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PROFESSOR: Hi.

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00:00:32,820 --> 00:00:35,490
Today is the day we've
all been waiting for.

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00:00:35,490 --> 00:00:41,080
We've come to the last lecture
of our course, and it's, I

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00:00:41,080 --> 00:00:44,600
think, a rather satisfying
lecture today, not just

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00:00:44,600 --> 00:00:47,450
because it is the last one,
but content-wise too.

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00:00:47,450 --> 00:00:49,540
Today we're going to
talk about uniform

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00:00:49,540 --> 00:00:51,550
convergence of series.

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00:00:51,550 --> 00:00:54,850
Remember, a series can be
viewed as a sequence of

16
00:00:54,850 --> 00:00:58,790
partial sums, and consequently
our discussion on uniform

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00:00:58,790 --> 00:01:00,580
convergence applies here.

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00:01:00,580 --> 00:01:02,050
And what we're going to do--

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00:01:02,050 --> 00:01:04,200
you see, again, it's
the same old story.

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00:01:04,200 --> 00:01:07,200
Now that we know what the
concept means, is there an

21
00:01:07,200 --> 00:01:11,110
easy way to tell when we
have the property?

22
00:01:11,110 --> 00:01:13,870
And I figure again that, this
being the last lecture, I

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00:01:13,870 --> 00:01:16,220
should give you some big
names to remember.

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00:01:16,220 --> 00:01:20,330
And name-dropping-wise, I come
to our first concept today,

25
00:01:20,330 --> 00:01:21,400
which I call--

26
00:01:21,400 --> 00:01:22,470
I don't call it that.

27
00:01:22,470 --> 00:01:26,390
It's named the Weierstrauss
M-test.

28
00:01:26,390 --> 00:01:31,080
The Weierstrauss M-test is a
very, very convenient method

29
00:01:31,080 --> 00:01:35,390
for determining whether a
given series converges

30
00:01:35,390 --> 00:01:36,780
uniformly or not.

31
00:01:36,780 --> 00:01:39,170
By the way, let me just
make one little aside.

32
00:01:39,170 --> 00:01:43,030
Instead of saying the sequence
of partial sums converges

33
00:01:43,030 --> 00:01:47,440
uniformly to the infinite
series, we usually abbreviate

34
00:01:47,440 --> 00:01:51,400
that simply by saying, the
infinite series is uniformly

35
00:01:51,400 --> 00:01:52,610
convergent.

36
00:01:52,610 --> 00:01:53,880
Let me just say that
one more time.

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00:01:53,880 --> 00:01:57,440
If I say that the series is
uniformly convergent, that's

38
00:01:57,440 --> 00:02:00,890
an abbreviation for saying that
the sequence of partial

39
00:02:00,890 --> 00:02:03,950
sums converges uniformly
to the series.

40
00:02:03,950 --> 00:02:06,880
But at any rate, let me now
go over the so-called

41
00:02:06,880 --> 00:02:08,780
Weierstrauss M-test with you.

42
00:02:08,780 --> 00:02:10,250
It's a rather simple test.

43
00:02:10,250 --> 00:02:12,950
I will give you both the proof
of the test and some

44
00:02:12,950 --> 00:02:14,450
applications of it.

45
00:02:14,450 --> 00:02:16,160
The test simply says this--

46
00:02:16,160 --> 00:02:17,990
and this is where the name
M-test comes from.

47
00:02:17,990 --> 00:02:19,570
See, because you put an
'M' in over here.

48
00:02:19,570 --> 00:02:21,530
If I put a 'C' over here, they
probably would have called it

49
00:02:21,530 --> 00:02:23,040
the Weierstrass C-test.

50
00:02:23,040 --> 00:02:24,400
But the idea is this.

51
00:02:24,400 --> 00:02:27,790
Suppose that the series,
summation 'n' goes from 1 to

52
00:02:27,790 --> 00:02:30,740
infinity, capital
'M sub n', is a

53
00:02:30,740 --> 00:02:34,190
positive convergent series--

54
00:02:34,190 --> 00:02:36,540
this is going to be sort of
like the comparison test--

55
00:02:36,540 --> 00:02:41,010
and that the absolute value of
''f sub n' of x' is less than

56
00:02:41,010 --> 00:02:45,300
or equal to 'M sub n' for each
'n' and for each 'x' in some

57
00:02:45,300 --> 00:02:47,540
interval from 'a' to 'b'.

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00:02:47,540 --> 00:02:51,860
Then if this condition is
obeyed, the sequence of

59
00:02:51,860 --> 00:02:53,760
partial sums--

60
00:02:53,760 --> 00:02:56,980
some 'k' goes from 1 to 'n',
of ''f sub k' of x'--

61
00:02:56,980 --> 00:03:00,720
converges uniformly to its limit
function, namely what we

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00:03:00,720 --> 00:03:04,065
call the series 'n' goes from
1 to infinity, ''f sub n' of

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00:03:04,065 --> 00:03:06,990
x', on the interval
from 'a' to 'b'.

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00:03:06,990 --> 00:03:10,670
See, in other words, if I can
get this kind of a bound on

65
00:03:10,670 --> 00:03:15,110
what's going on, where these
form the terms of a positive

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00:03:15,110 --> 00:03:21,250
convergent series, then this
series converges uniformly.

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00:03:21,250 --> 00:03:23,820
Now, again, let me outline
the proof.

68
00:03:23,820 --> 00:03:27,340
Once again, let me remind you
all of these proofs are done

69
00:03:27,340 --> 00:03:33,340
rigorously and also with
supplementary wording in the

70
00:03:33,340 --> 00:03:36,480
notes, so that you can pick this
up again if you missed

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00:03:36,480 --> 00:03:38,580
the high points in
the lecture.

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00:03:38,580 --> 00:03:42,310
You see, to prove convergence
in the first place, I

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00:03:42,310 --> 00:03:43,460
have to show what?

74
00:03:43,460 --> 00:03:47,760
That the sequence of partial
sums can be made to differ

75
00:03:47,760 --> 00:03:51,510
from this limit, which I call
'f of x', by as small an

76
00:03:51,510 --> 00:03:55,030
amount as I want, just by taking
'n' sufficiently large.

77
00:03:55,030 --> 00:03:57,680
And the way I work this thing
is I say, OK, let's take a

78
00:03:57,680 --> 00:03:59,350
look at this difference.

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00:03:59,350 --> 00:04:03,660
Well, remembering that another
name for 'f of x' is the limit

80
00:04:03,660 --> 00:04:06,700
of this thing as 'k' goes from
one to infinity, I just

81
00:04:06,700 --> 00:04:11,070
substitute this in place
of 'f of x'.

82
00:04:11,070 --> 00:04:12,570
I then get this.

83
00:04:12,570 --> 00:04:16,670
And now you see, if I subtract
the first 'n' terms from this

84
00:04:16,670 --> 00:04:20,300
infinite series, what's left
is all the terms in the 'n

85
00:04:20,300 --> 00:04:23,280
plus first' on.

86
00:04:23,280 --> 00:04:26,450
In other words, that this is in
turn equal to the absolute

87
00:04:26,450 --> 00:04:29,880
value of the summation, 'k'
goes from 'n plus 1' to

88
00:04:29,880 --> 00:04:32,610
infinity, ''f sub k' of x'.

89
00:04:32,610 --> 00:04:36,590
Again, don't be upset by the
'k's replacing the 'n'.

90
00:04:36,590 --> 00:04:39,070
Remember that 'k' is
a dummy variable.

91
00:04:39,070 --> 00:04:42,390
The hardship would be that if
I use the subscript 'n' over

92
00:04:42,390 --> 00:04:44,720
here, I would have a logical
contradiction.

93
00:04:44,720 --> 00:04:49,020
How can you say 'n' goes from
'n plus 1' to infinity? 'n'

94
00:04:49,020 --> 00:04:51,510
can't equal 'n plus 1',
so I just change

95
00:04:51,510 --> 00:04:52,600
the subscript here.

96
00:04:52,600 --> 00:04:56,600
At any rate, if I now invoke
the triangle inequality for

97
00:04:56,600 --> 00:04:57,680
absolute values--

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00:04:57,680 --> 00:05:01,020
namely, that the absolute value
of a sum is less than or

99
00:05:01,020 --> 00:05:03,300
equal to the sum of the
absolute values--

100
00:05:03,300 --> 00:05:06,530
I then get that the thing that
I'm investigating is less than

101
00:05:06,530 --> 00:05:10,490
or equal to the sum, 'k' goes
from 'n plus 1' to infinity,

102
00:05:10,490 --> 00:05:13,530
absolute value ''f
sub k' of x'.

103
00:05:13,530 --> 00:05:16,260
Now, remember what the
Weierstrass condition was.

104
00:05:16,260 --> 00:05:19,920
It was that for each 'k', the
absolute value of ''f sub k'

105
00:05:19,920 --> 00:05:22,810
of x', for all 'x' in my
interval, was less than or

106
00:05:22,810 --> 00:05:25,230
equal to 'M sub k'.

107
00:05:25,230 --> 00:05:28,020
So I can now go from
here to here.

108
00:05:28,020 --> 00:05:31,600
But what do I know about the
series summation, 'M sub k'?

109
00:05:31,600 --> 00:05:34,650
I know that that's a positive
convergent series.

110
00:05:34,650 --> 00:05:37,630
In particular, because it's
convergent, that means that

111
00:05:37,630 --> 00:05:40,550
the tail end-- meaning from
'n plus 1' to infinity--

112
00:05:40,550 --> 00:05:45,920
that sum can be made smaller
than any prescribed epsilon,

113
00:05:45,920 --> 00:05:49,780
smaller than any given epsilon,
simply by picking 'n'

114
00:05:49,780 --> 00:05:51,220
sufficiently large.

115
00:05:51,220 --> 00:05:55,040
See, in other words, I can make
this difference less than

116
00:05:55,040 --> 00:05:57,790
epsilon for 'n' sufficiently
large.

117
00:05:57,790 --> 00:05:59,180
And here's the key step.

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00:05:59,180 --> 00:06:06,520
Since 'M sub k' does not depend
on 'x', this inequality

119
00:06:06,520 --> 00:06:10,460
was done independently
of the choice of 'x'.

120
00:06:10,460 --> 00:06:12,640
You see, in other words, not
only do I have that this

121
00:06:12,640 --> 00:06:16,730
converges to this, but since
this was done independently of

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00:06:16,730 --> 00:06:19,790
the choice of 'x', this is
uniform convergence.

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00:06:19,790 --> 00:06:22,500
And maybe I should write that
down, because that's important

124
00:06:22,500 --> 00:06:28,710
to remember, that this is
independently of 'x'.

125
00:06:28,710 --> 00:06:32,880
That's what makes the
convergence uniform.

126
00:06:32,880 --> 00:06:37,030
Well, let's continue on, I
think, by means of an example

127
00:06:37,030 --> 00:06:38,900
might come in handy now.

128
00:06:38,900 --> 00:06:41,450
I think after you do any of
these abstract things, I think

129
00:06:41,450 --> 00:06:44,890
a nice example shows what's
coming off very nicely.

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00:06:44,890 --> 00:06:48,480
And so I very creatively
call this 'An Example'.

131
00:06:48,480 --> 00:06:50,670
Let's look at the following
series.

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00:06:50,670 --> 00:06:56,600
'cosine x' plus 'cosine '4x over
4' plus et cetera 'cosine

133
00:06:56,600 --> 00:07:01,270
''n squared x' over 'n
squared'', where 'n' is the

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00:07:01,270 --> 00:07:02,730
number of the term over here.

135
00:07:02,730 --> 00:07:05,450
For example, the third
term would be 'cosine

136
00:07:05,450 --> 00:07:08,250
9x' over 9, et cetera.

137
00:07:08,250 --> 00:07:13,520
And I now want to look at
the limit function here.

138
00:07:13,520 --> 00:07:16,210
See, is this a convergent
series?

139
00:07:16,210 --> 00:07:17,860
Is it absolutely convergent?

140
00:07:17,860 --> 00:07:19,330
Is it uniformly convergent?

141
00:07:19,330 --> 00:07:20,630
What is it?

142
00:07:20,630 --> 00:07:23,870
And here's how the Weierstrauss
M-test works.

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00:07:23,870 --> 00:07:28,180
We look at the absolute value
of the term making up this

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00:07:28,180 --> 00:07:29,670
particular series.

145
00:07:29,670 --> 00:07:32,760
See, the absolute value of
''cosine 'n squared x'' over

146
00:07:32,760 --> 00:07:33,380
'n squared''.

147
00:07:33,380 --> 00:07:34,610
Now, let's take a
look at this.

148
00:07:34,610 --> 00:07:38,910
Since 'n' is an integer, 'n
squared' is positive, so the

149
00:07:38,910 --> 00:07:42,460
absolute value of 'n squared'
is still in squared.

150
00:07:42,460 --> 00:07:47,810
Since the cosine has to be
between minus 1 and 1, the

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00:07:47,810 --> 00:07:50,850
magnitude of the cosine,
regardless of the value of 'n

152
00:07:50,850 --> 00:07:53,280
squared x', is less than
or equal to 1.

153
00:07:53,280 --> 00:07:56,340
In particular, then, the
absolute value of ''cosine 'n

154
00:07:56,340 --> 00:07:59,400
squared x'' over 'n squared''
is less than or equal to '1

155
00:07:59,400 --> 00:08:03,990
over 'n squared''
for each 'n'.

156
00:08:03,990 --> 00:08:09,610
And I might add, and all 'x',
because no matter what 'x' is

157
00:08:09,610 --> 00:08:12,790
and no matter what 'n' is,
'cosine 'n squared x'' in

158
00:08:12,790 --> 00:08:15,170
magnitude cannot exceed 1.

159
00:08:15,170 --> 00:08:19,070
On the other hand, we already
know, in particular by the

160
00:08:19,070 --> 00:08:23,360
integral test, that summation
'n' goes from 1 to infinity or

161
00:08:23,360 --> 00:08:26,980
0 to infinity is a positive
convergent series.

162
00:08:26,980 --> 00:08:31,410
Consequently, by the
Weierstrauss M-test, this

163
00:08:31,410 --> 00:08:34,360
series here is uniformly
convergent,

164
00:08:34,360 --> 00:08:35,970
which again means what?

165
00:08:35,970 --> 00:08:39,500
That the sequence of partial
sums, namely sum 'k' goes from

166
00:08:39,500 --> 00:08:43,380
1 to 'n', ''cosine 'k squared
x'' over 'k squared'',

167
00:08:43,380 --> 00:08:47,280
converges uniformly to summation
'n' goes from 1 to

168
00:08:47,280 --> 00:08:51,120
infinity ''cosine 'n squared
x'' over 'n squared''.

169
00:08:51,120 --> 00:08:52,410
That's how the test works.

170
00:08:52,410 --> 00:08:54,440
But now, from an engineering
point, the

171
00:08:54,440 --> 00:08:55,850
more important question--

172
00:08:55,850 --> 00:08:57,240
what does it mean?

173
00:08:57,240 --> 00:08:58,870
How can we use it?

174
00:08:58,870 --> 00:09:02,790
Well, for example, let's suppose
that in working some

175
00:09:02,790 --> 00:09:07,470
particular applied problem,
one had to integrate this

176
00:09:07,470 --> 00:09:12,510
particular series, 'cosine x'
plus ''cosine 4x' over 4' plus

177
00:09:12,510 --> 00:09:16,140
''cosine 9x' over 9', et cetera,
the infinite series

178
00:09:16,140 --> 00:09:20,440
from 0 to some value 't'.

179
00:09:20,440 --> 00:09:21,410
Now here's the key point.

180
00:09:21,410 --> 00:09:23,500
And by the way, I just
write it this way--

181
00:09:23,500 --> 00:09:26,570
to get used to the ominous
notation, notice that what

182
00:09:26,570 --> 00:09:29,120
looks a little bit ominous here
is just another way of

183
00:09:29,120 --> 00:09:31,760
writing out, longhand, what
the terms inside the

184
00:09:31,760 --> 00:09:33,250
parentheses are.

185
00:09:33,250 --> 00:09:34,500
The idea is this.

186
00:09:34,500 --> 00:09:36,070
Notice what this says--

187
00:09:36,070 --> 00:09:40,060
this says first compute the
sum and then integrate.

188
00:09:40,060 --> 00:09:44,020
For example, we might not
know what limit this sum

189
00:09:44,020 --> 00:09:45,150
approaches.

190
00:09:45,150 --> 00:09:48,270
Or if we do know it, it may be
a particularly complicated

191
00:09:48,270 --> 00:09:49,880
thing to write down.

192
00:09:49,880 --> 00:09:52,640
You see, in other words, if we
follow the problem in the

193
00:09:52,640 --> 00:09:56,830
order that the operations are
given, we must first sum this

194
00:09:56,830 --> 00:09:59,510
series and then integrate it.

195
00:09:59,510 --> 00:10:01,370
But what have we
already shown?

196
00:10:01,370 --> 00:10:04,300
We've already shown that this
series is uniformly

197
00:10:04,300 --> 00:10:08,160
convergent, but for a uniform
convergent series, we saw last

198
00:10:08,160 --> 00:10:11,720
time that you can interchange
the order of summation and

199
00:10:11,720 --> 00:10:13,060
integration.

200
00:10:13,060 --> 00:10:17,200
In other words, by uniform
convergence, what I can now do

201
00:10:17,200 --> 00:10:21,370
is integrate this thing
here, term by term.

202
00:10:21,370 --> 00:10:24,960
See, 'sine x' plus ''sine
4x' over 16'.

203
00:10:24,960 --> 00:10:27,530
Just the usual technique
of integration.

204
00:10:27,530 --> 00:10:30,740
And now you see I can integrate
first and then

205
00:10:30,740 --> 00:10:31,890
compute the limit.

206
00:10:31,890 --> 00:10:34,170
In other words, I can say, look,
this integral is just

207
00:10:34,170 --> 00:10:37,920
'sine t' plus ''sine 4t'
over 16' plus et

208
00:10:37,920 --> 00:10:39,720
cetera, and so forth.

209
00:10:39,720 --> 00:10:42,520
Again, if you want to see this
from the abstract point of

210
00:10:42,520 --> 00:10:47,020
view, what I'm saying is that by
uniform convergence, I can

211
00:10:47,020 --> 00:10:49,170
switch these two symbols.

212
00:10:49,170 --> 00:10:52,460
Now, the important point is that
to integrate ''cosine 'n

213
00:10:52,460 --> 00:10:54,440
squared x'' over
'n squared''--

214
00:10:54,440 --> 00:10:55,660
remember, 'n' is a
constant here.

215
00:10:55,660 --> 00:10:57,600
This is a very easy thing
to integrate.

216
00:10:57,600 --> 00:10:59,270
In fact, it comes
out to be what?

217
00:10:59,270 --> 00:11:02,470
''Sine 'n squared x'' over
'n to the fourth''.

218
00:11:02,470 --> 00:11:04,960
You see the 'n squared' comes
down as an 'n to the fourth'.

219
00:11:04,960 --> 00:11:08,450
And when I evaluate that between
0 and 1, all I wind up

220
00:11:08,450 --> 00:11:12,400
with is summation 'n' goes from
one to infinity, ''sine

221
00:11:12,400 --> 00:11:14,910
'n squared t'' over 'n
to the fourth''.

222
00:11:14,910 --> 00:11:19,050
And this is a perfectly good,
bona fide convergent series.

223
00:11:19,050 --> 00:11:22,320
And, in real life, you see I
can approximate this to any

224
00:11:22,320 --> 00:11:25,790
degree of accuracy that I want
just by going out far enough

225
00:11:25,790 --> 00:11:28,490
before I chop off the
remaining terms.

226
00:11:28,490 --> 00:11:31,860
In other words, the beauty in
this case with series is that

227
00:11:31,860 --> 00:11:35,920
when they converge uniformly and
you have to integrate the

228
00:11:35,920 --> 00:11:38,840
limit function, you can
integrate the individual

229
00:11:38,840 --> 00:11:42,940
member of the sequence first and
then take the limit as n

230
00:11:42,940 --> 00:11:44,630
goes to infinity.

231
00:11:44,630 --> 00:11:45,380
OK.

232
00:11:45,380 --> 00:11:46,720
So far so good.

233
00:11:46,720 --> 00:11:50,640
Let's apply this now, in
particular, to power series.

234
00:11:50,640 --> 00:11:53,400
And again, let me keep reminding
you, all of this

235
00:11:53,400 --> 00:11:57,240
stuff is done much more slowly
in the supplementary notes.

236
00:11:57,240 --> 00:12:00,350
My main reason for repeating
what's in the supplementary

237
00:12:00,350 --> 00:12:04,140
notes is I think this material
is both difficult enough and

238
00:12:04,140 --> 00:12:07,770
new enough so I think you should
hear parts of it before

239
00:12:07,770 --> 00:12:09,470
you're sent out on your
own to read it.

240
00:12:09,470 --> 00:12:12,680
Hopefully my words will sound
familiar to you as you hear

241
00:12:12,680 --> 00:12:15,335
them repeated in the
supplementary notes.

242
00:12:17,930 --> 00:12:22,520
Let's suppose that a power
series summation, 'a n', 'x to

243
00:12:22,520 --> 00:12:25,550
the n', as 'n' goes from 0 to
infinity, converges for the

244
00:12:25,550 --> 00:12:27,570
absolute value of 'x'
less than 'R'.

245
00:12:27,570 --> 00:12:30,850
And let me pick a couple of
values of 'x', 'x sub 0' and

246
00:12:30,850 --> 00:12:36,410
'x sub 1', such that 'x sub 0'
in magnitude is less than 'x

247
00:12:36,410 --> 00:12:38,340
sub 1' in magnitude,
which in turn is

248
00:12:38,340 --> 00:12:40,190
less than 'R' in magnitude.

249
00:12:40,190 --> 00:12:42,370
To give you a hint as to what
I'm going to be driving at,

250
00:12:42,370 --> 00:12:45,590
I'm going to try to prove that
this series converges

251
00:12:45,590 --> 00:12:47,800
uniformly within 'R'.

252
00:12:47,800 --> 00:12:50,650
And I'm going to try to prove
it by setting up the

253
00:12:50,650 --> 00:12:52,920
Weierstrauss M-test.

254
00:12:52,920 --> 00:12:55,640
And I'm going to have to compare
this with a positive

255
00:12:55,640 --> 00:12:58,610
convergent series, and the
positive convergent series

256
00:12:58,610 --> 00:13:01,980
that I have in mind is going to
make use of the fact that

257
00:13:01,980 --> 00:13:06,360
the magnitude of 'x0' divided by
the magnitude of 'x1' is a

258
00:13:06,360 --> 00:13:08,960
positive constant less than 1.

259
00:13:08,960 --> 00:13:11,160
And I'm going to set this thing
up so that I can use a

260
00:13:11,160 --> 00:13:12,940
geometric series on it.

261
00:13:12,940 --> 00:13:15,610
And if that sounds confusing,
let me just go through this

262
00:13:15,610 --> 00:13:18,940
now in slow motion and show you
what I'm driving at here.

263
00:13:18,940 --> 00:13:23,230
See, first of all, since 'x1'
is within the radius of

264
00:13:23,230 --> 00:13:27,450
convergence, that means in
particular that summation 'a n

265
00:13:27,450 --> 00:13:29,980
''x sub 1' to the n'',
as 'n' goes from 0

266
00:13:29,980 --> 00:13:32,450
to infinity, converges.

267
00:13:32,450 --> 00:13:35,390
Since this converges, in
particular-- the very first

268
00:13:35,390 --> 00:13:36,410
test that we learned--

269
00:13:36,410 --> 00:13:39,990
in particular, the n-th
term must go to 0.

270
00:13:39,990 --> 00:13:42,060
Remember, for a convergent
series, the n-th

271
00:13:42,060 --> 00:13:43,770
term goes to 0.

272
00:13:43,770 --> 00:13:49,310
Now, since the limit of 'a n 'x1
to the n-th'' is 0, that

273
00:13:49,310 --> 00:13:53,170
means that for n large enough,
this will get smaller than any

274
00:13:53,170 --> 00:13:54,110
positive constant.

275
00:13:54,110 --> 00:13:56,700
Otherwise it couldn't
converge on 0.

276
00:13:56,700 --> 00:13:57,580
All right?

277
00:13:57,580 --> 00:14:01,470
So all I'm saying there, I
guess, is if this is 0 and

278
00:14:01,470 --> 00:14:04,940
this is 'M', if I can't make
this thing less than 'M', how

279
00:14:04,940 --> 00:14:06,110
can the limit ever be 0?

280
00:14:06,110 --> 00:14:08,440
In other words, if these things
can't get past 'M', how

281
00:14:08,440 --> 00:14:10,170
could the limit be 0?

282
00:14:10,170 --> 00:14:13,490
So all I'm saying is that, given
a positive 'M', I can

283
00:14:13,490 --> 00:14:16,735
always find an 'n' such that
when I go out far enough-- in

284
00:14:16,735 --> 00:14:18,570
other words, when the subscript
'n' is greater than

285
00:14:18,570 --> 00:14:23,340
capital 'N', the magnitude of
'a n 'x1 to the n'' is less

286
00:14:23,340 --> 00:14:25,180
than capital 'M'.

287
00:14:25,180 --> 00:14:26,910
Now the key idea is this.

288
00:14:26,910 --> 00:14:31,910
What I'm going to do is look at
summation 'a n 'x0 to the

289
00:14:31,910 --> 00:14:35,790
n'' where 'x0' is as
given over here.

290
00:14:35,790 --> 00:14:38,240
Now what I'm going to try to
do is set this up for the

291
00:14:38,240 --> 00:14:41,720
Weierstrauss M-test, and
the way I'm going to

292
00:14:41,720 --> 00:14:43,160
do that is as follows.

293
00:14:43,160 --> 00:14:47,060
I look at the absolute value
of 'a n 'x0 to the n''.

294
00:14:47,060 --> 00:14:51,590
Somehow or other, the thing I
have control over is not 'a n

295
00:14:51,590 --> 00:14:56,580
'x0 to the n'', but rather
'a n 'x1 to the n''.

296
00:14:56,580 --> 00:15:00,590
And now I use that very common
mathematical trick that gets

297
00:15:00,590 --> 00:15:03,940
us out of all sorts of binds,
and that is that, since I

298
00:15:03,940 --> 00:15:08,010
would like 'a n 'x1 to the n''
over here, I just put it in.

299
00:15:08,010 --> 00:15:10,470
And then I must cancel
it out again.

300
00:15:10,470 --> 00:15:15,460
In other words, all I do is I
rewrite 'a n 'x0 to the n'' as

301
00:15:15,460 --> 00:15:20,820
'a n' times 'x0' over 'x1 to the
n', times 'x1 to the n'.

302
00:15:20,820 --> 00:15:22,970
In other words, notice that this
just cancels and I have

303
00:15:22,970 --> 00:15:24,730
what I started with over here.

304
00:15:24,730 --> 00:15:28,830
The point is that the magnitude
of 'a n 'x1 to the

305
00:15:28,830 --> 00:15:30,350
n'' is less than 'M'.

306
00:15:30,350 --> 00:15:31,550
We already saw that.

307
00:15:31,550 --> 00:15:34,240
So what I do is I split
this thing up.

308
00:15:34,240 --> 00:15:36,730
Remember, the absolute value of
a product is the product of

309
00:15:36,730 --> 00:15:38,020
the absolute values.

310
00:15:38,020 --> 00:15:39,550
So I write it as what?

311
00:15:39,550 --> 00:15:43,530
The magnitude of this times
this, times the magnitude of

312
00:15:43,530 --> 00:15:45,770
''x0 over x1' to the n'.

313
00:15:45,770 --> 00:15:50,310
In other words, I guess what
I should have over here--

314
00:15:50,310 --> 00:15:51,840
this is still an equality.

315
00:15:51,840 --> 00:15:52,660
This is just rewriting.

316
00:15:52,660 --> 00:15:55,420
The absolute value of a product
is a product of the

317
00:15:55,420 --> 00:15:56,510
absolute values.

318
00:15:56,510 --> 00:15:59,620
Consequently, this is the
absolute value of 'a n 'x1 to

319
00:15:59,620 --> 00:16:02,400
the n'' times the absolute
value of ''x0

320
00:16:02,400 --> 00:16:03,770
over x1' to the n'.

321
00:16:03,770 --> 00:16:07,840
And the key point is that the
absolute value of 'a n 'x1 to

322
00:16:07,840 --> 00:16:11,780
the n'' is less than 'M' for
'n' sufficiently large.

323
00:16:11,780 --> 00:16:14,720
In other words, for 'n'
sufficiently large, the

324
00:16:14,720 --> 00:16:19,220
magnitude of 'a n 'x0 to the n''
is less than or equal to

325
00:16:19,220 --> 00:16:20,670
this expression here.

326
00:16:20,670 --> 00:16:22,510
But what is this expression?

327
00:16:22,510 --> 00:16:26,690
My claim is that this is the
n-th term of a convergent

328
00:16:26,690 --> 00:16:30,850
positive series, namely a
convergent geometric series.

329
00:16:30,850 --> 00:16:32,700
I'm going to to show you
this in more detail.

330
00:16:32,700 --> 00:16:37,620
All I'm saying is observe that
the magnitude of 'x0 over x1',

331
00:16:37,620 --> 00:16:40,860
since 'x0' is less than 'x1'
in magnitude, is a positive

332
00:16:40,860 --> 00:16:43,090
constant less than 1.

333
00:16:43,090 --> 00:16:44,750
Therefore, this is what?

334
00:16:44,750 --> 00:16:50,680
This is a geometric series whose
ratio is 'x0 over x1'.

335
00:16:50,680 --> 00:16:53,300
In other words, as you pass from
the n-th term to the 'n

336
00:16:53,300 --> 00:16:57,190
plus first' term, in each case
you just multiply by 'x0 over

337
00:16:57,190 --> 00:16:59,590
x1', which is a constant.

338
00:16:59,590 --> 00:17:04,290
Therefore, by the Weierstrauss
M-test, this given series is

339
00:17:04,290 --> 00:17:08,970
uniformly convergent inside the
interval of convergence.

340
00:17:08,970 --> 00:17:10,170
You see?

341
00:17:10,170 --> 00:17:13,680
In other words, what this now
means is that we can replace

342
00:17:13,680 --> 00:17:17,200
the condition that the series
was absolutely convergent

343
00:17:17,200 --> 00:17:20,910
inside the radius of convergence
by, it's uniformly

344
00:17:20,910 --> 00:17:23,540
convergent inside the radius
of convergence.

345
00:17:23,540 --> 00:17:26,930
And now you see the question is,
what does that help us do?

346
00:17:26,930 --> 00:17:30,200
And again, this is left in
great detail for the

347
00:17:30,200 --> 00:17:33,170
supplementary notes and the
exercises, but I thought that

348
00:17:33,170 --> 00:17:36,360
for a finale, we should do
a rather nice practical

349
00:17:36,360 --> 00:17:37,330
application.

350
00:17:37,330 --> 00:17:42,160
In particular, a problem that
we haven't tackled at all.

351
00:17:42,160 --> 00:17:44,760
We wern't able to tackle it, but
a problem that we've used

352
00:17:44,760 --> 00:17:49,130
as a scapegoat many times to
show, for example, why the

353
00:17:49,130 --> 00:17:52,570
First Fundamental Theorem of
integral calculus was not the

354
00:17:52,570 --> 00:17:54,700
cure-all it was supposed
to be, for example.

355
00:17:54,700 --> 00:17:56,280
Let me show you what example
I have in mind.

356
00:17:59,180 --> 00:18:01,120
I'm thinking of this example.

357
00:18:01,120 --> 00:18:05,130
Find the area of the region 'R',
where 'R' is the region

358
00:18:05,130 --> 00:18:08,020
bounded above by our old friend
'y' equals 'e to the

359
00:18:08,020 --> 00:18:12,590
minus 'x squared'', below by the
x-axis, on the left by the

360
00:18:12,590 --> 00:18:16,790
y-axis, and on the right by
the line 'x' equals 1.

361
00:18:16,790 --> 00:18:18,790
You see, what we used to
say before was what?

362
00:18:18,790 --> 00:18:21,830
We know by the definition of a
definite integral that the

363
00:18:21,830 --> 00:18:25,170
area of the region 'R' is the
integral from 0 to 1, 'e to

364
00:18:25,170 --> 00:18:27,070
the minus 'x squared'', 'dx'.

365
00:18:27,070 --> 00:18:30,230
But what we do not know
explicitly is a function, 'G

366
00:18:30,230 --> 00:18:32,670
of x', for which 'G prime
of x' is 'e to

367
00:18:32,670 --> 00:18:34,220
the minus 'x squared''.

368
00:18:34,220 --> 00:18:36,520
Remember, we could solve this
problem by trapezoidal

369
00:18:36,520 --> 00:18:40,380
approximations and things
of this type.

370
00:18:40,380 --> 00:18:43,490
We could approximate it, but we
couldn't get a good bound

371
00:18:43,490 --> 00:18:45,270
on the answer very
conveniently.

372
00:18:45,270 --> 00:18:49,140
What I thought I'd like to show
you now is how one uses

373
00:18:49,140 --> 00:18:51,950
power series to solve this
kind of a problem.

374
00:18:51,950 --> 00:18:55,310
If nothing else, this one
application should be enough

375
00:18:55,310 --> 00:18:58,910
impetus to show you the
power of power series.

376
00:18:58,910 --> 00:19:02,170
They are a very, very important
and powerful

377
00:19:02,170 --> 00:19:03,840
analytical technique.

378
00:19:03,840 --> 00:19:06,030
Let's see how we can
handle this.

379
00:19:06,030 --> 00:19:10,060
First of all, from our previous
material, we already

380
00:19:10,060 --> 00:19:15,250
know that 'e to the u' is
represented by '1 plus u plus

381
00:19:15,250 --> 00:19:17,930
''u squared' over '2 factorial''
plus' et cetera,

382
00:19:17,930 --> 00:19:20,600
''u to the n' over 'n
factorial'', et cetera, for

383
00:19:20,600 --> 00:19:22,540
all real numbers 'u'.

384
00:19:22,540 --> 00:19:24,450
OK, we already know that.

385
00:19:24,450 --> 00:19:28,270
In particular, if I now think
of-- see, since 'u' is

386
00:19:28,270 --> 00:19:31,610
generically just a name for any
real number, since minus

387
00:19:31,610 --> 00:19:34,690
'x squared' is a real number
also, I can replace 'u' by

388
00:19:34,690 --> 00:19:36,150
minus 'x squared'.

389
00:19:36,150 --> 00:19:39,740
And replacing 'u' by minus 'x
squared' leads to what? 'e to

390
00:19:39,740 --> 00:19:44,570
the minus 'x squared'' is
1, minus 'x squared'.

391
00:19:44,570 --> 00:19:47,440
Now minus 'x squared', squared,
is 'x to the fourth',

392
00:19:47,440 --> 00:19:49,060
over 2 factorial.

393
00:19:49,060 --> 00:19:54,990
And in general, we see when
I replace 'u' by minus 'x

394
00:19:54,990 --> 00:19:59,440
squared', this becomes plus or
minus 'x to the 2n' over 'n

395
00:19:59,440 --> 00:20:00,610
factorial'.

396
00:20:00,610 --> 00:20:03,980
And to handle the plus or minus,
I simply put in my sign

397
00:20:03,980 --> 00:20:07,040
alternator-- that's minus
1 to the n-th power.

398
00:20:07,040 --> 00:20:10,980
And the reason I use 'n' here,
rather than, say, 'n plus 1',

399
00:20:10,980 --> 00:20:14,200
is to keep in mind that in power
series, this is called

400
00:20:14,200 --> 00:20:16,200
the 0-th term.

401
00:20:16,200 --> 00:20:19,910
In other words, when 'n' is 0,
minus 1 to the 0 would be

402
00:20:19,910 --> 00:20:23,280
positive, and I want the first
term to be positive.

403
00:20:23,280 --> 00:20:26,360
At any rate, to make a
long story short--

404
00:20:26,360 --> 00:20:27,440
shorter--

405
00:20:27,440 --> 00:20:30,820
another way of writing this
is that 'e to the minus 'x

406
00:20:30,820 --> 00:20:34,740
squared'' is summation 'n' goes
from 0 to infinity, minus

407
00:20:34,740 --> 00:20:38,920
'1 to the n', 'x to the 2n'
over 'n factorial'.

408
00:20:38,920 --> 00:20:42,810
Consequently, since these two
are synonyms, to compute the

409
00:20:42,810 --> 00:20:46,570
integral from 0 to 1, ''e to the
minus 'x squared'' dx', I

410
00:20:46,570 --> 00:20:50,610
can replace 'e to the minus 'x
squared'' by the power series

411
00:20:50,610 --> 00:20:54,850
which converges to
it uniformly.

412
00:20:54,850 --> 00:20:58,920
Namely, I can now write
this as what?

413
00:20:58,920 --> 00:21:02,640
Integral from 0 to 1, summation
'n' goes from 0 to

414
00:21:02,640 --> 00:21:05,770
infinity, minus '1 to the n',
'x to the 2n' over 'n

415
00:21:05,770 --> 00:21:08,090
factorial' times 'dx'.

416
00:21:08,090 --> 00:21:09,510
Now notice what this says.

417
00:21:09,510 --> 00:21:13,250
This says, in the order of the
given operations, that I must

418
00:21:13,250 --> 00:21:15,890
form the sum of this series
first, which is not an easy

419
00:21:15,890 --> 00:21:18,840
job, and then integrate
that from 0 to 1.

420
00:21:18,840 --> 00:21:24,760
But the beauty is that, by the
Weierstrauss M-test, I know

421
00:21:24,760 --> 00:21:28,380
that this converges uniformly
wherever it converges

422
00:21:28,380 --> 00:21:29,400
absolutely.

423
00:21:29,400 --> 00:21:31,540
I already know that it does
converge absolutely

424
00:21:31,540 --> 00:21:32,540
for all real 'x'.

425
00:21:32,540 --> 00:21:35,150
At any rate, then, I now know
that it's uniformly

426
00:21:35,150 --> 00:21:36,100
convergent.

427
00:21:36,100 --> 00:21:38,750
One of the beauties of the fact
that this is uniformly

428
00:21:38,750 --> 00:21:41,430
convergent is I can interchange
the order of

429
00:21:41,430 --> 00:21:44,860
summation and integration,
which I do over here.

430
00:21:44,860 --> 00:21:46,380
But this is crucial.

431
00:21:46,380 --> 00:21:49,200
Let me make sure I
underline that.

432
00:21:49,200 --> 00:21:51,990
You see, if there wasn't uniform
convergence here,

433
00:21:51,990 --> 00:21:54,670
mechanically I can change the
order, but I may get a

434
00:21:54,670 --> 00:21:56,020
different answer.

435
00:21:56,020 --> 00:21:58,790
But since this is uniformly
convergent, these are still

436
00:21:58,790 --> 00:22:02,100
synonyms, and the beauty is, if
I just look at the typical

437
00:22:02,100 --> 00:22:04,620
term that I'm trying to
integrate here, that's a very

438
00:22:04,620 --> 00:22:06,020
easy thing to integrate.

439
00:22:06,020 --> 00:22:08,250
Namely, I just do what?

440
00:22:08,250 --> 00:22:10,290
Replace this by an exponent
one larger--

441
00:22:10,290 --> 00:22:11,830
that's '2n plus 1'.

442
00:22:11,830 --> 00:22:13,890
Divide through by '2n plus 1'.

443
00:22:13,890 --> 00:22:17,030
In other words, the integral
from 0 to 1, ''e to the minus

444
00:22:17,030 --> 00:22:20,180
'x squared'' dx', is just this
particular power series,

445
00:22:20,180 --> 00:22:24,790
namely summation 'n' goes from 0
to infinity, minus '1 to the

446
00:22:24,790 --> 00:22:30,190
n', 'x to the '2n plus 1'' over
''2n plus 1' times 'n

447
00:22:30,190 --> 00:22:31,300
factorial''.

448
00:22:31,300 --> 00:22:34,560
That must be evaluated
between 0 and 1.

449
00:22:34,560 --> 00:22:37,800
Since the lower limit gives me
0 and the upper limit here

450
00:22:37,800 --> 00:22:41,500
just gives me a 1, rewriting
this-- this is just what?

451
00:22:41,500 --> 00:22:45,840
Summation 'n' goes from 0 to
infinity, minus '1 to the n'

452
00:22:45,840 --> 00:22:49,430
over ''2n plus 1' times
'n factorial''.

453
00:22:49,430 --> 00:22:49,790
In other words--

454
00:22:49,790 --> 00:22:50,510
let's take a look at this.

455
00:22:50,510 --> 00:22:53,820
When 'n' is 0, this term is 1.

456
00:22:53,820 --> 00:22:59,440
When 'n' is 1, this is
minus 1 over what?

457
00:22:59,440 --> 00:23:02,910
3 times 1 is 3.

458
00:23:02,910 --> 00:23:08,350
When 'n' is 2, this is 5 times
2 factorial, which is 10.

459
00:23:08,350 --> 00:23:12,250
When 'n' is 3, this is
minus 1 to the third

460
00:23:12,250 --> 00:23:13,830
power, which is minus.

461
00:23:13,830 --> 00:23:17,200
This is 7 times 3 factorial.

462
00:23:17,200 --> 00:23:19,330
7 times 3 factorial is what?

463
00:23:19,330 --> 00:23:22,600
7 times 6, which is
42, et cetera.

464
00:23:22,600 --> 00:23:26,900
In other words, what we now know
is that 'A sub R' is 1

465
00:23:26,900 --> 00:23:31,330
minus 1/3 plus 1/10
minus 1/42.

466
00:23:31,330 --> 00:23:35,130
And if we keep on going this
way, the next term is 1/216,

467
00:23:35,130 --> 00:23:37,540
which we can compute
very easily.

468
00:23:37,540 --> 00:23:38,840
What is this, by the way?

469
00:23:38,840 --> 00:23:43,410
This is a convergent
alternating series.

470
00:23:43,410 --> 00:23:46,680
And as you recall from our
learning exercises and some of

471
00:23:46,680 --> 00:23:49,190
the material in our
supplementary notes, not only

472
00:23:49,190 --> 00:23:54,340
does this thing converge, but
in such a case the error is

473
00:23:54,340 --> 00:23:58,380
never greater than the magnitude
of the next term.

474
00:23:58,380 --> 00:24:02,820
So, you see, what I can do
over here is just take 1,

475
00:24:02,820 --> 00:24:08,610
subtract 1/3, add 1/10, subtract
1/42, and know that

476
00:24:08,610 --> 00:24:12,450
whatever answer I get when I
do this has to be exact to

477
00:24:12,450 --> 00:24:16,470
within no more of an
error then 1/216.

478
00:24:16,470 --> 00:24:18,920
And in fact, if I now
add on the 1/216,

479
00:24:18,920 --> 00:24:22,130
I think I get 0.748.

480
00:24:22,130 --> 00:24:25,620
And the next term is going to
be something like 1/1,300,

481
00:24:25,620 --> 00:24:28,850
which means that this must be
correct to within three

482
00:24:28,850 --> 00:24:30,750
decimal places.

483
00:24:30,750 --> 00:24:32,740
And this is the way the
so-called error function, the

484
00:24:32,740 --> 00:24:35,510
table of error function,
is actually computed.

485
00:24:35,510 --> 00:24:38,450
Now I'm going to end the
material of the course right

486
00:24:38,450 --> 00:24:40,280
at this particular point.

487
00:24:40,280 --> 00:24:43,260
In other words, I will leave the
remaining applications for

488
00:24:43,260 --> 00:24:45,980
the exercises and the
supplementary notes.

489
00:24:45,980 --> 00:24:49,330
What I would like to do is to
become subjective for a moment

490
00:24:49,330 --> 00:24:50,820
or so, if I may.

491
00:24:50,820 --> 00:24:53,450
And even though this course is
quite different from a regular

492
00:24:53,450 --> 00:24:56,400
classroom course, I would like
to end it like a regular

493
00:24:56,400 --> 00:24:57,230
classroom course.

494
00:24:57,230 --> 00:25:02,600
And that is to make a genuine
remark to the effect that this

495
00:25:02,600 --> 00:25:05,000
course has been a pleasure
for me to teach.

496
00:25:05,000 --> 00:25:07,020
It has been very enlightening.

497
00:25:07,020 --> 00:25:11,000
It has been soul-searching to
both plan the lectures and the

498
00:25:11,000 --> 00:25:14,390
supplementary notes and the
learning exercises, and I

499
00:25:14,390 --> 00:25:17,880
myself, to coin a cliche, am a
much better person for having

500
00:25:17,880 --> 00:25:19,650
gone through all of this.

501
00:25:19,650 --> 00:25:23,650
Many other people worked hard
with me on this also.

502
00:25:23,650 --> 00:25:26,620
And in particular, I would like
to single out three of my

503
00:25:26,620 --> 00:25:29,000
colleagues for recognition.

504
00:25:29,000 --> 00:25:30,360
Professor Harold Mickley,
who was the

505
00:25:30,360 --> 00:25:32,040
director of the center.

506
00:25:32,040 --> 00:25:36,200
My friend, John Fitch, who is
the manager of the self-study

507
00:25:36,200 --> 00:25:40,730
projects and, in particular,
was the director of the

508
00:25:40,730 --> 00:25:44,900
lectures, and also gave me
invaluable advice in many

509
00:25:44,900 --> 00:25:47,220
places as to how to simplify
the lectures and make them

510
00:25:47,220 --> 00:25:48,470
more meaningful.

511
00:25:48,470 --> 00:25:51,910
And finally, my friend Charles
Patton, who is our electronics

512
00:25:51,910 --> 00:25:55,120
wizard here and kept everything
going from a

513
00:25:55,120 --> 00:25:57,700
technological point of view,
and gave me, also, many

514
00:25:57,700 --> 00:25:58,980
valuable pointers.

515
00:25:58,980 --> 00:26:02,210
Now, from your point of view,
I am hoping that you too can

516
00:26:02,210 --> 00:26:05,510
sit back and feel content, now
that the course is over.

517
00:26:05,510 --> 00:26:09,090
I'm hoping that this long trip
back through Calculus

518
00:26:09,090 --> 00:26:13,290
Revisited has given you a new
insight to sets, functions,

519
00:26:13,290 --> 00:26:16,770
circular functions, hyperbolic
functions, differentiation and

520
00:26:16,770 --> 00:26:20,850
integration, power series, so
that you can go back enriched,

521
00:26:20,850 --> 00:26:25,510
revitalized, and tackle the
projects of your choice.

522
00:26:25,510 --> 00:26:28,930
I hate to see the course come to
an end, but in closing, it

523
00:26:28,930 --> 00:26:32,640
was my pleasure and I hope that
we will have the pleasure

524
00:26:32,640 --> 00:26:36,730
of getting together again soon
to tackle, together, Part Two

525
00:26:36,730 --> 00:26:38,380
of Calculus Revisited.

526
00:26:38,380 --> 00:26:42,270
And so, until that
time, so long.

527
00:26:42,270 --> 00:26:45,300
ANNOUNCER: Funding for the
publication of this video was

528
00:26:45,300 --> 00:26:50,020
provided by the Gabriella and
Paul Rosenbaum Foundation.

529
00:26:50,020 --> 00:26:54,190
Help OCW continue to provide
free and open access to MIT

530
00:26:54,190 --> 00:26:58,390
courses by making a donation
at ocw.mit.edu/donate.