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PROFESSOR: What we're going
to talk about today is a
continuation of last time.
I want to review Newton's method
because I want to talk
to you about its accuracy.
So if you remember, the way
Newton's method works is this.
If you have a curve and you
want to know whether it
crosses the axis.
And you don't know where this
point is, this point which
I'll call x here, what you
do is you take a guess.
Maybe you take a
point x0 here.
And then you go down to this
point on the graph. and you
draw the tangent line.
I'll draw these in a couple of
different colors so that you
can see the difference
between them.
So here's a tangent line.
It's coming out like that.
And that one is going to
get a little closer
to our target point.
But now the trick is, and this
is rather hard to see because
the scale gets small incredibly
fast, is that if
you go right up from that,
and you do this
same trick over again.
That is, this is your second
guess, x1, and now you draw
the second tangent line.
Which is going to come
down this way.
That's really close.
You can see here on the
chalkboard, it's practically
the same as the dot of x.
So that's the next guess.
Which is x2.
And I want to analyze, now,
how close it gets.
And just describe to
you how it works.
So let me just remind you
of the formulas, too.
It's worth having them
in your head.
So the formula for the
next one is this.
And then the idea is just
to repeat this process.
Which has a fancy name, which is
in algorithms, which is to
iterate, if you like.
So we repeat the process.
And that means, for example, we
generate x2 from x1 by the
same formula.
And we did this last time.
And, more generally, the n + 1st
is generated from the nth
guess, by this formula here.
So what I'd like to do is just
draw the picture of one step a
little bit more closely.
So I want to blow
up the picture,
which is above me there.
That's a little too high.
Where are my erasers?
Got to get it a little lower
than that, since I'm going to
depict everything above
the line here.
So here's my curve
coming down.
And suppose that x1 is here, so
this is directly above it
is this point here.
And then as I drew it,
this green tangent
coming down like that.
It's a little bit closer, and
this was the place, x2, and
then here was x, our target,
which is where the curve
crosses as opposed
to the straight
tangent line crossing.
So that's the picture that I
want you to keep in mind.
And now, we're just going to do
a very qualitative kind of
error analysis.
So here's our error analysis.
And we're starting out, the
distance between x1 and x is
what we want to measure.
In other words, how close we
are to where we're heading.
And so I've called that, I'm
going to call that Error 1.
That's x - x1.
In absolute value.
And then, E2 would be x -
x2, in absolute value.
And so forth.
And, last time, when I was
estimating the size of this,
so En would be whatever
it was.
Last time, remember, we did
it for a specific case.
So last time, I actually
wrote down the numbers.
And they were these numbers,
maybe you could call them En,
which was the absolute value
of square root of 5 - xn.
These are the sizes that I was
writing down last time.
And I just want to talk about
in general what to expect.
That worked amazingly well, and
I want to show you that
that's true pretty
much in general.
So the first distance, again, is
E1, is this distance here.
That's the E1.
And then this shorter distance,
here, this little
bit, which I'll mark maybe
in green, is E2.
So how much shorter
is E1 than E2?
Well, the idea is
pretty simple.
It's that if this distance in
this vertical, distances are
probably about the same as the
perpendicular distance.
And this is basically the
situation of a curve touching
a tangent line.
Then the separation is going
to be quadratic.
And that's basically what's
going to happen.
So, in other words the distance
E2 is going to be
about the square of
the distance E1.
And that's really the only
feature of this that I
want to point out.
So, approximately, this
is the situation that
we're going to get.
And so what that means is, and
maybe thinking from last time,
what we had was something
roughly like this.
You have an E0, you have an E1,
you have an E2, you have
an E3, and so forth.
Maybe I'll write down E4 here.
And last time, this was
about 10 ^ - 1.
So the expectation based on
this rule is that the next
error's the square of
the previous one.
So that's 10 ^ - 2.
The next one is the square
of the previous one.
So that's 10 ^ - 4.
And the next one is the square
of that, that's 10 ^ - 8.
And this one is 10 ^ - 16.
So the thing that's impressive
about this list of numbers is
you can see that the number of
digits of accuracy doubles at
each stage.
Accuracy doubles at each step.
The number of digits of accuracy
doubles at each step.
So, very, very quickly you get
past the accuracy of your
calculator, as you saw
on your problem set.
And this thing works
beautifully.
So, let me just summarize by
saying that Newton's method
works very well.
By which I mean this
kind of rate.
And I want to be just
slightly specific.
If, there's really two
conditions disguised in this,
that are going on.
One is that f' has
to be, not to be.
To be not small.
And f'' has to be not too big.
That's roughly speaking what's
going on. i'll explain these
in just a second.
And x0 starts nearby.
Nearby the target value x.
So that's really what's
going on here.
So let me just illustrate
to you.
So I'm not going to explain
this, except to say the reason
why this second derivative gets
involved is that it's how
curved the curve is, that
how far away you get.
If the second derivative were
0, that would be the best
possible case.
Then we would get
it on the nose.
If the second derivative is not
too big, that means the
quadratic part is not too big.
So we don't get away very far
from the green line to the
curve itself.
The other thing to say is,
as I said, that x0
needs to start nearby.
So I'll explain that
by explaining what
maybe could go wrong.
So the ways the method can fail,
and one example which
actually would have happened in
our example from last time,
which was y = x^2 -
5, is suppose we'd
started x0 over here.
Then this thing would have gone
off to the left, and we
would have landed on not
the square root of 5
but the other root.
So if it's too far away, then
we get the wrong root.
So that's an example, explaining
that the x0 needs
to start near the root that
we're talking about.
Otherwise, the method doesn't
know which root
you're asking for.
It only knows where
you started.
So it may go off to
the wrong place.
OK, it can't read your mind.
Yes, question.
STUDENT: [INAUDIBLE]
PROFESSOR: Oh, good question.
So the question was, what
if the first error
is larger than 1.
Are you in trouble?
And the answer is,
absolutely, yes.
If you have quadratic behavior,
you can see.
If you have a quadratic nearby,
it's pretty close to
the straight line.
But far away, a parabola is
miles from a straight line.
It's way, way, way far away.
So if you're foolish enough to
start over here, you may have
some trouble making progress.
Actually, it isn't - when I,
that little wiggle there just
meant proportional to.
In fact, in the particular case
of a parabola, it manages
to get back.
It saves itself.
But there's no guarantee
of that sort of thing.
You really do want to start
reasonably close.
Yep.
STUDENT: [INAUDIBLE]
PROFESSOR: What you have to do
is you have to watch out.
That is, it's hard to know
what assumptions
to make about x0.
You plug it into the machine
and you see what you get.
And either it works
or it doesn't.
You can tell that it's marching
toward a specific
place, and you can tell
that that place
probably is a 0, usually.
But maybe it's not the one
you were looking for.
So in other words, you have
to use your head.
You run the program and then
you see what it does.
And if you're lucky - the
problem is, if you have no
idea where the 0 is, you may
just wander around forever.
As we'll see in a second.
So the next example here
is the following.
I said that f' has to
be not too small.
There's a real catastrophe
hiding
just inside this picture.
Which is the transition between
when you find the
positive root and when you find
the negative root here.
Which is, if you're
right down here.
If you were foolish enough to
get 0, then what's going to
happen is your tangent
line is horizontal.
It doesn't even meet the axis.
So in the formula, you can
see that's a catastrophe.
Because there's an f'
in the denominator.
So that's 0.
That's undefined.
It's not surprising, it's
consistent that the parallel
line doesn't meet the axis.
And you have no x1.
So you had, so if you like,
another point here is that f'
= 0 is a disaster.
A disaster for the method.
Because the next, so
say, if f (x0) =
0, then x1 is undefined.
And finally, there's one other
weird thing that can happen.
Which is, which I'll just draw
a picture of schematically.
Which you can get
from a wiggle.
So this wiggle has
three roots.
The way I've drawn it.
And it can be that you can start
over here with your x0.
And draw your tangent line and
go over here to an x1.
And then that tangent
line will take you
right back to the x0.
I didn't draw it quite right,
but that's about right.
So it goes over like this.
So let me draw the two tangent
lines, so that
you can see it properly.
Sorry, I messed it up.
So here are the two
tangent lines.
This guy and this guy.
And it just goes back and forth.
x0 cycles to x1, and x1
goes back to x0.
We have a cycle.
And it never goes anywhere.
This is, the grass is
always greener.
It's over here, it thinks, oh, I
really would prefer to go to
this 0 and then it thinks
oh, I want to go back.
And it goes back and forth,
and back and forth.
Grass is always greener on the
other side of the fence.
Never, never gets anywhere.
So those are the sorts of things
that can go wrong with
Newton's method.
Nevertheless, it's fantastic.
It works very well, in
a lot of situations.
And solves basically any
equation that you can imagine,
numerically.
Next we're going to move on.
We're going to move on to
something which is a little
theoretical.
Which is the mean
value theorem.
And that will allow us in just a
day or so to launch into the
ideas of integration,
which is the whole
second half of the course.
So let's get started
with that.
The mean value theorem will
henceforth be abbreviated MVT.
So I don't have to write
quite as much every
time I refer to it.
The mean values theorem,
colloquially, says the following.
If you go from Boston to LA,
which I think a lot of Red Sox
fans are going to want to do
soon, so that's 3,000 miles.
In 6 hours, then at some
time you are going
at a certain speed.
The average of this speed.
Average, so speed, which
in this case is what?
So we're going at the
average speed.
That's 3,000 miles times
6 hours, so that's
500 miles per hour.
Exactly.
So sometime on your journey,
of course, some of the time
you're going more than 500 miles
an hour, sometimes you
are going less.
And some time you must've
been going 500
miles an hour exactly.
That's the mean value theorem.
The reason why it's called mean
value theorem is that
word mean is the same
as the word average.
So now I'm going to
state it in math
symbols, the same theorem.
And it's a formula.
It says that the difference
quotient, so this is the
distance traveled divided
by the time elapsed.
That's the average speed, is
equal to the infinitesimal
speed for some time
in between.
So some c, which is in
between a and b.
I'm not quite done.
It's a real theorem,
it has hypotheses.
I've told you the conclusion
first, but there are some
hypotheses, they're
straightforward hypotheses.
Provided f is differentiable;
that is, it has a derivative
in the interval a < x < b.
And continuous in a < or =
x, less than or equal to.
There has to be a sense that you
can make out of the speed,
or the rate of change of f at
each intermediate point.
And in order for the values at
the ends to make sense, it has
to be continuous.
There has to be a link between
the values at the ends and
what's going on in between.
If it were discontinuous, there
would be no relation
between the left and
right values and
the rest of the function.
So here's the theorem,
conclusion and its hypothesis.
And it means what I said more
colloquially up above.
Now, I'm going to prove this
theorem immediately.
At least, give a geometric
intuitive argument, which is
not very different from the
one that's given in a very
systematic treatment.
So here's the proof of the
mean value theorem.
It's really just a picture.
So here's a place, and here's
another place on the graph.
And the graph is going along
like this, let's say.
And this line here is
the secant line.
So this is (a, f (
a )) down here.
And this is (b, f
( b )) up there.
And this segment is the secant,
its slope is the slope
that we're aiming for.
The slope of that line is
the left-hand side of
this formula here.
So we need to find something
with that slope.
And what we need to find is a
tangent line with that slope,
because what's on the right-hand
side is the slip of
a tangent line.
So here's how we construct it.
We take a parallel
line, down here.
And then we just translate it
up, leaving it parallel, we
move it up.
Towards this one.
Until it touches.
And where it touches, at this
point of tangency, down there,
I've just found my value of c.
And you can see that if the
tangent line is parallel to
this line, that's exactly
the equation we want.
So this thing has
slope f' (c).
And this other one has slope
equal to this complicated
expression, f ( b) -
f (a) / (b - a).
That is almost the
end of the proof.
There's one problem.
So, again, we move a
parallel line up.
Move up the parallel line
until it touches.
There's a little subtlety
here, which I
just want to emphasize.
Which is that that dotted line
keeps on going here.
But when we bring it up, we're
going to ignore what's
happening outside of a.
And beyond b, alright?
So we're just going to ignore
the rest of the graph.
But there is one thing
that can go wrong.
So if it does not touch, then
the picture looks likes this.
Here are the same two points.
And the graph is all above.
And we brought up our thing.
And it went like that.
So we didn't construct
a tangent line.
If this happens.
So we're in trouble,
in that point.
In this situation, sorry.
But there's a trick, which is
a straightforward trick.
Then bring the tangent lines
down from the top.
So parallel lines, sorry,
not tangent lines.
Parallel lines.
From above.
So, that's the whole story.
That's how we cook up this
point c, with the right
properties.
I want to point out just one
more theoretical thing.
And then the rest, we're going
to be drawing conclusions.
So there's one more theoretical
remark about the
proof, which is something that
is fairly important to
understand.
When you understand a proof, you
should always be thinking
about why the hypotheses
are necessary.
Where do I use the hypothesis.
And I want to give you an
example where the proof
doesn't work to show you
that the hypothesis is
an important one.
So the example is
the following.
I'll just take a function
which is two
straight lines like this.
And if you try to perform this
trick with these things, then
it's going to come up and it's
going to touch here.
But the problem is that
the tangent line
is not defined here.
There are lots of tangents, and
there's no derivative at
this point.
So the derivative doesn't
exist here.
So this is the claim that one
bad point ruins the proof.
We need f' to exist at
all so, f' ( x ) to
exist at all x in between.
Can't get away even with
one defective point.
Now it's time to draw
some consequences.
And the main consequence is
going to have to do with
applications to graphing.
But we'll see tomorrow and for
the rest of the course that
this is even more
significance.
It's significant to all
the rest of Calculus.
I'm going to list three
consequences which you're
quite familiar with already.
So, the first one is if f'
is positive, then f is
increasing.
And the second one is if f'
is negative, then f is
decreasing.
And the last one seems like the
simplest. But even this
one alone is the key
to everything.
If f' = 0, then f is constant.
These are three consequences,
now, of
the mean value theorem.
And let me show you how
they're proved.
I just told you that they were
true, maybe a while ago.
And certainly I mentioned
the first two.
The last one was so simple that
we maybe just swept it
under the rug.
You did use it on a problem
set, once or twice.
But it turns out that this
actually requires proof, and
we're going to give the
proof right now.
The way that the proof goes is
simply to write down, to
rewrite star.
Rewrite our formula.
Which says that f (b) - f
(a) / (b - a) = f' (c).
And you see I've written it from
left to right here to say
that the right-hand side
information about the
derivative is going to be giving
the information about
the function.
That's the way I'm
going to read it.
In order to express this,
though, I'm going to just
rewrite it a couple
of times here.
So here's f ( a ), multiplying
through by the denominator.
And now I'm going to write it in
another customary form for
the mean value theorem.
Which is f ( b ) = f ( a
) + f' (c) ( b - a).
So here's another version.
I should probably have put this
one in the box to begin
with anyway.
And, just changing it around
algebraically,
it's this fact here.
They're the same thing.
And now with the formula written
in this form, I claim
that I can check these
three facts.
Let's start with
the first one.
I'm going to set things up
always so that a < b.
And that's the setup
of the theorem.
And so that means that
b - a is positive.
Which means that this
factor over here
is a positive number.
If f' is positive, which is
what happens in the first
case, that's the assumption that
we're making, then this
is a positive number.
And so f( b ) > f( a ).
Which means that it's
increasing.
It goes up as the
value goes up.
Similarly, if f' (c) is
negative, then this is a
positive times a negative
number, this is negative. f (
b ) < f(a).
So it goes the other way.
Maybe I'll write this way.
And finally, if f' (c) =
0, then f ( b ) = f(a).
Which if you apply it to all
possible ends means if you can
do it for every interval, which
you can't, then that
means that f is constant.
It never gets to
change values.
Well you might have believed
these facts already.
But I just want to emphasize to
you that this turns out to
be the one key link between
infinitesimals, between limits
and these actual differences.
Before, we were saying that the
difference quotient was
approximately equal
to the derivative.
Now we're saying that it's
exactly equal to a derivative.
Although we don't know exactly
which point to use.
It's some point in between.
I'm going to be deducing some
other consequences in a
second, but let me stop for
second to make sure that
everybody's on board.
Especially since I've finished
the blackboards here.
Before we, everybody happy?
One question.
STUDENT: [INAUDIBLE]
PROFESSOR: I'm just going to
repeat your question first.
I'm a little bit confused, you
said, about what guarantees
that there's a point
of tangency.
That's what you said.
So do you want to elaborate, or
do you want to want to stop
with what you just send?
What is it that confuses you?
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: So I'm not claiming
that there's only one point.
This could wiggle a lot of times
and it maybe touches at
ten places.
In other words, it's
OK with me if it
touches more than once.
Then I just have more,
the more the merrier.
In other words, I
don't want there
necessarily only to be one.
It could come down like this.
And touch a second time.
Is that what was
concerning you?
So in mathematics, when we claim
that this is true for
some point, we don't necessarily
mean that it
doesn't work for others.
In fact, if the function is
constant, this is 0 and in
fact this equation is
true for every c.
That satisfies your question?
The fact that this point exists
actually is a touchy point.
I just convinced you
of it visually.
It's a geometric issue,
whether you're
allowed to do this.
Indeed, it has to do with the
existence of tangent lines and
more analysis then we can
do in this class.
Yeah.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Pardon me.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
what's the difference between
this and the linear
approximation.
And I think, let me see if
I can describe that.
I'll leave the theorem
on the board.
I'm going to get rid
of the colloquial
version of the theorem.
And I'll try to describe to
you the difference between
this and the linear
approximation.
I was planning to do that in a
while, but we'll do it right
now since that's what
you're asking.
That's fine.
So here's the situation.
The linear approximation, so
let's say comparison with
linear approximation.
They're very closely related.
The linear approximation says
the change in f over the
change in x, that's the
left-hand side of this thing,
is approximately f' (a).
For b near a, and
b - a = delta x.
This statement, which is in
the box, which is sitting
right up there, is the statement
that this change in
f is actually equal
to something.
Not approximately equal to it.
It's equal to f' of some c.
And the problem here is that we
don't know exactly which c.
This is for some c.
Between a and b.
Right, so.
That's the difference
between the two.
And let me elaborate
a little bit.
If you're trying to understand
what f ( b ) - f ( a ) / (b -
a) is, the mean value theorem
is telling you for sure that
it's equal to this f' (c).
So that means it's less than
or equal to the largest
possible value on the largest
value you can get, for sure.
And this is on the
whole interval.
And I'm going to include the
ends, because when you take a
max it's sometimes achieved
at the ends.
And similarly, because it's f'
(c), it's definitely bigger
than the min on this
same interval here.
This is all you can say based
on the mean value theorem.
All you know is this.
And colloquially, what that
means is that the average
speed is between the maximum
and the minimum.
Not very surprising.
The mean value theorem is
supposed to be very
intuitively obvious.
It's saying the average speed is
trapped between the maximum
speed and the minimum speed.
For sure, that's something,
that's why, incidentally this
wasn't really proved
when Newton and
Leibniz were around.
But, let's write this so
that you can read it.
Average speed is between
the max and the min.
But nobody had any trouble,
they didn't disbelieve it
because it's a very
natural thing.
Now if, for example, I take
any kind of linear
approximation; say, for
instance, e ^ x is
approximately 1 + x.
Then I'm making the
guess, now, don't
want to say this yet.
That's not going to explain
it to you well enough.
What we're saying, so this
is the mean value here.
This is what the mean
value theorem says.
And here's the linear
approximation.
The linear approximation is
saying that the average speed
is approximately the
initial speed, or
possibly the final speed.
So if a is the left endpoint,
then it's the initial speed.
If it happens to be the right
endpoint, if the value of x is
to the left then it's
the final speed.
So those are the - so you can
see it's approximately right.
Because the speed, when you're
on a short interval, shouldn't
be varying very much.
The max and the min should
be pretty close together.
So that's why the linear
approximation is reasonable.
And this is telling you
absolutely, it's no less than
the min and no more
than the max.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The little kink?
STUDENT: [INAUDIBLE]
PROFESSOR: If you approach
from the top.
So if it's still under here
I can show you it again.
Oh yeah, it's still there.
Good.
STUDENT: [INAUDIBLE]
PROFESSOR: Oh, the one with
the wiggle on top?
Yeah, this one you can't.
Because there's nothing to touch
and it also fails from
the bottom because there's
this bad point.
From the top, it could work.
It can certainly
work both ways.
So, for example.
See if you're a machine, you
maybe don't have a way of
doing this.
But if you're a human being you
can spot all the places.
There are a bunch of spots
where the slope is right.
And it's perfectly OK.
All of them work.
STUDENT: [INAUDIBLE]
PROFESSOR: It's not that
the c is the same.
It's just we've now found one,
two, three, four, five c's for
which it works.
STUDENT: [INAUDIBLE]
PROFESSOR: If you're asked to
find a c, so first of all
that's kind of a
phony question.
There are some questions on your
problem set which ask you
to find a c.
That actually is struggling to
get you to understand what the
statement of the mean value
theorem is, but you should not
pay a lot of attention
to those questions.
They're not very impressive.
But, of course, you would have
to find all the - if it asked
you to find one, you find one.
If you can find some
more, fine.
You can pick whichever
one you want.
Mean value theorem just
doesn't care.
The mean value theorem doesn't
care because actually, the
mean value theorem is never
used except in real life,
except in this context here.
You can never nail down which c
it is, so the only thing you
can say is that you're going
slower than the maximum speed
and faster than the
minimum speed.
Sorry, say that again?
STUDENT: [INAUDIBLE]
PROFESSOR: If you're asked for a
specific c, you have to find
a specific c.
And it has to be in the range.
In between, it has
to be in here.
So now I want to tell you
about another kind of
application, which is really
just a consequence of what
I've described here.
I should emphasize, by the way,
this, probably, should be
doing this.
I guess we've never used
this color here.
This popular.
This is pink.
So this one is so good.
So since we're going
to do this.
So the reason why the
exclamation points are
temporary, this is such
an obvious fact.
But this is the way that you're
going to want to use
the mean value theorem, and this
is the only way you need
to understand the mean
value theorem.
On your test, or ever
in your whole life.
So this is the way
it will be used.
As I will make very clear when
we review for the exam.
In practice what happens is you
even forget about the mean
value theorem, and what you
remember is these three
properties here.
Which are themselves
consequences of
the mean value theorem.
So these are the ones that
I want to illustrate now.
In my next discussion here.
I'm just going to talk about
inequalities. inequalities are
relationships between
functions.
And I'm going to prove a couple
of them using the
properties over there, the
properties that functions with
positive derivatives
are increasing.
Here's an example. e ^ x
> 1 + x, where x > 0.
The proof is the following.
I consider, so here's a proof.
I consider the function f ( x ),
which is the difference. e
^ x - (1 + x).
I observe that it starts at f (
0 ) equal to, well, that's e
^ 0 - (1 + 0), which is 0.
And, it keeps on going.
f' (x) = e ^ x.
If I differentiate here,
the 1 goes away.
I get - 1.
That's the derivative
of the function.
And this function, because
e ^ x > 1, for
x positive is positive.
As x gets bigger and
bigger, this rate
of increase is positive.
And therefore, three dots,
that's therefore, f ( x ) is
bigger than its starting place.
For x > 0.
If it's increasing, then that's,
in particular, it's
increasing starting from 0.
So this is true.
Now, all I have to do is read
what this inequality says.
And what it says is that e ^ x,
just plug in for f ( x ),
which is right here. -( 1 + x)
is greater than the starting
value, which was 0.
Now, I put the thing that's
negative on the other side.
So that's the same thing
as e^x > 1 + x.
That's a typical inequality.
And now, we'll use this
principle again.
Oh gee, I erased the
wrong thing.
I erased the statement
and not the proof.
Well, hide the proof.
The next thing I want to prove
to you is that e ^ x > 1
+ x + (x^2 / 2).
So, how do I do that?
I introduce a function g (x),
which is e^x minus this.
And now, I'm just going
to do exactly the
same thing I did before.
Which is, I get started
with g ( 0 ).
Which is 1 - 1.
Which is 0.
And g' ( x ) is e ^ x minus -
now, look at what happens when
I differentiate this.
The 1 goes away.
The x gives me a 1, and the
x^2 / 2 gives me a + x.
And this one is positive for
x > 0, because of step 1.
Because of the previous
one that I did.
So this one is increasing.
g is increasing.
Which says that g
( x ) > g ( 0 ).
And if you just read that off,
it's exactly the same as our
inequality here. e^x >
1 + x + (x^2 / 2).
Now, you can keep on going with
this essentially forever.
And let me just write
down what you get.
You get e ^ x > 1
+ x + (x^2 / 2).
The next one turns out
to be (x^3 / 3 * 2) +
(x^4 / 4 * 3 * 2).
And you can do whatever
you want.
You can do others.
And this is like the tortoise
and the hare.
This is the tortoise,
and this is the
hare, it's always ahead.
But eventually, if you go
infinitely far, it catches up.
So this turns out to
be exactly equal to
e ^ x in the limit.
And we'll talk about
that maybe at
the end of the course.
