1 00:00:00,040 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,690 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 00:00:32,404 --> 00:00:33,210 HERBERT GROSS: Hi. 9 00:00:33,210 --> 00:00:35,210 Well, I guess we're in the home stretch of the course. 10 00:00:35,210 --> 00:00:39,210 We're down to our last concept, and it will take us 11 00:00:39,210 --> 00:00:43,150 essentially two lectures to cover this new concept. 12 00:00:43,150 --> 00:00:46,430 Today we'll do the concept in general and next time we'll 13 00:00:46,430 --> 00:00:50,640 apply it specifically to the concept of series. 14 00:00:50,640 --> 00:00:53,710 But the concept we want to talk about is something called 15 00:00:53,710 --> 00:00:55,420 'Uniform Convergence'. 16 00:00:55,420 --> 00:00:59,050 Let me say at the outset that this is a very subtle topic. 17 00:00:59,050 --> 00:01:00,510 It is difficult. 18 00:01:00,510 --> 00:01:03,620 It seems to be beyond the scope of our textbook, because 19 00:01:03,620 --> 00:01:04,900 it's not mentioned there. 20 00:01:04,900 --> 00:01:08,840 Consequently, I will try to give the highlights as I speak 21 00:01:08,840 --> 00:01:11,430 with you, but the supplementary notes will 22 00:01:11,430 --> 00:01:15,210 contain a more detailed explanation of the things that 23 00:01:15,210 --> 00:01:16,770 we're going to talk about. 24 00:01:16,770 --> 00:01:20,690 Now, to set the stage properly for our discussion of uniform 25 00:01:20,690 --> 00:01:24,980 convergence, I think it's wise that we at least review the 26 00:01:24,980 --> 00:01:27,670 concept of convergence in general. 27 00:01:27,670 --> 00:01:30,180 Well, let's take a look over here. 28 00:01:30,180 --> 00:01:34,320 Recall that when we write that the limit of ''f sub n' of x' 29 00:01:34,320 --> 00:01:39,360 as 'n' approaches infinity equals 'f of x' for all 'x' in 30 00:01:39,360 --> 00:01:40,980 the closed interval from 'a' to 'b'-- 31 00:01:40,980 --> 00:01:44,910 when this happens, another way of saying this is that the 32 00:01:44,910 --> 00:01:48,110 sequence 'f sub n', the sequence of functions 'f sub 33 00:01:48,110 --> 00:01:52,710 n', converges to the function 'f' on the closed interval 34 00:01:52,710 --> 00:01:54,060 from 'a' to 'b'. 35 00:01:54,060 --> 00:01:57,350 Now, again, these tend to be words unless you look at a 36 00:01:57,350 --> 00:01:58,520 specific example. 37 00:01:58,520 --> 00:02:00,210 Let's just pick one over here. 38 00:02:00,210 --> 00:02:03,630 Let ''f sub n' of x' be ''n' over '2n plus 39 00:02:03,630 --> 00:02:05,940 1'' times 'x squared'. 40 00:02:05,940 --> 00:02:10,880 Notice, of course, that the value of this particular 41 00:02:10,880 --> 00:02:14,880 number depends both on 'x' and 'n'. 42 00:02:14,880 --> 00:02:19,770 At any rate, let's pick a fixed 'x', hold it that way, 43 00:02:19,770 --> 00:02:22,240 and take the limit of ''f sub n' of x' as 44 00:02:22,240 --> 00:02:23,770 'n' approaches infinity. 45 00:02:23,770 --> 00:02:26,300 In other words, as we let 'n' approach infinity in this 46 00:02:26,300 --> 00:02:29,020 case, notice that the limit of 'n' over '2n 47 00:02:29,020 --> 00:02:31,260 plus 1' becomes 1/2. 48 00:02:31,260 --> 00:02:33,050 'x' has been chosen independently of 49 00:02:33,050 --> 00:02:34,150 the choice of 'n'. 50 00:02:34,150 --> 00:02:36,510 Consequently, the limit function in this case, 'f of 51 00:02:36,510 --> 00:02:39,880 x', is 1/2 'x squared'. 52 00:02:39,880 --> 00:02:43,150 And in this case, what we say is that the sequence of 53 00:02:43,150 --> 00:02:48,440 functions 'n 'x squared'' over '2n plus 1' converges to 1/2 54 00:02:48,440 --> 00:02:49,720 'x squared'. 55 00:02:49,720 --> 00:02:52,240 Now what does this mean more specifically? 56 00:02:52,240 --> 00:02:54,250 In other words, let's see if we can look at a few specific 57 00:02:54,250 --> 00:02:55,450 values of 'x'. 58 00:02:55,450 --> 00:02:59,390 For example, if I choose 'x' to be 2-- 59 00:02:59,390 --> 00:03:02,360 in other words, if I choose 'x' to be 2, if we look over 60 00:03:02,360 --> 00:03:06,430 here, this says that ''f sub n' of x' is '4n' 61 00:03:06,430 --> 00:03:08,490 over '2n plus 1'. 62 00:03:08,490 --> 00:03:11,550 If I now take the limit as 'n' approaches infinity, I'm going 63 00:03:11,550 --> 00:03:13,800 to wind up with what? 64 00:03:13,800 --> 00:03:18,560 'x' is replaced by 2, 2 squared is 4, 1/2 of 4 is 2. 65 00:03:18,560 --> 00:03:22,500 In other words, the limit of ''f sub n' of 2', as 'n' goes 66 00:03:22,500 --> 00:03:24,270 to infinity, is 2. 67 00:03:24,270 --> 00:03:29,270 In a similar way, if I replace 'x' by 4, 1/2 'x 68 00:03:29,270 --> 00:03:30,560 squared' becomes 8. 69 00:03:30,560 --> 00:03:33,870 And so what we have is, at the limit, as 'n' approaches 70 00:03:33,870 --> 00:03:36,790 infinity, ''f sub n' of 4' is 8. 71 00:03:36,790 --> 00:03:39,980 The key thing being that once you choose 'x', notice that 72 00:03:39,980 --> 00:03:43,870 for a fixed 'x', ''f sub n' of x' is a constant and you're 73 00:03:43,870 --> 00:03:47,690 now taking the limit of a sequence of numbers. 74 00:03:47,690 --> 00:03:49,990 At any rate, here's what the key point is. 75 00:03:49,990 --> 00:03:53,120 What does this mean by our basic definition? 76 00:03:53,120 --> 00:03:55,910 By our basic definition, it means that we can find the 77 00:03:55,910 --> 00:03:59,540 number, capital 'N sub 1', such that when 'n' is greater 78 00:03:59,540 --> 00:04:03,020 than capital 'N sub 1', the absolute value of ''f sub n' 79 00:04:03,020 --> 00:04:06,260 of 2', minus 2, is less than epsilon. 80 00:04:06,260 --> 00:04:10,170 In a similar way, this means that we can find the number 81 00:04:10,170 --> 00:04:13,400 capital 'N sub 2' such that for any 'n' greater than 82 00:04:13,400 --> 00:04:16,426 capital 'N sub 2', the absolute value of ''f sub n' 83 00:04:16,426 --> 00:04:19,680 of 4' minus 8 is also less than epsilon. 84 00:04:19,680 --> 00:04:24,060 The key point is that 'N1' and 'N2' can be different. 85 00:04:24,060 --> 00:04:27,590 In other words, you may have to go out further to make this 86 00:04:27,590 --> 00:04:30,500 difference less than epsilon than you do to make this 87 00:04:30,500 --> 00:04:32,190 difference less than epsilon. 88 00:04:32,190 --> 00:04:34,900 In other words, you see what's happening here-- and we're 89 00:04:34,900 --> 00:04:36,830 going to review this in writing in a few minutes, so 90 00:04:36,830 --> 00:04:38,310 that you'll see it in front you. 91 00:04:38,310 --> 00:04:42,350 What happens here is that you see that for different values 92 00:04:42,350 --> 00:04:45,460 of 'x', we get different values of 'n'. 93 00:04:45,460 --> 00:04:48,910 And since there are infinitely many values of 'x', it means 94 00:04:48,910 --> 00:04:51,150 that in general, we're going to be in a little bit of 95 00:04:51,150 --> 00:04:55,430 trouble trying to find one 'n' that works for everything. 96 00:04:55,430 --> 00:04:56,460 And let me show you what that means 97 00:04:56,460 --> 00:04:59,230 again, going more slowly. 98 00:04:59,230 --> 00:05:01,720 I simply call this two basic definitions. 99 00:05:01,720 --> 00:05:04,720 In other words, if we have a sequence of functions 'f sub 100 00:05:04,720 --> 00:05:07,670 n', each of which is defined on the closed interval from 101 00:05:07,670 --> 00:05:11,680 'a' to 'b', we say that that sequence of functions 102 00:05:11,680 --> 00:05:13,450 converges point-wise-- 103 00:05:13,450 --> 00:05:14,910 that means number by number-- 104 00:05:14,910 --> 00:05:17,370 to 'f' on [a, b] 105 00:05:17,370 --> 00:05:21,170 if this limit, ''f sub n' of x' as 'n' approaches infinity, 106 00:05:21,170 --> 00:05:25,340 equals 'f of x' for each 'x' in [a,b]. 107 00:05:25,340 --> 00:05:27,930 In other words, given epsilon greater than 0, we can find 108 00:05:27,930 --> 00:05:31,550 'N1' such that 'n' greater than 'N1' implies that the 109 00:05:31,550 --> 00:05:35,900 absolute value of ''f sub n' of x1' minus 'f of x1' is less 110 00:05:35,900 --> 00:05:39,720 than epsilon for a given 'x1' in [a,b]. 111 00:05:39,720 --> 00:05:43,860 In general, the choice of 'N1' depends on the choice of 'x1', 112 00:05:43,860 --> 00:05:47,450 and there are infinitely many such choices to make in [a,b]. 113 00:05:47,450 --> 00:05:50,080 Now the key point is this, and this is where uniform 114 00:05:50,080 --> 00:05:51,740 convergence comes in. 115 00:05:51,740 --> 00:05:56,130 If we can find one 'N' such that whenever little 'n' is 116 00:05:56,130 --> 00:06:00,090 greater than that capital 'N', ''f sub n' of x' minus 'f of 117 00:06:00,090 --> 00:06:07,320 x' in absolute value is less than epsilon for every 'x' in 118 00:06:07,320 --> 00:06:10,000 the closed interval, then we say that the 119 00:06:10,000 --> 00:06:12,530 convergence is uniform. 120 00:06:12,530 --> 00:06:16,220 In other words, if we can find one 'N' that makes that 121 00:06:16,220 --> 00:06:19,960 difference less than epsilon for the entire interval, then 122 00:06:19,960 --> 00:06:22,970 we call the convergence uniform. 123 00:06:22,970 --> 00:06:26,070 Now, you see, convergence in general is a tough topic. 124 00:06:26,070 --> 00:06:30,240 In particular, uniform convergence may seem even more 125 00:06:30,240 --> 00:06:33,620 remote, and therefore what I'd like to do now is-- saving the 126 00:06:33,620 --> 00:06:36,490 formal proofs for the supplementary notes, let me 127 00:06:36,490 --> 00:06:40,860 show you pictorially just what the concept of uniform 128 00:06:40,860 --> 00:06:42,200 convergence really is. 129 00:06:45,570 --> 00:06:48,810 So let me give you a pictorial representation. 130 00:06:48,810 --> 00:06:53,020 Let's suppose I have the curve 'y' equals 'f of x'. 131 00:06:53,020 --> 00:06:57,580 Now, to be within epsilon of 'f of x' means I retrace this 132 00:06:57,580 --> 00:07:01,650 curve displaced epsilon units above the original position, 133 00:07:01,650 --> 00:07:03,350 and epsilon units below. 134 00:07:03,350 --> 00:07:06,430 In other words, for a given an epsilon, I now draw the curve 135 00:07:06,430 --> 00:07:09,710 'y' equals ''f of x' plus epsilon' and 'y' equals ''f of 136 00:07:09,710 --> 00:07:11,730 x' minus epsilon'. 137 00:07:11,730 --> 00:07:15,390 Now, what uniform convergence means is this, that for this 138 00:07:15,390 --> 00:07:19,630 given epsilon, I can find a capital 'N' such that whenever 139 00:07:19,630 --> 00:07:23,290 'n' is greater than capital 'N', the curve 'y' equals ''f 140 00:07:23,290 --> 00:07:26,190 sub n' of x' lies in this shaded region. 141 00:07:26,190 --> 00:07:29,960 In other words, it can bounce around all over, but it can't 142 00:07:29,960 --> 00:07:31,180 get outside of this region. 143 00:07:31,180 --> 00:07:35,270 In other words, once I'm far enough out of my sequence, all 144 00:07:35,270 --> 00:07:41,120 of the curves lie in this particular region. 145 00:07:41,120 --> 00:07:43,590 Now, of course, the question is what does this all mean? 146 00:07:43,590 --> 00:07:44,820 And the answer is-- well, look. 147 00:07:44,820 --> 00:07:48,510 Let's take epsilon to be, and I put this in quotation marks, 148 00:07:48,510 --> 00:07:52,020 "very, very small." Let's take epsilon, for example, to be so 149 00:07:52,020 --> 00:07:55,290 small that it's within the thickness of our chalk. 150 00:07:55,290 --> 00:07:56,540 If I now do this-- 151 00:07:58,900 --> 00:08:01,540 see, I draw the curve 'y' equals 'f of x'. 152 00:08:01,540 --> 00:08:04,930 Notice now that the thickness of my curve itself is the band 153 00:08:04,930 --> 00:08:06,710 width 2 epsilon. 154 00:08:06,710 --> 00:08:10,410 All I'm saying is that for this very, very small epsilon, 155 00:08:10,410 --> 00:08:14,140 when 'n' is sufficiently large, the curve 'y' equals 156 00:08:14,140 --> 00:08:18,720 ''f sub n' of x' appears to lie inside of this curve. 157 00:08:18,720 --> 00:08:21,590 You see, in other words, what you're saying is that for a 158 00:08:21,590 --> 00:08:24,930 large enough 'n' and small enough epsilon-- 159 00:08:24,930 --> 00:08:28,330 loosely speaking what you saying is that 'y' equals ''f 160 00:08:28,330 --> 00:08:31,620 sub n' of x' looks like 'y' equals 'f of x', for a 161 00:08:31,620 --> 00:08:33,510 sufficiently large values of 'n'. 162 00:08:33,510 --> 00:08:38,220 In other words, it appears that we can't really tell the 163 00:08:38,220 --> 00:08:42,750 n-th curve in the sequence from the limit function. 164 00:08:42,750 --> 00:08:44,430 And I want to make a few key observations 165 00:08:44,430 --> 00:08:45,850 about what that means. 166 00:08:45,850 --> 00:08:48,370 I've written the whole thing out on the blackboard so that 167 00:08:48,370 --> 00:08:51,860 you can see this after I say it, but what I want you to see 168 00:08:51,860 --> 00:08:55,100 is, can you begin to get the feeling that with this kind of 169 00:08:55,100 --> 00:08:56,970 a condition, for example-- 170 00:08:56,970 --> 00:09:01,260 if each 'f sub n' happens to be continuous, in other words, 171 00:09:01,260 --> 00:09:05,200 if each member of my sequence is unbroken, then the limit 172 00:09:05,200 --> 00:09:07,430 function itself must also be unbroken. 173 00:09:07,430 --> 00:09:10,970 Because you see I can squeeze this thing down to such a 174 00:09:10,970 --> 00:09:14,950 narrow width that there's no room for a break in here. 175 00:09:14,950 --> 00:09:19,700 Also notice that if the curve 'y' equals ''f sub n' of x' is 176 00:09:19,700 --> 00:09:21,490 caught inside this curve-- 177 00:09:21,490 --> 00:09:24,840 if I, for example, were computing the area of the 178 00:09:24,840 --> 00:09:26,010 region 'R'-- 179 00:09:26,010 --> 00:09:28,970 for a large enough 'n', I couldn't tell the difference 180 00:09:28,970 --> 00:09:33,470 in area if I use 'y' equals 'f of x' for my top curve, or 181 00:09:33,470 --> 00:09:36,730 whether I use 'y' equals ''f sub n' of x'. 182 00:09:36,730 --> 00:09:40,560 Well, keep that in mind, and all I'm saying is this, that 183 00:09:40,560 --> 00:09:43,810 from our picture, it should seem clear that if the 184 00:09:43,810 --> 00:09:47,670 sequence 'f sub n' converges uniformly to 'f' on [a, b], 185 00:09:47,670 --> 00:09:51,420 and if each 'f sub n' is continuous on [a,b], then-- 186 00:09:51,420 --> 00:09:53,140 and this is a fundamental result. 187 00:09:53,140 --> 00:09:57,380 Fundamental result one, 'f' is also continuous on [a, b]. 188 00:10:01,230 --> 00:10:02,240 Now you say, look. 189 00:10:02,240 --> 00:10:04,300 That's what you'd expect, isn't it, if every member of 190 00:10:04,300 --> 00:10:05,620 the sequence is continuous? 191 00:10:05,620 --> 00:10:08,570 Why shouldn't the limit function also be continuous? 192 00:10:08,570 --> 00:10:12,030 The point is, well, maybe that's what you expect, but 193 00:10:12,030 --> 00:10:13,500 note this-- 194 00:10:13,500 --> 00:10:17,070 in one of our earlier lectures, we already saw that 195 00:10:17,070 --> 00:10:20,640 if we only had point-wise convergence, this did not need 196 00:10:20,640 --> 00:10:21,640 to be true. 197 00:10:21,640 --> 00:10:25,240 In particular, recall our example in which we defined 198 00:10:25,240 --> 00:10:29,480 ''f sub n' of x' to be 'x to the n', where the domain of 199 00:10:29,480 --> 00:10:32,650 'f' was the closed interval from 0 to 1. 200 00:10:32,650 --> 00:10:34,330 Remember what happened in that case? 201 00:10:34,330 --> 00:10:36,960 Each of these 'f sub n's is continuous. 202 00:10:36,960 --> 00:10:42,060 But the limit function, you may recall, is what? 203 00:10:42,060 --> 00:10:47,990 It's 0 if 'x' is less than 1, and 1 if 'x' equals 1. 204 00:10:47,990 --> 00:10:51,540 In other words, the limit function was discontinuous at 205 00:10:51,540 --> 00:10:52,890 'x' equals 1. 206 00:10:52,890 --> 00:10:55,360 By the way, you might like to see what this means from 207 00:10:55,360 --> 00:10:56,870 another point of view. 208 00:10:56,870 --> 00:10:58,750 And let me show you what it does mean from 209 00:10:58,750 --> 00:10:59,910 another point of view. 210 00:10:59,910 --> 00:11:04,190 Since 'f' continuous at 'x' equals 'x sub 1' means that 211 00:11:04,190 --> 00:11:08,110 the limit of 'f of x' as 'x' approaches 'x1' is 'f of x1', 212 00:11:08,110 --> 00:11:13,620 and since 'f of x' itself, by definition, is the limit of 213 00:11:13,620 --> 00:11:22,360 ''f sub n' of x' as 'n' approaches infinity, we may 214 00:11:22,360 --> 00:11:26,550 rewrite our first condition in the form-- what? 215 00:11:26,550 --> 00:11:32,100 We may rewrite our first equation, this equation here, 216 00:11:32,100 --> 00:11:33,320 in what form? 217 00:11:33,320 --> 00:11:38,820 Limit as 'x' approaches 'x1' of limit 'n' approaches 218 00:11:38,820 --> 00:11:41,270 infinity, ''f sub n' of x'. 219 00:11:41,270 --> 00:11:44,470 And that equals the limit as 'n' approaches infinity, ''f 220 00:11:44,470 --> 00:11:46,660 sub n' of x1'. 221 00:11:46,660 --> 00:11:50,330 Now keep in mind that since each 'f sub n' is given to be 222 00:11:50,330 --> 00:11:55,510 continuous, by definition of continuity ''f sub n' of x1' 223 00:11:55,510 --> 00:11:59,280 is the same as saying the limit as 'x' approaches 'x1', 224 00:11:59,280 --> 00:12:00,690 ''f sub n' of x'. 225 00:12:00,690 --> 00:12:05,060 The point I'm making is, if you now put this together with 226 00:12:05,060 --> 00:12:09,410 this, it says the rather remarkable thing that unless 227 00:12:09,410 --> 00:12:13,860 you have uniform convergence when you interchange the order 228 00:12:13,860 --> 00:12:16,770 in which you take the limits over here, you may very well 229 00:12:16,770 --> 00:12:18,200 get a different answer. 230 00:12:18,200 --> 00:12:20,920 In other words, when you're dealing with non-uniform 231 00:12:20,920 --> 00:12:25,550 convergence, you must be very, very careful to perform every 232 00:12:25,550 --> 00:12:27,650 operation in the given order. 233 00:12:27,650 --> 00:12:29,210 What we're saying is what? 234 00:12:29,210 --> 00:12:31,090 This will give you an answer. 235 00:12:31,090 --> 00:12:32,710 This will give you an answer. 236 00:12:32,710 --> 00:12:36,140 But if the convergence is not uniform, the answers may be 237 00:12:36,140 --> 00:12:39,730 different, and consequently by changing the order you destroy 238 00:12:39,730 --> 00:12:42,000 the whole physical meaning of the problem. 239 00:12:42,000 --> 00:12:44,590 Well, again, that's reemphasized in the 240 00:12:44,590 --> 00:12:45,890 supplementary notes. 241 00:12:45,890 --> 00:12:47,940 Let me continue on here. 242 00:12:47,940 --> 00:12:51,510 Let me tell you another interesting property of 243 00:12:51,510 --> 00:12:54,510 uniform convergence. 244 00:12:54,510 --> 00:12:58,420 Suppose the sequence ''f sub n' of x' converges uniformly 245 00:12:58,420 --> 00:13:00,590 to 'f of x' on [a, b]. 246 00:13:00,590 --> 00:13:03,220 The point that's rather interesting is that you can 247 00:13:03,220 --> 00:13:06,530 reverse the order of integration and taking the 248 00:13:06,530 --> 00:13:08,210 limit in this particular case. 249 00:13:08,210 --> 00:13:12,060 In other words, suppose you want to compute the integral 250 00:13:12,060 --> 00:13:14,800 of the limit function from 'a' to 'b'. 251 00:13:14,800 --> 00:13:19,790 What you can do instead is compute the integral of the 252 00:13:19,790 --> 00:13:23,440 n-th number of your sequence, and then take the limit as 'n' 253 00:13:23,440 --> 00:13:24,700 goes to infinity. 254 00:13:24,700 --> 00:13:27,820 In other words, rewriting this, it says that if you have 255 00:13:27,820 --> 00:13:32,080 uniform convergence, you can take the limit inside the 256 00:13:32,080 --> 00:13:33,470 integral sign. 257 00:13:33,470 --> 00:13:37,740 And again, these results are proven in our 258 00:13:37,740 --> 00:13:39,400 supplementary notes. 259 00:13:39,400 --> 00:13:43,040 We also say a few words about corresponding results for 260 00:13:43,040 --> 00:13:45,750 differentiation in our supplementary notes. 261 00:13:45,750 --> 00:13:50,100 And I should point out that differentiation is a far more 262 00:13:50,100 --> 00:13:52,000 subtle thing than integration. 263 00:13:52,000 --> 00:13:55,080 See, remember that for integration, all you need is 264 00:13:55,080 --> 00:13:56,520 continuity. 265 00:13:56,520 --> 00:13:59,340 For differentiation, you need smoothness. 266 00:13:59,340 --> 00:14:04,420 The point is that as you put a thin band around the function 267 00:14:04,420 --> 00:14:07,760 'y' equals 'f of x' when you have uniform convergence, 268 00:14:07,760 --> 00:14:10,410 that's enough to make sure that the limit function must 269 00:14:10,410 --> 00:14:13,200 be continuous if each of the members in the sequence is 270 00:14:13,200 --> 00:14:13,960 continuous. 271 00:14:13,960 --> 00:14:16,790 But without going into the details of this thing, it does 272 00:14:16,790 --> 00:14:20,000 turn out that for the degree of smoothness that you need, 273 00:14:20,000 --> 00:14:22,510 these things can jump around enough so that for 274 00:14:22,510 --> 00:14:25,860 differentiation, we do have to be a little bit more careful. 275 00:14:25,860 --> 00:14:28,650 Rather than to becloud the issue, I will stick with 276 00:14:28,650 --> 00:14:31,810 integration topics for our lecture today. 277 00:14:31,810 --> 00:14:34,880 Now, the other thing that I want to mention is, again, in 278 00:14:34,880 --> 00:14:38,830 terms of doing what comes naturally, I think we are 279 00:14:38,830 --> 00:14:41,850 tempted to look at something like this and say, look, all I 280 00:14:41,850 --> 00:14:43,270 did was bring the limit inside. 281 00:14:43,270 --> 00:14:45,360 Why can't I do that? 282 00:14:45,360 --> 00:14:48,120 And instead of saying, look, you can't, I think the best 283 00:14:48,120 --> 00:14:50,290 thing to show is that when you don't have uniform 284 00:14:50,290 --> 00:14:52,570 convergence, you get two different answers. 285 00:14:52,570 --> 00:14:53,980 Again, the idea being what? 286 00:14:53,980 --> 00:14:56,270 We are not saying that you can't do this. 287 00:14:56,270 --> 00:14:58,920 We are not saying that you can't compute this. 288 00:14:58,920 --> 00:15:01,530 All we're saying is that if the convergence is not 289 00:15:01,530 --> 00:15:08,050 uniform, these two expressions may very well 290 00:15:08,050 --> 00:15:10,880 name different numbers. 291 00:15:10,880 --> 00:15:13,180 Now, to show you what I have in mind here, let me give you 292 00:15:13,180 --> 00:15:15,110 an example. 293 00:15:15,110 --> 00:15:18,180 See, to show you that 2 need not be true if the convergence 294 00:15:18,180 --> 00:15:21,630 is not uniform, consider the following example. 295 00:15:21,630 --> 00:15:25,290 Now, in the supplementary notes, I repeat this example 296 00:15:25,290 --> 00:15:28,270 both the way I have it on the board and also from an 297 00:15:28,270 --> 00:15:30,880 algebraic point of view, without using the pictures. 298 00:15:30,880 --> 00:15:33,700 But in terms of the picture, here's what we do. 299 00:15:33,700 --> 00:15:37,020 We define a function on the closed interval from 0 to 2, 300 00:15:37,020 --> 00:15:40,160 which I'll call 'f sub n', as follows. 301 00:15:40,160 --> 00:15:45,460 For a given 'n', I will locate the points '1/n' and '2/n'. 302 00:15:45,460 --> 00:15:49,900 For example, if 'n' happened to be 50, this would be 1/50 303 00:15:49,900 --> 00:15:52,030 and this would be 2/50. 304 00:15:52,030 --> 00:15:57,570 Now what I do, is at the 'x' value '1/n', I take as the 305 00:15:57,570 --> 00:16:01,510 corresponding y-value, 'n squared'. 306 00:16:01,510 --> 00:16:05,060 And I draw the straight line that goes from the origin to 307 00:16:05,060 --> 00:16:08,110 this point, '1/n' comma 'n squared'. 308 00:16:08,110 --> 00:16:12,290 Then I draw the straight line that comes right back to the 309 00:16:12,290 --> 00:16:17,800 x-intercept, '2/n', and I finish off the curve by just 310 00:16:17,800 --> 00:16:22,180 letting it hug the x-axis till we get over to 'x' equals 2. 311 00:16:22,180 --> 00:16:24,660 I'll come back to this on the next board, to show you why I 312 00:16:24,660 --> 00:16:27,590 chose this, but let's make a few observations just to make 313 00:16:27,590 --> 00:16:30,860 sure that you understand what this function looks like. 314 00:16:30,860 --> 00:16:33,630 I'll just make a few arbitrary remarks about it. 315 00:16:33,630 --> 00:16:36,310 First of all, for each 'n', ''f sub n' of 316 00:16:36,310 --> 00:16:38,640 '1/n'' is 'n squared'. 317 00:16:38,640 --> 00:16:41,010 That's just another way of indicating a 318 00:16:41,010 --> 00:16:42,260 label for this point. 319 00:16:44,610 --> 00:16:50,990 Secondly, my claim is that for any number 'x sub 0', if 'x 320 00:16:50,990 --> 00:16:56,580 sub 0' is greater than '2/n', ''f sub n' of x0' must be 0. 321 00:16:56,580 --> 00:16:58,810 And the reason for that is quite simple. 322 00:16:58,810 --> 00:17:00,760 I'm just trying to show you how to read this picture. 323 00:17:00,760 --> 00:17:04,410 Namely, notice that as soon as 'x' gets to be as great as 324 00:17:04,410 --> 00:17:10,140 '2/n', the 'f' value is 0, because the function is 325 00:17:10,140 --> 00:17:11,960 hugging the x-axis. 326 00:17:11,960 --> 00:17:14,579 And just as a final observation, notice that when 327 00:17:14,579 --> 00:17:19,420 'x sub 0' is 0, ''f sub n' of 0' is 0 for every 'n', meaning 328 00:17:19,420 --> 00:17:23,660 that every member of my family of functions goes through this 329 00:17:23,660 --> 00:17:24,520 particular point. 330 00:17:24,520 --> 00:17:26,020 In other words, let me just label this. 331 00:17:26,020 --> 00:17:31,850 This is 'y' equals ''f sub n' of x'. 332 00:17:31,850 --> 00:17:35,110 Well, by the way if ''f sub n' of 0' is 0 for each 'n', in 333 00:17:35,110 --> 00:17:37,810 particular the limit of ''f sub n' of 0' as 'n' approaches 334 00:17:37,810 --> 00:17:39,920 infinity is 0. 335 00:17:39,920 --> 00:17:43,080 What happens if we pick a non-zero value? 336 00:17:43,080 --> 00:17:47,490 For example, suppose I pick 'x0' to be greater than 0 but 337 00:17:47,490 --> 00:17:49,190 less than or equal to 2? 338 00:17:49,190 --> 00:17:54,030 The key point is this, that since the limit of '2/n' as 339 00:17:54,030 --> 00:17:58,490 'n' approaches infinity is 0, given a value of 'x0' which is 340 00:17:58,490 --> 00:18:02,720 not 0, I can find the capital 'N' such that when 'n' is 341 00:18:02,720 --> 00:18:06,660 greater than capital 'N', '2/n' is less than 'x0'. 342 00:18:06,660 --> 00:18:10,020 In other words, if 'x0' is greater than 0, and '2/n' 343 00:18:10,020 --> 00:18:13,860 approaches 0, for large enough values of 'n', '2/n' 344 00:18:13,860 --> 00:18:15,340 be less than 'x0'. 345 00:18:15,340 --> 00:18:18,210 In particular, when that happens, if we couple this 346 00:18:18,210 --> 00:18:20,010 with our earlier observation-- 347 00:18:20,010 --> 00:18:21,510 what earlier observation? 348 00:18:21,510 --> 00:18:24,780 Well, this one. 349 00:18:24,780 --> 00:18:27,610 If we couple that with our earlier observation, we see 350 00:18:27,610 --> 00:18:30,150 that when 'n' is greater than capital 'N', ''f sub 351 00:18:30,150 --> 00:18:32,110 n' of x0' is 0. 352 00:18:32,110 --> 00:18:36,530 Correspondingly, then, the limit of ''f sub n' of x0' as 353 00:18:36,530 --> 00:18:39,870 'n' approaches infinity, by definition, is 0. 354 00:18:39,870 --> 00:18:40,260 In other words-- 355 00:18:40,260 --> 00:18:42,650 I'm going to reinforce this later, but notice that the 356 00:18:42,650 --> 00:18:45,650 limit function here is the function which 357 00:18:45,650 --> 00:18:47,480 is identically 0. 358 00:18:47,480 --> 00:18:51,600 Now, since this may look very abstract to you, let me take a 359 00:18:51,600 --> 00:18:52,450 few minutes-- 360 00:18:52,450 --> 00:18:54,960 and I hope this doesn't insult your intelligence, but let me 361 00:18:54,960 --> 00:18:58,490 just take a few minutes and redraw this for a couple of 362 00:18:58,490 --> 00:19:01,200 different values of 'n', just so that you can see what's 363 00:19:01,200 --> 00:19:02,450 starting to happen here. 364 00:19:04,870 --> 00:19:08,730 Keep that picture in mind, and now look what this means. 365 00:19:08,730 --> 00:19:15,460 For example, when 'n' is 1, '1/n' is 1, '2/n' is 2, 'n 366 00:19:15,460 --> 00:19:16,760 squared' is 1. 367 00:19:16,760 --> 00:19:20,920 In other words, the graph 'y' equals 'f1 of x' is just this 368 00:19:20,920 --> 00:19:23,876 triangular-- 369 00:19:23,876 --> 00:19:25,480 just this. 370 00:19:25,480 --> 00:19:27,300 Why give it a name? 371 00:19:27,300 --> 00:19:30,030 Well, let's try a tougher one. 372 00:19:30,030 --> 00:19:34,100 Let's see what the member 'f sub 20' looks like. 373 00:19:34,100 --> 00:19:35,590 Recall how you draw this, now. 374 00:19:35,590 --> 00:19:38,000 With the subscript 20, what do you do? 375 00:19:38,000 --> 00:19:43,670 You come in to the point 1/20, and at that 376 00:19:43,670 --> 00:19:44,930 point, you do what? 377 00:19:44,930 --> 00:19:49,900 You locate the point 1/20 comma 'n squared'. 378 00:19:49,900 --> 00:19:52,200 In this case, it's 400. 379 00:19:52,200 --> 00:19:54,700 And I have obviously haven't drawn this to scale, but you 380 00:19:54,700 --> 00:19:55,440 now do what? 381 00:19:55,440 --> 00:19:57,060 Draw the straight line that goes from the 382 00:19:57,060 --> 00:19:59,120 origin to this point. 383 00:19:59,120 --> 00:20:01,870 Then from this point, you draw the straight line that comes 384 00:20:01,870 --> 00:20:06,190 back to the x-axis, hitting it at 'x' equals 1/10. 385 00:20:06,190 --> 00:20:08,640 And then you come across the x-axis all the way 386 00:20:08,640 --> 00:20:10,050 to 'x' equals 2. 387 00:20:10,050 --> 00:20:13,810 This would be the graph of 'y' equals ''f sub 20' of x'. 388 00:20:13,810 --> 00:20:17,010 And by the way, do you sense what's happening over here? 389 00:20:17,010 --> 00:20:23,040 See, notice that as 'n' gets very, very large, the curve 390 00:20:23,040 --> 00:20:26,650 hugs the x-axis, starting in closer and 391 00:20:26,650 --> 00:20:28,220 closer to the y-axis. 392 00:20:28,220 --> 00:20:31,580 But what happens is someplace in here, no matter how close 393 00:20:31,580 --> 00:20:35,560 'x sub 0' is to 0, there comes a very high peak. 394 00:20:35,560 --> 00:20:37,220 In fact, what is that high peak? 395 00:20:37,220 --> 00:20:38,830 It's 'n squared'. 396 00:20:38,830 --> 00:20:41,470 In other words, when this number is very close to the 397 00:20:41,470 --> 00:20:45,220 y-axis, the peak is very, very high. 398 00:20:45,220 --> 00:20:47,580 In other words, no matter how you put the squeeze on over 399 00:20:47,580 --> 00:20:52,800 here, this particular peak jumps out. 400 00:20:52,800 --> 00:20:55,830 This is why this particular sequence of functions is not 401 00:20:55,830 --> 00:20:57,070 uniformly convergent. 402 00:20:57,070 --> 00:20:59,820 Again, this is done more slowly in the notes. 403 00:20:59,820 --> 00:21:01,880 But at any rate, let me show you an interesting thing that 404 00:21:01,880 --> 00:21:03,760 happens over here. 405 00:21:03,760 --> 00:21:07,290 Let me redraw this now for a general 'n'. 406 00:21:07,290 --> 00:21:09,930 In other words, let me draw 'y' equals ''f sub n' of x' 407 00:21:09,930 --> 00:21:11,430 for any old 'n'. 408 00:21:11,430 --> 00:21:14,140 Recall what our definition was, now, especially 409 00:21:14,140 --> 00:21:16,000 with this as review. 410 00:21:16,000 --> 00:21:18,960 We locate the point '1/n' comma 'n squared'. 411 00:21:18,960 --> 00:21:20,710 We then draw the line that goes from the 412 00:21:20,710 --> 00:21:22,180 origin to that point. 413 00:21:22,180 --> 00:21:26,220 Then we draw the line that goes from that point back to 414 00:21:26,220 --> 00:21:29,410 the x-axis at the point '2/n'. 415 00:21:29,410 --> 00:21:32,300 And then we come across to 'x' equals 2. 416 00:21:32,300 --> 00:21:37,770 Let's try to visualize what the integral from 0 to 2, ''f 417 00:21:37,770 --> 00:21:39,790 sub n' of x', 'dx', means. 418 00:21:39,790 --> 00:21:43,090 After all, in a case of a continuous curve, which this 419 00:21:43,090 --> 00:21:46,250 is, isn't the definite integral interpreted just as 420 00:21:46,250 --> 00:21:47,790 the area under the curve? 421 00:21:47,790 --> 00:21:51,370 Well, you see, the curve coincides with the x-axis from 422 00:21:51,370 --> 00:21:53,380 '2/n' on to 2. 423 00:21:53,380 --> 00:21:57,860 Consequently, this triangular region which I call 'R' is the 424 00:21:57,860 --> 00:21:59,210 area under the curve. 425 00:21:59,210 --> 00:22:02,440 In other words, the integral from 0 to 2, ''f sub n' of x', 426 00:22:02,440 --> 00:22:05,720 'dx', is the area of the region 'R'. 427 00:22:05,720 --> 00:22:06,750 But here's the point. 428 00:22:06,750 --> 00:22:09,870 We can compute the area of the region 'R' very easily. 429 00:22:09,870 --> 00:22:11,780 It's a triangle, right? 430 00:22:11,780 --> 00:22:13,300 What is the area of a triangle? 431 00:22:13,300 --> 00:22:17,040 Well, it's 1/2 times the base-- 432 00:22:17,040 --> 00:22:19,410 but the base is just '2/n'-- 433 00:22:19,410 --> 00:22:20,790 times the height. 434 00:22:20,790 --> 00:22:22,900 The height is 'n squared'. 435 00:22:22,900 --> 00:22:27,150 In other words, the area of the region 'R' simply is 'n'. 436 00:22:27,150 --> 00:22:28,530 And that's rather interesting. 437 00:22:28,530 --> 00:22:32,730 In other words, for each 'n', this particular integral just 438 00:22:32,730 --> 00:22:34,770 turns out to be 'n' itself. 439 00:22:34,770 --> 00:22:37,220 That's what's interesting about this particular diagram. 440 00:22:37,220 --> 00:22:40,620 In other words, this thing rises so high that even though 441 00:22:40,620 --> 00:22:44,800 the base gets very, very small as 'n' gets large, the height 442 00:22:44,800 --> 00:22:47,980 increases so rapidly that the area under this curve, 443 00:22:47,980 --> 00:22:52,270 numerically, is always equal to 'n'. 444 00:22:52,270 --> 00:22:54,300 In fact, we can check that if you'd like. 445 00:22:54,300 --> 00:22:56,160 Come back to this particular case. 446 00:22:56,160 --> 00:22:58,280 Look at this particular triangle. 447 00:22:58,280 --> 00:23:00,510 The base is 2, the height is 1. 448 00:23:00,510 --> 00:23:04,560 The area is 1 unit. 449 00:23:04,560 --> 00:23:07,490 Look at this particular triangle. 450 00:23:07,490 --> 00:23:12,520 The base is 1/10, the height is 400. 451 00:23:12,520 --> 00:23:17,540 400 times 1/10 is 40, and half of that is 20. 452 00:23:17,540 --> 00:23:20,470 The area of this triangle is 20, which exactly 453 00:23:20,470 --> 00:23:21,590 matches this subscript. 454 00:23:21,590 --> 00:23:23,870 That's what's going to happen here all the time. 455 00:23:23,870 --> 00:23:27,640 In particular, then, if we compute the integral from 0 to 456 00:23:27,640 --> 00:23:33,810 2, 'f of n', 'x dx', and then let the limit as 'n' goes to 457 00:23:33,810 --> 00:23:36,430 infinity be computed, what do we get for an answer? 458 00:23:36,430 --> 00:23:40,450 We get that this limit is the limit of 'n' as 'n' approaches 459 00:23:40,450 --> 00:23:43,690 infinity, and that of course is infinity. 460 00:23:43,690 --> 00:23:47,050 On the other hand, suppose we bring the limit inside? 461 00:23:47,050 --> 00:23:49,960 In other words, suppose we compute this. 462 00:23:49,960 --> 00:23:53,620 Well, the point is that we have already shown that this 463 00:23:53,620 --> 00:23:57,370 is identically 0 for all 'x'. 464 00:23:57,370 --> 00:24:01,800 Consequently, this integral is the integral from 0 to 2, 0 465 00:24:01,800 --> 00:24:03,990 'dx', which is 0. 466 00:24:03,990 --> 00:24:06,760 In other words, if you first take the limit and then 467 00:24:06,760 --> 00:24:09,530 integrate, you get 0. 468 00:24:09,530 --> 00:24:12,490 On the other hand, if you first integrate and then take 469 00:24:12,490 --> 00:24:14,490 the limit, you get infinity. 470 00:24:14,490 --> 00:24:17,820 And this should be a glaring example to show you that the 471 00:24:17,820 --> 00:24:20,990 answer that you get indeed does depend on the order in 472 00:24:20,990 --> 00:24:22,620 which you do the operation. 473 00:24:22,620 --> 00:24:24,260 Again, let me emphasize-- 474 00:24:24,260 --> 00:24:26,600 which of these two is wrong? 475 00:24:26,600 --> 00:24:29,210 The answer is, neither is wrong. 476 00:24:29,210 --> 00:24:32,150 All we're saying is that if you were supposed to solve 477 00:24:32,150 --> 00:24:35,710 this problem and by mistake you solve this one, you are 478 00:24:35,710 --> 00:24:39,150 going to get a drastically different answer. 479 00:24:39,150 --> 00:24:40,230 OK. 480 00:24:40,230 --> 00:24:41,670 Let's not beat this to death. 481 00:24:41,670 --> 00:24:43,170 So far, so good. 482 00:24:43,170 --> 00:24:47,340 Let me make one more remark, namely, what does all of this 483 00:24:47,340 --> 00:24:49,550 have to do with the study of series? 484 00:24:49,550 --> 00:24:53,710 See, now we're just talking about sequences of functions. 485 00:24:53,710 --> 00:24:56,630 And you see, the answer to this question is essentially 486 00:24:56,630 --> 00:25:02,020 going to be our last lecture of the course. 487 00:25:02,020 --> 00:25:04,140 But for now, what I'd like to do is to give you 488 00:25:04,140 --> 00:25:05,420 a preview of that. 489 00:25:05,420 --> 00:25:08,580 Namely, the application of uniform convergence to series 490 00:25:08,580 --> 00:25:09,750 is the following. 491 00:25:09,750 --> 00:25:13,300 Recall that when we write summation 'n' goes from 0 to 492 00:25:13,300 --> 00:25:16,310 infinity, 'a sub n', 'x to the n', that's an 493 00:25:16,310 --> 00:25:18,070 abbreviation for what? 494 00:25:18,070 --> 00:25:23,860 A polynomial, 'k' goes from 0 to 'n', 'a sub k', 'x sub k', 495 00:25:23,860 --> 00:25:25,550 as 'n' goes to infinity. 496 00:25:25,550 --> 00:25:30,390 In other words, recall that the sum of the series is a 497 00:25:30,390 --> 00:25:34,240 limit of a sequence of partial sums, and this is the n-th 498 00:25:34,240 --> 00:25:37,020 member of that sequence of partial sums. 499 00:25:37,020 --> 00:25:40,560 Again, if the sigma notation is throwing you off, all I'm 500 00:25:40,560 --> 00:25:46,800 saying is to observe that 'a0' plus 'a1 x' plus 'a2 'x 501 00:25:46,800 --> 00:25:50,160 squared'' plus-- et cetera, et cetera, et cetera, forever, 502 00:25:50,160 --> 00:25:53,500 just represents the limit of the following sequence. 503 00:25:53,500 --> 00:25:57,940 'a0', next member is 'a0 plus a1 x'. 504 00:25:57,940 --> 00:26:02,090 Next member is 'a0' plus 'a1 x' plus 'a2 'x squared''. 505 00:26:02,090 --> 00:26:06,310 The next member is 'a0' plus 'a1 x' plus 'a2 'x squared'' 506 00:26:06,310 --> 00:26:07,960 plus 'a3 'x cubed''. 507 00:26:07,960 --> 00:26:10,860 By the way, what is each member of the sequence? 508 00:26:10,860 --> 00:26:13,290 It's a polynomial. 509 00:26:13,290 --> 00:26:16,570 And polynomials have very nice properties, 510 00:26:16,570 --> 00:26:17,960 among which are what? 511 00:26:17,960 --> 00:26:20,780 Well, a polynomial is a continuous function. 512 00:26:20,780 --> 00:26:23,850 A polynomial is an integral function, et cetera. 513 00:26:23,850 --> 00:26:27,670 The idea, therefore, is that if this sequence of partial 514 00:26:27,670 --> 00:26:35,840 sums converges uniformly to the limit function, then, for 515 00:26:35,840 --> 00:26:39,800 example, the limit function, namely the power series, must 516 00:26:39,800 --> 00:26:44,640 be continuous since each partial sum that makes up the 517 00:26:44,640 --> 00:26:47,570 sequence of partial sums is also continuous. 518 00:26:47,570 --> 00:26:50,100 Namely, every polynomial is continuous. 519 00:26:50,100 --> 00:26:53,670 Also, if, for some reason or other, you want to integrate 520 00:26:53,670 --> 00:26:57,260 that power series from 'a' to 'b', if the convergence is 521 00:26:57,260 --> 00:26:59,800 uniform, I can then do what? 522 00:26:59,800 --> 00:27:03,880 I can then take the summation sign outside, integrate the 523 00:27:03,880 --> 00:27:08,100 n-th partial sum, and add these all up. 524 00:27:08,100 --> 00:27:09,610 You see, the idea being what? 525 00:27:09,610 --> 00:27:13,350 That the n-th partial sum is a polynomial, and a polynomial 526 00:27:13,350 --> 00:27:16,560 is a particularly simple thing to integrate. 527 00:27:16,560 --> 00:27:18,400 That's one of the easiest functions to 528 00:27:18,400 --> 00:27:20,150 integrate, in fact. 529 00:27:20,150 --> 00:27:21,380 OK. 530 00:27:21,380 --> 00:27:23,120 Now, here's the wrap up, then. 531 00:27:23,120 --> 00:27:26,760 What we shall show next time is that within the interval of 532 00:27:26,760 --> 00:27:30,780 absolute convergence, the sequence of partial sums, 533 00:27:30,780 --> 00:27:36,620 which we already know converges absolutely to the 534 00:27:36,620 --> 00:27:39,840 limit function, also converges uniformly. 535 00:27:39,840 --> 00:27:43,040 In other words, within the radius of convergence, the 536 00:27:43,040 --> 00:27:43,670 power series-- 537 00:27:43,670 --> 00:27:46,530 and I don't know how to say this other than to say, it 538 00:27:46,530 --> 00:27:50,750 enjoys the usual polynomial properties associated with a 539 00:27:50,750 --> 00:27:54,610 polynomial such as summation 'k' goes from 0 to 'n', 'a sub 540 00:27:54,610 --> 00:27:57,090 k', 'x to the k'. 541 00:27:57,090 --> 00:28:00,820 In other words, then, this about wraps up what our 542 00:28:00,820 --> 00:28:03,780 introductory lecture for today wanted to be, namely the 543 00:28:03,780 --> 00:28:06,250 concept of uniform convergence. 544 00:28:06,250 --> 00:28:09,320 What I would like you to do now is to study this material 545 00:28:09,320 --> 00:28:12,040 very carefully in the supplementary notes, go over 546 00:28:12,040 --> 00:28:15,260 the learning exercises so that you become familiar with this. 547 00:28:15,260 --> 00:28:18,080 Then we will wrap up our course in our next lecture, 548 00:28:18,080 --> 00:28:23,480 when we show what a very, very powerful tool this particular 549 00:28:23,480 --> 00:28:28,480 concept of absolute convergence is in the study of 550 00:28:28,480 --> 00:28:31,110 the mathematical concept of convergence. 551 00:28:31,110 --> 00:28:33,970 At any rate, until next time, then, goodbye. 552 00:28:36,550 --> 00:28:39,750 Funding for the publication of this video was provided by the 553 00:28:39,750 --> 00:28:43,810 Gabriella and Paul Rosenbaum Foundation. 554 00:28:43,810 --> 00:28:47,970 Help OCW continue to provide free and open access to MIT 555 00:28:47,970 --> 00:28:52,170 courses by making a donation at ocw.mit.edu/donate.