Past exponential functions?

Question

We have been taught that linear functions, usually expressed in the form $y=mx+b$, when given a input of 0,1,2,3, etc..., you can get from one output to the next by adding some constant (in this case, 1). $$ \begin{array}{c|l} \text{Input} & \text{Output} \\ \hline 0 & 1\\ 1 & 2\\ 2 & 3 \end{array} $$

But with exponential functions (which are usually expressed in the form $y=a\cdot b^x $), instead of adding a constant, you multiply by a constant. (In this case, 2) $$ \begin{array}{c|l} \text{Input} & \text{Output} \\ \hline 0 & 1\\ 1 & 2\\ 2 & 4\\ 3 & 8 \end{array} $$ But... we can keep going, can't we? $$ \begin{array}{c|l} \text{Input} & \text{Output} \\ \hline 0 & 1\\ 1 & 2\\ 2 & 4\\ 3 & 16\\ 4 & 256 \end{array} $$ In this example, you square the last output to get to the next one. I cannot find a 'general form' for such an equation, nor can I find much information online. Is there a name for these functions? Is there a general form for them? And can we keep going even past these 'super exponential functions'?

Answer 1

Note that your table can be written as

$$ \begin{array}{c|l} \text{Input} & \text{Output} \\ \hline 1 & 2^{2^{1-1}}\\ 2 & 2^{2^{2-1}}\\ 3 & 2^{2^{3-1}}\\ 4 & 2^{2^{4-1}} \end{array} $$

Tetration is the operation you could say that succeeds exponentiation. As an example, notationally, if we want to represent "a to the a to the a to the a"th power, we use

$$ ^4 a=a^{a^{a^a}}$$

Answer 2

Each of your tables is found by taking 2 to the power of the previous table (possibly with a shift by 1). As the formula for your second table is $2^x$, the formula for your third table is $2^{2^{x-1}}$ (for $x \geq 1$).

Answer 3

The map satisfies the recurrence relation $x_{n+1}={x_n}^2$ with base case $x_1=2$. By backtracking or induction one can show that $x_n=2^{2^{n-1}}$ for all $n\in\mathbb{N}$.

Answer 4

Your last table is incorrect, and as others have mentioned, tetration would be the next operation.

Let's try to figure out what these kinds of operations are and how they work, to begin with.

We will be considering $x+2$, $x\cdot2$, and $x^2$.

What is $x+2$? In simple terms, it means $x+1+1$ where we have $2$ ones.

What is $x\cdot2$? In terms of addition, it is $x+x$, where we add $x$ to itself $2$ times.

What is $x^2$? In terms of multiplication, it is $x\cdot x$, where we multiply $x$ to itself $2$ times.

From here, there is an operation $H_n$ which $n$ tells us how high up this tree of operations it is. For addition, $n=-1$. For multiplication, $n=0$. For exponentiation, $n=1$. We will be using base $2$.

But what about $H_2$? Well, as you can imagine, it is a repetition of exponentiation:

$$H_2(x)=x^x$$

It is simply $x$ raised to itself an amount of times.

Let us define our function a little bit better. $H_n(x,b)$, where $n$ is the 'level,' $x$ is the input, and $b$ is the base.

We now see that $H_2(x,2)=x^x$.

Also, we see that $H_2(x,3)=x^{x^x}$, and we can do this for any whole number for the base. This is called tetration.

What is beyond that is pentation:

$$H_3(x,2)=H_2(x,x)$$

It is difficult to express such a thing with normal notation, and even with my notation. However, I can define the general operation:

$$H_n(x,b)=H_{n-1}(x,H_{n-1}(x,H_{n-1}(x,H_{n-1}(x,\dots1))))$$

Where we apply the operation $H_{n-1}$ $b$ times. Very complicated!