FANDOM


Some months ago, or about 1 year ago? I calculated Moser's number, using up-arrow notation. Later I posted movie about Moser, to Nico-nico-douga:

http://www.nicovideo.jp/watch/sm21303901 http://www.nicovideo.jp/watch/sm21303970

However, those movies are in Japanese, so I want to write down the process of calculation, in English.

It needs some knowledges of approximation between up-arrow notation and Steinhaus-Moser notation.

Approximation section 1

we need some preparations to calculate moser.

First, let's consider about this.

  • (\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n})
^{(\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n})} when a and n is not small.

This can be calculated as follows:

  • =a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}
+\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}}

Here \underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}is so big, but \ll\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}

  • Therefore, a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}
+\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}}
\approx a^{a^{\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}}
=\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n+1}

If b=a, this can be described as follows, using tetration operator,

(a\uparrow\uparrow n)\uparrow\uparrow2\approx a\uparrow\uparrow(n+1)

using that step,

  • (a\uparrow\uparrow b)\uparrow\uparrow3\approx a\uparrow\uparrow(b+2)
  • (a\uparrow\uparrow b)\uparrow\uparrow4\approx a\uparrow\uparrow(b+3)
  • (a\uparrow\uparrow b)\uparrow\uparrow c\approx a\uparrow\uparrow(b+c-1)

and,

  • (a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow b)\approx a\uparrow\uparrow(b+(a\uparrow\uparrow b)-1)\approx a\uparrow\uparrow a\uparrow\uparrow b

Because of this approximation, the next can be also approximated as follows,

  • (a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b
  • (a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b

so, using pentation operator,

  • (a\uparrow^3 n)\uparrow^3 2\approx a\uparrow^3(n+1)

and also,

  • (a\uparrow^3 b)\uparrow^3 3\approx a\uparrow^3(b+2)
  • (a\uparrow^3 b)\uparrow^3 4\approx a\uparrow^3(b+3)
  • (a\uparrow^3 b)\uparrow^3 c\approx a\uparrow^3(b+c-1)
  • (a\uparrow^3 b)\uparrow^3 (a\uparrow^3 b)\approx a\uparrow^3 a\uparrow^3 b

Similarly and generally,

  • (a\uparrow^m b)\uparrow^m c\approx a\uparrow^m(b+c-1)
  • (a\uparrow^m b)\uparrow^m(a\uparrow^m b)\approx a\uparrow^m a\uparrow^m b

And of course,

  • (a\uparrow^m b)\uparrow^{m+1}(a\uparrow^m b)\approx a\uparrow^{m+1} a\uparrow^m b
  • (a\uparrow^m b)\uparrow^{m+c}(a\uparrow^m b)\approx a\uparrow^{m+c} a\uparrow^m b

Approximation section 2

For example, if n is not small,

  • 3\uparrow\uparrow n\approx(10\uparrow)^{n-1}1.1\approx10\uparrow\uparrow(n-1)
  • 10^{12}\uparrow\uparrow n\approx(10\uparrow)^{n+1}1.1\approx10\uparrow\uparrow(n+1)

The differences of exponential heights between tetration of 10, are only 1 or 2.

So if N>>0, for any a,

  • a\uparrow\uparrow N\approx10\uparrow\uparrow N

Nextly, for example,

  • a\uparrow\uparrow\uparrow3=a\uparrow\uparrow a\uparrow\uparrow a\approx10\uparrow\uparrow 10\uparrow\uparrow a' when a' is a, or a number near a.
  • a\uparrow\uparrow\uparrow4
=a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a
\approx10\uparrow\uparrow10\uparrow\uparrow10\uparrow\uparrow a'

So, if N>>0,

  • a\uparrow\uparrow\uparrow N\approx10\uparrow\uparrow\uparrow N

Similarly,

  • a\uparrow^m n\approx(10\uparrow^{m-1})^{n-1}a'
  • a\uparrow^m N\approx10\uparrow^m N

Those dispositions are necessary to calculate polygon notation.

Polygon notation

Now let's consider about n in p-gon, as n[p]. n is not small (at least 3 or more).

  • n[3]=n^n
  • n[3]_2=(n[3])[3]=(n^n)^{n^n}=n^{n\times n^n}=n^{n^{n+1}}
  • n[3]_3=(n[3]_2)[3]=(n^{n^{n+1}})^{n^{n^{n+1}}}=n^{n^{n+1}\times{n^{n^{n+1}}}}=n^{n^{n+1+{n^{n+1}}}}\approx{n^{n^{n^{n+1}}}} if n is not small
  • n[3]_4\approx(n^{n^{n^{n+1}}})^{n^{n^{n^{n+1}}}}
=n^{n^{n^{n+1}+n^{n^{n+1}}}}\approx{n^{n^{n^{n^{n+1}}}}}

Here we are passing the first steps of approximation.

So when it reach n in square,

  • n[4]=n[3]_n\approx (n\uparrow)^n(n+1)\lesssim(n+1)\uparrow\uparrow(n+1)
  • and if N>>0, N[4]\approx N\uparrow\uparrow N

if n' =: n+1,

  • n[4]\approx n'\uparrow\uparrow n'
  • n[4]_2=(n[4])[3]_{(n[4])}
\approx(n'\uparrow\uparrow n')\uparrow\uparrow(n'\uparrow\uparrow n')
\approx n'\uparrow\uparrow n'\uparrow\uparrow n' (using second steps of approximation)
  • n[4]_3\approx n'\uparrow\uparrow n'\uparrow\uparrow n'\uparrow\uparrow n'
  • n[5]=n[4]_n\approx n'\uparrow^3(n+1)=n'\uparrow^3 n'
  • n[5]_2=(n[5])[4]_{(n[5])}
\approx (n'\uparrow^3 n')\uparrow^3(n'\uparrow^3 n')
\approx n'\uparrow^3 n'\uparrow^3 n'
  • n[6]=n[5]_n\approx n'\uparrow^4(n+1)=n'\uparrow^4 n'

...Finally, in general,

  • n[p]\approx(n+1)\uparrow^{p-2}(n+1)

Calculation of 2[p]

2 is small, and so week, among up-arrow notation, so it's hard to apply n[p]\approx(n+1)\uparrow^{p-2}(n+1). However, 2[p]=(2[p-1])[p-1], and if p is not small, 2[p-1] is large number. So (2[p-1])[p-1] can be applied approximation of n[p].

\approx257\uparrow^3 257\uparrow^2 257
\approx10\uparrow^3 257\uparrow^2 257 = A-ooga

  • 2[7]=(2[6])[6]\approx(10\uparrow^3 257\uparrow^2 257)[6]

\approx(10\uparrow^3 257\uparrow^2 257)\uparrow^4(10\uparrow^3 257\uparrow^2 257)
\approx10\uparrow^4 10\uparrow^3 257\uparrow^2 257

  • 2[8]=(2[7])[7]\approx(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)[7]

\approx(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)\uparrow^5(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)

\approx10\uparrow^5 10\uparrow^4 10\uparrow^3 257\uparrow^2 257

Another approach to 2[p]

2[p]=2[p-1][p-1]=2[p-2][p-2][p-1]=2[p-3][p-3][p-2][p-1]

\dots=2^2[3][4][5]\dots[p-3][p-2][p-1]

now [3] is exponentiation, and [4] is approximately tetration, similarly [5] is pentation, [6] is hexation, and [p] is, approximately \uparrow^{p-2}.

So concidering pattern of previous section,

  • 2[p]=2^2[3][4][5]\dots[p-3][p-2][p-1]

\approx10\uparrow^{p-3}10\uparrow^{p-4}10\uparrow^{p-5}\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257

And Moser's number is,

  • 2[2[5]]\approx10\uparrow^{\text{Mega}-3}10\uparrow^{\text{Mega}-4}10\uparrow^{\text{Mega}-5}\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257

Can we approximate this more simply? Yes, there is two way to describe this structure.

  • 3\uparrow^{\text{Mega}-2}3

3\uparrow^n 3 can be written down as follows,

=3\uparrow^{n-1}3\uparrow^{n-2}\dots\uparrow^3 3\uparrow^2 3\uparrow(3\times(3+3+3)) It is similar to 2[p]

  • 2\uparrow^{\text{Mega}-2}4

2\uparrow^n 4(=2\uparrow^{n+1} 3)

=2\uparrow^{n-1}2\uparrow^{n-2}\dots\uparrow^3 2\uparrow^2 2\uparrow(2\times(2+2+2+2))

Compareing these and Moser, in fact,

  • 3\uparrow^{\text{Mega}-2}3\gtrsim2\uparrow^{\text{Mega}-2}4\gtrsim\text{Moser}

Because,

  • 3\uparrow^{\text{Mega}-2}3=\dots\uparrow^4 3\uparrow^3 3\uparrow^2 7625597484987

\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{7625597484986}1.1

  • 2\uparrow^{\text{Mega}-2}4=\dots\uparrow^4 2\uparrow^3 2\uparrow^2 65536

\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{65533}4.3

  • \text{Moser}\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{257}2.8

After all, However, Mega is already quite large number.

  • \text{Moser}\approx3\uparrow^{257\uparrow\uparrow257}3

This might be the most simple approximation, using up-arrow notation...

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.