1 00:00:00,040 --> 00:00:02,460 The following content is provided under a Creative 2 00:00:02,460 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,320 Your support will help MIT OpenCourseWare 4 00:00:06,320 --> 00:00:10,560 continue to offer high quality educational resources for free. 5 00:00:10,560 --> 00:00:13,300 To make a donation or view additional materials 6 00:00:13,300 --> 00:00:17,116 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,116 --> 00:00:17,740 at ocw.mit.edu. 8 00:00:32,145 --> 00:00:32,920 PROFESSOR: Hi. 9 00:00:32,920 --> 00:00:34,870 Today we, in a manner of speaking, 10 00:00:34,870 --> 00:00:39,150 start the second part of this part of our course, 11 00:00:39,150 --> 00:00:42,080 that what we have done up to now is, hopefully, 12 00:00:42,080 --> 00:00:46,270 to have presented the concept of partial derivatives 13 00:00:46,270 --> 00:00:47,620 with all the trimmings. 14 00:00:47,620 --> 00:00:50,250 And now what we would like to do is devote the remainder 15 00:00:50,250 --> 00:00:52,510 of the course to what we could call 16 00:00:52,510 --> 00:00:56,850 selected topics, in which the partial derivative plays 17 00:00:56,850 --> 00:01:00,580 a fundamental underlying thread. 18 00:01:00,580 --> 00:01:02,720 In other words, a situation in which 19 00:01:02,720 --> 00:01:06,730 we will now pick specific areas of calculus, 20 00:01:06,730 --> 00:01:09,370 and show how what we've learned about partial derivatives 21 00:01:09,370 --> 00:01:11,920 can be applied to these particular areas. 22 00:01:11,920 --> 00:01:15,590 Now, the first natural area, which would suggest itself, 23 00:01:15,590 --> 00:01:18,050 it seems to me, would be the concept 24 00:01:18,050 --> 00:01:21,480 of integrating a function of several variables. 25 00:01:21,480 --> 00:01:23,590 Since, after we study derivatives, 26 00:01:23,590 --> 00:01:26,790 we next associated the definite integral 27 00:01:26,790 --> 00:01:29,780 with an antiderivative, it might seem 28 00:01:29,780 --> 00:01:32,280 that a very natural inquiry now is 29 00:01:32,280 --> 00:01:35,900 to discuss the concept of infinite multiple sums 30 00:01:35,900 --> 00:01:39,090 as being the analog of the definite integral 31 00:01:39,090 --> 00:01:41,720 in our study of calculus of a single variable. 32 00:01:41,720 --> 00:01:44,370 Now, because these topics that we're picking 33 00:01:44,370 --> 00:01:48,180 have great application and have been studied in great depth, 34 00:01:48,180 --> 00:01:52,030 we by necessity will sort of skim the surface 35 00:01:52,030 --> 00:01:55,550 of many of these topics, so that we can present as many 36 00:01:55,550 --> 00:01:59,130 of the key points as we can in the time 37 00:01:59,130 --> 00:02:01,410 allotted for the course. 38 00:02:01,410 --> 00:02:03,600 We will make the lectures, hopefully, 39 00:02:03,600 --> 00:02:07,460 as short as possible, as pungent as possible, 40 00:02:07,460 --> 00:02:10,770 and leave most of the details to the learning exercises, 41 00:02:10,770 --> 00:02:13,130 and to the supplementary notes. 42 00:02:13,130 --> 00:02:16,310 At any rate, what our lecture today is called is "Double 43 00:02:16,310 --> 00:02:20,260 Sums, " or perhaps a better word would be multiple sums. 44 00:02:20,260 --> 00:02:22,030 But we will concentrate on the case 45 00:02:22,030 --> 00:02:24,370 of double sums for the usual reason 46 00:02:24,370 --> 00:02:26,840 that namely, in the case of two independent variables, 47 00:02:26,840 --> 00:02:30,240 we have convenient pictures still left at our convenience. 48 00:02:30,240 --> 00:02:33,930 Now the idea is, let's consider a region 49 00:02:33,930 --> 00:02:38,230 R. Say, the square, the unit square 50 00:02:38,230 --> 00:02:42,010 on the coordinate axes with vertices (0, 0), (1, 0), 51 00:02:42,010 --> 00:02:44,400 (1, 1), and (0, 1). 52 00:02:44,400 --> 00:02:46,590 Now, what we're going to do now is 53 00:02:46,590 --> 00:02:50,470 imagine that R is a thin plate. 54 00:02:50,470 --> 00:02:54,220 And what we would like to do is to find the mass of this plate 55 00:02:54,220 --> 00:02:59,410 R, assuming that the plate has a variable density. 56 00:02:59,410 --> 00:03:00,980 In other words, what we want to do 57 00:03:00,980 --> 00:03:05,320 is find the mass of the plate R, if its density rho at the point 58 00:03:05,320 --> 00:03:08,505 x comma y in R is given by rho of x, 59 00:03:08,505 --> 00:03:11,306 y equals x squared plus y squared. 60 00:03:11,306 --> 00:03:12,680 You see, by the way, the reason I 61 00:03:12,680 --> 00:03:15,920 picked this particular density distribution is, 62 00:03:15,920 --> 00:03:19,900 you may notice that x squared plus y squared, quite simply, 63 00:03:19,900 --> 00:03:22,510 is the square of the distance of the point 64 00:03:22,510 --> 00:03:25,550 x comma y from the origin. 65 00:03:25,550 --> 00:03:27,670 And this will make it very easy later 66 00:03:27,670 --> 00:03:31,300 for me to find where lowest densities occur, 67 00:03:31,300 --> 00:03:33,000 where greatest densities occur. 68 00:03:33,000 --> 00:03:36,310 Whereas this may not seem like a very applied problem. 69 00:03:36,310 --> 00:03:38,650 Notice that, if you want the geometric picture of what's 70 00:03:38,650 --> 00:03:42,390 going on here, the idea here is that the density 71 00:03:42,390 --> 00:03:46,970 is varying radially as we move away from the origin. 72 00:03:46,970 --> 00:03:49,420 In other words, the density is the square 73 00:03:49,420 --> 00:03:52,350 of the distance of the points from the origin, which 74 00:03:52,350 --> 00:03:55,640 means that the density is constant on any circle centered 75 00:03:55,640 --> 00:03:57,700 at the origin. 76 00:03:57,700 --> 00:04:00,740 Again, the key building block that we're going to use 77 00:04:00,740 --> 00:04:03,320 is again analogous to what we used in calculus 78 00:04:03,320 --> 00:04:04,690 of a single variable. 79 00:04:04,690 --> 00:04:06,960 For example, in calculus of a single variable, 80 00:04:06,960 --> 00:04:09,970 when we said that distance was equal to rate times time, 81 00:04:09,970 --> 00:04:12,930 it was assumed that the rate was constant. 82 00:04:12,930 --> 00:04:15,170 In a similar way, notice that if we 83 00:04:15,170 --> 00:04:18,010 assumed that the density were constant-- 84 00:04:18,010 --> 00:04:20,839 if the density were constant, then the mass 85 00:04:20,839 --> 00:04:24,930 would just be the density times the area. 86 00:04:24,930 --> 00:04:27,420 Again, most of us are used to thinking of density 87 00:04:27,420 --> 00:04:29,440 in terms of three dimensions, that the mass is 88 00:04:29,440 --> 00:04:31,610 the density times the volume. 89 00:04:31,610 --> 00:04:34,130 I prefer to pick a thin plate here, 90 00:04:34,130 --> 00:04:36,560 simply to utilize a simple diagram. 91 00:04:36,560 --> 00:04:38,240 But I did not want to get involved 92 00:04:38,240 --> 00:04:40,670 with three-dimensional diagrams. 93 00:04:40,670 --> 00:04:43,180 Another drawback, perhaps, to the system 94 00:04:43,180 --> 00:04:45,420 is that you may be wondering why we're not 95 00:04:45,420 --> 00:04:48,715 using the counterpart of what area was in calculus 96 00:04:48,715 --> 00:04:49,590 of a single variable. 97 00:04:49,590 --> 00:04:54,170 Why aren't we talking about the volume in two dimensions? 98 00:04:54,170 --> 00:04:57,060 Why couldn't we somehow visualize a volume problem 99 00:04:57,060 --> 00:04:57,560 here? 100 00:04:57,560 --> 00:04:59,950 Again, if that's what you'd like to do, 101 00:04:59,950 --> 00:05:02,780 notice that we could graph the density 102 00:05:02,780 --> 00:05:06,490 as a function of the point x comma y in the region R. 103 00:05:06,490 --> 00:05:09,030 Plot the density in the direction 104 00:05:09,030 --> 00:05:11,970 of the positive z-axis, in which case 105 00:05:11,970 --> 00:05:14,620 the problem that we would then be trying to solve 106 00:05:14,620 --> 00:05:17,490 would be to find the volume of a solid 107 00:05:17,490 --> 00:05:21,000 that you get by taking the parallelepiped, whose base is 108 00:05:21,000 --> 00:05:25,450 the region R, and intersect that with the surface z 109 00:05:25,450 --> 00:05:27,510 equals x squared plus y squared. 110 00:05:27,510 --> 00:05:31,090 Again, we will leave these details for the exercises. 111 00:05:31,090 --> 00:05:33,240 But the surface z equals x squared 112 00:05:33,240 --> 00:05:36,340 plus y squared, as you look in along the z-axis here, 113 00:05:36,340 --> 00:05:38,610 is a parabolic bowl opening up. 114 00:05:38,610 --> 00:05:40,470 In other words, z equals x squared 115 00:05:40,470 --> 00:05:44,300 plus y squared, if you take z equals a positive constant-- 116 00:05:44,300 --> 00:05:47,300 in other words, slice this parallel to the xy-plane-- 117 00:05:47,300 --> 00:05:49,430 if z is a positive constant, x squared 118 00:05:49,430 --> 00:05:53,140 plus y squared equals a positive constant is a circle. 119 00:05:53,140 --> 00:05:55,740 In any rate, just to summarize that, 120 00:05:55,740 --> 00:05:57,490 the geometric equivalent to our problem 121 00:05:57,490 --> 00:06:01,200 is find the volume of a solid S if S 122 00:06:01,200 --> 00:06:03,700 is the parallelepiped with the base R 123 00:06:03,700 --> 00:06:07,050 where R is just as given, and the top-- meaning 124 00:06:07,050 --> 00:06:09,340 it's intersected with the surface z 125 00:06:09,340 --> 00:06:11,640 equals x squared plus y squared. 126 00:06:11,640 --> 00:06:13,660 And again, let me mention as an aside 127 00:06:13,660 --> 00:06:17,970 that the same analogy appears in calculus of a single variable. 128 00:06:17,970 --> 00:06:22,200 Namely, if I write down the definite integral from 0 to 1 129 00:06:22,200 --> 00:06:26,100 x squared dx, I may view this, well, in many ways, 130 00:06:26,100 --> 00:06:27,660 but two of the major ways in which I 131 00:06:27,660 --> 00:06:29,618 may view this, the major one that we emphasized 132 00:06:29,618 --> 00:06:31,780 in our course was as an area. 133 00:06:31,780 --> 00:06:34,570 I can also view this as a mass. 134 00:06:34,570 --> 00:06:39,860 More specifically, let me take the region R as shown here. 135 00:06:39,860 --> 00:06:42,470 Then, you see, one interpretation 136 00:06:42,470 --> 00:06:45,130 of the integral from 0 to 1 x squared dx 137 00:06:45,130 --> 00:06:46,940 is that it's the area of the region 138 00:06:46,940 --> 00:06:50,320 R. Another interpretation is imagine 139 00:06:50,320 --> 00:06:55,300 that you have a very thin rod, geometrically of uniform shape. 140 00:06:55,300 --> 00:06:57,050 In other words, it's essentially a segment 141 00:06:57,050 --> 00:06:59,110 of the x-axis from 0 to 1. 142 00:06:59,110 --> 00:07:01,590 And suppose that the density of this rod 143 00:07:01,590 --> 00:07:05,860 is a function of position x, say rho of x is x squared. 144 00:07:05,860 --> 00:07:08,470 Then notice, in a manner analogous to what we were just 145 00:07:08,470 --> 00:07:11,280 talking about, the mass of this rod 146 00:07:11,280 --> 00:07:16,010 would be found by integrating x squared from 0 to 1. 147 00:07:16,010 --> 00:07:18,780 By the way, notice again a very important thing 148 00:07:18,780 --> 00:07:22,620 that comes up here, and that is that these two diagrams may 149 00:07:22,620 --> 00:07:24,670 be used for the same problem. 150 00:07:24,670 --> 00:07:27,800 You see, in many cases when one has a density 151 00:07:27,800 --> 00:07:30,530 distribution like this, one visualizes 152 00:07:30,530 --> 00:07:33,610 the density being plotted. 153 00:07:33,610 --> 00:07:36,560 In other words, the density rho of x equals x squared 154 00:07:36,560 --> 00:07:39,400 is plotted as the curve y equals rho of x, in this case, 155 00:07:39,400 --> 00:07:40,930 y equals x squared. 156 00:07:40,930 --> 00:07:43,800 And we then think of a way of being 157 00:07:43,800 --> 00:07:47,370 able to visualize what the density is doing here in terms 158 00:07:47,370 --> 00:07:49,130 of the shape of the curve here. 159 00:07:49,130 --> 00:07:52,250 That's exactly what we were doing earlier 160 00:07:52,250 --> 00:07:54,440 in the lecture, when we said that we can visualize 161 00:07:54,440 --> 00:07:59,860 the density rho of x, y as a function z equals rho of x, y. 162 00:07:59,860 --> 00:08:02,300 The analogy being that with two independent variables, 163 00:08:02,300 --> 00:08:06,530 we wouldn't have a curve, but rather we would have a surface. 164 00:08:06,530 --> 00:08:10,200 By the way, again, notice how this emphasizes the fact 165 00:08:10,200 --> 00:08:15,200 that the definite integral is in reality an infinite sum. 166 00:08:15,200 --> 00:08:17,490 That it is convenient in many cases 167 00:08:17,490 --> 00:08:21,870 to view this limit of a sum in terms of an area under a curve. 168 00:08:21,870 --> 00:08:25,600 But that the definite integral itself exists for more problems 169 00:08:25,600 --> 00:08:27,380 than just for computing areas. 170 00:08:27,380 --> 00:08:30,100 Now at any rate, all we're trying to point out here 171 00:08:30,100 --> 00:08:32,059 is that the same analogy will exist 172 00:08:32,059 --> 00:08:34,970 when we go to functions of several variables. 173 00:08:34,970 --> 00:08:38,340 If, in fact, we now return to the problem stated 174 00:08:38,340 --> 00:08:41,169 at the beginning of the lecture, we take our region 175 00:08:41,169 --> 00:08:43,990 R. Remember what this is, now. 176 00:08:43,990 --> 00:08:48,340 It's this square with vertices at these points. 177 00:08:48,340 --> 00:08:52,560 We know that the density is given at any point x comma y 178 00:08:52,560 --> 00:08:54,150 by x squared plus y squared. 179 00:08:54,150 --> 00:08:56,670 And what we'd like to do now is find the mass 180 00:08:56,670 --> 00:08:58,360 of this particular plate. 181 00:08:58,360 --> 00:09:00,880 Now, just as in calculus of a single variable, 182 00:09:00,880 --> 00:09:03,410 the argument that we use now is nothing more 183 00:09:03,410 --> 00:09:06,240 than if we took a small enough region here, 184 00:09:06,240 --> 00:09:08,910 we could assume that the variable density was 185 00:09:08,910 --> 00:09:10,095 essentially constant. 186 00:09:10,095 --> 00:09:13,160 In other words, this is how one makes physical approximations. 187 00:09:13,160 --> 00:09:15,930 We break this thing up into small pieces. 188 00:09:15,930 --> 00:09:19,250 Somehow or other find the mass of each small piece, 189 00:09:19,250 --> 00:09:20,790 and add those all up. 190 00:09:20,790 --> 00:09:23,470 The key point being, if the pieces are small enough 191 00:09:23,470 --> 00:09:26,530 for an approximation, we can say that the density 192 00:09:26,530 --> 00:09:28,170 is essentially constant. 193 00:09:28,170 --> 00:09:30,260 By the way, in the same way that we 194 00:09:30,260 --> 00:09:34,390 had to learn the sigma notation back in part one of our course, 195 00:09:34,390 --> 00:09:37,080 we will now have to somehow or other learn 196 00:09:37,080 --> 00:09:40,190 to master the notation for double sums. 197 00:09:40,190 --> 00:09:45,040 The idea, again, being this: if I partition this into n equal 198 00:09:45,040 --> 00:09:48,890 parts, and this segment into m equal parts, 199 00:09:48,890 --> 00:09:53,160 and if I call a typical x segment, delta x sub i, 200 00:09:53,160 --> 00:09:56,100 and a typical y segment, delta y_j, 201 00:09:56,100 --> 00:10:00,210 then to match up in coordinate fashion what the mass of this 202 00:10:00,210 --> 00:10:03,477 little piece would be, I would call that the (i, 203 00:10:03,477 --> 00:10:07,330 j)-th mass corresponding to the i-th x partition, 204 00:10:07,330 --> 00:10:09,980 and the j.th y partition. 205 00:10:09,980 --> 00:10:12,500 And the idea is, I'm going to add all of these up. 206 00:10:12,500 --> 00:10:16,280 That would motivate the notation called the double sum. 207 00:10:16,280 --> 00:10:18,590 Namely, the way you read this is you 208 00:10:18,590 --> 00:10:22,590 say, let me first pick a fixed value of j, 209 00:10:22,590 --> 00:10:27,460 and for that fixed value of j, I will sum up all these pieces 210 00:10:27,460 --> 00:10:29,380 as i goes from 1 to n. 211 00:10:29,380 --> 00:10:31,530 What that means pictorially is this. 212 00:10:31,530 --> 00:10:33,660 I will pick a fixed value of j. 213 00:10:33,660 --> 00:10:36,380 Well, since j controls the y-coordinate, 214 00:10:36,380 --> 00:10:39,500 a fixed value of j fixes your horizontal strip. 215 00:10:39,500 --> 00:10:41,000 And what you're saying is I will now 216 00:10:41,000 --> 00:10:44,210 add up all the masses along that strip. 217 00:10:44,210 --> 00:10:46,430 Then what you say is, and then as I 218 00:10:46,430 --> 00:10:51,780 do that, let me perform that for each j as j goes from 1 to m. 219 00:10:51,780 --> 00:10:53,370 In other words, what this would say 220 00:10:53,370 --> 00:10:58,010 is, add up all of the delta m's along each row, 221 00:10:58,010 --> 00:11:02,250 and then take the sum of all the possible values 222 00:11:02,250 --> 00:11:03,430 that the rows can take on. 223 00:11:03,430 --> 00:11:05,860 In other words, I'll add up these pieces, these pieces, 224 00:11:05,860 --> 00:11:08,990 these pieces, these pieces, then add those sums together. 225 00:11:08,990 --> 00:11:10,770 Notice, of course, I could have written 226 00:11:10,770 --> 00:11:12,700 this in the reverse order. 227 00:11:19,480 --> 00:11:20,910 Namely, this would say what? 228 00:11:20,910 --> 00:11:25,130 First hold i constant, and for a fixed i, sum these as j 229 00:11:25,130 --> 00:11:26,540 goes from 1 to m. 230 00:11:26,540 --> 00:11:31,300 In other words, for a fixed i, sum these as j goes from 1 231 00:11:31,300 --> 00:11:34,820 to m, and then sum over all the j's. 232 00:11:34,820 --> 00:11:36,520 Over all the i's. 233 00:11:36,520 --> 00:11:38,920 At any rate, the learning exercises 234 00:11:38,920 --> 00:11:41,790 will take care of making sure that you learn to manipulate 235 00:11:41,790 --> 00:11:43,520 this particular notation. 236 00:11:43,520 --> 00:11:45,560 The theory works as follows. 237 00:11:45,560 --> 00:11:49,650 What we do is, assuming that the density function is continuous, 238 00:11:49,650 --> 00:11:53,300 at each particular little rectangle that we have here we, 239 00:11:53,300 --> 00:11:56,400 we pick the point which has the smallest density, 240 00:11:56,400 --> 00:11:59,330 we picked the point which has the largest density. 241 00:11:59,330 --> 00:12:04,070 We then imagine that we had replaced this little segment 242 00:12:04,070 --> 00:12:07,780 by the same area, but with a material which 243 00:12:07,780 --> 00:12:12,380 has a constant smallest density, so that the true mass 244 00:12:12,380 --> 00:12:16,200 must be greater than the mass of that piece. 245 00:12:16,200 --> 00:12:19,050 And then we assume that this little piece was replaced 246 00:12:19,050 --> 00:12:20,930 by a new piece with the same shape, 247 00:12:20,930 --> 00:12:23,490 but whose density was constantly equal to the greatest 248 00:12:23,490 --> 00:12:24,610 density on this. 249 00:12:24,610 --> 00:12:26,940 So that this particular mass must be less 250 00:12:26,940 --> 00:12:29,210 than the mass of the new piece. 251 00:12:29,210 --> 00:12:34,480 And in that way, we squeeze the true mass between two extremes. 252 00:12:34,480 --> 00:12:37,740 Again, just as we did in calculus of a single variable. 253 00:12:37,740 --> 00:12:40,320 And again, to keep the notation going the same way, 254 00:12:40,320 --> 00:12:46,650 I let delta m sub i,j lower bar denote the mass of the piece 255 00:12:46,650 --> 00:12:51,100 that I get by replacing the density by the constant density 256 00:12:51,100 --> 00:12:55,960 equal to the smallest density, rho sub i,j lower bar. 257 00:12:55,960 --> 00:12:58,510 Treating that as a constant, multiplying that 258 00:12:58,510 --> 00:13:03,090 by the cross sectional area, delta x_i times delta y_j. 259 00:13:03,090 --> 00:13:08,160 I prefer to call that delta a sub i,j so we don't prejudice 260 00:13:08,160 --> 00:13:10,730 either the order in which we write these factors, 261 00:13:10,730 --> 00:13:14,400 or the particular coordinate system that one wants to use. 262 00:13:14,400 --> 00:13:16,240 But because I think you're more familiar 263 00:13:16,240 --> 00:13:18,900 with Cartesian coordinates, I will continue to write this. 264 00:13:18,900 --> 00:13:24,650 But the idea is, I compute the lowest, the smallest 265 00:13:24,650 --> 00:13:27,230 mass in other words, and underestimate. 266 00:13:27,230 --> 00:13:30,630 In a similar way, I compute and overestimate. 267 00:13:30,630 --> 00:13:33,535 I then know that the true mass of my (i, 268 00:13:33,535 --> 00:13:37,120 j)-th piece is caught between these two, 269 00:13:37,120 --> 00:13:40,820 therefore my total mass, which is the double sum of these over 270 00:13:40,820 --> 00:13:44,320 i and j, must be because must be, in turn, 271 00:13:44,320 --> 00:13:47,370 caught between these two double sums. 272 00:13:47,370 --> 00:13:49,420 In other words, what I have now done 273 00:13:49,420 --> 00:13:52,750 is caught the true mass between a lower 274 00:13:52,750 --> 00:13:55,450 bound and an upper bound. 275 00:13:55,450 --> 00:13:58,800 And again, to write this thing more symbolically 276 00:13:58,800 --> 00:14:01,850 so we can look at it here, what I'm saying is, on the one hand, 277 00:14:01,850 --> 00:14:05,370 my mass must be at least as great 278 00:14:05,370 --> 00:14:07,460 as the value of this double sum. 279 00:14:07,460 --> 00:14:09,330 On the other hand, it can be no greater 280 00:14:09,330 --> 00:14:11,630 than the value of this double sum. 281 00:14:11,630 --> 00:14:13,220 Somehow or other, the same as I did 282 00:14:13,220 --> 00:14:15,650 i calculus of a single variable, I would now 283 00:14:15,650 --> 00:14:19,830 like to put the squeeze on this as the sizes of the pieces 284 00:14:19,830 --> 00:14:23,400 delta x_i and delta y_j are allowed to go 285 00:14:23,400 --> 00:14:25,560 get arbitrarily close to 0 in size. 286 00:14:25,560 --> 00:14:29,280 What I hope happens is that in that limit, 287 00:14:29,280 --> 00:14:32,970 this difference goes to 0 so that M is caught between two 288 00:14:32,970 --> 00:14:35,770 equal things, and hence the true value of M 289 00:14:35,770 --> 00:14:38,740 must be that common value. 290 00:14:38,740 --> 00:14:41,190 You see, in a manner of speaking, theoretically, 291 00:14:41,190 --> 00:14:43,530 nothing new is happening that didn't already 292 00:14:43,530 --> 00:14:45,730 happen in calculus of a single variable, 293 00:14:45,730 --> 00:14:50,010 but from a computational point of view, limits of single sums 294 00:14:50,010 --> 00:14:53,150 have now been replaced by limits of double sums, 295 00:14:53,150 --> 00:14:56,790 so that computationally, there is a degree of difficulty 296 00:14:56,790 --> 00:14:59,830 more present in our present study 297 00:14:59,830 --> 00:15:02,810 than there was in our study of calculus of a single variable. 298 00:15:02,810 --> 00:15:04,920 The computation becomes more difficult, 299 00:15:04,920 --> 00:15:06,600 the theory stays the same. 300 00:15:06,600 --> 00:15:09,850 Now before we start getting into the idea of what limits mean, 301 00:15:09,850 --> 00:15:11,800 let me just summarize this for you 302 00:15:11,800 --> 00:15:14,450 in terms of a more concrete interpretation. 303 00:15:14,450 --> 00:15:19,450 Let me again take that square R with its vertices, 304 00:15:19,450 --> 00:15:23,540 the same one we dealt with, (0, 0), (1, 0), (1, 1), (0, 1). 305 00:15:23,540 --> 00:15:26,610 And let me now take as a special case the case 306 00:15:26,610 --> 00:15:31,310 in which I divide both the x region and the y region 307 00:15:31,310 --> 00:15:36,440 into two equal parts, so that my points of partition, you see, 308 00:15:36,440 --> 00:15:38,710 look like this. 309 00:15:38,710 --> 00:15:42,550 Now, you see, the advantage of picking my density 310 00:15:42,550 --> 00:15:45,500 to be x squared plus y squared is 311 00:15:45,500 --> 00:15:47,230 that since the density is radial, 312 00:15:47,230 --> 00:15:50,460 what this means is that in each of my partition 313 00:15:50,460 --> 00:15:54,320 rectangles-- in this case, each of my rectangles is a square. 314 00:15:54,320 --> 00:15:56,760 You see, keep in mind that first of all, I 315 00:15:56,760 --> 00:16:00,290 don't have to divide these two dimensions into the same number 316 00:16:00,290 --> 00:16:01,410 of equal parts. 317 00:16:01,410 --> 00:16:04,890 Secondly, they don't even have to be equal parts. 318 00:16:04,890 --> 00:16:08,052 But I've, just for symmetry here, elected to do this. 319 00:16:08,052 --> 00:16:10,510 Just so we get an idea of what's happening computationally, 320 00:16:10,510 --> 00:16:12,010 and the more difficult cases will be 321 00:16:12,010 --> 00:16:13,720 taken care of in the exercises. 322 00:16:13,720 --> 00:16:15,080 The idea now is what? 323 00:16:15,080 --> 00:16:19,380 That for each of these particular squares, the lowest 324 00:16:19,380 --> 00:16:22,380 density occurs at the point which makes up 325 00:16:22,380 --> 00:16:24,600 the lower left hand corner, because that 326 00:16:24,600 --> 00:16:28,120 will be the closest point to the origin in each 327 00:16:28,120 --> 00:16:29,150 of these squares. 328 00:16:29,150 --> 00:16:33,540 And the furthest point from the origin in each of these squares 329 00:16:33,540 --> 00:16:36,720 would be the point in the upper right hand corner. 330 00:16:36,720 --> 00:16:40,090 So you see in this particular case, 331 00:16:40,090 --> 00:16:41,600 let me do both of these at once. 332 00:16:41,600 --> 00:16:45,680 To find the smallest density in each of these regions, 333 00:16:45,680 --> 00:16:47,370 let me have the smallest one written 334 00:16:47,370 --> 00:16:50,960 in the black chalk on the lower part of this diagonal. 335 00:16:50,960 --> 00:16:54,470 You see the point (0, 0) corresponds to a density of 0. 336 00:16:54,470 --> 00:16:57,750 So the lowest possible density is 0 in this block. 337 00:16:57,750 --> 00:17:00,440 The point 1/2 comma 0 is the closest point 338 00:17:00,440 --> 00:17:02,780 to the origin in this little square. 339 00:17:02,780 --> 00:17:05,400 So x squared plus y squared in that case 340 00:17:05,400 --> 00:17:08,780 is simply 1/2 squared plus 0 squared, which is a quarter. 341 00:17:08,780 --> 00:17:12,630 The closest point here is 1/2 comma 1/2, 342 00:17:12,630 --> 00:17:15,150 so if we square the coordinates and add, 343 00:17:15,150 --> 00:17:17,760 we get 1/4 plus 1/4, which is 1/2. 344 00:17:17,760 --> 00:17:21,200 And in a similar way, the lowest density in this block 345 00:17:21,200 --> 00:17:23,829 is also 1/4. 346 00:17:23,829 --> 00:17:26,710 Now again, using the same argument, 347 00:17:26,710 --> 00:17:29,940 the point in the first block which 348 00:17:29,940 --> 00:17:33,260 is further from the origin is 1/2 comma 1/2. 349 00:17:33,260 --> 00:17:36,290 That density corresponds to 1/2. 350 00:17:36,290 --> 00:17:38,530 In other words, 1/2 squared plus 1/2 squared. 351 00:17:38,530 --> 00:17:40,880 So now putting in the biggest densities in each block, 352 00:17:40,880 --> 00:17:42,260 and carrying out the computations 353 00:17:42,260 --> 00:17:43,410 in a very trivial way. 354 00:17:43,410 --> 00:17:47,420 We see that the maximum densities are 1/2 here, 5/4 355 00:17:47,420 --> 00:17:52,020 here, 2 here, and 5/4 here. 356 00:17:52,020 --> 00:17:55,040 The idea is going to be now that what we can now do 357 00:17:55,040 --> 00:17:57,060 is we know that the area of each of these pieces 358 00:17:57,060 --> 00:17:59,770 is 1/2 by 1/2, which is a quarter. 359 00:17:59,770 --> 00:18:02,490 I will now assume two different plates. 360 00:18:02,490 --> 00:18:05,500 One will be the plate which has the same shape 361 00:18:05,500 --> 00:18:07,880 as this, but in this quadrant here, 362 00:18:07,880 --> 00:18:10,990 has a constant density of 0; in this quadrant here, 363 00:18:10,990 --> 00:18:14,130 has a constant density of 1/4; in this quadrant here, 364 00:18:14,130 --> 00:18:15,850 a constant density of 1/2; 365 00:18:15,850 --> 00:18:19,050 and in this quadrant, a constant density of 1/4. 366 00:18:19,050 --> 00:18:21,310 I can now compute that mass, and I 367 00:18:21,310 --> 00:18:24,960 know that mass must be less than the mass that I'm looking for, 368 00:18:24,960 --> 00:18:27,700 because I have put the lowest possible density 369 00:18:27,700 --> 00:18:28,830 in each quadrant. 370 00:18:28,830 --> 00:18:31,340 Correspondingly, if I now compute 371 00:18:31,340 --> 00:18:34,020 the mass of another plate which has the same shape, 372 00:18:34,020 --> 00:18:37,970 but in which each quadrant has the density of the greatest 373 00:18:37,970 --> 00:18:39,980 value of the density in each quadrant, 374 00:18:39,980 --> 00:18:42,930 that must give me an over-approximation. 375 00:18:42,930 --> 00:18:46,100 And so my mass is caught between these two. 376 00:18:46,100 --> 00:18:48,750 And just to let you get an idea of what this notation means, 377 00:18:48,750 --> 00:18:53,500 noticed the delta a sub i,j in this case is 1/4. 378 00:18:53,500 --> 00:18:56,810 For i equals 1, 2, and j equals 1 and 2-- in other words, 379 00:18:56,810 --> 00:19:03,060 our four elements of area are a sub 1, 1, a sub 1, 2, a sub 2, 380 00:19:03,060 --> 00:19:07,330 1, a sub 2, 2, where if you want to do these in order, 381 00:19:07,330 --> 00:19:11,330 just let i be 1 and 2 here, j be 1 and 2 here. 382 00:19:11,330 --> 00:19:17,010 For example, the maximum density in the first row, 383 00:19:17,010 --> 00:19:19,700 second column-- well, I'm reading these the wrong way 384 00:19:19,700 --> 00:19:20,200 here. 385 00:19:20,200 --> 00:19:21,290 The (i, j)-th. 386 00:19:21,290 --> 00:19:25,860 First row, second column, up this way, is 5/4. 387 00:19:25,860 --> 00:19:31,330 The minimum density in that same quadrant is 1/4, et cetera. 388 00:19:31,330 --> 00:19:33,300 At any rate, what I'm saying is if I now 389 00:19:33,300 --> 00:19:35,840 take the fact that for constant density, 390 00:19:35,840 --> 00:19:39,510 the mass is the density times the area, 391 00:19:39,510 --> 00:19:42,390 a lower approximation to my mass is 392 00:19:42,390 --> 00:19:45,520 obtained by adding up all of my constant densities 393 00:19:45,520 --> 00:19:47,680 and multiplying by the area. 394 00:19:47,680 --> 00:19:51,540 In other words, 0 times 1/4 plus 1/4 times 1/4 plus 1/2 times 395 00:19:51,540 --> 00:19:53,830 1/4 plus 1/4 times 1/4. 396 00:19:53,830 --> 00:19:56,320 That sum comes out to be 1/4. 397 00:19:56,320 --> 00:19:59,120 Correspondingly, the upper sum is 398 00:19:59,120 --> 00:20:01,770 what I get by taking each of the constant densities 399 00:20:01,770 --> 00:20:05,860 and multiplying it by the constant area, 1/4. 400 00:20:05,860 --> 00:20:08,810 That leads to a sum of 5/4. 401 00:20:08,810 --> 00:20:10,840 And therefore, no matter what else I 402 00:20:10,840 --> 00:20:13,400 know about the mass of my region R, 403 00:20:13,400 --> 00:20:19,150 I know that it must be more than 1/4, but less than 5/4. 404 00:20:19,150 --> 00:20:22,710 And the idea is that by putting on the squeeze, 405 00:20:22,710 --> 00:20:24,820 by making more and more subdivisions, 406 00:20:24,820 --> 00:20:29,720 I can hopefully find the exact value of M sub R. 407 00:20:29,720 --> 00:20:33,420 But finding this exact value involves a very, very 408 00:20:33,420 --> 00:20:35,430 difficult computation. 409 00:20:35,430 --> 00:20:39,030 Namely, I divide this up into m times 410 00:20:39,030 --> 00:20:41,880 n little elements of area. 411 00:20:41,880 --> 00:20:43,535 I pick a point in each region. 412 00:20:46,470 --> 00:20:50,290 The density at a particular point that I've picked out 413 00:20:50,290 --> 00:20:53,250 in that region, which I'll call c_(i, j) comma d_(i, 414 00:20:53,250 --> 00:20:58,240 j) is c_(i, j) squared plus d_(i, j) squared. 415 00:20:58,240 --> 00:21:03,660 This would give me the mass of a particular piece, 416 00:21:03,660 --> 00:21:05,650 and I add these up as i goes from 1 417 00:21:05,650 --> 00:21:08,010 to n, as j goes from 1 to n, and take 418 00:21:08,010 --> 00:21:11,620 the limit as the maximum delta X sub i and the maximum delta y 419 00:21:11,620 --> 00:21:13,740 sub j approach zero. 420 00:21:13,740 --> 00:21:17,340 And this, in most cases is very, very difficult. 421 00:21:17,340 --> 00:21:19,680 And in fact, even in this relatively simple case, 422 00:21:19,680 --> 00:21:21,780 what do I mean by relatively simple case? 423 00:21:21,780 --> 00:21:23,620 In terms of a region, what could be 424 00:21:23,620 --> 00:21:27,170 more straightforward than our little square, R, 425 00:21:27,170 --> 00:21:27,760 that we chose? 426 00:21:27,760 --> 00:21:30,790 In general, we'll have more complicated regions. 427 00:21:30,790 --> 00:21:34,860 Let me summarize today's lesson, essentially, by showing you, 428 00:21:34,860 --> 00:21:37,780 in particular, what it is that we're talking 429 00:21:37,780 --> 00:21:39,700 about in terms of double sums. 430 00:21:39,700 --> 00:21:41,570 Namely, what we're going to do is 431 00:21:41,570 --> 00:21:45,670 we'll assume that R is a reasonable subset of two 432 00:21:45,670 --> 00:21:46,860 dimensional space. 433 00:21:46,860 --> 00:21:50,180 And by reasonable, I mean that it's not actually 434 00:21:50,180 --> 00:21:53,320 a kinky type of thing that leads to pathological problems. 435 00:21:53,320 --> 00:21:56,740 The technical words are things like R is simply connected 436 00:21:56,740 --> 00:21:57,630 and things like this. 437 00:21:57,630 --> 00:21:59,470 We'll talk about that in more detail. 438 00:21:59,470 --> 00:22:01,330 What I'm saying is let me imagine 439 00:22:01,330 --> 00:22:06,780 that R is a fairly well defined region in the xy-plane. 440 00:22:06,780 --> 00:22:08,620 Let me assume that I have a mapping 441 00:22:08,620 --> 00:22:13,490 f that carries R into E. What does that mean? 442 00:22:13,490 --> 00:22:17,580 f is a real-valued function of two independent variables. 443 00:22:17,580 --> 00:22:20,600 Namely, R is a two-dimensional space. 444 00:22:20,600 --> 00:22:24,590 So a typical element in R is a two-tuple x comma 445 00:22:24,590 --> 00:22:27,350 y, and what we're saying that f maps x comma 446 00:22:27,350 --> 00:22:29,450 y into some number. 447 00:22:29,450 --> 00:22:32,760 Remember E, E^1, if you want to call it that, 448 00:22:32,760 --> 00:22:34,740 is the set of real numbers. 449 00:22:34,740 --> 00:22:38,280 So the idea is let's-- what we can say now is this. 450 00:22:38,280 --> 00:22:41,060 Let's take this region R. Let's partition it 451 00:22:41,060 --> 00:22:46,520 into a mesh of little rectangles here. 452 00:22:46,520 --> 00:22:50,390 Let's, in the (i, j)-th rectangle-- namely, 453 00:22:50,390 --> 00:22:54,190 the rectangle whose dimensions are delta x_i by delta y_j-- 454 00:22:54,190 --> 00:22:58,370 let's pick a particular point, and to identify it, 455 00:22:58,370 --> 00:23:01,900 we'll call it c sub i, j comma d sub i, j. 456 00:23:01,900 --> 00:23:05,260 Let's compute f at that point. 457 00:23:05,260 --> 00:23:09,290 Then we'll compute f times this area, 458 00:23:09,290 --> 00:23:12,790 and we'll sum this thing up as i goes from 1 to n, 459 00:23:12,790 --> 00:23:15,310 and j goes from 1 to m, and we're 460 00:23:15,310 --> 00:23:19,480 going to take a limit as this partition gets as fine as we 461 00:23:19,480 --> 00:23:21,060 wish, meaning the size of the pieces 462 00:23:21,060 --> 00:23:24,210 get arbitrarily close to zero in area. 463 00:23:24,210 --> 00:23:26,600 Both in the x and in the y direction. 464 00:23:26,600 --> 00:23:29,820 Notice, by the way, that because this is not 465 00:23:29,820 --> 00:23:33,790 a rectangular region, pieces of rectangles are left out. 466 00:23:33,790 --> 00:23:36,120 And that from a computational point of view, 467 00:23:36,120 --> 00:23:38,330 as you make this refinement greater and greater, 468 00:23:38,330 --> 00:23:41,530 this is why the theory is so voluminous on 469 00:23:41,530 --> 00:23:43,310 this particular type of topic. 470 00:23:43,310 --> 00:23:45,370 You must show that you're squeezing out all 471 00:23:45,370 --> 00:23:46,670 the error and things like this. 472 00:23:46,670 --> 00:23:49,760 But conceptually, all that we're really doing here 473 00:23:49,760 --> 00:23:50,940 is the following. 474 00:23:50,940 --> 00:23:53,690 What we do is we pick a particular point 475 00:23:53,690 --> 00:23:57,280 in a particular one of these rectangles. 476 00:23:57,280 --> 00:24:00,192 We evaluate f at that particular point, 477 00:24:00,192 --> 00:24:01,650 which exists because we're assuming 478 00:24:01,650 --> 00:24:03,214 that f is defined on this. 479 00:24:03,214 --> 00:24:05,130 And if you're having trouble visualizing this, 480 00:24:05,130 --> 00:24:07,510 f has nothing to do with this boundary. 481 00:24:07,510 --> 00:24:11,360 F is a function which maps each point into a number, 482 00:24:11,360 --> 00:24:15,240 so graphically, you can think of f as being a surface here. 483 00:24:15,240 --> 00:24:19,260 Namely if we visualize f of x comma y 484 00:24:19,260 --> 00:24:22,410 as being mapped with respect to the z-axis, 485 00:24:22,410 --> 00:24:24,750 we have a surface coming out here. 486 00:24:24,750 --> 00:24:26,830 And in a sense, what we're really trying to do 487 00:24:26,830 --> 00:24:30,860 is to find an approximation for the volume of the region cut 488 00:24:30,860 --> 00:24:33,000 off by this cylinder. 489 00:24:33,000 --> 00:24:37,825 And whose top is the particular surface z equals f of x, y. 490 00:24:37,825 --> 00:24:39,700 But forgetting about that for the time being, 491 00:24:39,700 --> 00:24:42,090 mechanically, all we do is we form 492 00:24:42,090 --> 00:24:44,200 this particular double sum. 493 00:24:44,200 --> 00:24:47,160 We form this double sum, and now I've put the brackets in here 494 00:24:47,160 --> 00:24:50,000 to emphasize that what this says is you first pick 495 00:24:50,000 --> 00:24:56,290 a fixed value of j, and sum this over all i from 1 to n. 496 00:24:56,290 --> 00:24:59,240 And when you're all through with that, the i no longer appears. 497 00:24:59,240 --> 00:25:02,340 You have something that depends only on j. 498 00:25:02,340 --> 00:25:03,290 And you now do what? 499 00:25:03,290 --> 00:25:06,290 Sum this result up over all values of j 500 00:25:06,290 --> 00:25:07,970 as j goes from 1 to m. 501 00:25:07,970 --> 00:25:11,460 And you then compute the limit as the maximum delta x_i 502 00:25:11,460 --> 00:25:14,960 and maximum delta y_j both approach 0. 503 00:25:14,960 --> 00:25:17,830 Now first of all, this limit may not exist. 504 00:25:17,830 --> 00:25:20,297 Secondly, the limit may exist, but it 505 00:25:20,297 --> 00:25:22,630 depends on the order in which you're adding these terms. 506 00:25:22,630 --> 00:25:24,850 If after all, we saw that we could change 507 00:25:24,850 --> 00:25:27,080 the sum of a simple infinite series 508 00:25:27,080 --> 00:25:28,880 by changing the order of the terms, 509 00:25:28,880 --> 00:25:30,850 certainly it shouldn't be surprising 510 00:25:30,850 --> 00:25:34,160 that an infinite double sum isn't affected-- 511 00:25:34,160 --> 00:25:35,910 or is affected, it shouldn't be surprising 512 00:25:35,910 --> 00:25:38,650 that it is affected if we change the order of the terms. 513 00:25:38,650 --> 00:25:41,560 So we have one of two ways of doing this, 514 00:25:41,560 --> 00:25:42,985 in terms of being systematized. 515 00:25:42,985 --> 00:25:45,920 Do we first sum in terms of the x's, and then 516 00:25:45,920 --> 00:25:48,110 add up the y increments? 517 00:25:48,110 --> 00:25:49,350 Or the other way around? 518 00:25:49,350 --> 00:25:51,840 The key result is this, though. 519 00:25:51,840 --> 00:25:56,020 It's that this particular limit will exist, and will 520 00:25:56,020 --> 00:25:59,180 be independent of what order you add these things in, 521 00:25:59,180 --> 00:26:02,360 provided only that the function f is 522 00:26:02,360 --> 00:26:04,780 at least piecewise continuous. 523 00:26:04,780 --> 00:26:07,070 In other words, that f is a continuous function. 524 00:26:07,070 --> 00:26:08,650 And if it's not continuous, it has 525 00:26:08,650 --> 00:26:12,020 breaks of only a finite number of places. 526 00:26:12,020 --> 00:26:16,520 And the idea is that not only will this limit exist, 527 00:26:16,520 --> 00:26:20,050 but in the manner analogous to that definite integral, when 528 00:26:20,050 --> 00:26:23,460 this limit exists, we denote it by what? 529 00:26:23,460 --> 00:26:25,830 We replace this by x and y. 530 00:26:25,830 --> 00:26:30,100 This gets replaced by dx and dy, and the double sum 531 00:26:30,100 --> 00:26:33,070 gets replaced by a double integral, 532 00:26:33,070 --> 00:26:38,360 and what we now do is indicate what the region R is over here. 533 00:26:38,360 --> 00:26:40,480 In other words, in terms of a summary, 534 00:26:40,480 --> 00:26:45,630 whenever I write down this expression, 535 00:26:45,630 --> 00:26:48,639 it's an abbreviation for a limit of a double sum. 536 00:26:48,639 --> 00:26:50,180 And what limit of a double sum is it? 537 00:26:50,180 --> 00:26:55,300 It's the particular limit of this double sum, if it exists, 538 00:26:55,300 --> 00:26:59,260 and that limit will exist if f is piecewise continuous. 539 00:26:59,260 --> 00:27:01,560 Now, the key point that I want to bring out here, 540 00:27:01,560 --> 00:27:05,240 and in fact, this will lead into the lecture of next time, also, 541 00:27:05,240 --> 00:27:08,450 is that if I had never heard of partial derivatives, 542 00:27:08,450 --> 00:27:11,150 it makes sense to talk about this kind of a sum. 543 00:27:11,150 --> 00:27:14,080 In other words, note that forming this limit 544 00:27:14,080 --> 00:27:17,550 does not require any knowledge of partial derivatives. 545 00:27:17,550 --> 00:27:22,250 In the same way that finding infinite sums in calculus one 546 00:27:22,250 --> 00:27:26,700 could be done independently of any knowledge of a derivative, 547 00:27:26,700 --> 00:27:30,460 this particular concept, this particular limits 548 00:27:30,460 --> 00:27:34,660 makes sense even if we've never heard of partial derivatives. 549 00:27:34,660 --> 00:27:37,690 And what does this double sum represent? 550 00:27:37,690 --> 00:27:40,100 Just by way of a quick summary again, 551 00:27:40,100 --> 00:27:43,310 there are two very specific physical interpretations 552 00:27:43,310 --> 00:27:44,540 we can give this. 553 00:27:44,540 --> 00:27:47,350 One is that this double sum represents 554 00:27:47,350 --> 00:27:50,660 the density of the plate whose shape is the region 555 00:27:50,660 --> 00:27:56,320 R-- the mass of the plate whose shape is the region R, 556 00:27:56,320 --> 00:27:59,340 and whose density at the point x comma y in R 557 00:27:59,340 --> 00:28:01,540 is given by f of x, y. 558 00:28:01,540 --> 00:28:05,070 In other words, this part here controls the density, 559 00:28:05,070 --> 00:28:08,270 and this part here controls the shape of the region. 560 00:28:08,270 --> 00:28:12,000 A second interpretation is that this limit, when it exists, 561 00:28:12,000 --> 00:28:14,620 represents the volume of the right cylinder. 562 00:28:14,620 --> 00:28:17,130 That means a cylinder is obtained 563 00:28:17,130 --> 00:28:20,200 by tracing over the region R with a line 564 00:28:20,200 --> 00:28:22,000 perpendicular to the xy-plane. 565 00:28:22,000 --> 00:28:26,990 That particular cylinder, whose top is the surface z 566 00:28:26,990 --> 00:28:30,230 equals f of x, y. 567 00:28:30,230 --> 00:28:32,770 Here are, then, the two interpretations 568 00:28:32,770 --> 00:28:33,800 that we give this. 569 00:28:33,800 --> 00:28:35,900 One is it can be a mass. 570 00:28:35,900 --> 00:28:38,400 The other is it can be a volume. 571 00:28:38,400 --> 00:28:41,880 The key point is that to evaluate these double sums, 572 00:28:41,880 --> 00:28:43,470 these limits of double sums, or if you 573 00:28:43,470 --> 00:28:45,940 want to call them infinite double sums, 574 00:28:45,940 --> 00:28:49,880 there is no necessity for knowing partial derivatives. 575 00:28:49,880 --> 00:28:53,450 However, one might expect that in the same way that there 576 00:28:53,450 --> 00:28:56,160 was a connection between single infinite 577 00:28:56,160 --> 00:29:00,590 sums and ordinary antiderivatives, 578 00:29:00,590 --> 00:29:04,740 there should be a connection between double or multiple sums 579 00:29:04,740 --> 00:29:08,610 and antiderivatives of partial-- 580 00:29:08,610 --> 00:29:11,630 Well, the antiderivative involving partial derivatives. 581 00:29:11,630 --> 00:29:13,810 It turns out that this is indeed the case. 582 00:29:13,810 --> 00:29:16,870 We will talk about this in more detail next time, 583 00:29:16,870 --> 00:29:20,310 but between now and next time, what I would like you to do 584 00:29:20,310 --> 00:29:23,630 is to drill particularly hard on these exercises, 585 00:29:23,630 --> 00:29:26,500 and make sure that you become familiar and feel 586 00:29:26,500 --> 00:29:29,800 at home with the notation that's used in denoting 587 00:29:29,800 --> 00:29:31,680 double and other multiple sums. 588 00:29:31,680 --> 00:29:34,260 At any rate then, until next time, good bye. 589 00:29:38,340 --> 00:29:40,730 Funding for the publication of this video 590 00:29:40,730 --> 00:29:45,580 was provided by the Gabriella and Paul Rosenbaum Foundation. 591 00:29:45,580 --> 00:29:49,760 Help OCW continue to provide free and open access to MIT 592 00:29:49,760 --> 00:29:54,178 courses by making a donation at ocw.mit.edu/donate.