n-Dimensional CubesDate: 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/ |
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