Polynomials can be used find the elements of a hypercube.

[W axis]

I found that using (x+2)^n will give you a polynomial with the n-dimensional elements that make up the n-dimensional hypercube.

Example a hendekeract is:



X^11+22x^10+220x^9+1320x^8+5280x^7+14784x^6+29568x^5+42240x^4+42240x^3+28160x^2+11266x+2048

That factored is (x+2)^11

[wendy]

It works with all of the regular polytopes, and all of the regular products (prism, tegum, pyramid, comb). With these, one supposes that any given polytope has a '1' at each end (eg cube = 1, 6, 12, 8, 1).

The four products then equate to including (*) or excluding (#) the first and last digit, eg prism product = *#. so

line = 1,2,1 -> 1,2,#. The prism-power of a line is (1,2)^n, #
= 1,2,1 => #,2,1. The tegum power of a line is #,(2,1)^n

point = 1,1 -> 1,1 The pyramid power of a point is eg (1,1)^[n+1]

comb product of a polygon 1,p,p,1 -> #,p,p,# -> (p,p)^[n-1].

You can multiply any polygon to get the appropriate product, eg

triangle-prism = 1,3,3,# * 1,3,3,# = 1,6,15,18,9,#
triangle-dodecahedron tegum = 1,3,3,# * 1,12,30,20,# = 1,15,69,146,150,60,#

Even the 4d torus (comb) of dodechedron * pentagon

#,12,30,20,# * #,5,5,# = #,60,210,250,100,#.