1 00:00:07,980 --> 00:00:09,430 PROFESSOR: Hi, and welcome. 2 00:00:09,430 --> 00:00:11,580 Today we're going to do a problem about least 3 00:00:11,580 --> 00:00:13,910 squares approximations. 4 00:00:13,910 --> 00:00:15,510 Here's the question. 5 00:00:15,510 --> 00:00:18,740 Find the quadratic equation through the origin that 6 00:00:18,740 --> 00:00:24,090 is a best fit for these three points: (1, 1), (2, 5), 7 00:00:24,090 --> 00:00:27,060 and (-1, -2). 8 00:00:27,060 --> 00:00:30,150 I'll give you a minute to work that out on your own. 9 00:00:30,150 --> 00:00:32,090 You can hit the pause button now, 10 00:00:32,090 --> 00:00:35,440 and we'll be back in a minute to work the problem out together. 11 00:00:47,430 --> 00:00:50,690 OK and we're back. 12 00:00:50,690 --> 00:00:53,790 So what's the first step in a problem like this? 13 00:00:53,790 --> 00:00:58,220 The first step is figuring out what our equation looks like, 14 00:00:58,220 --> 00:01:00,280 the one that we're going to find. 15 00:01:00,280 --> 00:01:07,740 So our equation is going to look like c*t plus d t squared 16 00:01:07,740 --> 00:01:10,150 equals y. 17 00:01:10,150 --> 00:01:11,790 So that's what we're looking for. 18 00:01:11,790 --> 00:01:15,780 We're looking for a quadratic equation through the origin. 19 00:01:15,780 --> 00:01:17,740 Now, if it were just any quadratic equation, 20 00:01:17,740 --> 00:01:19,530 then we would have a constant term, 21 00:01:19,530 --> 00:01:21,410 but through the origin just means 22 00:01:21,410 --> 00:01:25,730 that the constant term is 0. 23 00:01:25,730 --> 00:01:28,250 Now, what's the next step? 24 00:01:28,250 --> 00:01:31,980 Next we need to set up a matrix equation. 25 00:01:31,980 --> 00:01:43,890 So we need an A, a matrix A. And which 26 00:01:43,890 --> 00:01:46,740 matrix is that going to be? 27 00:01:46,740 --> 00:01:51,360 Well, let me start with the first coordinate of these three 28 00:01:51,360 --> 00:01:52,790 points. 29 00:01:52,790 --> 00:01:57,450 And I'm going to put them in the first column of this matrix, 1, 30 00:01:57,450 --> 00:01:59,200 2, -1. 31 00:01:59,200 --> 00:02:01,700 And then my second column is going 32 00:02:01,700 --> 00:02:08,410 to be the squares of these coordinates, 1, 4, and 1. 33 00:02:08,410 --> 00:02:14,490 And my x hat-- that's just c and d. 34 00:02:14,490 --> 00:02:21,960 And my b-- this is going to be the second coordinates, 1, 5, 35 00:02:21,960 --> 00:02:22,460 and -2. 36 00:02:25,820 --> 00:02:28,960 OK, why did I set this problem up like that? 37 00:02:28,960 --> 00:02:34,950 Well, multiply A times x hat. 38 00:02:34,950 --> 00:02:38,970 The first coordinate of A times x hat is 1 times 39 00:02:38,970 --> 00:02:41,720 c plus 1 times d. 40 00:02:41,720 --> 00:02:44,290 It's just the same as plugging in this first point 41 00:02:44,290 --> 00:02:48,670 into the left-hand side of this equation. 42 00:02:48,670 --> 00:02:53,060 And similarly, if I took the second point 43 00:02:53,060 --> 00:02:55,630 and plugged it in here, I would just get 2 times 44 00:02:55,630 --> 00:02:59,090 c plus 4 times d, which is just the same 45 00:02:59,090 --> 00:03:02,840 as the second coordinate in the multiplication A times x hat. 46 00:03:02,840 --> 00:03:06,490 2 times c plus 4 times d. 47 00:03:06,490 --> 00:03:08,120 OK, and where did the b come from? 48 00:03:08,120 --> 00:03:10,840 Well, b came from plugging in these points 49 00:03:10,840 --> 00:03:12,680 to the right-hand side. 50 00:03:12,680 --> 00:03:16,740 So 1, 5, and -2 are just the y-coordinates 51 00:03:16,740 --> 00:03:19,720 of these three points. 52 00:03:19,720 --> 00:03:27,070 So it would be great if we could solve A x hat equals b. 53 00:03:27,070 --> 00:03:33,680 But we can't solve A x hat equals 54 00:03:33,680 --> 00:03:38,610 b, because there isn't a quadratic equation 55 00:03:38,610 --> 00:03:41,500 through the origin that contains these three points. 56 00:03:41,500 --> 00:03:44,880 But we need to find the best approximation to that, 57 00:03:44,880 --> 00:03:48,340 so that's the same as solving A x hat equals 58 00:03:48,340 --> 00:03:51,660 the projection of b onto the column space of A. 59 00:03:51,660 --> 00:03:54,150 Because we only have a chance of solving A x hat equals 60 00:03:54,150 --> 00:03:57,150 something if it's in the column space of A. 61 00:03:57,150 --> 00:04:01,760 And remember from our projections part of the class 62 00:04:01,760 --> 00:04:13,600 that this is the same as solving A transpose A x hat equals 63 00:04:13,600 --> 00:04:14,340 A transpose b. 64 00:04:17,180 --> 00:04:19,540 So that's what we're really going to do. 65 00:04:19,540 --> 00:04:21,710 We're really going to solve A transpose 66 00:04:21,710 --> 00:04:24,640 A x hat equals A transpose b. 67 00:04:24,640 --> 00:04:27,840 And now all we have to do is just a computation. 68 00:04:27,840 --> 00:04:29,400 So let's do it. 69 00:04:32,190 --> 00:04:36,210 So what is A transpose A? 70 00:04:36,210 --> 00:04:37,850 I'm going to do this kind of quickly, 71 00:04:37,850 --> 00:04:41,400 because you should be good at multiplying matrices by now. 72 00:04:56,130 --> 00:05:02,350 Oh this-- I have the negative in the wrong place, thank you. 73 00:05:02,350 --> 00:05:03,475 Thanks for that correction. 74 00:05:03,475 --> 00:05:06,470 I have the negative backwards there. 75 00:05:10,390 --> 00:05:13,065 And what do I get when I multiply these? 76 00:05:13,065 --> 00:05:14,340 I'll let you check. 77 00:05:14,340 --> 00:05:17,952 This is 6, 8; 8, and 10. 78 00:05:22,560 --> 00:05:24,650 And what is A transpose b? 79 00:05:34,620 --> 00:05:37,970 Well, if you multiply this out, I'll 80 00:05:37,970 --> 00:05:43,480 let you check that you get 13 and 19. 81 00:05:46,540 --> 00:05:49,780 So this is just some computation. 82 00:05:49,780 --> 00:05:54,600 And so what we're solving here is [6, 8; 8, 10] 83 00:05:54,600 --> 00:05:59,720 times x hat equals [13; 19]. 84 00:05:59,720 --> 00:06:04,640 And we remember how to do this just by using elimination. 85 00:06:04,640 --> 00:06:08,690 We replace the second row by 3 times the second row 86 00:06:08,690 --> 00:06:11,090 minus 4 times the first row. 87 00:06:11,090 --> 00:06:13,900 Again, I'm going to do this quickly because you know this 88 00:06:13,900 --> 00:06:15,150 from other parts of the class. 89 00:06:18,280 --> 00:06:21,400 You similarly change the b vector, 90 00:06:21,400 --> 00:06:30,750 and we backsolve to get d is -5/2 91 00:06:30,750 --> 00:06:35,940 and c equals-- let's plug that in 92 00:06:35,940 --> 00:06:40,330 and see-- think you're going to get 11 over 2. 93 00:06:40,330 --> 00:06:44,330 So what's our final equation? 94 00:06:44,330 --> 00:06:56,365 Our final equation is y equals 11/2 t minus 5/2 t squared. 95 00:07:04,280 --> 00:07:04,780 OK. 96 00:07:04,780 --> 00:07:08,520 So this is our best fit quadratic equation 97 00:07:08,520 --> 00:07:10,150 through the origin. 98 00:07:10,150 --> 00:07:13,400 Now before I end, let me do a couple things. 99 00:07:13,400 --> 00:07:18,460 First, let me go back and review what the key steps were. 100 00:07:18,460 --> 00:07:23,380 Whenever you're faced with such a best fit equation, first, 101 00:07:23,380 --> 00:07:26,670 you want to see what the general form of the equation is. 102 00:07:26,670 --> 00:07:29,490 Next you want to write it in terms of matrices. 103 00:07:29,490 --> 00:07:33,780 Write down your matrix A and your vector b. 104 00:07:33,780 --> 00:07:38,750 And lastly, you set up your projection equation. 105 00:07:38,750 --> 00:07:43,480 And then all you have to do is just a computation. 106 00:07:43,480 --> 00:07:45,290 But it might also be worth noting 107 00:07:45,290 --> 00:07:48,230 that these three points certainly aren't 108 00:07:48,230 --> 00:07:51,210 on this quadratic equation. 109 00:07:51,210 --> 00:07:54,680 For instance, if I plug in 1 here, I don't get (1, 1), I get 110 00:07:54,680 --> 00:07:56,560 (1, 3). 111 00:07:56,560 --> 00:07:58,230 But that's OK. 112 00:07:58,230 --> 00:08:01,910 This is as close as we can do. 113 00:08:01,910 --> 00:08:03,318 Thanks.