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Title: Weierstrass M-Test
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Series: Weierstrass M-Test
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YouTube-Title: Weierstrass M-Test
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Quiz: Test your knowledge
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Subtitle on GitHub: wmt01_sub_eng.srt
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Definitions in the video: Weierstrass M-test
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Timestamps
00:00 Intro
00:28 Comparison for series of functions
00:56 Theorem about Weierstrass M-Test
03:19 Uniform convergence explanation
05:00 Proof of Weierstrass M-Test
09:46 Application of Weierstrass M-Test
11:25 Credits
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Subtitle in English
1 00:00:00,680 –> 00:00:02,310 Hello and welcome to this
2 00:00:02,350 –> 00:00:04,078 video about the Weierstrass
3 00:00:04,134 –> 00:00:04,970 M test.
4 00:00:05,350 –> 00:00:06,782 This is a topic from real
5 00:00:06,846 –> 00:00:08,222 analysis, so if you want
6 00:00:08,246 –> 00:00:09,390 to learn more, you should
7 00:00:09,430 –> 00:00:10,974 check out my real analysis
8 00:00:11,022 –> 00:00:11,610 course.
9 00:00:12,030 –> 00:00:13,374 However, before we start
10 00:00:13,422 –> 00:00:15,070 with this M test, I really
11 00:00:15,110 –> 00:00:16,270 want to thank all the nice
12 00:00:16,310 –> 00:00:17,462 people who support this channel
13 00:00:17,486 –> 00:00:18,926 on Steady here on YouTube
14 00:00:18,998 –> 00:00:20,326 or via other means.
15 00:00:20,518 –> 00:00:21,966 And please don’t forget with
16 00:00:21,998 –> 00:00:23,230 the link in the description
17 00:00:23,310 –> 00:00:24,918 you find additional material
18 00:00:25,014 –> 00:00:26,798 like PDF versions, quizzes
19 00:00:26,854 –> 00:00:27,970 and other stuff.
20 00:00:28,440 –> 00:00:28,968 Okay.
21 00:00:29,024 –> 00:00:30,328 With this I would say we
22 00:00:30,344 –> 00:00:32,184 can immediately start explaining
23 00:00:32,272 –> 00:00:34,100 the Weierstrass M test.
24 00:00:34,560 –> 00:00:35,808 Now what you should remember
25 00:00:35,904 –> 00:00:37,808 is that this M test is
26 00:00:37,824 –> 00:00:39,168 a comparison test for
27 00:00:39,224 –> 00:00:40,696 series, but for the
28 00:00:40,728 –> 00:00:42,552 uniform convergence for
29 00:00:42,576 –> 00:00:44,140 a sequence of functions.
30 00:00:44,560 –> 00:00:45,744 So it’s not so complicated
31 00:00:45,792 –> 00:00:46,320 at all.
32 00:00:46,400 –> 00:00:48,256 And it’s called m test simply
33 00:00:48,288 –> 00:00:49,728 because the constants in
34 00:00:49,744 –> 00:00:51,168 the theorem are called
35 00:00:51,264 –> 00:00:51,952 M.
36 00:00:52,136 –> 00:00:53,224 And there I would say we
37 00:00:53,232 –> 00:00:54,384 can immediately formulate
38 00:00:54,432 –> 00:00:55,340 this statement.
39 00:00:56,180 –> 00:00:57,436 And as already mentioned,
40 00:00:57,508 –> 00:00:59,060 the only thing you need here
41 00:00:59,140 –> 00:01:00,500 is to know what a series
42 00:01:00,620 –> 00:01:02,100 is and the uniform
43 00:01:02,140 –> 00:01:03,540 convergence for a sequence
44 00:01:03,580 –> 00:01:04,520 of functions.
45 00:01:04,940 –> 00:01:06,268 And these functions can be
46 00:01:06,284 –> 00:01:07,516 defined on any set.
47 00:01:07,588 –> 00:01:09,188 So let’s call it D.
48 00:01:09,364 –> 00:01:11,108 And indeed, often we have
49 00:01:11,164 –> 00:01:13,060 that D is a subset of the
50 00:01:13,100 –> 00:01:14,160 real numbers.
51 00:01:14,580 –> 00:01:16,452 Hence our functions fk
52 00:01:16,556 –> 00:01:18,100 should have D as their
53 00:01:18,140 –> 00:01:18,916 domain.
54 00:01:19,108 –> 00:01:20,356 And now the functions could
55 00:01:20,388 –> 00:01:22,164 be real valued or complex
56 00:01:22,212 –> 00:01:23,110 valued.
57 00:01:23,300 –> 00:01:24,786 So in general we could just
58 00:01:24,858 –> 00:01:26,386 write that they map into
59 00:01:26,418 –> 00:01:28,298 the complex numbers and
60 00:01:28,314 –> 00:01:29,594 then we have a function for
61 00:01:29,642 –> 00:01:31,218 each k in the natural
62 00:01:31,274 –> 00:01:33,026 numbers, so already
63 00:01:33,098 –> 00:01:34,882 a whole sequence of functions
64 00:01:34,946 –> 00:01:36,790 with the same domain D.
65 00:01:37,130 –> 00:01:38,570 However, for the Weierstrass
66 00:01:38,610 –> 00:01:40,050 M test we are interested
67 00:01:40,090 –> 00:01:41,866 in the partial sums, so the
68 00:01:41,898 –> 00:01:43,630 series of these functions.
69 00:01:44,010 –> 00:01:45,466 And to have the convergence
70 00:01:45,498 –> 00:01:46,994 of this series, we
71 00:01:47,042 –> 00:01:48,818 assume that we have constants
72 00:01:48,914 –> 00:01:50,732 with the name M.
73 00:01:50,906 –> 00:01:52,328 And as already mentioned,
74 00:01:52,424 –> 00:01:53,632 this is the whole reason
75 00:01:53,696 –> 00:01:55,540 we call it the M test.
76 00:01:56,000 –> 00:01:57,304 Now we have two assumptions
77 00:01:57,352 –> 00:01:58,400 for these constants.
78 00:01:58,520 –> 00:01:59,744 First, they should be non
79 00:01:59,792 –> 00:02:01,496 negative, and second, the
80 00:02:01,528 –> 00:02:02,480 series should be
81 00:02:02,520 –> 00:02:03,576 convergent.
82 00:02:03,768 –> 00:02:05,280 This means that the limit
83 00:02:05,320 –> 00:02:07,096 of the partial sums should
84 00:02:07,168 –> 00:02:07,900 exist.
85 00:02:08,440 –> 00:02:09,960 In other words, the value
86 00:02:10,040 –> 00:02:11,960 of the series is not equal
87 00:02:12,000 –> 00:02:13,096 to infinity.
88 00:02:13,288 –> 00:02:14,696 And now in addition, the
89 00:02:14,728 –> 00:02:16,256 constants should be always
90 00:02:16,368 –> 00:02:17,836 bigger than the values
91 00:02:17,868 –> 00:02:19,760 of the functions fk.
92 00:02:20,220 –> 00:02:21,844 More precisely, we look at
93 00:02:21,892 –> 00:02:23,684 f k of x and take
94 00:02:23,732 –> 00:02:25,640 the absolute value of this,
95 00:02:26,060 –> 00:02:27,444 and then this should be less
96 00:02:27,492 –> 00:02:28,676 or equal than
97 00:02:28,748 –> 00:02:29,560 M_k.
98 00:02:29,940 –> 00:02:31,580 And indeed we want that for
99 00:02:31,620 –> 00:02:33,492 all k in n and for
100 00:02:33,556 –> 00:02:35,092 all x in D.
101 00:02:35,276 –> 00:02:36,980 So the constants M_k
102 00:02:37,100 –> 00:02:39,044 form a uniform majorant
103 00:02:39,092 –> 00:02:40,160 for the functions.
104 00:02:40,620 –> 00:02:42,324 Ok, so these are the assumptions
105 00:02:42,372 –> 00:02:43,752 we need, and then we get
106 00:02:43,776 –> 00:02:45,320 the result for the series
107 00:02:45,440 –> 00:02:47,180 of the functions fk.
108 00:02:47,480 –> 00:02:48,872 Indeed, by the comparison
109 00:02:48,936 –> 00:02:50,472 test for series we
110 00:02:50,496 –> 00:02:51,576 immediately get that the
111 00:02:51,608 –> 00:02:53,208 sequence of partial sums
112 00:02:53,344 –> 00:02:55,336 converges for every x in
113 00:02:55,368 –> 00:02:56,016 D.
114 00:02:56,208 –> 00:02:57,928 However, we get even more
115 00:02:58,024 –> 00:02:59,736 because we deal with functions.
116 00:02:59,848 –> 00:03:01,120 The term uniform
117 00:03:01,160 –> 00:03:02,560 convergence also makes
118 00:03:02,600 –> 00:03:03,296 sense.
119 00:03:03,488 –> 00:03:05,392 So we get out that the series
120 00:03:05,496 –> 00:03:06,704 is uniformly
121 00:03:06,752 –> 00:03:07,740 convergent.
122 00:03:08,040 –> 00:03:09,808 Moreover, as a side note,
123 00:03:09,864 –> 00:03:11,488 I should also mention that
124 00:03:11,504 –> 00:03:12,888 we have the same result for
125 00:03:12,904 –> 00:03:14,192 the absolute value of the
126 00:03:14,216 –> 00:03:14,940 functions.
127 00:03:15,360 –> 00:03:17,168 So both nice things come
128 00:03:17,224 –> 00:03:18,640 out of the Weierstrass M-
129 00:03:18,680 –> 00:03:19,288 test.
130 00:03:19,464 –> 00:03:20,736 Now, in the case you don’t
131 00:03:20,768 –> 00:03:22,144 know exactly what the term
132 00:03:22,192 –> 00:03:23,856 uniform convergence means,
133 00:03:24,008 –> 00:03:25,616 let me refresh your memory
134 00:03:25,688 –> 00:03:26,208 here.
135 00:03:26,344 –> 00:03:27,528 Now, the first thing you
136 00:03:27,544 –> 00:03:28,808 should see here is that we
137 00:03:28,824 –> 00:03:30,248 get a well defined function
138 00:03:30,304 –> 00:03:32,100 we can call S,
139 00:03:32,440 –> 00:03:33,864 and this one has the same
140 00:03:33,912 –> 00:03:35,592 domain D and it sends
141 00:03:35,696 –> 00:03:37,460 x to this series.
142 00:03:38,250 –> 00:03:39,562 And what I mean there is
143 00:03:39,586 –> 00:03:41,506 that we just sum up f
144 00:03:41,578 –> 00:03:43,530 k of x and
145 00:03:43,570 –> 00:03:44,842 this limit should exist.
146 00:03:44,946 –> 00:03:46,266 So we get a well defined
147 00:03:46,338 –> 00:03:47,510 function S.
148 00:03:47,810 –> 00:03:49,538 And moreover, this capital
149 00:03:49,634 –> 00:03:51,426 S is the limit of the
150 00:03:51,458 –> 00:03:53,042 partial sums of the functions
151 00:03:53,106 –> 00:03:53,882 fk.
152 00:03:54,066 –> 00:03:55,954 And when I say limit, I mean
153 00:03:56,002 –> 00:03:57,650 with respect to the supremum
154 00:03:57,690 –> 00:03:58,378 norm.
155 00:03:58,554 –> 00:04:00,306 And please recall, the supremum
156 00:04:00,338 –> 00:04:02,210 norm is usually denoted with
157 00:04:02,250 –> 00:04:03,906 this infinity sign on the
158 00:04:03,938 –> 00:04:04,590 norm.
159 00:04:04,970 –> 00:04:06,626 Now, S is the uniform
160 00:04:06,698 –> 00:04:08,362 limit for the partial sums.
161 00:04:08,466 –> 00:04:10,178 So this goes to zero when
162 00:04:10,234 –> 00:04:11,710 n goes to infinity.
163 00:04:12,290 –> 00:04:14,026 Hence, this is exactly what
164 00:04:14,058 –> 00:04:15,354 we mean by the uniform
165 00:04:15,402 –> 00:04:17,306 convergence stated in the
166 00:04:17,338 –> 00:04:18,190 theorem.
167 00:04:18,690 –> 00:04:19,986 And maybe it’s always good
168 00:04:20,018 –> 00:04:21,586 to visualize this convergence
169 00:04:21,618 –> 00:04:22,402 in a picture.
170 00:04:22,506 –> 00:04:24,106 So let’s sketch the graph
171 00:04:24,178 –> 00:04:25,750 of the function S,
172 00:04:26,050 –> 00:04:27,634 and let’s simply say that
173 00:04:27,682 –> 00:04:29,482 here on the x axis, we have
174 00:04:29,506 –> 00:04:30,956 the domain D.
175 00:04:31,098 –> 00:04:32,864 Hence, for example, the graph
176 00:04:32,912 –> 00:04:34,488 of S could look like
177 00:04:34,544 –> 00:04:35,140 this.
178 00:04:35,560 –> 00:04:37,272 And now what one can do is
179 00:04:37,296 –> 00:04:38,768 to look at such an epsilon
180 00:04:38,824 –> 00:04:40,344 tube around this
181 00:04:40,392 –> 00:04:41,112 graph.
182 00:04:41,296 –> 00:04:42,992 And then the result is, no
183 00:04:43,016 –> 00:04:44,448 matter how small we choose
184 00:04:44,504 –> 00:04:46,384 this epsilon, the graph
185 00:04:46,472 –> 00:04:47,928 of the partial sums
186 00:04:48,064 –> 00:04:49,780 always lies inside.
187 00:04:50,280 –> 00:04:51,856 Obviously, this only has
188 00:04:51,888 –> 00:04:53,520 to hold eventually, which
189 00:04:53,560 –> 00:04:55,200 means n has to be big
190 00:04:55,240 –> 00:04:55,820 enough.
191 00:04:56,290 –> 00:04:57,970 So this is what you can remember.
192 00:04:58,090 –> 00:04:59,466 This is exactly what the
193 00:04:59,498 –> 00:05:01,310 uniform convergence means.
194 00:05:01,770 –> 00:05:03,178 Okay, then the next step
195 00:05:03,234 –> 00:05:05,010 would be to look at the proof
196 00:05:05,050 –> 00:05:06,110 of this statement.
197 00:05:06,610 –> 00:05:08,146 And in fact, it’s surprisingly
198 00:05:08,178 –> 00:05:08,706 easy.
199 00:05:08,818 –> 00:05:10,298 We only need the so called
200 00:05:10,354 –> 00:05:12,058 Cauchy criterion for
201 00:05:12,114 –> 00:05:12,710 series.
202 00:05:13,170 –> 00:05:14,482 And you can find that in
203 00:05:14,506 –> 00:05:16,378 my real analysis course in
204 00:05:16,434 –> 00:05:17,510 part 17.
205 00:05:18,050 –> 00:05:19,722 And it simply tells us that
206 00:05:19,746 –> 00:05:20,754 a series is
207 00:05:20,802 –> 00:05:22,790 convergent if and only
208 00:05:22,870 –> 00:05:24,534 if the partial sums form
209 00:05:24,582 –> 00:05:26,410 a Cauchy sequence.
210 00:05:26,750 –> 00:05:28,134 So you see, the Cauchy
211 00:05:28,182 –> 00:05:29,726 criterion simply
212 00:05:29,798 –> 00:05:31,430 uses the completeness of
213 00:05:31,470 –> 00:05:33,446 the real or complex numbers.
214 00:05:33,598 –> 00:05:35,014 Now, more precisely, this
215 00:05:35,062 –> 00:05:36,510 means that for all epsilon
216 00:05:36,550 –> 00:05:38,350 greater zero, we find an
217 00:05:38,390 –> 00:05:39,854 index capital N
218 00:05:40,022 –> 00:05:41,406 such that for all
219 00:05:41,478 –> 00:05:43,250 indices afterwards,
220 00:05:43,590 –> 00:05:45,014 and let’s simply call them
221 00:05:45,062 –> 00:05:46,462 lowercase n and
222 00:05:46,526 –> 00:05:48,376 m, we have that
223 00:05:48,408 –> 00:05:50,200 the sum in the absolute value
224 00:05:50,320 –> 00:05:52,020 is less than epsilon,
225 00:05:52,360 –> 00:05:54,040 and the sum has to go from
226 00:05:54,120 –> 00:05:55,688 m to the index
227 00:05:55,864 –> 00:05:56,560 n.
228 00:05:56,720 –> 00:05:58,240 So you could just say this
229 00:05:58,280 –> 00:05:59,624 partial sum is
230 00:05:59,672 –> 00:06:00,944 arbitrarily small
231 00:06:01,072 –> 00:06:02,020 eventually.
232 00:06:02,440 –> 00:06:04,168 And now please recall, in
233 00:06:04,184 –> 00:06:05,608 the assumption of the theorem,
234 00:06:05,664 –> 00:06:07,520 we assumed that we have the
235 00:06:07,560 –> 00:06:08,904 convergence for the series
236 00:06:08,992 –> 00:06:10,960 of Mk, which means
237 00:06:11,040 –> 00:06:12,800 we have the Cauchy criterion
238 00:06:12,880 –> 00:06:13,788 as well.
239 00:06:13,984 –> 00:06:15,340 Hence, for a given epsilon
240 00:06:15,380 –> 00:06:17,204 greater zero, we can choose
241 00:06:17,252 –> 00:06:18,644 a capital N as
242 00:06:18,692 –> 00:06:20,620 above, and then we immediately
243 00:06:20,660 –> 00:06:22,468 get a result for the series
244 00:06:22,564 –> 00:06:24,492 of the functions fk.
245 00:06:24,676 –> 00:06:25,972 So let’s simply look at the
246 00:06:25,996 –> 00:06:27,964 partial sums from m to
247 00:06:28,012 –> 00:06:29,076 n as well.
248 00:06:29,268 –> 00:06:30,644 And then we can just use
249 00:06:30,692 –> 00:06:32,252 the standard triangle
250 00:06:32,316 –> 00:06:33,240 inequality.
251 00:06:33,820 –> 00:06:35,508 It’s a finite sum, so we
252 00:06:35,524 –> 00:06:36,796 can just pull the absolute
253 00:06:36,828 –> 00:06:37,960 value inside.
254 00:06:38,300 –> 00:06:39,596 And then in the next step
255 00:06:39,668 –> 00:06:41,588 we can just estimate f
256 00:06:41,644 –> 00:06:43,360 k of x inside.
257 00:06:43,900 –> 00:06:45,308 And now the assumption comes
258 00:06:45,364 –> 00:06:47,092 in that we know that each
259 00:06:47,156 –> 00:06:49,028 one is less or equal than
260 00:06:49,084 –> 00:06:49,932 Mk.
261 00:06:50,116 –> 00:06:51,716 However, for the series of
262 00:06:51,748 –> 00:06:53,652 the non negative Mk’s,
263 00:06:53,756 –> 00:06:55,524 we already know it’s less
264 00:06:55,572 –> 00:06:57,204 than epsilon by the Cauchy
265 00:06:57,252 –> 00:06:58,120 criterion.
266 00:06:58,420 –> 00:06:59,924 In other words, the Cauchy
267 00:06:59,972 –> 00:07:01,660 criterion now also tells
268 00:07:01,700 –> 00:07:03,636 us that this series here
269 00:07:03,708 –> 00:07:05,276 is a convergent one,
270 00:07:05,468 –> 00:07:07,188 which already implies that
271 00:07:07,204 –> 00:07:08,762 the function S from above
272 00:07:08,876 –> 00:07:10,798 is a well defined function.
273 00:07:10,974 –> 00:07:12,894 In other words, it exists
274 00:07:13,022 –> 00:07:14,558 and it makes sense to talk
275 00:07:14,614 –> 00:07:15,342 about it.
276 00:07:15,486 –> 00:07:16,590 And thats what we will do
277 00:07:16,630 –> 00:07:18,494 now, because we want to talk
278 00:07:18,542 –> 00:07:20,326 about the supremum norm
279 00:07:20,518 –> 00:07:21,902 more concretely, we want
280 00:07:21,926 –> 00:07:23,390 to take the difference of
281 00:07:23,470 –> 00:07:25,134 S with the partial sums
282 00:07:25,262 –> 00:07:26,830 in the supremum norm.
283 00:07:26,990 –> 00:07:28,286 And exactly like before,
284 00:07:28,398 –> 00:07:29,918 lets fix an epsilon greater
285 00:07:29,974 –> 00:07:31,446 zero and lets choose the
286 00:07:31,478 –> 00:07:33,198 corresponding index capital
287 00:07:33,294 –> 00:07:33,910 N.
288 00:07:34,070 –> 00:07:35,582 And as always, the lowercase
289 00:07:35,646 –> 00:07:37,350 n should be greater or equal
290 00:07:37,390 –> 00:07:39,050 than this capital N.
291 00:07:39,200 –> 00:07:40,806 Now first, as a reminder,
292 00:07:40,878 –> 00:07:42,566 the supremum norm is given
293 00:07:42,638 –> 00:07:43,862 by the supremum.
294 00:07:44,006 –> 00:07:45,542 So instead of the norm, we
295 00:07:45,566 –> 00:07:47,182 can just write supremum
296 00:07:47,246 –> 00:07:49,090 over x in D.
297 00:07:49,390 –> 00:07:50,766 And then we just look at
298 00:07:50,798 –> 00:07:52,566 the absolute value of the
299 00:07:52,598 –> 00:07:54,170 value of the functions.
300 00:07:54,670 –> 00:07:56,374 In other words, we just put
301 00:07:56,422 –> 00:07:57,918 in x at the correct
302 00:07:57,974 –> 00:07:58,970 positions.
303 00:07:59,630 –> 00:08:01,118 And then in the next step
304 00:08:01,174 –> 00:08:02,638 we can simply substitute
305 00:08:02,694 –> 00:08:04,510 s with the limit
306 00:08:04,550 –> 00:08:05,370 definition.
307 00:08:05,850 –> 00:08:07,306 This means instead of the
308 00:08:07,338 –> 00:08:09,202 infinity symbol here we can
309 00:08:09,226 –> 00:08:11,050 just write limit of
310 00:08:11,090 –> 00:08:12,310 the partial sums.
311 00:08:12,610 –> 00:08:14,474 So maybe let’s use a number
312 00:08:14,562 –> 00:08:16,010 m that goes to
313 00:08:16,050 –> 00:08:16,830 infinity.
314 00:08:17,170 –> 00:08:18,674 And now we can simply use
315 00:08:18,722 –> 00:08:20,082 that the absolute value is
316 00:08:20,106 –> 00:08:21,930 a continuous function, which
317 00:08:21,970 –> 00:08:23,490 means we can pull out the
318 00:08:23,530 –> 00:08:24,258 limit.
319 00:08:24,434 –> 00:08:25,546 So it’s not a problem at
320 00:08:25,578 –> 00:08:27,082 all to write the limit in
321 00:08:27,106 –> 00:08:28,870 front of the absolute value.
322 00:08:29,250 –> 00:08:31,002 And then you should see inside
323 00:08:31,066 –> 00:08:32,026 the absolute value.
324 00:08:32,138 –> 00:08:33,274 It looks much simpler.
325 00:08:33,322 –> 00:08:35,132 Now we just subtract
326 00:08:35,196 –> 00:08:37,068 finite sums so we know
327 00:08:37,124 –> 00:08:38,920 a finite sum remains.
328 00:08:39,540 –> 00:08:40,884 However, maybe it’s not clear
329 00:08:40,932 –> 00:08:42,564 how to write this sum because
330 00:08:42,612 –> 00:08:44,436 we don’t know if m or n is
331 00:08:44,468 –> 00:08:45,076 bigger.
332 00:08:45,228 –> 00:08:46,572 But obviously in the case
333 00:08:46,636 –> 00:08:48,500 that m is bigger than n,
334 00:08:48,580 –> 00:08:50,156 we would say that k goes
335 00:08:50,188 –> 00:08:51,732 from n plus one to
336 00:08:51,796 –> 00:08:52,400 m.
337 00:08:52,740 –> 00:08:54,148 And of course, in the second
338 00:08:54,204 –> 00:08:55,580 case, we would just write
339 00:08:55,620 –> 00:08:56,960 it the other way around.
340 00:08:57,380 –> 00:08:58,604 It does not change so much,
341 00:08:58,652 –> 00:08:59,972 because we know that for
342 00:08:59,996 –> 00:09:01,740 both cases, if n and
343 00:09:01,780 –> 00:09:03,268 m are bigger than the capital
344 00:09:03,364 –> 00:09:04,964 N, then this whole
345 00:09:05,012 –> 00:09:06,300 sum is less than
346 00:09:06,340 –> 00:09:07,172 epsilon.
347 00:09:07,356 –> 00:09:08,748 This is simply the cauchy
348 00:09:08,804 –> 00:09:10,468 criterion we have at
349 00:09:10,524 –> 00:09:11,920 every point x.
350 00:09:12,620 –> 00:09:14,380 Hence, the conclusion is,
351 00:09:14,500 –> 00:09:16,332 even with the limit and with
352 00:09:16,356 –> 00:09:18,308 the supremum, we stay lower
353 00:09:18,364 –> 00:09:19,160 epsilon.
354 00:09:19,460 –> 00:09:21,012 The only thing we can lose
355 00:09:21,076 –> 00:09:22,412 here is the strict
356 00:09:22,476 –> 00:09:23,400 inequality.
357 00:09:23,940 –> 00:09:25,172 But obviously, this does
358 00:09:25,196 –> 00:09:26,840 not change anything, because
359 00:09:26,920 –> 00:09:28,880 epsilon was arbitrarily chosen
360 00:09:28,920 –> 00:09:30,112 from the beginning.
361 00:09:30,296 –> 00:09:31,704 And with that, we have it.
362 00:09:31,792 –> 00:09:33,472 This means that in the limit
363 00:09:33,536 –> 00:09:35,272 n to infinity, this
364 00:09:35,336 –> 00:09:37,140 norm here goes to zero.
365 00:09:37,520 –> 00:09:39,120 And since we have the supremum
366 00:09:39,160 –> 00:09:40,840 norm, this is exactly the
367 00:09:40,880 –> 00:09:42,540 uniform convergence.
368 00:09:42,920 –> 00:09:44,664 Therefore, this is already
369 00:09:44,752 –> 00:09:45,980 the whole proof.
370 00:09:46,280 –> 00:09:47,744 And then I would say, let’s
371 00:09:47,792 –> 00:09:49,392 look at a nice application
372 00:09:49,456 –> 00:09:51,040 of this Weierstrass m
373 00:09:51,080 –> 00:09:51,718 test.
374 00:09:51,904 –> 00:09:53,602 So let’s take an example
375 00:09:53,706 –> 00:09:55,658 of a series with a function
376 00:09:55,714 –> 00:09:56,586 inside.
377 00:09:56,778 –> 00:09:57,986 And if you know something
378 00:09:58,058 –> 00:09:59,986 about Fourier series, you
379 00:10:00,018 –> 00:10:01,370 might recognize that this
380 00:10:01,410 –> 00:10:02,538 one here is an important
381 00:10:02,634 –> 00:10:03,226 one.
382 00:10:03,378 –> 00:10:05,146 So let’s take f k from
383 00:10:05,218 –> 00:10:07,090 R to R, where f
384 00:10:07,130 –> 00:10:08,970 k of x is
385 00:10:09,010 –> 00:10:10,586 given as the cosine
386 00:10:10,658 –> 00:10:12,154 of kx
387 00:10:12,282 –> 00:10:13,790 over k squared.
388 00:10:14,170 –> 00:10:16,042 So this is our function depending
389 00:10:16,106 –> 00:10:16,746 on k.
390 00:10:16,858 –> 00:10:18,346 And now we can look at the
391 00:10:18,378 –> 00:10:19,430 absolute value,
392 00:10:20,090 –> 00:10:21,650 and we know that the cosine
393 00:10:21,690 –> 00:10:22,898 is bounded by one.
394 00:10:22,994 –> 00:10:24,602 So this is definitely less
395 00:10:24,666 –> 00:10:26,602 or equal than one over k
396 00:10:26,666 –> 00:10:27,350 squared.
397 00:10:27,850 –> 00:10:29,458 And this holds for every
398 00:10:29,554 –> 00:10:30,978 x in R.
399 00:10:31,154 –> 00:10:32,570 And now you only have to
400 00:10:32,610 –> 00:10:34,250 know that the series of one
401 00:10:34,290 –> 00:10:35,674 over k squared is
402 00:10:35,722 –> 00:10:37,522 convergent to apply the
403 00:10:37,546 –> 00:10:39,202 Weierstrass M test.
404 00:10:39,386 –> 00:10:40,938 And there are different possibilities
405 00:10:41,034 –> 00:10:42,562 to show this, but it’s
406 00:10:42,586 –> 00:10:44,578 definitely not too complicated.
407 00:10:44,714 –> 00:10:46,650 And a basic result you definitely
408 00:10:46,690 –> 00:10:47,670 should remember.
409 00:10:48,000 –> 00:10:49,304 And indeed, we can immediately
410 00:10:49,352 –> 00:10:50,912 make use of that with the
411 00:10:50,936 –> 00:10:52,460 Weierstrass M test.
412 00:10:52,880 –> 00:10:54,800 Our conclusion is that this
413 00:10:54,840 –> 00:10:56,832 series of functions is
414 00:10:56,896 –> 00:10:58,832 uniformly convergent.
415 00:10:59,016 –> 00:11:00,288 And indeed, this is really
416 00:11:00,344 –> 00:11:01,640 helpful as a result.
417 00:11:01,760 –> 00:11:03,152 For example, when you want
418 00:11:03,176 –> 00:11:04,736 to calculate the derivative
419 00:11:04,808 –> 00:11:06,800 of this function and
420 00:11:06,840 –> 00:11:07,888 something like that, you
421 00:11:07,904 –> 00:11:09,600 can find in my Fourier
422 00:11:09,680 –> 00:11:10,980 transform series.
423 00:11:11,520 –> 00:11:12,864 In fact, exactly.
424 00:11:12,912 –> 00:11:14,792 This result here plays a
425 00:11:14,816 –> 00:11:16,220 crucial role there.
426 00:11:16,680 –> 00:11:18,152 Okay, then, I think this
427 00:11:18,176 –> 00:11:19,680 is good enough for this video,
428 00:11:19,800 –> 00:11:21,096 and I really hope I meet
429 00:11:21,128 –> 00:11:22,352 you in another one.
430 00:11:22,456 –> 00:11:23,784 So, have a nice day, and
431 00:11:23,832 –> 00:11:24,700 bye bye.
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Quiz Content
Q1: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: I \rightarrow \mathbb{R}$. What is the correct definition for the uniform convergence to a function $f: I \rightarrow \mathbb{R}$?
A1: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \leq N ~ : ~ $ $$|f_n(x)-f(x)|<\varepsilon $$
A2: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~~ \forall x \in I ~ : ~$ $$ |f_n(x)-f(x)|<\varepsilon $$
A3: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~ : ~$ $$ |f_n(x)-f(x)|<\varepsilon $$
A4: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall x \in I ~~ \forall n \leq N ~ : ~$ $$ |f_n(x)-f(x)|<\varepsilon $$
Q2: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_k: \mathbb{R} \rightarrow \mathbb{R}$ such $|f_k(x)| < \frac{1}{k} $ for all $k \in \mathbb{N}$ and for all $x \in \mathbb{R}$. Is the Weierstrass M-Test applicable here?
A1: No, because $\sum_{k=1}^\infty \frac{1}{k}$ is not convergent.
A2: Yes, because $\sum_{k=1}^\infty \frac{1}{k} = \pi$
A3: Yes, because $\sum_{k=1}^\infty \frac{1}{k}$ is not convergent.
A4: One needs more information.
Q3: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_k: \mathbb{R} \rightarrow \mathbb{R}$ such $|f_k(x)| < \frac{1}{k^k} $ for all $k \in \mathbb{N}$ and for all $x \in \mathbb{R}$. What is not correct?
A1: The series of functions $\sum_{k=1}^\infty f_k$ is uniformly convergent.
A2: The series of functions $\sum_{k=1}^\infty f_k$ is pointwisely convergent.
A3: The series of functions $\sum_{k=1}^\infty |f_k|$ is uniformly convergent.
A4: The series of functions $\sum_{k=1}^\infty f_k$ is not well-defined.
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Last update: 2024-10
