Steinhaus–Moser notation

Template:Short description In mathematics, Steinhaus–Moser notation is a notation for expressing certain large numbers. It is an extension (devised by Leo Moser) of Hugo Steinhaus's polygon notation.<ref>Hugo Steinhaus, Mathematical Snapshots, Oxford University Press 1969³, Template:ISBN, pp. 28-29</ref>

DefinitionsEdit

n in a triangle a number Template:Math in a triangle means Template:Math.

n in a square a number Template:Math in a square is equivalent to "the number Template:Math inside Template:Math triangles, which are all nested."

n in a pentagon a number Template:Math in a pentagon is equivalent to "the number Template:Math inside Template:Math squares, which are all nested."

etc.: Template:Math written in an (Template:Math)-sided polygon is equivalent to "the number Template:Math inside Template:Math nested Template:Math-sided polygons". In a series of nested polygons, they are associated inward. The number Template:Math inside two triangles is equivalent to Template:Math inside one triangle, which is equivalent to Template:Math raised to the power of Template:Math.

Steinhaus defined only the triangle, the square, and the circle n in a circle, which is equivalent to the pentagon defined above.

Special valuesEdit

Steinhaus defined:

mega is the number equivalent to 2 in a circle: Template:Tooltip
megiston is the number equivalent to 10 in a circle: ⑩

Moser's number is the number represented by "2 in a megagon". Megagon is here the name of a polygon with "mega" sides (not to be confused with the polygon with one million sides).

Alternative notations:

use the functions square(x) and triangle(x)
let Template:Math be the number represented by the number Template:Math in Template:Math nested Template:Math-sided polygons; then the rules are:
- <math>M(n,1,3) = n^n</math>
- <math>M(n,1,p+1) = M(n,n,p)</math>
- <math>M(n,m+1,p) = M(M(n,1,p),m,p)</math>
and
- mega = <math>M(2,1,5)</math>
- megiston = <math>M(10,1,5)</math>
- moser = <math>M(2,1,M(2,1,5))</math>

MegaEdit

A mega, ②, is already a very large number, since ② = 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(3.2317 × 10⁶¹⁶)...))) [255 triangles] ...

Using the other notation:

mega = <math>M(2,1,5) = M(256,256,3)</math>

With the function <math>f(x)=x^x</math> we have mega = <math>f^{256}(256) = f^{258}(2)</math> where the superscript denotes a functional power, not a numerical power.

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

<math>M(256,2,3) =</math> <math>(256^{\,\!256})^{256^{256}}=256^{256^{257}}</math>
<math>M(256,3,3) =</math> <math>(256^{\,\!256^{257}})^{256^{256^{257}}}=256^{256^{257}\times 256^{256^{257}}}=256^{256^{257+256^{257}}}</math>≈<math>256^{\,\!256^{256^{257}}}</math>

Similarly:

<math>M(256,4,3) \approx</math> <math>{\,\!256^{256^{256^{256^{257}}}}}</math>
<math>M(256,5,3) \approx</math> <math>{\,\!256^{256^{256^{256^{256^{257}}}}}}</math>
<math>M(256,6,3) \approx</math> <math>{\,\!256^{256^{256^{256^{256^{256^{257}}}}}}}</math>

etc.

Thus:

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

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

After the first few steps the value of <math>n^n</math> is each time approximately equal to <math>256^n</math>. In fact, it is even approximately equal to <math>10^n</math> (see also approximate arithmetic for very large numbers). Using base 10 powers we get:

<math>M(256,1,3)\approx 3.23\times 10^{616}</math>
<math>M(256,2,3)\approx10^{\,\!1.99\times 10^{619}}</math> (<math>\log _{10} 616</math> is added to the 616)
<math>M(256,3,3)\approx10^{\,\!10^{1.99\times 10^{619}}}</math> (<math>619</math> is added to the <math>1.99\times 10^{619}</math>, which is negligible; therefore just a 10 is added at the bottom)
<math>M(256,4,3)\approx10^{\,\!10^{10^{1.99\times 10^{619}}}}</math>

...

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