1 00:00:00,000 --> 00:00:01,940 ANNOUNCER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons license. 3 00:00:03,690 --> 00:00:06,640 Your support will help MIT OpenCourseWare continue to 4 00:00:06,640 --> 00:00:09,980 offer high-quality educational resources for free. 5 00:00:09,980 --> 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 00:00:18,010 --> 00:00:31,555 9 00:00:31,555 --> 00:00:32,820 PROFESSOR: Hi. 10 00:00:32,820 --> 00:00:35,490 Today is the day we've all been waiting for. 11 00:00:35,490 --> 00:00:41,080 We've come to the last lecture of our course, and it's, I 12 00:00:41,080 --> 00:00:44,600 think, a rather satisfying lecture today, not just 13 00:00:44,600 --> 00:00:47,450 because it is the last one, but content-wise too. 14 00:00:47,450 --> 00:00:49,540 Today we're going to talk about uniform 15 00:00:49,540 --> 00:00:51,550 convergence of series. 16 00:00:51,550 --> 00:00:54,850 Remember, a series can be viewed as a sequence of 17 00:00:54,850 --> 00:00:58,790 partial sums, and consequently our discussion on uniform 18 00:00:58,790 --> 00:01:00,580 convergence applies here. 19 00:01:00,580 --> 00:01:02,050 And what we're going to do-- 20 00:01:02,050 --> 00:01:04,200 you see, again, it's the same old story. 21 00:01:04,200 --> 00:01:07,200 Now that we know what the concept means, is there an 22 00:01:07,200 --> 00:01:11,110 easy way to tell when we have the property? 23 00:01:11,110 --> 00:01:13,870 And I figure again that, this being the last lecture, I 24 00:01:13,870 --> 00:01:16,220 should give you some big names to remember. 25 00:01:16,220 --> 00:01:20,330 And name-dropping-wise, I come to our first concept today, 26 00:01:20,330 --> 00:01:21,400 which I call-- 27 00:01:21,400 --> 00:01:22,470 I don't call it that. 28 00:01:22,470 --> 00:01:26,390 It's named the Weierstrauss M-test. 29 00:01:26,390 --> 00:01:31,080 The Weierstrauss M-test is a very, very convenient method 30 00:01:31,080 --> 00:01:35,390 for determining whether a given series converges 31 00:01:35,390 --> 00:01:36,780 uniformly or not. 32 00:01:36,780 --> 00:01:39,170 By the way, let me just make one little aside. 33 00:01:39,170 --> 00:01:43,030 Instead of saying the sequence of partial sums converges 34 00:01:43,030 --> 00:01:47,440 uniformly to the infinite series, we usually abbreviate 35 00:01:47,440 --> 00:01:51,400 that simply by saying, the infinite series is uniformly 36 00:01:51,400 --> 00:01:52,610 convergent. 37 00:01:52,610 --> 00:01:53,880 Let me just say that one more time. 38 00:01:53,880 --> 00:01:57,440 If I say that the series is uniformly convergent, that's 39 00:01:57,440 --> 00:02:00,890 an abbreviation for saying that the sequence of partial 40 00:02:00,890 --> 00:02:03,950 sums converges uniformly to the series. 41 00:02:03,950 --> 00:02:06,880 But at any rate, let me now go over the so-called 42 00:02:06,880 --> 00:02:08,780 Weierstrauss M-test with you. 43 00:02:08,780 --> 00:02:10,250 It's a rather simple test. 44 00:02:10,250 --> 00:02:12,950 I will give you both the proof of the test and some 45 00:02:12,950 --> 00:02:14,450 applications of it. 46 00:02:14,450 --> 00:02:16,160 The test simply says this-- 47 00:02:16,160 --> 00:02:17,990 and this is where the name M-test comes from. 48 00:02:17,990 --> 00:02:19,570 See, because you put an 'M' in over here. 49 00:02:19,570 --> 00:02:21,530 If I put a 'C' over here, they probably would have called it 50 00:02:21,530 --> 00:02:23,040 the Weierstrass C-test. 51 00:02:23,040 --> 00:02:24,400 But the idea is this. 52 00:02:24,400 --> 00:02:27,790 Suppose that the series, summation 'n' goes from 1 to 53 00:02:27,790 --> 00:02:30,740 infinity, capital 'M sub n', is a 54 00:02:30,740 --> 00:02:34,190 positive convergent series-- 55 00:02:34,190 --> 00:02:36,540 this is going to be sort of like the comparison test-- 56 00:02:36,540 --> 00:02:41,010 and that the absolute value of ''f sub n' of x' is less than 57 00:02:41,010 --> 00:02:45,300 or equal to 'M sub n' for each 'n' and for each 'x' in some 58 00:02:45,300 --> 00:02:47,540 interval from 'a' to 'b'. 59 00:02:47,540 --> 00:02:51,860 Then if this condition is obeyed, the sequence of 60 00:02:51,860 --> 00:02:53,760 partial sums-- 61 00:02:53,760 --> 00:02:56,980 some 'k' goes from 1 to 'n', of ''f sub k' of x'-- 62 00:02:56,980 --> 00:03:00,720 converges uniformly to its limit function, namely what we 63 00:03:00,720 --> 00:03:04,065 call the series 'n' goes from 1 to infinity, ''f sub n' of 64 00:03:04,065 --> 00:03:06,990 x', on the interval from 'a' to 'b'. 65 00:03:06,990 --> 00:03:10,670 See, in other words, if I can get this kind of a bound on 66 00:03:10,670 --> 00:03:15,110 what's going on, where these form the terms of a positive 67 00:03:15,110 --> 00:03:21,250 convergent series, then this series converges uniformly. 68 00:03:21,250 --> 00:03:23,820 Now, again, let me outline the proof. 69 00:03:23,820 --> 00:03:27,340 Once again, let me remind you all of these proofs are done 70 00:03:27,340 --> 00:03:33,340 rigorously and also with supplementary wording in the 71 00:03:33,340 --> 00:03:36,480 notes, so that you can pick this up again if you missed 72 00:03:36,480 --> 00:03:38,580 the high points in the lecture. 73 00:03:38,580 --> 00:03:42,310 You see, to prove convergence in the first place, I 74 00:03:42,310 --> 00:03:43,460 have to show what? 75 00:03:43,460 --> 00:03:47,760 That the sequence of partial sums can be made to differ 76 00:03:47,760 --> 00:03:51,510 from this limit, which I call 'f of x', by as small an 77 00:03:51,510 --> 00:03:55,030 amount as I want, just by taking 'n' sufficiently large. 78 00:03:55,030 --> 00:03:57,680 And the way I work this thing is I say, OK, let's take a 79 00:03:57,680 --> 00:03:59,350 look at this difference. 80 00:03:59,350 --> 00:04:03,660 Well, remembering that another name for 'f of x' is the limit 81 00:04:03,660 --> 00:04:06,700 of this thing as 'k' goes from one to infinity, I just 82 00:04:06,700 --> 00:04:11,070 substitute this in place of 'f of x'. 83 00:04:11,070 --> 00:04:12,570 I then get this. 84 00:04:12,570 --> 00:04:16,670 And now you see, if I subtract the first 'n' terms from this 85 00:04:16,670 --> 00:04:20,300 infinite series, what's left is all the terms in the 'n 86 00:04:20,300 --> 00:04:23,280 plus first' on. 87 00:04:23,280 --> 00:04:26,450 In other words, that this is in turn equal to the absolute 88 00:04:26,450 --> 00:04:29,880 value of the summation, 'k' goes from 'n plus 1' to 89 00:04:29,880 --> 00:04:32,610 infinity, ''f sub k' of x'. 90 00:04:32,610 --> 00:04:36,590 Again, don't be upset by the 'k's replacing the 'n'. 91 00:04:36,590 --> 00:04:39,070 Remember that 'k' is a dummy variable. 92 00:04:39,070 --> 00:04:42,390 The hardship would be that if I use the subscript 'n' over 93 00:04:42,390 --> 00:04:44,720 here, I would have a logical contradiction. 94 00:04:44,720 --> 00:04:49,020 How can you say 'n' goes from 'n plus 1' to infinity? 'n' 95 00:04:49,020 --> 00:04:51,510 can't equal 'n plus 1', so I just change 96 00:04:51,510 --> 00:04:52,600 the subscript here. 97 00:04:52,600 --> 00:04:56,600 At any rate, if I now invoke the triangle inequality for 98 00:04:56,600 --> 00:04:57,680 absolute values-- 99 00:04:57,680 --> 00:05:01,020 namely, that the absolute value of a sum is less than or 100 00:05:01,020 --> 00:05:03,300 equal to the sum of the absolute values-- 101 00:05:03,300 --> 00:05:06,530 I then get that the thing that I'm investigating is less than 102 00:05:06,530 --> 00:05:10,490 or equal to the sum, 'k' goes from 'n plus 1' to infinity, 103 00:05:10,490 --> 00:05:13,530 absolute value ''f sub k' of x'. 104 00:05:13,530 --> 00:05:16,260 Now, remember what the Weierstrass condition was. 105 00:05:16,260 --> 00:05:19,920 It was that for each 'k', the absolute value of ''f sub k' 106 00:05:19,920 --> 00:05:22,810 of x', for all 'x' in my interval, was less than or 107 00:05:22,810 --> 00:05:25,230 equal to 'M sub k'. 108 00:05:25,230 --> 00:05:28,020 So I can now go from here to here. 109 00:05:28,020 --> 00:05:31,600 But what do I know about the series summation, 'M sub k'? 110 00:05:31,600 --> 00:05:34,650 I know that that's a positive convergent series. 111 00:05:34,650 --> 00:05:37,630 In particular, because it's convergent, that means that 112 00:05:37,630 --> 00:05:40,550 the tail end-- meaning from 'n plus 1' to infinity-- 113 00:05:40,550 --> 00:05:45,920 that sum can be made smaller than any prescribed epsilon, 114 00:05:45,920 --> 00:05:49,780 smaller than any given epsilon, simply by picking 'n' 115 00:05:49,780 --> 00:05:51,220 sufficiently large. 116 00:05:51,220 --> 00:05:55,040 See, in other words, I can make this difference less than 117 00:05:55,040 --> 00:05:57,790 epsilon for 'n' sufficiently large. 118 00:05:57,790 --> 00:05:59,180 And here's the key step. 119 00:05:59,180 --> 00:06:06,520 Since 'M sub k' does not depend on 'x', this inequality 120 00:06:06,520 --> 00:06:10,460 was done independently of the choice of 'x'. 121 00:06:10,460 --> 00:06:12,640 You see, in other words, not only do I have that this 122 00:06:12,640 --> 00:06:16,730 converges to this, but since this was done independently of 123 00:06:16,730 --> 00:06:19,790 the choice of 'x', this is uniform convergence. 124 00:06:19,790 --> 00:06:22,500 And maybe I should write that down, because that's important 125 00:06:22,500 --> 00:06:28,710 to remember, that this is independently of 'x'. 126 00:06:28,710 --> 00:06:32,880 That's what makes the convergence uniform. 127 00:06:32,880 --> 00:06:37,030 Well, let's continue on, I think, by means of an example 128 00:06:37,030 --> 00:06:38,900 might come in handy now. 129 00:06:38,900 --> 00:06:41,450 I think after you do any of these abstract things, I think 130 00:06:41,450 --> 00:06:44,890 a nice example shows what's coming off very nicely. 131 00:06:44,890 --> 00:06:48,480 And so I very creatively call this 'An Example'. 132 00:06:48,480 --> 00:06:50,670 Let's look at the following series. 133 00:06:50,670 --> 00:06:56,600 'cosine x' plus 'cosine '4x over 4' plus et cetera 'cosine 134 00:06:56,600 --> 00:07:01,270 ''n squared x' over 'n squared'', where 'n' is the 135 00:07:01,270 --> 00:07:02,730 number of the term over here. 136 00:07:02,730 --> 00:07:05,450 For example, the third term would be 'cosine 137 00:07:05,450 --> 00:07:08,250 9x' over 9, et cetera. 138 00:07:08,250 --> 00:07:13,520 And I now want to look at the limit function here. 139 00:07:13,520 --> 00:07:16,210 See, is this a convergent series? 140 00:07:16,210 --> 00:07:17,860 Is it absolutely convergent? 141 00:07:17,860 --> 00:07:19,330 Is it uniformly convergent? 142 00:07:19,330 --> 00:07:20,630 What is it? 143 00:07:20,630 --> 00:07:23,870 And here's how the Weierstrauss M-test works. 144 00:07:23,870 --> 00:07:28,180 We look at the absolute value of the term making up this 145 00:07:28,180 --> 00:07:29,670 particular series. 146 00:07:29,670 --> 00:07:32,760 See, the absolute value of ''cosine 'n squared x'' over 147 00:07:32,760 --> 00:07:33,380 'n squared''. 148 00:07:33,380 --> 00:07:34,610 Now, let's take a look at this. 149 00:07:34,610 --> 00:07:38,910 Since 'n' is an integer, 'n squared' is positive, so the 150 00:07:38,910 --> 00:07:42,460 absolute value of 'n squared' is still in squared. 151 00:07:42,460 --> 00:07:47,810 Since the cosine has to be between minus 1 and 1, the 152 00:07:47,810 --> 00:07:50,850 magnitude of the cosine, regardless of the value of 'n 153 00:07:50,850 --> 00:07:53,280 squared x', is less than or equal to 1. 154 00:07:53,280 --> 00:07:56,340 In particular, then, the absolute value of ''cosine 'n 155 00:07:56,340 --> 00:07:59,400 squared x'' over 'n squared'' is less than or equal to '1 156 00:07:59,400 --> 00:08:03,990 over 'n squared'' for each 'n'. 157 00:08:03,990 --> 00:08:09,610 And I might add, and all 'x', because no matter what 'x' is 158 00:08:09,610 --> 00:08:12,790 and no matter what 'n' is, 'cosine 'n squared x'' in 159 00:08:12,790 --> 00:08:15,170 magnitude cannot exceed 1. 160 00:08:15,170 --> 00:08:19,070 On the other hand, we already know, in particular by the 161 00:08:19,070 --> 00:08:23,360 integral test, that summation 'n' goes from 1 to infinity or 162 00:08:23,360 --> 00:08:26,980 0 to infinity is a positive convergent series. 163 00:08:26,980 --> 00:08:31,410 Consequently, by the Weierstrauss M-test, this 164 00:08:31,410 --> 00:08:34,360 series here is uniformly convergent, 165 00:08:34,360 --> 00:08:35,970 which again means what? 166 00:08:35,970 --> 00:08:39,500 That the sequence of partial sums, namely sum 'k' goes from 167 00:08:39,500 --> 00:08:43,380 1 to 'n', ''cosine 'k squared x'' over 'k squared'', 168 00:08:43,380 --> 00:08:47,280 converges uniformly to summation 'n' goes from 1 to 169 00:08:47,280 --> 00:08:51,120 infinity ''cosine 'n squared x'' over 'n squared''. 170 00:08:51,120 --> 00:08:52,410 That's how the test works. 171 00:08:52,410 --> 00:08:54,440 But now, from an engineering point, the 172 00:08:54,440 --> 00:08:55,850 more important question-- 173 00:08:55,850 --> 00:08:57,240 what does it mean? 174 00:08:57,240 --> 00:08:58,870 How can we use it? 175 00:08:58,870 --> 00:09:02,790 Well, for example, let's suppose that in working some 176 00:09:02,790 --> 00:09:07,470 particular applied problem, one had to integrate this 177 00:09:07,470 --> 00:09:12,510 particular series, 'cosine x' plus ''cosine 4x' over 4' plus 178 00:09:12,510 --> 00:09:16,140 ''cosine 9x' over 9', et cetera, the infinite series 179 00:09:16,140 --> 00:09:20,440 from 0 to some value 't'. 180 00:09:20,440 --> 00:09:21,410 Now here's the key point. 181 00:09:21,410 --> 00:09:23,500 And by the way, I just write it this way-- 182 00:09:23,500 --> 00:09:26,570 to get used to the ominous notation, notice that what 183 00:09:26,570 --> 00:09:29,120 looks a little bit ominous here is just another way of 184 00:09:29,120 --> 00:09:31,760 writing out, longhand, what the terms inside the 185 00:09:31,760 --> 00:09:33,250 parentheses are. 186 00:09:33,250 --> 00:09:34,500 The idea is this. 187 00:09:34,500 --> 00:09:36,070 Notice what this says-- 188 00:09:36,070 --> 00:09:40,060 this says first compute the sum and then integrate. 189 00:09:40,060 --> 00:09:44,020 For example, we might not know what limit this sum 190 00:09:44,020 --> 00:09:45,150 approaches. 191 00:09:45,150 --> 00:09:48,270 Or if we do know it, it may be a particularly complicated 192 00:09:48,270 --> 00:09:49,880 thing to write down. 193 00:09:49,880 --> 00:09:52,640 You see, in other words, if we follow the problem in the 194 00:09:52,640 --> 00:09:56,830 order that the operations are given, we must first sum this 195 00:09:56,830 --> 00:09:59,510 series and then integrate it. 196 00:09:59,510 --> 00:10:01,370 But what have we already shown? 197 00:10:01,370 --> 00:10:04,300 We've already shown that this series is uniformly 198 00:10:04,300 --> 00:10:08,160 convergent, but for a uniform convergent series, we saw last 199 00:10:08,160 --> 00:10:11,720 time that you can interchange the order of summation and 200 00:10:11,720 --> 00:10:13,060 integration. 201 00:10:13,060 --> 00:10:17,200 In other words, by uniform convergence, what I can now do 202 00:10:17,200 --> 00:10:21,370 is integrate this thing here, term by term. 203 00:10:21,370 --> 00:10:24,960 See, 'sine x' plus ''sine 4x' over 16'. 204 00:10:24,960 --> 00:10:27,530 Just the usual technique of integration. 205 00:10:27,530 --> 00:10:30,740 And now you see I can integrate first and then 206 00:10:30,740 --> 00:10:31,890 compute the limit. 207 00:10:31,890 --> 00:10:34,170 In other words, I can say, look, this integral is just 208 00:10:34,170 --> 00:10:37,920 'sine t' plus ''sine 4t' over 16' plus et 209 00:10:37,920 --> 00:10:39,720 cetera, and so forth. 210 00:10:39,720 --> 00:10:42,520 Again, if you want to see this from the abstract point of 211 00:10:42,520 --> 00:10:47,020 view, what I'm saying is that by uniform convergence, I can 212 00:10:47,020 --> 00:10:49,170 switch these two symbols. 213 00:10:49,170 --> 00:10:52,460 Now, the important point is that to integrate ''cosine 'n 214 00:10:52,460 --> 00:10:54,440 squared x'' over 'n squared''-- 215 00:10:54,440 --> 00:10:55,660 remember, 'n' is a constant here. 216 00:10:55,660 --> 00:10:57,600 This is a very easy thing to integrate. 217 00:10:57,600 --> 00:10:59,270 In fact, it comes out to be what? 218 00:10:59,270 --> 00:11:02,470 ''Sine 'n squared x'' over 'n to the fourth''. 219 00:11:02,470 --> 00:11:04,960 You see the 'n squared' comes down as an 'n to the fourth'. 220 00:11:04,960 --> 00:11:08,450 And when I evaluate that between 0 and 1, all I wind up 221 00:11:08,450 --> 00:11:12,400 with is summation 'n' goes from one to infinity, ''sine 222 00:11:12,400 --> 00:11:14,910 'n squared t'' over 'n to the fourth''. 223 00:11:14,910 --> 00:11:19,050 And this is a perfectly good, bona fide convergent series. 224 00:11:19,050 --> 00:11:22,320 And, in real life, you see I can approximate this to any 225 00:11:22,320 --> 00:11:25,790 degree of accuracy that I want just by going out far enough 226 00:11:25,790 --> 00:11:28,490 before I chop off the remaining terms. 227 00:11:28,490 --> 00:11:31,860 In other words, the beauty in this case with series is that 228 00:11:31,860 --> 00:11:35,920 when they converge uniformly and you have to integrate the 229 00:11:35,920 --> 00:11:38,840 limit function, you can integrate the individual 230 00:11:38,840 --> 00:11:42,940 member of the sequence first and then take the limit as n 231 00:11:42,940 --> 00:11:44,630 goes to infinity. 232 00:11:44,630 --> 00:11:45,380 OK. 233 00:11:45,380 --> 00:11:46,720 So far so good. 234 00:11:46,720 --> 00:11:50,640 Let's apply this now, in particular, to power series. 235 00:11:50,640 --> 00:11:53,400 And again, let me keep reminding you, all of this 236 00:11:53,400 --> 00:11:57,240 stuff is done much more slowly in the supplementary notes. 237 00:11:57,240 --> 00:12:00,350 My main reason for repeating what's in the supplementary 238 00:12:00,350 --> 00:12:04,140 notes is I think this material is both difficult enough and 239 00:12:04,140 --> 00:12:07,770 new enough so I think you should hear parts of it before 240 00:12:07,770 --> 00:12:09,470 you're sent out on your own to read it. 241 00:12:09,470 --> 00:12:12,680 Hopefully my words will sound familiar to you as you hear 242 00:12:12,680 --> 00:12:15,335 them repeated in the supplementary notes. 243 00:12:15,335 --> 00:12:17,930 244 00:12:17,930 --> 00:12:22,520 Let's suppose that a power series summation, 'a n', 'x to 245 00:12:22,520 --> 00:12:25,550 the n', as 'n' goes from 0 to infinity, converges for the 246 00:12:25,550 --> 00:12:27,570 absolute value of 'x' less than 'R'. 247 00:12:27,570 --> 00:12:30,850 And let me pick a couple of values of 'x', 'x sub 0' and 248 00:12:30,850 --> 00:12:36,410 'x sub 1', such that 'x sub 0' in magnitude is less than 'x 249 00:12:36,410 --> 00:12:38,340 sub 1' in magnitude, which in turn is 250 00:12:38,340 --> 00:12:40,190 less than 'R' in magnitude. 251 00:12:40,190 --> 00:12:42,370 To give you a hint as to what I'm going to be driving at, 252 00:12:42,370 --> 00:12:45,590 I'm going to try to prove that this series converges 253 00:12:45,590 --> 00:12:47,800 uniformly within 'R'. 254 00:12:47,800 --> 00:12:50,650 And I'm going to try to prove it by setting up the 255 00:12:50,650 --> 00:12:52,920 Weierstrauss M-test. 256 00:12:52,920 --> 00:12:55,640 And I'm going to have to compare this with a positive 257 00:12:55,640 --> 00:12:58,610 convergent series, and the positive convergent series 258 00:12:58,610 --> 00:13:01,980 that I have in mind is going to make use of the fact that 259 00:13:01,980 --> 00:13:06,360 the magnitude of 'x0' divided by the magnitude of 'x1' is a 260 00:13:06,360 --> 00:13:08,960 positive constant less than 1. 261 00:13:08,960 --> 00:13:11,160 And I'm going to set this thing up so that I can use a 262 00:13:11,160 --> 00:13:12,940 geometric series on it. 263 00:13:12,940 --> 00:13:15,610 And if that sounds confusing, let me just go through this 264 00:13:15,610 --> 00:13:18,940 now in slow motion and show you what I'm driving at here. 265 00:13:18,940 --> 00:13:23,230 See, first of all, since 'x1' is within the radius of 266 00:13:23,230 --> 00:13:27,450 convergence, that means in particular that summation 'a n 267 00:13:27,450 --> 00:13:29,980 ''x sub 1' to the n'', as 'n' goes from 0 268 00:13:29,980 --> 00:13:32,450 to infinity, converges. 269 00:13:32,450 --> 00:13:35,390 Since this converges, in particular-- the very first 270 00:13:35,390 --> 00:13:36,410 test that we learned-- 271 00:13:36,410 --> 00:13:39,990 in particular, the n-th term must go to 0. 272 00:13:39,990 --> 00:13:42,060 Remember, for a convergent series, the n-th 273 00:13:42,060 --> 00:13:43,770 term goes to 0. 274 00:13:43,770 --> 00:13:49,310 Now, since the limit of 'a n 'x1 to the n-th'' is 0, that 275 00:13:49,310 --> 00:13:53,170 means that for n large enough, this will get smaller than any 276 00:13:53,170 --> 00:13:54,110 positive constant. 277 00:13:54,110 --> 00:13:56,700 Otherwise it couldn't converge on 0. 278 00:13:56,700 --> 00:13:57,580 All right? 279 00:13:57,580 --> 00:14:01,470 So all I'm saying there, I guess, is if this is 0 and 280 00:14:01,470 --> 00:14:04,940 this is 'M', if I can't make this thing less than 'M', how 281 00:14:04,940 --> 00:14:06,110 can the limit ever be 0? 282 00:14:06,110 --> 00:14:08,440 In other words, if these things can't get past 'M', how 283 00:14:08,440 --> 00:14:10,170 could the limit be 0? 284 00:14:10,170 --> 00:14:13,490 So all I'm saying is that, given a positive 'M', I can 285 00:14:13,490 --> 00:14:16,735 always find an 'n' such that when I go out far enough-- in 286 00:14:16,735 --> 00:14:18,570 other words, when the subscript 'n' is greater than 287 00:14:18,570 --> 00:14:23,340 capital 'N', the magnitude of 'a n 'x1 to the n'' is less 288 00:14:23,340 --> 00:14:25,180 than capital 'M'. 289 00:14:25,180 --> 00:14:26,910 Now the key idea is this. 290 00:14:26,910 --> 00:14:31,910 What I'm going to do is look at summation 'a n 'x0 to the 291 00:14:31,910 --> 00:14:35,790 n'' where 'x0' is as given over here. 292 00:14:35,790 --> 00:14:38,240 Now what I'm going to try to do is set this up for the 293 00:14:38,240 --> 00:14:41,720 Weierstrauss M-test, and the way I'm going to 294 00:14:41,720 --> 00:14:43,160 do that is as follows. 295 00:14:43,160 --> 00:14:47,060 I look at the absolute value of 'a n 'x0 to the n''. 296 00:14:47,060 --> 00:14:51,590 Somehow or other, the thing I have control over is not 'a n 297 00:14:51,590 --> 00:14:56,580 'x0 to the n'', but rather 'a n 'x1 to the n''. 298 00:14:56,580 --> 00:15:00,590 And now I use that very common mathematical trick that gets 299 00:15:00,590 --> 00:15:03,940 us out of all sorts of binds, and that is that, since I 300 00:15:03,940 --> 00:15:08,010 would like 'a n 'x1 to the n'' over here, I just put it in. 301 00:15:08,010 --> 00:15:10,470 And then I must cancel it out again. 302 00:15:10,470 --> 00:15:15,460 In other words, all I do is I rewrite 'a n 'x0 to the n'' as 303 00:15:15,460 --> 00:15:20,820 'a n' times 'x0' over 'x1 to the n', times 'x1 to the n'. 304 00:15:20,820 --> 00:15:22,970 In other words, notice that this just cancels and I have 305 00:15:22,970 --> 00:15:24,730 what I started with over here. 306 00:15:24,730 --> 00:15:28,830 The point is that the magnitude of 'a n 'x1 to the 307 00:15:28,830 --> 00:15:30,350 n'' is less than 'M'. 308 00:15:30,350 --> 00:15:31,550 We already saw that. 309 00:15:31,550 --> 00:15:34,240 So what I do is I split this thing up. 310 00:15:34,240 --> 00:15:36,730 Remember, the absolute value of a product is the product of 311 00:15:36,730 --> 00:15:38,020 the absolute values. 312 00:15:38,020 --> 00:15:39,550 So I write it as what? 313 00:15:39,550 --> 00:15:43,530 The magnitude of this times this, times the magnitude of 314 00:15:43,530 --> 00:15:45,770 ''x0 over x1' to the n'. 315 00:15:45,770 --> 00:15:50,310 In other words, I guess what I should have over here-- 316 00:15:50,310 --> 00:15:51,840 this is still an equality. 317 00:15:51,840 --> 00:15:52,660 This is just rewriting. 318 00:15:52,660 --> 00:15:55,420 The absolute value of a product is a product of the 319 00:15:55,420 --> 00:15:56,510 absolute values. 320 00:15:56,510 --> 00:15:59,620 Consequently, this is the absolute value of 'a n 'x1 to 321 00:15:59,620 --> 00:16:02,400 the n'' times the absolute value of ''x0 322 00:16:02,400 --> 00:16:03,770 over x1' to the n'. 323 00:16:03,770 --> 00:16:07,840 And the key point is that the absolute value of 'a n 'x1 to 324 00:16:07,840 --> 00:16:11,780 the n'' is less than 'M' for 'n' sufficiently large. 325 00:16:11,780 --> 00:16:14,720 In other words, for 'n' sufficiently large, the 326 00:16:14,720 --> 00:16:19,220 magnitude of 'a n 'x0 to the n'' is less than or equal to 327 00:16:19,220 --> 00:16:20,670 this expression here. 328 00:16:20,670 --> 00:16:22,510 But what is this expression? 329 00:16:22,510 --> 00:16:26,690 My claim is that this is the n-th term of a convergent 330 00:16:26,690 --> 00:16:30,850 positive series, namely a convergent geometric series. 331 00:16:30,850 --> 00:16:32,700 I'm going to to show you this in more detail. 332 00:16:32,700 --> 00:16:37,620 All I'm saying is observe that the magnitude of 'x0 over x1', 333 00:16:37,620 --> 00:16:40,860 since 'x0' is less than 'x1' in magnitude, is a positive 334 00:16:40,860 --> 00:16:43,090 constant less than 1. 335 00:16:43,090 --> 00:16:44,750 Therefore, this is what? 336 00:16:44,750 --> 00:16:50,680 This is a geometric series whose ratio is 'x0 over x1'. 337 00:16:50,680 --> 00:16:53,300 In other words, as you pass from the n-th term to the 'n 338 00:16:53,300 --> 00:16:57,190 plus first' term, in each case you just multiply by 'x0 over 339 00:16:57,190 --> 00:16:59,590 x1', which is a constant. 340 00:16:59,590 --> 00:17:04,290 Therefore, by the Weierstrauss M-test, this given series is 341 00:17:04,290 --> 00:17:08,970 uniformly convergent inside the interval of convergence. 342 00:17:08,970 --> 00:17:10,170 You see? 343 00:17:10,170 --> 00:17:13,680 In other words, what this now means is that we can replace 344 00:17:13,680 --> 00:17:17,200 the condition that the series was absolutely convergent 345 00:17:17,200 --> 00:17:20,910 inside the radius of convergence by, it's uniformly 346 00:17:20,910 --> 00:17:23,540 convergent inside the radius of convergence. 347 00:17:23,540 --> 00:17:26,930 And now you see the question is, what does that help us do? 348 00:17:26,930 --> 00:17:30,200 And again, this is left in great detail for the 349 00:17:30,200 --> 00:17:33,170 supplementary notes and the exercises, but I thought that 350 00:17:33,170 --> 00:17:36,360 for a finale, we should do a rather nice practical 351 00:17:36,360 --> 00:17:37,330 application. 352 00:17:37,330 --> 00:17:42,160 In particular, a problem that we haven't tackled at all. 353 00:17:42,160 --> 00:17:44,760 We wern't able to tackle it, but a problem that we've used 354 00:17:44,760 --> 00:17:49,130 as a scapegoat many times to show, for example, why the 355 00:17:49,130 --> 00:17:52,570 First Fundamental Theorem of integral calculus was not the 356 00:17:52,570 --> 00:17:54,700 cure-all it was supposed to be, for example. 357 00:17:54,700 --> 00:17:56,280 Let me show you what example I have in mind. 358 00:17:56,280 --> 00:17:59,180 359 00:17:59,180 --> 00:18:01,120 I'm thinking of this example. 360 00:18:01,120 --> 00:18:05,130 Find the area of the region 'R', where 'R' is the region 361 00:18:05,130 --> 00:18:08,020 bounded above by our old friend 'y' equals 'e to the 362 00:18:08,020 --> 00:18:12,590 minus 'x squared'', below by the x-axis, on the left by the 363 00:18:12,590 --> 00:18:16,790 y-axis, and on the right by the line 'x' equals 1. 364 00:18:16,790 --> 00:18:18,790 You see, what we used to say before was what? 365 00:18:18,790 --> 00:18:21,830 We know by the definition of a definite integral that the 366 00:18:21,830 --> 00:18:25,170 area of the region 'R' is the integral from 0 to 1, 'e to 367 00:18:25,170 --> 00:18:27,070 the minus 'x squared'', 'dx'. 368 00:18:27,070 --> 00:18:30,230 But what we do not know explicitly is a function, 'G 369 00:18:30,230 --> 00:18:32,670 of x', for which 'G prime of x' is 'e to 370 00:18:32,670 --> 00:18:34,220 the minus 'x squared''. 371 00:18:34,220 --> 00:18:36,520 Remember, we could solve this problem by trapezoidal 372 00:18:36,520 --> 00:18:40,380 approximations and things of this type. 373 00:18:40,380 --> 00:18:43,490 We could approximate it, but we couldn't get a good bound 374 00:18:43,490 --> 00:18:45,270 on the answer very conveniently. 375 00:18:45,270 --> 00:18:49,140 What I thought I'd like to show you now is how one uses 376 00:18:49,140 --> 00:18:51,950 power series to solve this kind of a problem. 377 00:18:51,950 --> 00:18:55,310 If nothing else, this one application should be enough 378 00:18:55,310 --> 00:18:58,910 impetus to show you the power of power series. 379 00:18:58,910 --> 00:19:02,170 They are a very, very important and powerful 380 00:19:02,170 --> 00:19:03,840 analytical technique. 381 00:19:03,840 --> 00:19:06,030 Let's see how we can handle this. 382 00:19:06,030 --> 00:19:10,060 First of all, from our previous material, we already 383 00:19:10,060 --> 00:19:15,250 know that 'e to the u' is represented by '1 plus u plus 384 00:19:15,250 --> 00:19:17,930 ''u squared' over '2 factorial'' plus' et cetera, 385 00:19:17,930 --> 00:19:20,600 ''u to the n' over 'n factorial'', et cetera, for 386 00:19:20,600 --> 00:19:22,540 all real numbers 'u'. 387 00:19:22,540 --> 00:19:24,450 OK, we already know that. 388 00:19:24,450 --> 00:19:28,270 In particular, if I now think of-- see, since 'u' is 389 00:19:28,270 --> 00:19:31,610 generically just a name for any real number, since minus 390 00:19:31,610 --> 00:19:34,690 'x squared' is a real number also, I can replace 'u' by 391 00:19:34,690 --> 00:19:36,150 minus 'x squared'. 392 00:19:36,150 --> 00:19:39,740 And replacing 'u' by minus 'x squared' leads to what? 'e to 393 00:19:39,740 --> 00:19:44,570 the minus 'x squared'' is 1, minus 'x squared'. 394 00:19:44,570 --> 00:19:47,440 Now minus 'x squared', squared, is 'x to the fourth', 395 00:19:47,440 --> 00:19:49,060 over 2 factorial. 396 00:19:49,060 --> 00:19:54,990 And in general, we see when I replace 'u' by minus 'x 397 00:19:54,990 --> 00:19:59,440 squared', this becomes plus or minus 'x to the 2n' over 'n 398 00:19:59,440 --> 00:20:00,610 factorial'. 399 00:20:00,610 --> 00:20:03,980 And to handle the plus or minus, I simply put in my sign 400 00:20:03,980 --> 00:20:07,040 alternator-- that's minus 1 to the n-th power. 401 00:20:07,040 --> 00:20:10,980 And the reason I use 'n' here, rather than, say, 'n plus 1', 402 00:20:10,980 --> 00:20:14,200 is to keep in mind that in power series, this is called 403 00:20:14,200 --> 00:20:16,200 the 0-th term. 404 00:20:16,200 --> 00:20:19,910 In other words, when 'n' is 0, minus 1 to the 0 would be 405 00:20:19,910 --> 00:20:23,280 positive, and I want the first term to be positive. 406 00:20:23,280 --> 00:20:26,360 At any rate, to make a long story short-- 407 00:20:26,360 --> 00:20:27,440 shorter-- 408 00:20:27,440 --> 00:20:30,820 another way of writing this is that 'e to the minus 'x 409 00:20:30,820 --> 00:20:34,740 squared'' is summation 'n' goes from 0 to infinity, minus 410 00:20:34,740 --> 00:20:38,920 '1 to the n', 'x to the 2n' over 'n factorial'. 411 00:20:38,920 --> 00:20:42,810 Consequently, since these two are synonyms, to compute the 412 00:20:42,810 --> 00:20:46,570 integral from 0 to 1, ''e to the minus 'x squared'' dx', I 413 00:20:46,570 --> 00:20:50,610 can replace 'e to the minus 'x squared'' by the power series 414 00:20:50,610 --> 00:20:54,850 which converges to it uniformly. 415 00:20:54,850 --> 00:20:58,920 Namely, I can now write this as what? 416 00:20:58,920 --> 00:21:02,640 Integral from 0 to 1, summation 'n' goes from 0 to 417 00:21:02,640 --> 00:21:05,770 infinity, minus '1 to the n', 'x to the 2n' over 'n 418 00:21:05,770 --> 00:21:08,090 factorial' times 'dx'. 419 00:21:08,090 --> 00:21:09,510 Now notice what this says. 420 00:21:09,510 --> 00:21:13,250 This says, in the order of the given operations, that I must 421 00:21:13,250 --> 00:21:15,890 form the sum of this series first, which is not an easy 422 00:21:15,890 --> 00:21:18,840 job, and then integrate that from 0 to 1. 423 00:21:18,840 --> 00:21:24,760 But the beauty is that, by the Weierstrauss M-test, I know 424 00:21:24,760 --> 00:21:28,380 that this converges uniformly wherever it converges 425 00:21:28,380 --> 00:21:29,400 absolutely. 426 00:21:29,400 --> 00:21:31,540 I already know that it does converge absolutely 427 00:21:31,540 --> 00:21:32,540 for all real 'x'. 428 00:21:32,540 --> 00:21:35,150 At any rate, then, I now know that it's uniformly 429 00:21:35,150 --> 00:21:36,100 convergent. 430 00:21:36,100 --> 00:21:38,750 One of the beauties of the fact that this is uniformly 431 00:21:38,750 --> 00:21:41,430 convergent is I can interchange the order of 432 00:21:41,430 --> 00:21:44,860 summation and integration, which I do over here. 433 00:21:44,860 --> 00:21:46,380 But this is crucial. 434 00:21:46,380 --> 00:21:49,200 Let me make sure I underline that. 435 00:21:49,200 --> 00:21:51,990 You see, if there wasn't uniform convergence here, 436 00:21:51,990 --> 00:21:54,670 mechanically I can change the order, but I may get a 437 00:21:54,670 --> 00:21:56,020 different answer. 438 00:21:56,020 --> 00:21:58,790 But since this is uniformly convergent, these are still 439 00:21:58,790 --> 00:22:02,100 synonyms, and the beauty is, if I just look at the typical 440 00:22:02,100 --> 00:22:04,620 term that I'm trying to integrate here, that's a very 441 00:22:04,620 --> 00:22:06,020 easy thing to integrate. 442 00:22:06,020 --> 00:22:08,250 Namely, I just do what? 443 00:22:08,250 --> 00:22:10,290 Replace this by an exponent one larger-- 444 00:22:10,290 --> 00:22:11,830 that's '2n plus 1'. 445 00:22:11,830 --> 00:22:13,890 Divide through by '2n plus 1'. 446 00:22:13,890 --> 00:22:17,030 In other words, the integral from 0 to 1, ''e to the minus 447 00:22:17,030 --> 00:22:20,180 'x squared'' dx', is just this particular power series, 448 00:22:20,180 --> 00:22:24,790 namely summation 'n' goes from 0 to infinity, minus '1 to the 449 00:22:24,790 --> 00:22:30,190 n', 'x to the '2n plus 1'' over ''2n plus 1' times 'n 450 00:22:30,190 --> 00:22:31,300 factorial''. 451 00:22:31,300 --> 00:22:34,560 That must be evaluated between 0 and 1. 452 00:22:34,560 --> 00:22:37,800 Since the lower limit gives me 0 and the upper limit here 453 00:22:37,800 --> 00:22:41,500 just gives me a 1, rewriting this-- this is just what? 454 00:22:41,500 --> 00:22:45,840 Summation 'n' goes from 0 to infinity, minus '1 to the n' 455 00:22:45,840 --> 00:22:49,430 over ''2n plus 1' times 'n factorial''. 456 00:22:49,430 --> 00:22:49,790 In other words-- 457 00:22:49,790 --> 00:22:50,510 let's take a look at this. 458 00:22:50,510 --> 00:22:53,820 When 'n' is 0, this term is 1. 459 00:22:53,820 --> 00:22:59,440 When 'n' is 1, this is minus 1 over what? 460 00:22:59,440 --> 00:23:02,910 3 times 1 is 3. 461 00:23:02,910 --> 00:23:08,350 When 'n' is 2, this is 5 times 2 factorial, which is 10. 462 00:23:08,350 --> 00:23:12,250 When 'n' is 3, this is minus 1 to the third 463 00:23:12,250 --> 00:23:13,830 power, which is minus. 464 00:23:13,830 --> 00:23:17,200 This is 7 times 3 factorial. 465 00:23:17,200 --> 00:23:19,330 7 times 3 factorial is what? 466 00:23:19,330 --> 00:23:22,600 7 times 6, which is 42, et cetera. 467 00:23:22,600 --> 00:23:26,900 In other words, what we now know is that 'A sub R' is 1 468 00:23:26,900 --> 00:23:31,330 minus 1/3 plus 1/10 minus 1/42. 469 00:23:31,330 --> 00:23:35,130 And if we keep on going this way, the next term is 1/216, 470 00:23:35,130 --> 00:23:37,540 which we can compute very easily. 471 00:23:37,540 --> 00:23:38,840 What is this, by the way? 472 00:23:38,840 --> 00:23:43,410 This is a convergent alternating series. 473 00:23:43,410 --> 00:23:46,680 And as you recall from our learning exercises and some of 474 00:23:46,680 --> 00:23:49,190 the material in our supplementary notes, not only 475 00:23:49,190 --> 00:23:54,340 does this thing converge, but in such a case the error is 476 00:23:54,340 --> 00:23:58,380 never greater than the magnitude of the next term. 477 00:23:58,380 --> 00:24:02,820 So, you see, what I can do over here is just take 1, 478 00:24:02,820 --> 00:24:08,610 subtract 1/3, add 1/10, subtract 1/42, and know that 479 00:24:08,610 --> 00:24:12,450 whatever answer I get when I do this has to be exact to 480 00:24:12,450 --> 00:24:16,470 within no more of an error then 1/216. 481 00:24:16,470 --> 00:24:18,920 And in fact, if I now add on the 1/216, 482 00:24:18,920 --> 00:24:22,130 I think I get 0.748. 483 00:24:22,130 --> 00:24:25,620 And the next term is going to be something like 1/1,300, 484 00:24:25,620 --> 00:24:28,850 which means that this must be correct to within three 485 00:24:28,850 --> 00:24:30,750 decimal places. 486 00:24:30,750 --> 00:24:32,740 And this is the way the so-called error function, the 487 00:24:32,740 --> 00:24:35,510 table of error function, is actually computed. 488 00:24:35,510 --> 00:24:38,450 Now I'm going to end the material of the course right 489 00:24:38,450 --> 00:24:40,280 at this particular point. 490 00:24:40,280 --> 00:24:43,260 In other words, I will leave the remaining applications for 491 00:24:43,260 --> 00:24:45,980 the exercises and the supplementary notes. 492 00:24:45,980 --> 00:24:49,330 What I would like to do is to become subjective for a moment 493 00:24:49,330 --> 00:24:50,820 or so, if I may. 494 00:24:50,820 --> 00:24:53,450 And even though this course is quite different from a regular 495 00:24:53,450 --> 00:24:56,400 classroom course, I would like to end it like a regular 496 00:24:56,400 --> 00:24:57,230 classroom course. 497 00:24:57,230 --> 00:25:02,600 And that is to make a genuine remark to the effect that this 498 00:25:02,600 --> 00:25:05,000 course has been a pleasure for me to teach. 499 00:25:05,000 --> 00:25:07,020 It has been very enlightening. 500 00:25:07,020 --> 00:25:11,000 It has been soul-searching to both plan the lectures and the 501 00:25:11,000 --> 00:25:14,390 supplementary notes and the learning exercises, and I 502 00:25:14,390 --> 00:25:17,880 myself, to coin a cliche, am a much better person for having 503 00:25:17,880 --> 00:25:19,650 gone through all of this. 504 00:25:19,650 --> 00:25:23,650 Many other people worked hard with me on this also. 505 00:25:23,650 --> 00:25:26,620 And in particular, I would like to single out three of my 506 00:25:26,620 --> 00:25:29,000 colleagues for recognition. 507 00:25:29,000 --> 00:25:30,360 Professor Harold Mickley, who was the 508 00:25:30,360 --> 00:25:32,040 director of the center. 509 00:25:32,040 --> 00:25:36,200 My friend, John Fitch, who is the manager of the self-study 510 00:25:36,200 --> 00:25:40,730 projects and, in particular, was the director of the 511 00:25:40,730 --> 00:25:44,900 lectures, and also gave me invaluable advice in many 512 00:25:44,900 --> 00:25:47,220 places as to how to simplify the lectures and make them 513 00:25:47,220 --> 00:25:48,470 more meaningful. 514 00:25:48,470 --> 00:25:51,910 And finally, my friend Charles Patton, who is our electronics 515 00:25:51,910 --> 00:25:55,120 wizard here and kept everything going from a 516 00:25:55,120 --> 00:25:57,700 technological point of view, and gave me, also, many 517 00:25:57,700 --> 00:25:58,980 valuable pointers. 518 00:25:58,980 --> 00:26:02,210 Now, from your point of view, I am hoping that you too can 519 00:26:02,210 --> 00:26:05,510 sit back and feel content, now that the course is over. 520 00:26:05,510 --> 00:26:09,090 I'm hoping that this long trip back through Calculus 521 00:26:09,090 --> 00:26:13,290 Revisited has given you a new insight to sets, functions, 522 00:26:13,290 --> 00:26:16,770 circular functions, hyperbolic functions, differentiation and 523 00:26:16,770 --> 00:26:20,850 integration, power series, so that you can go back enriched, 524 00:26:20,850 --> 00:26:25,510 revitalized, and tackle the projects of your choice. 525 00:26:25,510 --> 00:26:28,930 I hate to see the course come to an end, but in closing, it 526 00:26:28,930 --> 00:26:32,640 was my pleasure and I hope that we will have the pleasure 527 00:26:32,640 --> 00:26:36,730 of getting together again soon to tackle, together, Part Two 528 00:26:36,730 --> 00:26:38,380 of Calculus Revisited. 529 00:26:38,380 --> 00:26:42,270 And so, until that time, so long. 530 00:26:42,270 --> 00:26:45,300 ANNOUNCER: Funding for the publication of this video was 531 00:26:45,300 --> 00:26:50,020 provided by the Gabriella and Paul Rosenbaum Foundation. 532 00:26:50,020 --> 00:26:54,190 Help OCW continue to provide free and open access to MIT 533 00:26:54,190 --> 00:26:58,390 courses by making a donation at ocw.mit.edu/donate. 534 00:26:58,390 --> 00:27:03,124