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What is the hypervolume of an n-dimensional hypercube?
Hint: What are the hypervolumes of a line segment, a square, and a cube?

How many vertices does an n-dimensional hypercube have?
Hint: How does the number of vertices change in constructing an n-dimensional hypercube from an (n – 1)-dimensional hypercube?

What is the number of edges contained in an n-dimensional hypercube in terms of the numbers of vertices and edges in an (n – 1)-dimensional hypercube?
Hint: How does the number of edges change in constructing an n-dimensional hypercube from an (n – 1)-dimensional hypercube?

What is bn, m for hypercubes in terms of previous entries?
Hint: In constructing an n-dimensional hypercube from an (n – 1)-dimensional hypercube, each (m – 1)-dimensional hypercube in the (n – 1)-dimensional hypercube contributes an m-dimensional hypercube in the n-dimensional hypercube.

What is bn, m for hypercubes?
Hint: The previous recurrence for bn, m must be solved explicitly. First find bn, 1 by repeatedly substituting the recurrence relation into itself until a closed-form expression is obtained. An expression for bn, 2 can be found similarly, etc.

What is the surface hyperarea of an n-cube?
Hint: How many (n – 1)-cubes are there in the n-cube, and what is the hypervolume of each?

How many vertices does an n-simplex have?
Hint: The natural generalization is the right one.

What is bn, m for simplices?
Hint: Notice that every set of three vertices in an n-simplex forms a triangle, every set of four forms a tetrahedron, etc. Binomial coefficients count the number of these objects.

What is the height of a regular n-simplex?
Hint: The hypervolume of an n-dimensional pyramid is h · A / n, where h is the height of the pyramid and A is the hyperarea of the base. What is the hypervolume of an n-simplex in terms of the apothem an and its (n – 1)-dimensional surface hyperarea? By equating the two, one can find the height in terms of n and an.

What is the hypervolume of a regular n-simplex in terms of s?
Hint: Find hn by considering right triangles.


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Eric Rowland