. Plotting it...

Plot3D[Arg[
Tetration[Log[3], SetPrecision [x + (I*y), \[Infinity]]]], {x, -4,
4}, {y, -4, 4}, PlotPoints -> 40, WorkingPrecision -> 60,
ColorFunction -> Hue, PlotLegends -> Automatic,  PlotRange -> All]


A bit of a counter-intuitive "property" of tetration and higher-order operations (pentation, hexation,...):

As we can see by plotting both functions, these are clearly not the same...

LogPlot[{x^x, x}, {x, 1, 9},
PlotLegends -> {"y = \!$$\*FractionBox[\(Log[x]$$, \
$$ProductLog[Log[x]]$$]\)", "y = x" }]
`