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n-Dimensional Cubes

Date: 09/06/2002 at 19:54:41
From: Greg
Subject: n-dimensional cubes

Define #(n,A) to be the number of corners of an n-dimensional cube 
whose distance to 0 is greater than A.

The limit (as n -> infinity) #(n,A)/(2^n) = 1.

How would I verify the limit statement?


Date: 10/26/2002 at 23:07:29
From: Doctor Nitrogen
Subject: Re: n-dimensional cubes

Hi, Greg:

Here is the framework of an argument for your answer:

Consider what happens when n = 2 and we have an ordinary square in 2D 
space. Place the center of the square at the origin, and let the 
length of the square's edge be L:

            (-L/2, L/2)  |-------------|(L/2, L/2)
                         |             |
                         |             |
                         |   (0, 0)    |
                         |             |
                         |             |
                         |             |
            (-L/2, -L/2) |-------------|(L/2, -L/2)

The square's four vertices are at (-L/2, L/2), (L/2, L/2),           
(-L/2, -L/2), and (L/2, -L/2). The distance from ANY vertex (corner) 
to the center of the square is equal to:

(1) sqrt(abs(+-L/2)^2 + abs(+-L/2)^2) = L*sqrt(2)/2,

where L*sqrt(2)is the length of the diagonal between any two opposite 
vertices.  

Now let A be ANY length LESS than L*sqrt(2)/2. This implies that 
  
   #(n, A) = 4.

Can you see this? The distance from any vertex of this square of edge 
length L to the origin is greater than A, provided 

   A < L*sqrt(2)/2.

Now, for your problem, notice for n = 2 the result 

   #(n, A)/2^2 = 4/4 = 1.

Suppose n = 3 and we are in 3D space. Consider a cube of edge length 
L. Place (0, 0, 0) at the center of the cube. Then the eight vertices 
will be at:

   (-L/2, L/2, -L/2), (-L/2, -L/2, L/2)

   (-L/2, -L/2, -L/2), (L/2, L/2, L/2)

   (L/2, L/2, -L/2), (L/2, -L/2, L/2)

   (L/2, -L/2, -L/2), (-L/2, L/2, L/2)


The length of any diagonal between opposite vertices will be 

   L*sqrt(3)

and the distance of any vertex from the center (0, 0, 0) will be

   L*sqrt(3)/2

which is half the length of the diagonal. So, for 

   L*sqrt(3)/2 > A

you will find that #(n, A)/2^3 = 8/8 = 1, for n = 3.

You can generalize this for higher dimensions. For example, for the 
tesseract (4-dimensional hypercube) with center at the origin

   (0, 0, 0, 0),

you will have a diagonal between any two opposite vertices of length

   sqrt(4) = 2,

   and for L*sqrt(4)/2 > A:

   #(n, A)/2^4 = 1 for n = 4, 

since all 2^4 = 16 vertices of the tesseract have the same distance 
from the center. 

The generalization for n > 4 would be, with the diagonals between 
opposite vertices equal in length to sqrt(n):

1. The distance from any vertex to the center 

   (0, 0, 0, 0, .............., 0), n zeros,

is L*sqrt(n)/2 in length.

2. If L*sqrt(n)/2 > A, then #(n, A)/2^n = 1. So, essentially, the 
limit should always be 1, as #(n, A) = 2^n for the n dimensional cube. 

Of course this is just a framework for an argument, so you might have 
to prove this formally to convince yourself.

You will find a link that has material on the diagonal used in this 
argument here at:

   Diagonals of Cubes in Different Dimensions
   http://mathforum.org/mam/00/master/essays/B3D/2/diagonals.html 

But please keep in mind that for the n cube, there are two KINDS of 
diagonals. Only one kind of diagonal is used in my above argument, 
that is, the diagonal between any two opposite vertices. 

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Higher-Dimensional Geometry

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