1 00:00:00,040 --> 00:00:01,940 FEMALE SPEAKER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 creative commons license. 3 00:00:03,690 --> 00:00:06,630 Your support will help MIT OpenCourseWare continue to 4 00:00:06,630 --> 00:00:09,990 offer high-quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation, or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:31,780 --> 00:00:32,770 PROFESSOR: Hi. 9 00:00:32,770 --> 00:00:35,880 Today, we're going to generalize our discussion of 10 00:00:35,880 --> 00:00:41,240 series to cover series of functions rather than series 11 00:00:41,240 --> 00:00:42,490 of just constants. 12 00:00:42,490 --> 00:00:45,130 In other words, you'll notice that, up until now, what we've 13 00:00:45,130 --> 00:00:47,790 been doing in our discussion of series was have a bunch of 14 00:00:47,790 --> 00:00:50,180 fixed numbers and then add them. 15 00:00:50,180 --> 00:00:52,850 You see, in the same way that one starts with constants and 16 00:00:52,850 --> 00:00:56,020 ordinary arithmetic and then starts talking about 17 00:00:56,020 --> 00:01:00,000 functions, the same thing occurs here, that maybe we are 18 00:01:00,000 --> 00:01:03,470 now interested in a series over a range of different 19 00:01:03,470 --> 00:01:04,780 values of 'x'. 20 00:01:04,780 --> 00:01:07,680 Now, rather than to talk abstractly about this, I'd 21 00:01:07,680 --> 00:01:12,130 like to return to a topic that we touched upon briefly much 22 00:01:12,130 --> 00:01:15,140 earlier in the course, and then come back to that topic 23 00:01:15,140 --> 00:01:17,760 with far greater power than we were able to 24 00:01:17,760 --> 00:01:19,580 exhibit up until now. 25 00:01:19,580 --> 00:01:24,160 And it's the idea of approximating a function by 26 00:01:24,160 --> 00:01:25,060 polynomials. 27 00:01:25,060 --> 00:01:27,930 In other words, this question of degree of contact. 28 00:01:27,930 --> 00:01:31,900 And again, rather than review that abstractly, let's talk in 29 00:01:31,900 --> 00:01:33,370 terms of a picture. 30 00:01:33,370 --> 00:01:35,240 You see, today's lecture is called polynomial 31 00:01:35,240 --> 00:01:37,550 approximations, and the idea is this. 32 00:01:37,550 --> 00:01:41,060 Let's suppose we have the curve, 'y' equals 'f of x'. 33 00:01:41,060 --> 00:01:42,140 It's smooth. 34 00:01:42,140 --> 00:01:44,520 It passes through the origin. 35 00:01:44,520 --> 00:01:46,020 And we're told the following. 36 00:01:46,020 --> 00:01:50,230 We say, look, we're interested in approximating this curve in 37 00:01:50,230 --> 00:01:52,960 a neighborhood of 'x' equals 0-- in other words, at this 38 00:01:52,960 --> 00:01:54,010 point here-- 39 00:01:54,010 --> 00:01:56,440 by various polynomials. 40 00:01:56,440 --> 00:01:58,590 Now, the simplest polynomial, of course, is a constant. 41 00:01:58,590 --> 00:02:01,290 In other words, the question is, how should we choose a 42 00:02:01,290 --> 00:02:04,790 straight line, 'y' equals a constant, 'y' equals ''P sub 43 00:02:04,790 --> 00:02:06,595 0' of x'-- in other words, a polynomial 44 00:02:06,595 --> 00:02:08,610 of degree 0, a constant-- 45 00:02:08,610 --> 00:02:12,880 if we want the best possible horizontal line that 46 00:02:12,880 --> 00:02:15,870 approximates the curve at this particular point? 47 00:02:15,870 --> 00:02:18,760 And I think you can see that it's rather trivial at this 48 00:02:18,760 --> 00:02:21,700 case to say, OK, let's take the horizontal line that 49 00:02:21,700 --> 00:02:23,240 passes through this point. 50 00:02:23,240 --> 00:02:26,610 See, any other horizontal line misses this point altogether. 51 00:02:26,610 --> 00:02:28,490 OK, so far, so good. 52 00:02:28,490 --> 00:02:30,750 Then somebody says, now, look it, if we remove the 53 00:02:30,750 --> 00:02:33,270 restriction that the line be horizontal, what about any 54 00:02:33,270 --> 00:02:33,990 straight line? 55 00:02:33,990 --> 00:02:37,720 What straight line has the best degree of contact with 56 00:02:37,720 --> 00:02:39,710 the curve at this particular point? 57 00:02:39,710 --> 00:02:42,280 In other words, what first degree polynomial has the 58 00:02:42,280 --> 00:02:44,410 highest degree of contact over here? 59 00:02:44,410 --> 00:02:46,760 Now again, sparing you the details-- 60 00:02:46,760 --> 00:02:50,070 because the details are supplied in the text-- 61 00:02:50,070 --> 00:02:52,450 the polynomial there is simply what? 62 00:02:52,450 --> 00:02:56,190 'f of 0' plus ''f prime of 0' times 'x''. 63 00:02:56,190 --> 00:02:58,330 In other words, it's a tangent line. 64 00:02:58,330 --> 00:03:02,080 The basic equation of a tangent line is 'y' equals 'mx 65 00:03:02,080 --> 00:03:06,480 plus b', where 'm' is the slope, which is 'f prime of 66 00:03:06,480 --> 00:03:10,060 0', and 'b' is the y-intercept, which is 67 00:03:10,060 --> 00:03:11,800 'f of 0' over here. 68 00:03:11,800 --> 00:03:13,400 Now, the next question comes up. 69 00:03:13,400 --> 00:03:17,380 What if we wanted a quadratic polynomial, a second degree 70 00:03:17,380 --> 00:03:20,290 polynomial that fits the curve even better, in other words, 71 00:03:20,290 --> 00:03:22,350 has a greater degree of contact at 72 00:03:22,350 --> 00:03:23,720 this particular point? 73 00:03:23,720 --> 00:03:27,310 Let's label that 'y' equals ''P sub 2' of x' to indicate a 74 00:03:27,310 --> 00:03:29,890 second degree polynomial. 75 00:03:29,890 --> 00:03:32,430 And to analyze what that polynomial must look like, 76 00:03:32,430 --> 00:03:35,910 let's simply write down some undetermined coefficients 77 00:03:35,910 --> 00:03:41,430 here, namely, let's let 'P2 of x' be 'a0 plus 'a1x' plus 'a2 78 00:03:41,430 --> 00:03:42,760 'x squared''. 79 00:03:42,760 --> 00:03:46,110 And the idea is to determine what 'a0', 'a1', and 'a2' must 80 00:03:46,110 --> 00:03:49,650 look like if this polynomial is to have a maximum degree of 81 00:03:49,650 --> 00:03:53,940 contact with the curve, 'y' equals 'f of x'. 82 00:03:53,940 --> 00:03:56,350 Now, what do you mean by maximum degree of contact? 83 00:03:56,350 --> 00:04:01,330 Well, what you mean is you want the function at 0 to 84 00:04:01,330 --> 00:04:03,620 equal the given function at 0. 85 00:04:03,620 --> 00:04:07,720 You want the first derivative of 'P2' evaluated at 0 to 86 00:04:07,720 --> 00:04:10,150 equal the derivative of 'f' evaluated at 0. 87 00:04:10,150 --> 00:04:13,590 And you want the second derivative of 'P2' evaluated 88 00:04:13,590 --> 00:04:16,000 at 0 equal to ''f double prime' of 0'. 89 00:04:16,000 --> 00:04:18,649 Well, let's go through this in slow motion. 90 00:04:18,649 --> 00:04:23,090 First of all, 'P2 of 0' is simply 'a0'. 91 00:04:23,090 --> 00:04:29,580 'P2 prime' is simply 'a1 plus '2 a2 x'', and evaluating that 92 00:04:29,580 --> 00:04:32,460 when 'x' equals 0, we just have 'a1'. 93 00:04:32,460 --> 00:04:35,700 So 'f of 0' must equal 'a0'. 94 00:04:35,700 --> 00:04:38,480 'a1' must equal 'f prime of 0'. 95 00:04:38,480 --> 00:04:40,250 Now, what about the second derivative? 96 00:04:40,250 --> 00:04:43,260 Notice that every time we differentiate this polynomial, 97 00:04:43,260 --> 00:04:44,780 another term drops out. 98 00:04:44,780 --> 00:04:48,780 'a0' drops out the first time, 'a1' the second time. 99 00:04:48,780 --> 00:04:51,580 And when you've differentiated this term twice, all you have 100 00:04:51,580 --> 00:04:53,320 left is '2 a2'. 101 00:04:53,320 --> 00:04:55,040 In other words, 'P2 double prime' 102 00:04:55,040 --> 00:04:57,570 evaluated at 0 is '2 a2'. 103 00:04:57,570 --> 00:05:00,210 That must be 'f double prime' of 0. 104 00:05:00,210 --> 00:05:06,450 Equating these two, 'a2' must be the second derivative of 105 00:05:06,450 --> 00:05:09,050 'f' evaluated at 0 over 2. 106 00:05:09,050 --> 00:05:12,140 I inadvertently wrote in 2 factorial here, which is the 107 00:05:12,140 --> 00:05:15,100 right answer, but I haven't led up to that yet, so if you 108 00:05:15,100 --> 00:05:18,050 want to call this 2, that's fine. 109 00:05:18,050 --> 00:05:19,670 At any rate, here's what I've shown. 110 00:05:19,670 --> 00:05:22,900 What I've shown is that, if you want the polynomial of 111 00:05:22,900 --> 00:05:27,830 degree 2 that fits the curve the best, in a neighborhood of 112 00:05:27,830 --> 00:05:30,480 0 comma 'f of 0', this is how the 113 00:05:30,480 --> 00:05:32,810 coefficients must be chosen. 114 00:05:32,810 --> 00:05:36,650 And in fact, this situation generalizes very nicely. 115 00:05:36,650 --> 00:05:39,690 You see, suppose you want the n-th degree polynomial 116 00:05:39,690 --> 00:05:40,950 expression. 117 00:05:40,950 --> 00:05:43,550 Differentiate this 'n' times. 118 00:05:43,550 --> 00:05:46,240 You see, each time you differentiate, one of these 119 00:05:46,240 --> 00:05:47,430 terms drops out. 120 00:05:47,430 --> 00:05:51,070 What happens to this term as you differentiate it? 121 00:05:51,070 --> 00:05:53,230 The first time you differentiate it, you bring 122 00:05:53,230 --> 00:05:54,370 down an 'n'. 123 00:05:54,370 --> 00:05:56,850 The next time you differentiate, you bring down 124 00:05:56,850 --> 00:05:58,710 an 'n minus 1'. 125 00:05:58,710 --> 00:06:01,470 By the time you've differentiated 'n' times, this 126 00:06:01,470 --> 00:06:06,380 is 'x to the 0', that's 1, 'a sub n' is still here, and 'n' 127 00:06:06,380 --> 00:06:09,640 times 'n minus 1' times 'n minus 2', et cetera, is just 128 00:06:09,640 --> 00:06:11,840 what we call 'n factorial'. 129 00:06:11,840 --> 00:06:15,270 In other words, the n-th derivative of ''P sub n' of x' 130 00:06:15,270 --> 00:06:17,530 is 'n factorial' times 'a sub n'. 131 00:06:17,530 --> 00:06:20,970 If you evaluate that at 0, that's still 'n factorial' 132 00:06:20,970 --> 00:06:22,820 times 'a sub n'. 133 00:06:22,820 --> 00:06:25,610 That must also equal the n-th derivative of the function 134 00:06:25,610 --> 00:06:29,050 evaluated at 0, by definition of what you mean an n-th 135 00:06:29,050 --> 00:06:30,600 degree of contact. 136 00:06:30,600 --> 00:06:33,910 At any rate, that says that the n-th coefficient, the 137 00:06:33,910 --> 00:06:37,050 coefficient of 'x to the n', must be the n-th derivative of 138 00:06:37,050 --> 00:06:40,930 'f' evaluated at 0 over 'n factorial'. 139 00:06:40,930 --> 00:06:43,040 By the way, notice what this says. 140 00:06:43,040 --> 00:06:45,630 And we'll come back to this much more strongly later. 141 00:06:45,630 --> 00:06:48,660 It says that, for this coefficient to exist, the n-th 142 00:06:48,660 --> 00:06:51,080 derivative of 'f' at 0 has to exist. 143 00:06:51,080 --> 00:06:54,970 In other words, if 'f' does not possess its n-th 144 00:06:54,970 --> 00:06:59,010 derivative, this particular equation doesn't make sense. 145 00:06:59,010 --> 00:07:01,490 And I'll show you what that means in a little while. 146 00:07:01,490 --> 00:07:04,950 But at any rate, notice that what we've now shown is that 147 00:07:04,950 --> 00:07:06,120 ''P sub n' of x'-- 148 00:07:06,120 --> 00:07:08,170 if you want to use the sigma notation-- 149 00:07:08,170 --> 00:07:09,530 can be written how? 150 00:07:09,530 --> 00:07:13,760 Well, it's the sum as 'k' goes from 0 to 'n', the k-th 151 00:07:13,760 --> 00:07:18,710 derivative of 'f' evaluated at 0 over 'k factorial', times 'x 152 00:07:18,710 --> 00:07:19,810 to the k-th' power. 153 00:07:19,810 --> 00:07:22,350 Which written out longhand is simply what? 154 00:07:22,350 --> 00:07:27,130 'a0' plus 'a1x' plus ''a2 'x squared'' over '2 factorial'' 155 00:07:27,130 --> 00:07:29,890 plus et cetera, the n-th derivative of 'f' evaluated at 156 00:07:29,890 --> 00:07:33,920 0 over 'n factorial' times 'x to the n'. 157 00:07:33,920 --> 00:07:38,630 And what we do is we let 'P of x' denote the limit of ''P sub 158 00:07:38,630 --> 00:07:40,170 n' of x' as 'n' goes to infinity. 159 00:07:40,170 --> 00:07:42,900 In other words, for a fixed 'x', we look at what the 160 00:07:42,900 --> 00:07:47,140 sequence converges to as 'n' goes to infinity, and we 161 00:07:47,140 --> 00:07:49,260 denote that as a series. 162 00:07:49,260 --> 00:07:53,330 You see, what happens is 'x' is a variable here, but if you 163 00:07:53,330 --> 00:07:56,370 replace 'x' by a specific number, what's inside the 164 00:07:56,370 --> 00:07:59,770 summation sign becomes a constant which depends only on 165 00:07:59,770 --> 00:08:01,630 'n', and we're back to our study of 166 00:08:01,630 --> 00:08:03,270 ordinary series again. 167 00:08:03,270 --> 00:08:05,980 Now, what does 'P of x' mean in this case? 168 00:08:05,980 --> 00:08:09,940 Going back to our diagram, it means that, as we let n get 169 00:08:09,940 --> 00:08:11,780 bigger and bigger and we determine more and more 170 00:08:11,780 --> 00:08:13,700 coefficients, we get what? 171 00:08:13,700 --> 00:08:17,310 Hopefully, what's a better fit to the curve over here, that 172 00:08:17,310 --> 00:08:20,680 we get a greater degree of contact every time we tack on 173 00:08:20,680 --> 00:08:23,070 the next term in the series. 174 00:08:23,070 --> 00:08:26,110 Now, because this may seem a little bit abstract, let's 175 00:08:26,110 --> 00:08:28,150 illustrate this thing numerically. 176 00:08:28,150 --> 00:08:30,490 Let's pick a specific example. 177 00:08:30,490 --> 00:08:32,799 And I'll pick an easy one to compute. 178 00:08:32,799 --> 00:08:35,159 Let 'f of x' be 'e to the x'. 179 00:08:35,159 --> 00:08:36,549 After all, what could be easier? 180 00:08:36,549 --> 00:08:38,770 Because the derivative of 'e to the x' with respect to 'x' 181 00:08:38,770 --> 00:08:41,740 is just 'e to the x', so the n-th derivative of 'e to the 182 00:08:41,740 --> 00:08:43,510 x' is always 'e to the x'. 183 00:08:43,510 --> 00:08:46,530 When 'x' is 0, 'e to the 0' is 1. 184 00:08:46,530 --> 00:08:49,260 So the n-th derivative of 'e to the x', 185 00:08:49,260 --> 00:08:51,250 evaluated at 0, is 1. 186 00:08:51,250 --> 00:08:54,090 If I divide that by 'n factorial', I just get '1 over 187 00:08:54,090 --> 00:08:55,370 'n factorial''. 188 00:08:55,370 --> 00:08:58,340 In other words, the sequence of polynomials that 189 00:08:58,340 --> 00:09:02,680 approximates 'e to the x' is given by what? 190 00:09:02,680 --> 00:09:06,900 Summation ''x to the k' over 'k factorial'' as 'k' goes 191 00:09:06,900 --> 00:09:08,580 from '1 to n'. 192 00:09:08,580 --> 00:09:11,930 And again, what this means is what? 193 00:09:11,930 --> 00:09:16,150 The first approximation is the line, 'y' equals 1. 194 00:09:16,150 --> 00:09:19,230 The straight line approximation for the highest 195 00:09:19,230 --> 00:09:21,420 degree of contact at the origin is 'y' 196 00:09:21,420 --> 00:09:23,060 equals '1 plus x'. 197 00:09:23,060 --> 00:09:25,990 The best quadratic approximation is 'y' equals '1 198 00:09:25,990 --> 00:09:28,680 plus x plus ''x squared' over 2'. 199 00:09:28,680 --> 00:09:33,420 The best cubic fit is when 'y' is '1 plus x plus ''x squared' 200 00:09:33,420 --> 00:09:38,090 over 2', plus 'x cubed over 6'', 6 being 3 factorial. 201 00:09:38,090 --> 00:09:41,400 And we can continue on this way, getting better fits as we 202 00:09:41,400 --> 00:09:42,810 add more and more terms. 203 00:09:42,810 --> 00:09:46,220 And again, I've spoken quite rapidly, the reason being that 204 00:09:46,220 --> 00:09:49,800 this part is done superbly in the textbook, complete with 205 00:09:49,800 --> 00:09:52,870 graphs, tables, and what have you. 206 00:09:52,870 --> 00:09:54,290 Now, here's the question. 207 00:09:54,290 --> 00:09:58,000 The question is, look it, we can compute 'Pn of x' 208 00:09:58,000 --> 00:09:59,500 for any given 'n'. 209 00:09:59,500 --> 00:10:02,280 And we intuitively get the feeling that, as 'n' 210 00:10:02,280 --> 00:10:07,080 increases, 'Pn' hopefully will look more like 'f of x'. 211 00:10:07,080 --> 00:10:09,990 In other words, what we would like to do is to be able to 212 00:10:09,990 --> 00:10:14,930 compare, or better still, to study 'f of x' by comparing it 213 00:10:14,930 --> 00:10:18,100 with the limit of 'Pn of x' as 'n' goes to infinity. 214 00:10:18,100 --> 00:10:21,320 In other words, what we hope will happen is that 'f of x' 215 00:10:21,320 --> 00:10:25,530 looks like ''P sub n' of x' for large values of 'n'. 216 00:10:25,530 --> 00:10:27,510 Now, we don't know where this is going to happen. 217 00:10:27,510 --> 00:10:32,150 In fact, the remainder of our course now revolves about 218 00:10:32,150 --> 00:10:33,480 three questions. 219 00:10:33,480 --> 00:10:35,210 The three questions are the following. 220 00:10:35,210 --> 00:10:38,450 First of all, does the limit, 'Pn of x' as 'n' approaches 221 00:10:38,450 --> 00:10:40,160 infinity, exist? 222 00:10:40,160 --> 00:10:45,600 In other words, does 'P of x' exist in the first place? 223 00:10:45,600 --> 00:10:48,250 Does this limit always exists for a given 'x'? 224 00:10:48,250 --> 00:10:51,200 Secondly, suppose the limit does exist. 225 00:10:51,200 --> 00:10:53,590 How do we know it's going to equal the given function, 'f 226 00:10:53,590 --> 00:10:55,630 of x', which it's trying to approximate? 227 00:10:55,630 --> 00:10:57,590 In other words, how do we know that we don't get an 228 00:10:57,590 --> 00:11:01,840 approximation which is nice near the point of contact, but 229 00:11:01,840 --> 00:11:04,630 then, no matter how far out we go beyond the point of 230 00:11:04,630 --> 00:11:06,470 contact, the approximation doesn't become 231 00:11:06,470 --> 00:11:08,010 sufficiently good? 232 00:11:08,010 --> 00:11:09,330 See? 233 00:11:09,330 --> 00:11:12,130 And the third question that comes up, assuming that the 234 00:11:12,130 --> 00:11:15,050 first have been answered, is suppose the limit does exist-- 235 00:11:15,050 --> 00:11:17,520 in other words, suppose this limit function, 'P of x', 236 00:11:17,520 --> 00:11:20,140 which is a limit of 'Pn of x' as 'n' approaches infinity, 237 00:11:20,140 --> 00:11:21,240 does exist-- 238 00:11:21,240 --> 00:11:24,630 the question is, does 'P' possess the polynomial 239 00:11:24,630 --> 00:11:27,150 properties that each piece of 'n' possesses? 240 00:11:27,150 --> 00:11:30,330 In other words, the question is, does the limit of a 241 00:11:30,330 --> 00:11:34,130 sequence of functions have the same property that each of the 242 00:11:34,130 --> 00:11:35,920 members of the sequence has? 243 00:11:35,920 --> 00:11:38,770 For example, every polynomial is continuous. 244 00:11:38,770 --> 00:11:41,340 The question, for example, that might come up is, is the 245 00:11:41,340 --> 00:11:44,590 limit of a sequence of continuous functions also 246 00:11:44,590 --> 00:11:46,040 continuous? 247 00:11:46,040 --> 00:11:49,020 Now, at first glance, it might seem as if these are rather 248 00:11:49,020 --> 00:11:50,250 trivial to answer. 249 00:11:50,250 --> 00:11:56,620 Let me proceed next by showing that the answer can be no to 250 00:11:56,620 --> 00:11:59,600 each of these three questions, after which we'll then try to 251 00:11:59,600 --> 00:12:02,420 show when the answer will be yes. 252 00:12:02,420 --> 00:12:03,440 The idea is this. 253 00:12:03,440 --> 00:12:06,300 Let me call these things counter-examples, even though 254 00:12:06,300 --> 00:12:09,770 I don't know what better word to use here. 255 00:12:09,770 --> 00:12:11,870 In other words, let me show you an example where the 256 00:12:11,870 --> 00:12:14,950 answer to each of questions one, two, and three 257 00:12:14,950 --> 00:12:17,170 happen to be false. 258 00:12:17,170 --> 00:12:21,500 For example, let 'f of x' be '1 over '1 minus x''. 259 00:12:21,500 --> 00:12:24,600 As a review in taking derivatives, 'f prime of x' 260 00:12:24,600 --> 00:12:28,000 would be '1 minus x' to the minus 2 power, ''f double 261 00:12:28,000 --> 00:12:30,620 prime' of x' would be 2 times '1 minus x' 262 00:12:30,620 --> 00:12:31,930 to the minus 3 power. 263 00:12:31,930 --> 00:12:34,280 See, in other words, you bring down the minus 2, but the 264 00:12:34,280 --> 00:12:36,830 derivative of what's inside with respect to 'x' is minus. 265 00:12:36,830 --> 00:12:39,090 So a minus times minus is plus. 266 00:12:39,090 --> 00:12:43,220 Similarly, the third derivative of 'f of x' is 3 267 00:12:43,220 --> 00:12:47,650 factorial, 2 times minus 3 times minus 1, '1 minus x' to 268 00:12:47,650 --> 00:12:48,740 the minus fourth. 269 00:12:48,740 --> 00:12:51,050 And in general, the n-th derivative of 'f', in this 270 00:12:51,050 --> 00:12:56,240 case, is 'n factorial' times '1 minus x' to the what? 271 00:12:56,240 --> 00:12:57,890 Minus 'n minus 1'. 272 00:12:57,890 --> 00:13:01,780 In other words, notice that the magnitude of the exponent 273 00:13:01,780 --> 00:13:06,160 is one greater than the number we're taking the factorial of. 274 00:13:06,160 --> 00:13:09,210 At any rate, if I now compute the n-th derivative of 'f' 275 00:13:09,210 --> 00:13:13,200 evaluated at 0, I just get 'n factorial'. 276 00:13:13,200 --> 00:13:16,950 And if I divide 'n factorial' by 'n factorial', I get 1. 277 00:13:16,950 --> 00:13:19,600 In other words, every one of my coefficients of the 278 00:13:19,600 --> 00:13:23,130 approximating sequence of polynomials is going to be 1. 279 00:13:23,130 --> 00:13:26,430 In other words, the approximation for 'f of x', 280 00:13:26,430 --> 00:13:30,970 where 'f of x' is '1 over '1 minus x'', is simply what? 281 00:13:30,970 --> 00:13:35,657 'Pn of x' is '1 plus x plus 'x squared'', et cetera, plus et 282 00:13:35,657 --> 00:13:37,700 cetera, 'x to the n-th' power. 283 00:13:37,700 --> 00:13:40,050 Now, here's the interesting point. 284 00:13:40,050 --> 00:13:41,090 This part is irrelevant. 285 00:13:41,090 --> 00:13:43,000 I just wanted to show you something over here. 286 00:13:43,000 --> 00:13:45,670 Notice that 'f of 2' is clearly minus 1. 287 00:13:45,670 --> 00:13:48,500 If I replace 'x' by 2 over here, 1 over 1 288 00:13:48,500 --> 00:13:50,420 minus 2 is minus 1. 289 00:13:50,420 --> 00:13:52,970 On the other hand, what I claim is that the limit of a 290 00:13:52,970 --> 00:13:56,740 sequence, ''P sub n' of x', as 'n' goes to infinity doesn't 291 00:13:56,740 --> 00:13:59,200 even exist when 'x' is 2, because look what 292 00:13:59,200 --> 00:14:00,440 happens over here. 293 00:14:00,440 --> 00:14:02,740 When 'x' is 2, this is what? 294 00:14:02,740 --> 00:14:07,820 1 plus 2 plus 2 squared plus et cetera '2 to the n'. 295 00:14:07,820 --> 00:14:11,840 That's 1 plus 2 plus 4 plus 8 plus 16, et cetera. 296 00:14:11,840 --> 00:14:15,170 And as I let 'n' go to infinity, that sum increases 297 00:14:15,170 --> 00:14:16,200 without bound. 298 00:14:16,200 --> 00:14:20,650 In other words, here's an example where the given limit 299 00:14:20,650 --> 00:14:21,760 didn't have to exist. 300 00:14:21,760 --> 00:14:24,550 In other words, the sequence of polynomials does not 301 00:14:24,550 --> 00:14:28,980 converge at all at 'x' equals 2. 302 00:14:28,980 --> 00:14:31,090 So much for showing that the answer to the first 303 00:14:31,090 --> 00:14:32,610 question can be no. 304 00:14:32,610 --> 00:14:35,900 As for the second question, let's define a function as 305 00:14:35,900 --> 00:14:39,170 follows, to show you why continuity is so important. 306 00:14:39,170 --> 00:14:41,880 Let 'f of x' be 'x squared' if 'x' is less 307 00:14:41,880 --> 00:14:43,270 than or equal to 2. 308 00:14:43,270 --> 00:14:46,130 And let it be 4 if 'x' is greater than 2. 309 00:14:46,130 --> 00:14:49,100 Now, if we compute the derivatives of 'f' evaluated 310 00:14:49,100 --> 00:14:52,320 at 0, we find that what? 'f of 0' is 0. 311 00:14:52,320 --> 00:14:54,970 See, 'x squared' at 0 is 0. 312 00:14:54,970 --> 00:14:57,600 '2x' at 0 is 0. 313 00:14:57,600 --> 00:15:00,650 The second derivative here is 2, so ''f double 314 00:15:00,650 --> 00:15:02,440 prime' of 0' is 2. 315 00:15:02,440 --> 00:15:05,640 And the third derivative and higher, since this is only 'x 316 00:15:05,640 --> 00:15:07,410 squared', are identically 0. 317 00:15:07,410 --> 00:15:09,960 In other words, notice that the second derivative of 'f' 318 00:15:09,960 --> 00:15:11,650 evaluated at 0 is 2. 319 00:15:11,650 --> 00:15:13,740 All other derivatives are 0. 320 00:15:13,740 --> 00:15:17,340 In other words, notice that 'Pn of x' is equal to 'x 321 00:15:17,340 --> 00:15:20,300 squared' when 'n' is greater than 2. 322 00:15:20,300 --> 00:15:22,730 And that's a straightforward computation. 323 00:15:22,730 --> 00:15:25,610 I will drill you on the homework problems to show you 324 00:15:25,610 --> 00:15:27,810 how to get more facility. 325 00:15:27,810 --> 00:15:28,700 We're doing these things. 326 00:15:28,700 --> 00:15:30,470 But here's the key point. 327 00:15:30,470 --> 00:15:33,840 Let's look to see what the limit of 'Pn of x' is as 'n' 328 00:15:33,840 --> 00:15:35,150 approaches infinity. 329 00:15:35,150 --> 00:15:38,710 Since 'Pn of x' is equal to 'x squared' for all 'n', in 330 00:15:38,710 --> 00:15:41,130 particular, then, since this doesn't depend on 'n', the 331 00:15:41,130 --> 00:15:43,090 limit off 'Pn of x' as 'n' goes to 332 00:15:43,090 --> 00:15:45,680 infinity is just 'x squared'. 333 00:15:45,680 --> 00:15:47,040 What this says is this. 334 00:15:47,040 --> 00:15:52,490 Our function, 'f of x', graphs like this. 335 00:15:52,490 --> 00:15:54,040 The sequence of polynomials-- 336 00:15:54,040 --> 00:15:56,230 which, by the way, converge to the limit 337 00:15:56,230 --> 00:15:58,120 function, 'x squared'-- 338 00:15:58,120 --> 00:15:59,630 look what they do. 339 00:15:59,630 --> 00:16:01,750 They're just 'y' equals 'x squared'. 340 00:16:01,750 --> 00:16:04,700 In other words, somehow or other, when we got to this 341 00:16:04,700 --> 00:16:07,740 sharp corner-- in other words, notice that 'f of x' is not 342 00:16:07,740 --> 00:16:09,660 differentiable at this point-- 343 00:16:09,660 --> 00:16:13,300 our sequence of polynomials blissfully went right on their 344 00:16:13,300 --> 00:16:16,260 merry way smoothly while the function itself 345 00:16:16,260 --> 00:16:17,630 leveled off like this. 346 00:16:17,630 --> 00:16:20,460 In other words, the limit function always exist, but as 347 00:16:20,460 --> 00:16:24,140 soon as 'x' is greater than 2, in this example, the function, 348 00:16:24,140 --> 00:16:27,480 'P of x', no longer is the same as 'f of x'. 349 00:16:27,480 --> 00:16:30,860 You see, 'P of x' is going up like this, 'f of x' has just 350 00:16:30,860 --> 00:16:32,170 leveled off this way. 351 00:16:32,170 --> 00:16:35,240 There are even more glaring examples, but I will leave 352 00:16:35,240 --> 00:16:38,500 those for the supplementary notes to go 353 00:16:38,500 --> 00:16:39,740 into in more detail. 354 00:16:39,740 --> 00:16:41,120 But so far, we've seen what? 355 00:16:41,120 --> 00:16:44,840 That the answers to questions one and two can be no. 356 00:16:44,840 --> 00:16:47,000 Let's now show that the answer to question 357 00:16:47,000 --> 00:16:48,250 three can also be no. 358 00:16:52,370 --> 00:16:55,880 Let's define a sequence of polynomials, ''P sub n' of x', 359 00:16:55,880 --> 00:16:59,620 to be 'x to the n', where the domain of 'P sub n' is the 360 00:16:59,620 --> 00:17:01,800 interval from 0 to 1. 361 00:17:01,800 --> 00:17:04,329 What that means is something like this. 362 00:17:04,329 --> 00:17:06,290 See, here's 0, 0. 363 00:17:06,290 --> 00:17:09,500 And I hope this reminds you of a homework problem that we had 364 00:17:09,500 --> 00:17:12,450 very early in the course, but that's irrelevant whether it 365 00:17:12,450 --> 00:17:13,250 does or not. 366 00:17:13,250 --> 00:17:15,490 'P1 of x' would look like this. 367 00:17:15,490 --> 00:17:18,660 'P2 of x' would look like this. 368 00:17:18,660 --> 00:17:20,390 'P3 of x' would look like this. 369 00:17:20,390 --> 00:17:22,510 In other words, all of these curves pass 370 00:17:22,510 --> 00:17:25,270 through 0, 0 and 1, 1. 371 00:17:25,270 --> 00:17:28,060 But as 'n' increases, the degree of 372 00:17:28,060 --> 00:17:31,070 contact here gets better. 373 00:17:31,070 --> 00:17:34,710 At any rate, let's take a look at the limit function. 374 00:17:34,710 --> 00:17:36,740 'P of x' is the limit of ''P sub n' of x' as 375 00:17:36,740 --> 00:17:38,550 'n' approaches infinity. 376 00:17:38,550 --> 00:17:41,810 'x to the n', as 'n' approaches infinity, well, if 377 00:17:41,810 --> 00:17:44,930 'x' is less than 1 in magnitude, 'x to the n-th' 378 00:17:44,930 --> 00:17:46,120 approach is 0. 379 00:17:46,120 --> 00:17:49,460 On the other hand, if 'x' is equal to 1, 1 to the n-th 380 00:17:49,460 --> 00:17:52,020 power is still 1 as 'n' approaches infinity. 381 00:17:52,020 --> 00:17:54,180 Therefore, 'P of x' would be what? 382 00:17:54,180 --> 00:18:00,450 It's 0 if 'x' is greater than or equal to 0 but less than 1. 383 00:18:00,450 --> 00:18:03,010 And it's 1 if 'x' equals 1. 384 00:18:03,010 --> 00:18:04,670 Now, here's the key point. 385 00:18:04,670 --> 00:18:08,130 Observe that, since 'Pn of x' is a polynomial, it's, in 386 00:18:08,130 --> 00:18:09,955 particular, continuous. 387 00:18:09,955 --> 00:18:12,450 In other words, each 'P sub n' is a 388 00:18:12,450 --> 00:18:15,370 continuous function, unbroken. 389 00:18:15,370 --> 00:18:18,210 On the other hand, look at the limit function. 390 00:18:18,210 --> 00:18:21,080 It's 0 all the way up to 1. 391 00:18:21,080 --> 00:18:24,085 And that 1, the function jumps to 1. 392 00:18:24,085 --> 00:18:29,060 In other words, graph-wise, the thing looks like this. 393 00:18:29,060 --> 00:18:31,620 This is a graph, 'y' equals 'P of x'. 394 00:18:31,620 --> 00:18:32,620 It's 0. 395 00:18:32,620 --> 00:18:34,420 All of a sudden, it jumps. 396 00:18:34,420 --> 00:18:37,060 Is the limit function, therefore, continuous 397 00:18:37,060 --> 00:18:38,320 with 'x' equals 1. 398 00:18:38,320 --> 00:18:40,140 And the answer is no. 399 00:18:40,140 --> 00:18:42,190 There's a jump discontinuity here. 400 00:18:42,190 --> 00:18:45,440 In other words, observe that each 'P sub n' is continuous 401 00:18:45,440 --> 00:18:46,940 when 'x' equals 1. 402 00:18:46,940 --> 00:18:49,790 The limit function exists, but it isn't continuous 403 00:18:49,790 --> 00:18:51,160 when 'x' equals 1. 404 00:18:51,160 --> 00:18:54,690 In other words, the properties of a sequence of convergent 405 00:18:54,690 --> 00:18:57,900 functions do not have to be inherited 406 00:18:57,900 --> 00:19:00,450 by the limit function. 407 00:19:00,450 --> 00:19:02,920 That shows, then, that the answer to our three 408 00:19:02,920 --> 00:19:04,610 questions can be no. 409 00:19:04,610 --> 00:19:07,620 What we would like to do is find out when and if the 410 00:19:07,620 --> 00:19:09,260 answer will be yes. 411 00:19:09,260 --> 00:19:12,460 What we shall do for the remainder of today's lesson is 412 00:19:12,460 --> 00:19:14,590 answer the first two questions. 413 00:19:14,590 --> 00:19:18,270 And this, by the way, is also done very nicely in the text. 414 00:19:18,270 --> 00:19:21,250 The third question, for some reason or other, is beyond the 415 00:19:21,250 --> 00:19:22,150 scope of the text. 416 00:19:22,150 --> 00:19:26,200 We will discuss this in our future lectures, plus the 417 00:19:26,200 --> 00:19:27,410 supplementary notes. 418 00:19:27,410 --> 00:19:32,660 In fact, that the answer to question number three will be 419 00:19:32,660 --> 00:19:35,710 the end of our course, when we finally answer 420 00:19:35,710 --> 00:19:36,910 that particular question. 421 00:19:36,910 --> 00:19:39,490 But at any rate, what I'm saying now is let's spend the 422 00:19:39,490 --> 00:19:42,260 remainder of today's lesson in showing how to answer 423 00:19:42,260 --> 00:19:45,110 questions one and two in the affirmative, and we'll save 424 00:19:45,110 --> 00:19:48,260 question three for the remaining lectures. 425 00:19:48,260 --> 00:19:50,790 The first thing I'd like to point out is the role of the 426 00:19:50,790 --> 00:19:53,940 ratio test and absolute convergence. 427 00:19:53,940 --> 00:19:57,480 Namely, given the series summation 'a n 'x to the n''-- 428 00:19:57,480 --> 00:19:59,990 And by the way, if you haven't noticed this by now, it's 429 00:19:59,990 --> 00:20:03,690 rather conventional sometimes, to save time, not to put the 430 00:20:03,690 --> 00:20:05,020 subscripts on here. 431 00:20:05,020 --> 00:20:07,980 But if it bothers you, I mean, I could do things like this. 432 00:20:07,980 --> 00:20:11,950 It's just that to save time and space, I sometimes will 433 00:20:11,950 --> 00:20:13,310 not put the subscripts on here. 434 00:20:13,310 --> 00:20:16,380 But at any rate, let's just talk about this for a moment. 435 00:20:16,380 --> 00:20:19,980 Notice that a series converges as soon as it converges 436 00:20:19,980 --> 00:20:21,780 absolutely. 437 00:20:21,780 --> 00:20:26,430 The point is that, for a positive series, I can use the 438 00:20:26,430 --> 00:20:28,700 ratio test or its equivalent. 439 00:20:28,700 --> 00:20:33,850 Namely, for example, what I can do is to test summation 440 00:20:33,850 --> 00:20:37,260 absolute value, 'a n 'x to the n'', for convergence by the 441 00:20:37,260 --> 00:20:40,170 ratio test, by the comparison m by the 442 00:20:40,170 --> 00:20:43,590 integral test, et cetera. 443 00:20:43,590 --> 00:20:46,150 And what I'm saying is, if that particular series 444 00:20:46,150 --> 00:20:48,960 converges, then the original series, namely, the one 445 00:20:48,960 --> 00:20:53,020 without the absolute value signs, converges absolutely 446 00:20:53,020 --> 00:20:54,620 for that same value of 'x'. 447 00:20:54,620 --> 00:20:57,300 Well, again, instead of talking about that, let's look 448 00:20:57,300 --> 00:20:58,160 at a particular example. 449 00:20:58,160 --> 00:20:59,870 And it looks like I've scraped the blackboard 450 00:20:59,870 --> 00:21:00,870 here a little bit. 451 00:21:00,870 --> 00:21:02,440 Let me just-- 452 00:21:02,440 --> 00:21:03,850 The first example is this. 453 00:21:03,850 --> 00:21:06,800 Let's suppose you look at summation n goes from 0 to 454 00:21:06,800 --> 00:21:09,990 infinity, absolute value of 'x to the n-th' power. 455 00:21:09,990 --> 00:21:14,340 By the way, notice that this is a geometric series with a 456 00:21:14,340 --> 00:21:16,280 ratio absolute value of 'x'. 457 00:21:16,280 --> 00:21:19,570 Consequently, the series converges if and only if the 458 00:21:19,570 --> 00:21:22,300 absolute value of 'x' is less than 1. 459 00:21:22,300 --> 00:21:24,900 By the way, let me make a little aside here. 460 00:21:24,900 --> 00:21:28,300 Many people would tackle this problem by the ratio test, as 461 00:21:28,300 --> 00:21:29,640 I hinted at over here. 462 00:21:29,640 --> 00:21:33,170 Namely, let the n-th term be the absolute value of 'x to 463 00:21:33,170 --> 00:21:34,270 the n-th' power. 464 00:21:34,270 --> 00:21:36,930 Then the 'n plus first' term would be the absolute value of 465 00:21:36,930 --> 00:21:39,070 'x to the 'n plus first'' power. 466 00:21:39,070 --> 00:21:42,150 Then the ratio between these two would be the absolute 467 00:21:42,150 --> 00:21:43,580 value of 'x'. 468 00:21:43,580 --> 00:21:47,690 Therefore, row, which is this limit, would be the limit of 469 00:21:47,690 --> 00:21:53,120 the absolute value of 'x' as 'n' goes to infinity. 470 00:21:53,120 --> 00:21:56,230 That's just equal to the absolute value of 'x' itself. 471 00:21:56,230 --> 00:21:59,920 And for convergence, notice that row must be less than 1, 472 00:21:59,920 --> 00:22:00,720 and that says what? 473 00:22:00,720 --> 00:22:03,820 The absolute value of 'x' must be less than 1, just 474 00:22:03,820 --> 00:22:05,050 as we did over here. 475 00:22:05,050 --> 00:22:08,580 I simply would like to make a little aside over here, and 476 00:22:08,580 --> 00:22:11,360 that is, if you use this particular approach, that 477 00:22:11,360 --> 00:22:14,430 comes under the heading of circular reasoning, because 478 00:22:14,430 --> 00:22:18,340 you may recall when we proved the ratio test, we did it by 479 00:22:18,340 --> 00:22:21,940 comparing the series with the geometric series, which meant 480 00:22:21,940 --> 00:22:23,570 that we already had to know that the 481 00:22:23,570 --> 00:22:25,400 geometric series converged. 482 00:22:25,400 --> 00:22:27,690 And this is a geometric series. 483 00:22:27,690 --> 00:22:30,380 But that's an aside which I just wanted to mention to you 484 00:22:30,380 --> 00:22:32,140 in forms of circular reasoning. 485 00:22:32,140 --> 00:22:34,600 The important point to observe is what? 486 00:22:34,600 --> 00:22:38,400 That this series converges if the absolute value of 'x' is 487 00:22:38,400 --> 00:22:39,600 less than 1. 488 00:22:39,600 --> 00:22:42,400 Consequently, that means that the series without the 489 00:22:42,400 --> 00:22:46,310 absolute value sign converges, in fact, absolutely, if the 490 00:22:46,310 --> 00:22:48,660 absolute value of 'x' is less than 1. 491 00:22:48,660 --> 00:22:53,220 In other words, for this particular series, '1 plus x 492 00:22:53,220 --> 00:22:56,160 plus 'x squared' plus 'x cubed'', et cetera, that will 493 00:22:56,160 --> 00:22:59,990 converge to a finite limit function if the absolute value 494 00:22:59,990 --> 00:23:02,590 of 'x' is less than 1. 495 00:23:02,590 --> 00:23:05,200 Let's look at another example. 496 00:23:05,200 --> 00:23:07,950 Suppose we look at the absolute value of 'x to the 497 00:23:07,950 --> 00:23:10,842 n-th' power over 'n factorial'. 498 00:23:10,842 --> 00:23:13,530 If we look at this, the nth term is the absolute value of 499 00:23:13,530 --> 00:23:16,210 'x to the n-th' over 'n factorial'. 500 00:23:16,210 --> 00:23:18,520 The 'n plus first' term, therefore, is the absolute 501 00:23:18,520 --> 00:23:21,450 value of 'x to the 'n plus 1'', over ''n plus 1' 502 00:23:21,450 --> 00:23:22,510 factorial'. 503 00:23:22,510 --> 00:23:25,280 Therefore, the ratio between the 'n plus first' term and 504 00:23:25,280 --> 00:23:30,980 the n-th term is the absolute value of 'x' over 'n plus 1'. 505 00:23:30,980 --> 00:23:32,960 Rho is this particular limit. 506 00:23:32,960 --> 00:23:35,730 Well, the limit of the absolute value of 'x' over 'n 507 00:23:35,730 --> 00:23:37,760 plus 1', as 'n' approaches infinity-- 508 00:23:37,760 --> 00:23:39,040 and here's the key point. 509 00:23:39,040 --> 00:23:42,330 Notice that 'x' is a fixed but finite number once we get 510 00:23:42,330 --> 00:23:42,830 started here. 511 00:23:42,830 --> 00:23:45,030 In other words, we pick an 'x' and fixed it. 512 00:23:45,030 --> 00:23:46,870 That means that that stays constant. 513 00:23:46,870 --> 00:23:49,620 But as 'n' goes to infinity, the denominator increases 514 00:23:49,620 --> 00:23:50,590 without bound. 515 00:23:50,590 --> 00:23:52,470 That makes the limit equal to 0. 516 00:23:52,470 --> 00:23:55,860 And 0 is obviously less than 1, regardless of what the 517 00:23:55,860 --> 00:23:57,760 value of 'x' happens to be here. 518 00:23:57,760 --> 00:24:00,380 In other words, in this particular case, this 519 00:24:00,380 --> 00:24:04,160 particular series converges for all real values of 'x'. 520 00:24:04,160 --> 00:24:06,920 And we sometimes abbreviate that by saying the absolute 521 00:24:06,920 --> 00:24:09,660 value of 'x' is less than infinity. 522 00:24:09,660 --> 00:24:12,130 And therefore, the original series, meaning the one 523 00:24:12,130 --> 00:24:14,770 without the absolute value sign, would, in particular, 524 00:24:14,770 --> 00:24:17,600 converge also for all real 'x'. 525 00:24:17,600 --> 00:24:22,240 The important point is that, in general, given any series-- 526 00:24:22,240 --> 00:24:24,650 in other words, limit of a sequence of polynomials of 527 00:24:24,650 --> 00:24:26,350 this particular type-- 528 00:24:26,350 --> 00:24:28,710 there always exists a number 'M'. 529 00:24:28,710 --> 00:24:30,560 Now, 'M' may be as small as 0. 530 00:24:30,560 --> 00:24:32,570 It may be as large as infinity. 531 00:24:32,570 --> 00:24:36,630 But there always exists an 'M', such that the particular 532 00:24:36,630 --> 00:24:40,520 series will converge absolutely if the magnitude of 533 00:24:40,520 --> 00:24:43,680 'x' is less than 'M', and diverge if the magnitude of 534 00:24:43,680 --> 00:24:45,500 'x' is greater than 'M'. 535 00:24:45,500 --> 00:24:48,410 In other words, this particular 'M', which is 536 00:24:48,410 --> 00:24:51,400 called the radius of convergence, governs the 537 00:24:51,400 --> 00:24:53,240 answer to the first question. 538 00:24:53,240 --> 00:24:57,460 In other words, it's this 'M', which we often find by the 539 00:24:57,460 --> 00:25:01,240 ratio test, that determines where the sequence of 540 00:25:01,240 --> 00:25:04,310 functions, ''P sub n' of x', for what values of 'x', that 541 00:25:04,310 --> 00:25:06,610 will converge to 'P of x'. 542 00:25:06,610 --> 00:25:09,410 And again, I think this is kind of difficult for you to 543 00:25:09,410 --> 00:25:10,370 see abstractly. 544 00:25:10,370 --> 00:25:11,890 The proof is done in the text. 545 00:25:11,890 --> 00:25:14,040 What I'd like to do is show you pictorially what's 546 00:25:14,040 --> 00:25:15,480 happening over here. 547 00:25:15,480 --> 00:25:18,600 See, let's suppose we have our series, summation 548 00:25:18,600 --> 00:25:20,390 'a n 'x to the n''. 549 00:25:20,390 --> 00:25:24,600 And we pick a particular value of 'x', say 'x sub 1'. 550 00:25:24,600 --> 00:25:26,480 Now, the idea is this. 551 00:25:26,480 --> 00:25:29,510 If this particular series converges-- 552 00:25:29,510 --> 00:25:32,320 and the proof is given in the text, and I won't repeat it 553 00:25:32,320 --> 00:25:34,350 here because I want you to see the overview here-- 554 00:25:34,350 --> 00:25:36,170 if this particular series converges-- 555 00:25:36,170 --> 00:25:39,000 in other words, if the series converges when 'x' is equal to 556 00:25:39,000 --> 00:25:40,180 'x sub 1'-- 557 00:25:40,180 --> 00:25:43,960 then it can be shown that it converges absolutely for any 558 00:25:43,960 --> 00:25:47,080 'x' which is smaller in magnitude then 'x sub 1'. 559 00:25:47,080 --> 00:25:48,750 In other words, the gist is this. 560 00:25:48,750 --> 00:25:51,950 If I know that the series converges over here, I can 561 00:25:51,950 --> 00:25:55,075 then draw this interval surrounding 0, and conclude 562 00:25:55,075 --> 00:25:58,750 that the series converges in this entire interval. 563 00:25:58,750 --> 00:26:02,800 Correspondingly, if this particular series diverges-- 564 00:26:02,800 --> 00:26:05,620 and by the way, the whole proof of this thing hinges on 565 00:26:05,620 --> 00:26:08,120 the comparison test, essentially. 566 00:26:08,120 --> 00:26:11,440 See, if this thing diverges, then certainly, if 'x' is 567 00:26:11,440 --> 00:26:15,670 larger than 'x1' in magnitude, this will also diverge. 568 00:26:15,670 --> 00:26:18,530 In other words, once I know that my series diverges at 569 00:26:18,530 --> 00:26:20,700 this particular value of 'x1'.. 570 00:26:20,700 --> 00:26:25,280 of 'x', I know it diverges for all values of 'x' which are 571 00:26:25,280 --> 00:26:28,750 greater in magnitude then 'x1', which means, in terms of 572 00:26:28,750 --> 00:26:33,440 a picture, for everything further away from 0, then 'x1' 573 00:26:33,440 --> 00:26:34,790 and minus 'x1'. 574 00:26:34,790 --> 00:26:37,600 In other words, if this converges, this converges. 575 00:26:37,600 --> 00:26:43,110 In here, if this diverges, we have divergence out here. 576 00:26:43,110 --> 00:26:44,290 Now, what does that mean? 577 00:26:44,290 --> 00:26:50,050 Let me show you again in terms of a extended diagram. 578 00:26:50,050 --> 00:26:52,680 See, the idea goes something like this. 579 00:26:52,680 --> 00:26:56,530 Given the series summation 'a n 'x to the n'', pick a number 580 00:26:56,530 --> 00:26:58,400 'x' equal to 'x1'. 581 00:26:58,400 --> 00:27:01,680 Suppose the series converges at 'x1'. 582 00:27:01,680 --> 00:27:05,800 Then we know that it converges every place in here, so 583 00:27:05,800 --> 00:27:07,520 there's no need to check this any further. 584 00:27:07,520 --> 00:27:10,430 What we do next is we pick a number, 'x2', 585 00:27:10,430 --> 00:27:12,020 outside of this interval. 586 00:27:12,020 --> 00:27:13,990 Say 'x2' is over here. 587 00:27:13,990 --> 00:27:16,180 Suppose, for the sake of argument-- see, what we're 588 00:27:16,180 --> 00:27:20,570 saying is 'a n 'x 1 to the n'' converges, so you 589 00:27:20,570 --> 00:27:23,130 just suppose this. 590 00:27:23,130 --> 00:27:24,850 That gives us this picture. 591 00:27:24,850 --> 00:27:28,890 Suppose 'a n 'x2 to the n'' diverges. 592 00:27:28,890 --> 00:27:31,690 Then we know that the series diverges for all 593 00:27:31,690 --> 00:27:33,850 larger values of 'x'. 594 00:27:33,850 --> 00:27:37,390 Now we know how the series behaves exactly, except in the 595 00:27:37,390 --> 00:27:39,620 interval between 'x1' and 'x2'. 596 00:27:39,620 --> 00:27:41,410 So what we do next is what? 597 00:27:41,410 --> 00:27:44,210 We pick a number which we'll call 'x3'. 598 00:27:44,210 --> 00:27:47,830 We then see whether this converges or diverges. 599 00:27:47,830 --> 00:27:51,140 If it converges at 'x3', we know that the radius of 600 00:27:51,140 --> 00:27:53,870 convergence extends all the way between minus 601 00:27:53,870 --> 00:27:56,280 'x3' and plus 'x3'. 602 00:27:56,280 --> 00:28:01,190 If it diverges, we know that, from 'x3' out and minus 'x3' 603 00:28:01,190 --> 00:28:03,090 out, the series diverges. 604 00:28:03,090 --> 00:28:06,700 In any event, each time we perform this operation, we cut 605 00:28:06,700 --> 00:28:08,470 the remaining space down. 606 00:28:08,470 --> 00:28:10,420 And we can keep on going this way. 607 00:28:10,420 --> 00:28:14,040 And you see what you're doing then, is you're zeroing in on 608 00:28:14,040 --> 00:28:16,860 this number that we called capital 'M' before. 609 00:28:16,860 --> 00:28:18,540 You see, it's just a series of refinements. 610 00:28:18,540 --> 00:28:20,580 And by the way, the formal proof of 611 00:28:20,580 --> 00:28:21,920 this looks very messy. 612 00:28:21,920 --> 00:28:24,300 And yet, when it's stripped of all embellishment, that's 613 00:28:24,300 --> 00:28:26,440 exactly what the formal proof says. 614 00:28:26,440 --> 00:28:31,280 Well, at any rate, I will reinforce the rest of this 615 00:28:31,280 --> 00:28:34,940 between the text, the notes, and the learning exercises. 616 00:28:34,940 --> 00:28:38,720 All I want to do now is talk about the second question. 617 00:28:38,720 --> 00:28:41,000 In other words, we've now talked about something called 618 00:28:41,000 --> 00:28:44,490 the radius of convergence, that tells us when the series 619 00:28:44,490 --> 00:28:47,140 of functions converges to the limit function. 620 00:28:47,140 --> 00:28:50,510 Now the question is, how can we measure whether that limit 621 00:28:50,510 --> 00:28:53,080 function converges to the original function that we're 622 00:28:53,080 --> 00:28:54,470 trying to approximate? 623 00:28:54,470 --> 00:28:57,160 In other words, how do we know whether 'P of x' actually 624 00:28:57,160 --> 00:28:59,010 equals 'f of x'? 625 00:28:59,010 --> 00:29:04,690 And again, this is done extremely well in the text. 626 00:29:04,690 --> 00:29:07,370 It's called Taylor's theorem with remainder. 627 00:29:07,370 --> 00:29:09,550 The proof is given very, very nicely. 628 00:29:09,550 --> 00:29:12,440 I think the hardest part is to show what the thing means. 629 00:29:12,440 --> 00:29:15,300 You see, using integration by parts repeatedly-- 630 00:29:15,300 --> 00:29:17,000 which is done in the text-- 631 00:29:17,000 --> 00:29:20,180 it follows that, if 'f' and its first 'n plus 1' 632 00:29:20,180 --> 00:29:23,750 derivatives exist at 'x' equals 0, it turns out-- and 633 00:29:23,750 --> 00:29:25,320 this is why this is called Taylor's theorem with 634 00:29:25,320 --> 00:29:26,250 remainder-- 635 00:29:26,250 --> 00:29:31,400 it turns out that 'f of x' can be written as 'Pn of x' plus a 636 00:29:31,400 --> 00:29:34,510 remainder term-- ''r sub n' of x'-- a remainder term. 637 00:29:34,510 --> 00:29:37,260 And then you see, computationally, the text 638 00:29:37,260 --> 00:29:39,900 shows what that remainder term looks like. 639 00:29:39,900 --> 00:29:44,005 It happens to be the integral from 0 to 'x', ''x minus t' to 640 00:29:44,005 --> 00:29:48,700 the n-th' over 'n factorial', 'n plus first' derivative of 641 00:29:48,700 --> 00:29:53,970 'f of t', 'dt', from 't' equals 0 to 't' equals 'x'. 642 00:29:53,970 --> 00:29:55,080 This looks messy. 643 00:29:55,080 --> 00:29:56,630 It is messy. 644 00:29:56,630 --> 00:30:01,035 There's no need to hammer this home in the lecture part of 645 00:30:01,035 --> 00:30:01,740 our course. 646 00:30:01,740 --> 00:30:05,660 What I prefer to do is to give several learning exercises 647 00:30:05,660 --> 00:30:07,840 that will give you drill in using this. 648 00:30:07,840 --> 00:30:11,250 But the point that I want you to understand is that, since 649 00:30:11,250 --> 00:30:14,220 this is all done well in the text, all I want you to do is 650 00:30:14,220 --> 00:30:15,890 to see where this thing fits in. 651 00:30:15,890 --> 00:30:19,510 In other words, what the text is showing is, look it, the 652 00:30:19,510 --> 00:30:21,900 difference between the function which we're 653 00:30:21,900 --> 00:30:26,400 approximating at 0 comma 'f of 0' by this sequence of 654 00:30:26,400 --> 00:30:30,560 polynomials, the difference between that function and the 655 00:30:30,560 --> 00:30:33,990 nth member of that sequence of polynomials is simply 656 00:30:33,990 --> 00:30:37,880 something called ''r sub n' of x', where numerically ''r sub 657 00:30:37,880 --> 00:30:39,830 n' of x' looks like this. 658 00:30:39,830 --> 00:30:43,000 And the point is that, if you now take the limit of this 659 00:30:43,000 --> 00:30:45,990 expression as 'n' goes to infinity, remember, 'P of x' 660 00:30:45,990 --> 00:30:47,100 is this limit. 661 00:30:47,100 --> 00:30:51,140 Therefore, the only way that 'P of x' can equal 'f of x' is 662 00:30:51,140 --> 00:30:56,530 if, for a given value of 'x', this remainder goes to 0 as 663 00:30:56,530 --> 00:30:58,070 'n' goes to infinity. 664 00:30:58,070 --> 00:31:01,360 In other words, leaving the computational details to the 665 00:31:01,360 --> 00:31:04,940 text, because it does it very nicely, what we're saying is 666 00:31:04,940 --> 00:31:08,290 that the significance of Taylor's remainder theorem is 667 00:31:08,290 --> 00:31:11,850 that, to find out where 'f of x' and 'P of x' are identical, 668 00:31:11,850 --> 00:31:12,820 we simply-- 669 00:31:12,820 --> 00:31:16,640 and I say simply meaning conceptually simply-- 670 00:31:16,640 --> 00:31:18,760 computationally, it might be difficult. 671 00:31:18,760 --> 00:31:22,750 What we do computationally is simply find all those values 672 00:31:22,750 --> 00:31:27,340 of 'x', for which ''r sub n' of x' goes to 0 in the limit 673 00:31:27,340 --> 00:31:29,370 as 'n' goes to infinity. 674 00:31:29,370 --> 00:31:32,270 You see, what we've done is we've now done two things. 675 00:31:32,270 --> 00:31:34,410 We've first shown when the sequence of 676 00:31:34,410 --> 00:31:36,220 polynomials has a limit. 677 00:31:36,220 --> 00:31:39,560 And secondly, we've shown when it does have a limit, when 678 00:31:39,560 --> 00:31:40,940 will it equal the given function that 679 00:31:40,940 --> 00:31:43,170 it's trying to represent. 680 00:31:43,170 --> 00:31:45,980 The question that still remains is, how do we know 681 00:31:45,980 --> 00:31:49,570 that the limit function has the same properties as each 682 00:31:49,570 --> 00:31:50,750 member of the sequence? 683 00:31:50,750 --> 00:31:54,330 In fact, we've just given an example where the limit 684 00:31:54,330 --> 00:31:56,470 function had different properties. 685 00:31:56,470 --> 00:31:59,600 And the question is, under what conditions can we be sure 686 00:31:59,600 --> 00:32:01,810 that certain nice results-- 687 00:32:01,810 --> 00:32:04,360 that we'd like to be true, because they're true about 688 00:32:04,360 --> 00:32:06,220 each polynomial in the sequence-- 689 00:32:06,220 --> 00:32:09,280 actually are true about the limit function. 690 00:32:09,280 --> 00:32:11,880 In fact, that shall be our topic for 691 00:32:11,880 --> 00:32:13,250 the next two lessons. 692 00:32:13,250 --> 00:32:16,140 And these next two lectures should complete the course. 693 00:32:16,140 --> 00:32:18,480 At any rate, until next time, goodbye. 694 00:32:21,300 --> 00:32:23,840 MALE SPEAKER: Funding for the publication of this video was 695 00:32:23,840 --> 00:32:28,560 provided by the Gabriella and Paul Rosenbaum Foundation. 696 00:32:28,560 --> 00:32:32,730 Help OCW continue to provide free and open access to MIT 697 00:32:32,730 --> 00:32:36,930 courses by making a donation at ocw.mit.edu/donate.