# Thread: Applications of general arrow notation?

1. ## Applications of general arrow notation?

I don't know where to put this, so I put it here cause it is so advanced...

Anyway, for those who knows what arrow notation is, I have some questions, and for those who don't know what it is, here's a brief explanaition:

The arrow notation $\displaystyle a\uparrow b$ is used to apply an operation multiple times ($\displaystyle b-1$ times) at variable $\displaystyle a$ with variable $\displaystyle a$. For example, $\displaystyle a\uparrow^n b$ means that operation $\displaystyle \uparrow^{n-1}$ is applied $\displaystyle b-1$ times at $\displaystyle a$, with $\displaystyle a$ itself. So since $\displaystyle \uparrow^0$ is defined as multiplication, $\displaystyle \uparrow^1$ means power since $\displaystyle a$ is multiplicated $\displaystyle b-1$ times with itself.

$\displaystyle a\uparrow^n b\ =\ \underbrace{a\uparrow^{n-1} a\uparrow^{n-1} a\uparrow^{n-1} ... \uparrow^{n-1} a}_{\uparrow^{n-1} \text{ comes } b-1 \text{ times}}$

$\displaystyle a\uparrow^1 b\ =\ \underbrace{a\uparrow^0 a\uparrow^0 a\uparrow^0 ... \uparrow^0 a}_{a \text{ comes } b \text{ times}}\ =\ a\cdot a\cdot a\cdot ... \cdot a\ =\ a^b$

All this also means that $\displaystyle a\uparrow^n b\ =\ a\uparrow^{n-1}(a\uparrow^n(b-1))$. For example, $\displaystyle a^b\ =\ a\uparrow^1b\ =\ a\uparrow^0(a\uparrow^1(b-1))\ =\ a\cdot a^{b-1}$.

This can go on forever, meaning that $\displaystyle n$ can grov from 1 to 2, from 2 to 3 and so on, creating new operators $\displaystyle \uparrow^n$ all the time. For example

$\displaystyle \begin{array}{ccccl}a\uparrow^0b&=&&=&a\cdot b\\ a\uparrow^1b&=&a\uparrow b&=&a^b\\ a\uparrow^2b&=&a\uparrow\uparrow b&=&\overbrace{a^{a^{a^{a^{a...}}}}} ^{\begin{array}{c}_{b\ \ a\text{'s}}\\^{b\!-\!1 \text{ exponents}}\end{array}}\\ a\uparrow^3b&=&a\uparrow\uparrow\uparrow b\\ \vdots\end{array}$

This can create very big numbers if we continue developing the operators. This is an very interesting invention made by Donald Knuth, but what is it good for? I recently saw an application of $\displaystyle \uparrow\uparrow$ when reading about power towers.

Cheers!

2. Originally Posted by TriKri
This is an very interesting invention made by Donald Knuth, but what is it good for? I recently saw an application of $\displaystyle \uparrow\uparrow$ when reading about power towers.
I think it is called Akerman function. But I do not see any purpose of it, perhaps people try to create a function that produces large numbers

3. Originally Posted by ThePerfectHacker
Perhaps people try to create a function that produces large numbers
Yes, but you can still use the old traditional way of creating a large number, like $\displaystyle n\ \epsilon\text{ the set of VERY LARGE NUMBERS}$

Another thing you can conclude from this is this, $\displaystyle a\uparrow^nb\ =\ a\uparrow^{n-1}(a\uparrow^n(b-1))$ can also be written as:

$\displaystyle a\uparrow^{n-1}(a\uparrow^n(b-1))\ =\ a\uparrow^nb$

For example we have $\displaystyle \uparrow^0\ =\ \cdot$

$\displaystyle a\uparrow^{-1}(a\cdot(b-1)) = a\cdot b$

Set temporarily the variable $\displaystyle x\ =\ a\cdot(b-1)$

$\displaystyle a\uparrow^{-1}x = a\cdot b\ =\ x+a$

Now we have no reason to conclude from this that $\displaystyle a\uparrow^{-1}x\ =\ x+a$ for all values of $\displaystyle x$, we only know it is true for $\displaystyle a|x$, but we can generalize so that it becomes true for all values of $\displaystyle x$.

So now when we know have $\displaystyle a\uparrow^{-1}x\ =\ a+x$, we can set up the equation

$\displaystyle a\uparrow^{-2}(a\uparrow^{-1}(b-1))\ =\ a\uparrow^{-1}b$
$\displaystyle a\uparrow^{-2}(a+(b-1))\ =\ a+b$

Set temporarily the variable $\displaystyle x\ =\ a+(b-1)$

$\displaystyle a\uparrow^{-2}x\ =\ x + 1$

Generalize this to make it true for all values of $\displaystyle x$.

$\displaystyle a\uparrow^{-3}(a\uparrow^{-2}(b-1))\ =\ a\uparrow^{-2}b$
$\displaystyle a\uparrow^{-3}((b-1)+1)\ =\ b+1$
$\displaystyle a\uparrow^{-3}b\ =\ b+1$

That is already true for all values of $\displaystyle b$ and doesn't have to be generalized.

But now $\displaystyle a\uparrow^{-3}b\ \equiv\ a\uparrow^{-2}b$. That means every operator $\displaystyle \uparrow^{n\leq-2}$ will do the same, namely return $\displaystyle b+1$.

Yet another thing is that the definitions of $\displaystyle a\uparrow^0b$ and $\displaystyle a\uparrow^1b$ doesn't directly lead to the conclusion that $\displaystyle a\uparrow^nb\ =\ a\uparrow^{n-1}(a\uparrow^n(b-1))$. It could as likely has been $\displaystyle a\uparrow^nb\ =\ (a\uparrow^n(b-1))\uparrow^{n-1}a$, that doesn't matter since $\displaystyle a\cdot b\ \equiv\ b\cdot a$, and $\displaystyle a+b\ \equiv\ b+a$. But $\displaystyle a^b \not \equiv\ b^a$, therefore it matters when $\displaystyle n \geq 2$, cause $\displaystyle a\uparrow^2b\ =\ a\uparrow\uparrow b$ would have looked like $\displaystyle (((...(a^a)^a)^a)...)^a$ and not like $\displaystyle a^{a^{a^{a...}}}$. And $\displaystyle b+1\ \not \equiv\ a+1$, thus it matters when n $\displaystyle n \leq -1$ as well.