Let's move on to a different shape, one that requires some calclus to understand.
This is the simplex, this is an in-dimensional generalization of a
triangle or a pyramid. The unit simplex is defined algebraically
as a subset of the unit cube that satisfies an additional constraint, this
being that the sum of the coordinates is less than or equal to 1.
But what does that mean in terms of our slider bar analogy?
This means that you can take any of the individual bars and slide it all the way
up to 1. However you can't do this independently.
If you want to move the other slider bars up you have to do so in a way but the sum
of the values does not excede the treshold of 1.
That means that this is a highly constrained set.
It's not a large subset of the n-dimensional cube.
It feels much small. We expect to see that reflected in the
volume. Let's see how that works.
First, lets explore a few properties and then we'll compute the [UNKNOWN] volume.
The number of corners of an n-dimensional simplex is much less than that of a cube.
The n-dimensional simplex has n plus 1 corners.
What is the volume? Well, we know, for a single simplex, it's
just a point. The number of points is one.
We know for a one dimensional simplex since its the same as a one dimensional
cube. We just get a length of 1.
Now, a triangle as we all know, gives us area one-half.
When we look at a three dimensional simplex, it's a cone over that triangle.
We know the volume of a cone is going to be 1 3rd, the height 1 times the the area
of the base, 1 half. Now, we start to see a little bit of a
pattern here. What if I told you that the 4 dimensional
simplex had 4 dimensional volume equal to 1 24th.
That's true. And knowing that you would be convinced
of the pattern, namely that the volume of the n dimensional simplex v sub n must be
one over n factorial. Now that's a good guess, let's see if we
can show it. Our strategy for computing volumes of the
n dimensional simplexes is the same as that Of a cone.
We're going to slice in a direction parallel to the base.
And what we're going to see is that when we slice and n+1 dimensional simplex,
what we'll get in an n-dimensional simplex whose size is rescaled.
By a factor of x in each coordinate, where x is the distance to the top on the
simplex. So for a one dimensional simplex the
appropriate volume element of the slice that's nothing more than dx.
In a two dimensional simplex, the appropriate area element is what?
It's simply x d x. In the three dimensional case, well,
we've done this before. This is going to be 1/2 x times x d x and
in general, the difficult step. Is to argue that the volume element, for
the n plus one simplex, is the volume of the base and simplex v sub n, times x to
the n. Since we're re scaling each coordinate.
By a factor of x. But once we have that, and then
multiplying by the thickness dx, we can compute this n plus one dimensional
volume as the integral of the volume form.
That is the integral of v sub n times x to the n d x.
Integrating as x goes from 0 to 1. This is a trivial integral since v sub n
is a constant. Routine x to the n plus 1 over n plus 1,
evaluated from 0 to 1. That gives us v sub n over n plus 1 and
so we can write down all of these volumes by induction and argue that v sub n is in
fact 1 over n factorial. That's a nice application of simple
integration. Let's move on to an n-dimensional ball of
radius 1. These are a little difficult to draw.
In 2D this is simply a disk of radius 1. In 1D it's a disk of radius 1, well it's
really an interval of length 2, and in 1D it is again a simple point.
Higher dimensional balls are not so easy to draw.
Now how do we define it rigorously? The unit ball is defined as those set of
points with coordinates x sub I between negative one and positive one satisfying
the additional constraint that the sum of the squares of the coordinates is also
less than or equal to 1. This is what we're used to in 2D.
When we say x squared plus y squared, less than or equal to 1, this is simply
the generalization of that. Now, in terms of a slider bar analogy.
Now, all of the individual bars can go from negative 1 to 1.
Each can go to the very top, or the very bottom.
But, in between well, you have some freedom to move the individual sliders up
and down. But you can't move them all past a
certain point where the sum of the squares is less than or equal to one.
Nevertheless, it feels like there's a lot of room inside of there to move around,
How do we compute the volume? Well again, for a radius 1 ball in
dimension n, what is volume going to be? In dimension 0, there is the single
point, volume 1. In dimension 1, this interval has length
2. In dimension 2, well, we know the
formula, pi r squared. In this case, r equals 1.
In dimension 3, volume is Four thirds pie.
Moving up to dimension n. Well what are we going to do here, lets
call that volume of the unit ball v sub n.
And to determine what that is lets consider what happens when the radius is
not one, but r in this case. The length of the one-dimensional is 2
times r. The area of the two-dimensional ball is
pi r squared. Volume, 4 3rds pi r cubed.
In general, having a ball of radius r and dimension n is going to give you the
volume. Of the uniball times r to the nth power.
That's going to be helpful for us, as we'll see.
