P12_lapl

3.3.2

The Left Associative Polyates Lemma

Introduction

Now that we've gone over the most common elementary recursive functions, the next natural question is how do we determine when one number is larger than another if they are defined using very different notations? We know that 10[5] must be greater than 2[5] by definition. However how do we know whether 2[5] is bigger or smaller than something like 10^^100? We can't simply compute 2[5] and 10^^100 and compare because both numbers are beyond our ability to compute or at least to actually write out their full decimal expansion. So how can we compare numbers if they are too large to compute?! This is where mathematical analysis comes in! In this chapter we will explore techniques we can use to determine, with mathematical certainty, which of two numbers is larger. Besides allowing us to order numbers of tremendous size without knowing their exact values, it will also give us a much greater understanding of large numbers and how they work. Large Numbers behave in very counter-intuitive ways, as you'll see. We can not always trust our intuition in these matters! Mathematical reasoning will quickly become indispensable when working with large numbers. We'll actually be able to prove some strange results, even though they seem "impossible".

In this article we are going to "prove" in an informal way, what I call the "Left Associative Polyates Lemma". This lemma is itself somewhat counter intuitive. It will form an important cornerstone for all our later analysis of large numbers in low end googology, and it has many many applications in the low level recursive range. For example, it can be used to bound expressions in Steinhaus-Moser polygon notation, it can be used to prove Moser < Graham, it can be used to make important inferences about the weak operators, and it can be used to resolve "disputes" between numbers in the primitive recursive range. "Proving" this lemma, or at least showing you why I am convinced it is true, will be excellent practice for the art of manipulating large numbers.

Impetus

The "Left Associative Polyates Lemma" (or LAPL for short), was motivated by several considerations of my own large number research. It's earliest occurrence had to do with a proof about comparing large numbers.

It is a simple matter to "prove" that:

giggol < grangol

gaggol < greagol

geegol < gigangol

etc.

This really just follows from the definitions. Since a giggol = E1#100 it must be less than E100#100 which is a grangol, since gaggol = E1#1#100 it must be less than E100#100#100 which is a greagol, and since geegol = E1#1#1#100 it must be less than E100#100#100#100 which is a gigangol. The next natural question is: How much larger? This is where it gets interesting. A giggol is normally defined as:

10^10^10^ ... ^10^10^10 w/100 10s

and a grangol can be defined as:

10^10^10^10^ ... ^10^10^10^100 w/100 10s

In other words, a giggol is simply a power tower of 10s 100 terms high, while a grangol is a power tower of 10s 100s terms high, topped off with a 100. It might seem then that giggol^100 is equal to a grangol, but this is in fact false, and the truth is far more counter-intuitive. In fact, it can easily be proven that a grangol is LARGER than giggol^giggol. That is:

Again we encounter something that seems to defy intuition. It might seem that if we raise a 100 term power tower to the power of another 100 term power tower, that we should get a power tower with about 200 terms. WRONG! It turns out it only adds about 1 or 2 extra terms to the larger of the power towers, no matter how large they get! As it so happens a giggol^giggol is between 10^^101, and 10^^102. The proof of this is, relatively speaking, not that difficult. First and foremost, the lower bound is very easy to explain. Since giggol^giggol must be greater than 10^giggol by definition, and 10^giggol = 10^(10^^100) = 10^^101, it follows that:

10^^101 < giggol^giggol

The upper bound is a little more difficult to prove, but not by much. It can be done by simply following the exponential laws. First interpret a giggol as 10^(10^^99). Let N = 10^^99, for ease of manipulation. Then it follows that:

giggol^giggol = (10^N)^(10^^100) = 10^(N*10^^100) <

10^(10^^100*10^^100) = 10^(10^N*10^N) = 10^10^(2N) <

10^10^(10N) = 10^10^10^(1+10^^98) <

10^10^10^10^(1+10^^97) <

10^10^10^10^10^(1+10^^96) <

...

< 10^10^10^...^10^10^(1+10^^1) w/100 10s before (1+10^^1) =

E(1+10^^1)#100 = E(1+10)#100 = E11#100 < E(10^10)#100 = E1#102 = 10^^102

Not only have we proven that giggol^giggol < 10^^102, but we've also shown that giggol^giggol < E11#100, which means it's less than a grangol. Incredibly, when we raise power towers to the power of other power towers, we find that they "collapse" dramatically. This takes some getting used to, but I've been practicing with large numbers for a while and you do begin to get a "supernumerary" sense of things eventually. For example, squaring a power tower does not significantly effect a power tower, no matter how large. In fact, the larger the initial power tower, the smaller the effect of squaring is going to seem! Strange.

Now I first encountered a form of LAPL when I tried to prove the next astounding result: that gaggol^^gaggol < greagol. Proving this is somewhat more difficult that the previous proof, because we must be adept not just with power towers, but now also with tetra-towers, which is a little more tricky. LAPL would actually simplify the proof to something on par with the previous proof, but this was before I had really properly formulated it. In any case, try to imagine this. A gaggol is not normally described as a power tower, but can be thought of as a "tetra-tower" of 10s 100 terms high (think of it as a left leaning tower):

gaggol = ^1010...1010 w/100 10s

Where ^pb is actually a power tower of b's p terms high! To describe gaggol as a power tower imagine this: Let "10" be the 1st tower, let "10^10^10^...^10" w/10 10s be the 2nd tower, let "10^10^...^10" w/"10^10^...^10 w/10 10s" 10s be the 3rd tower, and let each new tower have as many 10s as the previous tower. A gaggol is the 100th such tower! To describe a gaggol as a power tower we can say:

gaggol = 10^10^10^ ... ^10^10 w/10^^^99 10s

Let M = 10^^^99. We can then proceed with a proof that gaggol^^gaggol < greagol:

gaggol^^gaggol = (10^^M)^^(10^^M) =

gaggol^gaggol^gaggol^ ... ... ^gaggol^gaggol w/gaggol gaggols

This is somewhat tricky to imagine. It's a power tower of gaggols a gaggol terms high. Now imagine that each gaggol is replaced with a set of parenthesis containing a power tower of 10s 10^^^99 terms high. This is not an ordinary power tower, but a transcendent one. Now this power tower is itself part of a power tower, where there are a gaggol such power towers! We know from the previous example, that although the total number of 10s in the original expression would be M*gaggol, that it will collapse considerably. By how much would it actually collapse?! As it turns out, just the fact that gaggol^^gaggol < 10^^(M*gaggol), is enough to prove that it is less than a greagol. However at this point in my original proof I conjectured that (A^^B)^^C should be "approximately" A^^(B+C). So obviously (A^^B)^^C < A^^(BC). Although multiplying the "tetrates", B and C, was sufficient for the proof, I began to wonder: If (A^^B)^^C is approximately A^^(B+C), which is larger and under what conditions. This began my interest in this direction.

To finish up the proof. If we assume (A^^B)^^C < A^^(BC) then:

gaggol^^gaggol = (10^^M)^^(gaggol) < 10^^(M*gaggol) < 10^^(gaggol*gaggol) =

10^^(gaggol^2)

Again this is a bit difficult to imagine but, think of a gaggol as a HUGE power tower of 10s. Previously I said, that if we square it, it will barely make a difference. In fact, it won't even succeed in adding an extra 10 to the power tower. Since gaggol = 10^^M it follows that:

10^^(gaggol^2) = 10^^((10^^M)^2) < 10^^10^^(M+1) =

10^^10^^(1+10^^^99)

From here on, it's pretty much proof as usual. We simply increase the height of the tetra-tower, while simultaneously reducing the size of the leading term, using a technique I call the Ascending One (AO):

10^^10^^(1+10^^99) < 10^^10^^10^^(1+10^^^98) <

... etc. ...

10^^10^^10^^ ... ^^10^^10^^(1+10^^^1) w/100 10s before (1+10^^^1) =

E1#(1+10^^^1)#100 = E1#(1+10)#100 = E1#11#100 < E100#100#100 = greagol

So we conclude that gaggol^^gaggol < greagol. This is really quite an impressive feat when you stop to think about it. It means that we can compare these kinds of numbers even though they are far far FAR vaster than anything known to exist, even though we can't compute them, and know next to nothing about them. Yet the proof seems precarious. It's based on several leaps of supernumerical intuition. Could we somehow bring this chaos into order? LAPL is an attempt to do that by establishing that in fact (A^^B)^^C is not just approximately A^^(B+C), in fact, the sum of the tetrates is LARGER!

A Note on Terminology Used

Before we go to the general result for all "polyates", it is instructive to first prove the result only for the tetrates. Before that though, let me make a note on the terminology used here. You're probably familiar with the term exponent. It is used to denote the "p" in b^p. In such an expression we say "b" is the base and "p" is the exponent. We can also write b^p as b^p. But what would we call "p" in b^^p, or b^^^p, or b^^^^p, etc.

There is no universally accepted name for these, just as there is no universally accepted terms for the 4-dimensional analogies of polyhedra. This is probably because, for the most part, numbers like this serve almost no purpose in mathematics, and absolutely none in science. However we can easily develop terminology for our purposes here. I use the term polyponent for the generalized exponent. It simply combines the roots poly, meaning many, and exponent. Thus any example of a polyponent is a polyate.

For specific levels I use the following terms:

b^^p , p is called the tetraponent or tetrate

b^^^p , p is called the pentaponent or pentate

b^^^^p , p is called the hexaponent or hexate

b^^^^^p , p is called the heptaponent or heptate

b^^^^^^p , p is called the octaponent or octate

etc.

One can easily adapt the greek prefixes to extend this naming scheme, though we really won't have much use for it beyond tetrates and pentates. For larger cases it is simpler just to use the terms polyponent or polyates.

You'll notice many different variables being used throughout this article. This is not done in a purely haphazard manner. There are some general conventions about using variables that you should know.

Firstly, it is common to use the variables:

a,b,c,k,m,n,p,q

mainly for integer values. m,n usually indicate additionally that these are positive integers. The variables:

x,y,r

usually indicate real values. The variables:

z,c

indicate complex values, in the appropriate context. Complex numbers are the largest domain generally considered because they are the smallest domain needed to satisfy the fundamental theorem of algebra. Because I'm mainly interested in number extensions which arise naturally by extending the positive integers and addition, and complex numbers are the largest such set, I therefore conclude that complex numbers are the largest set of interest to classical arithmetic (the arithmetic involving only the integers under addition, subtraction and recursion).

We'll be exploring some of the consequences of this extension to complex numbers in this article.

The Left Associate Tetrates Conjecture (LATC)

We will begin by posing a unconditional conjecture for tetrates I'm calling the Left Associate Tetrates Conjecture:

What we want to determine is appropriate conditions for the following statement:

|(z^^m)^^n| < |z^^(m+n)|

This will allow us to prove the more general assertion, the left associative polyates Conjecture (LAPC):

(b<k>m)<k>n < b<k>(m+n)

Where b<k>p = b^^^...^^^p w/k ^s.

General Search Boundaries for LATC

We begin by defining appropriate general conditions for LATC. To keep our proof simple we will assume that p and q are integers. If we allowed p and q to be anything more general than integers we would get into the muddle of defining tetration with rational, real, or complex tetraponents. This is a complicated and open question best left for another discussion, so we will side step the issue here. To gain the full range of possible integer values for p and q we will need to reverse engineer tetration. From the basic definition we have:

b^^1 := b

b^^(n+1) := b^(b^^n)

To generalize to all integers we take the second property and invert it:

b^^(n+1) = b^(b^^n)

log_b(b^^(n+1)) = log_b(b^(b^^n))

log_b(b^^(n+1)) = b^^n

b^^n := log_b(b^^(n+1))

The last statement is take as a definition for extending tetration to the integers. This leads to some interesting results:

b^^0 := log_b(b^^1) = lob_b(b) = 1

b^^-1 := log_b(b^^0) = log_b(1) = 0

b^^-2 := log_b(b^^-1) = log_b(0) = undefined (or -infinity)

b^^-3 := log_b(b^^-2) = log_b(undefined) = undefined

b^^-4 := log_b(b^^-3) = log_b(undefined) = undefined

b^^-5 := log_b(b^^-4) = log_b(undefined) = undefined

etc.

As you can see, only b^^0 and b^^-1 return real results. The rest produce undefines. However, it isn't unreasonable to interpret b^^-2 as -infinity, since lim(x-->0+)log_b(x) = -oo , and lim(x-->-oo)b^x = 0. In any case, we therefore restrict the integer domain of b^^p to p>-2.

This extension will prove useful in our proof, but we really want to prove the condition for cases when m and n are positive integers. Weird things happen at the other values. For example, when m,n=0 we have that:

(b^^0)^^0 = b^^(0+0)

Since:

(b^^0)^^0 = (1)^^0 = 1

&

b^^(0+0) = b^^0 = 1

Negative tetrates are even more problematic. Consider these odd results:

(b^^-1)^^-1 = 0^^-1 = log_0(0^^0) = log_0(log_0(0^^1)) = log_0(log_0(0))

b^^-2 = -infinity

Due to these issues, we will only let m,n=Z+.

The next important question is the restriction of the base. Since we have restricted the tetraponent to positive integers, we can use any base on which iterated exponentiation can be computed. It should be obvious that for b>0 there is an obvious way to do this using real-valued exponentiation. The case for zero is cumbersome. Exploring the real exponentials for zero reveals unusual properties. We can begin with the definition that 0^1=0. This leads naturally to the following conclusions:

0^2 = 0*0 = 0

0^3 = 0*0*0 = 0

0^4 = 0*0*0*0 = 0

...

0^n = 0 : n=Z+

0^(1/2) = 0 since 0^2 =0

0^(1/3) = 0 since 0^3 = 0

0^(1/4) = 0 since 0^4 = 0

...

0^(1/n) = 0 : n=Z+

0^(2/3) = 0^(1/3) * 0^(1/3) = 0*0 = 0

...

0^(a/b) = 0 : a,b=Z+

0^sqrt(2) = 0^1 * 0^(1/3) * 0^(1/13) * 0^(1/253) * ... = 0*0*0*0*... = 0

Therefore:

Let 0^r := lim(a/b-->r)0^(a/b) where r is a positive real.

Thus we conclude:

0^r = 0 : r=R+

What about 0^0?

0^(n+1) = 0*0^n

So

0^n := 0^(n+1)/0

Thus:

0^0 = 0^1/0 = 0/0

0/0 is an indeterminate form, so in essence 0^0 is undefined. We can carry on to negative exponents in the following manner:

a^-n = 1/a^n

Therefore:

0^-1 = 1/0^1 = 1/0

0^-2 = 1/0^2 = 1/0

0^-3 = 1/0^3 = 1/0

etc.

We can now say that:

0^-x = 1/0^x = 1/0 : x>0

So we conclude that 0 raised to a negative real is "undefined" or "infinite". This causes a problem for tetrating 0, because it means 0^^2 = 0^0 = undefined. However there is a way that we can define 0^0 that has a natural explanation and will resolve an anomaly for our generalization.

Interestingly:

lim(x-->0+) x^x = 1

lim(x-->0+) x^x^x = 0

lim(x-->0+) x^x^x^x = 1

...

and in general

lim(x-->0+) x^^n = 1 : n is even

lim(x-->0+) x^^n = 0 : n is odd

This works quite well. If we let 0^0=1 by definition we can obtain the same definition recursively:

0^^1 = 0

0^^2 = 0^0 = 1

0^^3 = 0^0^0 = 0^1 = 0

0^^4 = 0^0^0^0 = 0^0 = 1

etc.

As erratic as this might seem, my TI-89 produces these results, with the minor warning that 0^0 has been replaced by 1. The reason for accepting this definition we'll become clear as we progress with our generalization. Now that we have a suitable definition for zero, what about negative real bases? These inevitably lead us into serious trouble, but not so obviously at first. Firstly we can let the following statements generally hold:

z^^1 := z

z^^(n+1) := z^(z^^n) : n=Z+

From this we can examine the results for z=-1:

(-1)^^1 = -1 (by definition)

(-1)^^2 = (-1)^(-1) = 1/(-1) = -1

Because (-1)^(-1) = -1, it sets up a recursion whereby (-1)^^n = -1. This seems fairly harmless. Let's look at -2:

(-2)^^1 = -2

(-2)^^2 = (-2)^(-2) = 1/((-2)^2) = 1/4

(-2)^^3 = (-2)^(-2)^(-2) = (-2)^(1/4) = 0.84+0.84i

(-2)^^4 = (-2)^(0.84+0.84i) = ????

This leads us straight into the problem of dealing with complex exponents. How do we resolve such expressions? It turns out that there is a very nice, commonly accepted definition, for how to extend to complex exponents. This leads to a beautiful theory about complex base tetration. The key to this is the idea that the natural exponential function exp(x) = e^x, can be extended to complex values via one of it's primary definitions. Namely:

Let:

e^x := lim(n-->oo) (1+x/n)^n

be generalized to:

e^z := lim(n-->oo) (1+z/n)^n

What this formula does essentially, is turn the problem of solving for a complex exponent into a problem of solving for a complex base. This can be done readily by extending the multiplicative property of complex numbers to exponentiation via the principle root.

Now when we substitute z=pi*i, we get the following result:

e^(pi*i) = -1

more generally:

e^(bi) = cos(b) +sin(b)i

where cosine and sine use radian measure

When combined with a real part we obtain this definition:

e^(a+bi) = e^a * e^(bi) = (e^a)(cos(b)+sin(b)i)

The result is that e^(a+bi) outputs a complex number whose magnitude is e^a, and whose angle from the real direction is "b" in radians. We can take this further to generalize to any complex base as well. Simply let:

(a+bi)^(c+di) = e^((c+di)ln(a+bi))

ln(a+bi) can be defined because any complex number can be expressed as a complex power of e, using it's magnitude and principle angle ( between -pi and pi). Using this definition we can extend our exploration, not just to the negative reals, but to the entire complex plane, with one caveat. Generally, tetrating non-real complex numbers will result in complex numbers. This raises the question of how to compare complex numbers. This issue can be taken care of by a simple consideration of what it means for a complex number to be "large". In the ordinary sense, complex numbers transcend ordering, and normally size relations are restricted to the set of real numbers (I think of googology as mainly being concerned with real values and nothing larger). However would we not consider something like:

googol+ googol*i

To be a "large" complex number. This is clear when both the imaginary part and the real part are large, but what about:

googol+i

1+googol*i

These are still "large" aren't they? The most natural definition that I've found is to interpret a complex numbers size as a function of it's distance from zero. In other words, it's magnitude or absolute value. There are two problems with this definition:

1) it means equidistant complex numbers are "equal" under this definition, ie. 1 = i = -1 = -i

2) large negatives are "larger" than small positives ie. -2 > 1

We have little to worry about in our generalization however, because the issue of negatives being larger than positives will not come into play over the interval of greatest interest, where z is a positive real. Note that if z is a positive real then:

z^x > 0 : z>1 , z=R+

z^x = 1 : z=1 , z=R+

z^x > 0 : 0< z < 1 , z=R+

Since a positive real raised to any real is a positive real, it follows that a positive real raised to a positive real is a positive real. This set's up an infinite recursion whereby z^^n is a positive real whenever z is a positive real. Thus for the interval of greatest concern, we find that choosing the absolute value over the "real ordering" makes no difference in the conclusion. So we will accept the absolute value definition for our generalization of LATC to the complex numbers. Our goal will be to come to some understanding of elements of the set:

S_LAT = {(z,m,n): |(z^^m)^^n| < |z^^(m+n)| & m,n=Z+ , z=C}

The part of the set of greatest interest is the subset S_LAT n R+, where "n" is the intersection of the sets. In other words, for what subset of positive reals does the statement hold? The more general complex question will also be considered. Studying this set will basically allow us to come up with a fairly general formulation of LATL to the complex numbers. This will take up the largest portion of the work here.

After we have done this, we will consider LAPL, but only for the cases of positive integer bases. We will avoid extending to complex bases because, again, this would lead to us having to consider the difficult question of complex tetrates.

Understanding the Iterated Exponential Map

Daniel Geisler is a well respected "tetrationalist" who has explored some of the interesting consequences of complex tetration. His main website on the subject is tetration.org. Here he presents the infamous tetration fractal that I used in the title image for this article. His research has direct bearing on our current research into the LAT Set.

Before we get to that, What's a tetrationalist? They are basically hobbyist mathematicians who are interested in the extension of tetration to complex tetrates. There is some degree of overlap between the tetrationalists and googologists, similar to the overlap between googologists and the recreational linguists. The issue of complex tetrates is not something we'll go into here. By it's very nature it's a part of continuous mathematics, and we generally can create really really large numbers without ever leaving a discrete framework. Continuous mathematics involves many higher topics of mathematics such as calculus and complex analysis. Despite the high level of mathematical sophistication of tetrationalists, they also find themselves in a similar situation as the googologists, namely, that their subject is obscure and lacks any practical applications (at least for the moment). Thus tetration is still considered something of "fringe" mathematics and is largely ignored by the professional mathematics community. The reason complex tetration has not garnered much interest however, is probably due more to the fact that no unique universally recognized definition of complex tetration has yet been devised. In order for a mathematical system to be "interesting" it must be more than a mere product of whim. It must be connected in as natural a way as possible to pre-existing structures, and it must not admit alternative systems or less it loses it's claim to uniqueness.

While there is no universally agreed way to extend to complex tetrates, it is generally accepted that the complex functions exp(z) = e^z and ln(z), define a system of complex exponents with all the desired properties. All the exponential laws can be extended to the complex without issue, and on the restricted domain of the reals we get the same values as with real-valued exponentiation. Because of this, z^^n is well defined for all n=Z+.

We know from experience in Chapter 3.2 that tetration rapidly produces huge values. We can say it "explodes" for all the integers greater than 1. It's also true that the values strictly increase towards infinity without bound. In calculus this is known as divergence. So we can say that tetration diverges over the positive integers greater than 1. But now we will want to consider a larger set of bases. Does tetration always explode for any given base? For real bases larger than 2, this is practically a given. Take for example 2.3:

2.3^^1 = 2.3

2.3^^2 = 2.3^2.3 = 6.7916...

2.3^^3 = 2.3^2.3^2.3 = 286.2342...

2.3^^4 = 2.3^286.2342 = 3.4587E+103

2.3^^5 = 2.3^(3.4587E+103) = 10^(1.2511E+103) ~ E103.09#2

etc.

So what happens when we chose a real base less than 2? Consider base 1:

1^^1 = 1

1^^2 = 1^1 = 1

1^^3 = 1^1^1 = 1

1^^4 = 1^1^1^1 = 1

1^^5 = 1^1^1^1^1 = 1

etc.

We see here that tetration at 1 does not explode. Rather it remains "fixed" at 1. This is known as a "fixed point" value. Technically a fixed point, on the exponential map, is some number c, for which:

z^c = c

When z=1, c=1. Other values produce other fixed points. Fixed points of this form are important for us here, because if z^^n = c, then z^^m = c for all m>n. Why? Consider:

If z^c = c

&

z^^n = c

Then

z^^(n+1) = z^(z^^n) = z^c = c

z^^(n+2) = z^(z^^(n+1)) = z^c = c

etc.

So an infinite loop is set up where by the value of all taller towers remains fixed at c. The fact that fixed points exist suggest that not all values explode under tetration. Consider the case of 1.1:

1.1^^1 = 1.1

1.1^^2 = 1.11053424105...

1.1^^3 = 1.11164880003...

1.1^^4 = 1.1117680015...

1.1^^5 = 1.11178052653...

1.1^^6 = 1.11178185374...

No matter how tall the power tower gets it never gets much larger than 1! Even though every time we increase the height of the power tower by 1, the result will be slightly larger, it is constantly slowing down so that while it never stops, it also never goes above a certain threshold. That threshold is the limit:

lim(n-->oo)1.1^^n ~ 1.11178201104

The power tower will never reach this value. Interestingly however, if we raise 1.1 to this threshold value we find that it returns itself:

1.1^1.11178201104... = 1.11178201104...

In other words, 1.11178201104 is a fixed point for 1.1. As the power tower is increasing it is approaching that fixed point, and hence it gets caught in an infinite loop of diminishing returns. When this happens the power tower is said to converge to a single fixed point.

So now we know that not all power towers explode. Some plod along and never really get anywhere. 1.1 never even gets close to 2 ever! So what about a value like 1.9? Will it diverge or converge? Let's see:

1.9^^1 = 1.9

1.9^^2 = 3.38557034392...

1.9^^3 = 8.7849627562...

1.9^^4 = 281.086509502...

1.9^^5 = 2.2587944E+79

1.9^^6 ~ 10^10^77.799 = E77.799#2

etc.

So we discover that 1.9 diverges while 1.1 converges. Is there some unique point at while it flips from converging to diverging? It turns out there is! It was first discovered in the 1700s by Leonhard Euler, the famous mathematician who gave us much of our modern notation, including the symbol for e and pi. He discovered that the "switch over" point occurs at approximately 1.4446678. More precisely we can say:

lim(n-->oo)z^^n converges : 1=<z=< e^e^(-1) ~ 1.4446678

lim(n-->oo)z^^n diverges : z> e^e^(-1) ~ 1.4446678

Interestingly when z=e^e^(-1), z^^n actually converges to the number e ~ 2.718281828. "e" is perhaps the second most famous mathematical constant after pi. This result will have important bearing on our later exploration of LATC over the positive reals. There is one other very important feature of tetration over the interval of 1 to 2. When the base, z, is only slightly larger than e^e^(-1) it behaves quite interestingly. The closer it is to e^e^(-1) from above the slower it grows towards the value of e. However once it surpasses e it begins to accelerate instead of decelerate and in short order it explodes into the tetrational range. Here is a tame example that doesn't take too long, z=1.48:

1.48^^1 = 1.48

1.48^^2 = 1.7864354932... < e

1.48^^3 = 2.01447355942... < e

1.48^^4 = 2.20286419405... < e

1.48^^5 = 2.37172011242... < e

1.48^^6 = 2.53403806754... < e

1.48^^7 = 2.70053411685... < e

1.48^^8 = 2.88268754189... > e

1.48^^9 = 3.0960739538... > e

1.48^^10 = 3.36622283991... > e

1.48^^11 = 3.74230237414... > e

1.48^^12 = 4.33681425458... > e

1.48^^13 = 5.47511737957... > e

1.48^^14 = 8.55466243457... > e

1.48^^15 = 28.6108987689... > e

1.48^^16 = 74,360.226...

BOOM!

1.48^^17 ~ 10^12,660

1.48^^18 ~ 10^10^12,659

...

The critical mass seems to be reached somewhere around 8 ~ 9. We can actually we get good estimate of the critical mass by simply solving 1.4446678^x >= x+10. The result is 7.83123. So after crossing this value an amplifying effect quickly takes over. We can actually make it take as long to cross this threshold as we like. Simply pick a number arbitrarily close to e^e^(-1) from above and you can make it take a 100 steps or a 1000 or more. Yet no matter how close the value is to e^e^(-1) eventually it will reach the critical mass and explode. This strange behavior is usually not discussed, but it is actually very relevant to our investigation of real and complex LATC as we'll see.

What happens when we look at real values smaller than 1 but greater than 0? Here things become even more counter-intuitive. It turns out that if z is between 0 and 1, than z^^n is also between 0 and 1. The reason is simple. A small number raised to any positive exponent is small. To get a result of 1 or greater would require a zero or negative exponent. However since a small number > 0 we get that small^small is still small, which results in an infinite loop of smallness. For most of the values in this range, the power tower simply approaches a unique fixed point between 0 and 1. However it doesn't approach it from one side, like the values from 1 to 1.44. Instead it oscillates from either side, closing in on the fixed point. We can use 0.6 as an illustration:

0.6^^1 = 0.6

0.6^^2 = 0.6^0.6 = 0.736021922818...

0.6^^3 = 0.6^0.7360... = 0.68661684946...

0.6^^4 = 0.6^0.6866... = 0.704165770971...

0.6^^5 = 0.6^0.7041... = 0.697881529461...

0.6^^6 = 0.6^0.6978... = 0.700125434673...

0.6^^7 = 0.6^0.7001... = 0.699323379662...

etc.

This strange oscillating behavior becomes more and more violent, with larger distances transversed between the oscillations, as we approach zero. Then something very odd happens at around 0.06. Instead of the oscillations zeroing in on a single fixed point, the oscillations branch off into two fixed points. These two values never converge to each other! Take 0.05 for example:

0.05^^1 = 0.05

0.05^^2 = 0.05^0.05 = 0.860891659332...

0.05^^3 = 0.05^0.8608... = 0.075849745546...

0.05^^4 = 0.05^0.0758... = 0.796741072476...

0.05^^5 = 0.05^0.7967... = 0.091921259401...

0.05^^6 = 0.05^0.0919... = 0.759290007184...

0.05^^7 = 0.05^0.7592... = 0.10283499357...

etc.

For values between 0 and 0.06 the oscillation never stops, and it keeps jumping from a value just shy of 1 to a value just over zero. This provides an interpretation for 0^0 = 1, 0^0^0=0, 0^0^0^0=1,etc. We can think of "zero" as also oscillating from two fixed points which are exactly equal to 1 and 0.

Extending this idea into the negative, as we saw, leads directly to complex exponentiation. So at this point we can just jump to the general behavior of complex valued tetration. It turns out that for most complex values the power tower will converge to a finite number of fixed points, but will stay within a certain radius. "Divergence" is actually rare in complex tetration! This is very counter-intuitive, since we are used to tetration exploding for the positive integer bases greater than 1. But it turns out that 1 is in good company and it's the positive reals greater than 1 than behave in an unusually divergent manner. Take the number "i" for example. Rather than explode, it spirals towards to a single complex value:

i^^1 = i

(1.000@0.500pi)

i^^2 = i^i = 0.20787...

(0.207@0.000pi)

i^^3 = i^0.2078... = 0.94715...+i0.32076...

(1.000@0.104pi)

i^^4 = i^(0.947...+i0.3207...) = 0.05009...+i0.60211...

(0.604@0.473pi)

i^^5 = i^(0.05009...+i0.60211...) = 0.38716...+i0.03052...

(0.388@0.025pi)

i^^6 = i^(0.38716...+i0.03052...) = 0.78227...+i0.54460...

(0.953@0.193pi)

etc.

Eventually i^^n settles down very close to the fixed point roughly at 0.438+i0.360. In polar form thats 0.567@0.219pi. That is by no means a large complex number. We'll define a large complex number as any complex number whose radius is greater than 1, and a small complex number as any complex number whose radius is less than 1. In the above example i never leaves a radius of 1. This is not atypical. Most complex numbers get "trapped" within a radius of 1.

At this point we turn to the important work of Daniel Geisler. On his website on tetration he has explored the complex iterated exponential map. Geisler has made a number of images of the "tetration fractal". He has identified many key areas on the map. Here is an image of the tetration fractal that I have made myself using some custom code:

Tetration By Escape Fractal

The yellow dot in the center represents the number 0. The cyan dots count out two units in each of the four directions. Black represents values which never reach the bailout value of 1E+6. All the other colors are based on the number of iterations needed to reach the escape value. If it takes less than 10 iterations the pixel is colored red. The red regions can therefore be thought of as the exploding regions and the black as the stagnant regions. It turns out that most of this area is quite stagnant.

This image gives us a bird's eye view, and a map of complex base tetration. Let's now turn our attention to the simplest case for LATC.

Examining S_LAT over reals greater than 2

There are some general properties that should hold for z in order for it to be included in S_LAT. The most important properties for the following proof are:

0 < 1 < z < z^z < z^z^z < z^z^z^z < ...

&

if z^^n < r < z^^(n+1)

then

r = E[z]c#n : 1<c<z

These properties hold for the "large numbers", that is to say, for all the real numbers greater than 1. We will now prove that if x>=2, then the LATC holds. Let LAT_[m,n] be the set of all x such that:

(x^^m)^^n < x^^(m+n)

I will prove that:

[2,oo) is a subset of Every LAT[m,n] set. Put another way, if x>=2, than the above condition holds for all positive integers m,n.

First we'll prove that [2,oo) is a subset of LAT[1,1].

(x^^1)^^1 = (x)^^1 = x

x^^2 = x^x

x^x is greater than x, for all x>1. Since exponentiation is a strictly increasing function such that:

x^a < x^b : a<b

It follows that:

x = x^1 < x^x , because 1<x.

Therefore (1,oo) is a subset of LAT[1,1] and therefore so is [2,oo)

Next we look at LAT[m,1]:

(x^^m)^^1 = x^^m

x^^(m+1) = x^(x^^m)

We already know that if x>1, then it has the property that x^^m < x^^(m+1).

Next we can look at LAT[1,n]:

(x^^1)^^n = (x)^^n = x^^n

x^^(1+n) = x^(x^^n)

Again if x>1 then x^^n < x^^(n+1).

This leaves us with the case LAT[m,n] where m,n>1. We can prove this in a general fashion. To do this let:

x' = x^^(m-1) >= 2^^1 = 2

x'' = x^^(m-2) >= 2^^0 = 1

So we have:

(x^^m)^^n =

(x^^m)^(x^^m)^ ... ^(x^^m) w/n x^^m's =

(x^x')^(x^x')^ ... ^(x^x') w/n x^x' =

x^(x' * x^(x' * ... x^(x' *x^x')) ... )) w/n x's =

x^x^(x''+x^(x''+x^ ... x^(x''+x^(x''+x^x'')) ... )) w/n+1 x's

Now we can observe that x^x'' >= 2^x'' > x'' for all real x''. Therefore:

< x^x^(2*x^(2*x^( ... 2*x^(x''+x^x'')) ... )) w/n+1 x's

<= x^x^x^(1+x^(1+x^( ... 1+x^(1+x''+x^x'')) ... )) w/n+1 x's

< x^x^x^x^(2+x^(...1+x^(1+x''+x^x'')) ... )) w/n+1 x's

<< x^x^x^x^x^ ... x^(2+x''+x^x'')) ... )) w/n x's

= E[x]2+x''+x^x''#n : Assume m=2 then we have:

E[x]2+1+x^1#n

############################################################################

############################################################################

We largely work with Counting numbers in googology, so this will still have a very broad applicability to the problems we'll consider. The next sensible condition would seem to be that b>1. If b=1 then we could simply observe that:

Since 1^n = 1 for all counting numbers, n

it follows that 1^^n = 1^1^1^...^1 w/n 1s = 1 for all counting numbers, n

Therefore

(1^^p)^^q = 1^^q = 1

&

1^^(p+q) = 1

thus:

(1^^p)^^q = 1^^(p+q)

So this refutes the claim we are trying to make. For this reason we agree that for the following proof that b>1. The next natural thing to do seems to be to check if there are any other anomalies that occur when either p or q = 1, or both. When both = 1 we find that:

(b^^1)^^1 = b^^1 = b

while...

b^^(1+1) = b^^2 = b^b

So the condition is satisfied in this case. What if p=1 and q does not:

(b^^1)^^q = b^^q

while...

b^^(1+q) = b^(b^^q)

Thus the condition is also satisfied. What if q=1 and p does not:

(b^^p)^^1 = b^^p

while...

b^^(p+1) = b^(b^^p)

So we can see that this only leaves us to prove whether the inequality holds when both p and q are greater than 1. So we now have a claim and the appropriate conditions:

Left-Associative Tetrates Conjecture

(b^^p)^^q < b^^(p+q) : b,p,q are Counting Numbers & b>1

So all that's left to do now is prove it for p,q>1. Begin by imagining a power tower q terms high, where each term is a power tower p terms high of b's:

Since b,p,q>1, we know that there has to at least be two b's for every power tower, and q set's of such power towers. The steps in the following proof come from the exponential laws. The most important here are:

(b^p)^q = b^(p*q)

(b^p)*(b^q) = b^(p+q)

The third principle which is essential to the proof is the following:

(b^x)+y < b^(x+1) : y < b^x & b>1

This is easy enough to prove. Assume b,x, and y are Counting Numbers and b>1. Furthermore assume that y< b^x. If so we can observe that:

(b^x)+y < (b^x)+(b^x) < 2*(b^x) =< b*(b^x) = b^(x+1)

implies...

(b^x)+y < b^(x+1)

So applying the product of exponents law to the expanded form of (b^^p)^^q we obtain:

This is admittedly not very clear looking. Every b^b^...^b is a power tower with p b's. It's get's multiplied by it's neighbor in it's second position. There is a total of q such towers in the above expression. This follows from the exponential law (b^p)^q = b^(p*q), except that here p and q are power towers. This can be rendered much more readable and comprehensible by observing that b^^p = b^(b^^(p-1)). Replacing each tower with this we obtain:

There are q copies of "p-1" in the above expression. Now remember that multiplication is communative, meaning that pq = qp. We will now perform just such a reversal ... for every asterisk in the expression! The result is a stream lined expression such as the following:

Note that there are q+1 "b"s in this complex power tower. Since there are q "p-1"s, but one additional b on the bottom, there is q+1 "b"s in total. If your not too sure how this expression works keep in mind that the exponents and multiplications are applied in alternative order. So first compute b^(b^^(p-1)), then multiply it by b^^(p-1), call the result X, then compute b^X, multiply that by b^^(p-1), ... and so on all the way through to the bottom. Now since b^^(p-1) = b^(b^^(p-2)), we can use the exponential law b^p*b^q = b^(b+q) to move up the terms by one level. The result is the following: