Welcome to calculus, I'm Professor Ghrist, and we're about to begin lecture 34, Bonus Material. In our main lesson, recall, we claimed that the volume of a unit radius and dimensional ball, Vs of N, is given by the following formulae. When N is even, that is, N equals 2 times K, then V sub n is pi to the k over k factorial. When n is odd, that is, 2 k plus 1, then V sub n is pi to the k, k factorial, 2 to the n over n factorial. Most students would suspect that formulae this complicated must be very, very difficult to derive. That is not exactly the case. We're going to do it right now. Recall how we computed the volume of a three dimensional ball. We sliced it by a two dimensional plane, obtaining, when we did so, a two dimensional ball, whose radius varied according to where the slicing took place. We're going to do the same thing in this context, but instead of a three-dimensional ball, with a two-dimensional slice, we'll begin with an n-dimensional unit ball, and obtain an n minus 1-dimensional ball as the slice. And the subtlety is in figuring out what is the radius, that n minus one dimensional slice. Our volume element for the n-dimensional unit ball, d V sub n is an n minus 1 dimensional ball of some radius, r. That has volume V sub n minus 1 times r to the n minus 1. We multiply by the thickness, dx, choosing some coordinate direction and calling it x. Now, what is that radius? Well, the same computation that we used in the three-dimensional case gives it to us easily. What do we do? We build a right triangle whose base is of length X, and whose hypotenuse is, of course, one, since this is a unit ball. What is the height of this triangle? It is r, and that is equal, by Pythagoras to the square root of 1 minus x squared. We therefore compute the volume V sub n by integrating the volume element. That is we have the integral of V sub n minus 1 times quantity root 1 minus x squared to the n minus first power dx. What are the limits of integration? This is a unit, n-dimensional ball, and so we integrate as x goes from negative one to one. Now what do you notice about this formula? Well, first of all, we need to know v sub-n minus 1 in order to compute it, but that's okay. We could proceed inductively. What is leftover when you pull out that constant is simply a single variable integral, one that is entirely doable in the context of a first semester calculus course. Let us compute the value of this integral, or rather, integrals, since there's one for each n. For notational convenience, let us call this integral I sub n. This is clearly a prime candidate for trigonometric substitution. And so, if we said, x equal to sin of theta, and dx equal to cosine theta, d theta. Then, I'm sure you see that that square root is going to be cleared away and we'll obtain the integral of cosine theta to the N minus one times cosine theta D theta. What are the limits? Well, when X is negative 1, theta is negative pi over 2. When X is positive 1, theta is pi over 2. And so we obtain the integral from negative pi over 2 to pi over 2 of cosine to the n. This is an integral that we have computed before. In the bonus material to Lecture 28, we showed that this is equal to, well, there are two cases, one where n is even, the other where n is odd. The n is even case. This has the value of 1 times 3 times 5, all the way up to n minus 1. Over 2 times 4 times 6, all the way up til n. This value times pi. When n is odd, the evens and odds are flipped. We have a product of even numbers, up to n minus 1, in the numerator. In the denominator, we have the product of odds, up to and including n, all of this times an extra factor of two. Notice the even odd dichotomy and the change between an extra factor of pi, and an extra factor of two. This is significant. Why? Well, when we go to compute the volume, V sub n, we showed that this is equal to V sub n minus one times I sub n. Therefore, we can proceed inductively. Let's build a table and begin with small values of n. Now we know the values of I sub n, we can list those and we know the values of V sub n. When n is zero, one, two, certainly, the zero dimensional volume of a ball is one. The one one dimensional volume of a one dimensional ball is two, and the two dimensional volume of a unit two dimensional ball is pi. Translating those items over to the V sub n minus one column, what do we obtain from this formula? We simply multiply across I sub n times V sub n minus 1, and indeed we see that this works the values that we already know. What about dimension three? I sub n is 4 3rd, V sub n minus 1 is pi. This tells us that V sub 3 is four 3rd pi. That is something that you certainly know. Now we get to something that most of us don't know. What is the volume of a four-dimensional ball? Well, according to this formula, we multiply I sub 4, that is, 3 8ths pi, times V sub 3, that is, 4 3rds pi. And we obtain pi squared over 2, so the volume of four-dimensional ball of radius r is pi squared over 2 times r to the fourth. And now we can continue inductively for as long as our curiosity, and pencil hold out, obtaining various values of V sub n. At this point, one wants to look for patterns such as the fact that there are powers of pi that increase, not at every dimension, but every other dimension. And also, there are other patterns in the coefficients if we look at what happens in the even case, it seems to be particularly nice looking. In fact, we get that pattern that we claimed at the beginning of this lesson. Namely that in the even case you get Pi to the K over K factorial or N equals 2K. And we see now the reason for the complexity in the odd case comes from the complexities that are in these integrals I sub n. Now again, there are lots of things that we could say about these volumes. We can prove these general formulae by a formal induction step, but I'm going to leave that one to you. I'm also going to leave to you working out the implications of why the volume is going to zero as the dimension goes to infinity. What I want to focus on is the question of who cares, I mean come on you've got volumes of 'n' dimensional walls And who cares about what a ball in dimension 100 might look like? You might be surprised to learn how many high dimensional spaces are all around you. Let's take, for example, robotics. What happens when we look at a reasonably sophisticated robot? Well, hopefully it moves, and each of its movements is tracing out some sort of dimension in its configuration space. We can try to measure all of the different joint angles. Let's say, each of these mechanical degrees of freedom leads to an extra dimension in the robot's configuration space. And if you have a highly articulated robot, one that can do lots of fine motion control, then in order to effect that control, you have to work in a potentially high dimensional space. Things get even more interesting if we consider what happens when you have, say a robot that is flying around say a quad rotor. How many dimensions are implicated here? Well, let's see, we have an X a Y and Z coordinate for its position. We have velocities in the x, y, and z direction. For these objects we also have angles that determine their roll, their pitch, their yaw, how they are situated in space and how they are moving about. And if you have several of these that you are trying to coordinate into say a swarm to accomplish some task, then you have a lot of dimensions to worry about. And again, if one is trying to do any sort of control or differential equations on these spaces in these systems, you are going to have to worry about things like volume in high dimensions. But really, these aren't that high. Where the dimension gets high is in data, anything associated with large data sets, let's say stock prices or imagery data, social networks. All of these implicate many, many, many variables, and in order to provide a mathematical analysis of data sets, you're going to need to know how to work in high dimensions. Now the tools that we have learned in this lesson are still pretty simplistic, and it's wonderful that we can reduce things like volumes of balls down to a simple, one-dimensional integral. For more general problems involving high dimensions, you're going to have to take some multivariable calculus to get the right tools for solving those problems.