Understanding Fourth Dimension FiguresDate: 07/05/98 at 20:06:22 From: Heidi Subject: fourth dimension equations I am doing a senior research project on the progression of the fourth dimension, gearing toward teaching a high school level geometry course. I have located the standard form for equations of a hypersphere, but my instructor asked me to include equations for several more figures including the tesseract and hypertetrahedron. I have searched your archives and have found some very useful material. As I mentioned earlier, I am a pre-service teacher and cannot wait until I can use this site as a reference for my classroom. Thanks for providing such a valuable service for students, parents and educators. Heidi Date: 07/05/98 at 22:52:21 From: Doctor Tom Subject: Re: fourth dimension equations Hello Heidi, The easiest way to think about 4 (or more) dimensions is to see what is going on in 1, 2, and 3 dimensions, and see what the pattern is. At least that will work for your examples. For example, the tesseract is just a 4D cube. Well, what does a 3D, 2D, and 1D "cube" look like? If we're interested in the points that lie inside the cube, a 1D cube is just a line segment. If in all cases we're looking at a cube with length 1, in 1D, the "cube" is all the points such that 0 <= x <= 1. In 2D, it is a square, so the interior is all the points (x, y) satisfying 0 <= x <= 1 and at the same time, 0 <= y <= 1. In 3D, it is the same as the square, except the interior is the set of all points (x, y, z) satisfying 0 <= x <= 1 and 0 <= y <= 1 and 0 <= z <= 1. To save space, this is often written as: 0 <= x, y, z <= 1. So in 4D, the interior of a tesseract is the set of all points (x, y, z, w) such that 0 <= x, y, z, w <= 1. In other words, points that have all their 4 coordinates between 0 and 1. The hyper-tetrahedron should be examined in the same way. In 1D, it is a line segment. In 2D it is a triangle. In 3D, it is a tetrahedron. I think the easiest way to look at these things is in terms of barycentric coordinates. For a triangle with the three vertices P0 = (x0, y0), P1 = (x1, y1), and P2 = (x2, y2), all the interior points are those that have the form a*P1 + b*P2 + c*P3, where a + b + c = 1 and a, b, and c are bigger than zero. In other words, all points of the following form: (a*x0 + b*x1 + c*x2, a*y0 + b*y1 + c*y2) For a line going from x0 to x1, it is all the points a*x0 + b*x1 where a + b = 1 and a and b are non-negative. You can work out the 3D form yourself, but here is the answer in 4-D: Given 5 points P0 = (x0, y0, z0, w0), P1 = (x1, y1, z1, w1), P2 = (x2, y2, z2, w2), and so on for P3 and P4, the interior of a hyper-tetrahedron (of dimension 4) is given by all the points satisfying: a*P0 + b*P1 + c*P2 + d*P3 + e*P4 where a + b + c + d + e = 1 and all of a, b, c, d, and e are positive. Or: (a*x0+b*x1+c*x2+d*x3+e*x4, a*y0+b*y1+c*y2+d*y3+e*y4, ...) If this doesn't make much sense, do look up barycentric coordinates. They're often very useful. Good luck in your future career. - Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/