WEBVTT 00:00:07.470 --> 00:00:10.720 GILBERT STRANG: OK, what I want to do today is show you 00:00:10.720 --> 00:00:13.735 two different ways that derivatives are used. 00:00:16.860 --> 00:00:21.920 In one of them, the problem is to find a close approximation 00:00:21.920 --> 00:00:25.110 to the value f at a point x. 00:00:25.110 --> 00:00:26.730 f of x. 00:00:26.730 --> 00:00:29.830 The second application is to solve an 00:00:29.830 --> 00:00:31.800 equation, where often-- 00:00:31.800 --> 00:00:35.420 and I use a different letter, capital F, just because it's a 00:00:35.420 --> 00:00:37.360 different function and there will be different 00:00:37.360 --> 00:00:39.890 examples for this one. 00:00:39.890 --> 00:00:43.690 So I just chose capital F to keep them separate. 00:00:43.690 --> 00:00:47.420 This is the problem of solving an equation. 00:00:47.420 --> 00:00:53.060 And Newton had an idea and it's survived all these 00:00:53.060 --> 00:00:55.880 centuries and it's still the good way. 00:00:55.880 --> 00:00:59.690 OK, what's this based on, both of them? 00:00:59.690 --> 00:01:00.460 That's the point. 00:01:00.460 --> 00:01:05.510 They're both based on the same idea. 00:01:05.510 --> 00:01:13.070 Suppose at a point, which is near the x we want, or near 00:01:13.070 --> 00:01:17.400 the solution to this problem, at some point, let me call it 00:01:17.400 --> 00:01:24.690 a, suppose we know the slope, the derivative, at that point. 00:01:24.690 --> 00:01:29.320 So I'm using f prime for the derivative. 00:01:29.320 --> 00:01:34.160 At that point, well, we know what the definition is. 00:01:34.160 --> 00:01:39.270 But I'm also supposing that we've got that number. 00:01:39.270 --> 00:01:45.850 And now I want to use this knowledge of the slope at that 00:01:45.850 --> 00:01:52.580 nearby point a to come close to the solution, the f of x, 00:01:52.580 --> 00:01:54.660 or the x there. 00:01:54.660 --> 00:02:00.660 OK, well you remember this is delta f divided by delta x. 00:02:00.660 --> 00:02:06.160 You recognize that before we take the limit this is two 00:02:06.160 --> 00:02:09.509 nearby points, delta x apart. 00:02:09.509 --> 00:02:13.570 Their values, the values of f at those points. 00:02:13.570 --> 00:02:16.970 It's the change in f divided by the change in x-- 00:02:16.970 --> 00:02:18.930 that's what the derivative is-- 00:02:18.930 --> 00:02:23.840 as one point approaches the point a 00:02:23.840 --> 00:02:26.740 where we know the slope. 00:02:26.740 --> 00:02:28.060 Here's the idea. 00:02:28.060 --> 00:02:33.620 I'm just going to erase that stuff. 00:02:33.620 --> 00:02:37.840 Well, now I won't have an equal sign anymore. 00:02:37.840 --> 00:02:41.680 I'll have an approximately equal. 00:02:41.680 --> 00:02:43.850 So the slope-- 00:02:43.850 --> 00:02:48.300 this is delta f over delta x-- 00:02:48.300 --> 00:02:52.410 is not the same as df dx, which is this. 00:02:52.410 --> 00:02:58.790 But if x is close to a this will normally be close to the 00:02:58.790 --> 00:03:01.930 correct slope, the instant slope at the point. 00:03:01.930 --> 00:03:08.820 I'm just going to use this approximation to find a good 00:03:08.820 --> 00:03:11.050 approximation for f of x. 00:03:11.050 --> 00:03:17.410 So in this application on the left I know the x and I want 00:03:17.410 --> 00:03:20.090 to find what is f at that point? 00:03:20.090 --> 00:03:23.390 So I look at this. 00:03:23.390 --> 00:03:27.140 I multiply up by x minus a. 00:03:27.140 --> 00:03:30.380 I move the f of a to the other side. 00:03:30.380 --> 00:03:31.630 And what do I have? 00:03:33.840 --> 00:03:37.570 I still have an approximation sign. 00:03:37.570 --> 00:03:42.240 So when I move the f of a to the other side, there it is. 00:03:42.240 --> 00:03:52.770 And then, the other part is x minus a times f prime at a. 00:03:52.770 --> 00:03:57.300 That's my formula. 00:03:57.300 --> 00:04:00.200 Let me just talk about that formula for a minute and then 00:04:00.200 --> 00:04:03.190 give examples. 00:04:03.190 --> 00:04:06.970 What that says is that if I want to find the value of f at 00:04:06.970 --> 00:04:14.360 a nearby point, a good thing to do is use this linear 00:04:14.360 --> 00:04:15.270 approximation. 00:04:15.270 --> 00:04:17.320 I use the word "linear" because the graph 00:04:17.320 --> 00:04:19.480 of that is a line. 00:04:19.480 --> 00:04:22.100 I'm following a straight line instead of following the 00:04:22.100 --> 00:04:23.970 curved graph. 00:04:23.970 --> 00:04:25.800 That's the message of today's lecture. 00:04:25.800 --> 00:04:28.510 Follow the line. 00:04:28.510 --> 00:04:30.170 So it's a line. 00:04:30.170 --> 00:04:33.680 It starts out at the correct point at x equal a. 00:04:33.680 --> 00:04:36.670 It has the corrects slope f prime of a. 00:04:36.670 --> 00:04:39.630 And if I don't go too far that line won't be too 00:04:39.630 --> 00:04:41.140 far from the curve. 00:04:41.140 --> 00:04:41.720 Good. 00:04:41.720 --> 00:04:44.040 You'll see it now in examples. 00:04:44.040 --> 00:04:47.200 Let me get the corresponding idea here. 00:04:47.200 --> 00:04:54.720 Now I have to remember to use capital F. So let me create 00:04:54.720 --> 00:05:00.240 the formula that helps to solve, approximately solves, 00:05:00.240 --> 00:05:02.030 this equation. 00:05:02.030 --> 00:05:03.480 What's the difference? 00:05:03.480 --> 00:05:06.950 Now x is what I'm looking for. 00:05:06.950 --> 00:05:10.930 x is what I'm looking for and F of x is what I know. 00:05:10.930 --> 00:05:12.930 I know it's to be 0. 00:05:12.930 --> 00:05:18.440 So I'm going to use this with capital F. I know F of x 00:05:18.440 --> 00:05:20.340 should be 0 and x is the unknown. 00:05:20.340 --> 00:05:25.110 So can I just move that equation around again to get 00:05:25.110 --> 00:05:26.000 an equation? 00:05:26.000 --> 00:05:30.550 I'll bring the x minus a up as I did before. 00:05:30.550 --> 00:05:34.290 And I have to use an approximation sign as always. 00:05:34.290 --> 00:05:42.330 The F of x is 0, so I have minus F of a, and then I'm 00:05:42.330 --> 00:05:44.630 dividing by F prime of a. 00:05:48.910 --> 00:05:50.670 There you have it. 00:05:50.670 --> 00:05:53.360 That is Newton's insight. 00:05:53.360 --> 00:05:59.160 Newton and then somebody named Raphson helped out, made it 00:05:59.160 --> 00:06:00.600 work for general functions. 00:06:00.600 --> 00:06:03.230 And once again, I forgot. 00:06:03.230 --> 00:06:05.190 I should be using capital F there. 00:06:07.750 --> 00:06:11.260 Because in the right side of the board I'm calling the 00:06:11.260 --> 00:06:19.220 function capital F. This is the little movement of x away 00:06:19.220 --> 00:06:22.130 from a, which will bring us closer to the answer. 00:06:22.130 --> 00:06:25.960 You'll see is this formula work in a picture. 00:06:25.960 --> 00:06:32.150 So I'm ready for an example starting with approximation. 00:06:32.150 --> 00:06:33.600 So here's my problem. 00:06:33.600 --> 00:06:34.850 Here's my example. 00:06:37.180 --> 00:06:43.260 Find the square root of-- 00:06:43.260 --> 00:06:48.210 so my problem is going to be find the square root or 00:06:48.210 --> 00:06:51.210 approximate square root of 9. 00:06:51.210 --> 00:06:54.120 I'm going to shift away from 9 a little bit. 00:06:54.120 --> 00:06:58.570 9.06. 00:06:58.570 --> 00:07:01.150 OK, how does that fit this example? 00:07:01.150 --> 00:07:08.760 My function is the square root function, x to the 1/2. 00:07:08.760 --> 00:07:12.310 It's derivative, f prime. 00:07:12.310 --> 00:07:16.990 I know the derivative of that is 1/2 x to one 00:07:16.990 --> 00:07:20.100 lower power minus 1/2. 00:07:20.100 --> 00:07:23.810 So that's 1 over 2 square root of x. 00:07:23.810 --> 00:07:25.590 Good. 00:07:25.590 --> 00:07:26.600 I know my function. 00:07:26.600 --> 00:07:28.590 I know its derivative. 00:07:28.590 --> 00:07:36.090 Now I pick a point a, which is close and easy. 00:07:36.090 --> 00:07:42.760 So close, I'm looking for a point a, which is near 9.06 00:07:42.760 --> 00:07:44.880 and has a nice square root. 00:07:44.880 --> 00:07:47.200 Well, 9. 00:07:47.200 --> 00:07:55.230 So I'll choose a to be 9. 00:07:55.230 --> 00:07:59.680 The correct value f at the point a is the 00:07:59.680 --> 00:08:01.730 square root of 9. 00:08:01.730 --> 00:08:02.470 3. 00:08:02.470 --> 00:08:03.410 That's the point. 00:08:03.410 --> 00:08:08.001 We can evaluate the function easily at 9. 00:08:08.001 --> 00:08:19.620 And f prime at a is 1 over 2 square root of a. 00:08:19.620 --> 00:08:20.180 9. 00:08:20.180 --> 00:08:21.450 Square root of 9. 00:08:21.450 --> 00:08:22.740 So what do I have for that? 00:08:22.740 --> 00:08:24.700 That's the easy number, 3. 00:08:24.700 --> 00:08:25.950 That's 1/6. 00:08:28.660 --> 00:08:33.799 In other words, I know what's happening at x equal 9. 00:08:33.799 --> 00:08:37.710 And that's a particular point I'm calling a and I'm working 00:08:37.710 --> 00:08:38.630 from there. 00:08:38.630 --> 00:08:41.460 Then what does my approximation say? 00:08:41.460 --> 00:08:46.000 It says that f at this nearby point. 00:08:46.000 --> 00:08:47.950 So x is 9.06. 00:08:47.950 --> 00:08:54.040 This is the x where I want to know the square root. 00:08:54.040 --> 00:08:58.010 So this is saying that the square root of 9.06-- 00:08:58.010 --> 00:09:00.560 that's my f of x-- 00:09:00.560 --> 00:09:06.120 is approximately f at a. 00:09:06.120 --> 00:09:07.970 That's the square root of 9. 00:09:07.970 --> 00:09:09.990 That's the 3. 00:09:09.990 --> 00:09:11.570 So far reasonable. 00:09:11.570 --> 00:09:16.090 That's the approximation I started with just using the 9. 00:09:16.090 --> 00:09:20.650 But now I'm improving it by the difference 00:09:20.650 --> 00:09:22.710 between x minus 3. 00:09:22.710 --> 00:09:25.290 Oh, what is x? 00:09:25.290 --> 00:09:28.250 So I'm plugging in x is 9.06. 00:09:28.250 --> 00:09:31.090 That's where I really want the square root. 00:09:31.090 --> 00:09:33.120 That's my x. 00:09:33.120 --> 00:09:37.150 Minus a, which is the 9. 00:09:37.150 --> 00:09:40.390 Times f prime, which we figured out as 1/6. 00:09:43.730 --> 00:09:47.960 That's the linear approximation following this 00:09:47.960 --> 00:09:52.540 line to this number, which is what? 00:09:52.540 --> 00:09:54.330 3 plus-- 00:09:54.330 --> 00:09:56.880 That's 0.06 divided by 6. 00:09:56.880 --> 00:09:57.780 That's 3. 00:09:57.780 --> 00:09:59.870 What do I have there? 00:09:59.870 --> 00:10:02.880 This difference is 0.06. 00:10:02.880 --> 00:10:07.550 Divided by 6 and its 0.01. 00:10:07.550 --> 00:10:10.150 That's the approximation. 00:10:10.150 --> 00:10:12.000 Closer than 3. 00:10:12.000 --> 00:10:13.920 That comes from following the line. 00:10:13.920 --> 00:10:16.300 Let me draw a graph to show you what I mean by 00:10:16.300 --> 00:10:18.150 following the line. 00:10:18.150 --> 00:10:19.895 Here is my square root function. 00:10:23.120 --> 00:10:25.860 The square root looks something like that. 00:10:25.860 --> 00:10:32.430 And here is the point x equals 9, where I know that the 00:10:32.430 --> 00:10:33.680 height is 3. 00:10:36.110 --> 00:10:38.350 What else do I know? 00:10:38.350 --> 00:10:42.550 Here is 9.06, a little further over, and I'm 00:10:42.550 --> 00:10:45.660 looking for that point. 00:10:45.660 --> 00:10:49.910 I'm looking for the square root of 9.06. 00:10:49.910 --> 00:10:55.290 So this was 9 here and this was 9.06. 00:10:55.290 --> 00:11:00.040 And how am I getting close to that point? 00:11:00.040 --> 00:11:04.440 Well, I'm not going to follow the curve to do square root of 00:11:04.440 --> 00:11:07.300 9.06 exactly, any more than your 00:11:07.300 --> 00:11:09.700 calculator or computer does. 00:11:09.700 --> 00:11:13.320 I'm going to follow this line. 00:11:13.320 --> 00:11:17.050 So that line is the tangent line. 00:11:17.050 --> 00:11:23.300 It's the line that goes through the right thing at a 00:11:23.300 --> 00:11:25.550 with the right slope. 00:11:25.550 --> 00:11:30.670 So you can see that by following the line 00:11:30.670 --> 00:11:31.526 that gave me the 3. 00:11:31.526 --> 00:11:32.776 And then here is the little-- you see what I'm-- 00:11:36.690 --> 00:11:39.760 I'm missing by a very small amount, practically 00:11:39.760 --> 00:11:42.020 too small to see. 00:11:42.020 --> 00:11:49.185 I'm picking the point on the line and that was this across, 00:11:49.185 --> 00:11:53.220 this delta x was the 0.06. 00:11:53.220 --> 00:11:56.560 The little tiny bit here is the 0.06. 00:11:56.560 --> 00:12:03.840 And this delta f is the little piece that I 00:12:03.840 --> 00:12:09.790 added on the 0.01. 00:12:09.790 --> 00:12:11.120 How did that come from? 00:12:11.120 --> 00:12:14.570 It came from the fact that the correct slope is 1/6. 00:12:14.570 --> 00:12:18.330 If I go over 0.06, I should go up 0.01. 00:12:18.330 --> 00:12:19.520 So you see what I'm doing? 00:12:19.520 --> 00:12:28.210 I'm taking that point as my close approximation to the 00:12:28.210 --> 00:12:28.760 square root. 00:12:28.760 --> 00:12:34.830 Closer than 3 to the square root of 9.06. 00:12:34.830 --> 00:12:34.950 OK. 00:12:34.950 --> 00:12:35.990 That's a first example. 00:12:35.990 --> 00:12:37.870 I'll give a second one. 00:12:37.870 --> 00:12:44.180 But first, I'd like to give an example that's like this one 00:12:44.180 --> 00:12:46.430 of Newton's method. 00:12:46.430 --> 00:12:48.120 May I change now to Newton's method? 00:12:48.120 --> 00:12:51.030 So I want to create an equation. 00:12:51.030 --> 00:12:54.270 And actually, I want it to solve the same problem. 00:12:54.270 --> 00:12:58.440 So I'm going to take my function to be x squared minus 00:12:58.440 --> 00:13:05.870 9.06 and I'll set that to 0. 00:13:05.870 --> 00:13:09.670 I'm just keeping my two examples 00:13:09.670 --> 00:13:12.775 close because the answer-- 00:13:16.440 --> 00:13:22.450 in both problems, I'm looking for the square root of 9.06. 00:13:22.450 --> 00:13:25.300 Of course, that's the solution to this equation, the square 00:13:25.300 --> 00:13:27.190 root of 9.06. 00:13:27.190 --> 00:13:34.210 OK again, I pick a point a close to the solution. 00:13:34.210 --> 00:13:39.830 And again, I'm going to take a to be 3. 00:13:39.830 --> 00:13:42.070 So 3 is close to the correct solution. 00:13:46.110 --> 00:13:49.370 The correct x is the square root of 9.06. 00:13:49.370 --> 00:13:55.100 But I'm starting close to it. 00:13:55.100 --> 00:14:01.040 OK, at that point I figure out F of a. 00:14:01.040 --> 00:14:03.700 Newton wants to know F of a, the value. 00:14:03.700 --> 00:14:06.740 And Newton also wants to know the slope. 00:14:06.740 --> 00:14:09.360 So the value at a is-- 00:14:09.360 --> 00:14:11.200 3 squared is 9. 00:14:11.200 --> 00:14:20.200 9 minus 9.06 is minus 0.06. 00:14:20.200 --> 00:14:29.160 F prime of a is-- what's the derivative of F? 00:14:29.160 --> 00:14:33.980 Capital F. Well, the derivative is certainly 2x. 00:14:33.980 --> 00:14:38.710 And at the point a it's 6. 00:14:38.710 --> 00:14:40.220 This is the 2x. 00:14:40.220 --> 00:14:42.300 2a. 00:14:42.300 --> 00:14:46.340 This is the 2a because I'm evaluating the slope 00:14:46.340 --> 00:14:47.140 at the point a. 00:14:47.140 --> 00:14:51.060 Actually, if you want a picture, it's quite 00:14:51.060 --> 00:14:53.390 interesting to see the picture. 00:14:53.390 --> 00:14:57.610 Let me graph this f of x now. 00:14:57.610 --> 00:15:00.170 What does my function f of x look like? 00:15:00.170 --> 00:15:00.920 x squared. 00:15:00.920 --> 00:15:03.150 That's a parabola going up. 00:15:03.150 --> 00:15:04.900 It starts below here. 00:15:04.900 --> 00:15:10.335 It starts somewhere down here and it curves up like so. 00:15:13.930 --> 00:15:15.270 So this is the point I want. 00:15:18.110 --> 00:15:21.370 There is the square root of 9.06. 00:15:21.370 --> 00:15:26.060 I've graphed a function x squared minus 9.06. 00:15:26.060 --> 00:15:29.140 That's the point I'd like to find. 00:15:29.140 --> 00:15:32.740 I'm not expecting to find it exactly. 00:15:32.740 --> 00:15:37.200 What I do is I have a nearby point 3. 00:15:40.490 --> 00:15:47.130 At 3 I do know the exact value and I know the slope. 00:15:52.080 --> 00:15:52.175 Ha. 00:15:52.175 --> 00:15:57.750 Even my bad art is telling me what's going to happen. 00:15:57.750 --> 00:15:59.980 Let me write that 3 better. 00:15:59.980 --> 00:16:02.490 a equals 3. 00:16:02.490 --> 00:16:04.770 This is where I know what's happening. 00:16:04.770 --> 00:16:11.320 I know the value of F. It's actually pretty small, but 00:16:11.320 --> 00:16:12.780 it's negative. 00:16:12.780 --> 00:16:14.510 I know the value of the slope. 00:16:14.510 --> 00:16:20.010 So this is like I've blown up this picture here. 00:16:23.060 --> 00:16:31.640 This F is the negative 0.06. 00:16:31.640 --> 00:16:37.550 Is the value of F. The slope is 6. 00:16:37.550 --> 00:16:41.060 And what's the x? 00:16:41.060 --> 00:16:46.160 What is my improved guess at the solution? 00:16:46.160 --> 00:16:49.610 My improved guess, I don't follow that curve because that 00:16:49.610 --> 00:16:51.120 would be perfect. 00:16:51.120 --> 00:16:53.240 But curves are hard to follow. 00:16:53.240 --> 00:17:01.140 I'll follow the straight line and that's my next x. 00:17:01.140 --> 00:17:03.460 My better x. 00:17:03.460 --> 00:17:07.170 You see, that's a lot better. 00:17:07.170 --> 00:17:09.089 Let me do the numbers here and you'll see 00:17:09.089 --> 00:17:10.329 that it's a lot better. 00:17:10.329 --> 00:17:12.930 What is the x that Newton's method gives us? 00:17:12.930 --> 00:17:16.109 I'm using Newton's formula here. 00:17:16.109 --> 00:17:22.710 So Newton's formula says that x minus the a is 3. 00:17:22.710 --> 00:17:27.589 And Newton's formula is going to take equal to get an 00:17:27.589 --> 00:17:29.440 approximate x. 00:17:29.440 --> 00:17:30.920 Minus F of a. 00:17:30.920 --> 00:17:32.810 So what's minus F of a? 00:17:32.810 --> 00:17:36.860 That will be plus 0.06. 00:17:36.860 --> 00:17:42.250 Divided by F prime of a, the slope, which was 6. 00:17:42.250 --> 00:17:43.010 What do I get? 00:17:43.010 --> 00:17:46.680 I did 0.01. 00:17:46.680 --> 00:17:51.270 Actually, the two examples, of course, are parallel. 00:17:51.270 --> 00:17:55.880 In a way, the graph of that square root function kind of 00:17:55.880 --> 00:18:01.390 got just flipped to this graph of the x squared function. 00:18:01.390 --> 00:18:03.930 The square root function and the x squared function are 00:18:03.930 --> 00:18:06.980 just sort of inverses to each other. 00:18:06.980 --> 00:18:14.370 Their graphs flip and so now the slope is 6 instead of 1/6. 00:18:14.370 --> 00:18:22.940 And do you see that that distance, this is the x minus 00:18:22.940 --> 00:18:29.320 a distance that has to be 0.01. 00:18:29.320 --> 00:18:31.930 That's what I concluded. 00:18:31.930 --> 00:18:38.360 I go across 1 if I'm going up/down by 6 because 00:18:38.360 --> 00:18:42.910 the slope is 6. 00:18:42.910 --> 00:18:45.250 OK, good. 00:18:45.250 --> 00:18:47.390 Now, how close am I? 00:18:47.390 --> 00:18:50.440 Well, I can't resist asking. 00:18:50.440 --> 00:18:58.010 Suppose I multiply 3.01 by 3.01. 00:18:58.010 --> 00:19:04.450 Am I close to the 9.06 that was my whole goal? 00:19:04.450 --> 00:19:05.680 And how close? 00:19:05.680 --> 00:19:07.550 Of course, I'm not going to be exact. 00:19:07.550 --> 00:19:12.530 If I do that multiplication I get 301, 903. 00:19:12.530 --> 00:19:21.810 Combined I get 9 and the decimal put it in there, 0601. 00:19:21.810 --> 00:19:28.940 So by that method or by that method I ended up with 3.01 as 00:19:28.940 --> 00:19:34.590 much closer to the square root of 9.06. 00:19:34.590 --> 00:19:40.300 You see when I square 3.01, it slightly overshot. 00:19:40.300 --> 00:19:43.800 It slightly overshot the 9.06. 00:19:43.800 --> 00:19:50.030 This little error there, this is the overshoot error. 00:19:50.030 --> 00:19:55.000 And when I square it, it's way out in the ten thousandths 00:19:55.000 --> 00:19:59.890 place, the fourth decimal place. 00:19:59.890 --> 00:20:00.010 OK. 00:20:00.010 --> 00:20:02.240 So those are the two parallel examples. 00:20:02.240 --> 00:20:07.700 May I go back to the linear approximation? 00:20:07.700 --> 00:20:11.370 Because I think another example's appropriate there. 00:20:11.370 --> 00:20:14.450 And then I'll come back and give another Newton example. 00:20:14.450 --> 00:20:17.250 So that's my plan here. 00:20:17.250 --> 00:20:19.620 Two examples of each. 00:20:19.620 --> 00:20:22.480 So those examples were parallel, the next two 00:20:22.480 --> 00:20:25.520 examples will be a little different. 00:20:25.520 --> 00:20:29.420 So let me show you example two. 00:20:29.420 --> 00:20:42.620 So linear approximation example two. 00:20:49.160 --> 00:20:49.298 So I want-- 00:20:49.298 --> 00:20:50.440 OK, I'm going to look for something that 00:20:50.440 --> 00:20:52.770 I don't know exactly. 00:20:52.770 --> 00:20:58.980 Let me take e to the 0.01. 00:20:58.980 --> 00:21:06.160 What's the value of e to the power of 0.01? 00:21:06.160 --> 00:21:11.930 So the function is e to the x. 00:21:17.080 --> 00:21:24.080 And I'm looking at x is 0.01. 00:21:24.080 --> 00:21:25.700 That's what I want. 00:21:25.700 --> 00:21:29.550 I am not going to get an exact number here. 00:21:29.550 --> 00:21:32.960 I'm going to follow the tangent line. 00:21:32.960 --> 00:21:35.710 I'm going to get a close number. 00:21:35.710 --> 00:21:38.120 So where shall I start? 00:21:38.120 --> 00:21:43.510 I'll start with a number that's close to that and where 00:21:43.510 --> 00:21:45.530 I do you know the correct e. 00:21:45.530 --> 00:21:48.030 And a number close to that is take-- 00:21:48.030 --> 00:21:52.630 I'll choose a to be 0. 00:21:52.630 --> 00:21:55.110 That's close to 0.01. 00:21:55.110 --> 00:22:00.770 Then f of a is e to the 0 power. 00:22:00.770 --> 00:22:03.790 So that's 1. 00:22:03.790 --> 00:22:07.260 Now for the straight line approximation, I also need the 00:22:07.260 --> 00:22:09.020 correct slope. 00:22:09.020 --> 00:22:12.360 Correct slope at a? 00:22:12.360 --> 00:22:14.610 Well, I do know the slope of this. 00:22:14.610 --> 00:22:18.460 f prime at 0. 00:22:18.460 --> 00:22:19.970 That's my a. 00:22:19.970 --> 00:22:21.410 f prime at 0. 00:22:21.410 --> 00:22:23.320 0 is my a. 00:22:23.320 --> 00:22:25.640 Well, I know the derivative of e to the x. 00:22:25.640 --> 00:22:27.130 That's one thing I like. 00:22:27.130 --> 00:22:29.510 It's e to the x. 00:22:29.510 --> 00:22:34.500 So at x equals 0, again I get a 1 for the derivative, which 00:22:34.500 --> 00:22:36.410 is the same, I get e to the 0. 00:22:36.410 --> 00:22:38.460 I get also a 1. 00:22:38.460 --> 00:22:42.620 So now I know what's happening at a equals 0. 00:22:42.620 --> 00:22:46.740 I want to know approximately what's happening at the nearby 00:22:46.740 --> 00:22:53.360 point, 0.01. 00:22:53.360 --> 00:22:56.780 e to the 0.01, e to the x. 00:22:56.780 --> 00:22:59.940 This is the e to the x. 00:22:59.940 --> 00:23:02.070 That's my function. 00:23:02.070 --> 00:23:06.740 And I'm only going to get it approximately, is the value at 00:23:06.740 --> 00:23:09.930 this known point. 00:23:09.930 --> 00:23:18.880 The value at the known point 0, the exact exponential is 1. 00:23:18.880 --> 00:23:26.900 Plus x minus a times the slope, the corrects slope, at 00:23:26.900 --> 00:23:30.640 this not quite perfect point, 0. 00:23:30.640 --> 00:23:32.090 And the correct slope is 1. 00:23:34.880 --> 00:23:38.660 And of course, a was 0. 00:23:38.660 --> 00:23:42.310 So you see I'm using all the facts at a to get an 00:23:42.310 --> 00:23:44.720 approximate fact at x. 00:23:44.720 --> 00:23:46.340 And what have I got here? 00:23:46.340 --> 00:23:47.590 I've just got 1 plus x. 00:23:51.990 --> 00:23:54.310 You know, in a way, that's perfect. 00:23:54.310 --> 00:23:58.540 Because it shows what the linear approximation is doing. 00:23:58.540 --> 00:24:00.930 You remember the series for e to the x? 00:24:00.930 --> 00:24:03.320 My correct function is e to the x. 00:24:03.320 --> 00:24:06.900 My approximate function is 1 plus x. 00:24:06.900 --> 00:24:08.450 What's the connection? 00:24:08.450 --> 00:24:12.750 You remember that e to the x, the series for e to the x 00:24:12.750 --> 00:24:21.290 started out 1 plus x plus 1/2 x squared, 1/6 x cubed. 00:24:21.290 --> 00:24:25.270 Those are the guys, those are the higher order corrections 00:24:25.270 --> 00:24:30.440 that following the line misses. 00:24:30.440 --> 00:24:35.510 Those are the parts where the function, the curve e to the 00:24:35.510 --> 00:24:38.020 x, has left the line. 00:24:44.340 --> 00:24:48.410 But if I don't go to far--So this is 1.01. 00:24:48.410 --> 00:24:49.980 x is 0.01. 00:24:49.980 --> 00:24:51.240 That's my approximation. 00:24:51.240 --> 00:24:52.390 1 plus x. 00:24:52.390 --> 00:24:55.320 That's the thing to notice. 00:24:55.320 --> 00:25:00.410 That linear approximations and you could come back to that-- 00:25:00.410 --> 00:25:04.200 the formula for any f and any a. 00:25:04.200 --> 00:25:11.030 Linear approximations are just like those power series. 00:25:11.030 --> 00:25:14.940 Just like the e to the x equal 1 plus x plus 1/2 x squared 00:25:14.940 --> 00:25:17.110 plus so on. 00:25:17.110 --> 00:25:22.810 Except we cut them off after just the constant term and the 00:25:22.810 --> 00:25:23.810 linear term. 00:25:23.810 --> 00:25:28.150 That's what this linear approximation is about. 00:25:28.150 --> 00:25:31.000 And you might say, what's the next term? 00:25:31.000 --> 00:25:35.980 And of course we know that the next term is 1/2 x squared and 00:25:35.980 --> 00:25:39.310 you could ask, what's the next term in this? 00:25:39.310 --> 00:25:43.430 In the general case, actually let me tell you the next term. 00:25:43.430 --> 00:25:45.630 Next would be-- 00:25:45.630 --> 00:25:47.630 but we're not using it. 00:25:47.630 --> 00:25:49.660 Would be the 1/2. 00:25:49.660 --> 00:25:54.200 It would be the x minus a squared and it would be the 00:25:54.200 --> 00:25:55.950 second derivative at a. 00:25:59.010 --> 00:26:02.670 If we kept it that's what we would keep. 00:26:02.670 --> 00:26:04.980 OK, good. 00:26:04.980 --> 00:26:08.600 You saw the main point of linear approximation stop at 00:26:08.600 --> 00:26:09.670 the linear term. 00:26:09.670 --> 00:26:14.690 Finally, I go back to an example of Newton's method. 00:26:14.690 --> 00:26:16.900 A second example of Newton's method. 00:26:16.900 --> 00:26:19.990 So I've been thinking, what should I do? 00:26:19.990 --> 00:26:24.520 Let me use Newton's method the way it's really used. 00:26:24.520 --> 00:26:29.840 The way you use Newton's method is you do this to come 00:26:29.840 --> 00:26:32.360 close to the solution. 00:26:32.360 --> 00:26:35.360 And then, you do it again. 00:26:35.360 --> 00:26:40.140 You do it again starting at 3.01. 00:26:40.140 --> 00:26:44.170 So I planned to do just the same thing for the next step 00:26:44.170 --> 00:26:48.135 of Newton's method, except the a. 00:26:52.210 --> 00:26:59.810 I'm still aiming to solve this same equation, but I'm going 00:26:59.810 --> 00:27:02.130 to get closer than 3.01. 00:27:02.130 --> 00:27:05.090 I got closer than 3 to 3.01. 00:27:05.090 --> 00:27:07.660 Now I'm going to restart there. 00:27:07.660 --> 00:27:11.150 a is now going to be 3.01. 00:27:11.150 --> 00:27:16.350 I need to compute F of a for Newton's method. 00:27:16.350 --> 00:27:20.900 So I have to do 3.01 squared and take away that to see how 00:27:20.900 --> 00:27:22.350 wrong I am. 00:27:22.350 --> 00:27:22.470 HA. 00:27:22.470 --> 00:27:23.980 We did 3.01 squared. 00:27:23.980 --> 00:27:26.300 Actually, right there. 00:27:26.300 --> 00:27:30.970 So if I take away the 9.06, the F of a is-- well, that was 00:27:30.970 --> 00:27:32.330 the whole point. 00:27:32.330 --> 00:27:34.910 That it was pretty darn close. 00:27:34.910 --> 00:27:37.750 But nothing compared to the closeness we're going to get 00:27:37.750 --> 00:27:39.600 at the second term. 00:27:39.600 --> 00:27:40.850 And what's F prime of a? 00:27:44.420 --> 00:27:46.650 Take the derivative 2x. 00:27:46.650 --> 00:27:49.350 At the point a it's 2a. 00:27:49.350 --> 00:27:52.560 And a is now 3.01. 00:27:52.560 --> 00:27:56.030 So the slope is 6.02. 00:27:56.030 --> 00:27:57.720 You see what I'm doing. 00:27:57.720 --> 00:28:05.310 I'm just moving over to this point, which that has become 00:28:05.310 --> 00:28:10.470 now the a in the second try. 00:28:10.470 --> 00:28:13.230 a in the second cycle of Newton's method. 00:28:13.230 --> 00:28:15.760 This is how Newton's method really is used. 00:28:15.760 --> 00:28:21.170 And now let's find the new x, the highly improved 00:28:21.170 --> 00:28:24.470 x, better than 3.01. 00:28:24.470 --> 00:28:29.190 So Newton's method says x, the new x, minus a. 00:28:29.190 --> 00:28:32.160 I'm just using Newton's formula. 00:28:32.160 --> 00:28:37.280 x minus the a is supposed to be minus the F of a. 00:28:37.280 --> 00:28:47.340 So that's 0.0001 divided by F prime of a, 6.02. 00:28:47.340 --> 00:28:50.330 This is the delta x you could say. 00:28:50.330 --> 00:28:52.000 This is the little correction. 00:28:52.000 --> 00:28:52.810 It's negative. 00:28:52.810 --> 00:28:55.890 It means that we need to pull back a little. 00:28:55.890 --> 00:28:57.130 And you see that. 00:28:57.130 --> 00:29:02.260 We slightly, slightly overshot by following the tangent line. 00:29:02.260 --> 00:29:05.050 The curve went up a little across 0 a little before the 00:29:05.050 --> 00:29:07.020 tangent line. 00:29:07.020 --> 00:29:08.770 This is extremely close. 00:29:08.770 --> 00:29:17.980 So now this gives me the new x right here, 3.01 minus this 00:29:17.980 --> 00:29:20.360 tiny little bit. 00:29:20.360 --> 00:29:25.170 And so that's what the calculator will do. 00:29:25.170 --> 00:29:27.120 I hope you'll do it on a calculator. 00:29:27.120 --> 00:29:31.740 Just make the calculator take that quantity, then 00:29:31.740 --> 00:29:33.030 make it square it. 00:29:35.610 --> 00:29:38.740 Then just go through this. 00:29:38.740 --> 00:29:39.910 Find the new x. 00:29:39.910 --> 00:29:40.530 Square it. 00:29:40.530 --> 00:29:41.840 Subtract 9.06. 00:29:41.840 --> 00:29:43.560 Let's see how close it is. 00:29:43.560 --> 00:29:45.260 I believe that the error-- 00:29:48.450 --> 00:29:51.430 I don't know what it is. 00:29:51.430 --> 00:29:55.930 That's more than I can do in my head, squaring that 3.01 00:29:55.930 --> 00:29:57.880 minus this little tiny bit. 00:29:57.880 --> 00:30:07.590 But I am confident that the error, the x new squared. 00:30:07.590 --> 00:30:13.010 This is the formula for x new, the second 00:30:13.010 --> 00:30:15.570 cycle of Newton's method. 00:30:15.570 --> 00:30:21.340 I think that minus the 9.06, I don't know-- 00:30:21.340 --> 00:30:22.915 can I just put a bunch of zeros? 00:30:28.010 --> 00:30:30.270 Somebody will want me to put in a 1 here, so 00:30:30.270 --> 00:30:32.410 I'll put in a 1. 00:30:32.410 --> 00:30:35.320 I bet it's way out there. 00:30:35.320 --> 00:30:38.480 So Newton's method is really a terrific success. 00:30:38.480 --> 00:30:41.370 Follow the line, then follow the next tangent line. 00:30:41.370 --> 00:30:46.260 Then follow the next one and you home in very, very quickly 00:30:46.260 --> 00:30:48.560 on the exact answer. 00:30:48.560 --> 00:30:52.400 You get more and more decimal places correct. 00:30:52.400 --> 00:30:55.630 OK, that's two uses of calculus coming 00:30:55.630 --> 00:30:58.080 from the same idea. 00:30:58.080 --> 00:31:02.120 The same idea delta f over delta x. 00:31:02.120 --> 00:31:06.070 In one case, it was f that we didn't know. 00:31:06.070 --> 00:31:09.440 In this case, it was x that we didn't know. 00:31:09.440 --> 00:31:13.390 In both cases that formula gives a terrific and a 00:31:13.390 --> 00:31:18.010 terrifically simple and close approximation 00:31:18.010 --> 00:31:20.030 to the exact answer. 00:31:20.030 --> 00:31:21.210 Good. 00:31:21.210 --> 00:31:23.290 Thank you very much. 00:31:23.290 --> 00:31:25.100 ANNOUNCER: This has been a production of MIT 00:31:25.100 --> 00:31:27.490 OpenCourseWare and Gilbert Strang. 00:31:27.490 --> 00:31:29.760 Funding for this video was provided by the Lord 00:31:29.760 --> 00:31:30.980 Foundation. 00:31:30.980 --> 00:31:34.110 To help OCW continue to provide free and open access 00:31:34.110 --> 00:31:37.190 to MIT courses, please make a donation at 00:31:37.190 --> 00:31:38.750 ocw.mit.edu/donate.