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 A square squared (Posted on 2004-03-20)
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.

 See The Solution Submitted by Tristan Rating: 4.2000 (5 votes)

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As the lengths, areas, volumes, etc. are all of unit size, the total value of each of these will just be a count of the number of edges, faces, cubic pieces, etc.

When transitioning from a given dimensional space to the next, say from a square to a cube, this can be said to be done by moving the original shape perpendicular to itself, during which process, each k-dimensional element traces out one element of k+1-dimension, but also leaves an initial and final impression of itself.  To illustrate, as a square is moved through the third dimension to trace out a cube, each of its vertices traces out a new edge, making 4 edges to the cube so far considered.  Each vertex also leaves an initial and final impression of itself, so the 4 vertices of a square lead to 8 vertices of the cube.  Meanwhile, the 4 edges of the square also leave their initial and final impressions; adding these 8 to the 4 edges traced by the vertices makes 12 vertices all together.  Finally, each edge traces out a square, making 4 faces, plus the original square leaves its initial and final impression, making a total of 6 faces on the cube.

The points (vertices) themselves are never generated by lower-dimensional pieces, and so just double each successive dimension: two end-points of a line segment, 4 vertices of a square, 8 vertices of a cube, 16 vertices of a hypercube.

Here's a chart for dimensions 1 through 7:
V   E   F   S
2   1   0   0   0   0   0   0 line segment
4   4   1   0   0   0   0   0 square
8  12   6   1   0   0   0   0 cube
16  32  24   8   1   0   0   0 hypercube
32  80  80  40  10   1   0   0
64 192 240 160  60  12   1   0
128 448 672 560 280  84  14   1

Where V, E, F, and S represent vertices, edges, faces and solids.

So a 4-dimensional hypercube has 32 edges, and if they are of unit length, the total length is 32. There are 24 2-d faces having a total area of 24.  There are 8 cubic surface volumes, each of unit volume, making the total volume of the hypersurface of the hypercube 8.

The above presents a recursive way of getting the figures for successive dimensions.  As for a direct formula for a given spot on the chart (where each number is twice the number just above it plus just 1 times the number up and to the left from it), that will take some more thought, but probably is related to combinations, as it is quite similar to the formation of Pascal's triangle.

 Posted by Charlie on 2004-03-20 11:23:31

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