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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

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to show that
something that you are

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going to be
repeatedly doing when

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you use quadratic approximations
is, in fact, true.

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So I'm going to
explain the situation,

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give a quick example,
and then show you

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what we're setting out to do.

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So the situation is
as follows: we're

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

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going to represent the
quadratic approximation to f

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at x equals 0.

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So what I've done
is say, Q of f I'm

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going to define to be the thing
on the right, which is exactly

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the formula you
were given in class

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for the quadratic approximation
of a 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

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to take the quadratic
approximation

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of a product of two
functions, that I

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can do 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, but is,

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in fact, easier in reality.

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So let me explain
what's happening

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and then I'll give
you an example.

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If I wanted to take the
quadratic approximation

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of a product of two
functions, what I want to show

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is that instead, I could take
the quadratic approximation

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of each individual function,
multiply those together,

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and then take the quadratic
approximation of what

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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

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e to the x and let's
let 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

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at x equals 0.

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

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your already knew this.

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And g, the quadratic
approximation of sine x,

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is just x.

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So if I wanted to find the
quadratic approximation

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to e to the x sine
x, what I could

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do-- what this is
claiming I can do--

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is instead I can take the
quadratic approximation

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of this function
times this function.

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So instead I can take the
quadratic approximation

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of-- f was the e to the
x-- 1 plus x plus x squared

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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

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is the quadratic
approximation to x plus x

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squared plus x cubed over 2.

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

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the quadratic approximation
to a polynomial at x equals 0

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is just all the terms up
to the quadratic term.

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So I drop off
higher-order terms.

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So I just 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,

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I know their individual
quadratic approximations,

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and so what I want
to do is I want

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to find the quadratic
approximation of this product

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by exploiting the
fact that I already

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know their individual
ones, and explain

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the fact that quadratic
approximation of polynomials

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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

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going to refer back to.

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

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to derive this for you, but
we have the product rule:

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f*g prime is equal to
what's on the right,

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and f*g double prime is
equal to what's on the right.

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So my goal here is
to show, remember,

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that the quadratic
approximation-- let

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me point over here again.

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The goal is to show the
quadratic approximation of f*g

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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

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because that's a little nicer.

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And then we'll show
the right-hand side

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and then we'll show
the left-hand side

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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

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and then 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 over 2

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x squared.

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

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x plus g 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,

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

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because it's going
to be a little long.

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And I'm going to group
them carefully so

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that I have all 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

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by 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

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know that anywhere
we see an f or a g

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

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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 f*g.

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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.

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And then I have f prime g--
so I'll just evaluate it

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at 0 at the end of the product--
plus f g 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

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are f of 0 times g
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.

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I get f g double prime at 0 over
2-- from those two-- plus g f

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double prime at 0 over
2-- from those two-- plus

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

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So there's an x times an x
there, 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

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mention the higher-order
terms, but I'm not

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going to write them all
the way out because of what

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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

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and an x cubed term from here.

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

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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.

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And where do the 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 magic

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on the board.

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This is a quadratic
approximation

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of f times a quadratic
approximation of g.

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Let me come over
here and remind you

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that I want the quadratic
approximation of that product.

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So what I'm going
to do is go back

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and look at what
I need from there,

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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

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to this thing,
which means then I'm

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applying it to this whole giant
thing, what do I actually get?

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This is actually a polynomial.

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I have something-- I have
a linear term, I have an x,

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I have an x squared,
I have an x cubed,

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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 want

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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-- or I
have constant, I'm sorry,

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I think I called this maybe
linear earlier-- constant,

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linear, quadratic term.

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And what I need is just
those if I'm looking

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for a quadratic approximation.

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So constant, linear,
quadratic term.

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I immediately drop the
cubic and the quartic term

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when I'm looking at a
quadratic approximation

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of a polynomial at 0.

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So if I want the
right-hand side,

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I just need what's
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 f*g, let me write
out the definition and then

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we'll use the cheat sheet.

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It's f*g at 0-- again,
this is f at 0, g at 0,

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that's what this
notation means--

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plus f*g quantity prime at
0 times x plus quantity f*g

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double prime at 0 over
2 times x squared.

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And now what we're
hoping, remember,

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is that what's in 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

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and I took all the derivatives.

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And this, in fact,
even though it

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looks more confusing
in evaluating

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such a quadratic
approximation, this way

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would be the easier way.

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

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what would be the easier way.

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So I get f*g 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

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walk over here-- comes
from the cheat sheet.

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The first thing.

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f*g 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

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to our quadratic of quadratic
of f times quadratic of g,

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it looks 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

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is this doesn't have
an over 2, but if you

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were paying attention
to the cheat sheet

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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 f*g double prime,
using the cheat sheet,

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is f double prime g plus g
double prime f plus 2 g prime f

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prime.

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I should have put 0's in there.

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00:11:41,450 --> 00:11:44,000
Just to be consistent let
me put these 0's in there.

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00:11:49,540 --> 00:11:52,890
2 f prime g prime at 0.

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00:11:52,890 --> 00:11:55,540
And then I have to divide
the whole thing by 2

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00:11:55,540 --> 00:11:57,800
because there's a
divided by 2 there,

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00:11:57,800 --> 00:11:59,332
and then times x squared.

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00:11:59,332 --> 00:12:01,730
So let me move out of
the way for a moment.

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00:12:01,730 --> 00:12:05,340
So this numerator came
from the cheat sheet

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00:12:05,340 --> 00:12:07,710
for the second 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 --> 00:12:12,666
You can work it
out for yourself.

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00:12:12,666 --> 00:12:15,040
You can just take the derivative
of the first derivative.

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00:12:15,040 --> 00:12:16,835
But that's where
this comes from.

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00:12:16,835 --> 00:12:19,210
So let me go back and we're
almost done with the problem.

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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 f*g at 0.

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00:12:29,630 --> 00:12:32,810
We get the second term
we want, f prime g at 0

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00:12:32,810 --> 00:12:38,090
plus g prime f at 0 times x.

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00:12:38,090 --> 00:12:39,590
And then the third
term is, in fact,

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00:12:39,590 --> 00:12:45,080
exactly what we want because
we get f double prime g at 0

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00:12:45,080 --> 00:12:52,810
over 2 plus g double prime f at
0 over 2 plus-- the 2's divide

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00:12:52,810 --> 00:12:58,280
out-- and I get f prime g
prime at 0 time x squared.

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00:12:58,280 --> 00:13:01,190
And if we look at
this last term and we

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00:13:01,190 --> 00:13:04,130
look at the squared term
in the box we see, in fact,

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00:13:04,130 --> 00:13:06,250
they 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 --> 00:13:10,580
So I'm going to go
back to the beginning,

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00:13:10,580 --> 00:13:12,829
give you the example, and
tell you what we were really

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00:13:12,829 --> 00:13:14,382
trying to do here.

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00:13:14,382 --> 00:13:15,590
So let's come back over here.

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00:13:15,590 --> 00:13:18,000
And let me remind
you, the goal was

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00:13:18,000 --> 00:13:22,060
to show that if I wanted to
take a quadratic approximation

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00:13:22,060 --> 00:13:24,780
of a product of two
functions, if I already

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00:13:24,780 --> 00:13:27,890
knew their individual
quadratic approximations,

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00:13:27,890 --> 00:13:31,670
you were told that you
could take those two

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00:13:31,670 --> 00:13:34,020
quadratic approximations,
multiply them, and drop off

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00:13:34,020 --> 00:13:36,230
the higher-order terms.

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00:13:36,230 --> 00:13:38,180
The higher order than 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 --> 00:13:44,370
And you've been told that
quadratic approximation

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00:13:44,370 --> 00:13:47,300
of their product is just
the quadratic approximation

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00:13:47,300 --> 00:13:50,000
of the product of their
quadratic approximations.

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00:13:50,000 --> 00:13:52,220
And so our goal
today was to show

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00:13:52,220 --> 00:13:54,000
that you don't drop
any of the terms

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00:13:54,000 --> 00:13:58,130
that you get if you do it by
this method or by 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 --> 00:14:01,208
So I think I'll stop there.