This is calculus in a single variable and I'm Robert Ghrist, professor of
mathematics and electrical and systems engineering, at the University of
Pennsylvania. We're about to begin lecture eleven, Bonus Material. What have we
looked at? In lecture eleven we considered the various rules for computing
derivatives. Linearity, product rule, chain rule. Other kinds of rules. Why
spend so much time focusing on the rules? One reason is they help with computations.
But another deeper reason is that there are structures in mathematics and related
areas that obey similar types of rules. And when we find them and recognize them,
we can come to a deeper understanding by knowing how calculus works. Let's look at
two specific examples. One example involves spaces or domains over which you
might do calculus. Here are some specific examples, one type of space might be a
interval or a line segment. Another might be a circular disk. One thing that's
interesting to talk about when you're looking at the geometry of spaces is the
boundary. And we're going to think of boundary as an operator, not unlike a
derivative. In fact, we're going to give it a symbol that is evocative of the
differentiation symbol, called the script-e, d, dow, for the boundary
operator. Now what is the boundary of an interval or a line segment? It consists
precisely of two end points. What is the boundary of a circular disc? It consists
precisely of those points that are on the boundary the circle. Now, observe how the
boundary behaves with respect to other structures on spaces. There's a type of
product called the Cartesian product that is somewhat familiar even if you've never
seen it formally before. We denote it by a times sign. For example, a rectangle,
solid, filled-in rectangle can be considered as the Cartesian product of two
intervals. A solid cylinder can be considered as a product of an interval
with a circular disc. I'm sure you recognize many other shapes as well that
can be similarly decomposed. Now, what happens when we take the boundary of a
product o f spaces? For a rectangle, what we get are four edges with those corners.
Now notice that this perimeter of the rectangle can be decomposed, into a union
of two pieces. The two edges on the side, and the two edges on the top. We can think
of that as the boundary of the first interval times the second union, the
boundary of the second interval times the first. Similarly with the solid cylinder,
if we look at the boundary of that, we can decompose it into a couple of pieces.
There's the left and right hand sides of the cylinder, the, the, the round disk
part, that can be thought of as the disk times the boundary of the interval. And
then there's that portion of the boundary of the cylinder that wraps around the
sides. That can be thought of as the interval cross the boundary, of the disk.
It is a general fact that the boundary of the product of two spaces, A times B, is
equal to the boundary of A times B union A times the boundary of B. And if you look
at that formula carefully, you'll see what looks just like a product rule. But with
different operations involved. And indeed, there's a close and deep connection
between the boundary operator, Dell, and the derivative operator D, that we know
and love. Let's consider another example, this time coming from lists and computer
science. Consider a list with five elements. I'm going to represent each
element abstractly by a variable, x, so that the list of five elements might
suggestively be called x to the fifth. Now consider the following operator, a
deletion operator, d, that acts by deleting one of the terms in the list.
What happens? Well, I could delete the first term, or I could have deleted the
second term, or the third term, or the fourth term or the fifth. In this setting
I get a list of four elements. Or another list of four elements or another or
another or another. And if we call those lists of four elements algebraically as X
to the fourth and if we represent the logical or as a formal addition, then what
do we have? We really have that the deletion operat or D applied to X to the
fifth equals five times X to the fourth. Now where have you seen something like
that before? It turns out that there's an entire calculus for lists. Let's look at a
simple example of how this works. Let us denote by x to the n a list of n elements.
What would x to the zero be? A list of zero elements. Oh, that would be an empty
list. And what should we call that mathematically? Let's call that one,
because x to the zero really ought to be one.
And now consider the set L of all finite lists. We'd like to know, what does L look
like, symbolically. Here's a statement that I think you'll agree is true. Any
list is either empty or it has a first entry. What does that translate to you,
algebraically? Any list means L, is, connotes an equal sign, empty. Oh, we've
seen what the empty list is, that's one. we've seen that OR, corresponds to a
formal addition. And what does it mean, for a list to have a first entry? That
means, you've got an X followed by some other finite list, or X times L. Now,
let's just take that algebraic equation and start manipulating it. Forget about
what these things mean grammatically, for the moment. If I move X, x L to the other
side, factor out L and divide through by 1-X, then we'll get that L is equal to one
over one minus x. Now what does that mean? I've come up with an algebraic expression
for all finite lists and I recognize one over one minus X as the geometric series
one plus X plus X squared plus X cubed etcetera. One way to interpret that, is
that we have derived the fact. That the set of all finite lists consists of the
empty list, or a list of one element, or a list of two elements, or a list of three
elements, etc. That's a pretty remarkable derivation. It's using a type of calculus,
that is very different from what we're learning in this class but has some
strange similarities. And as you go on in your calculus education, you'll see that
calculus applies not just to, problems in physics or economics, but in all kinds of
fields and in all kinds of manners, which, a t first, might not have been so obvious.