00:02
In part a, we're asked to find a formula for the number of vertices of an n pyramid in terms of the number of j faces of an n -minus -1 polytope q.
00:34
So recall that the n -piramid is formed by taking the convex whole of an n -1 polytope q, and a point x which is not in affine space of q so in part a it's pretty clear that f0 of pvn this is the number of vertices we're really only adding one more vertex which is that vertex outside the affine space of q this is going to be f0 of q plus one now in part b we're asked to find the number of k face of an n pyramid where k ranges from 1 all the way up to n minus 2.
01:53
So consider the number of two faces or the number of square faces.
02:01
I guess we should start with number of one faces.
02:03
These are the number of edges of a n pyramid.
02:13
Well, we have all the edges from polytoppe q.
02:20
And then in addition to that, for each one, vertex in the polytopic u i'm going to add exactly one more edge in the n pyramid so it follows that f1 of p vn is equal to f1 of q plus f0 of q now consider the number of square faces of p to n well this is going to be the number of square faces in the polytope q, in addition to that, for each edge in the polytope q, we're going to be adding exactly one more square face in p to the n.
03:23
So the edge in that third point outside the f1 space of q forms a new square face.
03:31
So we have that f2 of p to the n is equal to f1 of q, or sorry, f2 of q plus f1 of q.
03:50
And in fact, you might see a pattern here.
03:53
So you might guess that in general, fk of p to the n is going to be fk of q plus fk minus 1 of q.
04:11
And to see why, well, we have the number of k faces of p to the n.
04:18
This is going to be, we have all the k faces from polytope q.
04:23
And then for each of the k minus 1 faces of q, the addition of the new point in p the n leads to exactly one more k face in pvn.
04:43
So we have one more k face for each k minus one face in q...