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PROFESSOR: Welcome back
to recitation.

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Today what I want to do is
something maybe a little bit

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more theoretical, but the goal
is to show that something that

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you are going to be repeatedly
doing when you use quadratic

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approximations is,
in fact, true.

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So I'm going to explain the
situation, give a quick

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example, and then show you what
we're setting out to do.

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So the situation is as follows
we're going to--

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any time you see a Q of f,
that's going to represent the

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quadratic approximation
to f at x equals 0.

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So what I've done is say, Q of
f I'm going to define to be

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the thing on the right, which
is exactly the formula you

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were given in class for the
quadratic approximation of a

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function f at x equals 0.

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So f is approximately the thing
on the right near 0.

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Our goal is to show that if I
want to take the quadratic

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approximation of a product of
two functions, that I can do

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it in a different way.

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I can do it in the way written
on the right hand side, which

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actually looks more complicated
in this notation,

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but is, in fact, easier
in reality.

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So let me explain what's
happening and then I'll give

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you an example.

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If I wanted to take the
quadratic approximation of a

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product of two functions, what
I want to show is that

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instead, I could take the
quadratic approximation of

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each individual function,
multiply those together, and

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then take the quadratic
approximation of

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what I get as a result.

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So let me give you
an easy example.

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For instance, let's let f of x
equal e to the x and let's let

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g of x equal sine x.

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Then what is Q of f?

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Q of f is the quadratic
approximation to e to the x at

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x equals 0.

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And that's going to be 1 plus x
plus x squared over 2-- your

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already knew this.

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And g, the quadratic
approximation of

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sine x, is just x.

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So if I wanted to find the
quadratic approximation to e

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to the x sine x, what
I could do--

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what this is claiming
I can do--

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is instead I can take the
quadratic approximation of

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this function times
this function.

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So instead I can take the
quadratic approximation of--

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f was the e to the x--

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1 plus x plus x squared
over 2 times x.

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That's not a 4, sorry.

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That's a parentheses.

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1 plus x plus x squared
over 2 times x.

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And what is that?

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The quadratic approximation
to that is the quadratic

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approximation to x plus x
squared plus x cubed over 2.

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And at x equals 0, if I have
a polynomial, the quadratic

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approximation to a polynomial at
x equals 0 is just all the

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terms up to the quadratic
term.

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So I drop off higher order
terms. So I just

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get x plus x squared.

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So that's the idea.

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The idea is I have a product of
two functions, I know their

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individual quadratic
approximations, and so what I

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want to do is I want to find the
quadratic approximation of

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this product by exploiting the
fact that I already know their

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individual ones, and explain
the fact that quadratic

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approximation of polynomials
at x equals 0 is very easy.

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So that's the example,
that's the idea.

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So now let's see if
we can do it.

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OK.

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So we have a cheat sheet
up here that I'm going

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to refer back to.

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I didn't want to use it again
and I didn't want to have to

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derive this for you, but we
have the product rule-- fg

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prime is equal to what's on the
right, and fg double prime

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is equal to what's
on the right.

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So my goal here is to show,
remember, that the quadratic

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approximation--

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let me point over here again.

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The goal is to show the
quadratic approximation of fg

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is equal to the quadratic
approximation of quadratic

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approximation of f times the
quadratic approximation of g.

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So let's do, well let's do the
right hand side first because

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that's a little nicer.

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And then we'll show the right
hand side and then we'll show

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the left hand side and we'll
show they're equal.

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So let me start here with
the right hand side.

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OK?

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So, let's look at what's the
quadratic approximation of f

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and what's the quadratic
approximation of g and then

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we'll take their final quadratic
approximation.

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So Q of f, we have exactly
what we need there.

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f of 0 plus f prime of 0 times
x plus f double prime of 0

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over 2 x squared.

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Q of g is equal to g of 0 plus
g prime of 0 times x plus g

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double prime at 0 over
2 x squared.

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So now what I'm going to do is
multiply those two together.

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And I'm actually going to swing
this way a tiny bit, if

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that's OK, to write Q of f
times Q of g because it's

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going to be a little long.

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And I'm going to group them
carefully so that I have all

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the higher order terms
at the end.

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OK?

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So I'm going to get
f of 0 g of 0 by

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multiplying these two together.

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And then I'm going to get two
terms involving an x.

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I'm going to get an f prime
times g and a g prime times f.

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Actually, if you'll allow me,
we'll know that anywhere we

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see an f or a g or an f prime
or a g prime, or an f double

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prime or g double prime, they're
all evaluated at 0.

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So I'm going to drop the
0's from here on out.

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Anywhere you see those, I'm
evaluating them at 0.

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Otherwise this will
be way too long.

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So let me write this
just as fg.

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I can even write just a single
one evaluated at 0.

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It's the product evaluated at 0
and then I have f prime g--

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so I'll just evaluate it at 0
at the end of the product--

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plus fg prime evaluated at 0.

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This whole thing is times x.

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I get an x here, I
get an x here.

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Now I need to figure out what
terms give me an x squared.

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OK?

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So the terms that give me an x
squared are f of 0 times g

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double prime over 2.

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That gives me an x squared.

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f double prime times g gives
me an x squared.

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And f prime g prime gives
me an x squared.

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So let's write out those terms.
I get fg double prime

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at 0 over 2--

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from those two--

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plus g f double prime
at 0 over 2--

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from those two--

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plus f prime g prime at
0 all times x squared.

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So there's an x times
an x there--

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gives you an x squared.

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x squared there, x
squared there.

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Now I could keep going, and I
will mention the higher order

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terms, but I'm not going to
write them all the way out

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because of what we're
about to do.

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Let me show you where
they come from.

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You get an x cubed term
from here and an x

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cubed term from here.

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Maybe I'll write the x cubes,
but I won't write

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the x to the fourth.

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So the x cubed terms are f prime
g double prime 0 over 2

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plus g prime f double
prime at 0 over 2.

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And those are my
x cubed terms.

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And then I got some x to the
fourth terms. And where do the

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x to the fourth terms
come from?

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They come from this product.

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Right?

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But I want to point
out something.

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What I'm going to do, I'm
going to work some

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magic on the board.

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This is a quadratic
approximation of f times a

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quadratic approximation of g.

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Let me come over here and remind
you that I want the

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quadratic approximation
of that product.

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So what I'm going to do is go
back and look at what I need

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from there to be the quadratic
approximation of that product.

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So we come back over here.

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If I apply the quadratic
approximation to this thing,

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which means then I'm applying it
to this whole giant thing,

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what do I actually get?

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This is actually a polynomial.

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I have something--

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I have a linear term, I have an
x, I have an x squared, I

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have an x cubed, I have
an x to the fourth.

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So if I apply that quadratic
approximation,

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let's see what stays.

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This term stays.

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This term stays.

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This term stays.

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I'm going to erase what
disappears because I don't

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want us to get confused
by that.

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So these two terms disappear.

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Why?

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Because again, this
is a polynomial.

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I have linear--

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or I have constant, I'm sorry,
I think I called this maybe

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linear earlier-- constant
linear quadratic term.

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And what I need is just those if
I'm looking for a quadratic

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approximation.

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So constant linear quadratic
term, I immediately drop the

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cubic and the quartic term
when I'm looking at a

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quadratic approximation
of a polynomial at 0.

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So if I want the right hand
side, I just need what's

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underlined in blue.

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So now I'm going to put a big
box around that because that's

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going to be important.

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Whatever else happens,
we don't lose that.

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So now we've done the
right hand side.

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And now let's write out what
is the left hand side.

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And that's just going
to be plugging it

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straight into the formula.

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And using our cheat sheet.

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So Q of fg, let me write out the
definition and then we'll

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use the cheat sheet.

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It's fg at 0--

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again, this is f at 0,
g at 0, that's what

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this notation means--

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plus fg quantity prime at 0
times x plus quantity fg

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double prime at 0 over
2 times x squared.

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And now what we're hoping,
remember, is that what's in

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the box is what shows
up over here.

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Because this is the long
way to do the problem.

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This would be if I took either
the x sine x and I took all

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the derivatives.

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And this, in fact, even though
it looks more confusing in

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evaluating such a quadratic
approximation, this way would

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be the easier way.

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We just want to show we don't
lose anything by doing what

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would be the easier way.

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So I get fg at 0.

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And that's good-- we can see we
already have one of those,

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so that's nice.

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What do I get here?

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I get f prime g at 0 plus
g prime f at 0 times x.

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That comes-- let me remind you,
I'll walk over here--

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comes from the cheat sheet.

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The first thing.

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fg prime is f prime
g plus g prime f.

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We know that one pretty well,
but just to remind you.

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So that's where this
term comes from.

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This looks promising because
if we come back to our

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quadratic of quadratic of f
times quadratic of g, it looks

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exactly like the second
term here.

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So now we're hoping that the x
squared term looks like this.

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The only thing that might make
you nervous is this doesn't

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have an over 2, but if you were
paying attention to the

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cheat sheet you'll probably
see where that's coming.

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And I'll point it out
in one moment.

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So fg double prime--

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using the cheat sheet--

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is f double prime g plus
g double prime f

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plus 2 g prime f prime.

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I should have put
0's in there.

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Just to be consistent let me
put these 0's in there.

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00:11:44,000 --> 00:11:49,540

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2 f prime g prime at 0.

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And then I have to divide the
whole thing by 2 because

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there's a divided by 2 there,
and then times x squared.

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So let me move out of the
way for a moment.

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00:12:01,730 --> 00:12:06,670
So this numerator came from the
cheat sheet for the second

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00:12:06,670 --> 00:12:07,710
derivative.

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00:12:07,710 --> 00:12:09,280
And if you need,
we can go back.

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00:12:09,280 --> 00:12:11,260
Let me just remind
you, here it is.

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00:12:11,260 --> 00:12:12,690
You can work it out
for yourself.

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00:12:12,690 --> 00:12:14,270
You can just take the derivative
of the first

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00:12:14,270 --> 00:12:15,040
derivative.

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00:12:15,040 --> 00:12:17,140
But that's where this
comes from.

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00:12:17,140 --> 00:12:19,210
So let me go back and we're
almost done with the problem.

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00:12:19,210 --> 00:12:21,800

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00:12:21,800 --> 00:12:23,160
So what do we see?

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00:12:23,160 --> 00:12:29,630
Well, we see that
we get fg at 0.

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00:12:29,630 --> 00:12:33,390
We get the second term we want,
f prime g at 0 plus g

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00:12:33,390 --> 00:12:38,090
prime f at 0 times x.

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00:12:38,090 --> 00:12:41,190
And then the third term is, in
fact, exactly what we want

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00:12:41,190 --> 00:12:47,530
because we get f double prime
g at 0 over 2 plus g double

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00:12:47,530 --> 00:12:51,940
prime f at 0 over 2 plus--

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00:12:51,940 --> 00:12:53,420
the 2's divide out--

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00:12:53,420 --> 00:12:58,280
and I get f prime g prime
at 0 time x squared.

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00:12:58,280 --> 00:13:01,750
And if we look at this last
term and we look at the

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00:13:01,750 --> 00:13:04,230
squared term in the box
we see, in fact, they

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00:13:04,230 --> 00:13:06,250
are exactly the same.

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00:13:06,250 --> 00:13:08,850
So let me summarize because this
was kind of a long video.

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00:13:08,850 --> 00:13:10,910
So I'm going to go back to the
beginning, give you the

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00:13:10,910 --> 00:13:14,500
example, and tell you what we
were really trying to do here.

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00:13:14,500 --> 00:13:15,590
So let's come back over here.

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00:13:15,590 --> 00:13:20,030
And let me remind you, the goal
was to show that if I

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00:13:20,030 --> 00:13:22,670
wanted to take a quadratic
approximation of a product of

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00:13:22,670 --> 00:13:25,860
two functions, if I already
knew their individual

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00:13:25,860 --> 00:13:30,360
quadratic approximations, you
were told that you could take

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00:13:30,360 --> 00:13:33,380
those two quadratic
approximations, multiply them,

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00:13:33,380 --> 00:13:36,290
and drop off the higher
order terms. The

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00:13:36,290 --> 00:13:38,180
higher order, then 2.

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00:13:38,180 --> 00:13:39,680
So we had an example.

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00:13:39,680 --> 00:13:42,210
We knew these two quadratic
approximations.

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00:13:42,210 --> 00:13:44,450
And you've been told that
quadratic approximation of

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00:13:44,450 --> 00:13:47,600
their product is just the
quadratic approximation of the

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00:13:47,600 --> 00:13:50,000
product of their quadratic
approximations.

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00:13:50,000 --> 00:13:53,320
And so our goal today was to
show that you don't drop any

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00:13:53,320 --> 00:13:57,050
of the terms that you get if you
do it by this method or by

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00:13:57,050 --> 00:13:58,130
this method.

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00:13:58,130 --> 00:13:59,630
And we've done that.

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00:13:59,630 --> 00:14:01,560
So I think I'll stop there.

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00:14:01,560 --> 00:14:01,708