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Suppose $\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}$ are linearly independent vectors in $\mathbb{R}^{n}(1 \leq k \leq n) .$ Then the set $X^{k}=\operatorname{conv}\left\{ \pm \mathbf{v}_{1}, \ldots, \pm \mathbf{v}_{k}\right\}$ is called a $k$ -crosspolytope. a. Sketch $X^{1}$ and $X^{2}$ b. Determine the number of $k$ -faces of the 3 -dimensional crosspolytope $X^{3}$ for $k=0,1,2 .$ What is another name for $X^{3} ?$ c. Determine the number of $k$ -faces of the 4 -dimensional crosspolytope $X^{4}$ for $k=0,1,2,3 .$ Verify that your answer satisfies Euler's formula.d. Find a formula for $f_{k}\left(X^{n}\right),$ the number of $k$ -faces of $X^{n},$ for $0 \leq k \leq n-1$

a) The sketch of $X^{1}$ is a line segment as shown below:The sketch of $X^{2}$ is a parallelogram as shown below:b) Octahedronc) since $8-24+32-16=0$ , these numbers satisfies Euler formula.d) $f_{k}\left(X^{n}\right)=2^{k+1}\left(\begin{array}{c}{n} \\ {k+1}\end{array}\right)$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 5

Polytopes

Vectors

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rest sketch a one cross polito in a to cross Polito in part a. So to do this, all that press Polito contains linearly independent vectors. So in particular the one cross polito X one This is going to contain one the nearly independent sector. And then it will be the convex hole of this specter and the opposite of the specter. And we see that it's drawing a sketch, started the origin and draw a vector to the point. The one and we're also going to have the point negative B one over here included. And, of course, all the points in between. Since this is the convex hole, so it follows that X one is a line segment over. Now find X two I want to do is find a vector which is literally independent from the vectors. Next one. So from the one in particular so way to think of this as well we just find a vector that lies off the line segment or the line drawn through X one. So once again, we have our origin. Zero you ever vector be one and we have the opposite of new one. That's all included. But now We're also going to include a second linearly independent vectors to, and we could just arbitrarily say that he, too, is somewhere down here. And we're also going to include its opposite and we're going to take the convex hole. So not only going to include the line between the two and it's opposite one segment there. We're also going to include this entire geometric figure here, plane figure, and we see it. This figure is actually a parallelogram, and the reason it's a parallelogram or one reason is because he knows that the diagonals bisect each other. That's why it's not just a quadrille lateral, So the picture here is not a great doesn't look like. Besides, air parallel should be now in part B were asked to determine the number of K faces of the three dimensional cross polito X three, where K ranges from zero to and then toe identify what X three is. Well, first of all, call. We saw that X two is a parallelogram, so the number of zero faces of X three, first of all, how are you going to obtain X three? I wouldn't have to have another. The nearly independent vector from the one and two. The specter of the three is going to be coming out of the plane of the page. It's about a three dimensional figure, and we'll also have its opposite, which will be below the page if the D three is above the page. So we see that zero number of zero faces of X three. This is the number of Vergis is this is simply going to be four. Vergis is from X two, plus the two new Vergis is which makes F zero of x three b six. Now consider the one faces of X three thes. They're going to be the edges of X three. Now we see that we have four edges next to and in addition to that, we're going to have two times or mawr edges. Because for each of the four corners of parallelogram next to each of those four, Vergis is yeah, we're going to have a line segment from that vertex to the three, and also from that Vertex to the opposition, please. That's two times four is eight more edges and therefore a total of 12 edges. So F one, the next three is 12 now, looking at f two of x three. The Coghlan's Again Well, f two. This these air the number of faces of the three dimensional Figure x three. So you recognize that the parallelogram really doesn't contribute any faces. Here. It's on the inside of the figures, however we do have for each pair of Vergis is the next to for each adjacent pair of er disease. We will obtain two more faces, so we have one face above the plane and then one face below the plane. That pair. So we're going to have to plus two plus two plus two for a total of eight. There's four pairs. Always prepare. The Vergis is so we have that F two of X three is going to be eight. Eight faces my next three. You recognize that either from these numbers or from just intuitive idea of what extra should look like. X three is an octahedron now in part, C were asked to determine the number of K faces of the four dimensional cross polito X four, where K ranges from zero up to three and were asked to verify the answer satisfies Oilers formula. This is a little more difficult because now we're working with a four dimensional object X four, Harder to visualize. However, First we find f zero x four. This is the number of verjus ease of export. So I call that next three. We found that there were six for disease in the next four. What we're doing is we're adding another linearly independent vector the four plus its opposite and then formally convex hole. So it follows that were actually adding two more. Vergis is, and therefore the total number of er disease F zero of X four is going to be six plus two, which is eight. Yeah. Now, to find out for one of x four, this is going to be the number of edges and x four. So a call from previous part that the number of edges of x three this was 12. So we had each of the edges from that form and then we have for each vertex in x three. We're going to add to mawr edges. So one for each of the new verses. Next four. So we have a total of 12 edges from next three plus two times six, which is 12 again. So 12 plus 12 or 24 edges. So the F one of X four is going to be 24 edges trying number of faces of X four. We're going to use the number of faces of X three. So we saw that in the previous part, Member of faces of X three was eight. So we're going to include these faces. However, we're actually going to have Hey, for each Vertex and X three, we're going to have Forem or faces. So actually, we could say that for each pair of Vergis ease in x three, we're going to be adding two more faces. So one for each of the new, literally independent vectors, the three and its opposite. So we have the eight faces from x three, and adding to that, we're going to have well, how many ways are there to choose? Adjacent pair from six, Virtus is in an octahedron 12345678 and then nine, 10 11 12 Ways says 12 ways to pick pair of Vergis is from an octahedron. This is another way of saying there's 12 edges in the bedroom. If you already knew So for each edge. We're going to have two more faces and therefore we have eight plus two times 12 which is a plus 24 which is going to be 32 faces. So F two of X four is 32 To find the number of cubic faces of export. Well, we have that ultimately x three from which explores built. This is not going to contribute a cubic face because it lies in sight of export. However you can think of you scratched so they calculate the number of faces in x three. We saw that each edge contributed to Well, yeah, each edge of X two contributed to faces, so might suspect that next four each face each square face to face, you could say are X three is going to contribute to cubic faces. In fact, this is true something about the square face and then you have the to new linearly independent points. Well, if you take a square face and then you add this new linearly any point, you're going to obtain the three dimensional cubic face it is determined by those. Mhm is five points. So you're going to have the number of faces of x three times to which is 16. So we have the F three of X four is going to be two times eight, which is 16 cubic faces, and now he's Oilers formula. We see that eight minus 24 plus 32 minus 16. This is going to be equal to see negative 16 plus 30 to minus 16 0, which is the same as one plus negative one to the third power, which is what Oilers formula predicts. So it follows that this satisfies boilers formula. Finally, in Part D were asked to find a formula for F k f x d n the number of K faces of the N dimensional Cross polito X to the end, where K ranges from zero up to n minus one. Now we could draw a table as we did in previous problems. I suggest that you do this on your own, but to recognize is that this sort of follows a pattern again a bit like binomial pyramid binomial triangle. And so you see that if you draw this table in the cake column, you can divide by two D k plus one and obtain a number from the binomial triangle. So we have that f k of X to the end is equal to to to the K plus one and looking back. So, for example, we have that f three of X to the fore that that case K is equal to three. And so we see that 16 is divisible by two to the fourth and then also say we're looking at at zero of X to the three. Well, this was six and K is zero. In this case, six is indeed divisible by two. And this is what most by two. So we leave a three behind when we divide by two. And then once you divide this to the K plus one, you'll notice. But the term is simply going to be n choose K plus one the term from the binomial treat. What? So, for example, looking at F two of X to the fourth? This is 32 que is too. So we see that 32 divided by eight is four. We're sorry. Yeah, 32 divided by eight is four and then we have that four choose. Three is four. So satisfies the equation. Likewise, sort of the rest of these, and in fact this could be proved by induction, that this is the formula for F k F X to the end.

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