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August 15

Division by planes (generally)

This is based off of the pizza cutting question above. Will the maximum number of regions of a 3d object with n cuts (http://mathworld.wolfram.com/CubeDivisionbyPlanes.html, http://mathworld.wolfram.com/TorusCutting.html) always be a cubic polynomial? How can these polynomials be generated from an arbitrary shape? 70.162.10.166 (talk) 04:00, 15 August 2012 (UTC)[reply]

Expected value of half-normal sum?

I'm reading a paper with the following:

$\phi _{K}=E[{\text{max}}(\sum _{k=1}^{K}X_{k},0)]={\sqrt {\frac {K}{2\pi }}}$

where we know that all the $X_{k}$ are i.i.d. $N(0,\sigma ^{2})$

I would have thought that $\sum _{k=1}^{K}X_{k}$ would have a variance of $K\sigma ^{2}$ and that ${\text{max}}(\sum _{k=1}^{K}X_{k},0)$ would be a half-normal distribution meaning that $E[{\text{max}}(\sum _{k=1}^{K}X_{k},0)]$ would be $\sigma {\sqrt {\frac {2K}{\pi }}}$ rather than $\sigma {\sqrt {\frac {K}{2\pi }}}$

what am I missing? it's stated as though it is obvious but I'm not sure how it follows.

It's not a half-normal distribution. It's a mixture variable with 50% chance of being half-normal and 50% chance of being 0. Hence, its mean is only half that of a half-normal variable. -- Meni Rosenfeld (talk) 08:40, 15 August 2012 (UTC)[reply]

ah, I see. Thank you. — Preceding unsigned comment added by 202.14.156.14 (talk) 22:31, 15 August 2012 (UTC)[reply]

Maximum number of k-planes in n-space with the same mutual angle

What is the maximum number of 2-planes in Euclidean 3-space that have the same nontrivial* angle between each pair of planes? Okay, I know the answer to this, but does this generalize readily to k-planes in n-space? — Quondum^☏ 13:56, 15 August 2012 (UTC)[reply]

I think that it does generalise for k = 1 or n − 1, by considering the (n - 1)-faces of a n-simplex and the vertices of the same shape. Difficult to prove directly but each time you go up a dimension you can add one more plane/edge but not any more.

In other cases though, the first of which is 2-planes in 4-space, does it even make sense to talk about the angle between? I.e. I'm not sure it is well defined, or you at least need to say what the definition is first.--JohnBlackburne^words_deeds 15:12, 15 August 2012 (UTC)[reply]

IMHO the first case (i.e. the lowest dimension) would be 1-planes (which are lines...) in 3-space, and the answer would be 'at least 6', eg. lines joining centers of a cube's opposite edges. --CiaPan (talk) 22:05, 15 August 2012 (UTC)[reply]

They aren't equally spaced. If the cube has vertices {±1, ±1, ±1} then consider the lines through {1, 1, 0}, {1, -1, 0}, {1, 0, 1}. The angle between the first two is 90°, the angle between the second two is less.

One way of thinking of it: if you have an collection of 1-planes/lines you should be able to form a convex polytope from the convex shell of unit-length segments of these lines, with the points a distance 1 along the lines as vertices. As all the points subtend the same angle at the origin they must be the same distance apart. So any three of them form an equilateral triangle. Any four form a tetrahedron. Etc. The only n-polytope with triangles for faces, tetrahedrons for polyhedra etc. is a simplex, so the answer is the n + 1 vertices of a simplex. The (n − 1)-planes are then just the duals of those lines.--JohnBlackburne^words_deeds 23:13, 15 August 2012 (UTC)[reply]

You're thinking along the right lines, but the answer using n-simplexes is not correct (though it would be for oriented lines). You've given me food for thought on the definition of angle between k-planes in n-space: the concept becomes pretty complicated for n ≥ 4, so let's confine it to lines or n−1 planes in n-space. The duality of lines and n−1 planes in n-space has a nice symmetry, so we only need to consider k = 1 (lines in n-space). — Quondum^☏ 01:58, 16 August 2012 (UTC)[reply]

You're right, JohnBlackburne. I missed that, shame on me. --CiaPan (talk) 07:52, 16 August 2012 (UTC)[reply]

Okay, let me put some figures to the answer, to move into the generalization:

n = 1: 1 line

n = 2: 3 lines with the pairwise mutual angle of π/3

n = 3: 6 lines with the pairwise mutual angle of arctan(2)

~~n ≥ 4: I have no idea.~~

n ≥ 4: ≥ n + 3 (moving up one dimension, one can add at least one line) — Quondum^☏ 21:15, 17 August 2012 (UTC)[reply]

— Quondum^☏ 01:45, 17 August 2012 (UTC)[reply]

For n = 4 you may not be able to do better than 4 lines. You can think of the lines as pairs of antipodal points on a 3-sphere, so they form some convex polychoron, and I think it has to be regular as well (is there any way around this?). There are only 6 Convex regular polychoron. The simplex is ruled out since it doesn't have antipodal pairs of vertices. We have the additional requirement that any particular vertex must be adjacent to at least one vertex from each of the other antipodal pairs. Of the rest, only the 16-cell has this property (since the rest have too few vertices adjacent to a given vertex).

If this is right for n = 4 then for all n ≥ 4 you can only get n lines, since the the only convex regular n-polytopes for n > 4 are the simplex, the cross polytope (corresponding to the solution) and the n-cube which has 2ⁿ vertices but each vertex has only n neighbors. Rckrone (talk) 04:42, 17 August 2012 (UTC)[reply]

The claim I made that the polytope has to be regular is false. An easy counter-example in 3 dimensions is to take the arrangement of 6 lines and delete one (or more). Also my claim that you can only get 4 lines in 4 dimensions is obviously false since you can get at least as many as in 3 dimensions. Rckrone (talk) 06:32, 17 August 2012 (UTC)[reply]

We have ≥n+1 using the n vertices from JohnBlackburne's n-simplex arguments, with ≥6 in the case of n = 4 from your argument here. — Quondum^☏ 10:27, 17 August 2012 (UTC)[reply]