WEBVTT 00:00:00.000 --> 00:00:12.760 align:middle line:90% 00:00:12.760 --> 00:00:15.190 align:middle line:84% CASEY RODRIGUEZ: So last time we finished 00:00:15.190 --> 00:00:18.310 align:middle line:84% by proving Bolzano's intermediate value 00:00:18.310 --> 00:00:34.690 align:middle line:84% theorem, which states that a continuous function achieves 00:00:34.690 --> 00:00:38.020 align:middle line:84% all values in between the function evaluated 00:00:38.020 --> 00:00:38.840 align:middle line:90% at the endpoint. 00:00:38.840 --> 00:00:55.625 align:middle line:84% So if I have a continuous function, neither-- 00:00:55.625 --> 00:01:00.500 align:middle line:90% 00:01:00.500 --> 00:01:04.790 align:middle line:84% I take a value y between f of a and f of b, so either f of a 00:01:04.790 --> 00:01:07.370 align:middle line:84% is less than f of b and y's in between 00:01:07.370 --> 00:01:12.780 align:middle line:84% is bigger than f of a and less than f of b, or f of a 00:01:12.780 --> 00:01:17.510 align:middle line:84% is bigger than f of b, but y is still in between them. 00:01:17.510 --> 00:01:20.880 align:middle line:90% 00:01:20.880 --> 00:01:30.770 align:middle line:84% Then there exists c in the interval a, b so that f of c 00:01:30.770 --> 00:01:31.670 align:middle line:90% equals y. 00:01:31.670 --> 00:01:36.800 align:middle line:84% So every value in between f evaluated at the endpoints 00:01:36.800 --> 00:01:38.150 align:middle line:90% is achieved. 00:01:38.150 --> 00:01:39.800 align:middle line:84% I drew the picture that went along 00:01:39.800 --> 00:01:55.425 align:middle line:84% with this a, b, f of a, f of b, and that could be-- 00:01:55.425 --> 00:01:59.220 align:middle line:90% 00:01:59.220 --> 00:02:05.120 align:middle line:84% and if we take anything in between, 00:02:05.120 --> 00:02:13.322 align:middle line:84% then there has to be some c so that f of c equals y. 00:02:13.322 --> 00:02:14.780 align:middle line:84% At least in the picture, I've drawn 00:02:14.780 --> 00:02:16.780 align:middle line:84% there's three different guys, but there's always 00:02:16.780 --> 00:02:18.750 align:middle line:90% at least one. 00:02:18.750 --> 00:02:22.800 align:middle line:84% And then the last thing we talked 00:02:22.800 --> 00:02:24.360 align:middle line:84% about at the end of last lecture was 00:02:24.360 --> 00:02:32.390 align:middle line:84% that the image of a closed and bounded interval 00:02:32.390 --> 00:02:36.650 align:middle line:84% by a continuous function is, again, a closed and bounded 00:02:36.650 --> 00:02:51.190 align:middle line:84% interval, where e corresponds to the absolute max of f, 00:02:51.190 --> 00:02:52.680 align:middle line:90% and f corresponds so that-- 00:02:52.680 --> 00:02:55.180 align:middle line:84% I shouldn't call this f since we already have f being used-- 00:02:55.180 --> 00:03:00.500 align:middle line:90% 00:03:00.500 --> 00:03:03.200 align:middle line:90% let's make it d-- 00:03:03.200 --> 00:03:05.690 align:middle line:84% corresponds to the absolute min of f. 00:03:05.690 --> 00:03:29.050 align:middle line:84% So it's what we proved last time is that if f is continuous, 00:03:29.050 --> 00:03:37.800 align:middle line:84% then there exists e, d and r, such 00:03:37.800 --> 00:03:42.060 align:middle line:84% that the range of this closed and bounded interval 00:03:42.060 --> 00:03:46.080 align:middle line:84% is, again, a closed and bounded interval. 00:03:46.080 --> 00:03:51.330 align:middle line:84% So f is very well-behaved on closed and bounded intervals. 00:03:51.330 --> 00:03:54.072 align:middle line:84% And we'll see another way in which continuous functions are 00:03:54.072 --> 00:03:56.280 align:middle line:84% well behaved on a closed and bounded interval in just 00:03:56.280 --> 00:03:56.780 align:middle line:90% a second. 00:03:56.780 --> 00:04:00.184 align:middle line:90% 00:04:00.184 --> 00:04:08.820 align:middle line:84% Now, a simple application of the Bolzano intermediate value 00:04:08.820 --> 00:04:13.680 align:middle line:90% theorem is the following-- 00:04:13.680 --> 00:04:15.420 align:middle line:84% really is an application of that theorem 00:04:15.420 --> 00:04:17.019 align:middle line:84% that I labeled the bisection method. 00:04:17.019 --> 00:04:21.540 align:middle line:84% But let's chalk that up to the Bolzano intermediate value 00:04:21.540 --> 00:04:29.220 align:middle line:84% theorem, which is the following that if f of x 00:04:29.220 --> 00:04:49.520 align:middle line:84% is any odd polynomial, then f has at least one real root. 00:04:49.520 --> 00:04:57.230 align:middle line:90% 00:04:57.230 --> 00:04:59.750 align:middle line:84% So any odd polynomial has at least one real root. 00:04:59.750 --> 00:05:01.670 align:middle line:84% So by the fundamental theorem of algebra, 00:05:01.670 --> 00:05:06.740 align:middle line:84% every polynomial has exactly n roots, n corresponding 00:05:06.740 --> 00:05:09.230 align:middle line:84% to the degree of the polynomial, but these roots 00:05:09.230 --> 00:05:12.650 align:middle line:90% can be complex valued. 00:05:12.650 --> 00:05:15.800 align:middle line:84% But for odd polynomials, there must be at least one real root, 00:05:15.800 --> 00:05:18.870 align:middle line:84% and this is a consequence of the Bolzano intermediate value 00:05:18.870 --> 00:05:19.370 align:middle line:90% theorem. 00:05:19.370 --> 00:05:22.580 align:middle line:84% But instead of proving this theorem in its full generality, 00:05:22.580 --> 00:05:24.860 align:middle line:84% let me just give you a representative example. 00:05:24.860 --> 00:05:32.060 align:middle line:90% 00:05:32.060 --> 00:05:43.800 align:middle line:84% So let's take f of x equals x to the 2021 minus 00:05:43.800 --> 00:05:48.750 align:middle line:90% x to the 2020 plus-- 00:05:48.750 --> 00:05:49.620 align:middle line:90% I don't know. 00:05:49.620 --> 00:05:50.970 align:middle line:90% What's today? 00:05:50.970 --> 00:05:53.128 align:middle line:90% I think the 25 or 24. 00:05:53.128 --> 00:05:53.670 align:middle line:90% I don't know. 00:05:53.670 --> 00:06:04.830 align:middle line:90% --1025x minus 300. 00:06:04.830 --> 00:06:05.715 align:middle line:90% So what's the point? 00:06:05.715 --> 00:06:10.170 align:middle line:90% 00:06:10.170 --> 00:06:16.530 align:middle line:84% As long as I plug in a big enough x, 00:06:16.530 --> 00:06:20.000 align:middle line:90% this will swamp everything here. 00:06:20.000 --> 00:06:22.850 align:middle line:90% 00:06:22.850 --> 00:06:32.490 align:middle line:84% So, for example, if I stick in f of 100,000, 00:06:32.490 --> 00:06:35.850 align:middle line:84% I leave it to you to check that this is positive 00:06:35.850 --> 00:06:40.260 align:middle line:84% because this will be 100,000 to the 2020 minus 100,000 00:06:40.260 --> 00:06:46.320 align:middle line:84% to the 2021, and then plus this, which 00:06:46.320 --> 00:06:47.940 align:middle line:90% will be positive minus 300. 00:06:47.940 --> 00:06:48.880 align:middle line:90% This doesn't matter. 00:06:48.880 --> 00:06:55.440 align:middle line:84% So what really matters is this guy with the largest power 00:06:55.440 --> 00:06:58.830 align:middle line:84% will swamp all the other ones and eventually 00:06:58.830 --> 00:07:02.130 align:middle line:84% be positive if I stick in a large enough positive value. 00:07:02.130 --> 00:07:07.890 align:middle line:90% 00:07:07.890 --> 00:07:16.750 align:middle line:84% In fact, for this guy, if I stick in a negative value, 00:07:16.750 --> 00:07:19.260 align:middle line:84% then I get something very big, but now 00:07:19.260 --> 00:07:20.940 align:middle line:84% with a minus sign, which will again 00:07:20.940 --> 00:07:22.770 align:middle line:90% swamp all these earlier guys. 00:07:22.770 --> 00:07:25.200 align:middle line:84% In fact, I don't have to choose minus 100,000. 00:07:25.200 --> 00:07:27.992 align:middle line:84% If I just stick in 0, I will get minus 300, 00:07:27.992 --> 00:07:28.950 align:middle line:90% which will be negative. 00:07:28.950 --> 00:07:33.930 align:middle line:84% But you can check that if I stick in minus 100,000, 00:07:33.930 --> 00:07:35.100 align:middle line:90% this will still be negative. 00:07:35.100 --> 00:07:39.850 align:middle line:90% 00:07:39.850 --> 00:07:41.810 align:middle line:84% Thus, by the intermediate value theorem, 00:07:41.810 --> 00:07:59.380 align:middle line:84% there exists a c in 100,000 so that f of c 00:07:59.380 --> 00:08:03.880 align:middle line:84% equals 0 because 0 is something in between f evaluated 00:08:03.880 --> 00:08:05.230 align:middle line:90% at these two endpoints. 00:08:05.230 --> 00:08:12.960 align:middle line:90% 00:08:12.960 --> 00:08:16.500 align:middle line:84% And the idea is the same is that as long as you go far enough 00:08:16.500 --> 00:08:21.700 align:middle line:84% out and positive, then the polynomial will be positive 00:08:21.700 --> 00:08:24.580 align:middle line:84% if the coefficient out in front is positive; 00:08:24.580 --> 00:08:26.970 align:middle line:84% and if you go far enough out negative, 00:08:26.970 --> 00:08:33.380 align:middle line:84% then you'll get something that's negative. 00:08:33.380 --> 00:08:35.480 align:middle line:84% And then you apply the intermediate value theorem 00:08:35.480 --> 00:08:38.360 align:middle line:90% to be able to find a root. 00:08:38.360 --> 00:08:41.980 align:middle line:90% 00:08:41.980 --> 00:08:43.690 align:middle line:84% Now, of course, one other thing we 00:08:43.690 --> 00:08:49.525 align:middle line:84% did with the statement of the min-max theorem was, 00:08:49.525 --> 00:08:59.790 align:middle line:84% can we drop any of the assumptions on this theorem? 00:08:59.790 --> 00:09:03.910 align:middle line:84% And you can see that the example is no. 00:09:03.910 --> 00:09:12.410 align:middle line:84% So the question is, if f from a closed and bounded interval 00:09:12.410 --> 00:09:24.750 align:middle line:84% is not continuous and, let's say, f of 0 is less than 0, 00:09:24.750 --> 00:09:33.870 align:middle line:84% f of a, f of b is bigger than 0, then 00:09:33.870 --> 00:09:42.495 align:middle line:84% does there exist c, a, b, such that f of c equals 0? 00:09:42.495 --> 00:09:45.862 align:middle line:90% 00:09:45.862 --> 00:09:47.570 align:middle line:84% Let me put that question mark at the end. 00:09:47.570 --> 00:09:52.450 align:middle line:90% 00:09:52.450 --> 00:09:56.680 align:middle line:84% And it's easy to come up with an example of a function that 00:09:56.680 --> 00:09:59.560 align:middle line:84% is not continuous that satisfies these conclusions, 00:09:59.560 --> 00:10:02.830 align:middle line:84% and there does not exist a point in between so that f of c 00:10:02.830 --> 00:10:03.790 align:middle line:90% equals 0. 00:10:03.790 --> 00:10:07.190 align:middle line:84% Simplest being, I'll just draw a picture. 00:10:07.190 --> 00:10:17.370 align:middle line:90% Let's go from minus 1, 2. 00:10:17.370 --> 00:10:32.390 align:middle line:84% So then the function is f of x equals x minus 1. 00:10:32.390 --> 00:10:41.200 align:middle line:84% And this is for x not equal to 1, and let's make it a 1/2 00:10:41.200 --> 00:10:41.700 align:middle line:90% there. 00:10:41.700 --> 00:10:45.440 align:middle line:90% 00:10:45.440 --> 00:10:55.000 align:middle line:84% So this function looks like 1/2 there. 00:10:55.000 --> 00:10:58.195 align:middle line:84% So there is no value, there is no c in between. 00:10:58.195 --> 00:11:01.600 align:middle line:90% 00:11:01.600 --> 00:11:07.720 align:middle line:90% So x is between 0 and 2. 00:11:07.720 --> 00:11:11.860 align:middle line:84% So you see f of 0 is equal to minus 1, that's negative; 00:11:11.860 --> 00:11:14.200 align:middle line:84% f of 2 is equal to 1, it's positive; 00:11:14.200 --> 00:11:19.180 align:middle line:84% but there exists no c in between, so that f of c 00:11:19.180 --> 00:11:21.970 align:middle line:90% equals 0. 00:11:21.970 --> 00:11:24.460 align:middle line:90% So what's the point I'm making? 00:11:24.460 --> 00:11:29.140 align:middle line:84% We really do need this hypothesis 00:11:29.140 --> 00:11:34.120 align:middle line:84% that the function is continuous for this theorem to be true. 00:11:34.120 --> 00:11:36.610 align:middle line:84% If you drop that hypothesis, then the theorem 00:11:36.610 --> 00:11:37.600 align:middle line:90% is no longer true. 00:11:37.600 --> 00:12:02.450 align:middle line:90% 00:12:02.450 --> 00:12:11.610 align:middle line:84% So now we're going to move on to a new concept called 00:12:11.610 --> 00:12:15.330 align:middle line:90% uniform continuity. 00:12:15.330 --> 00:12:20.760 align:middle line:84% And before I do this, I'm going to rewrite 00:12:20.760 --> 00:12:22.350 align:middle line:90% the definition of continuity. 00:12:22.350 --> 00:12:48.740 align:middle line:90% 00:12:48.740 --> 00:12:51.920 align:middle line:84% So this is an old definition, but I 00:12:51.920 --> 00:12:55.320 align:middle line:84% want to highlight a specific aspect of it. 00:12:55.320 --> 00:12:58.010 align:middle line:84% So this is me writing the definition 00:12:58.010 --> 00:12:59.495 align:middle line:90% of a continuous function again. 00:12:59.495 --> 00:13:04.830 align:middle line:90% 00:13:04.830 --> 00:13:17.400 align:middle line:84% So function f from a set s to r is continuous. 00:13:17.400 --> 00:13:21.200 align:middle line:84% So we said that means it's continuous at every point, 00:13:21.200 --> 00:13:28.850 align:middle line:84% meaning if for all c And s, it's continuous at c, 00:13:28.850 --> 00:13:33.215 align:middle line:84% which means for all epsilon positive, there exists a delta. 00:13:33.215 --> 00:13:37.340 align:middle line:90% 00:13:37.340 --> 00:13:44.210 align:middle line:84% And we saw this in examples that we did proving continuity. 00:13:44.210 --> 00:13:51.980 align:middle line:84% This usually depends on epsilon and c, the point where 00:13:51.980 --> 00:14:00.520 align:middle line:84% I'm looking at, such that for all x and s, 00:14:00.520 --> 00:14:03.620 align:middle line:84% x minus c less than delta implies 00:14:03.620 --> 00:14:09.170 align:middle line:84% that f of x minus f of c is less than epsilon. 00:14:09.170 --> 00:14:16.930 align:middle line:90% 00:14:16.930 --> 00:14:18.770 align:middle line:84% So this is, again, a bit of review. 00:14:18.770 --> 00:14:21.400 align:middle line:84% But I just want to show you that, in general, 00:14:21.400 --> 00:14:26.510 align:middle line:84% when we verify that a function is continuous on a set, 00:14:26.510 --> 00:14:29.060 align:middle line:84% we have to show it's continuous at every point in the set. 00:14:29.060 --> 00:14:31.520 align:middle line:84% And that means for every epsilon, there exist delta. 00:14:31.520 --> 00:14:34.880 align:middle line:84% But this delta usually depends on epsilon, 00:14:34.880 --> 00:14:37.520 align:middle line:84% but it can also depend on the point c, which is where 00:14:37.520 --> 00:14:40.700 align:middle line:90% we're checking continuity at. 00:14:40.700 --> 00:14:45.890 align:middle line:84% So for simple example, let's take 00:14:45.890 --> 00:14:54.920 align:middle line:84% f of x equals 1 over x on the open interval 0, 1. 00:14:54.920 --> 00:15:07.680 align:middle line:84% So I claim f of x equals 1 over x is continuous. 00:15:07.680 --> 00:15:12.590 align:middle line:84% So proof we have to show that for every c and 0, 1 00:15:12.590 --> 00:15:14.520 align:middle line:84% for all epsilon, there exists that. 00:15:14.520 --> 00:15:25.190 align:middle line:84% So let's see b and 0, 1, but epsilon be positive. 00:15:25.190 --> 00:15:32.150 align:middle line:90% 00:15:32.150 --> 00:15:34.770 align:middle line:84% Again, since this kind of review, 00:15:34.770 --> 00:15:37.820 align:middle line:84% I'm not going to go through the process of really the thinking 00:15:37.820 --> 00:15:40.610 align:middle line:90% behind choosing this delta. 00:15:40.610 --> 00:15:43.310 align:middle line:84% I mean, there's calculations you do off to the side 00:15:43.310 --> 00:15:44.810 align:middle line:90% to try and get this delta. 00:15:44.810 --> 00:15:52.710 align:middle line:84% But let's choose delta to be the minimum of c over 2 00:15:52.710 --> 00:15:58.690 align:middle line:84% and c squared over 2 times epsilon. 00:15:58.690 --> 00:16:02.160 align:middle line:84% So what I want to again emphasize 00:16:02.160 --> 00:16:05.890 align:middle line:84% is this delta depends on c, the point we're looking at, 00:16:05.890 --> 00:16:06.660 align:middle line:90% and epsilon. 00:16:06.660 --> 00:16:10.110 align:middle line:90% 00:16:10.110 --> 00:16:11.430 align:middle line:90% So I claim this delta works. 00:16:11.430 --> 00:16:15.640 align:middle line:90% 00:16:15.640 --> 00:16:21.495 align:middle line:84% Suppose x is in 0, 1, and x minus c is less than delta. 00:16:21.495 --> 00:16:29.920 align:middle line:90% 00:16:29.920 --> 00:16:39.920 align:middle line:84% Then absolute value of x, this is by the triangle inequality. 00:16:39.920 --> 00:16:42.898 align:middle line:90% 00:16:42.898 --> 00:16:43.940 align:middle line:90% Let me write it this way. 00:16:43.940 --> 00:16:47.540 align:middle line:90% 00:16:47.540 --> 00:16:54.620 align:middle line:84% c absolute value is equal to c minus x 00:16:54.620 --> 00:16:57.800 align:middle line:84% plus x, which is less than or equal to c 00:16:57.800 --> 00:17:06.710 align:middle line:84% minus x plus x, which is less than delta, which is less than 00:17:06.710 --> 00:17:08.450 align:middle line:84% or equal to the minimum of these two. 00:17:08.450 --> 00:17:15.400 align:middle line:84% So it's certainly less than or equal to c over 2 plus x. 00:17:15.400 --> 00:17:24.520 align:middle line:84% And now c and x, these are in 0, 1, so they're all positive, 00:17:24.520 --> 00:17:26.170 align:middle line:90% and therefore I get that-- 00:17:26.170 --> 00:17:30.160 align:middle line:90% 00:17:30.160 --> 00:17:31.570 align:middle line:90% Let's go to the next board. 00:17:31.570 --> 00:17:38.260 align:middle line:90% 00:17:38.260 --> 00:17:43.300 align:middle line:84% So if I subtract the c over 2 over to the other side, 00:17:43.300 --> 00:17:49.170 align:middle line:84% I get that c over 2 is less than x. 00:17:49.170 --> 00:17:52.740 align:middle line:84% So this is as long as x is in 0 and x minus c 00:17:52.740 --> 00:17:55.380 align:middle line:90% is less than delta. 00:17:55.380 --> 00:17:58.230 align:middle line:90% This should not be a shock. 00:17:58.230 --> 00:18:05.520 align:middle line:84% There's 0, 1, c, c over 2 3c over 2. 00:18:05.520 --> 00:18:09.890 align:middle line:90% 00:18:09.890 --> 00:18:12.890 align:middle line:84% And so if x is in this interval, it certainly 00:18:12.890 --> 00:18:15.170 align:middle line:90% bounded away by c over 2. 00:18:15.170 --> 00:18:17.860 align:middle line:90% 00:18:17.860 --> 00:18:21.930 align:middle line:84% Or if x is in here, then x is certainly bigger than c over 2. 00:18:21.930 --> 00:18:24.550 align:middle line:90% 00:18:24.550 --> 00:18:31.280 align:middle line:84% And now we look at f of x minus f of c. 00:18:31.280 --> 00:18:34.220 align:middle line:84% This is equal to 1 over x minus 1 00:18:34.220 --> 00:18:43.360 align:middle line:84% over c equals c minus x over x times c, which is 00:18:43.360 --> 00:18:46.315 align:middle line:90% less than delta over x times c. 00:18:46.315 --> 00:18:59.030 align:middle line:90% 00:18:59.030 --> 00:19:04.340 align:middle line:84% And now x is bigger than c over 2, so if I take 1 over that, 00:19:04.340 --> 00:19:08.120 align:middle line:84% I get 1 over x is less than 2 over c. 00:19:08.120 --> 00:19:12.990 align:middle line:84% So this is less than or equal to 2 delta over c squared, 00:19:12.990 --> 00:19:15.770 align:middle line:84% and delta is the minimum of now c over 2 00:19:15.770 --> 00:19:18.600 align:middle line:84% and c squared over 2 times epsilon. 00:19:18.600 --> 00:19:20.340 align:middle line:84% So this is less than or equal to 2 00:19:20.340 --> 00:19:26.843 align:middle line:84% over c squared times c squared over 2 epsilon equals epsilon. 00:19:26.843 --> 00:19:28.760 align:middle line:84% So I just went through the proof of continuity 00:19:28.760 --> 00:19:34.800 align:middle line:84% for 1 over x, which using better machinery we could have proven. 00:19:34.800 --> 00:19:41.150 align:middle line:84% But I did this to highlight this point that's 00:19:41.150 --> 00:19:48.490 align:middle line:84% showing continuity of 1 over x on 0, 1, 00:19:48.490 --> 00:19:52.990 align:middle line:84% that in showing that, this delta that you are choosing 00:19:52.990 --> 00:19:55.800 align:middle line:84% depends on the point you're looking at. 00:19:55.800 --> 00:20:04.700 align:middle line:90% 00:20:04.700 --> 00:20:10.380 align:middle line:84% So what the definition of continuity says is for all c, 00:20:10.380 --> 00:20:13.800 align:middle line:84% for all epsilon, you can find a delta 00:20:13.800 --> 00:20:15.990 align:middle line:90% depending on epsilon and c. 00:20:15.990 --> 00:20:22.380 align:middle line:84% Now, uniform continuity removes that necessity of delta 00:20:22.380 --> 00:20:24.300 align:middle line:90% depending on c. 00:20:24.300 --> 00:20:28.370 align:middle line:84% And I'm going to make a possibly bad analogy in just a minute, 00:20:28.370 --> 00:20:30.120 align:middle line:84% but first let me write down the definition 00:20:30.120 --> 00:20:31.500 align:middle line:90% of uniform continuity. 00:20:31.500 --> 00:20:38.360 align:middle line:90% 00:20:38.360 --> 00:20:43.050 align:middle line:84% Let s be subset of r, f going from s to r. 00:20:43.050 --> 00:20:47.810 align:middle line:90% 00:20:47.810 --> 00:21:04.180 align:middle line:84% We say f is uniformly continuous on s 00:21:04.180 --> 00:21:10.980 align:middle line:84% if for all epsilon positive, there exists a delta depending 00:21:10.980 --> 00:21:19.520 align:middle line:84% only on epsilon, such that where all x, c and s, 00:21:19.520 --> 00:21:27.255 align:middle line:84% with x minus c less than delta, we get that f of x minus f of c 00:21:27.255 --> 00:21:28.130 align:middle line:90% is less than epsilon. 00:21:28.130 --> 00:21:30.890 align:middle line:90% 00:21:30.890 --> 00:21:37.090 align:middle line:84% So a function being continuous means that for all epsilon 00:21:37.090 --> 00:21:39.670 align:middle line:84% you can find delta depending on one of these points, 00:21:39.670 --> 00:21:43.810 align:middle line:84% depending on c so that if I had this, then I have this. 00:21:43.810 --> 00:21:47.140 align:middle line:84% Uniform continuity says you can find delta 00:21:47.140 --> 00:21:49.105 align:middle line:90% that works across the board. 00:21:49.105 --> 00:21:59.330 align:middle line:90% 00:21:59.330 --> 00:22:03.000 align:middle line:84% So I'm going to try and make an analogy that hopefully I 00:22:03.000 --> 00:22:06.540 align:middle line:84% don't have to cut out of this lecture. 00:22:06.540 --> 00:22:08.440 align:middle line:90% But it's a bit like this. 00:22:08.440 --> 00:22:11.700 align:middle line:84% Let's take a trip down memory lane 00:22:11.700 --> 00:22:13.395 align:middle line:90% and imagine being at a party. 00:22:13.395 --> 00:22:15.936 align:middle line:90% 00:22:15.936 --> 00:22:17.730 align:middle line:84% So you know everybody's in the party 00:22:17.730 --> 00:22:22.230 align:middle line:84% and let's say that everybody can hear, 00:22:22.230 --> 00:22:24.730 align:middle line:84% their ears work just about the same, 00:22:24.730 --> 00:22:26.970 align:middle line:84% and the goal is to try and understand 00:22:26.970 --> 00:22:29.370 align:middle line:84% each other's conversation, try and hear 00:22:29.370 --> 00:22:32.490 align:middle line:90% each other's conversations. 00:22:32.490 --> 00:22:39.940 align:middle line:84% Now, in order for me to hear this person's conversation, 00:22:39.940 --> 00:22:42.250 align:middle line:84% depending on how loud they talk, I 00:22:42.250 --> 00:22:45.550 align:middle line:84% have to be close enough to be able to hear them. 00:22:45.550 --> 00:22:48.520 align:middle line:84% And this changes from person to person in the room. 00:22:48.520 --> 00:22:53.390 align:middle line:84% Maybe I can stand a foot away from this person over here. 00:22:53.390 --> 00:22:56.440 align:middle line:84% Maybe this person over here is a low talker 00:22:56.440 --> 00:22:58.360 align:middle line:84% and I have to be almost up in their face 00:22:58.360 --> 00:22:59.590 align:middle line:90% to be able to hear them. 00:22:59.590 --> 00:23:01.390 align:middle line:84% Maybe that person just whispers and I 00:23:01.390 --> 00:23:04.900 align:middle line:84% have to be almost nose to nose to be able to hear them. 00:23:04.900 --> 00:23:07.250 align:middle line:84% That's a bit like a function being continuous. 00:23:07.250 --> 00:23:10.044 align:middle line:90% 00:23:10.044 --> 00:23:15.450 align:middle line:84% The threshold of hearing what they're saying is the epsilon 00:23:15.450 --> 00:23:18.030 align:middle line:84% and I have to change my delta depending on what point 00:23:18.030 --> 00:23:20.220 align:middle line:90% I'm at so that I can hear them. 00:23:20.220 --> 00:23:29.080 align:middle line:84% Now, uniform continuity, is kind of like everybody's 00:23:29.080 --> 00:23:33.950 align:middle line:84% more or less speaking around the same level, 00:23:33.950 --> 00:23:36.290 align:middle line:90% speaking at the same loudness. 00:23:36.290 --> 00:23:40.640 align:middle line:84% So I just can be within, say, one foot of everybody 00:23:40.640 --> 00:23:42.260 align:middle line:90% and be able to understand them. 00:23:42.260 --> 00:23:45.550 align:middle line:84% Go to this person if I'm within one foot I can hear them well. 00:23:45.550 --> 00:23:48.650 align:middle line:84% If I go to this person and within one foot 00:23:48.650 --> 00:23:51.500 align:middle line:90% I can hear them well, and so on. 00:23:51.500 --> 00:23:56.900 align:middle line:84% So that's kind of at least a loose analogy 00:23:56.900 --> 00:23:59.510 align:middle line:84% of the difference between continuity and uniform 00:23:59.510 --> 00:24:00.260 align:middle line:90% continuity. 00:24:00.260 --> 00:24:06.640 align:middle line:90% 00:24:06.640 --> 00:24:11.330 align:middle line:84% So this is a reasonably interesting definition. 00:24:11.330 --> 00:24:13.720 align:middle line:84% So we should see examples and negate it. 00:24:13.720 --> 00:24:16.930 align:middle line:90% 00:24:16.930 --> 00:24:21.340 align:middle line:84% First off, you should be able to convince yourself 00:24:21.340 --> 00:24:23.200 align:middle line:84% that a function that's uniformly continuous 00:24:23.200 --> 00:24:24.730 align:middle line:90% is actually continuous. 00:24:24.730 --> 00:24:27.030 align:middle line:84% I'm not going to write the proof of that down. 00:24:27.030 --> 00:24:30.280 align:middle line:84% It follows essentially straight from the definition. 00:24:30.280 --> 00:24:33.600 align:middle line:90% 00:24:33.600 --> 00:24:45.030 align:middle line:84% But let's look at a function, which is uniformly continuous. 00:24:45.030 --> 00:24:54.140 align:middle line:84% So the function f of x equals x squared on the interval 0, 00:24:54.140 --> 00:25:01.830 align:middle line:90% 1 is uniformly continuous. 00:25:01.830 --> 00:25:04.550 align:middle line:90% 00:25:04.550 --> 00:25:07.570 align:middle line:90% So what's the proof? 00:25:07.570 --> 00:25:09.320 align:middle line:84% I want you to notice I'm going to use kind 00:25:09.320 --> 00:25:14.150 align:middle line:84% of in a very essential way that we're 00:25:14.150 --> 00:25:15.680 align:middle line:84% on this closed and bounded interval. 00:25:15.680 --> 00:25:19.140 align:middle line:90% 00:25:19.140 --> 00:25:21.230 align:middle line:84% So we need to show that for every epsilon, 00:25:21.230 --> 00:25:24.355 align:middle line:84% there exists a delta so that no matter what two points-- 00:25:24.355 --> 00:25:27.340 align:middle line:90% 00:25:27.340 --> 00:25:31.320 align:middle line:84% So if you like, here's a picture that goes along with this-- 00:25:31.320 --> 00:25:37.240 align:middle line:90% 00:25:37.240 --> 00:25:40.020 align:middle line:84% so that if I look at maybe two points that 00:25:40.020 --> 00:25:43.020 align:middle line:84% are close to each other less than delta, 00:25:43.020 --> 00:25:52.930 align:middle line:84% their function's value is within epsilon. 00:25:52.930 --> 00:25:56.200 align:middle line:84% And so let's say I'm on an interval a, b. 00:25:56.200 --> 00:25:58.150 align:middle line:84% But not just this point, I could go over here 00:25:58.150 --> 00:26:02.600 align:middle line:84% to two other points that are within delta of each other. 00:26:02.600 --> 00:26:13.300 align:middle line:84% And if I look at the function evaluated at the two points, 00:26:13.300 --> 00:26:15.340 align:middle line:84% then the difference between these two guys 00:26:15.340 --> 00:26:16.465 align:middle line:90% is also less than epsilon. 00:26:16.465 --> 00:26:21.690 align:middle line:90% 00:26:21.690 --> 00:26:25.740 align:middle line:84% So I can move through the party and as long 00:26:25.740 --> 00:26:30.390 align:middle line:84% as I'm close enough within a uniform distance to somebody, 00:26:30.390 --> 00:26:33.490 align:middle line:90% I can hear them just fine. 00:26:33.490 --> 00:26:39.130 align:middle line:84% So what a nice trip down memory lane when we 00:26:39.130 --> 00:26:42.020 align:middle line:90% could be around other people. 00:26:42.020 --> 00:26:44.380 align:middle line:84% So let's show this function is uniformly continuous. 00:26:44.380 --> 00:26:45.790 align:middle line:90% Let epsilon be positive. 00:26:45.790 --> 00:26:48.970 align:middle line:90% 00:26:48.970 --> 00:26:56.110 align:middle line:84% Choose delta to be epsilon over 2. 00:26:56.110 --> 00:27:03.050 align:middle line:84% Now, you see this delta just depends on epsilon, 00:27:03.050 --> 00:27:12.920 align:middle line:84% then if x and c are in 0, 1, and x minus c is less than delta, 00:27:12.920 --> 00:27:16.690 align:middle line:84% then if I compute the difference, x 00:27:16.690 --> 00:27:22.000 align:middle line:84% squared minus c squared, I can write 00:27:22.000 --> 00:27:26.920 align:middle line:90% as x plus c times x minus c. 00:27:26.920 --> 00:27:28.660 align:middle line:84% And now I use the triangle inequality, 00:27:28.660 --> 00:27:34.660 align:middle line:84% this is less than or equal to x minus c. 00:27:34.660 --> 00:27:38.520 align:middle line:84% Now, these guys are in 0, 1, so their absolute values always 00:27:38.520 --> 00:27:40.650 align:middle line:90% bounded by 1. 00:27:40.650 --> 00:27:44.790 align:middle line:84% So this is less than or equal to 1 plus 1, 00:27:44.790 --> 00:27:49.590 align:middle line:84% and then x minus c is bounded by delta, 00:27:49.590 --> 00:27:51.690 align:middle line:84% which equals 2 times delta, and delta 00:27:51.690 --> 00:27:53.280 align:middle line:90% is chosen to be epsilon over 2. 00:27:53.280 --> 00:27:55.860 align:middle line:90% So I get epsilon. 00:27:55.860 --> 00:27:58.200 align:middle line:84% Now, again, how would you choose delta? 00:27:58.200 --> 00:28:00.930 align:middle line:84% So it looks like I just chose this delta out of nowhere. 00:28:00.930 --> 00:28:03.330 align:middle line:84% In practice, how do you choose delta? 00:28:03.330 --> 00:28:05.340 align:middle line:90% This is the computation you do. 00:28:05.340 --> 00:28:08.220 align:middle line:84% You take x squared minus c squared, split it, 00:28:08.220 --> 00:28:10.120 align:middle line:90% you have this. 00:28:10.120 --> 00:28:14.370 align:middle line:84% You can estimate with what you know above by 2 here 00:28:14.370 --> 00:28:15.750 align:middle line:90% times delta. 00:28:15.750 --> 00:28:19.200 align:middle line:84% So how do you choose delta so that 2 times delta 00:28:19.200 --> 00:28:21.420 align:middle line:90% is, say, equal to epsilon? 00:28:21.420 --> 00:28:23.940 align:middle line:84% You choose delta to be epsilon over 2. 00:28:23.940 --> 00:28:31.790 align:middle line:90% 00:28:31.790 --> 00:28:39.660 align:middle line:84% So let's negate this definition and then look at 1 00:28:39.660 --> 00:28:43.050 align:middle line:84% over x once again and check to see 00:28:43.050 --> 00:28:44.550 align:middle line:90% if it's uniformly continuous. 00:28:44.550 --> 00:28:45.840 align:middle line:90% We'll show it's not. 00:28:45.840 --> 00:29:00.150 align:middle line:84% So the negation f from s to r is not uniformly continuous. 00:29:00.150 --> 00:29:18.880 align:middle line:84% So I saw a little bit of the new Borat movie that came out. 00:29:18.880 --> 00:29:22.720 align:middle line:84% So I almost wanted to say f is uniformly continuous. 00:29:22.720 --> 00:29:23.380 align:middle line:90% Not. 00:29:23.380 --> 00:29:24.730 align:middle line:90% But I didn't. 00:29:24.730 --> 00:29:27.610 align:middle line:90% It's not uniformly continuous. 00:29:27.610 --> 00:29:30.850 align:middle line:84% And now you negate the rest of the definition. 00:29:30.850 --> 00:29:36.280 align:middle line:84% If there exists some bad epsilon, such that no matter 00:29:36.280 --> 00:29:45.820 align:middle line:84% what delta you choose, there exists some bad x and c 00:29:45.820 --> 00:29:51.370 align:middle line:84% and s, such that they're close to each other, 00:29:51.370 --> 00:30:00.610 align:middle line:84% but the values are not bigger than or equal to. 00:30:00.610 --> 00:30:09.490 align:middle line:90% 00:30:09.490 --> 00:30:15.730 align:middle line:90% So let's revisit 1 over x. 00:30:15.730 --> 00:30:19.890 align:middle line:90% 00:30:19.890 --> 00:30:30.820 align:middle line:84% This is not uniformly continuous on the open interval 0, 1. 00:30:30.820 --> 00:30:33.040 align:middle line:84% So you see that this function is continuous 00:30:33.040 --> 00:30:37.230 align:middle line:84% but it's not uniformly continuous. 00:30:37.230 --> 00:30:39.480 align:middle line:84% We just proved that a minute ago that it's continuous. 00:30:39.480 --> 00:30:43.460 align:middle line:84% So let me now give you the proof that this function is 00:30:43.460 --> 00:30:44.930 align:middle line:90% not uniformly continuous. 00:30:44.930 --> 00:30:49.880 align:middle line:90% And what's the idea behind this? 00:30:49.880 --> 00:30:52.440 align:middle line:90% Here's one. 00:30:52.440 --> 00:30:54.950 align:middle line:84% This is not to scale, but there's one 00:30:54.950 --> 00:31:04.670 align:middle line:84% as well, and 1 over x shoots up to plus infinity 00:31:04.670 --> 00:31:05.930 align:middle line:90% as x approaches 0. 00:31:05.930 --> 00:31:09.770 align:middle line:90% 00:31:09.770 --> 00:31:11.660 align:middle line:84% So what uniform continuity says is 00:31:11.660 --> 00:31:14.510 align:middle line:84% that if I take any two points in the interval that 00:31:14.510 --> 00:31:16.730 align:middle line:84% are close to each other, then the outputs 00:31:16.730 --> 00:31:20.900 align:middle line:90% have to be close to each other. 00:31:20.900 --> 00:31:24.450 align:middle line:84% Continuity, remember, says if I fix a point, 00:31:24.450 --> 00:31:26.340 align:middle line:84% then as long as I'm close to that point, 00:31:26.340 --> 00:31:28.440 align:middle line:84% the outputs will be close to each other. 00:31:28.440 --> 00:31:30.840 align:middle line:84% Uniform says, no matter what two points 00:31:30.840 --> 00:31:33.660 align:middle line:84% you choose, as long as they're close together, 00:31:33.660 --> 00:31:37.170 align:middle line:84% the outputs will be close together in a controlled way. 00:31:37.170 --> 00:31:42.420 align:middle line:84% Now, if I take two points very close to each other here, 00:31:42.420 --> 00:31:51.170 align:middle line:84% very close to 0, they can be pretty close to each other. 00:31:51.170 --> 00:32:00.690 align:middle line:84% I mean, their values can be quite large. 00:32:00.690 --> 00:32:06.350 align:middle line:84% Even though the difference between the arguments 00:32:06.350 --> 00:32:08.960 align:middle line:84% are small, the values can be quite large 00:32:08.960 --> 00:32:11.840 align:middle line:90% just because 1 over x shoots up. 00:32:11.840 --> 00:32:14.640 align:middle line:90% So let's quantify this. 00:32:14.640 --> 00:32:17.520 align:middle line:84% I have to tell you what the bad epsilon not is. 00:32:17.520 --> 00:32:21.620 align:middle line:90% 00:32:21.620 --> 00:32:23.510 align:middle line:90% Choose epsilon not equals 2. 00:32:23.510 --> 00:32:27.575 align:middle line:84% Now, any epsilon not will work just fine. 00:32:27.575 --> 00:32:31.050 align:middle line:90% 00:32:31.050 --> 00:32:34.690 align:middle line:90% Let delta be positive. 00:32:34.690 --> 00:32:40.080 align:middle line:84% So now we have to find x and c and s within distance delta 00:32:40.080 --> 00:32:53.790 align:middle line:84% to each other whose values vary by at least two. 00:32:53.790 --> 00:33:07.290 align:middle line:84% So let's choose c to be the minimum of delta and 1/2, 00:33:07.290 --> 00:33:10.560 align:middle line:84% and I'm going to choose x to be c over 2. 00:33:10.560 --> 00:33:14.820 align:middle line:90% 00:33:14.820 --> 00:33:19.230 align:middle line:84% So just ignoring this half for a minute, c is equal to delta, 00:33:19.230 --> 00:33:22.500 align:middle line:84% x is equal to delta over 2, and then 00:33:22.500 --> 00:33:26.340 align:middle line:84% 1 over c minus 1 over x is going to be 2 over delta. 00:33:26.340 --> 00:33:28.080 align:middle line:84% And if delta is very small, that's 00:33:28.080 --> 00:33:29.610 align:middle line:84% certainly bigger than or equal to 2. 00:33:29.610 --> 00:33:32.550 align:middle line:84% But I had to throw in this 1/2 here because I chose epsilon 00:33:32.550 --> 00:33:35.610 align:middle line:84% not to be 2, and delta's arbitrary. 00:33:35.610 --> 00:33:38.960 align:middle line:90% 00:33:38.960 --> 00:33:44.000 align:middle line:84% So then we see that x minus c, this 00:33:44.000 --> 00:33:47.900 align:middle line:84% equals c over 2, which is certainly less than or equal to 00:33:47.900 --> 00:33:52.130 align:middle line:84% since c is the minimum of delta and 1 over 1/2 00:33:52.130 --> 00:33:54.410 align:middle line:84% is less than delta over 2, which is less than delta. 00:33:54.410 --> 00:33:59.430 align:middle line:90% 00:33:59.430 --> 00:34:05.490 align:middle line:84% And if we compute 1 over x minus 1 over c, this is-- 00:34:05.490 --> 00:34:13.290 align:middle line:90% 00:34:13.290 --> 00:34:19.830 align:middle line:84% remember x is c over 2, so this 1 over x is 2 over c, so 1 00:34:19.830 --> 00:34:20.489 align:middle line:90% over c. 00:34:20.489 --> 00:34:23.850 align:middle line:90% 00:34:23.850 --> 00:34:29.969 align:middle line:84% And now from c being less than or equal to 1 over 1/2m 00:34:29.969 --> 00:34:33.020 align:middle line:84% I get that 1 over c is going to be bigger than or equal to 2. 00:34:33.020 --> 00:34:51.590 align:middle line:90% 00:34:51.590 --> 00:34:53.520 align:middle line:90% And that's the end of the proof. 00:34:53.520 --> 00:34:59.340 align:middle line:84% So we see that even though these two arguments are 00:34:59.340 --> 00:35:04.020 align:middle line:84% very close to each other, the outputs are, in fact, separated 00:35:04.020 --> 00:35:06.510 align:middle line:90% by some fixed distance. 00:35:06.510 --> 00:35:09.180 align:middle line:90% At least two. 00:35:09.180 --> 00:35:12.570 align:middle line:84% But we could actually make it bigger than 3, bigger than 4, 00:35:12.570 --> 00:35:13.395 align:middle line:90% whatever you like. 00:35:13.395 --> 00:35:27.320 align:middle line:90% 00:35:27.320 --> 00:35:29.070 align:middle line:84% And I'm not going to go over this example. 00:35:29.070 --> 00:35:31.960 align:middle line:90% 00:35:31.960 --> 00:35:36.790 align:middle line:84% But in fact, I'll leave it to you or maybe I'll 00:35:36.790 --> 00:35:38.500 align:middle line:84% put it on the assignment, f of x equals 00:35:38.500 --> 00:35:48.970 align:middle line:84% x squared is not uniformly continuous on r. 00:35:48.970 --> 00:35:52.660 align:middle line:84% So we saw that it's uniformly continuous on a closed interval 00:35:52.660 --> 00:35:59.650 align:middle line:84% 0, 1, but it's not uniformly continuous on all of r. 00:35:59.650 --> 00:36:02.920 align:middle line:84% The reason being again kind of because x squared 00:36:02.920 --> 00:36:04.370 align:middle line:90% is starting to get big. 00:36:04.370 --> 00:36:07.670 align:middle line:84% So I can take two things close to each other, 00:36:07.670 --> 00:36:10.570 align:middle line:84% stick it into x squared, and their outputs 00:36:10.570 --> 00:36:13.540 align:middle line:84% could be very different from each other. 00:36:13.540 --> 00:36:16.795 align:middle line:90% 00:36:16.795 --> 00:36:18.420 align:middle line:84% Didn't write out the proof of this one, 00:36:18.420 --> 00:36:26.720 align:middle line:84% so we're going to go by a little bit by the seat of our pants. 00:36:26.720 --> 00:36:32.990 align:middle line:90% 00:36:32.990 --> 00:36:35.820 align:middle line:84% So we want to show this is not uniformly continuous. 00:36:35.820 --> 00:36:44.250 align:middle line:84% So I think epsilon 0 equals 1 is going to work just fine. 00:36:44.250 --> 00:36:50.373 align:middle line:90% 00:36:50.373 --> 00:36:53.730 align:middle line:84% Let delta be positive, and the idea 00:36:53.730 --> 00:37:00.120 align:middle line:84% is we want to choose x to be some number. 00:37:00.120 --> 00:37:03.720 align:middle line:90% Let's call it a plus-- 00:37:03.720 --> 00:37:06.830 align:middle line:90% 00:37:06.830 --> 00:37:15.570 align:middle line:84% well, c will be some number, and c equals some number, 00:37:15.570 --> 00:37:20.310 align:middle line:90% x equals c plus delta over 2. 00:37:20.310 --> 00:37:22.590 align:middle line:84% So this is kind of scratch work off to the side. 00:37:22.590 --> 00:37:26.700 align:middle line:84% And let's see if we can somehow choose c. 00:37:26.700 --> 00:37:35.190 align:middle line:84% So first off, x minus c will be delta over 2, 00:37:35.190 --> 00:37:36.870 align:middle line:90% which is less than delta. 00:37:36.870 --> 00:37:43.050 align:middle line:84% And now let's see if we can choose c 00:37:43.050 --> 00:37:47.520 align:middle line:84% so that x squared minus c squared 00:37:47.520 --> 00:37:51.260 align:middle line:90% is bigger than or equal to 1. 00:37:51.260 --> 00:37:58.790 align:middle line:90% 00:37:58.790 --> 00:38:03.500 align:middle line:84% So c is kind of the thing that we get to play with here. 00:38:03.500 --> 00:38:05.720 align:middle line:84% I mean, this is how it actually goes 00:38:05.720 --> 00:38:09.590 align:middle line:84% if you want to figure out the proof before you write it down, 00:38:09.590 --> 00:38:12.830 align:middle line:90% the scratch work to the side. 00:38:12.830 --> 00:38:15.350 align:middle line:90% So choose c so that this-- 00:38:15.350 --> 00:38:18.908 align:middle line:84% now, let's start off with the thing 00:38:18.908 --> 00:38:20.450 align:middle line:84% that we're trying to bound from below 00:38:20.450 --> 00:38:22.310 align:middle line:90% and start computing some stuff. 00:38:22.310 --> 00:38:24.500 align:middle line:90% This is equal to x plus c-- 00:38:24.500 --> 00:38:26.120 align:middle line:90% and c will be positive. 00:38:26.120 --> 00:38:30.690 align:middle line:84% So this will be c, and x will be positive. 00:38:30.690 --> 00:38:38.360 align:middle line:84% So x plus c times x minus c; now x 00:38:38.360 --> 00:38:50.980 align:middle line:84% plus c will be 2c plus delta over 2; x minus c 00:38:50.980 --> 00:38:54.430 align:middle line:90% will be delta over 2. 00:38:54.430 --> 00:38:58.210 align:middle line:90% 00:38:58.210 --> 00:39:02.130 align:middle line:84% So far all of these things are equality, 00:39:02.130 --> 00:39:09.460 align:middle line:84% and we get this is equal to 2c plus-- 00:39:09.460 --> 00:39:14.650 align:middle line:84% so delta is 2c plus delta squared over 2, 00:39:14.650 --> 00:39:17.230 align:middle line:84% and here we're going to choose c positive. 00:39:17.230 --> 00:39:20.740 align:middle line:84% So we're trying to show it's not uniformly continuous on r. 00:39:20.740 --> 00:39:27.618 align:middle line:84% So we just need to find a value of c and a value of x so 00:39:27.618 --> 00:39:29.660 align:middle line:84% that they're within delta distance to each other, 00:39:29.660 --> 00:39:33.350 align:middle line:84% but their outputs are bigger than or equal to 1. 00:39:33.350 --> 00:39:36.560 align:middle line:84% So far if c is just something positive, 00:39:36.560 --> 00:39:39.962 align:middle line:84% x is c plus delta over 2, then they're close. 00:39:39.962 --> 00:39:41.420 align:middle line:84% Their difference, however, is going 00:39:41.420 --> 00:39:46.850 align:middle line:84% to be just by this computation equal to 2c plus delta over 2 00:39:46.850 --> 00:39:50.250 align:middle line:90% times delta over 2 squared. 00:39:50.250 --> 00:39:53.540 align:middle line:84% Now delta squared over 2 is always non-negative, 00:39:53.540 --> 00:39:57.170 align:middle line:84% so it's certainly bigger than or equal to delta times c. 00:39:57.170 --> 00:39:59.960 align:middle line:84% And now remember we wanted this thing 00:39:59.960 --> 00:40:01.550 align:middle line:90% to be bigger than or equal to 1. 00:40:01.550 --> 00:40:04.370 align:middle line:84% Right now we have it's bigger than or equal to delta c. 00:40:04.370 --> 00:40:12.810 align:middle line:84% So let's choose c to be 1 over delta, 00:40:12.810 --> 00:40:14.820 align:middle line:90% and let's see that that works. 00:40:14.820 --> 00:40:17.670 align:middle line:90% 00:40:17.670 --> 00:40:24.060 align:middle line:84% Choose c to be 1 over delta, x to be 1 over delta. 00:40:24.060 --> 00:40:27.390 align:middle line:90% So c plus delta over 2. 00:40:27.390 --> 00:40:30.480 align:middle line:90% 00:40:30.480 --> 00:40:32.730 align:middle line:90% Then these two values are close. 00:40:32.730 --> 00:40:38.610 align:middle line:84% x minus c equals delta over 2 is less than delta, 00:40:38.610 --> 00:40:45.840 align:middle line:84% and x squared minus c squared equals x plus c times 00:40:45.840 --> 00:40:50.704 align:middle line:90% x minus c. 00:40:50.704 --> 00:40:58.870 align:middle line:84% x plus c as we did a minute ago is going to be 2 over delta 00:40:58.870 --> 00:41:00.250 align:middle line:90% plus delta over 2. 00:41:00.250 --> 00:41:01.825 align:middle line:90% That's x plus c. 00:41:01.825 --> 00:41:06.554 align:middle line:84% x minus c is equal to delta over 2, 00:41:06.554 --> 00:41:10.690 align:middle line:84% which equals 1, plus delta squared over 4, 00:41:10.690 --> 00:41:14.500 align:middle line:84% which is bigger than or equal to 1, our bad epsilon. 00:41:14.500 --> 00:41:20.530 align:middle line:84% So f of x equals x squared is not uniformly continuous on r, 00:41:20.530 --> 00:41:22.483 align:middle line:84% even though it is uniformly continuous on this 00:41:22.483 --> 00:41:23.650 align:middle line:90% closed and bounded interval. 00:41:23.650 --> 00:41:27.200 align:middle line:90% 00:41:27.200 --> 00:41:35.540 align:middle line:84% So uniform continuity is really a relationship 00:41:35.540 --> 00:41:41.390 align:middle line:84% between a function and the domain on which it's defined. 00:41:41.390 --> 00:41:45.970 align:middle line:90% 00:41:45.970 --> 00:41:49.420 align:middle line:84% So this is kind of to wrap up our theme 00:41:49.420 --> 00:41:52.630 align:middle line:84% of continuous functions and how they're 00:41:52.630 --> 00:41:54.640 align:middle line:84% behaved on closed and bounded intervals. 00:41:54.640 --> 00:41:57.340 align:middle line:84% And we see here that uniform continuity 00:41:57.340 --> 00:42:01.570 align:middle line:84% is a much stronger notion than continuity, as I've said. 00:42:01.570 --> 00:42:04.240 align:middle line:84% I mean, you can easily prove from the definition 00:42:04.240 --> 00:42:07.480 align:middle line:84% that if a function is uniformly continuous on s, 00:42:07.480 --> 00:42:09.180 align:middle line:90% then it's continuous on s. 00:42:09.180 --> 00:42:14.740 align:middle line:90% 00:42:14.740 --> 00:42:19.000 align:middle line:84% And we've seen that in general the reverse inclusion does not 00:42:19.000 --> 00:42:19.580 align:middle line:90% hold. 00:42:19.580 --> 00:42:22.855 align:middle line:84% So let me just write this up as a remark. 00:42:22.855 --> 00:42:28.351 align:middle line:90% 00:42:28.351 --> 00:42:40.610 align:middle line:84% In general, f uniformly continuous 00:42:40.610 --> 00:42:46.320 align:middle line:90% implies that f is continuous. 00:42:46.320 --> 00:42:48.165 align:middle line:84% This just follows from the definition. 00:42:48.165 --> 00:42:49.415 align:middle line:90% This is not difficult to show. 00:42:49.415 --> 00:42:56.050 align:middle line:90% 00:42:56.050 --> 00:43:13.030 align:middle line:84% But the converse does not hold, as we've just seen 00:43:13.030 --> 00:43:15.160 align:middle line:84% We have two examples here. f of x 00:43:15.160 --> 00:43:18.550 align:middle line:84% equals 1 over x on the open interval 0, 1, which 00:43:18.550 --> 00:43:21.610 align:middle line:84% is continuous, but it's not uniformly continuous; 00:43:21.610 --> 00:43:24.560 align:middle line:84% and then we have the function f of x equals x squared, 00:43:24.560 --> 00:43:28.600 align:middle line:84% which is not uniformly continuous on r. 00:43:28.600 --> 00:43:31.450 align:middle line:84% But in fact, it is uniformly continuous 00:43:31.450 --> 00:43:34.750 align:middle line:84% on this closed and bounded interval 0, 1. 00:43:34.750 --> 00:43:36.730 align:middle line:84% This example here, though, is, in fact, 00:43:36.730 --> 00:43:44.620 align:middle line:84% the rule; that if I'm looking at a function 00:43:44.620 --> 00:43:48.070 align:middle line:84% on a closed and bounded interval, then, in fact, 00:43:48.070 --> 00:43:49.760 align:middle line:84% these two notions are equivalent. 00:43:49.760 --> 00:43:51.310 align:middle line:90% So that's the following theorem. 00:43:51.310 --> 00:43:55.980 align:middle line:90% 00:43:55.980 --> 00:44:00.830 align:middle line:84% Let f be a function from a to b to r. 00:44:00.830 --> 00:44:06.950 align:middle line:84% So closed and bounded interval now, not just an arbitrary set. 00:44:06.950 --> 00:44:14.240 align:middle line:84% Then f is continuous on a, b, meaning 00:44:14.240 --> 00:44:18.400 align:middle line:84% it's continuous at every point, if and only 00:44:18.400 --> 00:44:28.390 align:middle line:84% if f is uniformly continuous on a, b. 00:44:28.390 --> 00:44:33.010 align:middle line:84% And the necessity of working on a closed and bounded interval 00:44:33.010 --> 00:44:36.640 align:middle line:84% is seen again by this example we had 00:44:36.640 --> 00:44:39.310 align:middle line:84% here of f of x equals 1 over x and f 00:44:39.310 --> 00:44:44.160 align:middle line:84% of x equals x squared on r and 0, 1. 00:44:44.160 --> 00:44:48.580 align:middle line:84% But as long as we're looking on a closed and bounded interval, 00:44:48.580 --> 00:44:51.760 align:middle line:84% continuous is equivalent to uniform continuity. 00:44:51.760 --> 00:44:54.355 align:middle line:84% Now, this is also not the sharpest thing you can say. 00:44:54.355 --> 00:44:57.880 align:middle line:90% 00:44:57.880 --> 00:45:01.810 align:middle line:84% In fact, what you can say is that if I replace 00:45:01.810 --> 00:45:05.470 align:middle line:84% this with what's called a compact set, which 00:45:05.470 --> 00:45:07.305 align:middle line:84% I may put in the assignment, I may not, 00:45:07.305 --> 00:45:08.680 align:middle line:84% then the statement is still true. 00:45:08.680 --> 00:45:11.800 align:middle line:84% And that's in some sense the sharpest statement. 00:45:11.800 --> 00:45:16.530 align:middle line:84% But I think this is interesting enough 00:45:16.530 --> 00:45:20.750 align:middle line:84% to just work on a closed and bounded interval. 00:45:20.750 --> 00:45:22.655 align:middle line:90% So let's prove this. 00:45:22.655 --> 00:45:38.940 align:middle line:90% 00:45:38.940 --> 00:45:44.040 align:middle line:84% I leave this to you as an exercise, uniform continuity 00:45:44.040 --> 00:45:46.620 align:middle line:90% implying continuity. 00:45:46.620 --> 00:45:47.490 align:middle line:90% I leave it to you. 00:45:47.490 --> 00:45:50.670 align:middle line:90% 00:45:50.670 --> 00:45:56.170 align:middle line:84% And I will do more difficult one, 00:45:56.170 --> 00:46:01.360 align:middle line:84% and we'll use kind of a philosophy 00:46:01.360 --> 00:46:03.880 align:middle line:84% like we did in the previous lecture 00:46:03.880 --> 00:46:05.770 align:middle line:84% where we use the definition of continuity 00:46:05.770 --> 00:46:10.150 align:middle line:84% or this theorem about continuity in terms of sequences and also 00:46:10.150 --> 00:46:12.940 align:middle line:90% the Bolzano-Weierstrass theorem. 00:46:12.940 --> 00:46:18.320 align:middle line:90% 00:46:18.320 --> 00:46:18.820 align:middle line:90% Good. 00:46:18.820 --> 00:46:20.680 align:middle line:84% We still have the negation of the definition 00:46:20.680 --> 00:46:25.820 align:middle line:84% of uniform continuity, and that's what we'll need. 00:46:25.820 --> 00:46:28.850 align:middle line:90% 00:46:28.850 --> 00:46:33.860 align:middle line:84% So we're going to prove f continuous 00:46:33.860 --> 00:46:35.840 align:middle line:84% implies uniformly continuous, and we're going 00:46:35.840 --> 00:46:37.170 align:middle line:90% to do this by contradiction. 00:46:37.170 --> 00:46:43.760 align:middle line:84% So suppose f is continuous on a, b, 00:46:43.760 --> 00:46:45.470 align:middle line:90% but not uniformly continuous. 00:46:45.470 --> 00:47:00.820 align:middle line:90% 00:47:00.820 --> 00:47:10.590 align:middle line:84% And then we're going to break the universe some way, 00:47:10.590 --> 00:47:13.620 align:middle line:84% which shows that our assumption that f is not uniformly 00:47:13.620 --> 00:47:16.260 align:middle line:90% continuous does not hold. 00:47:16.260 --> 00:47:22.130 align:middle line:90% 00:47:22.130 --> 00:47:25.280 align:middle line:84% So we use this negation of the definition 00:47:25.280 --> 00:47:30.350 align:middle line:84% of uniform continuity, which says that there exists delta 0. 00:47:30.350 --> 00:47:31.820 align:middle line:90% So let me write this out. 00:47:31.820 --> 00:47:38.890 align:middle line:90% 00:47:38.890 --> 00:47:40.390 align:middle line:84% I'm just going to rewrite this here, 00:47:40.390 --> 00:47:43.660 align:middle line:84% then there exists a delta 0, such that for all delta 00:47:43.660 --> 00:47:50.770 align:middle line:84% positive, there exists x, c and a, b, 00:47:50.770 --> 00:47:58.800 align:middle line:84% which depend on delta, such that x minus c is less than delta 00:47:58.800 --> 00:48:07.770 align:middle line:84% and f of x minus f of c is greater than 00:48:07.770 --> 00:48:09.000 align:middle line:90% or equal to epsilon not. 00:48:09.000 --> 00:48:12.610 align:middle line:90% 00:48:12.610 --> 00:48:16.930 align:middle line:90% Now, this holds for every delta. 00:48:16.930 --> 00:48:21.970 align:middle line:84% You can find x and c so that x minus c is less than delta 00:48:21.970 --> 00:48:24.010 align:middle line:84% and f of x minus f of c is greater 00:48:24.010 --> 00:48:25.690 align:middle line:90% than or equal to epsilon not. 00:48:25.690 --> 00:48:30.490 align:middle line:84% So let's choose delta to be 1 over n for each natural number 00:48:30.490 --> 00:48:41.210 align:middle line:84% n, and for all natural numbers n, there exists 00:48:41.210 --> 00:48:46.970 align:middle line:90% xn, cn, and a, b, such that-- 00:48:46.970 --> 00:48:50.160 align:middle line:84% so this is the statement now with delta equals 1 over n. 00:48:50.160 --> 00:48:55.190 align:middle line:84% xn minus cn is less than 1 over n, 00:48:55.190 --> 00:49:06.070 align:middle line:84% and f of x sub n minus f of c sub n is bigger than or equal 00:49:06.070 --> 00:49:07.030 align:middle line:90% to epsilon 0. 00:49:07.030 --> 00:49:10.970 align:middle line:90% 00:49:10.970 --> 00:49:19.490 align:middle line:84% Now, I have two sequences here, and they're 00:49:19.490 --> 00:49:22.560 align:middle line:84% bounded because they're in this closed and bounded interval. 00:49:22.560 --> 00:49:27.350 align:middle line:84% So I can pass to a subsequence of one of them to start 00:49:27.350 --> 00:49:32.410 align:middle line:84% by the Bolzano-Weierstrass theorem-- 00:49:32.410 --> 00:49:34.420 align:middle line:90% not B-Z, Bolzano-Weierstrass. 00:49:34.420 --> 00:49:37.330 align:middle line:84% I think that's probably somewhere else in the lecture 00:49:37.330 --> 00:49:40.420 align:middle line:84% or the previous I wrote B-Z, but I meant B-W 00:49:40.420 --> 00:49:42.145 align:middle line:90% by Bolzano-Weierstrass-- 00:49:42.145 --> 00:49:44.820 align:middle line:90% 00:49:44.820 --> 00:50:05.110 align:middle line:84% there exists subsequent xn sub k of x sub n and x and a, b, such 00:50:05.110 --> 00:50:13.030 align:middle line:84% that limit as k goes to infinity of x sub n sub k equals x. 00:50:13.030 --> 00:50:16.630 align:middle line:84% So Bolzano-Weierstrass from any bounded sequence 00:50:16.630 --> 00:50:18.580 align:middle line:84% we can find a convergent subsequence. 00:50:18.580 --> 00:50:22.800 align:middle line:84% And because this subsequence is always between a and b, 00:50:22.800 --> 00:50:24.960 align:middle line:84% the limit x will be between a and b. 00:50:24.960 --> 00:50:28.820 align:middle line:90% 00:50:28.820 --> 00:50:42.190 align:middle line:84% Now, since the subsequence cn sub k 00:50:42.190 --> 00:50:52.990 align:middle line:90% is bounded between a and b-- 00:50:52.990 --> 00:50:55.960 align:middle line:84% so I have a subsequence of x sub n sub k, 00:50:55.960 --> 00:50:59.500 align:middle line:84% and I then obtain a subsequence of the c sub n 00:50:59.500 --> 00:51:03.400 align:middle line:84% by just picking these integers to be the same. 00:51:03.400 --> 00:51:09.742 align:middle line:84% And so now that subsequence is of the c sub 00:51:09.742 --> 00:51:12.840 align:middle line:84% n is they're all coming from a, b. 00:51:12.840 --> 00:51:14.970 align:middle line:90% So that's bounded by a, b. 00:51:14.970 --> 00:51:15.900 align:middle line:90% So it's bounded. 00:51:15.900 --> 00:51:22.020 align:middle line:84% So by Bolzano-Weierstrass applied now 00:51:22.020 --> 00:51:32.130 align:middle line:84% to this subsequence, there exists a subsequence c sub n 00:51:32.130 --> 00:51:45.700 align:middle line:84% sub k of j, of c sub n sub k, and an element c 00:51:45.700 --> 00:51:53.870 align:middle line:84% in a, b, such that the limit as j 00:51:53.870 --> 00:52:00.100 align:middle line:84% goes to infinity of c sub n sub k sub j equals c. 00:52:00.100 --> 00:52:08.100 align:middle line:84% So I found a subsequence of the x of n's, and now I 00:52:08.100 --> 00:52:11.700 align:middle line:84% look at a corresponding subsequence of the c sub 00:52:11.700 --> 00:52:16.897 align:middle line:84% n's where now the subsequent cn sub k's are coming from the n 00:52:16.897 --> 00:52:19.920 align:middle line:84% sub k's chosen for this subsequence, that's 00:52:19.920 --> 00:52:21.790 align:middle line:90% still a bounded sequence. 00:52:21.790 --> 00:52:23.970 align:middle line:84% So by Bolzano-Weierstrass, I can take 00:52:23.970 --> 00:52:27.210 align:middle line:84% a subsequence of that and an element 00:52:27.210 --> 00:52:29.580 align:middle line:90% c so that I have convergence. 00:52:29.580 --> 00:52:36.040 align:middle line:90% 00:52:36.040 --> 00:52:39.580 align:middle line:90% So now let me just summarize. 00:52:39.580 --> 00:52:55.360 align:middle line:90% 00:52:55.360 --> 00:53:15.900 align:middle line:84% So in summary, the sequences x sub n sub k of j-- 00:53:15.900 --> 00:53:19.620 align:middle line:84% so now this is even a subsequence of x sub n sub k 00:53:19.620 --> 00:53:22.440 align:middle line:90% and c sub n sub k sub j. 00:53:22.440 --> 00:53:27.220 align:middle line:90% 00:53:27.220 --> 00:53:36.060 align:middle line:84% These are subsequences of the original sequences x sub 00:53:36.060 --> 00:53:39.751 align:middle line:90% n and c sub n. 00:53:39.751 --> 00:53:51.330 align:middle line:90% 00:53:51.330 --> 00:54:00.860 align:middle line:84% And this x and c and a, b, such that the limit as j 00:54:00.860 --> 00:54:08.870 align:middle line:84% goes to infinity of x sub n sub j equals x, and limit as j 00:54:08.870 --> 00:54:19.000 align:middle line:84% goes to infinity of z sub n sub kj equals c. 00:54:19.000 --> 00:54:25.340 align:middle line:90% 00:54:25.340 --> 00:54:27.260 align:middle line:90% Now, I claim x equals c. 00:54:27.260 --> 00:54:37.530 align:middle line:90% 00:54:37.530 --> 00:54:45.830 align:middle line:84% Now, 0 is certainly bigger than or equal to x sub n sub k sub 00:54:45.830 --> 00:54:50.660 align:middle line:84% j minus c sub n sub k sub j, which is less 00:54:50.660 --> 00:54:54.050 align:middle line:84% than or equal to by how we've defined 00:54:54.050 --> 00:54:59.600 align:middle line:84% these sequences is less than or equal to 1 over n sub k sub j. 00:54:59.600 --> 00:55:06.120 align:middle line:90% 00:55:06.120 --> 00:55:13.160 align:middle line:84% So first off, by the definition of subsequences, 00:55:13.160 --> 00:55:16.430 align:middle line:84% j is always bigger than or equal to n sub k-- 00:55:16.430 --> 00:55:18.980 align:middle line:84% n sub k sub j is always bigger than or equal to j. 00:55:18.980 --> 00:55:22.730 align:middle line:84% So this is always less than or equal to 1 over j. 00:55:22.730 --> 00:55:26.450 align:middle line:84% This is just a consequence of how some sequences are defined. 00:55:26.450 --> 00:55:27.950 align:middle line:84% You always have to move to the right 00:55:27.950 --> 00:55:31.910 align:middle line:84% after you've made a choice of entry in the sequence 00:55:31.910 --> 00:55:35.540 align:middle line:84% that you're going to take as an element of your sequence. 00:55:35.540 --> 00:55:41.250 align:middle line:84% So 0 is bounded, bounds this from below, 00:55:41.250 --> 00:55:44.940 align:middle line:84% and is bounded above by 1 over j, which converges to 0. 00:55:44.940 --> 00:55:49.810 align:middle line:84% So the middle thing has to converge to 0 by the squeeze 00:55:49.810 --> 00:55:50.310 align:middle line:90% theorem. 00:55:50.310 --> 00:56:05.110 align:middle line:90% 00:56:05.110 --> 00:56:08.320 align:middle line:84% But these two things, that converges to x, this 00:56:08.320 --> 00:56:11.290 align:middle line:90% converges to c. 00:56:11.290 --> 00:56:15.310 align:middle line:84% And so the absolute value of x minus c equals 0, 00:56:15.310 --> 00:56:17.890 align:middle line:90% and therefore x equals c. 00:56:17.890 --> 00:56:29.753 align:middle line:90% 00:56:29.753 --> 00:56:30.920 align:middle line:90% Now, we're almost home free. 00:56:30.920 --> 00:56:40.440 align:middle line:90% 00:56:40.440 --> 00:56:47.040 align:middle line:84% This limit as j goes to infinity of x sub n sub k sub j equals 00:56:47.040 --> 00:56:56.200 align:middle line:84% c, and limit as j goes to infinity c sub n sub k 00:56:56.200 --> 00:56:58.705 align:middle line:90% sub j equals c. 00:56:58.705 --> 00:57:11.270 align:middle line:90% 00:57:11.270 --> 00:57:13.430 align:middle line:90% Now, where's the kicker? 00:57:13.430 --> 00:57:17.840 align:middle line:90% 00:57:17.840 --> 00:57:19.490 align:middle line:90% This. 00:57:19.490 --> 00:57:22.700 align:middle line:84% So these are subsequences of x sub n and c sub n. 00:57:22.700 --> 00:57:27.840 align:middle line:90% 00:57:27.840 --> 00:57:31.800 align:middle line:84% But each of those sequences is converging to a number c. 00:57:31.800 --> 00:57:37.530 align:middle line:84% So by the theorem about continuity, 00:57:37.530 --> 00:57:43.410 align:middle line:84% since f is continuous at c, this implies 00:57:43.410 --> 00:57:49.670 align:middle line:84% that 0 equals the limit as j goes to infinity 00:57:49.670 --> 00:57:53.120 align:middle line:90% of f of x sub n sub-- 00:57:53.120 --> 00:57:57.350 align:middle line:90% 00:57:57.350 --> 00:58:01.160 align:middle line:84% write it this way. f of c minus f of c 00:58:01.160 --> 00:58:17.920 align:middle line:84% equals limit j goes to infinity of f of n's of c sub n sub k 00:58:17.920 --> 00:58:21.797 align:middle line:84% sub j, which is bigger than or equal to epsilon 0. 00:58:21.797 --> 00:58:23.380 align:middle line:84% All of these are supposed to be bigger 00:58:23.380 --> 00:58:26.560 align:middle line:84% than or equal to some fixed epsilon 0, which is positive. 00:58:26.560 --> 00:58:29.650 align:middle line:90% 00:58:29.650 --> 00:58:31.330 align:middle line:84% But we've just shown that's equal to 0. 00:58:31.330 --> 00:58:34.700 align:middle line:90% 00:58:34.700 --> 00:58:36.302 align:middle line:90% And that's the contradiction. 00:58:36.302 --> 00:58:42.700 align:middle line:90% 00:58:42.700 --> 00:58:50.426 align:middle line:84% Thus, f is uniformly continuous on a, b. 00:58:50.426 --> 00:58:55.570 align:middle line:90% 00:58:55.570 --> 00:59:01.130 align:middle line:84% So that concludes what we'll say about continuity. 00:59:01.130 --> 00:59:03.590 align:middle line:84% And as you'll note kind of going forward, 00:59:03.590 --> 00:59:05.210 align:middle line:84% we're going to cover topics faster. 00:59:05.210 --> 00:59:08.270 align:middle line:84% That's just because we have more machinery to work with. 00:59:08.270 --> 00:59:14.820 align:middle line:84% At the start it was very slow going just because one, 00:59:14.820 --> 00:59:17.200 align:middle line:84% it's the start of your first proof based class. 00:59:17.200 --> 00:59:19.445 align:middle line:84% So I'm going to go slow; and two, 00:59:19.445 --> 00:59:20.820 align:middle line:84% we didn't have much to work with. 00:59:20.820 --> 00:59:24.180 align:middle line:84% We just had the real numbers are in 00:59:24.180 --> 00:59:26.350 align:middle line:84% ordered field with the least upper bound property. 00:59:26.350 --> 00:59:31.260 align:middle line:84% But now we've covered sequences, continuity we know more, 00:59:31.260 --> 00:59:37.455 align:middle line:84% and it'll be easier to prove new things and also 00:59:37.455 --> 00:59:41.580 align:middle line:84% hopefully familiarity with proofs 00:59:41.580 --> 00:59:45.810 align:middle line:84% is getting better so that my arguments can 00:59:45.810 --> 00:59:51.750 align:middle line:84% be a little bit shorter and not so involved. 00:59:51.750 --> 00:59:55.850 align:middle line:84% So we're moving on to something now. 00:59:55.850 --> 00:59:58.460 align:middle line:84% This is supposed to be about calculus. 00:59:58.460 --> 01:00:02.060 align:middle line:84% We've covered limits, continuity. 01:00:02.060 --> 01:00:07.870 align:middle line:84% The next big topic has to be differentiability. 01:00:07.870 --> 01:00:19.560 align:middle line:84% So this is a new chapter about the derivative. 01:00:19.560 --> 01:00:22.530 align:middle line:90% 01:00:22.530 --> 01:00:29.410 align:middle line:84% So let me first give you the precise definition. 01:00:29.410 --> 01:00:48.180 align:middle line:84% So first off, I'll be saying, let I be an interval, 01:00:48.180 --> 01:00:54.310 align:middle line:84% f goes from I to R, and c an element of I. 01:00:54.310 --> 01:00:55.900 align:middle line:84% Now, when I say an interval, I don't 01:00:55.900 --> 01:00:58.690 align:middle line:84% mean it has to be closed and bounded like ones 01:00:58.690 --> 01:00:59.830 align:middle line:90% we've been working on. 01:00:59.830 --> 01:01:01.945 align:middle line:90% I don't mean it has to be open. 01:01:01.945 --> 01:01:05.350 align:middle line:84% It doesn't have to be a bounded interval like from 0 to 1. 01:01:05.350 --> 01:01:07.420 align:middle line:90% It could be 0 to infinity. 01:01:07.420 --> 01:01:10.780 align:middle line:84% It can include the endpoints, except, of course, infinity. 01:01:10.780 --> 01:01:12.430 align:middle line:84% But I think what I mean and I don't 01:01:12.430 --> 01:01:15.670 align:middle line:84% have to write all this out for what exactly an interval is. 01:01:15.670 --> 01:01:20.320 align:middle line:84% So let I be an interval and f be a function from I to R and c 01:01:20.320 --> 01:01:23.700 align:middle line:90% a point of I. 01:01:23.700 --> 01:01:30.047 align:middle line:84% So first off, convince yourselves of this, 01:01:30.047 --> 01:01:31.630 align:middle line:84% but I think it should be pretty clear. 01:01:31.630 --> 01:01:41.090 align:middle line:90% 01:01:41.090 --> 01:01:44.810 align:middle line:84% For an interval, every point in the interval 01:01:44.810 --> 01:01:47.850 align:middle line:84% is a cluster point of the interval. 01:01:47.850 --> 01:01:53.700 align:middle line:84% This is not too difficult to prove. 01:01:53.700 --> 01:01:55.470 align:middle line:84% And I will not require you to prove it, 01:01:55.470 --> 01:01:56.970 align:middle line:84% but I'm just saying this because I'm 01:01:56.970 --> 01:01:58.350 align:middle line:90% about to define a limit at c. 01:01:58.350 --> 01:02:09.650 align:middle line:84% So it had better be a cluster point of a set 01:02:09.650 --> 01:02:11.390 align:middle line:84% that this limit is defined using. 01:02:11.390 --> 01:02:17.060 align:middle line:90% So of I, take away c. 01:02:17.060 --> 01:02:20.820 align:middle line:90% 01:02:20.820 --> 01:02:22.320 align:middle line:84% And if you don't like the statement, 01:02:22.320 --> 01:02:26.010 align:middle line:84% just forget I even wrote it and we'll continue on. 01:02:26.010 --> 01:02:29.314 align:middle line:90% 01:02:29.314 --> 01:02:39.360 align:middle line:84% You say f is differentiable at c-- 01:02:39.360 --> 01:02:46.200 align:middle line:84% so this is the new bit of terminology. 01:02:46.200 --> 01:03:01.510 align:middle line:84% If the limit x goes to c, f of x minus f of c over x minus c 01:03:01.510 --> 01:03:03.910 align:middle line:90% exists. 01:03:03.910 --> 01:03:07.450 align:middle line:84% If this limit exists, we say the functions differentiable at c, 01:03:07.450 --> 01:03:19.490 align:middle line:84% and we use a notation f prime f apostrophe c for this limit. 01:03:19.490 --> 01:03:39.650 align:middle line:90% 01:03:39.650 --> 01:03:48.710 align:middle line:84% If f is differentiable at every point of the interval I, 01:03:48.710 --> 01:03:51.870 align:middle line:84% we'd get a new function, this function 01:03:51.870 --> 01:04:15.670 align:middle line:84% f prime of c, which we denote the function that's 01:04:15.670 --> 01:04:18.100 align:middle line:84% a derivative of f as f prime or dfdx. 01:04:18.100 --> 01:04:24.550 align:middle line:90% 01:04:24.550 --> 01:04:27.370 align:middle line:84% So the simplest guys that are differential are, of course, 01:04:27.370 --> 01:04:28.090 align:middle line:90% polynomials. 01:04:28.090 --> 01:04:30.340 align:middle line:84% I think this is one of the first things you learn back 01:04:30.340 --> 01:04:32.260 align:middle line:90% in calculus. 01:04:32.260 --> 01:04:37.530 align:middle line:84% And I'm not going to go through what 01:04:37.530 --> 01:04:38.970 align:middle line:90% this is supposed to represent. 01:04:38.970 --> 01:04:41.370 align:middle line:84% You know this from calculus, the best way 01:04:41.370 --> 01:04:43.620 align:middle line:84% to say what this number represents 01:04:43.620 --> 01:04:46.110 align:middle line:84% is the instantaneous rate of change of the function 01:04:46.110 --> 01:04:47.340 align:middle line:90% f at this point. 01:04:47.340 --> 01:04:52.980 align:middle line:84% How is f changing at that point in the sense of increasing, 01:04:52.980 --> 01:04:54.150 align:middle line:90% decreasing, and so on? 01:04:54.150 --> 01:04:57.340 align:middle line:90% 01:04:57.340 --> 01:04:59.320 align:middle line:84% Or if you like, you interpret this 01:04:59.320 --> 01:05:01.720 align:middle line:84% as being the slope of the line that 01:05:01.720 --> 01:05:04.810 align:middle line:84% is perfectly tangent to the curve 01:05:04.810 --> 01:05:06.550 align:middle line:90% to the graph of the function f. 01:05:06.550 --> 01:05:11.800 align:middle line:84% But what does that mean to be tangent to the graph 01:05:11.800 --> 01:05:13.220 align:middle line:90% of a function? 01:05:13.220 --> 01:05:16.010 align:middle line:84% So that takes a bit of explaining. 01:05:16.010 --> 01:05:21.710 align:middle line:84% So I'm not going to get into what the actual interpretation 01:05:21.710 --> 01:05:22.993 align:middle line:90% of this number is. 01:05:22.993 --> 01:05:25.160 align:middle line:84% I think you have a pretty good idea of what that is. 01:05:25.160 --> 01:05:29.950 align:middle line:84% We're just going to start proving some properties of it. 01:05:29.950 --> 01:05:32.790 align:middle line:84% So the simplest example of a function 01:05:32.790 --> 01:05:40.830 align:middle line:84% which is differentiable is that of a monomial. 01:05:40.830 --> 01:05:50.925 align:middle line:90% So f of x equals x to the n. 01:05:50.925 --> 01:05:53.580 align:middle line:84% Let's put even a number in there. 01:05:53.580 --> 01:05:57.010 align:middle line:90% So let's write it this way. 01:05:57.010 --> 01:06:07.270 align:middle line:84% Let f have been R in the natural number 0, 01:06:07.270 --> 01:06:16.340 align:middle line:84% then function f of x equals alpha times x to the n 01:06:16.340 --> 01:06:17.470 align:middle line:90% is differentiable. 01:06:17.470 --> 01:06:22.176 align:middle line:90% 01:06:22.176 --> 01:06:24.090 align:middle line:84% So I should have said here as well 01:06:24.090 --> 01:06:26.190 align:middle line:84% if a function is differentiable at every point 01:06:26.190 --> 01:06:29.790 align:middle line:84% where it's defined, or if a function is 01:06:29.790 --> 01:06:32.550 align:middle line:84% differentiable at every point in its domain I, 01:06:32.550 --> 01:06:35.220 align:middle line:84% then we just say the function is differentiable. 01:06:35.220 --> 01:06:45.510 align:middle line:84% So then this monomial is differential and for all 01:06:45.510 --> 01:06:54.225 align:middle line:84% c f prime of c equals n times alpha x to the n minus 1, c. 01:06:54.225 --> 01:07:02.790 align:middle line:90% 01:07:02.790 --> 01:07:06.120 align:middle line:84% So let's give a proof of that, and we'll 01:07:06.120 --> 01:07:10.995 align:middle line:90% use this kind of simple formula. 01:07:10.995 --> 01:07:15.710 align:middle line:90% 01:07:15.710 --> 01:07:25.770 align:middle line:84% So let's take a look at x minus c times 01:07:25.770 --> 01:07:33.680 align:middle line:84% this kind of weird looking sum to x, n minus 1 minus j, 01:07:33.680 --> 01:07:35.460 align:middle line:90% c to the j. 01:07:35.460 --> 01:07:39.350 align:middle line:90% 01:07:39.350 --> 01:07:44.930 align:middle line:84% So multiply the x through, I get sum from j 01:07:44.930 --> 01:07:54.140 align:middle line:84% equals 0 to n minus 1, x to the n minus j, c to the j minus-- 01:07:54.140 --> 01:07:56.990 align:middle line:84% and I'm not going to use j again in the sum when 01:07:56.990 --> 01:07:58.790 align:middle line:90% I carry this c through. 01:07:58.790 --> 01:08:01.250 align:middle line:90% I'm going to use l-- 01:08:01.250 --> 01:08:11.000 align:middle line:84% sum from l equals 0 to n minus 1, x to the n minus-- 01:08:11.000 --> 01:08:21.479 align:middle line:84% I'll put it this way-- j plus 1, c to the j plus 1, l, 01:08:21.479 --> 01:08:23.850 align:middle line:84% and now I'm going to make a change of variables 01:08:23.850 --> 01:08:28.410 align:middle line:84% in this second sum and let j equals l plus 1. 01:08:28.410 --> 01:08:32.910 align:middle line:90% 01:08:32.910 --> 01:08:36.720 align:middle line:84% So in that case, when l equals 0, j starts at 1. 01:08:36.720 --> 01:08:41.100 align:middle line:84% When l hits n minus 1, j equals n. 01:08:41.100 --> 01:08:46.410 align:middle line:84% So this equals the sum from j equals 0, n minus 1, x, 01:08:46.410 --> 01:08:57.400 align:middle line:84% n minus j, c to the j minus sum j equals 1 to n, x, n minus j, 01:08:57.400 --> 01:08:59.750 align:middle line:90% c to the j. 01:08:59.750 --> 01:09:03.670 align:middle line:84% So it's the same terms in the sum, 01:09:03.670 --> 01:09:05.590 align:middle line:84% except starting at different places. 01:09:05.590 --> 01:09:08.069 align:middle line:84% So what doesn't get killed here is 01:09:08.069 --> 01:09:12.779 align:middle line:84% the j equals 0 term because this one starts at j equals 1. 01:09:12.779 --> 01:09:14.279 align:middle line:90% This one ends at n. 01:09:14.279 --> 01:09:16.390 align:middle line:84% And so it kills the n minus 1 term. 01:09:16.390 --> 01:09:18.300 align:middle line:84% So the only thing that survives from this sum 01:09:18.300 --> 01:09:24.000 align:middle line:84% is the j equals 0 term, x to the n minus 0, c to the 0 minus. 01:09:24.000 --> 01:09:27.305 align:middle line:90% 01:09:27.305 --> 01:09:32.490 align:middle line:84% And now from here, everything gets killed except the n terms. 01:09:32.490 --> 01:09:36.120 align:middle line:84% So x to the n minus n, c to the n 01:09:36.120 --> 01:09:39.580 align:middle line:84% equals x to the n minus c to the n. 01:09:39.580 --> 01:09:46.320 align:middle line:84% So this is therefore equal to this. 01:09:46.320 --> 01:09:50.130 align:middle line:90% 01:09:50.130 --> 01:09:52.450 align:middle line:84% So we use this to compute this limit. 01:09:52.450 --> 01:09:56.730 align:middle line:84% So then f prime of c is equal to the limit 01:09:56.730 --> 01:10:02.850 align:middle line:84% as x goes to c of x to the n alpha 01:10:02.850 --> 01:10:07.060 align:middle line:84% times x to the n times alpha times c to the n, 01:10:07.060 --> 01:10:08.370 align:middle line:90% where x minus c-- 01:10:08.370 --> 01:10:09.760 align:middle line:90% now, alpha is just a number. 01:10:09.760 --> 01:10:11.450 align:middle line:90% It can pop out-- 01:10:11.450 --> 01:10:15.560 align:middle line:84% x to the n minus c to the n divided by x minus c 01:10:15.560 --> 01:10:17.990 align:middle line:84% is this thing divided by x minus c. 01:10:17.990 --> 01:10:23.840 align:middle line:84% So this is equal to alpha times the limit as x goes to c of sum 01:10:23.840 --> 01:10:33.050 align:middle line:84% from j equals 0 to n minus 1, x to the n minus 1 minus j, c 01:10:33.050 --> 01:10:35.540 align:middle line:90% to the j. 01:10:35.540 --> 01:10:38.960 align:middle line:84% Now, this is just a polynomial in x. 01:10:38.960 --> 01:10:40.860 align:middle line:84% Polynomials are continuous functions. 01:10:40.860 --> 01:10:44.990 align:middle line:84% So the limit as x goes to c is just to plug in c to x. 01:10:44.990 --> 01:10:48.530 align:middle line:84% And so this equals alpha times sum from j 01:10:48.530 --> 01:10:55.400 align:middle line:84% equals 0 to n minus 1, c to the n minus 1 minus j times 01:10:55.400 --> 01:10:57.230 align:middle line:90% j equals-- 01:10:57.230 --> 01:10:59.900 align:middle line:84% so c to the n minus 1 minus j times c to the j 01:10:59.900 --> 01:11:02.210 align:middle line:90% gives me c to the n minus 1-- 01:11:02.210 --> 01:11:03.980 align:middle line:90% I'm summing in j. 01:11:03.980 --> 01:11:06.380 align:middle line:84% There's no j here, so that just pops out-- 01:11:06.380 --> 01:11:11.000 align:middle line:90% 01:11:11.000 --> 01:11:16.480 align:middle line:84% times the sum j equals 0, n minus 1. 01:11:16.480 --> 01:11:19.510 align:middle line:90% And there's n of the times 1. 01:11:19.510 --> 01:11:20.830 align:middle line:90% So this is just n. 01:11:20.830 --> 01:11:24.275 align:middle line:90% 01:11:24.275 --> 01:11:25.150 align:middle line:90% And that's the proof. 01:11:25.150 --> 01:11:27.960 align:middle line:90% 01:11:27.960 --> 01:11:32.010 align:middle line:84% So I'm sure you saw that at some point. 01:11:32.010 --> 01:11:35.850 align:middle line:84% Next time we'll give an example of a function which is not 01:11:35.850 --> 01:11:37.440 align:middle line:90% differentiable out of a point. 01:11:37.440 --> 01:11:39.660 align:middle line:84% Of course, I think you probably already know 01:11:39.660 --> 01:11:42.780 align:middle line:84% this, f of x equals the absolute value of x. 01:11:42.780 --> 01:11:45.270 align:middle line:84% If there's time at some point, I'll 01:11:45.270 --> 01:11:50.570 align:middle line:84% go over an example of a function, which 01:11:50.570 --> 01:11:52.780 align:middle line:84% was very surprising to people at one point. 01:11:52.780 --> 01:11:54.350 align:middle line:84% So now that we have differentiability 01:11:54.350 --> 01:12:01.790 align:middle line:84% and continuity on the board, for a long time since dead people 01:12:01.790 --> 01:12:06.970 align:middle line:84% believed that if you have a continuous function, 01:12:06.970 --> 01:12:08.550 align:middle line:84% any continuous function, then there 01:12:08.550 --> 01:12:13.170 align:middle line:84% has to be at least one point where it's differentiable. 01:12:13.170 --> 01:12:16.430 align:middle line:84% There must be at least one point where it's differentiable. 01:12:16.430 --> 01:12:18.560 align:middle line:84% And people tried to prove that for a long time 01:12:18.560 --> 01:12:20.870 align:middle line:84% and they couldn't because it's false. 01:12:20.870 --> 01:12:23.810 align:middle line:84% Weierstrass, who was, again kind of the godfather of everything 01:12:23.810 --> 01:12:28.700 align:middle line:84% we're doing, came up with a whole class of examples 01:12:28.700 --> 01:12:31.520 align:middle line:84% of functions that are continuous, 01:12:31.520 --> 01:12:34.470 align:middle line:90% but are differentiable nowhere. 01:12:34.470 --> 01:12:38.480 align:middle line:84% They're not differentiable at a single point. 01:12:38.480 --> 01:12:45.150 align:middle line:84% And perhaps depending on time, will go over this example. 01:12:45.150 --> 01:12:48.110 align:middle line:90% And think I'll stop there. 01:12:48.110 --> 01:12:55.000 align:middle line:90%