1 00:00:00,000 --> 00:00:02,330 The following content is provided under a Creative 2 00:00:02,330 --> 00:00:03,610 Commons license. 3 00:00:03,610 --> 00:00:05,990 Your support will help MIT OpenCourseWare 4 00:00:05,990 --> 00:00:10,010 continue to offer high quality educational resources for free. 5 00:00:10,010 --> 00:00:12,540 To make a donation, or to view additional materials 6 00:00:12,540 --> 00:00:15,870 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:15,870 --> 00:00:21,422 at ocw.mit.edu. 8 00:00:21,422 --> 00:00:23,380 PROFESSOR: What we're going to talk about today 9 00:00:23,380 --> 00:00:26,750 is a continuation of last time. 10 00:00:26,750 --> 00:00:29,600 I want to review Newton's method because I want to talk 11 00:00:29,600 --> 00:00:41,220 to you about its accuracy. 12 00:00:41,220 --> 00:00:46,090 So if you remember, the way Newton's method works is this. 13 00:00:46,090 --> 00:00:49,280 If you have a curve and you want to know 14 00:00:49,280 --> 00:00:52,642 whether it crosses the axis. 15 00:00:52,642 --> 00:00:54,100 And you don't know where this point 16 00:00:54,100 --> 00:01:00,100 is, this point which I'll call x here, what you do 17 00:01:00,100 --> 00:01:01,250 is you take a guess. 18 00:01:01,250 --> 00:01:03,590 Maybe you take a point x_0 here. 19 00:01:03,590 --> 00:01:06,640 And then you go down to this point on the graph, 20 00:01:06,640 --> 00:01:08,602 and you draw the tangent line. 21 00:01:08,602 --> 00:01:10,560 I'll draw these in a couple of different colors 22 00:01:10,560 --> 00:01:13,180 so that you can see the difference between them. 23 00:01:13,180 --> 00:01:14,410 So here's a tangent line. 24 00:01:14,410 --> 00:01:18,280 It's coming out like that. 25 00:01:18,280 --> 00:01:20,750 And that one is going to get a little closer 26 00:01:20,750 --> 00:01:23,090 to our target point. 27 00:01:23,090 --> 00:01:26,550 But now the trick is, and this is rather hard 28 00:01:26,550 --> 00:01:30,060 to see because the scale gets small incredibly fast, 29 00:01:30,060 --> 00:01:32,500 is that if you go right up from that, 30 00:01:32,500 --> 00:01:34,420 and you do this same trick over again. 31 00:01:34,420 --> 00:01:38,400 That is, this is your second guess, x_1, and now 32 00:01:38,400 --> 00:01:41,360 you draw the second tangent line. 33 00:01:41,360 --> 00:01:44,680 Which is going to come down this way. 34 00:01:44,680 --> 00:01:46,290 That's really close. 35 00:01:46,290 --> 00:01:48,570 You can see here on the chalkboard, 36 00:01:48,570 --> 00:01:51,740 it's practically the same as the dot of x. 37 00:01:51,740 --> 00:01:54,610 So that's the next guess. 38 00:01:54,610 --> 00:01:56,860 Which is x_2. 39 00:01:56,860 --> 00:02:03,620 And I want to analyze, now, how close it gets. 40 00:02:03,620 --> 00:02:05,990 And just describe to you how it works. 41 00:02:05,990 --> 00:02:09,060 So let me just remind you of the formulas, too. 42 00:02:09,060 --> 00:02:11,150 It's worth having them in your head. 43 00:02:11,150 --> 00:02:19,240 So the formula for the next one is this. 44 00:02:19,240 --> 00:02:23,570 And then the idea is just to repeat this process. 45 00:02:23,570 --> 00:02:28,720 Which has a fancy name, in algorithms, which 46 00:02:28,720 --> 00:02:31,210 is to iterate, if you like. 47 00:02:31,210 --> 00:02:32,530 So we repeat the process. 48 00:02:32,530 --> 00:02:35,720 And that means, for example, we generate x_2 from x_1 49 00:02:35,720 --> 00:02:41,730 by the same formula. 50 00:02:41,730 --> 00:02:43,170 And we did this last time. 51 00:02:43,170 --> 00:02:47,930 And, more generally, the (n+1)st is generated from the nth 52 00:02:47,930 --> 00:02:55,150 guess, by this formula here. 53 00:02:55,150 --> 00:03:02,820 So what I'd like to do is just draw the picture of one step 54 00:03:02,820 --> 00:03:03,920 a little bit more closely. 55 00:03:03,920 --> 00:03:05,580 So I want to blow up the picture, which 56 00:03:05,580 --> 00:03:10,880 is above me there. 57 00:03:10,880 --> 00:03:11,950 That's a little too high. 58 00:03:11,950 --> 00:03:16,750 Where are my erasers? 59 00:03:16,750 --> 00:03:18,390 Got to get it a little lower than that, 60 00:03:18,390 --> 00:03:21,890 since I'm going to depict everything above the line here. 61 00:03:21,890 --> 00:03:24,180 So here's my curve coming down. 62 00:03:24,180 --> 00:03:30,620 And suppose that x_1 is here, so this is directly 63 00:03:30,620 --> 00:03:32,170 above it is this point here. 64 00:03:32,170 --> 00:03:35,040 And then as I drew it, this green 65 00:03:35,040 --> 00:03:38,530 tangent coming down like that. 66 00:03:38,530 --> 00:03:42,970 It's a little bit closer, and this was the place, x_2, 67 00:03:42,970 --> 00:03:46,440 and then here was x, our target, which 68 00:03:46,440 --> 00:03:49,390 is where the curve crosses as opposed to the straight tangent 69 00:03:49,390 --> 00:03:52,600 line crossing. 70 00:03:52,600 --> 00:03:56,310 So that's the picture that I want you to keep in mind. 71 00:03:56,310 --> 00:04:00,440 And now, we're just going to do a very qualitative kind 72 00:04:00,440 --> 00:04:03,450 of error analysis. 73 00:04:03,450 --> 00:04:12,960 So here's our error analysis. 74 00:04:12,960 --> 00:04:19,355 And we're starting out, the distance between x_1 and x 75 00:04:19,355 --> 00:04:20,480 is what we want to measure. 76 00:04:20,480 --> 00:04:23,530 In other words, how close we are to where we're heading. 77 00:04:23,530 --> 00:04:26,600 And so I've called that, I'm going to call that Error 1. 78 00:04:26,600 --> 00:04:31,440 That's x - x1, in absolute value. 79 00:04:31,440 --> 00:04:37,010 And then, E_2 would be x - x_2, in absolute value. 80 00:04:37,010 --> 00:04:39,100 And so forth. 81 00:04:39,100 --> 00:04:44,190 And, last time, when I was estimating the size of this-- 82 00:04:44,190 --> 00:04:45,870 So E_n would be whatever it was. 83 00:04:45,870 --> 00:04:51,900 Last time, remember, we did it for a specific case. 84 00:04:51,900 --> 00:04:56,020 So last time, I actually wrote down the numbers. 85 00:04:56,020 --> 00:04:57,700 And they were these numbers, maybe 86 00:04:57,700 --> 00:05:00,365 you could call them E_n, which was the absolute value 87 00:05:00,365 --> 00:05:03,270 of square root of 5 minus x_n. 88 00:05:03,270 --> 00:05:06,610 These are the sizes that I was writing down last time. 89 00:05:06,610 --> 00:05:11,620 And I just want to talk about in general what to expect. 90 00:05:11,620 --> 00:05:13,480 That worked amazingly well, and I 91 00:05:13,480 --> 00:05:16,990 want to show you that that's true pretty much in general. 92 00:05:16,990 --> 00:05:23,660 So the first distance, again, is E_1, is this distance here. 93 00:05:23,660 --> 00:05:25,190 That's the E_1. 94 00:05:25,190 --> 00:05:27,570 And then this shorter distance, here, 95 00:05:27,570 --> 00:05:37,410 this little bit, which I'll mark maybe in green, is E_2. 96 00:05:37,410 --> 00:05:40,970 So how much shorter is E_1 than E_2? 97 00:05:40,970 --> 00:05:44,080 Well, the idea is pretty simple. 98 00:05:44,080 --> 00:05:46,895 It's that if this distance and this vertical distance, 99 00:05:46,895 --> 00:05:49,520 they are probably about the same as the perpendicular distance. 100 00:05:49,520 --> 00:05:51,550 And this is basically the situation 101 00:05:51,550 --> 00:05:54,140 of a curve touching a tangent line. 102 00:05:54,140 --> 00:05:58,010 Then the separation is going to be quadratic. 103 00:05:58,010 --> 00:06:00,150 And that's basically what's going to happen. 104 00:06:00,150 --> 00:06:03,780 So, in other words the distance E_2 105 00:06:03,780 --> 00:06:06,530 is going to be about the square of the distance E_1. 106 00:06:06,530 --> 00:06:10,579 And that's really the only feature of this 107 00:06:10,579 --> 00:06:11,620 that I want to point out. 108 00:06:11,620 --> 00:06:14,175 So, approximately, this is the situation 109 00:06:14,175 --> 00:06:17,880 that we're going to get. 110 00:06:17,880 --> 00:06:21,690 And so what that means is, and maybe 111 00:06:21,690 --> 00:06:24,380 thinking from last time, what we had was 112 00:06:24,380 --> 00:06:25,760 something roughly like this. 113 00:06:25,760 --> 00:06:29,060 You have an E_0, you have an E_1, you have an E_2, 114 00:06:29,060 --> 00:06:31,590 you have an E_3, and so forth. 115 00:06:31,590 --> 00:06:34,370 Maybe I'll write down E_4 here. 116 00:06:34,370 --> 00:06:37,330 And last time, this was about 10^(-1). 117 00:06:37,330 --> 00:06:39,650 So the expectation based on this rule 118 00:06:39,650 --> 00:06:42,710 is that the next error's the square of the previous one. 119 00:06:42,710 --> 00:06:44,500 So that's 10^(-2). 120 00:06:44,500 --> 00:06:47,140 The next one is the square of the previous one. 121 00:06:47,140 --> 00:06:49,090 So that's 10^(-4). 122 00:06:49,090 --> 00:06:52,069 And the next one is the square of that, that's 10^(-8). 123 00:06:52,069 --> 00:06:53,110 And this one is 10^(-16). 124 00:06:55,980 --> 00:06:59,550 So the thing that's impressive about this list of numbers 125 00:06:59,550 --> 00:07:02,780 is you can see that the number of digits of accuracy 126 00:07:02,780 --> 00:07:09,590 doubles at each stage. 127 00:07:09,590 --> 00:07:20,760 Accuracy doubles at each step. 128 00:07:20,760 --> 00:07:23,710 The number of digits of accuracy doubles at each step. 129 00:07:23,710 --> 00:07:28,150 So, very, very quickly you get past the accuracy 130 00:07:28,150 --> 00:07:31,150 of your calculator, as you saw on your problem set. 131 00:07:31,150 --> 00:07:34,220 And this thing works beautifully. 132 00:07:34,220 --> 00:07:42,700 So, let me just summarize by saying that Newton's method 133 00:07:42,700 --> 00:07:49,680 works very well. 134 00:07:49,680 --> 00:07:52,250 By which I mean this kind of rate. 135 00:07:52,250 --> 00:07:55,450 And I want to be just slightly specific. 136 00:07:55,450 --> 00:08:00,490 If-- there are really two conditions disguised in this, 137 00:08:00,490 --> 00:08:01,480 that are going on. 138 00:08:01,480 --> 00:08:09,890 One is that f' has to be, not to big-- to be not small. 139 00:08:09,890 --> 00:08:19,624 And f'' has to be not too big. 140 00:08:19,624 --> 00:08:21,290 That's roughly speaking what's going on. 141 00:08:21,290 --> 00:08:23,320 I'll explain these in just a second. 142 00:08:23,320 --> 00:08:34,050 And x_0 starts nearby. 143 00:08:34,050 --> 00:08:37,490 Nearby the target value x. 144 00:08:37,490 --> 00:08:42,050 So that's really what's going on here. 145 00:08:42,050 --> 00:08:44,610 So let me just illustrate to you. 146 00:08:44,610 --> 00:08:48,920 So I'm not going to explain this, except to say the reason 147 00:08:48,920 --> 00:08:51,120 why this second derivative gets involved 148 00:08:51,120 --> 00:08:53,360 is that it's how curved the curve is, 149 00:08:53,360 --> 00:08:55,330 that how far away you get. 150 00:08:55,330 --> 00:08:57,507 If the second derivative were 0, that 151 00:08:57,507 --> 00:08:58,840 would be the best possible case. 152 00:08:58,840 --> 00:09:00,660 Then we would get it on the nose. 153 00:09:00,660 --> 00:09:02,890 If the second derivative is not too big, 154 00:09:02,890 --> 00:09:05,360 that means the quadratic part is not too big. 155 00:09:05,360 --> 00:09:07,820 So we don't get away very far from the green line 156 00:09:07,820 --> 00:09:14,110 to the curve itself. 157 00:09:14,110 --> 00:09:17,460 The other thing to say is, as I said, 158 00:09:17,460 --> 00:09:19,050 that x_0 needs to start nearby. 159 00:09:19,050 --> 00:09:21,520 So I'll explain that by explaining 160 00:09:21,520 --> 00:09:23,310 what maybe could go wrong. 161 00:09:23,310 --> 00:09:40,080 So the ways the method can fail, and one example which actually 162 00:09:40,080 --> 00:09:45,080 would have happened in our example from last time, 163 00:09:45,080 --> 00:09:51,750 which was y = x^2 - 5, is suppose we'd started x_0 over 164 00:09:51,750 --> 00:09:54,840 here. 165 00:09:54,840 --> 00:09:57,430 Then this thing would have gone off to the left, 166 00:09:57,430 --> 00:10:00,713 and we would have landed on not the square root 167 00:10:00,713 --> 00:10:03,030 of 5 but the other root. 168 00:10:03,030 --> 00:10:11,750 So if it's too far away, then we get the wrong root. 169 00:10:11,750 --> 00:10:15,590 So that's an example, explaining that the x_0 needs 170 00:10:15,590 --> 00:10:18,710 to start near the root that we're talking about. 171 00:10:18,710 --> 00:10:21,460 Otherwise, the method doesn't know 172 00:10:21,460 --> 00:10:22,940 which root you're asking for. 173 00:10:22,940 --> 00:10:24,580 It only knows where you started. 174 00:10:24,580 --> 00:10:27,310 So it may go off to the wrong place. 175 00:10:27,310 --> 00:10:34,430 OK, it can't read your mind. 176 00:10:34,430 --> 00:10:35,350 Yes, question. 177 00:10:35,350 --> 00:10:42,610 STUDENT: [INAUDIBLE] 178 00:10:42,610 --> 00:10:44,460 PROFESSOR: Oh, good question. 179 00:10:44,460 --> 00:10:49,550 So the question was, what if the first error is larger than 1. 180 00:10:49,550 --> 00:10:51,220 Are you in trouble? 181 00:10:51,220 --> 00:10:56,000 And the answer is, absolutely, yes. 182 00:10:56,000 --> 00:10:58,110 If you have quadratic behavior, you can see. 183 00:10:58,110 --> 00:11:00,800 If you have a quadratic nearby, it's 184 00:11:00,800 --> 00:11:02,820 pretty close to the straight line. 185 00:11:02,820 --> 00:11:05,900 But far away, a parabola is miles from a straight line. 186 00:11:05,900 --> 00:11:08,180 It's way, way, way far away. 187 00:11:08,180 --> 00:11:14,010 So if you're foolish enough to start over here, 188 00:11:14,010 --> 00:11:17,150 you may have some trouble making progress. 189 00:11:17,150 --> 00:11:20,430 Actually, it isn't, when I-- that little wiggle 190 00:11:20,430 --> 00:11:22,010 there just meant proportional to. 191 00:11:22,010 --> 00:11:24,570 In fact, in the particular case of a parabola, 192 00:11:24,570 --> 00:11:25,810 it manages to get back. 193 00:11:25,810 --> 00:11:27,150 It saves itself. 194 00:11:27,150 --> 00:11:30,010 But there's no guarantee of that sort of thing. 195 00:11:30,010 --> 00:11:32,790 You really do want to start reasonably close. 196 00:11:32,790 --> 00:11:33,350 Yep. 197 00:11:33,350 --> 00:11:40,140 STUDENT: [INAUDIBLE] 198 00:11:40,140 --> 00:11:42,580 PROFESSOR: What you have to do is you have to watch out. 199 00:11:42,580 --> 00:11:46,790 That is, it's hard to know what assumptions to make about x_0. 200 00:11:46,790 --> 00:11:50,250 You plug it into the machine and you see what you get. 201 00:11:50,250 --> 00:11:53,227 And either it works or it doesn't. 202 00:11:53,227 --> 00:11:55,560 You can tell that it's marching toward a specific place, 203 00:11:55,560 --> 00:11:58,400 and you can tell that that place probably is a zero, usually. 204 00:11:58,400 --> 00:12:00,510 But maybe it's not the one you were looking for. 205 00:12:00,510 --> 00:12:02,384 So in other words, you have to use your head. 206 00:12:02,384 --> 00:12:05,230 You run the program and then you see what it does. 207 00:12:05,230 --> 00:12:08,000 And if you're lucky-- the problem is, if you have no idea 208 00:12:08,000 --> 00:12:12,270 where the zero is, you may just wander around forever. 209 00:12:12,270 --> 00:12:14,720 As we'll see in a second. 210 00:12:14,720 --> 00:12:20,870 So the next example here is the following. 211 00:12:20,870 --> 00:12:24,590 I said that f' has to be not too small. 212 00:12:24,590 --> 00:12:26,970 There's a real catastrophe hiding just 213 00:12:26,970 --> 00:12:28,250 inside this picture. 214 00:12:28,250 --> 00:12:30,935 Which is the transition between when you find the positive root 215 00:12:30,935 --> 00:12:32,890 and when you find the negative root here. 216 00:12:32,890 --> 00:12:35,480 Which is, if you're right down here. 217 00:12:35,480 --> 00:12:37,860 If you were foolish enough to get 0, 218 00:12:37,860 --> 00:12:41,060 then what's going to happen is your tangent line 219 00:12:41,060 --> 00:12:41,940 is horizontal. 220 00:12:41,940 --> 00:12:44,610 It doesn't even meet the axis. 221 00:12:44,610 --> 00:12:47,840 So in the formula, you can see that's a catastrophe. 222 00:12:47,840 --> 00:12:53,030 Because there's an f' in the denominator. 223 00:12:53,030 --> 00:12:53,530 So that's 0. 224 00:12:53,530 --> 00:12:54,791 That's undefined. 225 00:12:54,791 --> 00:12:56,290 It's not surprising, it's consistent 226 00:12:56,290 --> 00:12:59,480 that the parallel line doesn't meet the axis. 227 00:12:59,480 --> 00:13:01,940 And you have no x_1. 228 00:13:01,940 --> 00:13:06,370 So you had-- So if you like, another point here 229 00:13:06,370 --> 00:13:12,870 is that f' = 0 is a disaster. 230 00:13:12,870 --> 00:13:24,310 A disaster for the method. 231 00:13:24,310 --> 00:13:28,620 Because the next-- So say, if f(x_0) = 0, 232 00:13:28,620 --> 00:13:34,700 then x_1 is undefined. 233 00:13:34,700 --> 00:13:39,170 And finally, there's one other weird thing that can happen. 234 00:13:39,170 --> 00:13:43,460 Which is, which I'll just draw a picture of schematically. 235 00:13:43,460 --> 00:13:47,150 Which you can get from a wiggle. 236 00:13:47,150 --> 00:13:49,260 So this wiggle has three roots. 237 00:13:49,260 --> 00:13:52,060 The way I've drawn it. 238 00:13:52,060 --> 00:13:56,720 And it can be that you can start over here with your x_0. 239 00:13:56,720 --> 00:14:01,200 And draw your tangent line and go over here to an x_1. 240 00:14:01,200 --> 00:14:05,510 And then that tangent line will take you right back to the x_0. 241 00:14:05,510 --> 00:14:09,560 I didn't draw it quite right, but that's about right. 242 00:14:09,560 --> 00:14:11,330 So it goes over like this. 243 00:14:11,330 --> 00:14:13,520 So let me draw the two tangent lines, so 244 00:14:13,520 --> 00:14:14,760 that you can see it properly. 245 00:14:14,760 --> 00:14:16,614 Sorry, I messed it up. 246 00:14:16,614 --> 00:14:18,030 So here are the two tangent lines. 247 00:14:18,030 --> 00:14:20,870 This guy and this guy. 248 00:14:20,870 --> 00:14:25,950 And it just goes back and forth. x_0 cycles to x_1, 249 00:14:25,950 --> 00:14:27,520 and x_1 goes back to x_0. 250 00:14:27,520 --> 00:14:30,200 We have a cycle. 251 00:14:30,200 --> 00:14:32,090 And it never goes anywhere. 252 00:14:32,090 --> 00:14:34,915 This is, the grass is always greener. 253 00:14:34,915 --> 00:14:36,540 It's over here, it thinks, oh, I really 254 00:14:36,540 --> 00:14:39,000 would prefer to go to this zero and then it 255 00:14:39,000 --> 00:14:40,840 thinks oh, I want to go back. 256 00:14:40,840 --> 00:14:43,650 And it goes back and forth, and back and forth. 257 00:14:43,650 --> 00:14:47,060 Grass is always greener on the other side of the fence. 258 00:14:47,060 --> 00:14:50,090 Never, never gets anywhere. 259 00:14:50,090 --> 00:14:52,510 So those are the sorts of things that can go wrong 260 00:14:52,510 --> 00:14:53,490 with Newton's method. 261 00:14:53,490 --> 00:14:55,210 Nevertheless, it's fantastic. 262 00:14:55,210 --> 00:14:59,740 It works very well, in a lot of situations. 263 00:14:59,740 --> 00:15:03,510 And solves basically any equation that you can imagine, 264 00:15:03,510 --> 00:15:10,894 numerically. 265 00:15:10,894 --> 00:15:12,060 Next we're going to move on. 266 00:15:12,060 --> 00:15:13,518 We're going to move on to something 267 00:15:13,518 --> 00:15:15,730 which is a little theoretical. 268 00:15:15,730 --> 00:15:18,170 Which is the mean value theorem. 269 00:15:18,170 --> 00:15:22,410 And that will allow us in just a day or so 270 00:15:22,410 --> 00:15:26,150 to launch into the ideas of integration, 271 00:15:26,150 --> 00:15:31,140 which is the whole second half of the course. 272 00:15:31,140 --> 00:15:50,200 So let's get started with that. 273 00:15:50,200 --> 00:15:57,030 The mean value theorem will henceforth be abbreviated MVT. 274 00:15:57,030 --> 00:15:58,630 So I don't have to write quite as 275 00:15:58,630 --> 00:16:03,750 much every time I refer to it. 276 00:16:03,750 --> 00:16:07,290 The mean value theorem, colloquially, 277 00:16:07,290 --> 00:16:09,530 says the following. 278 00:16:09,530 --> 00:16:23,040 If you go from Boston to LA, which I think a lot of Red Sox 279 00:16:23,040 --> 00:16:31,110 fans are going to want to do soon, so that's 3,000 miles. 280 00:16:31,110 --> 00:16:47,800 In 6 hours, then at some time you 281 00:16:47,800 --> 00:16:55,640 are going at a certain speed. 282 00:16:55,640 --> 00:17:01,360 The average of this speed. 283 00:17:01,360 --> 00:17:08,850 Average, so, speed, which in this case is what? 284 00:17:08,850 --> 00:17:10,770 So we're going at the average speed. 285 00:17:10,770 --> 00:17:16,890 That's 3,000 miles times 6 hours, 286 00:17:16,890 --> 00:17:21,420 so that's 500 miles per hour. 287 00:17:21,420 --> 00:17:23,340 Exactly. 288 00:17:23,340 --> 00:17:26,442 So some time on your journey-- of course, some of the time 289 00:17:26,442 --> 00:17:28,150 you're going more than 500 miles an hour, 290 00:17:28,150 --> 00:17:29,610 sometimes you are going less. 291 00:17:29,610 --> 00:17:33,220 And some time you must've been going 500 miles an hour 292 00:17:33,220 --> 00:17:35,110 exactly. 293 00:17:35,110 --> 00:17:37,520 That's the mean value theorem. 294 00:17:37,520 --> 00:17:39,890 The reason why it's called mean value theorem 295 00:17:39,890 --> 00:17:55,440 is that word mean is the same as the word average. 296 00:17:55,440 --> 00:18:08,160 So now I'm going to state it in math symbols, the same theorem. 297 00:18:08,160 --> 00:18:10,470 And it's a formula. 298 00:18:10,470 --> 00:18:16,275 It says that the difference quotient 299 00:18:16,275 --> 00:18:20,410 - so this is the distance traveled 300 00:18:20,410 --> 00:18:25,570 divided by the time elapsed, that's the average speed - 301 00:18:25,570 --> 00:18:35,320 is equal to the infinitesimal speed for some time in between. 302 00:18:35,320 --> 00:18:48,590 So some c, which is in between a and b-- I'm not quite done. 303 00:18:48,590 --> 00:18:53,130 It's a real theorem, it has hypotheses. 304 00:18:53,130 --> 00:18:56,230 I've told you the conclusion first, 305 00:18:56,230 --> 00:18:58,030 but there are some hypotheses, they're 306 00:18:58,030 --> 00:18:59,650 straightforward hypotheses. 307 00:18:59,650 --> 00:19:03,320 Provided f is differentiable, that is, 308 00:19:03,320 --> 00:19:12,820 it has a derivative in the interval a < x < b. 309 00:19:12,820 --> 00:19:21,560 And continuous in a <= x <= b. 310 00:19:29,110 --> 00:19:32,080 There has to be a sense that you can make out of the speed, 311 00:19:32,080 --> 00:19:36,140 or the rate of change of f at each intermediate point. 312 00:19:36,140 --> 00:19:40,200 And in order for the values at the ends to make sense, 313 00:19:40,200 --> 00:19:41,200 it has to be continuous. 314 00:19:41,200 --> 00:19:44,470 There has to be a link between the values at the ends 315 00:19:44,470 --> 00:19:47,409 and what's going on in between. 316 00:19:47,409 --> 00:19:48,950 If it were discontinuous, there would 317 00:19:48,950 --> 00:19:52,330 be no relation between the left and right values 318 00:19:52,330 --> 00:19:55,970 and the rest of the function. 319 00:19:55,970 --> 00:20:00,160 So here's the theorem, conclusion and its hypothesis. 320 00:20:00,160 --> 00:20:11,780 And it means what I said more colloquially up above. 321 00:20:11,780 --> 00:20:14,540 Now, I'm going to prove this theorem immediately. 322 00:20:14,540 --> 00:20:18,110 At least, give a geometric intuitive argument, 323 00:20:18,110 --> 00:20:21,220 which is not very different from the one that's given 324 00:20:21,220 --> 00:20:26,410 in a very systematic treatment. 325 00:20:26,410 --> 00:20:34,280 So here's the proof of the mean value theorem. 326 00:20:34,280 --> 00:20:36,860 It's really just a picture. 327 00:20:36,860 --> 00:20:42,880 So here's a place, and here's another place on the graph. 328 00:20:42,880 --> 00:20:47,180 And the graph is going along like this, let's say. 329 00:20:47,180 --> 00:20:50,700 And this line here is the secant line. 330 00:20:50,700 --> 00:20:55,310 So this is (a, f(a)) down here. 331 00:20:55,310 --> 00:20:59,320 And this is (b, f(b)) up there. 332 00:20:59,320 --> 00:21:02,720 And this segment is the secant, its slope 333 00:21:02,720 --> 00:21:04,480 is the slope that we're aiming for. 334 00:21:04,480 --> 00:21:09,400 The slope of that line is the left-hand side of this formula 335 00:21:09,400 --> 00:21:11,190 here. 336 00:21:11,190 --> 00:21:14,062 So we need to find something with that slope. 337 00:21:14,062 --> 00:21:16,520 And what we need to find is a tangent line with that slope, 338 00:21:16,520 --> 00:21:18,340 because what's on the right-hand side 339 00:21:18,340 --> 00:21:20,220 is the slope of a tangent line. 340 00:21:20,220 --> 00:21:22,340 So here's how we construct it. 341 00:21:22,340 --> 00:21:27,240 We take a parallel line, down here. 342 00:21:27,240 --> 00:21:31,125 And then we just translate it up, leaving it parallel, 343 00:21:31,125 --> 00:21:32,750 we move it up. 344 00:21:32,750 --> 00:21:34,080 Towards this one. 345 00:21:34,080 --> 00:21:38,410 Until it touches. 346 00:21:38,410 --> 00:21:43,040 And where it touches, at this point of tangency, down there, 347 00:21:43,040 --> 00:21:47,490 I've just found my value of c. 348 00:21:47,490 --> 00:21:49,830 And you can see that if the tangent line is parallel 349 00:21:49,830 --> 00:21:53,180 to this line, that's exactly the equation we want. 350 00:21:53,180 --> 00:21:59,910 So this thing has slope f'(c). 351 00:21:59,910 --> 00:22:07,150 And this other one has slope equal to this complicated 352 00:22:07,150 --> 00:22:15,970 expression, (f(b) - f(a)) / (b - a). 353 00:22:15,970 --> 00:22:19,840 That is almost the end of the proof. 354 00:22:19,840 --> 00:22:25,330 There's one problem. 355 00:22:25,330 --> 00:22:29,880 So, again, we move a parallel line up. 356 00:22:29,880 --> 00:22:43,310 Move up the parallel line until it touches. 357 00:22:43,310 --> 00:22:46,300 There's a little subtlety here, which I just want to emphasize. 358 00:22:46,300 --> 00:22:49,210 Which is that that dotted line keeps on going here. 359 00:22:49,210 --> 00:22:51,220 But when we bring it up, we're going 360 00:22:51,220 --> 00:22:54,500 to ignore what's happening outside of a. 361 00:22:54,500 --> 00:22:56,870 And beyond b, alright? 362 00:22:56,870 --> 00:23:01,590 So we're just going to ignore the rest of the graph. 363 00:23:01,590 --> 00:23:06,090 But there is one thing that can go wrong. 364 00:23:06,090 --> 00:23:16,800 So if it does not touch, then the picture looks likes this. 365 00:23:16,800 --> 00:23:18,230 Here are the same two points. 366 00:23:18,230 --> 00:23:20,240 And the graph is all above. 367 00:23:20,240 --> 00:23:22,060 And we brought up our thing. 368 00:23:22,060 --> 00:23:23,960 And it went like that. 369 00:23:23,960 --> 00:23:27,480 So we didn't construct a tangent line. 370 00:23:27,480 --> 00:23:29,520 If this happens. 371 00:23:29,520 --> 00:23:31,980 So we're in trouble, in that point. 372 00:23:31,980 --> 00:23:37,010 In this situation, sorry. 373 00:23:37,010 --> 00:23:40,860 But there's a trick, which is a straightforward trick. 374 00:23:40,860 --> 00:23:55,810 Then bring the tangent lines down from the top. 375 00:23:55,810 --> 00:23:58,040 So parallel lines, sorry, not tangent lines. 376 00:23:58,040 --> 00:24:06,580 Parallel lines. 377 00:24:06,580 --> 00:24:11,600 From above. 378 00:24:11,600 --> 00:24:16,380 So, that's the whole story. 379 00:24:16,380 --> 00:24:43,460 That's how we cook up this point c, with the right properties. 380 00:24:43,460 --> 00:24:46,640 I want to point out just one more theoretical thing. 381 00:24:46,640 --> 00:24:50,087 And then the rest, we're going to be drawing conclusions. 382 00:24:50,087 --> 00:24:51,670 So there's one more theoretical remark 383 00:24:51,670 --> 00:24:55,490 about the proof, which is something that is 384 00:24:55,490 --> 00:24:59,130 fairly important to understand. 385 00:24:59,130 --> 00:25:01,640 When you understand a proof, you should always 386 00:25:01,640 --> 00:25:06,170 be thinking about why the hypotheses are necessary. 387 00:25:06,170 --> 00:25:08,280 Where do I use the hypothesis. 388 00:25:08,280 --> 00:25:10,890 And I want to give you an example the proof doesn't 389 00:25:10,890 --> 00:25:15,350 work to show you that the hypothesis is an important one. 390 00:25:15,350 --> 00:25:17,230 So the example is the following. 391 00:25:17,230 --> 00:25:18,710 I'll just take a function which is 392 00:25:18,710 --> 00:25:22,400 two straight lines like this. 393 00:25:22,400 --> 00:25:27,920 And if you try to perform this trick with these things, 394 00:25:27,920 --> 00:25:32,380 then it's going to come up and it's going to touch here. 395 00:25:32,380 --> 00:25:36,070 But the problem is that the tangent line is not 396 00:25:36,070 --> 00:25:36,740 defined here. 397 00:25:36,740 --> 00:25:38,660 There are lots of tangents, and there's 398 00:25:38,660 --> 00:25:40,610 no derivative at this point. 399 00:25:40,610 --> 00:25:44,860 So the derivative doesn't exist here. 400 00:25:44,860 --> 00:25:57,230 So this is the claim that one bad point ruins the proof. 401 00:25:57,230 --> 00:26:08,730 We need f' to exist at all-- so, f'(x) to exist at all x 402 00:26:08,730 --> 00:26:14,340 in between. 403 00:26:14,340 --> 00:26:30,980 Can't get away even with one defective point. 404 00:26:30,980 --> 00:26:40,620 Now it's time to draw some consequences. 405 00:26:40,620 --> 00:26:45,670 And the main consequence is going 406 00:26:45,670 --> 00:26:57,880 to have to do with applications to graphing. 407 00:26:57,880 --> 00:27:01,090 But we'll see tomorrow and for the rest of the course 408 00:27:01,090 --> 00:27:03,920 that this is even more significant. 409 00:27:03,920 --> 00:27:09,790 It's significant to all the rest of calculus. 410 00:27:09,790 --> 00:27:12,440 I'm going to list three consequences which you're 411 00:27:12,440 --> 00:27:14,890 quite familiar with already. 412 00:27:14,890 --> 00:27:27,550 So, the first one is if f' is positive, then f is increasing. 413 00:27:27,550 --> 00:27:40,430 And the second one is if f' is negative, then f is decreasing. 414 00:27:40,430 --> 00:27:44,130 And the last one seems like the simplest. 415 00:27:44,130 --> 00:27:48,460 But even this one alone is the key to everything. 416 00:27:48,460 --> 00:28:03,340 If f' = 0, then f is constant. 417 00:28:03,340 --> 00:28:13,490 These are three consequences, now, of the mean value theorem. 418 00:28:13,490 --> 00:28:17,130 And let me show you how they're proved. 419 00:28:17,130 --> 00:28:22,870 I just told you that they were true, maybe, a while ago. 420 00:28:22,870 --> 00:28:27,030 And certainly I mentioned the first two. 421 00:28:27,030 --> 00:28:30,200 The last one was so simple that we maybe just swept it 422 00:28:30,200 --> 00:28:30,810 under the rug. 423 00:28:30,810 --> 00:28:36,570 You did use it on a problem set, once or twice. 424 00:28:36,570 --> 00:28:39,140 But it turns out that this actually requires proof, 425 00:28:39,140 --> 00:28:48,010 and we're going to give the proof right now. 426 00:28:48,010 --> 00:28:52,190 The way that the proof goes is simply to write down, 427 00:28:52,190 --> 00:28:54,500 to rewrite star. 428 00:28:54,500 --> 00:28:59,440 Rewrite our formula. 429 00:28:59,440 --> 00:29:05,440 Which says that (f(b) - f(a)) / (b - a) = f'(c). 430 00:29:10,050 --> 00:29:13,730 And you see I've written it from left to right here 431 00:29:13,730 --> 00:29:16,110 to say that the right-hand side information 432 00:29:16,110 --> 00:29:17,590 about the derivative is going to be 433 00:29:17,590 --> 00:29:19,340 giving the information about the function. 434 00:29:19,340 --> 00:29:22,230 That's the way I'm going to read it. 435 00:29:22,230 --> 00:29:26,360 In order to express this, though, I'm 436 00:29:26,360 --> 00:29:30,760 going to just rewrite it a couple of times here. 437 00:29:30,760 --> 00:29:37,360 So here's f(a), multiplying through by the denominator. 438 00:29:37,360 --> 00:29:40,470 And now I'm going to write it in another customary form 439 00:29:40,470 --> 00:29:42,300 for the mean value theorem. 440 00:29:42,300 --> 00:29:46,120 Which is f(b) = f(a) + f'(c) (b-a). 441 00:29:51,300 --> 00:29:53,524 So here's another version. 442 00:29:53,524 --> 00:29:55,440 I should probably have put this one in the box 443 00:29:55,440 --> 00:29:59,870 to begin with anyway. 444 00:29:59,870 --> 00:30:03,800 And, just changing it around algebraically, 445 00:30:03,800 --> 00:30:07,160 it's this fact here. 446 00:30:07,160 --> 00:30:13,310 They're the same thing. 447 00:30:13,310 --> 00:30:17,760 And now with the formula written in this form, 448 00:30:17,760 --> 00:30:24,060 I claim that I can check these three facts. 449 00:30:24,060 --> 00:30:26,680 Let's start with the first one. 450 00:30:26,680 --> 00:30:33,580 I'm going to set things up always so that a < b. 451 00:30:33,580 --> 00:30:36,840 And that's the setup of the theorem. 452 00:30:36,840 --> 00:30:42,010 And so that means that b - a is positive. 453 00:30:42,010 --> 00:30:48,190 Which means that this factor over here is a positive number. 454 00:30:48,190 --> 00:30:56,000 If f' is positive, which is what happens 455 00:30:56,000 --> 00:30:58,830 in the first case, that's the assumption that we're making, 456 00:30:58,830 --> 00:31:01,210 then this is a positive number. 457 00:31:01,210 --> 00:31:02,750 And so f(b) > f(a). 458 00:31:08,190 --> 00:31:09,565 Which means that it's increasing. 459 00:31:09,565 --> 00:31:13,900 It goes up as the value goes up. 460 00:31:13,900 --> 00:31:20,780 Similarly, if f'(c) is negative, then this is a positive times 461 00:31:20,780 --> 00:31:22,830 a negative number, this is negative. 462 00:31:22,830 --> 00:31:25,610 f(b) < f(a). 463 00:31:25,610 --> 00:31:37,850 So it goes the other way. 464 00:31:37,850 --> 00:31:39,520 Maybe I'll write this way. 465 00:31:39,520 --> 00:31:49,910 And finally, if f'(c) = 0, then f(b) = f(a). 466 00:31:49,910 --> 00:31:52,110 Which if you apply it to all possible ends, 467 00:31:52,110 --> 00:31:56,110 means if you can do it for every interval, which you can, 468 00:31:56,110 --> 00:31:57,940 then that means that f is constant. 469 00:31:57,940 --> 00:32:12,640 It never gets to change values. 470 00:32:12,640 --> 00:32:17,260 Well you might have believed these facts already. 471 00:32:17,260 --> 00:32:20,670 But I just want to emphasize to you that this turns out 472 00:32:20,670 --> 00:32:23,670 to be the one key link between infinitesimals, 473 00:32:23,670 --> 00:32:27,610 between limits and these actual differences. 474 00:32:27,610 --> 00:32:30,240 Before, we were saying that the difference quotient 475 00:32:30,240 --> 00:32:32,530 was approximately equal to the derivative. 476 00:32:32,530 --> 00:32:35,520 Now we're saying that it's exactly equal to a derivative. 477 00:32:35,520 --> 00:32:38,930 Although we don't know exactly which point to use. 478 00:32:38,930 --> 00:32:47,150 It's some point in between. 479 00:32:47,150 --> 00:32:49,900 I'm going to be deducing some other consequences in a second, 480 00:32:49,900 --> 00:32:52,187 but let me stop for second to make sure 481 00:32:52,187 --> 00:32:53,270 that everybody's on board. 482 00:32:53,270 --> 00:32:56,880 Especially since I've finished the blackboards here. 483 00:32:56,880 --> 00:32:59,510 Before we-- everybody happy? 484 00:32:59,510 --> 00:33:00,100 One question. 485 00:33:00,100 --> 00:33:08,617 STUDENT: [INAUDIBLE] 486 00:33:08,617 --> 00:33:10,950 PROFESSOR: I'm just going to repeat your question first. 487 00:33:10,950 --> 00:33:12,950 I'm a little bit confused, you said, 488 00:33:12,950 --> 00:33:16,230 about what guarantees that there's a point of tangency. 489 00:33:16,230 --> 00:33:19,030 That's what you said. 490 00:33:19,030 --> 00:33:20,780 So do you want to elaborate, or do you 491 00:33:20,780 --> 00:33:23,152 want to want to stop with what you just said? 492 00:33:23,152 --> 00:33:24,360 What is it that confuses you? 493 00:33:24,360 --> 00:33:28,764 STUDENT: [INAUDIBLE] 494 00:33:28,764 --> 00:33:29,430 PROFESSOR: Yeah. 495 00:33:29,430 --> 00:33:43,500 STUDENT: [INAUDIBLE] 496 00:33:43,500 --> 00:33:46,305 PROFESSOR: So I'm not claiming that there's only one point. 497 00:33:46,305 --> 00:33:48,180 This could wiggle a lot of times and it maybe 498 00:33:48,180 --> 00:33:49,990 touches at ten places. 499 00:33:49,990 --> 00:33:54,640 In other words, it's OK with me if it touches more than once. 500 00:33:54,640 --> 00:33:56,830 Then I just have more, the more the merrier. 501 00:33:56,830 --> 00:33:59,550 In other words, I don't want there necessarily only 502 00:33:59,550 --> 00:34:00,050 to be one. 503 00:34:00,050 --> 00:34:02,720 It could come down like this. 504 00:34:02,720 --> 00:34:05,140 And touch a second time. 505 00:34:05,140 --> 00:34:09,290 Is that what was concerning you? 506 00:34:09,290 --> 00:34:11,580 So in mathematics, when we claim that this 507 00:34:11,580 --> 00:34:14,410 is true for some point, we don't necessarily 508 00:34:14,410 --> 00:34:16,210 mean that it doesn't work for others. 509 00:34:16,210 --> 00:34:18,310 In fact, if the function is constant, 510 00:34:18,310 --> 00:34:25,960 this is 0 and in fact this equation is true for every c. 511 00:34:25,960 --> 00:34:30,840 That satisfies your question? 512 00:34:30,840 --> 00:34:33,620 The fact that this point exists actually is a touchy point. 513 00:34:33,620 --> 00:34:35,320 I just convinced you of it visually. 514 00:34:35,320 --> 00:34:39,030 It's a geometric issue, whether you're allowed to do this. 515 00:34:39,030 --> 00:34:41,840 Indeed, it has to do with the existence 516 00:34:41,840 --> 00:34:44,150 of tangent lines and more analysis 517 00:34:44,150 --> 00:34:46,120 than we can do in this class. 518 00:34:46,120 --> 00:34:46,620 Yeah. 519 00:34:46,620 --> 00:34:47,328 Another question. 520 00:34:47,328 --> 00:34:48,880 STUDENT: [INAUDIBLE] 521 00:34:48,880 --> 00:34:50,437 PROFESSOR: Pardon me. 522 00:34:50,437 --> 00:34:51,270 STUDENT: [INAUDIBLE] 523 00:34:51,270 --> 00:34:52,686 PROFESSOR: The question is, what's 524 00:34:52,686 --> 00:34:56,780 the difference between this and the linear approximation. 525 00:34:56,780 --> 00:35:11,010 And I think, let me see if I can describe that. 526 00:35:11,010 --> 00:35:12,520 I'll leave the theorem on the board. 527 00:35:12,520 --> 00:35:14,490 I'm going to get rid of the colloquial version 528 00:35:14,490 --> 00:35:19,380 of the theorem. 529 00:35:19,380 --> 00:35:26,200 And I'll try to describe to you the difference between this 530 00:35:26,200 --> 00:35:32,130 and the linear approximation. 531 00:35:32,130 --> 00:35:34,220 I was planning to do that in a while, 532 00:35:34,220 --> 00:35:36,890 but we'll do it right now since that's what you're asking. 533 00:35:36,890 --> 00:35:37,660 That's fine. 534 00:35:37,660 --> 00:35:45,570 So here's the situation. 535 00:35:45,570 --> 00:35:51,660 The linear approximation, so let's say comparison 536 00:35:51,660 --> 00:35:57,390 with linear approximation. 537 00:35:57,390 --> 00:35:59,710 They're very closely related. 538 00:35:59,710 --> 00:36:02,710 The linear approximation says the change in f over the change 539 00:36:02,710 --> 00:36:06,290 in x, that's the left-hand side of this thing, 540 00:36:06,290 --> 00:36:09,980 is approximately f'(a). 541 00:36:09,980 --> 00:36:21,200 For b near a, and b - a = delta x. 542 00:36:21,200 --> 00:36:23,110 This statement, which is in the box, which 543 00:36:23,110 --> 00:36:25,980 is sitting right up there, is the statement 544 00:36:25,980 --> 00:36:31,330 that this change in f is actually equal to something. 545 00:36:31,330 --> 00:36:33,180 Not approximately equal to it. 546 00:36:33,180 --> 00:36:37,670 It's equal to f' of some c. 547 00:36:37,670 --> 00:36:41,330 And the problem here is that we don't know exactly which c. 548 00:36:41,330 --> 00:36:43,930 This is for some c. 549 00:36:43,930 --> 00:36:53,190 Between a and b. 550 00:36:53,190 --> 00:36:54,810 Right, so. 551 00:36:54,810 --> 00:36:59,190 That's the difference between the two. 552 00:36:59,190 --> 00:37:19,700 And let me elaborate a little bit. 553 00:37:19,700 --> 00:37:27,050 If you're trying to understand what (f(b) - f(a)) / (b - 554 00:37:27,050 --> 00:37:32,840 a) is, the mean value theorem is telling you for sure that 555 00:37:32,840 --> 00:37:35,530 it's equal to this f'(c). 556 00:37:35,530 --> 00:37:38,230 So that means it's less than or equal to the largest 557 00:37:38,230 --> 00:37:44,280 possible value on the-- largest value you can get, for sure. 558 00:37:44,280 --> 00:37:48,110 And this is on the whole interval. 559 00:37:48,110 --> 00:37:49,980 And I'm going to include the ends, 560 00:37:49,980 --> 00:37:54,050 because when you take a max it's sometimes achieved at the ends. 561 00:37:54,050 --> 00:37:58,770 And similarly, because it's f'(c), it's definitely bigger 562 00:37:58,770 --> 00:38:07,030 than the min on this same interval here. 563 00:38:07,030 --> 00:38:13,420 This is all you can say based on the mean value theorem. 564 00:38:13,420 --> 00:38:14,620 All you know is this. 565 00:38:14,620 --> 00:38:17,710 And colloquially, what that means 566 00:38:17,710 --> 00:38:25,690 is that the average speed is between the maximum 567 00:38:25,690 --> 00:38:29,000 and the minimum. 568 00:38:29,000 --> 00:38:31,450 Not very surprising. 569 00:38:31,450 --> 00:38:33,120 The mean value theorem is supposed 570 00:38:33,120 --> 00:38:36,480 to be very intuitively obvious. 571 00:38:36,480 --> 00:38:39,370 It's saying the average speed is trapped 572 00:38:39,370 --> 00:38:42,850 between the maximum speed and the minimum speed. 573 00:38:42,850 --> 00:38:44,830 For sure, that's something, that's 574 00:38:44,830 --> 00:38:48,340 why-- incidentally this wasn't really proved when 575 00:38:48,340 --> 00:38:50,350 Newton and Leibniz were around. 576 00:38:50,350 --> 00:38:52,860 But, let's write this so that you can read it. 577 00:38:52,860 --> 00:39:01,350 Average speed is between the max and the min. 578 00:39:01,350 --> 00:39:04,960 But nobody had any trouble, they didn't disbelieve it 579 00:39:04,960 --> 00:39:09,750 because it's a very natural thing. 580 00:39:09,750 --> 00:39:16,980 Now if, for example, I take any kind of linear approximation, 581 00:39:16,980 --> 00:39:25,670 say, for instance, e^x is approximately 1 + x. 582 00:39:25,670 --> 00:39:30,620 Then I'm making the guess-- no, I don't want to say this yet. 583 00:39:30,620 --> 00:39:35,830 That's not going to explain it to you well enough. 584 00:39:35,830 --> 00:39:38,690 What we're saying, so this is the mean value here. 585 00:39:38,690 --> 00:39:40,840 This is what the mean value theorem says. 586 00:39:40,840 --> 00:39:47,270 And here's the linear approximation. 587 00:39:47,270 --> 00:39:52,140 The linear approximation is saying that the average speed 588 00:39:52,140 --> 00:39:59,550 is approximately the initial speed, or possibly 589 00:39:59,550 --> 00:40:01,900 the final speed. 590 00:40:01,900 --> 00:40:07,140 So if a is the left endpoint, then it's the initial speed. 591 00:40:07,140 --> 00:40:09,630 If it happens to be the right endpoint, if the value of x 592 00:40:09,630 --> 00:40:13,460 is to the left then it's the final speed. 593 00:40:13,460 --> 00:40:16,640 So those are the-- so you can see it's approximately right. 594 00:40:16,640 --> 00:40:19,180 Because the speed, when you're on a short interval, 595 00:40:19,180 --> 00:40:20,530 shouldn't be varying very much. 596 00:40:20,530 --> 00:40:22,696 The max and the min should be pretty close together. 597 00:40:22,696 --> 00:40:25,630 So that's why the linear approximation is reasonable. 598 00:40:25,630 --> 00:40:27,800 And this is telling you absolutely, 599 00:40:27,800 --> 00:40:34,140 it's no less than the min and no more than the max. 600 00:40:34,140 --> 00:40:34,640 Yeah. 601 00:40:34,640 --> 00:40:40,965 STUDENT: [INAUDIBLE] 602 00:40:40,965 --> 00:40:42,090 PROFESSOR: The little kink? 603 00:40:42,090 --> 00:40:46,074 STUDENT: [INAUDIBLE] 604 00:40:46,074 --> 00:40:47,740 PROFESSOR: If you approach from the top. 605 00:40:47,740 --> 00:40:50,507 So if it's still under here I can show you it again. 606 00:40:50,507 --> 00:40:51,590 Oh yeah, it's still there. 607 00:40:51,590 --> 00:40:52,090 Good. 608 00:40:52,090 --> 00:40:54,630 STUDENT: [INAUDIBLE] 609 00:40:54,630 --> 00:40:56,830 PROFESSOR: Oh, the one with the wiggle on top? 610 00:40:56,830 --> 00:40:57,872 Yeah, this one you can't. 611 00:40:57,872 --> 00:40:59,704 Because there's nothing to touch and it also 612 00:40:59,704 --> 00:41:02,160 fails from the bottom because there's this bad point. 613 00:41:02,160 --> 00:41:04,217 From the top, it could work. 614 00:41:04,217 --> 00:41:05,550 It can certainly work both ways. 615 00:41:05,550 --> 00:41:07,760 So, for example. 616 00:41:07,760 --> 00:41:09,940 See if you're a machine, you maybe 617 00:41:09,940 --> 00:41:11,310 don't have a way of doing this. 618 00:41:11,310 --> 00:41:14,330 But if you're a human being you can spot all the places. 619 00:41:14,330 --> 00:41:17,990 There are a bunch of spots where the slope is right. 620 00:41:17,990 --> 00:41:20,770 And it's perfectly OK. 621 00:41:20,770 --> 00:41:21,780 All of them work. 622 00:41:21,780 --> 00:41:25,139 STUDENT: [INAUDIBLE] 623 00:41:25,139 --> 00:41:26,930 PROFESSOR: It's not that the c is the same. 624 00:41:26,930 --> 00:41:29,470 It's just we've now found one, two, three, four, 625 00:41:29,470 --> 00:41:31,160 five c's for which it works. 626 00:41:31,160 --> 00:41:40,390 STUDENT: [INAUDIBLE] PROFESSOR: If you're asked to find a c, 627 00:41:40,390 --> 00:41:44,897 so first of all that's kind of a phony question. 628 00:41:44,897 --> 00:41:46,730 There are some questions on your problem set 629 00:41:46,730 --> 00:41:48,340 which ask you to find a c. 630 00:41:48,340 --> 00:41:51,850 That actually is struggling to get you to understand what 631 00:41:51,850 --> 00:41:54,310 the statement of the mean value theorem is, 632 00:41:54,310 --> 00:41:58,890 but you should not pay a lot of attention to those questions. 633 00:41:58,890 --> 00:42:01,870 They're not very impressive. 634 00:42:01,870 --> 00:42:04,860 But, of course, you would have to find all the-- if it asked 635 00:42:04,860 --> 00:42:06,110 you to find one, you find one. 636 00:42:06,110 --> 00:42:10,950 If you can find some more, fine. 637 00:42:10,950 --> 00:42:13,470 You can pick whichever one you want. 638 00:42:13,470 --> 00:42:16,560 Mean value theorem just doesn't care. 639 00:42:16,560 --> 00:42:18,190 The mean value theorem doesn't care 640 00:42:18,190 --> 00:42:21,060 because actually, the mean value theorem is never 641 00:42:21,060 --> 00:42:28,020 used except to-- in real life, except in this context here. 642 00:42:28,020 --> 00:42:31,225 You can never nail down which c it 643 00:42:31,225 --> 00:42:33,140 is, so the only thing you can say 644 00:42:33,140 --> 00:42:35,980 is that you're going slower than the maximum speed 645 00:42:35,980 --> 00:42:40,414 and faster than the minimum speed. 646 00:42:40,414 --> 00:42:41,330 Sorry, say that again? 647 00:42:41,330 --> 00:42:47,057 STUDENT: [INAUDIBLE] 648 00:42:47,057 --> 00:42:48,890 PROFESSOR: If you're asked for a specific c, 649 00:42:48,890 --> 00:42:51,080 you have to find a specific c. 650 00:42:51,080 --> 00:42:53,070 And it has to be in the range. 651 00:42:53,070 --> 00:43:04,420 In between, it has to be in here. 652 00:43:04,420 --> 00:43:07,880 So now I want to tell you about another kind of application, 653 00:43:07,880 --> 00:43:11,100 which is really just a consequence of what 654 00:43:11,100 --> 00:43:22,070 I've described here. 655 00:43:22,070 --> 00:43:26,050 I should emphasize, by the way, this, probably, 656 00:43:26,050 --> 00:43:27,420 should be doing this. 657 00:43:27,420 --> 00:43:32,120 I guess we've never used this color here. 658 00:43:32,120 --> 00:43:32,950 It's popular. 659 00:43:32,950 --> 00:43:33,640 This is pink. 660 00:43:33,640 --> 00:43:35,480 So this one is so good. 661 00:43:35,480 --> 00:43:47,360 So since we're going to do this. 662 00:43:47,360 --> 00:43:50,960 So the reason why the exclamation points 663 00:43:50,960 --> 00:43:54,150 are temporary, this is such an obvious fact. 664 00:43:54,150 --> 00:43:57,857 But this is the way that you're going 665 00:43:57,857 --> 00:43:59,440 to want to use the mean value theorem, 666 00:43:59,440 --> 00:44:01,650 and this is the only way you need to understand 667 00:44:01,650 --> 00:44:02,670 the mean value theorem. 668 00:44:02,670 --> 00:44:06,520 On your test, or ever in your whole life. 669 00:44:06,520 --> 00:44:10,630 So this is the way it will be used. 670 00:44:10,630 --> 00:44:17,140 As I will make very clear when we review for the exam. 671 00:44:17,140 --> 00:44:18,714 In practice what happens is you even 672 00:44:18,714 --> 00:44:21,130 forget about the mean value theorem, and what you remember 673 00:44:21,130 --> 00:44:24,360 is these three properties here. 674 00:44:24,360 --> 00:44:26,000 Which are themselves consequences 675 00:44:26,000 --> 00:44:27,720 of the mean value theorem. 676 00:44:27,720 --> 00:44:31,950 So these are the ones that I want to illustrate now. 677 00:44:31,950 --> 00:44:35,710 In my next discussion here. 678 00:44:35,710 --> 00:44:42,420 I'm just going to talk about inequalities. 679 00:44:42,420 --> 00:44:46,930 Inequalities are relationships between functions. 680 00:44:46,930 --> 00:44:50,170 And I'm going to prove a couple of them using the properties 681 00:44:50,170 --> 00:44:52,300 over there, the properties that functions 682 00:44:52,300 --> 00:44:56,900 with positive derivatives are increasing. 683 00:44:56,900 --> 00:44:58,900 Here's an example. 684 00:44:58,900 --> 00:45:08,390 e^x > 1 + x, where x > 0. 685 00:45:08,390 --> 00:45:10,940 The proof is the following. 686 00:45:10,940 --> 00:45:16,070 I consider-- So here's a proof. 687 00:45:16,070 --> 00:45:21,550 I consider the function f(x), which is the difference. 688 00:45:21,550 --> 00:45:22,490 e^x - (1+x). 689 00:45:27,360 --> 00:45:37,140 I observe that it starts at f(0) equal to, well, that's e^0 - 690 00:45:37,140 --> 00:45:42,110 (1+0), which is 0. 691 00:45:42,110 --> 00:45:45,510 And, it keeps on going. 692 00:45:45,510 --> 00:45:50,310 f'(x) is e^x, if I differentiate here, the 1 goes away. 693 00:45:50,310 --> 00:45:54,117 I get minus 1. 694 00:45:54,117 --> 00:45:55,700 That's the derivative of the function. 695 00:45:55,700 --> 00:46:03,750 And this function, because e^x > 1 for x positive, is positive. 696 00:46:03,750 --> 00:46:07,095 As x gets bigger and bigger, this rate of increase 697 00:46:07,095 --> 00:46:08,290 is positive. 698 00:46:08,290 --> 00:46:13,940 And therefore, three dots, that's therefore, 699 00:46:13,940 --> 00:46:23,350 f(x) is bigger than its starting place, for x > 0. 700 00:46:23,350 --> 00:46:27,100 If it's increasing, then that's-- in particular, 701 00:46:27,100 --> 00:46:28,730 it's increasing starting from 0. 702 00:46:28,730 --> 00:46:30,340 So this is true. 703 00:46:30,340 --> 00:46:36,000 Now, all I have to do is read what this inequality says. 704 00:46:36,000 --> 00:46:39,860 And what it says is that e^x, just plug in for f(x), 705 00:46:39,860 --> 00:46:45,130 which is right here, minus (1+x) is greater than the starting 706 00:46:45,130 --> 00:46:48,590 value, which was 0. 707 00:46:48,590 --> 00:46:52,050 Now, I put the thing that's negative on the other side. 708 00:46:52,050 --> 00:47:01,550 So that's the same thing as e^x > 1 + x. 709 00:47:01,550 --> 00:47:04,330 That's a typical inequality. 710 00:47:04,330 --> 00:47:11,150 And now, we'll use this principle again. 711 00:47:11,150 --> 00:47:12,900 Oh gee, I erased the wrong thing. 712 00:47:12,900 --> 00:47:15,490 I erased the statement and not the proof. 713 00:47:15,490 --> 00:47:23,340 Well, hide the proof. 714 00:47:23,340 --> 00:47:27,500 The next thing I want to prove to you is that e^x > 1 + x + 715 00:47:27,500 --> 00:47:28,000 x^2 / 2. 716 00:47:33,090 --> 00:47:34,300 So, how do I do that? 717 00:47:34,300 --> 00:47:42,174 I introduce a function g(x), which is e^x minus this. 718 00:47:42,174 --> 00:47:44,340 And now, I'm just going to do exactly the same thing 719 00:47:44,340 --> 00:47:45,780 I did before. 720 00:47:45,780 --> 00:47:51,870 Which is, I get started with g(0), which is 1 - 1. 721 00:47:51,870 --> 00:47:53,680 Which is 0. 722 00:47:53,680 --> 00:48:00,690 And g'(x) is e^x minus - now, look at what happens when I 723 00:48:00,690 --> 00:48:03,590 differentiate this. 724 00:48:03,590 --> 00:48:04,870 The 1 goes away. 725 00:48:04,870 --> 00:48:10,600 The x gives me a 1, and the x^2 / 2 gives me a plus x. 726 00:48:10,600 --> 00:48:18,190 And this one is positive for x > 0, because of step 1. 727 00:48:18,190 --> 00:48:21,840 Because of the previous one that I did. 728 00:48:21,840 --> 00:48:28,340 So this one is increasing. g is increasing. 729 00:48:28,340 --> 00:48:33,350 Which says that g(x) > g(0). 730 00:48:33,350 --> 00:48:36,620 And if you just read that off, it's exactly the same 731 00:48:36,620 --> 00:48:41,690 as our inequality here. e^x > 1 + x + x^2 / 2. 732 00:48:48,270 --> 00:48:54,010 Now, you can keep on going with this essentially forever. 733 00:48:54,010 --> 00:48:58,510 And let me just write down what you get. 734 00:48:58,510 --> 00:49:04,772 You get e^x is greater than 1 plus x plus x^2 / 2, 735 00:49:04,772 --> 00:49:06,480 the next one turns out to be x^3 / (3*2). 736 00:49:10,460 --> 00:49:11,730 x^4 / (4*3*2). 737 00:49:14,850 --> 00:49:16,820 And you can do whatever you want. 738 00:49:16,820 --> 00:49:19,490 You can do others. 739 00:49:19,490 --> 00:49:22,860 And this is like the tortoise and the hare. 740 00:49:22,860 --> 00:49:27,740 This is the tortoise, and this is the hare, it's always ahead. 741 00:49:27,740 --> 00:49:31,910 But eventually, if you go infinitely far, it catches up. 742 00:49:31,910 --> 00:49:38,190 So this turns out to be exactly equal to e^x in the limit. 743 00:49:38,190 --> 00:49:41,280 And we'll talk about that maybe at the end of the course.