
Title: Taylor’s Theorem

Series: Real Analysis

Chapter: Differentiable Functions

YouTubeTitle: Real Analysis 45  Taylor’s Theorem

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1 00:00:00,430 –> 00:00:02,250 Hello and welcome back to
2 00:00:02,259 –> 00:00:03,819 real analysis.
3 00:00:04,360 –> 00:00:06,130 And you already know as always
4 00:00:06,139 –> 00:00:07,329 first, I want to thank all
5 00:00:07,340 –> 00:00:08,449 the nice people that support
6 00:00:08,460 –> 00:00:09,630 this channel on study or
7 00:00:09,640 –> 00:00:10,289 paypal.
8 00:00:10,710 –> 00:00:12,569 Now, in today’s part 45
9 00:00:12,579 –> 00:00:13,689 we will talk about a very
10 00:00:13,699 –> 00:00:15,670 famous theorem namely
11 00:00:15,680 –> 00:00:16,930 Taylor’s theorem.
12 00:00:17,860 –> 00:00:19,329 This one is very important
13 00:00:19,340 –> 00:00:20,809 for a lot of applications
14 00:00:20,819 –> 00:00:21,930 and you might have already
15 00:00:21,940 –> 00:00:23,450 heard the term Taylor
16 00:00:23,459 –> 00:00:24,409 polynomial.
17 00:00:25,200 –> 00:00:26,639 In fact, the overall idea
18 00:00:26,649 –> 00:00:28,440 here is not hard to understand
19 00:00:28,629 –> 00:00:29,780 because it’s about an
20 00:00:29,790 –> 00:00:31,200 approximation method.
21 00:00:31,909 –> 00:00:33,049 Therefore, what you should
22 00:00:33,060 –> 00:00:34,790 imagine here is the graph
23 00:00:34,799 –> 00:00:36,689 of a function that is a few
24 00:00:36,700 –> 00:00:38,090 times differentiable.
25 00:00:39,099 –> 00:00:41,040 And then one will fix one
26 00:00:41,049 –> 00:00:42,240 point in the domain.
27 00:00:42,330 –> 00:00:43,720 And as often it’s called
28 00:00:43,729 –> 00:00:44,560 X zero.
29 00:00:45,319 –> 00:00:47,060 Now around this 0.1
30 00:00:47,069 –> 00:00:48,639 wants a local approximation
31 00:00:48,650 –> 00:00:49,500 for the function.
32 00:00:50,220 –> 00:00:51,799 And for this reason, this
33 00:00:51,810 –> 00:00:53,790 point X zero is often called
34 00:00:53,799 –> 00:00:55,000 the expansion point.
35 00:00:55,889 –> 00:00:56,389 OK.
36 00:00:56,400 –> 00:00:58,189 Now I already told you we
37 00:00:58,200 –> 00:00:59,900 want an approximation around
38 00:00:59,909 –> 00:01:01,549 this point and this should
39 00:01:01,560 –> 00:01:03,270 be given by a polynomial.
40 00:01:03,990 –> 00:01:05,379 In fact, you already know
41 00:01:05,388 –> 00:01:07,209 the simplest one given by
42 00:01:07,220 –> 00:01:07,989 the tangent.
43 00:01:09,050 –> 00:01:10,599 Please recall this is the
44 00:01:10,610 –> 00:01:12,169 linear approximation we
45 00:01:12,180 –> 00:01:13,629 introduced in the definition
46 00:01:13,639 –> 00:01:14,730 of the derivative.
47 00:01:15,739 –> 00:01:17,550 In other words, here we have
48 00:01:17,559 –> 00:01:19,220 a polynomial with decree
49 00:01:19,230 –> 00:01:19,699 one.
50 00:01:20,580 –> 00:01:21,199 Indeed.
51 00:01:21,209 –> 00:01:22,739 Now Taylor’s theorem
52 00:01:22,750 –> 00:01:24,519 generalizes this fact to
53 00:01:24,529 –> 00:01:26,400 polynomials with higher decrease.
54 00:01:27,099 –> 00:01:28,440 Hence, in the picture, with
55 00:01:28,449 –> 00:01:30,120 the next step, we would have
56 00:01:30,129 –> 00:01:30,959 a parabola.
57 00:01:31,709 –> 00:01:33,190 And in the same way as the
58 00:01:33,199 –> 00:01:34,739 tangent was the best linear
59 00:01:34,750 –> 00:01:36,080 approximation, this
60 00:01:36,089 –> 00:01:37,709 parabola should be the best
61 00:01:37,720 –> 00:01:39,120 quadratic approximation.
62 00:01:40,169 –> 00:01:41,639 Now what this exactly
63 00:01:41,650 –> 00:01:43,470 means we have to describe
64 00:01:43,480 –> 00:01:44,470 with a formula.
65 00:01:45,269 –> 00:01:47,019 Therefore, let’s first recall
66 00:01:47,029 –> 00:01:48,669 the best linear approximation
67 00:01:48,680 –> 00:01:50,389 we defined with the derivative.
68 00:01:51,059 –> 00:01:52,410 In order to do this, we will
69 00:01:52,419 –> 00:01:53,989 introduce a new variable,
70 00:01:54,000 –> 00:01:55,949 we call h, should be
71 00:01:55,959 –> 00:01:57,650 a small number we add to
72 00:01:57,660 –> 00:01:58,459 X zero.
73 00:01:59,209 –> 00:02:00,610 In this sense, the point
74 00:02:00,620 –> 00:02:02,480 X zero plus H will be
75 00:02:02,489 –> 00:02:03,739 our point X.
76 00:02:04,690 –> 00:02:06,669 Indeed often this h makes
77 00:02:06,680 –> 00:02:08,000 the whole formulation a little
78 00:02:08,008 –> 00:02:08,800 bit simpler.
79 00:02:09,500 –> 00:02:10,679 Now, what we find by the
80 00:02:10,690 –> 00:02:12,000 definition of the derivative
81 00:02:12,009 –> 00:02:13,720 is that F of X
82 00:02:13,729 –> 00:02:15,380 zero plus h
83 00:02:15,389 –> 00:02:17,350 given by F of X zero
84 00:02:17,970 –> 00:02:19,919 plus F prime of X
85 00:02:19,929 –> 00:02:20,929 zero times h
86 00:02:22,119 –> 00:02:23,979 plus a remainder term
87 00:02:23,990 –> 00:02:25,600 R of h times h.
88 00:02:27,110 –> 00:02:28,559 Now here when you compare
89 00:02:28,570 –> 00:02:29,990 to our original definition
90 00:02:30,000 –> 00:02:31,479 of the derivative, you see that
91 00:02:31,490 –> 00:02:33,179 h now plays the role of
92 00:02:33,190 –> 00:02:34,630 X minus X zero.
93 00:02:35,330 –> 00:02:36,830 Hence, in this formulation,
94 00:02:36,839 –> 00:02:38,580 the linear term here is
95 00:02:38,589 –> 00:02:39,949 easy to recognize.
96 00:02:40,949 –> 00:02:41,330 OK.
97 00:02:41,339 –> 00:02:42,710 Now, one important property
98 00:02:42,720 –> 00:02:44,360 missing here is that R of
99 00:02:44,369 –> 00:02:46,339 H goes to zero
100 00:02:46,350 –> 00:02:48,039 when H goes to zero.
101 00:02:48,899 –> 00:02:50,279 In fact, this is the proper,
102 00:02:50,380 –> 00:02:51,800 that makes the tangent the
103 00:02:51,809 –> 00:02:53,399 best linear approximation.
104 00:02:54,300 –> 00:02:54,720 OK.
105 00:02:54,729 –> 00:02:56,500 Now, maybe not so surprising
106 00:02:56,509 –> 00:02:58,039 in a similar way, we can
107 00:02:58,050 –> 00:02:59,190 write down a quadratic
108 00:02:59,199 –> 00:03:00,119 approximation.
109 00:03:00,649 –> 00:03:02,210 Of course, the linear term
110 00:03:02,220 –> 00:03:03,440 here should not change.
111 00:03:03,449 –> 00:03:04,669 But now we want to add a
112 00:03:04,679 –> 00:03:06,309 term with squared.
113 00:03:07,160 –> 00:03:08,759 And as we will see soon,
114 00:03:08,770 –> 00:03:10,020 the best quadratic
115 00:03:10,029 –> 00:03:11,580 approximation is then given
116 00:03:11,589 –> 00:03:13,580 with one half times F
117 00:03:13,589 –> 00:03:15,419 prime prime of X
118 00:03:15,429 –> 00:03:17,369 zero times H squared.
119 00:03:18,509 –> 00:03:20,270 And in a similar way as before,
120 00:03:20,279 –> 00:03:21,770 we have a remainder term,
121 00:03:21,779 –> 00:03:22,929 which is not the same, but
122 00:03:22,940 –> 00:03:24,020 we still call it r.
123 00:03:25,089 –> 00:03:26,869 But now we have it here
124 00:03:26,919 –> 00:03:28,190 with h squared.
125 00:03:29,070 –> 00:03:30,809 Moreover, please don’t forget
126 00:03:30,839 –> 00:03:31,050 that.
127 00:03:31,059 –> 00:03:32,729 Speaking of an approximation,
128 00:03:32,740 –> 00:03:34,160 this still means that this
129 00:03:34,169 –> 00:03:35,729 function r goes to
130 00:03:35,740 –> 00:03:37,250 zero when h goes to
131 00:03:37,259 –> 00:03:37,800 zero.
132 00:03:38,600 –> 00:03:39,039 OK.
133 00:03:39,050 –> 00:03:40,470 And now it might not surprise
134 00:03:40,479 –> 00:03:42,009 you that we can generalize
135 00:03:42,020 –> 00:03:43,770 this approximation to
136 00:03:43,779 –> 00:03:45,750 polynomials with higher degree
137 00:03:45,860 –> 00:03:47,690 when we have enough derivatives.
138 00:03:48,500 –> 00:03:50,240 And exactly this idea
139 00:03:50,250 –> 00:03:52,080 leads to Taylor’s theorem.
140 00:03:53,080 –> 00:03:54,970 Therefore, let’s formulate
141 00:03:54,979 –> 00:03:55,289 it.
142 00:03:56,250 –> 00:03:57,600 So the only thing we need
143 00:03:57,630 –> 00:03:59,369 is a function defined on
144 00:03:59,380 –> 00:04:00,089 an interval.
145 00:04:00,100 –> 00:04:02,000 I hence,
146 00:04:02,009 –> 00:04:03,399 let’s call it F as
147 00:04:03,410 –> 00:04:04,039 before.
148 00:04:05,149 –> 00:04:06,809 Now, we already know we need
149 00:04:06,820 –> 00:04:08,139 enough derivatives.
150 00:04:08,250 –> 00:04:09,839 Therefore, let’s assume that
151 00:04:09,850 –> 00:04:11,580 the function F is N plus
152 00:04:11,589 –> 00:04:12,720 one times different.
153 00:04:13,839 –> 00:04:15,100 The reason why this plus
154 00:04:15,110 –> 00:04:16,928 one is helpful you will see
155 00:04:16,940 –> 00:04:17,619 in a minute.
156 00:04:18,510 –> 00:04:18,910 OK.
157 00:04:18,920 –> 00:04:20,820 And as before we also fix
158 00:04:20,829 –> 00:04:22,779 an expansion point X zero.
159 00:04:23,829 –> 00:04:25,170 Now, the claim here is that
160 00:04:25,179 –> 00:04:27,160 for any number H we have
161 00:04:27,170 –> 00:04:28,720 such an approximation formula,
162 00:04:29,369 –> 00:04:30,609 namely this means that we
163 00:04:30,619 –> 00:04:32,529 take any from the real number
164 00:04:32,540 –> 00:04:34,529 line such that X zero plus
165 00:04:34,540 –> 00:04:36,369 H still lies in I,
166 00:04:37,170 –> 00:04:38,559 this is important because
167 00:04:38,570 –> 00:04:40,350 only then we can put this
168 00:04:40,359 –> 00:04:41,940 number into the function
169 00:04:41,950 –> 00:04:42,390 F.
170 00:04:43,160 –> 00:04:45,010 Hence, in this case, F of
171 00:04:45,019 –> 00:04:46,880 X zero plus H is given
172 00:04:46,890 –> 00:04:48,059 as a whole sum
173 00:04:49,149 –> 00:04:50,850 of all the derives of the
174 00:04:50,859 –> 00:04:52,589 function F at the point
175 00:04:52,600 –> 00:04:53,339 X zero.
176 00:04:54,140 –> 00:04:55,750 So please keep in mind this
177 00:04:55,760 –> 00:04:57,670 is a real number and we divide
178 00:04:57,679 –> 00:04:59,220 it by K factorial.
179 00:05:00,100 –> 00:05:00,589 OK.
180 00:05:00,600 –> 00:05:01,880 So here you should see we
181 00:05:01,890 –> 00:05:03,239 start with case equal to
182 00:05:03,250 –> 00:05:04,760 zero, which is just a
183 00:05:04,769 –> 00:05:06,600 function F at the point X
184 00:05:06,609 –> 00:05:06,970 zero.
185 00:05:07,750 –> 00:05:09,079 And then we go through all
186 00:05:09,089 –> 00:05:10,540 the derivatives until we
187 00:05:10,549 –> 00:05:11,760 reach the nth one.
188 00:05:12,619 –> 00:05:13,829 Also, you should note that
189 00:05:13,839 –> 00:05:15,720 we don’t see the K factorial
190 00:05:15,730 –> 00:05:17,209 at the first two terms here
191 00:05:17,679 –> 00:05:19,170 because for K is equal to
192 00:05:19,179 –> 00:05:20,690 zero and K is equal to
193 00:05:20,700 –> 00:05:22,429 one, we just get the factor
194 00:05:22,440 –> 00:05:22,890 one.
195 00:05:23,329 –> 00:05:25,119 However, at K is equal to
196 00:05:25,130 –> 00:05:26,950 two, we see one half
197 00:05:27,649 –> 00:05:29,359 and this means the next factor
198 00:05:29,369 –> 00:05:30,660 will be 1/6.
199 00:05:31,369 –> 00:05:31,769 OK?
200 00:05:31,779 –> 00:05:33,089 Now you might ask how this
201 00:05:33,100 –> 00:05:34,660 factor comes in and I can
202 00:05:34,670 –> 00:05:36,040 tell you we will see it in
203 00:05:36,049 –> 00:05:37,010 the proof soon.
204 00:05:37,640 –> 00:05:39,170 However, more importantly,
205 00:05:39,179 –> 00:05:40,320 here, I should tell you that
206 00:05:40,329 –> 00:05:41,899 we also have our variable
207 00:05:42,160 –> 00:05:44,109 in namely with H to the
208 00:05:44,119 –> 00:05:44,899 power K.
209 00:05:45,619 –> 00:05:47,049 Hence what we get here is
210 00:05:47,059 –> 00:05:48,799 indeed a polynomial with
211 00:05:48,809 –> 00:05:50,260 degree at most N.
212 00:05:51,130 –> 00:05:53,070 And as we have seen before,
213 00:05:53,079 –> 00:05:54,329 we also have a remainder
214 00:05:54,339 –> 00:05:55,470 term at the end.
215 00:05:56,420 –> 00:05:57,679 And this one, I just want
216 00:05:57,690 –> 00:05:59,200 to denote with a capital
217 00:05:59,209 –> 00:06:01,070 R and in
218 00:06:01,079 –> 00:06:02,660 addition, it gets an index
219 00:06:02,670 –> 00:06:04,429 N, of course, the
220 00:06:04,440 –> 00:06:05,799 whole remainder term also
221 00:06:05,809 –> 00:06:07,260 depends on X zero.
222 00:06:07,269 –> 00:06:09,260 But for us, mainly the functional
223 00:06:09,269 –> 00:06:11,200 relation to h is important.
224 00:06:12,109 –> 00:06:13,679 Therefore, you see, we often
225 00:06:13,690 –> 00:06:15,410 just say ah and of H
226 00:06:15,420 –> 00:06:17,010 for the whole remainder term.
227 00:06:17,769 –> 00:06:19,380 However, the more important
228 00:06:19,390 –> 00:06:21,049 part is the front part.
229 00:06:21,059 –> 00:06:22,510 So the whole polynomial,
230 00:06:23,660 –> 00:06:25,470 in fact, this one is called
231 00:06:25,480 –> 00:06:27,239 the Taylor polynomial or
232 00:06:27,250 –> 00:06:28,980 more concretely the nth
233 00:06:28,989 –> 00:06:30,619 order Taylor polynomial.
234 00:06:31,459 –> 00:06:31,839 OK.
235 00:06:31,850 –> 00:06:33,000 Now, the whole claim here
236 00:06:33,010 –> 00:06:34,299 is not finished yet
237 00:06:34,450 –> 00:06:35,890 because of course, as
238 00:06:35,899 –> 00:06:37,630 before, we can say
239 00:06:37,640 –> 00:06:38,859 something about the remainder
240 00:06:38,869 –> 00:06:39,519 term here.
241 00:06:40,170 –> 00:06:41,600 Indeed, the remainder term
242 00:06:41,609 –> 00:06:43,279 gets better when we have
243 00:06:43,290 –> 00:06:45,000 one derivative more than
244 00:06:45,010 –> 00:06:46,829 we need for the Taylor polynomial.
245 00:06:47,390 –> 00:06:48,079 From this.
246 00:06:48,089 –> 00:06:49,380 It follows that we find an
247 00:06:49,390 –> 00:06:51,179 intermediate point we call
248 00:06:51,209 –> 00:06:52,970 C, this is a
249 00:06:52,980 –> 00:06:54,559 lower case Greek letter
250 00:06:54,570 –> 00:06:56,540 also often called Xi.
251 00:06:57,260 –> 00:06:58,470 Now, the important part is
252 00:06:58,480 –> 00:07:00,239 that this number lies in
253 00:07:00,250 –> 00:07:02,200 between X zero and X
254 00:07:02,209 –> 00:07:03,160 zero plus H.
255 00:07:04,029 –> 00:07:05,459 This means that if we want
256 00:07:05,470 –> 00:07:07,279 to write this as an interval,
257 00:07:07,290 –> 00:07:08,559 we have to distinguish two
258 00:07:08,570 –> 00:07:09,200 cases.
259 00:07:10,000 –> 00:07:11,450 It just depends which of
260 00:07:11,459 –> 00:07:13,309 the two numbers here is bigger
261 00:07:13,320 –> 00:07:14,140 than the other one.
262 00:07:14,920 –> 00:07:16,239 So we don’t know the exact
263 00:07:16,250 –> 00:07:18,000 value of the number C here,
264 00:07:18,040 –> 00:07:19,350 but we know the range of
265 00:07:19,359 –> 00:07:19,549 it.
266 00:07:20,390 –> 00:07:21,980 However often this is
267 00:07:21,989 –> 00:07:23,660 enough for an estimate of
268 00:07:23,670 –> 00:07:25,399 the remainder term are and
269 00:07:25,410 –> 00:07:26,019 of h.
270 00:07:26,880 –> 00:07:28,429 And in fact, the formula
271 00:07:28,440 –> 00:07:29,880 for the remainder term is
272 00:07:29,890 –> 00:07:31,420 so easy to remember
273 00:07:31,429 –> 00:07:33,140 because it looks exactly
274 00:07:33,149 –> 00:07:34,540 like the last term in the
275 00:07:34,549 –> 00:07:35,679 Taylor polynomial.
276 00:07:36,109 –> 00:07:37,760 More concretely, the number
277 00:07:37,769 –> 00:07:39,209 of the derivative is N plus
278 00:07:39,220 –> 00:07:41,040 one and we divide by
279 00:07:41,049 –> 00:07:42,579 N plus one factorial.
280 00:07:43,329 –> 00:07:44,730 In addition, at the end,
281 00:07:44,739 –> 00:07:46,730 we have to the power N
282 00:07:46,739 –> 00:07:47,380 plus one.
283 00:07:48,059 –> 00:07:49,500 However, we find the
284 00:07:49,510 –> 00:07:50,890 difference and this is the
285 00:07:50,899 –> 00:07:52,309 only one we don’t
286 00:07:52,320 –> 00:07:53,649 evaluate the function at
287 00:07:53,660 –> 00:07:54,809 the point X zero.
288 00:07:54,920 –> 00:07:56,369 But now at the point
289 00:07:56,429 –> 00:07:58,309 C OK,
290 00:07:58,339 –> 00:07:59,230 there we have it.
291 00:07:59,250 –> 00:08:01,170 This whole thing is the famous
292 00:08:01,179 –> 00:08:02,959 Taylor formula which holds
293 00:08:02,970 –> 00:08:04,750 for every function F
294 00:08:04,760 –> 00:08:06,170 which is often enough
295 00:08:06,549 –> 00:08:06,910 differentiable.
296 00:08:07,640 –> 00:08:08,899 And maybe I should tell you
297 00:08:09,000 –> 00:08:10,779 sometimes you also see it
298 00:08:10,790 –> 00:08:11,769 in a different form.
299 00:08:12,649 –> 00:08:13,959 This happens when one is
300 00:08:13,970 –> 00:08:15,940 interested in a Taylor polynomial
301 00:08:15,950 –> 00:08:17,420 but not in the explicit
302 00:08:17,429 –> 00:08:18,929 calculation for the remainder
303 00:08:18,940 –> 00:08:19,380 term.
304 00:08:20,109 –> 00:08:21,589 Then one just writes
305 00:08:21,600 –> 00:08:23,440 plus big O of
306 00:08:23,450 –> 00:08:24,929 h to the power and plus
307 00:08:24,940 –> 00:08:26,589 one, it just
308 00:08:26,600 –> 00:08:28,269 reminds us that here in the
309 00:08:28,279 –> 00:08:29,600 remainder term
310 00:08:29,869 –> 00:08:31,429 occurs with the power N plus
311 00:08:31,440 –> 00:08:31,910 one.
312 00:08:32,349 –> 00:08:34,030 But we don’t care how big
313 00:08:34,039 –> 00:08:35,229 the constant here is.
314 00:08:36,299 –> 00:08:38,169 Now, this curved O here is
315 00:08:38,179 –> 00:08:40,020 what we call a Landau symbol.
316 00:08:40,619 –> 00:08:42,080 And at the moment, we just
317 00:08:42,090 –> 00:08:43,469 use it as a shortcut for
318 00:08:43,479 –> 00:08:44,770 the whole remainder term
319 00:08:44,780 –> 00:08:45,169 here.
320 00:08:46,049 –> 00:08:47,750 However, later, I will tell
321 00:08:47,760 –> 00:08:48,989 you a little bit more about
322 00:08:49,000 –> 00:08:50,650 this symbol and related
323 00:08:50,659 –> 00:08:51,070 ones.
324 00:08:51,849 –> 00:08:52,349 OK.
325 00:08:52,359 –> 00:08:53,950 Now, for the end of the theorem,
326 00:08:53,960 –> 00:08:55,830 let me show you once again
327 00:08:55,840 –> 00:08:57,780 another formulation for this
328 00:08:57,789 –> 00:08:58,669 formula here.
329 00:08:59,500 –> 00:09:00,700 This happens if you don’t
330 00:09:00,710 –> 00:09:01,869 want to use the variable,
331 00:09:02,340 –> 00:09:03,719 but rather just to point
332 00:09:03,729 –> 00:09:05,390 X in the interval I,
333 00:09:06,219 –> 00:09:07,679 then of course, all the
334 00:09:07,690 –> 00:09:09,330 constants are the same.
335 00:09:09,340 –> 00:09:10,820 But now we have the factor
336 00:09:10,830 –> 00:09:12,320 X minus X zero.
337 00:09:13,080 –> 00:09:14,789 And obviously we also need
338 00:09:14,799 –> 00:09:15,929 the power K here.
339 00:09:16,859 –> 00:09:18,349 So now you see there is no
340 00:09:18,359 –> 00:09:19,479 difference at all.
341 00:09:19,530 –> 00:09:21,520 It just depends which variables
342 00:09:21,530 –> 00:09:23,080 you want to use in your problem.
343 00:09:23,900 –> 00:09:25,320 However, here please note
344 00:09:25,330 –> 00:09:26,880 both things, either with
345 00:09:26,890 –> 00:09:28,659 variable H or with variable
346 00:09:28,669 –> 00:09:30,650 X are called the nth order
347 00:09:30,659 –> 00:09:31,719 Taylor polynomial.
348 00:09:32,599 –> 00:09:33,099 OK.
349 00:09:33,109 –> 00:09:34,659 Then I would say in the next
350 00:09:34,669 –> 00:09:35,880 videos we can discuss
351 00:09:35,890 –> 00:09:37,690 examples applications
352 00:09:37,770 –> 00:09:39,469 and also the proof of this
353 00:09:39,479 –> 00:09:40,780 nice theorem here.
354 00:09:41,409 –> 00:09:42,570 Therefore, I hope I see you
355 00:09:42,580 –> 00:09:44,059 there and have a nice day.
356 00:09:44,190 –> 00:09:44,940 Bye.

Quiz Content
Q1: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. Then one can write for every $x_0, h \in \mathbb{R}$: $$f(x_0 +h) = f(x_0) + f^\prime(x_0) \cdot h + r(h) \cdot h$$ What does $r(h)$ satisfy?
A1: $r(h) = h$
A2: $r(h) = 1$
A3: $\frac{r(h)}{h} \xrightarrow{h \rightarrow 0} 0 $.
A4: $r(h) \xrightarrow{h \rightarrow 0} 0 $.
Q2: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a twotimes differentiable function. Then one can write for every $x_0, h \in \mathbb{R}$: $$f(x_0 +h) = f(x_0) + f^\prime(x_0) \cdot h + R(h)$$ What does $R(h)$, in general, not satisfy?
A1: $$R(h) \xrightarrow{h \rightarrow 0} 0 $$.
A2: $$\frac{R(h)}{h} \xrightarrow{h \rightarrow 0} 0 $$.
A3: $$\frac{R(h)}{h^2} \xrightarrow{h \rightarrow 0} \frac{1}{2} f^{\prime \prime}(x_0) $$.
A4: $$\frac{R(h)}{h^3} \xrightarrow{h \rightarrow 0} 0 $$.
Q3: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = x^5$. How does the $3$th order Taylor polynomial $T_3(h)$ with expansion point $x_0 = 0$ look like?
A1: $T_3(h) = h^5$
A2: $T_3(h) = h^4$
A3: $T_3(h) = h^3$
A4: $T_3(h) = h^2$
A5: $T_3(h) = h$
A6: $T_3(h) = 0$
Q4: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \sin(x)$. How does the $2$th order Taylor polynomial $T_2(h)$ with expansion point $x_0 = 0$ look like?
A1: $T_2(h) = 0$
A2: $T_2(h) = 1$
A3: $T_2(h) = 1 + h$
A4: $T_2(h) = h$
A5: $T_2(h) = h + h^2$
A6: $T_2(h) = 0$