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Math Help - Applications of general arrow notation?

  1. #1
    Senior Member TriKri's Avatar
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    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 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!
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  2. #2
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    Quote Originally Posted by TriKri View Post
    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
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  3. #3
    Senior Member TriKri's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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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