Applications of general arrow notation?

**TriKri** · December 7th 2006, 03:50 PM

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

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

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

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

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

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

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 $\uparrow\uparrow$ when reading about power towers.

Cheers!

**ThePerfectHacker** · December 7th 2006, 06:55 PM

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 $\uparrow\uparrow$ 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

**TriKri** · December 8th 2006, 06:00 AM

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 $n\ \epsilon\text{ the set of VERY LARGE NUMBERS}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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