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PROFESSOR: All right.
9
00:00:23,620 --> 00:00:25,960
So last time we
talked about methods
10
00:00:25,960 --> 00:00:28,040
for solving recurrences,
and we spent
11
00:00:28,040 --> 00:00:31,624
most of our time talking about
divide-and-conquer recurrences.
12
00:00:31,624 --> 00:00:33,790
These are recurrences where
you break the problem up
13
00:00:33,790 --> 00:00:37,210
into much smaller sub-problems,
like half the size
14
00:00:37,210 --> 00:00:39,570
or 2/3 the size.
15
00:00:39,570 --> 00:00:41,690
And they come up quite a
bit in computer science
16
00:00:41,690 --> 00:00:45,150
when you're doing algorithm
design and algorithm analysis.
17
00:00:45,150 --> 00:00:46,720
Today, we're going
to spend our time
18
00:00:46,720 --> 00:00:48,880
talking about a different
kind of recurrence that's
19
00:00:48,880 --> 00:00:50,770
called a linear recurrence.
20
00:00:50,770 --> 00:00:52,610
They also come up
in computer science
21
00:00:52,610 --> 00:00:55,360
and a lot of other fields.
22
00:00:55,360 --> 00:00:57,390
Now, I'm going to give
you the formal definition
23
00:00:57,390 --> 00:00:59,650
of a linear recurrence later.
24
00:00:59,650 --> 00:01:02,990
First I want to start
with an example,
25
00:01:02,990 --> 00:01:05,480
and this is an example of a
linear recurrence that comes up
26
00:01:05,480 --> 00:01:06,800
in population modeling.
27
00:01:06,800 --> 00:01:10,260
In fact, it comes up
in a lot of places.
28
00:01:10,260 --> 00:01:13,460
And to start it off, we're going
to analyze a particular problem
29
00:01:13,460 --> 00:01:16,760
that we call the
graduate student job
30
00:01:16,760 --> 00:01:21,810
problem or the graduate
student job nightmare.
31
00:01:21,810 --> 00:01:26,150
And the question here
is, will your TAs
32
00:01:26,150 --> 00:01:28,350
be able to get a job as
a professor somewhere
33
00:01:28,350 --> 00:01:29,425
when they get their PhD?
34
00:01:36,100 --> 00:01:39,040
So this is a problem
they worry about a lot.
35
00:01:39,040 --> 00:01:43,280
And the idea in this problem is
that there's some discipline,
36
00:01:43,280 --> 00:01:45,270
say computer science.
37
00:01:45,270 --> 00:01:48,310
And if you look in all the
universities in the world,
38
00:01:48,310 --> 00:01:51,640
there's M faculty lines.
39
00:01:51,640 --> 00:02:02,790
So the total number of jobs is
M, and that's a fixed value.
40
00:02:02,790 --> 00:02:04,690
Because of budgetary
constraints,
41
00:02:04,690 --> 00:02:07,330
universities aren't
growing, and so they're not
42
00:02:07,330 --> 00:02:10,680
going to create more computer
science faculty positions.
43
00:02:10,680 --> 00:02:12,680
It's going to stay
fixed over time.
44
00:02:19,900 --> 00:02:22,990
So it's a constant.
45
00:02:22,990 --> 00:02:29,610
And every year, professors
generate more graduates
46
00:02:29,610 --> 00:02:31,680
who become professors.
47
00:02:31,680 --> 00:02:35,830
And in particular, we're going
to assume that in this field
48
00:02:35,830 --> 00:02:39,130
every professor
graduates 1 student who
49
00:02:39,130 --> 00:02:43,650
goes on to become a professor,
or tries to if there's jobs.
50
00:02:43,650 --> 00:02:54,910
So each professor
generates 1 graduate
51
00:02:54,910 --> 00:03:00,740
who becomes a new professor, as
long as there's jobs, per year.
52
00:03:03,460 --> 00:03:07,280
With one exception, and
that's first-year professors
53
00:03:07,280 --> 00:03:09,760
because they're too busy
learning how to teach,
54
00:03:09,760 --> 00:03:12,540
getting grants, doing
administrative stuff,
55
00:03:12,540 --> 00:03:14,682
just figuring out
how it all works.
56
00:03:14,682 --> 00:03:16,890
So they don't have time to
produce any grad students.
57
00:03:23,510 --> 00:03:27,830
So first year or new professors
don't produce anything.
58
00:03:33,340 --> 00:03:37,105
Except first-year profs--
oops, let me correct that.
59
00:03:43,280 --> 00:03:44,750
So first-year profs have 0.
60
00:03:53,340 --> 00:03:56,130
Now, matters are made
worse by the fact
61
00:03:56,130 --> 00:03:59,280
that Congress passed a
law, and this is true,
62
00:03:59,280 --> 00:04:04,730
that more or less bars mandatory
retirements in colleges.
63
00:04:04,730 --> 00:04:09,780
And so that means there are no
retirements in this problem,
64
00:04:09,780 --> 00:04:14,150
and we're going to assume
the faculty live forever.
65
00:04:14,150 --> 00:04:16,870
And so once you fill a
position, it's filled forever.
66
00:04:16,870 --> 00:04:18,160
It doesn't exist anymore.
67
00:04:18,160 --> 00:04:21,149
In fact, if you walk
around the math department,
68
00:04:21,149 --> 00:04:23,150
you can see the impact of this.
69
00:04:23,150 --> 00:04:23,650
[LAUGHTER]
70
00:04:23,650 --> 00:04:29,347
I think the median age
is now well into the 70s.
71
00:04:29,347 --> 00:04:31,680
And there will be actually a
phenomenon over the next 10
72
00:04:31,680 --> 00:04:35,150
years as the math faculty
progress into their 80s
73
00:04:35,150 --> 00:04:38,044
where they actually, in reality,
probably do start to retire.
74
00:04:38,044 --> 00:04:39,460
And there will be
a whole new wave
75
00:04:39,460 --> 00:04:42,460
of people hired in mathematics,
just as happened back
76
00:04:42,460 --> 00:04:45,720
in the '50s and '60s after
the Sputnik crisis where
77
00:04:45,720 --> 00:04:47,810
a lot of mathematicians
were hired.
78
00:04:47,810 --> 00:04:50,700
And then they stayed in those
positions, for our purposes,
79
00:04:50,700 --> 00:04:52,530
forever.
80
00:04:52,530 --> 00:04:56,380
Now, the question is, when
do all the jobs get filled?
81
00:05:04,640 --> 00:05:09,130
So when are all M jobs
filled by this process?
82
00:05:12,830 --> 00:05:14,600
Now, to be able to
answer this question,
83
00:05:14,600 --> 00:05:18,790
we need one more
piece of information.
84
00:05:18,790 --> 00:05:21,296
Can anybody think
about one more fact
85
00:05:21,296 --> 00:05:23,670
here that we need before we
start going off and answering
86
00:05:23,670 --> 00:05:26,380
this question, analyzing it?
87
00:05:26,380 --> 00:05:26,880
Yeah?
88
00:05:26,880 --> 00:05:28,580
AUDIENCE: How many
professors do we start with?
89
00:05:28,580 --> 00:05:30,580
PROFESSOR: How many
professors do we start with?
90
00:05:30,580 --> 00:05:32,320
What's the boundary condition?
91
00:05:32,320 --> 00:05:35,300
What's the base case if we
were doing an induction?
92
00:05:35,300 --> 00:05:35,800
All right.
93
00:05:35,800 --> 00:05:45,280
So let's say the
boundary condition,
94
00:05:45,280 --> 00:05:48,170
and this always is
important with recurrences,
95
00:05:48,170 --> 00:05:53,346
is that the first professor
is hired in year 1,
96
00:05:53,346 --> 00:05:54,970
and there were none
before that person.
97
00:05:59,190 --> 00:05:59,690
All right?
98
00:05:59,690 --> 00:06:01,339
So in year 1
there's 1 professor,
99
00:06:01,339 --> 00:06:02,380
and that professor's new.
100
00:06:04,680 --> 00:06:05,180
OK.
101
00:06:05,180 --> 00:06:06,650
So now we have all
the information
102
00:06:06,650 --> 00:06:08,990
necessary to solve the problem.
103
00:06:08,990 --> 00:06:11,025
So let's do that and
set up a recurrence.
104
00:06:16,290 --> 00:06:25,770
We're going to define f of n
to be the number of professors
105
00:06:25,770 --> 00:06:26,570
during year n.
106
00:06:31,880 --> 00:06:35,700
And we know from our boundary
condition that in year 0
107
00:06:35,700 --> 00:06:43,330
there were none, and in year
1 there was 1, a new one.
108
00:06:43,330 --> 00:06:47,230
What is f of 2?
109
00:06:47,230 --> 00:06:50,360
How many professors
are there in year 2?
110
00:06:50,360 --> 00:06:51,327
AUDIENCE: 1.
111
00:06:51,327 --> 00:06:53,160
PROFESSOR: 1, because
the one that was there
112
00:06:53,160 --> 00:06:54,368
was too young to do anything.
113
00:06:54,368 --> 00:06:59,540
So he or she is the only one
left, all right, in year 2.
114
00:06:59,540 --> 00:07:01,370
What is f of 3?
115
00:07:01,370 --> 00:07:05,000
How many profs are
there in year 3?
116
00:07:05,000 --> 00:07:08,870
2-- the one you had, and by that
point that one is old enough
117
00:07:08,870 --> 00:07:12,045
to produce a new one.
118
00:07:12,045 --> 00:07:12,590
All right.
119
00:07:12,590 --> 00:07:13,910
What's f of 4?
120
00:07:16,542 --> 00:07:18,360
AUDIENCE: 3.
121
00:07:18,360 --> 00:07:20,890
PROFESSOR: 3, the 2
you had, and there
122
00:07:20,890 --> 00:07:23,520
was one who's been there 2
years to produce a new one.
123
00:07:27,901 --> 00:07:28,400
All right.
124
00:07:28,400 --> 00:07:33,008
And let's do one more, f of 5?
125
00:07:33,008 --> 00:07:33,507
AUDIENCE: 5.
126
00:07:33,507 --> 00:07:35,050
PROFESSOR: 5.
127
00:07:35,050 --> 00:07:41,120
These guys produced 1 each,
and you had 3 existing.
128
00:07:41,120 --> 00:07:41,770
All right.
129
00:07:41,770 --> 00:07:44,680
So we can actually write
down the recurrence now
130
00:07:44,680 --> 00:07:49,840
by sort of the process
we just went through.
131
00:07:49,840 --> 00:07:55,850
For years 2 and beyond, the
number of professors in year n
132
00:07:55,850 --> 00:07:59,240
is the number that we
had last year-- that's
133
00:07:59,240 --> 00:08:05,190
the previous ones-- plus the
number that were generated,
134
00:08:05,190 --> 00:08:08,900
new professors, the graduate
students that graduated.
135
00:08:08,900 --> 00:08:11,250
In terms of f, how many
new ones are there?
136
00:08:11,250 --> 00:08:12,830
AUDIENCE: f of n minus 2.
137
00:08:12,830 --> 00:08:14,580
PROFESSOR: f of n
minus 2, because that's
138
00:08:14,580 --> 00:08:17,080
the number of professors that
were there a couple years ago,
139
00:08:17,080 --> 00:08:19,150
and they are now
generating them.
140
00:08:19,150 --> 00:08:23,200
So that's the new ones there.
141
00:08:23,200 --> 00:08:26,720
Do people recognize
that recurrence?
142
00:08:26,720 --> 00:08:28,390
Yeah, it's pretty famous.
143
00:08:28,390 --> 00:08:29,810
How many people have not seen?
144
00:08:29,810 --> 00:08:31,525
This is called the
Fibonacci recurrence.
145
00:08:31,525 --> 00:08:33,720
It produces the
Fibonacci numbers.
146
00:08:33,720 --> 00:08:36,480
How many people
have not seen it?
147
00:08:36,480 --> 00:08:37,640
Yeah, very famous.
148
00:08:37,640 --> 00:08:40,780
Yeah, pretty much
everybody has seen that.
149
00:08:40,780 --> 00:08:43,280
Actually, this is
the first recurrence
150
00:08:43,280 --> 00:08:47,020
that was known to be
studied of all recurrences.
151
00:08:47,020 --> 00:08:51,260
It was published by
Leonardo Fibonacci of Pisa
152
00:08:51,260 --> 00:08:55,620
in 1202, all right,
so over 800 years ago,
153
00:08:55,620 --> 00:08:59,120
and he studied it for
modeling the population
154
00:08:59,120 --> 00:09:00,680
growth of rabbits.
155
00:09:00,680 --> 00:09:04,450
And the idea is that you
have a pair of rabbits,
156
00:09:04,450 --> 00:09:06,880
and in every month
after their first year
157
00:09:06,880 --> 00:09:08,910
of life-- sorry,
first month of life--
158
00:09:08,910 --> 00:09:11,221
they produce two new rabbits.
159
00:09:11,221 --> 00:09:11,720
All right?
160
00:09:11,720 --> 00:09:13,053
So it's the same notion as here.
161
00:09:13,053 --> 00:09:15,630
The first pair you do
nothing, but after that you're
162
00:09:15,630 --> 00:09:18,450
reproducing one for one.
163
00:09:18,450 --> 00:09:20,480
And that's an
abstraction, but it
164
00:09:20,480 --> 00:09:24,170
produces the same recurrence.
165
00:09:24,170 --> 00:09:27,440
Now, Fibonacci is credited
with discovering it.
166
00:09:27,440 --> 00:09:30,010
Really that means he's the
one that told the Europeans
167
00:09:30,010 --> 00:09:31,680
about it back then.
168
00:09:31,680 --> 00:09:35,800
And in fact, it's now been
traced back in to about 200 BC.
169
00:09:35,800 --> 00:09:37,900
The Indian
mathematicians knew all
170
00:09:37,900 --> 00:09:40,950
about Fibonacci's
recurrence, and they
171
00:09:40,950 --> 00:09:44,410
were using it to study certain
properties of grammar and music
172
00:09:44,410 --> 00:09:47,190
way back at 200 BC.
173
00:09:47,190 --> 00:09:50,150
This recurrence comes up in
all sorts of applications.
174
00:09:50,150 --> 00:09:52,710
Kepler used it in
the 16th century
175
00:09:52,710 --> 00:09:55,900
while studying how
the leaves of a flower
176
00:09:55,900 --> 00:09:58,460
are arranged around the
stem-- how many leaves
177
00:09:58,460 --> 00:10:02,270
you have in sort of each level
coming out around a stem.
178
00:10:02,270 --> 00:10:05,820
The first solution was
discovered by de Moivre
179
00:10:05,820 --> 00:10:06,654
in the 18th century.
180
00:10:06,654 --> 00:10:09,153
And we're going to talk about
how to solve this in a minute.
181
00:10:09,153 --> 00:10:11,500
But he was the first one
to figure out a closed form
182
00:10:11,500 --> 00:10:13,840
expression for f of n.
183
00:10:13,840 --> 00:10:16,270
Lame used it in the
19th century when
184
00:10:16,270 --> 00:10:19,170
he was studying the
Euclidean GCD algorithm.
185
00:10:19,170 --> 00:10:21,990
You know that pulverizer
thing and doing GCDs?
186
00:10:21,990 --> 00:10:25,690
It turns out that if you want
to analyze the running time,
187
00:10:25,690 --> 00:10:27,920
well, you get a Fibonacci
recurrence comes
188
00:10:27,920 --> 00:10:31,720
into play there, and that was
discovered in the 19th century.
189
00:10:31,720 --> 00:10:34,980
In the 20th century, it was
used in the study of optics,
190
00:10:34,980 --> 00:10:38,470
economics, and algorithms, and
it was named for Fibonacci.
191
00:10:38,470 --> 00:10:41,540
It got a name in
the 19th century.
192
00:10:41,540 --> 00:10:43,570
In fact, this is
so popular and used
193
00:10:43,570 --> 00:10:47,200
in so many places there is a
journal in mathematics called
194
00:10:47,200 --> 00:10:49,560
the Fibonacci Quarterly
Journal where they
195
00:10:49,560 --> 00:10:53,087
study these kinds of things.
196
00:10:53,087 --> 00:10:55,420
So today what we're going to
do is solve this recurrence
197
00:10:55,420 --> 00:10:58,060
and actually solve a much
broader family of recurrences
198
00:10:58,060 --> 00:11:01,370
called linear recurrences.
199
00:11:01,370 --> 00:11:02,980
And we're going to
get a closed form.
200
00:11:02,980 --> 00:11:05,021
I mean, you can produce
the Fibonacci numbers one
201
00:11:05,021 --> 00:11:07,880
after another, but we're
going to derive a formula
202
00:11:07,880 --> 00:11:11,041
for the n-th Fibonacci number.
203
00:11:11,041 --> 00:11:12,540
And when we're going
to do it, we're
204
00:11:12,540 --> 00:11:16,710
going to do it more broadly for
a class of linear recurrences.
205
00:11:16,710 --> 00:11:18,155
So let me define what that is.
206
00:11:23,950 --> 00:11:32,740
So a recurrence is
said to be linear
207
00:11:32,740 --> 00:11:47,500
if it is of the form f of n
equals a constant a1 times
208
00:11:47,500 --> 00:11:55,570
f of n minus 1 plus a2 times
f of n minus 2 plus dot dot
209
00:11:55,570 --> 00:12:02,800
dot d'th constant ad
times f of n minus d.
210
00:12:02,800 --> 00:12:06,440
And we could simplify
that as the sum
211
00:12:06,440 --> 00:12:14,590
i equals 1 to d of a sub
i times f of n minus i.
212
00:12:14,590 --> 00:12:19,190
And the constants
are fixed here,
213
00:12:19,190 --> 00:12:25,280
so for fixed a sub i and d.
214
00:12:25,280 --> 00:12:27,420
So the number of terms
has to be a constant,
215
00:12:27,420 --> 00:12:30,510
and each coefficient
has to be a constant.
216
00:12:30,510 --> 00:12:32,030
Can't vary.
217
00:12:32,030 --> 00:12:36,870
And we define d to be the
order of the recurrence.
218
00:12:36,870 --> 00:12:40,300
So d is the order
of the recurrence.
219
00:12:42,820 --> 00:12:44,790
OK?
220
00:12:44,790 --> 00:12:47,095
And you can see, of course,
that Fibonacci's recurrence
221
00:12:47,095 --> 00:12:48,660
is linear.
222
00:12:48,660 --> 00:12:51,080
What's its order?
223
00:12:51,080 --> 00:12:52,190
2.
224
00:12:52,190 --> 00:12:54,880
And the coefficients,
the a's, are just 1.
225
00:12:54,880 --> 00:12:58,410
So it's a simple
linear recurrence.
226
00:12:58,410 --> 00:12:58,910
All right.
227
00:12:58,910 --> 00:13:00,790
So let's see how to solve it.
228
00:13:00,790 --> 00:13:02,330
Well, actually,
before we do that,
229
00:13:02,330 --> 00:13:05,390
can you see the difference
between this recurrence,
230
00:13:05,390 --> 00:13:08,140
linear, and divide-and-conquer
recurrences?
231
00:13:11,180 --> 00:13:13,050
Right?
232
00:13:13,050 --> 00:13:16,120
What do I have here inside for
a divide-and-conquer recurrence?
233
00:13:19,400 --> 00:13:21,640
I get a fraction of n, right?
234
00:13:21,640 --> 00:13:23,630
And here, I'm
subtracting a constant
235
00:13:23,630 --> 00:13:26,730
from n, usually like 1,
2, 3, an integer from n.
236
00:13:26,730 --> 00:13:30,940
So linear is when inside you
have n minus 1, n minus 2.
237
00:13:30,940 --> 00:13:33,945
Divide and conquer,
you got n/2 or 3/4 n.
238
00:13:33,945 --> 00:13:37,992
And it makes a huge
difference in the solution.
239
00:13:37,992 --> 00:13:38,780
All right.
240
00:13:38,780 --> 00:13:41,529
So what I'm going to do is
give you a closed-form solution
241
00:13:41,529 --> 00:13:42,820
for these kinds of recurrences.
242
00:13:42,820 --> 00:13:47,690
And it won't be completely
easy because it took Europeans
243
00:13:47,690 --> 00:13:51,501
six centuries to find
the solution to this.
244
00:13:51,501 --> 00:13:52,000
Right?
245
00:13:52,000 --> 00:13:53,720
Fibonacci discovers
the thing in 1200
246
00:13:53,720 --> 00:13:55,210
and tells everybody about it.
247
00:13:55,210 --> 00:13:58,305
And it wasn't until
600 years later
248
00:13:58,305 --> 00:14:00,620
that they figured out
a closed-form solution.
249
00:14:03,824 --> 00:14:04,920
So let's do that.
250
00:14:08,040 --> 00:14:11,480
Now, what we're going to
do to do it this first time
251
00:14:11,480 --> 00:14:13,610
ourselves, and we
don't have the formula,
252
00:14:13,610 --> 00:14:15,750
is to use guess and verify.
253
00:14:15,750 --> 00:14:17,620
So we're going to
guess a solution
254
00:14:17,620 --> 00:14:19,750
and check that it works.
255
00:14:19,750 --> 00:14:23,030
And we're going to guess,
really, a class of solutions.
256
00:14:23,030 --> 00:14:29,966
We're going to try f of n is
an exponential in n, alpha
257
00:14:29,966 --> 00:14:32,180
to the n for some
constant alpha.
258
00:14:38,310 --> 00:14:40,590
Now, we're going
to figure out what
259
00:14:40,590 --> 00:14:45,130
alpha is as we go along during
the verification process.
260
00:14:45,130 --> 00:14:45,630
All right?
261
00:14:45,630 --> 00:14:50,830
So let's try to verify
this guess and plug it in.
262
00:14:50,830 --> 00:14:56,595
We know that f of n is f of n
minus 1 plus f of n minus 2.
263
00:14:56,595 --> 00:15:00,120
Well, let's plug that
in and see what we get.
264
00:15:00,120 --> 00:15:03,310
That gives us alpha to the
n equals alpha to the n
265
00:15:03,310 --> 00:15:08,880
minus 1 plus alpha
to the n minus 2.
266
00:15:08,880 --> 00:15:09,710
All right?
267
00:15:09,710 --> 00:15:15,300
Now, that means I can divide
by alpha to the n minus 2,
268
00:15:15,300 --> 00:15:20,920
and I get alpha squared
equals alpha plus 1.
269
00:15:20,920 --> 00:15:25,220
And now I can use the quadratic
formula to solve for alpha.
270
00:15:25,220 --> 00:15:25,720
All right?
271
00:15:25,720 --> 00:15:30,550
That means that alpha
squared minus alpha minus 1
272
00:15:30,550 --> 00:15:38,760
equals 0, which means that
alpha equals-- minus minus 1
273
00:15:38,760 --> 00:15:42,680
is 1 plus or minus the square
root of-- minus 1 squared
274
00:15:42,680 --> 00:15:49,961
is 1 minus 4ac plus 4 over 2.
275
00:15:49,961 --> 00:15:50,460
OK?
276
00:15:50,460 --> 00:15:54,990
So there's two possible
solutions here.
277
00:15:54,990 --> 00:15:57,910
This is 1 plus or minus the
square root of 5 over 2.
278
00:16:00,670 --> 00:16:05,670
So it works if f of n
equals either-- well, we'll
279
00:16:05,670 --> 00:16:08,450
call the roots here
alpha 1 and alpha 2.
280
00:16:14,870 --> 00:16:18,370
Alpha 1 will be the
positive case, 1
281
00:16:18,370 --> 00:16:21,176
plus square root of 5 over 2.
282
00:16:21,176 --> 00:16:23,830
And alpha 2 will be
the negative case, 1
283
00:16:23,830 --> 00:16:28,580
minus square root of 5 over 2.
284
00:16:28,580 --> 00:16:29,080
All right?
285
00:16:29,080 --> 00:16:35,700
So guess and verify works so
far if we have an exponential
286
00:16:35,700 --> 00:16:39,260
with either one of those bases.
287
00:16:39,260 --> 00:16:41,570
All right?
288
00:16:41,570 --> 00:16:45,905
By the way, does anybody
recognize that number?
289
00:16:49,220 --> 00:16:50,742
It's a famous number.
290
00:16:50,742 --> 00:16:51,700
AUDIENCE: Golden ratio.
291
00:16:51,700 --> 00:16:53,360
PROFESSOR: The
golden ratio, which
292
00:16:53,360 --> 00:16:56,230
is supposed to have all these
magical mystical properties.
293
00:16:56,230 --> 00:16:58,000
That when you look
at a building,
294
00:16:58,000 --> 00:17:02,060
if its aspect ratio is that,
it's perfect to the human eye.
295
00:17:02,060 --> 00:17:02,930
I don't know.
296
00:17:02,930 --> 00:17:05,180
But there's a lot of stuff
about the golden ratio that
297
00:17:05,180 --> 00:17:07,235
happens to come up here.
298
00:17:07,235 --> 00:17:07,735
OK?
299
00:17:10,240 --> 00:17:14,790
Now in fact, there's more
than just these two solutions.
300
00:17:14,790 --> 00:17:18,819
It turns out that whenever
you have a linear recurrence
301
00:17:18,819 --> 00:17:21,810
and you've got two or
more solutions like that,
302
00:17:21,810 --> 00:17:25,430
any linear combination
is also a solution.
303
00:17:25,430 --> 00:17:25,930
All right?
304
00:17:25,930 --> 00:17:29,160
So we're just going to
state that as a fact.
305
00:17:29,160 --> 00:17:31,630
It's not too hard to prove,
but we won't prove it in class.
306
00:17:35,110 --> 00:17:39,420
So if f of n equals
alpha 1 to the n--
307
00:17:39,420 --> 00:17:44,590
and this is true for any alpha
1 and alpha 2-- and f of n
308
00:17:44,590 --> 00:17:53,200
equals alpha 2 to
the n are solutions
309
00:17:53,200 --> 00:18:04,030
to a linear recurrence--
and here I mean
310
00:18:04,030 --> 00:18:06,680
without yet applying
the boundary conditions.
311
00:18:06,680 --> 00:18:08,860
So far, we've ignored
the boundary conditions,
312
00:18:08,860 --> 00:18:11,580
and we'll just do that
for a little longer.
313
00:18:11,580 --> 00:18:13,660
So if there are solutions
without worrying
314
00:18:13,660 --> 00:18:24,710
about the boundary
conditions, then f of n
315
00:18:24,710 --> 00:18:31,260
equals c1 times alpha 1 to
the n plus c2 alpha 2 to the n
316
00:18:31,260 --> 00:18:34,550
is also a solution for
any constants c1 and c2.
317
00:18:56,130 --> 00:18:56,630
All right?
318
00:18:56,630 --> 00:19:01,320
So any linear combination
of our solutions also works.
319
00:19:01,320 --> 00:19:05,580
If you plugged it in to do
verify, it would be fine.
320
00:19:05,580 --> 00:19:14,380
So that means that f
of n equals c1 times
321
00:19:14,380 --> 00:19:18,110
1 plus square root
of 5 over 2 to the n
322
00:19:18,110 --> 00:19:25,090
plus c2 1 minus square
root of 5 over 2 to the n
323
00:19:25,090 --> 00:19:33,500
is a solution-- oops--
to Fibonacci's recurrence
324
00:19:33,500 --> 00:19:35,320
without boundary
conditions again.
325
00:19:41,410 --> 00:19:41,910
All right?
326
00:19:41,910 --> 00:19:46,750
So I could plug this
expression into the recurrence,
327
00:19:46,750 --> 00:19:49,885
and it would satisfy it.
328
00:19:49,885 --> 00:19:50,910
I won't do that.
329
00:19:54,180 --> 00:19:57,790
But we haven't dealt with
the boundary conditions yet.
330
00:19:57,790 --> 00:19:59,970
And in fact, dealing with
the boundary conditions
331
00:19:59,970 --> 00:20:04,270
is what determines the
values of these constants.
332
00:20:04,270 --> 00:20:09,010
I could have Fibonacci's
recurrence where f of 0 was 10
333
00:20:09,010 --> 00:20:12,040
and f of 1 was 20
if I wanted to,
334
00:20:12,040 --> 00:20:15,860
and then the recurrence
would be the same afterwards.
335
00:20:15,860 --> 00:20:19,680
And it will turn out I get
different constants here.
336
00:20:19,680 --> 00:20:20,180
All right?
337
00:20:20,180 --> 00:20:22,880
But otherwise, the form is
going to look like this.
338
00:20:22,880 --> 00:20:24,340
So let's see how
to make that work.
339
00:20:24,340 --> 00:20:26,340
Let's see how to determine
the constant factors.
340
00:20:48,770 --> 00:20:49,390
OK.
341
00:20:49,390 --> 00:21:05,360
So to determine the
constant factors,
342
00:21:05,360 --> 00:21:08,180
we plug in the
boundary conditions.
343
00:21:08,180 --> 00:21:14,590
So we have f of 0 equals 0
from the boundary condition.
344
00:21:14,590 --> 00:21:17,190
And now we plug that into
our formula over there.
345
00:21:20,050 --> 00:21:26,740
That's c1 times alpha 1 to the
0 plus c2 alpha 2 to the 0.
346
00:21:26,740 --> 00:21:29,600
Of course, anything
to the 0 is just 1.
347
00:21:32,480 --> 00:21:38,471
And that means that
c2 equals minus c1.
348
00:21:38,471 --> 00:21:39,902
All right?
349
00:21:39,902 --> 00:21:42,280
And now I'll use the
next boundary condition
350
00:21:42,280 --> 00:21:44,180
to nail them down.
351
00:21:50,750 --> 00:21:51,250
All right?
352
00:21:51,250 --> 00:21:56,350
So I know also that
f(1) is 1, and that
353
00:21:56,350 --> 00:22:02,380
equals c1 1 plus square root
of 5 over 2 to the first power
354
00:22:02,380 --> 00:22:10,050
plus c2 1 minus the square root
of 5 over 2 to the first power.
355
00:22:10,050 --> 00:22:13,720
I plug in c2 as minus c1.
356
00:22:13,720 --> 00:22:19,480
So I get c1 1 plus square
root of 5 over 2 minus c1 1
357
00:22:19,480 --> 00:22:23,790
minus square root of 5 over 2.
358
00:22:23,790 --> 00:22:27,250
And now that I can
factor out the c1,
359
00:22:27,250 --> 00:22:33,890
I get 1 minus 1 square root of
5 minus minus square root of 5,
360
00:22:33,890 --> 00:22:40,370
which gives me c1 2
square root of 5 over 2.
361
00:22:40,370 --> 00:22:42,790
These cancel.
362
00:22:42,790 --> 00:22:47,480
This was all equal to 1, so
that means that c1 equals 1
363
00:22:47,480 --> 00:22:50,370
over the square root of 5.
364
00:22:50,370 --> 00:22:53,840
And of course, c2 is minus that.
365
00:22:53,840 --> 00:22:57,640
c2 is minus 1 over
the square root of 5.
366
00:23:00,560 --> 00:23:02,815
Any questions so far?
367
00:23:05,585 --> 00:23:06,563
All right.
368
00:23:12,930 --> 00:23:15,660
Now I could write
out the formula
369
00:23:15,660 --> 00:23:16,750
for the Fibonacci numbers.
370
00:23:22,630 --> 00:23:23,130
All right?
371
00:23:23,130 --> 00:23:32,380
So the solution is f
of n equals c1, which
372
00:23:32,380 --> 00:23:36,810
is 1 over square root of
5, times the n-th power
373
00:23:36,810 --> 00:23:42,800
of the first root plus
c2, which is minus 1
374
00:23:42,800 --> 00:23:46,180
over square root of 5, times the
n-th power of the second root.
375
00:23:51,030 --> 00:23:54,900
And that is the formula for
the n-th Fibonacci number.
376
00:23:57,580 --> 00:24:00,590
You wouldn't have guessed
that to start with obviously.
377
00:24:00,590 --> 00:24:02,900
That would require pretty
divine inspiration.
378
00:24:02,900 --> 00:24:06,140
And you can sort of see why it
took them 600 years in Europe
379
00:24:06,140 --> 00:24:07,660
to figure out the answer.
380
00:24:07,660 --> 00:24:08,160
All right?
381
00:24:08,160 --> 00:24:09,610
It's not the first
thing you'd think about.
382
00:24:09,610 --> 00:24:12,240
In fact, if somebody told you
the answer and said this is it,
383
00:24:12,240 --> 00:24:14,230
you'd go, oh, give me a break.
384
00:24:14,230 --> 00:24:14,730
All right?
385
00:24:14,730 --> 00:24:17,450
It does not look like-- I
mean, what are the chances that
386
00:24:17,450 --> 00:24:20,350
evaluates to an integer?
387
00:24:20,350 --> 00:24:20,850
All right?
388
00:24:20,850 --> 00:24:25,080
It's got square root of
5's all over the place.
389
00:24:25,080 --> 00:24:26,140
Right?
390
00:24:26,140 --> 00:24:30,470
And here I'm telling
you that f of 6,
391
00:24:30,470 --> 00:24:34,240
the sixth Fibonacci number,
which is 8, 3 plus 5,
392
00:24:34,240 --> 00:24:38,070
I'm telling you that is equal
to 1 over square root of 5 1
393
00:24:38,070 --> 00:24:42,250
plus the square root of 5 over
2 to the sixth power minus 1
394
00:24:42,250 --> 00:24:46,290
over square root of 5 1 minus
the square root of 5 over 2
395
00:24:46,290 --> 00:24:48,350
to the sixth power.
396
00:24:48,350 --> 00:24:51,350
I mean, would you believe
me if I told you that?
397
00:24:51,350 --> 00:24:52,970
Probably not.
398
00:24:52,970 --> 00:24:54,560
What are the
chances that's true?
399
00:24:54,560 --> 00:24:58,450
Somehow, magically, all those
square root of 5's all go away,
400
00:24:58,450 --> 00:25:02,890
and you're just left
with 8, a simple integer.
401
00:25:02,890 --> 00:25:03,390
OK?
402
00:25:06,130 --> 00:25:08,180
All right.
403
00:25:08,180 --> 00:25:13,895
Yeah, there's sort of some more
interesting things about this.
404
00:25:13,895 --> 00:25:17,890
What happens to this value
here as n gets large?
405
00:25:20,450 --> 00:25:22,938
What does it do as n gets large?
406
00:25:22,938 --> 00:25:26,260
AUDIENCE: Gets small.
407
00:25:26,260 --> 00:25:30,210
PROFESSOR: It gets small Because
1 minus the square root of 5
408
00:25:30,210 --> 00:25:34,480
over 2 is about 0.6.
409
00:25:34,480 --> 00:25:39,120
This is about, here, this is
about 0.618 something or other.
410
00:25:39,120 --> 00:25:42,050
And if I take a fraction less
than 1 to the n-th power,
411
00:25:42,050 --> 00:25:44,440
it goes to 0.
412
00:25:44,440 --> 00:25:46,430
It goes away.
413
00:25:46,430 --> 00:25:50,140
So in fact, as n
gets large, this
414
00:25:50,140 --> 00:25:54,050
is what the n-th Fibonacci
number starts to look like.
415
00:25:54,050 --> 00:25:58,250
In particular, f of n equals
just that first one-- 1
416
00:25:58,250 --> 00:26:05,530
over square root of 5 times the
golden ratio to the n-th power
417
00:26:05,530 --> 00:26:09,260
plus some error term, delta n.
418
00:26:09,260 --> 00:26:17,770
And delta n is less than a
1/10 for n greater than 4.
419
00:26:20,300 --> 00:26:27,330
And it is little l of 1
in general going to 0.
420
00:26:27,330 --> 00:26:29,070
So in fact, the n-th
Fibonacci number
421
00:26:29,070 --> 00:26:31,720
is about the n-th power of
the golden ratio divided
422
00:26:31,720 --> 00:26:35,020
by square root of 5.
423
00:26:35,020 --> 00:26:35,922
Yeah?
424
00:26:35,922 --> 00:26:38,064
AUDIENCE: [INAUDIBLE]
top of the right board--
425
00:26:38,064 --> 00:26:38,730
PROFESSOR: Yeah?
426
00:26:38,730 --> 00:26:43,158
AUDIENCE: --you say that you can
pick sort of any constants that
427
00:26:43,158 --> 00:26:44,009
satisfy it?
428
00:26:44,009 --> 00:26:44,634
PROFESSOR: Yep.
429
00:26:44,634 --> 00:26:46,602
AUDIENCE: Does that
shift the sequence
430
00:26:46,602 --> 00:26:48,570
over, or what does that do?
431
00:26:48,570 --> 00:26:50,046
I guess I don't understand.
432
00:26:50,046 --> 00:26:51,810
PROFESSOR: OK.
433
00:26:51,810 --> 00:26:55,609
You're asking, what are
these things doing here?
434
00:26:55,609 --> 00:26:56,234
AUDIENCE: Yeah.
435
00:26:56,234 --> 00:26:58,570
And why [? isn't there-- ?]
why are they also solutions?
436
00:26:58,570 --> 00:27:00,125
Because this is the right ones?
437
00:27:00,125 --> 00:27:01,166
PROFESSOR: The right one.
438
00:27:01,166 --> 00:27:01,760
OK.
439
00:27:01,760 --> 00:27:05,120
So there's two parts
to a recurrence.
440
00:27:05,120 --> 00:27:11,120
There's this part,
OK, and there's
441
00:27:11,120 --> 00:27:17,250
the boundary conditions-- f(0)
equals 0 and f(1) equals 1.
442
00:27:17,250 --> 00:27:19,450
There are really
two parts to it.
443
00:27:19,450 --> 00:27:21,610
If you change the
boundary conditions,
444
00:27:21,610 --> 00:27:26,240
you would change the
rest of the terms, right?
445
00:27:26,240 --> 00:27:28,460
If I started with
10 and 20, f of 2
446
00:27:28,460 --> 00:27:34,510
would be 30, OK, if I'm using
this recurrence form for n
447
00:27:34,510 --> 00:27:36,522
bigger than or equal to 2.
448
00:27:36,522 --> 00:27:40,200
Now, so when I'm computing a
solution, first I'm saying,
449
00:27:40,200 --> 00:27:41,860
let's ignore the
boundary conditions
450
00:27:41,860 --> 00:27:45,770
and look at all the recurrences
that share this part.
451
00:27:45,770 --> 00:27:49,960
And that family of
recurrences, I'm claiming,
452
00:27:49,960 --> 00:27:51,580
have these solutions.
453
00:27:51,580 --> 00:27:54,130
This is the solution
space if I just
454
00:27:54,130 --> 00:27:59,460
worry about this part and
not the boundary conditions.
455
00:27:59,460 --> 00:28:03,160
Now, once I plug the
boundary conditions in,
456
00:28:03,160 --> 00:28:08,220
that determines which
values of c1 and c2 I get.
457
00:28:08,220 --> 00:28:11,740
And so if I take the version
of the recurrence where f(0)
458
00:28:11,740 --> 00:28:16,400
is 0 and f(1) is 1, then c1
is going to be 1 over root 5,
459
00:28:16,400 --> 00:28:19,720
and c2 is going to be
negative 1 over root 5.
460
00:28:19,720 --> 00:28:23,660
If I use this
recurrence with f(0)
461
00:28:23,660 --> 00:28:28,430
equals 10 and f(1) equals 20,
I get different constants,
462
00:28:28,430 --> 00:28:31,362
but these powers stay the same.
463
00:28:31,362 --> 00:28:33,320
In fact, maybe we'll make
an exercise like that
464
00:28:33,320 --> 00:28:35,069
from the problem set
to figure out what it
465
00:28:35,069 --> 00:28:37,450
is if I started with 10 and 20.
466
00:28:37,450 --> 00:28:38,020
OK?
467
00:28:38,020 --> 00:28:40,240
So that's what
the constants have
468
00:28:40,240 --> 00:28:43,160
to do-- they come from the
boundary condition, which
469
00:28:43,160 --> 00:28:45,591
is sort of a cool fact.
470
00:28:45,591 --> 00:28:47,280
Does that make sense?
471
00:28:47,280 --> 00:28:48,060
AUDIENCE: Yeah.
472
00:28:48,060 --> 00:28:51,570
PROFESSOR: Any other questions
about what I was doing?
473
00:28:51,570 --> 00:28:56,590
So we used guess and verify, but
I sort of guessed a form first.
474
00:28:56,590 --> 00:29:01,390
And as I verified, I sort of
revised my guess along the way.
475
00:29:01,390 --> 00:29:01,890
All right?
476
00:29:01,890 --> 00:29:03,160
So I started guessing this.
477
00:29:03,160 --> 00:29:05,290
And then I said,
well, let's refine
478
00:29:05,290 --> 00:29:11,307
that guess so that really
it's one of these two guys.
479
00:29:11,307 --> 00:29:12,890
And then I said-- I
used a fact that I
480
00:29:12,890 --> 00:29:14,681
didn't prove-- that
said I could use really
481
00:29:14,681 --> 00:29:17,570
any linear combination,
plugged it into the base cases,
482
00:29:17,570 --> 00:29:20,301
and got my constants.
483
00:29:20,301 --> 00:29:20,800
All right?
484
00:29:20,800 --> 00:29:23,269
Now, this will lead up to
a formula or an approach
485
00:29:23,269 --> 00:29:25,310
so you don't have to go
through this on your own.
486
00:29:25,310 --> 00:29:27,710
You just plug in the
formula in general.
487
00:29:30,300 --> 00:29:30,800
OK.
488
00:29:33,402 --> 00:29:35,360
Let's get back to the
original question of when
489
00:29:35,360 --> 00:29:37,660
all the jobs are filled.
490
00:29:37,660 --> 00:29:43,091
For what value of n is
f of n bigger than M?
491
00:29:43,091 --> 00:29:43,590
All right?
492
00:29:48,790 --> 00:29:49,420
Let's do that.
493
00:30:03,940 --> 00:30:04,450
OK.
494
00:30:04,450 --> 00:30:20,130
To see when all the jobs
are filled, all M jobs,
495
00:30:20,130 --> 00:30:24,876
for the n when f of n is
bigger and equal to M,
496
00:30:24,876 --> 00:30:28,300
when we got a Fibonacci
number that's bigger than M.
497
00:30:28,300 --> 00:30:31,770
And we just showed
that basically f of n,
498
00:30:31,770 --> 00:30:35,090
well, it equals this
where that's a tiny thing.
499
00:30:35,090 --> 00:30:35,590
All right?
500
00:30:35,590 --> 00:30:40,020
So we need to figure out when
is 1 over square root of 5 times
501
00:30:40,020 --> 00:30:44,080
the golden ratio to
the n-th power-- plus
502
00:30:44,080 --> 00:30:47,430
I have this tiny little
thing that doesn't matter--
503
00:30:47,430 --> 00:30:50,200
is bigger than or equal to M?
504
00:30:50,200 --> 00:30:54,790
So I can solve for n now by
subtracting the delta term,
505
00:30:54,790 --> 00:30:57,300
multiplying by square root of 5.
506
00:30:57,300 --> 00:31:02,260
That gives me the golden
ratio to the n-th power
507
00:31:02,260 --> 00:31:06,240
bigger than or equal
to square root 5 times
508
00:31:06,240 --> 00:31:08,050
M minus the tiny thing.
509
00:31:10,800 --> 00:31:14,040
Now I take logs.
510
00:31:14,040 --> 00:31:21,450
And I'm in trouble when n is
bigger than or equal to log
511
00:31:21,450 --> 00:31:30,825
of this over log of the base.
512
00:31:38,710 --> 00:31:41,070
In other words, this is theta.
513
00:31:41,070 --> 00:31:42,550
What's this theta of here?
514
00:31:45,830 --> 00:31:48,500
That goes away.
515
00:31:48,500 --> 00:31:50,434
I can skip all the
constant factors.
516
00:31:50,434 --> 00:31:52,225
What's this theta of
here, this expression,
517
00:31:52,225 --> 00:31:56,224
in terms of theta of something
to do with some function of M?
518
00:31:56,224 --> 00:31:57,220
AUDIENCE: Log of M?
519
00:31:57,220 --> 00:32:02,910
PROFESSOR: Log of M. So the jobs
get filled in log of M years.
520
00:32:02,910 --> 00:32:05,250
And in fact, if I
plugged in, for example,
521
00:32:05,250 --> 00:32:07,810
M equals 10,000 into
this expression,
522
00:32:07,810 --> 00:32:09,290
I would find that
all the jobs are
523
00:32:09,290 --> 00:32:12,540
filled in 20 years, which
is more or less what
524
00:32:12,540 --> 00:32:14,780
happened in computer science.
525
00:32:14,780 --> 00:32:18,221
Ballpark, those numbers
are roughly correct.
526
00:32:18,221 --> 00:32:18,720
All right?
527
00:32:18,720 --> 00:32:21,530
But you can solve it exactly now
by just plugging in whatever M
528
00:32:21,530 --> 00:32:22,490
you want.
529
00:32:22,490 --> 00:32:23,870
That's a fraction
less than 1/10.
530
00:32:23,870 --> 00:32:25,286
And you can solve
to find out when
531
00:32:25,286 --> 00:32:26,973
your population gets that big.
532
00:32:29,851 --> 00:32:30,350
OK.
533
00:32:30,350 --> 00:32:30,933
Any questions?
534
00:32:36,410 --> 00:32:36,910
OK.
535
00:32:36,910 --> 00:32:39,250
So now what I want
to do is show you
536
00:32:39,250 --> 00:32:43,495
how to use this same idea to
solve any linear recurrence,
537
00:32:43,495 --> 00:32:45,620
and this is the process
you'll go through to do it.
538
00:33:01,625 --> 00:33:02,310
All right.
539
00:33:02,310 --> 00:33:06,420
So we're going to solve a
general linear recurrence.
540
00:33:20,040 --> 00:33:27,440
So in this case, we've got
f of n is the sum from i
541
00:33:27,440 --> 00:33:30,590
equals 1 to d, for an
order d recurrence,
542
00:33:30,590 --> 00:33:34,530
a sub i f of n minus i.
543
00:33:34,530 --> 00:33:38,280
And we've got to have
boundary conditions.
544
00:33:38,280 --> 00:33:46,570
So we'll have f of 0
is b0, f of 1 is b1.
545
00:33:46,570 --> 00:33:48,100
And how many boundary
conditions do
546
00:33:48,100 --> 00:33:49,599
you think I'm going
to need to have?
547
00:33:49,599 --> 00:33:51,790
Any guesses?
548
00:33:51,790 --> 00:33:54,242
I had two for Fibonacci.
549
00:33:54,242 --> 00:33:56,320
In this case, I've got d terms.
550
00:33:56,320 --> 00:33:57,059
AUDIENCE: d.
551
00:33:57,059 --> 00:33:57,600
PROFESSOR: d.
552
00:33:57,600 --> 00:33:59,266
I'm going to need d
boundary conditions.
553
00:33:59,266 --> 00:34:03,650
So we'll go all the
way to f of d minus 1
554
00:34:03,650 --> 00:34:05,370
equal to b of d minus 1.
555
00:34:07,992 --> 00:34:08,780
All right?
556
00:34:08,780 --> 00:34:16,839
And now I'm going to try
f of n is alpha to the n.
557
00:34:16,839 --> 00:34:20,989
We're going to plug it
into this expression.
558
00:34:20,989 --> 00:34:25,090
And then when we do,
we get alpha to the n
559
00:34:25,090 --> 00:34:31,480
equals a1 alpha to
the n minus 1 plus
560
00:34:31,480 --> 00:34:37,360
a2 alpha to the n minus 2
all the way down to a sub
561
00:34:37,360 --> 00:34:44,310
d alpha to the n minus d.
562
00:34:44,310 --> 00:34:44,810
All right?
563
00:34:44,810 --> 00:34:47,280
Just plugging into there.
564
00:34:47,280 --> 00:34:50,790
I can divide everything
by alpha to the n minus d,
565
00:34:50,790 --> 00:34:54,290
and that gives me
alpha to the d equals
566
00:34:54,290 --> 00:35:04,320
a1 alpha to the d minus 1 a2
alpha to the d minus 2 ad times
567
00:35:04,320 --> 00:35:08,760
alpha to the 0, which is just 1.
568
00:35:08,760 --> 00:35:12,805
And I can rewrite this as
a polynomial equal to 0.
569
00:35:22,800 --> 00:35:26,255
So that means that
alpha to the d minus
570
00:35:26,255 --> 00:35:34,420
a1 alpha to the d minus 1
minus a2 alpha to the d minus 2
571
00:35:34,420 --> 00:35:38,600
minus ad is 0.
572
00:35:38,600 --> 00:35:41,460
This is called the
characteristic equation
573
00:35:41,460 --> 00:35:42,210
of the recurrence.
574
00:35:56,770 --> 00:35:58,490
OK?
575
00:35:58,490 --> 00:36:03,590
Now, what you need to do is
compute the roots of this,
576
00:36:03,590 --> 00:36:08,610
and we'll use the roots to get
the solution to the recurrence.
577
00:36:08,610 --> 00:36:10,780
Now, the simple case
is when all the roots
578
00:36:10,780 --> 00:36:15,120
are different-- so let's
do that first-- like there
579
00:36:15,120 --> 00:36:16,530
was in the Fibonacci example.
580
00:36:21,440 --> 00:36:30,280
All d roots are
different, and let's call
581
00:36:30,280 --> 00:36:33,440
them alpha 1 to alpha d.
582
00:36:39,610 --> 00:36:44,070
In that case, the solution
is all linear combinations
583
00:36:44,070 --> 00:36:49,490
of their n-th
powers, just like it
584
00:36:49,490 --> 00:36:56,180
was with Fibonacci without
the boundary conditions.
585
00:36:56,180 --> 00:37:00,040
c1 alpha 1 to the
n plus c2 alpha 2
586
00:37:00,040 --> 00:37:08,580
to the n plus cd
alpha d to the n.
587
00:37:08,580 --> 00:37:09,080
All right?
588
00:37:09,080 --> 00:37:11,340
Same thing happens that
happened in Fibonacci,
589
00:37:11,340 --> 00:37:14,780
that this becomes the
solution before the boundary
590
00:37:14,780 --> 00:37:16,690
conditions are applied.
591
00:37:16,690 --> 00:37:17,190
All right?
592
00:37:30,500 --> 00:37:33,730
And by the way, if your
characteristic equation has
593
00:37:33,730 --> 00:37:36,930
imaginary roots, that's fine.
594
00:37:36,930 --> 00:37:39,480
It doesn't matter
because the imaginary i
595
00:37:39,480 --> 00:37:41,900
term will disappear just
like the square root of 5
596
00:37:41,900 --> 00:37:43,350
disappeared.
597
00:37:43,350 --> 00:37:49,130
Has to because we know f of n,
in this case, is a real number.
598
00:37:49,130 --> 00:37:51,110
All right?
599
00:37:51,110 --> 00:37:51,610
OK.
600
00:37:51,610 --> 00:37:54,600
Now, to find out the values
of the coefficients, what
601
00:37:54,600 --> 00:37:57,710
are we going to do?
602
00:37:57,710 --> 00:37:58,710
What is it?
603
00:37:58,710 --> 00:37:59,960
AUDIENCE: Boundary conditions.
604
00:37:59,960 --> 00:38:01,490
PROFESSOR: Boundary
conditions, yep.
605
00:38:01,490 --> 00:38:13,870
So now we solve for c1, c2, cd
from the boundary conditions.
606
00:38:13,870 --> 00:38:20,270
That is, f of i is b
of i for i from 0 to d.
607
00:38:23,030 --> 00:38:28,960
So for example, we
know that f of 0
608
00:38:28,960 --> 00:38:35,650
is c1 times alpha 1 to the
0, which is 1, c2 alpha 2
609
00:38:35,650 --> 00:38:40,480
to the 0 plus cd.
610
00:38:40,480 --> 00:38:45,950
And that's going to be
equal to b0 and so forth.
611
00:38:45,950 --> 00:38:48,600
So you get a system of
equations, d equations
612
00:38:48,600 --> 00:38:53,820
in d variables, that you solve
to find the coefficients.
613
00:38:53,820 --> 00:38:57,170
And that will give the
unique and correct solution
614
00:38:57,170 --> 00:39:00,360
to the recurrence.
615
00:39:00,360 --> 00:39:00,860
All right?
616
00:39:00,860 --> 00:39:05,620
Now, turns out this system of
equations is never degenerate.
617
00:39:05,620 --> 00:39:07,550
It always has a solution.
618
00:39:07,550 --> 00:39:11,860
And you can prove that,
but we won't do that.
619
00:39:11,860 --> 00:39:17,760
Any questions about
how to do that?
620
00:39:17,760 --> 00:39:18,420
All right.
621
00:39:18,420 --> 00:39:19,209
Yeah?
622
00:39:19,209 --> 00:39:21,083
AUDIENCE: Sorry, I just
had a quick question.
623
00:39:21,083 --> 00:39:25,700
But what exactly was the
distance there between n and d?
624
00:39:25,700 --> 00:39:29,710
d is like the degree of--
which the terms you go back to.
625
00:39:29,710 --> 00:39:35,260
PROFESSOR: Yes, Yeah. d is
how far back you're going.
626
00:39:35,260 --> 00:39:39,339
You go as far back
as f of n minus d.
627
00:39:39,339 --> 00:39:39,880
AUDIENCE: OK.
628
00:39:39,880 --> 00:39:40,340
Cool.
629
00:39:40,340 --> 00:39:41,965
PROFESSOR: And that
becomes the degree,
630
00:39:41,965 --> 00:39:45,177
and that is a
constant-- 2, 3, 4.
631
00:39:45,177 --> 00:39:47,510
Well, I don't think we'll
ever ask you anything beyond 4
632
00:39:47,510 --> 00:39:49,660
because it gets to
be a pain to do.
633
00:39:49,660 --> 00:39:53,450
Typically, it's 2 or 3,
sometimes even just 1.
634
00:39:53,450 --> 00:39:55,450
And then that
becomes the power of
635
00:39:55,450 --> 00:39:58,010
your characteristic equation.
636
00:39:58,010 --> 00:40:01,797
It's the order of the
characteristic equation.
637
00:40:01,797 --> 00:40:02,630
Any other questions?
638
00:40:05,240 --> 00:40:05,740
All right.
639
00:40:05,740 --> 00:40:08,550
That was the nice case
where all the roots
640
00:40:08,550 --> 00:40:11,660
of your characteristic
equation were different.
641
00:40:11,660 --> 00:40:14,625
If they're not all different,
it's a little more complicated.
642
00:40:33,380 --> 00:40:42,010
So the tricky case
is repeated roots.
643
00:40:49,320 --> 00:40:54,550
Now, the theorem, which we won't
prove but tells you what to do,
644
00:40:54,550 --> 00:41:10,460
is that if alpha is a root of
the characteristic equation
645
00:41:10,460 --> 00:41:16,530
and it is repeated r times-- so
x minus alpha to the r-th power
646
00:41:16,530 --> 00:41:39,576
is a factor-- then alpha to
the n, n times alpha to the n,
647
00:41:39,576 --> 00:41:43,350
n squared times alpha
to the n, all the way up
648
00:41:43,350 --> 00:41:47,790
to n to the r minus 1
times alpha to the n
649
00:41:47,790 --> 00:41:49,570
are all solutions
to the recurrence.
650
00:42:04,460 --> 00:42:04,960
All right?
651
00:42:04,960 --> 00:42:07,090
And then you would
treat them just
652
00:42:07,090 --> 00:42:08,420
as you would the other roots.
653
00:42:08,420 --> 00:42:10,380
You take linear
combinations, just
654
00:42:10,380 --> 00:42:13,560
like you did with the other
roots to put it all together.
655
00:42:17,470 --> 00:42:22,020
By the way, is anybody starting
to recognize a similarity
656
00:42:22,020 --> 00:42:24,615
with something else that
you've studied in the past?
657
00:42:24,615 --> 00:42:25,990
AUDIENCE: Differential
equations.
658
00:42:25,990 --> 00:42:27,910
PROFESSOR:
Differential equations.
659
00:42:27,910 --> 00:42:32,550
This is the discrete analog
of differential equations.
660
00:42:32,550 --> 00:42:33,050
All right?
661
00:42:33,050 --> 00:42:35,390
Recurrences is the same thing.
662
00:42:35,390 --> 00:42:37,450
All the math we're
going to do henceforth
663
00:42:37,450 --> 00:42:39,840
is going to look just
like what you did
664
00:42:39,840 --> 00:42:42,409
with differential equations.
665
00:42:42,409 --> 00:42:44,950
So that's sort of good news and,
I guess, bad news, depending
666
00:42:44,950 --> 00:42:45,820
on whether you like that stuff.
667
00:42:45,820 --> 00:42:46,612
AUDIENCE: Bad news.
668
00:42:46,612 --> 00:42:47,445
PROFESSOR: Bad news.
669
00:42:47,445 --> 00:42:48,020
Yeah, OK.
670
00:42:51,810 --> 00:42:54,520
All right, let's do an
example that uses maybe
671
00:42:54,520 --> 00:42:57,480
the repeated roots case.
672
00:42:57,480 --> 00:43:00,140
Suppose there's a
plant out there,
673
00:43:00,140 --> 00:43:04,510
and this plant lives
forever but only reproduces
674
00:43:04,510 --> 00:43:08,070
in the first year of life
and then never again.
675
00:43:08,070 --> 00:43:11,000
And it reproduces one for one.
676
00:43:11,000 --> 00:43:11,500
All right?
677
00:43:11,500 --> 00:43:13,530
Let's see what
happens in that case.
678
00:43:19,890 --> 00:43:22,310
Actually, there's a plant
sort of like this in Hawaii.
679
00:43:22,310 --> 00:43:27,517
I think it's called the
century plant, and it's rare.
680
00:43:27,517 --> 00:43:29,225
We'll see why when we
solve this problem.
681
00:43:31,990 --> 00:43:46,290
So it reproduces one for one
during the first year of life,
682
00:43:46,290 --> 00:43:56,825
then never again, and
the plant lives forever.
683
00:44:05,500 --> 00:44:10,951
So our question is, how fast
does the plant population grow?
684
00:44:10,951 --> 00:44:11,450
All right?
685
00:44:11,450 --> 00:44:15,010
In year n, how many
plants there are.
686
00:44:15,010 --> 00:44:18,035
So to figure that out, we're
going to set up a recurrence.
687
00:44:21,200 --> 00:44:26,330
We're going to let f of n be
the number of plants in year n.
688
00:44:33,150 --> 00:44:36,610
And we're going to say that the
first plant miraculously comes
689
00:44:36,610 --> 00:44:39,090
into existence in year 1.
690
00:44:39,090 --> 00:44:43,060
So that in year 0 there's
none of them, and in year 1
691
00:44:43,060 --> 00:44:45,512
there's 1.
692
00:44:45,512 --> 00:44:47,720
And now we want to know how
many there are in year n.
693
00:45:05,540 --> 00:45:06,070
OK.
694
00:45:06,070 --> 00:45:07,361
So let's set up the recurrence.
695
00:45:10,190 --> 00:45:15,200
f of n equals-- well,
the previous year,
696
00:45:15,200 --> 00:45:20,155
how many plants were
there in terms of f?
697
00:45:20,155 --> 00:45:21,030
AUDIENCE: [INAUDIBLE]
698
00:45:21,030 --> 00:45:23,300
PROFESSOR: f of n minus 1.
699
00:45:23,300 --> 00:45:23,800
All right?
700
00:45:23,800 --> 00:45:25,341
That's how many
there were last year,
701
00:45:25,341 --> 00:45:27,120
and they're all
still alive, plus
702
00:45:27,120 --> 00:45:31,360
we've got to add the
number of new plants.
703
00:45:31,360 --> 00:45:34,810
Well, that would
be-- yeah, well,
704
00:45:34,810 --> 00:45:36,330
how many new plants are there?
705
00:45:36,330 --> 00:45:40,560
It's all the plants that
were one-year-old last year.
706
00:45:40,560 --> 00:45:44,650
So that's all the
ones that were alive
707
00:45:44,650 --> 00:45:47,370
last year minus the ones that
were alive the year before.
708
00:45:50,450 --> 00:45:52,850
That's how many new plants
there were last year,
709
00:45:52,850 --> 00:45:56,186
and they each produce
1 new one this year.
710
00:45:56,186 --> 00:45:56,915
Is that OK?
711
00:45:56,915 --> 00:45:57,706
Everybody buy that?
712
00:46:00,265 --> 00:46:00,765
OK.
713
00:46:03,500 --> 00:46:10,020
So that equals 2 f of n
minus 1 minus f of n minus 2.
714
00:46:14,840 --> 00:46:16,280
OK.
715
00:46:16,280 --> 00:46:19,148
What's my characteristic
equation for this recurrence?
716
00:46:24,140 --> 00:46:24,680
Yeah?
717
00:46:24,680 --> 00:46:29,440
AUDIENCE: Alpha squared
minus 2 alpha plus 1.
718
00:46:29,440 --> 00:46:31,344
PROFESSOR: Yes.
719
00:46:31,344 --> 00:46:32,180
OK?
720
00:46:32,180 --> 00:46:39,931
alpha squared minus
2 alpha plus 1.
721
00:46:39,931 --> 00:46:40,430
All right?
722
00:46:40,430 --> 00:46:41,730
We jumped to that answer.
723
00:46:41,730 --> 00:46:45,863
I could have written down alpha
to the n minus 2 times alpha
724
00:46:45,863 --> 00:46:49,139
to the n minus 1 plus
alpha to the n minus 2
725
00:46:49,139 --> 00:46:50,930
and then divided by
alpha to the n minus 2.
726
00:46:50,930 --> 00:46:54,570
But pretty quickly, you want
to start just reading it off.
727
00:46:54,570 --> 00:46:56,420
That's the
characteristic equation.
728
00:46:56,420 --> 00:46:58,760
I get alpha squared, and
then bringing this over,
729
00:46:58,760 --> 00:47:03,000
a minus 2 alpha, bring
this over, plus 1 equals 0.
730
00:47:05,680 --> 00:47:10,000
Any questions on that?
731
00:47:10,000 --> 00:47:11,790
This is the process
now you will always
732
00:47:11,790 --> 00:47:13,420
use that we're going through.
733
00:47:13,420 --> 00:47:13,920
All right.
734
00:47:13,920 --> 00:47:15,294
What are the roots
to this thing?
735
00:47:18,117 --> 00:47:20,200
What are the roots to the
characteristic equation?
736
00:47:20,200 --> 00:47:24,184
AUDIENCE: Alpha
minus 1 [INAUDIBLE].
737
00:47:24,184 --> 00:47:24,850
PROFESSOR: Yeah.
738
00:47:24,850 --> 00:47:26,850
Alpha minus 1
times alpha minus 1
739
00:47:26,850 --> 00:47:29,300
is 0, which means alpha equals?
740
00:47:29,300 --> 00:47:29,800
AUDIENCE: 1.
741
00:47:29,800 --> 00:47:30,830
PROFESSOR: 1.
742
00:47:30,830 --> 00:47:37,590
So the roots are alpha equals
1, and it's a double root.
743
00:47:37,590 --> 00:47:39,070
OK?
744
00:47:39,070 --> 00:47:40,040
It's a double root.
745
00:47:40,040 --> 00:47:48,110
It occurs twice because this
is alpha minus 1 squared.
746
00:47:48,110 --> 00:47:49,550
OK.
747
00:47:49,550 --> 00:47:54,590
So what solutions am I going
to use to my recurrence?
748
00:48:00,330 --> 00:48:04,350
What's f of n going
to be now before I
749
00:48:04,350 --> 00:48:07,466
put in the boundary conditions?
750
00:48:07,466 --> 00:48:13,516
Well, it's going to be
c1 times what to the n?
751
00:48:13,516 --> 00:48:14,448
AUDIENCE: 1.
752
00:48:14,448 --> 00:48:15,140
PROFESSOR: 1.
753
00:48:15,140 --> 00:48:16,960
That's the first root.
754
00:48:16,960 --> 00:48:17,600
What's here?
755
00:48:17,600 --> 00:48:21,780
What's the next thing
I'm going to put here?
756
00:48:21,780 --> 00:48:22,740
AUDIENCE: [INAUDIBLE].
757
00:48:22,740 --> 00:48:24,555
PROFESSOR: n times 1 to the n.
758
00:48:24,555 --> 00:48:27,540
All right?
759
00:48:27,540 --> 00:48:30,150
Because I've got a root
that's repeated twice,
760
00:48:30,150 --> 00:48:34,790
r equals 2, so I have alpha
to the n and n alpha to the n.
761
00:48:34,790 --> 00:48:37,435
Just happens alpha's 1,
which makes it really easy.
762
00:48:39,770 --> 00:48:40,270
All right?
763
00:48:40,270 --> 00:48:43,000
So this is now my solution
before the boundary conditions.
764
00:48:46,270 --> 00:48:48,630
In fact, that gets even simpler.
765
00:48:48,630 --> 00:48:49,130
That's 1.
766
00:48:49,130 --> 00:48:51,308
That's c1 plus c2 times n.
767
00:48:54,520 --> 00:48:56,025
And now all that's
left is to plug
768
00:48:56,025 --> 00:48:57,150
in the boundary conditions.
769
00:49:10,680 --> 00:49:14,140
So let's do that.
770
00:49:14,140 --> 00:49:17,225
Well, f of 0 equals 0.
771
00:49:21,330 --> 00:49:23,514
If I plug it into here,
what happens for f of 0?
772
00:49:23,514 --> 00:49:24,430
What does that become?
773
00:49:24,430 --> 00:49:25,800
AUDIENCE: [INAUDIBLE].
774
00:49:25,800 --> 00:49:27,430
PROFESSOR: C1.
775
00:49:27,430 --> 00:49:28,690
So c1 equals 0.
776
00:49:28,690 --> 00:49:29,990
That's good.
777
00:49:29,990 --> 00:49:32,990
f of 1 equals 1.
778
00:49:32,990 --> 00:49:37,060
And now that's going
to be c1 plus c2.
779
00:49:37,060 --> 00:49:40,570
c1 is 0, so it
means c2 equals 1.
780
00:49:43,031 --> 00:49:43,530
All right.
781
00:49:43,530 --> 00:49:45,390
So this is really easy.
782
00:49:45,390 --> 00:49:48,430
What's f of n equal?
783
00:49:48,430 --> 00:49:50,520
n.
784
00:49:50,520 --> 00:49:53,593
So we went through a lot of work
to get an answer that probably
785
00:49:53,593 --> 00:49:55,940
we could have guessed
pretty easily by just
786
00:49:55,940 --> 00:49:58,107
plugging in a few examples.
787
00:49:58,107 --> 00:50:00,190
But the nice thing is this
works for all the cases
788
00:50:00,190 --> 00:50:05,890
when it's not so easy to guess
by plugging in the examples.
789
00:50:05,890 --> 00:50:09,850
Yeah, now you can see why this
is a relatively rare plant.
790
00:50:09,850 --> 00:50:12,080
Its population is
growing very slowly.
791
00:50:14,990 --> 00:50:15,540
OK.
792
00:50:15,540 --> 00:50:17,140
Any questions about that?
793
00:50:17,140 --> 00:50:18,140
That had repeated roots.
794
00:50:21,790 --> 00:50:22,520
All right.
795
00:50:22,520 --> 00:50:23,020
Yeah?
796
00:50:23,020 --> 00:50:24,700
AUDIENCE: So we just
guessed that f of n
797
00:50:24,700 --> 00:50:26,890
is equal to alpha
to the n, and we
798
00:50:26,890 --> 00:50:29,680
don't know if that works for
like different [INAUDIBLE].
799
00:50:29,680 --> 00:50:32,170
But how do you know that it
works until you plug it in?
800
00:50:32,170 --> 00:50:35,170
If you tried it out [INAUDIBLE]?
801
00:50:35,170 --> 00:50:37,294
PROFESSOR: Yeah.
802
00:50:37,294 --> 00:50:38,710
We didn't go through
and prove it.
803
00:50:38,710 --> 00:50:42,750
We used some facts along the
way that we didn't prove,
804
00:50:42,750 --> 00:50:43,590
which you can do.
805
00:50:43,590 --> 00:50:47,100
Like, in particular, the fact we
didn't prove that theorem there
806
00:50:47,100 --> 00:50:48,800
that those will be the roots.
807
00:50:48,800 --> 00:50:50,940
And we didn't prove that
any linear combination
808
00:50:50,940 --> 00:50:52,781
would be a solution.
809
00:50:52,781 --> 00:50:55,280
But those are facts you could
take from-- you could actually
810
00:50:55,280 --> 00:50:55,779
prove them.
811
00:50:55,779 --> 00:50:57,490
They're not too hard to prove.
812
00:50:57,490 --> 00:50:59,760
But otherwise, I think we
went through every step
813
00:50:59,760 --> 00:51:01,031
and narrowed it in.
814
00:51:01,031 --> 00:51:02,530
If we wanted to be
really sure, we'd
815
00:51:02,530 --> 00:51:03,990
go back and prove
it by induction
816
00:51:03,990 --> 00:51:06,020
against the original recurrence.
817
00:51:06,020 --> 00:51:10,030
And we'd say this would be
the induction hypothesis.
818
00:51:10,030 --> 00:51:12,930
We'd cover the base
cases-- f of 0 and f of 1.
819
00:51:12,930 --> 00:51:16,610
And then we'd plug this into
the recurrence up there,
820
00:51:16,610 --> 00:51:22,050
which is, does n equal twice
n minus 1 minus n minus 2,
821
00:51:22,050 --> 00:51:23,100
is what we would do.
822
00:51:23,100 --> 00:51:24,790
And you see that it's true.
823
00:51:24,790 --> 00:51:29,000
So you could really
prove it by induction
824
00:51:29,000 --> 00:51:30,720
for any of the
solutions we ever get.
825
00:51:30,720 --> 00:51:33,740
And if you worked this
way to get the solution,
826
00:51:33,740 --> 00:51:35,390
it will always work.
827
00:51:35,390 --> 00:51:38,172
This method never fails,
gives you the wrong answer.
828
00:51:38,172 --> 00:51:39,630
In fact, that's an
important point.
829
00:51:39,630 --> 00:51:41,963
When we tell you to solve a
recurrence using this method
830
00:51:41,963 --> 00:51:45,450
on an exam or for
homework, you don't
831
00:51:45,450 --> 00:51:47,460
have to verify it by
induction unless we
832
00:51:47,460 --> 00:51:49,550
say verify by induction.
833
00:51:49,550 --> 00:51:51,130
Because we're
giving you a method
834
00:51:51,130 --> 00:51:52,540
that is guaranteed to work.
835
00:51:52,540 --> 00:51:55,220
And if you do the method
right, you're fine.
836
00:51:55,220 --> 00:51:57,960
There's no guessing here so far.
837
00:51:57,960 --> 00:52:00,440
In a minute, we're going
to do a little guessing,
838
00:52:00,440 --> 00:52:03,573
but not so far.
839
00:52:03,573 --> 00:52:04,406
Any other questions?
840
00:52:07,670 --> 00:52:10,100
OK.
841
00:52:10,100 --> 00:52:17,301
This works for all homogeneous
linear recurrences.
842
00:52:17,301 --> 00:52:17,800
All right?
843
00:52:17,800 --> 00:52:19,615
Remember homogeneous
and inhomogeneous
844
00:52:19,615 --> 00:52:22,450
from differential equations?
845
00:52:22,450 --> 00:52:23,940
It's the same
thing happens here.
846
00:52:31,600 --> 00:52:34,037
And now we're going to talk
about the inhomogeneous case.
847
00:52:41,950 --> 00:52:42,450
All right.
848
00:52:42,450 --> 00:52:48,100
So we've been looking at
linear recurrences, which
849
00:52:48,100 --> 00:52:54,450
means you have something like
f of n minus a1 f of n minus 1
850
00:52:54,450 --> 00:53:01,892
minus ad f of n
minus d equals 0,
851
00:53:01,892 --> 00:53:03,225
and that means it's homogeneous.
852
00:53:09,170 --> 00:53:11,930
Now, instead of 0, I might
have had something else
853
00:53:11,930 --> 00:53:14,930
here, might have
been equal to 1,
854
00:53:14,930 --> 00:53:21,040
maybe n squared, or some
general function g of n.
855
00:53:21,040 --> 00:53:22,953
These cases are
all inhomogeneous.
856
00:53:30,430 --> 00:53:31,800
OK?
857
00:53:31,800 --> 00:53:35,570
Now, solving inhomogeneous
linear equations
858
00:53:35,570 --> 00:53:38,250
is just one step harder
than homogeneous.
859
00:53:38,250 --> 00:53:38,750
Yeah?
860
00:53:38,750 --> 00:53:39,583
AUDIENCE: All right.
861
00:53:39,583 --> 00:53:41,577
Could you define homogeneous
and inhomogeneous
862
00:53:41,577 --> 00:53:43,660
for people who didn't take
differential equations?
863
00:53:43,660 --> 00:53:44,326
PROFESSOR: Yeah.
864
00:53:44,326 --> 00:53:49,170
Homogeneous means
you have the 0 here.
865
00:53:49,170 --> 00:53:52,720
You got all these f
terms here, right,
866
00:53:52,720 --> 00:53:54,380
and there's nothing else.
867
00:53:54,380 --> 00:53:57,940
f of n minus a1 f n
minus 1 dot dot dot
868
00:53:57,940 --> 00:54:01,590
minus ad f n minus d equals 0.
869
00:54:01,590 --> 00:54:05,610
Like with Fibonacci
numbers-- f of n minus f n
870
00:54:05,610 --> 00:54:09,640
minus 1 minus f n minus 2 is 0.
871
00:54:09,640 --> 00:54:11,250
OK?
872
00:54:11,250 --> 00:54:14,710
I could consider
other recurrences
873
00:54:14,710 --> 00:54:17,770
where there's something else
out here that it equals.
874
00:54:17,770 --> 00:54:19,550
Like I could have a
Fibonacci recurrence
875
00:54:19,550 --> 00:54:23,300
where I have f(n) equals f of
n minus 1 plus f(n) minus 2
876
00:54:23,300 --> 00:54:26,780
plus g of n, like n cubed.
877
00:54:26,780 --> 00:54:29,820
That would give me a
Fibonacci-like recurrence.
878
00:54:29,820 --> 00:54:32,680
And as soon as you put
a non-zero out here,
879
00:54:32,680 --> 00:54:34,775
then it's inhomogeneous.
880
00:54:34,775 --> 00:54:36,150
And now I'm going
to tell you how
881
00:54:36,150 --> 00:54:37,960
to solve that kind
of recurrence,
882
00:54:37,960 --> 00:54:41,460
which is more general.
883
00:54:41,460 --> 00:54:41,960
OK?
884
00:54:44,642 --> 00:54:45,142
OK.
885
00:54:48,905 --> 00:54:51,363
So let me outline the method,
and then we'll do an example.
886
00:55:02,220 --> 00:55:02,840
All right.
887
00:55:02,840 --> 00:55:23,330
So the general inhomogeneous
recurrence is exactly this.
888
00:55:23,330 --> 00:55:32,280
It's f of n minus a1 f of n
minus 1 minus a sub d f of n
889
00:55:32,280 --> 00:55:38,270
minus d equals g of n, where
that's some fixed function
890
00:55:38,270 --> 00:55:39,770
of n, nothing to do with f.
891
00:55:42,550 --> 00:55:46,240
And we solve it according to
the following three-step method.
892
00:55:49,200 --> 00:55:59,280
In step 1, we
replace g of n by 0,
893
00:55:59,280 --> 00:56:03,464
thereby creating the situation
we already know how to solve.
894
00:56:06,330 --> 00:56:11,450
And we solve the
homogeneous recurrence--
895
00:56:11,450 --> 00:56:25,220
because you got a 0 there now--
but we don't go all the way.
896
00:56:25,220 --> 00:56:27,720
We ignore the boundary
conditions for now.
897
00:56:35,321 --> 00:56:37,320
So you don't get all the
way to the final answer
898
00:56:37,320 --> 00:56:39,420
where you plugged in
the boundary conditions
899
00:56:39,420 --> 00:56:41,510
to get the constant
coefficients.
900
00:56:41,510 --> 00:56:45,510
Leave the constant
coefficients undecided.
901
00:56:45,510 --> 00:56:55,420
Then in step 2, we
put back in g of n,
902
00:56:55,420 --> 00:57:05,880
and we find any what's called
particular solution, again
903
00:57:05,880 --> 00:57:08,126
ignoring boundary conditions.
904
00:57:08,126 --> 00:57:10,392
So you'll have the
constant factors there too.
905
00:57:20,141 --> 00:57:20,640
All right?
906
00:57:20,640 --> 00:57:24,990
And then step 3 is going to be
to put the whole thing together
907
00:57:24,990 --> 00:57:27,371
and plug in the boundary
conditions and solve it.
908
00:57:35,330 --> 00:57:41,675
So we add the homogeneous
and particular--
909
00:57:41,675 --> 00:57:54,420
and I'll show you how to
do step 2 in a minute--
910
00:57:54,420 --> 00:58:05,930
so we add the homogeneous and
particular solutions together
911
00:58:05,930 --> 00:58:08,570
and then use the boundary
conditions to get the answer.
912
00:58:30,560 --> 00:58:31,060
OK.
913
00:58:31,060 --> 00:58:32,110
So let's do an example.
914
00:58:35,750 --> 00:58:38,370
Let's go down here and do an
example of the three steps.
915
00:58:57,480 --> 00:59:00,780
So let's say that
our recurrence is
916
00:59:00,780 --> 00:59:09,190
this nasty-looking thing-- f
of n equals 4 f of n minus 1
917
00:59:09,190 --> 00:59:12,850
plus 3 to the n.
918
00:59:12,850 --> 00:59:19,240
And the boundary condition
is that f of 1 is 1.
919
00:59:19,240 --> 00:59:24,800
So step 1 says, ignore
this 3 to the n thing
920
00:59:24,800 --> 00:59:28,130
and get back to just the
homogeneous form, which
921
00:59:28,130 --> 00:59:32,490
is f of n minus 4 f
of n minus 1 is 0.
922
00:59:32,490 --> 00:59:33,720
So let's solve that.
923
00:59:33,720 --> 00:59:36,720
What's the
characteristic equation
924
00:59:36,720 --> 00:59:41,062
for the homogeneous part?
925
00:59:41,062 --> 00:59:42,520
What's the
characteristic equation?
926
00:59:45,789 --> 00:59:46,723
What is it?
927
00:59:46,723 --> 00:59:49,012
AUDIENCE: Alpha [INAUDIBLE].
928
00:59:49,012 --> 00:59:49,720
PROFESSOR: Close.
929
00:59:49,720 --> 00:59:52,873
It's-- you could either say
alpha to the n minus 4 alpha
930
00:59:52,873 --> 00:59:54,780
to the n minus 1.
931
00:59:54,780 --> 00:59:59,080
But better to simplify it
into an order 1 polynomial.
932
00:59:59,080 --> 01:00:04,470
So it would be alpha minus 4,
really simple in this case.
933
01:00:04,470 --> 01:00:06,950
So the characteristic
equation is alpha minus 4,
934
01:00:06,950 --> 01:00:12,850
alpha minus 4, all
right, equals 0.
935
01:00:12,850 --> 01:00:16,600
And it's really easy to
find the root of this thing.
936
01:00:16,600 --> 01:00:20,840
It's just alpha equals
4 is your only root.
937
01:00:20,840 --> 01:00:27,040
And that means the
homogeneous solution
938
01:00:27,040 --> 01:00:32,134
is f of n equals a
constant times 4 to the n.
939
01:00:34,761 --> 01:00:35,260
All right?
940
01:00:35,260 --> 01:00:36,150
So that's step 1.
941
01:00:43,670 --> 01:00:44,360
OK.
942
01:00:44,360 --> 01:00:48,155
Let's do step 2, which
we've not tried before.
943
01:00:51,910 --> 01:00:56,830
We need to find a
particular solution,
944
01:00:56,830 --> 01:01:04,520
and this just means
any old solution
945
01:01:04,520 --> 01:01:13,590
to f of n minus 4 f of n
minus 1 equals 3 to the n,
946
01:01:13,590 --> 01:01:15,970
without worrying about
boundary conditions.
947
01:01:15,970 --> 01:01:21,395
Now, there's basic rules
to use to figure out
948
01:01:21,395 --> 01:01:23,690
what to guess here.
949
01:01:23,690 --> 01:01:30,210
And basically, the idea is that
you guess something for f of n
950
01:01:30,210 --> 01:01:34,800
that looks a whole lot
like this g term out here.
951
01:01:34,800 --> 01:01:38,070
And in particular, if
this g term is 3 to the n,
952
01:01:38,070 --> 01:01:41,270
you guess a constant
times 3 to the n.
953
01:01:41,270 --> 01:01:44,680
If it's 5 to the n, you guess
a constant times 5 to the n.
954
01:01:44,680 --> 01:01:49,420
If it's n squared, you guess a
polynomial that's of degree 2
955
01:01:49,420 --> 01:01:51,590
and so forth.
956
01:01:51,590 --> 01:01:52,090
OK?
957
01:01:52,090 --> 01:01:58,920
So let's guess a
constant times 3 to the n
958
01:01:58,920 --> 01:02:01,560
and see if we can
make it work, not
959
01:02:01,560 --> 01:02:03,452
worrying about the
boundary conditions.
960
01:02:13,890 --> 01:02:16,346
All right.
961
01:02:16,346 --> 01:02:20,220
And this is just like
differential equations, right?
962
01:02:20,220 --> 01:02:22,400
Same guessing
strategy exactly you
963
01:02:22,400 --> 01:02:26,250
use in differential equations,
those of you who've had that.
964
01:02:26,250 --> 01:02:31,840
So we guess f of n equals a
constant times 3 to the n,
965
01:02:31,840 --> 01:02:34,320
and let's plug it in.
966
01:02:34,320 --> 01:02:37,640
So we plug that in up there.
967
01:02:37,640 --> 01:02:44,300
We get c3 to the n minus
c3 to the n minus 1
968
01:02:44,300 --> 01:02:47,601
equals 3 to the n.
969
01:02:47,601 --> 01:02:48,100
All right?
970
01:02:48,100 --> 01:02:50,240
So let's divide by
3 to the n here.
971
01:02:50,240 --> 01:02:56,710
We get 3c minus c equals 3.
972
01:02:56,710 --> 01:02:58,570
Did I do that right?
973
01:02:58,570 --> 01:03:01,815
No, I left off my 4 here,
right, left off that.
974
01:03:01,815 --> 01:03:02,830
So there's a 4 here.
975
01:03:06,150 --> 01:03:10,900
And that means
that c is minus 3.
976
01:03:10,900 --> 01:03:11,400
OK?
977
01:03:11,400 --> 01:03:15,600
So I got minus c equals
3, so c is minus 3.
978
01:03:15,600 --> 01:03:16,100
All right.
979
01:03:16,100 --> 01:03:27,860
That means that the particular
solution is just f of n
980
01:03:27,860 --> 01:03:36,060
equals-- c is minus 3--
minus 3 to the n plus 1.
981
01:03:36,060 --> 01:03:36,560
All right?
982
01:03:36,560 --> 01:03:41,370
So now I found a solution where
there's no constants involved.
983
01:03:41,370 --> 01:03:43,780
This time, the
constant went away just
984
01:03:43,780 --> 01:03:46,830
plugging into the
recurrence formula.
985
01:03:46,830 --> 01:03:49,130
I didn't use base cases yet.
986
01:03:49,130 --> 01:03:51,050
I just found a
particular case that
987
01:03:51,050 --> 01:03:56,980
works for the recurrence--
minus 3 to the n plus 1.
988
01:03:56,980 --> 01:03:59,492
OK?
989
01:03:59,492 --> 01:04:01,430
All right.
990
01:04:01,430 --> 01:04:02,546
Now we go to step 3.
991
01:04:19,300 --> 01:04:19,800
All right.
992
01:04:19,800 --> 01:04:30,670
Step 3, we find the
general solution
993
01:04:30,670 --> 01:04:33,520
by adding the homogeneous
and the particular solution.
994
01:04:37,850 --> 01:04:43,810
So we take f of
n equals c1 times
995
01:04:43,810 --> 01:04:50,790
4 to the n plus negative
3 to the n plus 1.
996
01:04:55,350 --> 01:04:58,660
And all I've got to
do is determine c1,
997
01:04:58,660 --> 01:05:01,894
and I do that from the
boundary conditions.
998
01:05:01,894 --> 01:05:02,560
Yeah, let's see.
999
01:05:02,560 --> 01:05:05,800
Where was my boundary
condition here?
1000
01:05:05,800 --> 01:05:09,580
Ah, up there-- f of 1 is 1.
1001
01:05:09,580 --> 01:05:12,690
So f of 1 equals 1.
1002
01:05:12,690 --> 01:05:15,760
Plugging in 1 here,
I get c1 times 4
1003
01:05:15,760 --> 01:05:21,430
to the 1 minus 3 squared.
1004
01:05:21,430 --> 01:05:21,930
All right?
1005
01:05:21,930 --> 01:05:25,120
So I have 4c1 minus 9.
1006
01:05:25,120 --> 01:05:30,100
Put the 9 over here,
I get 10 equals 4c1.
1007
01:05:30,100 --> 01:05:32,905
And that means that c1 is 5/2.
1008
01:05:37,570 --> 01:05:39,870
And now I know the
final solution for f(n).
1009
01:05:44,170 --> 01:05:49,845
It's 5/2 4 to the n
minus 3 to the n plus 1.
1010
01:05:56,650 --> 01:05:58,244
Now, when you're
all done with this,
1011
01:05:58,244 --> 01:05:59,660
you don't have--
we won't make you
1012
01:05:59,660 --> 01:06:03,579
do an inductive proof, which you
could do to verify it's right.
1013
01:06:03,579 --> 01:06:05,370
If you wanted to be
really, really careful,
1014
01:06:05,370 --> 01:06:06,950
you should do that.
1015
01:06:06,950 --> 01:06:10,952
But it is a good idea just
to check a couple values of n
1016
01:06:10,952 --> 01:06:13,535
to make sure you didn't make a
mistake because you might have.
1017
01:06:13,535 --> 01:06:16,440
With all these calculations,
you might have made a mistake.
1018
01:06:16,440 --> 01:06:20,200
So let's check, for
example, f of 2.
1019
01:06:23,000 --> 01:06:30,300
So by the recurrence,
f of 2 is 4 times
1020
01:06:30,300 --> 01:06:39,960
f of 1-- it's 4 times 1-- plus
3 squared, which is 9, is 13.
1021
01:06:39,960 --> 01:06:43,790
And let's just check
when we plug in 2 here.
1022
01:06:43,790 --> 01:06:52,590
f of 2 is 5/2 times 16 minus 27.
1023
01:06:52,590 --> 01:06:57,210
5 times 8 is 40, minus 27 is 13.
1024
01:06:57,210 --> 01:06:58,920
Just as a sanity check.
1025
01:06:58,920 --> 01:07:01,170
Because there's a decent
chance if you made a mistake,
1026
01:07:01,170 --> 01:07:04,240
it'll catch it pretty
quick with n equals 2 or 3.
1027
01:07:04,240 --> 01:07:04,740
All right?
1028
01:07:04,740 --> 01:07:07,880
Just to make sure
you got it right.
1029
01:07:07,880 --> 01:07:14,280
Any questions about that?
1030
01:07:14,280 --> 01:07:16,630
So the tricky part
here is guessing
1031
01:07:16,630 --> 01:07:18,730
the particular solution.
1032
01:07:18,730 --> 01:07:24,250
So let me give you the rules
for that, just write those down,
1033
01:07:24,250 --> 01:07:26,100
and maybe I'll do
one last example.
1034
01:07:40,320 --> 01:07:40,820
All right.
1035
01:07:40,820 --> 01:07:43,950
So we're guessing a
particular solution.
1036
01:07:56,740 --> 01:08:03,570
So if g-- that's saying
the non-homogeneous part--
1037
01:08:03,570 --> 01:08:14,660
is exponential, you should
try guessing an exponential
1038
01:08:14,660 --> 01:08:15,880
of the same type.
1039
01:08:20,740 --> 01:08:31,050
So for example, say that g is
2 to the n plus 3 to the n.
1040
01:08:31,050 --> 01:08:37,050
What you should do is guess
some constant a times 2 to the n
1041
01:08:37,050 --> 01:08:40,279
plus some constant
b times 3 to the n.
1042
01:08:40,279 --> 01:08:43,340
Plug it in-- not to the
boundary conditions.
1043
01:08:43,340 --> 01:08:47,520
Plug it into the recurrence
equation and solve for a and b.
1044
01:08:47,520 --> 01:08:50,939
And it generally will work.
1045
01:08:50,939 --> 01:09:00,450
If g is polynomial, you
should guess a polynomial
1046
01:09:00,450 --> 01:09:01,420
of the same degree.
1047
01:09:07,840 --> 01:09:21,109
So for example, say g of
n is n squared minus 1.
1048
01:09:21,109 --> 01:09:33,050
You should guess-- let's
see-- guess an squared
1049
01:09:33,050 --> 01:09:36,229
plus bn plus c.
1050
01:09:36,229 --> 01:09:38,951
Plug that in for f of n.
1051
01:09:38,951 --> 01:09:39,450
All right?
1052
01:09:39,450 --> 01:09:44,186
So when we're doing these
guesses, we're guessing f of n.
1053
01:09:44,186 --> 01:09:45,668
Same thing up here.
1054
01:10:07,930 --> 01:10:10,800
Now, say you mixed two together.
1055
01:10:10,800 --> 01:10:15,130
Suppose you have an
example where g of n
1056
01:10:15,130 --> 01:10:18,710
equals 2 to the n plus n.
1057
01:10:18,710 --> 01:10:22,359
What do you suppose
you do in that case?
1058
01:10:22,359 --> 01:10:24,190
AUDIENCE: [INAUDIBLE]
to guess [INAUDIBLE].
1059
01:10:24,190 --> 01:10:27,660
PROFESSOR: Yeah, guess each one
separately, add them together.
1060
01:10:27,660 --> 01:10:33,550
So you're going to guess
f of n equals a times 2
1061
01:10:33,550 --> 01:10:37,530
to the n plus bn plus c.
1062
01:10:37,530 --> 01:10:38,110
All right?
1063
01:10:38,110 --> 01:10:40,610
Because you take the guess for
that plus the guess for that.
1064
01:10:44,050 --> 01:10:44,560
All right.
1065
01:10:44,560 --> 01:10:48,540
Now, there's one last thing
that can go wrong here.
1066
01:10:48,540 --> 01:10:53,841
And that is you can try your
guess, and it doesn't work.
1067
01:10:53,841 --> 01:10:54,340
All right?
1068
01:10:54,340 --> 01:10:55,940
So there's rules for that too.
1069
01:10:58,460 --> 01:11:06,140
For example, if, say,
g(n) is 2 to the n
1070
01:11:06,140 --> 01:11:13,440
and your guess of a times 2 to
the n, where a is a constant,
1071
01:11:13,440 --> 01:11:14,575
fails.
1072
01:11:14,575 --> 01:11:17,800
And we'll do an example in
a minute where it fails.
1073
01:11:17,800 --> 01:11:22,900
What you do then is you guess
a polynomial times 2 to the n.
1074
01:11:22,900 --> 01:11:28,630
So you'd guess an plus
b times 2 to the n.
1075
01:11:28,630 --> 01:11:32,370
And if that fails,
your next guess
1076
01:11:32,370 --> 01:11:39,130
would be an squared plus
bn plus c times 2 to the n.
1077
01:11:39,130 --> 01:11:41,670
And it won't happen,
but if that failed,
1078
01:11:41,670 --> 01:11:43,820
you'd guess a cubic
times 2 to the n.
1079
01:11:43,820 --> 01:11:47,480
You keep pounding it with
another factor of n in front
1080
01:11:47,480 --> 01:11:50,030
if the guess fails.
1081
01:11:50,030 --> 01:11:52,260
And that's true
for anything at all
1082
01:11:52,260 --> 01:11:55,167
you would be doing like
this, and that'll work.
1083
01:11:55,167 --> 01:11:57,500
And I don't think I've ever
encountered an example where
1084
01:11:57,500 --> 01:12:00,760
you have to go very
far to make that fly.
1085
01:12:04,180 --> 01:12:06,260
Same thing we had
for repeated roots
1086
01:12:06,260 --> 01:12:09,310
in the characteristic equation.
1087
01:12:09,310 --> 01:12:11,610
Multiply those factors
of n in front of it
1088
01:12:11,610 --> 01:12:13,811
to get to the answer.
1089
01:12:13,811 --> 01:12:14,310
All right.
1090
01:12:14,310 --> 01:12:17,210
Let's do one more
example where it fails,
1091
01:12:17,210 --> 01:12:18,832
so we get to see what happens.
1092
01:12:39,970 --> 01:12:48,470
Let's try this recurrence--
f of n is 2 f of n minus 1
1093
01:12:48,470 --> 01:12:51,400
plus 2 to the n.
1094
01:12:51,400 --> 01:12:56,870
And the boundary
condition is f of 0 is 1.
1095
01:12:56,870 --> 01:13:02,090
What's the first
thing I do for this?
1096
01:13:02,090 --> 01:13:04,450
AUDIENCE: Set it equal to 0.
1097
01:13:04,450 --> 01:13:06,515
PROFESSOR: Set it equal
to 0 and solve it.
1098
01:13:06,515 --> 01:13:07,962
Get the homogeneous solution.
1099
01:13:13,240 --> 01:13:13,820
OK.
1100
01:13:13,820 --> 01:13:16,740
So what's the
characteristic polynomial?
1101
01:13:20,290 --> 01:13:24,520
Alpha minus 2 is 0.
1102
01:13:24,520 --> 01:13:26,541
What's my root?
1103
01:13:26,541 --> 01:13:27,040
2.
1104
01:13:29,790 --> 01:13:33,550
And therefore, my
homogeneous solution
1105
01:13:33,550 --> 01:13:36,460
is c1 times 2 to the n.
1106
01:13:36,460 --> 01:13:38,900
Well, that's pretty simple.
1107
01:13:38,900 --> 01:13:39,892
What's the next step?
1108
01:13:43,860 --> 01:13:45,983
What do I have to find next?
1109
01:13:45,983 --> 01:13:46,858
AUDIENCE: Particular.
1110
01:13:46,858 --> 01:13:48,334
PROFESSOR: Particular solution.
1111
01:13:52,762 --> 01:13:53,270
All right.
1112
01:13:53,270 --> 01:13:55,282
So what am I going to guess?
1113
01:14:00,210 --> 01:14:01,690
a times 2 to the n.
1114
01:14:04,260 --> 01:14:05,340
All right?
1115
01:14:05,340 --> 01:14:08,110
So let's plug that in,
a times 2 to the n.
1116
01:14:08,110 --> 01:14:10,720
I'm going to plug it in to here.
1117
01:14:10,720 --> 01:14:18,100
So I get a 2 to the n
equals 2 times a times 2
1118
01:14:18,100 --> 01:14:23,200
to the n minus 1
plus 2 to the n.
1119
01:14:23,200 --> 01:14:24,760
Did I do that right?
1120
01:14:24,760 --> 01:14:27,190
Think so.
1121
01:14:27,190 --> 01:14:29,962
So I get 2 to the n, 2
to the n, 2 to the n.
1122
01:14:29,962 --> 01:14:36,040
I get a equals a plus 1.
1123
01:14:36,040 --> 01:14:38,510
Not so good.
1124
01:14:38,510 --> 01:14:39,650
All right?
1125
01:14:39,650 --> 01:14:42,586
There's no solution for a.
1126
01:14:42,586 --> 01:14:44,181
Well, that's bad.
1127
01:14:44,181 --> 01:14:44,680
All right.
1128
01:14:44,680 --> 01:14:45,960
So what do I do?
1129
01:14:52,701 --> 01:14:54,576
Any thoughts about what
I'm going to do next?
1130
01:14:57,414 --> 01:14:58,360
AUDIENCE: [INAUDIBLE].
1131
01:14:58,360 --> 01:14:58,940
PROFESSOR: What is it?
1132
01:14:58,940 --> 01:15:00,240
AUDIENCE: Change your guess.
1133
01:15:00,240 --> 01:15:01,200
PROFESSOR: Change the guess.
1134
01:15:01,200 --> 01:15:02,616
What's the next
guess going to be?
1135
01:15:06,780 --> 01:15:07,580
Yeah, all right.
1136
01:15:07,580 --> 01:15:18,630
So now I'm going to guess f
of n is an plus b 2 to the n.
1137
01:15:18,630 --> 01:15:19,840
We'll hope for better luck.
1138
01:15:19,840 --> 01:15:23,020
So let's plug that in.
1139
01:15:23,020 --> 01:15:32,100
an plus b 2 to the n into--
where am I, up there--
1140
01:15:32,100 --> 01:15:42,013
equals 2 a n minus 1 plus b
times 2 to the n minus 1 plus 2
1141
01:15:42,013 --> 01:15:44,435
to the n.
1142
01:15:44,435 --> 01:15:45,130
All right.
1143
01:15:45,130 --> 01:15:48,325
Now, I can divide out the 2 to
the n's here like I did before.
1144
01:15:51,000 --> 01:16:04,510
And I get an plus b equals
an minus a plus b plus 1.
1145
01:16:04,510 --> 01:16:09,240
That cancels here, b cancels.
1146
01:16:09,240 --> 01:16:13,511
So I got a solution--
a equals 1.
1147
01:16:13,511 --> 01:16:15,500
And I don't care what b is.
1148
01:16:15,500 --> 01:16:17,910
I didn't need to
set it to anything.
1149
01:16:17,910 --> 01:16:20,880
So I'll make it 0 just
to get it out of the way
1150
01:16:20,880 --> 01:16:22,960
because it didn't matter.
1151
01:16:22,960 --> 01:16:31,375
So my particular solution
then is f of n-- a is 1.
1152
01:16:31,375 --> 01:16:34,370
b I can just make 0 because
it didn't matter what I used,
1153
01:16:34,370 --> 01:16:35,790
so I'll make it simpler.
1154
01:16:35,790 --> 01:16:38,634
So I get n 2 to the n.
1155
01:16:38,634 --> 01:16:39,884
That's my particular solution.
1156
01:16:42,810 --> 01:16:46,190
And I've got my general
solution up there as c1 times 2
1157
01:16:46,190 --> 01:16:48,830
to the n-- sorry, the
homogeneous solution.
1158
01:16:51,530 --> 01:16:53,339
What's the next step?
1159
01:16:59,207 --> 01:17:00,300
What's step 3?
1160
01:17:04,570 --> 01:17:07,955
What do I do with these
guys, this solution
1161
01:17:07,955 --> 01:17:09,387
and that solution?
1162
01:17:09,387 --> 01:17:10,532
[? AUDIENCE: Add them. ?]
1163
01:17:10,532 --> 01:17:11,740
PROFESSOR: Add them together.
1164
01:17:11,740 --> 01:17:12,240
Good.
1165
01:17:12,240 --> 01:17:23,380
So the general solution is I
have f of n is the sum of c1 2
1166
01:17:23,380 --> 01:17:27,700
to the n plus the particular
solutions n 2 to the n.
1167
01:17:27,700 --> 01:17:31,189
How do I figure out what c1 is?
1168
01:17:31,189 --> 01:17:32,730
AUDIENCE: Plugging
in to [INAUDIBLE].
1169
01:17:32,730 --> 01:17:37,050
PROFESSOR: Plugging in the
boundary condition-- f(0)
1170
01:17:37,050 --> 01:17:38,275
equals 1.
1171
01:17:38,275 --> 01:17:40,400
f(0) equals 1.
1172
01:17:40,400 --> 01:17:47,240
Plug in 0, I get c1
2 to the 0 plus 0.
1173
01:17:47,240 --> 01:17:51,390
Well, that's pretty
easy. c1 equals 1.
1174
01:17:51,390 --> 01:17:54,230
So I now have the final answer.
1175
01:17:54,230 --> 01:17:59,940
f of n equals 2 to the
n plus n 2 to the n.
1176
01:17:59,940 --> 01:18:00,777
All right?
1177
01:18:00,777 --> 01:18:01,360
Any questions?
1178
01:18:05,730 --> 01:18:06,230
OK.
1179
01:18:06,230 --> 01:18:09,740
So it's a little
tedious to do this.
1180
01:18:09,740 --> 01:18:12,560
But the really nice thing
is any linear recurrence
1181
01:18:12,560 --> 01:18:17,390
you ever see, this method
always works, which is handy.
1182
01:18:17,390 --> 01:18:17,890
OK.
1183
01:18:17,890 --> 01:18:20,040
That's it for today.