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CHRISTINE BREINER: Welcome
back to recitation.

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In this video I want to explain
to you where the

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coefficients we saw
in Simpson's rule

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actually come from.

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So what I'm going to do is take
this curve that I have

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and show you what the picture
of Simpson's rule does.

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And then I'm actually going to
determine explicitly where the

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coefficients come from.

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So let's look at this picture
first. The picture I have

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here, the white curve
is going to be my y

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equals f of x function.

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And so, if you remember, what
Simpson's rule is saying, you

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have to have two delta x's.

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So in this case h is
equal to delta x.

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So what Simpson's rule is
saying is I can find,

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approximate, the area under this
curve from minus h to h

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by a certain expression.

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And the expression is, I said
delta x is equal to h, so let

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me write h over 3 times y0 plus
4 y1 plus y2-- oh, y2.

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So that's what you got, that's
what you saw in the lecture,

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that this is what the
coefficients are.

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So I want to explain why this
is a 1, why that's a 4, why

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that's a 1, and where
that 3 comes from.

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The picture is not going to
explain it, but the picture

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will help us understand
how to go.

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So what Simpson's rule does is
it takes those three points,

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so the x0, y0, the x1, y1 and
x2, y2 that are on the curve y

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equals f of x.

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And it finds a parabola through
those three points.

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So I'm going to do my best
to draw what looks like a

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parabola through those
three points.

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Something--

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I'll draw it lightly first
and then I'll fill it in.

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Something along these lines.

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Something like that.

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That's a sort of looks
like a parabola.

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And os the idea is Simpson's
rule is saying I can find the

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area under the blue curve.

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So I can find the area
under the blue curve.

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So this is a function, we'll
call this y equal P of x.

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And that's what you
were actually

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told about in lecture.

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That you're approximating your
function y equals f of x by a

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quadratic function that goes
through the values x0, y0, x1,

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y1, and x2, y2.

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And then you find the area
under that parabola.

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Between minus h and h.

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Now this picture, you might
say, Christine, this looks

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really like a bad estimate.

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This looks kind of stupid.

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Maybe you should have
picked a different

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function to estimate this.

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And I did this one because I
wanted to be zoomed out far

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and to show you, to give you a
little intuition about what

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happens when we make
h smaller.

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Notice that if I cut the size of
h in half, so if I started

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with here was minus h to h, what
would my three points be?

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My three points would be this
point, y1, and y2 would be

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right here.

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Well the quadratic through those
points is much closer to

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the actual function.

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And if I cut that in half again,
I'd have points here,

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here, and here.

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Something like that, and that
starts to look almost exactly

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like a quadratic.

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So if I found the area
under this--

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or if I wanted to estimate the
area under this piece of the

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curve using a quadratic through
those three points,

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they would be very close.

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So don't be alarmed by
Simpson's rule as an

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approximation based on
this large picture.

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So now, what do we have to do?

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Remember, what's our goal?

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Our goal is to figure
out where the

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coefficients come from.

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So what I actually need to do
is I need to evaluate a

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certain integral.

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Or I need to integrate
a certain function.

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I need to integrate P of x.

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So what I'm going to be finding
through the rest of

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this video, is the integral
from minus h

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to h of P of x dx.

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And my goal is to show that this
integral is equal to this

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expression.

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I want to show that
these are equal.

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That's my goal.

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So let's get down to it.

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What do I know?

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What's the only thing I know
right away about P of x?

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I know I'm choosing it to be a
quadratic function and I know

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it goes through three
certain points.

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Right?

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I'll write down what the three
points are when we need them.

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But first let's just say, in
general, let's take this

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integral for a general quadratic
and see how much

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information we actually need.

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So let me come over here.

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Actually, let me say
first, what do I

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mean my general quadratic?

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I mean something of the form big
A, capital Ax squared plus

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Bx plus C. So I'm not filling
in values for these

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coefficients yet.

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Those coefficients are coming
exactly from my three points.

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x0, y0, x1, y1 and x2, y2.

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So let's just integrate that
from minus h to h first and

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see what kind of information
we need.

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So if I come over here.

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So what do I get when I actually
integrate this?

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Well, I get Ax to the third over
3, plus Bx squared over

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2, plus Cx, and then I have to
evaluate for minus h to h.

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Well, if you were thinking about
this, it shouldn't be

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surprising that when I
do this, there's not

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going to be a B term.

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Why is that?

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Well, these two functions
are even.

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Ax squared and C are
even functions.

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And I'm integrating over
something that's symmetric

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about the y-axis.

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I'm going from minus h to h.

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So if you think about Ax
squared, and I'm going from

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minus h to h, I'm finding the
area under a quadratic, from

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minus h to h, it's going to be
twice the area from 0 to h.

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But Bx, that's a line with
slope B. If I wanted to

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integrate Bx from minus h to h,
that's the area, the signed

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area under the curve of a line
Bx, from minus h to h.

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If I just draw a quick
picture, what

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does that look like?

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It's symmetric with respect
to a rotation there.

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I'd have this signed area.

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This is the curve y equals Bx.

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From minus h to h, I get this
area is negative and this are

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is positive and they're equal.

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So, what I'm about to do, you
shouldn't be surprised there

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won't be a B term.

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So what do we actually get
when we evaluate this?

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We get 2A h cubed, over
3, plus 2C h.

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That's what we get.

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You can actually work it all,
put in and all the h's and see

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that, but I knew I was going to
have double what was here

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when I evaluated at h, and
nothing from the B term.

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So we're getting closer.

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We're getting closer.

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We might not look like we're
getting closer, but we're

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getting closer.

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So let's simplify this
expression a little bit.

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Actually, what I ultimately need
to do is I need to figure

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out C and I need to figure
out something over here.

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I need to actually figure
out Ah squared.

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And let me explain why I need
to figure out Ah squared.

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In the end, if I come over back
to what I want to show--

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let me even box it,
so we see it--

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I want to show that this
integral, which I've started

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to calculate, is equal to h over
3 times this quantity.

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So I want to keep one around,
but I want the other, any

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other powers of h to not
be there when I'm done.

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Right?

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I need one power of h there.

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In fact I need an
h over 3 there.

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So I think what I'll do is I'll
start off and I'll pull

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out an h over 3, and then I'll
try and figure out if I can

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get the rest of my expression
to look like what's in the

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parentheses.

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That's really, ultimately,
what I'd like to do.

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So let's come back over here.

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I'm going to pull out an h over
3 from both of these and

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we're going to see what
we end up with.

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OK so if I pull out an h over 3
here, what do I end up with?

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This is easy.

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That's 2A h squared.

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And this one's a little
bit trickier.

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But I have to multiply by 3
over 3 to get a 3 there.

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So I end up with 6C.

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OK.

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Let's make sure I didn't
make any mistakes.

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If I multiply through here
I get 2A h cubed over 3.

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If I multiply through here I get
2h C. Looking good so far.

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Now I have to figure out A and
C, or Ah squared and C. Well,

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C is pretty easy to find.

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Let's think about why that is.

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I have this polynomial.

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The polynomial was equal to--
if we come over here and we

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look again, it's Ax squared
plus Bx plus C. And the

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polynomial has to go through
those three points I had.

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So when x is 0, P of 0 is C. So
let's come back and look at

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our picture.

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When x is 0, what's the output
on the white curve?

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It's y1.

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So C is exactly y1.

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How am I going to
find Ah squared?

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Well, what I need to do is use
these other two points.

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I'm going to evaluate the
function P of x at minus h.

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And it's output has to be y0.

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And I'm going to evaluate the
function P of x at h and it's

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output has to be y2.

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So we're going to come over to
the other side, but that's

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really what we're doing next.

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So let's come over here and
let's evaluate P of h

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and P of minus h.

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So P of h is Ah squared plus Bh
plus C. And P of minus h is

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Ah squared minus
Bh plus C. OK.

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What else?

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Again we said this one has to
be y2, the output, and this

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one has to be y0.

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Because the output at h has to
agree with the output of the

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function f of x.

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And its output at h was y2.

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The output at minus h of P has
to be the same as the output

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at minus h of f.

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And that was y0.

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Now this might not look fun yet
but what if I add these

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two things together.

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What happens?

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I get 2A h squared.

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These drop out.

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And then I get plus 2C is
equal to y0 plus y2.

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I'm getting closer.

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I'm getting much closer.

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Because now notice what I have.
I have 2A h squared, I

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can isolate that.

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And that's something that
I want to figure out.

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I want to figure 2A h squared.

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So let's figure out
what that is.

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2A h squared is equal to y0 plus
y2 minus, well what did

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we say C was?

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C was the output
at x equals 0.

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Which is y1.

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So it's minus 2C, which
is minus 2 y1.

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So now we're very close,
we're very close to

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getting what we want.

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And that's good, because
I'm almost out of room.

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So let's take that expression,
we were working on this

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expression right here,
and let's start to

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fill in what we get.

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We get h over 3 times 2A h
squared which is y0 plus y2

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minus 2 y1, and then
I have to add 6C.

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Now what's C?

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C is y1.

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So I'm going to add 6 y1.

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And if I simplify
that, I get y0--

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00:12:15,950 --> 00:12:16,840
bingo--

250
00:12:16,840 --> 00:12:21,770
plus 4 y1 plus y2.

251
00:12:21,770 --> 00:12:23,290
And that's what we wanted.

252
00:12:23,290 --> 00:12:26,260
So let me take us through kind
of where all this came from

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00:12:26,260 --> 00:12:29,920
again, what the big pieces
were and we'll see now we

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00:12:29,920 --> 00:12:33,330
understand how we do Simpson's
rule, and these small chunks

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00:12:33,330 --> 00:12:34,680
of Simpson's rule.

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00:12:34,680 --> 00:12:35,930
So let's come back to
the very beginning.

257
00:12:35,930 --> 00:12:40,580

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00:12:40,580 --> 00:12:43,690
OK, in the very beginning, what
we had was a function f

259
00:12:43,690 --> 00:12:45,250
of x, that was the white curve.
y equals f of x was the

260
00:12:45,250 --> 00:12:47,270
white curve.

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00:12:47,270 --> 00:12:52,600
And then I found a quadratic
that went through minus h, y0,

262
00:12:52,600 --> 00:12:54,900
0, y1, and h, y2.

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00:12:54,900 --> 00:12:55,930
Went through those
three points.

264
00:12:55,930 --> 00:12:59,340
And I called that quadratic
P of x.

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00:12:59,340 --> 00:13:02,200
And then what I was doing, we
know Simpson's rule says find

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00:13:02,200 --> 00:13:06,190
the area under the curve of P
of x between minus h and h.

267
00:13:06,190 --> 00:13:10,110
So what I did was I came over
here and I said OK, P of x,

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00:13:10,110 --> 00:13:12,640
integral of P of x from
minus h to h.

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00:13:12,640 --> 00:13:16,200
I wrote P of x in a
very general form.

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00:13:16,200 --> 00:13:17,900
And then I actually
found an integral.

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00:13:17,900 --> 00:13:21,700
I came through and after writing
it out, calculating

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00:13:21,700 --> 00:13:23,400
the integral, I got to
this expression.

273
00:13:23,400 --> 00:13:28,180
This is actually the integral
of P of x from minus h to h.

274
00:13:28,180 --> 00:13:30,620
So it's the area under the curve
of P of x from minus h

275
00:13:30,620 --> 00:13:32,270
to h is that.

276
00:13:32,270 --> 00:13:35,060
And now I had to figure this
out, how to write this in

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00:13:35,060 --> 00:13:38,250
terms of the outputs
of f of x.

278
00:13:38,250 --> 00:13:40,340
Which were y0, y1, and y2.

279
00:13:40,340 --> 00:13:42,550
Those were the ones we
were interested in.

280
00:13:42,550 --> 00:13:45,270
So I ultimately knew I wanted
an h over three in front.

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00:13:45,270 --> 00:13:48,800
I knew I wanted my integral in
my quadratic to be h over 3

282
00:13:48,800 --> 00:13:53,330
times something, so I pulled
out an h over 3, and then I

283
00:13:53,330 --> 00:13:55,340
looked at what this
expression was.

284
00:13:55,340 --> 00:13:57,760
And if I could get this
expression to look like the

285
00:13:57,760 --> 00:14:01,710
inside of what I wanted, which
was y0 plus 4 y1 plus y2, I

286
00:14:01,710 --> 00:14:03,460
was in business.

287
00:14:03,460 --> 00:14:08,890
And so then I used outputs that
I knew for p of x to find

288
00:14:08,890 --> 00:14:14,190
2A h squared and to find C. I
know P of h is the output of

289
00:14:14,190 --> 00:14:17,560
the f of x function at the
far right, which was y2.

290
00:14:17,560 --> 00:14:19,990
And I knew P of minus h was
the output of the f of x

291
00:14:19,990 --> 00:14:21,070
function at the far left.

292
00:14:21,070 --> 00:14:22,500
That's y0.

293
00:14:22,500 --> 00:14:26,210
I actually then evaluate P at h
and minus h, and when I add

294
00:14:26,210 --> 00:14:29,770
those together, I get my
2Ah squared in terms

295
00:14:29,770 --> 00:14:31,600
of y0, y1, and y2.

296
00:14:31,600 --> 00:14:34,320
Let me do this in terms
of y0, y1, and y2.

297
00:14:34,320 --> 00:14:37,030
Because I also knew C was y1.

298
00:14:37,030 --> 00:14:38,180
We talked about that as well.

299
00:14:38,180 --> 00:14:39,820
C had to be y1.

300
00:14:39,820 --> 00:14:43,000
So I can then do the
substitution I need right here

301
00:14:43,000 --> 00:14:44,380
and show where the
coefficients in

302
00:14:44,380 --> 00:14:46,750
Simpson's rule come from.

303
00:14:46,750 --> 00:14:48,390
So hopefully that was
informative for you.

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00:14:48,390 --> 00:14:50,370
And I think I'll stop there.

305
00:14:50,370 --> 00:14:50,490