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1. plus 2x cubed plus
n
minus 1x to the
n
, plus nx to the
n
plus 1.
2. proportional to
N
squared, which is natural.
3. you need, the
n
minus 1 and
n
minus 2.
4. to the
n
plus
n
-- or minus nx to the
n
plus 1.
5. are an integer divided by
n
.
6. So by increasing
N
, basically you
7.
N
-Not again...
8. have some number of equations, say
n
equations and
n
unknowns.
9. Derivative of this is now minus
n
plus 1, x to the
n
.
10. As
n
gets big, the square root of
n
11. I will get
n
. Very good.
12. all these first
n
minus 1 rectangles,
13. It's like an
N
1, 5.
14. is f of
n
minus 4 f of
n
minus 1 is 0.
15. linearly increase with
n
.
16. Minus nx to the
n
plus 1.
17. equals 3 to the
n
.
18. AUDIENCE: f of
n
minus 2.
19. So we guess f of
n
equals a constant times 3 to the
n
,
20. in these bounds as
n
gets big?
21. And that's basically the limit as
n
22. And that's basically the limit as
n
23. to a finite, positive solution, as you let
n
approach infinity, but that a sub
n
and
24. What's f of
n
equal?
25. We're going to let f of
n
be the number of plants in year
n
.
26. AUDIENCE: 2 to the
n
.
27. I've got 0, 1, 2, 3,
n
minus 2,
n
minus 1,
n
,
28.
n
squared times alpha to the
n
, all the way up
29. PROFESSOR:
n
times 1 to the
n
.
30. And delta
n
is less than a 1/10 for
n
greater than 4.
31. We're going to let f of
n
be the number of plants in year
n
.
32. That the
n
that depends on t, there's actually one largest
n
33. m to result
n
times.
34. so now I add the f
n
.
35. limit as
n
approaches infinity of a sub
n
is equal to 0, so this may actually converge.
36. It's a tree on
n
vertices.
37. that the
n
-th sum exceeds
n
times some fixed quantity a is
38. there. A sub
n
is always positive. B sub
n
is always positive. We know that simpler series
39. d alpha to the
n
minus d.
40. being
n
, this ends up being VE.
41. Might be something like
n
cubed.
42. every such tree has
n
edges.
43. What happens when
n
goes to infinity?
44. PROFESSOR:
N
0, sigma squared, right.
45. like, well, there are
n
minus 1 edges in any
n
vertex tree.
46. So that means that
n
is prime.
47. root of
n
that will evenly divide
n
.
48. So there's order
n
choices, so this is order
n
.
49. has
n
minus 1 edges.
50. as
n
goes to infinity.
51. And now I'm going to try f of
n
is alpha to the
n
.
52. r equals 2, so I have alpha to the
n
and
n
alpha to the
n
.
53. So for example, say that g is 2 to the
n
plus 3 to the
n
.
54. expression for f of
n
.
55. where
n
goes from 1 to infinity, of
n
to the k – 1 power over
n
^k + 7, where k is greater
56. One over
n
, right?
57. to f of
n
minus 4 f of
n
minus 1 equals 3 to the
n
,
58. What is f of
n
?
59. And
n
minus 1 down to 0.
60. It'll get
n
different answers from the
n
right-hand sides,
61. I hope it's only
n
.
62. So that equals 2 f of
n
minus 1 minus f of
n
minus 2.
63. That's the square root of
n
.
64. It's 5/2 4 to the
n
minus 3 to the
n
plus 1.
65. So for example, say g of
n
is
n
squared minus 1.
66. that the sum is at most f of
n
plus the integral from 1 to
n
.
67. goes to some
N
0, 1.
68. If it's 5 to the
n
, you guess a constant times 5 to the
n
.
69. We know that f of
n
is f of
n
minus 1 plus f of
n
minus 2.
70. That's c1 plus c2 times
n
.
71. the f of
n
is what we--
72. f of
n
minus 1 plus a2 times f of
n
minus 2 plus dot dot
73. to
n
times a?
74. S-C-I-F-U-
N
-- .org.
75. this nasty-looking thing-- f of
n
equals 4 f of
n
minus 1
76. That gives us alpha to the
n
equals alpha to the
n
77. So that's all together exactly
n
edges.
78. If I have
n
words, there's 2 to the
n
different splits.
79. to the
n
plus cd alpha d to the
n
.
80. to the
n
for some constant alpha.
81. We get c3 to the
n
minus c3 to the
n
minus 1
82. plus 1, x to the
n
plus nx to the
n
plus 1 over 1
83. by
n
, because sigma hat is equal to 1 over
n
sum
84. So for example, take
n
equals 100.
85. of
n
, nothing to do with f.
86. times 4^1, raised to the
n
power, we could bring that
n
down here and say it was 3^
n
87. We know that's a tree on
n
vertices has
n
minus 1 edges.
88. In general, the identity matrix in size
n
x
n
is an
n
x
n
matrix
89. and the result would be
n
by
n
.
90. So a tree with
n
vertices has
n
minus 1 edges.
91. of freedom divided by
n
.
92. so the gap here, square root of
n
minus 1, and
n
equals 100.
93. and r^
n
sin(
n
*theta).
94. because A could be m by
n
.
95. So that's for
n
large.
96. So for example, say that g is 2 to the
n
plus 3 to the
n
.
97. That's 2 to the
n
.
98. AUDIENCE:
n
times
n
plus 1 times 2n plus 1, all over 6.
99. equal 1 to
n
of these indicators.
100. so its area is f of
n
minus 1, and then f of
n
.
101. So we have 0, 1, 2, 3,
n
minus 2,
n
minus 1,
n
, and now f of
n
102. I get a fraction of
n
, right?
103. That's the square root of
n
.
104. growing as a function of
n
.
105. What is f of
n
?
106. is f of
n
equals a constant times 4 to the
n
.
107. as dividing here by
n
minus 1.
108. And that's why you see i minus 1 over
n
, or i over
n
.
109. done this one-- that's just
n
times
n
plus 1 over 2,
110. So for example, say g of
n
is
n
squared minus 1.
111. is a factor-- then alpha to the
n
,
n
times alpha to the
n
,
112. f of
n
?
113. 1 because
n
^k divided by
n
^k, which, of course is 1, + 7 over
n
^k. (Sorry, I can't seem to
114. plus g of
n
, like
n
cubed.
115. It's f of
n
minus a1 f of
n
minus 1 minus a sub d f of
n
116. That's getting bigger as
n
gets big,
117. bound to the random variable S sub
n
.
118. or
n
by
n
or m by
n
array of special numbers.
119. of
n
minus 2/3.
120. f of
n
minus 1, and f of
n
.
121. maybe
n
squared, or some general function g of
n
.
122. Divide and conquer, you got
n
/2 or 3/4
n
.
123.
n
equals 2, maybe even-- we'll make
124. here it's 2 over
n
, 3 over
n
, 4 over
n
, 5 over
n
.
125. your confidence intervals, is going to be not
n
-1 but
n
-2. The best way to remember
126. for the
n
when f of
n
is bigger and equal to M,
127. So if f of
n
equals alpha 1 to the
n
--
128. E has successfully
n
minus 1 edges.
129. is proportional only to
N
. Therefore,
130. And there's only
n
suffixes, remember that.
131. square root of
n
?
132. to the
n
minus 1.
133. Derivative of this is now minus
n
plus 1, x to the
n
.
134. Like with Fibonacci numbers-- f of
n
minus f
n
135. I get a fraction of
n
, right?
136. So that equals 2 f of
n
minus 1 minus f of
n
minus 2.
137.
n
.
138.
n
the comments section below.
139. matrix. It must have size
n
by
n
.
140. actually... here, right? 3^
n
over 4^
n
—I can rewrite that as three-fourths to the
n
141.
n
equals 2.
142. So it's
N
0, 1, right?
143. means you have something like f of
n
minus a1 f of
n
minus 1
144. so I'm going to have
n
of them about
n
squared.
145. it moves into
n
-dimensional space,
146. from
n
, usually like 1, 2, 3, an integer from
n
.
147. One minus x to the
n
.
148. to the
n
minus 1 plus alpha to the
n
minus 2
149. So there were
n
different sub-problems.
150. as a chi square
n
minus 1 divided by
n
.
151. of my chi square distribution with
n
152. So if f of
n
equals alpha 1 to the
n
--
153. that i equals 0 to
n
, x to the i equals 1 minus x to the
n
154. But I'll take
n
to be three.
155. Square root of
n
over
n
to the 3/2, that goes to 0.
156. and you've got
n
minus 1,000.
157. during year
n
.
158. goes to
N
distribution as
n
goes to infinity.
159. of
n
plus 1 times m.
160. to
n
minus 1.
161. There exists an
n
and a delta.
162. For what value of
n
is f of
n
bigger than M?
163. on
n
.
164. so I get square root of
n
over 2/3
n
to the 3/2.
165. It's the
n
-th payment, so it's m dollars in
n
minus 1 years.
166. We're going to try f of
n
is an exponential in
n
, alpha
167. Is it an
N
0, 1?
168. why basically, square root of
N
over
N
is small.
169. Plug that in for f of
n
.
170. plus 3 to the
n
.
171. that multiplication of 1 over
n
^4. But, ultimately, the limit as
n
approaches infinity is gonna
172.
n
over b sub
n
, flipping that up, cleaning up with some algebra—we got a limit which
173. So we have
n
vertices.
174. equals c1 times alpha 1 to the
n
plus c2 alpha 2 to the
n
175. which is order
n
, which is order
n
squared.
176. to show every single algebraic step. We have 3^
n
times 4^
n
, and we have 4^
n
times 3^
n
.
177. for the
n
-th Fibonacci number.
178. at most,
n
.
179. QUITE APPLY FOR FOOD
N
-- QUITE
180. where I have f(
n
) equals f of
n
minus 1 plus f(
n
) minus 2
181. which is, does
n
equal twice
n
minus 1 minus
n
minus 2,
182. So I'm multiplying by a positive
n
.
183. It is r^
n
-- for the nth pair, r^
n
times cos(
n
*theta),
184. Minus x to the
n
.
185. So I always think, T-H-I-
N
-K, T-H-I-
N
-K, remember that.
186. I didn't actually define DP of
n
.
187. PROFESSOR: f of
n
minus 1.
188. ix to the i equals x minus
n
plus 1, x to the
n
plus 1,
189. and this is now delta
n
is at most square root of
n
,
190. So linear is when inside you have
n
minus 1,
n
minus 2.
191. So for each value of
n
, we look at
n
observations.
192. One plus delta
n
over 2/3
n
to the 3/2.
193. PROFESSOR: 2 the
n
?
194. I should not see
n
here, but I should
n
minus 1.
195. AUDIENCE:
N
0, sigma squared?
196. what this
n
goes to infinity means.
197. and so forth, up to
n
x to the
n
.
198. plus some error term, delta
n
.
199. the exponents. Well, now we have the limit as
n
approaches infinity of
n
to the k – 1
200. What happens when
n
goes to infinity?
201. the maximum from 1 to
n
.
202.
n
.
203. The only
n
that works is infinity.
204. to
N
, where
N
is some vector that's
205. of the first
n
square roots.
206. There's only
n
choices for that.
207. Initially, we have
n
words to do.
208.
n
.
209. f of
n
minus a1 f
n
minus 1 dot dot dot
210. of a sub
n
over b sub
n
is a positive, finite value. And, a sub
n
is going to be positive.
211. to infinity, of just simply 3^
n
over 4^
n
. Okay, well... let's see here... Is a sub
n
212. to linearly decrease as we increase
n
.
213. to the
n
minus 1.
214. diverge? Well, we can look at the limit as
n
approaches infinity of 3^
n
+ 2 over 4^
n
215. as
n
approaches infinity of
n
to the k – 1 over
n
^k + 7. Now,
n
is that positive integer—the
216. That says u_(
n
+1) is zero.
217. So we suppose P of
n
.
218. And
n
minus 1 down to 0.
219. It's 5/2 4 to the
n
minus 3 to the
n
plus 1.
220. at month
n
- 2.
221. 4 to the
n
plus negative 3 to the
n
plus 1.
222. A limit as
n
goes to infinity of 2/3
n
to the 3/2
223. to
N
, not to
N
squared.
224. and it lasts for
n
years.
225. an
n
-year $m-dollar payment annuity,
226. S sub
n
is a sum of the first
n
of
227. So the average-- average goes like
N
.
228. infinity of
n
to the k – 1 over
n
^k over—now, distributing the 1 over
n
^k in the denominator—we have
229. has area f of
n
minus 1 here, and then, lastly, area f of
n
.
230. to
n
to the r minus 1 times alpha to the
n
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