Polynomials can be used find the elements of a hypercube.

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Polynomials can be used find the elements of a hypercube.

Postby W axis » Tue Nov 11, 2008 9:00 pm

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
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Re: Polynomials can be used find the elements of a hypercube.

Postby wendy » Wed Nov 12, 2008 8:03 am

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,#.
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Re: Polynomials can be used find the elements of a hypercube

Postby Keiji » Sat May 25, 2013 9:20 am

Hi wendy,

Although this is an old topic, I just found it while moving the forums around. Can you give a little more explanation on this, in particular, how a pyramid has two arguments?

Am I right in saying

pyramid = **
prism = *#
tegum = #*
comb = ##

Can you give some examples of these with say, all three dimensional arguments?
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Re: Polynomials can be used find the elements of a hypercube

Postby Klitzing » Sat May 25, 2013 10:37 am

Coming back to the original polynomials:

  • Simplex product "polynomial": (1/x)*[(x+1)^(D+1)],
    e.g. for D=2 (triangle): x^2+3x+3+1x^(-1)
    i.e. 1 body term (x^2) and 1 nulloid term (x^(-1))
  • Prism product polynomial: (x+2)^D,
    e.g. for D=3 (cube): x^3+6x^2+12x+8
    i.e. 1 body term (x^3) but no nulloid term (x^(-1))
  • Tegum product "polynomial": (1/x)*[(2x+1)^D],
    e.g. for D=3 (octahedron): 8x^2+12x+6+1x^(-1)
    i.e. no body term (x^3) but 1 nulloid term (x^(-1))
  • Honeycomb product polynomial: [N*(x+1)]^D, where N-> infinity
    e.g. for D=3 (cubical honeycomb): N^3*(x^3+3x^2+3x+1)
    i.e. no body term (as a horochoron, when used in hyperbolic tesselations: x^4), no nulloid (x^(-1))

--- rk
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