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PROFESSOR: Hi.
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Today, we're going to generalize
our discussion of
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series to cover series of
functions rather than series
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00:00:41,240 --> 00:00:42,490
of just constants.
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In other words, you'll notice
that, up until now, what we've
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been doing in our discussion of
series was have a bunch of
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00:00:47,790 --> 00:00:50,180
fixed numbers and
then add them.
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You see, in the same way that
one starts with constants and
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ordinary arithmetic and then
starts talking about
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00:00:56,020 --> 00:01:00,000
functions, the same thing occurs
here, that maybe we are
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00:01:00,000 --> 00:01:03,470
now interested in a series
over a range of different
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values of 'x'.
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00:01:04,780 --> 00:01:07,680
Now, rather than to talk
abstractly about this, I'd
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00:01:07,680 --> 00:01:12,130
like to return to a topic that
we touched upon briefly much
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00:01:12,130 --> 00:01:15,140
earlier in the course, and then
come back to that topic
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00:01:15,140 --> 00:01:17,760
with far greater power
than we were able to
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00:01:17,760 --> 00:01:19,580
exhibit up until now.
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00:01:19,580 --> 00:01:24,160
And it's the idea of
approximating a function by
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00:01:24,160 --> 00:01:25,060
polynomials.
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00:01:25,060 --> 00:01:27,930
In other words, this question
of degree of contact.
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00:01:27,930 --> 00:01:31,900
And again, rather than review
that abstractly, let's talk in
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00:01:31,900 --> 00:01:33,370
terms of a picture.
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00:01:33,370 --> 00:01:35,240
You see, today's lecture
is called polynomial
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00:01:35,240 --> 00:01:37,550
approximations, and
the idea is this.
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00:01:37,550 --> 00:01:41,060
Let's suppose we have the curve,
'y' equals 'f of x'.
35
00:01:41,060 --> 00:01:42,140
It's smooth.
36
00:01:42,140 --> 00:01:44,520
It passes through the origin.
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00:01:44,520 --> 00:01:46,020
And we're told the following.
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00:01:46,020 --> 00:01:50,230
We say, look, we're interested
in approximating this curve in
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00:01:50,230 --> 00:01:52,960
a neighborhood of 'x' equals
0-- in other words, at this
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00:01:52,960 --> 00:01:54,010
point here--
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00:01:54,010 --> 00:01:56,440
by various polynomials.
42
00:01:56,440 --> 00:01:58,590
Now, the simplest polynomial,
of course, is a constant.
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00:01:58,590 --> 00:02:01,290
In other words, the question
is, how should we choose a
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00:02:01,290 --> 00:02:04,790
straight line, 'y' equals a
constant, 'y' equals ''P sub
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00:02:04,790 --> 00:02:06,595
0' of x'-- in other words,
a polynomial
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00:02:06,595 --> 00:02:08,610
of degree 0, a constant--
47
00:02:08,610 --> 00:02:12,880
if we want the best possible
horizontal line that
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00:02:12,880 --> 00:02:15,870
approximates the curve at
this particular point?
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00:02:15,870 --> 00:02:18,760
And I think you can see that
it's rather trivial at this
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00:02:18,760 --> 00:02:21,700
case to say, OK, let's take
the horizontal line that
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00:02:21,700 --> 00:02:23,240
passes through this point.
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00:02:23,240 --> 00:02:26,610
See, any other horizontal line
misses this point altogether.
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00:02:26,610 --> 00:02:28,490
OK, so far, so good.
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00:02:28,490 --> 00:02:30,750
Then somebody says, now, look
it, if we remove the
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restriction that the line be
horizontal, what about any
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00:02:33,270 --> 00:02:33,990
straight line?
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00:02:33,990 --> 00:02:37,720
What straight line has the best
degree of contact with
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00:02:37,720 --> 00:02:39,710
the curve at this particular
point?
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00:02:39,710 --> 00:02:42,280
In other words, what first
degree polynomial has the
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00:02:42,280 --> 00:02:44,410
highest degree of contact
over here?
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00:02:44,410 --> 00:02:46,760
Now again, sparing
you the details--
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because the details are
supplied in the text--
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00:02:50,070 --> 00:02:52,450
the polynomial there
is simply what?
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00:02:52,450 --> 00:02:56,190
'f of 0' plus ''f prime
of 0' times 'x''.
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00:02:56,190 --> 00:02:58,330
In other words, it's
a tangent line.
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00:02:58,330 --> 00:03:02,080
The basic equation of a tangent
line is 'y' equals 'mx
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00:03:02,080 --> 00:03:06,480
plus b', where 'm' is the slope,
which is 'f prime of
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00:03:06,480 --> 00:03:10,060
0', and 'b' is the y-intercept,
which is
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00:03:10,060 --> 00:03:11,800
'f of 0' over here.
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Now, the next question
comes up.
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00:03:13,400 --> 00:03:17,380
What if we wanted a quadratic
polynomial, a second degree
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00:03:17,380 --> 00:03:20,290
polynomial that fits the curve
even better, in other words,
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00:03:20,290 --> 00:03:22,350
has a greater degree
of contact at
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00:03:22,350 --> 00:03:23,720
this particular point?
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Let's label that 'y' equals ''P
sub 2' of x' to indicate a
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00:03:27,310 --> 00:03:29,890
second degree polynomial.
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And to analyze what that
polynomial must look like,
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let's simply write down some
undetermined coefficients
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00:03:35,910 --> 00:03:41,430
here, namely, let's let 'P2 of
x' be 'a0 plus 'a1x' plus 'a2
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00:03:41,430 --> 00:03:42,760
'x squared''.
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00:03:42,760 --> 00:03:46,110
And the idea is to determine
what 'a0', 'a1', and 'a2' must
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00:03:46,110 --> 00:03:49,650
look like if this polynomial is
to have a maximum degree of
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00:03:49,650 --> 00:03:53,940
contact with the curve,
'y' equals 'f of x'.
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00:03:53,940 --> 00:03:56,350
Now, what do you mean by maximum
degree of contact?
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00:03:56,350 --> 00:04:01,330
Well, what you mean is you
want the function at 0 to
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00:04:01,330 --> 00:04:03,620
equal the given function at 0.
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00:04:03,620 --> 00:04:07,720
You want the first derivative
of 'P2' evaluated at 0 to
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00:04:07,720 --> 00:04:10,150
equal the derivative of
'f' evaluated at 0.
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00:04:10,150 --> 00:04:13,590
And you want the second
derivative of 'P2' evaluated
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00:04:13,590 --> 00:04:16,000
at 0 equal to ''f double
prime' of 0'.
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00:04:16,000 --> 00:04:18,649
Well, let's go through
this in slow motion.
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00:04:18,649 --> 00:04:23,090
First of all, 'P2 of
0' is simply 'a0'.
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00:04:23,090 --> 00:04:29,580
'P2 prime' is simply 'a1 plus '2
a2 x'', and evaluating that
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00:04:29,580 --> 00:04:32,460
when 'x' equals 0, we
just have 'a1'.
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00:04:32,460 --> 00:04:35,700
So 'f of 0' must equal 'a0'.
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00:04:35,700 --> 00:04:38,480
'a1' must equal 'f
prime of 0'.
97
00:04:38,480 --> 00:04:40,250
Now, what about the
second derivative?
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00:04:40,250 --> 00:04:43,260
Notice that every time we
differentiate this polynomial,
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00:04:43,260 --> 00:04:44,780
another term drops out.
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00:04:44,780 --> 00:04:48,780
'a0' drops out the first time,
'a1' the second time.
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00:04:48,780 --> 00:04:51,580
And when you've differentiated
this term twice, all you have
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00:04:51,580 --> 00:04:53,320
left is '2 a2'.
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00:04:53,320 --> 00:04:55,040
In other words, 'P2
double prime'
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00:04:55,040 --> 00:04:57,570
evaluated at 0 is '2 a2'.
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00:04:57,570 --> 00:05:00,210
That must be 'f double
prime' of 0.
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00:05:00,210 --> 00:05:06,450
Equating these two, 'a2' must
be the second derivative of
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00:05:06,450 --> 00:05:09,050
'f' evaluated at 0 over 2.
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00:05:09,050 --> 00:05:12,140
I inadvertently wrote in 2
factorial here, which is the
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00:05:12,140 --> 00:05:15,100
right answer, but I haven't led
up to that yet, so if you
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00:05:15,100 --> 00:05:18,050
want to call this
2, that's fine.
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00:05:18,050 --> 00:05:19,670
At any rate, here's
what I've shown.
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00:05:19,670 --> 00:05:22,900
What I've shown is that, if
you want the polynomial of
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00:05:22,900 --> 00:05:27,830
degree 2 that fits the curve the
best, in a neighborhood of
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00:05:27,830 --> 00:05:30,480
0 comma 'f of 0',
this is how the
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00:05:30,480 --> 00:05:32,810
coefficients must be chosen.
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00:05:32,810 --> 00:05:36,650
And in fact, this situation
generalizes very nicely.
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00:05:36,650 --> 00:05:39,690
You see, suppose you want the
n-th degree polynomial
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00:05:39,690 --> 00:05:40,950
expression.
119
00:05:40,950 --> 00:05:43,550
Differentiate this 'n' times.
120
00:05:43,550 --> 00:05:46,240
You see, each time you
differentiate, one of these
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00:05:46,240 --> 00:05:47,430
terms drops out.
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00:05:47,430 --> 00:05:51,070
What happens to this term
as you differentiate it?
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00:05:51,070 --> 00:05:53,230
The first time you differentiate
it, you bring
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00:05:53,230 --> 00:05:54,370
down an 'n'.
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00:05:54,370 --> 00:05:56,850
The next time you differentiate,
you bring down
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00:05:56,850 --> 00:05:58,710
an 'n minus 1'.
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00:05:58,710 --> 00:06:01,470
By the time you've
differentiated 'n' times, this
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00:06:01,470 --> 00:06:06,380
is 'x to the 0', that's 1, 'a
sub n' is still here, and 'n'
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00:06:06,380 --> 00:06:09,640
times 'n minus 1' times 'n minus
2', et cetera, is just
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00:06:09,640 --> 00:06:11,840
what we call 'n factorial'.
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00:06:11,840 --> 00:06:15,270
In other words, the n-th
derivative of ''P sub n' of x'
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00:06:15,270 --> 00:06:17,530
is 'n factorial' times
'a sub n'.
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00:06:17,530 --> 00:06:20,970
If you evaluate that at 0,
that's still 'n factorial'
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00:06:20,970 --> 00:06:22,820
times 'a sub n'.
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00:06:22,820 --> 00:06:25,610
That must also equal the n-th
derivative of the function
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00:06:25,610 --> 00:06:29,050
evaluated at 0, by definition
of what you mean an n-th
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00:06:29,050 --> 00:06:30,600
degree of contact.
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00:06:30,600 --> 00:06:33,910
At any rate, that says that
the n-th coefficient, the
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00:06:33,910 --> 00:06:37,050
coefficient of 'x to the n',
must be the n-th derivative of
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00:06:37,050 --> 00:06:40,930
'f' evaluated at 0 over
'n factorial'.
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00:06:40,930 --> 00:06:43,040
By the way, notice
what this says.
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00:06:43,040 --> 00:06:45,630
And we'll come back to this
much more strongly later.
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00:06:45,630 --> 00:06:48,660
It says that, for this
coefficient to exist, the n-th
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00:06:48,660 --> 00:06:51,080
derivative of 'f' at
0 has to exist.
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00:06:51,080 --> 00:06:54,970
In other words, if 'f' does
not possess its n-th
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00:06:54,970 --> 00:06:59,010
derivative, this particular
equation doesn't make sense.
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00:06:59,010 --> 00:07:01,490
And I'll show you what that
means in a little while.
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00:07:01,490 --> 00:07:04,950
But at any rate, notice that
what we've now shown is that
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00:07:04,950 --> 00:07:06,120
''P sub n' of x'--
150
00:07:06,120 --> 00:07:08,170
if you want to use the
sigma notation--
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00:07:08,170 --> 00:07:09,530
can be written how?
152
00:07:09,530 --> 00:07:13,760
Well, it's the sum as 'k' goes
from 0 to 'n', the k-th
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00:07:13,760 --> 00:07:18,710
derivative of 'f' evaluated at 0
over 'k factorial', times 'x
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00:07:18,710 --> 00:07:19,810
to the k-th' power.
155
00:07:19,810 --> 00:07:22,350
Which written out longhand
is simply what?
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00:07:22,350 --> 00:07:27,130
'a0' plus 'a1x' plus ''a2 'x
squared'' over '2 factorial''
157
00:07:27,130 --> 00:07:29,890
plus et cetera, the n-th
derivative of 'f' evaluated at
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00:07:29,890 --> 00:07:33,920
0 over 'n factorial'
times 'x to the n'.
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00:07:33,920 --> 00:07:38,630
And what we do is we let 'P of
x' denote the limit of ''P sub
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00:07:38,630 --> 00:07:40,170
n' of x' as 'n' goes
to infinity.
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00:07:40,170 --> 00:07:42,900
In other words, for a fixed
'x', we look at what the
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00:07:42,900 --> 00:07:47,140
sequence converges to as 'n'
goes to infinity, and we
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00:07:47,140 --> 00:07:49,260
denote that as a series.
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00:07:49,260 --> 00:07:53,330
You see, what happens is 'x' is
a variable here, but if you
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00:07:53,330 --> 00:07:56,370
replace 'x' by a specific
number, what's inside the
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00:07:56,370 --> 00:07:59,770
summation sign becomes a
constant which depends only on
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00:07:59,770 --> 00:08:01,630
'n', and we're back
to our study of
168
00:08:01,630 --> 00:08:03,270
ordinary series again.
169
00:08:03,270 --> 00:08:05,980
Now, what does 'P of x'
mean in this case?
170
00:08:05,980 --> 00:08:09,940
Going back to our diagram, it
means that, as we let n get
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00:08:09,940 --> 00:08:11,780
bigger and bigger and we
determine more and more
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00:08:11,780 --> 00:08:13,700
coefficients, we get what?
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00:08:13,700 --> 00:08:17,310
Hopefully, what's a better fit
to the curve over here, that
174
00:08:17,310 --> 00:08:20,680
we get a greater degree of
contact every time we tack on
175
00:08:20,680 --> 00:08:23,070
the next term in the series.
176
00:08:23,070 --> 00:08:26,110
Now, because this may seem a
little bit abstract, let's
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00:08:26,110 --> 00:08:28,150
illustrate this thing
numerically.
178
00:08:28,150 --> 00:08:30,490
Let's pick a specific example.
179
00:08:30,490 --> 00:08:32,799
And I'll pick an easy
one to compute.
180
00:08:32,799 --> 00:08:35,159
Let 'f of x' be 'e to the x'.
181
00:08:35,159 --> 00:08:36,549
After all, what could
be easier?
182
00:08:36,549 --> 00:08:38,770
Because the derivative of 'e to
the x' with respect to 'x'
183
00:08:38,770 --> 00:08:41,740
is just 'e to the x', so the
n-th derivative of 'e to the
184
00:08:41,740 --> 00:08:43,510
x' is always 'e to the x'.
185
00:08:43,510 --> 00:08:46,530
When 'x' is 0, 'e
to the 0' is 1.
186
00:08:46,530 --> 00:08:49,260
So the n-th derivative
of 'e to the x',
187
00:08:49,260 --> 00:08:51,250
evaluated at 0, is 1.
188
00:08:51,250 --> 00:08:54,090
If I divide that by 'n
factorial', I just get '1 over
189
00:08:54,090 --> 00:08:55,370
'n factorial''.
190
00:08:55,370 --> 00:08:58,340
In other words, the sequence
of polynomials that
191
00:08:58,340 --> 00:09:02,680
approximates 'e to the
x' is given by what?
192
00:09:02,680 --> 00:09:06,900
Summation ''x to the k' over
'k factorial'' as 'k' goes
193
00:09:06,900 --> 00:09:08,580
from '1 to n'.
194
00:09:08,580 --> 00:09:11,930
And again, what this
means is what?
195
00:09:11,930 --> 00:09:16,150
The first approximation is
the line, 'y' equals 1.
196
00:09:16,150 --> 00:09:19,230
The straight line approximation
for the highest
197
00:09:19,230 --> 00:09:21,420
degree of contact at
the origin is 'y'
198
00:09:21,420 --> 00:09:23,060
equals '1 plus x'.
199
00:09:23,060 --> 00:09:25,990
The best quadratic approximation
is 'y' equals '1
200
00:09:25,990 --> 00:09:28,680
plus x plus ''x squared'
over 2'.
201
00:09:28,680 --> 00:09:33,420
The best cubic fit is when 'y'
is '1 plus x plus ''x squared'
202
00:09:33,420 --> 00:09:38,090
over 2', plus 'x cubed over
6'', 6 being 3 factorial.
203
00:09:38,090 --> 00:09:41,400
And we can continue on this way,
getting better fits as we
204
00:09:41,400 --> 00:09:42,810
add more and more terms.
205
00:09:42,810 --> 00:09:46,220
And again, I've spoken quite
rapidly, the reason being that
206
00:09:46,220 --> 00:09:49,800
this part is done superbly in
the textbook, complete with
207
00:09:49,800 --> 00:09:52,870
graphs, tables, and
what have you.
208
00:09:52,870 --> 00:09:54,290
Now, here's the question.
209
00:09:54,290 --> 00:09:58,000
The question is, look it,
we can compute 'Pn of x'
210
00:09:58,000 --> 00:09:59,500
for any given 'n'.
211
00:09:59,500 --> 00:10:02,280
And we intuitively get the
feeling that, as 'n'
212
00:10:02,280 --> 00:10:07,080
increases, 'Pn' hopefully will
look more like 'f of x'.
213
00:10:07,080 --> 00:10:09,990
In other words, what we would
like to do is to be able to
214
00:10:09,990 --> 00:10:14,930
compare, or better still, to
study 'f of x' by comparing it
215
00:10:14,930 --> 00:10:18,100
with the limit of 'Pn of x'
as 'n' goes to infinity.
216
00:10:18,100 --> 00:10:21,320
In other words, what we hope
will happen is that 'f of x'
217
00:10:21,320 --> 00:10:25,530
looks like ''P sub n' of x'
for large values of 'n'.
218
00:10:25,530 --> 00:10:27,510
Now, we don't know where this
is going to happen.
219
00:10:27,510 --> 00:10:32,150
In fact, the remainder of our
course now revolves about
220
00:10:32,150 --> 00:10:33,480
three questions.
221
00:10:33,480 --> 00:10:35,210
The three questions
are the following.
222
00:10:35,210 --> 00:10:38,450
First of all, does the limit,
'Pn of x' as 'n' approaches
223
00:10:38,450 --> 00:10:40,160
infinity, exist?
224
00:10:40,160 --> 00:10:45,600
In other words, does 'P of x'
exist in the first place?
225
00:10:45,600 --> 00:10:48,250
Does this limit always exists
for a given 'x'?
226
00:10:48,250 --> 00:10:51,200
Secondly, suppose the
limit does exist.
227
00:10:51,200 --> 00:10:53,590
How do we know it's going to
equal the given function, 'f
228
00:10:53,590 --> 00:10:55,630
of x', which it's trying
to approximate?
229
00:10:55,630 --> 00:10:57,590
In other words, how do we know
that we don't get an
230
00:10:57,590 --> 00:11:01,840
approximation which is nice near
the point of contact, but
231
00:11:01,840 --> 00:11:04,630
then, no matter how far out
we go beyond the point of
232
00:11:04,630 --> 00:11:06,470
contact, the approximation
doesn't become
233
00:11:06,470 --> 00:11:08,010
sufficiently good?
234
00:11:08,010 --> 00:11:09,330
See?
235
00:11:09,330 --> 00:11:12,130
And the third question that
comes up, assuming that the
236
00:11:12,130 --> 00:11:15,050
first have been answered, is
suppose the limit does exist--
237
00:11:15,050 --> 00:11:17,520
in other words, suppose this
limit function, 'P of x',
238
00:11:17,520 --> 00:11:20,140
which is a limit of 'Pn of x'
as 'n' approaches infinity,
239
00:11:20,140 --> 00:11:21,240
does exist--
240
00:11:21,240 --> 00:11:24,630
the question is, does 'P'
possess the polynomial
241
00:11:24,630 --> 00:11:27,150
properties that each piece
of 'n' possesses?
242
00:11:27,150 --> 00:11:30,330
In other words, the question
is, does the limit of a
243
00:11:30,330 --> 00:11:34,130
sequence of functions have the
same property that each of the
244
00:11:34,130 --> 00:11:35,920
members of the sequence has?
245
00:11:35,920 --> 00:11:38,770
For example, every polynomial
is continuous.
246
00:11:38,770 --> 00:11:41,340
The question, for example, that
might come up is, is the
247
00:11:41,340 --> 00:11:44,590
limit of a sequence of
continuous functions also
248
00:11:44,590 --> 00:11:46,040
continuous?
249
00:11:46,040 --> 00:11:49,020
Now, at first glance, it might
seem as if these are rather
250
00:11:49,020 --> 00:11:50,250
trivial to answer.
251
00:11:50,250 --> 00:11:56,620
Let me proceed next by showing
that the answer can be no to
252
00:11:56,620 --> 00:11:59,600
each of these three questions,
after which we'll then try to
253
00:11:59,600 --> 00:12:02,420
show when the answer
will be yes.
254
00:12:02,420 --> 00:12:03,440
The idea is this.
255
00:12:03,440 --> 00:12:06,300
Let me call these things
counter-examples, even though
256
00:12:06,300 --> 00:12:09,770
I don't know what better
word to use here.
257
00:12:09,770 --> 00:12:11,870
In other words, let me show
you an example where the
258
00:12:11,870 --> 00:12:14,950
answer to each of questions
one, two, and three
259
00:12:14,950 --> 00:12:17,170
happen to be false.
260
00:12:17,170 --> 00:12:21,500
For example, let 'f of x'
be '1 over '1 minus x''.
261
00:12:21,500 --> 00:12:24,600
As a review in taking
derivatives, 'f prime of x'
262
00:12:24,600 --> 00:12:28,000
would be '1 minus x' to the
minus 2 power, ''f double
263
00:12:28,000 --> 00:12:30,620
prime' of x' would be
2 times '1 minus x'
264
00:12:30,620 --> 00:12:31,930
to the minus 3 power.
265
00:12:31,930 --> 00:12:34,280
See, in other words, you bring
down the minus 2, but the
266
00:12:34,280 --> 00:12:36,830
derivative of what's inside with
respect to 'x' is minus.
267
00:12:36,830 --> 00:12:39,090
So a minus times
minus is plus.
268
00:12:39,090 --> 00:12:43,220
Similarly, the third derivative
of 'f of x' is 3
269
00:12:43,220 --> 00:12:47,650
factorial, 2 times minus 3 times
minus 1, '1 minus x' to
270
00:12:47,650 --> 00:12:48,740
the minus fourth.
271
00:12:48,740 --> 00:12:51,050
And in general, the n-th
derivative of 'f', in this
272
00:12:51,050 --> 00:12:56,240
case, is 'n factorial' times
'1 minus x' to the what?
273
00:12:56,240 --> 00:12:57,890
Minus 'n minus 1'.
274
00:12:57,890 --> 00:13:01,780
In other words, notice that the
magnitude of the exponent
275
00:13:01,780 --> 00:13:06,160
is one greater than the number
we're taking the factorial of.
276
00:13:06,160 --> 00:13:09,210
At any rate, if I now compute
the n-th derivative of 'f'
277
00:13:09,210 --> 00:13:13,200
evaluated at 0, I just
get 'n factorial'.
278
00:13:13,200 --> 00:13:16,950
And if I divide 'n factorial'
by 'n factorial', I get 1.
279
00:13:16,950 --> 00:13:19,600
In other words, every one of
my coefficients of the
280
00:13:19,600 --> 00:13:23,130
approximating sequence of
polynomials is going to be 1.
281
00:13:23,130 --> 00:13:26,430
In other words, the
approximation for 'f of x',
282
00:13:26,430 --> 00:13:30,970
where 'f of x' is '1 over '1
minus x'', is simply what?
283
00:13:30,970 --> 00:13:35,657
'Pn of x' is '1 plus x plus 'x
squared'', et cetera, plus et
284
00:13:35,657 --> 00:13:37,700
cetera, 'x to the n-th' power.
285
00:13:37,700 --> 00:13:40,050
Now, here's the interesting
point.
286
00:13:40,050 --> 00:13:41,090
This part is irrelevant.
287
00:13:41,090 --> 00:13:43,000
I just wanted to show you
something over here.
288
00:13:43,000 --> 00:13:45,670
Notice that 'f of 2'
is clearly minus 1.
289
00:13:45,670 --> 00:13:48,500
If I replace 'x' by 2
over here, 1 over 1
290
00:13:48,500 --> 00:13:50,420
minus 2 is minus 1.
291
00:13:50,420 --> 00:13:52,970
On the other hand, what I claim
is that the limit of a
292
00:13:52,970 --> 00:13:56,740
sequence, ''P sub n' of x', as
'n' goes to infinity doesn't
293
00:13:56,740 --> 00:13:59,200
even exist when 'x' is
2, because look what
294
00:13:59,200 --> 00:14:00,440
happens over here.
295
00:14:00,440 --> 00:14:02,740
When 'x' is 2, this is what?
296
00:14:02,740 --> 00:14:07,820
1 plus 2 plus 2 squared plus
et cetera '2 to the n'.
297
00:14:07,820 --> 00:14:11,840
That's 1 plus 2 plus 4 plus
8 plus 16, et cetera.
298
00:14:11,840 --> 00:14:15,170
And as I let 'n' go to infinity,
that sum increases
299
00:14:15,170 --> 00:14:16,200
without bound.
300
00:14:16,200 --> 00:14:20,650
In other words, here's an
example where the given limit
301
00:14:20,650 --> 00:14:21,760
didn't have to exist.
302
00:14:21,760 --> 00:14:24,550
In other words, the sequence
of polynomials does not
303
00:14:24,550 --> 00:14:28,980
converge at all at
'x' equals 2.
304
00:14:28,980 --> 00:14:31,090
So much for showing that
the answer to the first
305
00:14:31,090 --> 00:14:32,610
question can be no.
306
00:14:32,610 --> 00:14:35,900
As for the second question,
let's define a function as
307
00:14:35,900 --> 00:14:39,170
follows, to show you why
continuity is so important.
308
00:14:39,170 --> 00:14:41,880
Let 'f of x' be 'x squared'
if 'x' is less
309
00:14:41,880 --> 00:14:43,270
than or equal to 2.
310
00:14:43,270 --> 00:14:46,130
And let it be 4 if 'x'
is greater than 2.
311
00:14:46,130 --> 00:14:49,100
Now, if we compute the
derivatives of 'f' evaluated
312
00:14:49,100 --> 00:14:52,320
at 0, we find that what?
'f of 0' is 0.
313
00:14:52,320 --> 00:14:54,970
See, 'x squared' at 0 is 0.
314
00:14:54,970 --> 00:14:57,600
'2x' at 0 is 0.
315
00:14:57,600 --> 00:15:00,650
The second derivative here
is 2, so ''f double
316
00:15:00,650 --> 00:15:02,440
prime' of 0' is 2.
317
00:15:02,440 --> 00:15:05,640
And the third derivative and
higher, since this is only 'x
318
00:15:05,640 --> 00:15:07,410
squared', are identically 0.
319
00:15:07,410 --> 00:15:09,960
In other words, notice that the
second derivative of 'f'
320
00:15:09,960 --> 00:15:11,650
evaluated at 0 is 2.
321
00:15:11,650 --> 00:15:13,740
All other derivatives are 0.
322
00:15:13,740 --> 00:15:17,340
In other words, notice that
'Pn of x' is equal to 'x
323
00:15:17,340 --> 00:15:20,300
squared' when 'n' is
greater than 2.
324
00:15:20,300 --> 00:15:22,730
And that's a straightforward
computation.
325
00:15:22,730 --> 00:15:25,610
I will drill you on the homework
problems to show you
326
00:15:25,610 --> 00:15:27,810
how to get more facility.
327
00:15:27,810 --> 00:15:28,700
We're doing these things.
328
00:15:28,700 --> 00:15:30,470
But here's the key point.
329
00:15:30,470 --> 00:15:33,840
Let's look to see what the limit
of 'Pn of x' is as 'n'
330
00:15:33,840 --> 00:15:35,150
approaches infinity.
331
00:15:35,150 --> 00:15:38,710
Since 'Pn of x' is equal to 'x
squared' for all 'n', in
332
00:15:38,710 --> 00:15:41,130
particular, then, since this
doesn't depend on 'n', the
333
00:15:41,130 --> 00:15:43,090
limit off 'Pn of x'
as 'n' goes to
334
00:15:43,090 --> 00:15:45,680
infinity is just 'x squared'.
335
00:15:45,680 --> 00:15:47,040
What this says is this.
336
00:15:47,040 --> 00:15:52,490
Our function, 'f of x',
graphs like this.
337
00:15:52,490 --> 00:15:54,040
The sequence of polynomials--
338
00:15:54,040 --> 00:15:56,230
which, by the way, converge
to the limit
339
00:15:56,230 --> 00:15:58,120
function, 'x squared'--
340
00:15:58,120 --> 00:15:59,630
look what they do.
341
00:15:59,630 --> 00:16:01,750
They're just 'y' equals
'x squared'.
342
00:16:01,750 --> 00:16:04,700
In other words, somehow or
other, when we got to this
343
00:16:04,700 --> 00:16:07,740
sharp corner-- in other words,
notice that 'f of x' is not
344
00:16:07,740 --> 00:16:09,660
differentiable at this point--
345
00:16:09,660 --> 00:16:13,300
our sequence of polynomials
blissfully went right on their
346
00:16:13,300 --> 00:16:16,260
merry way smoothly while
the function itself
347
00:16:16,260 --> 00:16:17,630
leveled off like this.
348
00:16:17,630 --> 00:16:20,460
In other words, the limit
function always exist, but as
349
00:16:20,460 --> 00:16:24,140
soon as 'x' is greater than 2,
in this example, the function,
350
00:16:24,140 --> 00:16:27,480
'P of x', no longer is
the same as 'f of x'.
351
00:16:27,480 --> 00:16:30,860
You see, 'P of x' is going up
like this, 'f of x' has just
352
00:16:30,860 --> 00:16:32,170
leveled off this way.
353
00:16:32,170 --> 00:16:35,240
There are even more glaring
examples, but I will leave
354
00:16:35,240 --> 00:16:38,500
those for the supplementary
notes to go
355
00:16:38,500 --> 00:16:39,740
into in more detail.
356
00:16:39,740 --> 00:16:41,120
But so far, we've seen what?
357
00:16:41,120 --> 00:16:44,840
That the answers to questions
one and two can be no.
358
00:16:44,840 --> 00:16:47,000
Let's now show that the
answer to question
359
00:16:47,000 --> 00:16:48,250
three can also be no.
360
00:16:48,250 --> 00:16:52,370
361
00:16:52,370 --> 00:16:55,880
Let's define a sequence of
polynomials, ''P sub n' of x',
362
00:16:55,880 --> 00:16:59,620
to be 'x to the n', where the
domain of 'P sub n' is the
363
00:16:59,620 --> 00:17:01,800
interval from 0 to 1.
364
00:17:01,800 --> 00:17:04,329
What that means is something
like this.
365
00:17:04,329 --> 00:17:06,290
See, here's 0, 0.
366
00:17:06,290 --> 00:17:09,500
And I hope this reminds you of a
homework problem that we had
367
00:17:09,500 --> 00:17:12,450
very early in the course, but
that's irrelevant whether it
368
00:17:12,450 --> 00:17:13,250
does or not.
369
00:17:13,250 --> 00:17:15,490
'P1 of x' would look
like this.
370
00:17:15,490 --> 00:17:18,660
'P2 of x' would look
like this.
371
00:17:18,660 --> 00:17:20,390
'P3 of x' would look
like this.
372
00:17:20,390 --> 00:17:22,510
In other words, all of
these curves pass
373
00:17:22,510 --> 00:17:25,270
through 0, 0 and 1, 1.
374
00:17:25,270 --> 00:17:28,060
But as 'n' increases,
the degree of
375
00:17:28,060 --> 00:17:31,070
contact here gets better.
376
00:17:31,070 --> 00:17:34,710
At any rate, let's take a look
at the limit function.
377
00:17:34,710 --> 00:17:36,740
'P of x' is the limit of
''P sub n' of x' as
378
00:17:36,740 --> 00:17:38,550
'n' approaches infinity.
379
00:17:38,550 --> 00:17:41,810
'x to the n', as 'n' approaches
infinity, well, if
380
00:17:41,810 --> 00:17:44,930
'x' is less than 1 in magnitude,
'x to the n-th'
381
00:17:44,930 --> 00:17:46,120
approach is 0.
382
00:17:46,120 --> 00:17:49,460
On the other hand, if 'x' is
equal to 1, 1 to the n-th
383
00:17:49,460 --> 00:17:52,020
power is still 1 as 'n'
approaches infinity.
384
00:17:52,020 --> 00:17:54,180
Therefore, 'P of x'
would be what?
385
00:17:54,180 --> 00:18:00,450
It's 0 if 'x' is greater than or
equal to 0 but less than 1.
386
00:18:00,450 --> 00:18:03,010
And it's 1 if 'x' equals 1.
387
00:18:03,010 --> 00:18:04,670
Now, here's the key point.
388
00:18:04,670 --> 00:18:08,130
Observe that, since 'Pn of x'
is a polynomial, it's, in
389
00:18:08,130 --> 00:18:09,955
particular, continuous.
390
00:18:09,955 --> 00:18:12,450
In other words, each
'P sub n' is a
391
00:18:12,450 --> 00:18:15,370
continuous function, unbroken.
392
00:18:15,370 --> 00:18:18,210
On the other hand, look
at the limit function.
393
00:18:18,210 --> 00:18:21,080
It's 0 all the way up to 1.
394
00:18:21,080 --> 00:18:24,085
And that 1, the function
jumps to 1.
395
00:18:24,085 --> 00:18:29,060
In other words, graph-wise,
the thing looks like this.
396
00:18:29,060 --> 00:18:31,620
This is a graph, 'y'
equals 'P of x'.
397
00:18:31,620 --> 00:18:32,620
It's 0.
398
00:18:32,620 --> 00:18:34,420
All of a sudden, it jumps.
399
00:18:34,420 --> 00:18:37,060
Is the limit function,
therefore, continuous
400
00:18:37,060 --> 00:18:38,320
with 'x' equals 1.
401
00:18:38,320 --> 00:18:40,140
And the answer is no.
402
00:18:40,140 --> 00:18:42,190
There's a jump discontinuity
here.
403
00:18:42,190 --> 00:18:45,440
In other words, observe that
each 'P sub n' is continuous
404
00:18:45,440 --> 00:18:46,940
when 'x' equals 1.
405
00:18:46,940 --> 00:18:49,790
The limit function exists,
but it isn't continuous
406
00:18:49,790 --> 00:18:51,160
when 'x' equals 1.
407
00:18:51,160 --> 00:18:54,690
In other words, the properties
of a sequence of convergent
408
00:18:54,690 --> 00:18:57,900
functions do not have
to be inherited
409
00:18:57,900 --> 00:19:00,450
by the limit function.
410
00:19:00,450 --> 00:19:02,920
That shows, then, that the
answer to our three
411
00:19:02,920 --> 00:19:04,610
questions can be no.
412
00:19:04,610 --> 00:19:07,620
What we would like to do is
find out when and if the
413
00:19:07,620 --> 00:19:09,260
answer will be yes.
414
00:19:09,260 --> 00:19:12,460
What we shall do for the
remainder of today's lesson is
415
00:19:12,460 --> 00:19:14,590
answer the first
two questions.
416
00:19:14,590 --> 00:19:18,270
And this, by the way, is also
done very nicely in the text.
417
00:19:18,270 --> 00:19:21,250
The third question, for some
reason or other, is beyond the
418
00:19:21,250 --> 00:19:22,150
scope of the text.
419
00:19:22,150 --> 00:19:26,200
We will discuss this in our
future lectures, plus the
420
00:19:26,200 --> 00:19:27,410
supplementary notes.
421
00:19:27,410 --> 00:19:32,660
In fact, that the answer to
question number three will be
422
00:19:32,660 --> 00:19:35,710
the end of our course, when
we finally answer
423
00:19:35,710 --> 00:19:36,910
that particular question.
424
00:19:36,910 --> 00:19:39,490
But at any rate, what I'm saying
now is let's spend the
425
00:19:39,490 --> 00:19:42,260
remainder of today's lesson
in showing how to answer
426
00:19:42,260 --> 00:19:45,110
questions one and two in the
affirmative, and we'll save
427
00:19:45,110 --> 00:19:48,260
question three for the
remaining lectures.
428
00:19:48,260 --> 00:19:50,790
The first thing I'd like to
point out is the role of the
429
00:19:50,790 --> 00:19:53,940
ratio test and absolute
convergence.
430
00:19:53,940 --> 00:19:57,480
Namely, given the series
summation 'a n 'x to the n''--
431
00:19:57,480 --> 00:19:59,990
And by the way, if you haven't
noticed this by now, it's
432
00:19:59,990 --> 00:20:03,690
rather conventional sometimes,
to save time, not to put the
433
00:20:03,690 --> 00:20:05,020
subscripts on here.
434
00:20:05,020 --> 00:20:07,980
But if it bothers you, I mean,
I could do things like this.
435
00:20:07,980 --> 00:20:11,950
It's just that to save time and
space, I sometimes will
436
00:20:11,950 --> 00:20:13,310
not put the subscripts
on here.
437
00:20:13,310 --> 00:20:16,380
But at any rate, let's just talk
about this for a moment.
438
00:20:16,380 --> 00:20:19,980
Notice that a series converges
as soon as it converges
439
00:20:19,980 --> 00:20:21,780
absolutely.
440
00:20:21,780 --> 00:20:26,430
The point is that, for a
positive series, I can use the
441
00:20:26,430 --> 00:20:28,700
ratio test or its equivalent.
442
00:20:28,700 --> 00:20:33,850
Namely, for example, what I can
do is to test summation
443
00:20:33,850 --> 00:20:37,260
absolute value, 'a n 'x to the
n'', for convergence by the
444
00:20:37,260 --> 00:20:40,170
ratio test, by the comparison
m by the
445
00:20:40,170 --> 00:20:43,590
integral test, et cetera.
446
00:20:43,590 --> 00:20:46,150
And what I'm saying is, if
that particular series
447
00:20:46,150 --> 00:20:48,960
converges, then the original
series, namely, the one
448
00:20:48,960 --> 00:20:53,020
without the absolute value
signs, converges absolutely
449
00:20:53,020 --> 00:20:54,620
for that same value of 'x'.
450
00:20:54,620 --> 00:20:57,300
Well, again, instead of talking
about that, let's look
451
00:20:57,300 --> 00:20:58,160
at a particular example.
452
00:20:58,160 --> 00:20:59,870
And it looks like I've scraped
the blackboard
453
00:20:59,870 --> 00:21:00,870
here a little bit.
454
00:21:00,870 --> 00:21:02,440
Let me just--
455
00:21:02,440 --> 00:21:03,850
The first example is this.
456
00:21:03,850 --> 00:21:06,800
Let's suppose you look at
summation n goes from 0 to
457
00:21:06,800 --> 00:21:09,990
infinity, absolute value of
'x to the n-th' power.
458
00:21:09,990 --> 00:21:14,340
By the way, notice that this is
a geometric series with a
459
00:21:14,340 --> 00:21:16,280
ratio absolute value of 'x'.
460
00:21:16,280 --> 00:21:19,570
Consequently, the series
converges if and only if the
461
00:21:19,570 --> 00:21:22,300
absolute value of 'x'
is less than 1.
462
00:21:22,300 --> 00:21:24,900
By the way, let me make
a little aside here.
463
00:21:24,900 --> 00:21:28,300
Many people would tackle this
problem by the ratio test, as
464
00:21:28,300 --> 00:21:29,640
I hinted at over here.
465
00:21:29,640 --> 00:21:33,170
Namely, let the n-th term be
the absolute value of 'x to
466
00:21:33,170 --> 00:21:34,270
the n-th' power.
467
00:21:34,270 --> 00:21:36,930
Then the 'n plus first' term
would be the absolute value of
468
00:21:36,930 --> 00:21:39,070
'x to the 'n plus
first'' power.
469
00:21:39,070 --> 00:21:42,150
Then the ratio between these
two would be the absolute
470
00:21:42,150 --> 00:21:43,580
value of 'x'.
471
00:21:43,580 --> 00:21:47,690
Therefore, row, which is this
limit, would be the limit of
472
00:21:47,690 --> 00:21:53,120
the absolute value of 'x'
as 'n' goes to infinity.
473
00:21:53,120 --> 00:21:56,230
That's just equal to the
absolute value of 'x' itself.
474
00:21:56,230 --> 00:21:59,920
And for convergence, notice that
row must be less than 1,
475
00:21:59,920 --> 00:22:00,720
and that says what?
476
00:22:00,720 --> 00:22:03,820
The absolute value of 'x' must
be less than 1, just
477
00:22:03,820 --> 00:22:05,050
as we did over here.
478
00:22:05,050 --> 00:22:08,580
I simply would like to make a
little aside over here, and
479
00:22:08,580 --> 00:22:11,360
that is, if you use this
particular approach, that
480
00:22:11,360 --> 00:22:14,430
comes under the heading of
circular reasoning, because
481
00:22:14,430 --> 00:22:18,340
you may recall when we proved
the ratio test, we did it by
482
00:22:18,340 --> 00:22:21,940
comparing the series with the
geometric series, which meant
483
00:22:21,940 --> 00:22:23,570
that we already had
to know that the
484
00:22:23,570 --> 00:22:25,400
geometric series converged.
485
00:22:25,400 --> 00:22:27,690
And this is a geometric
series.
486
00:22:27,690 --> 00:22:30,380
But that's an aside which I just
wanted to mention to you
487
00:22:30,380 --> 00:22:32,140
in forms of circular
reasoning.
488
00:22:32,140 --> 00:22:34,600
The important point to
observe is what?
489
00:22:34,600 --> 00:22:38,400
That this series converges if
the absolute value of 'x' is
490
00:22:38,400 --> 00:22:39,600
less than 1.
491
00:22:39,600 --> 00:22:42,400
Consequently, that means that
the series without the
492
00:22:42,400 --> 00:22:46,310
absolute value sign converges,
in fact, absolutely, if the
493
00:22:46,310 --> 00:22:48,660
absolute value of 'x'
is less than 1.
494
00:22:48,660 --> 00:22:53,220
In other words, for this
particular series, '1 plus x
495
00:22:53,220 --> 00:22:56,160
plus 'x squared' plus 'x
cubed'', et cetera, that will
496
00:22:56,160 --> 00:22:59,990
converge to a finite limit
function if the absolute value
497
00:22:59,990 --> 00:23:02,590
of 'x' is less than 1.
498
00:23:02,590 --> 00:23:05,200
Let's look at another example.
499
00:23:05,200 --> 00:23:07,950
Suppose we look at the absolute
value of 'x to the
500
00:23:07,950 --> 00:23:10,842
n-th' power over
'n factorial'.
501
00:23:10,842 --> 00:23:13,530
If we look at this, the nth term
is the absolute value of
502
00:23:13,530 --> 00:23:16,210
'x to the n-th' over
'n factorial'.
503
00:23:16,210 --> 00:23:18,520
The 'n plus first' term,
therefore, is the absolute
504
00:23:18,520 --> 00:23:21,450
value of 'x to the 'n plus
1'', over ''n plus 1'
505
00:23:21,450 --> 00:23:22,510
factorial'.
506
00:23:22,510 --> 00:23:25,280
Therefore, the ratio between
the 'n plus first' term and
507
00:23:25,280 --> 00:23:30,980
the n-th term is the absolute
value of 'x' over 'n plus 1'.
508
00:23:30,980 --> 00:23:32,960
Rho is this particular limit.
509
00:23:32,960 --> 00:23:35,730
Well, the limit of the absolute
value of 'x' over 'n
510
00:23:35,730 --> 00:23:37,760
plus 1', as 'n' approaches
infinity--
511
00:23:37,760 --> 00:23:39,040
and here's the key point.
512
00:23:39,040 --> 00:23:42,330
Notice that 'x' is a fixed but
finite number once we get
513
00:23:42,330 --> 00:23:42,830
started here.
514
00:23:42,830 --> 00:23:45,030
In other words, we pick
an 'x' and fixed it.
515
00:23:45,030 --> 00:23:46,870
That means that that
stays constant.
516
00:23:46,870 --> 00:23:49,620
But as 'n' goes to infinity,
the denominator increases
517
00:23:49,620 --> 00:23:50,590
without bound.
518
00:23:50,590 --> 00:23:52,470
That makes the limit
equal to 0.
519
00:23:52,470 --> 00:23:55,860
And 0 is obviously less than
1, regardless of what the
520
00:23:55,860 --> 00:23:57,760
value of 'x' happens
to be here.
521
00:23:57,760 --> 00:24:00,380
In other words, in this
particular case, this
522
00:24:00,380 --> 00:24:04,160
particular series converges for
all real values of 'x'.
523
00:24:04,160 --> 00:24:06,920
And we sometimes abbreviate that
by saying the absolute
524
00:24:06,920 --> 00:24:09,660
value of 'x' is less
than infinity.
525
00:24:09,660 --> 00:24:12,130
And therefore, the original
series, meaning the one
526
00:24:12,130 --> 00:24:14,770
without the absolute value sign,
would, in particular,
527
00:24:14,770 --> 00:24:17,600
converge also for
all real 'x'.
528
00:24:17,600 --> 00:24:22,240
The important point is that, in
general, given any series--
529
00:24:22,240 --> 00:24:24,650
in other words, limit of a
sequence of polynomials of
530
00:24:24,650 --> 00:24:26,350
this particular type--
531
00:24:26,350 --> 00:24:28,710
there always exists
a number 'M'.
532
00:24:28,710 --> 00:24:30,560
Now, 'M' may be as small as 0.
533
00:24:30,560 --> 00:24:32,570
It may be as large
as infinity.
534
00:24:32,570 --> 00:24:36,630
But there always exists an 'M',
such that the particular
535
00:24:36,630 --> 00:24:40,520
series will converge absolutely
if the magnitude of
536
00:24:40,520 --> 00:24:43,680
'x' is less than 'M', and
diverge if the magnitude of
537
00:24:43,680 --> 00:24:45,500
'x' is greater than 'M'.
538
00:24:45,500 --> 00:24:48,410
In other words, this particular
'M', which is
539
00:24:48,410 --> 00:24:51,400
called the radius of
convergence, governs the
540
00:24:51,400 --> 00:24:53,240
answer to the first question.
541
00:24:53,240 --> 00:24:57,460
In other words, it's this 'M',
which we often find by the
542
00:24:57,460 --> 00:25:01,240
ratio test, that determines
where the sequence of
543
00:25:01,240 --> 00:25:04,310
functions, ''P sub n' of x', for
what values of 'x', that
544
00:25:04,310 --> 00:25:06,610
will converge to 'P of x'.
545
00:25:06,610 --> 00:25:09,410
And again, I think this is kind
of difficult for you to
546
00:25:09,410 --> 00:25:10,370
see abstractly.
547
00:25:10,370 --> 00:25:11,890
The proof is done in the text.
548
00:25:11,890 --> 00:25:14,040
What I'd like to do is show
you pictorially what's
549
00:25:14,040 --> 00:25:15,480
happening over here.
550
00:25:15,480 --> 00:25:18,600
See, let's suppose we have
our series, summation
551
00:25:18,600 --> 00:25:20,390
'a n 'x to the n''.
552
00:25:20,390 --> 00:25:24,600
And we pick a particular value
of 'x', say 'x sub 1'.
553
00:25:24,600 --> 00:25:26,480
Now, the idea is this.
554
00:25:26,480 --> 00:25:29,510
If this particular series
converges--
555
00:25:29,510 --> 00:25:32,320
and the proof is given in the
text, and I won't repeat it
556
00:25:32,320 --> 00:25:34,350
here because I want you to
see the overview here--
557
00:25:34,350 --> 00:25:36,170
if this particular series
converges--
558
00:25:36,170 --> 00:25:39,000
in other words, if the series
converges when 'x' is equal to
559
00:25:39,000 --> 00:25:40,180
'x sub 1'--
560
00:25:40,180 --> 00:25:43,960
then it can be shown that it
converges absolutely for any
561
00:25:43,960 --> 00:25:47,080
'x' which is smaller in
magnitude then 'x sub 1'.
562
00:25:47,080 --> 00:25:48,750
In other words, the
gist is this.
563
00:25:48,750 --> 00:25:51,950
If I know that the series
converges over here, I can
564
00:25:51,950 --> 00:25:55,075
then draw this interval
surrounding 0, and conclude
565
00:25:55,075 --> 00:25:58,750
that the series converges
in this entire interval.
566
00:25:58,750 --> 00:26:02,800
Correspondingly, if this
particular series diverges--
567
00:26:02,800 --> 00:26:05,620
and by the way, the whole proof
of this thing hinges on
568
00:26:05,620 --> 00:26:08,120
the comparison test,
essentially.
569
00:26:08,120 --> 00:26:11,440
See, if this thing diverges,
then certainly, if 'x' is
570
00:26:11,440 --> 00:26:15,670
larger than 'x1' in magnitude,
this will also diverge.
571
00:26:15,670 --> 00:26:18,530
In other words, once I know
that my series diverges at
572
00:26:18,530 --> 00:26:20,700
this particular value
of 'x1'..
573
00:26:20,700 --> 00:26:25,280
of 'x', I know it diverges for
all values of 'x' which are
574
00:26:25,280 --> 00:26:28,750
greater in magnitude then 'x1',
which means, in terms of
575
00:26:28,750 --> 00:26:33,440
a picture, for everything
further away from 0, then 'x1'
576
00:26:33,440 --> 00:26:34,790
and minus 'x1'.
577
00:26:34,790 --> 00:26:37,600
In other words, if this
converges, this converges.
578
00:26:37,600 --> 00:26:43,110
In here, if this diverges, we
have divergence out here.
579
00:26:43,110 --> 00:26:44,290
Now, what does that mean?
580
00:26:44,290 --> 00:26:50,050
Let me show you again in terms
of a extended diagram.
581
00:26:50,050 --> 00:26:52,680
See, the idea goes something
like this.
582
00:26:52,680 --> 00:26:56,530
Given the series summation 'a n
'x to the n'', pick a number
583
00:26:56,530 --> 00:26:58,400
'x' equal to 'x1'.
584
00:26:58,400 --> 00:27:01,680
Suppose the series converges
at 'x1'.
585
00:27:01,680 --> 00:27:05,800
Then we know that it converges
every place in here, so
586
00:27:05,800 --> 00:27:07,520
there's no need to check
this any further.
587
00:27:07,520 --> 00:27:10,430
What we do next is we
pick a number, 'x2',
588
00:27:10,430 --> 00:27:12,020
outside of this interval.
589
00:27:12,020 --> 00:27:13,990
Say 'x2' is over here.
590
00:27:13,990 --> 00:27:16,180
Suppose, for the sake of
argument-- see, what we're
591
00:27:16,180 --> 00:27:20,570
saying is 'a n 'x 1 to the
n'' converges, so you
592
00:27:20,570 --> 00:27:23,130
just suppose this.
593
00:27:23,130 --> 00:27:24,850
That gives us this picture.
594
00:27:24,850 --> 00:27:28,890
Suppose 'a n 'x2 to
the n'' diverges.
595
00:27:28,890 --> 00:27:31,690
Then we know that the series
diverges for all
596
00:27:31,690 --> 00:27:33,850
larger values of 'x'.
597
00:27:33,850 --> 00:27:37,390
Now we know how the series
behaves exactly, except in the
598
00:27:37,390 --> 00:27:39,620
interval between
'x1' and 'x2'.
599
00:27:39,620 --> 00:27:41,410
So what we do next is what?
600
00:27:41,410 --> 00:27:44,210
We pick a number which
we'll call 'x3'.
601
00:27:44,210 --> 00:27:47,830
We then see whether this
converges or diverges.
602
00:27:47,830 --> 00:27:51,140
If it converges at 'x3', we
know that the radius of
603
00:27:51,140 --> 00:27:53,870
convergence extends all
the way between minus
604
00:27:53,870 --> 00:27:56,280
'x3' and plus 'x3'.
605
00:27:56,280 --> 00:28:01,190
If it diverges, we know that,
from 'x3' out and minus 'x3'
606
00:28:01,190 --> 00:28:03,090
out, the series diverges.
607
00:28:03,090 --> 00:28:06,700
In any event, each time we
perform this operation, we cut
608
00:28:06,700 --> 00:28:08,470
the remaining space down.
609
00:28:08,470 --> 00:28:10,420
And we can keep on
going this way.
610
00:28:10,420 --> 00:28:14,040
And you see what you're doing
then, is you're zeroing in on
611
00:28:14,040 --> 00:28:16,860
this number that we called
capital 'M' before.
612
00:28:16,860 --> 00:28:18,540
You see, it's just a series
of refinements.
613
00:28:18,540 --> 00:28:20,580
And by the way, the
formal proof of
614
00:28:20,580 --> 00:28:21,920
this looks very messy.
615
00:28:21,920 --> 00:28:24,300
And yet, when it's stripped of
all embellishment, that's
616
00:28:24,300 --> 00:28:26,440
exactly what the formal
proof says.
617
00:28:26,440 --> 00:28:31,280
Well, at any rate, I will
reinforce the rest of this
618
00:28:31,280 --> 00:28:34,940
between the text, the notes,
and the learning exercises.
619
00:28:34,940 --> 00:28:38,720
All I want to do now is talk
about the second question.
620
00:28:38,720 --> 00:28:41,000
In other words, we've now talked
about something called
621
00:28:41,000 --> 00:28:44,490
the radius of convergence, that
tells us when the series
622
00:28:44,490 --> 00:28:47,140
of functions converges to
the limit function.
623
00:28:47,140 --> 00:28:50,510
Now the question is, how can we
measure whether that limit
624
00:28:50,510 --> 00:28:53,080
function converges to the
original function that we're
625
00:28:53,080 --> 00:28:54,470
trying to approximate?
626
00:28:54,470 --> 00:28:57,160
In other words, how do we know
whether 'P of x' actually
627
00:28:57,160 --> 00:28:59,010
equals 'f of x'?
628
00:28:59,010 --> 00:29:04,690
And again, this is done
extremely well in the text.
629
00:29:04,690 --> 00:29:07,370
It's called Taylor's theorem
with remainder.
630
00:29:07,370 --> 00:29:09,550
The proof is given very,
very nicely.
631
00:29:09,550 --> 00:29:12,440
I think the hardest part is to
show what the thing means.
632
00:29:12,440 --> 00:29:15,300
You see, using integration
by parts repeatedly--
633
00:29:15,300 --> 00:29:17,000
which is done in the text--
634
00:29:17,000 --> 00:29:20,180
it follows that, if 'f' and
its first 'n plus 1'
635
00:29:20,180 --> 00:29:23,750
derivatives exist at 'x' equals
0, it turns out-- and
636
00:29:23,750 --> 00:29:25,320
this is why this is called
Taylor's theorem with
637
00:29:25,320 --> 00:29:26,250
remainder--
638
00:29:26,250 --> 00:29:31,400
it turns out that 'f of x' can
be written as 'Pn of x' plus a
639
00:29:31,400 --> 00:29:34,510
remainder term-- ''r sub n'
of x'-- a remainder term.
640
00:29:34,510 --> 00:29:37,260
And then you see,
computationally, the text
641
00:29:37,260 --> 00:29:39,900
shows what that remainder
term looks like.
642
00:29:39,900 --> 00:29:44,005
It happens to be the integral
from 0 to 'x', ''x minus t' to
643
00:29:44,005 --> 00:29:48,700
the n-th' over 'n factorial',
'n plus first' derivative of
644
00:29:48,700 --> 00:29:53,970
'f of t', 'dt', from 't' equals
0 to 't' equals 'x'.
645
00:29:53,970 --> 00:29:55,080
This looks messy.
646
00:29:55,080 --> 00:29:56,630
It is messy.
647
00:29:56,630 --> 00:30:01,035
There's no need to hammer this
home in the lecture part of
648
00:30:01,035 --> 00:30:01,740
our course.
649
00:30:01,740 --> 00:30:05,660
What I prefer to do is to give
several learning exercises
650
00:30:05,660 --> 00:30:07,840
that will give you drill
in using this.
651
00:30:07,840 --> 00:30:11,250
But the point that I want you
to understand is that, since
652
00:30:11,250 --> 00:30:14,220
this is all done well in the
text, all I want you to do is
653
00:30:14,220 --> 00:30:15,890
to see where this
thing fits in.
654
00:30:15,890 --> 00:30:19,510
In other words, what the text
is showing is, look it, the
655
00:30:19,510 --> 00:30:21,900
difference between the
function which we're
656
00:30:21,900 --> 00:30:26,400
approximating at 0 comma 'f
of 0' by this sequence of
657
00:30:26,400 --> 00:30:30,560
polynomials, the difference
between that function and the
658
00:30:30,560 --> 00:30:33,990
nth member of that sequence
of polynomials is simply
659
00:30:33,990 --> 00:30:37,880
something called ''r sub n' of
x', where numerically ''r sub
660
00:30:37,880 --> 00:30:39,830
n' of x' looks like this.
661
00:30:39,830 --> 00:30:43,000
And the point is that, if you
now take the limit of this
662
00:30:43,000 --> 00:30:45,990
expression as 'n' goes to
infinity, remember, 'P of x'
663
00:30:45,990 --> 00:30:47,100
is this limit.
664
00:30:47,100 --> 00:30:51,140
Therefore, the only way that 'P
of x' can equal 'f of x' is
665
00:30:51,140 --> 00:30:56,530
if, for a given value of 'x',
this remainder goes to 0 as
666
00:30:56,530 --> 00:30:58,070
'n' goes to infinity.
667
00:30:58,070 --> 00:31:01,360
In other words, leaving the
computational details to the
668
00:31:01,360 --> 00:31:04,940
text, because it does it very
nicely, what we're saying is
669
00:31:04,940 --> 00:31:08,290
that the significance of
Taylor's remainder theorem is
670
00:31:08,290 --> 00:31:11,850
that, to find out where 'f of x'
and 'P of x' are identical,
671
00:31:11,850 --> 00:31:12,820
we simply--
672
00:31:12,820 --> 00:31:16,640
and I say simply meaning
conceptually simply--
673
00:31:16,640 --> 00:31:18,760
computationally, it might
be difficult.
674
00:31:18,760 --> 00:31:22,750
What we do computationally is
simply find all those values
675
00:31:22,750 --> 00:31:27,340
of 'x', for which ''r sub n' of
x' goes to 0 in the limit
676
00:31:27,340 --> 00:31:29,370
as 'n' goes to infinity.
677
00:31:29,370 --> 00:31:32,270
You see, what we've done is
we've now done two things.
678
00:31:32,270 --> 00:31:34,410
We've first shown when
the sequence of
679
00:31:34,410 --> 00:31:36,220
polynomials has a limit.
680
00:31:36,220 --> 00:31:39,560
And secondly, we've shown when
it does have a limit, when
681
00:31:39,560 --> 00:31:40,940
will it equal the given
function that
682
00:31:40,940 --> 00:31:43,170
it's trying to represent.
683
00:31:43,170 --> 00:31:45,980
The question that still remains
is, how do we know
684
00:31:45,980 --> 00:31:49,570
that the limit function has the
same properties as each
685
00:31:49,570 --> 00:31:50,750
member of the sequence?
686
00:31:50,750 --> 00:31:54,330
In fact, we've just given an
example where the limit
687
00:31:54,330 --> 00:31:56,470
function had different
properties.
688
00:31:56,470 --> 00:31:59,600
And the question is, under what
conditions can we be sure
689
00:31:59,600 --> 00:32:01,810
that certain nice results--
690
00:32:01,810 --> 00:32:04,360
that we'd like to be true,
because they're true about
691
00:32:04,360 --> 00:32:06,220
each polynomial in
the sequence--
692
00:32:06,220 --> 00:32:09,280
actually are true about
the limit function.
693
00:32:09,280 --> 00:32:11,880
In fact, that shall
be our topic for
694
00:32:11,880 --> 00:32:13,250
the next two lessons.
695
00:32:13,250 --> 00:32:16,140
And these next two lectures
should complete the course.
696
00:32:16,140 --> 00:32:18,480
At any rate, until next
time, goodbye.
697
00:32:18,480 --> 00:32:21,300
698
00:32:21,300 --> 00:32:23,840
MALE SPEAKER: Funding for the
publication of this video was
699
00:32:23,840 --> 00:32:28,560
provided by the Gabriella and
Paul Rosenbaum Foundation.
700
00:32:28,560 --> 00:32:32,730
Help OCW continue to provide
free and open access to MIT
701
00:32:32,730 --> 00:32:36,930
courses by making a donation
at ocw.mit.edu/donate.
702
00:32:36,930 --> 00:32:41,663