1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,690 continue to offer high-quality educational resources for free. 5 00:00:10,690 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,270 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,270 --> 00:00:18,220 at ocw.mit.edu. 8 00:00:30,954 --> 00:00:31,800 HERBERT GROSS: Hi. 9 00:00:31,800 --> 00:00:35,730 Welcome to our final lecture in part three of calculus. 10 00:00:35,730 --> 00:00:38,850 And I suppose it would be nice to have an appropriate story 11 00:00:38,850 --> 00:00:39,660 to tell. 12 00:00:39,660 --> 00:00:41,700 And the only story I can think of 13 00:00:41,700 --> 00:00:45,780 is, on the transatlantic flight, the stewardess coming in 14 00:00:45,780 --> 00:00:49,320 to the passenger's cabin midway over the Atlantic Ocean 15 00:00:49,320 --> 00:00:51,150 and saying to the passengers, well, 16 00:00:51,150 --> 00:00:54,100 I have some good news and some bad news for you. 17 00:00:54,100 --> 00:00:58,380 The bad news is that we're almost out of gas. 18 00:00:58,380 --> 00:01:01,780 And the good news is that we're making wonderful time. 19 00:01:01,780 --> 00:01:03,870 And the reason I thought of this particular story 20 00:01:03,870 --> 00:01:06,960 is that, in a sense, we are not almost out of gas, 21 00:01:06,960 --> 00:01:10,420 but in terms of this being a fundamentals type of course 22 00:01:10,420 --> 00:01:15,120 in which we're trying to give the ingredients of what 23 00:01:15,120 --> 00:01:18,870 goes into all advanced mathematics courses, 24 00:01:18,870 --> 00:01:21,510 we have sort of come to the end of the road. 25 00:01:21,510 --> 00:01:26,250 We have finished the so-called ABC's of mathematics. 26 00:01:26,250 --> 00:01:28,860 As far as making wonderful time is concerned, 27 00:01:28,860 --> 00:01:32,550 I would say we've done deliberately just the opposite, 28 00:01:32,550 --> 00:01:36,210 that we have essentially taken one or two central themes 29 00:01:36,210 --> 00:01:40,590 and tried to unify all of elementary calculus 30 00:01:40,590 --> 00:01:43,530 in terms of these central themes. 31 00:01:43,530 --> 00:01:45,030 And when you do something like this, 32 00:01:45,030 --> 00:01:47,760 it's very difficult to come to the end of a course. 33 00:01:47,760 --> 00:01:49,712 You sort of don't know where to stop. 34 00:01:49,712 --> 00:01:51,420 So you go back to a little trick that you 35 00:01:51,420 --> 00:01:54,352 learned in about the third grade when you wrote compositions 36 00:01:54,352 --> 00:01:56,310 where you learned that you could write anything 37 00:01:56,310 --> 00:01:59,040 that you wanted as long as the last line was, 38 00:01:59,040 --> 00:02:01,410 we all had a very nice time. 39 00:02:01,410 --> 00:02:03,420 So we now must look for a line along 40 00:02:03,420 --> 00:02:05,970 those particular thoughts. 41 00:02:05,970 --> 00:02:10,350 And if we recall that we began as our major theme 42 00:02:10,350 --> 00:02:14,790 in this course the idea of mathematical structure, 43 00:02:14,790 --> 00:02:19,980 the idea of what new math really meant, not new meaning brand 44 00:02:19,980 --> 00:02:24,820 new, but new meaning the opposite of passe, new meaning 45 00:02:24,820 --> 00:02:27,900 meaningful, I can think of no nicer way 46 00:02:27,900 --> 00:02:31,440 of ending the course than by picking as an application 47 00:02:31,440 --> 00:02:36,810 a topic that existed long before the so-called new math was 48 00:02:36,810 --> 00:02:39,630 born, and which exists in an even 49 00:02:39,630 --> 00:02:44,160 more enlightened way in the spirit of the new mathematics. 50 00:02:44,160 --> 00:02:47,530 And the topic that I have in mind to conclude with-- 51 00:02:47,530 --> 00:02:49,890 and it also ties in with our lecture of last time 52 00:02:49,890 --> 00:02:51,060 on the dot product-- 53 00:02:51,060 --> 00:02:54,090 is the subject called orthogonal functions, 54 00:02:54,090 --> 00:02:58,470 which has as a special subtopic, Fourier series. 55 00:02:58,470 --> 00:03:01,200 Now the way this comes about is the following. 56 00:03:01,200 --> 00:03:04,110 And you have to know nothing about vector spaces 57 00:03:04,110 --> 00:03:07,260 to appreciate what we're going to do now, that in a sense 58 00:03:07,260 --> 00:03:10,950 we imitate what Fourier actually did in trying to solve 59 00:03:10,950 --> 00:03:12,690 a particular type of problem. 60 00:03:12,690 --> 00:03:16,620 We suppose that we have an infinite sequence of, functions 61 00:03:16,620 --> 00:03:18,615 which I'll denote by the family phi sub 62 00:03:18,615 --> 00:03:21,780 n of x, and a particular closed interval 63 00:03:21,780 --> 00:03:25,510 from a to b such that each member of my family phi sub n 64 00:03:25,510 --> 00:03:29,310 of x is integral on the interval from a to b. 65 00:03:29,310 --> 00:03:32,370 And we'll also assume that the integral from a 66 00:03:32,370 --> 00:03:36,870 to b of the product of any two of these functions, two 67 00:03:36,870 --> 00:03:40,170 different members of the set, that that integral is 0. 68 00:03:40,170 --> 00:03:42,050 In other words, the interval from a to b 69 00:03:42,050 --> 00:03:45,210 phi sub i of x times phi sub j of x dx 70 00:03:45,210 --> 00:03:50,190 is 0, provided i is unequal to j. 71 00:03:50,190 --> 00:03:52,920 So we're assuming this particular property, 72 00:03:52,920 --> 00:03:54,870 remembering of course that the product of two 73 00:03:54,870 --> 00:03:56,790 integral functions is, again, integrable 74 00:03:56,790 --> 00:03:58,680 so that this thing makes sense. 75 00:03:58,680 --> 00:04:01,920 Now traditionally, when a family of functions 76 00:04:01,920 --> 00:04:04,650 had this particular property, that family 77 00:04:04,650 --> 00:04:09,360 was set to be orthogonal on the interval from a to b. 78 00:04:09,360 --> 00:04:11,310 And I will leave it for a little bit 79 00:04:11,310 --> 00:04:15,360 later to show you how this jibes with the meaning of orthogonal 80 00:04:15,360 --> 00:04:18,839 as we defined it in terms of dot products last time. 81 00:04:18,839 --> 00:04:20,750 Let me show you how, for example, Fourier 82 00:04:20,750 --> 00:04:23,400 decided to use this type of information. 83 00:04:23,400 --> 00:04:25,230 Suppose we had a given function f 84 00:04:25,230 --> 00:04:28,500 of x that also happened to be integrable on the interval 85 00:04:28,500 --> 00:04:29,520 from a to b. 86 00:04:29,520 --> 00:04:31,590 Or written more succinctly, we assume 87 00:04:31,590 --> 00:04:35,160 that the integral from a to b f of x dx exists. 88 00:04:35,160 --> 00:04:38,430 Let's also suppose that f of x happens 89 00:04:38,430 --> 00:04:41,670 to be a linear combination of the members 90 00:04:41,670 --> 00:04:44,270 of the orthogonal family phi sub n of x, in other words, 91 00:04:44,270 --> 00:04:46,800 that for the given orthogonal family, f of x 92 00:04:46,800 --> 00:04:51,990 is summation Cn phi n of x as n goes from 1 to infinity. 93 00:04:51,990 --> 00:04:55,090 And the question is, under-- 94 00:04:55,090 --> 00:04:58,370 or with the knowledge that this is an orthogonal family, 95 00:04:58,370 --> 00:05:01,400 how can we determine each coefficient C sub 96 00:05:01,400 --> 00:05:04,310 n if this condition here is the whole? 97 00:05:04,310 --> 00:05:07,100 Or to make this more concrete and keep 98 00:05:07,100 --> 00:05:09,350 it as non-symbolic as possible, let's 99 00:05:09,350 --> 00:05:12,320 suppose I wanted to figure out here what C sub 3 100 00:05:12,320 --> 00:05:14,000 happened to be. 101 00:05:14,000 --> 00:05:15,290 The trick is this. 102 00:05:15,290 --> 00:05:19,440 I recognize that, from this basic definition, 103 00:05:19,440 --> 00:05:25,170 if I multiply phi sub 3 of x by any other member of this family 104 00:05:25,170 --> 00:05:28,140 other than phi sub 3 itself and integrate from a 105 00:05:28,140 --> 00:05:30,540 to b, that term will be 0. 106 00:05:30,540 --> 00:05:33,930 So what I do is I come back to this relationship. 107 00:05:33,930 --> 00:05:37,980 And I multiply both sides of this equation by phi sub 3 108 00:05:37,980 --> 00:05:41,880 of x, keeping in mind if I wanted to find C sub k, 109 00:05:41,880 --> 00:05:44,490 I would have multiplied both sides by phi sub k of x. 110 00:05:44,490 --> 00:05:46,940 I'm just picking k equals 3 here. 111 00:05:46,940 --> 00:05:49,560 I multiply both sides by phi sub 3 of x. 112 00:05:49,560 --> 00:05:55,830 And because phi sub 3 of x is a constant subscript, meaning 113 00:05:55,830 --> 00:05:58,600 it doesn't depend on the summation n which only affects 114 00:05:58,600 --> 00:06:00,810 the subscript here, I can bring phi sub up 3 115 00:06:00,810 --> 00:06:03,120 of x inside the summation sign. 116 00:06:03,120 --> 00:06:08,100 Now the trick is I integrate both sides of this equality 117 00:06:08,100 --> 00:06:09,750 from a to b. 118 00:06:09,750 --> 00:06:11,160 In other words, I put an integral 119 00:06:11,160 --> 00:06:14,670 from a to b on both sides here, augment this with a dx. 120 00:06:14,670 --> 00:06:17,730 And I now know that this is the case. 121 00:06:17,730 --> 00:06:20,407 My next gimmick, if you want to call it that-- 122 00:06:20,407 --> 00:06:21,990 and if you don't want to call it that, 123 00:06:21,990 --> 00:06:23,790 you should, because it is a gimmick, 124 00:06:23,790 --> 00:06:25,260 as we'll explain in the moment-- 125 00:06:25,260 --> 00:06:30,240 I simply interchange the order of summation and integration, 126 00:06:30,240 --> 00:06:32,430 remembering that the integral of the sum 127 00:06:32,430 --> 00:06:34,300 is the sum of the integrals. 128 00:06:34,300 --> 00:06:36,060 I interchange this order. 129 00:06:36,060 --> 00:06:38,820 And I now have the summation n goes from 1 130 00:06:38,820 --> 00:06:45,720 to infinity integral from a to b Cn phi n of x phi 3 of x dx. 131 00:06:45,720 --> 00:06:48,660 And now remembering that the integral 132 00:06:48,660 --> 00:06:52,920 from a to b, phi sub n of x times phi sub 3 of x dx, 133 00:06:52,920 --> 00:06:58,290 by definition of orthogonality is 0 unless n equals 3, 134 00:06:58,290 --> 00:07:02,070 all of these terms drop out except when n equals 3. 135 00:07:02,070 --> 00:07:03,270 Now what happens here? 136 00:07:03,270 --> 00:07:06,960 When n equals 3, this term here is C sub 3. 137 00:07:06,960 --> 00:07:11,910 This is phi sub 3 squared x dx integrated from a to b. 138 00:07:11,910 --> 00:07:15,750 And all the remaining terms are 0, because the integrand is 0. 139 00:07:15,750 --> 00:07:17,970 In other words, the right-hand side simply becomes-- 140 00:07:17,970 --> 00:07:20,460 well, since C3 is a constant, I can take it 141 00:07:20,460 --> 00:07:21,840 outside the integral sign. 142 00:07:21,840 --> 00:07:27,180 It's C3 integral from a to b C sub 3 squared of x dx. 143 00:07:27,180 --> 00:07:30,640 And now you see from this relationship here, 144 00:07:30,640 --> 00:07:32,850 I can solve for C sub 3. 145 00:07:32,850 --> 00:07:36,150 Namely, I simply divide both sides of this equation 146 00:07:36,150 --> 00:07:38,410 by this number here. 147 00:07:38,410 --> 00:07:42,750 Now what I said by gimmick is simply this. 148 00:07:42,750 --> 00:07:45,000 When we said that the integral of a sum 149 00:07:45,000 --> 00:07:48,210 is the sum of the integrals, that was only for a finite sum. 150 00:07:48,210 --> 00:07:50,310 You may remember in part one of our course 151 00:07:50,310 --> 00:07:53,310 when we talked about uniform convergence, one of the things 152 00:07:53,310 --> 00:07:56,010 that we pointed out was that for infinite sums, 153 00:07:56,010 --> 00:07:59,490 you weren't always guaranteed that you could interchange 154 00:07:59,490 --> 00:08:02,010 the order of summation and integration. 155 00:08:02,010 --> 00:08:05,350 Or as I usually put it, you could always interchange them, 156 00:08:05,350 --> 00:08:07,782 but you may not get the same answer. 157 00:08:07,782 --> 00:08:09,240 So the first question that comes up 158 00:08:09,240 --> 00:08:11,850 is, how do I know that I am allowed 159 00:08:11,850 --> 00:08:14,670 to interchange this order? 160 00:08:14,670 --> 00:08:17,292 By the way, the second question that comes up is-- 161 00:08:17,292 --> 00:08:18,750 and this is a more subtle question, 162 00:08:18,750 --> 00:08:20,980 but going back to what we're talking about before-- 163 00:08:20,980 --> 00:08:24,950 how do I even know that there is a linear combination of the phi 164 00:08:24,950 --> 00:08:28,560 sub n's that comes out to be the given f of x? 165 00:08:28,560 --> 00:08:30,600 You see you notice that our demonstration here 166 00:08:30,600 --> 00:08:33,456 presupposes that f of x can be written this way. 167 00:08:33,456 --> 00:08:35,039 You see the two questions that come up 168 00:08:35,039 --> 00:08:38,130 is, given an f of x, first of all, 169 00:08:38,130 --> 00:08:40,000 can it be written this way? 170 00:08:40,000 --> 00:08:42,539 And secondly, even if it can, do we 171 00:08:42,539 --> 00:08:45,930 have the right to interchange the order of integration 172 00:08:45,930 --> 00:08:46,872 and summation? 173 00:08:46,872 --> 00:08:48,330 By the way, assuming for the moment 174 00:08:48,330 --> 00:08:50,700 that we can do this mechanically at least, 175 00:08:50,700 --> 00:08:53,430 notice that what C3 turns out to be 176 00:08:53,430 --> 00:08:56,610 is the integral from a to b f of x phi sub 177 00:08:56,610 --> 00:09:00,600 3 of x dx divided by the integral form a to b phi sub 3 178 00:09:00,600 --> 00:09:05,430 of x squared times dx, and again, replacing the 3 by a k. 179 00:09:05,430 --> 00:09:07,620 This is how every one of these coefficients 180 00:09:07,620 --> 00:09:09,240 could have been determined. 181 00:09:09,240 --> 00:09:11,070 And now you see, as a very, very brief 182 00:09:11,070 --> 00:09:15,660 aside which I will go into much more detail in the study guide 183 00:09:15,660 --> 00:09:18,270 when we talk about the exercises and the various discussion 184 00:09:18,270 --> 00:09:22,770 of the exercises, given a series of functions which 185 00:09:22,770 --> 00:09:25,520 is integrable, a sequence of functions 186 00:09:25,520 --> 00:09:28,240 integrable on a given interval from a to b, 187 00:09:28,240 --> 00:09:32,700 one usually defines the dot product of these two functions 188 00:09:32,700 --> 00:09:36,930 to be the integral from a to b the product of the two 189 00:09:36,930 --> 00:09:38,910 functions times dx. 190 00:09:38,910 --> 00:09:40,490 You see, remember last time we showed 191 00:09:40,490 --> 00:09:42,570 that all you needed to have a dot product 192 00:09:42,570 --> 00:09:44,700 was a rule that showed you how to combine 193 00:09:44,700 --> 00:09:46,950 two members of your set to give you a number? 194 00:09:46,950 --> 00:09:49,170 And certainly, this thing here is a number. 195 00:09:49,170 --> 00:09:53,470 And it also had to have three or four basic properties. 196 00:09:53,470 --> 00:09:55,260 And as one of the exercises, we will 197 00:09:55,260 --> 00:09:58,560 show that those properties for a dot product are obeyed here. 198 00:09:58,560 --> 00:10:00,630 But this is where the word "orthogonal functions" 199 00:10:00,630 --> 00:10:01,690 come from. 200 00:10:01,690 --> 00:10:04,830 Notice that in terms of this definition of a dot product, 201 00:10:04,830 --> 00:10:08,995 two vectors are orthogonal if their dot product is 0. 202 00:10:08,995 --> 00:10:10,620 And now they're saying that this equals 203 00:10:10,620 --> 00:10:13,590 0 is the same as saying that this integral is 0. 204 00:10:13,590 --> 00:10:17,130 And that's precisely what the definition of orthogonal was. 205 00:10:17,130 --> 00:10:19,110 So you see, this is the connection 206 00:10:19,110 --> 00:10:22,770 between the vector space interpretation of orthogonal 207 00:10:22,770 --> 00:10:26,828 and the traditional interpretation of orthogonal. 208 00:10:26,828 --> 00:10:28,620 And as I say, I'll leave further discussion 209 00:10:28,620 --> 00:10:30,030 of that for the notes. 210 00:10:30,030 --> 00:10:31,900 But the important point is this. 211 00:10:31,900 --> 00:10:34,800 If I mechanically solve for these coefficients-- 212 00:10:34,800 --> 00:10:38,410 in other words, suppose I'm given f of x. 213 00:10:38,410 --> 00:10:40,500 And I'm given a family of functions 214 00:10:40,500 --> 00:10:42,840 phi sub n of x which is orthogonal on the interval 215 00:10:42,840 --> 00:10:44,130 from a to b. 216 00:10:44,130 --> 00:10:47,280 Suppose that without even knowing whether f of x 217 00:10:47,280 --> 00:10:50,133 can be represented as a linear combination of the phis, 218 00:10:50,133 --> 00:10:52,425 and that without even knowing whether I can interchange 219 00:10:52,425 --> 00:10:56,020 these two operations, the order in which I perform these two 220 00:10:56,020 --> 00:10:58,990 operations, let's suppose that I mechanically 221 00:10:58,990 --> 00:11:03,770 invent as my coefficients the C's in this particular way. 222 00:11:03,770 --> 00:11:05,740 In other words, regardless of whether it 223 00:11:05,740 --> 00:11:07,540 means anything or not, I can always 224 00:11:07,540 --> 00:11:11,720 multiply f of x by phi sub k of x, integrate from a to b. 225 00:11:11,720 --> 00:11:13,720 You see, the product of two integrable functions 226 00:11:13,720 --> 00:11:15,010 is always integrable. 227 00:11:15,010 --> 00:11:19,760 I can always divide that by the square of the function phi sub 228 00:11:19,760 --> 00:11:23,770 k of x from a to b, keeping in mind, by the way, that 229 00:11:23,770 --> 00:11:28,750 as long as phi sub 3 is not identically 0 230 00:11:28,750 --> 00:11:35,550 or if it's not 0 other than on a set of points of measure 231 00:11:35,550 --> 00:11:39,030 0, what have you, notice that the square of a number 232 00:11:39,030 --> 00:11:41,170 can't be negative. 233 00:11:41,170 --> 00:11:44,310 So if phi sub 3 of x is different from 0 234 00:11:44,310 --> 00:11:46,170 in enough places, by squaring it, 235 00:11:46,170 --> 00:11:47,920 I have a positive integrand. 236 00:11:47,920 --> 00:11:49,920 And when I integrate a positive integrand 237 00:11:49,920 --> 00:11:51,720 I can't get 0 for an answer. 238 00:11:51,720 --> 00:11:53,640 So I don't have a zero denominator here. 239 00:11:53,640 --> 00:11:55,120 But the point is this. 240 00:11:55,120 --> 00:11:59,250 If I choose the C's as I did here, 241 00:11:59,250 --> 00:12:04,710 see, with the C's as just defined, we write that f is x-- 242 00:12:04,710 --> 00:12:06,640 and then we put a little squiggle here. 243 00:12:06,640 --> 00:12:08,490 See, don't know if we had the right 244 00:12:08,490 --> 00:12:11,760 to assume that this series here is equal to f of x, 245 00:12:11,760 --> 00:12:14,220 but we write this little squiggle mark here. 246 00:12:14,220 --> 00:12:18,780 We call the resulting infinite series capital F of x. 247 00:12:18,780 --> 00:12:21,120 And what we say is that capital F 248 00:12:21,120 --> 00:12:25,920 of x is the Fourier representation of little f of x 249 00:12:25,920 --> 00:12:28,530 with respect to the family phi sub n of x. 250 00:12:28,530 --> 00:12:31,200 You see, we don't say that these two are equal. 251 00:12:31,200 --> 00:12:33,570 But when we choose these coefficients the way I just 252 00:12:33,570 --> 00:12:36,090 mentioned, this is, in any event, 253 00:12:36,090 --> 00:12:39,990 called the Fourier representation of the function 254 00:12:39,990 --> 00:12:40,800 little f of x. 255 00:12:40,800 --> 00:12:43,140 I see there are many questions that come up. 256 00:12:43,140 --> 00:12:46,180 And in the form of an overview, let me just state them. 257 00:12:46,180 --> 00:12:48,720 First of all, how do we know that capital F of x 258 00:12:48,720 --> 00:12:50,130 is even a convergent series? 259 00:12:50,130 --> 00:12:51,930 After all, this is an infinite series. 260 00:12:51,930 --> 00:12:54,060 How do we know it even converges? 261 00:12:54,060 --> 00:12:56,910 Secondly, if it does converge, how do we know 262 00:12:56,910 --> 00:12:59,580 that it converges to f of x? 263 00:12:59,580 --> 00:13:01,440 The answer is we don't know. 264 00:13:01,440 --> 00:13:02,520 We really don't know. 265 00:13:02,520 --> 00:13:05,640 But there is a rather remarkable result. 266 00:13:05,640 --> 00:13:08,100 And the result is not that difficult to prove. 267 00:13:08,100 --> 00:13:09,960 But in the form of an overview, I 268 00:13:09,960 --> 00:13:11,820 prefer to omit the proof here. 269 00:13:11,820 --> 00:13:15,000 It will be an exercise in the study guide, 270 00:13:15,000 --> 00:13:17,010 but I think it's important to see what 271 00:13:17,010 --> 00:13:19,710 the statement of the result is. 272 00:13:19,710 --> 00:13:22,530 And then we can talk about the ramifications of that. 273 00:13:22,530 --> 00:13:24,540 But the proof would actually obscure 274 00:13:24,540 --> 00:13:26,340 the beauty of the result. 275 00:13:26,340 --> 00:13:30,390 And the aside, which I leave as an exercise, is simply this. 276 00:13:30,390 --> 00:13:32,820 Suppose you took a linear combination 277 00:13:32,820 --> 00:13:39,780 of the first n of the terms phi 1 of x up to phi n of x. 278 00:13:39,780 --> 00:13:43,890 If you choose the C sub k's to be the so-called Fourier 279 00:13:43,890 --> 00:13:46,170 coefficients the way I just mentioned, 280 00:13:46,170 --> 00:13:50,010 what is interesting is this, that even if the Fourier 281 00:13:50,010 --> 00:13:53,760 representation doesn't converge to f of x, 282 00:13:53,760 --> 00:13:57,480 it has the property that it gives you the best 283 00:13:57,480 --> 00:14:00,250 least square approximation. 284 00:14:00,250 --> 00:14:02,550 In other words, if I were to compute 285 00:14:02,550 --> 00:14:07,080 the difference between f of x and the sum of the first n 286 00:14:07,080 --> 00:14:10,350 terms in the Fourier representation 287 00:14:10,350 --> 00:14:12,840 where the C's were chosen according to the method I just 288 00:14:12,840 --> 00:14:15,548 mentioned to you before and square 289 00:14:15,548 --> 00:14:17,340 that difference-- you see, this is why it's 290 00:14:17,340 --> 00:14:19,410 called the mean square idea. 291 00:14:19,410 --> 00:14:23,140 You see, large errors and small errors, in other words, 292 00:14:23,140 --> 00:14:27,330 a large negative error can cancel a large positive error 293 00:14:27,330 --> 00:14:30,360 because the algebraic signs tend to cancel one another. 294 00:14:30,360 --> 00:14:35,590 But by squaring large errors, all of these become what? 295 00:14:35,590 --> 00:14:40,420 Squares of large magnitudes are large positive magnitudes. 296 00:14:40,420 --> 00:14:43,810 By squaring, I really magnify the errors. 297 00:14:43,810 --> 00:14:46,150 The positives can't cancel the negatives. 298 00:14:46,150 --> 00:14:48,730 And what this says is that if I use 299 00:14:48,730 --> 00:14:52,780 this linear combination of the phi sub n's, 300 00:14:52,780 --> 00:14:58,610 that the square approximation error is less 301 00:14:58,610 --> 00:15:01,760 using these coefficients than if I had used 302 00:15:01,760 --> 00:15:05,180 any other possible coefficients, that if I picked 303 00:15:05,180 --> 00:15:07,790 any other coefficients in the whole world other than the ones 304 00:15:07,790 --> 00:15:10,610 that I found by the method indicated just a little while 305 00:15:10,610 --> 00:15:15,770 ago, if I summed these terms, subtracted that 306 00:15:15,770 --> 00:15:19,790 from f of x squared and integrated the results from a 307 00:15:19,790 --> 00:15:23,510 to b, I would get a number which was at least as large. 308 00:15:23,510 --> 00:15:26,000 And in most cases in fact, unless the d sub k's 309 00:15:26,000 --> 00:15:28,040 were chosen to equal the C sub k's, I 310 00:15:28,040 --> 00:15:30,950 would get a number which was larger than this. 311 00:15:30,950 --> 00:15:32,930 Now you see what this thing says? 312 00:15:32,930 --> 00:15:37,220 What this says is that there can't be any real interval 313 00:15:37,220 --> 00:15:40,370 where this gets very, very far away 314 00:15:40,370 --> 00:15:44,550 from f of x, because you see, if it did, if it did get very, 315 00:15:44,550 --> 00:15:47,630 very far away, this would be a large magnitude. 316 00:15:47,630 --> 00:15:51,020 Squaring it would give me a positive large magnitude. 317 00:15:51,020 --> 00:15:53,810 Integrating that from a to b would make this thing still 318 00:15:53,810 --> 00:15:54,980 very, very large. 319 00:15:54,980 --> 00:15:57,740 And why should it stay smaller than any other linear 320 00:15:57,740 --> 00:15:59,600 combination that I can get? 321 00:15:59,600 --> 00:16:02,330 And I'll emphasize this more from a concrete point 322 00:16:02,330 --> 00:16:03,960 of view in a few moments. 323 00:16:03,960 --> 00:16:07,910 But for the time being, what the impact of this beautiful result 324 00:16:07,910 --> 00:16:10,648 is-- and notice, it doesn't look like much when 325 00:16:10,648 --> 00:16:11,690 you look at it like this. 326 00:16:11,690 --> 00:16:13,640 It just looks like some abstract remark. 327 00:16:13,640 --> 00:16:16,910 But the beauty of this remark the impact of it 328 00:16:16,910 --> 00:16:19,370 is that, when you write the so-called Fourier 329 00:16:19,370 --> 00:16:23,187 representation, as opposed to power series, which I'll also 330 00:16:23,187 --> 00:16:24,770 mention at the end of the lecture what 331 00:16:24,770 --> 00:16:29,660 the contrast really is, the Fourier representation, if it 332 00:16:29,660 --> 00:16:33,740 does converge at all, converges to the function, what we call 333 00:16:33,740 --> 00:16:36,170 in the large, that over the whole interval 334 00:16:36,170 --> 00:16:38,940 it starts to fit the function very nicely. 335 00:16:38,940 --> 00:16:40,790 Now I'll come back to that in terms 336 00:16:40,790 --> 00:16:43,310 of a more specific example in a few moments. 337 00:16:43,310 --> 00:16:46,670 Let me point out that, Fourier, being basically an engineer 338 00:16:46,670 --> 00:16:50,150 and a scientist, worked with one particular set 339 00:16:50,150 --> 00:16:54,410 of orthogonal functions, a set of orthogonal functions that 340 00:16:54,410 --> 00:16:57,110 were very, very natural to anybody working 341 00:16:57,110 --> 00:17:00,110 with wave motion or things of this type, 342 00:17:00,110 --> 00:17:01,280 as you'll see in a moment. 343 00:17:01,280 --> 00:17:03,750 The special case that I have in mind is this. 344 00:17:03,750 --> 00:17:06,050 Consider the family of functions-- 345 00:17:06,050 --> 00:17:08,003 well, I can call this cosine 0x, which 346 00:17:08,003 --> 00:17:09,170 is another way of writing 1. 347 00:17:09,170 --> 00:17:14,819 But I have 1 cosine x, cosine 2x, cosine 3x, cosine 4x, 348 00:17:14,819 --> 00:17:18,920 et cetera, and also included in this family, 349 00:17:18,920 --> 00:17:22,339 sine x, sine 2x, sine 3x, et cetera. 350 00:17:22,339 --> 00:17:25,609 I claim that this family of functions 351 00:17:25,609 --> 00:17:29,780 is orthogonal on the interval from negative pi to pi. 352 00:17:29,780 --> 00:17:31,190 What I mean by that is what? 353 00:17:31,190 --> 00:17:34,760 That if I take any two different members of this family, 354 00:17:34,760 --> 00:17:39,620 multiply them together, integrate from minus pi to pi, 355 00:17:39,620 --> 00:17:41,453 the result will always be 0. 356 00:17:41,453 --> 00:17:43,370 Now the-- I won't go through this whole thing. 357 00:17:43,370 --> 00:17:46,700 That will also be one of the exercises in the study guide 358 00:17:46,700 --> 00:17:47,750 will be to verify this. 359 00:17:47,750 --> 00:17:49,850 But let me give you a head start in doing this. 360 00:17:49,850 --> 00:17:52,850 Let me pick two members of the cosine, two different numbers 361 00:17:52,850 --> 00:17:56,240 of the cosine family here, and multiply them together, 362 00:17:56,240 --> 00:17:57,890 and integrate from minus pi to pi. 363 00:17:57,890 --> 00:18:00,410 In other words, let me take integral from minus pi 364 00:18:00,410 --> 00:18:07,350 to pi cosine mx cosine nx dx where m is unequal to n. 365 00:18:07,350 --> 00:18:10,430 Now again, without belaboring this point, 366 00:18:10,430 --> 00:18:14,090 remember that the formula for cosine a cosine b 367 00:18:14,090 --> 00:18:17,600 comes from looking at the formulas for cosine a plus b 368 00:18:17,600 --> 00:18:22,070 cosine of a minus b and adding these two results. 369 00:18:22,070 --> 00:18:24,650 And the sine a sine b terms drop out. 370 00:18:24,650 --> 00:18:27,680 To make a long story short, or somewhat shorter, 371 00:18:27,680 --> 00:18:31,430 cosine mx times cosine nx is another way 372 00:18:31,430 --> 00:18:35,780 of writing one half the sum of cosine m 373 00:18:35,780 --> 00:18:40,820 plus nx plus cosine m minus n x. 374 00:18:40,820 --> 00:18:43,160 Now you see, to integrate this, very simply, 375 00:18:43,160 --> 00:18:44,320 this just gives me what? 376 00:18:44,320 --> 00:18:45,530 A 1/2 here. 377 00:18:45,530 --> 00:18:49,100 This is sine m plus n x over m plus n. 378 00:18:49,100 --> 00:18:52,020 This is sine m minus x-- 379 00:18:52,020 --> 00:18:55,010 m minus n times x over m minus n. 380 00:18:55,010 --> 00:18:58,010 And those, by the way, the fact that m is unequal to n, 381 00:18:58,010 --> 00:19:02,290 guarantees me that my denominator here can't be zero. 382 00:19:02,290 --> 00:19:04,970 And at any rate, noticing that when I evaluate this 383 00:19:04,970 --> 00:19:10,620 from x equals minus pi to pi, that these are integral, 384 00:19:10,620 --> 00:19:13,410 meaning whole-number multiples of pi, 385 00:19:13,410 --> 00:19:15,180 both these terms are zero. 386 00:19:15,180 --> 00:19:18,240 And we've established the fact that this is zero. 387 00:19:18,240 --> 00:19:22,500 So at least the members of the cosine family are orthogonal. 388 00:19:22,500 --> 00:19:26,550 One can also show the members of the sine family are orthogonal. 389 00:19:26,550 --> 00:19:30,420 And one can also show that any member of the cosine family 390 00:19:30,420 --> 00:19:31,770 taken-- 391 00:19:31,770 --> 00:19:34,950 multiplied by any member of the sine family integrated 392 00:19:34,950 --> 00:19:38,220 from minus pi to pi also gives zero. 393 00:19:38,220 --> 00:19:40,620 So the point then is, fair enough, 394 00:19:40,620 --> 00:19:44,640 that this family here is an orthogonal family 395 00:19:44,640 --> 00:19:47,310 on the interval from minus pi to pi. 396 00:19:47,310 --> 00:19:48,700 And the beauty of this result-- 397 00:19:48,700 --> 00:19:50,850 and let me point out that in textbooks 398 00:19:50,850 --> 00:19:55,560 on advanced mathematics, the multitude of pages 399 00:19:55,560 --> 00:19:59,800 I use to prove the results, but the statements 400 00:19:59,800 --> 00:20:01,710 of the terms that you prove are not 401 00:20:01,710 --> 00:20:04,170 that difficult to keep track of. 402 00:20:04,170 --> 00:20:08,100 Let me just give one major result, a result that 403 00:20:08,100 --> 00:20:10,290 is very, very crucial, and one that I will 404 00:20:10,290 --> 00:20:12,390 emphasize in the study guide. 405 00:20:12,390 --> 00:20:13,620 And that is this. 406 00:20:13,620 --> 00:20:16,620 Suppose that I am given a particular function little 407 00:20:16,620 --> 00:20:20,250 f of x, which is piecewise smooth on the interval 408 00:20:20,250 --> 00:20:22,035 from minus pi to pi. 409 00:20:22,035 --> 00:20:23,910 And remember what I mean by piecewise smooth. 410 00:20:23,910 --> 00:20:26,460 I mean it's a differentiable function except that it 411 00:20:26,460 --> 00:20:29,880 may have a finite number of jump discontinuities, 412 00:20:29,880 --> 00:20:32,730 that if I plot the graph y equals f of x, 413 00:20:32,730 --> 00:20:36,630 I get a smooth curve with the possible exception 414 00:20:36,630 --> 00:20:41,070 that there may be a few jumps in it, by few meaning what? 415 00:20:41,070 --> 00:20:44,550 A not-too-great infinite number, a set of measures 416 00:20:44,550 --> 00:20:45,400 zero type thing. 417 00:20:45,400 --> 00:20:48,130 But we won't belabor that point right now. 418 00:20:48,130 --> 00:20:49,500 But the idea is this. 419 00:20:49,500 --> 00:20:52,620 Suppose that I have this piecewise smooth function 420 00:20:52,620 --> 00:20:53,880 little f of x. 421 00:20:53,880 --> 00:20:57,270 And suppose capital F of x is the Fourier 422 00:20:57,270 --> 00:21:00,630 representation of little f of x relative to my sines 423 00:21:00,630 --> 00:21:01,290 and cosines. 424 00:21:01,290 --> 00:21:06,000 In other words, capital F of x is the linear combinations 425 00:21:06,000 --> 00:21:08,460 of the cosine terms and the sine terms 426 00:21:08,460 --> 00:21:11,127 where the coefficients are chosen by that little recipe 427 00:21:11,127 --> 00:21:13,210 that I showed you at the beginning of the lecture, 428 00:21:13,210 --> 00:21:14,940 and which, as I say, I'll reinforce 429 00:21:14,940 --> 00:21:17,790 as we do the exercises. 430 00:21:17,790 --> 00:21:21,127 At any rate, let's call this capital F of x. 431 00:21:21,127 --> 00:21:22,960 And in this particular case, in other words, 432 00:21:22,960 --> 00:21:24,050 what particular case? 433 00:21:24,050 --> 00:21:27,150 If little f of x is piecewise smooth 434 00:21:27,150 --> 00:21:30,300 and the orthogonal functions are the sines and cosines 435 00:21:30,300 --> 00:21:33,420 on the interval from minus pi the pi, then 436 00:21:33,420 --> 00:21:37,800 the Fourier representation of f of x, capital F of x, 437 00:21:37,800 --> 00:21:39,430 is given by the following. 438 00:21:39,430 --> 00:21:41,280 And this is a remarkable result. It 439 00:21:41,280 --> 00:21:44,910 says that capital F of x at any value of x 440 00:21:44,910 --> 00:21:47,250 in the interval from minus pi to pi 441 00:21:47,250 --> 00:21:52,860 is little f of x plus plus little f of x minus over 2. 442 00:21:52,860 --> 00:21:55,030 Now, what does f of x plus mean? 443 00:21:55,030 --> 00:21:59,280 That means the limit of f of x as you approach the given value 444 00:21:59,280 --> 00:22:01,140 of x from the positive side. 445 00:22:01,140 --> 00:22:03,870 And this f is x minus is the limit 446 00:22:03,870 --> 00:22:06,510 of f of x as you approach the given value of x 447 00:22:06,510 --> 00:22:08,040 from the negative side. 448 00:22:08,040 --> 00:22:10,900 In fact, to keep this easier to understand 449 00:22:10,900 --> 00:22:14,100 so I don't use x in two different ways, what I'm saying 450 00:22:14,100 --> 00:22:18,030 is capital F of x sub 0 is computed as follows. 451 00:22:18,030 --> 00:22:22,470 See what little f of x sub 0 is as x approaches x0 452 00:22:22,470 --> 00:22:23,550 from the left. 453 00:22:23,550 --> 00:22:28,035 See what f of x0 of f of x approaches as x approaches x0 454 00:22:28,035 --> 00:22:28,660 from the right. 455 00:22:31,170 --> 00:22:34,140 Add these two results and divide by 2. 456 00:22:34,140 --> 00:22:36,930 I hope that you can see that what I'm doing here 457 00:22:36,930 --> 00:22:39,330 is taking the average of these two results. 458 00:22:39,330 --> 00:22:41,610 Half the sum is the average. 459 00:22:41,610 --> 00:22:46,460 Also observe that if f happens to be continuous at x0, 460 00:22:46,460 --> 00:22:50,190 f of x0 plus is the same as f of x0 minus. 461 00:22:50,190 --> 00:22:52,710 You see, the only time that these two things can 462 00:22:52,710 --> 00:22:59,220 be different is if there is a jump in the function at x0, 463 00:22:59,220 --> 00:23:03,440 because if there is no jump, as I approach x0 from the left 464 00:23:03,440 --> 00:23:06,510 and approach x0 from the right, the two values I get 465 00:23:06,510 --> 00:23:07,870 must be the same. 466 00:23:07,870 --> 00:23:11,460 In other words, notice that at the points where 467 00:23:11,460 --> 00:23:16,590 f is continuous, this is nothing more than f of x0 itself. 468 00:23:16,590 --> 00:23:18,960 And what this famous result says is 469 00:23:18,960 --> 00:23:23,130 that, except at the points at which little f of x 470 00:23:23,130 --> 00:23:27,240 is discontinuous, capital F, the Fourier representation, 471 00:23:27,240 --> 00:23:30,450 is little f itself, that the Fourier 472 00:23:30,450 --> 00:23:34,150 representation converges to the function itself. 473 00:23:34,150 --> 00:23:36,090 And what it does that's remarkable 474 00:23:36,090 --> 00:23:39,960 is that where there is a jump in the function, 475 00:23:39,960 --> 00:23:44,310 it equals the average of the jump. 476 00:23:44,310 --> 00:23:47,010 Well, let me show you that in terms of an example. 477 00:23:47,010 --> 00:23:49,440 Let me pick a very simple function to begin with. 478 00:23:49,440 --> 00:23:50,970 Let me define the function little f 479 00:23:50,970 --> 00:23:56,460 of x to be negative 1 when x is in the interval from minus pi 480 00:23:56,460 --> 00:23:57,510 to 0. 481 00:23:57,510 --> 00:24:02,820 And let it be, say, positive 1 when x is 482 00:24:02,820 --> 00:24:05,080 in the interval from 0 to pi. 483 00:24:05,080 --> 00:24:07,830 So graphically, you see how this thing is going to look. 484 00:24:07,830 --> 00:24:09,670 It goes like this. 485 00:24:12,460 --> 00:24:14,950 By the way, notice that this is what mathematically we 486 00:24:14,950 --> 00:24:17,070 call an odd function. 487 00:24:17,070 --> 00:24:20,710 F of x is the negative of f of minus x. 488 00:24:20,710 --> 00:24:23,650 Notice that the sine terms are odd. 489 00:24:23,650 --> 00:24:25,300 The cosine terms are even. 490 00:24:25,300 --> 00:24:28,690 You see, cosine of x is the same as cosine of minus x. 491 00:24:28,690 --> 00:24:33,130 But sine of x is minus sine minus x. 492 00:24:33,130 --> 00:24:35,230 So the sine terms are odd. 493 00:24:35,230 --> 00:24:37,210 The cosine terms are even. 494 00:24:37,210 --> 00:24:40,120 So the first hint or the first nice simplification 495 00:24:40,120 --> 00:24:42,310 that takes place by choosing this example 496 00:24:42,310 --> 00:24:45,110 is that because this is an odd function, 497 00:24:45,110 --> 00:24:47,550 the cosine terms will all be absent. 498 00:24:47,550 --> 00:24:51,430 And it turns out that if I use the procedure indicated earlier 499 00:24:51,430 --> 00:24:52,370 in the lecture-- 500 00:24:52,370 --> 00:24:54,162 and again that will be one of the exercises 501 00:24:54,162 --> 00:24:56,350 so that you get the drill that's necessary here, 502 00:24:56,350 --> 00:24:58,810 again, I only want to give you the overview-- 503 00:24:58,810 --> 00:25:02,290 it turns out that the Fourier series representation 504 00:25:02,290 --> 00:25:05,020 of this particular function, capital F of x, 505 00:25:05,020 --> 00:25:10,000 turns out to be 4 over pi times the summation sine 506 00:25:10,000 --> 00:25:12,340 n x over n for n odd. 507 00:25:12,340 --> 00:25:15,730 And what that means is it's 4 over pi times 508 00:25:15,730 --> 00:25:24,490 sine x plus sine 3x over 3 plus sine 5x over 5, et cetera. 509 00:25:24,490 --> 00:25:27,670 By the way, these are very difficult things to draw. 510 00:25:27,670 --> 00:25:29,740 You're not used to working with these. 511 00:25:29,740 --> 00:25:33,010 I should point out this, that if there is anything that 512 00:25:33,010 --> 00:25:36,145 works nicely in the laboratory, it's sine and cosine terms, 513 00:25:36,145 --> 00:25:38,020 because those are things that you can produce 514 00:25:38,020 --> 00:25:40,360 very nicely on an oscilloscope. 515 00:25:40,360 --> 00:25:43,630 Analytically, these are very nasty terms to handle. 516 00:25:43,630 --> 00:25:47,470 But electronically, it's a very nice way 517 00:25:47,470 --> 00:25:49,540 of getting a feeling for Fourier series. 518 00:25:49,540 --> 00:25:52,100 But I don't have electronic chalk here, 519 00:25:52,100 --> 00:25:54,580 so I just talk about it and draw these things for you. 520 00:25:54,580 --> 00:25:56,590 But for example, to show you what I mean 521 00:25:56,590 --> 00:25:59,560 here, let me just take the first two terms. 522 00:25:59,560 --> 00:26:04,180 Let me plot the curve y equals 4 over pi times 523 00:26:04,180 --> 00:26:08,170 the quantity sine x plus sine 3x over 3. 524 00:26:08,170 --> 00:26:10,370 And if I plot that curve, which I've done here 525 00:26:10,370 --> 00:26:13,810 in the accentuated chalk, look at how nicely 526 00:26:13,810 --> 00:26:19,150 that curve starts already to fit the graph y equals f of x. 527 00:26:19,150 --> 00:26:20,650 You see how it fits over here? 528 00:26:20,650 --> 00:26:25,750 No perfect fit any place, but notice that, by and large, it 529 00:26:25,750 --> 00:26:30,100 fits the curve very nicely without too big an error all 530 00:26:30,100 --> 00:26:31,015 the way along here. 531 00:26:31,015 --> 00:26:32,890 It has roughly the right shape. 532 00:26:32,890 --> 00:26:37,210 And notice also it goes through the origin, 533 00:26:37,210 --> 00:26:40,030 because when x is 0, capital F of x is 0. 534 00:26:40,030 --> 00:26:43,130 And that's exactly the average of the jump over here. 535 00:26:43,130 --> 00:26:45,110 So you jump from minus 1 to 1. 536 00:26:45,110 --> 00:26:47,590 So the origin is the average over here. 537 00:26:47,590 --> 00:26:50,320 By the way, even though I don't draw very well, what 538 00:26:50,320 --> 00:26:52,630 I have tried to do here is show what 539 00:26:52,630 --> 00:26:55,390 happens if I tack on just a couple of more terms. 540 00:26:55,390 --> 00:26:59,530 I've tacked onto this the term sine 5x over 5 541 00:26:59,530 --> 00:27:01,690 and sine 7x over 7. 542 00:27:01,690 --> 00:27:05,240 And then the graph would look something like this. 543 00:27:05,240 --> 00:27:07,330 You see how nicely this starts to fit? 544 00:27:07,330 --> 00:27:09,100 It fits very nicely along here. 545 00:27:09,100 --> 00:27:13,060 But wherever there's a jump, it jumps to-- it averages 546 00:27:13,060 --> 00:27:14,740 out this way. 547 00:27:14,740 --> 00:27:17,350 By the way, one other observation-- 548 00:27:17,350 --> 00:27:20,830 obviously, the function is I'm constructing here 549 00:27:20,830 --> 00:27:22,960 is periodic with period 2 pi. 550 00:27:22,960 --> 00:27:27,400 If I replace x by x plus 2 pi, I don't change the sine any. 551 00:27:27,400 --> 00:27:29,020 And so obviously, the function which 552 00:27:29,020 --> 00:27:32,230 I'm calling capital F of x represents more 553 00:27:32,230 --> 00:27:33,670 than this given function. 554 00:27:33,670 --> 00:27:38,200 Rather, it represents the given function reproduced infinitely 555 00:27:38,200 --> 00:27:41,260 in both directions with period 2 pi. 556 00:27:41,260 --> 00:27:44,140 But again, I'll leave that for the exercises. 557 00:27:44,140 --> 00:27:46,700 To summarize what's happened over here, 558 00:27:46,700 --> 00:27:48,370 you see how nicely this thing fits? 559 00:27:48,370 --> 00:27:52,390 If I actually took capital F of x and computed this limit 560 00:27:52,390 --> 00:27:55,300 for each value of x, what the graph would look 561 00:27:55,300 --> 00:27:57,000 like-- now obviously, I can't get this 562 00:27:57,000 --> 00:27:59,860 on the oscilloscope, because no matter how many terms I use, 563 00:27:59,860 --> 00:28:02,800 no matter how large a number, n is still finite. 564 00:28:02,800 --> 00:28:05,080 But if I actually took the infinite sum, 565 00:28:05,080 --> 00:28:07,810 the graph that I would get mathematically 566 00:28:07,810 --> 00:28:12,680 is I would actually get the straight line x equals-- 567 00:28:12,680 --> 00:28:16,900 I'm sorry, y equals negative from minus pi to 0. 568 00:28:16,900 --> 00:28:23,260 It would be y equals positive 1 from 0 to pi. 569 00:28:23,260 --> 00:28:27,400 And at 0 itself, the function would be 0. 570 00:28:27,400 --> 00:28:30,340 You see, it splits the gap. 571 00:28:30,340 --> 00:28:32,260 It splits the jump. 572 00:28:32,260 --> 00:28:34,780 And then of course, the function would repeat itself 573 00:28:34,780 --> 00:28:37,270 at a regular interval of-- 574 00:28:37,270 --> 00:28:41,020 regular period of 2 pi, that this is what capital F of x 575 00:28:41,020 --> 00:28:43,000 would look like if you graphed it. 576 00:28:43,000 --> 00:28:46,730 By the way, to give you an example of an even function, 577 00:28:46,730 --> 00:28:49,870 if we were to use this same technique for getting 578 00:28:49,870 --> 00:28:53,740 the Fourier series for f of x equals the absolute value of x 579 00:28:53,740 --> 00:28:55,930 on the interval from minus pi to pi, 580 00:28:55,930 --> 00:29:01,430 it turns out that capital F of x would be pi over 2 minus 4 581 00:29:01,430 --> 00:29:07,180 over pi summation over n odd cosine nx over n squared. 582 00:29:07,180 --> 00:29:10,810 By the way, if this looks a little bit 583 00:29:10,810 --> 00:29:13,630 like the previous problem, notice that this 584 00:29:13,630 --> 00:29:15,700 is more than just coincidence. 585 00:29:15,700 --> 00:29:18,687 If you differentiate the square root of x-- 586 00:29:18,687 --> 00:29:20,770 I'm sorry, if you differentiate the absolute value 587 00:29:20,770 --> 00:29:22,600 of x, what is the derivative? 588 00:29:22,600 --> 00:29:27,460 It's minus 1 for x negative and positive 1 for x positive. 589 00:29:27,460 --> 00:29:30,430 It's the function that we were just talking about before, 590 00:29:30,430 --> 00:29:34,120 only it doesn't exist at x equals 0, you see? 591 00:29:34,120 --> 00:29:36,940 What I'm driving at over here is that it's not surprising 592 00:29:36,940 --> 00:29:42,460 that the terms in here are the integrals of the terms that 593 00:29:42,460 --> 00:29:44,320 were in our previous summation. 594 00:29:44,320 --> 00:29:47,380 But notice that the absolute value of x is an even function. 595 00:29:47,380 --> 00:29:50,200 Consequently, it's the cosine terms which appear here 596 00:29:50,200 --> 00:29:52,430 rather than the sine terms. 597 00:29:52,430 --> 00:29:56,410 And again, I've run the risk of drawing my own diagrams here. 598 00:29:56,410 --> 00:30:00,010 I have plotted this for n equals 3, 599 00:30:00,010 --> 00:30:05,950 meaning I've plotted y equals pi over 2 minus 4 over pi times 600 00:30:05,950 --> 00:30:10,570 the quantity cosine x plus cosine 3x over 9. 601 00:30:10,570 --> 00:30:12,760 And that's represented by this curve 602 00:30:12,760 --> 00:30:15,890 drawn in the accentuated chalk over here. 603 00:30:15,890 --> 00:30:19,870 And again, notice how even for just a small number of terms, 604 00:30:19,870 --> 00:30:24,580 we have already captured in the large the shape of the curve y 605 00:30:24,580 --> 00:30:27,820 equals the absolute value of x, which by the way 606 00:30:27,820 --> 00:30:31,240 contrasts very sharply with power series. 607 00:30:31,240 --> 00:30:33,130 Remember how power series worked? 608 00:30:33,130 --> 00:30:36,310 First of all, you had to require that the function that you 609 00:30:36,310 --> 00:30:40,240 were talking about possessed derivatives of all orders. 610 00:30:40,240 --> 00:30:44,440 Secondly, when you wanted to fit the function by its power 611 00:30:44,440 --> 00:30:47,063 series, you picked a particular point. 612 00:30:47,063 --> 00:30:47,980 Remember what you got? 613 00:30:47,980 --> 00:30:50,950 You first got a tangent line approximation. 614 00:30:50,950 --> 00:30:53,290 Then you got a quadratic approximation 615 00:30:53,290 --> 00:30:56,500 that fit the curve in a neighborhood at this point 616 00:30:56,500 --> 00:30:58,780 better than the straight line did. 617 00:30:58,780 --> 00:31:01,720 But after a while, the thing would sag off. 618 00:31:01,720 --> 00:31:06,580 Then you added a cubic term. 619 00:31:06,580 --> 00:31:09,330 And that fit even better for a while. 620 00:31:09,330 --> 00:31:11,590 And then again something weird might happen. 621 00:31:11,590 --> 00:31:14,450 And you kept on going like this so that ultimately, 622 00:31:14,450 --> 00:31:16,750 even though you had uniform convergence, what you were 623 00:31:16,750 --> 00:31:20,710 doing was you were getting a splendid fit near one 624 00:31:20,710 --> 00:31:26,770 particular point and slowly but surely, and actually very, very 625 00:31:26,770 --> 00:31:29,860 slowly, and also actually, very surely, but you 626 00:31:29,860 --> 00:31:32,890 may not have noticed it, these approximations were 627 00:31:32,890 --> 00:31:36,620 starting to fit better and better as you went out here. 628 00:31:36,620 --> 00:31:38,560 But the least square approximations 629 00:31:38,560 --> 00:31:39,730 would not be too good. 630 00:31:39,730 --> 00:31:41,210 In other words, out here someplace, 631 00:31:41,210 --> 00:31:44,440 there was starting to become some drastically large errors. 632 00:31:44,440 --> 00:31:49,030 On the other hand, the Fourier representation does what? 633 00:31:49,030 --> 00:31:53,950 It fits the curve that you're talking about very, very 634 00:31:53,950 --> 00:31:56,960 nicely very, very quickly. 635 00:31:56,960 --> 00:31:59,200 And this is one reason why Fourier series 636 00:31:59,200 --> 00:32:01,570 are as powerful as they are. 637 00:32:01,570 --> 00:32:04,570 And by the way, again, to end on a nice theme, 638 00:32:04,570 --> 00:32:07,060 Fourier series probably have as much space 639 00:32:07,060 --> 00:32:09,100 devoted to them in pure mathematics 640 00:32:09,100 --> 00:32:11,270 as they do in applied mathematics. 641 00:32:11,270 --> 00:32:14,270 It's a beautiful topic from both points of view. 642 00:32:14,270 --> 00:32:16,360 But I felt that for our lesson today, 643 00:32:16,360 --> 00:32:20,350 since it is a finishing touch to essentially the equivalent 644 00:32:20,350 --> 00:32:23,230 of three-plus semesters of calculus, 645 00:32:23,230 --> 00:32:26,770 that we should juxtaposition the old and the new 646 00:32:26,770 --> 00:32:29,920 and show the peaceful coexistence versus 647 00:32:29,920 --> 00:32:32,290 this idea of polarization where you either 648 00:32:32,290 --> 00:32:35,800 have to be a traditionalist or a modernist, 649 00:32:35,800 --> 00:32:39,100 where you either have to be an applied man or a purist, 650 00:32:39,100 --> 00:32:42,370 that all of these ideas mingle side by side. 651 00:32:42,370 --> 00:32:44,260 And about the only choice you have as a human 652 00:32:44,260 --> 00:32:46,570 being depending on what your interests are 653 00:32:46,570 --> 00:32:49,240 is what proportion you're going to mix them in. 654 00:32:49,240 --> 00:32:50,860 Well, at any rate, this brings us 655 00:32:50,860 --> 00:32:53,380 to the part of the course that always gives me 656 00:32:53,380 --> 00:32:54,820 a little bit of remorse. 657 00:32:54,820 --> 00:32:57,400 And that is, I hate to let my captive audience go. 658 00:32:57,400 --> 00:32:59,230 We have now been through, as I say, 659 00:32:59,230 --> 00:33:02,840 the equivalent of a couple of years of mathematics together. 660 00:33:02,840 --> 00:33:04,730 It has been a great pleasure for me. 661 00:33:04,730 --> 00:33:06,560 I hope it has been for you. 662 00:33:06,560 --> 00:33:10,270 And once again, I would like to thank everybody who has worked 663 00:33:10,270 --> 00:33:12,400 with me on this, but in particular, 664 00:33:12,400 --> 00:33:15,250 the usual three people who have been with me through thick 665 00:33:15,250 --> 00:33:18,790 and thin on this, John [? Fitchar, ?] project 666 00:33:18,790 --> 00:33:24,660 self-study manager, who has also done a lot of work with me 667 00:33:24,660 --> 00:33:26,340 in helping me write the study guide, 668 00:33:26,340 --> 00:33:29,790 who has given me a lot of advice on how to give these lectures 669 00:33:29,790 --> 00:33:32,340 and how to prepare them for the taping, 670 00:33:32,340 --> 00:33:35,220 and has also been our director for the taping, 671 00:33:35,220 --> 00:33:39,030 to Miss Elise [? Pelletier, ?] who in addition to her usual 672 00:33:39,030 --> 00:33:42,480 secretarial work, has typed much of our study guide and has 673 00:33:42,480 --> 00:33:49,170 worked as a video technician for us, and to our chief-- 674 00:33:49,170 --> 00:33:53,340 I'll just call him chief of audio visual effects 675 00:33:53,340 --> 00:33:55,848 at the center here, Charles [? Patton. ?] 676 00:33:55,848 --> 00:33:57,890 I am particularly indebted to these three people, 677 00:33:57,890 --> 00:33:59,307 indebted to a lot of other people. 678 00:33:59,307 --> 00:34:02,250 I thank you all for being with me this long. 679 00:34:02,250 --> 00:34:04,450 And I hope that our paths will cross again. 680 00:34:09,290 --> 00:34:11,690 Funding for the publication of this video 681 00:34:11,690 --> 00:34:16,550 was provided by the Gabriella and Paul Rosenbaum Foundation. 682 00:34:16,550 --> 00:34:20,719 Help OCW continue to provide free and open access to MIT 683 00:34:20,719 --> 00:34:26,154 courses by making a donation at ocw.mit.edu/donate.