Steinhaus-Moser notation

(Redirected from Steinhaus polygon notation)

In mathematics, Moser's polygon notation is a means of expressing certain extremely large numbers. It is an extension of Steinhaus's polygon notation.

(a number n in a triangle) means nⁿ

(a number n in a square) is equivalent with "the number n inside n triangles, which are all nested"

(a number n in a pentagon) is equivalent with "the number n inside n squares, which are all nested"

etc.: n written in an (m+1)-sided polygon is equivalent with "the number n inside n m-sided polygons, which are all nested"

Steinhaus only defined the triangle, the square, and a circle , equivalent to the pentagon defined above.

Steinhaus defined:

"mega" is the number equivalent to 2 in a circle:
"megistron" is the number equivalent to 10 in a circle:

Moser's number is the number represented by "2 in a megagon", where a "megagon" is a polygon with "mega" sides.

Alternative notations:

use the functions square(x) and triangle(x)
let M(n,m,p) be the number represented by the number n in m nested p-sided polygons; then the rules are:
- $M (n,1,3) = n n$
- $M (n,1, p + 1) = M (n, n, p)$
- $M(n,m+1,p) = M\big(M(n,1,p),m,p\big)$

and

- mega = $M (2,1,5)$
- moser = $M\big(2,1,M(2,1,5)\big)$

Contents

Mega

Note that is already a very large number, since = square(square(2)) = square(triangle(triangle(2))) = square(triangle(2²)) = square(triangle(4)) = square(4⁴) = square(256) = triangle(triangle(triangle(...triangle(256)...))) [256 triangles] = triangle(triangle(triangle(...triangle(256²⁵⁶)...))) [255 triangles] = triangle(triangle(triangle(...triangle(3.2 × 10⁶¹⁶)...))) [254 triangles] = ...

Using the other notation:

mega = M(2,1,5) = M(256,256,3)

With the function $f (x) = x x$ we have mega = $f 256 (256) = f 258 (2)$ where the superscript denotes a functional power, not a numerical power.

We have (note the convention that powers are evaluated from right to left):

M(256,2,3) = $(256^{\,\!256})^{256^{256}}=256^{256^{257}}$
M(256,3,3) = $(256^{\,\!256^{257}})^{256^{256^{257}}}=256^{256^{257}\times 256^{256^{257}}}=256^{256^{257+256^{257}}}$ ≈ $256^{\,\!256^{256^{257}}}$

Similarly:

M(256,4,3) ≈ ${\,\!256^{256^{256^{256^{257}}}}}$
M(256,5,3) ≈ ${\,\!256^{256^{256^{256^{256^{257}}}}}}$

etc.

Thus:

mega = $M(256,256,3)\approx(256\uparrow)^{256}257$ , where $(256\uparrow)^{256}$ denotes a functional power of the function $f (n) = 256 n$ .

Rounding more crudely (replacing the 257 at the end by 256), we get mega ≈ $256\uparrow\uparrow 257$ , using Knuth's up-arrow notation.

Note that after the first few steps the value of $n n$ is each time approximately equal to $256 n$ . In fact, it is even approximately equal to $10 n$ (see also approximate arithmetic for very large numbers). Using base 10 powers we get:

$M(256,1,3)\approx 3.23\times 10^{616}$
$M(256,2,3)\approx10^{\,\!1.99\times 10^{619}}$ ( $log 10616$ is added to the 616)
$M(256,3,3)\approx10^{\,\!10^{1.99\times 10^{619}}}$ ( $619$ is added to the $1.99\times 10^{619}$ , which is negligible; therefore just a 10 is added at the bottom)

$M(256,4,3)\approx10^{\,\!10^{10^{1.99\times 10^{619}}}}$

...

mega = $M(256,256,3)\approx(10\uparrow)^{255}1.99\times 10^{619}$ , where $(10\uparrow)^{255}$ denotes a functional power of the function $f (n) = 10 n$ . Hence $10\uparrow\uparrow 257 < \mbox{mega} < 10\uparrow\uparrow 258$