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