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A square squared (Posted on 2004-03-20) Difficulty: 3 of 5
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: 4.6000 (5 votes)

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Solution The Rest of a Solution | Comment 2 of 5 |
(In reply to Most of a solution by Charlie)

To find the formula for the number of s-dimensional components in a cube of c dimensions, you need to consider how the components of varying dimensions relate to each other.

As is shown in the chart in the first comment, the number of 0-dimensional components (vertices) is given by 2c.

There are c dimensions at right angles to each vertex, so c edges can be drawn from each. Each edge connects 2 vertices, so divide by 2 to find the number of edges.

There are (c-1) faces at right angles to each edge, and each face has 4 edges, therefore the number of faces is the number of edges multiplied by (c-1) and divided by 4.

For cubes, there are (c-2) from each face, and each cube has six faces, so multiply the faces by (c-2) and divide by 6.

To calculate the number of

Here you can see the resemblance to the Pascal sequence (which defines the terms of Pascal's triangle). A slight variation is all that is necessary. The general equation is:

<BLOCKQUOTE dir=ltr style=\"MARGIN-RIGHT: 0px\">

2^(c-s) c!

 s! (c-s)!

</BLOCKQUOTE>

This is the formula for the number of edges / faces / cubes etc. Since the c-dimensional cube has side length of one unit, the formula holds for this question. For a c-dimensional cube with side length n units, the s-dimensional surface area is given by the number of s-dimensional components multiplied by n^s.

 


  Posted by Iain on 2004-03-20 17:46:16
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