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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
as 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 that 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 some
constant times e^(-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
r_1, and this other radius
is some longer radius, r_2.
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 that a bit
so that it's a little bit easier
to read here.
So here we go.
So here I have radius r_1,
and here I have radius r_2.
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
the 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 r_1 and r_2.
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 r_1 and r_2.
And the range here is that--
Because this is, if you like,
a solid of revolution.
So the variable r is the same
as what we used to call x.
And it's ranging
between r_1 and r_2,
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
2 pi times the radius,
when we sweep the circle around.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: So the
question is, why
is what 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 chunk.
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. r_1
and r_2 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 r_1 and the r_2.
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^(-r^2),
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 this?
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 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 r e^(-r^2).
So that's the antiderivative.
And we're evaluating
it at r_1 and r_2.
So with the minus sign
that's going to get reversed.
The answer is going to
be pi times e^(-r_1^2),
that's the bottom one, minus
e^(r_2^2), 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,
times e^(-r_1^2) - e^(-r_2^2).
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 r_1, right, this is the
role that r_1 plays is this,
and the r_2 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 r_1 and r_2,
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^(-r_1^2) - e^(-r_2^2).
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
computation 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))^2.
Because it's really (2a)^2
is 4-- the quantity is 4a^2,
and then I bring
that exponent out.
Minus e^(-a^2) to the
9th power, 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) is equal to-- 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
the 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,
that it looks like these
places in the middle
are the most likely.
But that's the plot 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 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
than 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 sums.
So this is 1, this
is Riemann sums.
We start out with a, which
is a point we call x_0,
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. y_0 = f(x_0),
that's the point above a.
Then y_1 = f(x_1), that's
the point above x_1.
And so forth, all the way
up to y_n, which is f(x_n).
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 y_0 plus y_1
plus... up to y_(n-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 y_1 to y_n.
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 y_2
and this is y_3, if you
pick one of these shapes,
this height here is y_2,
and this height is y_3,
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 (y_0 + y_1) / 2.
That's the first trapezoid.
The next one is (y_1 + y_2) / 2.
And this keeps on going.
And at the end, I have
(y_(n-2) + y_(n-1)) / 2.
And then last of all, I
have (y_(n-1) + y_n) / 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
y_0 / 2 to start out with.
And now, y_1 got
mentioned twice.
Each time with a factor of 1/2.
So we get a whole y_1 in here.
And the same thing
is going to be
true of all the middle terms.
You're going to get y_2 and
all the way up to y_(n-1).
But then, the last one is
unmatched. y_n 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, a half of this
plus a half of that,
there's an overlap.
The y_1 through y_n
things are listed in both.
But the y_0 only
gets counted 1/2
and the y_n 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 plus the right
Riemann sum divided by 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
y_0, this is y_2,
and this middle segment is y_1.
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, y_0, y_1, and y_2.
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 they are 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
(y_0 + 4y_1 + y_2) / 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 (y_0 +
4y_1 + y_2) / 6 plus--
That's the first chunk.
Now, the second chunk, maybe
I'll just put it in here,
starts-- This is x_2.
This is x_0.
And it goes all the way to x_4.
So x_2, x_3, x_4.
So the next one involves
the indices 2, 3 and 4.
So this is (y_2 +
4y_3 + y_4) / 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) plus 2y_(n-1)-- sorry,
plus 4y_(n-1) plus y_n.
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 get 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) + y_n.
So again, 1 and a 4 to start.
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.
