1
00:00:01,060 --> 00:00:03,375
So, now we'll look
at third kind of sum

2
00:00:03,375 --> 00:00:06,310
that comes up all the
time called harmonic sums.

3
00:00:06,310 --> 00:00:08,850
And we'll begin by
examining an example where

4
00:00:08,850 --> 00:00:10,520
they come up really directly.

5
00:00:10,520 --> 00:00:12,500
So, here's the puzzle.

6
00:00:12,500 --> 00:00:16,560
Suppose that I'm trying to stack
a bunch of books on a table.

7
00:00:16,560 --> 00:00:19,450
Assume all the books are
the same size and weight

8
00:00:19,450 --> 00:00:23,180
and uniform, and I'd like
to stack them up one on top

9
00:00:23,180 --> 00:00:24,860
of the other in some way.

10
00:00:24,860 --> 00:00:28,720
And try to get them as far
out past the end of the table

11
00:00:28,720 --> 00:00:29,910
as I can manage.

12
00:00:29,910 --> 00:00:32,570
Now, notice in this picture,
it seems kind of paradoxical.

13
00:00:32,570 --> 00:00:36,170
The top book, the back
end of the top book

14
00:00:36,170 --> 00:00:38,440
is past the edge of the table.

15
00:00:38,440 --> 00:00:40,060
Is it possible to do that?

16
00:00:40,060 --> 00:00:44,840
Is it possible to get the top
book, the back of the top book

17
00:00:44,840 --> 00:00:46,090
past the edge of the table?

18
00:00:46,090 --> 00:00:50,670
And how far out can you get the
further most book to the right?

19
00:00:50,670 --> 00:00:52,920
That's the question
we want to ask.

20
00:00:52,920 --> 00:00:56,730
Well, let's go back and
do it for the simplest

21
00:00:56,730 --> 00:00:58,010
case, which is one book.

22
00:00:58,010 --> 00:01:01,200
So, this amount will be a
function of how many books

23
00:01:01,200 --> 00:01:02,280
we have.

24
00:01:02,280 --> 00:01:04,709
We're interested in the
overhang using n books.

25
00:01:04,709 --> 00:01:08,040
Overhang is the amount
past the edge of the table

26
00:01:08,040 --> 00:01:13,360
that the rightmost end
of any book can be.

27
00:01:13,360 --> 00:01:14,569
What do you do with one book?

28
00:01:14,569 --> 00:01:16,526
Well, with one book,
assuming that the thing is

29
00:01:16,526 --> 00:01:18,380
uniform, the center
of mass in the middle.

30
00:01:18,380 --> 00:01:20,990
Let's assume the
book is of length 1.

31
00:01:20,990 --> 00:01:22,770
So, the center of
mass of the book

32
00:01:22,770 --> 00:01:24,090
is at halfway down the book.

33
00:01:24,090 --> 00:01:27,260
And if that center of mass
is not over the table,

34
00:01:27,260 --> 00:01:30,970
then you're going to have torque
and the book is going to fall.

35
00:01:30,970 --> 00:01:34,550
So, you've got to keep the
center of mass supported.

36
00:01:34,550 --> 00:01:36,660
And the way to get
the largest overhang

37
00:01:36,660 --> 00:01:39,470
is to have the center of mass
right at the edge of the table

38
00:01:39,470 --> 00:01:40,500
here.

39
00:01:40,500 --> 00:01:42,650
And in that case,
you can get the book

40
00:01:42,650 --> 00:01:46,030
to stick out half a book
length without falling.

41
00:01:46,030 --> 00:01:51,290
And what that tells us is that
the one book overhang is 1/2.

42
00:01:51,290 --> 00:01:54,720
It'll balance with
the furthest end

43
00:01:54,720 --> 00:01:58,120
out exactly if the center
of mass is on the edge,

44
00:01:58,120 --> 00:02:01,950
and I get a half a book
length for unit overhang.

45
00:02:01,950 --> 00:02:04,870
Let's proceed recursively
or inductively.

46
00:02:04,870 --> 00:02:06,520
Suppose I have n books.

47
00:02:06,520 --> 00:02:08,310
How am I going to
get them to balance?

48
00:02:08,310 --> 00:02:10,180
Well, let's assume
that I figured out

49
00:02:10,180 --> 00:02:13,180
how to get a so-called stable
stack of n books, which

50
00:02:13,180 --> 00:02:15,910
if I completely supported
it flat on the table,

51
00:02:15,910 --> 00:02:17,742
it wouldn't fall over.

52
00:02:17,742 --> 00:02:19,200
And I'm going to
show you how to go

53
00:02:19,200 --> 00:02:21,500
from n to n plus 1, which
is how you construct

54
00:02:21,500 --> 00:02:25,330
an arbitrarily large stack of
books that won't fall over.

55
00:02:25,330 --> 00:02:28,570
Well, if the stack completely
resting on the table

56
00:02:28,570 --> 00:02:32,060
won't fall over,
that means that if I

57
00:02:32,060 --> 00:02:36,020
have the center of mass of it
past the edge of the table,

58
00:02:36,020 --> 00:02:38,056
by definition of
the center of mass,

59
00:02:38,056 --> 00:02:39,930
there's going to be an
equal amount of weight

60
00:02:39,930 --> 00:02:41,467
on both sides of
the center of mass,

61
00:02:41,467 --> 00:02:43,800
and the thing is going to
fall off the edge of the table

62
00:02:43,800 --> 00:02:46,620
by the same reasoning
as we did for one book.

63
00:02:46,620 --> 00:02:50,030
So, the stable and n
stack-- stable in the sense

64
00:02:50,030 --> 00:02:54,460
that it won't fall over of
self if it was lying completely

65
00:02:54,460 --> 00:02:55,300
over the table.

66
00:02:55,300 --> 00:02:59,060
In fact, it won't fall over
as long as its center of mass

67
00:02:59,060 --> 00:03:00,260
is over the table.

68
00:03:00,260 --> 00:03:02,270
And to get it out
the furthest amount

69
00:03:02,270 --> 00:03:04,040
to the right, what
I'm going to do

70
00:03:04,040 --> 00:03:07,500
is put it at the
edge of the table.

71
00:03:07,500 --> 00:03:12,170
So, now I know how to place
a stable stack of n books

72
00:03:12,170 --> 00:03:14,850
to get the largest
overhang out of it.

73
00:03:14,850 --> 00:03:17,050
What next?

74
00:03:17,050 --> 00:03:19,980
Well, let's consider
n plus 1 books.

75
00:03:19,980 --> 00:03:21,072
And what do I have to do?

76
00:03:21,072 --> 00:03:22,530
So, I'm trying to
do the same deal.

77
00:03:22,530 --> 00:03:24,590
Suppose that I have a
nice stack of n books

78
00:03:24,590 --> 00:03:28,490
and I know how to support
it so it won't tip over.

79
00:03:28,490 --> 00:03:31,900
And I now have n plus 1 books
and I want to get out further.

80
00:03:31,900 --> 00:03:33,170
What do I have to do?

81
00:03:33,170 --> 00:03:36,480
Well, by the basic reasoning
that we just went through,

82
00:03:36,480 --> 00:03:39,380
now the center of mass of the
whole stack of n plus 1 books

83
00:03:39,380 --> 00:03:40,945
has to be over the
edge of the table.

84
00:03:40,945 --> 00:03:42,820
That's the way I'm going
to get out furthest.

85
00:03:42,820 --> 00:03:45,820
So, I know where the center
of mass of n plus books

86
00:03:45,820 --> 00:03:49,210
is going to be, at
the edge of the table.

87
00:03:49,210 --> 00:03:52,310
What about the center of
mass of the top n books?

88
00:03:52,310 --> 00:03:54,100
Well, I need them
to be supported.

89
00:03:54,100 --> 00:03:56,650
I need their center of
mass to be supported.

90
00:03:56,650 --> 00:03:59,020
They'll be supported,
providing their center of mass

91
00:03:59,020 --> 00:04:01,180
is over the bottom
book somewhere.

92
00:04:01,180 --> 00:04:02,810
And the way to get
it out furthest

93
00:04:02,810 --> 00:04:06,300
is to have it over the right
edge of the bottom book.

94
00:04:06,300 --> 00:04:08,920
So, I'm going to put
the center of mass

95
00:04:08,920 --> 00:04:16,320
of the top n books at the edge
of the n plus first book here.

96
00:04:16,320 --> 00:04:19,140
And that means that the
incremental overhang that I

97
00:04:19,140 --> 00:04:20,890
get, the increase
in overhang that I

98
00:04:20,890 --> 00:04:25,260
get by adding one more book,
we can call the delta overhang.

99
00:04:25,260 --> 00:04:28,730
And it's the distance between
the center of mass of n

100
00:04:28,730 --> 00:04:32,460
plus 1 books and the
center of mass of n books.

101
00:04:32,460 --> 00:04:35,460
N here, and n plus 1 here.

102
00:04:35,460 --> 00:04:36,930
Well, let's see what's going on.

103
00:04:36,930 --> 00:04:41,330
The center of mass of the n
books is at some location here.

104
00:04:41,330 --> 00:04:44,540
And the center of mass
of the bottom book

105
00:04:44,540 --> 00:04:47,580
is halfway away,
half a book length

106
00:04:47,580 --> 00:04:50,480
away from where the
n books are balanced

107
00:04:50,480 --> 00:04:52,580
on the edge of the bottom book.

108
00:04:52,580 --> 00:04:54,970
So, the center of mass
of the n books is here.

109
00:04:54,970 --> 00:04:57,260
The center of mass of
the bottom book is there.

110
00:04:57,260 --> 00:05:00,060
The distance
between them is 1/2.

111
00:05:00,060 --> 00:05:04,870
And I need the table to be at
the balance point between the n

112
00:05:04,870 --> 00:05:06,370
books and the one book.

113
00:05:06,370 --> 00:05:09,400
That's where the center of mass
of the n plus 1 books will be.

114
00:05:09,400 --> 00:05:11,670
So, I need to calculate
amount that's going to be,

115
00:05:11,670 --> 00:05:12,980
the increase in overhang.

116
00:05:12,980 --> 00:05:15,040
So, let's abstract
it a little bit.

117
00:05:15,040 --> 00:05:16,560
The delta overhang
is the distance

118
00:05:16,560 --> 00:05:20,230
from the n book to the n
plus 1 book centers of mass.

119
00:05:20,230 --> 00:05:22,980
And if we think of this
as a balancing diagram,

120
00:05:22,980 --> 00:05:24,019
there's the n books.

121
00:05:24,019 --> 00:05:26,310
Or at least, there's the
center of mass of the n books.

122
00:05:26,310 --> 00:05:29,220
There's the center of mass of
the 1 book, their distance 1/2

123
00:05:29,220 --> 00:05:30,730
apart, which we said.

124
00:05:30,730 --> 00:05:33,320
And they have to balance
at the edge of the table.

125
00:05:33,320 --> 00:05:36,470
So, think of the edge of
the table as the pivot point

126
00:05:36,470 --> 00:05:37,410
and it's there.

127
00:05:37,410 --> 00:05:40,260
And I need to calculate,
where is that pivot point?

128
00:05:40,260 --> 00:05:44,650
How do I get this fulcrum
or this balance beam

129
00:05:44,650 --> 00:05:47,820
to balance with weight n
here and weight 1 there,

130
00:05:47,820 --> 00:05:50,120
when their total
length apart is 1/2.

131
00:05:50,120 --> 00:05:52,000
What's this distance?

132
00:05:52,000 --> 00:05:54,550
That distance is the delta
that I'm trying to calculate.

133
00:05:54,550 --> 00:05:58,480
Well, you just know from
physics that the balance point

134
00:05:58,480 --> 00:06:03,030
is going to be the
distance 1/2 divided

135
00:06:03,030 --> 00:06:07,380
by the sum of n and n plus 1.

136
00:06:07,380 --> 00:06:12,060
I need the n times this amount
to equal 1 times that amount.

137
00:06:12,060 --> 00:06:16,410
And if you check that out, it
means that delta is 1/2 over n

138
00:06:16,410 --> 00:06:17,330
plus 1.

139
00:06:17,330 --> 00:06:20,740
Or simplifying, 1
over twice n plus 1.

140
00:06:20,740 --> 00:06:22,660
You should stare at that
diagram a little bit

141
00:06:22,660 --> 00:06:24,240
and remember your
elementary physics

142
00:06:24,240 --> 00:06:29,960
to realize the reasoning
behind the formula for delta.

143
00:06:29,960 --> 00:06:31,979
Well, now I'm done
because basically, I've

144
00:06:31,979 --> 00:06:33,520
just figured out
that the increase is

145
00:06:33,520 --> 00:06:34,610
this delta overhang.

146
00:06:34,610 --> 00:06:35,780
And now I know what it is.

147
00:06:35,780 --> 00:06:38,720
It's 1 over twice n plus 1.

148
00:06:38,720 --> 00:06:41,430
And so, what I can conclude
is that the overhang

149
00:06:41,430 --> 00:06:45,670
of n books, B1 is
1/2 and Bn plus 1

150
00:06:45,670 --> 00:06:49,560
is Bn plus 1 over
twice n plus 1.

151
00:06:49,560 --> 00:06:51,890
So this is a recursive
definition of Bn,

152
00:06:51,890 --> 00:06:54,070
but it's easy to
see how it unwinds.

153
00:06:54,070 --> 00:07:05,130
It means that Bn is 1/2 plus
1/2 of 1 plus 1 plus 1/2 of 2

154
00:07:05,130 --> 00:07:07,800
plus 1 plus 1/2 of
3 plus 1 and so on.

155
00:07:07,800 --> 00:07:12,090
If I factor out the 1/2,
Bn is 1/2 times 1 plus 1/2

156
00:07:12,090 --> 00:07:16,710
plus 1/3 out through 1 over n.

157
00:07:16,710 --> 00:07:18,770
That sum is the harmonic sum.

158
00:07:18,770 --> 00:07:21,450
The sum 1 plus 1/2
up through 1 over n

159
00:07:21,450 --> 00:07:23,800
is called Hn, or
the harmonic sum.

160
00:07:23,800 --> 00:07:26,570
And what we figured out, or
really the harmonic number,

161
00:07:26,570 --> 00:07:28,310
is the value of that sum.

162
00:07:28,310 --> 00:07:31,590
And what we figured
out is that Bn,

163
00:07:31,590 --> 00:07:34,820
the amount that I
can get n books out

164
00:07:34,820 --> 00:07:39,741
up past the edge of
the table is Hn over 2.