Triangular numbers and n-simplices

Higher-dimensional geometry (previously "Polyshapes").

Triangular numbers and n-simplices

Postby quickfur » Mon Dec 01, 2008 8:23 pm

I've found the following curious relation between the triangular numbers of old and the coordinates of the n-simplex.

Let B(i) be the square root of the reciprocal triangular numbers, that is to say, B(i) = sqrt(2 / (i(i+1))).
Let A(i) = -sqrt(2i / (i+1)) = -sqrt(i^2 * (2 / (i(i+1)))). That is to say, A(i) is the negative square root of the square numbers divided by the triangular numbers.

Define the following set of points in n-space:

Code: Select all
(B(1), B(2), B(3), ... , B(n-2), B(n-1), B(n))
(A(1), B(2), B(3), ... , B(n-2), B(n-1), B(n))
(   0, A(2), B(3), ... , B(n-2), B(n-1), B(n))
(   0,    0, A(3), ... , B(n-2), B(n-1), B(n))
(   0,    0,    0, ... ,      0, A(n-1), B(n))
(   0,    0,    0, ... ,      0,      0, A(n))

The convex hull of these points form a regular n-simplex centered on the origin with edge length 2. The apex lies along the last coordinate axis (so in a sense the simplex is "upright") and the base has a plane of symmetry, with these two properties recursively true for the base (n-1)-simplex. So this is no accidental coincidence... the triangular numbers give rise to a "nice" orientation of the n-simplices.

The Greeks were on to something, I tell ya! ;)
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Re: Triangular numbers and n-simplices

Postby Keiji » Tue Dec 02, 2008 12:01 pm

Nice find! :D
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