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PROFESSOR: Hi.

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Well, this is sort of a summary
lecture for the big

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group about differential
calculus.

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And it's got a fancy title, Six
Functions, that we know.

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Well, five of them
that we know.

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And a new one-- of course,
there has to

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be something new--

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Six Rules, and Six Theorems.
So I haven't emphasized

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theorems, but it seemed like
this was an occasion where we

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could see the main points of the
math behind the functions

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and the rules.

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OK.

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So, here are my first five
functions, all familiar.

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And, what I'm happy about is
that, if we understand those

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five and the rules to create
more out of them, we get

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practically everything,
everything we frequently use.

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OK.

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So, I wrote down function
one, power of x, and its

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derivative, function
two and its

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derivative, function three.

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Function four has,
a little bit,

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something that is important.

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If it's e to the x,
then we know the

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derivative is e to the x.

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But, if it's e to the c, x
of factor c, comes down.

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Important case, you could
say the chain rule.

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The derivative is that times
the derivative of what's

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inside, which is the c.

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And, finally, the natural
logarithm with the great

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derivative of 1/x.

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And now, oh, I left space
to go from function one

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backwards, to remember the
function that came before it.

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So, what function has
this derivative?

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I'm looking here at the other
generation, the older

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generation.

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Well, the function with that
derivative is we need the

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power to be one higher, right?

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And then, the derivative of
that, we need to divide by n

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plus 1 so that, when we take the
derivative, the n plus 1

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comes down, cancels this, and
gives us x to the n-th.

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The function that comes before
sine x will be--

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oh, there was cos x
in that direction.

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In this direction, we need
minus cos x because the

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derivative of minus cos
x is plus sine x.

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But, for this guy, cosine x,
that came from sine x.

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And, what about this one?

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What function has
this derivative?

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Well, with exponentials, we
expect to see that exponential

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always, e to the c, x again,
but, since this would bring

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down a c and here we don't
want it, we'd better

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divide by that c.

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So then, if I take that, that's
e to the c, x divided

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by c, so the c will come
down, cancel the c,

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just the way here.

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And, oh, we've never
figured out log x.

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That'll be something novel to
do for integral calculus.

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But, I think, if I write down
the answer, I think it's x

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times log x minus x.

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I believe that works.

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I would use the product
rule on that.

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x times the derivative of that
would be a 1 minus that.

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And the derivative of that
would be a 1, so two ones

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would cancel, and the product
rule would leave me with log x

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times the derivative of that.

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It works.

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It works.

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And notice the one beautiful
thing in this list, that the

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case here is great unless
I'm dividing by 0.

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If n is minus 1,
I'm in trouble.

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If n is minus 1, I don't have
here something whose--

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if n is minus 1, I can't
divide by 0.

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I don't get x to the minus
1 out of x to the 0.

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That rule fails at
n equal minus 1.

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But look, here, is exactly
fills in that whole.

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Wonderful.

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Here is the minus 1 power, and
here is where it comes from.

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So that log just filled in the
one hole that was left there.

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OK.

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Otherwise, you know
these guys.

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But here's a new one:
a step function.

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A step function, it's 0 and it
jumps up to 1 at x equals 0.

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So, here's x.

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The function is 0 until
it gets to that point.

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So it's level, then it takes
a step up, a jump up, to 1.

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And let's say it's 1 at that
point, so it takes that jump.

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All right.

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OK.

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That's a function that's
actually quite important.

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And it's sort of like a two-part
function, it's got a

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part to the left and a
part to the right.

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And they don't meet, it's a
non-continuous function.

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Can I figure out what is it
that-- so here will be the old

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generation, what graph do
I put there so that the

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derivative is 0 and then 1?

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Well, that's not too hard.

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If I put 0's here, the
derivative will be 0.

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And now, over here, I want the
derivative to be a constant 1.

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And we know that the derivative
of x is what I

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need, so this is 0 and then
x, two parts again.

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And the derivatives of those
parts are 0 and then 1.

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And I often call that a ramp
function because it looks a

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little like a ramp.

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OK.

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What about going this way?

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Ah, that's a little more
interesting because what's the

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derivative of a step function?

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What's the slope of
a step function?

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Well, the slope here is
certainly 0, and the slope

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along here is certainly 0,
so, is the answer 0?

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Well, of course not.

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All the action is
at this jump.

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And what's the derivative
there?

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Now, a careful person
would say there is

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no derivative there.

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The limit of delta f/delta x,
you don't get a correct answer

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there because delta f jumps
by one, and delta x

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could be very small.

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And, as delta x goes to 0, we
have 1/0, we have infinite.

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Well, I say, what?

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Let's go for infinite.

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So my derivative is
0 and 0, and, at

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this point, it's infinite.

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It's a spike, or sometimes
called a delta function.

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It's 0, and then infinite at
one point, and then 0.

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And the oddball thing is that
the area under that one-point

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tower, spike, is supposed
to be 1.

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Because, do you remember--

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and we'll do more areas if we
get to integral calculus--

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but, the area under
this function is

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supposed to be this one.

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The area under the cosine
function is sine x.

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The area under this function
should be this one, so the

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area is 0 here.

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Run along here.

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No area under it.

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Then, I have a one-point
spike, and the area is

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supposed to jump to 1
under that spike,

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at that single point.

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That spike is infinitely tall,
and it actually has a little

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area under it.

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Ah, well, your teacher may say
get that function out of here.

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That's not a function.

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And I'm afraid that's
a true fact that

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it's not a real function.

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So you could say I don't
want to see this

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thing, clear it out.

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But, actually, that's
very useful.

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It's a model for something that
happens very quickly: an

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instant, an impulse, so
I'll leave it there.

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I'll leave it there,
but I'll go on.

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So, if you don't like it, you
don't have to look at it.

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OK.

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So those were the
six functions,

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now for the six rules.

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Nothing too fancy here.

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I don't think I really
emphasized the most important

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and simplest rule that, if you
have as a combination, like

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you add two functions, then
the derivatives add.

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Or, if you multiply that
function by 2 and that

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function by 3 before you add,
then you multiply the

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derivatives by 2 and
3 before you add.

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It's that fact that
allowed us--

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I mean, you've used
it all the time.

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If you integrated x plus x
squared, you used the sum rule

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to integrate--

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ah, sorry--

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took the derivative.

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If you want the slope of x plus
x squared, you would say

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oh, no problem: 1 plus 2x.

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1 coming from the first
function, 2x from the x

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squared function.

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So the slope of a sum is just
the sum of the slopes.

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You constantly use that to build
many more functions out

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of the simple, anything, x
squared plus x cubed plus x 4,

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if you know its derivative and
you're using this rule.

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Now, the product rule,
we worked through.

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You've practiced that.

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The quotient rule is a little
messier with this minus sign

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and the division by g squared.

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It's a fraction.

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And then, a little more
complicated, was

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this inverse function.

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Do you remember that if you
start from y equals f of x--

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which is what we always
have been doing--

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and then you say all right,
switch it so that x isn't the

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input anymore, it's now
the output, and the

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input is the y.

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So you're reversing
the function.

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You're flipping the graph.

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We did this to get between
e to the x and log x.

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That was the most important
case of doing this flip

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between y equals e to the
x and x equals log of y.

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And the chain rule tells us that
the derivative of this

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inverse function is 1 over the
derivative of the original.

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Nice rule.

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And here's the full-scale
chain rule.

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Oh, that deserves to be put
inside a box or something

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because this is a really
great way to create new

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functions as a chain.

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You start with x.

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You do g of x, and then
that's the input to f.

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You will know that chain rule.

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And you remember that that
produces a product, the

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derivative of f times the
derivative of g, but there was

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this little trick, right?

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This g of x was the y.

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00:13:03,180 --> 00:13:07,210
I'll just remind you that this
g of x is the y, and you have

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to get y out of the answer.

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00:13:11,420 --> 00:13:15,590
Use this to get an answer
in terms of x.

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Wherever you see y, you
have to put in g of x.

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00:13:20,220 --> 00:13:21,560
So, that's the chain rule.

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And then the final rule that
I want to mention is this

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L'hopital rule about--

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well, a lot of calculus is about
a ratio of f of x to g

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00:13:34,620 --> 00:13:39,110
of x when it's going to 0/0.

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00:13:39,110 --> 00:13:41,900
What do you do about 0/0?

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Well, as we're going to some
point, like x equals a, if

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this is going to 0/0, then
you're allowed to look.

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The slopes will tell you how
quickly each one is going to

220
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0, and the ratio becomes a
ratio of the two slopes.

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00:14:08,830 --> 00:14:16,670
So, normally then, this answer
would be the derivative at a

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00:14:16,670 --> 00:14:19,000
divided by the derivative
at a.

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00:14:21,940 --> 00:14:28,130
If we're lucky, this 0/0 thing,
when we look at the

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slopes, isn't 0/0 any more.

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It's good numbers,
and L'hopital

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gets the answer right.

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OK.

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That's a review of L'hopital's
rule, just really remembering

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that that's an important rule
that came directly from the

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idea of the derivative.

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00:14:50,030 --> 00:14:56,440
We're using the important part
of the function because the

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constant term in that
function is 0.

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Good.

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00:15:00,390 --> 00:15:05,060
OK, are you ready for
six theorems?

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That is a handful, but
let's just tackle it.

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Why not?

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00:15:13,740 --> 00:15:14,290
Why not?

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OK.

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So, six functions were easy.

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00:15:18,310 --> 00:15:22,510
Well, we start with the big
theorem, the big theorem, the

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fundamental theorem
of calculus.

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00:15:24,440 --> 00:15:28,320
The fundamental theorem
of calculus, OK,

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that ought to be important.

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And what does it say?

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00:15:34,720 --> 00:15:38,630
It says that the two operations
of going from

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function one to two by taking
the derivative, the slope, the

247
00:15:44,030 --> 00:15:48,740
speed, is the reverse of
going the other way,

248
00:15:48,740 --> 00:15:50,170
from two back to one.

249
00:15:50,170 --> 00:15:58,295
It's really saying that, if
I start with a function--

250
00:15:58,295 --> 00:15:59,740
Here, this would be one way.

251
00:15:59,740 --> 00:16:05,540
If I start with a function, f,
I take the derivative to get

252
00:16:05,540 --> 00:16:08,490
function two, the speed,
the slope.

253
00:16:08,490 --> 00:16:10,130
Then, if I go backwards--

254
00:16:10,130 --> 00:16:16,560
which is this integrating that
integration symbol that's the

255
00:16:16,560 --> 00:16:19,220
core in integral calculus--

256
00:16:19,220 --> 00:16:22,000
if I take the derivative
and then take the

257
00:16:22,000 --> 00:16:26,080
integral, I'm back to f.

258
00:16:26,080 --> 00:16:30,990
And what you actually get
in this number is f at--

259
00:16:30,990 --> 00:16:31,940
it depends.

260
00:16:31,940 --> 00:16:36,000
It's like a delta f, really.

261
00:16:36,000 --> 00:16:44,190
It's the f at the end minus
the f at the start.

262
00:16:44,190 --> 00:16:45,770
Maybe you'll remember that.

263
00:16:45,770 --> 00:16:54,750
When we talked about it, there
was one lecture on big picture

264
00:16:54,750 --> 00:16:59,940
of the integral, and there may
be more coming, but that was

265
00:16:59,940 --> 00:17:03,490
the one where we had
this kind of thing.

266
00:17:03,490 --> 00:17:08,569
And, in the other direction, if
I start with function two,

267
00:17:08,569 --> 00:17:11,869
do its integral to get function
one, take the

268
00:17:11,869 --> 00:17:16,040
derivative of that, then I'm
back to function two.

269
00:17:16,040 --> 00:17:22,240
Actually, you're going to say
I knew that: function one to

270
00:17:22,240 --> 00:17:24,780
two, back to one.

271
00:17:24,780 --> 00:17:28,280
Or start with two, go to one,
then back to two, that's the

272
00:17:28,280 --> 00:17:29,800
fundamental theorem.

273
00:17:29,800 --> 00:17:34,030
That those two operations,
of taking the

274
00:17:34,030 --> 00:17:36,160
derivative, that limit--

275
00:17:36,160 --> 00:17:40,860
You remember what's tricky about
all that is that this

276
00:17:40,860 --> 00:17:45,940
d,f, d,x, involves a limit
as delta x goes to 0.

277
00:17:45,940 --> 00:17:49,010
And this integral will also
involve a limit as

278
00:17:49,010 --> 00:17:51,230
delta x goes to 0.

279
00:17:51,230 --> 00:17:54,690
So that's the point at which it
became calculus instead of

280
00:17:54,690 --> 00:17:56,870
just algebra.

281
00:17:56,870 --> 00:17:58,340
Well, important.

282
00:18:02,060 --> 00:18:07,160
I should say, let's assume here,
that these functions are

283
00:18:07,160 --> 00:18:09,240
all continuous functions.

284
00:18:09,240 --> 00:18:14,150
And I'm going to assume that
these theorems will apply to

285
00:18:14,150 --> 00:18:15,990
continuous functions.

286
00:18:15,990 --> 00:18:19,140
And do you remember
what that meant?

287
00:18:19,140 --> 00:18:22,880
Basically, it meant that that
jump function is not

288
00:18:22,880 --> 00:18:24,630
continuous.

289
00:18:24,630 --> 00:18:26,420
And that delta function is--

290
00:18:26,420 --> 00:18:29,280
well, that's not even
a function.

291
00:18:29,280 --> 00:18:33,650
The ramp function is continuous
but, of course, the

292
00:18:33,650 --> 00:18:35,240
derivative isn't.

293
00:18:35,240 --> 00:18:37,780
OK.

294
00:18:37,780 --> 00:18:38,750
All right.

295
00:18:38,750 --> 00:18:42,950
So, we've got functions that
we can draw without raising

296
00:18:42,950 --> 00:18:45,290
our pen, without lifting
the chalk.

297
00:18:45,290 --> 00:18:49,890
And here's the fact about
them, that if I have a

298
00:18:49,890 --> 00:18:53,340
continuous function on an
interval-- so, here is some

299
00:18:53,340 --> 00:18:56,450
point, a, and here is some
point, b, and my

300
00:18:56,450 --> 00:18:57,830
function goes like that.

301
00:18:57,830 --> 00:18:59,290
Oh, it doesn't do that.

302
00:18:59,290 --> 00:19:02,330
It goes like that.

303
00:19:02,330 --> 00:19:09,690
Then this thing says that this
maximum is actually reached,

304
00:19:09,690 --> 00:19:12,430
and this minimum is
actually reached.

305
00:19:12,430 --> 00:19:15,620
And any value in-between,
anywhere between this height

306
00:19:15,620 --> 00:19:20,590
and this height, there are
points where the function

307
00:19:20,590 --> 00:19:22,470
equals that.

308
00:19:22,470 --> 00:19:28,320
The continuous function hits its
maximum, hits its minimum,

309
00:19:28,320 --> 00:19:29,920
hits every point in-between.

310
00:19:29,920 --> 00:19:33,150
Where, if it wasn't continuous,
you see it could

311
00:19:33,150 --> 00:19:38,330
go up, and then, suddenly,
never reach that point,

312
00:19:38,330 --> 00:19:39,580
suddenly drop to there.

313
00:19:42,340 --> 00:19:45,910
There's a function not
continuous, of course, because

314
00:19:45,910 --> 00:19:48,410
it fell down there.

315
00:19:48,410 --> 00:19:53,560
And it never reached m because
it was this close, as close as

316
00:19:53,560 --> 00:19:54,120
it could be.

317
00:19:54,120 --> 00:19:57,290
But it never got there because,
at the last minute,

318
00:19:57,290 --> 00:19:58,780
it jumped down.

319
00:19:58,780 --> 00:20:00,740
OK.

320
00:20:00,740 --> 00:20:08,130
So, that's sort of a good
theoretical bit about

321
00:20:08,130 --> 00:20:09,550
continuous functions.

322
00:20:09,550 --> 00:20:10,800
OK.

323
00:20:20,510 --> 00:20:22,450
So, that's new.

324
00:20:22,450 --> 00:20:25,665
That was not mentioned before.

325
00:20:25,665 --> 00:20:34,500
But you can see it by just
drawing a picture where it

326
00:20:34,500 --> 00:20:38,610
hits the max, hits the min, hits
all values in-between.

327
00:20:38,610 --> 00:20:42,720
And then, you see the point,
y, continuous was needed

328
00:20:42,720 --> 00:20:47,250
because, if you let it jump,
the result doesn't work.

329
00:20:47,250 --> 00:20:47,620
OK.

330
00:20:47,620 --> 00:20:48,800
Here's another thing.

331
00:20:48,800 --> 00:20:51,470
This is now called the
mean value theorem.

332
00:20:54,040 --> 00:20:56,030
That's a neat theorem.

333
00:20:56,030 --> 00:20:57,100
OK.

334
00:20:57,100 --> 00:20:57,280
Oh.

335
00:20:57,280 --> 00:21:01,780
Now, here, our function is going
to have a derivative

336
00:21:01,780 --> 00:21:03,660
over some region.

337
00:21:03,660 --> 00:21:05,900
That function probably
had a derivative.

338
00:21:05,900 --> 00:21:06,820
OK.

339
00:21:06,820 --> 00:21:07,820
OK.

340
00:21:07,820 --> 00:21:16,110
So, that function, or this
function, f of x, here's the

341
00:21:16,110 --> 00:21:21,310
idea of the mean
value theorem.

342
00:21:21,310 --> 00:21:27,870
This is like delta f/delta
x for the whole

343
00:21:27,870 --> 00:21:29,810
interval from a to b.

344
00:21:29,810 --> 00:21:34,100
Delta x is b minus a,
the whole jump.

345
00:21:34,100 --> 00:21:38,270
Delta f is f at the end
minus f at this end.

346
00:21:38,270 --> 00:21:42,380
So that delta f/delta x is like
your average speed over

347
00:21:42,380 --> 00:21:44,630
the whole trip.

348
00:21:44,630 --> 00:21:47,800
Like you went on the
MassPike, right?

349
00:21:47,800 --> 00:21:50,560
And you entered at 1:00 o'clock
and came out at 4:00

350
00:21:50,560 --> 00:21:53,980
o'clock, so you were on the
pike for three hours.

351
00:21:53,980 --> 00:21:58,930
And your trip meter
shows 200 miles.

352
00:21:58,930 --> 00:22:07,640
So your average speed, average
speed, was 200 divided by 3,

353
00:22:07,640 --> 00:22:09,750
that number of miles per hour.

354
00:22:09,750 --> 00:22:13,480
Yeah, about 66 miles-- well,
probably illegal.

355
00:22:13,480 --> 00:22:15,670
OK.

356
00:22:15,670 --> 00:22:20,160
A little over 66 miles an
hour: 200/3, so you're

357
00:22:20,160 --> 00:22:21,710
slightly over the speed limit.

358
00:22:21,710 --> 00:22:28,900
Well, the mean value theorem
catches you because you could

359
00:22:28,900 --> 00:22:37,490
say well, but when did
I pass the limit?

360
00:22:37,490 --> 00:22:41,210
When was I going more than 65?

361
00:22:41,210 --> 00:22:45,190
And the mean value theorem says
there was a time, there

362
00:22:45,190 --> 00:22:52,080
was a moment when your speed,
when the speedometer, itself,

363
00:22:52,080 --> 00:22:53,600
was exactly.

364
00:22:53,600 --> 00:22:59,950
This instant speed equaled
the average speed.

365
00:22:59,950 --> 00:23:01,750
Shall I say that again?

366
00:23:01,750 --> 00:23:10,180
If you travel with a smooth
changes of speed, no jumps in

367
00:23:10,180 --> 00:23:18,770
speed, then, if I look at the
average speed over a delta t,

368
00:23:18,770 --> 00:23:23,400
there is some point inside that
one where the average

369
00:23:23,400 --> 00:23:29,530
speed agrees with the
instant speed.

370
00:23:29,530 --> 00:23:35,890
Or you could say, if
you prefer slope--

371
00:23:35,890 --> 00:23:41,530
Suppose the average slope, the
up over a cross, is 10, So in

372
00:23:41,530 --> 00:23:45,870
the time at cross, you
eventually got up 10.

373
00:23:45,870 --> 00:23:48,910
Then there will be some
point when your

374
00:23:48,910 --> 00:23:51,300
climbing rate was 10.

375
00:23:51,300 --> 00:23:56,650
There'd be some point when that
instant slope is also 10.

376
00:23:56,650 --> 00:23:59,340
OK.

377
00:23:59,340 --> 00:24:02,360
That's the mean value theorem.

378
00:24:02,360 --> 00:24:04,250
This is called the mean value.

379
00:24:04,250 --> 00:24:07,880
Mean value is another
word for average.

380
00:24:07,880 --> 00:24:12,200
So the mean value equals the
instant value at some point.

381
00:24:12,200 --> 00:24:16,900
But we don't know, that point
could be anywhere.

382
00:24:16,900 --> 00:24:17,430
OK.

383
00:24:17,430 --> 00:24:24,840
Now, I'm ready for the last two
theorems. And the first

384
00:24:24,840 --> 00:24:29,300
one is called Taylor Series,
the Taylor's theorem.

385
00:24:29,300 --> 00:24:33,800
And we have touched on that.

386
00:24:33,800 --> 00:24:38,110
And what is Taylor
Series about?

387
00:24:38,110 --> 00:24:44,840
Taylor Series is when you know
what's going on at some point

388
00:24:44,840 --> 00:24:50,060
x equal a, and you want to know
what the function is at

389
00:24:50,060 --> 00:24:53,990
some point x near a.

390
00:24:53,990 --> 00:24:57,230
So x is near a.

391
00:24:57,230 --> 00:25:03,780
And, to a very low
approximation, f of x is

392
00:25:03,780 --> 00:25:05,580
pretty close to f of a.

393
00:25:05,580 --> 00:25:08,970
This is the constant term.

394
00:25:08,970 --> 00:25:10,890
That's where the trip started.

395
00:25:10,890 --> 00:25:15,380
So this is like a trip meter
for a very short trip.

396
00:25:15,380 --> 00:25:19,620
The first thing would be to know
what was the trip meter

397
00:25:19,620 --> 00:25:22,650
reading at the start.

398
00:25:22,650 --> 00:25:30,900
But then the correction term, so
this is the calculus term,

399
00:25:30,900 --> 00:25:35,985
it's the speed at the start
times the time of the trip.

400
00:25:39,760 --> 00:25:43,910
If you only keep this, the
trip meter isn't moving.

401
00:25:43,910 --> 00:25:48,280
When you add on this, you're
like following a tangent line.

402
00:25:48,280 --> 00:25:55,050
If I try to describe it, you're
pretending the speed

403
00:25:55,050 --> 00:25:56,380
didn't change.

404
00:25:56,380 --> 00:25:59,030
Here, you're pretending the
trip meter didn't change.

405
00:25:59,030 --> 00:26:00,190
Nothing happened.

406
00:26:00,190 --> 00:26:01,550
Here is the next term.

407
00:26:01,550 --> 00:26:08,490
But now, of course, this speed
normally changes too.

408
00:26:08,490 --> 00:26:12,630
So calculus says there is
a term from the second

409
00:26:12,630 --> 00:26:15,570
derivative, there's
a bending term.

410
00:26:15,570 --> 00:26:19,720
This, we would be correct to
stop right there on a straight

411
00:26:19,720 --> 00:26:23,170
line: constant speed.

412
00:26:23,170 --> 00:26:27,730
But now, if the speed is
increasing, your trip meter

413
00:26:27,730 --> 00:26:34,300
graph is bending upwards, you'd
better have a correction

414
00:26:34,300 --> 00:26:36,750
from the second derivative.

415
00:26:36,750 --> 00:26:41,880
That's the slope of the slope,
the rate of change

416
00:26:41,880 --> 00:26:45,310
of the rate of change.

417
00:26:45,310 --> 00:26:47,710
It's the acceleration.

418
00:26:47,710 --> 00:26:53,980
So, if I had constant
acceleration, like I drop this

419
00:26:53,980 --> 00:26:56,120
chalk, it accelerates.

420
00:26:56,120 --> 00:27:02,650
So, from where I drop it, that
gives me its original height.

421
00:27:02,650 --> 00:27:06,280
Its original speed might be
0, if I hold onto it.

422
00:27:06,280 --> 00:27:11,350
But then, this term would
account for the second

423
00:27:11,350 --> 00:27:12,890
derivative, the acceleration.

424
00:27:12,890 --> 00:27:20,660
And that would give me the right
answer, the right answer

425
00:27:20,660 --> 00:27:25,030
to the next term, but now I've
drawn the famous three dots.

426
00:27:25,030 --> 00:27:30,730
So three dots is the way to
say there are more terms

427
00:27:30,730 --> 00:27:34,660
because the acceleration
might not be constant.

428
00:27:34,660 --> 00:27:36,380
What's the next term?

429
00:27:36,380 --> 00:27:38,340
If you know the next
term, then you

430
00:27:38,340 --> 00:27:40,820
and Taylor are square.

431
00:27:40,820 --> 00:27:46,470
The next term will be
1/3 factorial, 1/6.

432
00:27:46,470 --> 00:27:51,470
It'll be a third derivative of
f at the known point times

433
00:27:51,470 --> 00:27:53,750
this x minus a cubed.

434
00:27:53,750 --> 00:27:57,680
You see that these terms
are getting,

435
00:27:57,680 --> 00:28:00,340
typically, for a nice function--

436
00:28:00,340 --> 00:28:03,150
and we saw this for
e to the x.

437
00:28:03,150 --> 00:28:06,370
We saw the Taylor Series
for e to the x.

438
00:28:06,370 --> 00:28:12,490
Can I remind you of the Taylor
Series for e to the x around

439
00:28:12,490 --> 00:28:18,610
the point 0 because e to the x
is the greatest function I've

440
00:28:18,610 --> 00:28:21,070
spoken about, at all?

441
00:28:21,070 --> 00:28:27,540
So, if this was e to the x, it
would start out at e to the 0,

442
00:28:27,540 --> 00:28:29,260
which is 1.

443
00:28:29,260 --> 00:28:36,530
Its slope is 1, so this
is 1 times x.

444
00:28:36,530 --> 00:28:40,790
Its second derivative
is, again, 1.

445
00:28:40,790 --> 00:28:44,640
And a is 0 here, so
this would be 1/2,

446
00:28:44,640 --> 00:28:47,670
1/2 factorial x squared.

447
00:28:47,670 --> 00:28:50,160
And then that next three-dot
term would be

448
00:28:50,160 --> 00:28:54,100
1/3 factorial x cube.

449
00:28:54,100 --> 00:28:56,660
And you remember what
it looks like.

450
00:28:56,660 --> 00:29:00,880
So the Taylor Series just
looks messy because I'm

451
00:29:00,880 --> 00:29:03,450
writing any old f.

452
00:29:03,450 --> 00:29:08,520
I'm allowing it to be the start
point, to be a, and not

453
00:29:08,520 --> 00:29:09,800
necessarily 0.

454
00:29:09,800 --> 00:29:12,390
But, typically, it's 0.

455
00:29:12,390 --> 00:29:16,930
And the e to the x series
is the best example.

456
00:29:16,930 --> 00:29:20,300
But I want to show you
one more example.

457
00:29:20,300 --> 00:29:23,090
That'll be my last theorem.

458
00:29:23,090 --> 00:29:26,510
I just mention it here because
it's just like

459
00:29:26,510 --> 00:29:29,240
the mean value theorem.

460
00:29:29,240 --> 00:29:37,220
If I do stop, suppose I stop
here and I don't include the x

461
00:29:37,220 --> 00:29:40,890
cube term, the third derivative
term, then I've

462
00:29:40,890 --> 00:29:42,570
made an error.

463
00:29:42,570 --> 00:29:45,700
And, of course, that error
depends on what the third

464
00:29:45,700 --> 00:29:50,970
derivative is, the one I
skipped, the x minus a cube,

465
00:29:50,970 --> 00:29:54,445
the thing I skipped, and
the 1/3 factorial.

466
00:29:57,810 --> 00:30:03,600
And this third derivative
is, at some point,

467
00:30:03,600 --> 00:30:04,850
between a and x.

468
00:30:07,465 --> 00:30:13,570
That's a lot to put in, but the
mean value theorem said

469
00:30:13,570 --> 00:30:16,870
you could take the derivative at
some point in-between, some

470
00:30:16,870 --> 00:30:19,360
point along the MassPike.

471
00:30:19,360 --> 00:30:22,990
And this is just the same thing,
but I'm keeping more

472
00:30:22,990 --> 00:30:28,060
terms. I'm quitting at any
point, and then I would take

473
00:30:28,060 --> 00:30:31,530
the next derivative at somewhere
along the MassPike.

474
00:30:34,410 --> 00:30:37,520
What should you learn
out of that?

475
00:30:37,520 --> 00:30:42,920
I think the idea is
Taylor Series.

476
00:30:42,920 --> 00:30:49,320
And, of course, we have
two possibilities.

477
00:30:49,320 --> 00:30:58,860
Either we cut the series off and
we make some error, but we

478
00:30:58,860 --> 00:31:04,440
get a pretty good answer, or we
let the series go forever.

479
00:31:04,440 --> 00:31:06,770
And then comes the question.

480
00:31:06,770 --> 00:31:09,750
Then we have an infinite number
of terms, and then the

481
00:31:09,750 --> 00:31:13,510
question is does that series add
up to a finite thing like

482
00:31:13,510 --> 00:31:15,200
e to the x?

483
00:31:15,200 --> 00:31:19,120
Or does it add up to a delta
function or something

484
00:31:19,120 --> 00:31:21,580
impossible?

485
00:31:21,580 --> 00:31:25,320
So that leads to the question
of learning

486
00:31:25,320 --> 00:31:29,340
about infinite series.

487
00:31:29,340 --> 00:31:32,710
In calculus, Taylor
Series is where

488
00:31:32,710 --> 00:31:35,280
infinite series come from.

489
00:31:35,280 --> 00:31:39,920
And, if we want to go all the
way with them, then we have to

490
00:31:39,920 --> 00:31:43,050
begin to think about what does
it mean for that infinite

491
00:31:43,050 --> 00:31:46,180
series to add up to a
number, or maybe it

492
00:31:46,180 --> 00:31:49,930
just goes off to infinity.

493
00:31:49,930 --> 00:31:53,890
Does it converge, or
does it diverge?

494
00:31:53,890 --> 00:31:58,410
Ah, that would be another
lecture or two.

495
00:31:58,410 --> 00:32:06,740
Let me complete today with one
more theorem, a famous one,

496
00:32:06,740 --> 00:32:07,970
the binomial theorem.

497
00:32:07,970 --> 00:32:10,750
So, what's the binomial
theorem about?

498
00:32:10,750 --> 00:32:14,870
The binomial theorem is about
powers of 1 plus x.

499
00:32:14,870 --> 00:32:20,690
1 plus x is a typical binomial:
two things, 1 and x.

500
00:32:20,690 --> 00:32:22,900
And we have various powers.

501
00:32:22,900 --> 00:32:26,960
Well, if the powers are the
first power, the second power,

502
00:32:26,960 --> 00:32:32,020
the third power, we can write
out, we can square 1 plus x,

503
00:32:32,020 --> 00:32:34,850
and we can get 1 plus x cubed.

504
00:32:34,850 --> 00:32:36,830
And, out of it, we get this.

505
00:32:36,830 --> 00:32:40,200
And there would be 1 plus
x to the 0-th power.

506
00:32:40,200 --> 00:32:43,110
And do you see that there's a
whole lot of ones in the neat

507
00:32:43,110 --> 00:32:44,440
pattern there?

508
00:32:44,440 --> 00:32:46,900
And then there's a
2, and a 3, 3.

509
00:32:46,900 --> 00:32:53,500
And if you'd like to know this
one, it would be 1, 4, 6, 4, 1

510
00:32:53,500 --> 00:32:56,430
would be the next
row of Pascal.

511
00:32:56,430 --> 00:33:03,630
Pascal really had a sense of
beauty or art in this triangle

512
00:33:03,630 --> 00:33:06,570
of numbers.

513
00:33:06,570 --> 00:33:11,550
And that's the triangle you
get, Pascal's triangle, if

514
00:33:11,550 --> 00:33:12,770
you're taking--

515
00:33:12,770 --> 00:33:17,790
A whole number, a power is 1
plus x to the third power,

516
00:33:17,790 --> 00:33:23,170
fourth power, fifth power,
sixth power, but what if

517
00:33:23,170 --> 00:33:27,390
you're taking to some other
power, any power, p?

518
00:33:27,390 --> 00:33:32,200
So now I'm interested in this
guy to a power of p that,

519
00:33:32,200 --> 00:33:34,250
maybe, is not two, three,
four, five.

520
00:33:34,250 --> 00:33:42,260
It could be 1/2, 1 plus x square
root to the 1/2 power,

521
00:33:42,260 --> 00:33:46,440
or 1 plus x to the
minus 1 power.

522
00:33:46,440 --> 00:33:54,480
All other powers are possible
and, for those,

523
00:33:54,480 --> 00:33:56,240
the Taylor's theorem.

524
00:33:56,240 --> 00:33:57,490
And here's my function.

525
00:34:01,540 --> 00:34:06,690
And I could apply Taylor's
theorem to find the--

526
00:34:06,690 --> 00:34:11,190
and I'll do it at x equals 0,
that's the place Taylor liked

527
00:34:11,190 --> 00:34:16,670
the best. So the
constant term--

528
00:34:16,670 --> 00:34:20,210
think of this Taylor expansion
that we just did--

529
00:34:20,210 --> 00:34:22,880
at x equals 0, this
thing is 1.

530
00:34:22,880 --> 00:34:27,500
So, the big theory starts
out with a 1 for

531
00:34:27,500 --> 00:34:29,080
the constant term.

532
00:34:29,080 --> 00:34:33,730
Then what I do for the next
term of the Taylor Series?

533
00:34:33,730 --> 00:34:37,840
I take the derivative and
I put x equals 0.

534
00:34:37,840 --> 00:34:39,719
And what do I get then?

535
00:34:39,719 --> 00:34:42,420
I get p times x.

536
00:34:42,420 --> 00:34:46,610
So this is the constant
term: f of 0.

537
00:34:46,610 --> 00:34:50,270
This is the derivative:
times x minus a

538
00:34:50,270 --> 00:34:52,270
divided by 1 factorial.

539
00:34:52,270 --> 00:34:55,389
Well, you didn't see all those
things because one factorial I

540
00:34:55,389 --> 00:34:56,530
didn't write.

541
00:34:56,530 --> 00:34:59,740
And then the next term would be
the next derivative, of p

542
00:34:59,740 --> 00:35:03,510
minus 1 will come down, so
you'll have p, p minus 1.

543
00:35:03,510 --> 00:35:05,600
You're supposed to divide
by 2 factorial.

544
00:35:05,600 --> 00:35:08,690
That multiplies x squared.

545
00:35:08,690 --> 00:35:17,070
Well, my point is just that
this binomial formula is

546
00:35:17,070 --> 00:35:19,090
Taylor's formula.

547
00:35:19,090 --> 00:35:24,260
The binomial theorem, with
these, this is called a

548
00:35:24,260 --> 00:35:26,100
binomial coefficient.

549
00:35:26,100 --> 00:35:31,480
Gamblers know all about
that, you know?

550
00:35:31,480 --> 00:35:34,650
If you've got p things and you
want to take two, how many

551
00:35:34,650 --> 00:35:36,140
ways to do it?

552
00:35:36,140 --> 00:35:39,000
You know, how many ways to get
two aces out of a deck, all

553
00:35:39,000 --> 00:35:45,230
these things are hidden in those
numbers, which gamblers

554
00:35:45,230 --> 00:35:47,820
learn or lose.

555
00:35:47,820 --> 00:35:49,190
OK.

556
00:35:49,190 --> 00:35:52,460
So, I'll make one last point
about the binomial theorem.

557
00:35:57,780 --> 00:36:00,050
Those were Taylor Series.

558
00:36:00,050 --> 00:36:01,490
This is a Taylor series.

559
00:36:01,490 --> 00:36:03,220
What's the difference?

560
00:36:03,220 --> 00:36:07,990
The difference is these
series stop.

561
00:36:07,990 --> 00:36:10,220
This is a series: 1
plus x squared.

562
00:36:10,220 --> 00:36:12,770
That's the Taylor Series,
but the third

563
00:36:12,770 --> 00:36:14,790
derivative is 0, right?

564
00:36:14,790 --> 00:36:18,480
The third derivative of that
function, because that

565
00:36:18,480 --> 00:36:21,700
function's only going up to x
squared, the third derivative

566
00:36:21,700 --> 00:36:25,170
is 0, so the rest of Taylor
Series has died.

567
00:36:25,170 --> 00:36:26,310
It's not there.

568
00:36:26,310 --> 00:36:27,830
So that's all there is.

569
00:36:30,810 --> 00:36:33,870
The derivative of any of those
powers, one, two, three, four,

570
00:36:33,870 --> 00:36:39,040
five powers, after I take enough
derivatives, gone.

571
00:36:39,040 --> 00:36:44,270
But, if I take a power like
minus 1, or 1/2, or pi, or

572
00:36:44,270 --> 00:36:48,910
anything, then I can take
derivatives forever

573
00:36:48,910 --> 00:36:54,080
without hitting 0.

574
00:36:54,080 --> 00:36:56,570
In other words, this series
goes on, and on, and on.

575
00:36:56,570 --> 00:36:57,650
Those three dots--

576
00:36:57,650 --> 00:37:01,160
let me move that eraser so you
see those three dots--

577
00:37:01,160 --> 00:37:07,040
that signals an infinite series
and the question of

578
00:37:07,040 --> 00:37:10,560
does it add up to a finite
number, what's going on with

579
00:37:10,560 --> 00:37:11,950
infinite series?

580
00:37:11,950 --> 00:37:19,440
But, for the moment, my point
is just this is what

581
00:37:19,440 --> 00:37:22,390
calculus can do.

582
00:37:22,390 --> 00:37:26,320
If you not only take that slope,
but the slope of the

583
00:37:26,320 --> 00:37:28,890
slope, and the third derivative,
and all higher

584
00:37:28,890 --> 00:37:33,340
derivatives, that's what Taylor
Series tells you.

585
00:37:33,340 --> 00:37:34,250
OK.

586
00:37:34,250 --> 00:37:40,590
So that's the, in some way, high
point of the highlights

587
00:37:40,590 --> 00:37:45,680
of calculus, and I sure hope
they're helpful to you.

588
00:37:45,680 --> 00:37:47,440
Thank you.

589
00:37:47,440 --> 00:37:49,650
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590
00:37:49,650 --> 00:37:52,040
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591
00:37:52,040 --> 00:37:54,310
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592
00:37:54,310 --> 00:37:55,530
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593
00:37:55,530 --> 00:37:58,660
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594
00:37:58,660 --> 00:38:02,050
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595
00:38:02,050 --> 00:38:03,610
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