# Tetration

Tetration is, for the most part, another name for iterated exponentiation. Iterated exponentiation is when you raise a number to the power of itself several times. The formal definition of tetration is

\[\large ^{ x }n \equiv n^{n^{n^{^{.^{.^{.}}}}}},\]

where there are \(x\) \(n\)'s on the right side of the equation. This definition allows tetration to represent huge numbers in a tiny amount of space.

#### Contents

## How to Calculate Tetration

In order to calculate a tetration, you must recall how to calculate a power tower.

What is \({ 3 }^{ { 3 }^{ 3 } }\) equal to in the simplest terms?

A common mistake is to solve for \({ 3 }^{ { 3 }^{ 3 } }\) as if it were \({ \big({ 3 }^{ 3 }\big) }^{ 3 }\). This gives us \({ 27 }^{ 3 }\), or 19,683. However, the true definition of a power tower says to calculate it starting from the top--that is, calculating it as \({ 3 }^{ ({ 3 }^{ 3 }) }\). Using this definition, we get \({ 3 }^{ 27 }\), or approximately \(\boxed{7.6\times { 10 }^{ 12 }},\) which is

muchlarger! \(_\square\)

Using this definition, equations containing tetration can be solved algebraically.

If \(\sqrt [ 4 ]{ ^{ 3 }x } =2\), what is \(x\) equal to?

Using the definition of tetration, \(\sqrt [ 4 ]{ ^{ 3 }x }\) can be simplified to \(\sqrt [ 4 ]{ { x }^{ { x }^{ x } } } \). We can then raise both sides of the equation to the fourth power to get \({ x }^{ { x }^{ x } }={ 2 }^{ { 4 } }\). Because \({ 2 }^{ 2 }=4\), we can rewrite this equation as \({ x }^{ { x }^{ x } }={ 2 }^{ { 2 }^{ 2 } }\), which is the same thing as saying that \(x=2.\ _\square\)

When computing a "tetra tower" of the form \(^{ ^{ ^{ \ddots }{ x }_{ 2 } }{ x }_{ 1 } }n\), similarly to exponents, you start at the top and work your way down.

What is \(^{ ^{ ^2 }2 }2\) in simplest terms?

Starting from the top, the equation can be expanded to \(^{ { 2 }^{ 2 } }2\), simplified to \(^{ 4 }2\), expanded again to \({ 2 }^{ { 2 }^{ { 2 }^{ 2 } } }\), and then simplified once more to \(65,536.\ _\square \)

More/better examples here.

## Algebra and Expanding the Definition

[TODO: Make this section follow a logical progression]

When dealing with problems containing a variable in the height of the tetration, other strategies are necessary.

If \(^{ ^{ 2 }x }3=27\), what is \({ x }^{ x }?\)

The equation can be expanded to say that \(^{ { x }^{ x } }3=27\). We can use an underbrace to express that there's a certain amount of something: using an underbrace, we can rewrite the equation as \(\underbrace {3^{3^{^{.^{.^{.}}}}}}_{{x^x}} =27.\) Since \(27={ 3 }^{ 3 }\), the right side of the equation can be expanded to say that \(\underbrace {3^{3^{^{.^{.^{.}}}}}}_{{x^x}}=3^3.\) But since \({ 3 }^{ 3 }\) is equivalent to \(\underbrace {3^{3^{^{.^{.^{.}}}}}}_{2},\) this means that \(\underbrace {3^{3^{^{.^{.^{.}}}}}}_{{x^x}} =\underbrace {3^{3^{^{.^{.^{.}}}}}}_{2},\) which is the same thing as saying that \({ x }^{ x }=2.\) If you want to know the approximate value of \(x,\) it's \({ x }\approx 1.559\). You'll learn how to calculate the exact value of \(x\) later. \(_\square\)

It's hard to define how to calculate a tetration with a base of 0. An easy way to figure out what the tetration might be is to expand it, then solve for a limit that approaches it.

What is \(^{ 4 }0\) in simplest terms?

First, we can start by expanding the expression. \(^{ 4 }0\) means a power tower with a size of 4, so it is equivalent to \({ 0 }^{ { 0 }^{ { 0 }^{ 0 } } }\). To simplify this value further, it can be rewritten as

\[ 0^{\Big(0^{^{\left(0^{{}^0}\right)}}\Big)}.\]

Although \({ 0 }^{ 0 }\) is technically indeterminate, it's generally agreed that the limit \(\lim _{ x\rightarrow 0 }{ { x }^{ x } } \) is equal to the value of \({ 0 }^{ 0 }\). If you don't know what a limit is, all it's asking is "what is the value of \({ x }^{ x }\) as \(x\) gets smaller and smaller to 0?" As it turns out, this value is 1. So, this means that \( 0^{\Big(0^{^{\left(0^{{}^0}\right)}}\Big)}\) is equivalent to \({ 0 }^{ { (0 }^{ 1 }) }\). \({ 0 }^{ 1 }\) is 0, so the expression simplifies to \({ 0 }^{ 0 }\). As we've established, \({ 0 }^{ 0 }\) is generally agreed upon to be 1, so the value of \(^{ 4 }0\) is equal to 1. \(_\square\)

As you may have noticed, the value of \(^{ x }0\) alternates between 0 and 1--that is, if \(x\) is even, the value is 1, and if \(x\) is odd, the value is 0. This means that the expression can be represented as either the piecewise function

\[ ^x0 = \begin{cases} 1 && \text{if } x= \text{even}\\ 0 && \text{if } x= \text{odd}, \end{cases} \]

or the explicit function \(-2\text{ mod}\big(\frac { x }{ 2 } ,1\big)+1\).

If the height of a tetration in an equation was infinite, it can still be simplified into a non-infinite value using algebra. There are several methods that can be used to simplify it.

What is \(^{ \infty }\sqrt { 2 } \) in simplest terms?

The easiest way to simplify the value is to see what \(\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}\) approaches as the size of the tower approaches infinity. \({ \sqrt { 2 } }^{ \sqrt { 2 } }\) is approximately 1.633, \({ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ \sqrt { 2 } } }\) is approximately 1.761, and as the size approaches infinity, the value approaches \(\boxed { 2 } \). The easiest way to prove this is by using basic substitution: if \(\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}=x,\) then that also means that \({ \sqrt { 2 } }^{ { x } }=x\). [TODO: Clean this mess up.] \(\log _{ \sqrt { 2 } }{ x } =x\) \(\log _{ \sqrt { 2 } }{ x } -x=0\) \(\log _{ \sqrt { 2 } }{ x } -\log _{ 2 }{ { 2 }^{ x } } =0\) \(\log _{ 2 }{ { 2 }^{ x } } =\frac { \log _{ \sqrt { 2 } }{ { 2 }^{ x } } }{ \log _{ \sqrt { 2 } }{ 2 } } \) \(\log _{ 2 }{ { 2 }^{ x } } =\frac { \log _{ \sqrt { 2 } }{ { 2 }^{ x } } }{ 2 } \) \(\log _{ \sqrt { 2 } }{ x } -\frac { \log _{ \sqrt { 2 } }{ { 2 }^{ x } } }{ 2 } =0\) \(2\log _{ \sqrt { 2 } }{ x } -\log _{ \sqrt { 2 } }{ { 2 }^{ x } } =0\) \(\log _{ \sqrt { 2 } }{ { x }^{ 2 } } -\log _{ \sqrt { 2 } }{ { 2 }^{ x } } =0\) \(\log _{ \sqrt { 2 } }{ { x }^{ 2 } } =\log _{ \sqrt { 2 } }{ { 2 }^{ x } } \) \({ x }^{ 2 }={ 2 }^{ x }\)

Because \(^{ x }n={ n }^{ { n }^{ { n }^{.^{.^{.}} } } }\) when there are \(x\) \(n\)'s on the right side of the equation, \(^{ x+1 }n={ n }^{ \big({ n }^{ { n }^{ { n }^{.^{.^{.}}} } }\big) }\) is true when there are \(x+1\) \(n\)'s on the right side. This equation can be rewritten to say that \(^{ x+1 }n={ n }^{ ^{ x }n }\). This allows us to extend the definition of tetration to zero.

What is \(^{ 0 }{ 4 }\) in simplest terms?

Replacing \(x\) with 0 in the identity \(^{ x+1 }n={ n }^{ ^{ x }n }\), we can show that \(^{ 1 }n={ n }^{ ^{ 0 }n }\). Because \(^{ 1 }n=n\), \(n={ n }^{ ^{ 0 }n }\). Taking the logarithm with base \(n\) of both sides, we get that \(1=^{ 0 }n\). This means that the \(0^\text{th}\) tetration of anything is always 1. Because of this, it can be said that \(^{ 0 }{ 4 }=1.\ _\square\)

## Tetra-root

Something about tetration

## Tetra-logarithm

Something about tetration

## Real Heights

Something about tetration

## Tetration with Complex Numbers

Something about tetration

## Advanced Equations

Something about tetration

## Notation

Although the form \(^{ x }n\) is very common when it comes to tetration, several other forms have been devised as well.

Usually, the most used of these forms is Knuth's up-arrow notation [TODO: Make a wiki page on the notation]. The form represents tetration with two up arrows--the base on the left of the two arrows, and the height on the right. Using this form, \(^{ x }n\) becomes \(n↑↑x\). Often, the arrows will get represented with two of the unicode character ^. This form can have advantages over the standard one, because instead of simply relying on position (which can get messy when you're dealing with exponents or roots), it uses a symbol which is easy to spot and isn't very confusing.

Another form, the Conway chained arrow notation[TODO: Make a wiki page on this notation too], is similar to up-arrow notation except that, instead of up arrows, it uses right arrows. There are two up arrows: the base on the left of the two and the height in the middle. On the right is a 2. This makes the value look like \(n→x→2\). To see why there's a 2 on the right, visit this wiki on the notation. [Someone please make this wiki cuz I don't want to do it]

More notations [note: include exp and stuff about tetra-root/log]

## Graphing Equations containing Tetration

Something about tetration

## Tetration in Higher Dimensions

Something about tetration