1 00:00:00,000 --> 00:00:00,499 2 00:00:00,499 --> 00:00:02,944 The following content is provided under a Creative 3 00:00:02,944 --> 00:00:03,610 Commons license. 4 00:00:03,610 --> 00:00:05,460 Your support will help MIT OpenCourseWare 5 00:00:05,460 --> 00:00:09,350 continue to offer high-quality educational resources for free. 6 00:00:09,350 --> 00:00:12,530 To make a donation, or to view additional materials 7 00:00:12,530 --> 00:00:16,160 from hundreds of MIT courses, visit MIT OpenCourseWare 8 00:00:16,160 --> 00:00:20,730 at ocw.mit.edu. 9 00:00:20,730 --> 00:00:24,520 PROFESSOR STRANG: OK, so I feel I should have brought 10 00:00:24,520 --> 00:00:26,700 pizza instead of math problems. 11 00:00:26,700 --> 00:00:34,190 I mean, but this is our final, final hour. 12 00:00:34,190 --> 00:00:36,820 Open for any questions, any discussion 13 00:00:36,820 --> 00:00:42,210 about Fourier topics. 14 00:00:42,210 --> 00:00:45,610 Looking toward tomorrow evening's exam. 15 00:00:45,610 --> 00:00:50,140 And, well, hoping you'll find the whole course 16 00:00:50,140 --> 00:00:54,190 useful at many times in the future. 17 00:00:54,190 --> 00:01:00,090 So, somebody emailed me that the posted solution, 18 00:01:00,090 --> 00:01:03,200 so I just posted a quick solution 19 00:01:03,200 --> 00:01:07,270 to the un-collected homework ten, 20 00:01:07,270 --> 00:01:09,590 and so I haven't even looked to see what 21 00:01:09,590 --> 00:01:11,590 did it say for this question. 22 00:01:11,590 --> 00:01:12,760 But what is the answer? 23 00:01:12,760 --> 00:01:17,320 If I convolve the delta function with the delta function, 24 00:01:17,320 --> 00:01:18,990 what do I get? 25 00:01:18,990 --> 00:01:20,070 Delta, right. 26 00:01:20,070 --> 00:01:21,930 If I convolve anything with delta 27 00:01:21,930 --> 00:01:26,850 I get delta because I'm over in the other space. 28 00:01:26,850 --> 00:01:28,760 I'm multiplying by one. 29 00:01:28,760 --> 00:01:33,430 Yeah, so if I use the convolution rule to go over 30 00:01:33,430 --> 00:01:37,000 to the other space, where this becomes just one times one 31 00:01:37,000 --> 00:01:46,080 I'm getting one, and transform back and I've got delta. 32 00:01:46,080 --> 00:01:49,150 Well that question won't be on tomorrow's exam. 33 00:01:49,150 --> 00:01:51,770 But I thought we could deal with that fast. 34 00:01:51,770 --> 00:01:56,080 Now I'm open to questions about any topic. 35 00:01:56,080 --> 00:01:56,850 Homework or not. 36 00:01:56,850 --> 00:01:57,940 Yeah, thanks. 37 00:01:57,940 --> 00:02:01,120 AUDIENCE: [INAUDIBLE] 38 00:02:01,120 --> 00:02:02,460 PROFESSOR STRANG: I could try. 39 00:02:02,460 --> 00:02:04,320 Did I make up the exam? 40 00:02:04,320 --> 00:02:05,210 Probably. 41 00:02:05,210 --> 00:02:06,530 Yeah, I'll recognize it. 42 00:02:06,530 --> 00:02:14,050 OK, alright, I'll read it out. 43 00:02:14,050 --> 00:02:25,180 OK, so this an exam from 2005, not that far back. 44 00:02:25,180 --> 00:02:31,617 OK, so it has some questions. 45 00:02:31,617 --> 00:02:33,450 Shall I tell you what the whole question is, 46 00:02:33,450 --> 00:02:35,320 but it looks like this Part D is sort 47 00:02:35,320 --> 00:02:36,930 of separate from the others. 48 00:02:36,930 --> 00:02:41,610 But why not, this gives us some questions to think about. 49 00:02:41,610 --> 00:02:44,440 So if I just read out the questions. 50 00:02:44,440 --> 00:02:52,170 One was, find the coefficients of-- The function is e^(-x). 51 00:02:52,170 --> 00:02:54,620 That would naturally occur to me as a function 52 00:02:54,620 --> 00:03:00,240 that you could-- So, from zero to 2pi. 53 00:03:00,240 --> 00:03:02,890 And then periodic. 54 00:03:02,890 --> 00:03:04,260 Always good to graph it. 55 00:03:04,260 --> 00:03:06,910 So will e to the minus x look like? 56 00:03:06,910 --> 00:03:12,060 It'll come down to, much steeper than that, 57 00:03:12,060 --> 00:03:14,140 but that would do it. 58 00:03:14,140 --> 00:03:15,710 So that's e^(-x). 59 00:03:15,710 --> 00:03:20,920 Here it's one, and here it's e^(-2pi). 60 00:03:20,920 --> 00:03:23,620 And then repeat, repeat, repeat. 61 00:03:23,620 --> 00:03:27,150 So find its Fourier coefficients. 62 00:03:27,150 --> 00:03:30,430 I'll just say what the question is and you can ask me. 63 00:03:30,430 --> 00:03:32,610 And then what's the decay rate. 64 00:03:32,610 --> 00:03:34,520 Oh, maybe you can tell me the decay rate. 65 00:03:34,520 --> 00:03:37,660 Even before you tell me the coefficient. 66 00:03:37,660 --> 00:03:41,840 What would the decay rate be if I expressed this 67 00:03:41,840 --> 00:03:49,140 as a combination of Fourier terms, harmonics. 68 00:03:49,140 --> 00:03:52,840 How quickly will those coefficients drop off? 69 00:03:52,840 --> 00:03:55,040 So 1/k, only. 70 00:03:55,040 --> 00:04:02,050 That's right, 1/k, because this function has a jump at zero. 71 00:04:02,050 --> 00:04:07,360 Well, at zero and again at 2pi, at every one of those points. 72 00:04:07,360 --> 00:04:10,540 It has a jump, I guess, of about this distance, 73 00:04:10,540 --> 00:04:12,670 between one and that number. 74 00:04:12,670 --> 00:04:16,880 Which would be the jump that we would see of course every time. 75 00:04:16,880 --> 00:04:20,930 Yeah, so the coefficients would decay like 1/k. 76 00:04:20,930 --> 00:04:27,800 Then it asked to compute the sum of those coefficients squared. 77 00:04:27,800 --> 00:04:32,970 OK, so how can we find the sum of those coefficients squared 78 00:04:32,970 --> 00:04:37,040 even before I know what they are? 79 00:04:37,040 --> 00:04:40,140 Well, you know the answer, right? 80 00:04:40,140 --> 00:04:43,940 How am I going to do this problem? 81 00:04:43,940 --> 00:04:48,740 Yeah, so I'm looking for the sum of the squares of coefficients 82 00:04:48,740 --> 00:04:51,710 and then there's this magic trick. 83 00:04:51,710 --> 00:04:56,390 What is it that helps me to find that sum of squares? 84 00:04:56,390 --> 00:05:00,970 How do I go about that? 85 00:05:00,970 --> 00:05:05,560 I connect it to the integral of the function squared. 86 00:05:05,560 --> 00:05:08,465 Integral zero to 2pi of the function squared. 87 00:05:08,465 --> 00:05:11,160 And you remember the reason, and I've 88 00:05:11,160 --> 00:05:13,950 left a little space for the fudge factor. 89 00:05:13,950 --> 00:05:18,070 2pi, or 1/(2pi), we can figure out what it is by taking 90 00:05:18,070 --> 00:05:19,190 a simple case. 91 00:05:19,190 --> 00:05:22,490 Yeah, why don't we take a simple case to figure out should there 92 00:05:22,490 --> 00:05:25,770 be a 2pi or a 1/(2pi) or what? 93 00:05:25,770 --> 00:05:28,530 What's a simple case here? 94 00:05:28,530 --> 00:05:31,860 Suppose I take f(x)=1. 95 00:05:31,860 --> 00:05:36,240 Suppose I take the function f(x)=1, then the right side-- 96 00:05:36,240 --> 00:05:42,200 So if f(x) is one, the right side would be 2pi, I guess, 97 00:05:42,200 --> 00:05:43,630 the integral of one. 98 00:05:43,630 --> 00:05:45,660 And the left side would be what? 99 00:05:45,660 --> 00:05:49,870 What would be the Fourier coefficients 100 00:05:49,870 --> 00:05:53,510 of the constant function? 101 00:05:53,510 --> 00:05:57,580 Well just c_0, the only one we would have would be c_0, 102 00:05:57,580 --> 00:06:02,150 the constant term, the c_0*e^0. 103 00:06:02,150 --> 00:06:03,550 And it would be one. 104 00:06:03,550 --> 00:06:07,890 So I'd have a one there, when k is zero, 105 00:06:07,890 --> 00:06:09,670 and otherwise I wouldn't have anything. 106 00:06:09,670 --> 00:06:12,690 So this would be one squared, this would be a one. 107 00:06:12,690 --> 00:06:16,600 So I do need to divide by 2pi. 108 00:06:16,600 --> 00:06:19,610 Do you agree with that? 109 00:06:19,610 --> 00:06:26,420 To make it correct I'd better have a 1/(2pi) to get that 110 00:06:26,420 --> 00:06:27,790 to be right. 111 00:06:27,790 --> 00:06:31,770 OK, and then this gives me the answer then, 112 00:06:31,770 --> 00:06:38,540 I just do this integral, e^(-2x), then f is e^(-x). 113 00:06:38,540 --> 00:06:43,240 So f squared would be e^(-2x), which I can certainly integrate 114 00:06:43,240 --> 00:06:45,710 and get some expression. 115 00:06:45,710 --> 00:06:46,890 Yeah, OK. 116 00:06:46,890 --> 00:06:49,210 So that's a straightforward part. 117 00:06:49,210 --> 00:06:56,960 And now to the Part D, which you asked about which is, oh wow. 118 00:06:56,960 --> 00:07:03,700 OK, so now I'm asking about an equation u, du/dx, 119 00:07:03,700 --> 00:07:08,830 plus u(x) equals a train of deltas. 120 00:07:08,830 --> 00:07:12,860 So delta(x) made periodic. 121 00:07:12,860 --> 00:07:14,360 Can I write it that way? 122 00:07:14,360 --> 00:07:16,530 So this is the periodic case. 123 00:07:16,530 --> 00:07:20,820 So I'm looking for a periodic answer. 124 00:07:20,820 --> 00:07:23,620 So I'm in the Fourier series world 125 00:07:23,620 --> 00:07:28,360 when I take-- I'll be looking for coefficients, not 126 00:07:28,360 --> 00:07:30,430 the integral transform. 127 00:07:30,430 --> 00:07:33,750 OK. 128 00:07:33,750 --> 00:07:36,140 So alright, so I'm making it clear. 129 00:07:36,140 --> 00:07:40,390 So how do I proceed now, with such a problem? 130 00:07:40,390 --> 00:07:45,100 So I've mentioned this morning that a problem like this 131 00:07:45,100 --> 00:07:45,800 could appear. 132 00:07:45,800 --> 00:07:48,470 So I'm happy to discuss it. 133 00:07:48,470 --> 00:07:50,730 What do I do? 134 00:07:50,730 --> 00:07:51,700 Transform. 135 00:07:51,700 --> 00:07:52,910 Absolutely. 136 00:07:52,910 --> 00:07:54,650 Take Fourier transform. 137 00:07:54,650 --> 00:07:58,440 OK, so what happens when I take Fourier transforms of u? 138 00:07:58,440 --> 00:08:01,410 Well, I guess, so we're doing periodic, 139 00:08:01,410 --> 00:08:03,660 so we're going to go to coefficients, right? 140 00:08:03,660 --> 00:08:09,710 I'll call them, shall I call, shall-- I think of u(x), 141 00:08:09,710 --> 00:08:12,620 shall I just call its coefficient c? 142 00:08:12,620 --> 00:08:15,700 It's the only function I have around so I might as well use 143 00:08:15,700 --> 00:08:17,580 c for it. 144 00:08:17,580 --> 00:08:21,820 So here's the function, and Fourier takes this 145 00:08:21,820 --> 00:08:25,370 to the set of coefficients, OK. 146 00:08:25,370 --> 00:08:29,610 And the point is that we can separate-- 147 00:08:29,610 --> 00:08:33,500 I mean this is the whole reason why Fourier's so great. 148 00:08:33,500 --> 00:08:37,350 That we can do each frequency separately. 149 00:08:37,350 --> 00:08:38,510 So let's see then. 150 00:08:38,510 --> 00:08:44,430 So this has coefficient c_k, and what are the coefficients, 151 00:08:44,430 --> 00:08:50,080 what's the k-th coefficient, so this is all the e^(ikx) term. 152 00:08:50,080 --> 00:09:00,100 What's the k-th coefficient of the derivative? ikc_k, right? 153 00:09:00,100 --> 00:09:04,530 And what's the k-th coefficient of the delta function? 154 00:09:04,530 --> 00:09:06,100 One, right? 155 00:09:06,100 --> 00:09:09,740 Is that right or is there 1/(2pi) or something? 156 00:09:09,740 --> 00:09:10,900 Is it 1/(2pi)? 157 00:09:10,900 --> 00:09:13,380 OK, 1/(2pi). 158 00:09:13,380 --> 00:09:14,560 Yeah, I guess that's right. 159 00:09:14,560 --> 00:09:17,610 Because the coefficient, I would have 160 00:09:17,610 --> 00:09:23,420 to take the integral times, yeah, the integral 161 00:09:23,420 --> 00:09:25,920 with the delta function will give me the one, 162 00:09:25,920 --> 00:09:27,670 and then there's a 1/2pi. 163 00:09:27,670 --> 00:09:28,840 Right, OK. 164 00:09:28,840 --> 00:09:33,040 So now I've got the answer. 165 00:09:33,040 --> 00:09:37,890 Now I have this equation at each k. 166 00:09:37,890 --> 00:09:40,060 Following each eigenvector, following 167 00:09:40,060 --> 00:09:42,410 each separate frequency. 168 00:09:42,410 --> 00:09:44,520 And it's easy to do. 169 00:09:44,520 --> 00:09:53,180 It's just 1/(2pi) And there's 1+ik, I think. 170 00:09:53,180 --> 00:09:57,300 OK, and now those are the c_k's. 171 00:09:57,300 --> 00:10:00,780 So I think maybe this is it. 172 00:10:00,780 --> 00:10:05,630 Sum from minus infinity, I know what the c_k is. 173 00:10:05,630 --> 00:10:10,930 2pi(1+ik), times e^(ikx). 174 00:10:10,930 --> 00:10:15,420 175 00:10:15,420 --> 00:10:18,620 That's my u(x), I think. 176 00:10:18,620 --> 00:10:19,830 Had to be, right? 177 00:10:19,830 --> 00:10:26,100 So I just did the sort of automatic steps. 178 00:10:26,100 --> 00:10:28,020 When I made up this exam, I don't 179 00:10:28,020 --> 00:10:33,370 know whether I was thinking I could find out-- Can 180 00:10:33,370 --> 00:10:45,560 I actually do that sum to produce some function u(x) 181 00:10:45,560 --> 00:10:47,910 that I am familiar with. 182 00:10:47,910 --> 00:10:52,890 I'm not sure. 183 00:10:52,890 --> 00:10:55,190 Anybody reading the question, so the question 184 00:10:55,190 --> 00:10:58,440 says solve this differential equation. 185 00:10:58,440 --> 00:11:02,610 Doesn't say what it means by solve. 186 00:11:02,610 --> 00:11:03,710 Anyway, I've done it. 187 00:11:03,710 --> 00:11:05,930 I guess. 188 00:11:05,930 --> 00:11:13,410 Shall we agree that I passed on this question? 189 00:11:13,410 --> 00:11:17,620 If we knew some tricky way to do this addition, 190 00:11:17,620 --> 00:11:19,670 but I don't immediately see it. 191 00:11:19,670 --> 00:11:21,330 So that's good. 192 00:11:21,330 --> 00:11:23,380 OK, you're welcome. 193 00:11:23,380 --> 00:11:24,210 Thanks. 194 00:11:24,210 --> 00:11:27,070 OK, so that was a very good question to ask. 195 00:11:27,070 --> 00:11:32,120 Because it got, it got the method 196 00:11:32,120 --> 00:11:37,340 of how do you deal with a differential equation. 197 00:11:37,340 --> 00:11:41,220 By the way, this could be a difference equation, too. 198 00:11:41,220 --> 00:11:46,510 That could be u at x+h minus u at x-h. 199 00:11:46,510 --> 00:11:52,650 Shall I just emphasize that that wouldn't have been any harder? 200 00:11:52,650 --> 00:11:59,910 u(x+h)-u(x-h) over 2h, let me take, for example, 201 00:11:59,910 --> 00:12:03,760 plus u equals delta. 202 00:12:03,760 --> 00:12:09,060 This is like I've taken a centered difference there, 203 00:12:09,060 --> 00:12:10,560 instead of a derivative. 204 00:12:10,560 --> 00:12:12,780 How would you write a solution to that? 205 00:12:12,780 --> 00:12:18,374 Same method. 206 00:12:18,374 --> 00:12:19,540 So what would the method be? 207 00:12:19,540 --> 00:12:21,700 I'd take transforms. 208 00:12:21,700 --> 00:12:25,280 So here this would have coefficient c_k, 209 00:12:25,280 --> 00:12:28,350 this would have coefficients 1/(2pi). 210 00:12:28,350 --> 00:12:30,670 And what would be the coefficients of these guys? 211 00:12:30,670 --> 00:12:33,100 Well, there's a 1/(2h). 212 00:12:33,100 --> 00:12:38,670 So what are the Fourier coefficients of u(x+h)? 213 00:12:38,670 --> 00:12:41,280 Just to see that all these things-- 214 00:12:41,280 --> 00:12:44,040 I've got constant coefficients here. 215 00:12:44,040 --> 00:12:46,570 Why shouldn't Fourier work? 216 00:12:46,570 --> 00:12:53,951 So if it's u(x+h), so there's an e to the something factor, 217 00:12:53,951 --> 00:12:54,450 right? 218 00:12:54,450 --> 00:12:58,780 When I've shifted a function then in the transform 219 00:12:58,780 --> 00:13:00,370 I see an e to the i thing. 220 00:13:00,370 --> 00:13:01,710 So let's just see. 221 00:13:01,710 --> 00:13:10,000 If u(x), so if u(x) is the sum of c_k*e^(ikx), 222 00:13:10,000 --> 00:13:21,220 then u(x+h) would be the sum of c_k*e^(ik(x+h)), right? 223 00:13:21,220 --> 00:13:23,300 No, nothing surprising there. 224 00:13:23,300 --> 00:13:25,220 So now I'm seeing the factor. 225 00:13:25,220 --> 00:13:29,780 There's c_k, it's multiplied by the factor e^(ikh). 226 00:13:29,780 --> 00:13:34,110 227 00:13:34,110 --> 00:13:43,210 So this is the same c_k, but an e^(ikh), from that one. 228 00:13:43,210 --> 00:13:47,030 And this would be the minus, and this would shift the other way. 229 00:13:47,030 --> 00:13:52,500 So it would be the same c_k, and probably e^(-ikh). 230 00:13:52,500 --> 00:13:54,470 If I did that right. 231 00:13:54,470 --> 00:13:55,420 Yeah. 232 00:13:55,420 --> 00:13:59,700 Anyway, once again, we have still linear problem. 233 00:13:59,700 --> 00:14:05,360 This is c_k times something equals 1/(2pi). 234 00:14:05,360 --> 00:14:08,060 235 00:14:08,060 --> 00:14:12,030 By the way, this thing is what I would call the transfer 236 00:14:12,030 --> 00:14:12,590 function. 237 00:14:12,590 --> 00:14:14,010 You see that language? 238 00:14:14,010 --> 00:14:16,640 We haven't used that language much. 239 00:14:16,640 --> 00:14:21,950 Or at all, it's sort of more of a systems theory, control 240 00:14:21,950 --> 00:14:23,200 theory word. 241 00:14:23,200 --> 00:14:29,520 But hey, we're always doing the same correct thing. 242 00:14:29,520 --> 00:14:35,670 That's the thing that transfers the input to the output. 243 00:14:35,670 --> 00:14:38,690 So maybe the transfer function is one over. 244 00:14:38,690 --> 00:14:41,940 Maybe I include the division in the transfer function. 245 00:14:41,940 --> 00:14:47,510 So finally I get c_k is this 1/(2pi) guy, 246 00:14:47,510 --> 00:14:52,410 divided by whatever that was, 1/(2h) times that number, 247 00:14:52,410 --> 00:14:55,140 minus 1/(2h) of that plus one. 248 00:14:55,140 --> 00:14:57,870 Whatever is multiplying c_k there. 249 00:14:57,870 --> 00:14:59,250 It's a transfer function. 250 00:14:59,250 --> 00:14:59,750 Yeah. 251 00:14:59,750 --> 00:15:04,910 So there's another word which we could have, and should have, 252 00:15:04,910 --> 00:15:06,930 used before today. 253 00:15:06,930 --> 00:15:09,870 But here it is. 254 00:15:09,870 --> 00:15:12,920 OK, so that's some thoughts about that question. 255 00:15:12,920 --> 00:15:14,660 Any other direction to go? 256 00:15:14,660 --> 00:15:15,737 AUDIENCE: [INAUDIBLE] 257 00:15:15,737 --> 00:15:17,070 PROFESSOR STRANG: Yes. go ahead. 258 00:15:17,070 --> 00:15:18,200 Yeah. 259 00:15:18,200 --> 00:15:22,560 AUDIENCE: [INAUDIBLE] 260 00:15:22,560 --> 00:15:24,420 PROFESSOR STRANG: Lab problem 12, OK. 261 00:15:24,420 --> 00:15:28,000 4.5, Fourier integrals, number 12. 262 00:15:28,000 --> 00:15:30,360 Oh yeah, maybe that was also. 263 00:15:30,360 --> 00:15:33,050 A question was raised about was it 264 00:15:33,050 --> 00:15:42,310 right on the posted solution. 265 00:15:42,310 --> 00:15:45,040 So maybe not. 266 00:15:45,040 --> 00:15:46,930 Let's see what we want to do. 267 00:15:46,930 --> 00:15:49,400 Yeah I just thought that was pretty-- I 268 00:15:49,400 --> 00:15:51,700 could remember the day I thought of this equation. 269 00:15:51,700 --> 00:15:55,910 Integral of u minus derivative of u equal delta(x). 270 00:15:55,910 --> 00:16:02,030 I thought that's kind of a cool looking equation. 271 00:16:02,030 --> 00:16:05,440 But I don't know that I've ever solved it, before. 272 00:16:05,440 --> 00:16:12,410 So integral of u, actually yeah, I look cheerful, 273 00:16:12,410 --> 00:16:17,570 but-- So I'm finishing the fourth edition 274 00:16:17,570 --> 00:16:19,640 of the linear algebra book. 275 00:16:19,640 --> 00:16:21,550 Introduction to Linear Algebra. 276 00:16:21,550 --> 00:16:23,610 So I love writing. 277 00:16:23,610 --> 00:16:25,910 I've done all the writing and now comes 278 00:16:25,910 --> 00:16:30,530 the only horrible part, solving all those stupid exercises. 279 00:16:30,530 --> 00:16:36,520 So that, you may think of me in sunny Singapore or somewhere, 280 00:16:36,520 --> 00:16:40,030 but what I'll be doing is not on the beach. 281 00:16:40,030 --> 00:16:46,740 It's in an office somewhere, doing Problem 1.1.1, 282 00:16:46,740 --> 00:16:50,870 and two, three, it's not life. 283 00:16:50,870 --> 00:16:55,540 I mean, it's. 284 00:16:55,540 --> 00:16:57,400 But, OK. 285 00:16:57,400 --> 00:16:59,540 Anyway, that looks like a pretty good equation. 286 00:16:59,540 --> 00:17:03,880 And certainly I'm going to transform it. 287 00:17:03,880 --> 00:17:06,800 So I can see a little question coming up. 288 00:17:06,800 --> 00:17:09,770 So am in the Fourier integral world, yeah. 289 00:17:09,770 --> 00:17:15,630 So I should you use this notation. u hat of k, right? 290 00:17:15,630 --> 00:17:22,510 Over ik, is that the right thing when I've integrated? 291 00:17:22,510 --> 00:17:23,010 Yeah. 292 00:17:23,010 --> 00:17:24,900 That's made it smoother. 293 00:17:24,900 --> 00:17:28,840 That's dragging the function down. 294 00:17:28,840 --> 00:17:34,480 And I guess the one point that's a little tricky there 295 00:17:34,480 --> 00:17:38,600 is always, are you dividing by zero. 296 00:17:38,600 --> 00:17:41,440 Because zero is one of the frequencies. 297 00:17:41,440 --> 00:17:46,240 And so I sort of need u hat of zero to be zero for this. 298 00:17:46,240 --> 00:17:49,190 So it's looking slightly dangerous at k=0, 299 00:17:49,190 --> 00:17:53,880 but otherwise it's certainly improving things by speeding up 300 00:17:53,880 --> 00:17:55,100 the decay rate. 301 00:17:55,100 --> 00:17:59,920 And then this would be minus ik u hat of k, as we just said. 302 00:17:59,920 --> 00:18:02,170 And this would be the one, maybe it's 303 00:18:02,170 --> 00:18:05,350 a one in the Fourier integral world, 304 00:18:05,350 --> 00:18:07,930 where the 2pi went in a different point. 305 00:18:07,930 --> 00:18:11,680 So I think that's, yeah. 306 00:18:11,680 --> 00:18:14,220 So I would now just, the usual thing 307 00:18:14,220 --> 00:18:15,940 was the transfer function. 308 00:18:15,940 --> 00:18:21,630 1/(ik)-ik, so that's my transfer function. 309 00:18:21,630 --> 00:18:23,640 That's multiplying u hat. 310 00:18:23,640 --> 00:18:26,000 Maybe I can simplify that by putting it 311 00:18:26,000 --> 00:18:30,620 all over ik, one minus ik. 312 00:18:30,620 --> 00:18:35,900 Is it plus k? 313 00:18:35,900 --> 00:18:41,010 Does that look good, or does that look bad? k squared? 314 00:18:41,010 --> 00:18:44,840 Does that look better? 315 00:18:44,840 --> 00:18:52,920 So I'm wondering, so over here is k squared over ik, 316 00:18:52,920 --> 00:18:57,830 is that the same as the minus ik? 317 00:18:57,830 --> 00:18:58,570 Yeah. 318 00:18:58,570 --> 00:19:01,390 Yeah, bring it up and the minus i squared is one 319 00:19:01,390 --> 00:19:02,650 and I've got k squared. 320 00:19:02,650 --> 00:19:04,560 Yeah, so this looks good. 321 00:19:04,560 --> 00:19:10,250 So that's then, dividing by it gives me, I'm dividing the one. 322 00:19:10,250 --> 00:19:14,750 So u hat k is one over this. 323 00:19:14,750 --> 00:19:17,140 So it's the transfer function. 324 00:19:17,140 --> 00:19:19,550 One plus k squared. 325 00:19:19,550 --> 00:19:25,110 OK. 326 00:19:25,110 --> 00:19:27,500 I don't know whether I had in mind 327 00:19:27,500 --> 00:19:29,960 to go any further than this. 328 00:19:29,960 --> 00:19:32,350 Did you get this far, or what? 329 00:19:32,350 --> 00:19:34,520 AUDIENCE: [INAUDIBLE] 330 00:19:34,520 --> 00:19:36,700 PROFESSOR STRANG: It said that? 331 00:19:36,700 --> 00:19:40,500 Oh jeez. 332 00:19:40,500 --> 00:19:43,650 I bitterly regret saying these things. 333 00:19:43,650 --> 00:19:45,440 Yeah, OK. 334 00:19:45,440 --> 00:19:46,760 I don't know u(x). 335 00:19:46,760 --> 00:19:48,420 Did anybody have any luck? 336 00:19:48,420 --> 00:19:50,940 You did. 337 00:19:50,940 --> 00:19:53,290 Did I say it with a derivative? 338 00:19:53,290 --> 00:19:56,980 Oh one, over one plus k squared, oh yeah. 339 00:19:56,980 --> 00:20:01,220 One over one plus k squared, we know what to do. 340 00:20:01,220 --> 00:20:06,090 And then ik will take its derivative. 341 00:20:06,090 --> 00:20:09,150 So what do we do if it's one over one plus k squared? 342 00:20:09,150 --> 00:20:15,240 That's the one where, that was the guy which was the e^(-x), 343 00:20:15,240 --> 00:20:18,400 right? 344 00:20:18,400 --> 00:20:20,450 Oh, it's just one side. 345 00:20:20,450 --> 00:20:22,480 No, two sides, isn't it? 346 00:20:22,480 --> 00:20:23,380 Two sides, yeah. 347 00:20:23,380 --> 00:20:25,550 Yeah, we just got a corner. 348 00:20:25,550 --> 00:20:26,890 Yeah, OK. 349 00:20:26,890 --> 00:20:28,620 That's looking good. 350 00:20:28,620 --> 00:20:33,700 And now maybe I need to divide by two. a is one here 351 00:20:33,700 --> 00:20:37,670 but there is a division by two. 352 00:20:37,670 --> 00:20:39,450 Yeah, that looks good. 353 00:20:39,450 --> 00:20:44,450 OK, and now I take-- This function, 354 00:20:44,450 --> 00:20:47,080 its transform has that. 355 00:20:47,080 --> 00:20:49,270 And now if I want to multiply by ik 356 00:20:49,270 --> 00:20:50,560 I have to take the derivative. 357 00:20:50,560 --> 00:20:53,400 So I believe with your help here, 358 00:20:53,400 --> 00:20:58,460 that the derivative of that is the same. 359 00:20:58,460 --> 00:21:03,880 Can I just take the derivative while we're looking at it? 360 00:21:03,880 --> 00:21:07,010 I mean e^x is the most-- You know, 361 00:21:07,010 --> 00:21:10,430 that was created to take its derivative, right? 362 00:21:10,430 --> 00:21:12,570 So its derivative is itself. 363 00:21:12,570 --> 00:21:15,780 The derivative of this is a minus. 364 00:21:15,780 --> 00:21:22,530 So I think that maybe that's the answer. e^x over two, 365 00:21:22,530 --> 00:21:25,630 coming up to 1/2, I guess. 366 00:21:25,630 --> 00:21:28,840 And then-- Oh, well the picture won't look, 367 00:21:28,840 --> 00:21:30,720 because of that minus sign. 368 00:21:30,720 --> 00:21:31,310 Yeah. 369 00:21:31,310 --> 00:21:31,860 Huh. 370 00:21:31,860 --> 00:21:39,720 Good, because that minus sign, it's minus e^(-x) over two. 371 00:21:39,720 --> 00:21:41,500 So it starts at minus 1/2. 372 00:21:41,500 --> 00:21:43,570 So there's a jump of one. 373 00:21:43,570 --> 00:21:45,130 Or drop of one. 374 00:21:45,130 --> 00:21:51,120 Is that right, there should be a drop of one in u hat? 375 00:21:51,120 --> 00:21:53,070 Probably. 376 00:21:53,070 --> 00:21:54,010 Yeah, yeah. 377 00:21:54,010 --> 00:21:57,480 The integral will be smooth at zero, 378 00:21:57,480 --> 00:22:00,470 where the delta function's hitting us. 379 00:22:00,470 --> 00:22:02,460 I have a minus sign and a derivative. 380 00:22:02,460 --> 00:22:06,250 But yeah, doesn't that look good? 381 00:22:06,250 --> 00:22:11,560 If the derivative has a delta, the function has a jump. 382 00:22:11,560 --> 00:22:14,530 And with that minus sign, and with one delta, 383 00:22:14,530 --> 00:22:17,000 the jump should be a drop of one. 384 00:22:17,000 --> 00:22:17,780 Which is that. 385 00:22:17,780 --> 00:22:18,480 Yeah. 386 00:22:18,480 --> 00:22:20,930 And the integral minus the derivative 387 00:22:20,930 --> 00:22:23,590 should be otherwise zero. 388 00:22:23,590 --> 00:22:29,510 So again this is not unrelated to exam questions. 389 00:22:29,510 --> 00:22:33,370 I did all this stuff, and I really should check, 390 00:22:33,370 --> 00:22:34,330 did I get it right? 391 00:22:34,330 --> 00:22:37,010 Does that solve the differential equation? 392 00:22:37,010 --> 00:22:39,386 I mean, you could say of course it does if you 393 00:22:39,386 --> 00:22:40,510 took all these steps right. 394 00:22:40,510 --> 00:22:42,960 But it's wise to check. 395 00:22:42,960 --> 00:22:44,350 Plug it in. 396 00:22:44,350 --> 00:22:46,610 OK, it has the drop of one. 397 00:22:46,610 --> 00:22:51,660 So it deals correctly with the delta because, the derivative 398 00:22:51,660 --> 00:22:52,590 has a delta. 399 00:22:52,590 --> 00:22:55,170 And things match. 400 00:22:55,170 --> 00:22:58,360 And now what else do I have to check? 401 00:22:58,360 --> 00:23:00,890 I have to check out all the other points, right? 402 00:23:00,890 --> 00:23:05,500 I've just checked that yep, this answer is good at the jump. 403 00:23:05,500 --> 00:23:06,710 The tricky point. 404 00:23:06,710 --> 00:23:09,120 But now what about all the other points? 405 00:23:09,120 --> 00:23:13,540 So where the delta is zero, I should just 406 00:23:13,540 --> 00:23:18,380 check that the integral of u, so what is the integral of u? 407 00:23:18,380 --> 00:23:24,720 Let me just check for the points left of the origin. 408 00:23:24,720 --> 00:23:29,150 So I'm just going to look at this part. 409 00:23:29,150 --> 00:23:31,620 Yeah, x negative. 410 00:23:31,620 --> 00:23:36,460 So its integral, what's the integral of e^x over two? 411 00:23:36,460 --> 00:23:42,680 I guess e^x over two, right? 412 00:23:42,680 --> 00:23:43,990 Minus? 413 00:23:43,990 --> 00:23:45,920 And what's the derivative of e^x? 414 00:23:45,920 --> 00:23:50,520 Oh, the derivative is also e^x over two and I get zero, 415 00:23:50,520 --> 00:23:51,800 or x negative. 416 00:23:51,800 --> 00:23:53,940 As I want to. 417 00:23:53,940 --> 00:23:57,370 Because the delta is zero in that left side. 418 00:23:57,370 --> 00:23:58,840 And similarly on the right side. 419 00:23:58,840 --> 00:23:59,730 It should be OK. 420 00:23:59,730 --> 00:24:03,590 And then the key point was that the jump was right. 421 00:24:03,590 --> 00:24:04,620 Yeah, so thank you. 422 00:24:04,620 --> 00:24:06,340 That's good. 423 00:24:06,340 --> 00:24:06,940 Yes please. 424 00:24:06,940 --> 00:24:09,610 AUDIENCE: [INAUDIBLE] 425 00:24:09,610 --> 00:24:11,690 PROFESSOR STRANG: What, sorry? 426 00:24:11,690 --> 00:24:14,820 Why did I take the derivative? 427 00:24:14,820 --> 00:24:18,550 What will be the derivative of this? 428 00:24:18,550 --> 00:24:19,540 Yes, there is. 429 00:24:19,540 --> 00:24:20,140 That's right. 430 00:24:20,140 --> 00:24:22,270 What would be the derivative of this? 431 00:24:22,270 --> 00:24:23,800 Huh, yeah, good question. 432 00:24:23,800 --> 00:24:27,220 What's the derivative of this answer. 433 00:24:27,220 --> 00:24:33,070 So it's that, on the left. and then there's a delta function, 434 00:24:33,070 --> 00:24:36,870 a minus delta, right, because it's dropped down. 435 00:24:36,870 --> 00:24:40,080 And then the derivative of that is e^(-x). 436 00:24:40,080 --> 00:24:43,910 Yeah, so if I graph the derivative, cool. 437 00:24:43,910 --> 00:24:50,750 I graph the derivative, it looks like the function. 438 00:24:50,750 --> 00:24:56,750 It has a spike going down there. 439 00:24:56,750 --> 00:25:00,540 At the origin, and then the derivative here is positive. 440 00:25:00,540 --> 00:25:01,040 Right? 441 00:25:01,040 --> 00:25:03,030 This function's coming up. 442 00:25:03,030 --> 00:25:05,310 It's e^(-x) over two. 443 00:25:05,310 --> 00:25:06,730 It's positive, it's there. 444 00:25:06,730 --> 00:25:08,050 Yeah. 445 00:25:08,050 --> 00:25:11,050 Oh wow, that's a nice graph. 446 00:25:11,050 --> 00:25:17,620 So that's the derivative, this is the graph of du/dx. 447 00:25:17,620 --> 00:25:19,040 Yeah, thanks. 448 00:25:19,040 --> 00:25:23,850 And similarly I could graph the integral, the integral-- Well, 449 00:25:23,850 --> 00:25:26,730 would you want to see the integral? 450 00:25:26,730 --> 00:25:28,590 Probably not. 451 00:25:28,590 --> 00:25:30,810 Maybe, yeah. 452 00:25:30,810 --> 00:25:34,020 The integral would probably look pretty much just like that 453 00:25:34,020 --> 00:25:37,320 but without the delta, yeah. 454 00:25:37,320 --> 00:25:40,860 Cool, isn't that nice? 455 00:25:40,860 --> 00:25:44,040 It's artistic. 456 00:25:44,040 --> 00:25:45,820 That's the derivative, and the integral 457 00:25:45,820 --> 00:25:48,030 is the same thing without the delta. 458 00:25:48,030 --> 00:25:52,230 And when you subtract, you get the delta. 459 00:25:52,230 --> 00:25:57,890 OK, you sure you guys don't want to come to Singapore 460 00:25:57,890 --> 00:26:01,050 and help with these problem solutions? 461 00:26:01,050 --> 00:26:03,380 There's plenty for everybody. 462 00:26:03,380 --> 00:26:04,820 OK. 463 00:26:04,820 --> 00:26:07,760 Alright, so another question. 464 00:26:07,760 --> 00:26:10,310 Yeah. 465 00:26:10,310 --> 00:26:10,960 2005. 466 00:26:10,960 --> 00:26:11,835 AUDIENCE: [INAUDIBLE] 467 00:26:11,835 --> 00:26:14,020 PROFESSOR STRANG: OK, alright. 468 00:26:14,020 --> 00:26:14,860 Yeah, which one? 469 00:26:14,860 --> 00:26:19,501 AUDIENCE: [INAUDIBLE] 470 00:26:19,501 --> 00:26:22,000 PROFESSOR STRANG: OK, I'll just read out the whole question. 471 00:26:22,000 --> 00:26:26,040 Yeah. 472 00:26:26,040 --> 00:26:32,870 So this is 2005, Problem 3. 473 00:26:32,870 --> 00:26:37,970 OK it's a half-hat function. 474 00:26:37,970 --> 00:26:41,340 This is Fourier integral, because it's on the whole line. 475 00:26:41,340 --> 00:26:46,150 And it's half a hat for-- Between zero and one, 476 00:26:46,150 --> 00:26:47,950 the function is coming down. 477 00:26:47,950 --> 00:26:53,150 And then otherwise it's zero. 478 00:26:53,150 --> 00:26:55,390 So its graph, done. 479 00:26:55,390 --> 00:26:56,560 Graph its derivative. 480 00:26:56,560 --> 00:27:01,260 OK, let's graph the derivative of that function. 481 00:27:01,260 --> 00:27:04,250 Derivative is certainly zero along there. 482 00:27:04,250 --> 00:27:07,040 Then the derivative is minus one here. 483 00:27:07,040 --> 00:27:09,200 And then the derivative is zero. 484 00:27:09,200 --> 00:27:16,720 So the derivative looks to me like it's that, OK. 485 00:27:16,720 --> 00:27:21,160 So yeah, that function-- Is that right? 486 00:27:21,160 --> 00:27:21,660 No. 487 00:27:21,660 --> 00:27:24,430 It's not right. 488 00:27:24,430 --> 00:27:25,640 There's a delta. 489 00:27:25,640 --> 00:27:28,150 That was the tricky part of the problem. 490 00:27:28,150 --> 00:27:32,270 Right, there's a delta in the derivative, 491 00:27:32,270 --> 00:27:35,220 right here because of that jump. 492 00:27:35,220 --> 00:27:37,211 There's a delta. 493 00:27:37,211 --> 00:27:37,710 OK. 494 00:27:37,710 --> 00:27:39,210 Good. 495 00:27:39,210 --> 00:27:42,920 Now, is that better? 496 00:27:42,920 --> 00:27:43,480 That's good. 497 00:27:43,480 --> 00:27:44,550 OK. 498 00:27:44,550 --> 00:27:47,120 What's its transform? 499 00:27:47,120 --> 00:27:52,870 The transform of this derivative. 500 00:27:52,870 --> 00:27:54,120 What's the Fourier transform? 501 00:27:54,120 --> 00:27:56,510 Well, I guess we got two parts. 502 00:27:56,510 --> 00:27:59,660 So this is my function u'(x). 503 00:27:59,660 --> 00:28:00,620 The derivative. 504 00:28:00,620 --> 00:28:04,530 Now, what's u' transform? 505 00:28:04,530 --> 00:28:07,460 So it should be a function of k, now. 506 00:28:07,460 --> 00:28:09,960 Alright, what do you think? 507 00:28:09,960 --> 00:28:12,240 So it's the sum of two things. 508 00:28:12,240 --> 00:28:17,260 That delta, which Fourier transforms to one. 509 00:28:17,260 --> 00:28:23,940 And this guy down below between zero and one, 510 00:28:23,940 --> 00:28:28,110 which was like the one we did today, 511 00:28:28,110 --> 00:28:30,300 we've got the integral from zero to one. 512 00:28:30,300 --> 00:28:34,120 Now it happens to be a minus one e^(-ikx) dx. 513 00:28:34,120 --> 00:28:37,150 514 00:28:37,150 --> 00:28:44,140 And that, of course, we can do. 515 00:28:44,140 --> 00:28:47,110 OK Oh, there's solutions here. 516 00:28:47,110 --> 00:28:49,360 AUDIENCE: [INAUDIBLE] 517 00:28:49,360 --> 00:28:55,710 PROFESSOR STRANG: Yeah. 518 00:28:55,710 --> 00:28:58,930 So OK, so I think we're doing alright. 519 00:28:58,930 --> 00:29:03,720 This expression is familiar. 520 00:29:03,720 --> 00:29:09,500 It's 1-e^(-ik) over the ik. 521 00:29:09,500 --> 00:29:12,840 Is that right? 522 00:29:12,840 --> 00:29:13,700 We have a minus. 523 00:29:13,700 --> 00:29:18,600 Is the minus, yeah, I think the minus looks good. 524 00:29:18,600 --> 00:29:23,610 And yeah, at least, that's here and it's probably 525 00:29:23,610 --> 00:29:26,520 got the minus signs correct. 526 00:29:26,520 --> 00:29:27,790 So why was it? 527 00:29:27,790 --> 00:29:32,390 I plugged in x=1, and that's why I got an e^(-ik). 528 00:29:32,390 --> 00:29:35,430 And then I plugged in x=0 and that's where that one came 529 00:29:35,430 --> 00:29:36,150 from. 530 00:29:36,150 --> 00:29:42,850 OK, so, but your question was about-- That was it? 531 00:29:42,850 --> 00:29:43,410 Oh, I see. 532 00:29:43,410 --> 00:29:45,840 OK. 533 00:29:45,840 --> 00:29:49,910 Right, and then the next part asked about the transform 534 00:29:49,910 --> 00:29:51,830 of the original guy. 535 00:29:51,830 --> 00:29:53,720 The transform of the original guy. 536 00:29:53,720 --> 00:30:00,280 Now, what would be the transform of the original guy? 537 00:30:00,280 --> 00:30:11,620 The original u. 538 00:30:11,620 --> 00:30:17,310 Right, OK, yeah. 539 00:30:17,310 --> 00:30:19,060 The reason I'm sort of stuttering 540 00:30:19,060 --> 00:30:22,760 is that if I'm going to integrate-- 541 00:30:22,760 --> 00:30:25,240 I mean, what are you going to tell me? 542 00:30:25,240 --> 00:30:28,330 What's a quick way to find the transform of the original u 543 00:30:28,330 --> 00:30:30,060 now? 544 00:30:30,060 --> 00:30:31,870 Divide by ik. 545 00:30:31,870 --> 00:30:33,820 Divide by ik. 546 00:30:33,820 --> 00:30:37,670 And then I'm all ready to do that except I'm worried that 547 00:30:37,670 --> 00:30:39,300 k=0. 548 00:30:39,300 --> 00:30:41,730 But it should come out alright, now. 549 00:30:41,730 --> 00:30:47,900 So I have to hope that this thing comes out to be zero 550 00:30:47,900 --> 00:30:51,080 at k=0. 551 00:30:51,080 --> 00:30:54,950 Can you see that it does? 552 00:30:54,950 --> 00:30:57,880 What is, yeah? 553 00:30:57,880 --> 00:30:59,210 This is a good point. 554 00:30:59,210 --> 00:31:03,490 What does u hat at zero represent? 555 00:31:03,490 --> 00:31:05,620 Have you thought about that? 556 00:31:05,620 --> 00:31:08,120 If I look at the Fourier transform, which I've got here. 557 00:31:08,120 --> 00:31:10,530 So let me write what I've got here. 558 00:31:10,530 --> 00:31:17,170 I've got here that Fourier transform of this function. 559 00:31:17,170 --> 00:31:20,690 And if I take the Fourier transform of any function, 560 00:31:20,690 --> 00:31:22,580 and I look at zero frequency. 561 00:31:22,580 --> 00:31:24,820 What am I seeing? 562 00:31:24,820 --> 00:31:25,800 The average, right. 563 00:31:25,800 --> 00:31:27,240 I'm saying the average. 564 00:31:27,240 --> 00:31:32,734 What's the average value of that function? 565 00:31:32,734 --> 00:31:33,900 We never thought about that. 566 00:31:33,900 --> 00:31:35,120 But that's fun. 567 00:31:35,120 --> 00:31:37,650 What's the average of this guy? 568 00:31:37,650 --> 00:31:40,960 If I integrate over the whole line. 569 00:31:40,960 --> 00:31:42,560 Do I get zero? 570 00:31:42,560 --> 00:31:43,690 Yes. 571 00:31:43,690 --> 00:31:48,170 Because I get the integral of the delta part gives me a one. 572 00:31:48,170 --> 00:31:51,480 But the integral of this box part gives me a minus one. 573 00:31:51,480 --> 00:31:55,630 So the integral, so this does equal zero at k=0, 574 00:31:55,630 --> 00:32:00,840 and I can safely do that division and get, 575 00:32:00,840 --> 00:32:07,800 so now u hat of k is this expression divided by ik. 576 00:32:07,800 --> 00:32:11,090 So it's one minus this thing. 577 00:32:11,090 --> 00:32:12,600 I'm just copying. 578 00:32:12,600 --> 00:32:16,990 Minus ik, over ik, all that divided by ik. 579 00:32:16,990 --> 00:32:20,460 And I can, of course, maneuver that some more 580 00:32:20,460 --> 00:32:23,750 to make it look better. 581 00:32:23,750 --> 00:32:25,750 OK, these are good questions, because it's 582 00:32:25,750 --> 00:32:31,680 giving me a few functions to do the standard rules. 583 00:32:31,680 --> 00:32:39,010 Shifting, integrating, differentiating. 584 00:32:39,010 --> 00:32:40,550 Oh, here's a Christmas present. 585 00:32:40,550 --> 00:32:42,790 What does that mean? 586 00:32:42,790 --> 00:32:47,740 Is the convolution of this function with itself 587 00:32:47,740 --> 00:32:52,100 the whole hat? 588 00:32:52,100 --> 00:32:53,960 Why is that a Christmas present? 589 00:32:53,960 --> 00:32:57,510 Is the convolution, the convolution 590 00:32:57,510 --> 00:33:00,430 of that with itself. 591 00:33:00,430 --> 00:33:05,070 So first point, we don't want to compute a convolution. 592 00:33:05,070 --> 00:33:07,180 You may have noticed, you have not 593 00:33:07,180 --> 00:33:09,450 computed convolutions of things like this. 594 00:33:09,450 --> 00:33:13,000 It's not a lot of fun. 595 00:33:13,000 --> 00:33:15,410 Because you have that integral to do. 596 00:33:15,410 --> 00:33:18,630 And you'd have to separate out the parts 597 00:33:18,630 --> 00:33:20,760 where it's this and the part where it's this 598 00:33:20,760 --> 00:33:22,010 and the parts where it's that. 599 00:33:22,010 --> 00:33:23,350 It's not nice. 600 00:33:23,350 --> 00:33:27,860 Much better to multiply in the transform domain. 601 00:33:27,860 --> 00:33:28,850 Much better. 602 00:33:28,850 --> 00:33:33,710 OK, so if I multiply in the transform domain, 603 00:33:33,710 --> 00:33:40,420 I don't think I get the Fourier transform of the hat. 604 00:33:40,420 --> 00:33:43,670 Of the whole hat. 605 00:33:43,670 --> 00:33:46,230 But can you give me a convincing argument 606 00:33:46,230 --> 00:33:50,880 for why-- This is, here I'm in the transform domain 607 00:33:50,880 --> 00:33:52,770 and I'm going to convolve with itself, 608 00:33:52,770 --> 00:33:59,970 so I'm just going to square it. 609 00:33:59,970 --> 00:34:05,500 With the decay rate, what's the decay rate as it is now? 610 00:34:05,500 --> 00:34:09,120 The decay rate is, yes. 611 00:34:09,120 --> 00:34:11,290 The decay rate is a good key. 612 00:34:11,290 --> 00:34:14,050 So what is the decay rate right now? 613 00:34:14,050 --> 00:34:15,690 Here's a function with a jump. 614 00:34:15,690 --> 00:34:18,250 So I know that even though has a slightly messy 615 00:34:18,250 --> 00:34:21,730 looking transform, I know the decay rate. 616 00:34:21,730 --> 00:34:25,570 It's what, with a jump is 1/k. 617 00:34:25,570 --> 00:34:27,690 OK, so this must be, and I sort of 618 00:34:27,690 --> 00:34:30,630 do see a k down here, so that's sensible. 619 00:34:30,630 --> 00:34:35,770 OK, now, if I convolve, then I multiply in this domain. 620 00:34:35,770 --> 00:34:38,590 So that would give me the k squared, as you said. 621 00:34:38,590 --> 00:34:43,380 So could that be the transform of the whole hat? 622 00:34:43,380 --> 00:34:50,060 Does the whole hat have a-- ooh, could be. 623 00:34:50,060 --> 00:34:53,310 I'm pretty sure the answer's no. 624 00:34:53,310 --> 00:34:54,740 Yeah, no way. 625 00:34:54,740 --> 00:34:59,290 Yeah, I think that that's the best answer. 626 00:34:59,290 --> 00:35:03,050 I've lost the reason. 627 00:35:03,050 --> 00:35:06,560 Because I'm getting the k squared decay rate. 628 00:35:06,560 --> 00:35:08,590 If I convolve something with itself, 629 00:35:08,590 --> 00:35:13,160 that would give me a k squared, and the decay rate 630 00:35:13,160 --> 00:35:16,070 for the transform of the hat is a k squared. 631 00:35:16,070 --> 00:35:18,840 But they wouldn't be the same, yeah. 632 00:35:18,840 --> 00:35:21,580 OK, yeah, I won't. 633 00:35:21,580 --> 00:35:22,240 Yes, please. 634 00:35:22,240 --> 00:35:29,990 AUDIENCE: [INAUDIBLE] 635 00:35:29,990 --> 00:35:30,990 PROFESSOR STRANG: Right. 636 00:35:30,990 --> 00:35:32,350 It's just convention. 637 00:35:32,350 --> 00:35:36,160 The difference was that that was the Fourier integral problem. 638 00:35:36,160 --> 00:35:37,920 Where we put, you might have noticed 639 00:35:37,920 --> 00:35:45,970 that the 2pi goes on the other-- And this was the Fourier series 640 00:35:45,970 --> 00:35:46,470 problem. 641 00:35:46,470 --> 00:35:47,870 That was its only difference. 642 00:35:47,870 --> 00:35:50,270 So it's just a convention. 643 00:35:50,270 --> 00:35:54,800 And maybe other people would choose a different convention. 644 00:35:54,800 --> 00:35:58,940 So that is a slight wiggle in the presentation 645 00:35:58,940 --> 00:36:03,770 that the 2pi in the Fourier integral section 646 00:36:03,770 --> 00:36:09,830 got moved to the transform in the other direction. 647 00:36:09,830 --> 00:36:12,450 Yeah, good. 648 00:36:12,450 --> 00:36:13,160 Yes, thanks. 649 00:36:13,160 --> 00:36:16,480 AUDIENCE: [INAUDIBLE] 650 00:36:16,480 --> 00:36:19,920 PROFESSOR STRANG: Oh, OK. 651 00:36:19,920 --> 00:36:24,310 I'm really just going by this rule that if there is a jump 652 00:36:24,310 --> 00:36:28,280 in the function, and nothing worse, 653 00:36:28,280 --> 00:36:32,520 then the decay rate is 1/k, and if there's a jump 654 00:36:32,520 --> 00:36:37,170 in the derivative, as there would be here -- 655 00:36:37,170 --> 00:36:40,400 so that has a jump in slope -- then the decay rate is one over 656 00:36:40,400 --> 00:36:42,960 k squared and so on. 657 00:36:42,960 --> 00:36:44,490 But those are the main cases. 658 00:36:44,490 --> 00:36:50,000 Yeah, so I'm not using anything deep there. 659 00:36:50,000 --> 00:36:54,010 And by the way, I realize here that one MATLAB problem 660 00:36:54,010 --> 00:36:58,710 that I didn't assign this semester but it's quite fun, 661 00:36:58,710 --> 00:37:03,120 is to actually see the Gibbs phenomenon. 662 00:37:03,120 --> 00:37:09,240 We know the terms in the Fourier series, with a jump. 663 00:37:09,240 --> 00:37:16,590 And if you computed-- Make it the periodic case, 664 00:37:16,590 --> 00:37:20,020 if you've computed the terms in the Fourier series 665 00:37:20,020 --> 00:37:24,140 they would stay near here and then 666 00:37:24,140 --> 00:37:27,320 I'd see the famous Gibbs stuff here. 667 00:37:27,320 --> 00:37:30,470 So that-- You know, it's quite pleasant 668 00:37:30,470 --> 00:37:37,370 to see the printout of the Gibbs phenomenon. 669 00:37:37,370 --> 00:37:40,950 Not getting any shorter, just moving 670 00:37:40,950 --> 00:37:44,810 closer to the jump as you increase the number of terms. 671 00:37:44,810 --> 00:37:48,920 Yeah, it's quite interesting. 672 00:37:48,920 --> 00:37:50,510 But couldn't do everything. 673 00:37:50,510 --> 00:37:51,760 Yeah, ready for more. 674 00:37:51,760 --> 00:37:54,150 Any questions? 675 00:37:54,150 --> 00:37:55,230 Well, these are good. 676 00:37:55,230 --> 00:37:57,130 Yes, thanks. 677 00:37:57,130 --> 00:37:58,800 AUDIENCE: [INAUDIBLE] 678 00:37:58,800 --> 00:38:00,591 PROFESSOR STRANG: With cyclic convolutions. 679 00:38:00,591 --> 00:38:13,670 OK, what could I do with cyclic convolutions? 680 00:38:13,670 --> 00:38:17,440 Let's see. 681 00:38:17,440 --> 00:38:20,770 What should I do with cyclic convolutions? 682 00:38:20,770 --> 00:38:30,240 I better make a little space. 683 00:38:30,240 --> 00:38:31,880 Let me make a little space and think. 684 00:38:31,880 --> 00:38:43,460 Anybody got a suggested problem for a cyclic convolution? 685 00:38:43,460 --> 00:38:47,450 I mean, one type of problem would certainly 686 00:38:47,450 --> 00:38:59,140 be if I gave you a cyclic convolution 687 00:38:59,140 --> 00:39:05,030 and asked for-- I mean, the direct way would be to say OK, 688 00:39:05,030 --> 00:39:10,430 what's the cyclic convolution of (1, 4, 2) cyclically 689 00:39:10,430 --> 00:39:14,070 with (2, 1, 3) or something. 690 00:39:14,070 --> 00:39:16,280 But you could-- That's just a calculation, 691 00:39:16,280 --> 00:39:19,050 so you would get that. 692 00:39:19,050 --> 00:39:22,170 I won't discuss that further. 693 00:39:22,170 --> 00:39:28,240 Now suppose I ask it with the unknown, the thing 694 00:39:28,240 --> 00:39:30,000 that I don't know here? 695 00:39:30,000 --> 00:39:31,440 So let me do that. 696 00:39:31,440 --> 00:39:40,790 Shall I say three, yeah. (3, 1, 1). 697 00:39:40,790 --> 00:39:48,320 Cyclically convolved with some unknown (x_0, x_1, x_2) 698 00:39:48,320 --> 00:39:51,140 equals some right-hand side, what 699 00:39:51,140 --> 00:39:55,010 should we take for the right-hand side? (1, 1, 1), 700 00:39:55,010 --> 00:39:57,250 for example? 701 00:39:57,250 --> 00:40:05,350 OK, I'll erase the top one. 702 00:40:05,350 --> 00:40:14,350 OK, how would you go with that? 703 00:40:14,350 --> 00:40:17,470 So that's three by three, three unknowns. 704 00:40:17,470 --> 00:40:20,620 It's linear, so somehow I could write that 705 00:40:20,620 --> 00:40:23,960 as three equations in three unknowns. 706 00:40:23,960 --> 00:40:28,430 The question is how do I get some insight into this. 707 00:40:28,430 --> 00:40:30,390 And when I'm seeing a convolution, what's 708 00:40:30,390 --> 00:40:34,690 my immediate thought? 709 00:40:34,690 --> 00:40:36,370 Transform. 710 00:40:36,370 --> 00:40:39,010 That's what this month is all about. 711 00:40:39,010 --> 00:40:42,710 November. 712 00:40:42,710 --> 00:40:46,010 So I'll take the DFT of everything. 713 00:40:46,010 --> 00:40:51,240 So what's the discrete Fourier transform of (1, 1, 1)? 714 00:40:51,240 --> 00:40:54,160 Up there is going to be the transform. 715 00:40:54,160 --> 00:40:59,830 So let me call it x_0 hat, x_1 hat, x_2 hat. 716 00:40:59,830 --> 00:41:01,160 Just to emphasize. 717 00:41:01,160 --> 00:41:07,800 So N is three here, in this problem. 718 00:41:07,800 --> 00:41:12,530 Or maybe c_0, c_1, c_2 you might prefer. 719 00:41:12,530 --> 00:41:13,830 Whatever. 720 00:41:13,830 --> 00:41:16,150 And now this is going to be a multiplication, 721 00:41:16,150 --> 00:41:20,620 so in MATLAB notation will be a dot star, right? 722 00:41:20,620 --> 00:41:24,660 And so now I have to-- These are the known ones, 723 00:41:24,660 --> 00:41:28,100 so what's the discrete Fourier transform of (1, 1, 1), 724 00:41:28,100 --> 00:41:31,270 the cyclic convolution of that? 725 00:41:31,270 --> 00:41:33,330 It's (1, 0, 0) is it? 726 00:41:33,330 --> 00:41:35,510 Or it's (3, 0, 0)? 727 00:41:35,510 --> 00:41:42,710 Is it three-- I guess when I asked that question, 728 00:41:42,710 --> 00:41:44,760 you're open to two answers. 729 00:41:44,760 --> 00:41:51,510 You could either be going from x space to frequency space 730 00:41:51,510 --> 00:41:52,760 or frequency to the other. 731 00:41:52,760 --> 00:41:57,080 Because I haven't told you what space we're in. 732 00:41:57,080 --> 00:41:59,870 OK, let's go with that one. 733 00:41:59,870 --> 00:42:07,830 How did you get to there? 734 00:42:07,830 --> 00:42:10,770 You multiplied by the three by three Fourier matrix, 735 00:42:10,770 --> 00:42:12,410 is that what you did? 736 00:42:12,410 --> 00:42:15,160 Alright, so now can you do that here? 737 00:42:15,160 --> 00:42:18,180 So separately I have to do a little multiplication 738 00:42:18,180 --> 00:42:20,240 of the three by three Fourier matrix. 739 00:42:20,240 --> 00:42:22,780 One, one-- Sorry, let me get it right. 740 00:42:22,780 --> 00:42:27,160 So I'm going to transform this. 741 00:42:27,160 --> 00:42:28,890 Make a little space here. 742 00:42:28,890 --> 00:42:29,800 Transform that. 743 00:42:29,800 --> 00:42:34,370 So it's 1, 1, 1, 1, 1 and this is 744 00:42:34,370 --> 00:42:38,190 w, w squared, w squared and w to the fourth, 745 00:42:38,190 --> 00:42:40,230 which is the same as w. 746 00:42:40,230 --> 00:42:44,760 OK, and now I want to multiply that by (3, 1, 1). 747 00:42:44,760 --> 00:42:47,780 And get the answer there. 748 00:42:47,780 --> 00:42:50,500 OK, so what do I have? 749 00:42:50,500 --> 00:42:52,050 Five. 750 00:42:52,050 --> 00:42:53,470 And what is that other one? 751 00:42:53,470 --> 00:42:54,100 Oh, boy. 752 00:42:54,100 --> 00:42:59,490 What is three plus w plus w squared? 753 00:42:59,490 --> 00:43:02,160 Anybody know that one? 754 00:43:02,160 --> 00:43:07,360 Well, I do know what one plus w plus w squared is. 755 00:43:07,360 --> 00:43:09,560 It is zero. 756 00:43:09,560 --> 00:43:12,570 Why is one plus w plus w squared zero? 757 00:43:12,570 --> 00:43:15,740 So I claim that if I have three plus-- I think the answer 758 00:43:15,740 --> 00:43:17,210 there is two. 759 00:43:17,210 --> 00:43:20,540 And here I think I've also got three, w, w squared, 760 00:43:20,540 --> 00:43:24,060 I think if it was a one the answer would be a zero. 761 00:43:24,060 --> 00:43:26,810 But it's a three, so I still have two there. 762 00:43:26,810 --> 00:43:29,400 I think that's probably right. 763 00:43:29,400 --> 00:43:32,410 Can I just remind you why? 764 00:43:32,410 --> 00:43:38,760 There's one, there's w, and there's w squared 765 00:43:38,760 --> 00:43:40,060 and they add to zero. 766 00:43:40,060 --> 00:43:42,930 Yeah, yeah. 767 00:43:42,930 --> 00:43:47,020 For many reasons. 768 00:43:47,020 --> 00:43:51,540 That's a very, very, handy property. 769 00:43:51,540 --> 00:43:55,450 That the sum of the nth roots of one 770 00:43:55,450 --> 00:43:58,710 add to zero, OK, so now where I am I? 771 00:43:58,710 --> 00:44:04,030 So I got (5, 2, 2). 772 00:44:04,030 --> 00:44:05,690 So I've transformed. 773 00:44:05,690 --> 00:44:09,980 Now what do I do next? 774 00:44:09,980 --> 00:44:11,290 I divide. 775 00:44:11,290 --> 00:44:18,070 So this is telling me that this is three separate equations. 776 00:44:18,070 --> 00:44:23,910 There's three, 2x_1 hat is zero. 777 00:44:23,910 --> 00:44:24,700 Right? 778 00:44:24,700 --> 00:44:26,110 No, x_2 hat. 779 00:44:26,110 --> 00:44:28,740 And 2x_2 hat. 780 00:44:28,740 --> 00:44:31,090 Sorry, one was right. 781 00:44:31,090 --> 00:44:35,120 And 2x_2 hat is zero. 782 00:44:35,120 --> 00:44:40,510 I'm just doing the multiplication 783 00:44:40,510 --> 00:44:41,780 as I'm supposed to. 784 00:44:41,780 --> 00:44:46,380 In the other space it's at each, for each component separately. 785 00:44:46,380 --> 00:44:52,890 So now I know x_0 hat is 3/5, and the others are zero. 786 00:44:52,890 --> 00:44:55,380 So now I know what these numbers are. 787 00:44:55,380 --> 00:45:01,270 Oh, yeah. (3/5, 0, 0). 788 00:45:01,270 --> 00:45:08,200 And now what? 789 00:45:08,200 --> 00:45:11,360 I mean, that statement's clearly true, right? 790 00:45:11,360 --> 00:45:14,320 Point star means five times this gives me three. 791 00:45:14,320 --> 00:45:17,700 Two times zero gives me zero, two times zero gives zero. 792 00:45:17,700 --> 00:45:21,590 So I'm golden, but I'm not finished. 793 00:45:21,590 --> 00:45:26,510 So that's not x, that's its transform. 794 00:45:26,510 --> 00:45:29,780 So what do I have to do? 795 00:45:29,780 --> 00:45:30,930 Go back. 796 00:45:30,930 --> 00:45:36,460 OK, so now can you tell me what this is? 797 00:45:36,460 --> 00:45:40,810 Up to a factor of 1/N, or N, which we may 798 00:45:40,810 --> 00:45:45,340 have to figure out separately. 799 00:45:45,340 --> 00:45:50,970 So apart from this possibly appearing, and probably 800 00:45:50,970 --> 00:45:55,460 appearing, give me an idea of what's 801 00:45:55,460 --> 00:46:01,710 the inverse transform of that? (3/5, 3/5, 3/5), good. 802 00:46:01,710 --> 00:46:06,400 3/5, all three times. 803 00:46:06,400 --> 00:46:16,140 OK, and now I can check to see if it's right. 804 00:46:16,140 --> 00:46:21,820 And again, I have not attempted to get the 1/N's right, 805 00:46:21,820 --> 00:46:23,420 I've just waited to the end. 806 00:46:23,420 --> 00:46:25,510 So now I'll just do this convolution 807 00:46:25,510 --> 00:46:28,150 and see if I really do get (1, 1, 1). 808 00:46:28,150 --> 00:46:30,990 So can we do that convolution? 809 00:46:30,990 --> 00:46:36,540 What's the typical, say, this guy. 810 00:46:36,540 --> 00:46:40,310 Where does that come from in the convolution? 811 00:46:40,310 --> 00:46:45,460 It comes from this, times what? 812 00:46:45,460 --> 00:46:46,570 This. 813 00:46:46,570 --> 00:46:48,280 That's the constant, right? 814 00:46:48,280 --> 00:46:49,700 And that's the constant. 815 00:46:49,700 --> 00:46:51,570 Contributes to the constant. 816 00:46:51,570 --> 00:46:55,050 And what other products contribute to the constant? 817 00:46:55,050 --> 00:47:00,860 So we're really seeing-- Thanks you for this question. 818 00:47:00,860 --> 00:47:13,450 So this one multiplies which one? 819 00:47:13,450 --> 00:47:16,790 The answer is, this one multiplies this guy. 820 00:47:16,790 --> 00:47:19,620 You may say don't worry me about it because they're the same. 821 00:47:19,620 --> 00:47:21,600 But it's nice to know. 822 00:47:21,600 --> 00:47:23,800 So this one multiplies this one. 823 00:47:23,800 --> 00:47:25,860 And this guy multiplies this guy. 824 00:47:25,860 --> 00:47:27,720 And we'll remember in a minute why. 825 00:47:27,720 --> 00:47:30,210 And then you add them up, that's what convolution is. 826 00:47:30,210 --> 00:47:38,100 So I'm getting 9/5+3/5+3/5 is 15/5, so that's three. 827 00:47:38,100 --> 00:47:43,470 So what's your conclusion here? 828 00:47:43,470 --> 00:47:45,010 I should divide by three, right? 829 00:47:45,010 --> 00:47:46,780 So I should divide by three. 830 00:47:46,780 --> 00:47:48,730 So that's easy to do. 831 00:47:48,730 --> 00:47:49,660 Those will all be one. 832 00:47:49,660 --> 00:47:52,130 Yeah, OK. 833 00:47:52,130 --> 00:47:54,350 So that's the right answer, right? 834 00:47:54,350 --> 00:47:59,620 And then I check that it's right. 835 00:47:59,620 --> 00:48:04,380 So let me just close by remembering why was it-- 836 00:48:04,380 --> 00:48:08,790 So I'm doing just a forward convolution here. 837 00:48:08,790 --> 00:48:11,610 Solving the equation was a deconvolution, 838 00:48:11,610 --> 00:48:12,830 figuring out that x. 839 00:48:12,830 --> 00:48:15,950 But now that I've got it and I'm just checking, 840 00:48:15,950 --> 00:48:17,580 that's just do it forward and see 841 00:48:17,580 --> 00:48:19,130 if you got the right answer. 842 00:48:19,130 --> 00:48:24,220 And so we're just remembering what 843 00:48:24,220 --> 00:48:30,520 it means to do a convolution. 844 00:48:30,520 --> 00:48:34,500 So you remember the point about convolutions. 845 00:48:34,500 --> 00:48:41,010 That if I'm convolving (a, b, c) cyclically with (d, e, 846 00:48:41,010 --> 00:48:45,450 f), that's a constant term. 847 00:48:45,450 --> 00:48:48,720 Just tell me again, now, what's the constant terms in this? 848 00:48:48,720 --> 00:48:55,080 It's ad plus b times what? 849 00:48:55,080 --> 00:49:01,830 This is the first term in the convolution. 850 00:49:01,830 --> 00:49:06,420 You remember that, the day we brought convolutions 851 00:49:06,420 --> 00:49:07,840 into the course? 852 00:49:07,840 --> 00:49:13,100 What does b multiply? f, why f. 853 00:49:13,100 --> 00:49:21,140 Because b is the coefficient of w, if I'm taking polynomials, 854 00:49:21,140 --> 00:49:24,920 this is a plus bw plus c w squared, 855 00:49:24,920 --> 00:49:29,740 and this corresponds to d plus ew plus f w squared. 856 00:49:29,740 --> 00:49:32,060 And I'm looking for the constant. 857 00:49:32,060 --> 00:49:35,440 So I get an a times a d gives me a constant. 858 00:49:35,440 --> 00:49:42,230 bw times what? times f w squared gives me 859 00:49:42,230 --> 00:49:46,240 the w cubed, which is the one, so that bf here 860 00:49:46,240 --> 00:49:48,860 and then there's a ce. 861 00:49:48,860 --> 00:49:53,910 Because c w squared multiplying ew 862 00:49:53,910 --> 00:49:57,650 gives me the w cubed, the one which folds in. 863 00:49:57,650 --> 00:50:01,670 Yeah, OK, so anyway. 864 00:50:01,670 --> 00:50:04,460 That's a quick reminder of cyclic convolutions. 865 00:50:04,460 --> 00:50:07,590 That was a good question to ask, yeah. 866 00:50:07,590 --> 00:50:12,440 You know, I fully realize this, especially this third topic, 867 00:50:12,440 --> 00:50:14,460 lots of things like this. 868 00:50:14,460 --> 00:50:16,870 Straightforward if you have lots of time, 869 00:50:16,870 --> 00:50:20,390 but we had to keep moving. 870 00:50:20,390 --> 00:50:25,950 So remembering what cyclic convolution was 871 00:50:25,950 --> 00:50:29,240 is a good chance to do it.