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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 

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 


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


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 

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 
Associated Topics:
College Higher-Dimensional Geometry

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