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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.