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PROFESSOR STRANG: OK,
thank you for coming today.

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00:00:23,020 --> 00:00:25,120
The day before Thanksgiving.

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00:00:25,120 --> 00:00:26,890
Day before my
birthday, actually.

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00:00:26,890 --> 00:00:29,680
So it's a special day.

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00:00:29,680 --> 00:00:31,350
Everybody gets an
A for showing up.

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00:00:31,350 --> 00:00:38,660
Even you.

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00:00:38,660 --> 00:00:40,950
So, let's see.

16
00:00:40,950 --> 00:00:44,950
Last time, I wrote down these
formulas for the Fourier

17
00:00:44,950 --> 00:00:46,530
integral transform.

18
00:00:46,530 --> 00:00:48,870
And I thought I'd
just write them again

19
00:00:48,870 --> 00:00:52,120
so you kind of photograph
them and remember them.

20
00:00:52,120 --> 00:00:54,580
They're easy to remember.

21
00:00:54,580 --> 00:00:56,460
As always, you
take the function,

22
00:00:56,460 --> 00:01:00,190
you multiply by e^(-ikx),
and you integrate.

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00:01:00,190 --> 00:01:05,290
To get the amount-- So k
is my frequency variable.

24
00:01:05,290 --> 00:01:08,240
It could well have been
omega or some other variable.

25
00:01:08,240 --> 00:01:13,020
I stayed with k because it
was k in the Fourier series.

26
00:01:13,020 --> 00:01:20,840
So that's the calculation
which as always, I mean,

27
00:01:20,840 --> 00:01:22,350
these are integrals
that we may be

28
00:01:22,350 --> 00:01:26,030
able to do if the function is
especially nice, or we may not.

29
00:01:26,030 --> 00:01:28,820
But that's the formula.

30
00:01:28,820 --> 00:01:31,580
And then to reconstruct
the function,

31
00:01:31,580 --> 00:01:39,220
we combine all they e^(ikx)'s
in that amount to get f(x) back.

32
00:01:39,220 --> 00:01:43,140
OK, nice formula.

33
00:01:43,140 --> 00:01:47,160
So I did one example last
time, and now could I just

34
00:01:47,160 --> 00:01:49,160
double it up?

35
00:01:49,160 --> 00:01:53,580
This is also in
the textbook and so

36
00:01:53,580 --> 00:01:55,970
this is now going to
be an even function.

37
00:01:55,970 --> 00:02:04,140
Last time the example I
did was zero up to x=0.

38
00:02:04,140 --> 00:02:06,570
This time I'll
make it symmetric,

39
00:02:06,570 --> 00:02:08,170
make the function even.

40
00:02:08,170 --> 00:02:12,740
And then I have two
pieces in the integral.

41
00:02:12,740 --> 00:02:16,580
And if you remember
what it was, you

42
00:02:16,580 --> 00:02:20,620
remember that this, I'll
just remind you what we did.

43
00:02:20,620 --> 00:02:24,630
We wrote that it's e^(-(a+ik)x).

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00:02:24,630 --> 00:02:27,620


45
00:02:27,620 --> 00:02:29,300
That was clear.

46
00:02:29,300 --> 00:02:33,630
And then when we integrated we
got that same function divided

47
00:02:33,630 --> 00:02:37,000
by -(a+ik).

48
00:02:37,000 --> 00:02:39,660
And then we put in the limits.

49
00:02:39,660 --> 00:02:43,480
And the answer was, let me maybe
write the answer down here.

50
00:02:43,480 --> 00:02:46,110
Was just at x equal
infinity the limit

51
00:02:46,110 --> 00:02:49,780
was zero because this
thing is tailing off.

52
00:02:49,780 --> 00:02:52,760
At x=0 this is one.

53
00:02:52,760 --> 00:02:55,700
It comes in with a minus
because that's the lower limit.

54
00:02:55,700 --> 00:03:04,330
So it was 1/(a+ik)
for the first half.

55
00:03:04,330 --> 00:03:07,550
And we were not
surprised to see this 1/k

56
00:03:07,550 --> 00:03:11,800
because the first
half all by itself

57
00:03:11,800 --> 00:03:15,140
has that jump, from zero to one.

58
00:03:15,140 --> 00:03:20,330
So we see that jump
reflected in slow decay.

59
00:03:20,330 --> 00:03:24,990
Alright, but now
I'm making it even.

60
00:03:24,990 --> 00:03:30,730
What are you going to guess for
the rate of decay of f hat of k

61
00:03:30,730 --> 00:03:32,790
for this function?

62
00:03:32,790 --> 00:03:37,290
This function no
longer has a jump.

63
00:03:37,290 --> 00:03:43,380
But it does have - I don't know
were we saying ramp, or corner?

64
00:03:43,380 --> 00:03:48,640
This is not a smooth point here,
because the derivative going up

65
00:03:48,640 --> 00:03:53,470
is plus a, so I'll just
put a circle right,

66
00:03:53,470 --> 00:03:58,320
the derivative is a*e^(ax),
and at x=0 that would be plus

67
00:03:58,320 --> 00:03:59,710
a going up.

68
00:03:59,710 --> 00:04:04,580
And here the derivative
is -a*e^(-ax).

69
00:04:04,580 --> 00:04:09,480
Put in x=0, and the derivative
coming down is minus x.

70
00:04:09,480 --> 00:04:12,160
So there's a jump
in the derivative.

71
00:04:12,160 --> 00:04:15,670
So what, just before
we see it, what

72
00:04:15,670 --> 00:04:23,070
will you expect for the rate
of decay of the transform?

73
00:04:23,070 --> 00:04:25,410
One over k to what power, now?

74
00:04:25,410 --> 00:04:30,190
So it didn't have a
jump, a jump was 1/k.

75
00:04:30,190 --> 00:04:32,090
This has a jump
in the derivative,

76
00:04:32,090 --> 00:04:36,510
so we're expecting one
over k squared. k squared,

77
00:04:36,510 --> 00:04:38,810
it'll be one order smoother.

78
00:04:38,810 --> 00:04:41,770
OK, you can easily
see that happen.

79
00:04:41,770 --> 00:04:54,080
Because this part, well
this is just e^((a-ik)x),

80
00:04:54,080 --> 00:04:59,070
which I'm going to integrate
to get this thing over a-ik.

81
00:04:59,070 --> 00:05:03,150
And I'm going to plug in the
limits minus infinity and zero.

82
00:05:03,150 --> 00:05:05,590
And at minus infinity
I'll get nothing,

83
00:05:05,590 --> 00:05:10,080
this e^(ax) at minus
infinity will be zero.

84
00:05:10,080 --> 00:05:12,460
Where the function starts.

85
00:05:12,460 --> 00:05:13,780
Way down at zero.

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00:05:13,780 --> 00:05:16,950
So and at x=0, this is a one.

87
00:05:16,950 --> 00:05:20,580
So I just get 1/(a+ik).

88
00:05:20,580 --> 00:05:24,370


89
00:05:24,370 --> 00:05:26,580
Over a minus, thank you.

90
00:05:26,580 --> 00:05:31,290
Right, over a minus.

91
00:05:31,290 --> 00:05:34,800
And somehow it
can't be an accident

92
00:05:34,800 --> 00:05:38,640
that this is the complex
conjugate of that somehow.

93
00:05:38,640 --> 00:05:39,980
That's not a surprise.

94
00:05:39,980 --> 00:05:46,150
OK, so let's put those
together into a single fraction

95
00:05:46,150 --> 00:05:47,880
and see what we have.

96
00:05:47,880 --> 00:05:50,030
So the denominator
of that fraction

97
00:05:50,030 --> 00:05:54,020
will be this times this.

98
00:05:54,020 --> 00:05:56,530
And that's the most
basic multiplication

99
00:05:56,530 --> 00:05:59,430
of complex numbers.

100
00:05:59,430 --> 00:06:02,490
That one times its
conjugate gives me what?

101
00:06:02,490 --> 00:06:05,440
It gives me an a squared.

102
00:06:05,440 --> 00:06:08,090
And what else?

103
00:06:08,090 --> 00:06:11,530
A plus k squared
because i times minus i

104
00:06:11,530 --> 00:06:18,900
is plus k squared,
and no imaginary part.

105
00:06:18,900 --> 00:06:22,650
There's a plus i k
a and a minus i k a,

106
00:06:22,650 --> 00:06:25,640
all we're seeing here
is the sum of squares.

107
00:06:25,640 --> 00:06:29,550
The usual z times z bar.

108
00:06:29,550 --> 00:06:32,470
And in the numerator, let's see.

109
00:06:32,470 --> 00:06:35,170
When I put it over
this-- So this

110
00:06:35,170 --> 00:06:37,600
was putting it over
this common denominator,

111
00:06:37,600 --> 00:06:43,520
so I should have an a-ik
going up on top there.

112
00:06:43,520 --> 00:06:48,750
And an a+ik going
up on top here.

113
00:06:48,750 --> 00:06:49,250
Right?

114
00:06:49,250 --> 00:06:50,670
Those are my two fractions.

115
00:06:50,670 --> 00:06:53,090
That over this,
and that over this.

116
00:06:53,090 --> 00:06:57,900
And now that numerator
simplifies, oh look it's great.

117
00:06:57,900 --> 00:07:02,160
I'm getting a real answer.

118
00:07:02,160 --> 00:07:05,270
And because the minus
ik and the plus ik

119
00:07:05,270 --> 00:07:08,330
cancel, and it's just the 2a.

120
00:07:08,330 --> 00:07:11,380
And probably no
surprise that somehow

121
00:07:11,380 --> 00:07:14,610
that that's the jump in slope.

122
00:07:14,610 --> 00:07:17,440
That must have something
to do with that 2a.

123
00:07:17,440 --> 00:07:27,460
So we got a real even, Fourier--
f hat from my real even f.

124
00:07:27,460 --> 00:07:30,830
And it decays like k squared.

125
00:07:30,830 --> 00:07:33,950
OK, so that's
another good example.

126
00:07:33,950 --> 00:07:35,580
A very useful example.

127
00:07:35,580 --> 00:07:36,080
Right.

128
00:07:36,080 --> 00:07:41,480
I could add other examples.

129
00:07:41,480 --> 00:07:48,970
One quite, before I use
that in application,

130
00:07:48,970 --> 00:07:52,510
let's do just a
few more examples.

131
00:07:52,510 --> 00:07:56,780
Suppose f(x) is
the delta function.

132
00:07:56,780 --> 00:08:00,590
What's f hat of k?

133
00:08:00,590 --> 00:08:07,120
Can you just plug in
f(x)=delta(x) here?

134
00:08:07,120 --> 00:08:11,770
Do that integration, and what
does f hat of k come out to be?

135
00:08:11,770 --> 00:08:12,660
One.

136
00:08:12,660 --> 00:08:16,480
Because if it's a delta
function there, at x,

137
00:08:16,480 --> 00:08:21,650
at x=0 the spike is at x=0,
so I plug in x=0, I get one.

138
00:08:21,650 --> 00:08:26,040
So we're kind of,
we'd be surprised

139
00:08:26,040 --> 00:08:28,850
if it wasn't a constant, right?

140
00:08:28,850 --> 00:08:32,370
A delta function
in physical space

141
00:08:32,370 --> 00:08:35,850
goes in-- has all
frequencies in equal amounts.

142
00:08:35,850 --> 00:08:39,930
And it's a constant
in frequency space.

143
00:08:39,930 --> 00:08:44,360
Then there's one more that
takes a little trick to do,

144
00:08:44,360 --> 00:08:49,730
but it's a very neat one.
f(x) is e to the minus

145
00:08:49,730 --> 00:08:53,780
x squared over two.

146
00:08:53,780 --> 00:08:58,180
Do you recognize that as
an important function,

147
00:08:58,180 --> 00:09:01,120
e to the minus x
squared or it usually

148
00:09:01,120 --> 00:09:03,510
has that e to the minus
x squared over two

149
00:09:03,510 --> 00:09:07,390
or sometimes an e to the
minus x squared over two sigma

150
00:09:07,390 --> 00:09:09,320
squared, a rescaling?

151
00:09:09,320 --> 00:09:15,400
But this would be the
bell-shaped curve.

152
00:09:15,400 --> 00:09:16,570
The bell-shaped curve.

153
00:09:16,570 --> 00:09:19,890
It decays very quickly.

154
00:09:19,890 --> 00:09:24,730
The variance, because
that's a two and not

155
00:09:24,730 --> 00:09:29,950
a two sigma squared, the
standard deviation is one here.

156
00:09:29,950 --> 00:09:31,300
The variance is one.

157
00:09:31,300 --> 00:09:34,990
So it's a bell shaped
curve that has about 2/3

158
00:09:34,990 --> 00:09:37,640
of its area between
minus one and one.

159
00:09:37,640 --> 00:09:43,090
This is the all important
function for probability.

160
00:09:43,090 --> 00:09:47,340
The normal distribution, the
Gaussian, both of those words

161
00:09:47,340 --> 00:09:51,300
are used, it's the most
important probability

162
00:09:51,300 --> 00:09:52,940
distribution.

163
00:09:52,940 --> 00:09:56,010
I need a one over
square root of 2pi

164
00:09:56,010 --> 00:09:59,610
to make the total
probability be one.

165
00:09:59,610 --> 00:10:02,060
But let me just leave it there.

166
00:10:02,060 --> 00:10:04,570
That's a very, very
important function.

167
00:10:04,570 --> 00:10:09,500
It's also going to be
important in the heat equation.

168
00:10:09,500 --> 00:10:13,760
In math finance, shows
up all over the place.

169
00:10:13,760 --> 00:10:19,920
And its integral would not be
easy to do from zero to one.

170
00:10:19,920 --> 00:10:22,820
The integral of that
function, from zero to one,

171
00:10:22,820 --> 00:10:24,700
we have tables of it.

172
00:10:24,700 --> 00:10:26,370
To the nth place.

173
00:10:26,370 --> 00:10:33,810
But so there's no simple,
elementary function

174
00:10:33,810 --> 00:10:35,670
whose derivative is this.

175
00:10:35,670 --> 00:10:39,760
That x squared is what's
making the integral tricky.

176
00:10:39,760 --> 00:10:44,480
So from zero to one, we
just have to give it a name.

177
00:10:44,480 --> 00:10:46,120
So, error function.

178
00:10:46,120 --> 00:10:48,530
This would be E-R-F,
error function,

179
00:10:48,530 --> 00:10:52,210
the integral of that thing
correctly normalized.

180
00:10:52,210 --> 00:10:57,640
I'm just saying, important,
important function.

181
00:10:57,640 --> 00:11:02,030
And it turns out that integrals
from minus infinity to infinity

182
00:11:02,030 --> 00:11:03,590
can be done.

183
00:11:03,590 --> 00:11:06,070
So, beautifully,
by some trickery,

184
00:11:06,070 --> 00:11:09,500
we can find the
transform of this.

185
00:11:09,500 --> 00:11:11,240
We can find the
transform of this,

186
00:11:11,240 --> 00:11:14,470
we can do this integral from
minus infinity to infinity,

187
00:11:14,470 --> 00:11:16,870
where we could not do
it from zero to one.

188
00:11:16,870 --> 00:11:22,970
So I'll just write down
the answer for this guy.

189
00:11:22,970 --> 00:11:25,980
Only because it's
such a key example.

190
00:11:25,980 --> 00:11:29,855
It's some constant that
involves 2pi times e

191
00:11:29,855 --> 00:11:34,880
to the minus k squared over two.

192
00:11:34,880 --> 00:11:36,950
Boy, that's pretty amazing.

193
00:11:36,950 --> 00:11:44,320
Right, the Fourier integral
transform, f hat of k,

194
00:11:44,320 --> 00:11:48,640
has the same form
as the function.

195
00:11:48,640 --> 00:11:53,120
And of course this function
is infinitely smooth.

196
00:11:53,120 --> 00:11:58,020
So its transform
decays infinitely fast.

197
00:11:58,020 --> 00:12:04,160
Yeah, there's no problems
like one over k squared here,

198
00:12:04,160 --> 00:12:10,230
there are no bumps in
the bell shaped curve.

199
00:12:10,230 --> 00:12:15,230
So I won't push that
example except I'll use it.

200
00:12:15,230 --> 00:12:18,150
What else should I
say just to, like,

201
00:12:18,150 --> 00:12:22,490
emphasize that this is such
an important distribution

202
00:12:22,490 --> 00:12:23,520
in probability?

203
00:12:23,520 --> 00:12:25,270
Why is it important
in probability?

204
00:12:25,270 --> 00:12:26,870
That's the question.

205
00:12:26,870 --> 00:12:32,230
Why does everybody assume,
if you can get away with it

206
00:12:32,230 --> 00:12:39,720
and don't have any natural
alternative, everybody

207
00:12:39,720 --> 00:12:42,240
assumes that noise,
whatever, is coming

208
00:12:42,240 --> 00:12:44,470
with a normal distribution.

209
00:12:44,470 --> 00:12:52,600
So, in other words, with
a sigma squared in there.

210
00:12:52,600 --> 00:12:55,500
So a normal distribution,
that has mean

211
00:12:55,500 --> 00:12:58,600
zero because it's absolutely
centered at the origin

212
00:12:58,600 --> 00:13:02,860
and it has variance one, but
I could change the variance

213
00:13:02,860 --> 00:13:07,060
and that would just
spread out or tighten

214
00:13:07,060 --> 00:13:08,810
the bell-shaped curve.

215
00:13:08,810 --> 00:13:12,020
Why is the bell-shaped
curve so important?

216
00:13:12,020 --> 00:13:14,495
That's certainly,
we're not going

217
00:13:14,495 --> 00:13:17,910
to launch into
theory of probability

218
00:13:17,910 --> 00:13:20,480
but it's the central
limit theorem.

219
00:13:20,480 --> 00:13:22,450
So let me just use those words.

220
00:13:22,450 --> 00:13:25,470
The central limit
theorem that says

221
00:13:25,470 --> 00:13:28,310
that if I start with other
probability distributions,

222
00:13:28,310 --> 00:13:30,800
like I'm flipping a coin.

223
00:13:30,800 --> 00:13:34,270
I flip a coin a million times.

224
00:13:34,270 --> 00:13:39,530
Then the mean, and let's say
zero for tails, one for heads.

225
00:13:39,530 --> 00:13:42,430
OK, so I flip, flip, flip.

226
00:13:42,430 --> 00:13:47,930
Well, the mean of
that, the expected mean

227
00:13:47,930 --> 00:13:50,720
is what, half a million, right?

228
00:13:50,720 --> 00:13:53,180
Half tails, half heads.

229
00:13:53,180 --> 00:13:55,720
So if I give zero for
tails, one for heads

230
00:13:55,720 --> 00:13:57,230
and flip a million
times the mean

231
00:13:57,230 --> 00:13:58,950
would be about half a million.

232
00:13:58,950 --> 00:14:04,410
And then, so let
me center the mean.

233
00:14:04,410 --> 00:14:07,360
I could have centered it by
taking minus one and one.

234
00:14:07,360 --> 00:14:08,660
That would have been smarter.

235
00:14:08,660 --> 00:14:11,380
Minus one and one.

236
00:14:11,380 --> 00:14:13,360
Minus one for
tails, one for heads

237
00:14:13,360 --> 00:14:16,070
would have had a mean of zero.

238
00:14:16,070 --> 00:14:19,840
And then it would be
natural if I have a million

239
00:14:19,840 --> 00:14:25,620
of these to divide by
a thousand, I think.

240
00:14:25,620 --> 00:14:27,450
Of course, the answer
won't be zero, right?

241
00:14:27,450 --> 00:14:29,350
If I do a million
flips it's not going

242
00:14:29,350 --> 00:14:34,170
to come out exactly half a
million and half a million.

243
00:14:34,170 --> 00:14:35,030
I'm remembering.

244
00:14:35,030 --> 00:14:41,220
I used to have a long discussion
with a nice guy in college.

245
00:14:41,220 --> 00:14:45,100
He ran for Mayor of
Boston, actually.

246
00:14:45,100 --> 00:14:49,600
But he had the idea that
after a million flips,

247
00:14:49,600 --> 00:14:56,520
suppose there had been
more heads than tails.

248
00:14:56,520 --> 00:15:03,210
Then the next flip, he figured,
was more likely to be tails.

249
00:15:03,210 --> 00:15:06,770
I couldn't convince them that
this was not mathematically

250
00:15:06,770 --> 00:15:09,520
the right thing to think about.

251
00:15:09,520 --> 00:15:12,730
And all I did was say
don't go to Las Vegas.

252
00:15:12,730 --> 00:15:16,790
I mean, if you're thinking
that way, save your money.

253
00:15:16,790 --> 00:15:18,950
So, anyway.

254
00:15:18,950 --> 00:15:20,940
But this is much studied.

255
00:15:20,940 --> 00:15:23,970
The variation, what
that curve looks like,

256
00:15:23,970 --> 00:15:26,020
that's quite interesting.

257
00:15:26,020 --> 00:15:30,790
But my point is, that as the
number gets bigger and bigger

258
00:15:30,790 --> 00:15:34,650
and we scale it properly,
the distribution

259
00:15:34,650 --> 00:15:38,150
will approach the normal.

260
00:15:38,150 --> 00:15:39,710
All sorts of distributions.

261
00:15:39,710 --> 00:15:44,360
If I just repeat and repeat
experiments and scale it,

262
00:15:44,360 --> 00:15:47,710
the central limit theorem
says you're always going

263
00:15:47,710 --> 00:15:49,340
to the normal distribution.

264
00:15:49,340 --> 00:15:51,830
So that's highly important.

265
00:15:51,830 --> 00:15:54,680
OK, and it comes up
different places.

266
00:15:54,680 --> 00:15:57,130
And it's quite a neat function.

267
00:15:57,130 --> 00:16:02,080
OK, so that's some examples.

268
00:16:02,080 --> 00:16:06,230
Now, let me use, like every
topics that I introduce,

269
00:16:06,230 --> 00:16:08,390
I want to find a use for.

270
00:16:08,390 --> 00:16:11,990
So now, can I start on this one?

271
00:16:11,990 --> 00:16:14,061
Constant coefficient
differential equations.

272
00:16:14,061 --> 00:16:16,310
I'm going to write down a
differential equation, which

273
00:16:16,310 --> 00:16:19,410
will look pretty
much like the ones we

274
00:16:19,410 --> 00:16:22,520
started this course with.

275
00:16:22,520 --> 00:16:27,470
And I could, well,
let me write it down.

276
00:16:27,470 --> 00:16:31,920
Minus d second u dx
squared, we're used to that.

277
00:16:31,920 --> 00:16:35,350
Now let me put in
an a squared u.

278
00:16:35,350 --> 00:16:39,330
Which is a lower lower order
term, we could deal with that.

279
00:16:39,330 --> 00:16:43,070
Equals some f(x).

280
00:16:43,070 --> 00:16:46,850
And now, because I want
to do Fourier integrals,

281
00:16:46,850 --> 00:16:50,670
I'm thinking all x.

282
00:16:50,670 --> 00:16:51,670
We're on the whole line.

283
00:16:51,670 --> 00:16:54,580
Instead of the
interval (0,1) where

284
00:16:54,580 --> 00:16:56,990
I might use Fourier
series and have

285
00:16:56,990 --> 00:17:00,580
sine series or cosine series,
depending on the boundary

286
00:17:00,580 --> 00:17:01,500
conditions.

287
00:17:01,500 --> 00:17:05,250
Here, the boundary condition
is just everything drops off

288
00:17:05,250 --> 00:17:06,420
at infinity.

289
00:17:06,420 --> 00:17:07,730
And minus infinity.

290
00:17:07,730 --> 00:17:13,180
So all these functions,
we can do these integrals.

291
00:17:13,180 --> 00:17:15,140
OK, so there's a good question.

292
00:17:15,140 --> 00:17:19,230
What's the solution?

293
00:17:19,230 --> 00:17:28,350
We could tackle it, but I want
to suggest to use Fourier.

294
00:17:28,350 --> 00:17:31,550
So it's not the only way,
but it's one way to see it.

295
00:17:31,550 --> 00:17:34,970
So now if I use, what do
I mean by using Fourier?

296
00:17:34,970 --> 00:17:38,050
It means I'm going to take
the Fourier integral transform

297
00:17:38,050 --> 00:17:41,030
of every term.

298
00:17:41,030 --> 00:17:43,080
So when I take the
Fourier transform

299
00:17:43,080 --> 00:17:45,470
of the right-hand side, I'm
going to get f hat of k,

300
00:17:45,470 --> 00:17:46,220
whatever.

301
00:17:46,220 --> 00:17:47,960
This is known, of course.

302
00:17:47,960 --> 00:17:50,360
This guy is given.

303
00:17:50,360 --> 00:17:51,820
That's the source term.

304
00:17:51,820 --> 00:17:54,210
And u is the unknown.

305
00:17:54,210 --> 00:17:58,430
OK, so I'm going
to take the Fourier

306
00:17:58,430 --> 00:18:02,710
transform of every term, well
this is, a is a constant. a

307
00:18:02,710 --> 00:18:07,370
had to be a constant,
or I couldn't do,

308
00:18:07,370 --> 00:18:09,680
you know if a depended
on x this would

309
00:18:09,680 --> 00:18:17,200
be some multiplication and
the transform would be a mess.

310
00:18:17,200 --> 00:18:20,630
Fourier applies when you've
got constant coefficients

311
00:18:20,630 --> 00:18:22,520
and nice boundary conditions.

312
00:18:22,520 --> 00:18:24,750
And here our boundary
conditions are nice,

313
00:18:24,750 --> 00:18:27,620
they just go to zero fast.

314
00:18:27,620 --> 00:18:34,730
OK, so the transform of this is,
that's a constant. u hat of k.

315
00:18:34,730 --> 00:18:37,960
And what's the
transform of that?

316
00:18:37,960 --> 00:18:40,320
So this is our chance
to use probably

317
00:18:40,320 --> 00:18:49,070
the most important rule
for Fourier integrals.

318
00:18:49,070 --> 00:18:52,470
Maybe you'll tell me what it is.

319
00:18:52,470 --> 00:18:53,970
You should think what it is.

320
00:18:53,970 --> 00:18:56,380
If I take a derivative,
that's the rule.

321
00:18:56,380 --> 00:19:00,060
If I take a derivative
of the function,

322
00:19:00,060 --> 00:19:03,720
what's happening in frequencies?

323
00:19:03,720 --> 00:19:05,710
I could make that happen here.

324
00:19:05,710 --> 00:19:09,570
If I took the derivative, yeah.

325
00:19:09,570 --> 00:19:12,710
So maybe if I take
the derivative here,

326
00:19:12,710 --> 00:19:15,490
so here it's just
remembering the rule.

327
00:19:15,490 --> 00:19:20,080
Suppose I take the
derivative of this equation.

328
00:19:20,080 --> 00:19:23,570
I get this integral,
and what would f' be?

329
00:19:23,570 --> 00:19:25,850
What would f hat, sorry.

330
00:19:25,850 --> 00:19:29,170
If I take the x
derivative of this,

331
00:19:29,170 --> 00:19:31,600
if I take the x derivative
of this equation,

332
00:19:31,600 --> 00:19:33,200
what happens on
the right-hand side

333
00:19:33,200 --> 00:19:36,010
when I take the x derivative?

334
00:19:36,010 --> 00:19:38,060
Down comes ik.

335
00:19:38,060 --> 00:19:40,280
Down comes ik.

336
00:19:40,280 --> 00:19:42,560
So when ik is coming
down, I won't even

337
00:19:42,560 --> 00:19:44,540
finish that equation.

338
00:19:44,540 --> 00:19:49,060
And ik is coming down,
when I take the derivative.

339
00:19:49,060 --> 00:19:52,490
So the derivative, the
transform, is multiplied by ik,

340
00:19:52,490 --> 00:19:55,760
higher frequencies
are emphasized now

341
00:19:55,760 --> 00:19:57,890
because of that k factor.

342
00:19:57,890 --> 00:20:01,780
And now if I take two
derivatives, I bring ik twice.

343
00:20:01,780 --> 00:20:04,690
So that's i squared k
squared, the i squared

344
00:20:04,690 --> 00:20:07,030
and the minus give me a plus.

345
00:20:07,030 --> 00:20:12,290
So that's just k
squared. u hat, of k.

346
00:20:12,290 --> 00:20:16,000
OK with that?

347
00:20:16,000 --> 00:20:19,730
And now, we get an
immediate formula

348
00:20:19,730 --> 00:20:25,550
for u hat of k, the solution.

349
00:20:25,550 --> 00:20:28,550
Well, it's the solution but
it's in frequency space.

350
00:20:28,550 --> 00:20:31,040
If we wanted to know it
in x space, as we do,

351
00:20:31,040 --> 00:20:33,090
we've got to transform back.

352
00:20:33,090 --> 00:20:34,620
But what do we get here?

353
00:20:34,620 --> 00:20:38,590
It's just f hat of k.

354
00:20:38,590 --> 00:20:42,740
Divided by, this is just
multiplied by a squared plus k

355
00:20:42,740 --> 00:20:47,590
squared.

356
00:20:47,590 --> 00:20:52,890
OK, so that's the answer.

357
00:20:52,890 --> 00:20:55,010
In frequency space.

358
00:20:55,010 --> 00:20:56,380
That was simple.

359
00:20:56,380 --> 00:21:00,080
And then if I wanted
it in x space,

360
00:21:00,080 --> 00:21:02,700
I take the reverse transform.

361
00:21:02,700 --> 00:21:08,010
Notice that this is, hidden
here are the same three steps

362
00:21:08,010 --> 00:21:12,720
that I emphasize all the
time about using eigenvectors

363
00:21:12,720 --> 00:21:13,830
and eigenvalues.

364
00:21:13,830 --> 00:21:15,580
Do you remember
those three steps

365
00:21:15,580 --> 00:21:17,490
for solving
differential equations?

366
00:21:17,490 --> 00:21:20,970
Difference equations,
linear equations, whatever?

367
00:21:20,970 --> 00:21:25,330
The three steps were,
find the coefficients,

368
00:21:25,330 --> 00:21:29,010
expand everything
in eigenfunctions.

369
00:21:29,010 --> 00:21:31,190
I won't write, I'll talk.

370
00:21:31,190 --> 00:21:34,870
The three steps were
expand in eigenfunctions,

371
00:21:34,870 --> 00:21:37,510
follow each eigenfunction
function separately,

372
00:21:37,510 --> 00:21:40,140
that was the trivial step
with just a division,

373
00:21:40,140 --> 00:21:41,750
like this division.

374
00:21:41,750 --> 00:21:48,500
And then use those coefficients
of the eigenfunctions,

375
00:21:48,500 --> 00:21:51,450
combine them all back
to get the answer.

376
00:21:51,450 --> 00:21:51,950
Right?

377
00:21:51,950 --> 00:21:55,730
Step one: write it in the
right basis, step two:

378
00:21:55,730 --> 00:21:57,510
easy in that basis.

379
00:21:57,510 --> 00:22:02,180
Step three: go back to
your physical space.

380
00:22:02,180 --> 00:22:04,540
We're doing exactly
the same thing here.

381
00:22:04,540 --> 00:22:09,840
These e^(ikx)'s are the
eigenfunctions of this thing.

382
00:22:09,840 --> 00:22:11,090
They're the eigenfunctions.

383
00:22:11,090 --> 00:22:14,120
And here the eigenvalue
of this stuff

384
00:22:14,120 --> 00:22:16,850
is k squared plus a squared.

385
00:22:16,850 --> 00:22:19,110
And that's what we divided by.

386
00:22:19,110 --> 00:22:22,990
And then the final job
of going back from u hat

387
00:22:22,990 --> 00:22:28,420
to u, by-- So write u there.

388
00:22:28,420 --> 00:22:33,640
Can I do, this f,
now it's really u

389
00:22:33,640 --> 00:22:39,090
that I'm wanting to bring
back to physical space.

390
00:22:39,090 --> 00:22:44,530
So just for the sake
of your eyes seeing it,

391
00:22:44,530 --> 00:22:47,970
let me put a u in.

392
00:22:47,970 --> 00:22:54,070
So that's the answer in a way.

393
00:22:54,070 --> 00:22:57,890
It's the answer, it's a
formula for the answer.

394
00:22:57,890 --> 00:23:03,220
It did depend on our being
able to do two integrals.

395
00:23:03,220 --> 00:23:07,640
The integral from f to f
hat may not have been easy,

396
00:23:07,640 --> 00:23:11,500
and then the integral from u
hat back to u, this integral,

397
00:23:11,500 --> 00:23:12,590
might not have been easy.

398
00:23:12,590 --> 00:23:14,290
So it's a formula.

399
00:23:14,290 --> 00:23:18,680
OK, now I want to go
with it a little longer.

400
00:23:18,680 --> 00:23:23,200
Because I want to show you how
the delta function pays off.

401
00:23:23,200 --> 00:23:29,040
So let me do the example where
f(x) is the delta function.

402
00:23:29,040 --> 00:23:35,210
So now we're really close
to where this course began.

403
00:23:35,210 --> 00:23:38,740
Differential equation
with a delta function.

404
00:23:38,740 --> 00:23:41,480
The only new thing is,
we're not on an interval,

405
00:23:41,480 --> 00:23:43,950
we're on the whole line.

406
00:23:43,950 --> 00:23:47,370
So I take transforms, so
what's the transform now

407
00:23:47,370 --> 00:23:51,850
of this specific f(x) is?

408
00:23:51,850 --> 00:23:52,570
One.

409
00:23:52,570 --> 00:23:54,220
We just saw.

410
00:23:54,220 --> 00:23:57,520
OK, so now we get a one there.

411
00:23:57,520 --> 00:23:59,710
Now we just divide by here.

412
00:23:59,710 --> 00:24:04,220
And we've got a one here.

413
00:24:04,220 --> 00:24:07,120
So we were able to go,
this was an integral

414
00:24:07,120 --> 00:24:12,280
we could easily do, to get
from delta to delta hat, which

415
00:24:12,280 --> 00:24:13,750
was just one.

416
00:24:13,750 --> 00:24:19,970
And fantastically, this is an
integral to go back to u(x),

417
00:24:19,970 --> 00:24:27,090
to go back to u(x), that's
an integral we can do.

418
00:24:27,090 --> 00:24:29,920
Well, you may ask
how can we do it.

419
00:24:29,920 --> 00:24:33,730
How do I find the u(x)
that has this transform?

420
00:24:33,730 --> 00:24:40,720
Well, I either use complex
variables to do integrals

421
00:24:40,720 --> 00:24:46,060
like this, residue methods
that are in Chapter 5,

422
00:24:46,060 --> 00:24:48,780
or I look in a table.

423
00:24:48,780 --> 00:24:50,890
Or I look at the
blackboard over there.

424
00:24:50,890 --> 00:24:52,700
I think that's the best way.

425
00:24:52,700 --> 00:24:54,451
Look at this blackboard.

426
00:24:54,451 --> 00:24:54,950
Right?

427
00:24:54,950 --> 00:24:58,570
Because this is
the answer we got.

428
00:24:58,570 --> 00:25:03,300
We got that same answer apart
from a constant factor 2a.

429
00:25:03,300 --> 00:25:05,130
So this is our function.

430
00:25:05,130 --> 00:25:08,190
This is our solution,
u(x) is this.

431
00:25:08,190 --> 00:25:10,130
What am I going to call that?

432
00:25:10,130 --> 00:25:12,570
Two-sided pulse?

433
00:25:12,570 --> 00:25:15,350
I'll call that the
two-sided pulse?

434
00:25:15,350 --> 00:25:19,870
Maybe I should give it a
name but I'll just write out

435
00:25:19,870 --> 00:25:21,840
those words two-sided pulse.

436
00:25:21,840 --> 00:25:25,550
That's it, divided by 2a.

437
00:25:25,550 --> 00:25:28,080
So we've got the answer.

438
00:25:28,080 --> 00:25:30,900
Let me just make a
little more space here.

439
00:25:30,900 --> 00:25:35,380
This was one over a
squared plus k squared,

440
00:25:35,380 --> 00:25:38,980
and now having seen that
already, I just say yep,

441
00:25:38,980 --> 00:25:40,150
that must be it.

442
00:25:40,150 --> 00:25:49,560
It's the two-sided pulse,
and I have to divide by 2a.

443
00:25:49,560 --> 00:25:53,140
Do you see that that's
the correct answer?

444
00:25:53,140 --> 00:25:57,570
We can substitute that in the
equation and see that it works.

445
00:25:57,570 --> 00:25:59,960
I mean, so we have
solved the problem.

446
00:25:59,960 --> 00:26:04,000
We have solved the problem
when the right side was delta.

447
00:26:04,000 --> 00:26:05,970
Let's put it into the equation.

448
00:26:05,970 --> 00:26:07,970
So this is just
because we did this,

449
00:26:07,970 --> 00:26:10,960
it's nice to do it again
after all this time.

450
00:26:10,960 --> 00:26:13,050
So I put it in the equation.

451
00:26:13,050 --> 00:26:15,850
This two-sided pulse over 2a.

452
00:26:15,850 --> 00:26:18,600
So what's my equation?

453
00:26:18,600 --> 00:26:22,250
Well, this is zero
most of the time.

454
00:26:22,250 --> 00:26:26,710
So I believe that if I
plug in this function,

455
00:26:26,710 --> 00:26:29,429
it will give me the zero.

456
00:26:29,429 --> 00:26:30,720
Do you want to just plug it in?

457
00:26:30,720 --> 00:26:36,060
I believe that if I plug in
a^(-x), or a^x, either one,

458
00:26:36,060 --> 00:26:41,090
can I just check that
u-- Try u=e^(-ax).

459
00:26:41,090 --> 00:26:44,590
Put it in and just
see that I get zero.

460
00:26:44,590 --> 00:26:47,380
Because yeah, two
derivatives bring down

461
00:26:47,380 --> 00:26:49,590
a squared with a minus.

462
00:26:49,590 --> 00:26:51,780
And there's a
squared with a plus.

463
00:26:51,780 --> 00:26:53,700
It works, right?

464
00:26:53,700 --> 00:26:59,620
Two derivatives of this function
bring down minus a twice,

465
00:26:59,620 --> 00:27:01,440
so that's a squared.

466
00:27:01,440 --> 00:27:04,160
So it's minus a
squared, plus a squared.

467
00:27:04,160 --> 00:27:05,750
Works.

468
00:27:05,750 --> 00:27:10,480
And then, of course, the
important point is x=0 where

469
00:27:10,480 --> 00:27:12,760
the spike is.

470
00:27:12,760 --> 00:27:18,470
What happens at the
spike, going back

471
00:27:18,470 --> 00:27:20,560
to the beginning of the course?

472
00:27:20,560 --> 00:27:23,250
This term is going
to be unimportant

473
00:27:23,250 --> 00:27:25,370
compared to this term.

474
00:27:25,370 --> 00:27:28,300
What do I see?

475
00:27:28,300 --> 00:27:34,630
With -u'' equal a spike, what
was the solution to that?

476
00:27:34,630 --> 00:27:36,690
u had a corner, right?

477
00:27:36,690 --> 00:27:41,100
The slope of u, what
did the slope of u do?

478
00:27:41,100 --> 00:27:43,630
It dropped by one,
was that right?

479
00:27:43,630 --> 00:27:46,140
The slope of u dropped by one.

480
00:27:46,140 --> 00:27:50,570
We used to have corners
going up and down

481
00:27:50,570 --> 00:27:53,540
and the difference between
the slopes was one.

482
00:27:53,540 --> 00:27:58,300
And here, the difference
between the slopes, ah, look.

483
00:27:58,300 --> 00:28:09,790
The slope has dropped by 2a,
and when we divided by the 2a,

484
00:28:09,790 --> 00:28:12,700
it was just right.

485
00:28:12,700 --> 00:28:18,450
And now, when I divide by the
2a, this has a slope of 1/2.

486
00:28:18,450 --> 00:28:21,670
This has a slope of minus
1/2, the drop is one.

487
00:28:21,670 --> 00:28:23,190
And we're right.

488
00:28:23,190 --> 00:28:25,520
It solves the equation.

489
00:28:25,520 --> 00:28:27,460
Nobody doubted that.

490
00:28:27,460 --> 00:28:29,500
OK, so that's great.

491
00:28:29,500 --> 00:28:34,820
We have found the
solution to this equation,

492
00:28:34,820 --> 00:28:39,050
when the right side is delta.

493
00:28:39,050 --> 00:28:40,420
Good.

494
00:28:40,420 --> 00:28:43,310
Now, can I ask you do
you remember the name?

495
00:28:43,310 --> 00:28:48,710
There's a special name for the
solution when the right side is

496
00:28:48,710 --> 00:28:51,110
a delta function.

497
00:28:51,110 --> 00:28:53,590
Whose name is
associated with that?

498
00:28:53,590 --> 00:28:56,085
So that this
particular u, I'm going

499
00:28:56,085 --> 00:28:57,840
to give it another letter.

500
00:28:57,840 --> 00:29:03,320
It's the particular
u, the special u

501
00:29:03,320 --> 00:29:06,410
when the right side is
delta, and whose name is

502
00:29:06,410 --> 00:29:09,880
associated with that solution?

503
00:29:09,880 --> 00:29:11,340
Green.

504
00:29:11,340 --> 00:29:13,180
It's the Green's function.

505
00:29:13,180 --> 00:29:14,000
Green's function.

506
00:29:14,000 --> 00:29:15,730
The famous Green's function.

507
00:29:15,730 --> 00:29:19,160
Green's function is just like
an inverse to the problem.

508
00:29:19,160 --> 00:29:22,570
This is like having an identity
on the right-hand side.

509
00:29:22,570 --> 00:29:24,450
It's like there it is.

510
00:29:24,450 --> 00:29:27,850
So let me just use G
for Green's function.

511
00:29:27,850 --> 00:29:31,220
So that's the Fourier transform
of the Green's function,

512
00:29:31,220 --> 00:29:33,880
and this is the
Green's function.

513
00:29:33,880 --> 00:29:35,990
This is the Green's function.

514
00:29:35,990 --> 00:29:41,780
Now I can give it its
name, Green's function,

515
00:29:41,780 --> 00:29:45,570
when I divide by the 2a.

516
00:29:45,570 --> 00:29:51,520
And now the slope is 1/2 going
up, and minus 1/2 coming down.

517
00:29:51,520 --> 00:29:53,600
And it's all right.

518
00:29:53,600 --> 00:29:56,630
OK, so we found the
Green's function.

519
00:29:56,630 --> 00:29:59,640
We found the fundamental
solution to the equation,

520
00:29:59,640 --> 00:30:03,680
and this is it.

521
00:30:03,680 --> 00:30:05,720
It was straight lines, right?

522
00:30:05,720 --> 00:30:12,430
It was straight lines in the
first weeks of the course.

523
00:30:12,430 --> 00:30:15,930
But now there's an
exponential drop-off

524
00:30:15,930 --> 00:30:20,510
caused by this additional term.

525
00:30:20,510 --> 00:30:25,320
OK, good.

526
00:30:25,320 --> 00:30:27,650
So that's straightforward.

527
00:30:27,650 --> 00:30:31,330
Depending on our being
able to recognize or do

528
00:30:31,330 --> 00:30:36,870
the transform back
to the x space.

529
00:30:36,870 --> 00:30:41,120
Now comes the question, what
about the original f(x)?

530
00:30:41,120 --> 00:30:43,640


531
00:30:43,640 --> 00:30:45,760
How can the Green's
function be used?

532
00:30:45,760 --> 00:30:51,340
So you're seeing now, what
use is this Green's function?

533
00:30:51,340 --> 00:30:54,630
With that right-hand side,
when the right-hand side

534
00:30:54,630 --> 00:30:56,260
is something different?

535
00:30:56,260 --> 00:30:58,690
When the right-hand side
is some different f(x)?

536
00:30:58,690 --> 00:31:08,260
So let me go back to
an f(x) on the right.

537
00:31:08,260 --> 00:31:14,520
And then there's an f hat
of k, after the transform.

538
00:31:14,520 --> 00:31:21,190
How can I use the Green's
function for a general source?

539
00:31:21,190 --> 00:31:24,810
The general source
term, a general load?

540
00:31:24,810 --> 00:31:30,780
This is a fundamental idea.

541
00:31:30,780 --> 00:31:33,530
I would say fundamental.

542
00:31:33,530 --> 00:31:37,370
How do you use the
Green's function?

543
00:31:37,370 --> 00:31:39,880
And remember, the
Green's function

544
00:31:39,880 --> 00:31:42,840
is like telling you
the inverse matrix.

545
00:31:42,840 --> 00:31:44,420
So it can't be too hard.

546
00:31:44,420 --> 00:31:47,200
It's like solving a
linear system when

547
00:31:47,200 --> 00:31:52,630
you know the inverse matrix.

548
00:31:52,630 --> 00:31:55,880
So that's the analogy,
but let's just focus

549
00:31:55,880 --> 00:32:00,120
on the particular question.

550
00:32:00,120 --> 00:32:03,670
I think the
intuition, you should

551
00:32:03,670 --> 00:32:07,030
have an intuition for how
the Green's function works.

552
00:32:07,030 --> 00:32:09,750
So the Green's function
was the solution when

553
00:32:09,750 --> 00:32:12,320
the source term was a delta.

554
00:32:12,320 --> 00:32:14,310
And here's the intuition.

555
00:32:14,310 --> 00:32:19,190
It's rough, but it works.

556
00:32:19,190 --> 00:32:25,920
Any source term,
f(x), is in some way

557
00:32:25,920 --> 00:32:29,220
a combination of delta.

558
00:32:29,220 --> 00:32:32,750
If f(x) is a
combination of deltas,

559
00:32:32,750 --> 00:32:37,300
then our answer u(x)
is the same combination

560
00:32:37,300 --> 00:32:39,230
of the Green's function, right?

561
00:32:39,230 --> 00:32:43,940
If the right-hand side
is some combination

562
00:32:43,940 --> 00:32:48,900
of this special delta, then
the solution will be the same.

563
00:32:48,900 --> 00:32:50,580
This is just linearity.

564
00:32:50,580 --> 00:32:55,900
Superposition, whatever short
or long word you like to use.

565
00:32:55,900 --> 00:32:59,550
So if I can make sense
of that statement,

566
00:32:59,550 --> 00:33:08,740
that f(x) is a combination
of deltas, then I'm in.

567
00:33:08,740 --> 00:33:12,780
Now, what do I mean by
a combination of deltas?

568
00:33:12,780 --> 00:33:18,070
I mean, well, those deltas are
going to be shifted deltas.

569
00:33:18,070 --> 00:33:20,630
Obviously the single
delta, delta(x),

570
00:33:20,630 --> 00:33:23,300
is a spike at the origin.

571
00:33:23,300 --> 00:33:26,790
That's only one point.

572
00:33:26,790 --> 00:33:30,770
I want to combine delta
of x and its shifts.

573
00:33:30,770 --> 00:33:34,230
So I'm going to have
to expect to be using

574
00:33:34,230 --> 00:33:37,590
G(x) and its shifts, right?

575
00:33:37,590 --> 00:33:40,390
OK so now I'll just
say this again.

576
00:33:40,390 --> 00:33:43,320
I'm thinking of f(x) as
a combination of delta

577
00:33:43,320 --> 00:33:47,290
and its shifts, and
then the solution u

578
00:33:47,290 --> 00:33:51,740
will be the same combination
of G(x) and its shifts.

579
00:33:51,740 --> 00:33:55,000
So now you just have to
tell me what combination.

580
00:33:55,000 --> 00:34:01,380
What combination of
delta and its shifts?

581
00:34:01,380 --> 00:34:03,100
Maybe you'll allow me.

582
00:34:03,100 --> 00:34:09,640
Let me just do this
maybe on that board.

583
00:34:09,640 --> 00:34:15,820
I just can't help writing
down the discrete case.

584
00:34:15,820 --> 00:34:19,520
So, in the discrete
case, the delta vector

585
00:34:19,520 --> 00:34:22,841
corresponds to something
like [1, 0, 0, 0].

586
00:34:22,841 --> 00:34:23,340
Right?

587
00:34:23,340 --> 00:34:26,040
That was a typical
delta vector with a one

588
00:34:26,040 --> 00:34:28,060
in the zeroth position.

589
00:34:28,060 --> 00:34:33,750
Then its shifts would be [0, 1,
 0, 0], that'd be a shift.

590
00:34:33,750 --> 00:34:37,160
And another shift
would be [0,  0, 1, 0].

591
00:34:37,160 --> 00:34:41,690
And another shift
would be [0,  0, 0, 1].

592
00:34:41,690 --> 00:34:44,800
So now there is the delta
vector and its shifts.

593
00:34:44,800 --> 00:34:46,190
These four guys.

594
00:34:46,190 --> 00:34:50,610
OK, now suppose my f,
my right-hand side,

595
00:34:50,610 --> 00:34:54,520
is [1, 2, 3, 7].

596
00:34:54,520 --> 00:34:59,070
I want to write that as a
combination of those deltas.

597
00:34:59,070 --> 00:35:03,280
This would be in the case when
if I know the solution for each

598
00:35:03,280 --> 00:35:07,330
of these guys, the Green's
function, the inverse matrix.

599
00:35:07,330 --> 00:35:11,600
Everybody sees if I know
the solution to those four,

600
00:35:11,600 --> 00:35:13,260
I know the inverse matrix.

601
00:35:13,260 --> 00:35:14,240
Right?

602
00:35:14,240 --> 00:35:16,880
Because if I can solve with
those four right-hand sides,

603
00:35:16,880 --> 00:35:21,520
those four solutions are the
columns of the inverse matrix.

604
00:35:21,520 --> 00:35:22,050
Right?

605
00:35:22,050 --> 00:35:24,970
You remember that
if I had a matrix A

606
00:35:24,970 --> 00:35:27,040
and I was looking
for its inverse,

607
00:35:27,040 --> 00:35:35,490
I solve A A inverse equal I.
And I is just these four guys.

608
00:35:35,490 --> 00:35:42,510
So A inverse is the solution
from these four guys.

609
00:35:42,510 --> 00:35:44,770
OK, now everybody's
going to tell me

610
00:35:44,770 --> 00:35:51,130
what's the solution for this
right-hand side [1, 2, 3, 7]?

611
00:35:51,130 --> 00:35:56,920
Suppose this guy has
solution, so A inverse,

612
00:35:56,920 --> 00:36:02,410
the columns of A inverse
are this Green's function.

613
00:36:02,410 --> 00:36:05,820
This Green's function
with a shift.

614
00:36:05,820 --> 00:36:10,120
Maybe SG, Green's
function with a shift.

615
00:36:10,120 --> 00:36:14,130
Yeah, S squared G, Green's
function with a double shift.

616
00:36:14,130 --> 00:36:17,490
S cubed G. I'm just
cooking up, I never

617
00:36:17,490 --> 00:36:18,760
used these letters before.

618
00:36:18,760 --> 00:36:20,600
But what's the answer?

619
00:36:20,600 --> 00:36:24,680
Then u is what?

620
00:36:24,680 --> 00:36:28,920
It's one times the
Green's function.

621
00:36:28,920 --> 00:36:34,830
And it's two times
the guy, right?

622
00:36:34,830 --> 00:36:38,670
This f is just,
this f is just, is

623
00:36:38,670 --> 00:36:42,440
one times the Green's function,
two times the shifted.

624
00:36:42,440 --> 00:36:46,130
Three times the double
shifted, and seven times

625
00:36:46,130 --> 00:36:48,620
the triple shift.

626
00:36:48,620 --> 00:36:49,120
Right?

627
00:36:49,120 --> 00:36:55,890
Just taking four minutes to
do something simple because

628
00:36:55,890 --> 00:37:00,050
over there, when I use
the continuous case

629
00:37:00,050 --> 00:37:04,740
it'll look a little strange,
but here it's so easy.

630
00:37:04,740 --> 00:37:08,100
That'll involve integrals,
this involves a sum.

631
00:37:08,100 --> 00:37:09,030
So what is it?

632
00:37:09,030 --> 00:37:14,020
I have G, twice the
shift of G. Three

633
00:37:14,020 --> 00:37:19,170
times the double shift of G,
and seven times the triple shift

634
00:37:19,170 --> 00:37:21,450
of G. Right?

635
00:37:21,450 --> 00:37:29,880
By linearity, by superposition,
if this is my f, this is my u.

636
00:37:29,880 --> 00:37:31,690
Everybody's with me here, right?

637
00:37:31,690 --> 00:37:35,380
That if I take a combination
of f's, the answer is

638
00:37:35,380 --> 00:37:37,960
the corresponding
combination of G's.

639
00:37:37,960 --> 00:37:42,890
OK, good.

640
00:37:42,890 --> 00:37:45,700
I didn't mean that.

641
00:37:45,700 --> 00:37:49,630
That wasn't so great.

642
00:37:49,630 --> 00:37:53,930
That shouldn't
have been called, I

643
00:37:53,930 --> 00:37:56,690
didn't mean to call those
the Green's functions.

644
00:37:56,690 --> 00:37:59,070
I meant to call those
the deltas, right?

645
00:37:59,070 --> 00:38:01,070
And the Green's functions
were the answers.

646
00:38:01,070 --> 00:38:05,040
So this a shifted delta, this
is a doubly shifted delta,

647
00:38:05,040 --> 00:38:07,650
this is a triply
shifted delta, and now

648
00:38:07,650 --> 00:38:14,380
this is one of the delta, two
of the shifted delta, three

649
00:38:14,380 --> 00:38:17,370
of the doubly shifted delta.

650
00:38:17,370 --> 00:38:21,150
The f is a combination
of deltas and its shifts.

651
00:38:21,150 --> 00:38:23,490
The u is a combination
of Green's function,

652
00:38:23,490 --> 00:38:25,470
and its shifts.

653
00:38:25,470 --> 00:38:27,650
Apologies for
making that mistake,

654
00:38:27,650 --> 00:38:31,920
but maybe it's brought
us back to the point.

655
00:38:31,920 --> 00:38:39,220
So the point is, express
your function f, your source,

656
00:38:39,220 --> 00:38:43,610
as a combination of deltas,
just exactly our plan.

657
00:38:43,610 --> 00:38:48,830
Then u is the same combination
of the G's with the same shift.

658
00:38:48,830 --> 00:38:52,260
Alright, now back here.

659
00:38:52,260 --> 00:38:58,440
How do I express
f(x), in what way is

660
00:38:58,440 --> 00:39:03,260
f(x) a combination of deltas?

661
00:39:03,260 --> 00:39:08,310
I mean, slow down just
to see that point.

662
00:39:08,310 --> 00:39:18,970
How much of delta, of the spike
at x=3, how much of delta(x-3),

663
00:39:18,970 --> 00:39:24,630
so that's a spike
at three, right?

664
00:39:24,630 --> 00:39:34,980
How much of that do you
think I need in f(x)? f of?

665
00:39:34,980 --> 00:39:36,480
f(3).

666
00:39:36,480 --> 00:39:39,800
Whatever f is at
that point, x=3,

667
00:39:39,800 --> 00:39:42,500
that's the amount I need there.

668
00:39:42,500 --> 00:39:45,290
So this would be f(3).

669
00:39:45,290 --> 00:39:49,330
So that that part
would sort of have

670
00:39:49,330 --> 00:39:56,530
the right pep, the right
punch, at the point three.

671
00:39:56,530 --> 00:40:01,570
Now, three could be
any point, that's

672
00:40:01,570 --> 00:40:03,220
any point along the line.

673
00:40:03,220 --> 00:40:07,020
Let me, I can't use x for
other points on the line

674
00:40:07,020 --> 00:40:09,010
because I've got an
x in the formula.

675
00:40:09,010 --> 00:40:12,390
Let me use t.

676
00:40:12,390 --> 00:40:17,720
So this was t equal to three.

677
00:40:17,720 --> 00:40:20,470
Now I want to do
it at all points.

678
00:40:20,470 --> 00:40:24,380
I want to take f(2), and
f(pi), and f at everything

679
00:40:24,380 --> 00:40:25,650
and put them together.

680
00:40:25,650 --> 00:40:28,600
And of course putting them
together in the continuous case

681
00:40:28,600 --> 00:40:32,820
means not sum, as I did
there, but integrate.

682
00:40:32,820 --> 00:40:38,080
So I have to, now I'm going
to change three to a t,

683
00:40:38,080 --> 00:40:49,450
because this is
the amount of f of,

684
00:40:49,450 --> 00:40:55,850
so that's the amount, that's
the delta functions which spike

685
00:40:55,850 --> 00:40:59,340
at t, multiplied by
f(t), and now how

686
00:40:59,340 --> 00:41:01,780
do I get f(x) out of this?

687
00:41:01,780 --> 00:41:05,950
I put them together. dt.

688
00:41:05,950 --> 00:41:08,190
I add up, this is
the combination

689
00:41:08,190 --> 00:41:10,850
I've been talking about.

690
00:41:10,850 --> 00:41:13,070
So this is like any f.

691
00:41:13,070 --> 00:41:20,680
This is like a crazy
delta function identity.

692
00:41:20,680 --> 00:41:23,460
Actually, it's not
crazy, it's the identity

693
00:41:23,460 --> 00:41:25,560
we've used our whole lives.

694
00:41:25,560 --> 00:41:28,380
Or at least our whole
18.085 lives, which

695
00:41:28,380 --> 00:41:31,480
is all that matters, right?

696
00:41:31,480 --> 00:41:35,910
OK, so I'm integrating
a delta function.

697
00:41:35,910 --> 00:41:39,650
And I want to see,
do I get that answer?

698
00:41:39,650 --> 00:41:42,020
And you're going to say
absolutely, clearly you

699
00:41:42,020 --> 00:41:43,280
get that answer.

700
00:41:43,280 --> 00:41:44,000
No.

701
00:41:44,000 --> 00:41:46,850
Everybody knows that if I
integrate something times

702
00:41:46,850 --> 00:41:51,520
a delta function that I
plug in at the point t=x,

703
00:41:51,520 --> 00:41:57,240
where that spike happens, I
put t=x, I get f(x), correct.

704
00:41:57,240 --> 00:41:59,900
So that's like any,
that's an identity

705
00:41:59,900 --> 00:42:02,360
or whatever you would
like to call it.

706
00:42:02,360 --> 00:42:06,560
And now just tell
me the final answer.

707
00:42:06,560 --> 00:42:09,600
Let me put it on
the board above.

708
00:42:09,600 --> 00:42:14,170
Now, so this was,
this expressed my f(x)

709
00:42:14,170 --> 00:42:16,200
as a combination of deltas.

710
00:42:16,200 --> 00:42:20,060
Just the way over here
I expressed [1, 2, 3, 7]

711
00:42:20,060 --> 00:42:22,300
as a combination of deltas.

712
00:42:22,300 --> 00:42:24,100
Now, what's that u?

713
00:42:24,100 --> 00:42:25,700
What's the function u?

714
00:42:25,700 --> 00:42:30,730
The solution that
comes from f(x).

715
00:42:30,730 --> 00:42:35,210
Just erase here so that
I can put all of them.

716
00:42:35,210 --> 00:42:38,150
The point of the
whole example now

717
00:42:38,150 --> 00:42:50,110
is for you to tell
me what's u(x)?

718
00:42:50,110 --> 00:42:51,110
Can you do it?

719
00:42:51,110 --> 00:42:53,330
You see what u(x) is
going to come out?

720
00:42:53,330 --> 00:42:56,420
It's going to come out nicely.

721
00:42:56,420 --> 00:43:01,480
I've written the right-hand side
f as a combination of deltas.

722
00:43:01,480 --> 00:43:05,580
I know the answer, for delta.

723
00:43:05,580 --> 00:43:12,610
It's G. What's the answer when
the delta is shifted along?

724
00:43:12,610 --> 00:43:17,520
What's the Green's function
when the spike is moved along

725
00:43:17,520 --> 00:43:20,940
to a point t?

726
00:43:20,940 --> 00:43:24,050
Just, because this
constant-coefficient,

727
00:43:24,050 --> 00:43:30,990
translation-invariant, LTI
problem is shift invariant,

728
00:43:30,990 --> 00:43:33,200
the answer, when the
spike is moved along,

729
00:43:33,200 --> 00:43:35,990
is just the answer
G moved along.

730
00:43:35,990 --> 00:43:38,600
So here's my answer.

731
00:43:38,600 --> 00:43:47,200
All this, this is my input and
my output is the same integral,

732
00:43:47,200 --> 00:43:52,610
this from minus infinity to
infinity, of -- where it was f,

733
00:43:52,610 --> 00:43:58,450
now it's G. Well
no, what do I want?

734
00:43:58,450 --> 00:44:01,210
Help me out here.

735
00:44:01,210 --> 00:44:06,130
No, f(t) is just the
amount of the delta,

736
00:44:06,130 --> 00:44:08,120
but now what's the solution?

737
00:44:08,120 --> 00:44:10,530
What do I write now?

738
00:44:10,530 --> 00:44:13,600
G of?

739
00:44:13,600 --> 00:44:24,270
x-t.

740
00:44:24,270 --> 00:44:27,040
That's it.

741
00:44:27,040 --> 00:44:29,680
You see why that works?

742
00:44:29,680 --> 00:44:32,920
Because this was the
input, G's the output.

743
00:44:32,920 --> 00:44:36,970
The problem was shift invariant,
so if I shifted the input

744
00:44:36,970 --> 00:44:38,990
I shifted the output.

745
00:44:38,990 --> 00:44:43,760
It was linear, so if I
add up a bunch of deltas,

746
00:44:43,760 --> 00:44:53,360
the solution is add up a
bunch of G. That's the answer.

747
00:44:53,360 --> 00:44:56,440
Oh, I could just
say one thing more.

748
00:44:56,440 --> 00:44:59,730
But you've got it
if you see math.

749
00:44:59,730 --> 00:45:03,750
So that's the point, that if
you know the Green's function,

750
00:45:03,750 --> 00:45:06,140
well yeah.

751
00:45:06,140 --> 00:45:11,660
Maybe from a practical point
of view, what have we done?

752
00:45:11,660 --> 00:45:17,400
The original way we did
it involved our computing

753
00:45:17,400 --> 00:45:18,760
two integrals.

754
00:45:18,760 --> 00:45:21,695
We had to, if we were
given an f(x), we

755
00:45:21,695 --> 00:45:23,970
had to find its
transform f hat of k,

756
00:45:23,970 --> 00:45:28,600
that we weren't sure we could
do with pencil and paper.

757
00:45:28,600 --> 00:45:31,840
And then we got an answer
with an f hat of k up here

758
00:45:31,840 --> 00:45:34,260
and then we had
to transform back.

759
00:45:34,260 --> 00:45:36,800
Step one and step
three, we had to do.

760
00:45:36,800 --> 00:45:40,980
Now this is better.

761
00:45:40,980 --> 00:45:44,210
This is like better,
because we were

762
00:45:44,210 --> 00:45:52,800
able to get an explicit answer,
G, when this was a delta.

763
00:45:52,800 --> 00:45:56,977
You could say I've got
it down to one integral.

764
00:45:56,977 --> 00:45:58,310
Well, for whatever that's worth.

765
00:45:58,310 --> 00:46:01,160
I was going to say,
probably can't do that one

766
00:46:01,160 --> 00:46:03,640
either depending what f(x) is.

767
00:46:03,640 --> 00:46:10,460
But that's a nice way to see
the answer, you have to admit.

768
00:46:10,460 --> 00:46:14,710
And it's because the
problem is shift-invariant

769
00:46:14,710 --> 00:46:16,830
and I can write the
answer that way.

770
00:46:16,830 --> 00:46:22,350
OK, and now one more
thing about this.

771
00:46:22,350 --> 00:46:26,000
And I've written the word,
the key word, down here.

772
00:46:26,000 --> 00:46:30,980
Convolution.

773
00:46:30,980 --> 00:46:37,540
Do you see the nice way
to write that answer?

774
00:46:37,540 --> 00:46:41,620
It's the convolution
of f with G.

775
00:46:41,620 --> 00:46:43,990
We didn't do integral
convolutions,

776
00:46:43,990 --> 00:46:47,280
we just did the discrete
sums, but I mentioned

777
00:46:47,280 --> 00:46:52,610
that in the integral case,
you have this same thing. t

778
00:46:52,610 --> 00:46:56,610
and x-t adding up
to x, just the way

779
00:46:56,610 --> 00:47:00,930
we had k and n-k adding up to n.

780
00:47:00,930 --> 00:47:03,900
In other words, this
is just notation.

781
00:47:03,900 --> 00:47:10,000
But I'm just going to write
the answer in a nice way.

782
00:47:10,000 --> 00:47:14,300
So after all that
lecture, the answer

783
00:47:14,300 --> 00:47:20,820
to the differential equations
is in three symbols f star G.

784
00:47:20,820 --> 00:47:23,840
Where this just
simply means this.

785
00:47:23,840 --> 00:47:29,210
This is the continuous
convolution, not cyclic,

786
00:47:29,210 --> 00:47:38,070
and there's the answer.

787
00:47:38,070 --> 00:47:41,870
I'll allow myself
one more thing.

788
00:47:41,870 --> 00:47:46,650
Here we had a convolution
for the right-hand side.

789
00:47:46,650 --> 00:47:48,870
We started with this,
this is a convolution.

790
00:47:48,870 --> 00:47:55,430
Now, what are the three
symbols that I write down?

791
00:47:55,430 --> 00:47:59,390
For the shorthand
for this equation?

792
00:47:59,390 --> 00:48:01,840
So this was to be
true for any f.

793
00:48:01,840 --> 00:48:04,800
And now can I write
down, how do I write?

794
00:48:04,800 --> 00:48:14,970
f(x) is equal to, what
is this right-hand side?

795
00:48:14,970 --> 00:48:19,950
It's f convolved with?

796
00:48:19,950 --> 00:48:29,860
So any f is the same f
convolved with delta.

797
00:48:29,860 --> 00:48:33,840
In convolution, delta is one.

798
00:48:33,840 --> 00:48:39,190
Because when I go
to the other space

799
00:48:39,190 --> 00:48:44,710
and I get a multiplication,
it really is one, right?

800
00:48:44,710 --> 00:48:46,220
In the other space.

801
00:48:46,220 --> 00:48:52,000
So in the other space, so
this convolution in x space

802
00:48:52,000 --> 00:48:54,820
turns into multiplication
in frequency space.

803
00:48:54,820 --> 00:48:59,000
And it just tells me that
f hat is f hat times one.

804
00:48:59,000 --> 00:49:04,200
So that's the way to look
at it in physical space.

805
00:49:04,200 --> 00:49:11,620
And this is the way to
look at the solution.

806
00:49:11,620 --> 00:49:16,860
So, one more thought and
I'll come back to that.

807
00:49:16,860 --> 00:49:20,700
So this G, the
Green's function, this

808
00:49:20,700 --> 00:49:25,110
is what a CAT scan does,
what an X-ray telescope does.

809
00:49:25,110 --> 00:49:27,460
What all sorts of
physical things do.

810
00:49:27,460 --> 00:49:31,330
Provided we can assume this
translation invariance,

811
00:49:31,330 --> 00:49:34,970
which is never perfectly
true because the telescope is

812
00:49:34,970 --> 00:49:35,910
finite.

813
00:49:35,910 --> 00:49:40,960
But a telescope takes the
star, takes the light signal,

814
00:49:40,960 --> 00:49:44,440
convolves it with the
telescope's own little Green's

815
00:49:44,440 --> 00:49:45,860
function.

816
00:49:45,860 --> 00:49:48,940
It blurs it, it's the
point spread function.

817
00:49:48,940 --> 00:49:52,560
It's the blurring function,
G. The Green's function

818
00:49:52,560 --> 00:49:55,190
on a telescope is
somehow, that's what's

819
00:49:55,190 --> 00:49:57,670
you're convolving with.

820
00:49:57,670 --> 00:50:03,920
And if you want to find that
star as a bright, single point.

821
00:50:03,920 --> 00:50:06,390
You've got to do deconvolution.

822
00:50:06,390 --> 00:50:09,530
You've got to do a
division to get the G out.

823
00:50:09,530 --> 00:50:13,800
May I just say those words
and then it's Thanksgiving?

824
00:50:13,800 --> 00:50:22,010
That a machine, a sensor which
is translation invariant,

825
00:50:22,010 --> 00:50:27,000
or you could say close
enough to pretend it is,

826
00:50:27,000 --> 00:50:29,340
because nothing is going
to be perfectly translation

827
00:50:29,340 --> 00:50:31,660
invariant all the
way out to infinity.

828
00:50:31,660 --> 00:50:34,950
But if it's near the star.

829
00:50:34,950 --> 00:50:40,130
So I look at this point star
in the telescope I see a blur.

830
00:50:40,130 --> 00:50:46,880
That's because the telescope
has convolved the correct thing,

831
00:50:46,880 --> 00:50:52,480
that I should have seen,
with G. It's blurred it

832
00:50:52,480 --> 00:50:54,660
by its point spread function.

833
00:50:54,660 --> 00:50:57,330
So what if the
person, the factory

834
00:50:57,330 --> 00:51:00,320
where the telescope was built
can test this whole thing

835
00:51:00,320 --> 00:51:01,420
on points.

836
00:51:01,420 --> 00:51:04,210
And it can find the
point spread function.

837
00:51:04,210 --> 00:51:07,810
And if I knew G,
then I could undo it.

838
00:51:07,810 --> 00:51:10,310
And get a clear picture.

839
00:51:10,310 --> 00:51:16,970
So it's that step that won the
Nobel Prize for the CAT scan,

840
00:51:16,970 --> 00:51:21,740
and I'm sure is winning
Nobel Prizes for astronomers.

841
00:51:21,740 --> 00:51:24,650
OK, have a great Thanksgiving,
I'll see you Monday.

842
00:51:24,650 --> 00:51:25,788
Good.

843
00:51:25,788 --> 00:51:26,287