hypervolume of truncated tesseract?

[Jonathan Post]

Hypervolume ~ 33.803896 as below:

Example: n = 7

Since 7 = 2^2 + 1^2 + 1^2 + 1^2, we have 4 permutations, and 16 sign
permutations, for 4*16 = 64 vertices.

(+-2, +-1, +-1, +-1) is a truncated tesseract.

In geometry, a truncated tesseract is a uniform polychoron
(4-dimensional uniform polytope) which is bounded by 24 cells: 8
truncated cubes, and 16 tetrahedra.  It has 64 vertices, 128 edges, 88
faces (64 triangles and 24 octagons).

We know that: "The truncated tesseract may be constructed by
truncating the vertices of the tesseract at 1 / (2 + sqrt(2)) of the
edge length."

1 / (2 + sqrt(2)) = 0.292893219

http://en.wikipedia.org/wiki/Truncated_tesseract

Don Reble  djr at nk.ca comments (used with permission):

 "In other words, the truncated squares are regular octagons."

> the hypervolume of a truncated tesseract is ... ?

"In N dimensions, a rectangular simplex is a simplex where N of the
edges are mutually orthogonal line segments of the same length L. It's
size is L^N / N! ."

"From the unit hypercube, each truncated corner loses one such
simplex, of hypervolume 1 / (1632 + 1152 sqrt2). There are 16 corners,
so the total loss is 1 / (102 + 72 sqrt2) = 0.00491..."

"That may be surprisingly tiny, but remember, those corners are small
to the fourth power, divided by 4!."

In this sum-of-squares example, we are not starting with "unit
hypercube" but, rather, ending after truncation with unit edge.  Don
Reble is absolutely correct to invoke the octagon as the basis for
normalization.

We know that:

An octagon is an eight-sided polygon. The inradius r, circumradius R,
and area A of the regular octagon can be computed directly from the
formulas for a general regular polygon with side length s and n = 8
sides as:

r = (1/2) s cot (pi/8) = (1/2) (1 + sqrt 2) s

R = (1/2) s csc (pi/8) = (1/2) sqrt[(4 + 2 sqrt 2)] s

A = (1/4) n s^2 cot (pi/8) =  2 (1 + sqrt 2) s^2

In our example, defined by (+-2, +-1, +-1, +-1), we have edge length:

d[(2, 1, 1, 1), (1, 2, 1, 1)] = sqrt[(2-1)^2) + (1-2)^2 + (1-1)^2 +
(1-1)^2] = sqrt 2.

There are diagonals of various lengths:

d[(2, 1, 1, 1), (-2, 1, 1, 1)] = sqrt[4^2 + 0^2 + 0^2 + 0^2] = 4.

d[(2, 1, 1, 1), (-2, 1, 1, -11)] = sqrt[0^2 + 0^2 + 0^2 + 2^2] = 2.

Our edge length of sqrt 2 combined with Don Reble's analysis gives:

L = 1 + sqrt 2, so the tesseract (before truncation) has hypervolume
(1 + sqrt 2)^4 ~ 33.9705627
After we remove the 16 rectangular tetrahedral, we have:

(1 + sqrt 2)^4   *   (1 – 1/(102 + 72 * sqrt 2)) ~ 33.9705627  * (1 -
0.00490620859)
~ 33.803896.




* Mohammed BOUAYOUN <Mohammed.BOUAYOUN at sanef.com> [Mar 29. 2007 07:34]:
>  
> Last day  was 27/2007,
> If we take the numbers 27,207,2007,20007, ... we have :
> the sum of digits 2 and 7 are 2+7=9 = 3^2
> and
> 27 = (3^2)*3
> 207=(3^2)*23
> 2007=(3^2)*223
> 20007=(3^2)*2223
> ...
> 200..007 = (3^2)*22..223
>  
> Your comment

Please do not submit any sequence based on this  ;-)

Observations of this type can be obtained by polynomial identities:
write a base R number as a0+a1*R*a2*R^2+...+an*R^n.
All sum of digits observation are obtained by
reducing the polynomial modulo _something_.
In your case something equals R-1 (nine) which
gives the plain old sum-of-digits.


>  
> Thanks
>  
>  

regards,   jj