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Welcome to calculus, I'm Professor Ghrist, and we're about to begin lecture

Â 34, Bonus Material. In our main lesson, recall, we claimed

Â that the volume of a unit radius and dimensional ball, Vs of N, is given by

Â the following formulae. When N is even, that is, N equals 2 times

Â K, then V sub n is pi to the k over k factorial.

Â When n is odd, that is, 2 k plus 1, then V sub n is pi to the k, k factorial, 2 to

Â the n over n factorial. Most students would suspect that formulae

Â this complicated must be very, very difficult to derive.

Â That is not exactly the case. We're going to do it right now.

Â Recall how we computed the volume of a three dimensional ball.

Â We sliced it by a two dimensional plane, obtaining, when we did so, a two

Â dimensional ball, whose radius varied according to where the slicing took

Â place. We're going to do the same thing in this

Â context, but instead of a three-dimensional ball, with a

Â two-dimensional slice, we'll begin with an n-dimensional unit ball, and obtain an

Â n minus 1-dimensional ball as the slice. And the subtlety is in figuring out what

Â is the radius, that n minus one dimensional slice.

Â Our volume element for the n-dimensional unit ball, d V sub n is an n minus 1

Â dimensional ball of some radius, r. That has volume V sub n minus 1 times r

Â to the n minus 1. We multiply by the thickness, dx,

Â choosing some coordinate direction and calling it x.

Â Now, what is that radius? Well, the same computation that we used

Â in the three-dimensional case gives it to us easily.

Â What do we do? We build a right triangle whose base is

Â of length X, and whose hypotenuse is, of course, one, since this is a unit ball.

Â What is the height of this triangle? It is r, and that is equal, by Pythagoras

Â to the square root of 1 minus x squared. We therefore compute the volume V sub n

Â by integrating the volume element. That is we have the integral of V sub n

Â minus 1 times quantity root 1 minus x squared to the n minus first power dx.

Â What are the limits of integration? This is a unit, n-dimensional ball, and

Â so we integrate as x goes from negative one to one.

Â Now what do you notice about this formula?

Â Well, first of all, we need to know v sub-n minus 1 in order to compute it, but

Â that's okay. We could proceed inductively.

Â What is leftover when you pull out that constant is simply a single variable

Â integral, one that is entirely doable in the context of a first semester calculus

Â course. Let us compute the value of this

Â integral, or rather, integrals, since there's one for each n.

Â For notational convenience, let us call this integral I sub n.

Â This is clearly a prime candidate for trigonometric substitution.

Â And so, if we said, x equal to sin of theta, and dx equal to cosine theta, d

Â theta. Then, I'm sure you see that that square

Â root is going to be cleared away and we'll obtain the integral of cosine theta

Â to the N minus one times cosine theta D theta.

Â What are the limits? Well, when X is negative 1, theta is

Â negative pi over 2. When X is positive 1, theta is pi over 2.

Â And so we obtain the integral from negative pi over 2 to pi over 2 of cosine

Â to the n. This is an integral that we have computed

Â before. In the bonus material to Lecture 28, we

Â showed that this is equal to, well, there are two cases, one where n is even, the

Â other where n is odd. The n is even case.

Â This has the value of 1 times 3 times 5, all the way up to n minus 1.

Â Over 2 times 4 times 6, all the way up til n.

Â This value times pi. When n is odd, the evens and odds are

Â flipped. We have a product of even numbers, up to

Â n minus 1, in the numerator. In the denominator, we have the product

Â of odds, up to and including n, all of this times an extra factor of two.

Â Notice the even odd dichotomy and the change between an extra factor of pi, and

Â an extra factor of two. This is significant.

Â Why? Well, when we go to compute the volume, V

Â sub n, we showed that this is equal to V sub n minus one times I sub n.

Â Therefore, we can proceed inductively. Let's build a table and begin with small

Â values of n. Now we know the values of I sub n, we can

Â list those and we know the values of V sub n.

Â When n is zero, one, two, certainly, the zero dimensional volume of a ball is one.

Â The one one dimensional volume of a one dimensional ball is two, and the two

Â dimensional volume of a unit two dimensional ball is pi.

Â Translating those items over to the V sub n minus one column, what do we obtain

Â from this formula? We simply multiply across I sub n times V

Â sub n minus 1, and indeed we see that this works the values that we already

Â know. What about dimension three?

Â I sub n is 4 3rd, V sub n minus 1 is pi. This tells us that V sub 3 is four 3rd

Â pi. That is something that you certainly

Â know. Now we get to something that most of us

Â don't know. What is the volume of a four-dimensional

Â ball? Well, according to this formula, we

Â multiply I sub 4, that is, 3 8ths pi, times V sub 3, that is, 4 3rds pi.

Â And we obtain pi squared over 2, so the volume of four-dimensional ball of radius

Â r is pi squared over 2 times r to the fourth.

Â And now we can continue inductively for as long as our curiosity, and pencil hold

Â out, obtaining various values of V sub n. At this point, one wants to look for

Â patterns such as the fact that there are powers of pi that increase, not at every

Â dimension, but every other dimension. And also, there are other patterns in the

Â coefficients if we look at what happens in the even case, it seems to be

Â particularly nice looking. In fact, we get that pattern that we

Â claimed at the beginning of this lesson. Namely that in the even case you get Pi

Â to the K over K factorial or N equals 2K. And we see now the reason for the

Â complexity in the odd case comes from the complexities that are in these integrals

Â I sub n. Now again, there are lots of things that

Â we could say about these volumes. We can prove these general formulae by a

Â formal induction step, but I'm going to leave that one to you.

Â I'm also going to leave to you working out the implications of why the volume is

Â going to zero as the dimension goes to infinity.

Â What I want to focus on is the question of who cares, I mean come on you've got

Â volumes of 'n' dimensional walls And who cares about what a ball in dimension 100

Â might look like? You might be surprised to learn how many

Â high dimensional spaces are all around you.

Â Let's take, for example, robotics. What happens when we look at a reasonably

Â sophisticated robot? Well, hopefully it moves, and each of its

Â movements is tracing out some sort of dimension in its configuration space.

Â We can try to measure all of the different joint angles.

Â Let's say, each of these mechanical degrees of freedom leads to an extra

Â dimension in the robot's configuration space.

Â And if you have a highly articulated robot, one that can do lots of fine

Â motion control, then in order to effect that control, you have to work in a

Â potentially high dimensional space. Things get even more interesting if we

Â consider what happens when you have, say a robot that is flying around say a quad

Â rotor. How many dimensions are implicated here?

Â Well, let's see, we have an X a Y and Z coordinate for its position.

Â We have velocities in the x, y, and z direction.

Â For these objects we also have angles that determine their roll, their pitch,

Â their yaw, how they are situated in space and how they are moving about.

Â And if you have several of these that you are trying to coordinate into say a swarm

Â to accomplish some task, then you have a lot of dimensions to worry about.

Â And again, if one is trying to do any sort of control or differential equations

Â on these spaces in these systems, you are going to have to worry about things like

Â volume in high dimensions. But really, these aren't that high.

Â Where the dimension gets high is in data, anything associated with large data sets,

Â let's say stock prices or imagery data, social networks.

Â All of these implicate many, many, many variables, and in order to provide a

Â mathematical analysis of data sets, you're going to need to know how to work

Â in high dimensions. Now the tools that we have learned in

Â this lesson are still pretty simplistic, and it's wonderful that we can reduce

Â things like volumes of balls down to a simple, one-dimensional integral.

Â For more general problems involving high dimensions, you're going to have to take

Â some multivariable calculus to get the right tools for solving those problems.

Â