4-Dimensional Objects

[argo]

Right. I'm not a stupid person, or at least I don't think so, when it comes to these things, and yet I have been left by most of this thread puzzled as to why anyone would think that a "4D" object would have any more sides than a "3D" object when the 4th dimension is an expression of transition through TIME rather than spacial dimensionality......?

Here's your problem; you are confusing physics with mathmatics. We can derive a mathematical formula to describe an object in one dimension, or two dimensions or three dimensions. Those dimensions do not have to have real-world analogues such as "height" or "width" they are idealized mathematical constructs. The question this thread is asking is what is the formula(s) for describing an object in 4 arbitrary, idealized dimensions (or in n arbitrary, idealized dimensions where n is a variable) regardless of what its real-world analogue may be.

I'll leave the discussion of how hyperdimensions may or may not translate into real-world physical analogues to people with a more developed frontal lobe than I. N-dimensional math always gave me a migrane

 

[Cor Azer]

There is nothing arbitrary about it. 0! = 1 is perfectly consistent with the Gamma function.

True, I guess "arbitrarily" was the wrong word to use.

Still, if a person is unfamiliar with factorials, I have a strong suspicion they wouldn't fully understand the Gamma function.

 

[Tarril Wolfeye]

True, I guess "arbitrarily" was the wrong word to use.

Still, if a person is unfamiliar with factorials, I have a strong suspicion they wouldn't fully understand the Gamma function.

You don't need the Gamma function or any arbitrarity (is that a word?)

If you define n! = n * (n-1)! and say n = 1 then 1! = 1 * 0! and so 0! = 1! = 1
Of course this doesn't work for (-1)!, because you have to divide by zero.

 

[Tarril Wolfeye]

Anyway, for 3D solids, S = E - V + 2. A cube has six sides, 12 edges, and 8 vertices. 6=12-8+2. For the triangular-based pyramid (D4) it's 4=6-4+2. For the square-based pyramid it's 5=8-5+2. It would be interesting to derive the formula for polychorons... if I could get this formula to kick out the right numbers!

The formula for Polychorons is V + S = E + C.
V = vertices (corners), E = edges, S = sides, C = cells

http://mathworld.wolfram.com/PolyhedralFormula.html