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PROFESSOR: Hi.
9
00:00:33,710 --> 00:00:38,740
Our lecture for today I've
entitled 'Logarithms Without
10
00:00:38,740 --> 00:00:39,800
Exponents'.
11
00:00:39,800 --> 00:00:43,160
And before I try to clarify
that, let me make a couple of
12
00:00:43,160 --> 00:00:44,390
general remarks.
13
00:00:44,390 --> 00:00:48,720
First of all, we have now
completed the rudiments, the
14
00:00:48,720 --> 00:00:51,240
basics so to speak, of
both differential
15
00:00:51,240 --> 00:00:52,550
and integral calculus.
16
00:00:52,550 --> 00:00:56,340
Consequently, what our next
ambition will be is to take
17
00:00:56,340 --> 00:01:00,020
these principles and apply
them to special functions
18
00:01:00,020 --> 00:01:02,260
which are worthy of
investigation.
19
00:01:02,260 --> 00:01:04,890
And you see, in connection with
this, the first function
20
00:01:04,890 --> 00:01:08,670
that I've chosen to talk about
is called the logarithm.
21
00:01:08,670 --> 00:01:11,630
Now the other interesting thing
is that we are used to
22
00:01:11,630 --> 00:01:14,570
logarithms from high school
where they were viewed as
23
00:01:14,570 --> 00:01:16,980
being a different kind
of notation for
24
00:01:16,980 --> 00:01:18,840
talking about exponents.
25
00:01:18,840 --> 00:01:21,500
Now, in the same way when
we treated the circular
26
00:01:21,500 --> 00:01:25,030
functions, we mentioned that
one did not have to invent
27
00:01:25,030 --> 00:01:27,670
triangles to talk about
trigonometry.
28
00:01:27,670 --> 00:01:30,490
The interesting point is that
one does not have to talk
29
00:01:30,490 --> 00:01:33,960
about exponents to invent
the logarithm function.
30
00:01:33,960 --> 00:01:37,130
This is why I have entitled
the lesson, as I say,
31
00:01:37,130 --> 00:01:40,010
'Logarithms without
Exponents'.
32
00:01:40,010 --> 00:01:43,100
You see, what I would like to do
is try to mimic, especially
33
00:01:43,100 --> 00:01:45,930
from an engineering point of
view, how many of these
34
00:01:45,930 --> 00:01:48,610
mathematical topics
originated.
35
00:01:48,610 --> 00:01:51,790
Let's look at a rather
straightforward physical
36
00:01:51,790 --> 00:01:56,050
principle, a principle that I'm
sure we believe permeates
37
00:01:56,050 --> 00:01:58,340
many real life situations.
38
00:01:58,340 --> 00:02:01,990
It's what I call the rule
of compound interest.
39
00:02:01,990 --> 00:02:06,000
Namely, let's suppose we have a
physical situation in which
40
00:02:06,000 --> 00:02:07,490
the quantity that
we're measuring,
41
00:02:07,490 --> 00:02:08,590
which we'll call 'm'.
42
00:02:08,590 --> 00:02:12,310
In other words, the rate of
change of the quantity is
43
00:02:12,310 --> 00:02:18,520
proportional to the amount
present, 'dm/ dt' equals 'km'.
44
00:02:18,520 --> 00:02:22,750
Now notice that this is a pretty
harmless statement.
45
00:02:22,750 --> 00:02:24,750
The rate of change
is proportional
46
00:02:24,750 --> 00:02:25,920
to the amount present.
47
00:02:25,920 --> 00:02:29,520
We certainly would expect that
many experiments would run
48
00:02:29,520 --> 00:02:32,820
this way, and there seems to be
nothing at all supernatural
49
00:02:32,820 --> 00:02:34,270
about this kind of
an assumption.
50
00:02:34,270 --> 00:02:38,090
At any rate, let's apply our
previous principles, and see
51
00:02:38,090 --> 00:02:41,130
if we can't, from this
differential equation, figure
52
00:02:41,130 --> 00:02:43,360
out what 'm' has to be.
53
00:02:43,360 --> 00:02:46,990
Separating the variables, we
get that 'dm' over 'm' is
54
00:02:46,990 --> 00:02:48,860
equal to 'kdt'.
55
00:02:48,860 --> 00:02:53,130
And now integrating, meaning
taking the inverse derivative,
56
00:02:53,130 --> 00:02:54,130
we have what?
57
00:02:54,130 --> 00:02:57,310
That the integral of 'dm'
over 'm' is equal
58
00:02:57,310 --> 00:02:59,830
to 'kt' plus a constant.
59
00:02:59,830 --> 00:03:03,300
Now you see, the interesting
point is at this stage of the
60
00:03:03,300 --> 00:03:08,400
game, we do not know explicitly
how to find a
61
00:03:08,400 --> 00:03:12,970
function whose derivative with
respect to 'm' is '1 over m'.
62
00:03:12,970 --> 00:03:14,710
In fact, that's the
problem that
63
00:03:14,710 --> 00:03:16,700
underlies today's lecture.
64
00:03:16,700 --> 00:03:20,130
And we will try to answer this
question, showing how once we
65
00:03:20,130 --> 00:03:23,760
tackle this from a philosophic
point of view, the rest of the
66
00:03:23,760 --> 00:03:27,530
details follow from material
that we've already learned.
67
00:03:27,530 --> 00:03:30,590
At any rate, focusing our
attention on the problem, the
68
00:03:30,590 --> 00:03:34,140
question is this: to determine
a function, which I'll call
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00:03:34,140 --> 00:03:36,240
capital 'L of x'--
70
00:03:36,240 --> 00:03:40,820
the 'L' sort of to forewarn us
of the fact that we will
71
00:03:40,820 --> 00:03:43,220
somehow get a logarithm
out of this--
72
00:03:43,220 --> 00:03:47,050
such that 'L prime of
x' is '1 over x'.
73
00:03:47,050 --> 00:03:50,690
And you see, what I want you to
notice is that if you write
74
00:03:50,690 --> 00:03:53,750
'1 over x' in exponential form,
in other words, if you
75
00:03:53,750 --> 00:03:57,340
try to write '1 over x' as 'x
to the n', notice that to
76
00:03:57,340 --> 00:04:00,460
solve this problem, 'L
of x' would be what?
77
00:04:00,460 --> 00:04:04,570
The integral of 'x' to
the 'minus 1 dx'.
78
00:04:04,570 --> 00:04:07,930
In other words, this is the case
of integral ''x to the n'
79
00:04:07,930 --> 00:04:11,290
dx', with 'n' equal
to minus 1.
80
00:04:11,290 --> 00:04:14,350
And you'll recall that when
we studied the recipe--
81
00:04:14,350 --> 00:04:16,430
let me just write that over
here-- when we studied the
82
00:04:16,430 --> 00:04:20,740
recipe of how you integrate 'x
to the n', we saw that that
83
00:04:20,740 --> 00:04:27,240
was 'x to the 'n + 1'' over
'n + 1' plus a constant.
84
00:04:27,240 --> 00:04:30,810
And we then observed that this
doesn't even make sense when
85
00:04:30,810 --> 00:04:33,910
'n' is equal to minus 1,
because we have a 0
86
00:04:33,910 --> 00:04:35,060
denominator.
87
00:04:35,060 --> 00:04:37,140
From the other point of view,
the way to look at
88
00:04:37,140 --> 00:04:38,700
this is we said what?
89
00:04:38,700 --> 00:04:42,490
That when you differentiate, you
lower the exponent by 1.
90
00:04:42,490 --> 00:04:47,160
Notice that to wind up with an
exponent of minus 1, we would
91
00:04:47,160 --> 00:04:51,340
have had to start with
an exponent of 0.
92
00:04:51,340 --> 00:04:53,950
But when you differentiate 'x'
to the 0, that being a
93
00:04:53,950 --> 00:05:00,010
constant, the derivative
of 'x' to the 0 is 0.
94
00:05:00,010 --> 00:05:03,870
It's not '1 over x'.
95
00:05:03,870 --> 00:05:07,620
You see, in other words, this
particular recipe that we're
96
00:05:07,620 --> 00:05:10,590
talking about, we don't
have working for us.
97
00:05:10,590 --> 00:05:13,180
In other words, this is an
interesting point again.
98
00:05:13,180 --> 00:05:15,770
You say gee whiz, if the recipe
for the integral ''x to
99
00:05:15,770 --> 00:05:18,650
the n' dx' works for everything
except 'n' equals
100
00:05:18,650 --> 00:05:20,730
minus 1, nothing's perfect.
101
00:05:20,730 --> 00:05:22,180
Let's be content.
102
00:05:22,180 --> 00:05:23,700
Since it works for every
number except
103
00:05:23,700 --> 00:05:25,710
that one, why worry?
104
00:05:25,710 --> 00:05:28,040
This is analogous to when
we talked about taking
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00:05:28,040 --> 00:05:30,220
derivatives and saw that
we get a 0 over
106
00:05:30,220 --> 00:05:31,850
0 form every time.
107
00:05:31,850 --> 00:05:34,260
You see, if you pick a function
at random, the
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00:05:34,260 --> 00:05:36,890
likelihood that the limit of
'f of x' as 'x' approaches
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00:05:36,890 --> 00:05:40,820
'a', the likelihood of that
being 0 over 0 is very small,
110
00:05:40,820 --> 00:05:43,760
except when you form
the derivative, you
111
00:05:43,760 --> 00:05:45,860
always get 0 over 0.
112
00:05:45,860 --> 00:05:49,560
And in a similar way, granted
that integral ''x' to the
113
00:05:49,560 --> 00:05:52,810
minus 1 dx' is one very
special case.
114
00:05:52,810 --> 00:05:56,990
What we have seen is that this
comes up every single time
115
00:05:56,990 --> 00:06:00,620
that we want to solve a problem
of the form 'dm/ dt'
116
00:06:00,620 --> 00:06:02,040
equals 'km'.
117
00:06:02,040 --> 00:06:05,330
In other words then, what it
really boils down to is that
118
00:06:05,330 --> 00:06:09,350
unless we can come up with
a function 'L of x' whose
119
00:06:09,350 --> 00:06:13,400
derivative is '1 over x', we
cannot solve the problem that
120
00:06:13,400 --> 00:06:15,230
we started off today's
lesson with.
121
00:06:15,230 --> 00:06:18,610
In other words, we must leave
it in the form integral 'dm
122
00:06:18,610 --> 00:06:21,760
over m' equals 'kt'
plus a constant.
123
00:06:21,760 --> 00:06:25,900
Well at any rate, let's see what
such a function capital
124
00:06:25,900 --> 00:06:27,640
'L of x' must look like.
125
00:06:27,640 --> 00:06:31,050
I'm going to utilize both the
differential calculus approach
126
00:06:31,050 --> 00:06:34,450
and the integral calculus
approach.
127
00:06:34,450 --> 00:06:37,540
By differential calculus,
assuming that 'y' equals 'L of
128
00:06:37,540 --> 00:06:40,510
x', what do we know about
the function 'y'?
129
00:06:40,510 --> 00:06:44,110
By definition, we've defined it
to be that function whose
130
00:06:44,110 --> 00:06:46,330
derivative is '1 over x'.
131
00:06:46,330 --> 00:06:48,850
So we know its derivative
is '1 over x'.
132
00:06:48,850 --> 00:06:53,150
The next point is that knowing
that 'y prime' is '1 over x',
133
00:06:53,150 --> 00:06:55,490
and even though we can't
integrate '1 over x', we can
134
00:06:55,490 --> 00:06:57,390
certainly differentiate
'1 over x'.
135
00:06:57,390 --> 00:06:59,640
We can now differentiate
'1 over x'.
136
00:06:59,640 --> 00:07:02,760
The derivative of '1 over x' is
minus '1 over 'x squared'',
137
00:07:02,760 --> 00:07:04,870
and we find that 'y
double prime' is
138
00:07:04,870 --> 00:07:06,840
minus '1 over 'x squared''.
139
00:07:06,840 --> 00:07:11,100
By the way, we should observe
that we must shy away from 'x'
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00:07:11,100 --> 00:07:14,400
equaling 0, because you see, we
would have a 0 denominator
141
00:07:14,400 --> 00:07:15,360
in that case.
142
00:07:15,360 --> 00:07:19,780
If we think of most physical
situations, notice that we
143
00:07:19,780 --> 00:07:21,240
talk about the amount--
144
00:07:21,240 --> 00:07:23,160
the rate of change
is proportional
145
00:07:23,160 --> 00:07:24,380
to the amount present.
146
00:07:24,380 --> 00:07:27,760
From a physical point of view,
the amount present can't be
147
00:07:27,760 --> 00:07:31,110
negative, so let's put the
restriction on here that the
148
00:07:31,110 --> 00:07:36,790
domain of 'L' will be all
positive numbers.
149
00:07:39,330 --> 00:07:41,060
Now here's the interesting
thing.
150
00:07:41,060 --> 00:07:46,660
Intuitively, I can now visualize
how the function 'L
151
00:07:46,660 --> 00:07:47,950
of x' must look.
152
00:07:47,950 --> 00:07:51,100
Namely, since its first
derivative is '1 over x' and
153
00:07:51,100 --> 00:07:53,620
'x' is greater than 0, that
tells me that the curve is
154
00:07:53,620 --> 00:07:55,050
always rising.
155
00:07:55,050 --> 00:07:58,750
And secondly, since '1 over 'x
squared'' is always positive,
156
00:07:58,750 --> 00:08:01,780
minus '1 over 'x squared'' is
always negative, that tells me
157
00:08:01,780 --> 00:08:04,650
that my curve is always
spilling water.
158
00:08:04,650 --> 00:08:07,140
In essence then, what
must the curve do?
159
00:08:07,140 --> 00:08:10,110
The curve must be rising
but spilling water.
160
00:08:10,110 --> 00:08:13,100
In other words, the curve
'L of x' belongs to this
161
00:08:13,100 --> 00:08:14,410
particular family.
162
00:08:14,410 --> 00:08:17,940
Notice that once we have one
curve which we call 'y' equals
163
00:08:17,940 --> 00:08:21,800
'L of x', by any displacement
parallel--
164
00:08:21,800 --> 00:08:24,430
vertical displacement,
we get what?
165
00:08:24,430 --> 00:08:28,820
A member called 'y' equals
'L of x' plus 'c'.
166
00:08:28,820 --> 00:08:31,360
All that changes is what?
167
00:08:31,360 --> 00:08:34,640
The point at which the curve
crosses the x-axis, but
168
00:08:34,640 --> 00:08:37,140
whichever member of the
family we pick, what
169
00:08:37,140 --> 00:08:38,789
typifies 'L of x'?
170
00:08:38,789 --> 00:08:40,870
Its derivative is '1 over x'.
171
00:08:40,870 --> 00:08:43,990
In other words, somehow or
other, we visualize that 'L of
172
00:08:43,990 --> 00:08:48,850
x' exists, and its graph
is something like this.
173
00:08:48,850 --> 00:08:51,860
Now suppose we want a more
tangible form, namely how do
174
00:08:51,860 --> 00:08:54,760
you compute 'L of x'
for a given 'x'?
175
00:08:54,760 --> 00:08:57,690
I thought what might be a nice
review now is to see how we
176
00:08:57,690 --> 00:09:01,300
can use our integral calculus
approach, and in particular
177
00:09:01,300 --> 00:09:03,730
the second fundamental theorem
of interval calculus.
178
00:09:06,920 --> 00:09:10,270
By the integral calculus
approach, we do is is we pick
179
00:09:10,270 --> 00:09:14,000
any positive number 'a', and
once picked, we fix it.
180
00:09:14,000 --> 00:09:18,650
You see, we have a great deal
of freedom in how we choose
181
00:09:18,650 --> 00:09:19,990
the positive number a.
182
00:09:19,990 --> 00:09:22,020
But let's pick one and
leave it here.
183
00:09:22,020 --> 00:09:25,480
Now what we'll do is in the
'y-t' plane, we'll draw the
184
00:09:25,480 --> 00:09:29,140
curve 'y' equals '1 over t'.
185
00:09:29,140 --> 00:09:32,550
And what we'll do now, we'll
study the area of the region
186
00:09:32,550 --> 00:09:36,910
'R' where 'R' is bounded above
by 'y' equals '1 over t', on
187
00:09:36,910 --> 00:09:41,310
the left by 't' equals 'a', on
the right by 't' equals 'x',
188
00:09:41,310 --> 00:09:44,420
and below by the t-axis.
189
00:09:44,420 --> 00:09:48,080
Now we've already seen that the
function that we get by
190
00:09:48,080 --> 00:09:52,750
taking the definite integral
from 'a' 'to x', 'dt over t',
191
00:09:52,750 --> 00:09:56,590
has the property that its
derivative is '1 over x'.
192
00:09:56,590 --> 00:09:59,310
In other words, recall from the
fundamental theorem, this
193
00:09:59,310 --> 00:10:03,120
is just a generalization of the
fact that if 'f of t' is a
194
00:10:03,120 --> 00:10:07,480
continuous function, then the
integral from 'a' to 'x', ''f
195
00:10:07,480 --> 00:10:11,120
of t' dt', is a function
of 'x'.
196
00:10:11,120 --> 00:10:13,730
And its derivative with
respect to 'x'
197
00:10:13,730 --> 00:10:16,600
is just 'f of x'.
198
00:10:16,600 --> 00:10:20,490
You see, in other words, I can
now view capital 'L of x' as
199
00:10:20,490 --> 00:10:24,920
being an area under the curve
'y' equals '1 over t'.
200
00:10:24,920 --> 00:10:28,610
Notice that I have a degree of
freedom here, namely what that
201
00:10:28,610 --> 00:10:32,000
area is depends on where
I choose 'a'.
202
00:10:32,000 --> 00:10:36,490
Notice that if I choose a
differently, changing 'a'--
203
00:10:36,490 --> 00:10:39,610
let's call this 'a
prime', say--
204
00:10:39,610 --> 00:10:43,880
changing 'a' changes the area
under the curve, but notice it
205
00:10:43,880 --> 00:10:46,150
changes it by a fixed amount.
206
00:10:46,150 --> 00:10:50,160
In other words, notice that
shifting 'a' just changes 'L
207
00:10:50,160 --> 00:10:52,440
of x' by a constant,
just like the
208
00:10:52,440 --> 00:10:54,500
ordinary in definite integral.
209
00:10:54,500 --> 00:10:57,540
At any rate, these are the two
approaches that we have.
210
00:10:57,540 --> 00:11:00,960
And if we superimpose them,
all we're saying is what?
211
00:11:00,960 --> 00:11:02,910
If you start with the
differential calculus
212
00:11:02,910 --> 00:11:06,970
approach, 'y' equals capital 'L
of x' must be a member of
213
00:11:06,970 --> 00:11:11,820
this family, crossing the axis
at some point (a,0) .
214
00:11:11,820 --> 00:11:14,770
And from the integral calculus
point of view, if you take the
215
00:11:14,770 --> 00:11:18,010
curve 'y' equals '1 over t'
and compute the area under
216
00:11:18,010 --> 00:11:22,420
that curve from 'a'
to 'x', that area
217
00:11:22,420 --> 00:11:25,690
function is 'L of x'.
218
00:11:25,690 --> 00:11:30,290
Now we'll let that rest for the
time being, and now have a
219
00:11:30,290 --> 00:11:32,180
brief digression.
220
00:11:32,180 --> 00:11:34,560
You see, sooner or later, I've
got to get to what a logarithm
221
00:11:34,560 --> 00:11:37,060
is, and we might just
as well do that now.
222
00:11:37,060 --> 00:11:40,620
So let's now take a look and
see what we mean by a
223
00:11:40,620 --> 00:11:42,280
logarithmic function.
224
00:11:42,280 --> 00:11:46,010
You see, quite frequently in
mathematics, what one does is
225
00:11:46,010 --> 00:11:49,950
one deals with a particular
special case in which one is
226
00:11:49,950 --> 00:11:51,220
interested.
227
00:11:51,220 --> 00:11:54,980
And having deduced very nice
results, he looks at this
228
00:11:54,980 --> 00:11:58,320
thing and says, you know, all
of these results came from a
229
00:11:58,320 --> 00:12:01,530
particular recipe, a particular
structure that this
230
00:12:01,530 --> 00:12:03,130
system obeyed.
231
00:12:03,130 --> 00:12:06,860
What he then does is he takes
this recipe or structure,
232
00:12:06,860 --> 00:12:11,280
gives it a general name, and
now is able to extend this
233
00:12:11,280 --> 00:12:14,870
property to a larger
class of objects.
234
00:12:14,870 --> 00:12:16,830
For example, let's
look at this way.
235
00:12:16,830 --> 00:12:19,990
What is the nice thing about
logarithms in the traditional
236
00:12:19,990 --> 00:12:21,270
sense of the word?
237
00:12:21,270 --> 00:12:24,030
Remember when we learned
logarithms in high school, by
238
00:12:24,030 --> 00:12:27,800
use of logarithms, we were able
to replace multiplication
239
00:12:27,800 --> 00:12:30,420
problems by addition problems,
et cetera.
240
00:12:30,420 --> 00:12:33,680
In other words, remember the
key structural property was
241
00:12:33,680 --> 00:12:36,610
that the logarithm of a product
was the sum of the
242
00:12:36,610 --> 00:12:37,490
logarithms.
243
00:12:37,490 --> 00:12:40,620
That was the structural thing
that we used over and over and
244
00:12:40,620 --> 00:12:41,650
over again.
245
00:12:41,650 --> 00:12:44,200
In fact, it's rather interesting
to point out that
246
00:12:44,200 --> 00:12:47,050
in many cases where we used the
properties of logarithms,
247
00:12:47,050 --> 00:12:50,370
we never really had to know
what a logarithm was.
248
00:12:50,370 --> 00:12:52,080
All we had to know was what?
249
00:12:52,080 --> 00:12:55,190
What properties did it obey,
and did we have a book of
250
00:12:55,190 --> 00:12:58,610
tables so we could look up the
logarithm when we had to.
251
00:12:58,610 --> 00:13:02,810
Well at any rate, using this as
motivation, we now define a
252
00:13:02,810 --> 00:13:06,250
logarithmic function,
specifically a function f is
253
00:13:06,250 --> 00:13:11,630
called logarithmic if for all
'x1', 'x2' in the domain of
254
00:13:11,630 --> 00:13:16,140
'f', 'f' of the product of 'x1'
and 'x2' is 'f of x1'
255
00:13:16,140 --> 00:13:17,720
plus 'f of x2'.
256
00:13:17,720 --> 00:13:20,440
In other words, since we know
what nice properties a
257
00:13:20,440 --> 00:13:23,670
logarithm has, let's define
any function 'f' to be
258
00:13:23,670 --> 00:13:27,460
logarithmic if 'f' of a
product is equal to
259
00:13:27,460 --> 00:13:31,380
the sum of the 'f's.
260
00:13:31,380 --> 00:13:34,400
Now you see, this may sound
like a pretty simple
261
00:13:34,400 --> 00:13:37,460
statement, but it's
quite demanding.
262
00:13:37,460 --> 00:13:39,000
In other words, notice--
263
00:13:39,000 --> 00:13:41,220
by the way, somebody says,
I wonder if there are any
264
00:13:41,220 --> 00:13:42,610
logarithmic functions?
265
00:13:42,610 --> 00:13:45,430
You see, notice that by the way
we began, there must be at
266
00:13:45,430 --> 00:13:47,260
least one logarithmic function,
267
00:13:47,260 --> 00:13:50,230
namely the usual logarithm.
268
00:13:50,230 --> 00:13:52,580
We know there's at least one
function which has these
269
00:13:52,580 --> 00:13:53,880
properties.
270
00:13:53,880 --> 00:13:56,550
So the set of all logarithmic
functions is certainly
271
00:13:56,550 --> 00:13:57,920
not the empty set.
272
00:13:57,920 --> 00:14:01,180
At any rate, let's see what
we can deduce about any
273
00:14:01,180 --> 00:14:02,820
logarithmic function.
274
00:14:02,820 --> 00:14:06,690
Well for example, I claim that
if 'f' is logarithmic, 'f of
275
00:14:06,690 --> 00:14:08,020
1' must be 0.
276
00:14:08,020 --> 00:14:09,150
Why is that?
277
00:14:09,150 --> 00:14:13,110
Well, notice that 1 can be
written as 1 times 1.
278
00:14:13,110 --> 00:14:16,530
And since 'f of 1' times
1 is 'f of 1' plus 'f
279
00:14:16,530 --> 00:14:18,610
of 1', we have what?
280
00:14:18,610 --> 00:14:23,040
That 'f of 1' is equal to 'f of
1' plus 'f of 1', therefore
281
00:14:23,040 --> 00:14:27,110
twice 'f of 1' is 'f of 1',
and therefore by algebraic
282
00:14:27,110 --> 00:14:29,780
manipulation, 'f of 1' is 0.
283
00:14:29,780 --> 00:14:33,330
Which, by the way, does check
out with the usual logarithm,
284
00:14:33,330 --> 00:14:37,570
the logarithm of 1 to
any base 'b' is 0.
285
00:14:37,570 --> 00:14:41,900
Secondly, if 'f of '1 over x''
is defined, you see in other
286
00:14:41,900 --> 00:14:45,170
words, I'm not sure whether '1
over x' is in the domain of
287
00:14:45,170 --> 00:14:48,620
'f' just because 'x' is, but
suppose '1 over x' is in the
288
00:14:48,620 --> 00:14:49,920
domain of 'f'.
289
00:14:49,920 --> 00:14:55,345
Then 'f of '1 over x'' is
equal to minus 'f of x'.
290
00:14:55,345 --> 00:15:00,050
In other words, 'f' of a number
is minus 'f' of the
291
00:15:00,050 --> 00:15:01,890
reciprocal of that number.
292
00:15:01,890 --> 00:15:05,570
Again, a familiar logarithmic
property, why does it follow
293
00:15:05,570 --> 00:15:07,050
from our basic definition?
294
00:15:07,050 --> 00:15:11,250
Well notice that we already
know that 'f of 1' is 0.
295
00:15:11,250 --> 00:15:16,120
We also know that 1 is equal
to 'x' times '1 over x',
296
00:15:16,120 --> 00:15:17,990
provided 'x' is not 0.
297
00:15:17,990 --> 00:15:22,280
By the way, if 'x' were 0, 1
over 0 wouldn't be defined, so
298
00:15:22,280 --> 00:15:23,930
we wouldn't be worrying
about that anyway.
299
00:15:23,930 --> 00:15:26,140
But notice now we have what?
300
00:15:26,140 --> 00:15:31,550
If 'x' is not 0, 'f of 'x times
'1 over x'' is 0, but by
301
00:15:31,550 --> 00:15:35,370
the logarithmic property, that's
also 'f of x' plus 'f
302
00:15:35,370 --> 00:15:36,750
of '1 over x''.
303
00:15:36,750 --> 00:15:41,880
Since these two together add up
to 0, 'f of '1 over x' must
304
00:15:41,880 --> 00:15:44,280
just be minus 'f of x'.
305
00:15:44,280 --> 00:15:46,780
And we'll just take a few more
properties like this just to
306
00:15:46,780 --> 00:15:49,130
make sure that you see what
properties logarithmic
307
00:15:49,130 --> 00:15:50,650
functions have.
308
00:15:50,650 --> 00:15:53,490
Remember again the ordinary
logarithm, the logarithm of a
309
00:15:53,490 --> 00:15:56,270
quotient was the difference
of the logarithms.
310
00:15:56,270 --> 00:15:59,400
Again, 'f of 'x over y''
is equal to 'f of
311
00:15:59,400 --> 00:16:01,130
x' minus 'f of y'.
312
00:16:01,130 --> 00:16:02,530
Why is that the case?
313
00:16:02,530 --> 00:16:04,710
Well, look at 'f
of 'x over y''.
314
00:16:04,710 --> 00:16:08,560
That says 'f of x'
times '1 over y'.
315
00:16:08,560 --> 00:16:12,040
But again, by logarithmic
properties, 'f' of a product
316
00:16:12,040 --> 00:16:13,440
is the sum of the 'f's.
317
00:16:13,440 --> 00:16:18,310
Therefore 'f of 'x times '1 over
y''' is 'f of x' plus 'f
318
00:16:18,310 --> 00:16:19,550
of '1 over y''.
319
00:16:19,550 --> 00:16:24,020
But we just saw that 'f of '1
over y' is minus 'f of y'.
320
00:16:24,020 --> 00:16:27,240
So we have this familiar
result now.
321
00:16:27,240 --> 00:16:30,640
And finally, by mathematical
induction, if 'n' is any
322
00:16:30,640 --> 00:16:35,300
positive integer, 'f of x' to
the n-th power is just 'n'
323
00:16:35,300 --> 00:16:36,810
times 'f of x'.
324
00:16:36,810 --> 00:16:38,630
And the proof is rather clear.
325
00:16:38,630 --> 00:16:42,070
Namely, 'f of x' to the n-th
power when 'n' is a positive
326
00:16:42,070 --> 00:16:46,160
integer means 'f of 'x'
times 'x' times 'x'
327
00:16:46,160 --> 00:16:48,440
times 'x'', 'n' times.
328
00:16:48,440 --> 00:16:52,140
And since the logarithmic
property says that 'f' of a
329
00:16:52,140 --> 00:16:56,120
product is a sum of the 'f's,
that's equal to 'f of x' plus
330
00:16:56,120 --> 00:17:00,360
'f of x' plus et cetera plus 'f
of x', 'n' times, and that
331
00:17:00,360 --> 00:17:03,290
precisely is just 'n'
times 'f of x'.
332
00:17:03,290 --> 00:17:05,730
In other words, notice that
our definition of a
333
00:17:05,730 --> 00:17:10,740
logarithmic function, simply
that 'f of 'x1 times x2'' is
334
00:17:10,740 --> 00:17:15,930
equal to 'f of x1' plus 'f of
x2' allows us to deduce all of
335
00:17:15,930 --> 00:17:19,970
the familiar properties that
were known to us in terms of
336
00:17:19,970 --> 00:17:23,369
the traditional meaning
of logarithm.
337
00:17:23,369 --> 00:17:26,420
And now we come to the most
important question of today's
338
00:17:26,420 --> 00:17:30,200
lecture, and that is what does
all of this discussion have to
339
00:17:30,200 --> 00:17:34,220
do with the function that we've
called capital 'L of x'?
340
00:17:34,220 --> 00:17:35,830
And that's what will
tackle next.
341
00:17:35,830 --> 00:17:39,670
Let's take a look and see
whether capital 'L of x' is
342
00:17:39,670 --> 00:17:43,220
indeed a logarithmic function.
343
00:17:43,220 --> 00:17:48,540
Well, if 'L' is to be a
logarithmic function, if 'b'
344
00:17:48,540 --> 00:17:53,220
is any constant, in particular
'L of 'b times x'' had better
345
00:17:53,220 --> 00:17:55,830
be equal to 'L of b'
plus 'L of x'.
346
00:17:55,830 --> 00:17:58,550
That's the definition
of logarithmic.
347
00:17:58,550 --> 00:18:00,210
Now we don't know
if that's true.
348
00:18:00,210 --> 00:18:04,390
What do we have at our disposal
to be able to see
349
00:18:04,390 --> 00:18:06,190
whether we can solve this
problem or not?
350
00:18:06,190 --> 00:18:09,260
How do we, using calculus,
check to see whether two
351
00:18:09,260 --> 00:18:10,750
functions are equal?
352
00:18:10,750 --> 00:18:12,860
Remember the standard
approach is what?
353
00:18:12,860 --> 00:18:15,730
Take the derivative
of both sides.
354
00:18:15,730 --> 00:18:17,970
If the derivatives are
equal, then the
355
00:18:17,970 --> 00:18:19,930
functions differ by a constant.
356
00:18:19,930 --> 00:18:22,530
And if we can show that that
constant is 0, then the two
357
00:18:22,530 --> 00:18:24,380
functions are equal.
358
00:18:24,380 --> 00:18:26,990
So let's take a look and see
how that works over here.
359
00:18:26,990 --> 00:18:31,230
First of all, let's see what the
derivative of 'L of bx' is
360
00:18:31,230 --> 00:18:32,650
with respect to 'x'.
361
00:18:32,650 --> 00:18:35,430
And by the way, here again is
the beauty of what we mean
362
00:18:35,430 --> 00:18:39,280
when we say that all of the
study that we're making allows
363
00:18:39,280 --> 00:18:42,340
us to utilize every fundamental
result that we've
364
00:18:42,340 --> 00:18:43,450
learned before.
365
00:18:43,450 --> 00:18:46,630
We've learned that the basic
definition of 'L' is what?
366
00:18:46,630 --> 00:18:49,980
That if you differentiate 'L'
with respect to the given
367
00:18:49,980 --> 00:18:52,960
variable, you get 1 over
that variable.
368
00:18:52,960 --> 00:18:56,940
We want the derivative of 'L
of bx' with respect to 'x'.
369
00:18:56,940 --> 00:19:00,140
Now, you see the idea is the
derivative of 'L of u' with
370
00:19:00,140 --> 00:19:02,590
respect to 'u' is '1 over u'.
371
00:19:02,590 --> 00:19:06,330
You see by the chain rule, if we
were differentiating 'L of
372
00:19:06,330 --> 00:19:12,050
bx' with respect to 'bx', that
would be '1 over bx'.
373
00:19:12,050 --> 00:19:13,290
I shouldn't say by the
chain rule yet.
374
00:19:13,290 --> 00:19:16,200
What I'm saying is if we
differentiate 'L of bx' with
375
00:19:16,200 --> 00:19:19,680
respect to 'bx', we would
have '1 over bx'.
376
00:19:19,680 --> 00:19:22,150
But we're not differentiating
with respect to 'bx', we're
377
00:19:22,150 --> 00:19:24,410
differentiating with
respect to 'x'.
378
00:19:24,410 --> 00:19:26,410
And this is where the
chain rule comes in.
379
00:19:26,410 --> 00:19:31,660
Namely, we rewrite the
derivative of 'L of bx' with
380
00:19:31,660 --> 00:19:35,730
respect to 'x' as the derivative
'L of bx' with
381
00:19:35,730 --> 00:19:39,020
respect to 'bx' times the
derivative of 'bx' with
382
00:19:39,020 --> 00:19:40,170
respect to 'b'.
383
00:19:40,170 --> 00:19:41,700
And in this way we get what?
384
00:19:41,700 --> 00:19:43,860
The first factor
is '1 over bx'.
385
00:19:43,860 --> 00:19:45,990
The second factor
is 'b' itself--
386
00:19:45,990 --> 00:19:47,310
remember 'b' is a constant--
387
00:19:47,310 --> 00:19:52,200
and therefore the derivative
of 'L of bx' is '1 over x'.
388
00:19:52,200 --> 00:19:54,860
On the other hand, what is
the derivative of 'L of
389
00:19:54,860 --> 00:19:56,440
b' plus 'L of x'?
390
00:19:56,440 --> 00:20:00,460
Since 'b' is a constant, the
derivative 'L of b' is 0.
391
00:20:00,460 --> 00:20:03,250
And by definition, the
derivative of 'L of x' with
392
00:20:03,250 --> 00:20:05,440
respect to 'x' is '1 over x'.
393
00:20:05,440 --> 00:20:09,710
And now you see, comparing these
two results, we see that
394
00:20:09,710 --> 00:20:14,430
'L of bx' and 'L of b' plus 'L
of x have the same derivative,
395
00:20:14,430 --> 00:20:18,220
hence they differ by a constant
which I'll call 'c'.
396
00:20:18,220 --> 00:20:21,830
In other words, notice that
capital 'L' is what I call
397
00:20:21,830 --> 00:20:23,300
almost logarithmic.
398
00:20:23,300 --> 00:20:26,670
If it weren't for this factor
constant 'c' in here, it would
399
00:20:26,670 --> 00:20:28,810
be a logarithmic function.
400
00:20:28,810 --> 00:20:30,820
Well at any rate,
what do we have?
401
00:20:30,820 --> 00:20:34,730
We know that capital 'L of bx'
is equal to capital 'L of b'
402
00:20:34,730 --> 00:20:38,450
plus capital 'L of x' plus
some constant 'c'.
403
00:20:38,450 --> 00:20:42,190
To evaluate the constant, we
only have to evaluate it at
404
00:20:42,190 --> 00:20:46,030
one element in the domain, let's
pick 'x' equal to 1.
405
00:20:46,030 --> 00:20:49,100
The motif being that if 'x' is
1, notice that on the left
406
00:20:49,100 --> 00:20:52,070
hand side you have an 'L of b'
term, on the right hand side
407
00:20:52,070 --> 00:20:54,590
you have an 'L of b' term,
and they will cancel.
408
00:20:54,590 --> 00:20:58,570
In other words, if 'x' is 1,
this equation becomes 'L of b'
409
00:20:58,570 --> 00:21:02,500
equals 'L of b' plus 'L of 1'
plus 'c', from which it
410
00:21:02,500 --> 00:21:06,180
follows that 'c' is
minus 'L of 1'.
411
00:21:06,180 --> 00:21:08,060
'c' is minus 'L of 1'.
412
00:21:08,060 --> 00:21:13,010
Now what do we want to 'c' to
be if 'c' were equal to 0?
413
00:21:13,010 --> 00:21:17,510
If 'c' where equal to 0, this
would be logarithmic, and all
414
00:21:17,510 --> 00:21:24,350
we need to make 'c' equal to 0
is to set 'L of 1' equal to 0.
415
00:21:24,350 --> 00:21:28,105
In other words, summarizing this
result, capital 'L of x'
416
00:21:28,105 --> 00:21:31,830
is logarithmic if capital
'L of 1' is 0.
417
00:21:31,830 --> 00:21:35,530
Now remember, we have a whole
family of 'L's that work.
418
00:21:35,530 --> 00:21:39,180
What we're saying now is let's
pick the member of the family
419
00:21:39,180 --> 00:21:42,660
of 'L's that passes through
the point (1,0) .
420
00:21:42,660 --> 00:21:45,320
And because that's logarithmic,
let's give that a
421
00:21:45,320 --> 00:21:46,320
special name.
422
00:21:46,320 --> 00:21:50,160
In essence, we will define
this symbol.
423
00:21:50,160 --> 00:21:51,750
It's written 'ln of x'.
424
00:21:51,750 --> 00:21:55,200
It will later be called the
natural logarithm of 'x'.
425
00:21:55,200 --> 00:21:58,120
I'm trying to avoid the word
logarithm here as much as
426
00:21:58,120 --> 00:22:01,400
possible, because I want you to
see that we haven't had to
427
00:22:01,400 --> 00:22:04,880
use exponents at all in making
this kind of a definition.
428
00:22:04,880 --> 00:22:09,340
But let's define 'ln of x' to
be the member of the family
429
00:22:09,340 --> 00:22:12,080
capital 'L of x' plus
'c' which passes
430
00:22:12,080 --> 00:22:14,320
through the point (1,0).
431
00:22:14,320 --> 00:22:17,950
In essence then, what
is 'ln of x'?
432
00:22:17,950 --> 00:22:20,800
I'll call it natural log, it's
easier for me to say.
433
00:22:20,800 --> 00:22:24,390
The natural 'log of x' is that
function whose derivative with
434
00:22:24,390 --> 00:22:28,470
respect to 'x' is '1 over
x', and characterized by
435
00:22:28,470 --> 00:22:30,980
the fact that what?
436
00:22:30,980 --> 00:22:33,620
If the input is 1,
the output is 0.
437
00:22:33,620 --> 00:22:35,260
In other words, the graph
passes through
438
00:22:35,260 --> 00:22:37,660
the point (1 , 0).
439
00:22:37,660 --> 00:22:41,650
Again, notice that in terms
of what we said earlier in
440
00:22:41,650 --> 00:22:46,190
today's lesson, we talked about
the functions 'L of x',
441
00:22:46,190 --> 00:22:48,880
and talked about, in the
differential calculus
442
00:22:48,880 --> 00:22:52,290
approach, the curve could
go through any
443
00:22:52,290 --> 00:22:53,470
point on the x-axis.
444
00:22:53,470 --> 00:22:56,440
What we're saying is now if the
point on the x-axis that
445
00:22:56,440 --> 00:23:00,860
the curve crosses at is (1 ,
0), that's what 'lnx' will
446
00:23:00,860 --> 00:23:03,140
mean, the natural
log function.
447
00:23:03,140 --> 00:23:06,910
In terms of the integral
calculus approach, if the
448
00:23:06,910 --> 00:23:11,010
particular 'a' value that we
choose, the fixed value that
449
00:23:11,010 --> 00:23:14,310
we're going to study the area
under, notice that all we're
450
00:23:14,310 --> 00:23:16,730
saying is pick 'a' to be 1.
451
00:23:16,730 --> 00:23:19,190
And by the way, as a
quick check, all
452
00:23:19,190 --> 00:23:19,940
we're saying is what?
453
00:23:19,940 --> 00:23:23,090
The 'natural log of x' is
defined to be the area under
454
00:23:23,090 --> 00:23:25,200
this curve as a function
of 'x'.
455
00:23:25,200 --> 00:23:28,090
In terms of the definite
integral, that's the integral
456
00:23:28,090 --> 00:23:30,760
from '1 to x,' 'dt over t'.
457
00:23:30,760 --> 00:23:33,880
And notice that if you pick 'x'
to be 1, the natural log
458
00:23:33,880 --> 00:23:39,330
of 1 is the integral from 1 to
1 'dt over t', and that's 0,
459
00:23:39,330 --> 00:23:42,030
just as it should be.
460
00:23:42,030 --> 00:23:44,340
So at any rate now, that
tells me how to
461
00:23:44,340 --> 00:23:46,170
define the natural log.
462
00:23:46,170 --> 00:23:48,260
Am I doing this in terms
of exponents?
463
00:23:48,260 --> 00:23:48,790
No.
464
00:23:48,790 --> 00:23:50,270
I am trying to do what?
465
00:23:50,270 --> 00:23:53,830
Find a function whose derivative
with respect to 'x'
466
00:23:53,830 --> 00:23:57,430
is '1 over x', and I'd also like
that function, since it's
467
00:23:57,430 --> 00:24:01,010
that close to being a
logarithmic function, to be a
468
00:24:01,010 --> 00:24:02,970
logarithmic function.
469
00:24:02,970 --> 00:24:07,250
In other words, this is how I
invented the function 'ln of
470
00:24:07,250 --> 00:24:09,190
x', the 'natural log of x'.
471
00:24:09,190 --> 00:24:12,310
It's derivative with respect to
'x' is '1 over x', and the
472
00:24:12,310 --> 00:24:16,480
natural log of a product is the
sum of the natural logs.
473
00:24:16,480 --> 00:24:20,050
And by the way, that's exactly
how we use this material.
474
00:24:20,050 --> 00:24:22,970
Let me just take a few minutes
and go over something that we
475
00:24:22,970 --> 00:24:25,380
already had an answer
for, just to show
476
00:24:25,380 --> 00:24:26,680
you how this works.
477
00:24:26,680 --> 00:24:30,650
Let me try to rederive the
product rule using logarithms.
478
00:24:30,650 --> 00:24:33,150
Suppose 'u' and 'v' are
differentiable functions of
479
00:24:33,150 --> 00:24:37,360
'x', and I want to find the
derivative of 'u' times 'v'
480
00:24:37,360 --> 00:24:38,440
with respect to 'x'.
481
00:24:38,440 --> 00:24:41,290
In other words, I'd like
to find 'dy dx'.
482
00:24:41,290 --> 00:24:44,480
OK, I take the natural
log of both sides.
483
00:24:44,480 --> 00:24:47,770
In other words, I say if 'y'
equals 'u' times 'v', the
484
00:24:47,770 --> 00:24:53,780
natural log 'y' equals natural
log 'u times v'.
485
00:24:53,780 --> 00:24:56,550
Now what is the property that
the natural log function has?
486
00:24:56,550 --> 00:24:59,840
I deliberately chose it to be
that member of the l family
487
00:24:59,840 --> 00:25:01,180
that was logarithmic.
488
00:25:01,180 --> 00:25:05,480
The natural log of 'u times
v' is 'natural log u' plus
489
00:25:05,480 --> 00:25:07,590
'natural log v'.
490
00:25:07,590 --> 00:25:11,700
Now, can I differentiate
this implicitly
491
00:25:11,700 --> 00:25:12,760
with respect to 'x'?
492
00:25:12,760 --> 00:25:15,500
You see how all of our
old stuff keeps
493
00:25:15,500 --> 00:25:17,040
coming up in new context.
494
00:25:17,040 --> 00:25:20,110
I take this equation and I
differentiate it implicitly
495
00:25:20,110 --> 00:25:21,690
with respect to 'x'.
496
00:25:21,690 --> 00:25:23,800
The derivative of the
left hand side is '1
497
00:25:23,800 --> 00:25:26,160
over y', 'dy dx'.
498
00:25:26,160 --> 00:25:29,020
The derivative of the right
hand side is what?
499
00:25:29,020 --> 00:25:31,380
Well, the derivative of 'log u'
with respect to 'u' is '1
500
00:25:31,380 --> 00:25:33,260
over u', but I'm differentiating
it with
501
00:25:33,260 --> 00:25:36,740
respect to 'x', so I must use
the correction factor by the
502
00:25:36,740 --> 00:25:38,530
chain rule 'du dx'.
503
00:25:38,530 --> 00:25:41,670
And similarly, the derivative of
'log v' with respect to 'x'
504
00:25:41,670 --> 00:25:44,020
is '1 over v', 'dv dx'.
505
00:25:44,020 --> 00:25:47,790
And now multiplying through by
'y', I have obtained that 'dy
506
00:25:47,790 --> 00:25:53,170
dx' is ''y over u' du dx'
plus ''y over v' dv dx'.
507
00:25:53,170 --> 00:25:56,470
But remembering that 'y' is
equal to 'u' times 'v', this
508
00:25:56,470 --> 00:26:02,060
becomes 'v 'du dx'' plus 'u 'dv
dx', and we have arrived
509
00:26:02,060 --> 00:26:05,870
at the familiar product
rule using logarithmic
510
00:26:05,870 --> 00:26:07,120
differentiation.
511
00:26:07,120 --> 00:26:09,290
You see the point that's really
important to stress
512
00:26:09,290 --> 00:26:12,190
here it that it wasn't important
whether the natural
513
00:26:12,190 --> 00:26:13,950
log was an exponent or not.
514
00:26:13,950 --> 00:26:16,440
The important thing that we used
about logarithms, whether
515
00:26:16,440 --> 00:26:18,700
it was our high school course,
whether it's going to be in
516
00:26:18,700 --> 00:26:22,140
our college calculus course, the
important thing was what?
517
00:26:22,140 --> 00:26:25,580
That when we wanted to write the
log of a product, it was
518
00:26:25,580 --> 00:26:27,580
the sum of the logs.
519
00:26:27,580 --> 00:26:29,720
And now the only additional
fact that we have from the
520
00:26:29,720 --> 00:26:32,820
calculus approach is that the
derivative of 'log x' with
521
00:26:32,820 --> 00:26:36,800
respect to 'x' was defined
to be '1 over x'.
522
00:26:36,800 --> 00:26:41,140
By the way, since there is
a tendency to think of
523
00:26:41,140 --> 00:26:44,970
traditional logarithms when one
uses the word logarithm,
524
00:26:44,970 --> 00:26:46,780
if we so wanted--
525
00:26:46,780 --> 00:26:48,850
and there's no reason why
we have to do this--
526
00:26:48,850 --> 00:26:51,940
if we wanted to associate
a base with
527
00:26:51,940 --> 00:26:53,540
the natural log system--
528
00:26:53,540 --> 00:26:56,650
this is just a little aside, and
we'll talk about this more
529
00:26:56,650 --> 00:26:59,720
perhaps in the learning
exercises or in the
530
00:26:59,720 --> 00:27:01,780
supplementary notes
as need be--
531
00:27:01,780 --> 00:27:03,090
but the idea is this.
532
00:27:03,090 --> 00:27:05,970
Notice that in a traditional
logarithm system, if the base
533
00:27:05,970 --> 00:27:09,380
is 'b', the base is
characterized by the fact that
534
00:27:09,380 --> 00:27:12,400
the 'log of 'b to the
base b'' is 1.
535
00:27:12,400 --> 00:27:18,100
Thus, if we use 'e' to denote
the base for the natural log,
536
00:27:18,100 --> 00:27:21,820
whatever 'e' is, it must be
characterized by the fact that
537
00:27:21,820 --> 00:27:24,050
the 'natural log of e' is 1.
538
00:27:24,050 --> 00:27:27,430
And the reason that I put the
word base in quotation marks
539
00:27:27,430 --> 00:27:31,150
here is that I would like you
to observe that again, if I
540
00:27:31,150 --> 00:27:34,640
have never heard of the word
exponent, it still makes sense
541
00:27:34,640 --> 00:27:38,650
to say find the number e
such that the 'natural
542
00:27:38,650 --> 00:27:40,220
log of e' is 1.
543
00:27:40,220 --> 00:27:43,290
In fact, I can give you two
interpretations for that
544
00:27:43,290 --> 00:27:47,510
number 'e', and as a byproduct,
even show you why
545
00:27:47,510 --> 00:27:50,430
the second fundamental theorem
of integral calculus is as
546
00:27:50,430 --> 00:27:53,346
powerful as it really is.
547
00:27:53,346 --> 00:27:56,160
See, the idea is this.
548
00:27:56,160 --> 00:27:59,300
To find what 'e' is, all we're
saying is take the curve 'y'
549
00:27:59,300 --> 00:28:02,620
equals 'natural log x'.
550
00:28:02,620 --> 00:28:05,000
Look to see where the
y-coordinate is 1.
551
00:28:05,000 --> 00:28:08,130
In other words, draw the
line 'y' equals 1.
552
00:28:08,130 --> 00:28:12,020
Where that line intercepts the
curve 'y' equals 'log x', that
553
00:28:12,020 --> 00:28:13,160
x-coordinate is 'e'.
554
00:28:13,160 --> 00:28:16,700
In other words, the 'natural log
of e' must be 1, so this
555
00:28:16,700 --> 00:28:19,650
is geometrically how I would
locate 'e' using
556
00:28:19,650 --> 00:28:21,130
differential calculus.
557
00:28:21,130 --> 00:28:26,250
If I wanted to use integral
calculus, notice that the 'log
558
00:28:26,250 --> 00:28:28,440
of e' by definition is what?
559
00:28:28,440 --> 00:28:31,960
The integral from 1 to
'e', 'dt over t'.
560
00:28:31,960 --> 00:28:34,750
That's the area of
this region 'R'.
561
00:28:34,750 --> 00:28:36,600
Now what do I want 'e' to be?
562
00:28:36,600 --> 00:28:41,420
I want 'e' to be that number
that makes this area 1.
563
00:28:41,420 --> 00:28:45,650
In other words, I want the
'natural log of e' to be 1.
564
00:28:45,650 --> 00:28:50,170
So again, in the same way that
one can think of pi as being
565
00:28:50,170 --> 00:28:53,490
geometrically constructed,
notice I can construct 'e'
566
00:28:53,490 --> 00:28:55,970
geometrically, namely
again what?
567
00:28:55,970 --> 00:29:00,450
I take the curve 'y' equals '1
over t' from 't' equals 1,
568
00:29:00,450 --> 00:29:04,120
bounded below by the t-axis,
and I keep shifting over to
569
00:29:04,120 --> 00:29:09,890
the right until the area of this
region 'R' is exactly 1.
570
00:29:09,890 --> 00:29:14,410
The 't' value that makes this
area 1 is called 'e'.
571
00:29:14,410 --> 00:29:17,880
And by the way, notice again
the power of what I mean by
572
00:29:17,880 --> 00:29:19,200
the area approach.
573
00:29:19,200 --> 00:29:22,430
Notice I can start now to
get estimates on what
574
00:29:22,430 --> 00:29:23,270
'e' must look like.
575
00:29:23,270 --> 00:29:25,530
Let me show you what I'm
driving at over here.
576
00:29:25,530 --> 00:29:28,840
Let's suppose we take the curve
'y' equals '1 over t'
577
00:29:28,840 --> 00:29:30,760
from 1 to 2.
578
00:29:30,760 --> 00:29:33,480
Now what do we mean
by natural log 2?
579
00:29:33,480 --> 00:29:38,280
Natural log 2 is by definition
the area of this region here.
580
00:29:38,280 --> 00:29:40,980
Now notice that the smallest
height of this region, since
581
00:29:40,980 --> 00:29:43,880
the curve is 'y' equals '1 over
t', the smallest height
582
00:29:43,880 --> 00:29:48,080
of this region is 1/2, and the
tallest height, the highest
583
00:29:48,080 --> 00:29:50,020
height in this region is 1.
584
00:29:50,020 --> 00:29:54,390
Consequently, whatever the area
of this region is, it's
585
00:29:54,390 --> 00:29:57,280
less than the area of the
inscribed rectangle--
586
00:29:57,280 --> 00:30:01,460
it's greater than the area of
the inscribed rectangle, and
587
00:30:01,460 --> 00:30:05,330
less than the area of the
circumscribed rectangle.
588
00:30:05,330 --> 00:30:08,620
Now notice that both of these
rectangles have base 1, and
589
00:30:08,620 --> 00:30:10,350
we're going from 1 to 2.
590
00:30:10,350 --> 00:30:13,490
The height of the inscribed
rectangle is 1/2, therefore
591
00:30:13,490 --> 00:30:16,200
the area of the small
rectangle is 1/2.
592
00:30:16,200 --> 00:30:20,930
The area of the big rectangle,
it's a 1 by 1 rectangle, is 1.
593
00:30:20,930 --> 00:30:25,290
And now what we have is that
whatever the natural log of 2
594
00:30:25,290 --> 00:30:29,250
is, it being the area of this
region, it must be what?
595
00:30:29,250 --> 00:30:32,600
Less than 1 but greater
than 1/2.
596
00:30:32,600 --> 00:30:35,020
We also know that whatever
'e' is, the 'log of
597
00:30:35,020 --> 00:30:37,530
e' is equal to 1.
598
00:30:37,530 --> 00:30:39,350
Well, let's go on a little
bit further.
599
00:30:39,350 --> 00:30:41,250
What about the log of 4?
600
00:30:41,250 --> 00:30:46,410
The log of 4, natural log of
4, is the log of 2 squared.
601
00:30:46,410 --> 00:30:51,290
But we've already seen that
one of the properties of a
602
00:30:51,290 --> 00:30:53,960
logarithmic function is that
you can bring the exponent
603
00:30:53,960 --> 00:30:56,020
down as a multiplier.
604
00:30:56,020 --> 00:30:59,340
See, 'f of 'x to the n''
is 'n 'f of x''.
605
00:30:59,340 --> 00:31:03,470
This becomes 2 log 2, and
because the natural log of 2
606
00:31:03,470 --> 00:31:06,680
is greater than 1/2, twice
the natural log of 2
607
00:31:06,680 --> 00:31:08,890
is more than 1.
608
00:31:08,890 --> 00:31:13,010
In other words, if log 2 is more
than 1/2, twice log 2 is
609
00:31:13,010 --> 00:31:14,400
more than 1.
610
00:31:14,400 --> 00:31:17,200
In other words, if we put these
three lines together, we
611
00:31:17,200 --> 00:31:20,890
have that the natural log of 2
is less than the 'natural log
612
00:31:20,890 --> 00:31:24,570
of e', which in turn is less
than the natural log of 4.
613
00:31:24,570 --> 00:31:28,490
By the way, notice that since
the derivative of 'log x' with
614
00:31:28,490 --> 00:31:32,660
respect to 'x' is '1 over x',
and that's positive, 'log x'
615
00:31:32,660 --> 00:31:34,580
is a one to one function.
616
00:31:34,580 --> 00:31:38,250
Therefore, if the log of 2 is
less than the 'log of e' is
617
00:31:38,250 --> 00:31:42,080
less than the log of 4, it
follows that 2 must be less
618
00:31:42,080 --> 00:31:45,230
than 'e', which in turn
must be less than 4.
619
00:31:45,230 --> 00:31:48,450
In other words, even with this
crude approximation of just
620
00:31:48,450 --> 00:31:52,260
inscribing and circumscribing
rectangles, I can show again
621
00:31:52,260 --> 00:31:56,410
without reference to exponents
that whatever the number 'e'
622
00:31:56,410 --> 00:32:02,330
is, the number e defined by the
fact that 'ln of e' is 1,
623
00:32:02,330 --> 00:32:06,400
that number 'e' is some number
between 2 and 4.
624
00:32:06,400 --> 00:32:08,830
But again, we won't dwell
on this too long.
625
00:32:08,830 --> 00:32:12,000
What I want to do now is to come
back to a summary point,
626
00:32:12,000 --> 00:32:14,570
and at the same time,
lead into the
627
00:32:14,570 --> 00:32:16,020
lecture for next time.
628
00:32:16,020 --> 00:32:20,790
Recall that we began this
lecture with the problem of
629
00:32:20,790 --> 00:32:24,740
solving the situation where
the rate of change was
630
00:32:24,740 --> 00:32:27,390
proportional to the
amount present.
631
00:32:27,390 --> 00:32:31,180
Given 'dm/ dt' equals 'km', we
separated variables and got
632
00:32:31,180 --> 00:32:34,920
down to the stage where the
integral of 'dm over m' was
633
00:32:34,920 --> 00:32:36,880
'kt' plus a constant.
634
00:32:36,880 --> 00:32:40,540
And all we did in the rest of
today's lecture that was at
635
00:32:40,540 --> 00:32:45,350
all different, was we invented
and constructed the particular
636
00:32:45,350 --> 00:32:48,970
function whose derivative would
be '1 over m', and which
637
00:32:48,970 --> 00:32:50,550
had the logarithmic property.
638
00:32:50,550 --> 00:32:52,920
In other words, we can now
say that this is the
639
00:32:52,920 --> 00:32:54,340
answer to the problem.
640
00:32:54,340 --> 00:32:58,240
This problem is explicitly
solved because the natural log
641
00:32:58,240 --> 00:33:00,690
function can be viewed as
an area under a curve.
642
00:33:00,690 --> 00:33:02,950
We can construct it
for each 'm'.
643
00:33:02,950 --> 00:33:05,400
This is then what we
managed to do.
644
00:33:05,400 --> 00:33:08,800
And by the way, all I'm saying
now is that since the log is a
645
00:33:08,800 --> 00:33:10,250
one to one function--
646
00:33:10,250 --> 00:33:12,250
see, its derivative is
always positive--
647
00:33:12,250 --> 00:33:14,670
can't we talk about the
inverse function?
648
00:33:14,670 --> 00:33:16,880
In other words, notice that
another way of writing this is
649
00:33:16,880 --> 00:33:21,720
that id 'natural log m' is 'kt'
plus 'c', 'm' itself must
650
00:33:21,720 --> 00:33:26,490
be the 'inverse log
of kt' plus 'c'.
651
00:33:26,490 --> 00:33:30,450
And you see, next time what
we're going to do is to
652
00:33:30,450 --> 00:33:35,610
explore what we mean
by the inverse log.
653
00:33:35,610 --> 00:33:39,300
At any rate, in summarizing
today's lecture, to make sure
654
00:33:39,300 --> 00:33:43,250
that we didn't fall victim to
all the computational details,
655
00:33:43,250 --> 00:33:46,500
notice that all we did was
physically motivated the
656
00:33:46,500 --> 00:33:50,070
necessity for inventing a
function whose derivative with
657
00:33:50,070 --> 00:33:51,920
respect to 'x' was '1 over x'.
658
00:33:51,920 --> 00:33:55,900
And from that point on,
everything else that we did
659
00:33:55,900 --> 00:33:59,730
followed as applications of
material that came before.
660
00:33:59,730 --> 00:34:03,980
It's in this sense that the
course now picks up in tempo.
661
00:34:03,980 --> 00:34:07,050
That as we go on now, we're
going to be able to cover
662
00:34:07,050 --> 00:34:10,679
larger globs of material in one
sitting, because we will
663
00:34:10,679 --> 00:34:14,130
find, at least for the next
several lectures, that every
664
00:34:14,130 --> 00:34:18,989
new topic is basically one new
idea together with all of the
665
00:34:18,989 --> 00:34:20,330
old recipes.
666
00:34:20,330 --> 00:34:22,330
At any rate, until next
time, goodbye.
667
00:34:25,489 --> 00:34:28,030
ANNOUNCER: Funding for the
publication of this video was
668
00:34:28,030 --> 00:34:32,750
provided by the Gabriella and
Paul Rosenbaum Foundation.
669
00:34:32,750 --> 00:34:36,920
Help OCW continue to provide
free and open access to MIT
670
00:34:36,920 --> 00:34:41,120
courses by making a donation
at ocw.mit.edu/donate.