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Title: Properties of the Spectrum
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Series: Functional Analysis
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YouTube-Title: Functional Analysis 30 | Properties of the Spectrum
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Subtitle in English
1 00:00:00,439 –> 00:00:02,220 Hello and welcome back to
2 00:00:02,230 –> 00:00:03,890 functional analysis.
3 00:00:04,639 –> 00:00:05,909 And there’s always many,
4 00:00:05,920 –> 00:00:07,349 many thanks to all the nice
5 00:00:07,360 –> 00:00:08,449 people that support this
6 00:00:08,460 –> 00:00:10,340 channel on study or paypal.
7 00:00:11,020 –> 00:00:12,699 In today’s part 30 we are
8 00:00:12,710 –> 00:00:14,560 still talking about spectral
9 00:00:14,569 –> 00:00:15,260 theory.
10 00:00:15,779 –> 00:00:17,100 This means that we still
11 00:00:17,110 –> 00:00:18,940 analyze the set sigma of
12 00:00:18,950 –> 00:00:20,739 T which is the spectrum
13 00:00:20,750 –> 00:00:22,159 of the operator T.
14 00:00:22,850 –> 00:00:24,719 So we have all complex numbers
15 00:00:24,729 –> 00:00:26,420 lambda such that T
16 00:00:26,430 –> 00:00:28,209 minus lambda identity
17 00:00:28,219 –> 00:00:29,680 is not invertible.
18 00:00:30,299 –> 00:00:31,750 On the other hand, we have
19 00:00:31,760 –> 00:00:33,349 the complement the resolvent
20 00:00:33,360 –> 00:00:35,200 set of T which is denoted
21 00:00:35,209 –> 00:00:36,029 by a row.
22 00:00:36,720 –> 00:00:38,299 Now please remember in this
23 00:00:38,310 –> 00:00:40,220 context, we will always omit
24 00:00:40,229 –> 00:00:42,110 mentioning the identity operator
25 00:00:42,119 –> 00:00:42,450 here.
26 00:00:43,200 –> 00:00:44,810 It’s simply shorter in this
27 00:00:44,819 –> 00:00:45,159 way.
28 00:00:45,840 –> 00:00:47,830 Also don’t forget we do
29 00:00:47,840 –> 00:00:49,790 everything in a complex Banach
30 00:00:49,830 –> 00:00:50,779 space X
31 00:00:51,599 –> 00:00:53,459 and for a bounded linear
32 00:00:53,470 –> 00:00:55,159 operator T from
33 00:00:55,169 –> 00:00:56,180 X to X.
34 00:00:56,830 –> 00:00:57,330 OK.
35 00:00:57,340 –> 00:00:59,090 Now, in this context, we
36 00:00:59,099 –> 00:01:01,009 have some nice results about
37 00:01:01,020 –> 00:01:01,729 the spectrum.
38 00:01:02,360 –> 00:01:04,230 So let’s summarize some facts
39 00:01:04,239 –> 00:01:05,620 in one proposition.
40 00:01:06,540 –> 00:01:08,160 First, I can tell you that
41 00:01:08,169 –> 00:01:09,569 the resolvent set is
42 00:01:09,580 –> 00:01:11,290 always an open set.
43 00:01:12,050 –> 00:01:13,480 In conclusion, this means
44 00:01:13,489 –> 00:01:15,370 that the spectrum is always
45 00:01:15,379 –> 00:01:16,239 a close set.
46 00:01:16,919 –> 00:01:18,629 Therefore, the visualization
47 00:01:18,639 –> 00:01:20,069 should always be that you
48 00:01:20,080 –> 00:01:21,559 are in the complex plane
49 00:01:21,779 –> 00:01:23,639 and you have some close set,
50 00:01:23,650 –> 00:01:24,900 which is the spectrum of
51 00:01:24,910 –> 00:01:26,620 T and everything
52 00:01:26,629 –> 00:01:28,379 outside is an open set,
53 00:01:28,389 –> 00:01:30,050 which is the resolvent set.
54 00:01:30,919 –> 00:01:32,769 Later, we will also see that
55 00:01:32,779 –> 00:01:34,309 the spectrum of a bounded
56 00:01:34,319 –> 00:01:36,300 linear operator is also
57 00:01:36,309 –> 00:01:38,239 a bounded set or in
58 00:01:38,250 –> 00:01:39,800 other words, this spectrum
59 00:01:39,809 –> 00:01:40,800 is compact.
60 00:01:41,580 –> 00:01:41,980 OK.
61 00:01:41,989 –> 00:01:43,919 Now, for part B let’s take
62 00:01:43,930 –> 00:01:45,529 an element from the resolvent
63 00:01:45,540 –> 00:01:45,949 set.
64 00:01:46,660 –> 00:01:48,209 So here we just have a complex
65 00:01:48,220 –> 00:01:49,480 number lambda.
66 00:01:50,300 –> 00:01:51,639 Now, the good thing we know
67 00:01:51,650 –> 00:01:53,569 is that the operator T minus
68 00:01:53,610 –> 00:01:55,519 lambda has an inverse.
69 00:01:56,230 –> 00:01:57,690 So writing the power minus
70 00:01:57,699 –> 00:01:59,220 one here makes sense.
71 00:01:59,349 –> 00:02:00,849 And we can also calculate
72 00:02:00,860 –> 00:02:01,959 the operator norm
73 00:02:02,769 –> 00:02:04,610 which is always greater
74 00:02:04,620 –> 00:02:06,589 or equal than
75 00:02:06,599 –> 00:02:08,070 one divided by the
76 00:02:08,080 –> 00:02:09,910 distance of Lambda to the
77 00:02:09,919 –> 00:02:11,910 spectrum in the picture.
78 00:02:11,919 –> 00:02:13,759 This means what is the shortest
79 00:02:13,770 –> 00:02:15,710 distance we have to go until
80 00:02:15,720 –> 00:02:17,580 we reach a point in the spectrum.
81 00:02:18,380 –> 00:02:19,559 And this is simply what we
82 00:02:19,570 –> 00:02:20,910 call the distance from a
83 00:02:20,919 –> 00:02:22,279 point to a set.
84 00:02:22,899 –> 00:02:24,490 So you see the closer
85 00:02:24,500 –> 00:02:25,800 lambda is to the spectrum
86 00:02:25,809 –> 00:02:27,679 of T, the bigger this
87 00:02:27,690 –> 00:02:28,699 norm has to be
88 00:02:29,399 –> 00:02:29,740 OK.
89 00:02:29,750 –> 00:02:31,550 Here we observe that we can
90 00:02:31,559 –> 00:02:33,320 have different operators
91 00:02:33,330 –> 00:02:34,889 when we have different, values
92 00:02:34,899 –> 00:02:35,779 lambda.
93 00:02:36,500 –> 00:02:38,199 Hence, we can look at a map
94 00:02:38,210 –> 00:02:40,000 that sends lambda to this
95 00:02:40,009 –> 00:02:40,740 operator.
96 00:02:41,210 –> 00:02:42,710 Of course, lambda has to
97 00:02:42,720 –> 00:02:44,320 come from the resolvent set.
98 00:02:44,330 –> 00:02:46,199 But then we get a bounded
99 00:02:46,210 –> 00:02:47,580 operator on X.
100 00:02:48,160 –> 00:02:50,110 So here BX denotes
101 00:02:50,119 –> 00:02:51,770 the space of all bounded
102 00:02:51,779 –> 00:02:53,720 linear operators from X
103 00:02:53,729 –> 00:02:54,559 into X.
104 00:02:55,330 –> 00:02:57,250 And now our result
105 00:02:57,259 –> 00:02:59,160 is that this map is
106 00:02:59,169 –> 00:03:00,509 an analytical map
107 00:03:01,210 –> 00:03:02,990 in short, this nice map is
108 00:03:03,000 –> 00:03:04,619 often called the resolvent.
109 00:03:05,619 –> 00:03:07,009 Of course, at this point,
110 00:03:07,020 –> 00:03:08,410 you should ask yourself,
111 00:03:08,419 –> 00:03:10,050 what does it mean for a map
112 00:03:10,059 –> 00:03:11,339 like this to be
113 00:03:11,350 –> 00:03:12,250 analytical?
114 00:03:12,949 –> 00:03:14,289 The short explanation here
115 00:03:14,300 –> 00:03:15,970 would be just the definition
116 00:03:16,710 –> 00:03:18,570 which tells us this map
117 00:03:18,580 –> 00:03:19,929 locally can be
118 00:03:19,940 –> 00:03:21,429 expressed as a Taylor
119 00:03:21,440 –> 00:03:22,089 series.
120 00:03:22,809 –> 00:03:24,339 What this means exactly.
121 00:03:24,380 –> 00:03:25,850 I will show you in the proof
122 00:03:25,860 –> 00:03:26,820 of this statement.
123 00:03:27,669 –> 00:03:28,880 Therefore, let’s immediately
124 00:03:28,889 –> 00:03:30,440 start proving these
125 00:03:30,449 –> 00:03:31,000 things.
126 00:03:31,669 –> 00:03:33,179 First, you should note in
127 00:03:33,190 –> 00:03:34,500 the case that the resolvent
128 00:03:34,509 –> 00:03:35,669 set is empty.
129 00:03:35,720 –> 00:03:37,649 Everything here is trivially
130 00:03:37,660 –> 00:03:38,149 true.
131 00:03:38,809 –> 00:03:40,399 For this reason, let’s consider
132 00:03:40,410 –> 00:03:41,610 the case where we have at
133 00:03:41,619 –> 00:03:43,399 least one point lambda
134 00:03:43,410 –> 00:03:45,350 zero in the resolvent set.
135 00:03:46,070 –> 00:03:47,589 This means that the operator
136 00:03:47,600 –> 00:03:48,960 T minus lambda
137 00:03:48,970 –> 00:03:50,800 zero has an inverse.
138 00:03:51,449 –> 00:03:52,759 Then for the next part, let’s
139 00:03:52,770 –> 00:03:54,470 consider the operator norm
140 00:03:54,479 –> 00:03:55,750 of this operator,
141 00:03:56,570 –> 00:03:57,589 it’s a constant.
142 00:03:57,600 –> 00:03:58,990 I want to call C
143 00:03:59,889 –> 00:04:01,559 and the reciprocal I want
144 00:04:01,570 –> 00:04:02,860 to call epsilon.
145 00:04:03,539 –> 00:04:03,960 OK.
146 00:04:03,970 –> 00:04:05,289 Now, what we want to do I
147 00:04:05,300 –> 00:04:06,669 can visualize in the picture
148 00:04:06,679 –> 00:04:07,220 above.
149 00:04:07,229 –> 00:04:09,110 So here we have lambda zero
150 00:04:09,710 –> 00:04:11,350 and now I want to look at
151 00:04:11,360 –> 00:04:13,059 the epsilon ball around this
152 00:04:13,070 –> 00:04:14,880 point in the complex plane.
153 00:04:15,630 –> 00:04:17,000 Now, when we show that this
154 00:04:17,010 –> 00:04:18,660 epsilon ball is completely
155 00:04:18,670 –> 00:04:20,170 inside the resolvent set,
156 00:04:20,178 –> 00:04:21,820 we have shown that the resolvent
157 00:04:21,829 –> 00:04:23,730 set is indeed an open set.
158 00:04:24,480 –> 00:04:25,790 Therefore, let’s take an
159 00:04:25,799 –> 00:04:27,519 arbitrary element from this
160 00:04:27,529 –> 00:04:28,450 epsilon ball.
161 00:04:28,940 –> 00:04:30,329 So it’s a complex number
162 00:04:30,339 –> 00:04:32,049 lambda where the distance
163 00:04:32,059 –> 00:04:33,950 to lambda zero is less
164 00:04:33,959 –> 00:04:34,959 than epsilon.
165 00:04:35,709 –> 00:04:36,130 OK.
166 00:04:36,140 –> 00:04:37,880 Now, having this, our plan
167 00:04:37,890 –> 00:04:39,190 is to show that this
168 00:04:39,200 –> 00:04:41,049 operator T minus lambda
169 00:04:41,149 –> 00:04:42,730 is also invertible.
170 00:04:43,609 –> 00:04:45,269 Therefore, let’s calculate
171 00:04:45,279 –> 00:04:45,829 a little bit.
172 00:04:46,690 –> 00:04:48,149 Now we want to transform
173 00:04:48,160 –> 00:04:49,510 this to the operator we already
174 00:04:49,519 –> 00:04:49,829 know.
175 00:04:49,839 –> 00:04:51,750 So T minus lambda zero.
176 00:04:52,019 –> 00:04:53,450 Therefore, we have to add
177 00:04:53,459 –> 00:04:54,880 and subtract lambda zero
178 00:04:54,890 –> 00:04:56,750 here or for example,
179 00:04:56,760 –> 00:04:58,440 we can write the equality
180 00:04:58,450 –> 00:04:59,390 in this way.
181 00:05:00,190 –> 00:05:01,380 Then in the next step, the
182 00:05:01,390 –> 00:05:02,799 operator we already know
183 00:05:02,809 –> 00:05:03,880 is invertible.
184 00:05:03,890 –> 00:05:05,619 We can just factor out here.
185 00:05:06,220 –> 00:05:07,679 So here we have the identity
186 00:05:07,690 –> 00:05:09,109 operator minus
187 00:05:09,959 –> 00:05:11,899 lambda minus lambda zero
188 00:05:11,920 –> 00:05:13,540 times the inverse.
189 00:05:14,170 –> 00:05:14,579 OK.
190 00:05:14,589 –> 00:05:16,380 Now I can tell you why this
191 00:05:16,390 –> 00:05:18,269 form here is so helpful for
192 00:05:18,279 –> 00:05:18,609 us.
193 00:05:18,820 –> 00:05:20,720 For this reason, let’s call
194 00:05:20,730 –> 00:05:22,140 this operator here.
195 00:05:22,149 –> 00:05:23,089 Just s
196 00:05:23,100 –> 00:05:25,049 first, let’s look at
197 00:05:25,059 –> 00:05:26,769 the operator norm of S
198 00:05:27,399 –> 00:05:28,859 where we can just use that
199 00:05:28,869 –> 00:05:30,299 this one is less than
200 00:05:30,309 –> 00:05:32,279 epsilon and this one is in
201 00:05:32,290 –> 00:05:33,779 the operator norm C.
202 00:05:34,510 –> 00:05:36,010 So in summary, this whole
203 00:05:36,019 –> 00:05:37,739 thing is less than epsilon
204 00:05:37,750 –> 00:05:39,720 times C which is by
205 00:05:39,730 –> 00:05:41,140 the definition of epsilon
206 00:05:41,149 –> 00:05:42,279 just one.
207 00:05:43,190 –> 00:05:44,480 So here you should see we
208 00:05:44,489 –> 00:05:46,190 have an operator with operator
209 00:05:46,200 –> 00:05:48,000 norm exactly one where we
210 00:05:48,010 –> 00:05:49,739 subtract an operator with
211 00:05:49,750 –> 00:05:51,010 norm less than one.
212 00:05:51,690 –> 00:05:53,200 And the result is such a
213 00:05:53,209 –> 00:05:55,000 combination is always
214 00:05:55,010 –> 00:05:55,910 invertible.
215 00:05:56,399 –> 00:05:58,220 Indeed, this is called a
216 00:05:58,230 –> 00:05:59,579 Neumann series
217 00:06:00,359 –> 00:06:02,309 an operator I minus S
218 00:06:02,320 –> 00:06:04,299 where the norm of S is less
219 00:06:04,309 –> 00:06:05,579 than one is
220 00:06:05,589 –> 00:06:06,459 invertible.
221 00:06:06,929 –> 00:06:08,559 Because the reasoning goes
222 00:06:08,570 –> 00:06:10,239 like for the geometric series
223 00:06:10,250 –> 00:06:12,019 for numbers, this
224 00:06:12,029 –> 00:06:13,529 means that we multiply the
225 00:06:13,540 –> 00:06:15,380 operator I minus S
226 00:06:15,390 –> 00:06:17,070 with the sum of the powers
227 00:06:17,079 –> 00:06:17,660 of S.
228 00:06:18,549 –> 00:06:19,750 And then it does not matter
229 00:06:19,760 –> 00:06:21,010 if we multiply it from the
230 00:06:21,019 –> 00:06:22,290 right or the left,
231 00:06:22,829 –> 00:06:24,410 we just get the same result
232 00:06:24,420 –> 00:06:25,890 by telescoping the two
233 00:06:25,899 –> 00:06:27,829 sums, which means
234 00:06:27,839 –> 00:06:29,410 I minus S to the
235 00:06:29,420 –> 00:06:30,950 power N plus one.
236 00:06:31,450 –> 00:06:32,910 And now we can use that the
237 00:06:32,920 –> 00:06:34,549 operator norm of S is less
238 00:06:34,559 –> 00:06:36,410 than one, which means that
239 00:06:36,420 –> 00:06:37,809 this sequence here
240 00:06:37,820 –> 00:06:39,510 converges to zero
241 00:06:39,519 –> 00:06:41,359 when N goes to infinity.
242 00:06:41,850 –> 00:06:43,619 And of course, this convergence
243 00:06:43,630 –> 00:06:45,079 is measured with the operator
244 00:06:45,089 –> 00:06:45,589 norm.
245 00:06:45,910 –> 00:06:47,720 Therefore, we conclude if
246 00:06:47,730 –> 00:06:49,450 we have an infinite sum here.
247 00:06:49,459 –> 00:06:50,640 So a series,
248 00:06:51,350 –> 00:06:53,149 then this operator defines
249 00:06:53,160 –> 00:06:55,059 the inverse of I minus
250 00:06:55,070 –> 00:06:57,010 S indeed this
251 00:06:57,019 –> 00:06:58,790 series is well defined as
252 00:06:58,799 –> 00:07:00,730 a limit because the norm
253 00:07:00,739 –> 00:07:02,179 of S is less than one.
254 00:07:02,279 –> 00:07:03,890 And we call it the Neumann
255 00:07:03,899 –> 00:07:04,579 series.
256 00:07:05,130 –> 00:07:05,450 OK.
257 00:07:05,459 –> 00:07:07,079 Now, this nice general
258 00:07:07,089 –> 00:07:08,920 result we can use here
259 00:07:09,640 –> 00:07:11,339 because now we can conclude
260 00:07:11,350 –> 00:07:13,059 that T minus Lambda
261 00:07:13,070 –> 00:07:14,940 is also invertible.
262 00:07:15,299 –> 00:07:16,679 That’s simply because it’s
263 00:07:16,690 –> 00:07:18,540 the product of two invertible
264 00:07:18,549 –> 00:07:19,339 operators.
265 00:07:19,929 –> 00:07:21,799 And this is true for all
266 00:07:21,809 –> 00:07:23,440 Lambda with this proper T
267 00:07:23,500 –> 00:07:25,200 and also lambda zero
268 00:07:25,209 –> 00:07:26,839 was arbitrary chosen,
269 00:07:26,850 –> 00:07:28,640 therefore row of T
270 00:07:28,649 –> 00:07:29,500 is open.
271 00:07:30,309 –> 00:07:32,059 In conclusion, our spectrum
272 00:07:32,070 –> 00:07:33,970 sigma of T is closed.
273 00:07:34,429 –> 00:07:35,529 So the whole statement in
274 00:07:35,540 –> 00:07:36,790 A is proven
275 00:07:37,510 –> 00:07:39,220 also, you might already see
276 00:07:39,230 –> 00:07:40,640 that we have in equation
277 00:07:40,649 –> 00:07:42,619 above our Taylor series.
278 00:07:43,309 –> 00:07:44,570 More concretely, we have
279 00:07:44,579 –> 00:07:46,519 T minus Lambda is equal
280 00:07:46,529 –> 00:07:48,290 to T minus lambda zero
281 00:07:48,299 –> 00:07:49,890 times the Neumann series.
282 00:07:50,630 –> 00:07:51,260 Or to put it.
283 00:07:51,269 –> 00:07:52,489 In correct words, we get
284 00:07:52,500 –> 00:07:54,000 the Neumann series when we
285 00:07:54,010 –> 00:07:55,720 look at the inverses here.
286 00:07:56,269 –> 00:07:57,730 And don’t forget taking
287 00:07:57,739 –> 00:07:59,359 inverses changes the order
288 00:07:59,369 –> 00:08:00,390 of the operators.
289 00:08:01,200 –> 00:08:02,500 Therefore, we have the Neumann
290 00:08:02,510 –> 00:08:03,799 series in front.
291 00:08:04,549 –> 00:08:05,000 OK.
292 00:08:05,010 –> 00:08:06,410 At this point, we can just
293 00:08:06,420 –> 00:08:07,970 substitute our operator S
294 00:08:07,980 –> 00:08:08,679 again.
295 00:08:08,809 –> 00:08:10,709 And then we get the
296 00:08:10,720 –> 00:08:11,970 series where we have this
297 00:08:11,980 –> 00:08:13,640 product to the power K.
298 00:08:14,160 –> 00:08:15,459 So what you should see is
299 00:08:15,470 –> 00:08:17,079 lambda minus lambda zero
300 00:08:17,089 –> 00:08:18,880 to the power K which looks
301 00:08:18,890 –> 00:08:20,200 like a Taylor series.
302 00:08:20,750 –> 00:08:20,980 OK.
303 00:08:20,989 –> 00:08:22,399 Maybe when we combine these
304 00:08:22,410 –> 00:08:23,839 two operators and put them
305 00:08:23,850 –> 00:08:25,760 in formed, then you see this
306 00:08:25,769 –> 00:08:27,519 is exactly a Taylor series.
307 00:08:28,190 –> 00:08:29,709 So these are just constant
308 00:08:29,720 –> 00:08:31,369 coefficients but now they
309 00:08:31,380 –> 00:08:33,359 are operators but
310 00:08:33,369 –> 00:08:34,349 otherwise it looks like a
311 00:08:34,359 –> 00:08:36,340 totally common Taylor series.
312 00:08:37,020 –> 00:08:38,280 And there you see this is
313 00:08:38,289 –> 00:08:39,770 exactly what we call an
314 00:08:39,780 –> 00:08:40,940 analytical map.
315 00:08:40,950 –> 00:08:42,659 And therefore C is also
316 00:08:42,669 –> 00:08:43,280 finished.
317 00:08:43,929 –> 00:08:44,159 OK.
318 00:08:44,169 –> 00:08:45,669 Having this, let’s close
319 00:08:45,679 –> 00:08:47,520 the video by proving B.
320 00:08:48,250 –> 00:08:50,109 So if we now take a complex
321 00:08:50,119 –> 00:08:51,900 number lambda from the spectrum
322 00:08:51,909 –> 00:08:53,859 of T, then we know from
323 00:08:53,869 –> 00:08:55,349 above that there’s no other
324 00:08:55,359 –> 00:08:56,260 choice than that.
325 00:08:56,270 –> 00:08:57,869 The distance from LAMBDA
326 00:08:57,880 –> 00:08:59,859 to LAMBDA zero is
327 00:08:59,869 –> 00:09:01,349 greater or equal than
328 00:09:01,359 –> 00:09:02,140 epsilon.
329 00:09:02,840 –> 00:09:04,830 And then looking at the reciprocals,
330 00:09:04,840 –> 00:09:06,659 we get that one
331 00:09:06,669 –> 00:09:08,309 over the distance is less
332 00:09:08,320 –> 00:09:09,539 or equal than C.
333 00:09:10,210 –> 00:09:11,409 And they are, please recall
334 00:09:11,419 –> 00:09:12,770 this is simply the operator
335 00:09:12,780 –> 00:09:14,630 norm of T minus lambda
336 00:09:14,640 –> 00:09:16,020 zero inverse.
337 00:09:16,729 –> 00:09:18,059 And with this, we are almost
338 00:09:18,070 –> 00:09:18,510 finished.
339 00:09:18,520 –> 00:09:20,200 Now we just have to go through
340 00:09:20,210 –> 00:09:21,960 all the points in the spectrum.
341 00:09:22,520 –> 00:09:22,830 OK.
342 00:09:22,840 –> 00:09:24,020 Maybe let’s write it down
343 00:09:24,030 –> 00:09:25,229 with more details.
344 00:09:25,239 –> 00:09:27,049 So the distance here is just
345 00:09:27,059 –> 00:09:28,789 the infimum where we
346 00:09:28,799 –> 00:09:30,020 simply go through all the
347 00:09:30,030 –> 00:09:31,429 points lambda in the
348 00:09:31,440 –> 00:09:32,109 spectrum.
349 00:09:32,659 –> 00:09:34,349 And now if we pull the infimum
350 00:09:34,559 –> 00:09:36,309 out of the denominator, it
351 00:09:36,320 –> 00:09:37,359 gets a supremum.
352 00:09:38,039 –> 00:09:39,320 And at this point, we can
353 00:09:39,330 –> 00:09:40,989 just use the inequality from
354 00:09:41,000 –> 00:09:41,539 above.
355 00:09:41,719 –> 00:09:43,020 And then we have shown our
356 00:09:43,030 –> 00:09:44,840 claim or in other
357 00:09:44,849 –> 00:09:46,479 words, also be is
358 00:09:46,489 –> 00:09:48,049 shown OK.
359 00:09:48,059 –> 00:09:49,289 With this, I hope I see you
360 00:09:49,299 –> 00:09:50,440 in the next video where I
361 00:09:50,450 –> 00:09:52,250 show you how we can use all
362 00:09:52,260 –> 00:09:52,690 of this.
363 00:09:53,419 –> 00:09:54,799 Have a nice day and see you
364 00:09:54,809 –> 00:09:55,239 then.
365 00:09:55,250 –> 00:09:55,950 Bye.
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Last update: 2024-10
