[{"Name":"Uniform Continuity from the Definition","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Uniform Continuity, Definition","Duration":"2m 53s","ChapterTopicVideoID":29543,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.900","Text":"A new topic, Uniform Continuity."},{"Start":"00:03.900 ","End":"00:07.020","Text":"Let\u0027s say we\u0027re given a nonempty set I,"},{"Start":"00:07.020 ","End":"00:09.090","Text":"and usually, that\u0027s an interval."},{"Start":"00:09.090 ","End":"00:12.405","Text":"We\u0027re given a function f defined on"},{"Start":"00:12.405 ","End":"00:17.390","Text":"I. I\u0027m going to remind you first what regular continuity is."},{"Start":"00:17.390 ","End":"00:20.450","Text":"F is called continuous at a point x_naught."},{"Start":"00:20.450 ","End":"00:23.550","Text":"If for all Epsilon there exists Delta,"},{"Start":"00:23.550 ","End":"00:29.389","Text":"such that if x is close to x_naught within a distance of Delta,"},{"Start":"00:29.389 ","End":"00:34.415","Text":"then f(x) is close to f(x_naught) within a distance of Epsilon."},{"Start":"00:34.415 ","End":"00:36.170","Text":"There\u0027s a picture here."},{"Start":"00:36.170 ","End":"00:40.610","Text":"Delta is horizontal distance."},{"Start":"00:40.610 ","End":"00:45.145","Text":"Epsilon denotes vertical distance."},{"Start":"00:45.145 ","End":"00:48.020","Text":"That\u0027s continuity at a point x_naught."},{"Start":"00:48.020 ","End":"00:52.010","Text":"In general, f is called continuous on the set I,"},{"Start":"00:52.010 ","End":"00:55.825","Text":"if it\u0027s continuous at each x_naught in I."},{"Start":"00:55.825 ","End":"00:58.155","Text":"Note that in this definition,"},{"Start":"00:58.155 ","End":"01:01.160","Text":"Delta depends on Epsilon and on"},{"Start":"01:01.160 ","End":"01:06.245","Text":"x_naught which brings us to the definition of uniform continuity."},{"Start":"01:06.245 ","End":"01:09.610","Text":"F is called uniformly continuous from the set I."},{"Start":"01:09.610 ","End":"01:14.690","Text":"If for all Epsilon there exists Delta not depending on any x_naught,"},{"Start":"01:14.690 ","End":"01:16.100","Text":"such that for any x,"},{"Start":"01:16.100 ","End":"01:17.570","Text":"y in the interval,"},{"Start":"01:17.570 ","End":"01:20.419","Text":"if x and y are close to within Delta,"},{"Start":"01:20.419 ","End":"01:24.385","Text":"then f(x) and f(y) are close to within Epsilon."},{"Start":"01:24.385 ","End":"01:27.620","Text":"Here\u0027s an illustration I found on the Internet."},{"Start":"01:27.620 ","End":"01:30.980","Text":"It stresses the same Delta works for"},{"Start":"01:30.980 ","End":"01:35.240","Text":"a given Epsilon everywhere on the curve regardless of x."},{"Start":"01:35.240 ","End":"01:38.305","Text":"It shows you in 3 different positions."},{"Start":"01:38.305 ","End":"01:41.010","Text":"This is Delta, this is the Epsilon,"},{"Start":"01:41.010 ","End":"01:43.005","Text":"this is Delta same Epsilon,"},{"Start":"01:43.005 ","End":"01:45.565","Text":"this is Delta same Epsilon."},{"Start":"01:45.565 ","End":"01:49.085","Text":"Now reversing the logic or negating it,"},{"Start":"01:49.085 ","End":"01:53.870","Text":"we can say that f is not uniformly continuous on the set I,"},{"Start":"01:53.870 ","End":"01:57.275","Text":"and negating the logic here if there exists Epsilon,"},{"Start":"01:57.275 ","End":"01:58.625","Text":"which doesn\u0027t have a Delta,"},{"Start":"01:58.625 ","End":"02:01.130","Text":"meaning that for every Delta there are x,"},{"Start":"02:01.130 ","End":"02:03.950","Text":"y which don\u0027t satisfy the condition,"},{"Start":"02:03.950 ","End":"02:06.905","Text":"meaning that x minus y is less than Delta."},{"Start":"02:06.905 ","End":"02:10.670","Text":"Nevertheless, f(x) minus f(y) is not less than Epsilon,"},{"Start":"02:10.670 ","End":"02:12.499","Text":"meaning bigger or equal to Epsilon."},{"Start":"02:12.499 ","End":"02:16.085","Text":"We have an example of a non-continuous function."},{"Start":"02:16.085 ","End":"02:20.193","Text":"It\u0027s continuous but not uniformly continuous,"},{"Start":"02:20.193 ","End":"02:22.025","Text":"f(x) equals 1 over x."},{"Start":"02:22.025 ","End":"02:24.530","Text":"In fact, this will be one of the exercises to"},{"Start":"02:24.530 ","End":"02:27.565","Text":"prove that this is not uniformly continuous."},{"Start":"02:27.565 ","End":"02:31.160","Text":"You can see that if we\u0027re given a Delta here,"},{"Start":"02:31.160 ","End":"02:32.615","Text":"this Epsilon will do,"},{"Start":"02:32.615 ","End":"02:35.750","Text":"but if we want the same Epsilon to be good,"},{"Start":"02:35.750 ","End":"02:37.655","Text":"further to the left,"},{"Start":"02:37.655 ","End":"02:40.040","Text":"we have to make Delta smaller."},{"Start":"02:40.040 ","End":"02:44.270","Text":"Uniform continuity, we can have the same Epsilon Delta."},{"Start":"02:44.270 ","End":"02:46.955","Text":"Delta wouldn\u0027t shrink like this."},{"Start":"02:46.955 ","End":"02:49.835","Text":"That\u0027s all for the introductory clip."},{"Start":"02:49.835 ","End":"02:53.370","Text":"The rest is in the exercises that follow."}],"ID":31143},{"Watched":false,"Name":"Exercise 1","Duration":"1m 56s","ChapterTopicVideoID":29529,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.570","Text":"In this exercise, we\u0027re asked to prove that"},{"Start":"00:03.570 ","End":"00:10.860","Text":"the constant function where f(x) is always 7 is uniformly continuous."},{"Start":"00:10.860 ","End":"00:13.230","Text":"Using the Epsilon Delta method,"},{"Start":"00:13.230 ","End":"00:15.015","Text":"we have to prove the following."},{"Start":"00:15.015 ","End":"00:17.865","Text":"For any given Epsilon bigger than 0,"},{"Start":"00:17.865 ","End":"00:22.645","Text":"there exists a Delta bigger than 0 corresponding to this Epsilon,"},{"Start":"00:22.645 ","End":"00:30.795","Text":"such that whenever x and y as such that their distance is less than Delta,"},{"Start":"00:30.795 ","End":"00:35.790","Text":"then the distance between f(x) and f(y) is less than Epsilon."},{"Start":"00:35.790 ","End":"00:41.940","Text":"In more shorthand notation using quantifiers for all is this upside down, A,"},{"Start":"00:41.940 ","End":"00:46.640","Text":"there exists is this mirror flipped E,"},{"Start":"00:46.640 ","End":"00:53.290","Text":"and if then can be replaced by an arrow."},{"Start":"00:53.290 ","End":"01:01.595","Text":"We start out by someone giving us Epsilon and we have to find the corresponding Delta,"},{"Start":"01:01.595 ","End":"01:04.820","Text":"which satisfies the above that if this is true,"},{"Start":"01:04.820 ","End":"01:06.380","Text":"then this is true."},{"Start":"01:06.380 ","End":"01:08.930","Text":"Now, if we just look at this expression,"},{"Start":"01:08.930 ","End":"01:12.655","Text":"f(x) minus f(y) is 0 always."},{"Start":"01:12.655 ","End":"01:14.540","Text":"This is always going to be true."},{"Start":"01:14.540 ","End":"01:17.075","Text":"So we can choose any Delta,"},{"Start":"01:17.075 ","End":"01:22.115","Text":"let\u0027s just say Delta equals 1 because we have to give a specific Delta."},{"Start":"01:22.115 ","End":"01:23.975","Text":"Now, to be precise,"},{"Start":"01:23.975 ","End":"01:29.495","Text":"what we have to show is this part that if x minus y is less than Delta,"},{"Start":"01:29.495 ","End":"01:34.240","Text":"then absolute value f(x) minus f(y) is less than Epsilon."},{"Start":"01:34.240 ","End":"01:39.215","Text":"Now, this is always true because this is equal to 0."},{"Start":"01:39.215 ","End":"01:45.545","Text":"We don\u0027t need this part even because this is always true and from logic,"},{"Start":"01:45.545 ","End":"01:50.389","Text":"if the second part of an implication is true,"},{"Start":"01:50.389 ","End":"01:53.525","Text":"then the implication is true."},{"Start":"01:53.525 ","End":"01:56.940","Text":"That concludes this exercise."}],"ID":31144},{"Watched":false,"Name":"Exercise 2","Duration":"1m 57s","ChapterTopicVideoID":29530,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"In this exercise, we have to prove that the function f defined"},{"Start":"00:04.500 ","End":"00:09.490","Text":"by f of x equals 2x plus 3 is uniformly continuous."},{"Start":"00:09.490 ","End":"00:12.420","Text":"What we have to prove using Epsilon,"},{"Start":"00:12.420 ","End":"00:14.280","Text":"Delta is the following."},{"Start":"00:14.280 ","End":"00:16.770","Text":"For all Epsilon bigger than 0,"},{"Start":"00:16.770 ","End":"00:20.295","Text":"there\u0027s a Delta bigger than 0 such that"},{"Start":"00:20.295 ","End":"00:24.869","Text":"if x minus y and absolute value is less than Delta,"},{"Start":"00:24.869 ","End":"00:29.610","Text":"then absolute value of f of x minus f of y is less than Epsilon."},{"Start":"00:29.610 ","End":"00:34.005","Text":"We are given Epsilon bigger than 0 arbitrary."},{"Start":"00:34.005 ","End":"00:42.065","Text":"Our task is to find a corresponding Delta bigger than 0, which satisfies this."},{"Start":"00:42.065 ","End":"00:45.425","Text":"Let\u0027s just work on this expression for a moment."},{"Start":"00:45.425 ","End":"00:54.200","Text":"Absolute value of f of x minus f of y is absolute value of 2x plus 3 minus 2y plus 3."},{"Start":"00:54.200 ","End":"01:01.985","Text":"It simplifies to twice absolute value of x minus y because this is less than Delta,"},{"Start":"01:01.985 ","End":"01:05.885","Text":"we\u0027ve got that this is less than 2 Delta."},{"Start":"01:05.885 ","End":"01:08.480","Text":"What we wanted here is Epsilon."},{"Start":"01:08.480 ","End":"01:14.910","Text":"The obvious thing to do is to let 2 Delta equal Epsilon."},{"Start":"01:14.910 ","End":"01:18.495","Text":"Choose Delta equals Epsilon over 2."},{"Start":"01:18.495 ","End":"01:19.710","Text":"You could choose smaller,"},{"Start":"01:19.710 ","End":"01:22.110","Text":"you could choose Epsilon over 3."},{"Start":"01:22.110 ","End":"01:26.620","Text":"Then still 2 Delta would be less than Epsilon."},{"Start":"01:26.960 ","End":"01:29.205","Text":"Anyway, in our case,"},{"Start":"01:29.205 ","End":"01:30.930","Text":"choosing Delta is Epsilon over 2,"},{"Start":"01:30.930 ","End":"01:33.510","Text":"we get f(x) minus f(y) less than 2 Delta,"},{"Start":"01:33.510 ","End":"01:35.085","Text":"which is Epsilon."},{"Start":"01:35.085 ","End":"01:37.155","Text":"That\u0027s basically it."},{"Start":"01:37.155 ","End":"01:40.550","Text":"Just to summarize, we\u0027re given Epsilon bigger than 0,"},{"Start":"01:40.550 ","End":"01:44.895","Text":"we choose Delta is Epsilon over 2."},{"Start":"01:44.895 ","End":"01:49.459","Text":"Then when x minus y absolute value is less than Delta,"},{"Start":"01:49.459 ","End":"01:54.650","Text":"then the absolute f(x) minus f(y) will be less than Epsilon as desired."},{"Start":"01:54.650 ","End":"01:57.810","Text":"That concludes this exercise."}],"ID":31145},{"Watched":false,"Name":"Exercise 3","Duration":"2m 28s","ChapterTopicVideoID":29531,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.450","Text":"In this exercise we\u0027re going to prove that"},{"Start":"00:03.450 ","End":"00:07.695","Text":"the square root function is uniformly continuous."},{"Start":"00:07.695 ","End":"00:10.530","Text":"The square root function is defined on"},{"Start":"00:10.530 ","End":"00:17.565","Text":"the half infinite interval from 0 to infinity including the 0."},{"Start":"00:17.565 ","End":"00:22.275","Text":"We\u0027re given a hint that for positive numbers,"},{"Start":"00:22.275 ","End":"00:28.080","Text":"absolute value of a minus b is less than or equal to absolute value of a plus b,"},{"Start":"00:28.080 ","End":"00:30.585","Text":"the difference is less than the sum."},{"Start":"00:30.585 ","End":"00:34.830","Text":"Our interval I is from 0 to infinity."},{"Start":"00:34.830 ","End":"00:39.200","Text":"What we have to show is that for all Epsilon bigger than 0"},{"Start":"00:39.200 ","End":"00:44.360","Text":"that are given we can find for it corresponding Delta bigger than 0,"},{"Start":"00:44.360 ","End":"00:49.669","Text":"such that on this interval if x is close to y within Delta,"},{"Start":"00:49.669 ","End":"00:54.170","Text":"then f(x) is close to f(y) within Epsilon."},{"Start":"00:54.170 ","End":"00:57.380","Text":"Let Epsilon be given,"},{"Start":"00:57.380 ","End":"01:01.830","Text":"we need to find the Delta which satisfies the above."},{"Start":"01:02.470 ","End":"01:06.230","Text":"The absolute value of f(x) minus f(y),"},{"Start":"01:06.230 ","End":"01:08.150","Text":"which is this expression here,"},{"Start":"01:08.150 ","End":"01:10.550","Text":"is equal to square root of x minus"},{"Start":"01:10.550 ","End":"01:13.805","Text":"square root of y because f is the square root function."},{"Start":"01:13.805 ","End":"01:15.679","Text":"Now using this hint,"},{"Start":"01:15.679 ","End":"01:20.165","Text":"with a and b replaced by root x and root y,"},{"Start":"01:20.165 ","End":"01:23.400","Text":"we get the following inequality."},{"Start":"01:23.400 ","End":"01:25.820","Text":"If this is equal to this,"},{"Start":"01:25.820 ","End":"01:27.830","Text":"which is less than or equal to this,"},{"Start":"01:27.830 ","End":"01:30.150","Text":"we get that this is this."},{"Start":"01:30.550 ","End":"01:34.205","Text":"Now what we can do is multiply these two,"},{"Start":"01:34.205 ","End":"01:36.695","Text":"one\u0027s in equality, one is in inequality."},{"Start":"01:36.695 ","End":"01:38.870","Text":"You can think of both of them as inequalities"},{"Start":"01:38.870 ","End":"01:41.480","Text":"less than or equal to and everything is positive,"},{"Start":"01:41.480 ","End":"01:42.890","Text":"so we can multiply."},{"Start":"01:42.890 ","End":"01:49.640","Text":"We get absolute value of f(x) minus f(y)^2 is less than or equal to this times this."},{"Start":"01:49.640 ","End":"01:52.895","Text":"Then using the difference of squares formula,"},{"Start":"01:52.895 ","End":"01:56.090","Text":"what we get is square root of x^2,"},{"Start":"01:56.090 ","End":"01:59.890","Text":"which is x minus square root of y^2 which is y."},{"Start":"01:59.890 ","End":"02:02.430","Text":"This is less than Delta,"},{"Start":"02:02.430 ","End":"02:07.865","Text":"so the obvious thing to do now is to define Delta equals Epsilon^2."},{"Start":"02:07.865 ","End":"02:08.930","Text":"Epsilon is given,"},{"Start":"02:08.930 ","End":"02:10.940","Text":"so we can do this."},{"Start":"02:10.940 ","End":"02:16.370","Text":"Then we get that this is less than Epsilon^2."},{"Start":"02:16.370 ","End":"02:18.635","Text":"Then taking the square root,"},{"Start":"02:18.635 ","End":"02:25.085","Text":"we have absolute value of f(x) minus f(y) less than Epsilon as required."},{"Start":"02:25.085 ","End":"02:28.530","Text":"That completes this exercise."}],"ID":31146},{"Watched":false,"Name":"Exercise 4","Duration":"2m 30s","ChapterTopicVideoID":29532,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:05.550","Text":"In this exercise we\u0027re going to prove that the function f(x) equals"},{"Start":"00:05.550 ","End":"00:11.775","Text":"the square root of absolute value of x plus 1 is uniformly continuous."},{"Start":"00:11.775 ","End":"00:14.730","Text":"Most of these problems start the same way."},{"Start":"00:14.730 ","End":"00:17.730","Text":"We have to prove that for all Epsilon there exists"},{"Start":"00:17.730 ","End":"00:22.770","Text":"Delta such that if x is close to y within Delta,"},{"Start":"00:22.770 ","End":"00:26.490","Text":"then f(x) is close to f(y) within Epsilon."},{"Start":"00:26.490 ","End":"00:28.705","Text":"We\u0027re given the Epsilon,"},{"Start":"00:28.705 ","End":"00:31.760","Text":"we have to supply the Delta."},{"Start":"00:31.760 ","End":"00:34.400","Text":"Let\u0027s do a computation."},{"Start":"00:34.400 ","End":"00:38.975","Text":"Often the way of starting is to simplify this expression."},{"Start":"00:38.975 ","End":"00:42.610","Text":"This is equal to from the definition, this."},{"Start":"00:42.610 ","End":"00:45.680","Text":"Now we use the conjugate,"},{"Start":"00:45.680 ","End":"00:49.385","Text":"and we have a square root plus or minus another square root."},{"Start":"00:49.385 ","End":"00:55.305","Text":"Multiply top and bottom by the same thing with a plus, the conjugate."},{"Start":"00:55.305 ","End":"00:57.920","Text":"Then you want to simplify,"},{"Start":"00:57.920 ","End":"01:00.065","Text":"remember the difference of squares rule,"},{"Start":"01:00.065 ","End":"01:04.025","Text":"a minus b times a plus b is a squared minus b squared."},{"Start":"01:04.025 ","End":"01:07.460","Text":"Well, the square root squared is just the thing without the square root,"},{"Start":"01:07.460 ","End":"01:09.415","Text":"and the same here,"},{"Start":"01:09.415 ","End":"01:12.245","Text":"and this is a denominator."},{"Start":"01:12.245 ","End":"01:15.110","Text":"Now the numerator can be simplified to"},{"Start":"01:15.110 ","End":"01:21.005","Text":"just the absolute value of absolute value of x minus absolute value of y."},{"Start":"01:21.005 ","End":"01:22.310","Text":"In the denominator,"},{"Start":"01:22.310 ","End":"01:26.440","Text":"note that each of these two terms is bigger or equal to 1."},{"Start":"01:26.440 ","End":"01:28.370","Text":"If you increase the denominator,"},{"Start":"01:28.370 ","End":"01:30.500","Text":"then you decrease the fraction."},{"Start":"01:30.500 ","End":"01:36.425","Text":"This comes out to be less than what you get if you put 1 and 1 here."},{"Start":"01:36.425 ","End":"01:39.680","Text":"Now there\u0027s a reverse triangle inequality."},{"Start":"01:39.680 ","End":"01:40.880","Text":"I wrote it in big,"},{"Start":"01:40.880 ","End":"01:42.890","Text":"this is an important one."},{"Start":"01:42.890 ","End":"01:45.440","Text":"Absolute value of difference is bigger or equal to"},{"Start":"01:45.440 ","End":"01:49.130","Text":"the absolute value of the difference of the absolute values."},{"Start":"01:49.130 ","End":"01:51.005","Text":"So using that here,"},{"Start":"01:51.005 ","End":"01:54.920","Text":"we get absolute value of x minus y over 2."},{"Start":"01:54.920 ","End":"01:58.880","Text":"We know that absolute value of x minus y is less than Delta,"},{"Start":"01:58.880 ","End":"02:01.765","Text":"so we\u0027ve got to the point Delta over 2."},{"Start":"02:01.765 ","End":"02:03.510","Text":"We wanted epsilon here,"},{"Start":"02:03.510 ","End":"02:07.760","Text":"so it\u0027s clear that we can define Delta to be 2 Epsilon,"},{"Start":"02:07.760 ","End":"02:11.105","Text":"and then Delta over 2 will be Epsilon."},{"Start":"02:11.105 ","End":"02:16.260","Text":"This thing will be less than Epsilon."},{"Start":"02:16.270 ","End":"02:18.470","Text":"This should be less than,"},{"Start":"02:18.470 ","End":"02:19.700","Text":"not less than or equal to,"},{"Start":"02:19.700 ","End":"02:20.880","Text":"and this is equal to."},{"Start":"02:20.880 ","End":"02:22.850","Text":"But altogether we get equals,"},{"Start":"02:22.850 ","End":"02:24.260","Text":"less than, or equal to less than."},{"Start":"02:24.260 ","End":"02:25.430","Text":"As long as we have a less than,"},{"Start":"02:25.430 ","End":"02:28.385","Text":"and it\u0027s less than Epsilon as required."},{"Start":"02:28.385 ","End":"02:31.470","Text":"That concludes this exercise."}],"ID":31147},{"Watched":false,"Name":"Exercise 5","Duration":"2m 51s","ChapterTopicVideoID":29533,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.110","Text":"In this exercise, we\u0027re going to show that the composition"},{"Start":"00:04.110 ","End":"00:08.685","Text":"of uniformly continuous functions is also uniformly continuous."},{"Start":"00:08.685 ","End":"00:14.145","Text":"Specifically, suppose f and g are uniformly continuous."},{"Start":"00:14.145 ","End":"00:16.520","Text":"Then f composition g,"},{"Start":"00:16.520 ","End":"00:20.010","Text":"we\u0027re going to prove is also uniformly continuous."},{"Start":"00:20.010 ","End":"00:26.730","Text":"Just to remind you, the composition f compose g of x means first apply g(x),"},{"Start":"00:26.730 ","End":"00:29.400","Text":"and then apply f to the result of that."},{"Start":"00:29.400 ","End":"00:33.665","Text":"Now here\u0027s the usual bit about what we have to prove that for all Epsilon,"},{"Start":"00:33.665 ","End":"00:39.184","Text":"there exists Delta such that if x and y are close within Delta,"},{"Start":"00:39.184 ","End":"00:46.070","Text":"then f compose g(x) minus f compose g(y) is close within Epsilon."},{"Start":"00:46.070 ","End":"00:47.720","Text":"I\u0027m referring to this line here."},{"Start":"00:47.720 ","End":"00:50.270","Text":"Writing f compose g and writing what it is."},{"Start":"00:50.270 ","End":"00:51.875","Text":"We\u0027re given Epsilon,"},{"Start":"00:51.875 ","End":"00:53.860","Text":"and we have to find Delta."},{"Start":"00:53.860 ","End":"01:00.110","Text":"Now, recall that we\u0027re given that both f and g are uniformly continuous."},{"Start":"01:00.110 ","End":"01:04.040","Text":"Because f is uniformly continuous."},{"Start":"01:04.040 ","End":"01:07.580","Text":"For this Epsilon, there exists,"},{"Start":"01:07.580 ","End":"01:10.010","Text":"using a letter other than Delta,"},{"Start":"01:10.010 ","End":"01:11.030","Text":"I need another letter,"},{"Start":"01:11.030 ","End":"01:12.170","Text":"I use the letter Eta,"},{"Start":"01:12.170 ","End":"01:14.750","Text":"there exists Eta bigger than 0,"},{"Start":"01:14.750 ","End":"01:18.695","Text":"such that if x minus y is less than Eta,"},{"Start":"01:18.695 ","End":"01:22.580","Text":"then f(x) minus f(y) is less than Epsilon;"},{"Start":"01:22.580 ","End":"01:24.760","Text":"an absolute value I didn\u0027t say."},{"Start":"01:24.760 ","End":"01:27.350","Text":"Eta is like Delta here."},{"Start":"01:27.350 ","End":"01:30.305","Text":"Now, g is also uniformly continuous,"},{"Start":"01:30.305 ","End":"01:31.820","Text":"and to interpret that,"},{"Start":"01:31.820 ","End":"01:35.660","Text":"we could say that for this Eta,"},{"Start":"01:35.660 ","End":"01:37.745","Text":"in the role of Epsilon,"},{"Start":"01:37.745 ","End":"01:42.812","Text":"there exists a Delta such that if x is close to y within Delta,"},{"Start":"01:42.812 ","End":"01:45.835","Text":"then g(x) is close to g(y) within Eta."},{"Start":"01:45.835 ","End":"01:48.065","Text":"Once Eta takes the role of Delta,"},{"Start":"01:48.065 ","End":"01:51.095","Text":"and the other time it takes the role of Epsilon."},{"Start":"01:51.095 ","End":"01:53.720","Text":"Now we\u0027re going to put these two things together."},{"Start":"01:53.720 ","End":"01:56.700","Text":"We got from Epsilon to Delta in two stages."},{"Start":"01:56.700 ","End":"01:58.545","Text":"For Epsilon, there is Eta,"},{"Start":"01:58.545 ","End":"02:00.945","Text":"and for Eta, there is Delta."},{"Start":"02:00.945 ","End":"02:04.130","Text":"I claim that this Delta is good for"},{"Start":"02:04.130 ","End":"02:08.760","Text":"our Epsilon in regards to the composition of f and g,"},{"Start":"02:08.760 ","End":"02:11.555","Text":"and the proof now is in one line."},{"Start":"02:11.555 ","End":"02:15.455","Text":"Because x is close to y within Delta,"},{"Start":"02:15.455 ","End":"02:18.970","Text":"g(x) is close to g(y) within Eta."},{"Start":"02:18.970 ","End":"02:22.475","Text":"I think of g(x) as another kind of x called an x\u0027,"},{"Start":"02:22.475 ","End":"02:25.400","Text":"and this we\u0027ll call it y\u0027."},{"Start":"02:25.400 ","End":"02:29.425","Text":"x\u0027 is close to y\u0027 within Eta."},{"Start":"02:29.425 ","End":"02:34.700","Text":"Then because of the continuity uniform of f,"},{"Start":"02:34.700 ","End":"02:40.280","Text":"we have that f(x\u0027) is close to f(y\u0027) within Epsilon,"},{"Start":"02:40.280 ","End":"02:44.195","Text":"but x\u0027 is g(x) and y\u0027 is g(y)."},{"Start":"02:44.195 ","End":"02:48.845","Text":"We have this, and this is exactly what we were trying to show up here."},{"Start":"02:48.845 ","End":"02:51.450","Text":"So we are done."}],"ID":31148},{"Watched":false,"Name":"Exercise 6 Part 1","Duration":"6m 4s","ChapterTopicVideoID":29534,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:04.470","Text":"In this exercise, a is less than c is less than b,"},{"Start":"00:04.470 ","End":"00:05.940","Text":"like in the picture,"},{"Start":"00:05.940 ","End":"00:08.280","Text":"and we have a function from the interval a,"},{"Start":"00:08.280 ","End":"00:10.470","Text":"b to the reals."},{"Start":"00:10.470 ","End":"00:13.080","Text":"Suppose that both the restrictions,"},{"Start":"00:13.080 ","End":"00:14.700","Text":"f restricted to a,"},{"Start":"00:14.700 ","End":"00:17.640","Text":"c and f restricted to c,"},{"Start":"00:17.640 ","End":"00:21.795","Text":"b that both of these are uniformly continuous."},{"Start":"00:21.795 ","End":"00:26.850","Text":"To rephrase that, you could say that f is uniformly continuous on a,"},{"Start":"00:26.850 ","End":"00:30.255","Text":"c and f is uniformly continuous on c, b."},{"Start":"00:30.255 ","End":"00:34.200","Text":"In one sense, there\u0027s only one function f and in another sense there are"},{"Start":"00:34.200 ","End":"00:40.185","Text":"3 functions because the domain is part of the definition of the function."},{"Start":"00:40.185 ","End":"00:44.540","Text":"We have to prove that f itself is uniformly"},{"Start":"00:44.540 ","End":"00:49.100","Text":"continuous and we have to do this in 2 different ways; 1,"},{"Start":"00:49.100 ","End":"00:53.345","Text":"using the definition of uniform continuity, and b,"},{"Start":"00:53.345 ","End":"00:58.370","Text":"using Cantor\u0027s theorem which I\u0027ll remind you of when we get to Part b."},{"Start":"00:58.370 ","End":"01:01.820","Text":"But let\u0027s now just start with Part a."},{"Start":"01:01.820 ","End":"01:05.720","Text":"Just to repeat, f is uniformly continuous on a,"},{"Start":"01:05.720 ","End":"01:08.240","Text":"c so there is Epsilon Delta,"},{"Start":"01:08.240 ","End":"01:15.320","Text":"but we\u0027ll call it Epsilon 1 and Delta 1. f is uniformly continuous on c,"},{"Start":"01:15.320 ","End":"01:20.830","Text":"b, so the same thing but we\u0027ll call it Epsilon 2 and Delta 2."},{"Start":"01:20.830 ","End":"01:25.740","Text":"This symbol by the way is shorthand for such that."},{"Start":"01:25.740 ","End":"01:30.215","Text":"I just wrote that, so you learned a new symbol in case you didn\u0027t know it already."},{"Start":"01:30.215 ","End":"01:33.460","Text":"Now what we have to prove is,"},{"Start":"01:33.460 ","End":"01:35.705","Text":"as usual for all Epsilon,"},{"Start":"01:35.705 ","End":"01:40.940","Text":"there exists Delta such that if x and y but in the interval a,"},{"Start":"01:40.940 ","End":"01:43.380","Text":"b are close within Delta,"},{"Start":"01:43.380 ","End":"01:46.530","Text":"then f(x) and f(y) are close within Epsilon."},{"Start":"01:46.530 ","End":"01:51.095","Text":"Epsilon is now given and we have to find the corresponding Delta."},{"Start":"01:51.095 ","End":"01:53.015","Text":"We\u0027re going to use these 2 facts,"},{"Start":"01:53.015 ","End":"01:55.120","Text":"the uniform continuity on a,"},{"Start":"01:55.120 ","End":"01:56.920","Text":"c and on c, b."},{"Start":"01:56.920 ","End":"02:02.565","Text":"We\u0027ll start with Epsilon 1 and Epsilon 2 and set both of them equal to Epsilon over 2."},{"Start":"02:02.565 ","End":"02:05.960","Text":"You\u0027ll soon see why I chose Epsilon over 2"},{"Start":"02:05.960 ","End":"02:08.780","Text":"but you could just ignore that for the time being,"},{"Start":"02:08.780 ","End":"02:10.595","Text":"just leave it as Epsilon Epsilon 2."},{"Start":"02:10.595 ","End":"02:13.905","Text":"When we have to, we\u0027ll define them."},{"Start":"02:13.905 ","End":"02:16.590","Text":"Delta 1, Delta 2 or as above,"},{"Start":"02:16.590 ","End":"02:18.210","Text":"Delta 1 corresponds to Epsilon 1,"},{"Start":"02:18.210 ","End":"02:21.015","Text":"Delta 2 corresponds to Epsilon 2."},{"Start":"02:21.015 ","End":"02:26.685","Text":"Now let\u0027s define Delta to be the minimum of Delta 1 and Delta 2."},{"Start":"02:26.685 ","End":"02:30.800","Text":"This is something that we do often so that if x is close to y within"},{"Start":"02:30.800 ","End":"02:35.645","Text":"Delta then its close within Delta 1 and it\u0027s close within Delta 2."},{"Start":"02:35.645 ","End":"02:39.080","Text":"Here I\u0027m basically repeating what I wrote here that we"},{"Start":"02:39.080 ","End":"02:42.530","Text":"have to show that if x and y are in the big interval a,"},{"Start":"02:42.530 ","End":"02:45.570","Text":"b, and they\u0027re close within Delta,"},{"Start":"02:45.570 ","End":"02:48.990","Text":"then f(x) and f(y) are close within Epsilon."},{"Start":"02:48.990 ","End":"02:51.875","Text":"There are 3 cases we\u0027re going to distinguish."},{"Start":"02:51.875 ","End":"02:56.045","Text":"It could be that x and y are both in the interval a,"},{"Start":"02:56.045 ","End":"03:00.350","Text":"c, could be that x and y are both in the interval c, b."},{"Start":"03:00.350 ","End":"03:02.810","Text":"These two are the easy cases."},{"Start":"03:02.810 ","End":"03:05.720","Text":"The third case is when x is in a,"},{"Start":"03:05.720 ","End":"03:08.195","Text":"c and y is c, b."},{"Start":"03:08.195 ","End":"03:09.995","Text":"Something I forgot to say,"},{"Start":"03:09.995 ","End":"03:12.830","Text":"we\u0027re assuming that x is less than y. I mean,"},{"Start":"03:12.830 ","End":"03:14.975","Text":"there\u0027s a total symmetry between x and y."},{"Start":"03:14.975 ","End":"03:16.655","Text":"If it\u0027s not the case,"},{"Start":"03:16.655 ","End":"03:18.890","Text":"then just change the labeling,"},{"Start":"03:18.890 ","End":"03:21.955","Text":"you label x as y and label y as x."},{"Start":"03:21.955 ","End":"03:24.510","Text":"Also it could be that x equals y."},{"Start":"03:24.510 ","End":"03:26.550","Text":"Of course, if x equals y,"},{"Start":"03:26.550 ","End":"03:30.045","Text":"then f(x) is equal to f(y)."},{"Start":"03:30.045 ","End":"03:32.265","Text":"This is 0 and this is 0,"},{"Start":"03:32.265 ","End":"03:33.665","Text":"and that also applies."},{"Start":"03:33.665 ","End":"03:35.180","Text":"That\u0027s getting technical."},{"Start":"03:35.180 ","End":"03:38.945","Text":"Now let\u0027s treat each of the cases separately."},{"Start":"03:38.945 ","End":"03:41.540","Text":"Let\u0027s start with case 1,"},{"Start":"03:41.540 ","End":"03:44.660","Text":"where x and y are both in the interval a,"},{"Start":"03:44.660 ","End":"03:48.950","Text":"c, then absolute value of x minus y is less than Delta,"},{"Start":"03:48.950 ","End":"03:51.230","Text":"so it\u0027s less than Delta 1."},{"Start":"03:51.230 ","End":"03:56.735","Text":"Then f(x) minus f(y) in absolute value is less than Epsilon 1."},{"Start":"03:56.735 ","End":"03:59.075","Text":"Anything less than Epsilon we could have chosen."},{"Start":"03:59.075 ","End":"04:01.895","Text":"Then f(x) minus f(y) is less than Epsilon."},{"Start":"04:01.895 ","End":"04:04.565","Text":"In case 3, you\u0027ll see why this."},{"Start":"04:04.565 ","End":"04:06.610","Text":"Case 2 is very similar."},{"Start":"04:06.610 ","End":"04:08.270","Text":"I just leave it here."},{"Start":"04:08.270 ","End":"04:11.495","Text":"It\u0027s exactly the same thing but with subscript 2."},{"Start":"04:11.495 ","End":"04:12.800","Text":"Now the third case,"},{"Start":"04:12.800 ","End":"04:13.850","Text":"where x is in a,"},{"Start":"04:13.850 ","End":"04:16.685","Text":"c and y is in c, b."},{"Start":"04:16.685 ","End":"04:17.900","Text":"On the one hand,"},{"Start":"04:17.900 ","End":"04:22.235","Text":"x minus c is less than Delta because"},{"Start":"04:22.235 ","End":"04:27.740","Text":"x minus c in absolute value is less than x minus y. I mean,"},{"Start":"04:27.740 ","End":"04:30.640","Text":"x is closer to c than it is to y."},{"Start":"04:30.640 ","End":"04:33.243","Text":"The distance is less than Delta 1,"},{"Start":"04:33.243 ","End":"04:37.940","Text":"and so f(x) is close to f(c) within Epsilon 1."},{"Start":"04:37.940 ","End":"04:40.380","Text":"Similarly for the subscript 2,"},{"Start":"04:40.380 ","End":"04:45.095","Text":"c is closer to y than x is to y."},{"Start":"04:45.095 ","End":"04:47.375","Text":"From the triangle inequality,"},{"Start":"04:47.375 ","End":"04:48.855","Text":"we have the following."},{"Start":"04:48.855 ","End":"04:50.340","Text":"Think of this is a,"},{"Start":"04:50.340 ","End":"04:52.105","Text":"think of this is b,"},{"Start":"04:52.105 ","End":"04:53.660","Text":"and this is a plus b."},{"Start":"04:53.660 ","End":"04:55.730","Text":"Absolute value of a plus b is less than or"},{"Start":"04:55.730 ","End":"04:58.775","Text":"equal to absolute value of a plus absolute value of b."},{"Start":"04:58.775 ","End":"05:02.480","Text":"The reason I did this is because we have an estimate for this,"},{"Start":"05:02.480 ","End":"05:04.640","Text":"this is less than or equal to Epsilon 1,"},{"Start":"05:04.640 ","End":"05:07.660","Text":"and this we know is less than or equal to Epsilon 2."},{"Start":"05:07.660 ","End":"05:13.535","Text":"This should be strict inequality."},{"Start":"05:13.535 ","End":"05:19.045","Text":"f(x) minus f(y) is strictly less than Epsilon 1 plus Epsilon 2."},{"Start":"05:19.045 ","End":"05:24.425","Text":"Really it\u0027s only at this point where we had to assign Epsilon 1 and Epsilon 2."},{"Start":"05:24.425 ","End":"05:29.235","Text":"I said above that each of these is Epsilon over 2."},{"Start":"05:29.235 ","End":"05:31.335","Text":"We could have decided that here,"},{"Start":"05:31.335 ","End":"05:32.960","Text":"we could have also done it differently."},{"Start":"05:32.960 ","End":"05:37.150","Text":"I mean, this could be Epsilon over 3 and this would be 2 Epsilon over 3."},{"Start":"05:37.150 ","End":"05:40.420","Text":"As long as the sum is Epsilon or less,"},{"Start":"05:40.420 ","End":"05:43.270","Text":"it seems the most natural to choose Epsilon over 2."},{"Start":"05:43.270 ","End":"05:47.360","Text":"This works and this works because each of these is less than Epsilon,"},{"Start":"05:47.360 ","End":"05:49.580","Text":"and this is less than Epsilon."},{"Start":"05:49.580 ","End":"05:57.090","Text":"In all 3 cases when x minus y is less than Delta in absolute value,"},{"Start":"05:57.090 ","End":"06:00.290","Text":"then f(x) minus f(y) in absolute value is less than Epsilon."},{"Start":"06:00.290 ","End":"06:01.506","Text":"That\u0027s what we had to share,"},{"Start":"06:01.506 ","End":"06:04.440","Text":"we\u0027re done for Part a."}],"ID":31149},{"Watched":false,"Name":"Exercise 6 Part 2","Duration":"1m 49s","ChapterTopicVideoID":29535,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.372","Text":"Now we come to part b of the exercise."},{"Start":"00:03.372 ","End":"00:11.235","Text":"Just to remind you we\u0027re given that f is uniformly continuous on a, c,"},{"Start":"00:11.235 ","End":"00:12.840","Text":"and on c, b,"},{"Start":"00:12.840 ","End":"00:16.170","Text":"and we have to prove that it\u0027s uniformly continuous on a,"},{"Start":"00:16.170 ","End":"00:20.580","Text":"b, using Cantor\u0027s theorem."},{"Start":"00:20.580 ","End":"00:22.755","Text":"In the Wikipedia,"},{"Start":"00:22.755 ","End":"00:25.860","Text":"it\u0027s called the Heine-Cantor theorem."},{"Start":"00:25.860 ","End":"00:32.480","Text":"The important part is if a function is continuous on a closed interval a,"},{"Start":"00:32.480 ","End":"00:36.080","Text":"b, then it\u0027s also uniformly continuous,"},{"Start":"00:36.080 ","End":"00:40.880","Text":"so all we have to do is to prove continuity on a, b."},{"Start":"00:40.880 ","End":"00:46.190","Text":"Now, f is uniformly continuous on the interval a,"},{"Start":"00:46.190 ","End":"00:50.825","Text":"c, and if it\u0027s uniformly continuous, then it\u0027s continuous."},{"Start":"00:50.825 ","End":"00:52.610","Text":"Similarly, for c, b."},{"Start":"00:52.610 ","End":"00:56.030","Text":"We have that f is continuous here a,"},{"Start":"00:56.030 ","End":"00:59.695","Text":"c, and continuous here c, b."},{"Start":"00:59.695 ","End":"01:03.470","Text":"Does it automatically follow that f is continuous on a,"},{"Start":"01:03.470 ","End":"01:05.450","Text":"b? Well, not quite."},{"Start":"01:05.450 ","End":"01:08.075","Text":"There\u0027s a snag at the point c,"},{"Start":"01:08.075 ","End":"01:10.630","Text":"because if it\u0027s continuous on a, c,"},{"Start":"01:10.630 ","End":"01:13.640","Text":"it only needs to be left continuous at"},{"Start":"01:13.640 ","End":"01:17.360","Text":"c. The limit as x goes to c from the left of f(x) is f(c)."},{"Start":"01:17.360 ","End":"01:21.195","Text":"Similarly, on the interval c, b,"},{"Start":"01:21.195 ","End":"01:26.675","Text":"continuity there just means that f is right continuous at c. However,"},{"Start":"01:26.675 ","End":"01:28.640","Text":"if f is left continuous at c,"},{"Start":"01:28.640 ","End":"01:30.815","Text":"and right continuous at c,"},{"Start":"01:30.815 ","End":"01:35.675","Text":"then it\u0027s continuous at c. That basically completes the proof."},{"Start":"01:35.675 ","End":"01:39.610","Text":"Therefore, f is continuous on all of a, b,"},{"Start":"01:39.610 ","End":"01:43.575","Text":"and by Cantor\u0027s theorem,"},{"Start":"01:43.575 ","End":"01:45.210","Text":"continuous on a, b,"},{"Start":"01:45.210 ","End":"01:47.280","Text":"means uniformly continuous."},{"Start":"01:47.280 ","End":"01:49.810","Text":"We are done."}],"ID":31150},{"Watched":false,"Name":"Exercise 7","Duration":"3m 41s","ChapterTopicVideoID":29536,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.160","Text":"In this exercise, we\u0027re going to show that the sum of"},{"Start":"00:03.160 ","End":"00:07.630","Text":"2 uniformly continuous functions is also uniformly continuous."},{"Start":"00:07.630 ","End":"00:09.970","Text":"Specifically, would take f and g,"},{"Start":"00:09.970 ","End":"00:11.860","Text":"which are uniformly continuous."},{"Start":"00:11.860 ","End":"00:16.285","Text":"I\u0027m going to prove that about the sum f plus g. For convenience,"},{"Start":"00:16.285 ","End":"00:20.830","Text":"we\u0027ll call the sum h. So what we have to prove is the following,"},{"Start":"00:20.830 ","End":"00:23.495","Text":"just the usual, I won\u0027t even read it out."},{"Start":"00:23.495 ","End":"00:27.510","Text":"In our case, we\u0027ll assume that Epsilon bigger than 0 is given,"},{"Start":"00:27.510 ","End":"00:28.960","Text":"someone gives us Epsilon,"},{"Start":"00:28.960 ","End":"00:34.870","Text":"we have to provide them with the corresponding Delta which satisfies this same as"},{"Start":"00:34.870 ","End":"00:40.995","Text":"always with h instead of f. F is uniformly continuous,"},{"Start":"00:40.995 ","End":"00:42.975","Text":"that\u0027s the interpretation,"},{"Start":"00:42.975 ","End":"00:45.150","Text":"g is uniformly continuous,"},{"Start":"00:45.150 ","End":"00:46.650","Text":"so this is the interpretation."},{"Start":"00:46.650 ","End":"00:47.970","Text":"Same as for h,"},{"Start":"00:47.970 ","End":"00:50.505","Text":"except that we\u0027ll use subscripts,"},{"Start":"00:50.505 ","End":"00:53.400","Text":"subscript 1 for f and subscript 2 for"},{"Start":"00:53.400 ","End":"00:57.090","Text":"g. We have Epsilon_1 and Delta_1, Epsilon_2 and Delta_2."},{"Start":"00:57.090 ","End":"00:59.010","Text":"In case you don\u0027t know what this is,"},{"Start":"00:59.010 ","End":"01:01.610","Text":"it\u0027s such that not that common."},{"Start":"01:01.610 ","End":"01:04.790","Text":"Sometimes I just write St or just spell it out."},{"Start":"01:04.790 ","End":"01:06.745","Text":"First, get back to Epsilon."},{"Start":"01:06.745 ","End":"01:12.120","Text":"We\u0027re going to let Epsilon_1 and Epsilon_2 both be Epsilon over 2."},{"Start":"01:12.120 ","End":"01:15.390","Text":"It\u0027s like I just pulled this out of a hat."},{"Start":"01:15.410 ","End":"01:18.275","Text":"Let\u0027s not decide that yet."},{"Start":"01:18.275 ","End":"01:19.955","Text":"You\u0027ll see why later,"},{"Start":"01:19.955 ","End":"01:23.150","Text":"we\u0027ll figure out what they are and then we\u0027ll put it back here."},{"Start":"01:23.150 ","End":"01:24.740","Text":"That\u0027s how these things work."},{"Start":"01:24.740 ","End":"01:30.989","Text":"You get towards the end and then you see what Epsilon_2 should be reverse engineering."},{"Start":"01:30.989 ","End":"01:32.789","Text":"Delta_1 we know exists,"},{"Start":"01:32.789 ","End":"01:37.060","Text":"corresponding to Epsilon_1 and Delta_2 corresponding to Epsilon_2."},{"Start":"01:37.060 ","End":"01:39.980","Text":"What we often do is let Delta be the minimum of"},{"Start":"01:39.980 ","End":"01:43.415","Text":"2 Deltas and that way if something is less than Delta,"},{"Start":"01:43.415 ","End":"01:47.180","Text":"then it\u0027s automatically less than Delta_1 and less than Delta_2."},{"Start":"01:47.180 ","End":"01:52.265","Text":"Now let\u0027s get back to proving this part that if x minus y is less than Delta,"},{"Start":"01:52.265 ","End":"01:57.095","Text":"then h(x) minus h(y) absolute value is less than Epsilon."},{"Start":"01:57.095 ","End":"01:59.198","Text":"Just put this in words."},{"Start":"01:59.198 ","End":"02:00.955","Text":"Like I said,"},{"Start":"02:00.955 ","End":"02:03.285","Text":"if something is less than Delta,"},{"Start":"02:03.285 ","End":"02:08.345","Text":"it\u0027s less than Delta_1 and less than Delta_2 because of the minimum,"},{"Start":"02:08.345 ","End":"02:11.570","Text":"and because of what we have here and here,"},{"Start":"02:11.570 ","End":"02:13.325","Text":"these 2 implications,"},{"Start":"02:13.325 ","End":"02:19.655","Text":"we get that f(x) is close to f(y) to within Epsilon_1."},{"Start":"02:19.655 ","End":"02:22.790","Text":"Whoops, I did a copy paste the wrong thing,"},{"Start":"02:22.790 ","End":"02:27.245","Text":"it should be g(x) minus g(y) less than Epsilon_2."},{"Start":"02:27.245 ","End":"02:31.775","Text":"Now, we want to evaluate absolute value of h(x) minus h(y)."},{"Start":"02:31.775 ","End":"02:33.680","Text":"H is f plus g,"},{"Start":"02:33.680 ","End":"02:37.490","Text":"so we get f plus g(x) minus f plus g(y),"},{"Start":"02:37.490 ","End":"02:39.340","Text":"so it\u0027s minus minus."},{"Start":"02:39.340 ","End":"02:43.340","Text":"Now we can break this up into 2 parts."},{"Start":"02:43.340 ","End":"02:47.930","Text":"We can have f(x) minus f(y) plus g(x) minus g(y),"},{"Start":"02:47.930 ","End":"02:50.300","Text":"and then apply the triangle inequality,"},{"Start":"02:50.300 ","End":"02:55.717","Text":"so it\u0027s less than or equal to this pair plus this pair."},{"Start":"02:55.717 ","End":"02:58.175","Text":"This is less than Epsilon_1,"},{"Start":"02:58.175 ","End":"03:00.485","Text":"this is less than Epsilon_2,"},{"Start":"03:00.485 ","End":"03:05.230","Text":"and what we would like is to have Epsilon written here."},{"Start":"03:05.230 ","End":"03:07.895","Text":"This is the point at which we say, okay,"},{"Start":"03:07.895 ","End":"03:12.620","Text":"let\u0027s take Epsilon_1 and Epsilon_2 to be 1/2 of Epsilon,"},{"Start":"03:12.620 ","End":"03:16.110","Text":"then we\u0027ll get 1/2 of Epsilon plus 1/2 of Epsilon equals Epsilon,"},{"Start":"03:16.110 ","End":"03:18.155","Text":"just 1 way to do it."},{"Start":"03:18.155 ","End":"03:24.500","Text":"You could take, 0.7 Epsilon here and 0.3 Epsilon here,"},{"Start":"03:24.500 ","End":"03:28.235","Text":"or any 2 bits that add up to Epsilon or less."},{"Start":"03:28.235 ","End":"03:30.320","Text":"So we have what we wanted."},{"Start":"03:30.320 ","End":"03:35.390","Text":"Absolute value of h(x) minus h(y) is less than Epsilon."},{"Start":"03:35.390 ","End":"03:38.060","Text":"That\u0027s the missing piece of the puzzle,"},{"Start":"03:38.060 ","End":"03:41.700","Text":"this part and so we are done."}],"ID":31151},{"Watched":false,"Name":"Exercise 8","Duration":"7m 10s","ChapterTopicVideoID":29537,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.700","Text":"In this exercise f and g are real functions on an interval I."},{"Start":"00:05.700 ","End":"00:08.736","Text":"There are 5 claims here,"},{"Start":"00:08.736 ","End":"00:11.715","Text":"and we have to disprove all of them."},{"Start":"00:11.715 ","End":"00:15.900","Text":"In mathematics to disprove a claim all you need is 1 counterexample."},{"Start":"00:15.900 ","End":"00:18.450","Text":"That\u0027s how we\u0027ll disprove them."},{"Start":"00:18.450 ","End":"00:21.085","Text":"We\u0027ll read each one as we come to it."},{"Start":"00:21.085 ","End":"00:24.090","Text":"In part A the claim is that,"},{"Start":"00:24.090 ","End":"00:26.490","Text":"if f and g are uniformly continuous,"},{"Start":"00:26.490 ","End":"00:32.595","Text":"then so is the product f times g. Our counterexample is to take"},{"Start":"00:32.595 ","End":"00:38.870","Text":"both f(x) and g(x) to be x. I don\u0027t know why I wrote 0 infinity."},{"Start":"00:38.870 ","End":"00:42.065","Text":"That\u0027s okay. We can just make it."},{"Start":"00:42.065 ","End":"00:46.655","Text":"Both f and g are the same, have bounded derivatives."},{"Start":"00:46.655 ","End":"00:51.515","Text":"The derivative of f as well as g is 1, so it\u0027s bounded."},{"Start":"00:51.515 ","End":"00:57.530","Text":"They are uniformly continuous by a theorem or exercise that we haven\u0027t done yet,"},{"Start":"00:57.530 ","End":"00:58.820","Text":"we\u0027ll prove it in the future."},{"Start":"00:58.820 ","End":"01:01.985","Text":"Now it\u0027s easy enough to do with Epsilon and Delta."},{"Start":"01:01.985 ","End":"01:05.360","Text":"In fact, if you take Delta equals Epsilon it will work,"},{"Start":"01:05.360 ","End":"01:08.030","Text":"but we\u0027ll use this claim or theorem."},{"Start":"01:08.030 ","End":"01:12.095","Text":"Now, is f times g uniformly continuous?"},{"Start":"01:12.095 ","End":"01:18.677","Text":"Now recall, we let h equals f times g. f times g is x times x is x squared."},{"Start":"01:18.677 ","End":"01:24.650","Text":"This is not uniformly continuous and we\u0027ll prove it by contradiction."},{"Start":"01:24.650 ","End":"01:29.225","Text":"In other words, we\u0027ll assume that it is."},{"Start":"01:29.225 ","End":"01:33.935","Text":"I\u0027m going to give you an Epsilon for which there is no Delta."},{"Start":"01:33.935 ","End":"01:38.440","Text":"Suppose it was uniformly continuous and that Epsilon equals 2,"},{"Start":"01:38.440 ","End":"01:43.265","Text":"then there\u0027s a corresponding Delta as in the definition of uniform continuity."},{"Start":"01:43.265 ","End":"01:50.240","Text":"We can choose x equals Delta plus 1 over Delta and y equals 1 over Delta."},{"Start":"01:50.240 ","End":"01:55.925","Text":"What we\u0027d expect is that the absolute value of h(x) minus h(y),"},{"Start":"01:55.925 ","End":"01:58.070","Text":"which is x squared minus y squared,"},{"Start":"01:58.070 ","End":"01:59.675","Text":"would be less than Epsilon."},{"Start":"01:59.675 ","End":"02:05.925","Text":"However, in practice x-squared minus y-squared is this expression."},{"Start":"02:05.925 ","End":"02:09.930","Text":"If you expand it the 1 over Delta squared cancels,"},{"Start":"02:09.930 ","End":"02:14.420","Text":"we just get Delta squared plus twice this times this which is plus 2."},{"Start":"02:14.420 ","End":"02:16.370","Text":"That is not less than 2,"},{"Start":"02:16.370 ","End":"02:22.685","Text":"which is a contradiction showing that h is not uniformly continuous."},{"Start":"02:22.685 ","End":"02:24.360","Text":"Next, part B."},{"Start":"02:24.360 ","End":"02:28.205","Text":"Assuming that the product is uniformly continuous,"},{"Start":"02:28.205 ","End":"02:33.919","Text":"the claim is that at least 1 of them has to be uniformly continuous, and that\u0027s false."},{"Start":"02:33.919 ","End":"02:39.155","Text":"The counterexample will be to take both f and g(x) to be the function which is"},{"Start":"02:39.155 ","End":"02:44.390","Text":"1 on the rationals and minus 1 on the irrationals."},{"Start":"02:44.390 ","End":"02:46.970","Text":"Now, f and g are not continuous."},{"Start":"02:46.970 ","End":"02:48.770","Text":"Actually, it\u0027s even worse than that."},{"Start":"02:48.770 ","End":"02:51.155","Text":"They\u0027re not continuous at any point,"},{"Start":"02:51.155 ","End":"02:52.820","Text":"but the product is 1."},{"Start":"02:52.820 ","End":"02:56.005","Text":"It\u0027s either 1 squared or minus 1 squared."},{"Start":"02:56.005 ","End":"03:00.170","Text":"A constant function is certainly uniformly continuous."},{"Start":"03:00.170 ","End":"03:04.175","Text":"In part C the claim is that the quotient"},{"Start":"03:04.175 ","End":"03:08.260","Text":"of 2 uniformly continuous functions is continuous."},{"Start":"03:08.260 ","End":"03:15.135","Text":"Our counterexample will be to take f(x) as 1 and g(x) as x."},{"Start":"03:15.135 ","End":"03:18.290","Text":"We\u0027ll take it on the interval from 0 to infinity,"},{"Start":"03:18.290 ","End":"03:21.680","Text":"where certainly g is not 0."},{"Start":"03:21.680 ","End":"03:23.900","Text":"We know that f and g are uniformly continuous,"},{"Start":"03:23.900 ","End":"03:25.168","Text":"f is a constant function,"},{"Start":"03:25.168 ","End":"03:28.060","Text":"and g(x) was mentioned in part A."},{"Start":"03:28.060 ","End":"03:32.780","Text":"However, the quotient, which is 1/x, is not."},{"Start":"03:32.780 ","End":"03:36.409","Text":"We\u0027ll do it by using Epsilon Delta proof by contradiction."},{"Start":"03:36.409 ","End":"03:37.940","Text":"If it is uniform continuous,"},{"Start":"03:37.940 ","End":"03:43.345","Text":"choose Epsilon equals 1 and let Delta be the corresponding Delta."},{"Start":"03:43.345 ","End":"03:47.855","Text":"Now choose n big enough so that 1 is less than Delta."},{"Start":"03:47.855 ","End":"03:53.075","Text":"Let\u0027s define x to be 1 and y equals 1 plus 1."},{"Start":"03:53.075 ","End":"03:56.060","Text":"Absolute value of x minus y is less than"},{"Start":"03:56.060 ","End":"04:00.350","Text":"Delta because both x and y are between 0 and Delta."},{"Start":"04:00.350 ","End":"04:03.325","Text":"1 is less than Delta, so it\u0027s 1 plus 1."},{"Start":"04:03.325 ","End":"04:10.140","Text":"The absolute value of h(x) minus h(y) is equal to 1/x is n,"},{"Start":"04:10.140 ","End":"04:12.130","Text":"1/y is n plus 1."},{"Start":"04:12.130 ","End":"04:14.210","Text":"The absolute value of the difference is 1,"},{"Start":"04:14.210 ","End":"04:19.175","Text":"which is not less than Epsilon because 1 is not less than 1."},{"Start":"04:19.175 ","End":"04:23.505","Text":"That\u0027s part C. In part D,"},{"Start":"04:23.505 ","End":"04:25.850","Text":"the claim to disprove is that if f is"},{"Start":"04:25.850 ","End":"04:29.695","Text":"continuous and bounded then f is uniformly continuous."},{"Start":"04:29.695 ","End":"04:36.675","Text":"The counterexample will be f(x) equals cosine of Pi x^2."},{"Start":"04:36.675 ","End":"04:38.300","Text":"I\u0027ll show you what it looks like."},{"Start":"04:38.300 ","End":"04:39.860","Text":"Here\u0027s a Picture."},{"Start":"04:39.860 ","End":"04:44.010","Text":"It\u0027s a bit like the cosine of Pi x."},{"Start":"04:44.010 ","End":"04:45.995","Text":"Only because it\u0027s x^2,"},{"Start":"04:45.995 ","End":"04:51.350","Text":"it frequency keeps increasing and like a spring,"},{"Start":"04:51.350 ","End":"04:58.834","Text":"these peaks get closer and closer to the swings or the slope gets steeper and steeper."},{"Start":"04:58.834 ","End":"05:00.920","Text":"That\u0027s just for intuition."},{"Start":"05:00.920 ","End":"05:05.240","Text":"Let\u0027s show that it isn\u0027t uniformly continuous."},{"Start":"05:05.240 ","End":"05:10.885","Text":"Note that if we take f of the square root of an even number,"},{"Start":"05:10.885 ","End":"05:15.275","Text":"we get cosine of that even number times Pi."},{"Start":"05:15.275 ","End":"05:18.365","Text":"Cosine of 0, 2Pi,"},{"Start":"05:18.365 ","End":"05:20.645","Text":"4Pi, 6Pi, etc."},{"Start":"05:20.645 ","End":"05:22.075","Text":"is 1."},{"Start":"05:22.075 ","End":"05:25.320","Text":"On the other hand, cosine of Pi, 3Pi,"},{"Start":"05:25.320 ","End":"05:26.805","Text":"5Pi, 7Pi,"},{"Start":"05:26.805 ","End":"05:30.480","Text":"an odd number times Pi is minus 1."},{"Start":"05:30.480 ","End":"05:36.300","Text":"Choose Epsilon equals to or less and let\u0027s say that"},{"Start":"05:36.300 ","End":"05:42.260","Text":"Delta is the corresponding Delta for this Epsilon if it were uniformly continuous."},{"Start":"05:42.260 ","End":"05:47.690","Text":"Choose x equals the square root of 2k and y equals the square root of 2k plus 1,"},{"Start":"05:47.690 ","End":"05:52.910","Text":"where k is such that this difference is less than Delta."},{"Start":"05:52.910 ","End":"05:56.060","Text":"You might ask, how do you know there is such a k?"},{"Start":"05:56.060 ","End":"06:00.170","Text":"Well, it\u0027s easy enough to show that this expression,"},{"Start":"06:00.170 ","End":"06:03.710","Text":"this difference goes to 0 as k goes to infinity."},{"Start":"06:03.710 ","End":"06:07.265","Text":"1 way to show it would be to multiply and divide"},{"Start":"06:07.265 ","End":"06:11.165","Text":"by the conjugate square root plus the square root."},{"Start":"06:11.165 ","End":"06:17.720","Text":"We get 1 over this plus this and we get 1 over infinity is 0."},{"Start":"06:17.720 ","End":"06:19.895","Text":"I\u0027ll leave this as an exercise."},{"Start":"06:19.895 ","End":"06:22.565","Text":"The absolute value of x minus y less than Delta."},{"Start":"06:22.565 ","End":"06:27.435","Text":"I mean, this is x and this is y and less than Delta and bigger than 0."},{"Start":"06:27.435 ","End":"06:31.565","Text":"The absolute value, something between 0 and Delta, is less than Delta."},{"Start":"06:31.565 ","End":"06:36.755","Text":"But the absolute value of the difference is according to this,"},{"Start":"06:36.755 ","End":"06:39.815","Text":"1 minus minus 1,"},{"Start":"06:39.815 ","End":"06:41.990","Text":"which is absolute value of 2,"},{"Start":"06:41.990 ","End":"06:45.620","Text":"which is not less than Epsilon contradiction,"},{"Start":"06:45.620 ","End":"06:49.090","Text":"so f is not uniformly continuous."},{"Start":"06:49.090 ","End":"06:51.338","Text":"Last part, E,"},{"Start":"06:51.338 ","End":"06:57.230","Text":"says that if f is uniformly continuous then f is bounded. That\u0027s false."},{"Start":"06:57.230 ","End":"06:59.495","Text":"We already saw an example,"},{"Start":"06:59.495 ","End":"07:02.720","Text":"f(x)=x, obviously not bounded."},{"Start":"07:02.720 ","End":"07:06.275","Text":"We know that it is uniformly continuous."},{"Start":"07:06.275 ","End":"07:10.650","Text":"That concludes part E and this exercise."}],"ID":31152},{"Watched":false,"Name":"Exercise 9","Duration":"3m 21s","ChapterTopicVideoID":29538,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.590","Text":"In this exercise, we\u0027re given 2 functions, f and g,"},{"Start":"00:04.590 ","End":"00:10.920","Text":"which are uniformly continuous and bounded on an interval I."},{"Start":"00:10.920 ","End":"00:14.490","Text":"We have to prove that the product is also uniformly"},{"Start":"00:14.490 ","End":"00:19.305","Text":"continuous and automatically it will be bounded because the product is bounded."},{"Start":"00:19.305 ","End":"00:22.200","Text":"To say that f is bounded means that the absolute value"},{"Start":"00:22.200 ","End":"00:25.620","Text":"of f(x) is less than or equal to some constant K for"},{"Start":"00:25.620 ","End":"00:28.380","Text":"all x in the interval and similarly for g will use the"},{"Start":"00:28.380 ","End":"00:32.475","Text":"constant m. What we have to show is the usual given Epsilon,"},{"Start":"00:32.475 ","End":"00:37.515","Text":"there is Delta such that if x and y are close to within Delta,"},{"Start":"00:37.515 ","End":"00:42.230","Text":"then h(x) and h(y) are close to within Epsilon for any x,"},{"Start":"00:42.230 ","End":"00:44.180","Text":"y on the interval."},{"Start":"00:44.180 ","End":"00:50.140","Text":"Now, let\u0027s evaluate h(x) minus h(y) absolute value."},{"Start":"00:50.140 ","End":"00:53.737","Text":"This is equal to h is f times g,"},{"Start":"00:53.737 ","End":"00:54.935","Text":"so we have this."},{"Start":"00:54.935 ","End":"00:58.730","Text":"Now, we can subtract and add the same quantity,"},{"Start":"00:58.730 ","End":"01:01.220","Text":"which is f(x) g(y)."},{"Start":"01:01.220 ","End":"01:05.030","Text":"This allows us to separate this into 2."},{"Start":"01:05.030 ","End":"01:11.165","Text":"This is less than or equal to the absolute value of this plus the absolute value of this."},{"Start":"01:11.165 ","End":"01:12.500","Text":"But for the first pair,"},{"Start":"01:12.500 ","End":"01:15.200","Text":"we can take f(x) out the brackets,"},{"Start":"01:15.200 ","End":"01:16.580","Text":"out the absolute value."},{"Start":"01:16.580 ","End":"01:19.615","Text":"For the second pair, we can take g(y) out."},{"Start":"01:19.615 ","End":"01:22.220","Text":"We end up with this expression."},{"Start":"01:22.220 ","End":"01:29.530","Text":"Now, we know that f(x) is less than or equal to K and g(y) is less than or equal to M,"},{"Start":"01:29.530 ","End":"01:32.305","Text":"so we get this expression."},{"Start":"01:32.305 ","End":"01:34.820","Text":"We\u0027ll return to this in a moment."},{"Start":"01:34.820 ","End":"01:40.550","Text":"Let\u0027s just say what we mean by f is uniformly continuous and g is uniformly continuous."},{"Start":"01:40.550 ","End":"01:42.935","Text":"Well, the usual Epsilon Delta,"},{"Start":"01:42.935 ","End":"01:47.280","Text":"but here for f we\u0027ll use a subscript 1 on Epsilon and Delta,"},{"Start":"01:47.280 ","End":"01:51.860","Text":"for g we\u0027ll use a subscript 2 because we have actually 3 pairs."},{"Start":"01:51.860 ","End":"01:56.750","Text":"For h, we have Epsilon and Delta without subscript."},{"Start":"01:56.750 ","End":"01:58.355","Text":"Now, as we often do,"},{"Start":"01:58.355 ","End":"02:01.670","Text":"we let Delta be the minimum of the 2 Deltas so that"},{"Start":"02:01.670 ","End":"02:05.270","Text":"if absolute value of x minus y is less than Delta,"},{"Start":"02:05.270 ","End":"02:09.050","Text":"then it\u0027s less than Delta 1 and it\u0027s less than Delta 2,"},{"Start":"02:09.050 ","End":"02:12.410","Text":"so we get that f(x) minus f(y) is less than"},{"Start":"02:12.410 ","End":"02:17.580","Text":"Epsilon 1 and g(x) minus g(y) is less than Epsilon 2."},{"Start":"02:17.580 ","End":"02:22.280","Text":"That if x minus y is less than Delta,"},{"Start":"02:22.280 ","End":"02:27.350","Text":"then we have h(x) minus h(y) and returning here is"},{"Start":"02:27.350 ","End":"02:32.590","Text":"less than K times Epsilon 2, that\u0027s this."},{"Start":"02:32.590 ","End":"02:35.245","Text":"Here get Epsilon 1."},{"Start":"02:35.245 ","End":"02:39.230","Text":"For now it\u0027s clear what to take for Epsilon 1 and Epsilon 2."},{"Start":"02:39.230 ","End":"02:43.040","Text":"If we take Epsilon 2 to be Epsilon over 2K,"},{"Start":"02:43.040 ","End":"02:46.990","Text":"and Epsilon 1 equals Epsilon over 2m."},{"Start":"02:46.990 ","End":"02:49.025","Text":"Then if we put them here and here,"},{"Start":"02:49.025 ","End":"02:51.770","Text":"we get that if x minus y is less than Delta,"},{"Start":"02:51.770 ","End":"02:54.320","Text":"then h(x) minus h(y) is less than."},{"Start":"02:54.320 ","End":"02:57.060","Text":"This comes out to be Epsilon over 2,"},{"Start":"02:57.160 ","End":"03:01.145","Text":"this comes out to be Epsilon over 2, which is Epsilon. Just to summarize."},{"Start":"03:01.145 ","End":"03:02.960","Text":"If we\u0027re given Epsilon,"},{"Start":"03:02.960 ","End":"03:06.010","Text":"we choose Epsilon 1 and Epsilon 2 as follows."},{"Start":"03:06.010 ","End":"03:09.720","Text":"Then we get Delta 1 and Delta 2 from"},{"Start":"03:09.720 ","End":"03:13.800","Text":"here and then we let Delta be the minimum of the 2 Deltas,"},{"Start":"03:13.800 ","End":"03:17.655","Text":"so this delta will correspond to the given Epsilon."},{"Start":"03:17.655 ","End":"03:21.820","Text":"That completes this exercise."}],"ID":31153},{"Watched":false,"Name":"Exercise 10","Duration":"3m 38s","ChapterTopicVideoID":29539,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.145","Text":"In this exercise, we have a function defined on the open interval a, b,"},{"Start":"00:05.145 ","End":"00:08.010","Text":"and we\u0027re given that it\u0027s differentiable and"},{"Start":"00:08.010 ","End":"00:11.250","Text":"furthermore that the derivative is bounded on the interval."},{"Start":"00:11.250 ","End":"00:13.140","Text":"Then there are two parts, a and b."},{"Start":"00:13.140 ","End":"00:17.580","Text":"In part a, we have to show that there exists some constant M,"},{"Start":"00:17.580 ","End":"00:20.730","Text":"such that the absolute value of f(y) minus"},{"Start":"00:20.730 ","End":"00:26.550","Text":"f(x) is less than or equal to M times the absolute value of y minus x."},{"Start":"00:26.550 ","End":"00:27.945","Text":"This, by the way,"},{"Start":"00:27.945 ","End":"00:30.570","Text":"means that f is Lipschitz continuous."},{"Start":"00:30.570 ","End":"00:33.870","Text":"This is the definition of Lipschitz continuity, i.e."},{"Start":"00:33.870 ","End":"00:40.035","Text":"the existence of such an M means that f is Lipschitz continuous."},{"Start":"00:40.035 ","End":"00:41.590","Text":"In part b,"},{"Start":"00:41.590 ","End":"00:45.620","Text":"we\u0027ll prove that f is uniformly continuous."},{"Start":"00:45.620 ","End":"00:50.135","Text":"But we won\u0027t use all the details except that f is Lipschitz continuous."},{"Start":"00:50.135 ","End":"00:51.460","Text":"In other words, in part b,"},{"Start":"00:51.460 ","End":"00:56.600","Text":"we\u0027ll show that Lipschitz continuity implies uniform continuity."},{"Start":"00:56.600 ","End":"00:58.730","Text":"We\u0027ll start with part a."},{"Start":"00:58.730 ","End":"01:02.450","Text":"Suppose that x and y are in our interval a, b."},{"Start":"01:02.450 ","End":"01:08.120","Text":"There are two cases: either x equals y or x is not equal to y."},{"Start":"01:08.120 ","End":"01:09.979","Text":"If x equals y,"},{"Start":"01:09.979 ","End":"01:15.410","Text":"then the condition clearly holds because 0 is less than or equal to M times 0."},{"Start":"01:15.410 ","End":"01:17.180","Text":"If x is not equal to y,"},{"Start":"01:17.180 ","End":"01:19.730","Text":"then that\u0027s assumed that x is less than y."},{"Start":"01:19.730 ","End":"01:22.000","Text":"This means without loss of generality."},{"Start":"01:22.000 ","End":"01:23.420","Text":"If it\u0027s the other way around,"},{"Start":"01:23.420 ","End":"01:27.155","Text":"then relabel them, label y as x and label x as y."},{"Start":"01:27.155 ","End":"01:30.890","Text":"Note that because x and y are in the interval a,"},{"Start":"01:30.890 ","End":"01:32.510","Text":"b, the closed interval x,"},{"Start":"01:32.510 ","End":"01:35.510","Text":"y is the sub interval of a, b,"},{"Start":"01:35.510 ","End":"01:40.600","Text":"and so f is continuous on x, y,"},{"Start":"01:40.600 ","End":"01:44.220","Text":"and it\u0027s also differentiable on the open x,"},{"Start":"01:44.220 ","End":"01:46.730","Text":"y because this is a subset of this."},{"Start":"01:46.730 ","End":"01:50.270","Text":"This means that f satisfies the conditions of"},{"Start":"01:50.270 ","End":"01:54.920","Text":"the mean value theorem according to Lagrange"},{"Start":"01:54.920 ","End":"02:00.260","Text":"and this means that f(y) minus f(x) over y"},{"Start":"02:00.260 ","End":"02:06.815","Text":"minus x equals f′ of c for some c in the interval from x to y,"},{"Start":"02:06.815 ","End":"02:10.490","Text":"which is also in the interval from a to b."},{"Start":"02:10.490 ","End":"02:14.315","Text":"Now given that f′ is bounded on the interval,"},{"Start":"02:14.315 ","End":"02:19.655","Text":"and bounded means that the absolute value is less than or equal to some constant,"},{"Start":"02:19.655 ","End":"02:22.970","Text":"call that constant M. What we"},{"Start":"02:22.970 ","End":"02:27.820","Text":"get is that this in absolute value is less than or equal to M,"},{"Start":"02:27.820 ","End":"02:32.360","Text":"which means that the numerator is less than or equal to m times the denominator,"},{"Start":"02:32.360 ","End":"02:35.060","Text":"and this is exactly what we had to show."},{"Start":"02:35.060 ","End":"02:39.905","Text":"Where is it? Yeah, it\u0027s exactly what we had to show so that concludes part a."},{"Start":"02:39.905 ","End":"02:41.480","Text":"Now in part b,"},{"Start":"02:41.480 ","End":"02:44.780","Text":"we need to show that f is uniformly continuous."},{"Start":"02:44.780 ","End":"02:50.690","Text":"We\u0027ll use the Epsilon Delta definition and this is the usual thing that we have to prove."},{"Start":"02:50.690 ","End":"02:52.760","Text":"Given Epsilon, we have to find Delta."},{"Start":"02:52.760 ","End":"02:58.550","Text":"Like I said, all we\u0027re going to use from part a is the Lipschitz continuity."},{"Start":"02:58.550 ","End":"03:02.450","Text":"If y minus x is less than Delta,"},{"Start":"03:02.450 ","End":"03:07.640","Text":"then f(y) minus f(x) is less than or equal to M times y minus x,"},{"Start":"03:07.640 ","End":"03:09.685","Text":"which is M times Delta."},{"Start":"03:09.685 ","End":"03:14.635","Text":"If this is less than M Delta and we want this less than Epsilon,"},{"Start":"03:14.635 ","End":"03:19.550","Text":"then natural thing to do is to choose Delta equals Epsilon over"},{"Start":"03:19.550 ","End":"03:24.350","Text":"M or less and then we\u0027ll get that this is less than M Delta,"},{"Start":"03:24.350 ","End":"03:29.320","Text":"which is Epsilon for all y and x such that this is less than Delta."},{"Start":"03:29.320 ","End":"03:32.045","Text":"That proves what we had to show."},{"Start":"03:32.045 ","End":"03:35.570","Text":"In other words, that f is uniformly continuous on a,"},{"Start":"03:35.570 ","End":"03:38.430","Text":"b, and we are done."}],"ID":31154},{"Watched":false,"Name":"Exercise 11","Duration":"2m ","ChapterTopicVideoID":29540,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.749","Text":"In this exercise, we\u0027re given that f is uniformly continuous on I,"},{"Start":"00:06.749 ","End":"00:09.735","Text":"and we\u0027re also given that the absolute value of"},{"Start":"00:09.735 ","End":"00:13.155","Text":"f is bigger or equal to some positive constant"},{"Start":"00:13.155 ","End":"00:21.540","Text":"C. We have to prove that the reciprocal of f is also uniformly continuous on I."},{"Start":"00:21.540 ","End":"00:27.015","Text":"This is the usual Epsilon Delta definition of what we have to show."},{"Start":"00:27.015 ","End":"00:30.870","Text":"For the function g, f is uniformly continuous."},{"Start":"00:30.870 ","End":"00:34.020","Text":"You want to use a different letter than Epsilon."},{"Start":"00:34.020 ","End":"00:36.740","Text":"Given Eta bigger than 0,"},{"Start":"00:36.740 ","End":"00:39.970","Text":"there exists Delta bigger than 0,"},{"Start":"00:39.970 ","End":"00:43.430","Text":"such that if x is close to y within Delta,"},{"Start":"00:43.430 ","End":"00:46.295","Text":"then f(x) is close to f(y) within Eta."},{"Start":"00:46.295 ","End":"00:50.435","Text":"Take Eta to be Epsilon times c squared."},{"Start":"00:50.435 ","End":"00:53.990","Text":"You\u0027ll see why we chose this when we get to it."},{"Start":"00:53.990 ","End":"00:58.490","Text":"So choose the same Delta for G that we used"},{"Start":"00:58.490 ","End":"01:03.185","Text":"for f. If the absolute value of x minus y is less than Delta,"},{"Start":"01:03.185 ","End":"01:06.455","Text":"then absolute value of g(x) minus g(y),"},{"Start":"01:06.455 ","End":"01:12.335","Text":"which by definition is absolute value 1 over f(x) minus 1 over f(y)."},{"Start":"01:12.335 ","End":"01:15.305","Text":"Cross multiply, subtract, we get this."},{"Start":"01:15.305 ","End":"01:19.520","Text":"Now we know that absolute value of f (x) is bigger or equal to"},{"Start":"01:19.520 ","End":"01:24.560","Text":"c from the given and similarly for f(y)."},{"Start":"01:24.560 ","End":"01:28.340","Text":"If we decrease the denominator,"},{"Start":"01:28.340 ","End":"01:30.170","Text":"we can only increase the fraction."},{"Start":"01:30.170 ","End":"01:35.495","Text":"So this is less than or equal to f(x) minus f(y) absolute value over c squared."},{"Start":"01:35.495 ","End":"01:40.115","Text":"f(x) minus f(y) here is less than Eta."},{"Start":"01:40.115 ","End":"01:44.615","Text":"Eta over c^2 is equal to Epsilon."},{"Start":"01:44.615 ","End":"01:47.270","Text":"Which shows that this same Delta is good for"},{"Start":"01:47.270 ","End":"01:51.230","Text":"Epsilon in the uniform continuity of g. In other words,"},{"Start":"01:51.230 ","End":"01:52.790","Text":"if x minus y is less than Delta,"},{"Start":"01:52.790 ","End":"01:56.690","Text":"then g(x) minus g(y) less and Epsilon absolute value here and here."},{"Start":"01:56.690 ","End":"02:00.510","Text":"That concludes this exercise and we\u0027re done."}],"ID":31155},{"Watched":false,"Name":"Exercise 12","Duration":"4m 53s","ChapterTopicVideoID":29541,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.975","Text":"This exercise has 2 parts which are similar."},{"Start":"00:03.975 ","End":"00:10.574","Text":"In part a, we have the function f(x)=e^x defined on all the reals."},{"Start":"00:10.574 ","End":"00:13.650","Text":"We have to show this is not uniformly continuous."},{"Start":"00:13.650 ","End":"00:15.810","Text":"However, in part b,"},{"Start":"00:15.810 ","End":"00:20.070","Text":"function g is also e^x,"},{"Start":"00:20.070 ","End":"00:23.775","Text":"but only on the closed interval from 0-1."},{"Start":"00:23.775 ","End":"00:28.755","Text":"And in part b, we have to show that g is uniformly continuous."},{"Start":"00:28.755 ","End":"00:31.710","Text":"Changing the domain from the reals to the interval"},{"Start":"00:31.710 ","End":"00:35.969","Text":"0,1 changes whether it isn\u0027t or is uniformly continuous."},{"Start":"00:35.969 ","End":"00:39.210","Text":"In both cases, we\u0027re supposed to work directly from the definition,"},{"Start":"00:39.210 ","End":"00:42.120","Text":"meaning not to quote some theorems."},{"Start":"00:42.120 ","End":"00:43.470","Text":"Let\u0027s start with part a,"},{"Start":"00:43.470 ","End":"00:45.075","Text":"which we\u0027ll do by contradiction."},{"Start":"00:45.075 ","End":"00:49.445","Text":"Suppose that f is uniformly continuous."},{"Start":"00:49.445 ","End":"00:52.460","Text":"Then for every Epsilon there is a Delta."},{"Start":"00:52.460 ","End":"00:56.675","Text":"Let Delta correspond to Epsilon=1,"},{"Start":"00:56.675 ","End":"01:01.475","Text":"meaning that if absolute value of y minus x is less than Delta,"},{"Start":"01:01.475 ","End":"01:07.655","Text":"then f(y) minus f(x) is less than 1."},{"Start":"01:07.655 ","End":"01:13.315","Text":"Now, choose some Eta between 0 and Delta."},{"Start":"01:13.315 ","End":"01:22.160","Text":"Let x be a real number and let y equals x plus Eta so that absolute value of y minus x,"},{"Start":"01:22.160 ","End":"01:24.610","Text":"which is Eta, is less than Delta."},{"Start":"01:24.610 ","End":"01:30.048","Text":"Then we can apply this and get that e^y minus e^x is less than 1."},{"Start":"01:30.048 ","End":"01:34.475","Text":"But y is x plus Eta so we get this."},{"Start":"01:34.475 ","End":"01:37.100","Text":"Now, we can factorize this,"},{"Start":"01:37.100 ","End":"01:39.635","Text":"e^x can be factored out."},{"Start":"01:39.635 ","End":"01:46.265","Text":"Because it\u0027s positive and because e^Eta minus 1 is also positive,"},{"Start":"01:46.265 ","End":"01:48.470","Text":"because Eta is bigger than 0,"},{"Start":"01:48.470 ","End":"01:49.760","Text":"so e^Eta is bigger than 1,"},{"Start":"01:49.760 ","End":"01:54.305","Text":"that means we can drop the absolute value because each of these factors is positive."},{"Start":"01:54.305 ","End":"01:59.375","Text":"Now, divide both sides by e^Eta minus 1."},{"Start":"01:59.375 ","End":"02:03.140","Text":"This is positive. The inequality keeps the direction."},{"Start":"02:03.140 ","End":"02:07.580","Text":"We have that e^x is less than 1 over e^Eta minus 1."},{"Start":"02:07.580 ","End":"02:11.900","Text":"Now, this is true for any x and that\u0027s impossible."},{"Start":"02:11.900 ","End":"02:18.745","Text":"Because if you let x go to infinity then e^x goes to infinity."},{"Start":"02:18.745 ","End":"02:22.320","Text":"It can\u0027t stay less than some number."},{"Start":"02:22.320 ","End":"02:26.180","Text":"That\u0027s a contradiction which came from assuming that f is"},{"Start":"02:26.180 ","End":"02:31.190","Text":"uniformly continuous and so it isn\u0027t as required."},{"Start":"02:31.190 ","End":"02:33.620","Text":"Now, let\u0027s go on to part b,"},{"Start":"02:33.620 ","End":"02:38.990","Text":"where g is defined just on the interval from 0-1, the closed interval."},{"Start":"02:38.990 ","End":"02:41.240","Text":"We could use Cantor\u0027s theorem and say,"},{"Start":"02:41.240 ","End":"02:43.760","Text":"g is continuous on a closed interval,"},{"Start":"02:43.760 ","End":"02:46.715","Text":"but Cantor\u0027s theorem it\u0027s uniformly continuous."},{"Start":"02:46.715 ","End":"02:51.955","Text":"But the requirement was to use the Epsilon Delta techniques."},{"Start":"02:51.955 ","End":"02:53.990","Text":"Let\u0027s start again."},{"Start":"02:53.990 ","End":"03:01.400","Text":"G is continuous on the interval 0,1 and it\u0027s differentiable on the open interval 0,1."},{"Start":"03:01.400 ","End":"03:05.275","Text":"I mean, E^x is differentiable, of course."},{"Start":"03:05.275 ","End":"03:10.385","Text":"Then we can use the Lagrange\u0027s mean value theorem"},{"Start":"03:10.385 ","End":"03:15.844","Text":"and get that if x is less than or equal to y"},{"Start":"03:15.844 ","End":"03:19.535","Text":"then g(x) minus g(y) equals"},{"Start":"03:19.535 ","End":"03:27.695","Text":"g\u0027 of some number c times x minus y for some c. C is between x and y."},{"Start":"03:27.695 ","End":"03:31.670","Text":"Normally this is dated with x strictly less than y,"},{"Start":"03:31.670 ","End":"03:38.690","Text":"but it also works when x=y because then we get 0 equals something times 0, which is okay."},{"Start":"03:38.690 ","End":"03:41.540","Text":"Also, without loss of generality,"},{"Start":"03:41.540 ","End":"03:43.100","Text":"x is less than or equal to y."},{"Start":"03:43.100 ","End":"03:46.055","Text":"If x is bigger than y then just reverse the labeling."},{"Start":"03:46.055 ","End":"03:51.780","Text":"Yes, so I should have said without loss of generality x is less than or equal to y."},{"Start":"03:51.780 ","End":"03:55.490","Text":"Of course, if c is between x and y then c is also"},{"Start":"03:55.490 ","End":"03:59.495","Text":"between 0 and 1 because x and y are in the interval 0,1."},{"Start":"03:59.495 ","End":"04:08.120","Text":"We get the absolute value of g(x) minus g(y) is g\u0027(c) times absolute value of x minus y."},{"Start":"04:08.120 ","End":"04:09.755","Text":"This is also an absolute value."},{"Start":"04:09.755 ","End":"04:13.520","Text":"Note the absolute value of g\u0027(c) which is the absolute value of"},{"Start":"04:13.520 ","End":"04:17.765","Text":"e^c because the derivative of e^x is e^x."},{"Start":"04:17.765 ","End":"04:20.664","Text":"This is positive, so it\u0027s e^c."},{"Start":"04:20.664 ","End":"04:24.380","Text":"It\u0027s increasing so that\u0027s less than or equal to e^1."},{"Start":"04:24.380 ","End":"04:28.190","Text":"We get that g(x) minus g(y) and absolute value is"},{"Start":"04:28.190 ","End":"04:32.800","Text":"less than or equal to e times absolute value of x minus y."},{"Start":"04:32.800 ","End":"04:38.705","Text":"Given Epsilon bigger than 0 we can choose Delta equals Epsilon over e or less."},{"Start":"04:38.705 ","End":"04:43.190","Text":"Then we\u0027ll get that if x minus y is less than Delta then g(x)"},{"Start":"04:43.190 ","End":"04:47.780","Text":"minus g(y) absolute value is less than e times Epsilon over e,"},{"Start":"04:47.780 ","End":"04:50.105","Text":"which is Epsilon as required."},{"Start":"04:50.105 ","End":"04:54.030","Text":"That concludes part b of this exercise."}],"ID":31156},{"Watched":false,"Name":"Exercise 13 (Cantors Theorem)","Duration":"3m 41s","ChapterTopicVideoID":29542,"CourseChapterTopicPlaylistID":294556,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.810","Text":"In this exercise, we\u0027re going to prove Cantor\u0027s theorem,"},{"Start":"00:03.810 ","End":"00:06.675","Text":"also known as the Heine-Cantor theorem."},{"Start":"00:06.675 ","End":"00:10.560","Text":"If f from a,b to R is continuous,"},{"Start":"00:10.560 ","End":"00:12.990","Text":"and it\u0027s important this is the closed interval,"},{"Start":"00:12.990 ","End":"00:16.635","Text":"then f is uniformly continuous."},{"Start":"00:16.635 ","End":"00:19.575","Text":"We\u0027ll do a proof by contradiction."},{"Start":"00:19.575 ","End":"00:23.745","Text":"Suppose that f is not uniformly continuous."},{"Start":"00:23.745 ","End":"00:28.500","Text":"That means that some Epsilon doesn\u0027t have a Delta."},{"Start":"00:28.500 ","End":"00:32.010","Text":"In particular, for any natural number n,"},{"Start":"00:32.010 ","End":"00:34.560","Text":"1 over n is no good as a Delta,"},{"Start":"00:34.560 ","End":"00:37.350","Text":"there is no good Delta for this Epsilon."},{"Start":"00:37.350 ","End":"00:39.000","Text":"When I say no good,"},{"Start":"00:39.000 ","End":"00:42.860","Text":"I mean that there exists x and y, well,"},{"Start":"00:42.860 ","End":"00:44.540","Text":"depending on n, so it\u0027s x_n,"},{"Start":"00:44.540 ","End":"00:47.740","Text":"y_n, in the interval such that,"},{"Start":"00:47.740 ","End":"00:52.280","Text":"although x_n minus y_n absolute value is less than Delta,"},{"Start":"00:52.280 ","End":"00:53.735","Text":"which is 1 over n,"},{"Start":"00:53.735 ","End":"00:59.390","Text":"still, the absolute value of f of x_n minus f of y_n is not less than Epsilon,"},{"Start":"00:59.390 ","End":"01:01.610","Text":"it\u0027s bigger or equal to Epsilon."},{"Start":"01:01.610 ","End":"01:07.460","Text":"Since for all n we have an x_n and a y_n what we\u0027ve done is built 2 sequences,"},{"Start":"01:07.460 ","End":"01:10.365","Text":"sequence x_n and a sequence y_n."},{"Start":"01:10.365 ","End":"01:12.810","Text":"The sequences are in a closed interval,"},{"Start":"01:12.810 ","End":"01:18.567","Text":"so we can apply Bolzano-Weierstrass theorem to the sequence x_n."},{"Start":"01:18.567 ","End":"01:22.130","Text":"First of all, it has a convergent sub-sequence."},{"Start":"01:22.130 ","End":"01:27.240","Text":"Call that convergence sub-sequence x_n_k,"},{"Start":"01:27.240 ","End":"01:30.920","Text":"and convergent means it has a limit and let\u0027s say the limit is"},{"Start":"01:30.920 ","End":"01:35.690","Text":"L. I claim that the sub-sequence y_n_k,"},{"Start":"01:35.690 ","End":"01:40.280","Text":"also converges to L. This is equivalent to"},{"Start":"01:40.280 ","End":"01:45.995","Text":"showing that the absolute value of y_n_k minus L tends to 0."},{"Start":"01:45.995 ","End":"01:48.590","Text":"Note 3 things. First of all,"},{"Start":"01:48.590 ","End":"01:52.280","Text":"that the absolute value of y_n_k minus L by"},{"Start":"01:52.280 ","End":"01:54.770","Text":"the triangle inequality is less than"},{"Start":"01:54.770 ","End":"01:57.605","Text":"or equal to the absolute value of y_n_k minus x_n_k,"},{"Start":"01:57.605 ","End":"02:00.565","Text":"plus absolute value of x_n_k minus L."},{"Start":"02:00.565 ","End":"02:05.210","Text":"Because this plus this is this, triangle inequality."},{"Start":"02:05.210 ","End":"02:12.125","Text":"Second thing to note is that y_n_k minus x_n_k is less than 1 over n_k."},{"Start":"02:12.125 ","End":"02:15.060","Text":"Because it is true for any n,"},{"Start":"02:15.060 ","End":"02:17.550","Text":"in particular, the sub-sequence n_k."},{"Start":"02:17.550 ","End":"02:24.215","Text":"Thirdly, note that because x_n_k converges to L,"},{"Start":"02:24.215 ","End":"02:26.510","Text":"x_n_k minus L goes to 0,"},{"Start":"02:26.510 ","End":"02:29.800","Text":"I should have said as k goes to infinity."},{"Start":"02:29.800 ","End":"02:32.315","Text":"Now what we have, just to summarize,"},{"Start":"02:32.315 ","End":"02:35.240","Text":"this is less than or equal to this plus this,"},{"Start":"02:35.240 ","End":"02:37.895","Text":"this is less than 1 over n_k,"},{"Start":"02:37.895 ","End":"02:40.264","Text":"and this goes to 0."},{"Start":"02:40.264 ","End":"02:42.835","Text":"It follows that,"},{"Start":"02:42.835 ","End":"02:49.505","Text":"this goes to 0 because this being less than 1 over n_k goes to 0,"},{"Start":"02:49.505 ","End":"02:50.750","Text":"and this goes to 0,"},{"Start":"02:50.750 ","End":"02:52.610","Text":"if both of these go to 0,"},{"Start":"02:52.610 ","End":"02:56.600","Text":"and this is less than or equal to it and non-negative,"},{"Start":"02:56.600 ","End":"02:59.950","Text":"then it\u0027s some which between 0 and something that goes to 0,"},{"Start":"02:59.950 ","End":"03:02.000","Text":"so it also goes to 0."},{"Start":"03:02.000 ","End":"03:04.115","Text":"Now if we copy this here,"},{"Start":"03:04.115 ","End":"03:07.865","Text":"this minus this is bigger or equal to Epsilon."},{"Start":"03:07.865 ","End":"03:09.380","Text":"It\u0027s true for all n,"},{"Start":"03:09.380 ","End":"03:12.170","Text":"and in particular, it\u0027s true for n equals n_k."},{"Start":"03:12.170 ","End":"03:16.940","Text":"So for any k, this minus this absolute value bigger or equal to Epsilon."},{"Start":"03:16.940 ","End":"03:19.565","Text":"Now, let k go to infinity."},{"Start":"03:19.565 ","End":"03:21.755","Text":"This tends to L,"},{"Start":"03:21.755 ","End":"03:24.925","Text":"this tends to L. So,"},{"Start":"03:24.925 ","End":"03:28.835","Text":"we get that absolute value of L minus L or"},{"Start":"03:28.835 ","End":"03:33.145","Text":"0 is bigger or equal to Epsilon and that\u0027s a contradiction."},{"Start":"03:33.145 ","End":"03:38.030","Text":"The contradiction came from assuming that f is not uniformly continuous,"},{"Start":"03:38.030 ","End":"03:39.215","Text":"therefore it is,"},{"Start":"03:39.215 ","End":"03:41.970","Text":"and that concludes this proof."}],"ID":31157}],"Thumbnail":null,"ID":294556},{"Name":"Sufficient Conditions for Uniform Continuity","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Sufficient Condition for Uniform Continuity Part 1","Duration":"4m 1s","ChapterTopicVideoID":29515,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.305","Text":"In this section, we\u0027ll be concerned with sufficient conditions for uniform continuity."},{"Start":"00:07.305 ","End":"00:09.660","Text":"There will be 5 of them in this clip,"},{"Start":"00:09.660 ","End":"00:11.385","Text":"just the first one."},{"Start":"00:11.385 ","End":"00:18.390","Text":"Sufficient conditions are useful for proving that a function is uniformly continuous."},{"Start":"00:18.390 ","End":"00:20.190","Text":"In the following section,"},{"Start":"00:20.190 ","End":"00:24.780","Text":"we will have necessary conditions for uniform continuity."},{"Start":"00:24.780 ","End":"00:30.015","Text":"These will be useful to prove that a function is not uniformly continuous."},{"Start":"00:30.015 ","End":"00:33.405","Text":"Number 1, the first one of these is a theorem,"},{"Start":"00:33.405 ","End":"00:36.810","Text":"the Heine-Cantor theorem or just Cantor\u0027s theorem,"},{"Start":"00:36.810 ","End":"00:43.340","Text":"that if we have a function that\u0027s continuous on a closed and bounded interval I."},{"Start":"00:43.340 ","End":"00:47.870","Text":"In this case, f is uniformly continuous on the interval."},{"Start":"00:47.870 ","End":"00:50.345","Text":"So 2 things, f should be continuous,"},{"Start":"00:50.345 ","End":"00:52.505","Text":"the interval should be closed and bounded."},{"Start":"00:52.505 ","End":"00:54.985","Text":"Then we have uniform continuity."},{"Start":"00:54.985 ","End":"01:00.845","Text":"We actually prove this in the previous section in an exercise example."},{"Start":"01:00.845 ","End":"01:06.245","Text":"The function f (x) = x^2 is uniformly continuous"},{"Start":"01:06.245 ","End":"01:12.600","Text":"on the interval from 1 to 40, including the endpoints."},{"Start":"01:12.600 ","End":"01:20.030","Text":"The reason for that is f is continuous on it and I is closed and bounded."},{"Start":"01:20.030 ","End":"01:23.070","Text":"Now a useful lemma."},{"Start":"01:23.150 ","End":"01:26.995","Text":"Not only will it be useful, but in fact,"},{"Start":"01:26.995 ","End":"01:33.610","Text":"it\u0027s also a sufficient condition for proving uniform continuity."},{"Start":"01:33.610 ","End":"01:40.445","Text":"If you can prove uniform continuity on a larger interval of 2,"},{"Start":"01:40.445 ","End":"01:42.865","Text":"meaning that 1 is contained in the other,"},{"Start":"01:42.865 ","End":"01:47.380","Text":"then we automatically get uniform continuity for the smaller one."},{"Start":"01:47.380 ","End":"01:49.090","Text":"This is what it states."},{"Start":"01:49.090 ","End":"01:52.064","Text":"If f is uniformly continuous on the interval I,"},{"Start":"01:52.064 ","End":"01:53.410","Text":"doesn\u0027t have to be an interval,"},{"Start":"01:53.410 ","End":"01:54.885","Text":"but let\u0027s say,"},{"Start":"01:54.885 ","End":"01:58.720","Text":"and J is a subset or subinterval of I,"},{"Start":"01:58.720 ","End":"02:03.055","Text":"then f is also uniformly continuous on J."},{"Start":"02:03.055 ","End":"02:06.355","Text":"Strictly speaking, it\u0027s not the same f we should say"},{"Start":"02:06.355 ","End":"02:11.470","Text":"the restriction of f to J that we say it this way."},{"Start":"02:11.470 ","End":"02:15.540","Text":"Let\u0027s prove this lemma."},{"Start":"02:15.540 ","End":"02:18.395","Text":"That Epsilon bigger than 0 be given,"},{"Start":"02:18.395 ","End":"02:24.620","Text":"there exists Delta bigger than 0 such that for all x and y and I,"},{"Start":"02:24.620 ","End":"02:28.040","Text":"if x is close to y within Delta,"},{"Start":"02:28.040 ","End":"02:33.595","Text":"then f(x) is close to f(y) due within Epsilon."},{"Start":"02:33.595 ","End":"02:41.210","Text":"The same Epsilon Delta pair is good for showing the uniform continuity of f on J."},{"Start":"02:41.210 ","End":"02:45.375","Text":"Given Epsilon choose this same Delta here."},{"Start":"02:45.375 ","End":"02:48.795","Text":"Then if x and y are in J,"},{"Start":"02:48.795 ","End":"02:50.535","Text":"then they\u0027re also in I,"},{"Start":"02:50.535 ","End":"02:55.560","Text":"because J is a subset of I and then this implication that if this is true,"},{"Start":"02:55.560 ","End":"03:00.635","Text":"then this is true also follows because x and y are in I."},{"Start":"03:00.635 ","End":"03:03.665","Text":"That\u0027s the proof of the lemma."},{"Start":"03:03.665 ","End":"03:07.685","Text":"Now an example, the function f(x),"},{"Start":"03:07.685 ","End":"03:11.210","Text":"which is 1 over 1 plus e^x,"},{"Start":"03:11.210 ","End":"03:17.840","Text":"is uniformly continuous on the interval (1,4) open at both ends."},{"Start":"03:17.840 ","End":"03:24.390","Text":"This requires some explanation because the interval (1,4) is not closed."},{"Start":"03:24.390 ","End":"03:27.635","Text":"What we\u0027re going to do is use the lemma we just proved."},{"Start":"03:27.635 ","End":"03:30.065","Text":"I use it as a workaround so here are the steps."},{"Start":"03:30.065 ","End":"03:33.770","Text":"First of all, f is continuous on the closed interval (1,4),"},{"Start":"03:33.770 ","End":"03:40.520","Text":"and therefore it\u0027s uniformly continuous on (1,4) by Cantor\u0027s theorem,"},{"Start":"03:40.520 ","End":"03:43.990","Text":"let J be the open (0,4),"},{"Start":"03:43.990 ","End":"03:46.755","Text":"then J is a subset of I."},{"Start":"03:46.755 ","End":"03:50.239","Text":"We can use our lemma because f is uniformly"},{"Start":"03:50.239 ","End":"03:55.640","Text":"continuous on I and therefore it\u0027s uniformly continuous on J."},{"Start":"03:55.640 ","End":"03:57.325","Text":"That\u0027s what we wanted."},{"Start":"03:57.325 ","End":"04:01.800","Text":"That concludes this example as well as this clip."}],"ID":31127},{"Watched":false,"Name":"Sufficient Condition for Uniform Continuity Part 2","Duration":"3m 58s","ChapterTopicVideoID":29516,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.615","Text":"Now we come to Part 2 of this sufficient conditions for uniform continuity."},{"Start":"00:06.615 ","End":"00:07.950","Text":"There\u0027s 5 altogether."},{"Start":"00:07.950 ","End":"00:09.405","Text":"This is the 2nd."},{"Start":"00:09.405 ","End":"00:15.645","Text":"This time it\u0027s a proposition that if f is continuous on an interval I,"},{"Start":"00:15.645 ","End":"00:21.690","Text":"and if it has a finite limit of each of the 2 endpoints,"},{"Start":"00:21.690 ","End":"00:24.285","Text":"I\u0027ll return to this at the end,"},{"Start":"00:24.285 ","End":"00:27.600","Text":"then f is uniformly continuous on I."},{"Start":"00:27.600 ","End":"00:31.140","Text":"This basically means that you don\u0027t have to check it for closed endpoints."},{"Start":"00:31.140 ","End":"00:32.985","Text":"It\u0027s only at the open ones."},{"Start":"00:32.985 ","End":"00:35.340","Text":"Well, you\u0027ll see. I\u0027ll prove part of it."},{"Start":"00:35.340 ","End":"00:36.636","Text":"There\u0027s several cases,"},{"Start":"00:36.636 ","End":"00:38.910","Text":"I\u0027ll just prove it for the main case."},{"Start":"00:38.910 ","End":"00:41.885","Text":"Well, we have an interval (a,b)."},{"Start":"00:41.885 ","End":"00:47.210","Text":"It could be (a,b) and it\u0027s half-closed or half-closed on the other side or closed,"},{"Start":"00:47.210 ","End":"00:49.505","Text":"or it could be that 1 of these 2 is infinity."},{"Start":"00:49.505 ","End":"00:52.700","Text":"It would just be a bore to do all cases."},{"Start":"00:52.700 ","End":"00:56.300","Text":"In this case, by the given,"},{"Start":"00:56.300 ","End":"01:00.380","Text":"we have a finite limit at the left endpoint,"},{"Start":"01:00.380 ","End":"01:02.044","Text":"at the right endpoint."},{"Start":"01:02.044 ","End":"01:03.440","Text":"Now, the left endpoint, it\u0027s"},{"Start":"01:03.440 ","End":"01:07.190","Text":"only a 1-sided limit from the right and at the right endpoint,"},{"Start":"01:07.190 ","End":"01:09.245","Text":"it\u0027s a 1-sided limit from the left."},{"Start":"01:09.245 ","End":"01:12.260","Text":"What these 2 are basically saying is that the endpoints,"},{"Start":"01:12.260 ","End":"01:16.190","Text":"we have a removable discontinuity,"},{"Start":"01:16.190 ","End":"01:22.430","Text":"I mean, so the trick is now to extend it to the endpoints also."},{"Start":"01:22.430 ","End":"01:25.880","Text":"Commonly an extension of a function is written with a tilde,"},{"Start":"01:25.880 ","End":"01:27.520","Text":"a wave over it."},{"Start":"01:27.520 ","End":"01:33.320","Text":"We\u0027ll define f tilde of x to be the same as f(x) on the interval (a,b)."},{"Start":"01:33.320 ","End":"01:35.915","Text":"At the left endpoint a,"},{"Start":"01:35.915 ","End":"01:37.265","Text":"we\u0027ll define it as c,"},{"Start":"01:37.265 ","End":"01:39.020","Text":"at the right endpoint b,"},{"Start":"01:39.020 ","End":"01:43.190","Text":"we\u0027ll define it as d. This f tilde is continuous,"},{"Start":"01:43.190 ","End":"01:47.620","Text":"not only in the middle but at the endpoints also because of this."},{"Start":"01:47.620 ","End":"01:51.920","Text":"By Cantor\u0027s theorem, since f is continuous on"},{"Start":"01:51.920 ","End":"01:56.170","Text":"the closed interval (a,b), it\u0027s uniformly continuous."},{"Start":"01:56.170 ","End":"02:00.155","Text":"We had a lemma that if we take a restriction to a subset,"},{"Start":"02:00.155 ","End":"02:02.690","Text":"in this case to the open interval (a,b),"},{"Start":"02:02.690 ","End":"02:05.090","Text":"it\u0027s still uniformly continuous."},{"Start":"02:05.090 ","End":"02:08.885","Text":"That concludes the proof and now let\u0027s do an example."},{"Start":"02:08.885 ","End":"02:13.940","Text":"We have to show that the function f(x) equals sine x over x is"},{"Start":"02:13.940 ","End":"02:19.745","Text":"uniformly continuous on the open interval from 0 to infinity."},{"Start":"02:19.745 ","End":"02:25.025","Text":"Note there\u0027s no problem with the denominator because 0 is excluded from this interval."},{"Start":"02:25.025 ","End":"02:28.415","Text":"Now what we have to show by the proposition is that"},{"Start":"02:28.415 ","End":"02:33.011","Text":"the limit as x goes to 0 from the right and x goes to infinity,"},{"Start":"02:33.011 ","End":"02:37.340","Text":"well, from the left it has to be that these 2 exist in our finite,"},{"Start":"02:37.340 ","End":"02:41.615","Text":"some people say that a limit of infinity means the limit exists."},{"Start":"02:41.615 ","End":"02:43.310","Text":"Some people say it doesn\u0027t exist."},{"Start":"02:43.310 ","End":"02:44.870","Text":"But to remove doubt,"},{"Start":"02:44.870 ","End":"02:47.135","Text":"they exist in finite."},{"Start":"02:47.135 ","End":"02:54.800","Text":"Now, the limit as x goes to 0 of sine x over x is a well-known limit and it\u0027s equal to 1."},{"Start":"02:54.800 ","End":"02:57.185","Text":"It\u0027s actually equal on the left also,"},{"Start":"02:57.185 ","End":"03:00.030","Text":"and it\u0027s a removable singularity."},{"Start":"03:00.140 ","End":"03:02.330","Text":"For the limit at infinity,"},{"Start":"03:02.330 ","End":"03:06.065","Text":"we\u0027ll use Heine\u0027s definition of a limit using sequences."},{"Start":"03:06.065 ","End":"03:09.920","Text":"If x_n is a sequence that goes to infinity,"},{"Start":"03:09.920 ","End":"03:17.050","Text":"then the limit as n goes to infinity of sine x times 1 over x_n."},{"Start":"03:17.050 ","End":"03:20.640","Text":"Just broken this up into sine x times 1 over x."},{"Start":"03:20.640 ","End":"03:25.580","Text":"This is 0 because we have a bounded sequence times the null sequence."},{"Start":"03:25.580 ","End":"03:27.935","Text":"This is bounded by plus or minus 1,"},{"Start":"03:27.935 ","End":"03:29.860","Text":"and this goes to 0."},{"Start":"03:29.860 ","End":"03:33.520","Text":"It\u0027s well-known that bounded times null is null."},{"Start":"03:33.520 ","End":"03:35.780","Text":"Finally a remark."},{"Start":"03:35.780 ","End":"03:41.330","Text":"If you had to show that this function f(x) is continuous on say,"},{"Start":"03:41.330 ","End":"03:44.900","Text":"(0,1) where the endpoint 1 is included,"},{"Start":"03:44.900 ","End":"03:50.825","Text":"then you only have to show the limit as x goes to 0 from the right."},{"Start":"03:50.825 ","End":"03:55.234","Text":"On the closed side you don\u0027t have to prove anything."},{"Start":"03:55.234 ","End":"03:58.920","Text":"With that remark, we conclude this clip."}],"ID":31128},{"Watched":false,"Name":"Sufficient Condition for Uniform Continuity Part 3","Duration":"2m 26s","ChapterTopicVideoID":29501,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.870","Text":"In this clip we come to the third of 5 sufficient conditions for uniform continuity."},{"Start":"00:06.870 ","End":"00:10.694","Text":"If the condition is met then the function is uniformly continuous."},{"Start":"00:10.694 ","End":"00:14.100","Text":"In this case it\u0027s the following expressed to the proposition."},{"Start":"00:14.100 ","End":"00:16.470","Text":"Let f be differentiable on open interval,"},{"Start":"00:16.470 ","End":"00:19.245","Text":"I, and suppose that f\u0027,"},{"Start":"00:19.245 ","End":"00:25.105","Text":"the derivative, is bounded on I then f is uniformly continuous on I."},{"Start":"00:25.105 ","End":"00:27.840","Text":"Now in one of the previous exercises,"},{"Start":"00:27.840 ","End":"00:31.250","Text":"we proved this for a finite interval and"},{"Start":"00:31.250 ","End":"00:35.435","Text":"the infinite case is practically the same so we won\u0027t do that."},{"Start":"00:35.435 ","End":"00:37.805","Text":"We\u0027ll settle for the finite case proof."},{"Start":"00:37.805 ","End":"00:39.155","Text":"As an example,"},{"Start":"00:39.155 ","End":"00:42.440","Text":"we\u0027ll take the function f(x) equals 4x plus sine x and"},{"Start":"00:42.440 ","End":"00:46.040","Text":"show that it\u0027s uniformly continuous on the whole real line."},{"Start":"00:46.040 ","End":"00:48.350","Text":"We\u0027ll prove it by this proposition."},{"Start":"00:48.350 ","End":"00:54.200","Text":"If we just show that f\u0027 is bounded then we\u0027re done. Now, what is f\u0027?"},{"Start":"00:54.200 ","End":"00:56.240","Text":"Well, derivative of x is 4,"},{"Start":"00:56.240 ","End":"01:00.440","Text":"derivative of sine x is cosine x and clearly this is"},{"Start":"01:00.440 ","End":"01:05.815","Text":"between 3 and 5 because cosine x is between minus 1 and 1."},{"Start":"01:05.815 ","End":"01:08.910","Text":"So if you take 4 plus or minus 1,"},{"Start":"01:08.910 ","End":"01:10.705","Text":"that\u0027s between 3 and 5,"},{"Start":"01:10.705 ","End":"01:13.760","Text":"which means if you want to express it in terms of absolute value,"},{"Start":"01:13.760 ","End":"01:18.875","Text":"that the absolute value of f\u0027 of x is less than or equal to the larger of this 2,"},{"Start":"01:18.875 ","End":"01:21.530","Text":"the 5 so f\u0027 is bounded."},{"Start":"01:21.530 ","End":"01:24.245","Text":"Its absolute value is less than or equal to some constant."},{"Start":"01:24.245 ","End":"01:25.805","Text":"That\u0027s all we had to do."},{"Start":"01:25.805 ","End":"01:30.365","Text":"That completes the example and now I\u0027d like to make a remark."},{"Start":"01:30.365 ","End":"01:34.070","Text":"In some books this proposition is stated to hold on"},{"Start":"01:34.070 ","End":"01:37.310","Text":"an open interval because it\u0027s not always"},{"Start":"01:37.310 ","End":"01:40.700","Text":"clear what differentiable means if it\u0027s not on an open interval."},{"Start":"01:40.700 ","End":"01:45.680","Text":"But the proposition is also true if we remove the word open as long as we understand that"},{"Start":"01:45.680 ","End":"01:51.445","Text":"differentiable at an endpoint means that it has a 1-sided derivative there."},{"Start":"01:51.445 ","End":"01:56.585","Text":"For our purposes, we can just drop the word open in this proposition."},{"Start":"01:56.585 ","End":"01:59.280","Text":"To be safe, I\u0027m going to make another change."},{"Start":"01:59.280 ","End":"02:02.210","Text":"Let\u0027s change the interval I,"},{"Start":"02:02.210 ","End":"02:04.625","Text":"here to the interior of I."},{"Start":"02:04.625 ","End":"02:09.695","Text":"What I mean is f\u0027 surely exists on the interior,"},{"Start":"02:09.695 ","End":"02:13.145","Text":"meaning the interval without any endpoints."},{"Start":"02:13.145 ","End":"02:14.750","Text":"For this proposition to work,"},{"Start":"02:14.750 ","End":"02:21.080","Text":"we really only need f\u0027 to exist on the interior and to be bounded there."},{"Start":"02:21.080 ","End":"02:24.530","Text":"I wrote the little circle there means interior."},{"Start":"02:24.530 ","End":"02:27.180","Text":"That concludes this clip."}],"ID":31129},{"Watched":false,"Name":"Sufficient Condition for Uniform Continuity Part 4","Duration":"3m 6s","ChapterTopicVideoID":29502,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.245","Text":"Now another condition for uniform continuity,"},{"Start":"00:04.245 ","End":"00:08.894","Text":"a sufficient condition which you use to prove that a function is uniform continuous."},{"Start":"00:08.894 ","End":"00:12.255","Text":"If a function is continuous and periodic,"},{"Start":"00:12.255 ","End":"00:17.850","Text":"then f is uniformly continuous and we\u0027ll prove it in the following exercise."},{"Start":"00:17.850 ","End":"00:19.710","Text":"Now before the examples,"},{"Start":"00:19.710 ","End":"00:22.140","Text":"a useful claim it\u0027s in 2 parts,"},{"Start":"00:22.140 ","End":"00:24.810","Text":"but really part b) is important."},{"Start":"00:24.810 ","End":"00:32.235","Text":"If we have a periodic function with some period p and g is any function,"},{"Start":"00:32.235 ","End":"00:36.630","Text":"then g(f(x)) is periodic with the same period."},{"Start":"00:36.630 ","End":"00:41.490","Text":"It should be clear that if we put x plus p instead of x,"},{"Start":"00:41.490 ","End":"00:44.120","Text":"and f(x) plus p is the same as f(x) so"},{"Start":"00:44.120 ","End":"00:48.425","Text":"the result comes out the same even if you apply g afterwards."},{"Start":"00:48.425 ","End":"00:50.885","Text":"Part b) is the one that\u0027s useful for us."},{"Start":"00:50.885 ","End":"00:54.950","Text":"If you add continuous here and continuous here,"},{"Start":"00:54.950 ","End":"00:58.635","Text":"what we get is that if f is continuous and periodic,"},{"Start":"00:58.635 ","End":"01:02.795","Text":"the period p and g is not any function but a continuous function,"},{"Start":"01:02.795 ","End":"01:07.100","Text":"then g(f(x)) is continuous and periodic with"},{"Start":"01:07.100 ","End":"01:11.390","Text":"period p. This implies that it\u0027s uniformly continuous."},{"Start":"01:11.390 ","End":"01:13.865","Text":"Now some examples."},{"Start":"01:13.865 ","End":"01:19.070","Text":"The sine function is periodic with period 2 Pi,"},{"Start":"01:19.070 ","End":"01:20.900","Text":"and it\u0027s also continuous,"},{"Start":"01:20.900 ","End":"01:24.115","Text":"and therefore it\u0027s uniformly continuous."},{"Start":"01:24.115 ","End":"01:30.710","Text":"Cosine of 4x is periodic and continuous and therefore uniformly continuous."},{"Start":"01:30.710 ","End":"01:33.530","Text":"By the way, there\u0027s a way of computing the period."},{"Start":"01:33.530 ","End":"01:39.695","Text":"In general cosine of ax plus b or sine of ax plus b are both"},{"Start":"01:39.695 ","End":"01:47.090","Text":"periodic with period 2 Pi divided by a and hence uniformly continuous."},{"Start":"01:47.090 ","End":"01:50.420","Text":"Now, this one is a composition of functions."},{"Start":"01:50.420 ","End":"01:54.800","Text":"We take sine of 10x and then we take the function e to"},{"Start":"01:54.800 ","End":"02:00.065","Text":"the power of applied to this function so by what we said,"},{"Start":"02:00.065 ","End":"02:08.345","Text":"this is periodic with the same period as sine 10x which is Pi over 5 or 2 Pi over 10."},{"Start":"02:08.345 ","End":"02:11.680","Text":"Next another composition,"},{"Start":"02:11.680 ","End":"02:17.340","Text":"cosine 2x is periodic with period Pi so if we take the cube root,"},{"Start":"02:17.340 ","End":"02:20.120","Text":"it\u0027s also periodic with period Pi and it\u0027s"},{"Start":"02:20.120 ","End":"02:24.050","Text":"a continuous function of a continuous function so it\u0027s continuous,"},{"Start":"02:24.050 ","End":"02:26.585","Text":"therefore it\u0027s uniformly continuous."},{"Start":"02:26.585 ","End":"02:29.155","Text":"Now the last example,"},{"Start":"02:29.155 ","End":"02:34.590","Text":"z(x) is natural log of sine squared x plus 4."},{"Start":"02:34.590 ","End":"02:38.870","Text":"This is actually a composition of a composition of functions."},{"Start":"02:38.870 ","End":"02:42.665","Text":"Sine x is continuous with period 2 Pi."},{"Start":"02:42.665 ","End":"02:48.680","Text":"If I take the function x^2 plus 4 and combine it with sine x,"},{"Start":"02:48.680 ","End":"02:50.840","Text":"I get sine squared x plus 4,"},{"Start":"02:50.840 ","End":"02:54.020","Text":"and then take the function natural log of and compose it"},{"Start":"02:54.020 ","End":"02:57.290","Text":"with that so this is a composition of functions,"},{"Start":"02:57.290 ","End":"02:59.750","Text":"but the innermost one is sine x,"},{"Start":"02:59.750 ","End":"03:03.330","Text":"and this is also uniformly continuous."},{"Start":"03:03.330 ","End":"03:07.600","Text":"With that example, we\u0027ll conclude this clip."}],"ID":31130},{"Watched":false,"Name":"Sufficient Condition for Uniform Continuity Part 5","Duration":"3m 6s","ChapterTopicVideoID":29503,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:07.440","Text":"This is our 5th and last of the sufficient conditions for uniform continuity."},{"Start":"00:07.440 ","End":"00:11.400","Text":"What this says informally is that the sum of"},{"Start":"00:11.400 ","End":"00:17.130","Text":"2 uniformly continuous functions is also uniformly continuous."},{"Start":"00:17.130 ","End":"00:20.595","Text":"We already proved this in a previous exercise,"},{"Start":"00:20.595 ","End":"00:23.970","Text":"so let\u0027s just do an example to show how to use this."},{"Start":"00:23.970 ","End":"00:26.805","Text":"We\u0027ll show that this function f(x),"},{"Start":"00:26.805 ","End":"00:29.805","Text":"which is x^3 over 1 plus x plus x^2,"},{"Start":"00:29.805 ","End":"00:35.510","Text":"is uniformly continuous on the interval from 0 to infinity,"},{"Start":"00:35.510 ","End":"00:38.195","Text":"the open interval, the positive numbers."},{"Start":"00:38.195 ","End":"00:43.430","Text":"Now note that the denominator is not 0 on this interval,"},{"Start":"00:43.430 ","End":"00:45.445","Text":"if x is positive,"},{"Start":"00:45.445 ","End":"00:49.920","Text":"then so is 1 plus x plus x^2 and so it\u0027s not going to be 0,"},{"Start":"00:49.920 ","End":"00:55.790","Text":"so this is defined everywhere on this interval and now let\u0027s break it up into 2."},{"Start":"00:55.790 ","End":"00:58.760","Text":"If we do division of polynomials,"},{"Start":"00:58.760 ","End":"01:00.320","Text":"and if you remember how to do that,"},{"Start":"01:00.320 ","End":"01:06.655","Text":"then the result comes out that this goes into this x minus 1 times with remainder 1,"},{"Start":"01:06.655 ","End":"01:09.560","Text":"and this is the expression that we get."},{"Start":"01:09.560 ","End":"01:14.780","Text":"You can also do it without division of polynomials as follows."},{"Start":"01:14.780 ","End":"01:19.870","Text":"Break the numerator up into x^3 minus 1 plus 1,"},{"Start":"01:19.870 ","End":"01:22.955","Text":"and we do this because we\u0027re going to use the formula from"},{"Start":"01:22.955 ","End":"01:26.480","Text":"algebra for difference of cubes, which is this,"},{"Start":"01:26.480 ","End":"01:30.425","Text":"we can apply that here because this is x^3 minus 1^3,"},{"Start":"01:30.425 ","End":"01:36.825","Text":"so we get that this is x minus 1 times, following this pattern,"},{"Start":"01:36.825 ","End":"01:40.545","Text":"x^2 plus 1 times x plus 1^2,"},{"Start":"01:40.545 ","End":"01:44.270","Text":"and this happens to be the same as what\u0027s on the denominator,"},{"Start":"01:44.270 ","End":"01:46.355","Text":"so this cancels with this,"},{"Start":"01:46.355 ","End":"01:50.225","Text":"and we\u0027re left with x minus 1 here plus this bit."},{"Start":"01:50.225 ","End":"01:54.730","Text":"Now let\u0027s call this function f(x), the whole thing."},{"Start":"01:54.730 ","End":"01:59.235","Text":"This is going to be our g(x) and this is h(x),"},{"Start":"01:59.235 ","End":"02:02.360","Text":"and we\u0027re going to use our sufficient condition that if"},{"Start":"02:02.360 ","End":"02:05.750","Text":"this is uniformly continuous and this is uniformly continuous,"},{"Start":"02:05.750 ","End":"02:07.450","Text":"and so is this."},{"Start":"02:07.450 ","End":"02:10.470","Text":"This is just in words what I just said,"},{"Start":"02:10.470 ","End":"02:15.120","Text":"so g(x) has a derivative of 1,"},{"Start":"02:15.120 ","End":"02:18.935","Text":"so we can use the condition on the bounded derivative."},{"Start":"02:18.935 ","End":"02:23.825","Text":"So g is uniformly continuous on the interval, now what about h?"},{"Start":"02:23.825 ","End":"02:26.810","Text":"What I\u0027m going to do is show that h is uniformly"},{"Start":"02:26.810 ","End":"02:29.540","Text":"continuous not only on the open interval,"},{"Start":"02:29.540 ","End":"02:33.295","Text":"but even on the half-closed interval."},{"Start":"02:33.295 ","End":"02:36.840","Text":"When the function is continuous on an interval,"},{"Start":"02:36.840 ","End":"02:39.020","Text":"to show that it\u0027s uniformly continuous,"},{"Start":"02:39.020 ","End":"02:44.180","Text":"we just have to show that at the open endpoints it has a finite limit."},{"Start":"02:44.180 ","End":"02:48.215","Text":"I just was being lazy when I added the 0 here,"},{"Start":"02:48.215 ","End":"02:51.230","Text":"because otherwise we\u0027d have to show 2 limits and I could"},{"Start":"02:51.230 ","End":"02:54.430","Text":"get by with just showing the 1 limit at infinity."},{"Start":"02:54.430 ","End":"02:57.090","Text":"Clearly, the denominator goes to infinity,"},{"Start":"02:57.090 ","End":"03:01.445","Text":"so we get then 1 over infinity situation which is 0,"},{"Start":"03:01.445 ","End":"03:04.100","Text":"and that\u0027s the last piece of the puzzle,"},{"Start":"03:04.100 ","End":"03:06.660","Text":"so we are done."}],"ID":31131},{"Watched":false,"Name":"Exercise 1","Duration":"2m 1s","ChapterTopicVideoID":29504,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.930","Text":"In this exercise, we\u0027ll show that the function f(x),"},{"Start":"00:03.930 ","End":"00:06.630","Text":"which is equal to 1 minus cosine x over x squared,"},{"Start":"00:06.630 ","End":"00:10.260","Text":"is uniformly continuous on the interval 0,"},{"Start":"00:10.260 ","End":"00:13.365","Text":"1, open at 0 and closed at 1."},{"Start":"00:13.365 ","End":"00:15.750","Text":"F is an elementary function,"},{"Start":"00:15.750 ","End":"00:19.380","Text":"and there\u0027s a well-known fact that"},{"Start":"00:19.380 ","End":"00:23.880","Text":"all elementary functions are continuous everywhere that they are defined."},{"Start":"00:23.880 ","End":"00:27.194","Text":"Now, this is defined everywhere on this interval."},{"Start":"00:27.194 ","End":"00:28.860","Text":"It isn\u0027t defined at 0,"},{"Start":"00:28.860 ","End":"00:32.275","Text":"but that\u0027s not part of our interval, so we\u0027re okay."},{"Start":"00:32.275 ","End":"00:37.160","Text":"We\u0027ll use 1 of the sufficient conditions we learned about wrapped up as"},{"Start":"00:37.160 ","End":"00:43.550","Text":"a proposition that if a function f is continuous on an interval and it has a limit,"},{"Start":"00:43.550 ","End":"00:48.485","Text":"which automatically implies finite limit at each of the endpoints of I,"},{"Start":"00:48.485 ","End":"00:50.890","Text":"then f is uniformly continuous on I."},{"Start":"00:50.890 ","End":"00:53.555","Text":"We don\u0027t have to check both end points,"},{"Start":"00:53.555 ","End":"00:55.980","Text":"only the ones that are open,"},{"Start":"00:55.980 ","End":"00:59.365","Text":"and in our case we only have to check at 0."},{"Start":"00:59.365 ","End":"01:05.390","Text":"We have to show that the limit as x goes to 0 from the right of f exists."},{"Start":"01:05.390 ","End":"01:09.335","Text":"In this course, exists means and is finite."},{"Start":"01:09.335 ","End":"01:13.925","Text":"This is actually a famous limit and it\u0027s known to be a half."},{"Start":"01:13.925 ","End":"01:15.350","Text":"But if you didn\u0027t know that,"},{"Start":"01:15.350 ","End":"01:21.994","Text":"you could also do it by L\u0027Hopital\u0027s rule because we have a situation of 0 over 0."},{"Start":"01:21.994 ","End":"01:24.590","Text":"Because when x goes to 0 of cosine x goes to 1,"},{"Start":"01:24.590 ","End":"01:28.250","Text":"so we have 1 minus 1 over 0 squared 0 over 0,"},{"Start":"01:28.250 ","End":"01:30.230","Text":"so that\u0027s the case for L\u0027Hopital,"},{"Start":"01:30.230 ","End":"01:36.080","Text":"so we can differentiate separately numerator and denominator and get this."},{"Start":"01:36.080 ","End":"01:37.895","Text":"We could differentiate again,"},{"Start":"01:37.895 ","End":"01:41.750","Text":"but this is equal to a famous limit,"},{"Start":"01:41.750 ","End":"01:42.890","Text":"sine x over x,"},{"Start":"01:42.890 ","End":"01:44.660","Text":"which is more famous than this."},{"Start":"01:44.660 ","End":"01:47.330","Text":"Still, if you differentiate it, again,"},{"Start":"01:47.330 ","End":"01:50.405","Text":"you would get cosine x over 1,"},{"Start":"01:50.405 ","End":"01:53.120","Text":"and when x goes to 0, that goes to 1."},{"Start":"01:53.120 ","End":"01:56.510","Text":"But we have the 1/2 here also, so it\u0027s 1/2."},{"Start":"01:56.510 ","End":"01:59.330","Text":"Everything is fine this exists and is finite,"},{"Start":"01:59.330 ","End":"02:02.190","Text":"and that concludes the exercise."}],"ID":31132},{"Watched":false,"Name":"Exercise 2","Duration":"1m 42s","ChapterTopicVideoID":29505,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In this exercise, we have to show that the function f(x) which is equal to"},{"Start":"00:04.830 ","End":"00:10.230","Text":"xe to the -x^2 is uniformly continuous on the real line."},{"Start":"00:10.230 ","End":"00:19.305","Text":"This is an elementary function and it\u0027s defined everywhere, so it\u0027s continuous."},{"Start":"00:19.305 ","End":"00:22.950","Text":"We\u0027re going to use the proposition that if we have"},{"Start":"00:22.950 ","End":"00:27.893","Text":"a continuous function on an interval and has a limit at each of the endpoints,"},{"Start":"00:27.893 ","End":"00:30.558","Text":"we only have to check at the open endpoints,"},{"Start":"00:30.558 ","End":"00:33.870","Text":"then f is uniformly continuous on the interval."},{"Start":"00:33.870 ","End":"00:36.300","Text":"The real line can be expressed as an interval,"},{"Start":"00:36.300 ","End":"00:38.271","Text":"interval minus infinity to infinity,"},{"Start":"00:38.271 ","End":"00:41.164","Text":"so we have to check at both endpoints."},{"Start":"00:41.164 ","End":"00:44.240","Text":"In other words, we have to show that both the limits as x goes to"},{"Start":"00:44.240 ","End":"00:47.755","Text":"infinity or minus infinity of f(x) exist."},{"Start":"00:47.755 ","End":"00:50.000","Text":"Limit as x goes to infinity,"},{"Start":"00:50.000 ","End":"00:54.155","Text":"we can write it as x over e to the x^2,"},{"Start":"00:54.155 ","End":"00:57.890","Text":"and we get something of the form infinity over infinity,"},{"Start":"00:57.890 ","End":"01:02.150","Text":"which is a sign that we should use L\u0027Hopital\u0027s rule."},{"Start":"01:02.150 ","End":"01:03.800","Text":"Also for the other side,"},{"Start":"01:03.800 ","End":"01:09.589","Text":"for minus infinity, we get similarly minus infinity over infinity."},{"Start":"01:09.589 ","End":"01:12.215","Text":"So we\u0027re going to use L\u0027Hopital\u0027s rule for each of them."},{"Start":"01:12.215 ","End":"01:13.580","Text":"For the infinity case,"},{"Start":"01:13.580 ","End":"01:18.035","Text":"we differentiate top and bottom and get 1 over 2xe to the x^2."},{"Start":"01:18.035 ","End":"01:21.121","Text":"This now becomes 1 over infinity, which is 0."},{"Start":"01:21.121 ","End":"01:24.615","Text":"This say, mean symbolically,"},{"Start":"01:24.615 ","End":"01:28.780","Text":"1 over something that tends to infinity, 0."},{"Start":"01:28.780 ","End":"01:32.730","Text":"The other one comes out 1 over minus infinity, which is also 0."},{"Start":"01:32.730 ","End":"01:34.670","Text":"Minus 0 is the same as 0."},{"Start":"01:34.670 ","End":"01:38.030","Text":"Both of these exist on our finite."},{"Start":"01:38.030 ","End":"01:39.605","Text":"That\u0027s all that we had to show."},{"Start":"01:39.605 ","End":"01:42.780","Text":"So that concludes this exercise."}],"ID":31133},{"Watched":false,"Name":"Exercise 3","Duration":"1m 28s","ChapterTopicVideoID":29506,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.675","Text":"In this exercise, we have to show that the function f(x),"},{"Start":"00:03.675 ","End":"00:07.185","Text":"which is equal to 1 over 1 plus e to the 1 over x,"},{"Start":"00:07.185 ","End":"00:14.380","Text":"is uniformly continuous on the interval from 0 to infinity open interval."},{"Start":"00:14.960 ","End":"00:19.250","Text":"When you use the fact that all elementary functions are continuous,"},{"Start":"00:19.250 ","End":"00:22.730","Text":"and this is defined everywhere at 0 to infinity,"},{"Start":"00:22.730 ","End":"00:24.440","Text":"we just have to exclude 0,"},{"Start":"00:24.440 ","End":"00:27.095","Text":"and 0 is excluded from this interval."},{"Start":"00:27.095 ","End":"00:29.390","Text":"So it\u0027s continuous. Now,"},{"Start":"00:29.390 ","End":"00:32.960","Text":"we\u0027ll use the proposition that if we have a function that\u0027s continuous on"},{"Start":"00:32.960 ","End":"00:38.285","Text":"an interval and it has a limit at each of the endpoints,"},{"Start":"00:38.285 ","End":"00:41.480","Text":"then it\u0027s uniformly continuous on the interval."},{"Start":"00:41.480 ","End":"00:44.825","Text":"We have 2 end points, 0 and infinity,"},{"Start":"00:44.825 ","End":"00:45.965","Text":"0 from the right,"},{"Start":"00:45.965 ","End":"00:50.215","Text":"we have to show that both of these exist and are finite."},{"Start":"00:50.215 ","End":"00:53.160","Text":"Well, let\u0027s take this one first."},{"Start":"00:53.160 ","End":"00:55.620","Text":"When x goes to 0 from the right,"},{"Start":"00:55.620 ","End":"01:00.014","Text":"we get 1 over 1 plus e to the 1 over positive 0,"},{"Start":"01:00.014 ","End":"01:03.170","Text":"which is 1 over 1 plus e to the infinity."},{"Start":"01:03.170 ","End":"01:06.079","Text":"1 over infinity is 0,"},{"Start":"01:06.079 ","End":"01:08.215","Text":"exists and is finite."},{"Start":"01:08.215 ","End":"01:09.930","Text":"As for the other one,"},{"Start":"01:09.930 ","End":"01:11.180","Text":"instead of 0 plus,"},{"Start":"01:11.180 ","End":"01:12.500","Text":"we have infinity here."},{"Start":"01:12.500 ","End":"01:14.480","Text":"1 over infinity is 0,"},{"Start":"01:14.480 ","End":"01:18.050","Text":"1 over 1 plus e^0 is 1 over 1 plus 1,"},{"Start":"01:18.050 ","End":"01:21.530","Text":"which is 1/2, also exists and is finite."},{"Start":"01:21.530 ","End":"01:24.320","Text":"So both limits exist and are finite,"},{"Start":"01:24.320 ","End":"01:29.220","Text":"and that\u0027s all we had to do to complete this exercise. So we\u0027re done."}],"ID":31134},{"Watched":false,"Name":"Exercise 4","Duration":"1m 7s","ChapterTopicVideoID":29507,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.660","Text":"In this exercise, we\u0027re going to show that the arctangent function"},{"Start":"00:03.660 ","End":"00:07.290","Text":"is uniformly continuous on the whole real line."},{"Start":"00:07.290 ","End":"00:11.895","Text":"As usual, we\u0027ll note that this is an elementary function."},{"Start":"00:11.895 ","End":"00:15.345","Text":"It\u0027s an inverse trigonometric function."},{"Start":"00:15.345 ","End":"00:20.025","Text":"It\u0027s defined everywhere on the interval, so it\u0027s continuous."},{"Start":"00:20.025 ","End":"00:25.080","Text":"We\u0027ll use the proposition that we always use that if we have a function"},{"Start":"00:25.080 ","End":"00:29.910","Text":"that\u0027s continuous on an interval and it has a finite limit at each of the end points,"},{"Start":"00:29.910 ","End":"00:33.075","Text":"then it\u0027s uniformly continuous on the interval."},{"Start":"00:33.075 ","End":"00:36.535","Text":"We have to show that at the end point,"},{"Start":"00:36.535 ","End":"00:38.735","Text":"which are minus infinity and infinity,"},{"Start":"00:38.735 ","End":"00:40.865","Text":"that these limits exist."},{"Start":"00:40.865 ","End":"00:45.380","Text":"Now it\u0027s just known from trigonometry that when x goes to infinity,"},{"Start":"00:45.380 ","End":"00:48.200","Text":"the arctangent is Pi over 2."},{"Start":"00:48.200 ","End":"00:51.950","Text":"This follows from the fact that when an angle goes to Pi over 2,"},{"Start":"00:51.950 ","End":"00:54.490","Text":"the tangent goes to infinity."},{"Start":"00:54.490 ","End":"00:56.315","Text":"Similarly on the other side,"},{"Start":"00:56.315 ","End":"00:59.090","Text":"it\u0027s minus Pi over 2."},{"Start":"00:59.090 ","End":"01:02.450","Text":"Both of these limits exist on our finite,"},{"Start":"01:02.450 ","End":"01:04.835","Text":"and that\u0027s what we had to show."},{"Start":"01:04.835 ","End":"01:08.010","Text":"That concludes this exercise."}],"ID":31135},{"Watched":false,"Name":"Exercise 5","Duration":"1m 39s","ChapterTopicVideoID":29508,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"In this exercise, we\u0027re asked to prove that the function f(x),"},{"Start":"00:05.790 ","End":"00:07.920","Text":"which is the natural log of x,"},{"Start":"00:07.920 ","End":"00:13.185","Text":"is uniformly continuous for x bigger or equal to 1."},{"Start":"00:13.185 ","End":"00:18.060","Text":"We\u0027re going to use one of the sufficiency conditions from the tutorial."},{"Start":"00:18.060 ","End":"00:19.350","Text":"There were 5 of them."},{"Start":"00:19.350 ","End":"00:22.185","Text":"I think this was Number 3."},{"Start":"00:22.185 ","End":"00:27.245","Text":"Anyway, it says that if f is differentiable on an interval"},{"Start":"00:27.245 ","End":"00:33.530","Text":"and f\u0027 is bounded on the interior of the interval,"},{"Start":"00:33.530 ","End":"00:37.160","Text":"meaning the interval throw away any endpoints,"},{"Start":"00:37.160 ","End":"00:40.819","Text":"then f is uniformly continuous on the interval."},{"Start":"00:40.819 ","End":"00:46.615","Text":"The interior of this interval is the open interval from 1 to infinity."},{"Start":"00:46.615 ","End":"00:51.590","Text":"The derivative of natural log of x is 1 over x."},{"Start":"00:51.590 ","End":"00:53.945","Text":"So let\u0027s prove the boundedness."},{"Start":"00:53.945 ","End":"00:56.405","Text":"But if x is in this interval,"},{"Start":"00:56.405 ","End":"01:00.590","Text":"then it means that x is bigger than 1."},{"Start":"01:00.590 ","End":"01:03.190","Text":"If x is bigger than 1,"},{"Start":"01:03.190 ","End":"01:07.920","Text":"then 1 over x is between 0 and 1."},{"Start":"01:07.920 ","End":"01:09.620","Text":"Because this is bigger than 1,"},{"Start":"01:09.620 ","End":"01:12.320","Text":"we can take the reciprocal and get that the reciprocal of"},{"Start":"01:12.320 ","End":"01:15.185","Text":"this switches direction of inequality,"},{"Start":"01:15.185 ","End":"01:17.045","Text":"so less than 1."},{"Start":"01:17.045 ","End":"01:20.555","Text":"We also get an extra thing that is bigger than 0,"},{"Start":"01:20.555 ","End":"01:22.035","Text":"because if it\u0027s bigger than 1,"},{"Start":"01:22.035 ","End":"01:27.310","Text":"then it\u0027s bigger than 0, and the reciprocal of a positive number is a positive number."},{"Start":"01:27.310 ","End":"01:31.460","Text":"From this, we can actually get double inequality."},{"Start":"01:31.460 ","End":"01:33.800","Text":"This shows that 1 over x is bounded,"},{"Start":"01:33.800 ","End":"01:39.570","Text":"meaning that f\u0027(x) is bounded and that concludes this exercise."}],"ID":31136},{"Watched":false,"Name":"Exercise 6","Duration":"1m 44s","ChapterTopicVideoID":29509,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.350","Text":"In this exercise, we\u0027re going to show that the square root of x function is"},{"Start":"00:04.350 ","End":"00:09.270","Text":"uniformly continuous for x bigger or equal to 1."},{"Start":"00:09.270 ","End":"00:15.690","Text":"We already did this or something similar in the first section using Epsilon Delta,"},{"Start":"00:15.690 ","End":"00:17.700","Text":"but here we\u0027ll prove it using one of"},{"Start":"00:17.700 ","End":"00:21.060","Text":"these sufficiency conditions for absolute continuity."},{"Start":"00:21.060 ","End":"00:22.815","Text":"In fact, we\u0027ll use this one."},{"Start":"00:22.815 ","End":"00:26.759","Text":"That if f is continuous on an interval,"},{"Start":"00:26.759 ","End":"00:28.380","Text":"in this case, this,"},{"Start":"00:28.380 ","End":"00:32.745","Text":"and if we show that f\u0027 is bounded on the interior,"},{"Start":"00:32.745 ","End":"00:35.130","Text":"which is the open interval 1 to infinity,"},{"Start":"00:35.130 ","End":"00:40.275","Text":"then we\u0027ve got that f is uniformly continuous on the original interval."},{"Start":"00:40.275 ","End":"00:43.977","Text":"The derivative is 1 over 2 root x."},{"Start":"00:43.977 ","End":"00:48.770","Text":"So we have to show that this is bounded on the open interval from 1 to infinity."},{"Start":"00:48.770 ","End":"00:51.610","Text":"If x is in this interval,"},{"Start":"00:51.610 ","End":"00:57.290","Text":"it means that x is bigger than 1 and so the square root of"},{"Start":"00:57.290 ","End":"01:03.860","Text":"x is bigger than 1 also and so twice root to x is bigger than 2."},{"Start":"01:03.860 ","End":"01:08.045","Text":"Now we want to take a reciprocal and we can leave that for a moment."},{"Start":"01:08.045 ","End":"01:13.250","Text":"We can say that 1 over 2 root x is less than 1/2."},{"Start":"01:13.250 ","End":"01:16.100","Text":"When you take the reciprocal of positive numbers,"},{"Start":"01:16.100 ","End":"01:19.385","Text":"you reverse the direction of inequality."},{"Start":"01:19.385 ","End":"01:26.000","Text":"We also get another inequality because this 2 root x is positive,"},{"Start":"01:26.000 ","End":"01:28.340","Text":"its reciprocal is also positive."},{"Start":"01:28.340 ","End":"01:30.440","Text":"In general, if a is a positive number,"},{"Start":"01:30.440 ","End":"01:32.275","Text":"then so is its reciprocal."},{"Start":"01:32.275 ","End":"01:35.150","Text":"This shows that 1 over 2 root x is bounded."},{"Start":"01:35.150 ","End":"01:36.950","Text":"It\u0027s sandwiched between 0 and 1/2."},{"Start":"01:36.950 ","End":"01:38.450","Text":"So certainly bounded."},{"Start":"01:38.450 ","End":"01:40.565","Text":"This is f\u0027(x)."},{"Start":"01:40.565 ","End":"01:45.270","Text":"We\u0027ve shown what we needed to show and that concludes this exercise."}],"ID":31137},{"Watched":false,"Name":"Exercise 7","Duration":"57s","ChapterTopicVideoID":29510,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.580","Text":"In this exercise, we\u0027re going to prove that"},{"Start":"00:02.580 ","End":"00:07.515","Text":"the arc tangent function is uniformly continuous on all the real line."},{"Start":"00:07.515 ","End":"00:10.590","Text":"We\u0027ve actually proved this in the previous exercise,"},{"Start":"00:10.590 ","End":"00:12.840","Text":"but we\u0027re going to do it differently here."},{"Start":"00:12.840 ","End":"00:19.095","Text":"We\u0027re going to use the proposition that if the derivative of the function is bounded,"},{"Start":"00:19.095 ","End":"00:23.205","Text":"then we have uniform continuity, won\u0027t read it out."},{"Start":"00:23.205 ","End":"00:26.970","Text":"The derivative of the arc tangent is well known."},{"Start":"00:26.970 ","End":"00:29.430","Text":"It\u0027s 1 over 1 plus x squared."},{"Start":"00:29.430 ","End":"00:31.895","Text":"We have to show this is bounded."},{"Start":"00:31.895 ","End":"00:36.700","Text":"First of all, it\u0027s a positive number because everything here is positive."},{"Start":"00:36.700 ","End":"00:40.790","Text":"Also the numerator is less than the denominator."},{"Start":"00:40.790 ","End":"00:42.245","Text":"If you have a fraction,"},{"Start":"00:42.245 ","End":"00:44.285","Text":"the numerator and denominator are positive,"},{"Start":"00:44.285 ","End":"00:47.380","Text":"but the numerator is smaller then obviously it\u0027s less than 1."},{"Start":"00:47.380 ","End":"00:52.835","Text":"F\u0027(x) is bounded between 0 and 1, and it\u0027s bounded."},{"Start":"00:52.835 ","End":"00:57.900","Text":"That\u0027s what we had to show and that concludes this exercise."}],"ID":31138},{"Watched":false,"Name":"Exercise 8","Duration":"1m 17s","ChapterTopicVideoID":29511,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.795","Text":"In this exercise, we\u0027re going to prove that the function f(x) equals"},{"Start":"00:03.795 ","End":"00:08.235","Text":"x^2 over x plus 1 is uniformly continuous on this interval,"},{"Start":"00:08.235 ","End":"00:11.445","Text":"which means for positive x."},{"Start":"00:11.445 ","End":"00:13.800","Text":"We\u0027re going to use 1 of"},{"Start":"00:13.800 ","End":"00:17.940","Text":"the sufficient conditions from the tutorial I think it\u0027s the third of"},{"Start":"00:17.940 ","End":"00:25.095","Text":"5 which basically says that if f has a bounded derivative,"},{"Start":"00:25.095 ","End":"00:29.430","Text":"then we can deduce uniform continuity."},{"Start":"00:29.430 ","End":"00:34.320","Text":"Now the only place this function is not differentiable is at x equals minus 1"},{"Start":"00:34.320 ","End":"00:39.290","Text":"which is totally outside our interval so we don\u0027t have to worry about that."},{"Start":"00:39.290 ","End":"00:43.225","Text":"Just to show now that it\u0027s bounded on this interval,"},{"Start":"00:43.225 ","End":"00:45.575","Text":"first of all, let\u0027s compute what it is."},{"Start":"00:45.575 ","End":"00:47.945","Text":"Using the quotient rule,"},{"Start":"00:47.945 ","End":"00:49.535","Text":"we get the following,"},{"Start":"00:49.535 ","End":"00:52.325","Text":"which simplifies to this."},{"Start":"00:52.325 ","End":"00:57.680","Text":"Note that the denominator is 1 more than the numerator and in general,"},{"Start":"00:57.680 ","End":"01:03.515","Text":"if a is a positive number then a over a plus 1 is between 0 and 1."},{"Start":"01:03.515 ","End":"01:05.300","Text":"Here, a is this numerator,"},{"Start":"01:05.300 ","End":"01:08.570","Text":"which is certainly positive when x is positive and"},{"Start":"01:08.570 ","End":"01:12.095","Text":"so our derivative f prime is between 0 and 1,"},{"Start":"01:12.095 ","End":"01:17.640","Text":"meaning that it\u0027s bounded and that\u0027s what we had to show so that concludes this clip."}],"ID":31139},{"Watched":false,"Name":"Exercise 9","Duration":"2m 54s","ChapterTopicVideoID":29512,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this exercise, we\u0027re going to prove that the function f(x) equals root x"},{"Start":"00:05.340 ","End":"00:11.400","Text":"sine root x is uniformly continuous on the interval from 0 to infinity,"},{"Start":"00:11.400 ","End":"00:15.000","Text":"including the 0, meaning the non-negative numbers."},{"Start":"00:15.000 ","End":"00:21.170","Text":"We\u0027d like to use the sufficient condition that if the derivative is bounded,"},{"Start":"00:21.170 ","End":"00:23.165","Text":"then it\u0027s uniformly continuous."},{"Start":"00:23.165 ","End":"00:26.240","Text":"But not clear if f is differentiable at 0,"},{"Start":"00:26.240 ","End":"00:29.128","Text":"not even if it has a right derivative"},{"Start":"00:29.128 ","End":"00:33.200","Text":"because the square root of x as the derivative of 1 over 2 root x."},{"Start":"00:33.200 ","End":"00:38.270","Text":"It\u0027s not clear if f is differentiable at 0 or even if it has a right derivative there."},{"Start":"00:38.270 ","End":"00:45.695","Text":"What we\u0027re going to do is break it up into the union of 2 intervals which fit together."},{"Start":"00:45.695 ","End":"00:49.408","Text":"This interval is closed on the right and this interval is closed on the left,"},{"Start":"00:49.408 ","End":"00:51.980","Text":"and we had an exercise to show that if a function is"},{"Start":"00:51.980 ","End":"00:54.755","Text":"uniformly continuous on here and on here,"},{"Start":"00:54.755 ","End":"00:58.460","Text":"then it\u0027s uniformly continuous on the union, which is this."},{"Start":"00:58.460 ","End":"01:00.540","Text":"I wrote that out."},{"Start":"01:00.540 ","End":"01:01.810","Text":"Now, as a matter of fact,"},{"Start":"01:01.810 ","End":"01:05.485","Text":"it turns out that this is right differentiable at 0."},{"Start":"01:05.485 ","End":"01:06.790","Text":"Just in case you\u0027re wondering,"},{"Start":"01:06.790 ","End":"01:08.995","Text":"the function starts out looking like this."},{"Start":"01:08.995 ","End":"01:12.190","Text":"In fact, the right derivative exists here and is equal to 1,"},{"Start":"01:12.190 ","End":"01:14.680","Text":"but it\u0027s the whole exercise to prove that."},{"Start":"01:14.680 ","End":"01:17.290","Text":"We use this technique of breaking it up into 2."},{"Start":"01:17.290 ","End":"01:20.590","Text":"On 0,1, there is no problem because this is"},{"Start":"01:20.590 ","End":"01:24.280","Text":"a closed and bounded interval and f is continuous on it,"},{"Start":"01:24.280 ","End":"01:27.545","Text":"and so it\u0027s uniformly continuous."},{"Start":"01:27.545 ","End":"01:33.040","Text":"We just have to concentrate now on the interval from 1 to infinity and we\u0027ll"},{"Start":"01:33.040 ","End":"01:38.391","Text":"show that the derivative is bounded on the interior of the interval."},{"Start":"01:38.391 ","End":"01:42.235","Text":"Actually, it\u0027s bounded on all of the positive x, so let\u0027s show that."},{"Start":"01:42.235 ","End":"01:44.294","Text":"Let x positive,"},{"Start":"01:44.294 ","End":"01:47.600","Text":"then the derivative is the following,"},{"Start":"01:47.600 ","End":"01:50.740","Text":"using the product and chain rules."},{"Start":"01:50.740 ","End":"01:53.150","Text":"This simplifies to this."},{"Start":"01:53.150 ","End":"01:57.005","Text":"Now, we\u0027ll need a result which is well-known."},{"Start":"01:57.005 ","End":"02:00.800","Text":"The absolute value of sine of Alpha is less than or"},{"Start":"02:00.800 ","End":"02:04.730","Text":"equal to the absolute value of Alpha for any Alpha."},{"Start":"02:04.730 ","End":"02:10.370","Text":"I\u0027ll present here a proof and you can pause and refer to it if you want."},{"Start":"02:10.370 ","End":"02:12.140","Text":"Anyway, back to our exercise,"},{"Start":"02:12.140 ","End":"02:15.335","Text":"so we have this inequality and of course,"},{"Start":"02:15.335 ","End":"02:20.090","Text":"the cosine is also bounded by 1 in absolute value,"},{"Start":"02:20.090 ","End":"02:23.105","Text":"so we have these 2 inequalities."},{"Start":"02:23.105 ","End":"02:30.065","Text":"That will give us that absolute value of f\u0027 is less than or equal to."},{"Start":"02:30.065 ","End":"02:31.790","Text":"Here from this inequality,"},{"Start":"02:31.790 ","End":"02:37.190","Text":"we replace sine root x by root x and here we\u0027ll replace cosine root x by 1,"},{"Start":"02:37.190 ","End":"02:38.576","Text":"so we get this."},{"Start":"02:38.576 ","End":"02:43.820","Text":"Root x cancels, we get a 1/2 plus 1/2, which is 1."},{"Start":"02:43.820 ","End":"02:48.229","Text":"F\u0027 is bounded, it\u0027s absolute value is less than some constant,"},{"Start":"02:48.229 ","End":"02:49.670","Text":"1 in this case."},{"Start":"02:49.670 ","End":"02:51.190","Text":"That\u0027s all we had to show."},{"Start":"02:51.190 ","End":"02:54.580","Text":"That concludes this exercise."}],"ID":31140},{"Watched":false,"Name":"Exercise 10","Duration":"3m 54s","ChapterTopicVideoID":29513,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this exercise, we\u0027re going to prove that the function"},{"Start":"00:03.480 ","End":"00:06.990","Text":"f(x) equals x cosine 1 over x is uniformly"},{"Start":"00:06.990 ","End":"00:13.890","Text":"continuous for positive x that is on the interval from 0 to infinity, open."},{"Start":"00:13.890 ","End":"00:18.900","Text":"Now, this function is problematic when x goes to 0."},{"Start":"00:18.900 ","End":"00:21.000","Text":"It\u0027s not defined at 0 even."},{"Start":"00:21.000 ","End":"00:26.445","Text":"To be safe, we\u0027ll split up the interval from 0 to infinity into two pieces,"},{"Start":"00:26.445 ","End":"00:28.995","Text":"from 0-1, including 1,"},{"Start":"00:28.995 ","End":"00:32.035","Text":"and from 1 to infinity including the 1."},{"Start":"00:32.035 ","End":"00:37.265","Text":"We\u0027ve been shown previously that if we have union of 2 intervals with a common endpoint,"},{"Start":"00:37.265 ","End":"00:40.460","Text":"then uniform continuity on one together with"},{"Start":"00:40.460 ","End":"00:45.257","Text":"uniform continuity on the other gives uniform continuity on the union."},{"Start":"00:45.257 ","End":"00:50.165","Text":"We\u0027ll use different techniques for each of the intervals."},{"Start":"00:50.165 ","End":"00:54.330","Text":"Let\u0027s start with the interval 0,1. f is"},{"Start":"00:54.330 ","End":"01:00.170","Text":"an elementary function and is defined everywhere on this interval, so it\u0027s continuous."},{"Start":"01:00.170 ","End":"01:05.540","Text":"Then we\u0027ll use the proposition that if a function is continuous on"},{"Start":"01:05.540 ","End":"01:11.825","Text":"an interval and the limit exists and is finite at the end points where it\u0027s open,"},{"Start":"01:11.825 ","End":"01:14.150","Text":"just that the 0 end point,"},{"Start":"01:14.150 ","End":"01:17.045","Text":"then that\u0027s good for uniform continuity."},{"Start":"01:17.045 ","End":"01:19.895","Text":"We just have to compute 1 limit,"},{"Start":"01:19.895 ","End":"01:23.130","Text":"the right limit at 0 and there\u0027s no need to"},{"Start":"01:23.130 ","End":"01:26.555","Text":"check at 1 though it wouldn\u0027t hurt if you did that also."},{"Start":"01:26.555 ","End":"01:30.265","Text":"For a change, we\u0027ll use Heine\u0027s definition of a limit."},{"Start":"01:30.265 ","End":"01:36.140","Text":"Instead of taking x to 0 plus meaning right limit,"},{"Start":"01:36.140 ","End":"01:39.590","Text":"we\u0027ll take a sequence that goes to 0 from the right,"},{"Start":"01:39.590 ","End":"01:44.095","Text":"and then we\u0027ll compute the limit of x cosine 1 over x,"},{"Start":"01:44.095 ","End":"01:46.095","Text":"replace x by x_n,"},{"Start":"01:46.095 ","End":"01:48.050","Text":"and we\u0027ll show that for any such sequence,"},{"Start":"01:48.050 ","End":"01:50.880","Text":"this limit is 0."},{"Start":"01:50.880 ","End":"01:54.545","Text":"Because what we have is a bounded sequence,"},{"Start":"01:54.545 ","End":"01:57.320","Text":"that\u0027s the cosine, it\u0027s between plus and minus 1,"},{"Start":"01:57.320 ","End":"01:58.865","Text":"and a null sequence,"},{"Start":"01:58.865 ","End":"02:01.145","Text":"x_n that goes to 0,"},{"Start":"02:01.145 ","End":"02:03.350","Text":"and the product will also be a null sequence,"},{"Start":"02:03.350 ","End":"02:05.045","Text":"so it will go to 0."},{"Start":"02:05.045 ","End":"02:07.295","Text":"By the Heine\u0027s definition,"},{"Start":"02:07.295 ","End":"02:14.405","Text":"the limit of f(x) or x cosine 1 over x is 0 as x goes to 0 from the right."},{"Start":"02:14.405 ","End":"02:18.110","Text":"That takes care of the 0,1 interval."},{"Start":"02:18.110 ","End":"02:21.455","Text":"Now we need to do the 1 infinity interval."},{"Start":"02:21.455 ","End":"02:25.595","Text":"If you don\u0027t like Heine\u0027s definition of a limit,"},{"Start":"02:25.595 ","End":"02:27.650","Text":"you could use the sandwich theorem."},{"Start":"02:27.650 ","End":"02:34.820","Text":"We could say that x cosine 1 over x is sandwiched between minus x and x for positive x,"},{"Start":"02:34.820 ","End":"02:37.700","Text":"because this is between minus 1 and 1."},{"Start":"02:37.700 ","End":"02:42.080","Text":"Then you could apply the sandwich theorem as x goes to 0 and this will go to 0,"},{"Start":"02:42.080 ","End":"02:43.310","Text":"this will go to minus 0,"},{"Start":"02:43.310 ","End":"02:44.480","Text":"which is also 0,"},{"Start":"02:44.480 ","End":"02:46.885","Text":"and this is forced to go to 0."},{"Start":"02:46.885 ","End":"02:50.420","Text":"Getting back to the 1 infinity interval,"},{"Start":"02:50.420 ","End":"02:54.560","Text":"we\u0027ll use another proposition that says we just have to show that"},{"Start":"02:54.560 ","End":"03:00.150","Text":"the derivative is bounded and you don\u0027t have to check it that at the end points."},{"Start":"03:00.740 ","End":"03:04.520","Text":"f(x) is x cosine 1 over x,"},{"Start":"03:04.520 ","End":"03:08.045","Text":"and using the product rule and the chain rule,"},{"Start":"03:08.045 ","End":"03:10.790","Text":"we get that f\u0027(x) is this,"},{"Start":"03:10.790 ","End":"03:16.580","Text":"and let\u0027s simplify that a bit and we get cosine 1 over x plus 1 over x sine 1 over x."},{"Start":"03:16.580 ","End":"03:18.140","Text":"Show it\u0027s bounded."},{"Start":"03:18.140 ","End":"03:22.445","Text":"We\u0027ll take the absolute value of the derivative and we get,"},{"Start":"03:22.445 ","End":"03:26.180","Text":"using the triangle inequality, the following,"},{"Start":"03:26.180 ","End":"03:30.290","Text":"and then cosine of anything is less than or equal to 1."},{"Start":"03:30.290 ","End":"03:31.985","Text":"Similarly for sine,"},{"Start":"03:31.985 ","End":"03:38.610","Text":"and 1 over x is less than or equal to 1 for x bigger than 1."},{"Start":"03:38.610 ","End":"03:44.330","Text":"All this comes to less than or equal to 1 plus 1 times 1, so it\u0027s 2."},{"Start":"03:44.330 ","End":"03:46.744","Text":"Since the absolute value is bounded,"},{"Start":"03:46.744 ","End":"03:48.935","Text":"the derivative of f is bounded there,"},{"Start":"03:48.935 ","End":"03:51.455","Text":"and that\u0027s all we had left to show."},{"Start":"03:51.455 ","End":"03:54.540","Text":"That concludes this exercise."}],"ID":31141},{"Watched":false,"Name":"Exercise 11","Duration":"3m 25s","ChapterTopicVideoID":29514,"CourseChapterTopicPlaylistID":294557,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.000","Text":"In this exercise, we\u0027re going to prove the claim we made in"},{"Start":"00:04.000 ","End":"00:09.235","Text":"the tutorial that if a function is continuous and periodic,"},{"Start":"00:09.235 ","End":"00:11.815","Text":"then it\u0027s uniformly continuous."},{"Start":"00:11.815 ","End":"00:13.135","Text":"So here goes,"},{"Start":"00:13.135 ","End":"00:14.740","Text":"for if it\u0027s periodic,"},{"Start":"00:14.740 ","End":"00:17.125","Text":"it has some period p,"},{"Start":"00:17.125 ","End":"00:20.410","Text":"which means that if we take f(x),"},{"Start":"00:20.410 ","End":"00:28.180","Text":"will be the same as f(x) plus any whole number times p. Because f is continuous,"},{"Start":"00:28.180 ","End":"00:32.650","Text":"it\u0027s going to be uniformly continuous on the closed interval in R,"},{"Start":"00:32.650 ","End":"00:38.695","Text":"and we\u0027ll take the interval 0 to p. If we translate that into Epsilon Delta,"},{"Start":"00:38.695 ","End":"00:40.630","Text":"for any Epsilon bigger than 0,"},{"Start":"00:40.630 ","End":"00:44.120","Text":"there exists a Delta such that for all x,"},{"Start":"00:44.120 ","End":"00:45.935","Text":"y in this interval,"},{"Start":"00:45.935 ","End":"00:48.455","Text":"if x is close to y within Delta,"},{"Start":"00:48.455 ","End":"00:52.535","Text":"then f(x) will be close to f(y) within Epsilon."},{"Start":"00:52.535 ","End":"00:57.245","Text":"We can assume that Delta is less than or equal to p, if it isn\u0027t,"},{"Start":"00:57.245 ","End":"00:59.105","Text":"just shrink Delta,"},{"Start":"00:59.105 ","End":"01:02.030","Text":"make it p. Because if you take a smaller Delta,"},{"Start":"01:02.030 ","End":"01:04.000","Text":"it\u0027s also going to work."},{"Start":"01:04.000 ","End":"01:09.890","Text":"Now, I will show that this Delta is good for this Epsilon in all of the reals."},{"Start":"01:09.890 ","End":"01:11.120","Text":"In other words, for any 2x,"},{"Start":"01:11.120 ","End":"01:12.785","Text":"y on the real line,"},{"Start":"01:12.785 ","End":"01:15.220","Text":"not just from 0 to p,"},{"Start":"01:15.220 ","End":"01:17.260","Text":"if x is close to y within Delta,"},{"Start":"01:17.260 ","End":"01:19.650","Text":"f(x) is close to f(y) within Epsilon."},{"Start":"01:19.650 ","End":"01:23.300","Text":"We can always assume that x is less than or equal to y,"},{"Start":"01:23.300 ","End":"01:26.630","Text":"if not, then just swap x and y around."},{"Start":"01:26.630 ","End":"01:29.705","Text":"So given x and y,"},{"Start":"01:29.705 ","End":"01:32.570","Text":"there exists a whole number n,"},{"Start":"01:32.570 ","End":"01:35.455","Text":"such that if you move x,"},{"Start":"01:35.455 ","End":"01:37.665","Text":"you add np to it,"},{"Start":"01:37.665 ","End":"01:40.145","Text":"for some whole number n,"},{"Start":"01:40.145 ","End":"01:42.470","Text":"then we get within the interval 0,"},{"Start":"01:42.470 ","End":"01:49.430","Text":"p, we start from x and it\u0027s in some interval of length p. So just keep moving it."},{"Start":"01:49.430 ","End":"01:51.440","Text":"In this case, n would be negative,"},{"Start":"01:51.440 ","End":"01:54.050","Text":"or possibly this should be minus np."},{"Start":"01:54.050 ","End":"01:55.955","Text":"Anyway, you get the idea."},{"Start":"01:55.955 ","End":"01:59.660","Text":"So if we move x down by np will be in here."},{"Start":"01:59.660 ","End":"02:03.915","Text":"Now, y_1 might not be in this same interval,"},{"Start":"02:03.915 ","End":"02:09.560","Text":"we\u0027re adding Delta which is less than p. So it could still be in 0,"},{"Start":"02:09.560 ","End":"02:14.695","Text":"p, but at most it will be in the interval from 0 to 2p, let\u0027s write that."},{"Start":"02:14.695 ","End":"02:18.360","Text":"If we let y_1 equals y plus np,"},{"Start":"02:18.360 ","End":"02:20.380","Text":"again, it should be minus,"},{"Start":"02:20.380 ","End":"02:25.100","Text":"because the difference between x_1 and y_1 is the same"},{"Start":"02:25.100 ","End":"02:29.450","Text":"as the distance between x and y and that\u0027s less than Delta,"},{"Start":"02:29.450 ","End":"02:33.390","Text":"which means that it\u0027s less than p. So for most we\u0027re adding p,"},{"Start":"02:33.390 ","End":"02:36.540","Text":"we still have to stay within 0, 2P."},{"Start":"02:36.540 ","End":"02:42.750","Text":"Now that we have x_1 and y_1 both within the interval from 0 to 2p,"},{"Start":"02:42.750 ","End":"02:46.310","Text":"and we also have that distance, like I said,"},{"Start":"02:46.310 ","End":"02:48.620","Text":"is the same as the distance between x and y,"},{"Start":"02:48.620 ","End":"02:50.590","Text":"which is less than Delta,"},{"Start":"02:50.590 ","End":"02:57.020","Text":"what we get is that distance from f(x) to f(y) is the same as the distance"},{"Start":"02:57.020 ","End":"03:03.345","Text":"from f(x) plus np to f(y) plus np."},{"Start":"03:03.345 ","End":"03:13.355","Text":"By periodicity, this f(x) is the same as f(x) plus a whole number times p. Anyway,"},{"Start":"03:13.355 ","End":"03:16.370","Text":"this and this are just x_1 and y_1,"},{"Start":"03:16.370 ","End":"03:17.540","Text":"which is less than Epsilon."},{"Start":"03:17.540 ","End":"03:20.840","Text":"So we have that f(x) minus f(y) absolute value is less than"},{"Start":"03:20.840 ","End":"03:25.200","Text":"Epsilon as required and that completes the proof."}],"ID":31142}],"Thumbnail":null,"ID":294557},{"Name":"Necessary Conditions for Uniform Continuity","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Necessary Condition for Uniform Continuity Part 1","Duration":"7m 9s","ChapterTopicVideoID":29522,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29522.jpeg","UploadDate":"2022-07-21T15:50:45.4370000","DurationForVideoObject":"PT7M9S","Description":null,"MetaTitle":"Necessary Condition for Uniform Continuity Part 1: Video + Workbook | Proprep","MetaDescription":"Uniform Continuity - Necessary Conditions for Uniform Continuity. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/calculus-i%2c-ii-and-iii/uniform-continuity/necessary-conditions-for-uniform-continuity/vid31116","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.390","Text":"In this section, we\u0027ll talk about necessary conditions for uniform continuity."},{"Start":"00:06.390 ","End":"00:08.040","Text":"In the previous section,"},{"Start":"00:08.040 ","End":"00:11.865","Text":"we talked about sufficient conditions for uniform continuity."},{"Start":"00:11.865 ","End":"00:15.240","Text":"Let me just briefly review those 2 concepts."},{"Start":"00:15.240 ","End":"00:19.350","Text":"If a function satisfies the sufficient conditions, sufficient means enough,"},{"Start":"00:19.350 ","End":"00:21.675","Text":"it\u0027s enough for uniform continuity,"},{"Start":"00:21.675 ","End":"00:24.180","Text":"then it\u0027s uniformly continuous."},{"Start":"00:24.180 ","End":"00:30.495","Text":"A necessary condition is something that uniformly continuous function has to satisfy."},{"Start":"00:30.495 ","End":"00:32.760","Text":"In other words, if it doesn\u0027t satisfy"},{"Start":"00:32.760 ","End":"00:37.170","Text":"a necessary condition then it\u0027s not uniformly continuous."},{"Start":"00:37.170 ","End":"00:41.915","Text":"Usually, sufficient conditions are used for proving uniform continuity,"},{"Start":"00:41.915 ","End":"00:47.885","Text":"whereas necessary conditions are usually used for proving non-uniform continuity."},{"Start":"00:47.885 ","End":"00:51.680","Text":"I started numbering with 0 because after I\u0027d done a few,"},{"Start":"00:51.680 ","End":"00:57.000","Text":"I realized I\u0027d forgotten the trivial one so I put it at the beginning at number 0."},{"Start":"00:57.000 ","End":"01:02.045","Text":"That is, that if f is uniformly continuous, then it\u0027s continuous."},{"Start":"01:02.045 ","End":"01:04.100","Text":"The name implies it,"},{"Start":"01:04.100 ","End":"01:07.070","Text":"but it\u0027s not immediately obvious."},{"Start":"01:07.070 ","End":"01:12.230","Text":"We did prove it in the clip where we defined uniform continuity."},{"Start":"01:12.230 ","End":"01:14.240","Text":"The converse of this,"},{"Start":"01:14.240 ","End":"01:17.840","Text":"which is used to proving non-uniform continuity is that if"},{"Start":"01:17.840 ","End":"01:22.805","Text":"a function f is not continuous then it\u0027s not uniformly continuous."},{"Start":"01:22.805 ","End":"01:25.610","Text":"Let\u0027s do another condition here."},{"Start":"01:25.610 ","End":"01:30.035","Text":"Necessary condition number 1 can be stated as a proposition"},{"Start":"01:30.035 ","End":"01:34.669","Text":"that if f is uniformly continuous on a finite interval,"},{"Start":"01:34.669 ","End":"01:37.210","Text":"then it\u0027s bounded on that interval."},{"Start":"01:37.210 ","End":"01:40.925","Text":"If we take the logical converse,"},{"Start":"01:40.925 ","End":"01:44.600","Text":"then if f is unbounded on a finite interval,"},{"Start":"01:44.600 ","End":"01:48.305","Text":"then it\u0027s not uniformly continuous on that interval."},{"Start":"01:48.305 ","End":"01:50.795","Text":"I\u0027ll give an example,"},{"Start":"01:50.795 ","End":"01:54.005","Text":"f(x) equals natural log of x."},{"Start":"01:54.005 ","End":"02:02.175","Text":"It\u0027s not uniformly continuous on the open interval 0,1 because it\u0027s not bounded on 0,1."},{"Start":"02:02.175 ","End":"02:03.770","Text":"How do we know it\u0027s not bounded?"},{"Start":"02:03.770 ","End":"02:07.280","Text":"Because the limit as x goes to 0 from the right and"},{"Start":"02:07.280 ","End":"02:11.705","Text":"the natural log of x is minus infinity, so not bounded."},{"Start":"02:11.705 ","End":"02:20.750","Text":"Another example, f(x)=1/x is not uniformly continuous on the interval 0,3,"},{"Start":"02:20.750 ","End":"02:23.810","Text":"since it\u0027s not bounded on 0,3."},{"Start":"02:23.810 ","End":"02:26.625","Text":"3 is not the important thing, the 0 is."},{"Start":"02:26.625 ","End":"02:30.590","Text":"It\u0027s not bounded on 0,3 because the limit as x goes to"},{"Start":"02:30.590 ","End":"02:35.330","Text":"0(1/x) is infinity, so not bounded."},{"Start":"02:35.330 ","End":"02:39.095","Text":"The next example is like the first one,"},{"Start":"02:39.095 ","End":"02:41.765","Text":"only instead of 0,1,"},{"Start":"02:41.765 ","End":"02:44.560","Text":"we\u0027ll take 0 infinity."},{"Start":"02:44.560 ","End":"02:47.480","Text":"Now this time we can\u0027t apply"},{"Start":"02:47.480 ","End":"02:51.965","Text":"the proposition because the proposition talks about a finite interval."},{"Start":"02:51.965 ","End":"02:57.890","Text":"But we can get around that problem because in Example 1 we showed it\u0027s not uniformly"},{"Start":"02:57.890 ","End":"03:01.280","Text":"continuous on 0,1 and"},{"Start":"03:01.280 ","End":"03:08.310","Text":"0,1 is a subset of 0 infinity."},{"Start":"03:08.310 ","End":"03:13.535","Text":"We learned that if f is uniformly continuous on a set,"},{"Start":"03:13.535 ","End":"03:17.635","Text":"then it\u0027s also uniformly continuous on any subset."},{"Start":"03:17.635 ","End":"03:21.080","Text":"If it were uniformly continuous on 0 infinity,"},{"Start":"03:21.080 ","End":"03:22.790","Text":"it would be uniformly continuous on 0,1,"},{"Start":"03:22.790 ","End":"03:28.280","Text":"which it isn\u0027t, and therefore it isn\u0027t uniformly continuous on 0 infinity."},{"Start":"03:28.280 ","End":"03:30.440","Text":"Now a couple of remarks."},{"Start":"03:30.440 ","End":"03:33.335","Text":"The first remark, well, let me give a background."},{"Start":"03:33.335 ","End":"03:35.375","Text":"In this clip we had 2 conditions,"},{"Start":"03:35.375 ","End":"03:37.490","Text":"we call them number 0 and number 1,"},{"Start":"03:37.490 ","End":"03:40.715","Text":"that if f is uniformly continuous then f is continuous,"},{"Start":"03:40.715 ","End":"03:43.820","Text":"and the other ones, number 1 says that it\u0027s bounded."},{"Start":"03:43.820 ","End":"03:48.830","Text":"But the thing is that even if it\u0027s continuous and bounded on a finite interval,"},{"Start":"03:48.830 ","End":"03:52.400","Text":"it still might not be uniformly continuous."},{"Start":"03:52.400 ","End":"03:53.930","Text":"That\u0027s an example,"},{"Start":"03:53.930 ","End":"03:58.920","Text":"f(x)=sin(1/x) is continuous on"},{"Start":"03:58.920 ","End":"04:03.215","Text":"0,1 because it\u0027s an elementary function and it\u0027s defined on 0,1,"},{"Start":"04:03.215 ","End":"04:04.940","Text":"and it\u0027s also bounded."},{"Start":"04:04.940 ","End":"04:09.555","Text":"We know sine is bounded between plus and minus 1,"},{"Start":"04:09.555 ","End":"04:14.050","Text":"but it\u0027s still not uniformly continuous and we\u0027ll prove this,"},{"Start":"04:14.050 ","End":"04:16.475","Text":"I think it\u0027s in the following clip."},{"Start":"04:16.475 ","End":"04:23.626","Text":"The other remark is that we didn\u0027t just say finite interval for no reason,"},{"Start":"04:23.626 ","End":"04:27.065","Text":"it doesn\u0027t work on an infinite interval, this proposition."},{"Start":"04:27.065 ","End":"04:35.450","Text":"As an example, f(x)=x is uniformly continuous on the real line,"},{"Start":"04:35.450 ","End":"04:38.315","Text":"but it\u0027s not bounded obviously."},{"Start":"04:38.315 ","End":"04:41.320","Text":"All that remains now is the proof."},{"Start":"04:41.320 ","End":"04:44.929","Text":"For the proof, we\u0027ll need the following theorem."},{"Start":"04:44.929 ","End":"04:47.225","Text":"We won\u0027t prove this theorem,"},{"Start":"04:47.225 ","End":"04:48.950","Text":"we\u0027ll just borrow it."},{"Start":"04:48.950 ","End":"04:52.240","Text":"It\u0027s called the continuous extension theorem."},{"Start":"04:52.240 ","End":"04:56.510","Text":"What it says is that if we have an open interval a, b,"},{"Start":"04:56.510 ","End":"05:02.150","Text":"and we have a uniformly continuous function on this open interval,"},{"Start":"05:02.150 ","End":"05:07.145","Text":"then we can define it at the endpoints a and b,"},{"Start":"05:07.145 ","End":"05:11.192","Text":"such that f is continuous on the closed interval a, b."},{"Start":"05:11.192 ","End":"05:12.985","Text":"It\u0027s actually if and only if,"},{"Start":"05:12.985 ","End":"05:14.900","Text":"one direction is trivial,"},{"Start":"05:14.900 ","End":"05:17.870","Text":"if it\u0027s continuous on the closed interval a, b,"},{"Start":"05:17.870 ","End":"05:20.850","Text":"then it\u0027s uniformly continuous on the closed interval a, b,"},{"Start":"05:20.850 ","End":"05:23.855","Text":"and therefore it\u0027s uniformly continuous on the subset,"},{"Start":"05:23.855 ","End":"05:26.045","Text":"which is the open interval a, b."},{"Start":"05:26.045 ","End":"05:28.280","Text":"Anyway, the other direction we won\u0027t prove,"},{"Start":"05:28.280 ","End":"05:34.250","Text":"but we\u0027ll use this to prove that if f is uniformly continuous on a finite interval,"},{"Start":"05:34.250 ","End":"05:37.010","Text":"then it\u0027s bounded on that interval."},{"Start":"05:37.010 ","End":"05:39.750","Text":"The proof uses this theorem."},{"Start":"05:39.750 ","End":"05:44.300","Text":"If it\u0027s uniformly continuous on a, b,"},{"Start":"05:44.300 ","End":"05:49.640","Text":"then we extend it to a continuous function and call that f tilde on all of a,"},{"Start":"05:49.640 ","End":"05:53.465","Text":"b. F tilde is the same as f on the open interval,"},{"Start":"05:53.465 ","End":"05:56.740","Text":"but it\u0027s also defined on the endpoints."},{"Start":"05:56.740 ","End":"06:00.275","Text":"Now, f tilde is bounded because"},{"Start":"06:00.275 ","End":"06:04.310","Text":"continuous function on a closed finite interval is bounded."},{"Start":"06:04.310 ","End":"06:09.980","Text":"If it\u0027s bounded, then so is little f. It\u0027s bounded on all of a,"},{"Start":"06:09.980 ","End":"06:11.558","Text":"b with the endpoint,"},{"Start":"06:11.558 ","End":"06:14.030","Text":"it\u0027s bounded on a, b without the endpoints."},{"Start":"06:14.030 ","End":"06:19.670","Text":"So far we\u0027ve just proved the proposition for the case of an open interval a,"},{"Start":"06:19.670 ","End":"06:23.165","Text":"b, but that\u0027s not the only finite interval."},{"Start":"06:23.165 ","End":"06:25.430","Text":"It could be 3 other things."},{"Start":"06:25.430 ","End":"06:28.850","Text":"It could be open on the left and not on the right,"},{"Start":"06:28.850 ","End":"06:29.915","Text":"open on the right,"},{"Start":"06:29.915 ","End":"06:32.285","Text":"or not open on either end."},{"Start":"06:32.285 ","End":"06:36.995","Text":"We can deal with all these 3 cases simultaneously because in each of the 3 cases,"},{"Start":"06:36.995 ","End":"06:39.865","Text":"the open interval a, b is a subset."},{"Start":"06:39.865 ","End":"06:42.695","Text":"If f is bounded on the subset,"},{"Start":"06:42.695 ","End":"06:44.615","Text":"and it\u0027s bounded on the original,"},{"Start":"06:44.615 ","End":"06:47.390","Text":"because to get to one of these 3 from here,"},{"Start":"06:47.390 ","End":"06:49.730","Text":"we either add 1 point or 2 points,"},{"Start":"06:49.730 ","End":"06:52.715","Text":"either at b or at a or at both of them."},{"Start":"06:52.715 ","End":"06:56.360","Text":"Adding a couple of points won\u0027t change the boundedness."},{"Start":"06:56.360 ","End":"07:00.835","Text":"Something is bounded and then you have 2 more values it\u0027s still bounded."},{"Start":"07:00.835 ","End":"07:03.450","Text":"Here\u0027s that in writing."},{"Start":"07:03.450 ","End":"07:09.700","Text":"That concludes the proof of the proposition and this clip."}],"ID":31116},{"Watched":false,"Name":"Necessary Condition for Uniform Continuity Part 2","Duration":"5m 12s","ChapterTopicVideoID":29523,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.913","Text":"In this clip, we\u0027ll talk about the second necessary condition for uniform continuity."},{"Start":"00:05.913 ","End":"00:07.985","Text":"Expressed as a proposition,"},{"Start":"00:07.985 ","End":"00:13.035","Text":"the condition is that if we have 2 sequences"},{"Start":"00:13.035 ","End":"00:18.675","Text":"in the interval whose distance tends to 0 as n goes to infinity,"},{"Start":"00:18.675 ","End":"00:27.083","Text":"then the distance between f(x_n) and f(y_n) also goes to 0 when n goes to infinity,"},{"Start":"00:27.083 ","End":"00:29.910","Text":"but this is more useful in the converse,"},{"Start":"00:29.910 ","End":"00:33.180","Text":"so we\u0027re going to prove that a function is not uniformly continuous."},{"Start":"00:33.180 ","End":"00:35.985","Text":"If we have 2 sequences,"},{"Start":"00:35.985 ","End":"00:37.210","Text":"x and y_n,"},{"Start":"00:37.210 ","End":"00:40.760","Text":"in our interval, whose distance tends to 0,"},{"Start":"00:40.760 ","End":"00:47.030","Text":"but the distance between f(x_n) and f(y_n) does not go to 0,"},{"Start":"00:47.030 ","End":"00:50.945","Text":"then the function is not uniformly continuous."},{"Start":"00:50.945 ","End":"00:54.305","Text":"In other words, uniformly continuous has the property,"},{"Start":"00:54.305 ","End":"00:58.610","Text":"doesn\u0027t have the property, not uniformly continuous."},{"Start":"00:58.610 ","End":"01:04.780","Text":"Here\u0027s an example of the use of the converse part of the proposition."},{"Start":"01:04.780 ","End":"01:13.580","Text":"We\u0027ll show that f(x)=1/x is not uniformly continuous on the open interval 0,1."},{"Start":"01:13.580 ","End":"01:19.633","Text":"Now, we had similar example in the previous clip,"},{"Start":"01:19.633 ","End":"01:24.314","Text":"and there we showed that it\u0027s not uniformly continuous by saying that it\u0027s unbounded."},{"Start":"01:24.314 ","End":"01:27.950","Text":"It\u0027s unbounded because when x goes to 0 from the right,"},{"Start":"01:27.950 ","End":"01:29.960","Text":"1/x goes to infinity."},{"Start":"01:29.960 ","End":"01:34.130","Text":"I want to show it using the proposition here,"},{"Start":"01:34.130 ","End":"01:36.205","Text":"different way of proving it."},{"Start":"01:36.205 ","End":"01:39.090","Text":"We\u0027ll take 2 sequences,"},{"Start":"01:39.090 ","End":"01:44.275","Text":"x_n is 1 and y_n is 1 plus 1."},{"Start":"01:44.275 ","End":"01:50.190","Text":"Now, the claim is that the distance between x_n and y_n goes to 0,"},{"Start":"01:50.190 ","End":"01:54.345","Text":"but the distance between f(x_n) and f(y_n) does not go to 0."},{"Start":"01:54.345 ","End":"01:56.760","Text":"That\u0027s what this is saying here."},{"Start":"01:56.760 ","End":"01:59.925","Text":"Let\u0027s verify these 2 claims."},{"Start":"01:59.925 ","End":"02:07.850","Text":"The distance between x_n and y_n is 1 minus 1 plus 1 should be absolute value here,"},{"Start":"02:07.850 ","End":"02:09.050","Text":"but this is bigger than this,"},{"Start":"02:09.050 ","End":"02:12.515","Text":"so we don\u0027t need the absolute value and this is equal to,"},{"Start":"02:12.515 ","End":"02:16.275","Text":"by Algebra, 1(n+1)."},{"Start":"02:16.275 ","End":"02:18.980","Text":"Obviously, the denominator goes to infinity,"},{"Start":"02:18.980 ","End":"02:21.229","Text":"so this goes to 0."},{"Start":"02:21.229 ","End":"02:26.945","Text":"Now, the second thing we need to prove is that this absolute value doesn\u0027t go to 0."},{"Start":"02:26.945 ","End":"02:31.820","Text":"Well, 1/x_n is n,"},{"Start":"02:31.820 ","End":"02:34.610","Text":"1 is n, and similarly,"},{"Start":"02:34.610 ","End":"02:44.050","Text":"1+1 is n+1, and this difference and absolute value is 1 and so it doesn\u0027t go to 0."},{"Start":"02:44.050 ","End":"02:47.870","Text":"Our f is not uniformly continuous."},{"Start":"02:47.870 ","End":"02:50.390","Text":"Here, I just wrote out the remark I made"},{"Start":"02:50.390 ","End":"02:54.860","Text":"earlier that we could have done it in a more brief way."},{"Start":"02:54.860 ","End":"02:56.300","Text":"Now, a remark,"},{"Start":"02:56.300 ","End":"02:58.505","Text":"in this type of exercise,"},{"Start":"02:58.505 ","End":"03:01.850","Text":"the sequences we use often involve sines and"},{"Start":"03:01.850 ","End":"03:07.535","Text":"cosines and the following equalities might be useful."},{"Start":"03:07.535 ","End":"03:12.290","Text":"I won\u0027t read them out. You could always pause or even print this table."},{"Start":"03:12.290 ","End":"03:14.909","Text":"Let\u0027s just start on an example."},{"Start":"03:14.909 ","End":"03:18.995","Text":"This will be example 3 that uses trigonometric sequences."},{"Start":"03:18.995 ","End":"03:25.055","Text":"Let\u0027s take the example which we mentioned in the previous clip of f(x)=sine1/x."},{"Start":"03:25.055 ","End":"03:29.435","Text":"We mentioned that it\u0027s bounded and it\u0027s continuous,"},{"Start":"03:29.435 ","End":"03:32.390","Text":"but still, it\u0027s not uniformly continuous."},{"Start":"03:32.390 ","End":"03:34.625","Text":"We\u0027ll do this using sequences."},{"Start":"03:34.625 ","End":"03:39.405","Text":"We\u0027ll let x_n be the sequence 1 Pi,"},{"Start":"03:39.405 ","End":"03:40.500","Text":"where n goes 1,"},{"Start":"03:40.500 ","End":"03:41.580","Text":"2, 3, 4, 5,"},{"Start":"03:41.580 ","End":"03:46.560","Text":"etc, and y_n is 1 Pi+Pi/2,"},{"Start":"03:46.560 ","End":"03:48.965","Text":"n equals 1, 2, 3, 4, 5, etc."},{"Start":"03:48.965 ","End":"03:54.602","Text":"Now, the claim is that absolute value of x_n minus y_n goes to 0,"},{"Start":"03:54.602 ","End":"04:00.580","Text":"but absolute value of f(x_n) minus f(y_n) does not go to 0."},{"Start":"04:00.580 ","End":"04:02.380","Text":"In our proposition,"},{"Start":"04:02.380 ","End":"04:06.545","Text":"we showed that this means that f is not uniformly continuous."},{"Start":"04:06.545 ","End":"04:09.170","Text":"The first of the 2 claims,"},{"Start":"04:09.170 ","End":"04:15.330","Text":"absolute value of x_n minus y_n is absolute value of this minus this."},{"Start":"04:15.330 ","End":"04:18.815","Text":"We don\u0027t need the absolute value because this is bigger than this."},{"Start":"04:18.815 ","End":"04:22.920","Text":"We can cross multiply nPi"},{"Start":"04:22.920 ","End":"04:28.455","Text":"plus Pi/2 minus nPi is just Pi/2 over the product of the denominators."},{"Start":"04:28.455 ","End":"04:32.030","Text":"This is a constant and the denominator goes to infinity,"},{"Start":"04:32.030 ","End":"04:34.100","Text":"so this goes to 0."},{"Start":"04:34.100 ","End":"04:37.200","Text":"Now, the second half of the claim,"},{"Start":"04:37.200 ","End":"04:43.415","Text":"f(x_n) minus f(y_n) is the sine of 1 over this,"},{"Start":"04:43.415 ","End":"04:50.690","Text":"which is sine of nPi minus sine of nPi plus Pi/2 in absolute value."},{"Start":"04:50.690 ","End":"04:52.790","Text":"If you look back at that table,"},{"Start":"04:52.790 ","End":"04:54.980","Text":"we have that this is 0,"},{"Start":"04:54.980 ","End":"04:58.000","Text":"this is minus 1^n."},{"Start":"04:58.000 ","End":"05:01.070","Text":"The difference in any case is 1."},{"Start":"05:01.070 ","End":"05:02.600","Text":"It\u0027s either 1 or minus 1,"},{"Start":"05:02.600 ","End":"05:05.210","Text":"but the absolute value is 1,"},{"Start":"05:05.210 ","End":"05:07.280","Text":"and that doesn\u0027t tend to 0,"},{"Start":"05:07.280 ","End":"05:09.530","Text":"so f is not uniformly continuous."},{"Start":"05:09.530 ","End":"05:13.110","Text":"That concludes this example and this clip."}],"ID":31117},{"Watched":false,"Name":"Necessary Condition for Uniform Continuity Part 3","Duration":"5m 7s","ChapterTopicVideoID":29524,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this clip, we come to the third and last of"},{"Start":"00:03.690 ","End":"00:09.315","Text":"the necessary conditions for uniform continuity phases as a proposition."},{"Start":"00:09.315 ","End":"00:16.110","Text":"If f is uniformly continuous on a semi-infinite interval and differentiable,"},{"Start":"00:16.110 ","End":"00:21.735","Text":"then the limit as x goes to infinity of the derivative is not infinity,"},{"Start":"00:21.735 ","End":"00:25.425","Text":"meaning it\u0027s not true that this limit is infinity,"},{"Start":"00:25.425 ","End":"00:28.095","Text":"and we usually use it in the converse."},{"Start":"00:28.095 ","End":"00:31.185","Text":"If the limit is infinity,"},{"Start":"00:31.185 ","End":"00:34.915","Text":"then f is not uniformly continuous."},{"Start":"00:34.915 ","End":"00:38.855","Text":"We\u0027ll prove it, the converse part."},{"Start":"00:38.855 ","End":"00:40.190","Text":"Suppose, on the contrary,"},{"Start":"00:40.190 ","End":"00:42.080","Text":"this limit is infinity,"},{"Start":"00:42.080 ","End":"00:44.465","Text":"choose x and y in this interval,"},{"Start":"00:44.465 ","End":"00:46.535","Text":"x is the smaller of the 2."},{"Start":"00:46.535 ","End":"00:48.515","Text":"By the mean value theorem,"},{"Start":"00:48.515 ","End":"00:54.455","Text":"there exists a c between x and y such that f(x) minus"},{"Start":"00:54.455 ","End":"01:01.910","Text":"f(y) in absolute value is absolute value of f prime of c times x minus y absolute value."},{"Start":"01:01.910 ","End":"01:05.270","Text":"Well, the mean value theorem doesn\u0027t have the absolute values."},{"Start":"01:05.270 ","End":"01:09.379","Text":"Apply the mean value theorem and then take everything in absolute value."},{"Start":"01:09.379 ","End":"01:12.145","Text":"Now, let Delta be bigger than 0,"},{"Start":"01:12.145 ","End":"01:15.465","Text":"any Delta, and choose Eta,"},{"Start":"01:15.465 ","End":"01:17.460","Text":"I want to reserve Epsilon for later."},{"Start":"01:17.460 ","End":"01:22.805","Text":"Choose Eta such that Eta is less than Delta but still positive."},{"Start":"01:22.805 ","End":"01:25.909","Text":"You could take it as Delta over 2 if you want to be precise."},{"Start":"01:25.909 ","End":"01:29.900","Text":"Now, to say that f\u0027(x) goes to infinity in"},{"Start":"01:29.900 ","End":"01:34.070","Text":"absolute value means that there is some positive constant K,"},{"Start":"01:34.070 ","End":"01:37.820","Text":"and we can also make sure that K is bigger than a,"},{"Start":"01:37.820 ","End":"01:40.350","Text":"the a from the interval, a,"},{"Start":"01:40.350 ","End":"01:44.710","Text":"necessary increase K such that if x is bigger than K,"},{"Start":"01:44.710 ","End":"01:49.159","Text":"then f\u0027(x) is bigger than any positive number,"},{"Start":"01:49.159 ","End":"01:52.280","Text":"and we take that positive number to be 1 over Eta."},{"Start":"01:52.280 ","End":"01:55.675","Text":"Now this is for any x less than y in the interval."},{"Start":"01:55.675 ","End":"01:58.445","Text":"Now, let\u0027s choose something more specific."},{"Start":"01:58.445 ","End":"02:01.970","Text":"We\u0027ll choose x_0 bigger than K,"},{"Start":"02:01.970 ","End":"02:07.720","Text":"and having chosen x_0 will let y_0 equals x plus Eta."},{"Start":"02:07.720 ","End":"02:11.505","Text":"Then x_0 minus y_0 is Eta,"},{"Start":"02:11.505 ","End":"02:12.840","Text":"which is less than Delta,"},{"Start":"02:12.840 ","End":"02:17.765","Text":"so we have absolute value of x_0 minus y_0 that is less than Delta."},{"Start":"02:17.765 ","End":"02:24.070","Text":"Also note that c is bigger than K because x_0 is bigger than K and y_0 is bigger than K,"},{"Start":"02:24.070 ","End":"02:26.815","Text":"and c is between x_0 and y_0."},{"Start":"02:26.815 ","End":"02:31.865","Text":"It\u0027s also bigger than K. This is the c for x_0, y_0."},{"Start":"02:31.865 ","End":"02:37.560","Text":"Now, c could be the x here because c is bigger than K,"},{"Start":"02:37.560 ","End":"02:39.225","Text":"and for any x bigger than K,"},{"Start":"02:39.225 ","End":"02:43.710","Text":"we have this, so this is bigger than 1 over Eta,"},{"Start":"02:43.710 ","End":"02:47.880","Text":"and x_0 minus y_0 is exactly equal to Eta."},{"Start":"02:47.880 ","End":"02:50.270","Text":"We have that this is bigger than this,"},{"Start":"02:50.270 ","End":"02:52.475","Text":"which is equal to 1,"},{"Start":"02:52.475 ","End":"02:55.430","Text":"and now let\u0027s take this as our Epsilon,"},{"Start":"02:55.430 ","End":"03:02.880","Text":"so what we\u0027ve shown is that Epsilon equals 1 has no corresponding Delta."},{"Start":"03:02.880 ","End":"03:04.350","Text":"For any Delta,"},{"Start":"03:04.350 ","End":"03:12.425","Text":"we found x_0 and y_0 such that x_0 minus y_0 is less than Delta and absolute value,"},{"Start":"03:12.425 ","End":"03:17.540","Text":"and yet f(x_ 0) minus f(y_0) is bigger than Epsilon,"},{"Start":"03:17.540 ","End":"03:20.990","Text":"so this means that f is not uniformly continuous,"},{"Start":"03:20.990 ","End":"03:23.294","Text":"and that\u0027s the proof and,"},{"Start":"03:23.294 ","End":"03:25.700","Text":"won\u0027t give an example in this clip,"},{"Start":"03:25.700 ","End":"03:28.100","Text":"there will be some exercises following it."},{"Start":"03:28.100 ","End":"03:30.815","Text":"I will like to make a couple of remarks."},{"Start":"03:30.815 ","End":"03:35.480","Text":"First of all, the proposition does not work on a finite interval."},{"Start":"03:35.480 ","End":"03:39.620","Text":"We took our interval as a infinity,"},{"Start":"03:39.620 ","End":"03:43.415","Text":"but it won\u0027t work on a finite interval (a, b)."},{"Start":"03:43.415 ","End":"03:48.980","Text":"If you take f(x) equals the square root of 1 minus x on the interval (0,1),"},{"Start":"03:48.980 ","End":"03:53.635","Text":"and certainly f is differentiable on the open interval (0,1),"},{"Start":"03:53.635 ","End":"03:59.030","Text":"and the limit as x goes to 1 from the left of the derivative of f,"},{"Start":"03:59.030 ","End":"04:03.005","Text":"which is minus 1 over twice square root of 1 minus x."},{"Start":"04:03.005 ","End":"04:09.380","Text":"This is infinity because 1 minus x goes to 0 from the right,"},{"Start":"04:09.380 ","End":"04:11.764","Text":"square root of that goes to 0,"},{"Start":"04:11.764 ","End":"04:16.115","Text":"and the reciprocal of it goes to infinity."},{"Start":"04:16.115 ","End":"04:17.240","Text":"The minus doesn\u0027t matter,"},{"Start":"04:17.240 ","End":"04:18.725","Text":"it\u0027s an absolute value."},{"Start":"04:18.725 ","End":"04:24.365","Text":"However, f(x) is uniformly continuous on (0,1)."},{"Start":"04:24.365 ","End":"04:29.465","Text":"It\u0027s pretty much the mirror image of square root of x."},{"Start":"04:29.465 ","End":"04:31.580","Text":"In fact, you can make a substitution,"},{"Start":"04:31.580 ","End":"04:37.675","Text":"replace x by 1 minus x here and you get this."},{"Start":"04:37.675 ","End":"04:42.275","Text":"The mirror image of uniformly continuous is uniformly continuous."},{"Start":"04:42.275 ","End":"04:50.300","Text":"Now the second remark is that the proposition doesn\u0027t work if instead of the condition,"},{"Start":"04:50.300 ","End":"04:53.600","Text":"the limit of f\u0027(x) is infinity,"},{"Start":"04:53.600 ","End":"04:57.080","Text":"if you replace it with f(x) is nearly unbounded."},{"Start":"04:57.080 ","End":"05:00.919","Text":"It doesn\u0027t work, and there is a counter example,"},{"Start":"05:00.919 ","End":"05:04.955","Text":"which will be in 1 of the exercises following."},{"Start":"05:04.955 ","End":"05:07.800","Text":"That concludes this clip."}],"ID":31118},{"Watched":false,"Name":"Exercise 1","Duration":"2m 49s","ChapterTopicVideoID":29525,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"In this exercise, we\u0027re going to show that the function sine of"},{"Start":"00:03.330 ","End":"00:06.795","Text":"x squared is not uniformly continuous."},{"Start":"00:06.795 ","End":"00:09.615","Text":"Here\u0027s a picture of what it looks like."},{"Start":"00:09.615 ","End":"00:13.320","Text":"Notice that the oscillations get more and more frequent and hence"},{"Start":"00:13.320 ","End":"00:18.360","Text":"steeper and steeper and that\u0027s 1 of the reasons why it\u0027s not uniformly continuous."},{"Start":"00:18.360 ","End":"00:26.355","Text":"We\u0027re going to use the proposition that we had is that if we have 2 sequences,"},{"Start":"00:26.355 ","End":"00:27.675","Text":"x_n and y_n,"},{"Start":"00:27.675 ","End":"00:31.080","Text":"such that the difference between x_n and y_n goes to 0,"},{"Start":"00:31.080 ","End":"00:34.590","Text":"but the difference between f(x_n) and f(y_n) does not go to 0,"},{"Start":"00:34.590 ","End":"00:38.340","Text":"then we have non uniform continuity."},{"Start":"00:38.340 ","End":"00:39.770","Text":"Just as a by the way,"},{"Start":"00:39.770 ","End":"00:42.680","Text":"this function is continuous and bounded,"},{"Start":"00:42.680 ","End":"00:45.740","Text":"but that\u0027s not enough for uniform continuity."},{"Start":"00:45.740 ","End":"00:50.060","Text":"We\u0027re going to define x_n and y_n by trial and error and skill."},{"Start":"00:50.060 ","End":"00:52.735","Text":"This is the x_n and y_n we come up with."},{"Start":"00:52.735 ","End":"00:56.850","Text":"Note that f(x_n) is sine of x_n squared,"},{"Start":"00:56.850 ","End":"00:59.640","Text":"x_n squared is this without the square root and we know"},{"Start":"00:59.640 ","End":"01:02.745","Text":"that the sine of 2n Pi plus Pi/2 is 1."},{"Start":"01:02.745 ","End":"01:09.795","Text":"Similarly, sine of 2n Pi minus Pi/2 is minus 1,"},{"Start":"01:09.795 ","End":"01:16.610","Text":"so what we get is that the absolute value of the difference is equal to 2,"},{"Start":"01:16.610 ","End":"01:20.875","Text":"even without the absolute value of 1 minus minus 1."},{"Start":"01:20.875 ","End":"01:23.480","Text":"The sequence doesn\u0027t tend to 0."},{"Start":"01:23.480 ","End":"01:27.115","Text":"Something it\u0027s constantly equal to 2 does not go to 0,"},{"Start":"01:27.115 ","End":"01:30.365","Text":"so this relates to the second of these 2."},{"Start":"01:30.365 ","End":"01:35.210","Text":"Now we\u0027ll show that x_n minus y_n does go to 0 and then that"},{"Start":"01:35.210 ","End":"01:40.010","Text":"will mean that it\u0027s not uniformly continuous. Let\u0027s evaluate."},{"Start":"01:40.010 ","End":"01:43.060","Text":"The absolute value of x_n minus y_n."},{"Start":"01:43.060 ","End":"01:46.100","Text":"The square root of this minus the square root of this don\u0027t"},{"Start":"01:46.100 ","End":"01:49.265","Text":"need the absolute value because this is bigger than this."},{"Start":"01:49.265 ","End":"01:53.090","Text":"Now what we\u0027ll do when we have a square root plus or minus the square root is to"},{"Start":"01:53.090 ","End":"01:57.145","Text":"multiply and divide by the conjugate."},{"Start":"01:57.145 ","End":"02:01.610","Text":"Now the conjugate is the same thing, but with the plus."},{"Start":"02:01.610 ","End":"02:03.020","Text":"Now on the numerator,"},{"Start":"02:03.020 ","End":"02:06.785","Text":"if we take a plus b times a minus b,"},{"Start":"02:06.785 ","End":"02:09.230","Text":"we get a squared minus b squared."},{"Start":"02:09.230 ","End":"02:12.230","Text":"A is the whole square root, b is the whole square root,"},{"Start":"02:12.230 ","End":"02:16.175","Text":"so the a squared is without the square root and b squared is without the square root."},{"Start":"02:16.175 ","End":"02:22.310","Text":"Now, this minus this is Pi/2 minus minus Pi/2, which is Pi."},{"Start":"02:22.310 ","End":"02:25.700","Text":"Now, when we take the limit as n goes to infinity,"},{"Start":"02:25.700 ","End":"02:28.535","Text":"the numerator is a constant Pi."},{"Start":"02:28.535 ","End":"02:30.830","Text":"This goes to infinity,"},{"Start":"02:30.830 ","End":"02:32.120","Text":"what\u0027s under the square root,"},{"Start":"02:32.120 ","End":"02:33.620","Text":"and so does this."},{"Start":"02:33.620 ","End":"02:35.960","Text":"Symbolically we have Pi over"},{"Start":"02:35.960 ","End":"02:39.040","Text":"the square root of infinity plus the square root of infinity."},{"Start":"02:39.040 ","End":"02:42.500","Text":"Just Pi over infinity, which is 0."},{"Start":"02:42.500 ","End":"02:49.470","Text":"That proves the part about the 0 and that concludes this exercise."}],"ID":31119},{"Watched":false,"Name":"Exercise 2","Duration":"2m 10s","ChapterTopicVideoID":29526,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.070","Text":"In this exercise, we\u0027re going to show that e^x cosine 1 over x"},{"Start":"00:05.070 ","End":"00:10.350","Text":"is not uniformly continuous on the open interval 0,1."},{"Start":"00:10.350 ","End":"00:14.430","Text":"We\u0027re going to use the usual proposition with"},{"Start":"00:14.430 ","End":"00:17.460","Text":"2 sequences if we can find x_n"},{"Start":"00:17.460 ","End":"00:21.548","Text":"and y_n such that the difference between x_n and y_n goes to 0."},{"Start":"00:21.548 ","End":"00:24.794","Text":"But the difference between fx and fy_n does not go to 0,"},{"Start":"00:24.794 ","End":"00:27.905","Text":"then we have non-uniform continuity."},{"Start":"00:27.905 ","End":"00:36.380","Text":"We\u0027ll take our x_n and y_n here to be 1 over 2n Pi and 1 over 2n plus 1Pi."},{"Start":"00:36.380 ","End":"00:40.100","Text":"Now we\u0027re going to show these 2 things that this goes to 0,"},{"Start":"00:40.100 ","End":"00:41.915","Text":"but this doesn\u0027t go to 0."},{"Start":"00:41.915 ","End":"00:44.816","Text":"Start with the x_n minus y_n,"},{"Start":"00:44.816 ","End":"00:48.230","Text":"you have 1 over 2n Pi minus 1 over 2n plus 1Pi."},{"Start":"00:48.230 ","End":"00:50.620","Text":"No need for absolute value."},{"Start":"00:50.620 ","End":"00:52.458","Text":"This is equal to,"},{"Start":"00:52.458 ","End":"00:56.330","Text":"take the 1 over Pi outside the brackets and then use"},{"Start":"00:56.330 ","End":"01:01.320","Text":"cross multiplication so we get 2n plus 1 minus 2n,"},{"Start":"01:01.320 ","End":"01:03.870","Text":"which is 1 over 2n(2n plus 1)."},{"Start":"01:03.870 ","End":"01:06.340","Text":"Obviously, this goes to 0, is a constant,"},{"Start":"01:06.340 ","End":"01:07.850","Text":"this is constant,"},{"Start":"01:07.850 ","End":"01:11.664","Text":"denominator here goes to infinity, so this goes to 0."},{"Start":"01:11.664 ","End":"01:13.060","Text":"That\u0027s the first part."},{"Start":"01:13.060 ","End":"01:14.960","Text":"Now we have to show the second part,"},{"Start":"01:14.960 ","End":"01:17.715","Text":"so f of x_n minus f of y_n."},{"Start":"01:17.715 ","End":"01:25.230","Text":"Function was e^x cosine 1 over x and if x is 1 over 2n Pi,"},{"Start":"01:25.230 ","End":"01:28.130","Text":"then 1 over x is just 2n Pi,"},{"Start":"01:28.130 ","End":"01:29.665","Text":"so we have the following."},{"Start":"01:29.665 ","End":"01:32.745","Text":"Now, cosine of 2n Pi is 1."},{"Start":"01:32.745 ","End":"01:36.495","Text":"Cosine of 2n plus 1Pi is minus 1."},{"Start":"01:36.495 ","End":"01:38.685","Text":"Simplifies to this."},{"Start":"01:38.685 ","End":"01:47.840","Text":"Now, this exponent goes to 0 because 1 over 2n Pi is 1 over infinity like 0."},{"Start":"01:47.840 ","End":"01:51.260","Text":"Similarly here, we get 1 over infinity, which is 0."},{"Start":"01:51.260 ","End":"01:53.149","Text":"If both of these is 0,"},{"Start":"01:53.149 ","End":"01:55.640","Text":"then this goes to 1, e^0,"},{"Start":"01:55.640 ","End":"02:02.555","Text":"and this goes to 1 so we end up with absolute value of 1 plus 1, which is 2."},{"Start":"02:02.555 ","End":"02:03.920","Text":"If this goes to 2,"},{"Start":"02:03.920 ","End":"02:05.994","Text":"it doesn\u0027t go to 0."},{"Start":"02:05.994 ","End":"02:10.590","Text":"That\u0027s all we needed to complete this clip and we\u0027re done."}],"ID":31120},{"Watched":false,"Name":"Exercise 3","Duration":"2m 20s","ChapterTopicVideoID":29527,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:06.210","Text":"In this exercise, we\u0027re going to show that the function x sine x is not"},{"Start":"00:06.210 ","End":"00:12.470","Text":"uniformly continuous on the interval 0 to infinity, half close."},{"Start":"00:12.470 ","End":"00:15.600","Text":"We\u0027ll use the proposition with 2 sequences,"},{"Start":"00:15.600 ","End":"00:18.090","Text":"and in our case,"},{"Start":"00:18.090 ","End":"00:24.240","Text":"we\u0027ll let x_n be 2nPi and y_n be 2nPi plus 1."},{"Start":"00:24.240 ","End":"00:26.120","Text":"So we have to show 2 things,"},{"Start":"00:26.120 ","End":"00:29.255","Text":"that the difference goes to 0,"},{"Start":"00:29.255 ","End":"00:34.010","Text":"but f(1) minus f of the other doesn\u0027t go to 0."},{"Start":"00:34.010 ","End":"00:37.730","Text":"We\u0027ll start with the one that goes to 0,"},{"Start":"00:37.730 ","End":"00:40.655","Text":"x_n minus y_n in absolute value,"},{"Start":"00:40.655 ","End":"00:46.100","Text":"this minus this, but plus 1, that goes to 0."},{"Start":"00:46.100 ","End":"00:47.495","Text":"So that\u0027s this part."},{"Start":"00:47.495 ","End":"00:49.370","Text":"Now the other part,"},{"Start":"00:49.370 ","End":"00:52.920","Text":"f(x_n) minus f(y_n) is,"},{"Start":"00:52.920 ","End":"00:54.900","Text":"the function is x sine x,"},{"Start":"00:54.900 ","End":"00:58.380","Text":"so it\u0027s 2nPi sine 2nPi minus,"},{"Start":"00:58.380 ","End":"01:01.605","Text":"and then the same thing with 2nPi plus 1."},{"Start":"01:01.605 ","End":"01:04.215","Text":"Sine 2nPi is 0."},{"Start":"01:04.215 ","End":"01:10.040","Text":"Here we can simplify because the sine of an angle,"},{"Start":"01:10.040 ","End":"01:15.370","Text":"if you add 2nPi to the angle is the same thing as sine of the angle."},{"Start":"01:15.370 ","End":"01:19.620","Text":"We can subtract the 2nPi from the sine to remove it,"},{"Start":"01:19.620 ","End":"01:23.580","Text":"and here we get 0."},{"Start":"01:23.580 ","End":"01:26.405","Text":"What we have now is,"},{"Start":"01:26.405 ","End":"01:28.550","Text":"forget about the minus because it\u0027s an absolute value."},{"Start":"01:28.550 ","End":"01:32.480","Text":"So we have 2nPi plus 1 times sine 1."},{"Start":"01:32.480 ","End":"01:39.245","Text":"We can pull n out of this brackets and put it as 1 in the denominator,"},{"Start":"01:39.245 ","End":"01:42.230","Text":"so we end up with this."},{"Start":"01:42.230 ","End":"01:44.915","Text":"Now, 2nPi plus 1 squared,"},{"Start":"01:44.915 ","End":"01:48.390","Text":"the limit of that is 2nPi."},{"Start":"01:48.390 ","End":"01:51.200","Text":"We still have to figure out the limit of the second factor,"},{"Start":"01:51.200 ","End":"01:58.565","Text":"this sine 1 over 1/ n. We can do that using the limit of sine t/t,"},{"Start":"01:58.565 ","End":"02:00.720","Text":"which is 1 as t goes to 0,"},{"Start":"02:00.720 ","End":"02:04.730","Text":"and if we replace t by 1, then that goes to 0."},{"Start":"02:04.730 ","End":"02:08.255","Text":"So this limit will also be equal to 1."},{"Start":"02:08.255 ","End":"02:11.885","Text":"We\u0027ll have 2nPi times 1, which is 2nPi."},{"Start":"02:11.885 ","End":"02:15.420","Text":"In short, if this thing goes to 2nPi,"},{"Start":"02:15.420 ","End":"02:17.415","Text":"it does not go to 0."},{"Start":"02:17.415 ","End":"02:20.830","Text":"That concludes this exercise."}],"ID":31121},{"Watched":false,"Name":"Exercise 4","Duration":"1m 33s","ChapterTopicVideoID":29528,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"This exercise we\u0027re going to show that the natural log of x on"},{"Start":"00:04.050 ","End":"00:08.295","Text":"the interval from 0 to 1 is not uniformly continuous."},{"Start":"00:08.295 ","End":"00:13.065","Text":"We\u0027re going to use the proposition with 2 sequences,"},{"Start":"00:13.065 ","End":"00:19.740","Text":"although it could be made easier if we use the proposition about unboundedness."},{"Start":"00:19.740 ","End":"00:24.180","Text":"This function f is unbounded as x goes to 0 from the right,"},{"Start":"00:24.180 ","End":"00:25.620","Text":"it goes to minus infinity,"},{"Start":"00:25.620 ","End":"00:28.305","Text":"so we could say straight away not uniformly continuous,"},{"Start":"00:28.305 ","End":"00:31.725","Text":"but we want to practice using this proposition."},{"Start":"00:31.725 ","End":"00:39.630","Text":"Let x_n be e to the minus n and y_n be e to the minus n minus 1."},{"Start":"00:39.630 ","End":"00:42.300","Text":"We have to show these 2 things."},{"Start":"00:42.300 ","End":"00:45.405","Text":"This goes to 0, but this doesn\u0027t go to 0."},{"Start":"00:45.405 ","End":"00:52.025","Text":"X_n minus y_n absolute value is e to the minus n minus e to the minus n minus 1."},{"Start":"00:52.025 ","End":"00:55.415","Text":"This is bigger than this, we can drop the absolute value."},{"Start":"00:55.415 ","End":"01:00.590","Text":"Take e to the minus n outside the brackets and we have 1 minus e to the minus 1."},{"Start":"01:00.590 ","End":"01:03.800","Text":"E to the minus n goes to e to the minus infinity,"},{"Start":"01:03.800 ","End":"01:07.830","Text":"which is 0, so all this goes to 0."},{"Start":"01:07.990 ","End":"01:13.310","Text":"Now we have the other part to check that this doesn\u0027t go to 0."},{"Start":"01:13.310 ","End":"01:17.450","Text":"We have natural log of e to the minus n minus natural log of e to"},{"Start":"01:17.450 ","End":"01:24.500","Text":"the minus n minus 1 and this is equal to minus n. This is minus n minus 1."},{"Start":"01:24.500 ","End":"01:26.400","Text":"When you subtract them,"},{"Start":"01:26.400 ","End":"01:27.990","Text":"we just get 1,"},{"Start":"01:27.990 ","End":"01:30.405","Text":"which doesn\u0027t go to 0."},{"Start":"01:30.405 ","End":"01:34.360","Text":"That proved the second part, and we\u0027re done."}],"ID":31122},{"Watched":false,"Name":"Exercise 5","Duration":"3m 11s","ChapterTopicVideoID":29518,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.160","Text":"This exercise has 2 parts which don\u0027t appear to be connected,"},{"Start":"00:05.160 ","End":"00:08.010","Text":"but actually they are, as you\u0027ll see."},{"Start":"00:08.010 ","End":"00:11.910","Text":"In part a we have to proof that the limit as n goes to infinity of"},{"Start":"00:11.910 ","End":"00:15.990","Text":"this expression in n is 0 and in part b,"},{"Start":"00:15.990 ","End":"00:21.150","Text":"we have to prove that the function sine of e^x is not uniformly continuous."},{"Start":"00:21.150 ","End":"00:28.740","Text":"We\u0027ll start with part a and try and simplify the expression and then compute the limit."},{"Start":"00:28.740 ","End":"00:33.330","Text":"Natural log minus the natural log is"},{"Start":"00:33.330 ","End":"00:38.690","Text":"the natural log of the quotient and each of these is a positive number,"},{"Start":"00:38.690 ","End":"00:41.735","Text":"so the expression is well-defined."},{"Start":"00:41.735 ","End":"00:44.480","Text":"Well, I\u0027m assuming n is a positive integer."},{"Start":"00:44.480 ","End":"00:50.550","Text":"Now this, we can simplify to natural log of 1 plus 1/4n,"},{"Start":"00:50.550 ","End":"00:57.440","Text":"just divide top and bottom by 2Pi n. Then limit as n goes to infinity,"},{"Start":"00:57.440 ","End":"01:00.065","Text":"the limit of 1 plus the limit of 1/4n,"},{"Start":"01:00.065 ","End":"01:04.325","Text":"which is symbolically 1/4 times infinity."},{"Start":"01:04.325 ","End":"01:07.065","Text":"But that is equal to 1,"},{"Start":"01:07.065 ","End":"01:10.980","Text":"the bracket part, and natural log of 1 is 0."},{"Start":"01:10.980 ","End":"01:12.810","Text":"That\u0027s what we had to show."},{"Start":"01:12.810 ","End":"01:15.015","Text":"That\u0027s part a."},{"Start":"01:15.015 ","End":"01:17.415","Text":"Next in part b,"},{"Start":"01:17.415 ","End":"01:24.280","Text":"we\u0027re going to prove that f(x) equals sine of e^x is not uniformly continuous."},{"Start":"01:24.320 ","End":"01:28.620","Text":"We\u0027re going to use the method with 2 sequences and you probably guessed that"},{"Start":"01:28.620 ","End":"01:33.085","Text":"the 2 sequences will be this and this."},{"Start":"01:33.085 ","End":"01:36.170","Text":"Going to use the proposition that I mentioned."},{"Start":"01:36.170 ","End":"01:38.645","Text":"I won\u0027t read it out, should be familiar."},{"Start":"01:38.645 ","End":"01:42.125","Text":"We\u0027ll let the 2 sequences be, like we said,"},{"Start":"01:42.125 ","End":"01:45.905","Text":"natural log of 2Pi n plus Pi/2,"},{"Start":"01:45.905 ","End":"01:47.225","Text":"and the other one,"},{"Start":"01:47.225 ","End":"01:51.125","Text":"natural log of 2Pi n. Now we showed in part"},{"Start":"01:51.125 ","End":"01:55.765","Text":"a that the absolute value of x_n minus y_n is 0."},{"Start":"01:55.765 ","End":"01:59.150","Text":"All that remains is the second part to show"},{"Start":"01:59.150 ","End":"02:03.125","Text":"that the absolute value of f(x_n) minus f(y_n),"},{"Start":"02:03.125 ","End":"02:05.975","Text":"this sequence doesn\u0027t go to 0."},{"Start":"02:05.975 ","End":"02:10.265","Text":"Just a reminder, function f(x) is sine of e to the x."},{"Start":"02:10.265 ","End":"02:12.650","Text":"Of course we\u0027ll use the formula that e to"},{"Start":"02:12.650 ","End":"02:16.430","Text":"the natural log of something is that something itself."},{"Start":"02:16.430 ","End":"02:21.050","Text":"We\u0027re going to let a equals 2Pi n plus Pi/2 once,"},{"Start":"02:21.050 ","End":"02:22.835","Text":"and the other time we\u0027re going to let a equals"},{"Start":"02:22.835 ","End":"02:27.665","Text":"2Pi n. That way we\u0027ll get e to the power of x_n,"},{"Start":"02:27.665 ","End":"02:31.190","Text":"e to the power of natural log of something, is that something."},{"Start":"02:31.190 ","End":"02:40.820","Text":"Similarly here we have a 2Pi n. What we get is that f(x_n) is sine of e to the x_n,"},{"Start":"02:40.820 ","End":"02:42.110","Text":"which is sine of this,"},{"Start":"02:42.110 ","End":"02:43.950","Text":"which is 1,"},{"Start":"02:43.950 ","End":"02:50.640","Text":"sine of 2Pi n plus something is that something because sine is periodic in 2Pi."},{"Start":"02:50.640 ","End":"02:54.330","Text":"The other one\u0027s f(y_n) is sine of each of the y_n,"},{"Start":"02:54.330 ","End":"02:57.975","Text":"which is sine of 2n Pi, which is 0."},{"Start":"02:57.975 ","End":"03:03.575","Text":"This expression becomes the absolute value of 1 minus 0, which is 1."},{"Start":"03:03.575 ","End":"03:07.205","Text":"This sequence does not tend to 0,"},{"Start":"03:07.205 ","End":"03:11.580","Text":"which is all that remained to show and we are done."}],"ID":31123},{"Watched":false,"Name":"Exercise 6","Duration":"3m 25s","ChapterTopicVideoID":29688,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31317},{"Watched":false,"Name":"Exercise 7","Duration":"2m 22s","ChapterTopicVideoID":29519,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.615","Text":"In this exercise, f(x) is e^1/x and it\u0027s defined everywhere except at 0."},{"Start":"00:06.615 ","End":"00:12.885","Text":"In part a, we have to prove that f is uniformly continuous for negative x."},{"Start":"00:12.885 ","End":"00:15.930","Text":"That\u0027s this part and in part b,"},{"Start":"00:15.930 ","End":"00:23.595","Text":"we\u0027re going to prove that f is not uniformly continuous for positive x, that\u0027s here."},{"Start":"00:23.595 ","End":"00:27.630","Text":"Now, f is continuous on each of the 2 parts,"},{"Start":"00:27.630 ","End":"00:31.350","Text":"on minus infinity 0 and also on 0 infinity."},{"Start":"00:31.350 ","End":"00:36.000","Text":"In order to prove that it\u0027s uniformly continuous,"},{"Start":"00:36.000 ","End":"00:38.040","Text":"we just have to show that f has"},{"Start":"00:38.040 ","End":"00:43.324","Text":"a finite limit at each of the open endpoints of the interval."},{"Start":"00:43.324 ","End":"00:47.060","Text":"This interval has 2 open-end points at minus infinity and at 0,"},{"Start":"00:47.060 ","End":"00:48.875","Text":"so we need to prove 2 limits."},{"Start":"00:48.875 ","End":"00:53.450","Text":"Let\u0027s take the minus infinity first, e^1/x."},{"Start":"00:53.450 ","End":"01:00.200","Text":"When x goes to minus infinity is like e^0 or e to the minus 0, which is 1."},{"Start":"01:00.200 ","End":"01:04.145","Text":"When x goes to 0 from the left,"},{"Start":"01:04.145 ","End":"01:12.140","Text":"1/x goes to minus infinity and we get symbolically to the minus infinity, which is 0."},{"Start":"01:12.140 ","End":"01:16.760","Text":"Each of these limits exist and is finite so that\u0027s part a."},{"Start":"01:16.760 ","End":"01:18.690","Text":"Now in part b,"},{"Start":"01:18.690 ","End":"01:22.535","Text":"we\u0027ll use the proposition that if a function is unbounded,"},{"Start":"01:22.535 ","End":"01:24.845","Text":"then it\u0027s not uniformly continuous,"},{"Start":"01:24.845 ","End":"01:27.950","Text":"but this only applies to a finite interval."},{"Start":"01:27.950 ","End":"01:30.095","Text":"What we\u0027ll do first of all,"},{"Start":"01:30.095 ","End":"01:32.660","Text":"is not take the whole 0 to infinity,"},{"Start":"01:32.660 ","End":"01:35.580","Text":"but just take 0-1."},{"Start":"01:35.580 ","End":"01:43.400","Text":"This we show by taking the limit as x goes to 0, it\u0027s infinity."},{"Start":"01:43.400 ","End":"01:46.310","Text":"It\u0027s unbounded on 0,1."},{"Start":"01:46.310 ","End":"01:47.840","Text":"I show you in the picture,"},{"Start":"01:47.840 ","End":"01:49.385","Text":"that\u0027s this part here."},{"Start":"01:49.385 ","End":"01:51.410","Text":"This part goes to infinity."},{"Start":"01:51.410 ","End":"01:53.860","Text":"On 0,1 it\u0027s unbounded."},{"Start":"01:53.860 ","End":"01:58.605","Text":"But note that 0,1 is subset of 0 infinity."},{"Start":"01:58.605 ","End":"02:04.370","Text":"We can apply the principle if a function is uniformly continuous on an interval,"},{"Start":"02:04.370 ","End":"02:07.490","Text":"and it\u0027s also uniformly continuous on a sub-interval."},{"Start":"02:07.490 ","End":"02:12.080","Text":"Therefore, f is not uniformly continuous on 0 infinity because if it was,"},{"Start":"02:12.080 ","End":"02:16.235","Text":"then it would be uniformly continuous on 0,1, which it\u0027s not."},{"Start":"02:16.235 ","End":"02:22.170","Text":"That concludes part b and this clip."}],"ID":31124},{"Watched":false,"Name":"Exercise 8","Duration":"4m 29s","ChapterTopicVideoID":29520,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"This exercise has 2 parts which are seemingly unrelated."},{"Start":"00:04.530 ","End":"00:07.035","Text":"Let me give you some context."},{"Start":"00:07.035 ","End":"00:11.370","Text":"We\u0027re talking about the proposition which is 1 of"},{"Start":"00:11.370 ","End":"00:15.810","Text":"the necessary conditions for uniform continuity."},{"Start":"00:15.810 ","End":"00:18.480","Text":"Which basically says that if f is continuous on"},{"Start":"00:18.480 ","End":"00:21.805","Text":"a semi-infinite interval and differentiable"},{"Start":"00:21.805 ","End":"00:28.924","Text":"and the limit as x goes to infinity of the derivative is infinity in absolute value,"},{"Start":"00:28.924 ","End":"00:32.500","Text":"then the function is not uniformly continuous."},{"Start":"00:32.500 ","End":"00:36.140","Text":"By the way, this would work also on the semi-infinite"},{"Start":"00:36.140 ","End":"00:40.490","Text":"interval in the other direction if it was from minus infinity to a."},{"Start":"00:40.490 ","End":"00:42.305","Text":"Now, Part A,"},{"Start":"00:42.305 ","End":"00:44.810","Text":"is just to prove that something is not uniformly"},{"Start":"00:44.810 ","End":"00:48.500","Text":"continuous using this proposition, Part B,"},{"Start":"00:48.500 ","End":"00:51.440","Text":"the idea is to show that if we replace"},{"Start":"00:51.440 ","End":"00:58.055","Text":"this condition by simply saying that f\u0027 is unbounded,"},{"Start":"00:58.055 ","End":"01:01.460","Text":"then that\u0027s not good enough because we have"},{"Start":"01:01.460 ","End":"01:04.970","Text":"a counter example that is uniformly continuous,"},{"Start":"01:04.970 ","End":"01:08.035","Text":"but still f\u0027 is unbounded,"},{"Start":"01:08.035 ","End":"01:10.975","Text":"so we really need it to go to infinity."},{"Start":"01:10.975 ","End":"01:12.275","Text":"In Part A,"},{"Start":"01:12.275 ","End":"01:18.895","Text":"f is differentiable on the interval from 0 to infinity and therefore also continuous."},{"Start":"01:18.895 ","End":"01:22.390","Text":"We\u0027re going to show that the limit as x goes to infinity"},{"Start":"01:22.390 ","End":"01:25.930","Text":"of the derivative in absolute value is infinity."},{"Start":"01:25.930 ","End":"01:27.430","Text":"There\u0027s a picture here."},{"Start":"01:27.430 ","End":"01:30.910","Text":"This is the function f(x),"},{"Start":"01:30.910 ","End":"01:37.285","Text":"which is x natural log of x and the other graph is the derivative f\u0027."},{"Start":"01:37.285 ","End":"01:40.780","Text":"We can see that derivative goes to infinity."},{"Start":"01:40.780 ","End":"01:43.600","Text":"Anyway, we\u0027re not going to rely on the visual proof,"},{"Start":"01:43.600 ","End":"01:46.225","Text":"we\u0027re going to prove it. Here it goes."},{"Start":"01:46.225 ","End":"01:50.420","Text":"The derivative using the product rule is the derivative of"},{"Start":"01:50.420 ","End":"01:56.285","Text":"1 times log x plus x times the derivative of log x natural log."},{"Start":"01:56.285 ","End":"02:01.565","Text":"Simplifying it, it is natural log of x plus 1."},{"Start":"02:01.565 ","End":"02:04.040","Text":"Since this goes to infinity,"},{"Start":"02:04.040 ","End":"02:07.415","Text":"adding 1 still goes to infinity."},{"Start":"02:07.415 ","End":"02:09.230","Text":"If we take the absolute value,"},{"Start":"02:09.230 ","End":"02:11.855","Text":"it also goes to infinity."},{"Start":"02:11.855 ","End":"02:14.230","Text":"That concludes Part A."},{"Start":"02:14.230 ","End":"02:16.530","Text":"Now on to Part B."},{"Start":"02:16.530 ","End":"02:19.400","Text":"There\u0027s a picture. There it is."},{"Start":"02:19.400 ","End":"02:25.415","Text":"This is a picture of the function and this is the picture of the derivative."},{"Start":"02:25.415 ","End":"02:29.275","Text":"It\u0027s not the same scale vertically as horizontally."},{"Start":"02:29.275 ","End":"02:31.640","Text":"We have to prove, first of all,"},{"Start":"02:31.640 ","End":"02:33.725","Text":"that this is uniformly continuous."},{"Start":"02:33.725 ","End":"02:41.690","Text":"We\u0027ll do that by showing that the limit as x goes to infinity exists and is finite."},{"Start":"02:41.690 ","End":"02:46.640","Text":"We don\u0027t need to check at the other end because we had a closed interval,"},{"Start":"02:46.640 ","End":"02:48.355","Text":"1 and infinity,"},{"Start":"02:48.355 ","End":"02:49.980","Text":"closed on the left."},{"Start":"02:49.980 ","End":"02:53.240","Text":"We only have to check the right hand, which is infinity."},{"Start":"02:53.240 ","End":"02:56.689","Text":"We\u0027ll do this using Sandwich Theorem."},{"Start":"02:56.689 ","End":"03:02.555","Text":"This absolute value is between 0 and1/x because cosine of anything"},{"Start":"03:02.555 ","End":"03:09.155","Text":"is less than or equal to 1 in absolute value,1/x goes to 0 as x goes to infinity."},{"Start":"03:09.155 ","End":"03:13.415","Text":"This is something between 0 and something that goes to 0."},{"Start":"03:13.415 ","End":"03:16.614","Text":"This also goes to 0."},{"Start":"03:16.614 ","End":"03:21.650","Text":"Poppy, we were also asked to show that f\u0027 is unbounded."},{"Start":"03:21.650 ","End":"03:24.920","Text":"F\u0027 using the quotient rule and simplifying,"},{"Start":"03:24.920 ","End":"03:29.690","Text":"I\u0027ll leave you to check this is this expression and it\u0027s unbounded."},{"Start":"03:29.690 ","End":"03:33.335","Text":"The way we\u0027ll show it\u0027s unbounded is by finding a sequence x_n"},{"Start":"03:33.335 ","End":"03:38.090","Text":"such that the absolute value of f\u0027 goes to infinity."},{"Start":"03:38.090 ","End":"03:41.720","Text":"In the picture, that will be these peaks."},{"Start":"03:41.720 ","End":"03:43.969","Text":"The peaks tend to infinity."},{"Start":"03:43.969 ","End":"03:48.125","Text":"Even though the function itself doesn\u0027t go to infinity, it oscillates."},{"Start":"03:48.125 ","End":"03:55.860","Text":"These peaks correspond to x_n equals the cube root of 2 n Pi plus Pi/2."},{"Start":"03:55.860 ","End":"04:05.100","Text":"Cosine(x_n)^3 is cosine 2nPi plus Pi/2 is 0 and sine of 2n Pi plus Pi/2 is 1."},{"Start":"04:05.100 ","End":"04:10.310","Text":"If we substitute x_n in this expression, this Part 1,"},{"Start":"04:10.310 ","End":"04:15.230","Text":"this part zeros of the whole term drops out and we get just the minus 3x,"},{"Start":"04:15.230 ","End":"04:18.725","Text":"which is minus 3x_n,"},{"Start":"04:18.725 ","End":"04:21.065","Text":"which goes to minus infinity,"},{"Start":"04:21.065 ","End":"04:24.980","Text":"but the absolute value goes to infinity."},{"Start":"04:24.980 ","End":"04:29.430","Text":"That\u0027s what we wanted and that concludes this exercise."}],"ID":31125},{"Watched":false,"Name":"Exercise 9","Duration":"5m 57s","ChapterTopicVideoID":29521,"CourseChapterTopicPlaylistID":294558,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.140","Text":"In this exercise, f is a differentiable function and we\u0027re given its derivative,"},{"Start":"00:07.140 ","End":"00:12.870","Text":"f\u0027(x) is equal to this expression though we\u0027re not given what f is itself."},{"Start":"00:12.870 ","End":"00:17.655","Text":"In any event, we can only determine it up to a constant if we can integrate this."},{"Start":"00:17.655 ","End":"00:20.280","Text":"There are 3 things to prove in part a,"},{"Start":"00:20.280 ","End":"00:26.010","Text":"we have to prove this inequality and the illustration relates to part a."},{"Start":"00:26.010 ","End":"00:30.555","Text":"This is the picture of sine^4 x plus cosine^4 x."},{"Start":"00:30.555 ","End":"00:36.140","Text":"In part b, we\u0027re to prove that f is not uniformly continuous for"},{"Start":"00:36.140 ","End":"00:43.115","Text":"positive x and in part c to prove that f is uniformly continuous for negative x."},{"Start":"00:43.115 ","End":"00:48.845","Text":"We\u0027ll start with part a sine^4 x plus cosine^4 x."},{"Start":"00:48.845 ","End":"00:52.205","Text":"Using this formula from algebra,"},{"Start":"00:52.205 ","End":"00:56.825","Text":"where a is sine^2 x and b is cosine^2 x."},{"Start":"00:56.825 ","End":"01:00.890","Text":"That will give us the following if you substitute,"},{"Start":"01:00.890 ","End":"01:03.290","Text":"and then we\u0027ll use a trigonometric formula."},{"Start":"01:03.290 ","End":"01:11.850","Text":"Sine^2 plus cosine^2 is 1 and that gives us that this is 1 so we get 1 minus."},{"Start":"01:11.850 ","End":"01:17.540","Text":"This we can rewrite this thing is sine x cosine x^2."},{"Start":"01:17.540 ","End":"01:18.790","Text":"If we put a 2 in,"},{"Start":"01:18.790 ","End":"01:21.500","Text":"that will give us 4 sine^2 cosine^2."},{"Start":"01:21.500 ","End":"01:24.339","Text":"We compensate by putting 1/2 in front."},{"Start":"01:24.339 ","End":"01:28.805","Text":"Then we can use another formula from trigonometry, this one,"},{"Start":"01:28.805 ","End":"01:35.165","Text":"to figure out that 2 sine x cosine x is sine^2 2x."},{"Start":"01:35.165 ","End":"01:40.280","Text":"Sine of 2x^2 then we put the squared in between the sine and the 2x."},{"Start":"01:40.280 ","End":"01:45.375","Text":"That\u0027s how we write it. That gives us this equals this."},{"Start":"01:45.375 ","End":"01:48.155","Text":"We have this equality,"},{"Start":"01:48.155 ","End":"01:53.330","Text":"but we\u0027re headed for an inequality and it\u0027s going to be based on the sine^2 2x,"},{"Start":"01:53.330 ","End":"01:55.100","Text":"the inequality for this."},{"Start":"01:55.100 ","End":"02:00.500","Text":"Sine^2 2x, sine^2 of anything is between 0 and 1,"},{"Start":"02:00.500 ","End":"02:03.290","Text":"because sine is between 1 and minus 1."},{"Start":"02:03.290 ","End":"02:05.390","Text":"If we multiply this by a 1/2,"},{"Start":"02:05.390 ","End":"02:10.180","Text":"which is positive, then we get the following inequality."},{"Start":"02:10.180 ","End":"02:13.000","Text":"Now we multiply by minus 1,"},{"Start":"02:13.000 ","End":"02:16.385","Text":"which will reverse the direction of the inequality."},{"Start":"02:16.385 ","End":"02:21.130","Text":"We\u0027ll read it from right to left after we multiply by minus 1."},{"Start":"02:21.130 ","End":"02:24.615","Text":"Next, we\u0027ll add 1 everywhere here, here,"},{"Start":"02:24.615 ","End":"02:28.355","Text":"and here and that will give us the following."},{"Start":"02:28.355 ","End":"02:32.410","Text":"Now, notice that we have here 1 minus 1/2 sine^2 x,"},{"Start":"02:32.410 ","End":"02:34.930","Text":"which is exactly the right-hand side of this."},{"Start":"02:34.930 ","End":"02:43.310","Text":"We can replace it by sine^4 x plus cosine^4 x and that gives us what we want for part a."},{"Start":"02:43.310 ","End":"02:44.880","Text":"Now part b,"},{"Start":"02:44.880 ","End":"02:52.314","Text":"where we had to prove that f is not uniformly continuous on 0 infinity, the positive x\u0027s."},{"Start":"02:52.314 ","End":"02:56.455","Text":"The remainder f\u0027(x) was this expression."},{"Start":"02:56.455 ","End":"03:03.235","Text":"We\u0027re going to use the proposition that if f is continuous and differentiable"},{"Start":"03:03.235 ","End":"03:10.550","Text":"on a infinity and the derivative in absolute value goes to infinity,"},{"Start":"03:10.550 ","End":"03:13.430","Text":"then f is not uniformly continuous."},{"Start":"03:13.430 ","End":"03:17.150","Text":"We\u0027re going to show that the limit as x"},{"Start":"03:17.150 ","End":"03:21.490","Text":"goes to infinity of the derivative from here is infinity."},{"Start":"03:21.490 ","End":"03:23.505","Text":"We\u0027re going to use part a."},{"Start":"03:23.505 ","End":"03:26.495","Text":"In part a, we had the following inequality."},{"Start":"03:26.495 ","End":"03:27.680","Text":"Well, not exactly."},{"Start":"03:27.680 ","End":"03:29.165","Text":"We didn\u0027t have the e^x,"},{"Start":"03:29.165 ","End":"03:30.710","Text":"but e^x is positive."},{"Start":"03:30.710 ","End":"03:34.924","Text":"Take the inequality from part a and multiply by e^x everywhere."},{"Start":"03:34.924 ","End":"03:42.380","Text":"Now, we can take the limit of this as x goes to infinity and that\u0027s infinity."},{"Start":"03:42.380 ","End":"03:47.000","Text":"It\u0027s 1/2 infinity, which is infinity and this goes to infinity."},{"Start":"03:47.000 ","End":"03:51.695","Text":"If the 2 ends go to infinity and this is sandwiched in-between,"},{"Start":"03:51.695 ","End":"03:57.920","Text":"then we can say by the sandwich theorem that this also goes to infinity."},{"Start":"03:57.920 ","End":"04:01.835","Text":"Something goes to infinity when you put it in absolute value,"},{"Start":"04:01.835 ","End":"04:04.000","Text":"it goes to infinity."},{"Start":"04:04.000 ","End":"04:08.210","Text":"That\u0027s what we had to show so that concludes part b."},{"Start":"04:08.210 ","End":"04:11.030","Text":"Now, third and last part c,"},{"Start":"04:11.030 ","End":"04:14.765","Text":"we have to show that f is uniformly continuous"},{"Start":"04:14.765 ","End":"04:20.029","Text":"on the interval minus infinity to 0 the negative numbers."},{"Start":"04:20.029 ","End":"04:24.125","Text":"Again reminder what f\u0027(x) is."},{"Start":"04:24.125 ","End":"04:28.085","Text":"This time we have to prove something is uniformly continuous."},{"Start":"04:28.085 ","End":"04:30.640","Text":"We\u0027ll use the following proposition."},{"Start":"04:30.640 ","End":"04:36.280","Text":"That if f is differentiable on an interval and has a bounded derivative there,"},{"Start":"04:36.280 ","End":"04:39.949","Text":"then it\u0027s uniformly continuous on the interval."},{"Start":"04:39.949 ","End":"04:44.580","Text":"What we\u0027re going to do in our case I is going to be this interval and we\u0027re going to show"},{"Start":"04:44.580 ","End":"04:49.875","Text":"that e^x sine^4 x plus cosine^4 x is bounded on this interval."},{"Start":"04:49.875 ","End":"04:53.570","Text":"Again, we\u0027ll be using the inequality from part a."},{"Start":"04:53.570 ","End":"04:56.550","Text":"If x belongs to this interval,"},{"Start":"04:56.550 ","End":"04:59.650","Text":"that just means that x is negative."},{"Start":"04:59.650 ","End":"05:01.375","Text":"If x is negative,"},{"Start":"05:01.375 ","End":"05:04.465","Text":"then e^x is between 0 and 1."},{"Start":"05:04.465 ","End":"05:08.110","Text":"Each of the x is always positive and if x is less than 0,"},{"Start":"05:08.110 ","End":"05:10.640","Text":"then e^x is less than e^0."},{"Start":"05:10.640 ","End":"05:15.410","Text":"Now, we had the inequality from part a, which was this."},{"Start":"05:15.410 ","End":"05:19.040","Text":"We can actually multiply 2 inequalities together."},{"Start":"05:19.040 ","End":"05:20.390","Text":"Everything is positive."},{"Start":"05:20.390 ","End":"05:29.195","Text":"We can multiply and we get that 0 is less than e^x times this,"},{"Start":"05:29.195 ","End":"05:32.885","Text":"which is less than 1 times 1, which is 1."},{"Start":"05:32.885 ","End":"05:34.910","Text":"Now this is positive,"},{"Start":"05:34.910 ","End":"05:36.650","Text":"so it\u0027s the same as its absolute value."},{"Start":"05:36.650 ","End":"05:39.610","Text":"The absolute value is less than 1."},{"Start":"05:39.610 ","End":"05:43.395","Text":"But this is exactly f\u0027(x) in here."},{"Start":"05:43.395 ","End":"05:50.270","Text":"We\u0027ve proved that f\u0027(x) is bounded by showing that the absolute value is less than 1."},{"Start":"05:50.270 ","End":"05:53.290","Text":"That\u0027s what we needed to show; the boundedness."},{"Start":"05:53.290 ","End":"05:57.720","Text":"That concludes part c and the whole exercise."}],"ID":31126}],"Thumbnail":null,"ID":294558}]

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1.1

1