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PROFESSOR: Today I want to get started by correcting a mistake
that I made last time.
And this was mistaken terminology.
I said that what we were computing, when we computed
in this candy bar example was energy and not heat.
But it's both.
They're the same thing.
And in fact, energy, heat and work are all the
same thing in physics.
I was foolishly considering the much more, what am I trying to
say, the intuitive feeling of heat is just being the
same as temperature.
But in physics, usually heat is measured in calories and energy
can be in lots of things.
Maybe kilowatt-hours or ergs, these are various of the units.
And work would be in things like foot-pounds.
That is, lifting some weight some distance.
And the amount of force you have to apply.
And these all have conversions between them.
They're all the same quantity, in different units.
OK, so these are the same quantity.
Different units.
So that's about as much physics as we'll do for today.
And sorry about that.
Now, the example that I was starting to discuss last time
and then I'm going to carry out today was this
dartboard example.
We have a dartboard, which is some kind of target.
And we have a person, your little brother, who's
standing over there.
And somebody is throwing darts.
And the question is, how likely is he to be hit.
So I want to describe to you how we're going to make this
problem into a math problem.
Yep.
STUDENT: [INAUDIBLE]
PROFESSOR: What topic is this that we're going over.
We're going over an example.
Which is a dartboard example.
And it has to do with probability.
OK.
So what is the probability that this guy, your little
brother, gets hit by a dart.
Now, we have to put some assumptions into this
problem in order to make it a math problem.
And I'm really going to try to make them pretty
realistic assumptions.
So the first assumption is that the number of hits is
proportional to ce ^ - r ^2.
So that's actually a kind of a normal distribution.
That's the bell curve.
But as a function of the radius.
So this is the assumption that I'm going to
make in this problem.
And a problem in probability is a problem of the ratio
of the part to the whole.
So the part is where this little guy is standing.
And the whole is all the possible places where
the dart might hit.
Which is maybe the whole blackboard or extending beyond,
depending on how good an aim you think that this
older child has.
So, if you like, the part is, we'll start simply.
I mean, this doesn't sweep all the way around.
But we're going to talk about some section.
Like this.
Where this is some radius r1, and this other radius is
some longer radius, r2.
And the part that we'll first keep track of is
everything around there.
That's not very well-centered, but it's supposed to be
two concentric circles.
Maybe I should fix a bit so that it's a little
bit easier to read here.
So here we go.
So here I have radius r1, and here I have radius r2.
And then the region in between is what we're going to
try to keep track of.
So I claim that we'll be able to get, so this is what I'm
calling a part, to start out with.
And then we'll take a section of it to get the place where
the person is standing.
Now, I want to take a side view of e ^ - r ^2.
The function that we're talking about.
Again, that's the bell curve.
And it sort of looks like this.
This is the top value, this is now the r axis.
And this is up, or at least, so you should think of this in
terms of the fact that the horizontal here is the
plane of the dartboard.
And the vertical is measuring how likely it is that there
will be darts piling up here.
So if they were balls tumbling down or something else
falling on, many of them would pile up here.
Many fewer of them would be piling up farther away.
And the chunk that we're keeping track of is the
chunk between r1 and r2.
That's the corresponding region here.
And in order to calculate this part, we have to calculate
this volume of revolution.
Sweeping around.
Because really, in disguise, this is a ring.
This is a side view, it's really a ring.
Because we're rotating it around this axis here.
So we're trying to figure out what the total
volume of that ring is.
And that's going to be our weighting, our likelihood for
whether the hits are occurring in this section or in this
ring versus the rest.
To set this up, I remind you we're going to use
the method of shells.
That's really the only one that's going to work here.
And we want to integrate between r1 and r2.
And the range here is that because this is, if you like,
a solid of revolutions.
So the variable r is the same as what we used to call x.
And it's ranging between r1 and r2, and then we're
sweeping it around.
And the circumference of a little piece, so at a
fixed distance r, here, the circumference is
going to be 2 pi r.
So 2 pi r is the circumference.
And then the height is the height of that
green stem there.
That's e ^ - r ^2.
And then we're multiplying by the thickness, which is dr. So
the thickness of the green is dr. The height of this little
green guy is e ^ - r^2, and the circumference is e pi times the
radius, when we sweep the circle around.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: So the question is, why is we're interested
in not this pink area.
And the reason is an interpretation of what
I meant by this.
What I meant by this is that if you wanted to add up what the
likelihood is that this thing will be here versus here, I
want it to be, really, the proportions are the number
of hits times the, if you like, I wanted it d area.
That's really what I meant here.
The number of hits in a little chunk.
So little, maybe I'll call it delta a.
A little chuck.
Is proportional to the chunk times that.
So there's already an area factor.
And there's a height.
So there are a total of three dimensions involved.
There's the area and then this height.
So it's a matter of the, what I was given, what I intended
to say the problem is.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, yeah.
Exactly.
The height is c times that, whatever this c is.
In fact, we don't know what the c is, but because we're going
to have a part and a whole, we'll divide, the c
will always cancel.
So I'm throwing the c out.
I don't know what it is, and in the end it won't matter.
That's a very important question.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Say it again?
STUDENT: PROFESSOR:
PROFESSOR: So what I mean is the number of
hits in some chunk.
That is, suppose you imagine, the question is, what does
this left-hand side mean.
That right?
Is that the question that's you're asking?
When I try to understand what the distribution of
dartboard hits is, I should imagine my dartboard.
And very often there'll be a whole bunch of
holes in some places.
And fewer holes else.
I'm trying to figure out what the whole distribution
of those marks is.
And so some places will have more hits on them
and some places will have fewer hits on them.
And so what I want to measure is, on average,
the number of hits.
So this would really be some, this constant of
proportionality is ambiguous because it depends on
how many times you try.
If you throw a thousand times, it'll be much
more densely packed.
And if you have only a hundred times it'll be fewer.
So that's where this constant comes in.
But given that you have a certain number of times that
you tried, say, a thousand times, there will be a whole
bunch more piled in the middle.
And fewer and fewer as you get farther away.
Assuming that the person's aim is reasonable.
So that's what we're saying.
So we're thinking in terms of the person's always
aiming for the center.
So it's most likely that the person will hit the center.
But on the other hand, it's a fallible person, so
the person may miss.
And so it's less and less likely as you go farther out.
And we're just counting how many times this gets hit, how
many time this and so on.
In proportion to the area.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah. r1 and r2 are arbitrary.
We're going to make this calculation in general.
We're going to calculate what the likelihood is that we
hit any possible band.
And I want to leave those as just letters for now.
The r1 and the r2.
Because I want to be able to try various different
possibilities.
STUDENT: [INAUDIBLE]
PROFESSOR: Say it again., why do we have to take volume.
So this is what we were addressing before.
It's a volume because it's number of hit per unit area.
So there's a height, that is, number of hits.
And then there's an area and the product of those is, so
this is, if you like, a histogram of the
number of hits.
But this should be measured per area.
Not per length of r.
Because on the real diagram, it's going all the way around.
There's a lot more area to this red band than just
the distance implies.
OK.
So, having discussed the setup, this is a pretty standard setup
-- oh, one more question.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah.
So the question is, why is this a realistic.
Why is this choice of function here e to the minus r squared
a realistic choice of function for the darts.
So I can answer this with an analogy.
When people were asking themselves where the V-2
rockets from Germany hit London, they used this model.
It turned out to be the one which was the most accurate.
So that gives you an idea that this is actually real.
The question, this makes it look like people are masters.
That is, that they'll all hit in the center more
often than elsewhere.
But that's actually somewhat deceptive.
There's a difference between the mode, that is, the most
likely spot, and what happens on average.
So in other words, the single most likely spot is the center.
But there's rather little area in here, and in fact the
likelihood of hitting that is some little tiny bit.
In here.
In fact, you're much more likely to be out here.
So if you take the total of the volume, you'll see that much of
the volume is contributed from out here.
And in fact, the person hits rather rarely near the center.
So this is not a ridiculous thing to do.
If you think of it in terms of somebody's aiming at the center
but there's some random thing which is throwing the person
off, then there is likely to be to left or to the right, or
they might even get lucky and all those errors cancel
themselves and they happen to hit pretty much
near the center.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: How does the little brother come into play?
The little brother is going to come into play as follows.
I'll tell you in advance.
So the thing is, the little brother was not so stupid as to
stand in front of the target.
I know.
He stood about twice the radius of the target away.
And so, we're going to approximate the location
by some sector here.
Which is just going to be some chunk of one of these things.
We'll just break off a piece of it.
And that's how we're going to capture.
So the point is, the target is here.
But there is the possibility that the seven-year-old who's
throwing the darts actually missed the target.
That actually happens a lot.
So, does that answer your question?
Alright. are we ready now to to do?
One more question.
STUDENT: [INAUDIBLE]
PROFESSOR: I'm giving you this property here.
I'm telling you that this is what's called a mathematical
model, when you give somebody something like this.
In fact, that requires further justification.
It's an interesting issue.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: I'm giving it to you for now.
And it's something which really has to be justified.
In certain circumstances it is justified.
But, OK.
So anyway, here's our part.
This is going to be our chunk, for now, that we're going to
estimate the the importance.
The relative importance of.
And now, this is something whose antiderivative we can
just do by substitution or by guessing.
It's just - pi e ^ - r^2.
If you differentiate that, you get a - 2r, which cancels
the minus sign here.
So you get 2pi re ^ - r ^2.
So that's the antiderivative.
And we're evaluating it at r1 and r2.
So with the minus sign that's going to get reversed.
The answer is going to be pi ( e ^ - r1^2, that's the bottom
one. - e ^ - r2 ^2), that's the top.
So this is what our part gives us.
And, more technically, if you wanted to multiply through by
c, it would be c times this.
I'll say that in just a second.
OK, now I want to work.
So, if you like, the part is equal to maybe even I'll call
it c pi (e^ -r1 ^2 - e ^ - r2 ^2).
That's what it really is if I put in this factor.
So now there's no prejudice as to how many attempts we make.
Whether it was a thousand attempts or a million
attempts at the target.
Now, the most important second feature here of these kinds of
modeling problems is, there is always some kind
of idealization.
And the next thing that I want to discuss with you is the
interpretation of the whole.
That is, what's the family of all possibilities.
And in this case, what I'm going to claim is that the
reasonable way to think of the whole is it's that r can range
all the way from 0 to infinity.
Now, you may not like this.
But these are maybe my first and third favorite number, my
second favorite number being 1.
So infinity is a really useful concept.
Of course, it's nonsense in the context of the darts.
Because if you think of the basement wall where the kid
might miss a target, he'd probably hit the wall.
He's probably not going to hit one of the walls to the right,
and anyway he's certainly not going to hit over there.
So there's something artificial about sending the possibility
of hits all the way out to infinity.
On the other hand, the shape of this curve is such that the
real tail ends here, because of this exponential
decrease, are tiny.
And that's negligible.
And the point is that actually the value, if you go all the
way out to infinity, is the easiest value to calculate.
So by doing this, I'm idealizing the problem but I'm
actually making the numbers come out much more cleanly.
And this is just always done in mathematics.
That's what we did when we went from differences to
differentials, to differentiation and
infinitesimals.
So we like that.
Because it makes things easier, not because it
makes things harder.
So we're just going to pretend the whole is
from 0 to infinity.
And now let's just see what it is.
It's c pi times the starting place is e^ - 0 ^2.
That's r1, right, this is the role that r1 plays is this,
and the r2 is this value.
Minus e ^ - infinity ^2.
Which is that negligibly small number, 0.
So this is just c pi.
Because this number is 1, and this other number is 0.
This is just (1 - 0), in the parentheses there.
And now I can tell you from these two numbers what
the probability is.
The probability that we landed on the target in a radius
between r1 and r2, so that's this annulus here, is the ratio
of the part to the whole.
Which in this case just cancels the c pi.
So it's (e ^ - r1 ^2 - e ^ - r2^2).
There's the formula for the probability.
So the c canceled and the pi canceled.
It's all gone.
Now, again, let me just emphasize the way this
formula is supposed to work.
The total probability of every possibility here is supposed
to be set up to be 1.
This is some fraction of 1.
If you like, it's a percent.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: This one is giving, the question is, doesn't this
just give the probability of the ring?
This gives you the probability of the ring, but this is
a very, very wide ring.
This is a ring starting with 0, nothing, on the inside.
And then going all the way out.
So that's everything.
So this corresponds to everything.
This corresponds to a ring.
So now, let's see.
Where do I want to go from here.
So in order to make progress here, I still have to give you
one more piece of information.
And this is, again, supposed to be realistic.
When I was three years old and my brother's friend Ralph
was seven, I watched him throwing darts a lot.
And I would say that for Ralph, so for Ralph, at age seven,
anyway, later on he got a little better at it.
But Ralph at age seven, the probability that he hit
the target was about 1/2.
Right?
So he hit the target about half the time.
And the other times, there was cement on the walls of the
basement, it wasn't that bad.
Just bounced off.
That also meant that the points got a little
blunter as time went on.
So it was a little less dangerous when they hit.
Alright.
So now, so here's the extra assumption that I want to make.
So a is going to be the radius of the target.
Now, the other realistic assumption that I want to
make is where this little kid would be standing.
And now, here, I want to get very specific and just do the
competition in one case.
We're going to imagine the target is here.
And the kid is standing, say, between, so we'll
just do a section of this.
This is between 3 o'clock and 5 o'clock.
There's more of him, but it's lower down and
maybe negligible here.
So this section is the part, the chunk that
we want to see about.
And this is a, and then this distance here is 2a.
And then the longest distance here is 3a.
So the longest distance is 3a.
So in other words, what I'm saying is that the probability,
if you're standing too close, the chance Ralph hits younger
brother is about 1/6, right?
Because 2/12 = 1/6.
1/6 of the probability that we're between a and 3a.
That's the number that we're looking for.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The 2/12 came from the fact that we assumed.
So we made a very, very bold assumption here.
We assumed that this human being, who is actually
standing, the floor is about down here.
Maybe he wasn't that tall.
But anyway, so he's really a little bigger than this.
That the part of him that was close to the target covered
about this section here, between radius 2a and 3a.
As you'll see, actually from the computation, because the
likelihood drops off pretty quickly, whatever of him was
standing outside there wouldn't have mattered anyway.
So we're just worried about the part that's closest
to the target here.
STUDENT: [INAUDIBLE]
PROFESSOR: Why is it out of 12?
Because I made it a clock.
And I made it from 3 o'clock to 5 o'clock, so it's 2 of
the 12 hours of a clock.
It's just a way of me making a section that
you can visibly see.
So now, so here's what we're trying to calculate.
And in order to figure this out, I need one
more item. here.
So maybe I'll leave myself a little bit of room.
I have to figure out something about what our
information gave us.
Which is that the probability, sorry, this
probability was 1/2.
So let's remember what this is.
This is going to be e ^ - 0 ^2 - e ^ - a ^2.
That's what's this probability is.
And that's equal to 1/2.
So that means that 1 - e ^ - a ^2 = 1/2.
Which means that e ^ - a ^2 = 1/2.
I'm not going to calculate a, this is the part about the
information about a that I want to use.
That's what I'll use.
And now I'm going to calculate this other probability here.
So the probability right up there is this.
And that's going to be the same as e ^ - (2a)^2 - e ^ - (3a)^2.
That's what we calculated.
And now I want to use some of the arithmetic of exponents.
This is (e ^ - a ^2)^4.
Because it's really (2a)^2 = 4.
The quantity is 4a^2, and then I bring that exponent out. -
(e ^ - a ^2) ^ 9 that's 3 ^2.
And so this comes out to be (1/2) ^ 4 - (1/2) ^9.
Which is approximately 1/16.
This is negligible.
This part here.
And this is actually why these tails, as you go out to
infinity, don't really matter that much.
So this is a much smaller number.
So the probability of the whole band is 1/16.
And now I can answer the question up here.
This is approximately 1/6 * 1/16.
Which is about 1/100, or about 1%.
So if I stood there for 100 attempts, then my chances of
getting hit were pretty high.
So that's the computation.
That's a typical example of a problem in probability.
And let me just make one more connection with
what we did before.
This is connected to weighted averages or integrals
over weights.
But the weight that's involved in this problem was w ( r) =,
so let's just look at what happened in all of
those integrals.
What happened in all the integrals was, we had
this factor here.
2 pi r.
And if I include this c, it was really 2 pi c r e ^ - r ^2.
This was the weight that we were using.
The relative importance of things.
Now, this is not the same as e ^ - r ^2 that
we started out with.
Because this is the one-dimensional histogram.
And that has to do with the method of shells that gave us
that extra factor of r here.
So that also connects with the question at the beginning.
Which had to do with this paradox, and it looks like
these places in the middle are the most likely.
But that's the plot of e ^ - r ^2.
If you actually look at this plot here, you see that as r
goes to 0, it's going to 0.
This is a different graph here.
And actually, so this is what's happening really.
In terms of how likely it is that you'll get within a
certain distance of the center of the target.
Again, it's not the area under that curve that we're doing.
It's that volume of revolution.
We're going to change subjects now.
OK, one more question.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, that's supposed to be the
graph of w (r).
STUDENT: [INAUDIBLE]
PROFESSOR: Well, so, the question is, wouldn't
the importance of the center be greatest?
It's a question of which variable you're using.
According according to pure radius, it's not.
It turns out that there are some bands in radius which are
more important, more likely for hits than others.
It really has to do with the fact that the center, or the
core, of the target is really tiny.
So it's harder to hit.
Whereas a whole band around the outside has a lot more area.
Many, many ways to hit that band.
So it's a much larger area.
So there's a competition there between those two things.
So we're going to move on.
And I want to talk now about a different kind
of weighted average.
These weighted averages are going to be much simpler.
And they come up in what's called numerical integration.
There are many, many methods of integrating numerically.
And they're important because many, many integrals
don't have formulas.
And so you have to compute them with a calculator or a machine.
So the first type that we've already done are Riemann sums.
They turned out to be incredibly inefficient.
They're lousy.
The next rule that I'm going to describe is
a little improvement.
It's called the trapezoidal rule.
And this one is much more reasonable then
the Riemann sum.
Unfortunately, it's actually also pretty lousy.
There's another rule which is just a slightly trickier rule.
And it's actually amazing that it exists.
And it's called Simpson's Rule.
And this one is actually pretty good.
It's clever.
So let's get started with these.
And the way I'll get started is by reminding you what
the Riemann sum is.
So this is a good review, because you need to know
all three of these.
And you're going to want to see them all laid out in parallel.
So here's the setup.
OK, so here we go.
We have our graph, we have our function it starts out
at a, it ends up at b.
Maybe goes on, but we're only paying attention
to this interval.
This is a function y = f(x).
And we split it up when we do Riemann's sum.
So this is 1, this is Riemann sums.
We start out with a, which is a point we call x0, and then
we go a certain distance.
We go all the way over to b.
And we subdivide this thing with these delta x's.
Which are the step sizes.
So we have these little steps of size delta x.
Corresponding to these x values, we have y values.
y0 = f(x0), that's the point above a.
Then y1 = f ( x1), that's the point above x1.
And so forth, all the way up to yn, which = f (xn).
In order to figure out the area, you really need to know
something about the function.
You need to be able to evaluate it.
So that's what we've done.
We've evaluated here at n + 1 points.
Enumerated 0 through n.
And those are the numbers out of which we're going to get
all of our approximations to the integral.
So somehow we want average these numbers.
So here's our goal.
Our goal is to average, or add, I'm using average
very loosely here.
But I was going to say add up these numbers to
get an approximation.
To average or add the y's.
To get an approximation to the integral.
Which we know is the area under the curve.
So here's what the Riemann sum is.
It's the following thing.
You take (y0 + y1 + ... up to yn - 1) And you
multiply by delta x.
That's it.
Now, this one is the one with left endpoints.
So the left-hand sum.
There's also a right one.
Which is, if you start at the right-hand ends.
And that will go from y1 to yn.
OK, so this one is the right-hand.
Right Riemann sum.
Those are the two that we did before.
Now I'm going to describe to you the next two.
They have a similar pattern to them.
And the one with trapezoids requires a picture.
Here's a shape.
And here's a bunch of values.
And we're trying to estimate the size of these chunks.
And now, instead of doing something stupid, which is to
draw horizontal lines in rectangles, we're going to do
something slightly more clever.
Which is to draw straight lines that are diagonal.
You see that many of them actually coincide probably
pretty closely with what I drew there.
Although if they're curved, they miss by a little bit.
So this is called the trapezoidal rule.
Because if you pick one of these shapes, say, this is
y2 and this is y3, if you pick one of these shapes,
this height here is y2.
And this height is y3, and this base is delta x.
This is a trapezoid.
So this being a trapezoid, I can figure out its area.
And what do I get?
I get the base times the average height.
If you think about if, you work out what happens when you do
something with a straight line on top, like that, you'll
get this average.
So this is the average height of the trapezoid.
And now I want to add up.
I want to add them all up to get my formula for
the trapezoidal rule.
So what do I do?
I have delta x times the first one.
Which is y0 + y1 / 2.
That's the first trapezoid.
The next one is y1 + y2 / 2.
And this keeps on going.
And at the end, I have yn - 2 + yn - 1 / 2.
And then last of all, I have yn - 1 + yn / 2.
That's a very long formula here.
We're going to simplify it quite a bit in just a second.
What's this equal to?
Well, notice that I get y0 / 2 to start out with.
And now, y1 got mentioned twice.
Each time with a factor of 1/2.
So we get a whole y1 in here.
And the same thing is going to be true of
all the middle terms.
You're going to get y2 and all the way up to yn - 1.
But then, the last one is unmatched. yn is only
1/2, only counts 1/2.
So here is what's known as the trapezoidal rule.
Now, I'd like to compare it for you to the Riemann sums,
which are sitting just to the left here.
Here's the left one, and here's the right one.
If you take the average of the left and the right, that is,
1/2 of this + 1/2 of that, there's an overlap.
The y1 through yn things are listed in both.
But the y0 only gets counted 1/2 and the yn only
gets counted 1/2.
So what this is, is this is the symmetric compromise between
the two Riemann sums.
This is actually equal to the left Riemann sum + the
right Riemann sum / 2.
It's the average of them.
Now, this would be great and it does look like it's closer.
But actually it's not as impressive as it looks.
If you actually do it in practice, it's
not very efficient.
Although it's way better than a Riemann sum, it's
still not good enough.
So now I need to describe to you the fancier rule.
Which is known as Simpson's Rule.
And so, this is, if you like, 3, Method 3.
The idea is again to divide things into chunks.
But now it always needs n to be even.
In other words, we're going to deal not with just one
box, we're going to deal with pairs of boxes.
Here's delta x, and here's delta x again.
And we're going to study the area of this piece here.
So let me focus just on that part.
Let's reproduce it over here.
And here's the delta x, here's delta x.
And of course there are various heights.
This starts out at y0, this is y2, and this
middle segment is y1.
Now, the approximating curve that we're going
to use is a parabola.
That is, we're going to fit a parabola through
these three points.
And then we're going to use that as the approximating area.
Now, it doesn't look like, this looks like it misses.
But actually, most functions mostly wiggle either
one way or the other.
They don't switch.
They don't have inflection points.
So, this is a lousy, at this scale.
But when we get to a smaller scale, this becomes
really fantastic.
As an approximation.
Now, I need to tell you what the arithmetic is.
And in order to save time, it's on your problem set
what the actual formula is.
But I'm going to tell you how to think about it.
I want you to think about it as follows.
So the area under the parabola is going to be a base times
some kind of average height.
And the base here, you can already see.
It's 2 delta x.
The base is 2 delta x.
Now, the average height is weird.
You have to work out what it is for a parabola, depending on
those three numbers. y0, y1, and y2.
And it turns out to be the following formula.
It has to be an average, but it's an interesting
weighted average.
So this was the punchline, if you like.
Is that there were such things as interesting
weighted averages.
This one's very simple, it just involves three numbers.
But it's still interesting.
It's the following.
It turns out to be y0 + 4 y1 + y2 / 6.
Why divided by 6?
Well, it's supposed to be an average.
So the total is 1 + 4 + 1 of these things.
And 6 is in the denominator.
So it emphasizes the middle more than the sides.
And that's what happens with a parabola.
So this is a computation which is on your homework.
And now we can put this together for the full
Simpson's Rule formula.
Which I'll put up over here.
We have here 2 delta x, and we divide by 6.
And then we have y0 + 4 y1 + y2 / 6 +...
that's the first chunk.
Now, the second chunk, maybe I'll just put it in
here, starts this is x2.
This is x0.
And it goes all the way to x4.
So x2, x3, x4.
So the next one involves the indices 2, 3 and 4.
So this is y2 + 4 y3 + y4 / 6.
Oh, oh, oh, oh, no.
I think I'll get rid of these 6's.
I have too many 6's.
Alright.
Let's get rid of them here.
Let's take them out.
Put them out here.
Thank you.
All the way to the end, which is y(n - 2) + 2 y(n - 1) -
sorry, + 4 y(n - 1) + yn.
I was about to divide by 6, but you saved me.
So here are all the chunks.
Now, what does this pattern come out to be?
This comes out to be the following.
This is 1, 4, 1, added to 1, 4, 1 added to 1, 4, 1.
You add them up.
You 1, 4, and then there's a repeat, so you get a 2 and a
4, and a 2 and a 4 and a 1.
So the pattern is that it starts out with
1's on the far ends.
And then 4's next in.
And then it alternates 2's and 4's in between.
So the full pattern of Simpson's Rule is delta x /
3, I have now succeeded in canceling this 2 with this
6 and getting out that factor of 2.
And then here I have y0 + 4 y1 + 2 y2 + 4 y3 + ... it keeps on
going and keeps on going and keeps on going.
And in the end it's 2 y(n - 2) + 4 y(n - 1) + yn.
So again, 1 and a 4 to start.
1 and a 4 and a 1 to end.
And then alternating 2's and 4's in the middle And this
weird weighted average is way better.
As I will show you next time.