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PROFESSOR: Hi.
Today, we're going to generalize our discussion of
series to cover series of functions rather than series
of just constants.
In other words, you'll notice that, up until now, what we've
been doing in our discussion of series was have a bunch of
fixed numbers and then add them.
You see, in the same way that one starts with constants and
ordinary arithmetic and then starts talking about
functions, the same thing occurs here, that maybe we are
now interested in a series over a range of different
values of 'x'.
Now, rather than to talk abstractly about this, I'd
like to return to a topic that we touched upon briefly much
earlier in the course, and then come back to that topic
with far greater power than we were able to
exhibit up until now.
And it's the idea of approximating a function by
polynomials.
In other words, this question of degree of contact.
And again, rather than review that abstractly, let's talk in
terms of a picture.
You see, today's lecture is called polynomial
approximations, and the idea is this.
Let's suppose we have the curve, 'y' equals 'f of x'.
It's smooth.
It passes through the origin.
And we're told the following.
We say, look, we're interested in approximating this curve in
a neighborhood of 'x' equals 0-- in other words, at this
point here--
by various polynomials.
Now, the simplest polynomial, of course, is a constant.
In other words, the question is, how should we choose a
straight line, 'y' equals a constant, 'y' equals ''P sub
0' of x'-- in other words, a polynomial
of degree 0, a constant--
if we want the best possible horizontal line that
approximates the curve at this particular point?
And I think you can see that it's rather trivial at this
case to say, OK, let's take the horizontal line that
passes through this point.
See, any other horizontal line misses this point altogether.
OK, so far, so good.
Then somebody says, now, look it, if we remove the
restriction that the line be horizontal, what about any
straight line?
What straight line has the best degree of contact with
the curve at this particular point?
In other words, what first degree polynomial has the
highest degree of contact over here?
Now again, sparing you the details--
because the details are supplied in the text--
the polynomial there is simply what?
'f of 0' plus ''f prime of 0' times 'x''.
In other words, it's a tangent line.
The basic equation of a tangent line is 'y' equals 'mx
plus b', where 'm' is the slope, which is 'f prime of
0', and 'b' is the y-intercept, which is
'f of 0' over here.
Now, the next question comes up.
What if we wanted a quadratic polynomial, a second degree
polynomial that fits the curve even better, in other words,
has a greater degree of contact at
this particular point?
Let's label that 'y' equals ''P sub 2' of x' to indicate a
second degree polynomial.
And to analyze what that polynomial must look like,
let's simply write down some undetermined coefficients
here, namely, let's let 'P2 of x' be 'a0 plus 'a1x' plus 'a2
'x squared''.
And the idea is to determine what 'a0', 'a1', and 'a2' must
look like if this polynomial is to have a maximum degree of
contact with the curve, 'y' equals 'f of x'.
Now, what do you mean by maximum degree of contact?
Well, what you mean is you want the function at 0 to
equal the given function at 0.
You want the first derivative of 'P2' evaluated at 0 to
equal the derivative of 'f' evaluated at 0.
And you want the second derivative of 'P2' evaluated
at 0 equal to ''f double prime' of 0'.
Well, let's go through this in slow motion.
First of all, 'P2 of 0' is simply 'a0'.
'P2 prime' is simply 'a1 plus '2 a2 x'', and evaluating that
when 'x' equals 0, we just have 'a1'.
So 'f of 0' must equal 'a0'.
'a1' must equal 'f prime of 0'.
Now, what about the second derivative?
Notice that every time we differentiate this polynomial,
another term drops out.
'a0' drops out the first time, 'a1' the second time.
And when you've differentiated this term twice, all you have
left is '2 a2'.
In other words, 'P2 double prime'
evaluated at 0 is '2 a2'.
That must be 'f double prime' of 0.
Equating these two, 'a2' must be the second derivative of
'f' evaluated at 0 over 2.
I inadvertently wrote in 2 factorial here, which is the
right answer, but I haven't led up to that yet, so if you
want to call this 2, that's fine.
At any rate, here's what I've shown.
What I've shown is that, if you want the polynomial of
degree 2 that fits the curve the best, in a neighborhood of
0 comma 'f of 0', this is how the
coefficients must be chosen.
And in fact, this situation generalizes very nicely.
You see, suppose you want the n-th degree polynomial
expression.
Differentiate this 'n' times.
You see, each time you differentiate, one of these
terms drops out.
What happens to this term as you differentiate it?
The first time you differentiate it, you bring
down an 'n'.
The next time you differentiate, you bring down
an 'n minus 1'.
By the time you've differentiated 'n' times, this
is 'x to the 0', that's 1, 'a sub n' is still here, and 'n'
times 'n minus 1' times 'n minus 2', et cetera, is just
what we call 'n factorial'.
In other words, the n-th derivative of ''P sub n' of x'
is 'n factorial' times 'a sub n'.
If you evaluate that at 0, that's still 'n factorial'
times 'a sub n'.
That must also equal the n-th derivative of the function
evaluated at 0, by definition of what you mean an n-th
degree of contact.
At any rate, that says that the n-th coefficient, the
coefficient of 'x to the n', must be the n-th derivative of
'f' evaluated at 0 over 'n factorial'.
By the way, notice what this says.
And we'll come back to this much more strongly later.
It says that, for this coefficient to exist, the n-th
derivative of 'f' at 0 has to exist.
In other words, if 'f' does not possess its n-th
derivative, this particular equation doesn't make sense.
And I'll show you what that means in a little while.
But at any rate, notice that what we've now shown is that
''P sub n' of x'--
if you want to use the sigma notation--
can be written how?
Well, it's the sum as 'k' goes from 0 to 'n', the k-th
derivative of 'f' evaluated at 0 over 'k factorial', times 'x
to the k-th' power.
Which written out longhand is simply what?
'a0' plus 'a1x' plus ''a2 'x squared'' over '2 factorial''
plus et cetera, the n-th derivative of 'f' evaluated at
0 over 'n factorial' times 'x to the n'.
And what we do is we let 'P of x' denote the limit of ''P sub
n' of x' as 'n' goes to infinity.
In other words, for a fixed 'x', we look at what the
sequence converges to as 'n' goes to infinity, and we
denote that as a series.
You see, what happens is 'x' is a variable here, but if you
replace 'x' by a specific number, what's inside the
summation sign becomes a constant which depends only on
'n', and we're back to our study of
ordinary series again.
Now, what does 'P of x' mean in this case?
Going back to our diagram, it means that, as we let n get
bigger and bigger and we determine more and more
coefficients, we get what?
Hopefully, what's a better fit to the curve over here, that
we get a greater degree of contact every time we tack on
the next term in the series.
Now, because this may seem a little bit abstract, let's
illustrate this thing numerically.
Let's pick a specific example.
And I'll pick an easy one to compute.
Let 'f of x' be 'e to the x'.
After all, what could be easier?
Because the derivative of 'e to the x' with respect to 'x'
is just 'e to the x', so the n-th derivative of 'e to the
x' is always 'e to the x'.
When 'x' is 0, 'e to the 0' is 1.
So the n-th derivative of 'e to the x',
evaluated at 0, is 1.
If I divide that by 'n factorial', I just get '1 over
'n factorial''.
In other words, the sequence of polynomials that
approximates 'e to the x' is given by what?
Summation ''x to the k' over 'k factorial'' as 'k' goes
from '1 to n'.
And again, what this means is what?
The first approximation is the line, 'y' equals 1.
The straight line approximation for the highest
degree of contact at the origin is 'y'
equals '1 plus x'.
The best quadratic approximation is 'y' equals '1
plus x plus ''x squared' over 2'.
The best cubic fit is when 'y' is '1 plus x plus ''x squared'
over 2', plus 'x cubed over 6'', 6 being 3 factorial.
And we can continue on this way, getting better fits as we
add more and more terms.
And again, I've spoken quite rapidly, the reason being that
this part is done superbly in the textbook, complete with
graphs, tables, and what have you.
Now, here's the question.
The question is, look it, we can compute 'Pn of x'
for any given 'n'.
And we intuitively get the feeling that, as 'n'
increases, 'Pn' hopefully will look more like 'f of x'.
In other words, what we would like to do is to be able to
compare, or better still, to study 'f of x' by comparing it
with the limit of 'Pn of x' as 'n' goes to infinity.
In other words, what we hope will happen is that 'f of x'
looks like ''P sub n' of x' for large values of 'n'.
Now, we don't know where this is going to happen.
In fact, the remainder of our course now revolves about
three questions.
The three questions are the following.
First of all, does the limit, 'Pn of x' as 'n' approaches
infinity, exist?
In other words, does 'P of x' exist in the first place?
Does this limit always exists for a given 'x'?
Secondly, suppose the limit does exist.
How do we know it's going to equal the given function, 'f
of x', which it's trying to approximate?
In other words, how do we know that we don't get an
approximation which is nice near the point of contact, but
then, no matter how far out we go beyond the point of
contact, the approximation doesn't become
sufficiently good?
See?
And the third question that comes up, assuming that the
first have been answered, is suppose the limit does exist--
in other words, suppose this limit function, 'P of x',
which is a limit of 'Pn of x' as 'n' approaches infinity,
does exist--
the question is, does 'P' possess the polynomial
properties that each piece of 'n' possesses?
In other words, the question is, does the limit of a
sequence of functions have the same property that each of the
members of the sequence has?
For example, every polynomial is continuous.
The question, for example, that might come up is, is the
limit of a sequence of continuous functions also
continuous?
Now, at first glance, it might seem as if these are rather
trivial to answer.
Let me proceed next by showing that the answer can be no to
each of these three questions, after which we'll then try to
show when the answer will be yes.
The idea is this.
Let me call these things counter-examples, even though
I don't know what better word to use here.
In other words, let me show you an example where the
answer to each of questions one, two, and three
happen to be false.
For example, let 'f of x' be '1 over '1 minus x''.
As a review in taking derivatives, 'f prime of x'
would be '1 minus x' to the minus 2 power, ''f double
prime' of x' would be 2 times '1 minus x'
to the minus 3 power.
See, in other words, you bring down the minus 2, but the
derivative of what's inside with respect to 'x' is minus.
So a minus times minus is plus.
Similarly, the third derivative of 'f of x' is 3
factorial, 2 times minus 3 times minus 1, '1 minus x' to
the minus fourth.
And in general, the n-th derivative of 'f', in this
case, is 'n factorial' times '1 minus x' to the what?
Minus 'n minus 1'.
In other words, notice that the magnitude of the exponent
is one greater than the number we're taking the factorial of.
At any rate, if I now compute the n-th derivative of 'f'
evaluated at 0, I just get 'n factorial'.
And if I divide 'n factorial' by 'n factorial', I get 1.
In other words, every one of my coefficients of the
approximating sequence of polynomials is going to be 1.
In other words, the approximation for 'f of x',
where 'f of x' is '1 over '1 minus x'', is simply what?
'Pn of x' is '1 plus x plus 'x squared'', et cetera, plus et
cetera, 'x to the n-th' power.
Now, here's the interesting point.
This part is irrelevant.
I just wanted to show you something over here.
Notice that 'f of 2' is clearly minus 1.
If I replace 'x' by 2 over here, 1 over 1
minus 2 is minus 1.
On the other hand, what I claim is that the limit of a
sequence, ''P sub n' of x', as 'n' goes to infinity doesn't
even exist when 'x' is 2, because look what
happens over here.
When 'x' is 2, this is what?
1 plus 2 plus 2 squared plus et cetera '2 to the n'.
That's 1 plus 2 plus 4 plus 8 plus 16, et cetera.
And as I let 'n' go to infinity, that sum increases
without bound.
In other words, here's an example where the given limit
didn't have to exist.
In other words, the sequence of polynomials does not
converge at all at 'x' equals 2.
So much for showing that the answer to the first
question can be no.
As for the second question, let's define a function as
follows, to show you why continuity is so important.
Let 'f of x' be 'x squared' if 'x' is less
than or equal to 2.
And let it be 4 if 'x' is greater than 2.
Now, if we compute the derivatives of 'f' evaluated
at 0, we find that what? 'f of 0' is 0.
See, 'x squared' at 0 is 0.
'2x' at 0 is 0.
The second derivative here is 2, so ''f double
prime' of 0' is 2.
And the third derivative and higher, since this is only 'x
squared', are identically 0.
In other words, notice that the second derivative of 'f'
evaluated at 0 is 2.
All other derivatives are 0.
In other words, notice that 'Pn of x' is equal to 'x
squared' when 'n' is greater than 2.
And that's a straightforward computation.
I will drill you on the homework problems to show you
how to get more facility.
We're doing these things.
But here's the key point.
Let's look to see what the limit of 'Pn of x' is as 'n'
approaches infinity.
Since 'Pn of x' is equal to 'x squared' for all 'n', in
particular, then, since this doesn't depend on 'n', the
limit off 'Pn of x' as 'n' goes to
infinity is just 'x squared'.
What this says is this.
Our function, 'f of x', graphs like this.
The sequence of polynomials--
which, by the way, converge to the limit
function, 'x squared'--
look what they do.
They're just 'y' equals 'x squared'.
In other words, somehow or other, when we got to this
sharp corner-- in other words, notice that 'f of x' is not
differentiable at this point--
our sequence of polynomials blissfully went right on their
merry way smoothly while the function itself
leveled off like this.
In other words, the limit function always exist, but as
soon as 'x' is greater than 2, in this example, the function,
'P of x', no longer is the same as 'f of x'.
You see, 'P of x' is going up like this, 'f of x' has just
leveled off this way.
There are even more glaring examples, but I will leave
those for the supplementary notes to go
into in more detail.
But so far, we've seen what?
That the answers to questions one and two can be no.
Let's now show that the answer to question
three can also be no.
Let's define a sequence of polynomials, ''P sub n' of x',
to be 'x to the n', where the domain of 'P sub n' is the
interval from 0 to 1.
What that means is something like this.
See, here's 0, 0.
And I hope this reminds you of a homework problem that we had
very early in the course, but that's irrelevant whether it
does or not.
'P1 of x' would look like this.
'P2 of x' would look like this.
'P3 of x' would look like this.
In other words, all of these curves pass
through 0, 0 and 1, 1.
But as 'n' increases, the degree of
contact here gets better.
At any rate, let's take a look at the limit function.
'P of x' is the limit of ''P sub n' of x' as
'n' approaches infinity.
'x to the n', as 'n' approaches infinity, well, if
'x' is less than 1 in magnitude, 'x to the n-th'
approach is 0.
On the other hand, if 'x' is equal to 1, 1 to the n-th
power is still 1 as 'n' approaches infinity.
Therefore, 'P of x' would be what?
It's 0 if 'x' is greater than or equal to 0 but less than 1.
And it's 1 if 'x' equals 1.
Now, here's the key point.
Observe that, since 'Pn of x' is a polynomial, it's, in
particular, continuous.
In other words, each 'P sub n' is a
continuous function, unbroken.
On the other hand, look at the limit function.
It's 0 all the way up to 1.
And that 1, the function jumps to 1.
In other words, graph-wise, the thing looks like this.
This is a graph, 'y' equals 'P of x'.
It's 0.
All of a sudden, it jumps.
Is the limit function, therefore, continuous
with 'x' equals 1.
And the answer is no.
There's a jump discontinuity here.
In other words, observe that each 'P sub n' is continuous
when 'x' equals 1.
The limit function exists, but it isn't continuous
when 'x' equals 1.
In other words, the properties of a sequence of convergent
functions do not have to be inherited
by the limit function.
That shows, then, that the answer to our three
questions can be no.
What we would like to do is find out when and if the
answer will be yes.
What we shall do for the remainder of today's lesson is
answer the first two questions.
And this, by the way, is also done very nicely in the text.
The third question, for some reason or other, is beyond the
scope of the text.
We will discuss this in our future lectures, plus the
supplementary notes.
In fact, that the answer to question number three will be
the end of our course, when we finally answer
that particular question.
But at any rate, what I'm saying now is let's spend the
remainder of today's lesson in showing how to answer
questions one and two in the affirmative, and we'll save
question three for the remaining lectures.
The first thing I'd like to point out is the role of the
ratio test and absolute convergence.
Namely, given the series summation 'a n 'x to the n''--
And by the way, if you haven't noticed this by now, it's
rather conventional sometimes, to save time, not to put the
subscripts on here.
But if it bothers you, I mean, I could do things like this.
It's just that to save time and space, I sometimes will
not put the subscripts on here.
But at any rate, let's just talk about this for a moment.
Notice that a series converges as soon as it converges
absolutely.
The point is that, for a positive series, I can use the
ratio test or its equivalent.
Namely, for example, what I can do is to test summation
absolute value, 'a n 'x to the n'', for convergence by the
ratio test, by the comparison m by the
integral test, et cetera.
And what I'm saying is, if that particular series
converges, then the original series, namely, the one
without the absolute value signs, converges absolutely
for that same value of 'x'.
Well, again, instead of talking about that, let's look
at a particular example.
And it looks like I've scraped the blackboard
here a little bit.
Let me just--
The first example is this.
Let's suppose you look at summation n goes from 0 to
infinity, absolute value of 'x to the n-th' power.
By the way, notice that this is a geometric series with a
ratio absolute value of 'x'.
Consequently, the series converges if and only if the
absolute value of 'x' is less than 1.
By the way, let me make a little aside here.
Many people would tackle this problem by the ratio test, as
I hinted at over here.
Namely, let the n-th term be the absolute value of 'x to
the n-th' power.
Then the 'n plus first' term would be the absolute value of
'x to the 'n plus first'' power.
Then the ratio between these two would be the absolute
value of 'x'.
Therefore, row, which is this limit, would be the limit of
the absolute value of 'x' as 'n' goes to infinity.
That's just equal to the absolute value of 'x' itself.
And for convergence, notice that row must be less than 1,
and that says what?
The absolute value of 'x' must be less than 1, just
as we did over here.
I simply would like to make a little aside over here, and
that is, if you use this particular approach, that
comes under the heading of circular reasoning, because
you may recall when we proved the ratio test, we did it by
comparing the series with the geometric series, which meant
that we already had to know that the
geometric series converged.
And this is a geometric series.
But that's an aside which I just wanted to mention to you
in forms of circular reasoning.
The important point to observe is what?
That this series converges if the absolute value of 'x' is
less than 1.
Consequently, that means that the series without the
absolute value sign converges, in fact, absolutely, if the
absolute value of 'x' is less than 1.
In other words, for this particular series, '1 plus x
plus 'x squared' plus 'x cubed'', et cetera, that will
converge to a finite limit function if the absolute value
of 'x' is less than 1.
Let's look at another example.
Suppose we look at the absolute value of 'x to the
n-th' power over 'n factorial'.
If we look at this, the nth term is the absolute value of
'x to the n-th' over 'n factorial'.
The 'n plus first' term, therefore, is the absolute
value of 'x to the 'n plus 1'', over ''n plus 1'
factorial'.
Therefore, the ratio between the 'n plus first' term and
the n-th term is the absolute value of 'x' over 'n plus 1'.
Rho is this particular limit.
Well, the limit of the absolute value of 'x' over 'n
plus 1', as 'n' approaches infinity--
and here's the key point.
Notice that 'x' is a fixed but finite number once we get
started here.
In other words, we pick an 'x' and fixed it.
That means that that stays constant.
But as 'n' goes to infinity, the denominator increases
without bound.
That makes the limit equal to 0.
And 0 is obviously less than 1, regardless of what the
value of 'x' happens to be here.
In other words, in this particular case, this
particular series converges for all real values of 'x'.
And we sometimes abbreviate that by saying the absolute
value of 'x' is less than infinity.
And therefore, the original series, meaning the one
without the absolute value sign, would, in particular,
converge also for all real 'x'.
The important point is that, in general, given any series--
in other words, limit of a sequence of polynomials of
this particular type--
there always exists a number 'M'.
Now, 'M' may be as small as 0.
It may be as large as infinity.
But there always exists an 'M', such that the particular
series will converge absolutely if the magnitude of
'x' is less than 'M', and diverge if the magnitude of
'x' is greater than 'M'.
In other words, this particular 'M', which is
called the radius of convergence, governs the
answer to the first question.
In other words, it's this 'M', which we often find by the
ratio test, that determines where the sequence of
functions, ''P sub n' of x', for what values of 'x', that
will converge to 'P of x'.
And again, I think this is kind of difficult for you to
see abstractly.
The proof is done in the text.
What I'd like to do is show you pictorially what's
happening over here.
See, let's suppose we have our series, summation
'a n 'x to the n''.
And we pick a particular value of 'x', say 'x sub 1'.
Now, the idea is this.
If this particular series converges--
and the proof is given in the text, and I won't repeat it
here because I want you to see the overview here--
if this particular series converges--
in other words, if the series converges when 'x' is equal to
'x sub 1'--
then it can be shown that it converges absolutely for any
'x' which is smaller in magnitude then 'x sub 1'.
In other words, the gist is this.
If I know that the series converges over here, I can
then draw this interval surrounding 0, and conclude
that the series converges in this entire interval.
Correspondingly, if this particular series diverges--
and by the way, the whole proof of this thing hinges on
the comparison test, essentially.
See, if this thing diverges, then certainly, if 'x' is
larger than 'x1' in magnitude, this will also diverge.
In other words, once I know that my series diverges at
this particular value of 'x1'..
of 'x', I know it diverges for all values of 'x' which are
greater in magnitude then 'x1', which means, in terms of
a picture, for everything further away from 0, then 'x1'
and minus 'x1'.
In other words, if this converges, this converges.
In here, if this diverges, we have divergence out here.
Now, what does that mean?
Let me show you again in terms of a extended diagram.
See, the idea goes something like this.
Given the series summation 'a n 'x to the n'', pick a number
'x' equal to 'x1'.
Suppose the series converges at 'x1'.
Then we know that it converges every place in here, so
there's no need to check this any further.
What we do next is we pick a number, 'x2',
outside of this interval.
Say 'x2' is over here.
Suppose, for the sake of argument-- see, what we're
saying is 'a n 'x 1 to the n'' converges, so you
just suppose this.
That gives us this picture.
Suppose 'a n 'x2 to the n'' diverges.
Then we know that the series diverges for all
larger values of 'x'.
Now we know how the series behaves exactly, except in the
interval between 'x1' and 'x2'.
So what we do next is what?
We pick a number which we'll call 'x3'.
We then see whether this converges or diverges.
If it converges at 'x3', we know that the radius of
convergence extends all the way between minus
'x3' and plus 'x3'.
If it diverges, we know that, from 'x3' out and minus 'x3'
out, the series diverges.
In any event, each time we perform this operation, we cut
the remaining space down.
And we can keep on going this way.
And you see what you're doing then, is you're zeroing in on
this number that we called capital 'M' before.
You see, it's just a series of refinements.
And by the way, the formal proof of
this looks very messy.
And yet, when it's stripped of all embellishment, that's
exactly what the formal proof says.
Well, at any rate, I will reinforce the rest of this
between the text, the notes, and the learning exercises.
All I want to do now is talk about the second question.
In other words, we've now talked about something called
the radius of convergence, that tells us when the series
of functions converges to the limit function.
Now the question is, how can we measure whether that limit
function converges to the original function that we're
trying to approximate?
In other words, how do we know whether 'P of x' actually
equals 'f of x'?
And again, this is done extremely well in the text.
It's called Taylor's theorem with remainder.
The proof is given very, very nicely.
I think the hardest part is to show what the thing means.
You see, using integration by parts repeatedly--
which is done in the text--
it follows that, if 'f' and its first 'n plus 1'
derivatives exist at 'x' equals 0, it turns out-- and
this is why this is called Taylor's theorem with
remainder--
it turns out that 'f of x' can be written as 'Pn of x' plus a
remainder term-- ''r sub n' of x'-- a remainder term.
And then you see, computationally, the text
shows what that remainder term looks like.
It happens to be the integral from 0 to 'x', ''x minus t' to
the n-th' over 'n factorial', 'n plus first' derivative of
'f of t', 'dt', from 't' equals 0 to 't' equals 'x'.
This looks messy.
It is messy.
There's no need to hammer this home in the lecture part of
our course.
What I prefer to do is to give several learning exercises
that will give you drill in using this.
But the point that I want you to understand is that, since
this is all done well in the text, all I want you to do is
to see where this thing fits in.
In other words, what the text is showing is, look it, the
difference between the function which we're
approximating at 0 comma 'f of 0' by this sequence of
polynomials, the difference between that function and the
nth member of that sequence of polynomials is simply
something called ''r sub n' of x', where numerically ''r sub
n' of x' looks like this.
And the point is that, if you now take the limit of this
expression as 'n' goes to infinity, remember, 'P of x'
is this limit.
Therefore, the only way that 'P of x' can equal 'f of x' is
if, for a given value of 'x', this remainder goes to 0 as
'n' goes to infinity.
In other words, leaving the computational details to the
text, because it does it very nicely, what we're saying is
that the significance of Taylor's remainder theorem is
that, to find out where 'f of x' and 'P of x' are identical,
we simply--
and I say simply meaning conceptually simply--
computationally, it might be difficult.
What we do computationally is simply find all those values
of 'x', for which ''r sub n' of x' goes to 0 in the limit
as 'n' goes to infinity.
You see, what we've done is we've now done two things.
We've first shown when the sequence of
polynomials has a limit.
And secondly, we've shown when it does have a limit, when
will it equal the given function that
it's trying to represent.
The question that still remains is, how do we know
that the limit function has the same properties as each
member of the sequence?
In fact, we've just given an example where the limit
function had different properties.
And the question is, under what conditions can we be sure
that certain nice results--
that we'd like to be true, because they're true about
each polynomial in the sequence--
actually are true about the limit function.
In fact, that shall be our topic for
the next two lessons.
And these next two lectures should complete the course.
At any rate, until next time, goodbye.
MALE SPEAKER: Funding for the publication of this video was
provided by the Gabriella and Paul Rosenbaum Foundation.
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