User blog:Aetonal/Calculating Moser's Number

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})}$$ when a and n is not small. This can be calculated as follows: +\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}$$ Here $$\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}$$is so big, but $$\ll\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}$$ +\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}$$
 * $$(\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n})
 * $$=a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}
 * Therefore, $$a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}

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,

so, using pentation operator,
 * $$(a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b$$
 * $$(a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b$$


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

and also,


 * $$(a\uparrow^3 b)\uparrow^3 3\approx a\uparrow^3(b+2)$$
 * $$(a\uparrow^3 b)\uparrow^3 4\approx a\uparrow^3(b+3)$$
 * $$(a\uparrow^3 b)\uparrow^3 c\approx a\uparrow^3(b+c-1)$$
 * $$(a\uparrow^3 b)\uparrow^3 (a\uparrow^3 n)\approx a\uparrow^3 a\uparrow^3 n$$

Similarly and generally,


 * $$(a\uparrow^m b)\uparrow^m c\approx a\uparrow^3(b+c-1)$$
 * $$(a\uparrow^m b)\uparrow^m(a\uparrow^m b)\approx a\uparrow^m a\uparrow^m b$$

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,

The differences of exponential heights between tetration of 10, are only 1 or 2.
 * $$3\uparrow\uparrow n\approx(10\uparrow)^{n-1}1.1\approx10\uparrow\uparrow(n-1)$$
 * $$10^{12}\uparrow\uparrow n\approx(10\uparrow)^{n+1}1.1\approx10\uparrow\uparrow(n+1)$$

So if N>>0, for any a,
 * $$a\uparrow\uparrow N\approx10\uparrow\uparrow N$$

Nextly, for example,

=a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a \approx10\uparrow\uparrow10\uparrow\uparrow10\uparrow\uparrow a'$$
 * $$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

So, if N>>0,


 * $$a\uparrow\uparrow\uparrow N\approx10\uparrow\uparrow\uparrow N$$

Similary,


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

Those dispositions are necessary to calculate polygon notation.

=Polygon notation=

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

=n^{n^{n^{n+1}+n^{n^{n+1}}}}\approx{n^{n^{n^{n^{n+1}}}}}$$ Here we are passing the first steps of approximation.
 * $$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}}}}

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,

\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]\approx n'\uparrow\uparrow n'$$
 * $$n[4]_2=(n[4])[3]_{(n[4])}
 * $$n[4]_3\approx n'\uparrow\uparrow n'\uparrow\uparrow n'\uparrow\uparrow n'$$

\approx (n'\uparrow^3 n')\uparrow^3(n'\uparrow^3 n') \approx n'\uparrow^3 n'\uparrow^3 n'$$
 * $$n[5]=n[4]_n\approx n'\uparrow^3(n+1)=n'\uparrow^3 n'$$
 * $$n[5]_2=(n[5])[4]_{(n[5])}


 * $$n[6]=n[5]_n\approx n'\uparrow^4(n+1)=n'\uparrow^4 n'$$

...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].

\approx(257\uparrow^2 257)\uparrow^3(257\uparrow^2 257)$$ $$\approx257\uparrow^3 257\uparrow^2 257 \approx10\uparrow^3 257\uparrow^2 257$$ = A-ooga $$\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$$ $$\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$$
 * $$2[3]=2^2=4$$
 * $$2[4]=(2[3])[3]=4[3]=4^4=256$$
 * $$2[5]=(2[4])[4]=256[4]\approx(256\uparrow)^{256} 257\lesssim257\uparrow\uparrow257$$ = Mega
 * $$2[6]=(2[5])[5]\approx(257\uparrow\uparrow257)[5]
 * $$2[7]=(2[6])[6]\approx(10\uparrow^3 257\uparrow^2 257)[6]$$
 * $$2[8]=(2[7])[7]\approx(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)[7]$$

=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,

$$\approx10\uparrow^{p-3}10\uparrow^{p-4}10\uparrow^{p-5}\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257$$
 * $$2[p]=2^2[3][4][5]\dots[p-3][p-2][p-1]$$

=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, Because,
 * $$3\uparrow^{\text{Mega}-2}3\gtrsim2\uparrow^{\text{Mega}-2}4\gtrsim\text{Moser}$$

$$\approx\dots\uparrow^4 10\uparrow^3 12\uparrow^2 7625597484986$$ $$\approx\dots\uparrow^4 10\uparrow^3 6\uparrow^2 65534$$
 * $$3\uparrow^{\text{Mega}-2}3=\dots\uparrow^4 3\uparrow^3 3\uparrow^2 7625597484987$$
 * $$2\uparrow^{\text{Mega}-2}4=\dots\uparrow^4 2\uparrow^3 2\uparrow^2 65536$$
 * $$\text{Moser}\approx\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257$$

After all, However, Mega is already quite large number. This might be the most simple approximation, using up-arrow notation...
 * $$\text{Moser}\approx3\uparrow^{257\uparrow\uparrow257}3$$