A four-dimensional hypercube has vertices connecting 4 edges, each edge 90 degrees apart from each other. Each edge is 1 unit long. Find the 3-d surface volume of this cube. Find the 2-d surface area. Find the sum of the lengths of all the edges.

Find a general equation for the s-dimensional surface area of an c-dimensional cube with one unit side length. For example, if s=1 and c=3, you're finding the sum of the length of the edges of a normal cube.

A c-dimensional unit cube may be described as the set of all vectors of the form a1*e1+a2*e2+...+ac*ec where e1,e2,...,ec are orthogonal unit vectors, one in the direction of each coordinate axis, and a1,a2,...,ac are real numbers that vary over the closed interval [0,1]. The vertices, edges, faces, cubic faces, ..., s-dimensional faces, ..., are formed by fixing the elements of subsets of size c-s from {a1,a2,...,ac} at values 0 or 1 and letting the others (s in number) vary over [0,1]. The number of such subsets is clearly (c,c-s)=(c,s), and since each subset member can be set to either 0 or 1, there are exactly (c,s)*2^(c-s) different s-dimensional faces, each of unit s-dimensional "area" (or "content" as it is usually called). This is the same result as Iain found, and gives the same numbers as Charlie's table.

Edited on March 21, 2004, 10:22 pm

Posted by Richard on 2004-03-20 20:02:54