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HERBERT GROSS: Hi.
Well, I guess we're in the home
stretch of the course.
We're down to our last concept,
and it will take us
essentially two lectures to
cover this new concept.
Today we'll do the concept in
general and next time we'll
apply it specifically to
the concept of series.
But the concept we want to talk
about is something called
'Uniform Convergence'.
Let me say at the outset that
this is a very subtle topic.
It is difficult.
It seems to be beyond the scope
of our textbook, because
it's not mentioned there.
Consequently, I will try to give
the highlights as I speak
with you, but the supplementary
notes will
contain a more detailed
explanation of the things that
we're going to talk about.
Now, to set the stage properly
for our discussion of uniform
convergence, I think it's wise
that we at least review the
concept of convergence
in general.
Well, let's take a
look over here.
Recall that when we write that
the limit of ''f sub n' of x'
as 'n' approaches infinity
equals 'f of x' for all 'x' in
the closed interval
from 'a' to 'b'--
when this happens, another way
of saying this is that the
sequence 'f sub n', the sequence
of functions 'f sub
n', converges to the function
'f' on the closed interval
from 'a' to 'b'.
Now, again, these tend to be
words unless you look at a
specific example.
Let's just pick one over here.
Let ''f sub n' of x' be
''n' over '2n plus
1'' times 'x squared'.
Notice, of course, that the
value of this particular
number depends both
on 'x' and 'n'.
At any rate, let's pick a fixed
'x', hold it that way,
and take the limit of
''f sub n' of x' as
'n' approaches infinity.
In other words, as we let 'n'
approach infinity in this
case, notice that the limit
of 'n' over '2n
plus 1' becomes 1/2.
'x' has been chosen
independently of
the choice of 'n'.
Consequently, the limit function
in this case, 'f of
x', is 1/2 'x squared'.
And in this case, what we say
is that the sequence of
functions 'n 'x squared'' over
'2n plus 1' converges to 1/2
'x squared'.
Now what does this mean
more specifically?
In other words, let's see if we
can look at a few specific
values of 'x'.
For example, if I choose
'x' to be 2--
in other words, if I choose 'x'
to be 2, if we look over
here, this says that ''f
sub n' of x' is '4n'
over '2n plus 1'.
If I now take the limit as 'n'
approaches infinity, I'm going
to wind up with what?
'x' is replaced by 2, 2 squared
is 4, 1/2 of 4 is 2.
In other words, the limit of ''f
sub n' of 2', as 'n' goes
to infinity, is 2.
In a similar way, if I replace
'x' by 4, 1/2 'x
squared' becomes 8.
And so what we have is, at the
limit, as 'n' approaches
infinity, ''f sub
n' of 4' is 8.
The key thing being that once
you choose 'x', notice that
for a fixed 'x', ''f sub n' of
x' is a constant and you're
now taking the limit of
a sequence of numbers.
At any rate, here's what
the key point is.
What does this mean by
our basic definition?
By our basic definition, it
means that we can find the
number, capital 'N sub 1', such
that when 'n' is greater
than capital 'N sub 1', the
absolute value of ''f sub n'
of 2', minus 2, is less
than epsilon.
In a similar way, this means
that we can find the number
capital 'N sub 2' such that
for any 'n' greater than
capital 'N sub 2', the absolute
value of ''f sub n'
of 4' minus 8 is also
less than epsilon.
The key point is that 'N1' and
'N2' can be different.
In other words, you may have to
go out further to make this
difference less than epsilon
than you do to make this
difference less than epsilon.
In other words, you see what's
happening here-- and we're
going to review this in writing
in a few minutes, so
that you'll see it
in front you.
What happens here is that you
see that for different values
of 'x', we get different
values of 'n'.
And since there are infinitely
many values of 'x', it means
that in general, we're going
to be in a little bit of
trouble trying to find one 'n'
that works for everything.
And let me show you
what that means
again, going more slowly.
I simply call this two
basic definitions.
In other words, if we have a
sequence of functions 'f sub
n', each of which is defined
on the closed interval from
'a' to 'b', we say that that
sequence of functions
converges point-wise--
that means number by number--
to 'f' on [a, b]
if this limit, ''f sub n' of x'
as 'n' approaches infinity,
equals 'f of x' for
each 'x' in [a,b].
In other words, given epsilon
greater than 0, we can find
'N1' such that 'n' greater than
'N1' implies that the
absolute value of ''f sub n' of
x1' minus 'f of x1' is less
than epsilon for a given
'x1' in [a,b].
In general, the choice of 'N1'
depends on the choice of 'x1',
and there are infinitely many
such choices to make in [a,b].
Now the key point is this,
and this is where uniform
convergence comes in.
If we can find one 'N' such that
whenever little 'n' is
greater than that capital 'N',
''f sub n' of x' minus 'f of
x' in absolute value is less
than epsilon for every 'x' in
the closed interval, then
we say that the
convergence is uniform.
In other words, if we can find
one 'N' that makes that
difference less than epsilon for
the entire interval, then
we call the convergence
uniform.
Now, you see, convergence in
general is a tough topic.
In particular, uniform
convergence may seem even more
remote, and therefore what I'd
like to do now is-- saving the
formal proofs for the
supplementary notes, let me
show you pictorially just what
the concept of uniform
convergence really is.
So let me give you a pictorial
representation.
Let's suppose I have the curve
'y' equals 'f of x'.
Now, to be within epsilon of 'f
of x' means I retrace this
curve displaced epsilon units
above the original position,
and epsilon units below.
In other words, for a given an
epsilon, I now draw the curve
'y' equals ''f of x' plus
epsilon' and 'y' equals ''f of
x' minus epsilon'.
Now, what uniform convergence
means is this, that for this
given epsilon, I can find a
capital 'N' such that whenever
'n' is greater than capital 'N',
the curve 'y' equals ''f
sub n' of x' lies in
this shaded region.
In other words, it can bounce
around all over, but it can't
get outside of this region.
In other words, once I'm far
enough out of my sequence, all
of the curves lie in this
particular region.
Now, of course, the question
is what does this all mean?
And the answer is--
well, look.
Let's take epsilon to be, and I
put this in quotation marks,
"very, very small." Let's take
epsilon, for example, to be so
small that it's within the
thickness of our chalk.
If I now do this--
see, I draw the curve
'y' equals 'f of x'.
Notice now that the thickness of
my curve itself is the band
width 2 epsilon.
All I'm saying is that for this
very, very small epsilon,
when 'n' is sufficiently large,
the curve 'y' equals
''f sub n' of x' appears to
lie inside of this curve.
You see, in other words, what
you're saying is that for a
large enough 'n' and small
enough epsilon--
loosely speaking what you saying
is that 'y' equals ''f
sub n' of x' looks like 'y'
equals 'f of x', for a
sufficiently large
values of 'n'.
In other words, it appears that
we can't really tell the
n-th curve in the sequence
from the limit function.
And I want to make a few
key observations
about what that means.
I've written the whole thing out
on the blackboard so that
you can see this after I say it,
but what I want you to see
is, can you begin to get the
feeling that with this kind of
a condition, for example--
if each 'f sub n' happens to be
continuous, in other words,
if each member of my sequence
is unbroken, then the limit
function itself must
also be unbroken.
Because you see I can squeeze
this thing down to such a
narrow width that there's no
room for a break in here.
Also notice that if the curve
'y' equals ''f sub n' of x' is
caught inside this curve--
if I, for example, were
computing the area of the
region 'R'--
for a large enough 'n', I
couldn't tell the difference
in area if I use 'y' equals 'f
of x' for my top curve, or
whether I use 'y' equals
''f sub n' of x'.
Well, keep that in mind, and all
I'm saying is this, that
from our picture, it should
seem clear that if the
sequence 'f sub n' converges
uniformly to 'f' on [a, b],
and if each 'f sub n' is
continuous on [a,b], then--
and this is a fundamental
result.
Fundamental result one, 'f' is
also continuous on [a, b].
Now you say, look.
That's what you'd expect, isn't
it, if every member of
the sequence is continuous?
Why shouldn't the limit function
also be continuous?
The point is, well, maybe that's
what you expect, but
note this--
in one of our earlier lectures,
we already saw that
if we only had point-wise
convergence, this did not need
to be true.
In particular, recall our
example in which we defined
''f sub n' of x' to be 'x to
the n', where the domain of
'f' was the closed interval
from 0 to 1.
Remember what happened
in that case?
Each of these 'f sub
n's is continuous.
But the limit function, you
may recall, is what?
It's 0 if 'x' is less than
1, and 1 if 'x' equals 1.
In other words, the limit
function was discontinuous at
'x' equals 1.
By the way, you might like to
see what this means from
another point of view.
And let me show you what
it does mean from
another point of view.
Since 'f' continuous at 'x'
equals 'x sub 1' means that
the limit of 'f of x' as 'x'
approaches 'x1' is 'f of x1',
and since 'f of x' itself, by
definition, is the limit of
''f sub n' of x' as 'n'
approaches infinity, we may
rewrite our first condition
in the form-- what?
We may rewrite our first
equation, this equation here,
in what form?
Limit as 'x' approaches 'x1'
of limit 'n' approaches
infinity, ''f sub n' of x'.
And that equals the limit as 'n'
approaches infinity, ''f
sub n' of x1'.
Now keep in mind that since each
'f sub n' is given to be
continuous, by definition of
continuity ''f sub n' of x1'
is the same as saying the limit
as 'x' approaches 'x1',
''f sub n' of x'.
The point I'm making is, if you
now put this together with
this, it says the rather
remarkable thing that unless
you have uniform convergence
when you interchange the order
in which you take the limits
over here, you may very well
get a different answer.
In other words, when you're
dealing with non-uniform
convergence, you must be very,
very careful to perform every
operation in the given order.
What we're saying is what?
This will give you an answer.
This will give you an answer.
But if the convergence is not
uniform, the answers may be
different, and consequently by
changing the order you destroy
the whole physical meaning
of the problem.
Well, again, that's reemphasized
in the
supplementary notes.
Let me continue on here.
Let me tell you another
interesting property of
uniform convergence.
Suppose the sequence ''f sub n'
of x' converges uniformly
to 'f of x' on [a, b].
The point that's rather
interesting is that you can
reverse the order of integration
and taking the
limit in this particular case.
In other words, suppose you want
to compute the integral
of the limit function
from 'a' to 'b'.
What you can do instead is
compute the integral of the
n-th number of your sequence,
and then take the limit as 'n'
goes to infinity.
In other words, rewriting this,
it says that if you have
uniform convergence, you can
take the limit inside the
integral sign.
And again, these results
are proven in our
supplementary notes.
We also say a few words about
corresponding results for
differentiation in our
supplementary notes.
And I should point out that
differentiation is a far more
subtle thing than integration.
See, remember that for
integration, all you need is
continuity.
For differentiation, you
need smoothness.
The point is that as you put a
thin band around the function
'y' equals 'f of x' when you
have uniform convergence,
that's enough to make sure that
the limit function must
be continuous if each of the
members in the sequence is
continuous.
But without going into the
details of this thing, it does
turn out that for the degree of
smoothness that you need,
these things can jump around
enough so that for
differentiation, we do have to
be a little bit more careful.
Rather than to becloud the
issue, I will stick with
integration topics for
our lecture today.
Now, the other thing that I want
to mention is, again, in
terms of doing what comes
naturally, I think we are
tempted to look at something
like this and say, look, all I
did was bring the
limit inside.
Why can't I do that?
And instead of saying, look,
you can't, I think the best
thing to show is that when
you don't have uniform
convergence, you get two
different answers.
Again, the idea being what?
We are not saying that
you can't do this.
We are not saying that you
can't compute this.
All we're saying is that if
the convergence is not
uniform, these two expressions
may very well
name different numbers.
Now, to show you what I have in
mind here, let me give you
an example.
See, to show you that 2 need not
be true if the convergence
is not uniform, consider
the following example.
Now, in the supplementary notes,
I repeat this example
both the way I have it on the
board and also from an
algebraic point of view, without
using the pictures.
But in terms of the picture,
here's what we do.
We define a function on the
closed interval from 0 to 2,
which I'll call 'f sub
n', as follows.
For a given 'n', I will locate
the points '1/n' and '2/n'.
For example, if 'n' happened to
be 50, this would be 1/50
and this would be 2/50.
Now what I do, is at the 'x'
value '1/n', I take as the
corresponding y-value,
'n squared'.
And I draw the straight line
that goes from the origin to
this point, '1/n' comma
'n squared'.
Then I draw the straight line
that comes right back to the
x-intercept, '2/n', and I finish
off the curve by just
letting it hug the x-axis till
we get over to 'x' equals 2.
I'll come back to this on the
next board, to show you why I
chose this, but let's make a few
observations just to make
sure that you understand what
this function looks like.
I'll just make a few arbitrary
remarks about it.
First of all, for each
'n', ''f sub n' of
'1/n'' is 'n squared'.
That's just another way
of indicating a
label for this point.
Secondly, my claim is that for
any number 'x sub 0', if 'x
sub 0' is greater than '2/n',
''f sub n' of x0' must be 0.
And the reason for that
is quite simple.
I'm just trying to show you
how to read this picture.
Namely, notice that as soon as
'x' gets to be as great as
'2/n', the 'f' value is 0,
because the function is
hugging the x-axis.
And just as a final observation,
notice that when
'x sub 0' is 0, ''f sub n' of 0'
is 0 for every 'n', meaning
that every member of my family
of functions goes through this
particular point.
In other words, let me
just label this.
This is 'y' equals
''f sub n' of x'.
Well, by the way if ''f sub n'
of 0' is 0 for each 'n', in
particular the limit of ''f sub
n' of 0' as 'n' approaches
infinity is 0.
What happens if we pick
a non-zero value?
For example, suppose I pick 'x0'
to be greater than 0 but
less than or equal to 2?
The key point is this, that
since the limit of '2/n' as
'n' approaches infinity is 0,
given a value of 'x0' which is
not 0, I can find the capital
'N' such that when 'n' is
greater than capital 'N',
'2/n' is less than 'x0'.
In other words, if 'x0' is
greater than 0, and '2/n'
approaches 0, for large enough
values of 'n', '2/n'
be less than 'x0'.
In particular, when that
happens, if we couple this
with our earlier observation--
what earlier observation?
Well, this one.
If we couple that with our
earlier observation, we see
that when 'n' is greater than
capital 'N', ''f sub
n' of x0' is 0.
Correspondingly, then, the limit
of ''f sub n' of x0' as
'n' approaches infinity,
by definition, is 0.
In other words--
I'm going to reinforce this
later, but notice that the
limit function here is
the function which
is identically 0.
Now, since this may look very
abstract to you, let me take a
few minutes--
and I hope this doesn't insult
your intelligence, but let me
just take a few minutes and
redraw this for a couple of
different values of 'n', just
so that you can see what's
starting to happen here.
Keep that picture in mind, and
now look what this means.
For example, when 'n' is 1,
'1/n' is 1, '2/n' is 2, 'n
squared' is 1.
In other words, the graph 'y'
equals 'f1 of x' is just this
triangular--
just this.
Why give it a name?
Well, let's try a tougher one.
Let's see what the member
'f sub 20' looks like.
Recall how you draw this, now.
With the subscript 20,
what do you do?
You come in to the point
1/20, and at that
point, you do what?
You locate the point 1/20
comma 'n squared'.
In this case, it's 400.
And I have obviously haven't
drawn this to scale, but you
now do what?
Draw the straight line
that goes from the
origin to this point.
Then from this point, you draw
the straight line that comes
back to the x-axis, hitting
it at 'x' equals 1/10.
And then you come across
the x-axis all the way
to 'x' equals 2.
This would be the graph of 'y'
equals ''f sub 20' of x'.
And by the way, do you sense
what's happening over here?
See, notice that as 'n' gets
very, very large, the curve
hugs the x-axis, starting
in closer and
closer to the y-axis.
But what happens is someplace
in here, no matter how close
'x sub 0' is to 0, there
comes a very high peak.
In fact, what is
that high peak?
It's 'n squared'.
In other words, when this number
is very close to the
y-axis, the peak is
very, very high.
In other words, no matter how
you put the squeeze on over
here, this particular
peak jumps out.
This is why this particular
sequence of functions is not
uniformly convergent.
Again, this is done more
slowly in the notes.
But at any rate, let me show you
an interesting thing that
happens over here.
Let me redraw this now
for a general 'n'.
In other words, let me draw 'y'
equals ''f sub n' of x'
for any old 'n'.
Recall what our definition
was, now, especially
with this as review.
We locate the point '1/n'
comma 'n squared'.
We then draw the line
that goes from the
origin to that point.
Then we draw the line that goes
from that point back to
the x-axis at the point '2/n'.
And then we come across
to 'x' equals 2.
Let's try to visualize what the
integral from 0 to 2, ''f
sub n' of x', 'dx', means.
After all, in a case of a
continuous curve, which this
is, isn't the definite integral
interpreted just as
the area under the curve?
Well, you see, the curve
coincides with the x-axis from
'2/n' on to 2.
Consequently, this triangular
region which I call 'R' is the
area under the curve.
In other words, the integral
from 0 to 2, ''f sub n' of x',
'dx', is the area of
the region 'R'.
But here's the point.
We can compute the area of the
region 'R' very easily.
It's a triangle, right?
What is the area
of a triangle?
Well, it's 1/2 times
the base--
but the base is just '2/n'--
times the height.
The height is 'n squared'.
In other words, the area of the
region 'R' simply is 'n'.
And that's rather interesting.
In other words, for each 'n',
this particular integral just
turns out to be 'n' itself.
That's what's interesting about
this particular diagram.
In other words, this thing rises
so high that even though
the base gets very, very small
as 'n' gets large, the height
increases so rapidly that the
area under this curve,
numerically, is always
equal to 'n'.
In fact, we can check
that if you'd like.
Come back to this
particular case.
Look at this particular
triangle.
The base is 2, the
height is 1.
The area is 1 unit.
Look at this particular
triangle.
The base is 1/10, the
height is 400.
400 times 1/10 is 40, and
half of that is 20.
The area of this triangle
is 20, which exactly
matches this subscript.
That's what's going to happen
here all the time.
In particular, then, if we
compute the integral from 0 to
2, 'f of n', 'x dx', and then
let the limit as 'n' goes to
infinity be computed, what
do we get for an answer?
We get that this limit is the
limit of 'n' as 'n' approaches
infinity, and that of
course is infinity.
On the other hand, suppose we
bring the limit inside?
In other words, suppose
we compute this.
Well, the point is that we have
already shown that this
is identically 0 for all 'x'.
Consequently, this integral is
the integral from 0 to 2, 0
'dx', which is 0.
In other words, if you first
take the limit and then
integrate, you get 0.
On the other hand, if you first
integrate and then take
the limit, you get infinity.
And this should be a glaring
example to show you that the
answer that you get indeed does
depend on the order in
which you do the operation.
Again, let me emphasize--
which of these two is wrong?
The answer is, neither
is wrong.
All we're saying is that if
you were supposed to solve
this problem and by mistake you
solve this one, you are
going to get a drastically
different answer.
OK.
Let's not beat this to death.
So far, so good.
Let me make one more remark,
namely, what does all of this
have to do with the
study of series?
See, now we're just talking
about sequences of functions.
And you see, the answer to this
question is essentially
going to be our last lecture
of the course.
But for now, what I'd like
to do is to give you
a preview of that.
Namely, the application of
uniform convergence to series
is the following.
Recall that when we write
summation 'n' goes from 0 to
infinity, 'a sub n', 'x
to the n', that's an
abbreviation for what?
A polynomial, 'k' goes from 0 to
'n', 'a sub k', 'x sub k',
as 'n' goes to infinity.
In other words, recall that
the sum of the series is a
limit of a sequence of partial
sums, and this is the n-th
member of that sequence
of partial sums.
Again, if the sigma notation is
throwing you off, all I'm
saying is to observe that 'a0'
plus 'a1 x' plus 'a2 'x
squared'' plus-- et cetera, et
cetera, et cetera, forever,
just represents the limit of
the following sequence.
'a0', next member is
'a0 plus a1 x'.
Next member is 'a0' plus 'a1
x' plus 'a2 'x squared''.
The next member is 'a0' plus 'a1
x' plus 'a2 'x squared''
plus 'a3 'x cubed''.
By the way, what is each
member of the sequence?
It's a polynomial.
And polynomials have very
nice properties,
among which are what?
Well, a polynomial is a
continuous function.
A polynomial is an integral
function, et cetera.
The idea, therefore, is that
if this sequence of partial
sums converges uniformly to the
limit function, then, for
example, the limit function,
namely the power series, must
be continuous since each partial
sum that makes up the
sequence of partial sums
is also continuous.
Namely, every polynomial
is continuous.
Also, if, for some reason or
other, you want to integrate
that power series from 'a' to
'b', if the convergence is
uniform, I can then do what?
I can then take the summation
sign outside, integrate the
n-th partial sum, and
add these all up.
You see, the idea being what?
That the n-th partial sum is a
polynomial, and a polynomial
is a particularly simple
thing to integrate.
That's one of the easiest
functions to
integrate, in fact.
OK.
Now, here's the wrap up, then.
What we shall show next time is
that within the interval of
absolute convergence, the
sequence of partial sums,
which we already know converges
absolutely to the
limit function, also converges
uniformly.
In other words, within the
radius of convergence, the
power series--
and I don't know how to say
this other than to say, it
enjoys the usual polynomial
properties associated with a
polynomial such as summation 'k'
goes from 0 to 'n', 'a sub
k', 'x to the k'.
In other words, then, this
about wraps up what our
introductory lecture for today
wanted to be, namely the
concept of uniform
convergence.
What I would like you to do now
is to study this material
very carefully in the
supplementary notes, go over
the learning exercises so that
you become familiar with this.
Then we will wrap up our course
in our next lecture,
when we show what a very, very
powerful tool this particular
concept of absolute convergence
is in the study of
the mathematical concept
of convergence.
At any rate, until next
time, then, goodbye.
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