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