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.

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?

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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 much larger! $_\square$

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

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

ANSWER

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?

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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.

[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 ${ x }^{ x }?$

ANSWER

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?

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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?

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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?

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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 $^{ 0 }{ 4 }=1.\ _\square$

Something about tetration

Something about tetration

Something about tetration

Something about tetration

Something about tetration

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]

Something about tetration

Something about tetration