1
00:00:10 --> 00:00:16
Today is going to be one of the
more difficult lectures of the

2
00:00:14 --> 00:00:20
term.
So, put on your thinking caps,

3
00:00:17 --> 00:00:23
as they would say in elementary
school.

4
00:00:20 --> 00:00:26
The topic is going to be what's
called a convolution.

5
00:00:25 --> 00:00:31
The convolution is something
very peculiar that you do to two

6
00:00:29 --> 00:00:35
functions to get a third
function.

7
00:00:32 --> 00:00:38
It has its own special symbol.
f of t asterisk is the

8
00:00:38 --> 00:00:44
universal symbol that's used for
that.

9
00:00:41 --> 00:00:47
So, this is a new function of
t, which bears very little

10
00:00:45 --> 00:00:51
resemblance to the ones,
f of t, that you started with.

11
00:00:50 --> 00:00:56
I'm going to give you the
formula for it,

12
00:00:53 --> 00:00:59
but first, there are two ways
of motivating it,

13
00:00:57 --> 00:01:03
and both are important.
There is a formal motivation,

14
00:01:02 --> 00:01:08
which is why it's tucked into
the section on Laplace

15
00:01:07 --> 00:01:13
transform.
And, the formal motivation is

16
00:01:10 --> 00:01:16
the following.
Suppose we start with the

17
00:01:13 --> 00:01:19
Laplace transform of those two
functions.

18
00:01:17 --> 00:01:23
Now, the most natural question
to ask is, since Laplace

19
00:01:22 --> 00:01:28
transforms are really a pain to
calculate is from old Laplace

20
00:01:27 --> 00:01:33
transforms, is it easy to get
new ones?

21
00:01:32 --> 00:01:38
And, the first thing,
of course, summing functions is

22
00:01:35 --> 00:01:41
easy.
That gives you the sum of the

23
00:01:37 --> 00:01:43
transforms.
But, a more natural question

24
00:01:40 --> 00:01:46
would be, suppose I want to
multiply F of t and G

25
00:01:43 --> 00:01:49
of t.
Is there, hopefully,

26
00:01:45 --> 00:01:51
some neat formula?
If I multiply the product of

27
00:01:48 --> 00:01:54
the, take the product of these
two, is there some neat formula

28
00:01:52 --> 00:01:58
for the Laplace transform of
that product?

29
00:01:55 --> 00:02:01
That would simply life greatly.
And, the answer is,

30
00:01:58 --> 00:02:04
there is no such formula.
And there never will be.

31
00:02:03 --> 00:02:09
Well, we will not give up
entirely.

32
00:02:05 --> 00:02:11
Suppose we ask the other
question.

33
00:02:08 --> 00:02:14
Suppose instead I multiply the
Laplace transforms.

34
00:02:11 --> 00:02:17
Could that be related to
something I cook up out of F of

35
00:02:15 --> 00:02:21
t and G of t?
Could it be the transform of

36
00:02:20 --> 00:02:26
something I cook up out of F of
t or G of t?

37
00:02:23 --> 00:02:29
And, that's what the
convolution is for.

38
00:02:26 --> 00:02:32
The answer is that F of s times
G of s turns out

39
00:02:31 --> 00:02:37
to be the Laplace transform of
the convolution.

40
00:02:36 --> 00:02:42
The convolution,
and that's one way of defining

41
00:02:40 --> 00:02:46
it, is the function of t you
should put it there in order

42
00:02:45 --> 00:02:51
that its Laplace transform turn
out to be the product of F of s

43
00:02:50 --> 00:02:56
times G of s.
Now, I'll give you,

44
00:02:54 --> 00:03:00
in a moment,
the formula for it.

45
00:02:57 --> 00:03:03
But, I'll give you one and a
quarter minutes,

46
00:03:01 --> 00:03:07
well, two minutes of motivation
as to why there should be such

47
00:03:07 --> 00:03:13
formula.
Now, I won't calculate this out

48
00:03:10 --> 00:03:16
to the end because I don't have
time.

49
00:03:13 --> 00:03:19
But, here's the reason why
there should be such formula.

50
00:03:16 --> 00:03:22
And, you might suspect,
and therefore it would be worth

51
00:03:19 --> 00:03:25
looking for.
It's because,

52
00:03:21 --> 00:03:27
remember, I told you where the
Laplace transform came from,

53
00:03:24 --> 00:03:30
that the Laplace transform was
the continuous analog of a power

54
00:03:28 --> 00:03:34
series.
So, when you ask a general

55
00:03:31 --> 00:03:37
question like that,
the place to look for is if you

56
00:03:34 --> 00:03:40
know an analogous idea,
say, does it work.

57
00:03:37 --> 00:03:43
something like that work there?
So, here I have a power series

58
00:03:41 --> 00:03:47
summation, (a)n x to the n.

59
00:03:43 --> 00:03:49
Remember, you can write this in
computer notation as a of n

60
00:03:47 --> 00:03:53
to make it look like f
of n,

61
00:03:50 --> 00:03:56
f of t.
And, the analog is turned into

62
00:03:53 --> 00:03:59
t when you turn a power series
into the Laplace transform,

63
00:03:57 --> 00:04:03
and x gets turned into e to the
negative s,

64
00:04:01 --> 00:04:07
and one formula just turns into
the other.

65
00:04:05 --> 00:04:11
Okay, so, there's a formula for
F of x.

66
00:04:08 --> 00:04:14
This is the analog of the
Laplace transform.

67
00:04:12 --> 00:04:18
And, similarly,
G of x here is

68
00:04:15 --> 00:04:21
summation (b)n x to the n.

69
00:04:18 --> 00:04:24
Now, again, the naīve question
would be, well,

70
00:04:21 --> 00:04:27
suppose I multiply the two
corresponding coefficients

71
00:04:25 --> 00:04:31
together, and add up that power
series, summation (a)n (b)n

72
00:04:30 --> 00:04:36
times x to the n.

73
00:04:34 --> 00:04:40
Is that somehow,
that sum related to F and G?

74
00:04:37 --> 00:04:43
And, of course,
everybody knows the answer to

75
00:04:40 --> 00:04:46
that is no.
It has no relation whatever.

76
00:04:44 --> 00:04:50
But, suppose instead I multiply
these two guys.

77
00:04:47 --> 00:04:53
In that case,
I'll get a new power series.

78
00:04:50 --> 00:04:56
I don't know what its
coefficients are,

79
00:04:53 --> 00:04:59
but let's write them down.
Let's just call them (c)n's.

80
00:04:58 --> 00:05:04
So, what I'm asking is,
this corresponds to the product

81
00:05:02 --> 00:05:08
of the two Laplace transforms.
And, what I want to know is,

82
00:05:08 --> 00:05:14
is there a formula which says
that (c)n is equal to something

83
00:05:14 --> 00:05:20
that can be calculated out of
the (a)i and the (b)j.

84
00:05:19 --> 00:05:25
Now, the answer to that is,
yes, there is.

85
00:05:24 --> 00:05:30
And, the formula for (c)n is
called the convolution.

86
00:05:30 --> 00:05:36
Now, you could figure out this
formula yourself.

87
00:05:33 --> 00:05:39
You figure it out.
Anyone who's smart enough to be

88
00:05:36 --> 00:05:42
interested in the question in
the first place is smart enough

89
00:05:41 --> 00:05:47
to figure out what that formula
is.

90
00:05:43 --> 00:05:49
And, it will give you great
pleasure to see that it's just

91
00:05:47 --> 00:05:53
like the formula for the
convolution of going to give you

92
00:05:51 --> 00:05:57
now.
So, what is that formula for

93
00:05:54 --> 00:06:00
the convolution?
Okay, hang on.

94
00:05:56 --> 00:06:02
Now, you are not going to like
it.

95
00:06:00 --> 00:06:06
But, you didn't like the
formula for the Laplace

96
00:06:03 --> 00:06:09
transform, either.
You felt wiser,

97
00:06:06 --> 00:06:12
grown-up getting it.
But it's a mouthful to swallow.

98
00:06:10 --> 00:06:16
It's something you get used to
slowly.

99
00:06:13 --> 00:06:19
And, you will get used to the
convolution equally slowly.

100
00:06:17 --> 00:06:23
So, what is the convolution of
f of t and g of t?

101
00:06:22 --> 00:06:28
It's a function calculated

102
00:06:25 --> 00:06:31
according to the corresponding
formula.

103
00:06:28 --> 00:06:34
It's a function of t.
It is the integral from zero to

104
00:06:32 --> 00:06:38
t of f of u, --
u is a dummy variable because

105
00:06:37 --> 00:06:43
it's going to be integrated out
when I do the integration,

106
00:06:43 --> 00:06:49
g of (t minus u) dt.

107
00:06:49 --> 00:06:55
That's it.

108
00:06:50 --> 00:06:56
I didn't make it up.
I'm just varying the bad news.

109
00:06:55 --> 00:07:01
Well, what do you do when you
see a formula?

110
00:07:00 --> 00:07:06
Well, the first thing to do,
of course, is try calculating

111
00:07:05 --> 00:07:11
just to get some feeling for
what kind of a thing,

112
00:07:10 --> 00:07:16
you know.
Let's try some examples.

113
00:07:14 --> 00:07:20
Let's see, let's calculate,
what would be a modest

114
00:07:18 --> 00:07:24
beginning?
Let's calculate the convolution

115
00:07:21 --> 00:07:27
of t with itself.
Or, better yet,

116
00:07:24 --> 00:07:30
let's calculate the convolution
just so that you could tell the

117
00:07:28 --> 00:07:34
difference, t with t squared,
t squared with t,

118
00:07:32 --> 00:07:38
to make it a little easier.
By the way, the convolution is

119
00:07:37 --> 00:07:43
symmetric.
f star g is the same thing as g

120
00:07:40 --> 00:07:46
star f.
Let's put that down explicitly.

121
00:07:43 --> 00:07:49
I forgot to last period.
So, tell all the guys who came

122
00:07:47 --> 00:07:53
to the one o'clock lecture that
you know something that they

123
00:07:51 --> 00:07:57
don't.
Now, that's a theory.

124
00:07:53 --> 00:07:59
It's commutative.
This operation is commutative,

125
00:07:57 --> 00:08:03
in other words.
Now, that has to be a theorem

126
00:08:00 --> 00:08:06
because the formula is not
symmetric.

127
00:08:04 --> 00:08:10
The formula does not treat f
and g equally.

128
00:08:07 --> 00:08:13
And therefore,
this is not obvious.

129
00:08:11 --> 00:08:17
It's at least not obvious if
you look at it that way,

130
00:08:15 --> 00:08:21
but it is obvious if you look
at it that way.

131
00:08:19 --> 00:08:25
Why?
In other words,

132
00:08:21 --> 00:08:27
f star g is the guy
whose Laplace transform is F of

133
00:08:27 --> 00:08:33
s times G of s.
Well, what would g star f?

134
00:08:33 --> 00:08:39
That would be the guy whose

135
00:08:35 --> 00:08:41
Laplace transform is G
times F.

136
00:08:37 --> 00:08:43
But capital F times capital G
is the same as capital G times

137
00:08:41 --> 00:08:47
capital F.
So, it's because the Laplace

138
00:08:45 --> 00:08:51
transforms are commutative.
Ordinary multiplication is

139
00:08:48 --> 00:08:54
commutative.
It follows that this has to be

140
00:08:51 --> 00:08:57
commutative, too.
So, I'll write that down,

141
00:08:54 --> 00:09:00
since F times G is equal to GF.
And, you have to understand

142
00:08:58 --> 00:09:04
that here, I mean that these are
the Laplace transforms of those

143
00:09:02 --> 00:09:08
guys.
But, it's not obvious from the

144
00:09:06 --> 00:09:12
formula.
Okay, let's calculate the

145
00:09:08 --> 00:09:14
Laplace transform of,
sorry, the convolution of t

146
00:09:11 --> 00:09:17
star,
let's do it by the formula.

147
00:09:14 --> 00:09:20
All right, by the formula,
I calculate integral zero to t.

148
00:09:18 --> 00:09:24
Now, I take the first function,
but I change its variable to

149
00:09:23 --> 00:09:29
the dummy variable,
u.

150
00:09:24 --> 00:09:30
So, that's u squared.
I take the second function and

151
00:09:28 --> 00:09:34
replace its variable by u minus
t.

152
00:09:33 --> 00:09:39
So, this is times t minus u,
sorry.

153
00:09:38 --> 00:09:44
Okay, do you see that to
calculate this is what I have to

154
00:09:45 --> 00:09:51
write down?
That's what the formula

155
00:09:49 --> 00:09:55
becomes.
Anything wrong?

156
00:09:51 --> 00:09:57
Oh, sorry, the du,
the integration's with expect

157
00:09:57 --> 00:10:03
to u, of course.
Thanks very much.

158
00:10:03 --> 00:10:09
Okay, let's do it.
So, it is, integral of u

159
00:10:06 --> 00:10:12
squared t is,
remember, it's integrated with

160
00:10:10 --> 00:10:16
respect to u.
So, it's u cubed over three

161
00:10:13 --> 00:10:19
times t.
The rest of it is the integral

162
00:10:17 --> 00:10:23
of u cubed,
which is u to the forth over

163
00:10:21 --> 00:10:27
four.
All this is to be evaluated

164
00:10:24 --> 00:10:30
between zero and t at the upper
limit.

165
00:10:27 --> 00:10:33
So, I put u equal t,
I get t to the forth over three

166
00:10:32 --> 00:10:38
minus t to the forth
over four.

167
00:10:38 --> 00:10:44
Of course, at the lower limit,
u is zero.

168
00:10:41 --> 00:10:47
So, both of these are terms of
zero.

169
00:10:43 --> 00:10:49
There's nothing there.
And, the answer is,

170
00:10:46 --> 00:10:52
therefore, t to the forth
divided by,

171
00:10:49 --> 00:10:55
a third minus a quarter is a
twelfth.

172
00:10:53 --> 00:10:59
So, that's doing it from the
formula.

173
00:10:56 --> 00:11:02
But, of course,
there is an easier way to do

174
00:10:59 --> 00:11:05
it.
We can cheat and use the

175
00:11:01 --> 00:11:07
Laplace transform instead.
If I Laplace transform it,

176
00:11:06 --> 00:11:12
the Laplace transform of t
squared is what?

177
00:11:10 --> 00:11:16
It's two factorial divided by s
cubed.

178
00:11:13 --> 00:11:19
The Laplace transform of t

179
00:11:16 --> 00:11:22
is one divided by s
squared.

180
00:11:20 --> 00:11:26
And so, because this is the
convolution of these,

181
00:11:24 --> 00:11:30
it should correspond to the
product of the Laplace

182
00:11:28 --> 00:11:34
transforms, which is two over s
to the 5th power.

183
00:11:34 --> 00:11:40
Well, is that the same as this?
What's the Laplace transform

184
00:11:38 --> 00:11:44
of, in other words,
what's the inverse Laplace

185
00:11:42 --> 00:11:48
transform of two over s to the
fifth?

186
00:11:47 --> 00:11:53
Well, the inverse Laplace
transform of four factorial over

187
00:11:51 --> 00:11:57
s to the fifth is how much?

188
00:11:55 --> 00:12:01
That's t to the forth,
right?

189
00:11:58 --> 00:12:04
Now, how does this differ?
Well, to turn that into that,

190
00:12:03 --> 00:12:09
I should divide by four times
three.

191
00:12:06 --> 00:12:12
So, this should be one twelfth
t to the forth,

192
00:12:10 --> 00:12:16
one over four times three
because this is 24,

193
00:12:13 --> 00:12:19
and that's two,
so, divide by 12 to determine

194
00:12:15 --> 00:12:21
what constant,
yeah.

195
00:12:17 --> 00:12:23
So, it works,
at least in that case.

196
00:12:19 --> 00:12:25
But now, notice that this is
not an ordinary product.

197
00:12:22 --> 00:12:28
The convolution of t squared
and t is not something

198
00:12:26 --> 00:12:32
like t cubed.
It's something like t to the

199
00:12:30 --> 00:12:36
forth, and there's a funny
constant in there,

200
00:12:33 --> 00:12:39
too, very unpredictable.
Let's look at the convolution.

201
00:12:38 --> 00:12:44
Let's take another example of
the convolution.

202
00:12:41 --> 00:12:47
Let's do something really
humble just assure you that

203
00:12:45 --> 00:12:51
this, even at the simplest
example, this is not trivial.

204
00:12:50 --> 00:12:56
Let's take the convolution of f
of t with one.

205
00:12:54 --> 00:13:00
Can you take,
yeah, one is a function just

206
00:12:57 --> 00:13:03
like any function.
But, you get something out of

207
00:13:01 --> 00:13:07
the convolution,
yes, yes.

208
00:13:03 --> 00:13:09
Let's just write down the
formula.

209
00:13:05 --> 00:13:11
Now, I can't use the Laplace
transform here because you won't

210
00:13:09 --> 00:13:15
know what to do with it.
You don't have that formula

211
00:13:13 --> 00:13:19
yet.
It's a secret one that only I

212
00:13:15 --> 00:13:21
know.
So, let's do it.

213
00:13:16 --> 00:13:22
Let's calculate it out the way
it was supposed to.

214
00:13:19 --> 00:13:25
So, it's the integral from zero
to t of f of u,

215
00:13:23 --> 00:13:29
and now, what do I do with that
one?

216
00:13:25 --> 00:13:31
I'm supposed to take,
one is the function g of t,

217
00:13:29 --> 00:13:35
and wherever I see a
t, I'm supposed to plug in t

218
00:13:32 --> 00:13:38
minus u.
Well, I don't see any t there.

219
00:13:39 --> 00:13:45
But that's something for
rejoicing.

220
00:13:43 --> 00:13:49
There's nothing to do to make
the substitution.

221
00:13:48 --> 00:13:54
It's just one.
So, the answer is,

222
00:13:52 --> 00:13:58
it's this curious thing.
The convolution of a function

223
00:13:58 --> 00:14:04
with one, you integrate it from
zero to t.

224
00:14:04 --> 00:14:10
Well, as they said in Alice in
Wonderland, things are getting

225
00:14:08 --> 00:14:14
curiouser and curiouser.
I mean, what is going on with

226
00:14:12 --> 00:14:18
this crazy function,
and where are we supposed to

227
00:14:16 --> 00:14:22
start with it?
Well, I'm going to prove this

228
00:14:19 --> 00:14:25
for you, mostly because the
proof is easy.

229
00:14:23 --> 00:14:29
In other words,
I'm going to prove that that's

230
00:14:26 --> 00:14:32
true.
And, as I give the proof,

231
00:14:29 --> 00:14:35
you'll see where the
convolution is coming from.

232
00:14:32 --> 00:14:38
That's number one.
And, number two,

233
00:14:34 --> 00:14:40
the real reason I'm giving you
the proof: because it's a

234
00:14:38 --> 00:14:44
marvelous exercise in changing
the variables in a double

235
00:14:42 --> 00:14:48
integral.
Now, that's something you all

236
00:14:44 --> 00:14:50
know how to do,
even the ones who are taking

237
00:14:47 --> 00:14:53
18.02 concurrently,
and I didn't advise you to do

238
00:14:50 --> 00:14:56
that.
But, I've arranged the course

239
00:14:52 --> 00:14:58
so it's possible to do.
But, I knew that by the time we

240
00:14:56 --> 00:15:02
got to this, you would already
know how to change variables at

241
00:15:00 --> 00:15:06
a double integral.
So, and in fact,

242
00:15:04 --> 00:15:10
you will have the advantage of
remembering how to do it because

243
00:15:10 --> 00:15:16
you just had it about a week or
two ago, whereas all the other

244
00:15:15 --> 00:15:21
guys, it's something dim in
their distance.

245
00:15:19 --> 00:15:25
So, I'm reviewing how to change
variables at a double integral.

246
00:15:25 --> 00:15:31
I'm showing you it's good for
something.

247
00:15:29 --> 00:15:35
So, what we are out to try to
prove is this formula.

248
00:15:33 --> 00:15:39
Let's put that down in,
so you understand.

249
00:15:39 --> 00:15:45
Okay, let's do it.
Now, we'll use the desert

250
00:15:41 --> 00:15:47
island method.
So, you have as much time as

251
00:15:44 --> 00:15:50
you want.
You're on a desert island.

252
00:15:46 --> 00:15:52
In fact, I'm going to even go
it the opposite way.

253
00:15:49 --> 00:15:55
I'm going to start with--
you've got a lot of time on your

254
00:15:53 --> 00:15:59
hands and say,
gee, I wonder if I take the

255
00:15:56 --> 00:16:02
product of the Laplace
transforms, I wonder if there's

256
00:15:59 --> 00:16:05
some crazy function I could put
in there, which would make

257
00:16:03 --> 00:16:09
things work.
You've never heard of the

258
00:16:06 --> 00:16:12
convolution.
You're going to discover it all

259
00:16:09 --> 00:16:15
by yourself.
Okay, so how do you begin?

260
00:16:11 --> 00:16:17
So, we'll start with the left
hand side.

261
00:16:14 --> 00:16:20
We're looking for some nice way
of calculating that as the

262
00:16:17 --> 00:16:23
Laplace transform of a single
function.

263
00:16:19 --> 00:16:25
So, the way to begin is by
writing out the definitions.

264
00:16:23 --> 00:16:29
We couldn't use anything else
since we don't have anything

265
00:16:26 --> 00:16:32
else to use.
Now, looking ahead,

266
00:16:28 --> 00:16:34
I'm going to not use t.
I'm going to use two neutral

267
00:16:32 --> 00:16:38
variables when I calculate.
After all, the t is just a

268
00:16:35 --> 00:16:41
dummy variable anyway.
You will see in a minute the

269
00:16:40 --> 00:16:46
wisdom of doing this.
So, it's this times the

270
00:16:44 --> 00:16:50
integral, which gives the
Laplace transform of g.

271
00:16:48 --> 00:16:54
So, that's e to the negative s
v, let's say,

272
00:16:52 --> 00:16:58
times g of v,
dv.

273
00:16:53 --> 00:16:59
Okay, everybody can get that
far.

274
00:16:56 --> 00:17:02
But now we have to start
looking.

275
00:17:00 --> 00:17:06
Well, this is a single
integral, an 18.01 integral

276
00:17:03 --> 00:17:09
involving u, and this is an
18.01 integral involving v.

277
00:17:07 --> 00:17:13
But when you take the product
of two integrals like that,

278
00:17:11 --> 00:17:17
remember when you evaluate a
double integral,

279
00:17:14 --> 00:17:20
there's an easy case where it's
much easier than any other case.

280
00:17:19 --> 00:17:25
If you could write the inside,
if you are integrating over a

281
00:17:23 --> 00:17:29
rectangle, for example,
and you can write the integral

282
00:17:27 --> 00:17:33
as a product of a function just
of u, and a product of a

283
00:17:31 --> 00:17:37
function just as v,
then the integral is very easy

284
00:17:34 --> 00:17:40
to evaluate.
You can forget all the rules.

285
00:17:38 --> 00:17:44
You just take all the u part
out, all the v part out,

286
00:17:42 --> 00:17:48
and integrate them separately,
a to b, c to d.

287
00:17:44 --> 00:17:50
That's the easy case of
evaluating a double integral.

288
00:17:48 --> 00:17:54
It's what everybody tries to
do, even when it's not

289
00:17:51 --> 00:17:57
appropriate.
Now, here it is appropriate,

290
00:17:53 --> 00:17:59
except I'm going to use it
backwards.

291
00:17:56 --> 00:18:02
This is the result of having
done that.

292
00:17:58 --> 00:18:04
If this is the result of having
done it, what was the step just

293
00:18:02 --> 00:18:08
before it?
Well, I must have been trying

294
00:18:06 --> 00:18:12
to evaluate a double integral as
u runs from zero to infinity and

295
00:18:10 --> 00:18:16
v runs from zero to infinity,
of what?

296
00:18:13 --> 00:18:19
Well, of the product of these
two functions.

297
00:18:16 --> 00:18:22
Now, what is that?
e to the minus s u times e to

298
00:18:20 --> 00:18:26
the minus s v.

299
00:18:22 --> 00:18:28
Well, I must surely want to
combine those.

300
00:18:25 --> 00:18:31
e to the minus s u times e to
the minus s v.

301
00:18:30 --> 00:18:36
And, what's left?
Well, what gets dragged along?

302
00:18:33 --> 00:18:39
du dv.
This is the same as that

303
00:18:35 --> 00:18:41
because of that law I just gave
you this is the product of a

304
00:18:39 --> 00:18:45
function just of u,
and a function just of v.

305
00:18:42 --> 00:18:48
And therefore,
it's okay to separate the two

306
00:18:45 --> 00:18:51
integrals out that way because
I'm integrating sort of a

307
00:18:49 --> 00:18:55
rectangle that goes to infinity
that way and infinity that way.

308
00:18:54 --> 00:19:00
But, what I'm integrating is
over the plane,

309
00:18:57 --> 00:19:03
in other words,
this region of the plane as u,

310
00:19:00 --> 00:19:06
v goes from zero to infinity,
zero to infinity.

311
00:19:05 --> 00:19:11
Now, let's take a look.
What are we looking for?

312
00:19:10 --> 00:19:16
Well, we're looking for,
we would be very happy if u

313
00:19:15 --> 00:19:21
plus v were t.
Let's make it t.

314
00:19:20 --> 00:19:26
In other words,
I'm introducing a new variable,

315
00:19:25 --> 00:19:31
t, u plus v,
and it's suggested by the form

316
00:19:30 --> 00:19:36
in which I'm looking for the
answer.

317
00:19:35 --> 00:19:41
Now, of course you then have
to, we need another variable.

318
00:19:39 --> 00:19:45
We could keep either u or v.
Let's keep u.

319
00:19:43 --> 00:19:49
That means v,
we just gave a musical chairs.

320
00:19:46 --> 00:19:52
v got dropped out.
Well, we can't have three

321
00:19:50 --> 00:19:56
variables.
We only have room for two.

322
00:19:53 --> 00:19:59
But, we will remember it.
Rest in peace,

323
00:19:57 --> 00:20:03
v was equal to t minus u 
in case we ever need him

324
00:20:02 --> 00:20:08
again.
Okay, let's now put in the

325
00:20:05 --> 00:20:11
limits.
Let's put in the integral,

326
00:20:08 --> 00:20:14
the rest of the change of
variable.

327
00:20:10 --> 00:20:16
So, I'm now changing it to
these new variables,

328
00:20:14 --> 00:20:20
t and u, so it's e to the
negative s t.

329
00:20:18 --> 00:20:24
Well, f of u I don't
have to do anything to.

330
00:20:22 --> 00:20:28
But, g of v,
I'm not allowed to keep v,

331
00:20:25 --> 00:20:31
so v has to be changed to t
minus u.

332
00:20:30 --> 00:20:36
Amazing things are happening.
Now, I want to change this to

333
00:20:34 --> 00:20:40
an integral du dt.
Now, for that,

334
00:20:37 --> 00:20:43
you have to be a little
careful.

335
00:20:40 --> 00:20:46
We have two things to do to
figure out this;

336
00:20:43 --> 00:20:49
what goes with that?
And, we have to put in the

337
00:20:47 --> 00:20:53
limits, also.
Now, those are the two

338
00:20:50 --> 00:20:56
nontrivial operations,
when you change variables in a

339
00:20:55 --> 00:21:01
double integral.
So, let's be really careful.

340
00:21:00 --> 00:21:06
Let's do the easier of the two,
first.

341
00:21:03 --> 00:21:09
I want to change from du dv to
du dt.

342
00:21:07 --> 00:21:13
And now, to do that,
you have to put in the Jacobian

343
00:21:12 --> 00:21:18
matrix, the Jacobian
determinant.

344
00:21:15 --> 00:21:21
Ah-ha!
How many of you forgot that?

345
00:21:19 --> 00:21:25
I won't even bother asking.
Oh, come on,

346
00:21:23 --> 00:21:29
you only lose two points.
It doesn't matter if you put it

347
00:21:28 --> 00:21:34
in the Jacobian.
As you see, you're going to

348
00:21:34 --> 00:21:40
forget something.
You will lose less credit for

349
00:21:39 --> 00:21:45
forgetting than anything else.
So, it's the Jacobian of u and

350
00:21:45 --> 00:21:51
v with respect to u and t.
So, to calculate that,

351
00:21:49 --> 00:21:55
you write u equals u,
v equals t minus u,

352
00:21:54 --> 00:22:00
and then the Jacobian is
the partial of the matrix,

353
00:22:00 --> 00:22:06
the determinant of partial
derivatives.

354
00:22:05 --> 00:22:11
So, it's the determinant whose
entries are the partial of u

355
00:22:09 --> 00:22:15
with respect to u,
the partial of u with respect

356
00:22:13 --> 00:22:19
to t, but these are independent
variables.

357
00:22:16 --> 00:22:22
So, that's zero.
The partial of v with respect

358
00:22:20 --> 00:22:26
to u is negative one.
The partial of v with respect

359
00:22:24 --> 00:22:30
to t is one.
So, the Jacobian is one.

360
00:22:27 --> 00:22:33
So, if you forgot it,
no harm.

361
00:22:31 --> 00:22:37
So, the Jacobian is one.
Now, more serious,

362
00:22:34 --> 00:22:40
and in some ways,
I think, for most of you,

363
00:22:37 --> 00:22:43
the most difficult part of the
operation, is putting in the new

364
00:22:41 --> 00:22:47
limits.
Now, for that,

365
00:22:43 --> 00:22:49
you look at the region over
which you're integrating.

366
00:22:46 --> 00:22:52
I think I'd better do that
carefully.

367
00:22:49 --> 00:22:55
I need a bigger picture.
That's really what I'm trying

368
00:22:53 --> 00:22:59
to say.
So, here's the (u,

369
00:22:54 --> 00:23:00
v) coordinates.
And, I want to change these to

370
00:22:58 --> 00:23:04
(u, t) coordinates.
The integration is over the

371
00:23:01 --> 00:23:07
first quadrant.
So, what you do is,

372
00:23:05 --> 00:23:11
when you do the integral,
the first step is u is varying,

373
00:23:10 --> 00:23:16
and t is held fixed.
So, in the first integration,

374
00:23:15 --> 00:23:21
u varies.
t is held fixed.

375
00:23:17 --> 00:23:23
Now, what is holding t fixed in
this picture mean?

376
00:23:22 --> 00:23:28
Well, t is equal to u plus v.

377
00:23:26 --> 00:23:32
So, u plus v is fixed,
is a constant,

378
00:23:29 --> 00:23:35
in other words.
Now, where are the curves along

379
00:23:34 --> 00:23:40
which u plus v is a
constant?

380
00:23:38 --> 00:23:44
Well, they are these lines.
These are the lines along which

381
00:23:43 --> 00:23:49
u plus v equals a constant,
or t is a constant.

382
00:23:47 --> 00:23:53
The reason I'm holding t a
constant is because the first

383
00:23:52 --> 00:23:58
integration only allows u to
change.

384
00:23:55 --> 00:24:01
t is held fixed.
Okay, you let u increase.

385
00:23:59 --> 00:24:05
As u increases,
and t is held fixed,

386
00:24:02 --> 00:24:08
I'm traversing these lines in
this direction.

387
00:24:08 --> 00:24:14
That's the direction on which u
is increasing.

388
00:24:11 --> 00:24:17
I integrate from the point,
from the u value where they

389
00:24:15 --> 00:24:21
leave the region.
And, to enter the region,

390
00:24:18 --> 00:24:24
what's the u value where they
enter the region?

391
00:24:21 --> 00:24:27
u is equal to zero.
Everybody would know that.

392
00:24:24 --> 00:24:30
Not so many people would be
able to figure out what to put

393
00:24:28 --> 00:24:34
for where it leaves the region.
What's the value of u when it

394
00:24:34 --> 00:24:40
leaves the region?
Well, this is the curve,

395
00:24:38 --> 00:24:44
v equals zero.
But, v equals zero is,

396
00:24:42 --> 00:24:48
in another language,
u equals t.

397
00:24:46 --> 00:24:52
t minus u equals zero,
or u equals t.

398
00:24:51 --> 00:24:57
In other words,
they enter the region where u

399
00:24:55 --> 00:25:01
equals zero,
and they leave where u is t,

400
00:25:00 --> 00:25:06
has the value of t.
And, how about the other guys?

401
00:25:05 --> 00:25:11
For which t's do I want to do
this?

402
00:25:07 --> 00:25:13
Well, I want to do it for all
these t values.

403
00:25:10 --> 00:25:16
Well, now, the t value here,
that's the starting one.

404
00:25:14 --> 00:25:20
Here, t is zero,
and here t is not zero.

405
00:25:16 --> 00:25:22
And, if I go out and cover the
whole first quadrant,

406
00:25:20 --> 00:25:26
I'll be letting t increase to
infinity.

407
00:25:23 --> 00:25:29
The sum of u and v,
I will be letting increase to

408
00:25:26 --> 00:25:32
infinity.
So, it's zero to infinity.

409
00:25:30 --> 00:25:36
So, all this is an exercise in
taking this double integral in

410
00:25:35 --> 00:25:41
(u, v) coordinates,
and changing it to this double

411
00:25:40 --> 00:25:46
integral, an equivalent double
integral over the same region,

412
00:25:46 --> 00:25:52
but now in (u,
t) coordinates.

413
00:25:48 --> 00:25:54
And now, that's the answer.
Somewhere here is the answer

414
00:25:54 --> 00:26:00
because, look,
since the first integration is

415
00:25:58 --> 00:26:04
with respect to u,
this guy can migrate outside

416
00:26:02 --> 00:26:08
because it doesn't involve u.
That only involves t,

417
00:26:08 --> 00:26:14
and t is only caught by the
second integration.

418
00:26:11 --> 00:26:17
So, I can put this outside.
And, what do I end up with?

419
00:26:15 --> 00:26:21
The integral from zero to
infinity of e to the negative s

420
00:26:18 --> 00:26:24
t times,

421
00:26:22 --> 00:26:28
what's left?
A funny expression,

422
00:26:24 --> 00:26:30
but you're on your desert
island and found it.

423
00:26:27 --> 00:26:33
This funny expression,
integral from zero to t,

424
00:26:30 --> 00:26:36
f of u, g of t minus u vu,

425
00:26:34 --> 00:26:40
in short,
the convolution,

426
00:26:37 --> 00:26:43
exactly the convolution.
So, all you have to do is get

427
00:26:42 --> 00:26:48
the idea that there might be a
formula, sit down,

428
00:26:45 --> 00:26:51
change variables and double
integral it, ego,

429
00:26:48 --> 00:26:54
you've got your formula.
Well, I would like to spend

430
00:26:52 --> 00:26:58
much of the rest of the
period--- in other words,

431
00:26:56 --> 00:27:02
that's how it relates to the
Laplace transform.

432
00:26:59 --> 00:27:05
That's how it comes out of the
Laplace transform.

433
00:27:04 --> 00:27:10
Here's how to use it,
calculate it either with the

434
00:27:07 --> 00:27:13
Laplace transform or directly
from the integral.

435
00:27:10 --> 00:27:16
And, of course,
you will solve problems,

436
00:27:13 --> 00:27:19
Laplace transform problems,
differential equations using

437
00:27:17 --> 00:27:23
the convolution.
But, I have to tell you that

438
00:27:20 --> 00:27:26
most people, convolution is very
important.

439
00:27:23 --> 00:27:29
And, most people who use it
don't use it in connection with

440
00:27:27 --> 00:27:33
the Laplace transform.
They use it for its own sake.

441
00:27:30 --> 00:27:36
The first place I learned that
outside of MIT people used a

442
00:27:34 --> 00:27:40
convolution was actually from my
daughter.

443
00:27:39 --> 00:27:45
She's an environmental
engineer, an environmental

444
00:27:41 --> 00:27:47
consultant.
She does risk assessment,

445
00:27:44 --> 00:27:50
and stuff like that.
But anyway, she had this paper

446
00:27:47 --> 00:27:53
on acid rain she was trying to
read for a client,

447
00:27:50 --> 00:27:56
and she said something about
calculating acid rain falls on

448
00:27:53 --> 00:27:59
soil.
And then, from there,

449
00:27:55 --> 00:28:01
the stuff leeches into a river.
But, things happen to it on the

450
00:27:58 --> 00:28:04
way.
Soil combines in various ways,

451
00:28:01 --> 00:28:07
reduces the acidity,
and things happen.

452
00:28:03 --> 00:28:09
Chemical reactions take place,
blah, blah, blah,

453
00:28:06 --> 00:28:12
blah.
Anyways, she said,

454
00:28:08 --> 00:28:14
well, then they calculated in
the end how much the river gets

455
00:28:11 --> 00:28:17
polluted.
But, she said it's convolution.

456
00:28:13 --> 00:28:19
She said, what's the
convolution?

457
00:28:15 --> 00:28:21
So, I told her she was too
young to learn about the

458
00:28:18 --> 00:28:24
convolution.
And she knows that I thought

459
00:28:20 --> 00:28:26
I'd better look it up first.
I mean, I, of course,

460
00:28:23 --> 00:28:29
knew the convolution was,
but I was a little puzzled at

461
00:28:26 --> 00:28:32
that application of it.
So, I read the paper.

462
00:28:29 --> 00:28:35
It was interesting.
And, in thinking about it,

463
00:28:33 --> 00:28:39
other people have come to me,
some guy with a problem about,

464
00:28:38 --> 00:28:44
they drilled ice cores in the
North Pole, and from the

465
00:28:41 --> 00:28:47
radioactive carbon and so on,
deducing various things about

466
00:28:46 --> 00:28:52
the climate 60 billion years
ago, and it was all convolution.

467
00:28:50 --> 00:28:56
He asked me if I could explain
that to him.

468
00:28:53 --> 00:28:59
So, let me give you sort of
all-purpose thing,

469
00:28:56 --> 00:29:02
a simple all-purpose model,
which can be adapted,

470
00:28:59 --> 00:29:05
which is very good way of
thinking of the convolution,

471
00:29:03 --> 00:29:09
in my opinion.
It's a problem of radioactive

472
00:29:08 --> 00:29:14
dumping.
It's in the notes,

473
00:29:11 --> 00:29:17
by the way.
So, I'm just,

474
00:29:13 --> 00:29:19
if you want to take a chance,
and just listen to what I'm

475
00:29:18 --> 00:29:24
saying rather that just
scribbling everything down,

476
00:29:23 --> 00:29:29
maybe you'll be able to figure
it out for the notes,

477
00:29:28 --> 00:29:34
also.
So, the problem is we have some

478
00:29:33 --> 00:29:39
factory, or a nuclear plant,
or some thing like that,

479
00:29:38 --> 00:29:44
is producing radioactive waste,
not always at the same rate.

480
00:29:44 --> 00:29:50
And then, it carts it,
dumps it on a pile somewhere.

481
00:29:49 --> 00:29:55
So, radioactive waste is
dumped, and there's a dumping

482
00:29:54 --> 00:30:00
function.
I'll call that f of t,

483
00:29:58 --> 00:30:04
the dump rate.
That's the dumping rate.

484
00:30:03 --> 00:30:09
Let's say t is in years.
You like to have units,

485
00:30:07 --> 00:30:13
and quantity,
kilograms, I don't know,

486
00:30:10 --> 00:30:16
whatever you want.
Now, what does the dumping rate

487
00:30:15 --> 00:30:21
mean?
The dumping rate means that if

488
00:30:18 --> 00:30:24
I have two times that are close
together, for example,

489
00:30:23 --> 00:30:29
two successive days,
midnight on two successive

490
00:30:27 --> 00:30:33
days, then there's a time
interval between them.

491
00:30:33 --> 00:30:39
I'll call that delta t.
To say the dumping rate is f of

492
00:30:38 --> 00:30:44
t means that the amount
dumped in this time interval,

493
00:30:43 --> 00:30:49
in the time interval from t1 to
t1 plus one is

494
00:30:49 --> 00:30:55
approximately,
not exactly,

495
00:30:52 --> 00:30:58
because the dumping rate isn't
even constant within this time

496
00:30:58 --> 00:31:04
interval.
But it's approximately the

497
00:31:02 --> 00:31:08
dumping rate times the time over
which the dumping is taking

498
00:31:09 --> 00:31:15
place.
That's what I mean by the dump

499
00:31:13 --> 00:31:19
rate.
And, it gets more and more

500
00:31:16 --> 00:31:22
accurate, the smaller the time
interval you take.

501
00:31:21 --> 00:31:27
Okay, now here's my problem.
The problem is,

502
00:31:26 --> 00:31:32
you start dumping at time t
equals zero.

503
00:31:33 --> 00:31:39
At time t equal t,
how much radioactive waste is

504
00:31:39 --> 00:31:45
in the pile?

505
00:31:41 --> 00:31:47

506
00:31:55 --> 00:32:01
Now, what makes that problem
slightly complicated is

507
00:31:58 --> 00:32:04
radioactive waste decays.
If I put some at a certain day,

508
00:32:02 --> 00:32:08
and then go back several months
later and nothing's happened in

509
00:32:07 --> 00:32:13
between, I don't have the same
amount that I dumps because a

510
00:32:11 --> 00:32:17
fraction of it decayed.
I have less.

511
00:32:14 --> 00:32:20
And, our answer to the problem
must take account of,

512
00:32:18 --> 00:32:24
for each piece of waste,
how long it has been in the

513
00:32:22 --> 00:32:28
pile because that takes account
of how long it had to decay,

514
00:32:27 --> 00:32:33
and what it ends up as.
So, the calculation,

515
00:32:32 --> 00:32:38
the essential part of the
calculation will be that if you

516
00:32:37 --> 00:32:43
have an initial amount of this
substance, and it decays for a

517
00:32:43 --> 00:32:49
time, t, this is the amount left
at time t.

518
00:32:47 --> 00:32:53
This is the law of radioactive
decay.

519
00:32:51 --> 00:32:57
You knew that coming into
18.03, although,

520
00:32:55 --> 00:33:01
it's, of course,
a simple differential equation

521
00:33:00 --> 00:33:06
which produces it,
but I'll assume you simply know

522
00:33:05 --> 00:33:11
the answer.
k depends on the material,

523
00:33:10 --> 00:33:16
so I'm going to assume that the
nuclear plant dumps the same

524
00:33:14 --> 00:33:20
radioactive substance each time.
It's only one substance I'm

525
00:33:19 --> 00:33:25
calculating, and k is it.
So, assume the k is fixed.

526
00:33:23 --> 00:33:29
I don't have to change from one
k from one material to a k for

527
00:33:27 --> 00:33:33
another because it's mixing up
the stuff, just one material.

528
00:33:33 --> 00:33:39
Okay, and now let's calculate
it.

529
00:33:35 --> 00:33:41
Here's the idea.
I'll take the t-axis,

530
00:33:38 --> 00:33:44
but now I'm going to change its
name to the u-axis.

531
00:33:43 --> 00:33:49
You will see why in just a
second.

532
00:33:45 --> 00:33:51
It starts at zero.
I'm interested in what's

533
00:33:49 --> 00:33:55
happening at the time,
t.

534
00:33:51 --> 00:33:57
How much is left at time t?
So, I'm going to divide up the

535
00:33:56 --> 00:34:02
interval from zero to t on this
time axis into,

536
00:34:00 --> 00:34:06
well, here's u0,
the starting point,

537
00:34:03 --> 00:34:09
u1, u2, let's make this u1.
Oh, curses!

538
00:34:08 --> 00:34:14
u1, u2, u3, and so on.
Let's call this (u)n.

539
00:34:13 --> 00:34:19
So they're u(n + 1),
not that it matters.

540
00:34:18 --> 00:34:24
It doesn't matter.
Okay, now, the amount,

541
00:34:23 --> 00:34:29
so, what I'm going to do is
look at the amount,

542
00:34:28 --> 00:34:34
take the time interval from ui
to ui plus one.

543
00:34:36 --> 00:34:42
This is a time interval,

544
00:34:40 --> 00:34:46
delta u.
Divide it up into equal time

545
00:34:43 --> 00:34:49
intervals.
So, the amount dumped in the

546
00:34:46 --> 00:34:52
time interval from u(i) to u(i
plus one)

547
00:34:51 --> 00:34:57
is equal to approximately f of
u(i),

548
00:34:55 --> 00:35:01
the dumping function there,
times delta u.

549
00:35:00 --> 00:35:06
We calculated that before.
That's what the meaning of the

550
00:35:06 --> 00:35:12
dumping rate is.
By time t, how much has it

551
00:35:11 --> 00:35:17
decayed to?
It has decayed.

552
00:35:14 --> 00:35:20
How much is left,
in other words?

553
00:35:18 --> 00:35:24
Well, this is the starting
amount.

554
00:35:21 --> 00:35:27
So, the answer is going to be
it's f of (u)i times delta u

555
00:35:28 --> 00:35:34
times this factor,
which tells how much it decays,

556
00:35:34 --> 00:35:40
so, time.
So, this is the starting amount

557
00:35:39 --> 00:35:45
at time (u)i.
That's when it was first

558
00:35:41 --> 00:35:47
dumped, and this is the amount
that was dumped,

559
00:35:45 --> 00:35:51
times, multiply that by e to
the minus k times,

560
00:35:49 --> 00:35:55
now, what should I put up in
there?

561
00:35:51 --> 00:35:57
I have to put the length of
time that it had to decay.

562
00:35:55 --> 00:36:01
What is the length of time that
it had to decay?

563
00:36:00 --> 00:36:06
It was dumped at u(i).
I'm looking at time,

564
00:36:08 --> 00:36:14
t, it decayed for time length t
minus u i,

565
00:36:19 --> 00:36:25
the length of time it had all
the pile.

566
00:36:28 --> 00:36:34

567
00:36:32 --> 00:36:38
So, the stuff that was dumped
in this time interval,

568
00:36:36 --> 00:36:42
at time t when I come to look
at it, this is how much of it is

569
00:36:41 --> 00:36:47
left.
And now, all I have to do is

570
00:36:44 --> 00:36:50
add up that quantity for this
time, the stuff that was dumped

571
00:36:49 --> 00:36:55
in this time interval plus the
stuff dumped in,

572
00:36:54 --> 00:37:00
and so on, all the way up to
the stuff that was dumped

573
00:36:58 --> 00:37:04
yesterday.
And, the answer will be the

574
00:37:01 --> 00:37:07
total amount left at time,
t, that is not yet decayed will

575
00:37:06 --> 00:37:12
be approximately,
you add up the amount coming

576
00:37:10 --> 00:37:16
from the first time interval
plus the amount coming,

577
00:37:15 --> 00:37:21
and so on.
So, it will be f of u(i),

578
00:37:19 --> 00:37:25
I'll save the delta u for the
end, times e to the minus k

579
00:37:23 --> 00:37:29
times t minus u(i) times
delta u.

580
00:37:27 --> 00:37:33
So, these two parts represent

581
00:37:29 --> 00:37:35
the amount dumped,
and this is the decay factor.

582
00:37:33 --> 00:37:39
And, I had those up as I runs
from, well, where did I start?

583
00:37:37 --> 00:37:43
From one to n,
let's say.

584
00:37:39 --> 00:37:45
And now, let delta t go to
zero, in other words,

585
00:37:42 --> 00:37:48
make this delta u go to zero,
make this more accurate by

586
00:37:46 --> 00:37:52
taking finer and finer
subdivisions.

587
00:37:48 --> 00:37:54
In other words,
instead of looking every month

588
00:37:51 --> 00:37:57
to see how much was dumped,
let's look every week,

589
00:37:55 --> 00:38:01
every day, and so on,
to make this calculation more

590
00:37:58 --> 00:38:04
accurate.
And, the answer is,

591
00:38:00 --> 00:38:06
this approach is the exact
amount, which will be the

592
00:38:04 --> 00:38:10
integral.
This sum is a Riemann sum.

593
00:38:08 --> 00:38:14
It approaches the integral from
zero to, well,

594
00:38:12 --> 00:38:18
I'm adding it up from u1 equals
zero to un equals t,

595
00:38:18 --> 00:38:24
the final value.
So, it will be the integral

596
00:38:22 --> 00:38:28
from the starting point to the
ending point of f of u e to the

597
00:38:28 --> 00:38:34
minus k times t minus u to u.

598
00:38:34 --> 00:38:40
That's the answer to the
problem.

599
00:38:36 --> 00:38:42
It's given by this rather funny
looking integral.

600
00:38:39 --> 00:38:45
But, from this point of view,
it's entirely natural.

601
00:38:42 --> 00:38:48
It's a combination of the
dumping function.

602
00:38:44 --> 00:38:50
This doesn't care what the
material was.

603
00:38:47 --> 00:38:53
It only wants to know how much
was put on everyday.

604
00:38:50 --> 00:38:56
And, this part,
which doesn't care how much was

605
00:38:53 --> 00:38:59
put on each day,
it just is an intrinsic

606
00:38:55 --> 00:39:01
constant of the material
involving its decay rate.

607
00:39:00 --> 00:39:06
And, this total thing
represents the total amount.

608
00:39:04 --> 00:39:10
And that is,
what is it?

609
00:39:06 --> 00:39:12
It's the convolution of f of t
with what function?

610
00:39:11 --> 00:39:17
e to the minus k t.
It's the convolution of the

611
00:39:16 --> 00:39:22
dumping function and the decay
function.

612
00:39:19 --> 00:39:25
And, the convolution is exactly
the operation that you have to

613
00:39:25 --> 00:39:31
have to do that.
Okay, so, I think this is the

614
00:39:28 --> 00:39:34
most intuitive physical approach
to the meaning of the

615
00:39:33 --> 00:39:39
convolution.
In this particular,

616
00:39:37 --> 00:39:43
you can say,
well, that's very special.

617
00:39:39 --> 00:39:45
Okay, so it tells you what the
meaning of the convolution with

618
00:39:43 --> 00:39:49
an exponential is.
But, what about the convolution

619
00:39:46 --> 00:39:52
with all the other functions
we're going to have to use in

620
00:39:50 --> 00:39:56
this course.
They can all be interpreted

621
00:39:52 --> 00:39:58
just by being a little flexible
in your approach.

622
00:39:55 --> 00:40:01
I'll give you two examples of
this, well, three.

623
00:39:58 --> 00:40:04
First of all,
I'll use it for,

624
00:40:00 --> 00:40:06
in the problem set I ask you
about a bank account.

625
00:40:05 --> 00:40:11
That's not something any of you
are interested in.

626
00:40:08 --> 00:40:14
Okay, so, suppose instead I
dumped garbage --

627
00:40:12 --> 00:40:18

628
00:40:16 --> 00:40:22
-- undecaying.
So, something that doesn't

629
00:40:19 --> 00:40:25
decay at all,
what's the answer going to be?

630
00:40:22 --> 00:40:28
Well, the calculation will be
exactly the same.

631
00:40:26 --> 00:40:32
It will be the convolution of
the dumping function.

632
00:40:30 --> 00:40:36
The only difference is that now
the garbage isn't going to

633
00:40:34 --> 00:40:40
decay.
So, no matter how long it's

634
00:40:37 --> 00:40:43
left, the same amount is going
to be left at the end.

635
00:40:40 --> 00:40:46
In other words,
I don't want to exponential

636
00:40:42 --> 00:40:48
decay function.
I want to function,

637
00:40:44 --> 00:40:50
one, the constant function,
one, because once I stick it on

638
00:40:48 --> 00:40:54
the pile, nothing happens to it.
It just stays there.

639
00:40:51 --> 00:40:57
So, it's going to be the
convolution of this one because

640
00:40:54 --> 00:41:00
this is constant.
It's undecaying --

641
00:40:57 --> 00:41:03

642
00:41:04 --> 00:41:10
-- by the identical reasoning.
And so, what's the answer going

643
00:41:07 --> 00:41:13
to be?
It's going to be the integral

644
00:41:09 --> 00:41:15
from zero to t of f of u du.

645
00:41:12 --> 00:41:18
Now, that's an 18.01 problem.

646
00:41:14 --> 00:41:20
If I dump with a dumping rate,
f of u,

647
00:41:17 --> 00:41:23
and I dump from time zero to
time t, how much is on the pile?

648
00:41:20 --> 00:41:26
They don't give it.
They always give velocity

649
00:41:23 --> 00:41:29
problems, and problems of how to
slice up bread loaves,

650
00:41:26 --> 00:41:32
and stuff like that.
But, this is a real life

651
00:41:28 --> 00:41:34
problem.
If that's the dumping rate,

652
00:41:32 --> 00:41:38
and you dump for t days from
zero to time t,

653
00:41:35 --> 00:41:41
how much do you have left at
the end?

654
00:41:37 --> 00:41:43
Answer: the integral of f of u
du from zero to t.

655
00:41:42 --> 00:41:48
I'll give you another example.

656
00:41:46 --> 00:41:52
Suppose I wanted a dumping
function, suppose I wanted a

657
00:41:50 --> 00:41:56
function, wanted to interpret
something which grows like t,

658
00:41:54 --> 00:42:00
for instance.
All I want is a physical

659
00:41:57 --> 00:42:03
interpretation.
Well, I have to think,

660
00:42:01 --> 00:42:07
I'm making a pile of something,
a metaphorical pile,

661
00:42:04 --> 00:42:10
we don't actually have to make
a physical pile.

662
00:42:07 --> 00:42:13
And, the thing should be
growing like t.

663
00:42:09 --> 00:42:15
Well, what grows like t?
Not bacteria,

664
00:42:11 --> 00:42:17
they grow exponentially.
Before the lecture,

665
00:42:14 --> 00:42:20
I was trying to think of
something.

666
00:42:16 --> 00:42:22
So, I came up with chickens on
a chicken farm.

667
00:42:19 --> 00:42:25
Little baby chickens grow
linearly.

668
00:42:21 --> 00:42:27
All little animals,
anyway, I've observed that

669
00:42:23 --> 00:42:29
babies grow linearly,
at least for a while,

670
00:42:26 --> 00:42:32
thank God.
After a while,

671
00:42:27 --> 00:42:33
they taper off.
But, at the beginning,

672
00:42:32 --> 00:42:38
they eat every four hours or
whatever.

673
00:42:35 --> 00:42:41
And they eat the same amount,
pretty much.

674
00:42:39 --> 00:42:45
And, that adds up.
So, let's suppose this

675
00:42:43 --> 00:42:49
represents the linear growth of
chickens, of baby chicks.

676
00:42:48 --> 00:42:54
That makes them sound cuter,
less offensive.

677
00:42:52 --> 00:42:58
Okay, so, a farmer,
chicken farmer,

678
00:42:56 --> 00:43:02
whatever they call them,
is starting a new brood.

679
00:43:02 --> 00:43:08
So anyway, the hens lay at a
certain rate,

680
00:43:05 --> 00:43:11
and each of those are
incubated.

681
00:43:08 --> 00:43:14
And after a while,
little baby chicks come out.

682
00:43:12 --> 00:43:18
So, this will be the production
rate for new chickens.

683
00:43:18 --> 00:43:24

684
00:43:23 --> 00:43:29
Okay, and it will be the
convolution which will tell you

685
00:43:26 --> 00:43:32
at time, t, the number of
kilograms.

686
00:43:29 --> 00:43:35
We'd better do this in
kilograms, I'm afraid.

687
00:43:32 --> 00:43:38
Now, that's not as heartless as
it seems.

688
00:43:35 --> 00:43:41
The number of kilograms of
chickens times t.

689
00:43:38 --> 00:43:44
[LAUGHTER] It really isn't
heartless because,

690
00:43:41 --> 00:43:47
after all, why would the farmer
want to know that?

691
00:43:44 --> 00:43:50
Well, because a certain number
of pounds of chicken eat a

692
00:43:48 --> 00:43:54
certain number of pounds of
chicken feed,

693
00:43:51 --> 00:43:57
and that's how much he has to
dump, must have to give them

694
00:43:55 --> 00:44:01
every day.
That's how he calculates his

695
00:43:57 --> 00:44:03
expenses.
So, he will have to know the

696
00:44:01 --> 00:44:07
convolution is,
or better yet,

697
00:44:03 --> 00:44:09
he will hire you,
who knows what the convolution

698
00:44:07 --> 00:44:13
is.
And you'll be able to tell him.

699
00:44:09 --> 00:44:15
Okay, why don't we stop there
and go to recitation tomorrow.

700
00:44:13 --> 00:44:19
I'll be doing important things.