WEBVTT 00:00:00.000 --> 00:00:01.984 align:middle line:90% [SQUEAKING] 00:00:01.984 --> 00:00:04.464 align:middle line:90% [RUSTLING] 00:00:04.464 --> 00:00:07.440 align:middle line:90% [CLICKING] 00:00:07.440 --> 00:00:10.437 align:middle line:90% 00:00:10.437 --> 00:00:11.520 align:middle line:90% TOBIAS COLDING: All right. 00:00:11.520 --> 00:00:15.360 align:middle line:84% So last time, we were looking at a function from, say, R, 00:00:15.360 --> 00:00:17.940 align:middle line:84% or it could be some interval in r. 00:00:17.940 --> 00:00:20.840 align:middle line:90% 00:00:20.840 --> 00:00:24.720 align:middle line:90% And then we have a point x in R. 00:00:24.720 --> 00:00:37.200 align:middle line:84% And we say that f is differentiable at x0 00:00:37.200 --> 00:00:41.120 align:middle line:84% if when we form this different quotient-- 00:00:41.120 --> 00:00:42.830 align:middle line:90% so this quotient here-- 00:00:42.830 --> 00:00:50.240 align:middle line:90% 00:00:50.240 --> 00:01:02.350 align:middle line:84% if this here has a limit x go to x0. 00:01:02.350 --> 00:01:06.710 align:middle line:84% And so, again, this means that there is some number so 00:01:06.710 --> 00:01:12.870 align:middle line:84% that if x is sufficiently close, is delta close to x0, 00:01:12.870 --> 00:01:16.730 align:middle line:84% then this quotient here is close to that number. 00:01:16.730 --> 00:01:19.190 align:middle line:84% And it's always assumed that x is not 00:01:19.190 --> 00:01:22.950 align:middle line:84% equal to so that the denominator makes sense. 00:01:22.950 --> 00:01:28.750 align:middle line:84% So that is what it means for the function to be differentiable. 00:01:28.750 --> 00:01:32.270 align:middle line:84% And then one of the first things we saw 00:01:32.270 --> 00:01:50.150 align:middle line:84% was that we had a lemma that if f is differentiable at x0, 00:01:50.150 --> 00:01:53.040 align:middle line:84% then it implies that it's also continuous at x0. 00:01:53.040 --> 00:01:59.630 align:middle line:90% 00:01:59.630 --> 00:02:09.110 align:middle line:84% And just very briefly, we had four rules that if you 00:02:09.110 --> 00:02:14.990 align:middle line:84% had two functions, f and g, and they are both differentiable 00:02:14.990 --> 00:02:19.430 align:middle line:84% and you're looking at where you're adding the two functions, 00:02:19.430 --> 00:02:22.150 align:middle line:84% then the sum here of the two functions 00:02:22.150 --> 00:02:24.030 align:middle line:90% is also differentiable. 00:02:24.030 --> 00:02:28.750 align:middle line:84% And the derivative is the sum of the derivative. 00:02:28.750 --> 00:02:32.150 align:middle line:84% Another thing we had was that if you take a constant times 00:02:32.150 --> 00:02:35.990 align:middle line:84% a function and this function is differentiable, 00:02:35.990 --> 00:02:40.050 align:middle line:84% then this new function is also differentiable. 00:02:40.050 --> 00:02:42.670 align:middle line:84% And the derivative is that constant times 00:02:42.670 --> 00:02:46.070 align:middle line:84% the derivative of the original function. 00:02:46.070 --> 00:02:49.720 align:middle line:84% And then the third thing was the Leibniz rule. 00:02:49.720 --> 00:02:55.350 align:middle line:90% 00:02:55.350 --> 00:02:59.140 align:middle line:84% And that was if you have these two functions that 00:02:59.140 --> 00:03:03.420 align:middle line:84% are differentiable and you take the product 00:03:03.420 --> 00:03:05.820 align:middle line:84% and you take the derivative, that 00:03:05.820 --> 00:03:09.340 align:middle line:84% exists and is equal to the derivative of the first times 00:03:09.340 --> 00:03:16.340 align:middle line:84% the second plus the first times the derivative of the second. 00:03:16.340 --> 00:03:20.150 align:middle line:84% And then the fourth of these was the quotient rule. 00:03:20.150 --> 00:03:25.820 align:middle line:90% 00:03:25.820 --> 00:03:28.080 align:middle line:84% And the quotient rule was that you have, 00:03:28.080 --> 00:03:29.660 align:middle line:90% again, these two functions. 00:03:29.660 --> 00:03:35.700 align:middle line:84% You assume that g here is not equal to 0 at any point. 00:03:35.700 --> 00:03:39.420 align:middle line:84% And then you're looking at the quotient here. 00:03:39.420 --> 00:03:42.300 align:middle line:84% And the quotient is also differentiable. 00:03:42.300 --> 00:03:44.790 align:middle line:84% And the derivative is given by this formula. 00:03:44.790 --> 00:03:50.820 align:middle line:90% 00:03:50.820 --> 00:03:52.940 align:middle line:84% This is the formula for the derivative, 00:03:52.940 --> 00:03:56.380 align:middle line:84% where this is the function multiplied by itself. 00:03:56.380 --> 00:04:03.140 align:middle line:84% And then there was a last rule that was the chain rule. 00:04:03.140 --> 00:04:04.090 align:middle line:90% So the chain rule-- 00:04:04.090 --> 00:04:11.300 align:middle line:90% 00:04:11.300 --> 00:04:14.780 align:middle line:84% and that was that if you take a function, 00:04:14.780 --> 00:04:23.500 align:middle line:84% say, from some interval into some set, let's call it-- 00:04:23.500 --> 00:04:26.980 align:middle line:84% let's say that the interval that the function is defined 00:04:26.980 --> 00:04:29.620 align:middle line:90% is some interval from a to b. 00:04:29.620 --> 00:04:35.780 align:middle line:84% And this function takes value in another interval from c to d. 00:04:35.780 --> 00:04:37.620 align:middle line:84% And then you have a function g that 00:04:37.620 --> 00:04:43.660 align:middle line:84% is defined on c to d into R. Then 00:04:43.660 --> 00:04:45.840 align:middle line:90% you can look at the composition. 00:04:45.840 --> 00:04:49.440 align:middle line:84% You take some x, and you first apply f to it. 00:04:49.440 --> 00:04:51.900 align:middle line:84% And then afterwards, you apply g. 00:04:51.900 --> 00:04:55.510 align:middle line:84% And this is typically written as this. 00:04:55.510 --> 00:04:58.650 align:middle line:90% 00:04:58.650 --> 00:05:02.850 align:middle line:84% And the chain rule was that if f and g are differentiable, 00:05:02.850 --> 00:05:07.410 align:middle line:84% then the composition here is also differentiable. 00:05:07.410 --> 00:05:12.450 align:middle line:84% And the derivative of the composition 00:05:12.450 --> 00:05:18.010 align:middle line:84% is the derivative of the second function evaluated 00:05:18.010 --> 00:05:27.690 align:middle line:84% at this point here times the derivative of the first in x. 00:05:27.690 --> 00:05:30.530 align:middle line:90% So this was the chain rule. 00:05:30.530 --> 00:05:31.450 align:middle line:90% OK. 00:05:31.450 --> 00:05:34.690 align:middle line:90% That was just a quick recap. 00:05:34.690 --> 00:05:40.050 align:middle line:84% And now the first thing we'll be interested in 00:05:40.050 --> 00:05:47.950 align:middle line:84% is if we have a-- so suppose that f here is a function, 00:05:47.950 --> 00:05:53.290 align:middle line:84% let's say defined on some interval from a to b into R. 00:05:53.290 --> 00:05:55.790 align:middle line:90% And f is differentiable. 00:05:55.790 --> 00:06:01.890 align:middle line:90% 00:06:01.890 --> 00:06:04.130 align:middle line:84% It doesn't need to be differentiable for this here 00:06:04.130 --> 00:06:06.130 align:middle line:90% to make sense. 00:06:06.130 --> 00:06:08.850 align:middle line:90% But we say that x0-- 00:06:08.850 --> 00:06:15.010 align:middle line:84% and x0 is now in the open interval here. 00:06:15.010 --> 00:06:25.510 align:middle line:90% So we say that x0 is-- 00:06:25.510 --> 00:06:48.250 align:middle line:84% so we say that f has a local maximum at x0 00:06:48.250 --> 00:07:01.080 align:middle line:84% if there exists delta greater than 0 such 00:07:01.080 --> 00:07:05.720 align:middle line:84% that if you restrict f and you just 00:07:05.720 --> 00:07:14.000 align:middle line:84% think about f on this little interval around x0-- 00:07:14.000 --> 00:07:18.840 align:middle line:84% so you're forgetting about how f looks like everywhere else. 00:07:18.840 --> 00:07:24.940 align:middle line:84% Then if you're looking at the maximum of this function-- 00:07:24.940 --> 00:07:28.200 align:middle line:84% so this is the same as just the maximum 00:07:28.200 --> 00:07:32.530 align:middle line:90% of f on this little interval. 00:07:32.530 --> 00:07:36.480 align:middle line:90% 00:07:36.480 --> 00:07:43.760 align:middle line:84% Then the maximum is achieved at x0 if this is the case. 00:07:43.760 --> 00:07:45.880 align:middle line:90% So you have the picture here. 00:07:45.880 --> 00:07:47.060 align:middle line:90% You have a function. 00:07:47.060 --> 00:07:49.280 align:middle line:84% We talked about this a little bit last time. 00:07:49.280 --> 00:07:50.250 align:middle line:90% You have a function. 00:07:50.250 --> 00:07:53.110 align:middle line:84% Maybe the function looks like this. 00:07:53.110 --> 00:07:57.070 align:middle line:90% And this point here is x0. 00:07:57.070 --> 00:07:59.910 align:middle line:84% And so this here is the value at x0. 00:07:59.910 --> 00:08:02.710 align:middle line:84% And you see that there's some little interval-- 00:08:02.710 --> 00:08:04.550 align:middle line:90% say this interval here. 00:08:04.550 --> 00:08:06.310 align:middle line:84% If you're looking at the function 00:08:06.310 --> 00:08:08.510 align:middle line:84% in this little interval, then it's 00:08:08.510 --> 00:08:12.310 align:middle line:90% clear that the maximum is at x0. 00:08:12.310 --> 00:08:15.470 align:middle line:84% But, of course, it's just a local maximum. 00:08:15.470 --> 00:08:18.850 align:middle line:84% If you were allowed to move further away over here, 00:08:18.850 --> 00:08:22.370 align:middle line:84% you see that the value of the function is much larger. 00:08:22.370 --> 00:08:24.270 align:middle line:90% So this is a local maximum. 00:08:24.270 --> 00:08:27.270 align:middle line:84% And for this to make sense, you don't need the function 00:08:27.270 --> 00:08:29.190 align:middle line:90% to be differentiable. 00:08:29.190 --> 00:08:33.110 align:middle line:84% But when we apply it, we will assume 00:08:33.110 --> 00:08:34.909 align:middle line:90% that it is differentiable. 00:08:34.909 --> 00:08:37.570 align:middle line:90% So this is the local maximum. 00:08:37.570 --> 00:08:45.430 align:middle line:84% And this is the picture, that is it's just a maximum if you're 00:08:45.430 --> 00:08:50.010 align:middle line:84% looking at values near x0, but it's not a global maximum. 00:08:50.010 --> 00:08:54.070 align:middle line:84% Of course, it could be a global maximum. 00:08:54.070 --> 00:09:03.000 align:middle line:84% And then similarly, for a local minimum-- 00:09:03.000 --> 00:09:08.270 align:middle line:90% 00:09:08.270 --> 00:09:11.470 align:middle line:84% and I'll just draw the picture because the definition 00:09:11.470 --> 00:09:12.490 align:middle line:90% is kind of obvious. 00:09:12.490 --> 00:09:19.970 align:middle line:84% It's the same, that you take a function here. 00:09:19.970 --> 00:09:22.140 align:middle line:84% And so the function maybe could look like this. 00:09:22.140 --> 00:09:26.230 align:middle line:90% 00:09:26.230 --> 00:09:30.350 align:middle line:84% And you see that if this here was x0, 00:09:30.350 --> 00:09:34.430 align:middle line:84% then in some little neighborhood around x0, 00:09:34.430 --> 00:09:38.870 align:middle line:84% the value here would indeed be the minimum. 00:09:38.870 --> 00:09:41.390 align:middle line:84% But if you're moving much further away, 00:09:41.390 --> 00:09:44.630 align:middle line:84% then there might be other points that are even smaller, where 00:09:44.630 --> 00:09:46.470 align:middle line:90% the value is even smaller. 00:09:46.470 --> 00:09:49.100 align:middle line:90% So this is like a local minimum. 00:09:49.100 --> 00:09:54.780 align:middle line:84% And so the first observation you have about this 00:09:54.780 --> 00:09:56.630 align:middle line:90% is the following lemma. 00:09:56.630 --> 00:10:00.020 align:middle line:90% 00:10:00.020 --> 00:10:07.580 align:middle line:84% If f here is a function, say, on an interval like this, 00:10:07.580 --> 00:10:32.820 align:middle line:84% it's differentiable and has a local maximum or local minimum 00:10:32.820 --> 00:10:34.460 align:middle line:90% at x0. 00:10:34.460 --> 00:10:38.500 align:middle line:84% And now here, if I'm thinking about a function defined 00:10:38.500 --> 00:10:40.740 align:middle line:84% on a closed interval like this, it's 00:10:40.740 --> 00:10:44.420 align:middle line:84% important that where the local max or min 00:10:44.420 --> 00:10:46.400 align:middle line:90% is in an interior point. 00:10:46.400 --> 00:10:51.020 align:middle line:84% It's not allowed to be one of the boundary points. 00:10:51.020 --> 00:10:58.380 align:middle line:84% So if it is differentiable and it has a local maximum 00:10:58.380 --> 00:11:01.460 align:middle line:84% or minimum at such an interior point, 00:11:01.460 --> 00:11:11.420 align:middle line:84% then the claim is that then the derivative at x0 is equal to 0. 00:11:11.420 --> 00:11:16.260 align:middle line:90% 00:11:16.260 --> 00:11:18.560 align:middle line:84% And so now let's try to prove that. 00:11:18.560 --> 00:11:22.100 align:middle line:90% 00:11:22.100 --> 00:11:24.040 align:middle line:90% So we have here-- 00:11:24.040 --> 00:11:26.460 align:middle line:84% so the picture is-- let's just assume. 00:11:26.460 --> 00:11:48.130 align:middle line:84% So proof-- assume that f has a local maximum at x0. 00:11:48.130 --> 00:11:50.210 align:middle line:84% If it's a local minimum, the argument 00:11:50.210 --> 00:11:53.510 align:middle line:84% is the same with obvious modifications. 00:11:53.510 --> 00:11:54.210 align:middle line:90% OK. 00:11:54.210 --> 00:11:56.850 align:middle line:84% So you have some picture, like this. 00:11:56.850 --> 00:11:58.930 align:middle line:90% Again, here, you have x0. 00:11:58.930 --> 00:12:02.250 align:middle line:84% And now this here is the function that-- oops, 00:12:02.250 --> 00:12:03.650 align:middle line:90% maybe this. 00:12:03.650 --> 00:12:06.030 align:middle line:90% So this is the function right. 00:12:06.030 --> 00:12:10.290 align:middle line:84% We see that at the value here, at x0, it's maximum. 00:12:10.290 --> 00:12:11.830 align:middle line:90% And then who cares? 00:12:11.830 --> 00:12:16.650 align:middle line:84% We don't care about what happen when you are further away. 00:12:16.650 --> 00:12:19.430 align:middle line:84% And so you divide into two cases. 00:12:19.430 --> 00:12:23.850 align:middle line:84% You're looking at where x here is bigger than x0, 00:12:23.850 --> 00:12:25.930 align:middle line:90% if x is bigger than x0. 00:12:25.930 --> 00:12:28.050 align:middle line:90% And now you're looking at-- 00:12:28.050 --> 00:12:32.830 align:middle line:84% so then you have that f of x0 minus f of x0. 00:12:32.830 --> 00:12:35.490 align:middle line:90% 00:12:35.490 --> 00:12:37.230 align:middle line:90% The value here is maximum. 00:12:37.230 --> 00:12:41.310 align:middle line:90% So this value here is-- 00:12:41.310 --> 00:12:57.370 align:middle line:84% so if this is the case and x is sufficiently close to x0-- 00:12:57.370 --> 00:13:01.990 align:middle line:84% if it's very close here, then it has a maximum at this point. 00:13:01.990 --> 00:13:05.050 align:middle line:84% So this means that this difference here is non-positive. 00:13:05.050 --> 00:13:09.090 align:middle line:90% 00:13:09.090 --> 00:13:12.550 align:middle line:84% This didn't really use that it was strictly bigger than, right? 00:13:12.550 --> 00:13:16.570 align:middle line:84% It just used that it was close to x0. 00:13:16.570 --> 00:13:20.330 align:middle line:84% But now this is where we're using that it's bigger than-- 00:13:20.330 --> 00:13:32.510 align:middle line:84% because you also have that x minus x0-- 00:13:32.510 --> 00:13:35.450 align:middle line:84% and, again, this is where this assumption's used-- 00:13:35.450 --> 00:13:39.170 align:middle line:90% this thing here is positive. 00:13:39.170 --> 00:13:42.880 align:middle line:84% So this means that if you take something nonpositive 00:13:42.880 --> 00:13:48.080 align:middle line:84% and you divide it by something positive, then what you get 00:13:48.080 --> 00:13:52.980 align:middle line:84% is something that is nonpositive. 00:13:52.980 --> 00:13:57.240 align:middle line:84% So this implies that for the difference quotient, 00:13:57.240 --> 00:14:05.080 align:middle line:84% f of x minus f of x0, x minus x zero-- 00:14:05.080 --> 00:14:07.200 align:middle line:90% again, this here is nonpositive. 00:14:07.200 --> 00:14:10.600 align:middle line:84% This is positive, which means that the whole thing is 00:14:10.600 --> 00:14:12.200 align:middle line:90% nonpositive. 00:14:12.200 --> 00:14:19.000 align:middle line:84% So as you're going towards x0 from the right, 00:14:19.000 --> 00:14:22.320 align:middle line:84% the difference quotient here is nonpositive. 00:14:22.320 --> 00:14:28.220 align:middle line:84% If you now go to x0 from the left, 00:14:28.220 --> 00:14:33.440 align:middle line:90% if you go in this direction-- 00:14:33.440 --> 00:14:38.880 align:middle line:84% so this means that now you have x here is less than x0 00:14:38.880 --> 00:14:41.830 align:middle line:90% and sufficiently close. 00:14:41.830 --> 00:14:51.280 align:middle line:90% 00:14:51.280 --> 00:14:56.560 align:middle line:84% So if you have that, then well, you again have-- 00:14:56.560 --> 00:14:59.120 align:middle line:84% this was just because it was a local maximum. 00:14:59.120 --> 00:15:04.520 align:middle line:84% You still have that this thing here is not positive. 00:15:04.520 --> 00:15:18.800 align:middle line:84% But now we have that x minus x0, this thing here now is-- 00:15:18.800 --> 00:15:21.920 align:middle line:84% because of this, this means that this thing here 00:15:21.920 --> 00:15:24.786 align:middle line:90% is strictly less than 0. 00:15:24.786 --> 00:15:27.880 align:middle line:84% So you take something that is nonpositive 00:15:27.880 --> 00:15:30.640 align:middle line:84% and you divide it by something negative, 00:15:30.640 --> 00:15:33.220 align:middle line:84% then you get something nonnegative. 00:15:33.220 --> 00:15:38.870 align:middle line:84% So therefore, this implies that for the difference quotient, f 00:15:38.870 --> 00:15:44.950 align:middle line:84% of x minus f of x0, for the difference quotient here, 00:15:44.950 --> 00:15:47.970 align:middle line:84% you get that this thing here, this is nonpositive. 00:15:47.970 --> 00:15:49.010 align:middle line:90% This is negative. 00:15:49.010 --> 00:15:52.630 align:middle line:84% This means that this here is nonnegative. 00:15:52.630 --> 00:16:01.630 align:middle line:84% So as you're coming in-- so you see that as you're approaching 00:16:01.630 --> 00:16:05.510 align:middle line:84% this direction to x0, then the difference quotient 00:16:05.510 --> 00:16:06.750 align:middle line:90% is not positive. 00:16:06.750 --> 00:16:09.830 align:middle line:84% When you're coming in here, it's nonnegative. 00:16:09.830 --> 00:16:11.830 align:middle line:90% But it has a limit. 00:16:11.830 --> 00:16:16.430 align:middle line:84% And so the limit from this side would have to be-- 00:16:16.430 --> 00:16:18.190 align:middle line:84% the limit from this side would have 00:16:18.190 --> 00:16:21.930 align:middle line:84% to be nonpositive and the limit from this one nonnegative. 00:16:21.930 --> 00:16:24.870 align:middle line:84% So therefore, the limit must be 0. 00:16:24.870 --> 00:16:26.740 align:middle line:84% So we conclude that the limit here-- 00:16:26.740 --> 00:16:31.070 align:middle line:90% 00:16:31.070 --> 00:16:36.660 align:middle line:84% hence, the limit here, as x goes to x0 00:16:36.660 --> 00:16:42.050 align:middle line:84% of f of x of the difference quotient must be 0. 00:16:42.050 --> 00:16:45.220 align:middle line:90% 00:16:45.220 --> 00:16:50.180 align:middle line:84% And so this is the derivative at that point-- 00:16:50.180 --> 00:16:52.790 align:middle line:84% and similarly if it was a local minimum. 00:16:52.790 --> 00:16:57.500 align:middle line:90% 00:16:57.500 --> 00:17:02.820 align:middle line:84% And at point where the derivative-- 00:17:02.820 --> 00:17:06.900 align:middle line:90% so maybe I'll just say that-- 00:17:06.900 --> 00:17:18.619 align:middle line:84% so if f prime at x0 is equal to 0, 00:17:18.619 --> 00:17:23.780 align:middle line:84% then we often say that then x0 is said 00:17:23.780 --> 00:17:38.560 align:middle line:90% to be a critical point for f. 00:17:38.560 --> 00:17:41.960 align:middle line:84% And let me just give you an example. 00:17:41.960 --> 00:17:44.860 align:middle line:90% Let me give you two examples. 00:17:44.860 --> 00:17:50.540 align:middle line:84% So suppose you take f of x equal to x squared. 00:17:50.540 --> 00:17:52.860 align:middle line:90% This here is a polynomial. 00:17:52.860 --> 00:17:56.360 align:middle line:84% We know how to compute the derivative of any polynomial. 00:17:56.360 --> 00:17:58.420 align:middle line:84% This is just using the Leibniz rule-- 00:17:58.420 --> 00:18:00.040 align:middle line:90% I mean, using these rules. 00:18:00.040 --> 00:18:02.680 align:middle line:84% In this case, you would only need the Leibniz rule together 00:18:02.680 --> 00:18:06.620 align:middle line:84% with-- we knew what the derivative of the function 00:18:06.620 --> 00:18:08.980 align:middle line:90% f of x equal to x is. 00:18:08.980 --> 00:18:15.260 align:middle line:84% So in this case, the derivative of this here is 2x. 00:18:15.260 --> 00:18:20.550 align:middle line:84% And so this means that x is equal to 0 is a critical point. 00:18:20.550 --> 00:18:24.580 align:middle line:90% 00:18:24.580 --> 00:18:28.340 align:middle line:84% It's actually the only critical point. 00:18:28.340 --> 00:18:30.700 align:middle line:84% And so you see, in this case, this 00:18:30.700 --> 00:18:32.580 align:middle line:90% is how the function looked like. 00:18:32.580 --> 00:18:35.310 align:middle line:84% This is how the function looks like. 00:18:35.310 --> 00:18:36.270 align:middle line:90% Here, you have 0. 00:18:36.270 --> 00:18:39.610 align:middle line:90% 00:18:39.610 --> 00:18:46.170 align:middle line:84% Another example is that if you take another polynomial, 00:18:46.170 --> 00:18:51.890 align:middle line:84% where instead of x to the power 2, it's x to the power 3, 00:18:51.890 --> 00:18:59.470 align:middle line:84% then the derivative of this function is 3x squared. 00:18:59.470 --> 00:19:05.630 align:middle line:84% And so you see again that x equal to 0 is a critical point. 00:19:05.630 --> 00:19:09.090 align:middle line:84% The derivative here, if you plug in 0, you get 0. 00:19:09.090 --> 00:19:10.720 align:middle line:90% So this is a critical point. 00:19:10.720 --> 00:19:17.930 align:middle line:90% 00:19:17.930 --> 00:19:25.850 align:middle line:84% But in this case, the graph of the function 00:19:25.850 --> 00:19:29.690 align:middle line:84% for this example here, the graph, 00:19:29.690 --> 00:19:32.570 align:middle line:84% it looks like something like this. 00:19:32.570 --> 00:19:34.550 align:middle line:84% It becomes very flat here, like this. 00:19:34.550 --> 00:19:37.190 align:middle line:90% 00:19:37.190 --> 00:19:40.530 align:middle line:84% So 0 is a critical point, but it's not 00:19:40.530 --> 00:19:42.590 align:middle line:84% like it has a local maximum or minimum. 00:19:42.590 --> 00:19:51.370 align:middle line:90% 00:19:51.370 --> 00:19:51.870 align:middle line:90% OK. 00:19:51.870 --> 00:19:56.730 align:middle line:84% So the next thing I want to do is, I want to now apply it. 00:19:56.730 --> 00:20:00.490 align:middle line:84% I want to apply this simple observation. 00:20:00.490 --> 00:20:07.010 align:middle line:84% And I want to apply it to get a consequence, 00:20:07.010 --> 00:20:08.270 align:middle line:90% a simple consequence. 00:20:08.270 --> 00:20:11.810 align:middle line:84% And then I want to apply that simple consequence to get 00:20:11.810 --> 00:20:15.690 align:middle line:90% a lot of other consequences. 00:20:15.690 --> 00:20:20.610 align:middle line:84% So the first thing I want to do is 00:20:20.610 --> 00:20:24.460 align:middle line:84% what's called Rolle's lemma or theorem. 00:20:24.460 --> 00:20:41.360 align:middle line:90% 00:20:41.360 --> 00:20:46.760 align:middle line:84% So this is the following, that I have a function 00:20:46.760 --> 00:20:52.390 align:middle line:90% f from a to b into. 00:20:52.390 --> 00:20:55.360 align:middle line:90% 00:20:55.360 --> 00:20:57.760 align:middle line:90% R. 00:20:57.760 --> 00:21:00.860 align:middle line:84% And I'm assuming that f is differentiable. 00:21:00.860 --> 00:21:08.920 align:middle line:90% 00:21:08.920 --> 00:21:14.080 align:middle line:84% And I'm also assuming that the value initially 00:21:14.080 --> 00:21:19.600 align:middle line:84% and the value at the end, that those two values are the same. 00:21:19.600 --> 00:21:22.340 align:middle line:84% Sometimes one set those value equal to 0, 00:21:22.340 --> 00:21:26.800 align:middle line:84% but they just need to be the same. 00:21:26.800 --> 00:21:31.950 align:middle line:84% And then the claim is that then there exist. 00:21:31.950 --> 00:21:36.188 align:middle line:90% 00:21:36.188 --> 00:21:47.080 align:middle line:84% a x0, where is strictly between a and b. 00:21:47.080 --> 00:21:54.410 align:middle line:84% And the derivative at this x0 is equal to 0. 00:21:54.410 --> 00:21:57.024 align:middle line:90% 00:21:57.024 --> 00:21:59.960 align:middle line:84% And so the idea is just very simple, 00:21:59.960 --> 00:22:16.080 align:middle line:90% that you have a function. 00:22:16.080 --> 00:22:18.160 align:middle line:90% And the function initially-- 00:22:18.160 --> 00:22:20.240 align:middle line:90% this is set a. 00:22:20.240 --> 00:22:22.000 align:middle line:90% Here, you have b. 00:22:22.000 --> 00:22:25.380 align:middle line:84% The function initially and at the end 00:22:25.380 --> 00:22:27.020 align:middle line:90% end up at the same value. 00:22:27.020 --> 00:22:35.250 align:middle line:84% So this here is f of a, and that's the same as f of b. 00:22:35.250 --> 00:22:38.230 align:middle line:90% 00:22:38.230 --> 00:22:42.670 align:middle line:84% And then who knows what happened in between? 00:22:42.670 --> 00:22:44.070 align:middle line:90% Maybe like this. 00:22:44.070 --> 00:22:46.830 align:middle line:84% And the claim is that then there's 00:22:46.830 --> 00:22:49.890 align:middle line:84% always some point where the derivative is 0. 00:22:49.890 --> 00:22:52.830 align:middle line:84% In this case, it looks like that those two 00:22:52.830 --> 00:22:56.830 align:middle line:84% would be the points where the derivative was 0. 00:22:56.830 --> 00:23:02.870 align:middle line:84% And the proof is just this proof that there's 00:23:02.870 --> 00:23:04.120 align:middle line:90% three cases to consider. 00:23:04.120 --> 00:23:14.910 align:middle line:90% 00:23:14.910 --> 00:23:16.870 align:middle line:84% One is-- I mean, you don't really 00:23:16.870 --> 00:23:19.270 align:middle line:84% have to divide it up by this, but let me divide it up 00:23:19.270 --> 00:23:20.650 align:middle line:90% into these three cases. 00:23:20.650 --> 00:23:27.260 align:middle line:84% One is that the function here is constant, everywhere 00:23:27.260 --> 00:23:30.180 align:middle line:90% equal to the same. 00:23:30.180 --> 00:23:32.340 align:middle line:84% Well, if the function's constant, 00:23:32.340 --> 00:23:34.600 align:middle line:90% then the derivative is 0. 00:23:34.600 --> 00:23:36.500 align:middle line:90% We have already proven that. 00:23:36.500 --> 00:23:52.620 align:middle line:84% Case 2, somewhere, there exist x such that the value at this x 00:23:52.620 --> 00:23:56.800 align:middle line:84% is strictly bigger than the value at the endpoints. 00:23:56.800 --> 00:23:59.500 align:middle line:90% 00:23:59.500 --> 00:24:03.020 align:middle line:90% And then case 3-- 00:24:03.020 --> 00:24:09.340 align:middle line:84% there exists an x such that the value at this x 00:24:09.340 --> 00:24:12.810 align:middle line:84% is strictly less than the values at the endpoint. 00:24:12.810 --> 00:24:15.340 align:middle line:90% 00:24:15.340 --> 00:24:20.000 align:middle line:84% It's clear that one of these three cases happen, right? 00:24:20.000 --> 00:24:25.140 align:middle line:84% Of course, it could be that case 2 and 3 could both happen. 00:24:25.140 --> 00:24:29.620 align:middle line:84% But at least one of the three cases would happen. 00:24:29.620 --> 00:24:34.260 align:middle line:84% In case 1, we already know that the derivative is 0 everywhere. 00:24:34.260 --> 00:24:36.180 align:middle line:90% So let's just look at case 2. 00:24:36.180 --> 00:24:38.420 align:middle line:90% Case 3 is similar. 00:24:38.420 --> 00:24:42.560 align:middle line:84% So if there exists some x where the value-- 00:24:42.560 --> 00:24:46.380 align:middle line:84% so in this picture here, this here is maybe x. 00:24:46.380 --> 00:24:51.900 align:middle line:84% The value here is larger than at the two endpoints. 00:24:51.900 --> 00:24:54.980 align:middle line:84% Well, by the extreme value theorem 00:24:54.980 --> 00:24:57.500 align:middle line:84% that we proved earlier-- just because if a function is 00:24:57.500 --> 00:24:59.620 align:middle line:84% differentiable, it's also continuous. 00:24:59.620 --> 00:25:01.780 align:middle line:84% By the extreme value theorem, there 00:25:01.780 --> 00:25:05.620 align:middle line:84% exists somewhere where, actually, the maximum 00:25:05.620 --> 00:25:07.166 align:middle line:90% is achieved. 00:25:07.166 --> 00:25:10.940 align:middle line:90% And since somewhere in between-- 00:25:10.940 --> 00:25:16.580 align:middle line:84% in this case here, the maximum is not 00:25:16.580 --> 00:25:19.240 align:middle line:90% achieved at the two endpoints. 00:25:19.240 --> 00:25:26.025 align:middle line:84% So the maximum must be achieved at an interior point. 00:25:26.025 --> 00:25:28.950 align:middle line:84% But now, it's not just a local maximum. 00:25:28.950 --> 00:25:30.590 align:middle line:90% It actually is a maximum. 00:25:30.590 --> 00:25:32.770 align:middle line:84% But already, if it was a local maximum, 00:25:32.770 --> 00:25:35.610 align:middle line:84% the derivative would be 0 at that point. 00:25:35.610 --> 00:25:37.330 align:middle line:90% So this proved case 2. 00:25:37.330 --> 00:25:41.090 align:middle line:90% And case 3 is similar. 00:25:41.090 --> 00:25:45.930 align:middle line:84% So this is Rolle's theorem Now, Rolle's theorem 00:25:45.930 --> 00:25:49.730 align:middle line:84% have then a bunch of applications. 00:25:49.730 --> 00:25:53.900 align:middle line:84% And the first one is to what's called the mean, value theorem. 00:25:53.900 --> 00:25:57.570 align:middle line:90% 00:25:57.570 --> 00:26:01.090 align:middle line:84% And there are two versions of the mean value theorem. 00:26:01.090 --> 00:26:04.810 align:middle line:84% One is the actual mean value theorem. 00:26:04.810 --> 00:26:08.530 align:middle line:84% And the other is like a fancy version of it, 00:26:08.530 --> 00:26:11.890 align:middle line:84% where you have two functions in play at the same time. 00:26:11.890 --> 00:26:15.810 align:middle line:84% And that's called the Cauchy mean value theorem. 00:26:15.810 --> 00:26:28.030 align:middle line:84% So first consequence of Rolle's theorem. 00:26:28.030 --> 00:26:33.530 align:middle line:90% 00:26:33.530 --> 00:26:35.880 align:middle line:84% And this is the mean value theorem. 00:26:35.880 --> 00:26:41.750 align:middle line:90% 00:26:41.750 --> 00:26:42.250 align:middle line:90% OK. 00:26:42.250 --> 00:26:45.760 align:middle line:84% And so the first one is the actual mean value theorem. 00:26:45.760 --> 00:26:55.530 align:middle line:90% 00:26:55.530 --> 00:26:57.370 align:middle line:84% And the actual mean value theorem 00:26:57.370 --> 00:27:04.050 align:middle line:84% says that if you take a function f from a 00:27:04.050 --> 00:27:08.370 align:middle line:84% to b on some interval like this and the function 00:27:08.370 --> 00:27:33.920 align:middle line:84% is differentiable, then there exists an x, where x is strictly 00:27:33.920 --> 00:27:42.840 align:middle line:84% between a and b and so that the derivative at x 00:27:42.840 --> 00:27:46.120 align:middle line:84% is equal to the difference quotient, where 00:27:46.120 --> 00:27:51.960 align:middle line:84% you do the difference quotient at the two endpoints. 00:27:51.960 --> 00:27:56.040 align:middle line:90% So this here is this side of it. 00:27:56.040 --> 00:27:58.400 align:middle line:84% That's what you often call the mean. 00:27:58.400 --> 00:28:01.240 align:middle line:84% And the mean is equal to the derivative 00:28:01.240 --> 00:28:02.860 align:middle line:90% at some intermediate point. 00:28:02.860 --> 00:28:05.620 align:middle line:90% This is what they say. 00:28:05.620 --> 00:28:12.800 align:middle line:84% And so this seems a lot better than Rolle's theorem. 00:28:12.800 --> 00:28:17.400 align:middle line:84% But it's an easy consequence of Rolle's theorem. 00:28:17.400 --> 00:28:19.200 align:middle line:90% Because if you look at-- 00:28:19.200 --> 00:28:24.680 align:middle line:90% 00:28:24.680 --> 00:28:31.008 align:middle line:84% so suppose you take the function-- so proof. 00:28:31.008 --> 00:28:34.920 align:middle line:90% 00:28:34.920 --> 00:28:40.600 align:middle line:84% Suppose you take the function h of x 00:28:40.600 --> 00:28:55.960 align:middle line:84% equal to f of x minus f of b minus f of a divided 00:28:55.960 --> 00:28:59.620 align:middle line:90% by b minus a times x minus a. 00:28:59.620 --> 00:29:04.120 align:middle line:84% Suppose you're looking at this function here, h. 00:29:04.120 --> 00:29:07.420 align:middle line:84% Then observe first that-- so this one, 00:29:07.420 --> 00:29:09.440 align:middle line:90% this function is differentiable. 00:29:09.440 --> 00:29:12.360 align:middle line:90% This is just a constant. 00:29:12.360 --> 00:29:16.400 align:middle line:90% This is just a linear function. 00:29:16.400 --> 00:29:18.120 align:middle line:90% a is a constant also. 00:29:18.120 --> 00:29:20.530 align:middle line:90% So this is differentiable. 00:29:20.530 --> 00:29:21.370 align:middle line:90% This is a constant. 00:29:21.370 --> 00:29:22.990 align:middle line:90% This is differentiable. 00:29:22.990 --> 00:29:25.730 align:middle line:84% And we're subtracting it from a differentiable function. 00:29:25.730 --> 00:29:30.012 align:middle line:90% So h is differentiable. 00:29:30.012 --> 00:29:34.990 align:middle line:90% h of a is f of a. 00:29:34.990 --> 00:29:43.390 align:middle line:84% And then this thing here, minus f of b minus f of a, b minus a-- 00:29:43.390 --> 00:29:45.550 align:middle line:90% and then this here is a minus a. 00:29:45.550 --> 00:29:46.690 align:middle line:90% So this is 0. 00:29:46.690 --> 00:29:49.590 align:middle line:90% So this is f of a. 00:29:49.590 --> 00:30:02.710 align:middle line:84% h of b is f of b minus and then f of b minus f of a, b minus a. 00:30:02.710 --> 00:30:07.110 align:middle line:84% And here, you plug in b, so b minus a. 00:30:07.110 --> 00:30:10.420 align:middle line:84% And so you see that these here cancel out. 00:30:10.420 --> 00:30:14.730 align:middle line:90% 00:30:14.730 --> 00:30:16.910 align:middle line:90% So you are subtracting this. 00:30:16.910 --> 00:30:20.430 align:middle line:84% But this is a minus minus, so you're adding. 00:30:20.430 --> 00:30:23.030 align:middle line:90% So this is f of a. 00:30:23.030 --> 00:30:28.510 align:middle line:84% So you see that for this function, the value at a and b 00:30:28.510 --> 00:30:31.150 align:middle line:90% are the same. 00:30:31.150 --> 00:30:34.260 align:middle line:90% So now we apply Rolle's theorem. 00:30:34.260 --> 00:30:53.310 align:middle line:90% 00:30:53.310 --> 00:31:13.510 align:middle line:84% So now we have, by Rolle's theorem applied to h, 00:31:13.510 --> 00:31:21.580 align:middle line:84% then we have that h is equal to 0-- the derivative of h 00:31:21.580 --> 00:31:32.372 align:middle line:84% is equal to 0 for some x, where x is strictly between a and b. 00:31:32.372 --> 00:31:33.580 align:middle line:90% This is what Rolle's theorem. 00:31:33.580 --> 00:31:36.220 align:middle line:84% This is because this function, h, has 00:31:36.220 --> 00:31:40.740 align:middle line:90% the same value at a as at b. 00:31:40.740 --> 00:31:45.730 align:middle line:84% Now, what is the derivative of h? 00:31:45.730 --> 00:31:53.940 align:middle line:84% So the derivative of h is the derivative of f. 00:31:53.940 --> 00:31:56.840 align:middle line:84% And, again, this is just a constant. 00:31:56.840 --> 00:32:06.320 align:middle line:84% So we have to keep that minus f of b minus f of a, b minus a. 00:32:06.320 --> 00:32:09.840 align:middle line:90% 00:32:09.840 --> 00:32:11.000 align:middle line:90% So this is just a constant. 00:32:11.000 --> 00:32:13.460 align:middle line:84% And then derivative of this is just 1. 00:32:13.460 --> 00:32:16.330 align:middle line:84% So this is what the derivative is. 00:32:16.330 --> 00:32:18.970 align:middle line:84% And so you see that-- so we have this x, 00:32:18.970 --> 00:32:20.970 align:middle line:90% where this derivative is 0. 00:32:20.970 --> 00:32:24.410 align:middle line:90% So this here is 0. 00:32:24.410 --> 00:32:27.650 align:middle line:84% But if this here is 0, it means that this thing here 00:32:27.650 --> 00:32:29.690 align:middle line:90% is equal to that. 00:32:29.690 --> 00:32:32.410 align:middle line:84% And so that was the mean value theorem. 00:32:32.410 --> 00:32:35.970 align:middle line:84% So this proved the mean value theorem. 00:32:35.970 --> 00:32:37.770 align:middle line:90% OK. 00:32:37.770 --> 00:32:40.330 align:middle line:84% There's a second mean value theorem. 00:32:40.330 --> 00:32:42.696 align:middle line:84% And the second one is the Cauchy mean value theorem. 00:32:42.696 --> 00:32:45.370 align:middle line:90% 00:32:45.370 --> 00:32:51.010 align:middle line:84% And so this is a little more fancy. 00:32:51.010 --> 00:32:52.760 align:middle line:84% So the Cauchy mean value theorem-- 00:32:52.760 --> 00:33:07.330 align:middle line:90% 00:33:07.330 --> 00:33:13.290 align:middle line:84% so this is the following, that you have two functions, f and g. 00:33:13.290 --> 00:33:22.450 align:middle line:84% So we have f and g on some interval from a to b into R. 00:33:22.450 --> 00:33:24.230 align:middle line:84% And they are both differentiable. 00:33:24.230 --> 00:33:36.090 align:middle line:90% 00:33:36.090 --> 00:33:38.530 align:middle line:90% They're both differentiable. 00:33:38.530 --> 00:33:51.010 align:middle line:84% And then the claim is that there exists a x-- 00:33:51.010 --> 00:33:53.090 align:middle line:90% let's call it x0. 00:33:53.090 --> 00:33:56.330 align:middle line:84% Maybe I should have called it x0 up there, it's just one. 00:33:56.330 --> 00:34:03.270 align:middle line:84% There exists x0 with x0 strictly between a and b. 00:34:03.270 --> 00:34:08.358 align:middle line:90% 00:34:08.358 --> 00:34:09.310 align:middle line:90% OK. 00:34:09.310 --> 00:34:16.040 align:middle line:84% And so it's something about that the derivative of both functions 00:34:16.040 --> 00:34:19.679 align:middle line:84% kind of behave like the mean value theorem at the same time. 00:34:19.679 --> 00:34:24.639 align:middle line:90% So it is that the derivative-- 00:34:24.639 --> 00:34:27.600 align:middle line:84% I will explain this in a second-- 00:34:27.600 --> 00:34:37.440 align:middle line:84% that the derivative at this point times g of b minus g of a 00:34:37.440 --> 00:34:52.239 align:middle line:84% is equal to the derivative of g at x0 f of b minus f of a. 00:34:52.239 --> 00:34:58.980 align:middle line:84% So this is the Cauchy mean value theorem. 00:34:58.980 --> 00:35:01.840 align:middle line:90% So let's say that for this here. 00:35:01.840 --> 00:35:05.560 align:middle line:84% And let's just look at that expression 00:35:05.560 --> 00:35:09.220 align:middle line:84% here before we prove the Cauchy mean value theorem. 00:35:09.220 --> 00:35:17.400 align:middle line:90% 00:35:17.400 --> 00:35:25.480 align:middle line:84% So as a consequence-- so observe that, 00:35:25.480 --> 00:35:48.070 align:middle line:84% if g of b is not equal to g of a and g prime at x0 is not equal 00:35:48.070 --> 00:35:48.570 align:middle line:90% to 0-- 00:35:48.570 --> 00:35:52.440 align:middle line:90% 00:35:52.440 --> 00:35:55.860 align:middle line:84% if you have that, then you could divide-- 00:35:55.860 --> 00:35:58.340 align:middle line:90% you could have this expression. 00:35:58.340 --> 00:36:03.040 align:middle line:84% Let's say that-- then, by just dividing this here 00:36:03.040 --> 00:36:05.600 align:middle line:84% on both sides and this here on both sides, 00:36:05.600 --> 00:36:12.590 align:middle line:84% you would get that f prime of x0 divided by g prime at x0 00:36:12.590 --> 00:36:24.450 align:middle line:84% is equal to f of b minus f of a g of b minus g of a. 00:36:24.450 --> 00:36:27.790 align:middle line:90% 00:36:27.790 --> 00:36:31.070 align:middle line:90% You would have that. 00:36:31.070 --> 00:36:35.630 align:middle line:84% Now, I could think about this thing here 00:36:35.630 --> 00:36:46.350 align:middle line:84% as f of b minus f of a divided by b minus a. 00:36:46.350 --> 00:36:48.910 align:middle line:84% So I could think about it like this 00:36:48.910 --> 00:36:57.750 align:middle line:84% and then times b minus a, g of b minus g of a. 00:36:57.750 --> 00:37:03.270 align:middle line:84% And so you see that this here, it's something about-- 00:37:03.270 --> 00:37:06.150 align:middle line:84% the actual mean value theorem say that this thing 00:37:06.150 --> 00:37:09.990 align:middle line:84% here, that there would be some point, x0, in between that 00:37:09.990 --> 00:37:12.190 align:middle line:90% was equal to this. 00:37:12.190 --> 00:37:15.590 align:middle line:84% That would be the mean value theorem applied to f. 00:37:15.590 --> 00:37:18.830 align:middle line:84% And the mean value theorem applied to g 00:37:18.830 --> 00:37:24.910 align:middle line:84% would be that there exists some x0 so that 1 over this 00:37:24.910 --> 00:37:28.630 align:middle line:90% is equal to this thing here. 00:37:28.630 --> 00:37:30.310 align:middle line:84% And the Cauchy mean value theorem 00:37:30.310 --> 00:37:34.030 align:middle line:84% say that you can arrange so that both of these things 00:37:34.030 --> 00:37:35.190 align:middle line:90% happen at the same time. 00:37:35.190 --> 00:37:38.270 align:middle line:90% 00:37:38.270 --> 00:37:40.348 align:middle line:84% So this seems actually pretty good. 00:37:40.348 --> 00:37:41.640 align:middle line:90% You can do it at the same time. 00:37:41.640 --> 00:37:43.590 align:middle line:84% It's a little bit surprising, in a way. 00:37:43.590 --> 00:37:50.190 align:middle line:90% But the proof is quite trivial. 00:37:50.190 --> 00:37:52.470 align:middle line:90% But it's useful. 00:37:52.470 --> 00:37:58.890 align:middle line:84% So the proof is the following, that you just look-- 00:37:58.890 --> 00:38:00.830 align:middle line:84% so this is what we want to prove. 00:38:00.830 --> 00:38:05.350 align:middle line:90% So you just define-- 00:38:05.350 --> 00:38:07.580 align:middle line:90% and the exterior is unknown. 00:38:07.580 --> 00:38:11.800 align:middle line:84% So if I take this expression that I want to prove, 00:38:11.800 --> 00:38:17.900 align:middle line:84% if I move this thing here over on the other side, 00:38:17.900 --> 00:38:19.000 align:middle line:90% then I would get-- 00:38:19.000 --> 00:38:30.780 align:middle line:84% so in this case, I would define h of x to be f of x g b minus g 00:38:30.780 --> 00:38:41.780 align:middle line:90% a minus g of x f b minus f a. 00:38:41.780 --> 00:38:44.480 align:middle line:90% So I would define it like this. 00:38:44.480 --> 00:38:51.860 align:middle line:84% And you see that when you now take the derivative of h, 00:38:51.860 --> 00:38:53.240 align:middle line:90% you just get this here. 00:38:53.240 --> 00:39:02.780 align:middle line:90% 00:39:02.780 --> 00:39:07.180 align:middle line:84% This is because everything else is constant. 00:39:07.180 --> 00:39:09.780 align:middle line:90% So this is what you get. 00:39:09.780 --> 00:39:14.800 align:middle line:84% And you want to find a x0 where you want to find-- 00:39:14.800 --> 00:39:25.560 align:middle line:84% so want to find x0 where this thing here is 0. 00:39:25.560 --> 00:39:28.620 align:middle line:84% To find such a x0, it's the same as saying 00:39:28.620 --> 00:39:31.020 align:middle line:90% that you find an x0 over here. 00:39:31.020 --> 00:39:34.620 align:middle line:90% 00:39:34.620 --> 00:39:35.540 align:middle line:90% So now let's see. 00:39:35.540 --> 00:39:39.260 align:middle line:90% What is the value of h at a? 00:39:39.260 --> 00:39:43.740 align:middle line:84% If you plug in a here, then you get 00:39:43.740 --> 00:39:51.820 align:middle line:84% f of a, g of b minus g of a minus 00:39:51.820 --> 00:39:57.316 align:middle line:90% g of a f of b minus f of a. 00:39:57.316 --> 00:39:59.900 align:middle line:90% 00:39:59.900 --> 00:40:10.090 align:middle line:84% And so now you see here, I have this one here. 00:40:10.090 --> 00:40:14.650 align:middle line:90% I have f of a times g of a. 00:40:14.650 --> 00:40:16.890 align:middle line:90% That comes with a minus sign. 00:40:16.890 --> 00:40:21.590 align:middle line:84% I have the same term over here, but that comes with a plus. 00:40:21.590 --> 00:40:23.850 align:middle line:90% So those two cancel out. 00:40:23.850 --> 00:40:36.130 align:middle line:84% So I just get f of a g of b minus g of a f of b. 00:40:36.130 --> 00:40:39.530 align:middle line:90% So that's what I get there. 00:40:39.530 --> 00:40:45.290 align:middle line:90% And now I look at h of b. 00:40:45.290 --> 00:40:52.050 align:middle line:84% So if I plug in b up here for x, then 00:40:52.050 --> 00:41:05.600 align:middle line:84% I get f of b g of b minus g of a minus g of b f 00:41:05.600 --> 00:41:08.220 align:middle line:90% of b minus f of a. 00:41:08.220 --> 00:41:12.010 align:middle line:90% 00:41:12.010 --> 00:41:16.200 align:middle line:84% And now you see this when you multiply 00:41:16.200 --> 00:41:22.080 align:middle line:84% this by this, that term, f of g of b, you have this over here 00:41:22.080 --> 00:41:24.500 align:middle line:84% also, but it comes with a minus sign. 00:41:24.500 --> 00:41:26.440 align:middle line:90% So those cancel out. 00:41:26.440 --> 00:41:28.680 align:middle line:90% And so you're getting a term-- 00:41:28.680 --> 00:41:30.880 align:middle line:84% the only term that lives is this one times 00:41:30.880 --> 00:41:35.720 align:middle line:90% that, so minus g of a f of b. 00:41:35.720 --> 00:41:38.320 align:middle line:84% And then you get this term, minus minus, 00:41:38.320 --> 00:41:42.500 align:middle line:90% so plus g of b, f of a. 00:41:42.500 --> 00:41:48.800 align:middle line:90% 00:41:48.800 --> 00:41:49.340 align:middle line:90% Right. 00:41:49.340 --> 00:41:49.882 align:middle line:90% That's right. 00:41:49.882 --> 00:41:50.400 align:middle line:90% Sorry. 00:41:50.400 --> 00:41:54.160 align:middle line:84% And now you see that this here is the same as that. 00:41:54.160 --> 00:41:58.420 align:middle line:90% And these two here are the same. 00:41:58.420 --> 00:42:01.700 align:middle line:90% 00:42:01.700 --> 00:42:11.340 align:middle line:84% So we from this, we conclude that h of a is equal to h of b. 00:42:11.340 --> 00:42:15.560 align:middle line:84% So again, it's satisfying the requirements of Rolle's theorem. 00:42:15.560 --> 00:42:17.960 align:middle line:84% And so there is this intermediate point 00:42:17.960 --> 00:42:20.040 align:middle line:90% where the derivative is 0. 00:42:20.040 --> 00:42:24.360 align:middle line:84% But if the derivative of h is 0, that's 00:42:24.360 --> 00:42:29.140 align:middle line:84% exactly the point we wanted to find, where you have this. 00:42:29.140 --> 00:42:31.480 align:middle line:90% OK. 00:42:31.480 --> 00:42:35.840 align:middle line:84% And so, again, the Cauchy mean value theorem 00:42:35.840 --> 00:42:40.720 align:middle line:90% is sort of a stronger statement. 00:42:40.720 --> 00:42:43.840 align:middle line:84% It's a stronger statement because it allows 00:42:43.840 --> 00:42:45.840 align:middle line:90% you to have two functions. 00:42:45.840 --> 00:42:51.480 align:middle line:84% And then you know this point, where, to some extent, 00:42:51.480 --> 00:42:55.520 align:middle line:84% the mean value for both functions are satisfied. 00:42:55.520 --> 00:42:58.543 align:middle line:90% That's the idea. 00:42:58.543 --> 00:43:03.850 align:middle line:84% But it's an example of where, actually, the proof, it, 00:43:03.850 --> 00:43:07.670 align:middle line:84% in fact, worked more generally than what 00:43:07.670 --> 00:43:10.230 align:middle line:90% you used it for initially. 00:43:10.230 --> 00:43:13.790 align:middle line:84% In the initial, you just used it for a single function 00:43:13.790 --> 00:43:16.496 align:middle line:90% to prove the mean value theorem. 00:43:16.496 --> 00:43:16.996 align:middle line:90% OK. 00:43:16.996 --> 00:43:20.950 align:middle line:90% 00:43:20.950 --> 00:43:25.350 align:middle line:84% So now we want to prove-- so now we have a number of consequences 00:43:25.350 --> 00:43:27.630 align:middle line:90% of this. 00:43:27.630 --> 00:43:30.760 align:middle line:90% And the first one-- 00:43:30.760 --> 00:43:52.190 align:middle line:90% 00:43:52.190 --> 00:43:59.130 align:middle line:84% let me start with another consequence of Rolle's theorem. 00:43:59.130 --> 00:44:03.390 align:middle line:84% And then I come back to the Cauchy mean value theorem. 00:44:03.390 --> 00:44:05.890 align:middle line:90% So Taylor, Taylor expansion-- 00:44:05.890 --> 00:44:11.950 align:middle line:90% 00:44:11.950 --> 00:44:16.060 align:middle line:84% so the idea here is that you have a function, f. 00:44:16.060 --> 00:44:18.990 align:middle line:90% 00:44:18.990 --> 00:44:20.100 align:middle line:90% And you have a function. 00:44:20.100 --> 00:44:22.710 align:middle line:90% 00:44:22.710 --> 00:44:25.830 align:middle line:84% Let's say that it's a function on R-- 00:44:25.830 --> 00:44:29.950 align:middle line:84% doesn't need to be defined on all of R-- into R. 00:44:29.950 --> 00:44:34.870 align:middle line:90% And f, I'm assuming-- 00:44:34.870 --> 00:44:38.320 align:middle line:84% so let me do a special case of this, so special case first. 00:44:38.320 --> 00:44:46.950 align:middle line:90% 00:44:46.950 --> 00:44:50.030 align:middle line:84% So I assume I have a function like this 00:44:50.030 --> 00:44:52.701 align:middle line:90% and that it's differentiable. 00:44:52.701 --> 00:44:59.580 align:middle line:90% 00:44:59.580 --> 00:45:03.060 align:middle line:84% And then I'm assuming that the derivative-- so the derivative 00:45:03.060 --> 00:45:06.500 align:middle line:84% is now a function from R into R. And I'm 00:45:06.500 --> 00:45:10.643 align:middle line:84% assuming that the derivative is also differentiable. 00:45:10.643 --> 00:45:19.820 align:middle line:90% 00:45:19.820 --> 00:45:25.640 align:middle line:84% And the idea is now that I'm looking at the function. 00:45:25.640 --> 00:45:28.340 align:middle line:90% So the idea is like this. 00:45:28.340 --> 00:45:30.840 align:middle line:84% Again, in a minute, I'll prove a more general statement. 00:45:30.840 --> 00:45:35.540 align:middle line:84% But I just proved this one first to make the idea clear. 00:45:35.540 --> 00:45:51.700 align:middle line:84% So the idea is that you want to say that f locally can be 00:45:51.700 --> 00:46:01.560 align:middle line:90% approximated by a polynomial. 00:46:01.560 --> 00:46:05.500 align:middle line:90% 00:46:05.500 --> 00:46:07.980 align:middle line:84% So that's the idea of Taylor expansion, 00:46:07.980 --> 00:46:12.780 align:middle line:90% that if you take a function-- 00:46:12.780 --> 00:46:13.420 align:middle line:90% oops. 00:46:13.420 --> 00:46:16.820 align:middle line:84% If you take a function like this, some function, 00:46:16.820 --> 00:46:20.640 align:middle line:84% and you take a point, then in a neighborhood-- 00:46:20.640 --> 00:46:24.100 align:middle line:84% let's say that this is at some point here, x0. 00:46:24.100 --> 00:46:27.220 align:middle line:84% Then in the neighborhood of this point, 00:46:27.220 --> 00:46:30.020 align:middle line:84% instead of looking at the function, 00:46:30.020 --> 00:46:32.820 align:middle line:84% you get could get a very good idea 00:46:32.820 --> 00:46:36.460 align:middle line:84% about how the function behaves in this small neighborhood 00:46:36.460 --> 00:46:39.400 align:middle line:84% if you look at a certain polynomial. 00:46:39.400 --> 00:46:42.060 align:middle line:84% And the higher degree the polynomial, 00:46:42.060 --> 00:46:45.460 align:middle line:84% the better the approximation is to the function. 00:46:45.460 --> 00:46:51.220 align:middle line:84% And so we'll just do a simple version of this first, where 00:46:51.220 --> 00:46:54.330 align:middle line:90% you're looking at the-- 00:46:54.330 --> 00:46:57.730 align:middle line:84% so in this case, I would look at the-- 00:46:57.730 --> 00:47:05.250 align:middle line:90% 00:47:05.250 --> 00:47:09.330 align:middle line:90% so we have f here. 00:47:09.330 --> 00:47:14.170 align:middle line:90% And I'm interested in f. 00:47:14.170 --> 00:47:21.770 align:middle line:84% So we will be interested in f near x0. 00:47:21.770 --> 00:47:24.690 align:middle line:84% And then you're looking at what's called the Taylor 00:47:24.690 --> 00:47:36.410 align:middle line:90% polynomial at x0. 00:47:36.410 --> 00:47:40.850 align:middle line:84% And the Taylor polynomial, it depends on how many times 00:47:40.850 --> 00:47:43.290 align:middle line:84% you're willing to take the derivative. 00:47:43.290 --> 00:47:47.790 align:middle line:84% And so in this case, I just look at the first Taylor polynomial. 00:47:47.790 --> 00:47:50.410 align:middle line:90% 00:47:50.410 --> 00:47:54.480 align:middle line:84% And now it is a Taylor polynomial at x0. 00:47:54.480 --> 00:47:57.920 align:middle line:84% So this thing here is, by definition, 00:47:57.920 --> 00:48:02.840 align:middle line:90% the value of f at x0 plus-- 00:48:02.840 --> 00:48:11.640 align:middle line:84% and then the derivative of f at x0 minus x0. 00:48:11.640 --> 00:48:15.800 align:middle line:84% So this here is the first-order Taylor expansion of-- 00:48:15.800 --> 00:48:30.600 align:middle line:84% so this here is often called the first-order Taylor expansion 00:48:30.600 --> 00:48:31.540 align:middle line:90% at x0. 00:48:31.540 --> 00:48:37.900 align:middle line:90% 00:48:37.900 --> 00:48:41.000 align:middle line:84% Now, what you're interested in is, 00:48:41.000 --> 00:48:45.740 align:middle line:84% you're interested in showing that this polynomial here-- 00:48:45.740 --> 00:48:56.120 align:middle line:84% so, again, the idea is that f of x 00:48:56.120 --> 00:49:09.760 align:middle line:84% should be almost equal to P1 of x if x is almost equal to x0. 00:49:09.760 --> 00:49:14.680 align:middle line:84% So near x0, the function is very well 00:49:14.680 --> 00:49:17.280 align:middle line:84% approximated by, in this case, this linear function. 00:49:17.280 --> 00:49:21.640 align:middle line:90% 00:49:21.640 --> 00:49:29.640 align:middle line:84% So now the key here is, how do you 00:49:29.640 --> 00:49:32.640 align:middle line:84% estimate how well it approximates? 00:49:32.640 --> 00:49:39.856 align:middle line:90% 00:49:39.856 --> 00:49:45.920 align:middle line:84% So the statement is then that-- so you have here x0. 00:49:45.920 --> 00:49:48.520 align:middle line:90% You have here x. 00:49:48.520 --> 00:50:02.510 align:middle line:84% And the claim is now that there exist some c between x0 and x, 00:50:02.510 --> 00:50:04.270 align:middle line:90% so strictly between-- 00:50:04.270 --> 00:50:09.270 align:middle line:84% so between, which means that it's strictly between. 00:50:09.270 --> 00:50:10.390 align:middle line:90% Let me just write that. 00:50:10.390 --> 00:50:15.890 align:middle line:90% 00:50:15.890 --> 00:50:19.750 align:middle line:84% x could also lie on the other side. 00:50:19.750 --> 00:50:23.670 align:middle line:84% It's so obvious what I mean, strictly between also. 00:50:23.670 --> 00:50:33.230 align:middle line:84% So there exists some c in between these two so that f of x 00:50:33.230 --> 00:50:41.070 align:middle line:84% is equal to this Taylor polynomial plus and then 00:50:41.070 --> 00:50:43.310 align:middle line:90% the second derivative. 00:50:43.310 --> 00:50:49.110 align:middle line:84% So this is the derivative of the derivative. 00:50:49.110 --> 00:50:54.030 align:middle line:84% So f2 here is the derivative, and then you 00:50:54.030 --> 00:50:57.790 align:middle line:90% take the derivative like that. 00:50:57.790 --> 00:51:02.870 align:middle line:84% And that is evaluated at c and then divided by 2. 00:51:02.870 --> 00:51:08.160 align:middle line:90% And then it's x x0 squared. 00:51:08.160 --> 00:51:12.118 align:middle line:90% 00:51:12.118 --> 00:51:15.010 align:middle line:84% And then in a minute, I will just do the general case of it. 00:51:15.010 --> 00:51:27.930 align:middle line:90% 00:51:27.930 --> 00:51:28.430 align:middle line:90% OK. 00:51:28.430 --> 00:51:35.710 align:middle line:90% 00:51:35.710 --> 00:51:40.780 align:middle line:84% And so now, how do you prove this? 00:51:40.780 --> 00:51:50.020 align:middle line:90% 00:51:50.020 --> 00:51:54.760 align:middle line:90% So what you do is-- 00:51:54.760 --> 00:51:56.280 align:middle line:90% so let's try to prove this. 00:51:56.280 --> 00:52:05.164 align:middle line:90% 00:52:05.164 --> 00:52:17.260 align:middle line:84% So I define a M. We define a M. M is just some real number. 00:52:17.260 --> 00:52:22.330 align:middle line:84% And it's defined so that f of x-- 00:52:22.330 --> 00:52:26.940 align:middle line:90% 00:52:26.940 --> 00:52:28.420 align:middle line:90% I wonder if I should-- 00:52:28.420 --> 00:52:37.703 align:middle line:84% in my notes here, I call it, just 00:52:37.703 --> 00:52:39.120 align:middle line:84% because I thought it was clearer-- 00:52:39.120 --> 00:52:41.300 align:middle line:84% and maybe I will do this here also. 00:52:41.300 --> 00:52:45.460 align:middle line:84% I'll call this point here a and this here b. 00:52:45.460 --> 00:52:49.380 align:middle line:84% And then c here is somewhere in between. 00:52:49.380 --> 00:52:55.460 align:middle line:84% So this means that this thing here is b, and this is b. 00:52:55.460 --> 00:52:58.740 align:middle line:90% And this here is then b. 00:52:58.740 --> 00:53:01.920 align:middle line:90% And this here is a. 00:53:01.920 --> 00:53:05.620 align:middle line:84% It's just to be clear that somehow, you're 00:53:05.620 --> 00:53:07.300 align:middle line:90% kind of fixing x. 00:53:07.300 --> 00:53:11.980 align:middle line:90% 00:53:11.980 --> 00:53:13.860 align:middle line:90% So to prove it-- 00:53:13.860 --> 00:53:20.580 align:middle line:90% so we look at this thing here. 00:53:20.580 --> 00:53:30.780 align:middle line:90% And you look at thing here. 00:53:30.780 --> 00:53:36.140 align:middle line:90% And we subtract P1 of b. 00:53:36.140 --> 00:53:41.940 align:middle line:84% And then we define M so that you have 00:53:41.940 --> 00:53:48.670 align:middle line:84% that this thing here is b minus a squared. 00:53:48.670 --> 00:53:54.610 align:middle line:90% So define M by this thing here. 00:53:54.610 --> 00:53:58.530 align:middle line:84% This here just gives you some number. 00:53:58.530 --> 00:53:59.710 align:middle line:90% a and b are different. 00:53:59.710 --> 00:54:00.820 align:middle line:90% So this here is nonzero. 00:54:00.820 --> 00:54:03.490 align:middle line:90% 00:54:03.490 --> 00:54:05.370 align:middle line:90% M is just given by-- 00:54:05.370 --> 00:54:10.910 align:middle line:84% that is this difference here divided by this here times 2. 00:54:10.910 --> 00:54:13.675 align:middle line:90% 00:54:13.675 --> 00:54:21.360 align:middle line:84% So that's how we defined M. But now we need to show-- 00:54:21.360 --> 00:54:24.370 align:middle line:90% 00:54:24.370 --> 00:54:26.910 align:middle line:84% if you do it like this, it's like M is OK. 00:54:26.910 --> 00:54:31.050 align:middle line:84% It's just, by definition, you have-- 00:54:31.050 --> 00:54:34.250 align:middle line:90% 00:54:34.250 --> 00:54:45.240 align:middle line:84% so by definition, we have that f of b 00:54:45.240 --> 00:54:54.090 align:middle line:84% again is equal to P1 of b plus M 2 b minus a squared. 00:54:54.090 --> 00:54:57.260 align:middle line:90% 00:54:57.260 --> 00:54:58.330 align:middle line:90% But we need to prove-- 00:54:58.330 --> 00:55:04.440 align:middle line:90% 00:55:04.440 --> 00:55:10.960 align:middle line:84% we need to show that M here is the second derivative 00:55:10.960 --> 00:55:14.920 align:middle line:84% of the original function at some intermediate point. 00:55:14.920 --> 00:55:16.200 align:middle line:90% That's what we need to prove. 00:55:16.200 --> 00:55:20.600 align:middle line:90% 00:55:20.600 --> 00:55:22.532 align:middle line:90% And so you're looking at-- 00:55:22.532 --> 00:55:36.700 align:middle line:90% 00:55:36.700 --> 00:55:39.200 align:middle line:84% so now we're looking at the function 00:55:39.200 --> 00:55:49.760 align:middle line:84% h of x equal to f of x minus P1 of x 00:55:49.760 --> 00:55:55.800 align:middle line:90% minus M/2 x minus a squared. 00:55:55.800 --> 00:55:58.880 align:middle line:84% So if we're looking at this function-- 00:55:58.880 --> 00:56:03.840 align:middle line:84% well, if we're looking at h of-- let me do b first. 00:56:03.840 --> 00:56:16.680 align:middle line:84% So if I do b, then this here is f of b minus P1 of b minus M/2 00:56:16.680 --> 00:56:19.040 align:middle line:90% b minus a squared. 00:56:19.040 --> 00:56:25.080 align:middle line:84% But M was defined so that this thing here is equal to that. 00:56:25.080 --> 00:56:30.740 align:middle line:84% So by definition-- this is by definition of M-- 00:56:30.740 --> 00:56:36.480 align:middle line:90% 00:56:36.480 --> 00:56:38.320 align:middle line:84% I have that this thing here is 0. 00:56:38.320 --> 00:56:42.150 align:middle line:90% 00:56:42.150 --> 00:56:46.890 align:middle line:90% If I'm now looking at h of a-- 00:56:46.890 --> 00:56:52.270 align:middle line:90% 00:56:52.270 --> 00:56:58.270 align:middle line:84% well, this thing here-- here, you get f of a here, 00:56:58.270 --> 00:57:00.896 align:middle line:90% you get minus P1 of a. 00:57:00.896 --> 00:57:04.770 align:middle line:84% If you plug in a, here then this is 0. 00:57:04.770 --> 00:57:07.390 align:middle line:90% So this is what you get. 00:57:07.390 --> 00:57:10.590 align:middle line:90% But remember how P1 was given. 00:57:10.590 --> 00:57:12.690 align:middle line:90% P1 was here. 00:57:12.690 --> 00:57:14.570 align:middle line:90% And we now do it at a. 00:57:14.570 --> 00:57:16.170 align:middle line:90% So this here is really a. 00:57:16.170 --> 00:57:19.510 align:middle line:90% 00:57:19.510 --> 00:57:24.390 align:middle line:84% So you see that when you plug in x here, this thing here is 0. 00:57:24.390 --> 00:57:27.390 align:middle line:90% So you just get f of a here. 00:57:27.390 --> 00:57:34.150 align:middle line:84% So this thing here is f of a minus f of a. 00:57:34.150 --> 00:57:35.110 align:middle line:90% So that's 0. 00:57:35.110 --> 00:57:38.550 align:middle line:90% 00:57:38.550 --> 00:57:41.830 align:middle line:84% So you see that if you're looking at this function 00:57:41.830 --> 00:57:45.750 align:middle line:90% here, then at a, it's 0. 00:57:45.750 --> 00:57:47.750 align:middle line:90% At b, it's 0. 00:57:47.750 --> 00:58:15.430 align:middle line:84% So by Rolle's theorem, there exist a c1 between a and b such 00:58:15.430 --> 00:58:22.350 align:middle line:90% that h prime of c is equal to 0. 00:58:22.350 --> 00:58:28.730 align:middle line:90% 00:58:28.730 --> 00:58:31.190 align:middle line:90% Sorry, this is the c1. 00:58:31.190 --> 00:58:33.650 align:middle line:84% This is not the c we're looking for. 00:58:33.650 --> 00:58:35.270 align:middle line:84% The C we're looking for is something 00:58:35.270 --> 00:58:38.220 align:middle line:84% where the second derivative is something. 00:58:38.220 --> 00:58:41.260 align:middle line:84% But now you see that now you have this function, h. 00:58:41.260 --> 00:58:44.684 align:middle line:90% And here, you have a. 00:58:44.684 --> 00:58:48.060 align:middle line:90% Here, you have c1. 00:58:48.060 --> 00:58:51.700 align:middle line:84% And then b is further away, like this. 00:58:51.700 --> 00:58:58.806 align:middle line:84% And you have that in this case, the derivative is 0 at a. 00:58:58.806 --> 00:59:05.720 align:middle line:84% And the derivative-- sorry, not at a. 00:59:05.720 --> 00:59:12.620 align:middle line:84% But the derivative at this point here, h prime, is equal to 0. 00:59:12.620 --> 00:59:17.820 align:middle line:84% Let's calculate what the derivative of h is at a. 00:59:17.820 --> 00:59:20.740 align:middle line:84% I want to apply Rolle's theorem a second time. 00:59:20.740 --> 00:59:25.100 align:middle line:84% But in order to apply Rolle's theorem a second time, 00:59:25.100 --> 00:59:30.860 align:middle line:84% I really need the derivative of h at a to be 0. 00:59:30.860 --> 00:59:32.840 align:middle line:84% But let's see what it's equal to. 00:59:32.840 --> 00:59:36.340 align:middle line:90% 00:59:36.340 --> 00:59:42.300 align:middle line:84% And so, again, we have h up there at the top. 00:59:42.300 --> 00:59:50.620 align:middle line:84% And so this means that h prime is equal to f prime minus-- 00:59:50.620 --> 00:59:55.540 align:middle line:84% and then the Taylor polynomial here was the Taylor polynomial. 00:59:55.540 --> 00:59:58.260 align:middle line:84% When you take the derivative, this is just a constant. 00:59:58.260 --> 01:00:03.160 align:middle line:90% You're getting f prime a. 01:00:03.160 --> 01:00:05.860 align:middle line:84% This is all it gives when you take the derivative of this 01:00:05.860 --> 01:00:08.140 align:middle line:90% here. 01:00:08.140 --> 01:00:11.420 align:middle line:90% And then you're getting minus. 01:00:11.420 --> 01:00:15.080 align:middle line:90% And you're getting M x minus a. 01:00:15.080 --> 01:00:18.406 align:middle line:84% That's when you take the derivative of the last term. 01:00:18.406 --> 01:00:20.220 align:middle line:90% So this is what you get. 01:00:20.220 --> 01:00:21.760 align:middle line:90% And this here is at x. 01:00:21.760 --> 01:00:25.540 align:middle line:90% 01:00:25.540 --> 01:00:35.210 align:middle line:84% But so now h prime at a, this is f prime at a minus f prime at a. 01:00:35.210 --> 01:00:37.530 align:middle line:90% This thing here is actually 0. 01:00:37.530 --> 01:00:40.670 align:middle line:84% So you see that this whole thing is 0. 01:00:40.670 --> 01:00:47.530 align:middle line:84% So what we have is that we have that h prime at c1 is 0. 01:00:47.530 --> 01:00:53.650 align:middle line:84% And this is also equal to the derivative of a. 01:00:53.650 --> 01:00:57.510 align:middle line:84% So we can apply Rolle's theorem a second time. 01:00:57.510 --> 01:01:21.250 align:middle line:84% So by applying Rolle's theorem a second time, we have a c2. 01:01:21.250 --> 01:01:26.890 align:middle line:84% And now this c2, that's really going to be our c in this case. 01:01:26.890 --> 01:01:36.430 align:middle line:84% So we have a second c, where the derivative of the derivative. 01:01:36.430 --> 01:01:39.770 align:middle line:90% S-- The second derivative of h-- 01:01:39.770 --> 01:01:42.620 align:middle line:90% at that point is equal to 0. 01:01:42.620 --> 01:01:45.394 align:middle line:90% 01:01:45.394 --> 01:01:48.210 align:middle line:84% And this one here, remember, this here 01:01:48.210 --> 01:01:51.330 align:middle line:84% is-- when you apply Rolle's theorem, you started with a. 01:01:51.330 --> 01:01:53.010 align:middle line:90% You had here b. 01:01:53.010 --> 01:01:55.930 align:middle line:84% Then you got c1 that was between these two. 01:01:55.930 --> 01:01:57.330 align:middle line:90% That was c1. 01:01:57.330 --> 01:02:02.010 align:middle line:84% But now you're looking at h prime in this interval. 01:02:02.010 --> 01:02:06.370 align:middle line:84% And so the second one is lying here. 01:02:06.370 --> 01:02:09.890 align:middle line:84% So it's certainly between a and b, 01:02:09.890 --> 01:02:13.650 align:middle line:84% but it's actually even between a and c1. 01:02:13.650 --> 01:02:15.650 align:middle line:90% Now, what does this here mean? 01:02:15.650 --> 01:02:20.770 align:middle line:84% So we hope that this is the point we're looking for. 01:02:20.770 --> 01:02:23.910 align:middle line:84% And so to calculate the second derivative-- 01:02:23.910 --> 01:02:35.560 align:middle line:90% 01:02:35.560 --> 01:02:41.080 align:middle line:84% so this here was the derivative of h. 01:02:41.080 --> 01:02:45.200 align:middle line:84% So we just need to take another derivative of h. 01:02:45.200 --> 01:02:50.120 align:middle line:84% So if we do this, then we get that-- 01:02:50.120 --> 01:02:58.520 align:middle line:84% so h double prime of x at some x is equal to f double prime at x. 01:02:58.520 --> 01:03:00.620 align:middle line:84% And then you see this thing here. 01:03:00.620 --> 01:03:02.600 align:middle line:90% This is just a constant. 01:03:02.600 --> 01:03:04.140 align:middle line:90% This is a linear function. 01:03:04.140 --> 01:03:06.240 align:middle line:84% So the derivative of this one here is 1. 01:03:06.240 --> 01:03:10.380 align:middle line:90% So you're getting this minus M. 01:03:10.380 --> 01:03:13.720 align:middle line:84% And now we know that h double prime 01:03:13.720 --> 01:03:16.640 align:middle line:84% at this intermediate point, c2, which we also 01:03:16.640 --> 01:03:20.640 align:middle line:90% call c, that's equal to 0. 01:03:20.640 --> 01:03:25.000 align:middle line:84% So this means that f double prime at the c minus M 01:03:25.000 --> 01:03:26.240 align:middle line:90% is equal to 0. 01:03:26.240 --> 01:03:31.962 align:middle line:84% So this means that M really was the second derivative 01:03:31.962 --> 01:03:35.150 align:middle line:90% of f at this point. 01:03:35.150 --> 01:03:37.330 align:middle line:90% So this proof's sort of clever. 01:03:37.330 --> 01:03:43.310 align:middle line:90% 01:03:43.310 --> 01:03:46.510 align:middle line:90% The proof seems very clever. 01:03:46.510 --> 01:03:48.410 align:middle line:90% But if you think about it-- 01:03:48.410 --> 01:03:51.870 align:middle line:90% 01:03:51.870 --> 01:03:54.470 align:middle line:84% so it's applying Rolle's theorem twice. 01:03:54.470 --> 01:03:58.030 align:middle line:84% But it's, in a way, the only thing it could be. 01:03:58.030 --> 01:04:02.310 align:middle line:90% But anyway, OK. 01:04:02.310 --> 01:04:04.310 align:middle line:90% So that was Taylor. 01:04:04.310 --> 01:04:07.450 align:middle line:84% This was just a special case of Taylor expansion. 01:04:07.450 --> 01:04:09.380 align:middle line:90% Let me state the general case. 01:04:09.380 --> 01:04:13.750 align:middle line:90% 01:04:13.750 --> 01:04:17.150 align:middle line:84% So for proving the special case, we 01:04:17.150 --> 01:04:19.310 align:middle line:90% applied Rolle's theorem twice. 01:04:19.310 --> 01:04:23.910 align:middle line:84% But you can actually apply Rolle's theorem k times. 01:04:23.910 --> 01:04:27.230 align:middle line:84% And so if you apply k times, then you're 01:04:27.230 --> 01:04:30.750 align:middle line:90% getting the k Taylor expansion. 01:04:30.750 --> 01:04:37.310 align:middle line:84% So this is the general case of the Taylor expansion, so general 01:04:37.310 --> 01:04:46.490 align:middle line:90% case of Taylor expansion. 01:04:46.490 --> 01:04:52.990 align:middle line:90% 01:04:52.990 --> 01:04:57.430 align:middle line:84% So the general case is that you take a function on R 01:04:57.430 --> 01:05:10.390 align:middle line:84% into R. You assume that f is differentiable 01:05:10.390 --> 01:05:14.750 align:middle line:84% but also that f prime is differentiable 01:05:14.750 --> 01:05:20.270 align:middle line:84% and all the way up to, say, the k-- 01:05:20.270 --> 01:05:20.770 align:middle line:90% right. 01:05:20.770 --> 01:05:25.110 align:middle line:84% You assume that it's differentiable, 01:05:25.110 --> 01:05:30.260 align:middle line:84% and all the way up to the k-minus-1 derivative 01:05:30.260 --> 01:05:31.920 align:middle line:90% is also differentiable. 01:05:31.920 --> 01:05:33.590 align:middle line:90% So this here exists. 01:05:33.590 --> 01:05:36.420 align:middle line:90% 01:05:36.420 --> 01:05:37.880 align:middle line:90% So all of these exist. 01:05:37.880 --> 01:05:40.900 align:middle line:90% All these derivative exist. 01:05:40.900 --> 01:05:45.340 align:middle line:84% And then the claim is that-- so now the Taylor polynomial, 01:05:45.340 --> 01:05:48.520 align:middle line:90% the k Taylor polynomial-- 01:05:48.520 --> 01:05:58.380 align:middle line:90% 01:05:58.380 --> 01:06:19.300 align:middle line:84% or sometimes called the Taylor polynomial of degree k minus 1 01:06:19.300 --> 01:06:21.090 align:middle line:90% is that you're looking at-- 01:06:21.090 --> 01:06:24.910 align:middle line:90% 01:06:24.910 --> 01:06:25.540 align:middle line:90% oh, sorry. 01:06:25.540 --> 01:06:28.160 align:middle line:84% This here should be the k minus 1 Taylor polynomial. 01:06:28.160 --> 01:06:33.740 align:middle line:90% 01:06:33.740 --> 01:06:36.480 align:middle line:90% So this is where you write-- 01:06:36.480 --> 01:06:46.260 align:middle line:84% so you take-- again, you have here a, and you have b. 01:06:46.260 --> 01:06:50.700 align:middle line:84% Let me write it with a and b instead of with x, because I 01:06:50.700 --> 01:06:52.040 align:middle line:90% think it's a little clearer. 01:06:52.040 --> 01:06:55.380 align:middle line:84% So the thing is that the polynomial is where 01:06:55.380 --> 01:06:59.820 align:middle line:90% you say you expand around a. 01:06:59.820 --> 01:07:07.020 align:middle line:84% And so the polynomial, this is the k-minus-1 polynomial. 01:07:07.020 --> 01:07:09.620 align:middle line:84% For that, we write it in terms of x. 01:07:09.620 --> 01:07:15.780 align:middle line:84% And this is the polynomial where the constant term 01:07:15.780 --> 01:07:17.900 align:middle line:90% is the value of f at a. 01:07:17.900 --> 01:07:22.720 align:middle line:90% 01:07:22.720 --> 01:07:28.010 align:middle line:84% The coefficient to the first-order term is f prime at a 01:07:28.010 --> 01:07:31.330 align:middle line:90% and then x minus a. 01:07:31.330 --> 01:07:35.790 align:middle line:84% The coefficient to the second is the second derivative at a. 01:07:35.790 --> 01:07:41.210 align:middle line:84% And now it's divided by 2 x minus a squared. 01:07:41.210 --> 01:07:46.730 align:middle line:84% And then this next one is f triple prime. 01:07:46.730 --> 01:07:52.130 align:middle line:84% And now it's really 3 times 2 times 1 times 01:07:52.130 --> 01:07:55.970 align:middle line:90% x minus a to the power 3. 01:07:55.970 --> 01:08:03.650 align:middle line:84% And then you continue like this up to the k minus 1 derivative 01:08:03.650 --> 01:08:13.170 align:middle line:84% and then this k minus 1 factorial, x minus a, k minus 1. 01:08:13.170 --> 01:08:17.630 align:middle line:84% So this is the k minus 1 Taylor polynomial. 01:08:17.630 --> 01:08:20.670 align:middle line:84% And so you observe that you can just write it-- 01:08:20.670 --> 01:08:23.529 align:middle line:84% so it's typically written like this, 01:08:23.529 --> 01:08:36.490 align:middle line:84% that if the i derivative in a divided by i factorial x minus a 01:08:36.490 --> 01:08:47.810 align:middle line:84% to the power i-- and i is now summing from 0 to k minus 1. 01:08:47.810 --> 01:08:50.050 align:middle line:84% This is by definition equal to that. 01:08:50.050 --> 01:08:53.649 align:middle line:90% 01:08:53.649 --> 01:08:59.050 align:middle line:84% So that's the Taylor polynomial, the k minus 1 Taylor polynomial. 01:08:59.050 --> 01:09:07.050 align:middle line:84% And now the general case of the Taylor expansion is then-- 01:09:07.050 --> 01:09:11.130 align:middle line:84% so the general case of the Taylor expansion is then. 01:09:11.130 --> 01:09:20.850 align:middle line:90% 01:09:20.850 --> 01:09:26.380 align:middle line:84% So this is like the theorem, Taylor expansion. 01:09:26.380 --> 01:09:33.880 align:middle line:90% 01:09:33.880 --> 01:09:38.000 align:middle line:84% So you have again this function on R into R. 01:09:38.000 --> 01:09:43.120 align:middle line:84% And you have that all of the derivatives up to 01:09:43.120 --> 01:09:47.520 align:middle line:84% and including the k derivative, they all exist. 01:09:47.520 --> 01:09:57.280 align:middle line:84% And then you have that f of b is equal to the k minus 1 Taylor 01:09:57.280 --> 01:10:00.880 align:middle line:90% polynomial at b plus-- 01:10:00.880 --> 01:10:06.220 align:middle line:84% and then this term is usually called the remainder term. 01:10:06.220 --> 01:10:08.440 align:middle line:84% And the key point here is that it's 01:10:08.440 --> 01:10:12.480 align:middle line:84% easy to estimate the remainder term because there's 01:10:12.480 --> 01:10:14.240 align:middle line:90% an expression for it. 01:10:14.240 --> 01:10:19.400 align:middle line:84% This is the k derivative of f in this point that 01:10:19.400 --> 01:10:28.360 align:middle line:84% is an intermediate point divided by k factorial 01:10:28.360 --> 01:10:39.240 align:middle line:84% b minus a to the power k and where c here 01:10:39.240 --> 01:10:42.236 align:middle line:90% is lying between a and b. 01:10:42.236 --> 01:10:47.360 align:middle line:90% 01:10:47.360 --> 01:10:54.600 align:middle line:84% And again, the idea is that you have a function, any function. 01:10:54.600 --> 01:10:59.080 align:middle line:84% It needs to be differentiable k times. 01:10:59.080 --> 01:11:04.240 align:middle line:84% But then, actually, it's very close to a polynomial 01:11:04.240 --> 01:11:06.520 align:middle line:90% of degree k minus 1. 01:11:06.520 --> 01:11:09.160 align:middle line:84% And you can estimate it, how close it 01:11:09.160 --> 01:11:13.720 align:middle line:90% is, in terms of this term here. 01:11:13.720 --> 01:11:16.380 align:middle line:90% And so the proof of this-- 01:11:16.380 --> 01:11:19.550 align:middle line:84% so the proof is really just what we did, but let's 01:11:19.550 --> 01:11:21.030 align:middle line:90% just go over it again. 01:11:21.030 --> 01:11:25.470 align:middle line:84% We only did the first, where we did it in the case 01:11:25.470 --> 01:11:27.670 align:middle line:90% where k was equal to 2. 01:11:27.670 --> 01:11:30.600 align:middle line:84% And now let me try to do this in general. 01:11:30.600 --> 01:11:38.350 align:middle line:90% 01:11:38.350 --> 01:11:45.290 align:middle line:84% So you're looking at f of b minus-- 01:11:45.290 --> 01:11:49.490 align:middle line:84% we have this Taylor polynomial, k minus 1 Taylor polynomial. 01:11:49.490 --> 01:11:53.750 align:middle line:90% 01:11:53.750 --> 01:12:05.749 align:middle line:84% And you define M by the equation, 01:12:05.749 --> 01:12:14.910 align:middle line:84% that this here is m over k factorial times b minus a 01:12:14.910 --> 01:12:17.420 align:middle line:90% to the power k. 01:12:17.420 --> 01:12:24.020 align:middle line:84% So this here is how M is defined. 01:12:24.020 --> 01:12:28.580 align:middle line:84% b is not equal to M. So this is just some positive. 01:12:28.580 --> 01:12:31.380 align:middle line:84% It's actually a positive constant. 01:12:31.380 --> 01:12:33.680 align:middle line:90% And so you can just write-- 01:12:33.680 --> 01:12:37.260 align:middle line:90% 01:12:37.260 --> 01:12:41.900 align:middle line:84% so M here is given as the difference of those two. 01:12:41.900 --> 01:12:45.700 align:middle line:84% And then divide it by this and multiply by k factorial. 01:12:45.700 --> 01:12:50.220 align:middle line:90% So this defined M. 01:12:50.220 --> 01:12:56.540 align:middle line:84% And now what you do is, you then define h as before or similarly 01:12:56.540 --> 01:12:58.980 align:middle line:90% to before. 01:12:58.980 --> 01:13:06.580 align:middle line:84% h is then going to be f of x minus P k minus 1 01:13:06.580 --> 01:13:16.564 align:middle line:84% of x minus M over k factorial x minus a to the power k. 01:13:16.564 --> 01:13:21.700 align:middle line:90% That's how you define h. 01:13:21.700 --> 01:13:32.140 align:middle line:84% And by definition of M, by definition of M, 01:13:32.140 --> 01:13:39.740 align:middle line:84% you have that h of b is equal to 0. 01:13:39.740 --> 01:13:46.400 align:middle line:84% This is another way of saying how we define M. 01:13:46.400 --> 01:13:50.980 align:middle line:90% So now h of a-- 01:13:50.980 --> 01:14:04.660 align:middle line:84% well, h of a is just f of a minus p k minus 1 of a minus M 01:14:04.660 --> 01:14:09.440 align:middle line:84% over k factorial x minus a to the power k. 01:14:09.440 --> 01:14:12.140 align:middle line:90% 01:14:12.140 --> 01:14:15.430 align:middle line:84% So this one here is clearly 0. a minus a 0. 01:14:15.430 --> 01:14:18.450 align:middle line:90% So that's clearly 0. 01:14:18.450 --> 01:14:21.890 align:middle line:84% And this thing here, that's this polynomial. 01:14:21.890 --> 01:14:23.730 align:middle line:90% I wrote it up here. 01:14:23.730 --> 01:14:28.010 align:middle line:84% So if you plug in here a, you see this term here 01:14:28.010 --> 01:14:30.110 align:middle line:84% and all the other terms drops out. 01:14:30.110 --> 01:14:33.290 align:middle line:90% It's only this one that lives. 01:14:33.290 --> 01:14:37.990 align:middle line:84% So this thing here, this is the same as f of a. 01:14:37.990 --> 01:14:40.890 align:middle line:90% 01:14:40.890 --> 01:14:45.770 align:middle line:84% And so you have that the whole thing is actually 0. 01:14:45.770 --> 01:14:48.530 align:middle line:84% So because we chose M this way, we 01:14:48.530 --> 01:14:51.630 align:middle line:84% see that now Rolle's theorem can get started. 01:14:51.630 --> 01:14:54.850 align:middle line:84% And Rolle's theorem can get started. 01:14:54.850 --> 01:14:59.510 align:middle line:84% And this means that there exists a c1 between a 01:14:59.510 --> 01:15:03.010 align:middle line:84% and b where the derivative of h is 0. 01:15:03.010 --> 01:15:06.850 align:middle line:90% Now you check that-- 01:15:06.850 --> 01:15:10.850 align:middle line:84% if you're looking at this function again 01:15:10.850 --> 01:15:17.610 align:middle line:84% and take the derivative at a, the derivative at a is also 0. 01:15:17.610 --> 01:15:23.850 align:middle line:84% In fact, the whole idea is that all these derivatives 01:15:23.850 --> 01:15:29.110 align:middle line:84% up to the k minus 1 derivative of-- 01:15:29.110 --> 01:15:31.490 align:middle line:84% if you're looking at the k minus 1 all 01:15:31.490 --> 01:15:34.610 align:middle line:84% the way up to the k minus 1 derivative of h, 01:15:34.610 --> 01:15:38.050 align:middle line:90% they are all 0 at a. 01:15:38.050 --> 01:15:41.330 align:middle line:84% So you can repeat the Rolle's theorem 01:15:41.330 --> 01:15:46.810 align:middle line:84% So first here, this thing here together with this allows 01:15:46.810 --> 01:15:53.210 align:middle line:84% you to find this c1 strictly between a and b, 01:15:53.210 --> 01:15:58.290 align:middle line:90% where the derivative is 0. 01:15:58.290 --> 01:16:03.210 align:middle line:84% Then you observe that the derivative of h at a is 0. 01:16:03.210 --> 01:16:06.650 align:middle line:84% This then allows you to find a point 01:16:06.650 --> 01:16:13.960 align:middle line:84% between a and a second point, c1, which was less than b, 01:16:13.960 --> 01:16:20.360 align:middle line:84% where the second derivative is 0. 01:16:20.360 --> 01:16:25.880 align:middle line:84% Now you use that this second derivative of h at a is also 0. 01:16:25.880 --> 01:16:29.780 align:middle line:84% So you apply Rolle's theorem one more time. 01:16:29.780 --> 01:16:33.240 align:middle line:84% Each time you apply Rolle's theorem, the c's is 01:16:33.240 --> 01:16:36.360 align:middle line:90% moving to the left towards a. 01:16:36.360 --> 01:16:39.320 align:middle line:90% And so you apply it. 01:16:39.320 --> 01:16:42.360 align:middle line:84% You apply Rolle's theorem k times. 01:16:42.360 --> 01:16:46.120 align:middle line:84% And once you do this, you get eventually 01:16:46.120 --> 01:16:49.880 align:middle line:84% that this m here is the derivative 01:16:49.880 --> 01:16:53.520 align:middle line:84% of the original function at this point. 01:16:53.520 --> 01:16:57.400 align:middle line:90% So I wrote this out. 01:16:57.400 --> 01:17:01.260 align:middle line:84% Now, I did the special case, and I try to explain this here. 01:17:01.260 --> 01:17:03.580 align:middle line:90% I wrote it out in the notes. 01:17:03.580 --> 01:17:06.600 align:middle line:84% It's written out in the general case only. 01:17:06.600 --> 01:17:10.400 align:middle line:84% But on the other hand, it's written out completely there. 01:17:10.400 --> 01:17:11.380 align:middle line:90% Any questions? 01:17:11.380 --> 01:17:14.000 align:middle line:90% 01:17:14.000 --> 01:17:15.840 align:middle line:84% There's two things we didn't get to. 01:17:15.840 --> 01:17:18.040 align:middle line:84% There's these two rules about how 01:17:18.040 --> 01:17:22.740 align:middle line:84% to evaluate fractions that uses the Cauchy mean value theorem. 01:17:22.740 --> 01:17:23.860 align:middle line:90% And I'll start with those. 01:17:23.860 --> 01:17:27.620 align:middle line:84% Those are in the notes, but I will start with that next time. 01:17:27.620 --> 01:17:28.120 align:middle line:90% Yeah? 01:17:28.120 --> 01:17:31.000 align:middle line:84% STUDENT: So M is like [INAUDIBLE] derivative? 01:17:31.000 --> 01:17:33.900 align:middle line:84% TOBIAS COLDING: You do up to k derivative. 01:17:33.900 --> 01:17:36.360 align:middle line:84% It's when you do the k derivative, 01:17:36.360 --> 01:17:41.360 align:middle line:84% then that's when you get the c that you're actually 01:17:41.360 --> 01:17:42.500 align:middle line:90% looking for. 01:17:42.500 --> 01:17:47.160 align:middle line:84% And so in the special case I looked at, k was equal to 2. 01:17:47.160 --> 01:17:50.620 align:middle line:84% And the Taylor expansion was a linear-- 01:17:50.620 --> 01:17:55.100 align:middle line:84% so it was a polynomial of degree 1. 01:17:55.100 --> 01:17:58.520 align:middle line:90% So k minus 1 was 1. 01:17:58.520 --> 01:18:04.720 align:middle line:84% And so in the special case, we applied Rolle's theorem twice. 01:18:04.720 --> 01:18:07.710 align:middle line:84% And in this case, we do it k time. 01:18:07.710 --> 01:18:10.000 align:middle line:90%