[{"Name":"Continuity and Limits","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"1m 51s","ChapterTopicVideoID":20139,"CourseChapterTopicPlaylistID":113410,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:09.025","Text":"In this exercise, we\u0027re told that the limit as x goes to 0 of fx over x squared equals 5,"},{"Start":"00:09.025 ","End":"00:15.855","Text":"and we have to show that the limit as x goes to 0 of fx over x is 0."},{"Start":"00:15.855 ","End":"00:20.160","Text":"I want to comment on this x goes to 0."},{"Start":"00:20.160 ","End":"00:24.629","Text":"Both these expressions are not defined at x equal 0,"},{"Start":"00:24.629 ","End":"00:26.819","Text":"but that doesn\u0027t make any difference."},{"Start":"00:26.819 ","End":"00:28.619","Text":"When we take the limit,"},{"Start":"00:28.619 ","End":"00:32.070","Text":"we avoid the 0 itself."},{"Start":"00:32.070 ","End":"00:34.154","Text":"X just tends to 0,"},{"Start":"00:34.154 ","End":"00:37.300","Text":"but it isn\u0027t equal to 0."},{"Start":"00:37.340 ","End":"00:42.880","Text":"Just a point I wanted to make clear."},{"Start":"00:44.120 ","End":"00:49.625","Text":"The limit is x goes to 0 of fx over x is"},{"Start":"00:49.625 ","End":"00:56.345","Text":"equal to the limit as x goes to 0 of fx over x squared times x."},{"Start":"00:56.345 ","End":"00:59.165","Text":"We don\u0027t know that this limit exists yet,"},{"Start":"00:59.165 ","End":"01:01.460","Text":"but if it does,"},{"Start":"01:01.460 ","End":"01:03.320","Text":"then so does this and vice versa."},{"Start":"01:03.320 ","End":"01:07.055","Text":"We can take it to be this."},{"Start":"01:07.055 ","End":"01:09.950","Text":"Now this we can write as a product."},{"Start":"01:09.950 ","End":"01:20.000","Text":"We\u0027re using the theorem that if the limit of each factor in the product exists,"},{"Start":"01:20.000 ","End":"01:23.584","Text":"then the limit of the product exists and it\u0027s the product of the limits."},{"Start":"01:23.584 ","End":"01:27.489","Text":"If both of these 2 limits exist, we\u0027re okay."},{"Start":"01:27.489 ","End":"01:29.660","Text":"Then we just multiply them."},{"Start":"01:29.660 ","End":"01:34.790","Text":"The first limit, we know it exists because it\u0027s given that\u0027s equal to 5,"},{"Start":"01:34.790 ","End":"01:37.490","Text":"and the second one\u0027s just a trivial 1,"},{"Start":"01:37.490 ","End":"01:39.715","Text":"the limit is 0."},{"Start":"01:39.715 ","End":"01:43.470","Text":"When x goes to 0, then x goes to 0, so to speak."},{"Start":"01:43.470 ","End":"01:46.175","Text":"5 times 0 is 0,"},{"Start":"01:46.175 ","End":"01:49.355","Text":"so the answer is 0."},{"Start":"01:49.355 ","End":"01:52.140","Text":"We\u0027ve shown it and we\u0027re done."}],"ID":31301},{"Watched":false,"Name":"Exercise 2","Duration":"6m 32s","ChapterTopicVideoID":20145,"CourseChapterTopicPlaylistID":113410,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Let me rephrase this exercise for you."},{"Start":"00:03.270 ","End":"00:06.885","Text":"Basically, we\u0027re talking about this function,"},{"Start":"00:06.885 ","End":"00:11.430","Text":"f of x equals the whole part of x squared"},{"Start":"00:11.430 ","End":"00:16.725","Text":"times sine Pi x and it\u0027s defined on the non-negatives,"},{"Start":"00:16.725 ","End":"00:19.575","Text":"and we\u0027re concerned with its continuity."},{"Start":"00:19.575 ","End":"00:23.535","Text":"Now numbers can be divided into 3 categories."},{"Start":"00:23.535 ","End":"00:29.835","Text":"A number x might not be the square root of any natural number,"},{"Start":"00:29.835 ","End":"00:32.640","Text":"and here we\u0027re including 0 as a natural number."},{"Start":"00:32.640 ","End":"00:36.310","Text":"In other words, x squared is not a whole number"},{"Start":"00:36.310 ","End":"00:41.975","Text":"and if x is the square root of a natural number,"},{"Start":"00:41.975 ","End":"00:43.820","Text":"there\u0027s 2 sub possibilities."},{"Start":"00:43.820 ","End":"00:49.355","Text":"Either it itself is a natural number or it isn\u0027t."},{"Start":"00:49.355 ","End":"00:51.710","Text":"That\u0027s the third category,"},{"Start":"00:51.710 ","End":"00:55.160","Text":"where x is the square root of something"},{"Start":"00:55.160 ","End":"00:59.860","Text":"whole but not itself whole like square root of 2 or square root of 3."},{"Start":"00:59.860 ","End":"01:01.760","Text":"But square root of 4, for example,"},{"Start":"01:01.760 ","End":"01:05.225","Text":"would be in category B because it is whole."},{"Start":"01:05.225 ","End":"01:08.705","Text":"We have to show that in this case it\u0027s continuous,"},{"Start":"01:08.705 ","End":"01:11.360","Text":"in this case it\u0027s continuous,"},{"Start":"01:11.360 ","End":"01:12.770","Text":"and in this case,"},{"Start":"01:12.770 ","End":"01:16.400","Text":"it\u0027s not continuous, the function I mean."},{"Start":"01:16.400 ","End":"01:19.370","Text":"Let\u0027s start with Part a where"},{"Start":"01:19.370 ","End":"01:23.380","Text":"x is not the square root of any whole number or if you like it,"},{"Start":"01:23.380 ","End":"01:25.325","Text":"x squared is not a whole number."},{"Start":"01:25.325 ","End":"01:30.320","Text":"If it\u0027s not a natural number then we"},{"Start":"01:30.320 ","End":"01:35.910","Text":"can sandwich it in between 2 consecutive natural numbers as follows;"},{"Start":"01:35.910 ","End":"01:41.075","Text":"It\u0027s between k and k plus 1 strictly between x-naught squared."},{"Start":"01:41.075 ","End":"01:44.720","Text":"Now, let\u0027s let Epsilon be"},{"Start":"01:44.720 ","End":"01:49.565","Text":"the closest distance either from here to here or from here to here,"},{"Start":"01:49.565 ","End":"01:53.230","Text":"take the minimum of those 2 distances."},{"Start":"01:53.230 ","End":"01:55.200","Text":"If we do that,"},{"Start":"01:55.200 ","End":"02:00.050","Text":"then we can find an interval of plus or minus Epsilon that\u0027s still in this interval."},{"Start":"02:00.050 ","End":"02:03.440","Text":"Since the square function is continuous,"},{"Start":"02:03.440 ","End":"02:12.620","Text":"this Epsilon has a Delta such whenever x is within plus or minus Delta of x-naught then x"},{"Start":"02:12.620 ","End":"02:16.730","Text":"squared is within plus or minus Epsilon of"},{"Start":"02:16.730 ","End":"02:22.580","Text":"x-naught squared and so will be in this interval k, k plus 1."},{"Start":"02:22.580 ","End":"02:26.740","Text":"Now for such x, meaning x in this interval,"},{"Start":"02:26.740 ","End":"02:34.185","Text":"f of x is this expression but the whole part of x squared since"},{"Start":"02:34.185 ","End":"02:41.500","Text":"we\u0027re in this interval is going to be exactly k, the integer part."},{"Start":"02:41.500 ","End":"02:47.160","Text":"So we get that f of x is equal to k sine Pi x."},{"Start":"02:47.160 ","End":"02:54.150","Text":"It\u0027s clear that if f of x equals k sine Pi x this is continuous at least in"},{"Start":"02:54.150 ","End":"02:58.305","Text":"the interval that we discussed and"},{"Start":"02:58.305 ","End":"03:03.830","Text":"x-naught is in this interval so f is continuous in particular at x naught."},{"Start":"03:03.830 ","End":"03:06.145","Text":"That\u0027s Part a."},{"Start":"03:06.145 ","End":"03:16.590","Text":"Next, we move on to Part b where x is an integer k. Note that the limit as x"},{"Start":"03:16.590 ","End":"03:24.770","Text":"goes to k from the right of the whole path of x squared is k squared because when x is"},{"Start":"03:24.770 ","End":"03:28.910","Text":"just a little bit more than k then x squared is"},{"Start":"03:28.910 ","End":"03:33.590","Text":"just a little bit more than k squared and the whole part is k squared."},{"Start":"03:33.590 ","End":"03:40.805","Text":"On the other hand, if x goes to k from the left and x is a little bit less than k,"},{"Start":"03:40.805 ","End":"03:44.990","Text":"then x squared will be a little bit less than k squared"},{"Start":"03:44.990 ","End":"03:49.675","Text":"but the whole part will take it all the way down to k squared minus 1."},{"Start":"03:49.675 ","End":"03:53.985","Text":"What we need to show is that when x goes to k,"},{"Start":"03:53.985 ","End":"03:56.910","Text":"f of x goes to f of k,"},{"Start":"03:56.910 ","End":"04:00.400","Text":"f of k happens to be 0."},{"Start":"04:01.010 ","End":"04:03.930","Text":"For example, we see f of x,"},{"Start":"04:03.930 ","End":"04:08.250","Text":"if x is k which is a whole number"},{"Start":"04:08.250 ","End":"04:15.135","Text":"then sine of Pi times a whole number is 0 so f of x comes out to be 0."},{"Start":"04:15.135 ","End":"04:18.545","Text":"Let\u0027s evaluate these limits."},{"Start":"04:18.545 ","End":"04:23.290","Text":"When x goes to k from the right as we said,"},{"Start":"04:23.290 ","End":"04:29.220","Text":"f of x is k squared sine Pi k, and that\u0027s 0."},{"Start":"04:29.220 ","End":"04:31.575","Text":"When x goes to k from the left,"},{"Start":"04:31.575 ","End":"04:36.965","Text":"when x is very near k but lower than we said that f of x"},{"Start":"04:36.965 ","End":"04:44.190","Text":"is k squared minus 1 sine Pi k. The whole part is less than k squared,"},{"Start":"04:44.190 ","End":"04:46.380","Text":"so it\u0027s k squared minus 1,"},{"Start":"04:46.380 ","End":"04:49.455","Text":"but at times 0 is still 0."},{"Start":"04:49.455 ","End":"04:52.565","Text":"So both on the left and on the right it\u0027s"},{"Start":"04:52.565 ","End":"04:56.745","Text":"0 which is f of k and this is what we wanted to show."},{"Start":"04:56.745 ","End":"05:03.630","Text":"We\u0027ve proved Part b continuity at k. Now in Part c,"},{"Start":"05:03.630 ","End":"05:07.875","Text":"we have x-naught being square root of 3,"},{"Start":"05:07.875 ","End":"05:09.260","Text":"square root of a whole number,"},{"Start":"05:09.260 ","End":"05:11.095","Text":"but not a whole number."},{"Start":"05:11.095 ","End":"05:16.740","Text":"Sine of Pi x naught is not 0 because the sine"},{"Start":"05:16.740 ","End":"05:22.175","Text":"of Pi times something is 0 only when this is a whole number."},{"Start":"05:22.175 ","End":"05:29.120","Text":"If x tends to square root of n from the right then x squared is"},{"Start":"05:29.120 ","End":"05:32.205","Text":"slightly bigger than n so"},{"Start":"05:32.205 ","End":"05:36.920","Text":"the whole part is n but if x approaches square root of n from the left,"},{"Start":"05:36.920 ","End":"05:45.360","Text":"x squared will be slightly under n and the whole part is n minus 1."},{"Start":"05:45.360 ","End":"05:52.055","Text":"If we take the limit of f of x from the left of root n"},{"Start":"05:52.055 ","End":"05:58.625","Text":"and from the right of root n then from the right,"},{"Start":"05:58.625 ","End":"06:01.315","Text":"we get that this limit is"},{"Start":"06:01.315 ","End":"06:06.515","Text":"sine Pi x naught from this part and from this part just n as we saw."},{"Start":"06:06.515 ","End":"06:10.600","Text":"Whereas here, same thing except with n minus 1."},{"Start":"06:10.600 ","End":"06:14.055","Text":"We said that sine of Pi x-naught is not 0."},{"Start":"06:14.055 ","End":"06:17.370","Text":"If it\u0027s not 0 then when you times it by n or n"},{"Start":"06:17.370 ","End":"06:21.690","Text":"minus 1 you\u0027re going to get different things and that\u0027s what I\u0027ve said here."},{"Start":"06:21.690 ","End":"06:26.330","Text":"The left limit is not equal to the right limit at x-naught and so"},{"Start":"06:26.330 ","End":"06:32.400","Text":"the function is not continuous and we are done."}],"ID":31302},{"Watched":false,"Name":"Exercise 3","Duration":"1m 46s","ChapterTopicVideoID":20148,"CourseChapterTopicPlaylistID":113410,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"In this exercise, we have f which is defined on"},{"Start":"00:03.510 ","End":"00:07.275","Text":"the non-negative numbers, and it\u0027s continuous."},{"Start":"00:07.275 ","End":"00:12.870","Text":"We\u0027re told that the limit as x goes to infinity of f of x exists."},{"Start":"00:12.870 ","End":"00:16.095","Text":"We have to show that f is bounded."},{"Start":"00:16.095 ","End":"00:21.100","Text":"Now we\u0027ll use the definition of the limit at infinity."},{"Start":"00:21.100 ","End":"00:23.720","Text":"Say this limit is L."},{"Start":"00:23.720 ","End":"00:28.760","Text":"Then for any Epsilon, in particular, Epsilon equals 1,"},{"Start":"00:28.760 ","End":"00:34.445","Text":"there is a number M such that whenever x is bigger than M,"},{"Start":"00:34.445 ","End":"00:38.329","Text":"f of x is close to the limit within Epsilon,"},{"Start":"00:38.329 ","End":"00:40.530","Text":"in this case, 1."},{"Start":"00:41.000 ","End":"00:45.650","Text":"That gives us, if you use the triangle inequality on this,"},{"Start":"00:45.650 ","End":"00:51.450","Text":"that f of x is less than the absolute value of L plus 1."},{"Start":"00:51.920 ","End":"01:00.110","Text":"This means that f is bounded for x bigger than M i.e on the interval from M to infinity,"},{"Start":"01:00.110 ","End":"01:02.700","Text":"it\u0027s bounded by this."},{"Start":"01:03.280 ","End":"01:10.800","Text":"Now also, f is bounded on the closed interval from 0 to M."},{"Start":"01:10.800 ","End":"01:14.525","Text":"On a closed interval, a continuous function is bounded."},{"Start":"01:14.525 ","End":"01:17.420","Text":"For example, the extreme value theorem tell you that"},{"Start":"01:17.420 ","End":"01:21.990","Text":"it has its maximum and its minimum on closed interval."},{"Start":"01:21.990 ","End":"01:27.605","Text":"Altogether, we have that f is bounded on here and on here."},{"Start":"01:27.605 ","End":"01:29.680","Text":"Just take the maximum of the 2 bounds,"},{"Start":"01:29.680 ","End":"01:32.405","Text":"and it\u0027s bounded on the union of both,"},{"Start":"01:32.405 ","End":"01:37.070","Text":"which is the interval from 0 to infinity,"},{"Start":"01:37.070 ","End":"01:41.555","Text":"the non-negatives, which is the whole domain."},{"Start":"01:41.555 ","End":"01:46.230","Text":"F is bounded, and we are done."}],"ID":31303},{"Watched":false,"Name":"Exercise 4","Duration":"2m 20s","ChapterTopicVideoID":20152,"CourseChapterTopicPlaylistID":113410,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.029","Text":"In this exercise, we have 2 functions,"},{"Start":"00:03.029 ","End":"00:06.945","Text":"f and g, which are continuous."},{"Start":"00:06.945 ","End":"00:09.510","Text":"At some point af is not equal to"},{"Start":"00:09.510 ","End":"00:14.715","Text":"g. We have to show that there is some delta bigger than 0,"},{"Start":"00:14.715 ","End":"00:20.370","Text":"such that f is not equal to g in some neighborhood of a,"},{"Start":"00:20.370 ","End":"00:26.070","Text":"meaning when x minus a in absolute value is less than delta,"},{"Start":"00:26.070 ","End":"00:28.320","Text":"so within plus or minus delta of a,"},{"Start":"00:28.320 ","End":"00:34.690","Text":"f is still different from g. What we\u0027re going to do, it\u0027s a common strategy,"},{"Start":"00:34.690 ","End":"00:38.990","Text":"is to let h be the difference of the 2 functions."},{"Start":"00:38.990 ","End":"00:42.680","Text":"H is certainly continuous since f and g are,"},{"Start":"00:42.680 ","End":"00:44.930","Text":"and h of a is not equal to 0,"},{"Start":"00:44.930 ","End":"00:48.320","Text":"because f is not equal to g. Now,"},{"Start":"00:48.320 ","End":"00:52.580","Text":"h is continuous and there\u0027s an epsilon delta definition of continuity."},{"Start":"00:52.580 ","End":"00:54.695","Text":"Let\u0027s choose first of all epsilon."},{"Start":"00:54.695 ","End":"00:59.100","Text":"We\u0027ll take it as h of a in absolute value,"},{"Start":"00:59.100 ","End":"01:01.085","Text":"so epsilon\u0027s positive."},{"Start":"01:01.085 ","End":"01:04.445","Text":"For this epsilon, there is some delta,"},{"Start":"01:04.445 ","End":"01:10.110","Text":"such that whenever x minus a absolute value less than delta,"},{"Start":"01:10.110 ","End":"01:15.670","Text":"then absolute value h of x minus h of a is less than epsilon."},{"Start":"01:18.200 ","End":"01:23.450","Text":"Now when h of x minus h of a absolute value less than epsilon,"},{"Start":"01:23.450 ","End":"01:26.850","Text":"then h of x is not 0."},{"Start":"01:27.050 ","End":"01:29.735","Text":"Little proof by contradiction."},{"Start":"01:29.735 ","End":"01:31.745","Text":"Suppose it is 0,"},{"Start":"01:31.745 ","End":"01:37.175","Text":"then what we get from here just replace h of x with 0,"},{"Start":"01:37.175 ","End":"01:42.760","Text":"get 0 minus h of a absolute value less than epsilon."},{"Start":"01:42.760 ","End":"01:46.010","Text":"But this is just absolute value of h of a,"},{"Start":"01:46.010 ","End":"01:47.240","Text":"on the right-hand side,"},{"Start":"01:47.240 ","End":"01:51.000","Text":"epsilon is also equal to absolute value of h of a,"},{"Start":"01:51.000 ","End":"01:53.575","Text":"so we have something that\u0027s less than itself."},{"Start":"01:53.575 ","End":"01:57.060","Text":"No, that can\u0027t be, That\u0027s a contradiction."},{"Start":"01:57.060 ","End":"02:03.244","Text":"That means that h of x is not equal to 0,"},{"Start":"02:03.244 ","End":"02:08.200","Text":"and it\u0027s not equal to 0, whenever x minus a less than delta."},{"Start":"02:08.200 ","End":"02:12.680","Text":"When h is not 0, then f is not equal to g,"},{"Start":"02:12.680 ","End":"02:19.970","Text":"by the definition of h as the difference and we are done."}],"ID":31304},{"Watched":false,"Name":"Exercise 5","Duration":"4m 25s","ChapterTopicVideoID":20153,"CourseChapterTopicPlaylistID":113410,"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":"In this exercise, we have a function f which is continuous."},{"Start":"00:04.050 ","End":"00:09.105","Text":"We know that f of 0 is minus 1 and f of 1 is 3."},{"Start":"00:09.105 ","End":"00:16.680","Text":"Now we\u0027re going to define a set S of all the numbers in the interval 0,"},{"Start":"00:16.680 ","End":"00:20.340","Text":"1, where f of x is 0."},{"Start":"00:20.340 ","End":"00:25.530","Text":"Then there are 3 parts and we\u0027ll read each 1 and do it."},{"Start":"00:25.530 ","End":"00:28.935","Text":"First 1 to show that S is non-empty."},{"Start":"00:28.935 ","End":"00:34.290","Text":"For that we\u0027re going to use the intermediate value property because f is continuous,"},{"Start":"00:34.290 ","End":"00:36.495","Text":"it has the IVP."},{"Start":"00:36.495 ","End":"00:39.345","Text":"F of 0 is negative,"},{"Start":"00:39.345 ","End":"00:41.655","Text":"f of 1 is positive,"},{"Start":"00:41.655 ","End":"00:44.695","Text":"so somewhere in between 0 and 1,"},{"Start":"00:44.695 ","End":"00:49.085","Text":"there is some c such that f of c is 0,"},{"Start":"00:49.085 ","End":"00:54.530","Text":"so c belongs to S. It\u0027s between 0 and 1 and f of it is 0."},{"Start":"00:54.530 ","End":"00:57.244","Text":"So S is not empty."},{"Start":"00:57.244 ","End":"01:04.400","Text":"Now on to part b. Alpha is the supremum of the set S. Now why does that make sense?"},{"Start":"01:04.400 ","End":"01:06.710","Text":"Because S is non-empty and bounded."},{"Start":"01:06.710 ","End":"01:09.470","Text":"It\u0027s bounded in 0, 1 and we\u0027ve shown is non-empty,"},{"Start":"01:09.470 ","End":"01:10.864","Text":"so it has a supremum."},{"Start":"01:10.864 ","End":"01:13.550","Text":"We\u0027re okay there. Call it Alpha."},{"Start":"01:13.550 ","End":"01:19.490","Text":"Now we have to show that Alpha is bigger than 0 and less than or equal to 1."},{"Start":"01:19.490 ","End":"01:25.290","Text":"Now, Alpha is the least upper bound for S and"},{"Start":"01:25.290 ","End":"01:30.710","Text":"1 is an upper bound because 1 is bigger than all the elements of S,"},{"Start":"01:30.710 ","End":"01:32.240","Text":"because S is a subset of 0,"},{"Start":"01:32.240 ","End":"01:36.445","Text":"1 and 1 is bigger or equal to everything in 0, 1."},{"Start":"01:36.445 ","End":"01:39.470","Text":"Alpha\u0027s less than or equal to 1 because the"},{"Start":"01:39.470 ","End":"01:42.290","Text":"least upper bound is less than or equal to an upper bound."},{"Start":"01:42.290 ","End":"01:44.390","Text":"That\u0027s 1/2 of the inequality,"},{"Start":"01:44.390 ","End":"01:46.205","Text":"now let\u0027s prove the other half."},{"Start":"01:46.205 ","End":"01:49.875","Text":"Consider the point c, the c from here."},{"Start":"01:49.875 ","End":"01:54.870","Text":"C is in S and so Alpha\u0027s bigger or equal to"},{"Start":"01:54.870 ","End":"02:01.760","Text":"c because it\u0027s bigger or equal to all the elements of S. That\u0027s what an upper bound is."},{"Start":"02:01.760 ","End":"02:06.895","Text":"C is bigger than 0, so combining these Alpha\u0027s bigger than 0."},{"Start":"02:06.895 ","End":"02:08.760","Text":"That\u0027s the other half of the inequality,"},{"Start":"02:08.760 ","End":"02:10.815","Text":"so that\u0027s part b."},{"Start":"02:10.815 ","End":"02:15.650","Text":"Part c is to show that f of Alpha is 0."},{"Start":"02:15.650 ","End":"02:17.720","Text":"We\u0027ll do this by contradiction."},{"Start":"02:17.720 ","End":"02:21.880","Text":"Suppose on the contrary that f of Alpha is not 0."},{"Start":"02:21.880 ","End":"02:24.810","Text":"Now f is continuous at Alpha,"},{"Start":"02:24.810 ","End":"02:27.090","Text":"that means continuous everywhere."},{"Start":"02:27.090 ","End":"02:29.555","Text":"In a previous exercise,"},{"Start":"02:29.555 ","End":"02:34.970","Text":"we showed that there exists some Delta such"},{"Start":"02:34.970 ","End":"02:41.380","Text":"that f of x is not equal to 0 in some neighborhood of Alpha,"},{"Start":"02:41.380 ","End":"02:45.185","Text":"ie, something of the form x minus Alpha less than Delta,"},{"Start":"02:45.185 ","End":"02:47.185","Text":"the absolute value here."},{"Start":"02:47.185 ","End":"02:53.205","Text":"Let\u0027s let y equal Alpha minus Delta."},{"Start":"02:53.205 ","End":"02:57.930","Text":"Y can\u0027t be an upper bound of S because y"},{"Start":"02:57.930 ","End":"03:03.309","Text":"is less than Alpha and Alpha is the least upper bound."},{"Start":"03:03.560 ","End":"03:06.295","Text":"If it\u0027s not an upper bound,"},{"Start":"03:06.295 ","End":"03:11.335","Text":"then there must be some element of S which is bigger than y."},{"Start":"03:11.335 ","End":"03:12.760","Text":"On the other hand,"},{"Start":"03:12.760 ","End":"03:17.800","Text":"S is less than or equal to Alpha because Alpha is an upper bound,"},{"Start":"03:17.800 ","End":"03:19.360","Text":"it\u0027s the least upper bound."},{"Start":"03:19.360 ","End":"03:24.340","Text":"Now I claim that we\u0027ve reached a contradiction because we found"},{"Start":"03:24.340 ","End":"03:30.600","Text":"this S which satisfies absolute value of S minus Alpha\u0027s less than Delta."},{"Start":"03:30.600 ","End":"03:32.635","Text":"Perhaps I should explain this."},{"Start":"03:32.635 ","End":"03:37.950","Text":"This absolute value is equal to Alpha minus S because of this,"},{"Start":"03:37.950 ","End":"03:43.555","Text":"and Alpha minus S is less than Alpha minus y because of this,"},{"Start":"03:43.555 ","End":"03:46.815","Text":"but Alpha minus y is just Delta."},{"Start":"03:46.815 ","End":"03:49.250","Text":"That\u0027s how we get this inequality."},{"Start":"03:49.250 ","End":"03:55.879","Text":"Also, f of S is 0 because S belongs to the set S,"},{"Start":"03:55.879 ","End":"03:58.085","Text":"so this contradicts the above."},{"Start":"03:58.085 ","End":"04:00.350","Text":"What does it contradict specifically?"},{"Start":"04:00.350 ","End":"04:06.130","Text":"It contradicts that f of x is not equal to 0 whenever x satisfies this"},{"Start":"04:06.130 ","End":"04:12.909","Text":"and S is an example of a contradiction because it satisfies this,"},{"Start":"04:12.909 ","End":"04:16.140","Text":"but f of S is 0."},{"Start":"04:16.140 ","End":"04:21.350","Text":"This contradiction came from supposing that f of Alpha is not 0,"},{"Start":"04:21.350 ","End":"04:26.160","Text":"so f of Alpha is 0. We\u0027re done."}],"ID":31305}],"Thumbnail":null,"ID":113410},{"Name":"Existence of Extrema, Intermediate Value Property","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"1m 3s","ChapterTopicVideoID":20155,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In this exercise, f is defined on the closed interval a, b."},{"Start":"00:04.380 ","End":"00:09.090","Text":"It\u0027s continuous and it\u0027s positive everywhere on that interval."},{"Start":"00:09.090 ","End":"00:13.244","Text":"We have to show that there is some positive number Alpha,"},{"Start":"00:13.244 ","End":"00:19.860","Text":"such that f of x is bigger or equal to Alpha for all x in the interval a, b."},{"Start":"00:19.860 ","End":"00:23.215","Text":"We\u0027re going to use the Extreme Value Theorem here."},{"Start":"00:23.215 ","End":"00:26.180","Text":"F is continuous on a closed interval,"},{"Start":"00:26.180 ","End":"00:30.030","Text":"so it attains its minimum and also its maximum,"},{"Start":"00:30.030 ","End":"00:32.280","Text":"but we don\u0027t care about the maximum."},{"Start":"00:32.280 ","End":"00:36.664","Text":"It has a minimum somewhere at some x naught in the interval."},{"Start":"00:36.664 ","End":"00:41.255","Text":"What that means is that f of x naught being the minimum,"},{"Start":"00:41.255 ","End":"00:46.745","Text":"is less than or equal to f of x for all the other x in a, b."},{"Start":"00:46.745 ","End":"00:49.520","Text":"Now if we call this Alpha,"},{"Start":"00:49.520 ","End":"00:55.335","Text":"then Alpha being f of something is bigger than 0."},{"Start":"00:55.335 ","End":"00:59.240","Text":"Here is our Alpha that\u0027s bigger than 0 and less than"},{"Start":"00:59.240 ","End":"01:03.720","Text":"or equal to every f of x. We are done."}],"ID":20948},{"Watched":false,"Name":"Exercise 2","Duration":"1m 43s","ChapterTopicVideoID":20156,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"In this exercise, f is a function from the closed interval 0,"},{"Start":"00:04.980 ","End":"00:09.210","Text":"1 onto the open interval 0, 1."},{"Start":"00:09.210 ","End":"00:13.095","Text":"We have to show that f is not continuous."},{"Start":"00:13.095 ","End":"00:17.580","Text":"At least at one point it\u0027s not continuous."},{"Start":"00:17.580 ","End":"00:24.540","Text":"We\u0027ll do it by contradiction and we\u0027ll suppose that f is continuous."},{"Start":"00:24.540 ","End":"00:27.570","Text":"Now since it\u0027s continuous on a closed interval,"},{"Start":"00:27.570 ","End":"00:29.220","Text":"by the extreme value theorem,"},{"Start":"00:29.220 ","End":"00:33.990","Text":"it attains its minimum at some point x naught in this interval."},{"Start":"00:33.990 ","End":"00:40.410","Text":"What this means is that if we let m equals f of x naught,"},{"Start":"00:40.410 ","End":"00:43.610","Text":"then m is positive,"},{"Start":"00:43.610 ","End":"00:47.540","Text":"because f goes to the open interval,"},{"Start":"00:47.540 ","End":"00:53.195","Text":"meaning bigger than 0 and less than 1 strictly, so bigger than 0."},{"Start":"00:53.195 ","End":"01:00.435","Text":"Now, m over 2 is between 0 and 1,"},{"Start":"01:00.435 ","End":"01:07.725","Text":"because m is between 0 and 1 and f is onto."},{"Start":"01:07.725 ","End":"01:10.730","Text":"There must be a source for m over 2."},{"Start":"01:10.730 ","End":"01:15.865","Text":"In other words, the next one such that f of x1 equals m over 2."},{"Start":"01:15.865 ","End":"01:20.125","Text":"What we get is that f of x1 is m over 2,"},{"Start":"01:20.125 ","End":"01:23.905","Text":"which is less than m, and m is f of x naught."},{"Start":"01:23.905 ","End":"01:27.500","Text":"But, f of x naught is the minimum value of f,"},{"Start":"01:27.500 ","End":"01:32.510","Text":"and here we have something that\u0027s lower than the minimum and that\u0027s a contradiction."},{"Start":"01:32.510 ","End":"01:39.090","Text":"This contradiction came from assuming that f is continuous,"},{"Start":"01:39.090 ","End":"01:43.810","Text":"and so, it is not continuous and we\u0027re done."}],"ID":20949},{"Watched":false,"Name":"Exercise 3","Duration":"4m 54s","ChapterTopicVideoID":20157,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"In this exercise, we\u0027re asked to give"},{"Start":"00:02.670 ","End":"00:05.880","Text":"an example of a function f on the closed interval 0,"},{"Start":"00:05.880 ","End":"00:08.730","Text":"1, which is not continuous."},{"Start":"00:08.730 ","End":"00:13.020","Text":"Yet it still has the intermediate value property."},{"Start":"00:13.020 ","End":"00:19.030","Text":"Here I\u0027ve just written a reminder in case you\u0027ve forgotten what that property is."},{"Start":"00:19.030 ","End":"00:21.815","Text":"Here\u0027s our example."},{"Start":"00:21.815 ","End":"00:25.430","Text":"Define f from 0,"},{"Start":"00:25.430 ","End":"00:28.140","Text":"1 to minus 1,"},{"Start":"00:28.140 ","End":"00:32.945","Text":"1 by f of x equals sine of 1 over x."},{"Start":"00:32.945 ","End":"00:35.540","Text":"Now, this is not defined at x equals 0."},{"Start":"00:35.540 ","End":"00:39.740","Text":"At 0, we define it separately to be 0."},{"Start":"00:39.740 ","End":"00:43.235","Text":"First we\u0027ll show it satisfies the IVP,"},{"Start":"00:43.235 ","End":"00:46.550","Text":"and then we\u0027ll show that it\u0027s not continuous."},{"Start":"00:46.550 ","End":"00:51.935","Text":"Let\u0027s say that a is less than b in the interval 0, 1."},{"Start":"00:51.935 ","End":"00:57.995","Text":"We\u0027ll choose some y between f of a and f of b,"},{"Start":"00:57.995 ","End":"01:04.355","Text":"we have to find a c between a and b such that f of c equals y."},{"Start":"01:04.355 ","End":"01:11.315","Text":"We\u0027ll distinguish 2 cases according to whether or not a equals 0 or a is bigger than 0."},{"Start":"01:11.315 ","End":"01:14.225","Text":"If a is not 0,"},{"Start":"01:14.225 ","End":"01:16.790","Text":"then the interval a,"},{"Start":"01:16.790 ","End":"01:22.590","Text":"b excludes the bad point 0 and f is just sine of 1 over x"},{"Start":"01:22.590 ","End":"01:30.615","Text":"there and that this part we know is continuous and since it\u0027s continuous, it has IVP."},{"Start":"01:30.615 ","End":"01:32.720","Text":"So c exists."},{"Start":"01:32.720 ","End":"01:34.700","Text":"That\u0027s the trivial case."},{"Start":"01:34.700 ","End":"01:38.945","Text":"The more difficult case is when a is 0,"},{"Start":"01:38.945 ","End":"01:41.505","Text":"then it\u0027s the bad point."},{"Start":"01:41.505 ","End":"01:45.335","Text":"How do we show the existence of such c?"},{"Start":"01:45.335 ","End":"01:50.305","Text":"Well, we can choose an integer n big enough so that"},{"Start":"01:50.305 ","End":"01:55.130","Text":"1 over 2 Pi n plus Pi over 2 is less than or"},{"Start":"01:55.130 ","End":"02:00.170","Text":"equal to b. I guess you could use the Archimedean property of numbers,"},{"Start":"02:00.170 ","End":"02:06.840","Text":"but it\u0027s clear that 1 over this thing goes to 0 because the denominator goes to infinity."},{"Start":"02:06.840 ","End":"02:10.280","Text":"At some point is going to be less than or equal to b,"},{"Start":"02:10.280 ","End":"02:13.680","Text":"which is certainly positive."},{"Start":"02:14.300 ","End":"02:19.169","Text":"Now define a_1 and b_1 as follows,"},{"Start":"02:19.169 ","End":"02:22.215","Text":"f of a_1 is,"},{"Start":"02:22.215 ","End":"02:24.465","Text":"we take 1 over and then we take the sine."},{"Start":"02:24.465 ","End":"02:25.985","Text":"We have sine of this,"},{"Start":"02:25.985 ","End":"02:27.680","Text":"which is minus 1,"},{"Start":"02:27.680 ","End":"02:33.534","Text":"and f of b_1 is sine of this, which is 1."},{"Start":"02:33.534 ","End":"02:38.205","Text":"Now look, a is 0 in this case,"},{"Start":"02:38.205 ","End":"02:46.685","Text":"a_1 is bigger than a. I mean it\u0027s bigger than 0 and that\u0027s less than b_1."},{"Start":"02:46.685 ","End":"02:48.370","Text":"This has a bigger denominator,"},{"Start":"02:48.370 ","End":"02:52.705","Text":"so it\u0027s smaller and b_1 is less than b."},{"Start":"02:52.705 ","End":"02:55.000","Text":"Let me just go back and correct this."},{"Start":"02:55.000 ","End":"02:57.740","Text":"Make it a strict inequality."},{"Start":"02:57.740 ","End":"02:59.820","Text":"Certainly, we can make it less than,"},{"Start":"02:59.820 ","End":"03:01.330","Text":"makes no big difference,"},{"Start":"03:01.330 ","End":"03:02.860","Text":"but just to be precise,"},{"Start":"03:02.860 ","End":"03:05.365","Text":"since I\u0027ve written less than here."},{"Start":"03:05.365 ","End":"03:13.480","Text":"What we have is that y belongs to the interval from f of a to f of b."},{"Start":"03:13.480 ","End":"03:14.935","Text":"That\u0027s we chose it."},{"Start":"03:14.935 ","End":"03:21.925","Text":"But this is contained in the interval from minus 1 to 1 closed interval,"},{"Start":"03:21.925 ","End":"03:24.705","Text":"which is f of a_1, f of b_1."},{"Start":"03:24.705 ","End":"03:27.290","Text":"So y is in this interval."},{"Start":"03:27.290 ","End":"03:28.700","Text":"It\u0027s like we\u0027re replacing a,"},{"Start":"03:28.700 ","End":"03:32.880","Text":"b by a_1, b_1."},{"Start":"03:32.890 ","End":"03:36.650","Text":"Now f is continuous on the interval a_1,"},{"Start":"03:36.650 ","End":"03:40.100","Text":"b_1, which excludes 0."},{"Start":"03:40.100 ","End":"03:43.835","Text":"That\u0027s why it\u0027s continuous. The bad point is not there."},{"Start":"03:43.835 ","End":"03:49.170","Text":"It has the IVP, intermediate value property."},{"Start":"03:49.170 ","End":"03:52.050","Text":"There is some c in a_1,"},{"Start":"03:52.050 ","End":"03:55.320","Text":"b_1, such that f of c equals y,"},{"Start":"03:55.320 ","End":"03:58.400","Text":"but a_1, b_1 is contained, it\u0027s a subset of a,"},{"Start":"03:58.400 ","End":"04:01.070","Text":"b so we found our c in a,"},{"Start":"04:01.070 ","End":"04:03.925","Text":"b with f of c equals y."},{"Start":"04:03.925 ","End":"04:07.924","Text":"Now at this point, we\u0027ve shown the intermediate value property."},{"Start":"04:07.924 ","End":"04:14.420","Text":"Now the 2nd part is to show that f is not continuous at 0. What happens at 0?"},{"Start":"04:14.420 ","End":"04:19.730","Text":"Well, we\u0027ll define a sequence and the sequence is inspired by this."},{"Start":"04:19.730 ","End":"04:25.590","Text":"These are the axis where f of x or f of x_n is 1,"},{"Start":"04:25.590 ","End":"04:27.750","Text":"just as it is here."},{"Start":"04:27.750 ","End":"04:29.819","Text":"When n goes to infinity,"},{"Start":"04:29.819 ","End":"04:32.960","Text":"x_n goes to 0 cause the denominator goes to infinity."},{"Start":"04:32.960 ","End":"04:37.580","Text":"But the limit of f of x_n is the limit of 1."},{"Start":"04:37.580 ","End":"04:38.680","Text":"It\u0027s a constant sequence,"},{"Start":"04:38.680 ","End":"04:42.240","Text":"so the limit is 1 which is not 0,"},{"Start":"04:42.240 ","End":"04:43.500","Text":"which is f of 0."},{"Start":"04:43.500 ","End":"04:46.005","Text":"Otherwise x_n goes to 0,"},{"Start":"04:46.005 ","End":"04:48.485","Text":"but f of x_n doesn\u0027t go to f of 0,"},{"Start":"04:48.485 ","End":"04:54.960","Text":"which violates the continuity and that\u0027s it. We\u0027re done."}],"ID":20950},{"Watched":false,"Name":"Exercise 4","Duration":"1m 44s","ChapterTopicVideoID":20158,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.540","Text":"In this exercise, f is continuous on the closed interval 0, 1."},{"Start":"00:06.540 ","End":"00:10.575","Text":"We have to show that there is some point x-naught in this interval,"},{"Start":"00:10.575 ","End":"00:13.860","Text":"such that f of x-naught is this."},{"Start":"00:13.860 ","End":"00:18.750","Text":"It\u0027s the average of the values of f at 1/4, 1/2, and 3/4."},{"Start":"00:18.750 ","End":"00:23.640","Text":"Now, since f is continuous on this closed interval,"},{"Start":"00:23.640 ","End":"00:30.105","Text":"it attains its minimum and it\u0027s maximum by the extreme value theorem."},{"Start":"00:30.105 ","End":"00:31.560","Text":"The minimum let\u0027s say,"},{"Start":"00:31.560 ","End":"00:35.385","Text":"X_1 and the maximum let\u0027s say X_2."},{"Start":"00:35.385 ","End":"00:42.145","Text":"Now in general, the minimum value is less than or equal to some average,"},{"Start":"00:42.145 ","End":"00:45.140","Text":"which is less than or equal to some maximum."},{"Start":"00:45.140 ","End":"00:49.550","Text":"We have that f of X_1 is less than or equal to this expression,"},{"Start":"00:49.550 ","End":"00:53.130","Text":"which is less than or equal to f of X_2."},{"Start":"00:53.600 ","End":"00:59.250","Text":"Then I distinguish two cases here that either X_1 equals X_2 or not."},{"Start":"00:59.250 ","End":"01:01.040","Text":"If X_1 equals X_2,"},{"Start":"01:01.040 ","End":"01:03.470","Text":"the minimum and the maximum occur at the same point,"},{"Start":"01:03.470 ","End":"01:05.255","Text":"so the minimum is the maximum,"},{"Start":"01:05.255 ","End":"01:07.285","Text":"so f is constant."},{"Start":"01:07.285 ","End":"01:09.340","Text":"You can choose any x naught,"},{"Start":"01:09.340 ","End":"01:11.780","Text":"because this is equal to the constant value."},{"Start":"01:11.780 ","End":"01:17.185","Text":"The average of three constants is the same constant and any f of x-naught will do."},{"Start":"01:17.185 ","End":"01:21.670","Text":"The more interesting case is when X_1 is not equal to X_2,"},{"Start":"01:21.670 ","End":"01:26.060","Text":"and in that case, we use the intermediate value property,"},{"Start":"01:26.060 ","End":"01:30.045","Text":"to find some X naught in the interval from X_1 to"},{"Start":"01:30.045 ","End":"01:36.155","Text":"X_2 such that f of x-naught equals this expression."},{"Start":"01:36.155 ","End":"01:39.460","Text":"In this case too, we have an x-naught,"},{"Start":"01:39.460 ","End":"01:42.885","Text":"that f of x-naught equals this expression,"},{"Start":"01:42.885 ","End":"01:45.430","Text":"and we are done."}],"ID":20951},{"Watched":false,"Name":"Exercise 5","Duration":"2m 43s","ChapterTopicVideoID":20159,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.030","Text":"In this exercise, f of x is a polynomial of degree 2n,"},{"Start":"00:06.030 ","End":"00:08.850","Text":"and the leading coefficient is 1."},{"Start":"00:08.850 ","End":"00:13.695","Text":"We have to show that f attains its infimum."},{"Start":"00:13.695 ","End":"00:17.835","Text":"Now, let\u0027s consider the behavior of f at infinity."},{"Start":"00:17.835 ","End":"00:21.000","Text":"This is determined by the leading term,"},{"Start":"00:21.000 ","End":"00:26.940","Text":"so that when x goes to infinity or when x goes to minus infinity,"},{"Start":"00:26.940 ","End":"00:30.780","Text":"then f of x tends to infinity."},{"Start":"00:30.780 ","End":"00:33.870","Text":"Basically just by considering this term,"},{"Start":"00:33.870 ","End":"00:36.915","Text":"this goes to infinity on both sides."},{"Start":"00:36.915 ","End":"00:39.270","Text":"Now this limit and infinity on both sides,"},{"Start":"00:39.270 ","End":"00:41.420","Text":"we can translate as follows."},{"Start":"00:41.420 ","End":"00:43.745","Text":"For any K bigger than 0,"},{"Start":"00:43.745 ","End":"00:47.060","Text":"there exists an M bigger than 0,"},{"Start":"00:47.060 ","End":"00:51.290","Text":"such that for all x whose absolute value is bigger than M,"},{"Start":"00:51.290 ","End":"00:57.079","Text":"f of x is bigger than K. Now actually this is a 2-in-1 that we combine."},{"Start":"00:57.079 ","End":"01:02.660","Text":"There should be an M bigger than 0 for x bigger than M,"},{"Start":"01:02.660 ","End":"01:09.260","Text":"and there should be an M less than 0 for x less than minus M. But you"},{"Start":"01:09.260 ","End":"01:11.630","Text":"take the maximum of the 2M\u0027s and then we"},{"Start":"01:11.630 ","End":"01:16.625","Text":"can combine them with an absolute value like this."},{"Start":"01:16.625 ","End":"01:23.180","Text":"Now choose K to be bigger than absolute value of f of 0."},{"Start":"01:23.180 ","End":"01:28.025","Text":"I\u0027m sorry, I didn\u0027t state it but we still choose the M for this K."},{"Start":"01:28.025 ","End":"01:34.674","Text":"Since f is continuous on the interval from minus M to M,"},{"Start":"01:34.674 ","End":"01:38.420","Text":"on a closed interval by the extreme value theorem,"},{"Start":"01:38.420 ","End":"01:45.035","Text":"there exists some x naught such that f of x naught is the infimum."},{"Start":"01:45.035 ","End":"01:51.170","Text":"In other words, f achieves its minimum on the interval from minus M to M. Next,"},{"Start":"01:51.170 ","End":"01:57.350","Text":"we\u0027ll show that this minimum is actually true for all x also outside this interval."},{"Start":"01:57.350 ","End":"01:59.585","Text":"If x is not in this interval,"},{"Start":"01:59.585 ","End":"02:01.960","Text":"absolute value of x is bigger than M,"},{"Start":"02:01.960 ","End":"02:10.745","Text":"then f of x is bigger than K. But K is bigger than f of 0 because of this,"},{"Start":"02:10.745 ","End":"02:12.320","Text":"and also this is"},{"Start":"02:12.320 ","End":"02:18.740","Text":"bigger or equal to f of x naught because this is the infimum, the minimum."},{"Start":"02:18.740 ","End":"02:23.420","Text":"We get that f of x is bigger or equal to f of x naught"},{"Start":"02:23.420 ","End":"02:27.920","Text":"in both cases when it\u0027s between minus M and M,"},{"Start":"02:27.920 ","End":"02:34.000","Text":"because this is the infimum and outside because of this inequality."},{"Start":"02:34.000 ","End":"02:37.180","Text":"In either case, we get this inequality,"},{"Start":"02:37.180 ","End":"02:44.060","Text":"and so f achieves its infimum at x naught. We are done."}],"ID":20952},{"Watched":false,"Name":"Exercise 6","Duration":"4m 32s","ChapterTopicVideoID":20160,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.410","Text":"In this exercise, f is a real-valued function defined on the reels and its continuous."},{"Start":"00:07.410 ","End":"00:13.440","Text":"We have to show that f is a constant function in 1 of 2 cases."},{"Start":"00:13.440 ","End":"00:14.850","Text":"I mean, in each of the cases."},{"Start":"00:14.850 ","End":"00:17.790","Text":"If f of x is rational,"},{"Start":"00:17.790 ","End":"00:22.125","Text":"for every x, then f is a constant."},{"Start":"00:22.125 ","End":"00:26.730","Text":"If f of x is an integer for each rational x,"},{"Start":"00:26.730 ","End":"00:29.790","Text":"then also f is a constant."},{"Start":"00:29.790 ","End":"00:31.920","Text":"Well, let\u0027s start with part a."},{"Start":"00:31.920 ","End":"00:40.290","Text":"In part a, we\u0027ll do it by contradiction and suppose that f isn\u0027t constant."},{"Start":"00:40.290 ","End":"00:49.280","Text":"The least pair of numbers a and b such that f of a is not equal to f of b."},{"Start":"00:49.280 ","End":"00:52.760","Text":"Of course, a is not equal to b because if a was equal to b,"},{"Start":"00:52.760 ","End":"00:55.070","Text":"then f of a would equal f of b."},{"Start":"00:55.070 ","End":"00:56.945","Text":"It\u0027s not equal."},{"Start":"00:56.945 ","End":"01:05.330","Text":"Now, choose an irrational number y in the interval from f of a to f of b."},{"Start":"01:05.330 ","End":"01:12.800","Text":"By the IVP, there is some c in the open interval a,"},{"Start":"01:12.800 ","End":"01:19.580","Text":"b, such that f of c equals y. I want to add something."},{"Start":"01:19.580 ","End":"01:21.245","Text":"It may be clear, maybe not."},{"Start":"01:21.245 ","End":"01:25.850","Text":"I didn\u0027t say if a is bigger or smaller than b, but when I write a,"},{"Start":"01:25.850 ","End":"01:31.340","Text":"b, the interval, it\u0027s okay to say 3, 1 or something."},{"Start":"01:31.340 ","End":"01:32.740","Text":"It\u0027s okay if a is bigger than b,"},{"Start":"01:32.740 ","End":"01:37.520","Text":"it just means that the interval goes the other way but because a is not equal to b,"},{"Start":"01:37.520 ","End":"01:40.310","Text":"then there is definitely an open interval."},{"Start":"01:40.310 ","End":"01:43.760","Text":"Now back here, because f of c equals y,"},{"Start":"01:43.760 ","End":"01:49.145","Text":"then it contradicts the fact that f of x is always rational because for x equals c,"},{"Start":"01:49.145 ","End":"01:52.625","Text":"it\u0027s not because y is irrational."},{"Start":"01:52.625 ","End":"01:57.460","Text":"This contradiction shows that f is constant."},{"Start":"01:57.460 ","End":"02:00.115","Text":"Now on to part b."},{"Start":"02:00.115 ","End":"02:03.640","Text":"Let Alpha be any irrational number."},{"Start":"02:03.640 ","End":"02:09.295","Text":"We can find a sequence of rational numbers which converges to Alpha."},{"Start":"02:09.295 ","End":"02:13.590","Text":"Now by continuity, if r_n tends to Alpha,"},{"Start":"02:13.590 ","End":"02:16.710","Text":"then f of r_n tends to f of Alpha."},{"Start":"02:16.710 ","End":"02:20.215","Text":"F of r_n is a convergent sequence,"},{"Start":"02:20.215 ","End":"02:22.905","Text":"and so it\u0027s also a Cauchy sequence."},{"Start":"02:22.905 ","End":"02:26.170","Text":"We can choose Epsilon equals 1 and"},{"Start":"02:26.170 ","End":"02:30.970","Text":"interpret what being a Cauchy sequence means for this Epsilon."},{"Start":"02:30.970 ","End":"02:36.490","Text":"Well, it means in simple terms that from a certain point onwards,"},{"Start":"02:36.490 ","End":"02:41.915","Text":"the distance between any 2 sequence members is less than 1."},{"Start":"02:41.915 ","End":"02:45.460","Text":"Here\u0027s a more precise phrasing."},{"Start":"02:46.070 ","End":"02:51.890","Text":"Since these terms are integers and their distance is less than 1,"},{"Start":"02:51.890 ","End":"02:55.230","Text":"then they must be equal to each other."},{"Start":"02:56.000 ","End":"03:07.485","Text":"The sequence f of r_n is eventually a constant from this point, big and onward."},{"Start":"03:07.485 ","End":"03:11.585","Text":"If we have a sequence that eventually a constant,"},{"Start":"03:11.585 ","End":"03:17.660","Text":"then the limit which is f of Alpha is that same constant,"},{"Start":"03:17.660 ","End":"03:20.300","Text":"which is an integer because these are all integers."},{"Start":"03:20.300 ","End":"03:22.775","Text":"So f of Alpha is an integer."},{"Start":"03:22.775 ","End":"03:25.085","Text":"Now Alpha was arbitrary."},{"Start":"03:25.085 ","End":"03:31.765","Text":"What we\u0027ve shown is that f is an integer-valued function and its continuous."},{"Start":"03:31.765 ","End":"03:37.040","Text":"By the IVP, it must be constant."},{"Start":"03:37.040 ","End":"03:41.330","Text":"It\u0027s fairly clear that an integer-valued continuous function is constant,"},{"Start":"03:41.330 ","End":"03:44.180","Text":"but I\u0027ll show you a little proof of this part."},{"Start":"03:44.180 ","End":"03:46.190","Text":"I\u0027ll do it by contradiction."},{"Start":"03:46.190 ","End":"03:47.960","Text":"Suppose it\u0027s not a constant."},{"Start":"03:47.960 ","End":"03:55.890","Text":"We have 2 different Alphas where Alpha_1 goes to n_1 and Alpha_2 goes to n_2,"},{"Start":"03:55.890 ","End":"03:58.030","Text":"I mean f of them."},{"Start":"03:58.490 ","End":"04:01.985","Text":"These 2 integers are different."},{"Start":"04:01.985 ","End":"04:03.770","Text":"Now if 2 integers are different,"},{"Start":"04:03.770 ","End":"04:06.365","Text":"we can choose some non-integer between them,"},{"Start":"04:06.365 ","End":"04:11.285","Text":"something and a 1/2 or whatever between the 2 of them."},{"Start":"04:11.285 ","End":"04:16.769","Text":"By the IVP, for some Gamma in this interval,"},{"Start":"04:16.769 ","End":"04:20.910","Text":"f of Gamma is equal to y. F is not"},{"Start":"04:20.910 ","End":"04:26.160","Text":"integer-valued because y is not an integer and that\u0027s a contradiction."},{"Start":"04:26.860 ","End":"04:33.000","Text":"F is constant. We are done."}],"ID":20953},{"Watched":false,"Name":"Exercise 7","Duration":"3m 8s","ChapterTopicVideoID":20161,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.590","Text":"In this exercise, we have a real-valued function defined on the interval from 0-1,"},{"Start":"00:08.840 ","End":"00:13.470","Text":"a closed interval, and we suppose that f of"},{"Start":"00:13.470 ","End":"00:17.850","Text":"x is rational whenever x is irrational and vice versa,"},{"Start":"00:17.850 ","End":"00:21.000","Text":"f of x is irrational when x is rational."},{"Start":"00:21.000 ","End":"00:24.360","Text":"We have to show that f cannot be continuous."},{"Start":"00:24.360 ","End":"00:29.115","Text":"Now there\u0027s a trick that combines these 2 reversals,"},{"Start":"00:29.115 ","End":"00:36.000","Text":"what we can do is define g of x equal f of x minus x."},{"Start":"00:36.000 ","End":"00:40.865","Text":"Then g of x is irrational for all x in the interval,"},{"Start":"00:40.865 ","End":"00:45.800","Text":"because we either have rational minus irrational or irrational minus rational,"},{"Start":"00:45.800 ","End":"00:47.135","Text":"but in both cases,"},{"Start":"00:47.135 ","End":"00:49.670","Text":"it gives us irrational."},{"Start":"00:49.670 ","End":"00:55.790","Text":"Now, I claim that g cannot be a constant function."},{"Start":"00:55.790 ","End":"00:57.710","Text":"I\u0027m going to use proof by contradiction."},{"Start":"00:57.710 ","End":"01:00.080","Text":"Suppose that g is a constant,"},{"Start":"01:00.080 ","End":"01:03.175","Text":"let\u0027s call this constant alpha."},{"Start":"01:03.175 ","End":"01:06.180","Text":"If g of x is equal to Alpha,"},{"Start":"01:06.180 ","End":"01:11.110","Text":"then f of x is equal to x plus alpha everywhere."},{"Start":"01:11.110 ","End":"01:16.580","Text":"Plug in 0 and we get f of 0 is 0 plus alpha, which is alpha,"},{"Start":"01:16.580 ","End":"01:20.120","Text":"so alpha is irrational because f of"},{"Start":"01:20.120 ","End":"01:25.210","Text":"a rational number like 0 has got to be irrational as what was given."},{"Start":"01:25.210 ","End":"01:28.230","Text":"Now, if we divide alpha by a big enough n,"},{"Start":"01:28.230 ","End":"01:30.550","Text":"we\u0027ll get it between 0 and 1."},{"Start":"01:30.550 ","End":"01:32.930","Text":"If alpha happens to be negative,"},{"Start":"01:32.930 ","End":"01:39.435","Text":"then we just choose a negative n and it is easily seen to be possible."},{"Start":"01:39.435 ","End":"01:41.250","Text":"Just pick 1 of these n\u0027s,"},{"Start":"01:41.250 ","End":"01:47.925","Text":"so what we get is f of alpha of n is alpha plus alpha over n from the x plus alpha."},{"Start":"01:47.925 ","End":"01:50.700","Text":"Well, it did it backwards, alpha plus x."},{"Start":"01:50.700 ","End":"01:54.325","Text":"This is equal to alpha times 1 plus 1 over n,"},{"Start":"01:54.325 ","End":"01:56.285","Text":"and this is irrational,"},{"Start":"01:56.285 ","End":"02:04.055","Text":"because an irrational times a non-zero rational, gives us irrational."},{"Start":"02:04.055 ","End":"02:10.325","Text":"Now, this is actually a contradiction from the given f of x is rational for irrational x,"},{"Start":"02:10.325 ","End":"02:11.840","Text":"so that\u0027s the contradiction."},{"Start":"02:11.840 ","End":"02:15.965","Text":"We can now rule out the possibility that g is constant,"},{"Start":"02:15.965 ","End":"02:19.455","Text":"and that means it has at least 2 different values,"},{"Start":"02:19.455 ","End":"02:24.425","Text":"and these are irrational because g has only got irrational values."},{"Start":"02:24.425 ","End":"02:26.780","Text":"Now between any 2 numbers in particular,"},{"Start":"02:26.780 ","End":"02:29.449","Text":"between any 2 irrational numbers,"},{"Start":"02:29.449 ","End":"02:38.060","Text":"there\u0027s a rational number and g can\u0027t attain this because g only gets irrational values."},{"Start":"02:38.060 ","End":"02:40.940","Text":"So g doesn\u0027t have the IVP,"},{"Start":"02:40.940 ","End":"02:43.475","Text":"so it isn\u0027t continuous."},{"Start":"02:43.475 ","End":"02:47.000","Text":"That\u0027s g, but the question was about f. Well,"},{"Start":"02:47.000 ","End":"02:52.185","Text":"if g is not continuous and neither is f. If f were continuous,"},{"Start":"02:52.185 ","End":"02:57.380","Text":"then f of x minus x would also be continuous because x is continuous."},{"Start":"02:57.380 ","End":"03:01.315","Text":"This is like a mini proof by contradiction for this,"},{"Start":"03:01.315 ","End":"03:04.159","Text":"and so f means not continuous."},{"Start":"03:04.159 ","End":"03:05.600","Text":"This is what we wanted to show,"},{"Start":"03:05.600 ","End":"03:08.310","Text":"and so we are done."}],"ID":20954},{"Watched":false,"Name":"Exercise 8","Duration":"1m 49s","ChapterTopicVideoID":20162,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.195","Text":"In this exercise, f is continuous and it\u0027s periodic with period 2Pi."},{"Start":"00:07.195 ","End":"00:10.630","Text":"We have to show that there is an x naught such"},{"Start":"00:10.630 ","End":"00:14.005","Text":"that f of x naught plus Pi equals f of x naught."},{"Start":"00:14.005 ","End":"00:18.490","Text":"The way we tackle this is you want to make something equals 0 so if we"},{"Start":"00:18.490 ","End":"00:25.030","Text":"define g of x as f of x plus Pi minus f of x,"},{"Start":"00:25.030 ","End":"00:28.160","Text":"then g is also continuous,"},{"Start":"00:28.160 ","End":"00:31.300","Text":"and we\u0027re looking for a place where g is 0."},{"Start":"00:31.300 ","End":"00:35.110","Text":"Now, notice that g of Pi, by definition,"},{"Start":"00:35.110 ","End":"00:38.255","Text":"is f of 2Pi minus f of Pi,"},{"Start":"00:38.255 ","End":"00:43.040","Text":"and this is equal to f of 0 minus f of Pi."},{"Start":"00:43.040 ","End":"00:47.660","Text":"These 2 are equal by the periodic property and this is"},{"Start":"00:47.660 ","End":"00:53.750","Text":"equal and take the minus out f of Pi minus f of 0."},{"Start":"00:53.750 ","End":"00:56.705","Text":"If you look at what\u0027s inside the square brackets,"},{"Start":"00:56.705 ","End":"01:01.075","Text":"that is exactly g of 0."},{"Start":"01:01.075 ","End":"01:05.085","Text":"We have g of Pi is minus g of 0,"},{"Start":"01:05.085 ","End":"01:13.010","Text":"so g of 0 and g of Pi are negatives to each other so in the middle between them is 0."},{"Start":"01:13.010 ","End":"01:15.620","Text":"Because g of 0 could equal 0,"},{"Start":"01:15.620 ","End":"01:17.120","Text":"then g of Pi is 0,"},{"Start":"01:17.120 ","End":"01:23.560","Text":"but still 0 is between 0 and 0 if you take less than or equal to [inaudible] Zero is"},{"Start":"01:23.560 ","End":"01:30.600","Text":"in the closed interval inclusive from g of 0 to g of Pi."},{"Start":"01:30.600 ","End":"01:36.960","Text":"By the IVP, there is some x_0 in the interval from 0 to Pi such that g of"},{"Start":"01:36.960 ","End":"01:43.700","Text":"x naught equals naught and if you translate that by looking at the definition of g,"},{"Start":"01:43.700 ","End":"01:50.039","Text":"it means that f of x naught plus Pi equals f of x naught as required."}],"ID":20955},{"Watched":false,"Name":"Exercise 9","Duration":"4m 23s","ChapterTopicVideoID":20163,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.775","Text":"In this exercise, we have a function defined on the closed interval 0, 12."},{"Start":"00:05.775 ","End":"00:09.090","Text":"It\u0027s continuous and it\u0027s equal at the endpoints,"},{"Start":"00:09.090 ","End":"00:11.445","Text":"f of 0 equals f of 12."},{"Start":"00:11.445 ","End":"00:17.220","Text":"We have to find 4 points, x_1, x_2, x_3, x_4 in this interval"},{"Start":"00:17.220 ","End":"00:22.020","Text":"such that x_2 minus x_1 is 6,"},{"Start":"00:22.020 ","End":"00:25.080","Text":"x_4 minus x_3 is 3,"},{"Start":"00:25.080 ","End":"00:27.210","Text":"f of x_1 equals f of x_2,"},{"Start":"00:27.210 ","End":"00:29.535","Text":"and f of x_3 equals f of x_4."},{"Start":"00:29.535 ","End":"00:32.970","Text":"Our strategy will be first to find x_1, x_2."},{"Start":"00:32.970 ","End":"00:41.960","Text":"We\u0027ll find 2 points which differ by 6 and f is equal at both of them."},{"Start":"00:41.960 ","End":"00:45.055","Text":"Very similar to what we had in the beginning,"},{"Start":"00:45.055 ","End":"00:52.485","Text":"that 12 and 0 differ by 12 and f is equal at both of them."},{"Start":"00:52.485 ","End":"00:54.090","Text":"Then we get it down to 6."},{"Start":"00:54.090 ","End":"00:55.935","Text":"When we\u0027ve done that,"},{"Start":"00:55.935 ","End":"00:59.370","Text":"then we\u0027ll repeat the process with x_1, x_2"},{"Start":"00:59.370 ","End":"01:04.050","Text":"instead of 0,12 and we\u0027ll get 2 numbers that differ by 3"},{"Start":"01:04.050 ","End":"01:08.110","Text":"and also f is equal at both of them."},{"Start":"01:08.450 ","End":"01:14.885","Text":"For the first part, we define g of x equals f of x plus 6 minus f of x."},{"Start":"01:14.885 ","End":"01:18.040","Text":"We chose 6 because that\u0027s half the difference."},{"Start":"01:18.040 ","End":"01:23.360","Text":"Note that g of 0 is f of 6 minus f of 0."},{"Start":"01:24.410 ","End":"01:29.935","Text":"Well, initially it\u0027s f of 12 minus f of 6 but f of 12 is f of 0,"},{"Start":"01:29.935 ","End":"01:32.155","Text":"so it\u0027s f of 0 minus f of 6,"},{"Start":"01:32.155 ","End":"01:34.030","Text":"which is the reverse of this."},{"Start":"01:34.030 ","End":"01:38.615","Text":"What we see is that g of 6 is minus g of 0."},{"Start":"01:38.615 ","End":"01:41.010","Text":"They\u0027re negatives of each other."},{"Start":"01:41.010 ","End":"01:45.900","Text":"0 lies between them, inclusive."},{"Start":"01:45.900 ","End":"01:51.780","Text":"There is some x_1 in the interval from 0-6,"},{"Start":"01:51.780 ","End":"01:54.795","Text":"such that g of x_1 is 0."},{"Start":"01:54.795 ","End":"01:57.820","Text":"If you translate that by the definition of g,"},{"Start":"01:57.820 ","End":"02:03.510","Text":"it means that f of x_1 plus 6 minus f of x_1 is 0."},{"Start":"02:03.510 ","End":"02:07.575","Text":"All we have to do now is let x_1 plus 6 equal x_2."},{"Start":"02:07.575 ","End":"02:11.340","Text":"Then we get that x_1 and x_2 are both in this interval."},{"Start":"02:11.340 ","End":"02:13.880","Text":"Well, x_1 is in the smaller interval,"},{"Start":"02:13.880 ","End":"02:15.215","Text":"so it\u0027s certainly here."},{"Start":"02:15.215 ","End":"02:21.320","Text":"Anything in here, if I add 6 to it will still be in the interval from 0-12,"},{"Start":"02:21.320 ","End":"02:23.090","Text":"it\u0027ll be from 6-12,"},{"Start":"02:23.090 ","End":"02:24.815","Text":"but still, it will be here."},{"Start":"02:24.815 ","End":"02:30.620","Text":"It satisfies that f of x_2 equals f of x_1 from here"},{"Start":"02:30.620 ","End":"02:34.280","Text":"and x_2 minus x_1 is 6."},{"Start":"02:34.280 ","End":"02:36.410","Text":"That\u0027s the first part."},{"Start":"02:36.410 ","End":"02:38.465","Text":"Second part is very similar,"},{"Start":"02:38.465 ","End":"02:43.370","Text":"except instead of 0,12 will take x_1 and x_2."},{"Start":"02:43.370 ","End":"02:52.635","Text":"We define h of x equals f of x plus 3 minus f of x, 3 is half the 6."},{"Start":"02:52.635 ","End":"03:01.380","Text":"Then we get h at x_1 is equal to this and, h of x_1 plus 3,"},{"Start":"03:01.380 ","End":"03:06.980","Text":"just like before, is going to turn out to equal minus this."},{"Start":"03:06.980 ","End":"03:09.290","Text":"We have a point in between."},{"Start":"03:09.290 ","End":"03:18.540","Text":"Then we call that x_3 and h of x_3 is 0,"},{"Start":"03:18.540 ","End":"03:23.535","Text":"and then x_4 is x_3 plus 3."},{"Start":"03:23.535 ","End":"03:27.525","Text":"We have x_4 minus x_3 is 3 from here."},{"Start":"03:27.525 ","End":"03:31.080","Text":"From here we get that f of x_3 equals f of x_4."},{"Start":"03:31.080 ","End":"03:33.630","Text":"Everything is defined okay because"},{"Start":"03:33.630 ","End":"03:39.060","Text":"x plus 3 is in the interval from 3-12."},{"Start":"03:39.060 ","End":"03:40.850","Text":"Anyway, everything is defined properly"},{"Start":"03:40.850 ","End":"03:46.230","Text":"as long as we restrict x to be between 0 and 9."},{"Start":"03:46.370 ","End":"03:49.480","Text":"Let\u0027s just continue with verifying."},{"Start":"03:49.480 ","End":"03:56.490","Text":"Now, x_1 is in 0,6 and that\u0027s what it says here."},{"Start":"03:56.490 ","End":"04:04.650","Text":"X_3 is between 0 and 9 because x_3 is in this interval"},{"Start":"04:04.650 ","End":"04:11.795","Text":"and x_4 being x_3 plus 3 is going to be between 3 and 12."},{"Start":"04:11.795 ","End":"04:13.175","Text":"But in any case,"},{"Start":"04:13.175 ","End":"04:16.560","Text":"all of them, x_1 and x_2 we checked already,"},{"Start":"04:16.560 ","End":"04:19.175","Text":"and x_3 and x_4 are also in this interval,"},{"Start":"04:19.175 ","End":"04:24.090","Text":"so all the conditions are satisfied and we are done."}],"ID":20956},{"Watched":false,"Name":"Exercise 10","Duration":"2m 23s","ChapterTopicVideoID":20164,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.715","Text":"In this exercise, p is a polynomial function of odd degree."},{"Start":"00:05.715 ","End":"00:09.315","Text":"We have to show that p is onto."},{"Start":"00:09.315 ","End":"00:17.775","Text":"In other words, p is of the form a_0 plus a_1x and so on and so on, up to a_nx^n."},{"Start":"00:17.775 ","End":"00:20.595","Text":"But the important thing is that n is odd,"},{"Start":"00:20.595 ","End":"00:22.800","Text":"like n is 2k plus 1,"},{"Start":"00:22.800 ","End":"00:26.265","Text":"and the leading coefficient has to be non-0, of course."},{"Start":"00:26.265 ","End":"00:33.870","Text":"Now let\u0027s take any y in R. We have to show that p of something equals y."},{"Start":"00:33.870 ","End":"00:35.985","Text":"Let\u0027s consider 2 cases,"},{"Start":"00:35.985 ","End":"00:39.330","Text":"a_n bigger than 0 and then a_n less than 0."},{"Start":"00:39.330 ","End":"00:41.575","Text":"If a_n is positive,"},{"Start":"00:41.575 ","End":"00:44.360","Text":"then because of the odd degree,"},{"Start":"00:44.360 ","End":"00:48.390","Text":"the limit at infinity is infinity,"},{"Start":"00:48.390 ","End":"00:51.650","Text":"because we\u0027re just looking at this leading term,"},{"Start":"00:51.650 ","End":"00:54.770","Text":"positive times something that goes to infinity,"},{"Start":"00:54.770 ","End":"00:56.975","Text":"to the power of n is infinity."},{"Start":"00:56.975 ","End":"01:00.920","Text":"But when x goes to minus infinity,"},{"Start":"01:00.920 ","End":"01:05.720","Text":"then we have minus infinity to the odd degree,"},{"Start":"01:05.720 ","End":"01:08.145","Text":"and so it\u0027s minus infinity,"},{"Start":"01:08.145 ","End":"01:09.980","Text":"and we just get the other way around."},{"Start":"01:09.980 ","End":"01:11.555","Text":"If a_n is negative,"},{"Start":"01:11.555 ","End":"01:18.140","Text":"then we get minus infinity on the left and plus infinity on the right."},{"Start":"01:18.140 ","End":"01:19.630","Text":"Sorry, other way around."},{"Start":"01:19.630 ","End":"01:24.300","Text":"Minus infinity on the right and infinity on the left."},{"Start":"01:24.300 ","End":"01:26.005","Text":"But in either case,"},{"Start":"01:26.005 ","End":"01:34.860","Text":"you can find x_1 and x_2 such that y is between p of x_1 and p of x_2."},{"Start":"01:34.860 ","End":"01:37.395","Text":"Because in either case,"},{"Start":"01:37.395 ","End":"01:40.689","Text":"in 1 of the directions we go to infinity."},{"Start":"01:40.689 ","End":"01:44.515","Text":"So we can certainly find p of x_2 as large as we like."},{"Start":"01:44.515 ","End":"01:49.805","Text":"In either case, and 1 of the directions we go to minus infinity,"},{"Start":"01:49.805 ","End":"01:54.410","Text":"so we can find p of x_1 small as we like."},{"Start":"01:54.410 ","End":"02:00.550","Text":"Now, the intermediate value property says there is some x_3 between x_1 and x_2."},{"Start":"02:00.550 ","End":"02:04.705","Text":"By the way, I\u0027m assuming that x_1 is bigger than x_2 or smaller than x_2."},{"Start":"02:04.705 ","End":"02:07.045","Text":"When I write this interval notation, it could be."},{"Start":"02:07.045 ","End":"02:10.330","Text":"But either way, there\u0027s an x_3 between x_1 and x_2,"},{"Start":"02:10.330 ","End":"02:13.700","Text":"such that p of x_3 is exactly y."},{"Start":"02:13.700 ","End":"02:18.530","Text":"So y is in the image of f for all y,"},{"Start":"02:18.530 ","End":"02:22.590","Text":"and that means that f is onto and we are done."}],"ID":20957},{"Watched":false,"Name":"Exercise 11","Duration":"4m 10s","ChapterTopicVideoID":20165,"CourseChapterTopicPlaylistID":113411,"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, we\u0027re given a real valued function on the closed interval 0, 2."},{"Start":"00:05.700 ","End":"00:08.100","Text":"The function is continuous,"},{"Start":"00:08.100 ","End":"00:11.610","Text":"and we\u0027re given that f of nought is less than f of 2,"},{"Start":"00:11.610 ","End":"00:14.020","Text":"is less than f of 1."},{"Start":"00:14.240 ","End":"00:17.580","Text":"Here\u0027s a diagram, this is is f of nought,"},{"Start":"00:17.580 ","End":"00:19.380","Text":"which is less than f of 2,"},{"Start":"00:19.380 ","End":"00:21.870","Text":"which is less than f of 1."},{"Start":"00:21.870 ","End":"00:26.640","Text":"Now we have to show that there exist points x_0, x_1,"},{"Start":"00:26.640 ","End":"00:30.765","Text":"and x_2 in this interval such that, well,"},{"Start":"00:30.765 ","End":"00:36.465","Text":"this equality holds and that f of x_2 equals f of x_1."},{"Start":"00:36.465 ","End":"00:40.665","Text":"Now we get x nought basically from this point."},{"Start":"00:40.665 ","End":"00:42.620","Text":"This would be x nought because"},{"Start":"00:42.620 ","End":"00:50.480","Text":"by the intermediate value property between f of 1 and f of 0,"},{"Start":"00:50.480 ","End":"00:55.760","Text":"there must be a point which is equal to f of 2 because of this inequality."},{"Start":"00:55.760 ","End":"01:00.605","Text":"This would be our x nought and f of x nought is f of 2."},{"Start":"01:00.605 ","End":"01:02.420","Text":"Now we define a new function,"},{"Start":"01:02.420 ","End":"01:05.405","Text":"g of x based on f of x,"},{"Start":"01:05.405 ","End":"01:09.350","Text":"just using this formula and of course,"},{"Start":"01:09.350 ","End":"01:12.140","Text":"g is also continuous."},{"Start":"01:12.140 ","End":"01:14.690","Text":"Let\u0027s compute g of x_0."},{"Start":"01:14.690 ","End":"01:18.605","Text":"Just by substituting, we get this."},{"Start":"01:18.605 ","End":"01:26.675","Text":"Look at the details later and g of 1 plus 1/2 x_0 comes out to be this."},{"Start":"01:26.675 ","End":"01:29.780","Text":"Notice that these 2 are the same,"},{"Start":"01:29.780 ","End":"01:33.940","Text":"except the terms are reversed so one\u0027s minus the other,"},{"Start":"01:33.940 ","End":"01:36.485","Text":"not only they\u0027re negatives of each other,"},{"Start":"01:36.485 ","End":"01:38.285","Text":"but it\u0027s not the same point,"},{"Start":"01:38.285 ","End":"01:41.930","Text":"x nought and 1 plus a 1/2 x nought are not equal because"},{"Start":"01:41.930 ","End":"01:45.805","Text":"1 of them is less than 1 and the other is bigger than 1."},{"Start":"01:45.805 ","End":"01:48.349","Text":"By the intermediate value property,"},{"Start":"01:48.349 ","End":"01:50.270","Text":"there\u0027s going to be a 0 in between."},{"Start":"01:50.270 ","End":"01:56.715","Text":"We have to have an x_1 between x nought and 1 plus 1/2 x nought,"},{"Start":"01:56.715 ","End":"02:01.540","Text":"such that g at that point x_1 is 0."},{"Start":"02:03.590 ","End":"02:11.095","Text":"If we return to the definition of g and we plug in instead of x,"},{"Start":"02:11.095 ","End":"02:16.345","Text":"x_1, then we get that this is equal to 0."},{"Start":"02:16.345 ","End":"02:21.805","Text":"Now all that remains is that what I\u0027ve colored here to define that as x_2."},{"Start":"02:21.805 ","End":"02:30.495","Text":"Then we get that f of x_1 equals f of x_2 and that satisfies the condition."},{"Start":"02:30.495 ","End":"02:33.850","Text":"We\u0027re done, but don\u0027t go yet."},{"Start":"02:33.850 ","End":"02:37.895","Text":"I have an alternative solution to this problem."},{"Start":"02:37.895 ","End":"02:41.260","Text":"Now, the alternative solution is"},{"Start":"02:41.260 ","End":"02:46.210","Text":"to let g of x not the same g of x as above, is a new solution."},{"Start":"02:46.210 ","End":"02:50.350","Text":"Let g of x equals f of 1 plus x minus f of x."},{"Start":"02:50.350 ","End":"02:53.010","Text":"By trial and error, and this is what works."},{"Start":"02:53.010 ","End":"02:59.135","Text":"G of nought is f of 1 minus f of nought by substituting,"},{"Start":"02:59.135 ","End":"03:03.340","Text":"and that\u0027s bigger than 0 because we have f of 1 bigger than f of nought."},{"Start":"03:03.340 ","End":"03:08.010","Text":"Also, g of 1 is f of 2 minus f of 1."},{"Start":"03:08.010 ","End":"03:14.720","Text":"That\u0027s negative because we\u0027re told that f of 2 is less than f of 1,"},{"Start":"03:14.720 ","End":"03:17.960","Text":"f of 1 is the biggest of the 3 values."},{"Start":"03:17.960 ","End":"03:25.125","Text":"If we have a positive 0 and a negative 1 and somewhere in between 0 and 1,"},{"Start":"03:25.125 ","End":"03:31.970","Text":"we have a point x_1 where g of x_1 is 0 and if you interpret that,"},{"Start":"03:31.970 ","End":"03:35.945","Text":"we get that f of 1 plus x_1 minus f of x_1 is 0."},{"Start":"03:35.945 ","End":"03:40.005","Text":"Now let 1 plus x_1 be x_2,"},{"Start":"03:40.005 ","End":"03:44.335","Text":"and we have f of x_2 equals f of x_1."},{"Start":"03:44.335 ","End":"03:49.415","Text":"Now, the condition talked about an x_0 also,"},{"Start":"03:49.415 ","End":"03:52.895","Text":"that x_2 is 1 plus x_1 minus 1/2 x_0."},{"Start":"03:52.895 ","End":"03:58.705","Text":"We can just arbitrarily define x_0 equals 0,"},{"Start":"03:58.705 ","End":"04:05.865","Text":"and then we have that x_2 is 1 plus x_1 minus 1/2 x_0."},{"Start":"04:05.865 ","End":"04:10.720","Text":"We solved it 2 ways and we are done."}],"ID":20958},{"Watched":false,"Name":"Exercise 12","Duration":"2m 9s","ChapterTopicVideoID":20166,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.390","Text":"In this exercise, we have a function on the interval 0, 1,"},{"Start":"00:06.390 ","End":"00:10.005","Text":"the closed interval, and f is continuous,"},{"Start":"00:10.005 ","End":"00:12.960","Text":"and f is equal at the end points."},{"Start":"00:12.960 ","End":"00:19.575","Text":"We have to show that there is some point x_0 between 0 and 1/2,"},{"Start":"00:19.575 ","End":"00:24.075","Text":"such that f of x_0 equals f of x_0 plus 1/2."},{"Start":"00:24.075 ","End":"00:26.640","Text":"Of course, it\u0027s defined because if x_0 is here,"},{"Start":"00:26.640 ","End":"00:28.020","Text":"then x_0 is 1/2,"},{"Start":"00:28.020 ","End":"00:31.750","Text":"is still within the interval 0, 1."},{"Start":"00:31.970 ","End":"00:39.080","Text":"For the solution, we\u0027ll define a function g of x. g is"},{"Start":"00:39.080 ","End":"00:46.150","Text":"defined on the interval 0-1. g of x equals f of x plus 1/2 minus f of x."},{"Start":"00:46.150 ","End":"00:50.765","Text":"Obviously, we are looking for a place where g is 0, and that will be the x_0."},{"Start":"00:50.765 ","End":"00:54.215","Text":"Now, g is also continuous clearly,"},{"Start":"00:54.215 ","End":"00:56.255","Text":"and g of 1/2,"},{"Start":"00:56.255 ","End":"00:59.620","Text":"if you plug in 1/2,"},{"Start":"00:59.620 ","End":"01:03.540","Text":"we get f of 1 minus f of 1/2,"},{"Start":"01:03.540 ","End":"01:12.225","Text":"and this is equal to f0 minus f of 1/2 because f1 equals f0 by the given."},{"Start":"01:12.225 ","End":"01:13.790","Text":"Then this is equal to,"},{"Start":"01:13.790 ","End":"01:16.925","Text":"you can switch the order and put a minus,"},{"Start":"01:16.925 ","End":"01:19.790","Text":"and this is the same expression as this,"},{"Start":"01:19.790 ","End":"01:21.755","Text":"but with the minus in front,"},{"Start":"01:21.755 ","End":"01:28.630","Text":"so what we\u0027ve concluded is that g of 1/2 is minus g of 0."},{"Start":"01:28.630 ","End":"01:32.060","Text":"If 2 numbers are the negatives of each other,"},{"Start":"01:32.060 ","End":"01:34.385","Text":"0 is somewhere in between,"},{"Start":"01:34.385 ","End":"01:40.560","Text":"so we have 0 in the interval from g of 0 to g of 1/2."},{"Start":"01:40.560 ","End":"01:42.120","Text":"We don\u0027t know which is bigger,"},{"Start":"01:42.120 ","End":"01:43.455","Text":"but it doesn\u0027t matter."},{"Start":"01:43.455 ","End":"01:44.880","Text":"The interval could be backwards,"},{"Start":"01:44.880 ","End":"01:47.045","Text":"but there\u0027s still 0 is in it."},{"Start":"01:47.045 ","End":"01:53.750","Text":"Then we can apply the IVP to say that there is some x_0 in 0,"},{"Start":"01:53.750 ","End":"01:57.500","Text":"1/2 such that g is 0 at x_0."},{"Start":"01:57.500 ","End":"02:00.410","Text":"Now, if you just see what that means,"},{"Start":"02:00.410 ","End":"02:03.935","Text":"just plug in x_0 here,"},{"Start":"02:03.935 ","End":"02:06.185","Text":"and we get exactly what\u0027s here,"},{"Start":"02:06.185 ","End":"02:08.270","Text":"which is what we wanted."},{"Start":"02:08.270 ","End":"02:10.710","Text":"We are done."}],"ID":20959},{"Watched":false,"Name":"Exercise 13","Duration":"2m 16s","ChapterTopicVideoID":20167,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"In this exercise, we have 2 continuous functions,"},{"Start":"00:03.630 ","End":"00:07.155","Text":"f and g on the unit interval."},{"Start":"00:07.155 ","End":"00:11.850","Text":"We\u0027re given that the greatest lower bound of f on"},{"Start":"00:11.850 ","End":"00:16.379","Text":"the interval is equal to the greatest lower bound of g on the interval."},{"Start":"00:16.379 ","End":"00:22.779","Text":"We have to show that there is some point x naught in the interval where f and g are equal."},{"Start":"00:22.779 ","End":"00:26.405","Text":"Let\u0027s apply the extreme value theorem."},{"Start":"00:26.405 ","End":"00:30.095","Text":"After all, we have continuous functions on a closed interval."},{"Start":"00:30.095 ","End":"00:37.205","Text":"This means that the infimum is really a minimum and it\u0027s attained in each case."},{"Start":"00:37.205 ","End":"00:41.390","Text":"In other words, there are points x_1 and x_2,"},{"Start":"00:41.390 ","End":"00:49.225","Text":"such that the minimum of f occurs at x_1 and the minimum of g occurs at x_2."},{"Start":"00:49.225 ","End":"00:54.150","Text":"Now we\u0027re given that f of x_1 equals g of x_2."},{"Start":"00:54.150 ","End":"00:56.650","Text":"If x_1 happens to equal x_2,"},{"Start":"00:56.650 ","End":"01:01.595","Text":"then we\u0027re done because we can take our x naught to be that common value."},{"Start":"01:01.595 ","End":"01:05.065","Text":"Let\u0027s assume that x_1 is not equal to x_2."},{"Start":"01:05.065 ","End":"01:08.010","Text":"Note that g of x_2,"},{"Start":"01:08.010 ","End":"01:10.950","Text":"because it\u0027s equal to f of x_1 and f of x_1"},{"Start":"01:10.950 ","End":"01:14.340","Text":"is the minimum so it\u0027s less than or equal to f of x_2."},{"Start":"01:14.340 ","End":"01:16.735","Text":"The other way around,"},{"Start":"01:16.735 ","End":"01:19.970","Text":"it seems natural to take a different function,"},{"Start":"01:19.970 ","End":"01:26.180","Text":"define h as f minus g and h is also continuous of course."},{"Start":"01:26.180 ","End":"01:32.830","Text":"But h of x_1 is less than or equal to 0 because here at x_1,"},{"Start":"01:32.830 ","End":"01:37.430","Text":"f is less than or equal to g and h of x_2 is bigger or equal to"},{"Start":"01:37.430 ","End":"01:44.320","Text":"0 because here f minus g is bigger or equal to 0."},{"Start":"01:44.320 ","End":"01:48.005","Text":"By the intermediate value property,"},{"Start":"01:48.005 ","End":"01:57.650","Text":"h of x naught equals 0 for some x naught in the interval from x_1 to x_2,"},{"Start":"01:57.650 ","End":"02:03.440","Text":"which is contained in the original interval 0,1 and if h of x naught equals 0,"},{"Start":"02:03.440 ","End":"02:06.050","Text":"then by the definition of h as f minus g,"},{"Start":"02:06.050 ","End":"02:11.300","Text":"we get that f of x naught minus g of x naught is 0 and the rest is easy."},{"Start":"02:11.300 ","End":"02:16.710","Text":"We get that f of x naught equals g of x naught. We\u0027re done"}],"ID":20960},{"Watched":false,"Name":"Exercise 14","Duration":"3m 15s","ChapterTopicVideoID":20168,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.299","Text":"This exercise is a word problem."},{"Start":"00:03.299 ","End":"00:07.440","Text":"Cross country runner runs continuously an 8 kilometers"},{"Start":"00:07.440 ","End":"00:12.070","Text":"course in 40 minutes without taking a rest."},{"Start":"00:12.320 ","End":"00:21.315","Text":"Think about it, then it\u0027s an average of 5 minutes for each kilometer."},{"Start":"00:21.315 ","End":"00:25.080","Text":"Now, we have to show that somewhere along the course,"},{"Start":"00:25.080 ","End":"00:31.240","Text":"the runner must have covered a distance of 1 kilometer in exactly 5 minutes."},{"Start":"00:31.550 ","End":"00:39.440","Text":"Solution. Let\u0027s set up a variable x to be the distance in kilometers along"},{"Start":"00:39.440 ","End":"00:49.659","Text":"the course and we\u0027ll define a function from 0,7 to Reals,"},{"Start":"00:49.659 ","End":"00:56.965","Text":"where f of x is the time taken in minutes to cover the distance from x to x plus 1."},{"Start":"00:56.965 ","End":"01:00.340","Text":"That\u0027s why we only go up to 7 here."},{"Start":"01:00.340 ","End":"01:04.760","Text":"You might be wondering why we were given that it\u0027s without"},{"Start":"01:04.760 ","End":"01:11.255","Text":"a rest and that is necessary for the function f to be well-defined."},{"Start":"01:11.255 ","End":"01:15.380","Text":"I mean, if at exactly distance x plus 1,"},{"Start":"01:15.380 ","End":"01:18.890","Text":"the runner was resting,"},{"Start":"01:18.890 ","End":"01:21.875","Text":"then you couldn\u0027t say the time exactly"},{"Start":"01:21.875 ","End":"01:27.060","Text":"because at the same distance the time would have been increasing."},{"Start":"01:27.230 ","End":"01:30.830","Text":"F is well-defined there because of this,"},{"Start":"01:30.830 ","End":"01:34.100","Text":"what was given that the runner didn\u0027t take a rest."},{"Start":"01:34.100 ","End":"01:39.175","Text":"Now, notice that if we add up f of i,"},{"Start":"01:39.175 ","End":"01:43.005","Text":"when i goes from 0-7,"},{"Start":"01:43.005 ","End":"01:45.920","Text":"the time taken for the first kilometer,"},{"Start":"01:45.920 ","End":"01:47.179","Text":"for the second kilometer,"},{"Start":"01:47.179 ","End":"01:49.699","Text":"and so on up to the 8th kilometer,"},{"Start":"01:49.699 ","End":"01:52.790","Text":"which was from 7 to 7 plus 1,"},{"Start":"01:52.790 ","End":"01:58.170","Text":"the sum of the times is 40."},{"Start":"01:58.530 ","End":"02:04.000","Text":"Now, you can\u0027t have f of i being bigger"},{"Start":"02:04.000 ","End":"02:12.570","Text":"than 5 for all the i all 8 of them because if f of i"},{"Start":"02:12.570 ","End":"02:18.635","Text":"was bigger than 5 for each of these 8 kilometers,"},{"Start":"02:18.635 ","End":"02:22.335","Text":"then the total time would have been bigger than 40."},{"Start":"02:22.335 ","End":"02:28.640","Text":"This can\u0027t be true. At least there\u0027s 1 i for which f of i is not bigger than 5,"},{"Start":"02:28.640 ","End":"02:31.430","Text":"meaning less than or equal to 5."},{"Start":"02:31.430 ","End":"02:41.920","Text":"Similarly f of j has to be bigger or equal to 5 for some j."},{"Start":"02:41.980 ","End":"02:46.820","Text":"Otherwise the total time would be less than 40."},{"Start":"02:46.820 ","End":"02:54.530","Text":"We have i and j and by the intermediate value property,"},{"Start":"02:54.530 ","End":"03:03.200","Text":"there is some c between i and j inclusive such that f of c is exactly equal to 5."},{"Start":"03:03.200 ","End":"03:05.150","Text":"If you interpret this,"},{"Start":"03:05.150 ","End":"03:12.200","Text":"it means that the kilometer from distance c to distance c plus 1 took exactly 5 minutes."},{"Start":"03:12.200 ","End":"03:16.230","Text":"That\u0027s what we have to show so we are done."}],"ID":20961},{"Watched":false,"Name":"Exercise 15","Duration":"3m 42s","ChapterTopicVideoID":20169,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.285","Text":"In this exercise, we have a real valued function which is continuous and 1-1."},{"Start":"00:06.285 ","End":"00:11.210","Text":"We have to show that f is strictly monotone,"},{"Start":"00:11.210 ","End":"00:16.110","Text":"in other words, it\u0027s either strictly increasing or strictly decreasing."},{"Start":"00:16.110 ","End":"00:23.940","Text":"Now I claim that we can drop the word strictly from the statement of the problem"},{"Start":"00:23.940 ","End":"00:27.140","Text":"because it follows from the fact that f is 1-1"},{"Start":"00:27.140 ","End":"00:30.863","Text":"so we just have to show that f is increasing or decreasing"},{"Start":"00:30.863 ","End":"00:32.955","Text":"and the strictly will follow."},{"Start":"00:32.955 ","End":"00:37.080","Text":"It can\u0027t stay constant for a while because like I said,"},{"Start":"00:37.080 ","End":"00:38.415","Text":"it\u0027s 1-1."},{"Start":"00:38.415 ","End":"00:41.910","Text":"Let\u0027s do a proof by contradiction."},{"Start":"00:41.910 ","End":"00:46.660","Text":"Suppose that f is neither increasing nor decreasing,"},{"Start":"00:46.660 ","End":"00:51.665","Text":"that means it has to go up and down or down and up."},{"Start":"00:51.665 ","End":"00:53.090","Text":"Be more precise."},{"Start":"00:53.090 ","End":"00:58.505","Text":"There exists 3 values of x, x_1, x_2, x_3 in this order,"},{"Start":"00:58.505 ","End":"01:03.560","Text":"such that the middle one is less than the other 2,"},{"Start":"01:03.560 ","End":"01:07.520","Text":"so it goes down and up or the other way around,"},{"Start":"01:07.520 ","End":"01:10.490","Text":"the middle one could be larger than the 2 side ones,"},{"Start":"01:10.490 ","End":"01:13.009","Text":"and it goes up and down."},{"Start":"01:13.009 ","End":"01:21.370","Text":"Note the strict inequalities because of the fact that we dropped the strictly here,"},{"Start":"01:21.370 ","End":"01:25.700","Text":"we get strictly here in the contracept position."},{"Start":"01:25.700 ","End":"01:27.245","Text":"I highlighted this one."},{"Start":"01:27.245 ","End":"01:30.500","Text":"The 2 are so similar that we\u0027ll just do one case."},{"Start":"01:30.500 ","End":"01:33.080","Text":"We\u0027ll do the less than case."},{"Start":"01:33.080 ","End":"01:36.245","Text":"Here\u0027s a picture, I should label it."},{"Start":"01:36.245 ","End":"01:41.180","Text":"This would be x_1, x_2, x_3."},{"Start":"01:41.180 ","End":"01:43.070","Text":"This is the graph of f."},{"Start":"01:43.070 ","End":"01:46.660","Text":"Here we have f of x_1,"},{"Start":"01:46.660 ","End":"01:49.500","Text":"which is bigger than f of x_2,"},{"Start":"01:49.500 ","End":"01:53.340","Text":"and f of x_3 is also bigger than f of x_2."},{"Start":"01:53.340 ","End":"01:56.865","Text":"We can choose a value Alpha,"},{"Start":"01:56.865 ","End":"02:04.580","Text":"which separates the f of x_2 from f of x_1 and f of x_3, they\u0027re both bigger."},{"Start":"02:04.580 ","End":"02:06.830","Text":"Choose the smaller one of these 2,"},{"Start":"02:06.830 ","End":"02:11.330","Text":"and then put a line between this y and this y as follows."},{"Start":"02:11.330 ","End":"02:17.755","Text":"The strategy in words is to say that Alpha\u0027s between f of x_1 and f of x_2,"},{"Start":"02:17.755 ","End":"02:20.449","Text":"so we can find this point here"},{"Start":"02:20.449 ","End":"02:23.810","Text":"somewhere between x_1 and x_2 where we get the value Alpha."},{"Start":"02:23.810 ","End":"02:25.850","Text":"Similarly between x_2 and x_3,"},{"Start":"02:25.850 ","End":"02:29.990","Text":"we can find an x which has f of it is Alpha."},{"Start":"02:29.990 ","End":"02:34.040","Text":"Alpha is the value of the function at both points"},{"Start":"02:34.040 ","End":"02:37.025","Text":"and that would contradict the 1-1."},{"Start":"02:37.025 ","End":"02:39.094","Text":"That\u0027s just the idea."},{"Start":"02:39.094 ","End":"02:42.965","Text":"More precisely, by the intermediate value property,"},{"Start":"02:42.965 ","End":"02:46.440","Text":"there exists some a."},{"Start":"02:46.520 ","End":"02:53.955","Text":"A would be say here and then also going to be a b"},{"Start":"02:53.955 ","End":"03:03.990","Text":"here such that f of a is Alpha and f of b is Alpha,"},{"Start":"03:03.990 ","End":"03:06.620","Text":"and of course, a is not equal to b,"},{"Start":"03:06.620 ","End":"03:07.790","Text":"not just because of the picture,"},{"Start":"03:07.790 ","End":"03:12.150","Text":"but we know that x_2 is between a and b."},{"Start":"03:12.150 ","End":"03:16.185","Text":"I mean, a is less than x_2 from here,"},{"Start":"03:16.185 ","End":"03:21.510","Text":"and b is greater than x_2 because of this."},{"Start":"03:21.510 ","End":"03:24.890","Text":"So a is less than b, so a is not equal to b"},{"Start":"03:24.890 ","End":"03:29.075","Text":"and yet f of a equals f of b equals Alpha,"},{"Start":"03:29.075 ","End":"03:30.515","Text":"these 2 red points."},{"Start":"03:30.515 ","End":"03:32.180","Text":"So f is not 1-1."},{"Start":"03:32.180 ","End":"03:35.480","Text":"That\u0027s a contradiction and the contradiction came from"},{"Start":"03:35.480 ","End":"03:39.155","Text":"the supposition that f isn\u0027t monotone,"},{"Start":"03:39.155 ","End":"03:40.580","Text":"and therefore it is."},{"Start":"03:40.580 ","End":"03:43.560","Text":"We are done."}],"ID":20962},{"Watched":false,"Name":"Exercise 16","Duration":"2m 9s","ChapterTopicVideoID":20170,"CourseChapterTopicPlaylistID":113411,"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, f is a bijective map from"},{"Start":"00:04.500 ","End":"00:10.560","Text":"the reals to the interval from 0 to infinity, including the 0."},{"Start":"00:10.560 ","End":"00:16.845","Text":"I\u0027ll remind you that a bijective map is just a 1 to 1 and onto function."},{"Start":"00:16.845 ","End":"00:20.785","Text":"We have to show that R is not continuous."},{"Start":"00:20.785 ","End":"00:22.940","Text":"I\u0027ll just say that by the way,"},{"Start":"00:22.940 ","End":"00:28.850","Text":"if you change this to an open interval from 0-infinity,"},{"Start":"00:28.850 ","End":"00:32.970","Text":"then it is possible that f is continuous."},{"Start":"00:33.230 ","End":"00:36.440","Text":"Let\u0027s prove this by contradiction."},{"Start":"00:36.440 ","End":"00:39.860","Text":"Suppose that f is continuous,"},{"Start":"00:39.860 ","End":"00:41.030","Text":"in the previous exercise,"},{"Start":"00:41.030 ","End":"00:45.364","Text":"we showed that if f is 1 to 1 function on the reals,"},{"Start":"00:45.364 ","End":"00:50.575","Text":"then it has to be strictly increasing or strictly decreasing."},{"Start":"00:50.575 ","End":"00:52.975","Text":"There are 2 cases,"},{"Start":"00:52.975 ","End":"00:55.910","Text":"f increasing or f decreasing."},{"Start":"00:55.910 ","End":"00:57.650","Text":"We\u0027ll focus on the increasing,"},{"Start":"00:57.650 ","End":"01:02.239","Text":"but I\u0027ll indicate how you might modify it for the decreasing case."},{"Start":"01:02.239 ","End":"01:04.745","Text":"Now, f is onto,"},{"Start":"01:04.745 ","End":"01:12.555","Text":"so it has to equal 0 for some x naught because that\u0027s the range."},{"Start":"01:12.555 ","End":"01:22.495","Text":"Then f of x has to be negative for all x bigger than x naught because f is increasing."},{"Start":"01:22.495 ","End":"01:25.745","Text":"If f of x naught is 0,"},{"Start":"01:25.745 ","End":"01:28.220","Text":"then as we get further to the right,"},{"Start":"01:28.220 ","End":"01:30.470","Text":"we must get lower down."},{"Start":"01:30.470 ","End":"01:34.520","Text":"F of x has to be less than f of x naught, which is not."},{"Start":"01:34.520 ","End":"01:37.980","Text":"Now, if we have the decreasing case,"},{"Start":"01:37.980 ","End":"01:40.845","Text":"you would just switch this with this."},{"Start":"01:40.845 ","End":"01:44.140","Text":"The reason would be the decreasing."},{"Start":"01:44.140 ","End":"01:46.535","Text":"On the other hand,"},{"Start":"01:46.535 ","End":"01:52.490","Text":"f of x belongs to the interval from 0-infinity, including 0."},{"Start":"01:52.490 ","End":"01:55.750","Text":"F of x has to be bigger or equal to 0."},{"Start":"01:55.750 ","End":"01:57.765","Text":"If we look at this and this,"},{"Start":"01:57.765 ","End":"01:59.550","Text":"the 1 hand is bigger or equal to the 1 hand,"},{"Start":"01:59.550 ","End":"02:02.400","Text":"less then, that\u0027s a contradiction."},{"Start":"02:02.400 ","End":"02:05.915","Text":"This contradiction came from the assumption that f is continuous,"},{"Start":"02:05.915 ","End":"02:10.140","Text":"and therefore it\u0027s not. We\u0027re done."}],"ID":20963},{"Watched":false,"Name":"Exercise 17","Duration":"6m 2s","ChapterTopicVideoID":20171,"CourseChapterTopicPlaylistID":113411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.630","Text":"In this exercise, we have a continuous function f and in part a,"},{"Start":"00:06.630 ","End":"00:11.865","Text":"we suppose that f attains each of its values exactly twice."},{"Start":"00:11.865 ","End":"00:20.520","Text":"We\u0027re given that f of x_1 equals f of x_2 equals Alpha for some x_1,"},{"Start":"00:20.520 ","End":"00:24.840","Text":"x_2 Alpha in a and the f of x-naught"},{"Start":"00:24.840 ","End":"00:29.880","Text":"is bigger than Alpha for some x-naught in the interval x_1 and x_2."},{"Start":"00:29.880 ","End":"00:31.650","Text":"I guess I forgot to say this,"},{"Start":"00:31.650 ","End":"00:35.715","Text":"that x_1 is not equal to x_2."},{"Start":"00:35.715 ","End":"00:39.800","Text":"Then in part b, we\u0027ll use part a to show that f"},{"Start":"00:39.800 ","End":"00:43.880","Text":"cannot attain each of its values exactly 2 times,"},{"Start":"00:43.880 ","End":"00:51.740","Text":"meaning that this supposition was entirely hypothetical and it can\u0027t occur in fact."},{"Start":"00:51.740 ","End":"00:53.885","Text":"I\u0027ll start with a diagram."},{"Start":"00:53.885 ","End":"00:57.810","Text":"Here is x_1, here is x_2,"},{"Start":"00:57.810 ","End":"01:00.010","Text":"here\u0027s the value Alpha,"},{"Start":"01:00.010 ","End":"01:04.430","Text":"and the rest of it will come to as we proceed."},{"Start":"01:04.430 ","End":"01:11.780","Text":"In part a, we can conclude that for every x between x_1 and x_2,"},{"Start":"01:11.780 ","End":"01:15.695","Text":"strictly in between, f of x is not equal to Alpha,"},{"Start":"01:15.695 ","End":"01:17.990","Text":"doesn\u0027t cross this line."},{"Start":"01:17.990 ","End":"01:21.740","Text":"By the intermediate value property,"},{"Start":"01:21.740 ","End":"01:25.400","Text":"it has to always be above or always below."},{"Start":"01:25.400 ","End":"01:27.770","Text":"In our case, it\u0027s above the line."},{"Start":"01:27.770 ","End":"01:30.380","Text":"The reason it can\u0027t be once above and once"},{"Start":"01:30.380 ","End":"01:33.650","Text":"below is there might be a point in the middle where it\u0027s equal,"},{"Start":"01:33.650 ","End":"01:39.320","Text":"and then we\u0027d have the value Alpha attained more than twice so that can\u0027t be,"},{"Start":"01:39.320 ","End":"01:41.765","Text":"so there\u0027s only 2 possibilities."},{"Start":"01:41.765 ","End":"01:43.460","Text":"But as part of the given,"},{"Start":"01:43.460 ","End":"01:46.970","Text":"we were told that there is some x naught here not illustrated,"},{"Start":"01:46.970 ","End":"01:49.625","Text":"such that f of x naught is bigger than Alpha."},{"Start":"01:49.625 ","End":"01:56.100","Text":"We\u0027ve got to have the 2nd case that for all x in this interval between x_1 and x_2,"},{"Start":"01:56.100 ","End":"02:01.770","Text":"f of x is higher than Alpha and now let"},{"Start":"02:01.770 ","End":"02:09.350","Text":"Beta equal to the supremum of f of x in the closed interval from x_1 to x_2."},{"Start":"02:09.350 ","End":"02:11.225","Text":"This is Beta."},{"Start":"02:11.225 ","End":"02:14.240","Text":"By the Extreme Value Theorem,"},{"Start":"02:14.240 ","End":"02:23.430","Text":"there must be some point x_3 such that the maximum occurs at that point f of x_3 is Beta."},{"Start":"02:23.870 ","End":"02:27.105","Text":"Since Beta\u0027s bigger than Alpha,"},{"Start":"02:27.105 ","End":"02:32.025","Text":"must be that x_3 is in the interval from x_1 to x_2,"},{"Start":"02:32.025 ","End":"02:35.790","Text":"meaning is not 1 of the endpoints it\u0027s strictly in between."},{"Start":"02:35.790 ","End":"02:39.170","Text":"We need to show that x_3 is unique."},{"Start":"02:39.170 ","End":"02:42.395","Text":"That\u0027s what we have to show here in the picture it isn\u0027t."},{"Start":"02:42.395 ","End":"02:45.100","Text":"That\u0027s what we\u0027re going to do. Proof by contradiction."},{"Start":"02:45.100 ","End":"02:48.984","Text":"Suppose that there are 2 points, x_3 and x_4,"},{"Start":"02:48.984 ","End":"02:52.770","Text":"and we\u0027ll take x_3 to be the leftmost 1 of the 2,"},{"Start":"02:52.770 ","End":"03:01.065","Text":"such that f of x_3 is beta and f of x_4 is Beta as shown here got a tie."},{"Start":"03:01.065 ","End":"03:08.370","Text":"Let\u0027s pick any x_5 in this interval from x_3 to x_4."},{"Start":"03:08.370 ","End":"03:14.475","Text":"F of x_5 is not equal to Beta because we attain each value only twice,"},{"Start":"03:14.475 ","End":"03:19.660","Text":"and Beta is the supremum or the maximum so f of x_5 is less than"},{"Start":"03:19.660 ","End":"03:25.075","Text":"or equal to Beta and therefore have to be strictly less because it can\u0027t equal."},{"Start":"03:25.075 ","End":"03:30.580","Text":"We\u0027ll let Gamma equal f of x_5 back to the picture again,"},{"Start":"03:30.580 ","End":"03:34.465","Text":"Gamma is the value at this point x_5."},{"Start":"03:34.465 ","End":"03:36.740","Text":"You can see where we\u0027re going with this."},{"Start":"03:36.740 ","End":"03:38.960","Text":"If we just look at the diagram,"},{"Start":"03:38.960 ","End":"03:43.505","Text":"we see that between x_1 and x_3,"},{"Start":"03:43.505 ","End":"03:49.015","Text":"there has to be a point x_6 where the value is equal to Gamma,"},{"Start":"03:49.015 ","End":"03:54.395","Text":"because Gamma is between Alpha and Beta so that\u0027s the intermediate value property."},{"Start":"03:54.395 ","End":"03:59.615","Text":"Similarly, there\u0027s also going to be an x_7 between x_4 and x_2,"},{"Start":"03:59.615 ","End":"04:03.930","Text":"where we achieve the value Gamma,"},{"Start":"04:03.930 ","End":"04:06.560","Text":"but we also have the original x_5,"},{"Start":"04:06.560 ","End":"04:10.160","Text":"which is between x_3 and x_4."},{"Start":"04:10.160 ","End":"04:14.100","Text":"We have Gamma attained at least 3 times,"},{"Start":"04:14.100 ","End":"04:16.110","Text":"and that will be a contradiction."},{"Start":"04:16.110 ","End":"04:19.575","Text":"Let\u0027s just write it formally."},{"Start":"04:19.575 ","End":"04:22.680","Text":"F of x_1, which is Alpha,"},{"Start":"04:22.680 ","End":"04:24.105","Text":"is less than Gamma,"},{"Start":"04:24.105 ","End":"04:25.680","Text":"and f of x_3, which is Beta,"},{"Start":"04:25.680 ","End":"04:27.450","Text":"is bigger than Gamma."},{"Start":"04:27.450 ","End":"04:32.520","Text":"F of x_6 equals Gamma for some x_6 in this interval."},{"Start":"04:32.520 ","End":"04:36.975","Text":"Similarly, between x_2 and x_4,"},{"Start":"04:36.975 ","End":"04:39.465","Text":"we have an x_7."},{"Start":"04:39.465 ","End":"04:45.825","Text":"Therefore Gamma is equal to f of 3 different things,"},{"Start":"04:45.825 ","End":"04:48.825","Text":"and they are 3 different values."},{"Start":"04:48.825 ","End":"04:52.050","Text":"This is a contradiction."},{"Start":"04:52.050 ","End":"04:58.205","Text":"The contradiction came from assuming that there was another point besides"},{"Start":"04:58.205 ","End":"05:04.780","Text":"x_3 where f attained its maximum and that concludes part a."},{"Start":"05:04.780 ","End":"05:09.960","Text":"Now part b, choose any Alpha in the image of"},{"Start":"05:09.960 ","End":"05:15.600","Text":"f. Each value in the images attained exactly twice."},{"Start":"05:15.600 ","End":"05:21.060","Text":"Let\u0027s say it\u0027s attained at x_1 and x_2 that\u0027s a different x-1 and x_2 from part a,"},{"Start":"05:21.060 ","End":"05:25.905","Text":"such that f of x_1 equals f of x_2, which is Alpha."},{"Start":"05:25.905 ","End":"05:28.565","Text":"We got 2 cases."},{"Start":"05:28.565 ","End":"05:32.990","Text":"Either for all the values between x_1 and x_2,"},{"Start":"05:32.990 ","End":"05:34.490","Text":"f of x is bigger than Alpha,"},{"Start":"05:34.490 ","End":"05:36.830","Text":"or for all values in between them,"},{"Start":"05:36.830 ","End":"05:38.540","Text":"it\u0027s less than Alpha."},{"Start":"05:38.540 ","End":"05:41.385","Text":"Let\u0027s assume k is 1."},{"Start":"05:41.385 ","End":"05:43.220","Text":"Then if you follow part a,"},{"Start":"05:43.220 ","End":"05:45.200","Text":"we reach a contradiction."},{"Start":"05:45.200 ","End":"05:49.085","Text":"The other case, case 2, is very similar,"},{"Start":"05:49.085 ","End":"05:54.415","Text":"won\u0027t repeat all the results and we also reach a contradiction."},{"Start":"05:54.415 ","End":"05:57.300","Text":"In both cases, we reach a contradiction."},{"Start":"05:57.300 ","End":"06:03.040","Text":"There can\u0027t be such an f and we are done."}],"ID":20964}],"Thumbnail":null,"ID":113411},{"Name":"Differentiability and Rolles Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"2m 46s","ChapterTopicVideoID":20099,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.545","Text":"In this exercise, f is a real-valued function on this interval."},{"Start":"00:04.545 ","End":"00:09.885","Text":"It\u0027s twice differentiable and second derivative at 0 is positive."},{"Start":"00:09.885 ","End":"00:15.075","Text":"We have to show that for at least 1 natural number n,"},{"Start":"00:15.075 ","End":"00:18.885","Text":"f of 1 over n is not equal to 1."},{"Start":"00:18.885 ","End":"00:21.735","Text":"Nothing magic about minus 1 and 2."},{"Start":"00:21.735 ","End":"00:25.590","Text":"It just has to be an open interval that contains 0 and all the 1 over"},{"Start":"00:25.590 ","End":"00:29.820","Text":"n. The proof will be by contradiction."},{"Start":"00:29.820 ","End":"00:34.875","Text":"The opposite of this statement is that for all n,"},{"Start":"00:34.875 ","End":"00:38.355","Text":"f of 1 over n is equal to 1."},{"Start":"00:38.355 ","End":"00:44.855","Text":"Now, because 1 over n tends to 0 and f is continuous,"},{"Start":"00:44.855 ","End":"00:51.700","Text":"then we get that f of the limit is also equal to 1."},{"Start":"00:52.070 ","End":"00:58.040","Text":"This is the expression we get for f prime of 0."},{"Start":"00:58.040 ","End":"01:01.705","Text":"Now, f of 1 over n is 1,"},{"Start":"01:01.705 ","End":"01:05.670","Text":"we say here, and f of 0 is 1,"},{"Start":"01:05.670 ","End":"01:07.395","Text":"so we get this limit."},{"Start":"01:07.395 ","End":"01:10.965","Text":"This is the constant 0 for all n,"},{"Start":"01:10.965 ","End":"01:14.620","Text":"so the limit is 0."},{"Start":"01:14.620 ","End":"01:17.329","Text":"Now, for all n,"},{"Start":"01:17.329 ","End":"01:22.880","Text":"f of 0 equals f of 1 over n, they happen to equal 1."},{"Start":"01:22.880 ","End":"01:24.830","Text":"It doesn\u0027t have to be 0 here."},{"Start":"01:24.830 ","End":"01:27.560","Text":"Even though Rolle\u0027s theorem sometimes it\u0027s stated with a 0,"},{"Start":"01:27.560 ","End":"01:29.270","Text":"none of these 2 are equal."},{"Start":"01:29.270 ","End":"01:32.900","Text":"There exists some point c_n between them,"},{"Start":"01:32.900 ","End":"01:36.380","Text":"where the derivative is 0."},{"Start":"01:36.380 ","End":"01:40.205","Text":"Now, let\u0027s compute f double prime of 0."},{"Start":"01:40.205 ","End":"01:49.440","Text":"It\u0027s the limit of f prime of c_n minus f prime of 0 over c_n."},{"Start":"01:49.440 ","End":"01:54.035","Text":"I should have said c_n clearly tends to 0."},{"Start":"01:54.035 ","End":"01:55.880","Text":"By the Sandwich theorem,"},{"Start":"01:55.880 ","End":"01:59.345","Text":"c_n is less than 1 over n and bigger than 0."},{"Start":"01:59.345 ","End":"02:03.265","Text":"In the limit, it has to be 0."},{"Start":"02:03.265 ","End":"02:05.190","Text":"I\u0027ll just mark that."},{"Start":"02:05.190 ","End":"02:09.630","Text":"Just what I said, that c_n goes to 0."},{"Start":"02:09.630 ","End":"02:14.540","Text":"This limit is the limit of 0 minus 0 over c_n."},{"Start":"02:14.540 ","End":"02:17.930","Text":"Now, this expression is 0."},{"Start":"02:17.930 ","End":"02:21.780","Text":"It\u0027s the limit of 0 and hence 0."},{"Start":"02:21.780 ","End":"02:25.414","Text":"What we have is at f double prime of 0 is 0."},{"Start":"02:25.414 ","End":"02:30.700","Text":"But we were given initially that f double prime of 0 is bigger than 0."},{"Start":"02:30.700 ","End":"02:35.200","Text":"That\u0027s a contradiction; can\u0027t be equal to 0 and bigger than 0."},{"Start":"02:35.200 ","End":"02:40.275","Text":"The contradiction came from supposing that f of 1 over n equals 1 and therefore,"},{"Start":"02:40.275 ","End":"02:41.645","Text":"at least for some n,"},{"Start":"02:41.645 ","End":"02:46.020","Text":"f of 1 over n is not equal to 1. We are done."}],"ID":20903},{"Watched":false,"Name":"Exercise 2","Duration":"3m 49s","ChapterTopicVideoID":20100,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"In this exercise, f is differentiable function"},{"Start":"00:04.080 ","End":"00:09.000","Text":"such that f of 0 and f of 1 are both 0."},{"Start":"00:09.000 ","End":"00:11.100","Text":"That\u0027s these 2 points here."},{"Start":"00:11.100 ","End":"00:15.150","Text":"Also, the derivative is positive at both those points."},{"Start":"00:15.150 ","End":"00:16.920","Text":"If we drew a tangent,"},{"Start":"00:16.920 ","End":"00:21.630","Text":"it\u0027s sloping upwards and similarly here."},{"Start":"00:21.630 ","End":"00:25.349","Text":"In part 1, we have to show that there is some positive Delta"},{"Start":"00:25.349 ","End":"00:33.960","Text":"such that f is negative on the interval from 1 minus Delta to 1,"},{"Start":"00:33.960 ","End":"00:37.710","Text":"so here, f is negative."},{"Start":"00:37.710 ","End":"00:43.370","Text":"In part 2, we want to show that there are point c_1 and c_2."},{"Start":"00:43.370 ","End":"00:48.695","Text":"That would be like c_1 and c_2,"},{"Start":"00:48.695 ","End":"00:55.319","Text":"such that the derivative is 0 at both these points."},{"Start":"00:55.370 ","End":"01:00.600","Text":"Now, f prime of 1 is bigger than 0."},{"Start":"01:00.600 ","End":"01:02.565","Text":"If we express that as the limit,"},{"Start":"01:02.565 ","End":"01:05.250","Text":"the limit as x goes to 1,"},{"Start":"01:05.250 ","End":"01:07.280","Text":"but we\u0027ll take it from the left because"},{"Start":"01:07.280 ","End":"01:11.780","Text":"we\u0027re only in the interval from 0 to 1, so this limit,"},{"Start":"01:11.780 ","End":"01:16.965","Text":"f of x minus f of 1 over x minus 1 is f prime of 1,"},{"Start":"01:16.965 ","End":"01:20.120","Text":"and we\u0027ll call it L is something positive."},{"Start":"01:20.120 ","End":"01:23.945","Text":"If we choose an epsilon less than or equal to L,"},{"Start":"01:23.945 ","End":"01:32.710","Text":"there is Delta bigger than 0 such that whenever x is close to 1 within Delta,"},{"Start":"01:32.710 ","End":"01:38.135","Text":"then this expression minus the limit is less than epsilon."},{"Start":"01:38.135 ","End":"01:40.790","Text":"From this, if we interpret it in our case,"},{"Start":"01:40.790 ","End":"01:45.020","Text":"it means that if x is in this interval,"},{"Start":"01:45.020 ","End":"01:46.640","Text":"then this is true."},{"Start":"01:46.640 ","End":"01:49.490","Text":"Then from this using triangle inequalities,"},{"Start":"01:49.490 ","End":"01:56.325","Text":"we can get that f of x over x minus 1 is bigger than L minus epsilon."},{"Start":"01:56.325 ","End":"01:59.610","Text":"The f of 1 disappears because that\u0027s 0."},{"Start":"01:59.610 ","End":"02:04.115","Text":"We get that f of x is negative."},{"Start":"02:04.115 ","End":"02:08.645","Text":"Because if this expression is positive and the denominator is negative,"},{"Start":"02:08.645 ","End":"02:12.035","Text":"then the numerator must also be negative,"},{"Start":"02:12.035 ","End":"02:18.470","Text":"so f is below the x-axis as here."},{"Start":"02:18.470 ","End":"02:21.155","Text":"Now we can do a similar thing here."},{"Start":"02:21.155 ","End":"02:23.820","Text":"The derivative is also positive."},{"Start":"02:23.820 ","End":"02:25.665","Text":"Doing a similar thing to here,"},{"Start":"02:25.665 ","End":"02:27.409","Text":"we can get a different Delta,"},{"Start":"02:27.409 ","End":"02:32.240","Text":"such that from 0 to, I\u0027ll call it Delta 1,"},{"Start":"02:32.240 ","End":"02:39.280","Text":"the function is positive above the axis, I write down as this."},{"Start":"02:39.530 ","End":"02:44.920","Text":"Let\u0027s assume that the Delta and the Delta 1 or less than a half,"},{"Start":"02:44.920 ","End":"02:49.195","Text":"if necessary, shrink them because for any Delta,"},{"Start":"02:49.195 ","End":"02:53.305","Text":"this will work for a smaller Delta also here and here,"},{"Start":"02:53.305 ","End":"02:58.540","Text":"and that way these 2 points don\u0027t overlap when we get separate intervals here."},{"Start":"02:58.540 ","End":"03:01.135","Text":"Now we can apply IVP."},{"Start":"03:01.135 ","End":"03:05.125","Text":"I mean, in this interval and it\u0027s positive,"},{"Start":"03:05.125 ","End":"03:10.040","Text":"and you choose a point in this and it\u0027s negative."},{"Start":"03:10.400 ","End":"03:13.030","Text":"Say plus and minus,"},{"Start":"03:13.030 ","End":"03:14.170","Text":"between plus and minus,"},{"Start":"03:14.170 ","End":"03:19.080","Text":"there has to be a place where it\u0027s 0, and we\u0027ll call that c."},{"Start":"03:19.080 ","End":"03:22.992","Text":"Now we have 3 points where f is 0."},{"Start":"03:22.992 ","End":"03:24.930","Text":"At 0, at 1, and at c,"},{"Start":"03:24.930 ","End":"03:29.075","Text":"so now we can use the theorem due to roll,"},{"Start":"03:29.075 ","End":"03:38.650","Text":"that there must be some c_1 between 0 and c and a c_2 between c and 1,"},{"Start":"03:38.650 ","End":"03:42.350","Text":"such that the derivative at c_1 is 0 as we\u0027ve already drawn,"},{"Start":"03:42.350 ","End":"03:46.400","Text":"and the derivative of c_2 is 0."},{"Start":"03:46.400 ","End":"03:50.190","Text":"We are done, we\u0027ve shown all that we had to."}],"ID":20904},{"Watched":false,"Name":"Exercise 3","Duration":"1m 26s","ChapterTopicVideoID":20101,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"In this exercise, we have a function on"},{"Start":"00:03.030 ","End":"00:07.110","Text":"the open unit interval and it\u0027s differentiable 3 times."},{"Start":"00:07.110 ","End":"00:09.060","Text":"Let\u0027s suppose that for all n,"},{"Start":"00:09.060 ","End":"00:12.720","Text":"f of 1 over n is 0."},{"Start":"00:12.720 ","End":"00:17.980","Text":"I guess n has to be at least 2 to fit into this interval anyway."},{"Start":"00:18.500 ","End":"00:24.450","Text":"We have to show that there is some point x naught in the interval,"},{"Start":"00:24.450 ","End":"00:29.260","Text":"such that the third derivative at that point is 0."},{"Start":"00:29.260 ","End":"00:32.960","Text":"Now we\u0027ll use Rolle\u0027s Theorem many times."},{"Start":"00:32.960 ","End":"00:35.105","Text":"We\u0027ll start off with 4 points,"},{"Start":"00:35.105 ","End":"00:37.220","Text":"fifth quarter or a third or a half."},{"Start":"00:37.220 ","End":"00:39.985","Text":"but these 4 points, f is 0."},{"Start":"00:39.985 ","End":"00:43.770","Text":"Between these points we can get 3 points,"},{"Start":"00:43.770 ","End":"00:45.510","Text":"call them c_1, c_2,"},{"Start":"00:45.510 ","End":"00:51.355","Text":"c_3, such that the first derivative is 0 in each of these three."},{"Start":"00:51.355 ","End":"00:54.960","Text":"From 3 points we can get down to 2 points."},{"Start":"00:54.960 ","End":"00:56.895","Text":"Again applying Rolle\u0027s Theorem,"},{"Start":"00:56.895 ","End":"01:00.540","Text":"we\u0027ve got d_1 and d_2 between c_1,"},{"Start":"01:00.540 ","End":"01:04.400","Text":"c_2, c_3 and the second derivative."},{"Start":"01:04.400 ","End":"01:08.945","Text":"The derivative of the derivative is 0 at each of these."},{"Start":"01:08.945 ","End":"01:13.970","Text":"Now last time we\u0027ll apply Rolle\u0027s Theorem and find a point between these 2,"},{"Start":"01:13.970 ","End":"01:19.700","Text":"call it x naught and then because f double prime here is 0 and here"},{"Start":"01:19.700 ","End":"01:26.610","Text":"the derivative of f double-prime i.e if triple prime is 0 here. We are done."}],"ID":20905},{"Watched":false,"Name":"Exercise 4","Duration":"5m 33s","ChapterTopicVideoID":20102,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"In this exercise, we have to give an example of"},{"Start":"00:02.760 ","End":"00:05.865","Text":"a real-valued function defined on all of the reals,"},{"Start":"00:05.865 ","End":"00:11.010","Text":"which is differentiable everywhere except at the point x equals 1."},{"Start":"00:11.010 ","End":"00:14.580","Text":"We\u0027re going to start with a well-known example of"},{"Start":"00:14.580 ","End":"00:17.760","Text":"a function which isn\u0027t differentiable anywhere,"},{"Start":"00:17.760 ","End":"00:21.775","Text":"and that\u0027s the characteristic function, chi sub q."},{"Start":"00:21.775 ","End":"00:23.300","Text":"If you don\u0027t know what a characteristic function is,"},{"Start":"00:23.300 ","End":"00:25.430","Text":"never mind because I\u0027m going to give the definition."},{"Start":"00:25.430 ","End":"00:32.100","Text":"What it is, is it\u0027s 1 when x is rational and 0 when x is irrational."},{"Start":"00:32.100 ","End":"00:36.700","Text":"We want to modify it to make it differentiable at x equals 1."},{"Start":"00:36.700 ","End":"00:43.325","Text":"The way we\u0027ll do that is to multiply it by the parabola x minus 1 squared."},{"Start":"00:43.325 ","End":"00:46.760","Text":"We\u0027re going to try and flatten this function, and hopefully,"},{"Start":"00:46.760 ","End":"00:53.220","Text":"that will make it differentiable at x equals 1 but won\u0027t spoil it elsewhere."},{"Start":"00:53.220 ","End":"00:56.850","Text":"This is our f of x. Now, we need to"},{"Start":"00:56.850 ","End":"01:00.815","Text":"show 2 things that\u0027s differentiable at 1 and not differentiable elsewhere."},{"Start":"01:00.815 ","End":"01:04.565","Text":"At 1, we have, let\u0027s see,"},{"Start":"01:04.565 ","End":"01:10.105","Text":"if h is rational,"},{"Start":"01:10.105 ","End":"01:13.380","Text":"then so is 1 plus h,"},{"Start":"01:13.380 ","End":"01:20.970","Text":"so the chi part is equal to 1 and this is 1 plus h minus 1 squared is h squared."},{"Start":"01:20.970 ","End":"01:24.905","Text":"This is what we get. If h is irrational,"},{"Start":"01:24.905 ","End":"01:27.785","Text":"then so is h plus 1,"},{"Start":"01:27.785 ","End":"01:32.440","Text":"and so this part is 0, so we get 0."},{"Start":"01:32.440 ","End":"01:35.300","Text":"Also, note that f of 1 is 0."},{"Start":"01:35.300 ","End":"01:36.875","Text":"I need both of these."},{"Start":"01:36.875 ","End":"01:41.360","Text":"The difference quotient is this minus this over h,"},{"Start":"01:41.360 ","End":"01:45.875","Text":"which comes out to be h if h is rational."},{"Start":"01:45.875 ","End":"01:51.440","Text":"I mean it\u0027s h squared over h. If h is irrational,"},{"Start":"01:51.440 ","End":"01:53.940","Text":"then we get here is 0,"},{"Start":"01:53.940 ","End":"01:57.050","Text":"and 0 minus 0 over h is 0."},{"Start":"01:57.050 ","End":"02:00.650","Text":"Now, the absolute value of this quotient,"},{"Start":"02:00.650 ","End":"02:05.140","Text":"we can estimate it between 0 of course."}],"ID":20906},{"Watched":false,"Name":"Exercise 5","Duration":"3m 25s","ChapterTopicVideoID":20103,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.955","Text":"In this exercise, we have a function f which is differentiable at a point x-nought."},{"Start":"00:05.955 ","End":"00:11.790","Text":"In part a, we have to show that if f of x nought is not 0,"},{"Start":"00:11.790 ","End":"00:17.250","Text":"then the absolute value of f is also differentiable at x nought."},{"Start":"00:17.250 ","End":"00:21.420","Text":"Now in the other case where f of x nought equals nought, we can\u0027t say."},{"Start":"00:21.420 ","End":"00:24.140","Text":"Otherwise we could find an example that"},{"Start":"00:24.140 ","End":"00:27.395","Text":"the absolute value of f is differentiable at x nought,"},{"Start":"00:27.395 ","End":"00:29.840","Text":"and an example where it isn\u0027t."},{"Start":"00:29.840 ","End":"00:32.750","Text":"Let\u0027s get started with part a."},{"Start":"00:32.750 ","End":"00:36.650","Text":"If it\u0027s not 0, then it\u0027s got to be bigger than 0 or less than 0."},{"Start":"00:36.650 ","End":"00:39.215","Text":"We\u0027ll first take the case bigger than 0."},{"Start":"00:39.215 ","End":"00:42.740","Text":"Now f is continuous at x nought because it\u0027s differentiable"},{"Start":"00:42.740 ","End":"00:50.030","Text":"there and so we can find a whole neighborhood of x nought where f is positive."},{"Start":"00:50.030 ","End":"00:54.290","Text":"We\u0027ve seen this before given an Epsilon, there\u0027s a delta."},{"Start":"00:54.290 ","End":"00:56.270","Text":"If f is positive,"},{"Start":"00:56.270 ","End":"01:02.075","Text":"then absolute value of f equals f and since f is differentiable,"},{"Start":"01:02.075 ","End":"01:04.675","Text":"so is absolute value of f because it\u0027s the same."},{"Start":"01:04.675 ","End":"01:08.825","Text":"That was the case where f of x nought is positive."},{"Start":"01:08.825 ","End":"01:13.210","Text":"I could just say similarly for the case where it\u0027s negative,"},{"Start":"01:13.210 ","End":"01:16.675","Text":"it\u0027s almost the same but I\u0027ll show you proof any way."},{"Start":"01:16.675 ","End":"01:18.355","Text":"Not by repeating this,"},{"Start":"01:18.355 ","End":"01:22.630","Text":"but by considering the function minus f,"},{"Start":"01:22.630 ","End":"01:25.165","Text":"and call that big F. Now,"},{"Start":"01:25.165 ","End":"01:29.485","Text":"big F is also differentiable at x nought, but this time,"},{"Start":"01:29.485 ","End":"01:32.500","Text":"as before, big F of x nought is bigger than"},{"Start":"01:32.500 ","End":"01:35.725","Text":"0 because it\u0027s minus something that\u0027s negative."},{"Start":"01:35.725 ","End":"01:37.750","Text":"We can use the result of a,"},{"Start":"01:37.750 ","End":"01:45.220","Text":"apply it to big F and get that the absolute value of f is differentiable at x nought,"},{"Start":"01:45.220 ","End":"01:50.080","Text":"but absolute value of big F is the same as absolute value of little f,"},{"Start":"01:50.080 ","End":"01:53.860","Text":"because negating something doesn\u0027t change the absolute value."},{"Start":"01:53.860 ","End":"02:00.055","Text":"That\u0027s the case where it\u0027s negative and that covers both cases."},{"Start":"02:00.055 ","End":"02:02.710","Text":"That\u0027s part a."},{"Start":"02:02.710 ","End":"02:06.370","Text":"Now on to part b."},{"Start":"02:06.370 ","End":"02:12.250","Text":"As an example, we\u0027ll take f_1 of x to be equal to x,"},{"Start":"02:12.250 ","End":"02:15.895","Text":"and we\u0027ll consider the point x nought is 0,"},{"Start":"02:15.895 ","End":"02:20.230","Text":"then f_1 is certainly differentiable everywhere."},{"Start":"02:20.230 ","End":"02:22.375","Text":"F of x equals x."},{"Start":"02:22.375 ","End":"02:27.370","Text":"We have that f_1 of x nought is 0."},{"Start":"02:27.370 ","End":"02:29.865","Text":"Is it differentiable there or not?"},{"Start":"02:29.865 ","End":"02:34.220","Text":"The answer is no. The absolute value of f_1 of x is"},{"Start":"02:34.220 ","End":"02:39.815","Text":"just absolute value of x and absolute value of x is a famous example of the V-shape,"},{"Start":"02:39.815 ","End":"02:44.255","Text":"and it has a point sharp corner at 0,"},{"Start":"02:44.255 ","End":"02:46.370","Text":"and it\u0027s not differentiable there."},{"Start":"02:46.370 ","End":"02:47.990","Text":"That\u0027s 1 example."},{"Start":"02:47.990 ","End":"02:52.420","Text":"Now we have to give an example of the case where it is."},{"Start":"02:52.420 ","End":"02:57.210","Text":"Let\u0027s take f_2 of x equals x squared."},{"Start":"02:57.210 ","End":"03:03.410","Text":"Certainly f_2 is differentiable at x nought and the value at x nought is 0."},{"Start":"03:03.410 ","End":"03:08.420","Text":"But the absolute value of f_2 is the absolute value of x squared,"},{"Start":"03:08.420 ","End":"03:10.130","Text":"which is the same as x squared."},{"Start":"03:10.130 ","End":"03:16.280","Text":"Taking the absolute value didn\u0027t change anything so it is differentiable at x nought."},{"Start":"03:16.280 ","End":"03:20.300","Text":"You can take any positive differentiable function and do this."},{"Start":"03:20.300 ","End":"03:23.930","Text":"We found an example of each and that\u0027s part b,"},{"Start":"03:23.930 ","End":"03:26.130","Text":"and we are done."}],"ID":20907},{"Watched":false,"Name":"Exercise 6","Duration":"1m 32s","ChapterTopicVideoID":20104,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"In this exercise, we have 2 functions,"},{"Start":"00:03.060 ","End":"00:07.485","Text":"f and g, which are differentiable at x naught."},{"Start":"00:07.485 ","End":"00:13.530","Text":"Now we define a third function h to be the greatest of the 2 functions,"},{"Start":"00:13.530 ","End":"00:20.325","Text":"f and g. We have to show that if x naught f is not equal to g,"},{"Start":"00:20.325 ","End":"00:24.645","Text":"then the function h is differentiable at x naught."},{"Start":"00:24.645 ","End":"00:27.780","Text":"Now, if f is not equal to g at x naught,"},{"Start":"00:27.780 ","End":"00:29.445","Text":"then there\u0027s 2 possibilities."},{"Start":"00:29.445 ","End":"00:32.280","Text":"It\u0027s either bigger than or smaller than."},{"Start":"00:32.280 ","End":"00:35.775","Text":"We\u0027ll first take the case where it\u0027s greater than."},{"Start":"00:35.775 ","End":"00:38.689","Text":"Now, if f is bigger than g at a point,"},{"Start":"00:38.689 ","End":"00:42.935","Text":"then it\u0027s bigger than g in the whole neighborhood of x naught."},{"Start":"00:42.935 ","End":"00:44.675","Text":"If you don\u0027t see this immediately,"},{"Start":"00:44.675 ","End":"00:48.315","Text":"just think the difference function f minus g,"},{"Start":"00:48.315 ","End":"00:53.315","Text":"it\u0027s positive at x naught and it\u0027s continuous,"},{"Start":"00:53.315 ","End":"00:55.430","Text":"I should have said, at x naught."},{"Start":"00:55.430 ","End":"00:57.860","Text":"It\u0027s positive near x-naught,"},{"Start":"00:57.860 ","End":"00:59.830","Text":"meaning in a neighborhood of x naught."},{"Start":"00:59.830 ","End":"01:01.985","Text":"Now, in this neighborhood of x naught,"},{"Start":"01:01.985 ","End":"01:04.595","Text":"h equals f, think of this,"},{"Start":"01:04.595 ","End":"01:06.320","Text":"if a is bigger than b,"},{"Start":"01:06.320 ","End":"01:09.770","Text":"then the maximum of a b is a."},{"Start":"01:09.770 ","End":"01:14.060","Text":"If h equals f and f is differentiable at x naught,"},{"Start":"01:14.060 ","End":"01:16.415","Text":"then h is differentiable at x naught."},{"Start":"01:16.415 ","End":"01:22.355","Text":"The same idea works in the other case where f of x naught is less than g of x naught."},{"Start":"01:22.355 ","End":"01:26.540","Text":"This case we get that h equals g in a neighborhood of x naught,"},{"Start":"01:26.540 ","End":"01:28.850","Text":"and also h is differentiable at x naught,"},{"Start":"01:28.850 ","End":"01:30.575","Text":"so in either case."},{"Start":"01:30.575 ","End":"01:33.270","Text":"We are done."}],"ID":20908},{"Watched":false,"Name":"Exercise 7","Duration":"2m 15s","ChapterTopicVideoID":20105,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.490","Text":"In this exercise, we have a function f which is differentiable at x equals 1,"},{"Start":"00:05.490 ","End":"00:08.565","Text":"and the value of the function there is 1,"},{"Start":"00:08.565 ","End":"00:11.130","Text":"k is a natural number."},{"Start":"00:11.130 ","End":"00:17.805","Text":"We have to show that this limit is equal to this, I won\u0027t read it out."},{"Start":"00:17.805 ","End":"00:20.190","Text":"I just copied this here,"},{"Start":"00:20.190 ","End":"00:22.365","Text":"and I\u0027ll do a bit of algebra."},{"Start":"00:22.365 ","End":"00:24.170","Text":"From here to here I did 2 things."},{"Start":"00:24.170 ","End":"00:31.685","Text":"First of all, I took this minus k and subtracted minus 1 from each of the terms,"},{"Start":"00:31.685 ","End":"00:34.850","Text":"because there are k of these."},{"Start":"00:34.850 ","End":"00:41.810","Text":"The second thing is to take this n and put it in the denominators here."},{"Start":"00:41.810 ","End":"00:44.300","Text":"Basically, I\u0027m multiplying each term by n,"},{"Start":"00:44.300 ","End":"00:45.440","Text":"instead of multiplying by n,"},{"Start":"00:45.440 ","End":"00:48.025","Text":"I\u0027m dividing by 1."},{"Start":"00:48.025 ","End":"00:57.090","Text":"The next adjustment is to put a 2 here and 2 here,"},{"Start":"00:57.090 ","End":"01:02.610","Text":"k here and k here."},{"Start":"01:02.610 ","End":"01:07.260","Text":"Well, 1 here and 1 here."},{"Start":"01:07.260 ","End":"01:11.040","Text":"No need to write the 1. You get the idea."},{"Start":"01:11.040 ","End":"01:13.620","Text":"Then we can use the linearity of"},{"Start":"01:13.620 ","End":"01:19.099","Text":"the limit to break the limit of this sum into sum of limits,"},{"Start":"01:19.099 ","End":"01:22.865","Text":"and we can also take a constant outside the limit,"},{"Start":"01:22.865 ","End":"01:25.640","Text":"so we end up with this expression."},{"Start":"01:25.640 ","End":"01:29.225","Text":"Now let\u0027s look at 1 of the terms, say this 1."},{"Start":"01:29.225 ","End":"01:32.260","Text":"If you think if 2 as h,"},{"Start":"01:32.260 ","End":"01:36.860","Text":"basically this is just f of 1 plus h minus f of 1 over h,"},{"Start":"01:36.860 ","End":"01:43.075","Text":"which is the definition of the derivative of f at 1."},{"Start":"01:43.075 ","End":"01:46.140","Text":"Similarly with the k of all throughout,"},{"Start":"01:46.140 ","End":"01:48.420","Text":"we just have the k is h,"},{"Start":"01:48.420 ","End":"01:50.660","Text":"or here 1 is h,"},{"Start":"01:50.660 ","End":"01:53.330","Text":"and we get the derivative, how many times?"},{"Start":"01:53.330 ","End":"01:55.955","Text":"1 plus 2 plus, plus, plus,"},{"Start":"01:55.955 ","End":"02:02.055","Text":"plus k. Just take out these constants 1 plus 2 up to k,"},{"Start":"02:02.055 ","End":"02:07.435","Text":"f prime of 1, and this is a known arithmetic progression."},{"Start":"02:07.435 ","End":"02:11.340","Text":"The sum is k k plus 1/2,"},{"Start":"02:11.340 ","End":"02:15.570","Text":"and that\u0027s what we had to show and we are d1."}],"ID":20909},{"Watched":false,"Name":"Exercise 8","Duration":"1m 31s","ChapterTopicVideoID":20106,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.770","Text":"In this exercise, f is a function on the closed interval 0,"},{"Start":"00:04.770 ","End":"00:10.095","Text":"1, and it\u0027s differentiable and f of 0 is 0 and f of 1 is 1."},{"Start":"00:10.095 ","End":"00:14.580","Text":"We have to show that the equation f prime of x equals"},{"Start":"00:14.580 ","End":"00:20.235","Text":"2x has a solution on the open interval 0, 1."},{"Start":"00:20.235 ","End":"00:25.250","Text":"Means that there is some x between 0 and 1 for which this is true."},{"Start":"00:25.250 ","End":"00:28.040","Text":"Just thinking if it\u0027s true for all x,"},{"Start":"00:28.040 ","End":"00:31.925","Text":"then we could take the integral and say that f of x is x squared,"},{"Start":"00:31.925 ","End":"00:33.515","Text":"but that\u0027s not the case here."},{"Start":"00:33.515 ","End":"00:36.560","Text":"However, it gives an idea that if we define"},{"Start":"00:36.560 ","End":"00:41.215","Text":"a function g on this interval by taking our f and subtracting x squared,"},{"Start":"00:41.215 ","End":"00:43.100","Text":"maybe that will help us."},{"Start":"00:43.100 ","End":"00:49.320","Text":"Notice that g of 0 is 0 and g of 1 is also 0,"},{"Start":"00:49.320 ","End":"00:53.340","Text":"because g and f agree at 0 and 1."},{"Start":"00:53.340 ","End":"00:56.325","Text":"Then we can apply Rolle\u0027s theorem."},{"Start":"00:56.325 ","End":"01:04.680","Text":"We get that g prime is 0 for some c in the interval 0, 1."},{"Start":"01:04.680 ","End":"01:07.360","Text":"Now g prime at a point x,"},{"Start":"01:07.360 ","End":"01:12.965","Text":"we get by differentiating this f prime of x minus 2x."},{"Start":"01:12.965 ","End":"01:14.300","Text":"But instead of X,"},{"Start":"01:14.300 ","End":"01:19.720","Text":"we replace it with c and we get f prime of c minus 2c is 0,"},{"Start":"01:19.720 ","End":"01:23.055","Text":"which we can write as f prime of c equals 2c."},{"Start":"01:23.055 ","End":"01:27.605","Text":"In other words, x equals c is a solution to this equation,"},{"Start":"01:27.605 ","End":"01:31.860","Text":"or c is a solution, and we\u0027re done."}],"ID":20910},{"Watched":false,"Name":"Exercise 9","Duration":"1m 55s","ChapterTopicVideoID":20107,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"In this exercise, f is defined on the closed interval a, b,"},{"Start":"00:04.080 ","End":"00:09.615","Text":"and it has a third derivative for all x in the interval."},{"Start":"00:09.615 ","End":"00:16.875","Text":"Now let\u0027s suppose that at the end points f and f prime are both 0."},{"Start":"00:16.875 ","End":"00:21.060","Text":"We have to show that this equation has a solution."},{"Start":"00:21.060 ","End":"00:24.720","Text":"In other words, at some point in the interval,"},{"Start":"00:24.720 ","End":"00:26.715","Text":"I should have said in the interval a, b,"},{"Start":"00:26.715 ","End":"00:29.970","Text":"f triple prime of x is 0."},{"Start":"00:29.970 ","End":"00:34.515","Text":"We\u0027ll solve this using Rolle\u0027s Theorem many times, as you\u0027ll see,"},{"Start":"00:34.515 ","End":"00:40.425","Text":"we\u0027ll start off by seeing that f of a equals f of b is 0."},{"Start":"00:40.425 ","End":"00:42.175","Text":"There is some point,"},{"Start":"00:42.175 ","End":"00:45.260","Text":"call it d in the open interval a,"},{"Start":"00:45.260 ","End":"00:50.170","Text":"b, such that f prime of d is 0."},{"Start":"00:50.170 ","End":"00:55.040","Text":"Notice that now we have 3 points where f prime is 0, at least 3,"},{"Start":"00:55.040 ","End":"00:57.005","Text":"we have f prime at a,"},{"Start":"00:57.005 ","End":"01:00.235","Text":"f prime at b, and at d,"},{"Start":"01:00.235 ","End":"01:02.624","Text":"are all 3 equal to 0,"},{"Start":"01:02.624 ","End":"01:04.515","Text":"d is the middle one,"},{"Start":"01:04.515 ","End":"01:10.060","Text":"and we can find 2 other points like between a and d,"},{"Start":"01:10.060 ","End":"01:12.595","Text":"because f prime of a and f prime of d is 0,"},{"Start":"01:12.595 ","End":"01:18.640","Text":"then we have a c_1 between a and d such that f double-prime of c_1 is 0."},{"Start":"01:18.640 ","End":"01:21.365","Text":"Similarly between d and b."},{"Start":"01:21.365 ","End":"01:23.640","Text":"We get another point, c_2,"},{"Start":"01:23.640 ","End":"01:26.595","Text":"where f double prime is 0."},{"Start":"01:26.595 ","End":"01:32.500","Text":"C_1 and c_2 are different because the interval from a to d,"},{"Start":"01:32.500 ","End":"01:35.550","Text":"and from d to b, they don\u0027t overlap."},{"Start":"01:35.550 ","End":"01:37.664","Text":"There\u0027s 2 distinct points,"},{"Start":"01:37.664 ","End":"01:39.735","Text":"which means that between them,"},{"Start":"01:39.735 ","End":"01:41.809","Text":"we have some point c,"},{"Start":"01:41.809 ","End":"01:44.835","Text":"such that f double-prime prime,"},{"Start":"01:44.835 ","End":"01:48.590","Text":"which is f triple prime, is 0."},{"Start":"01:48.590 ","End":"01:55.320","Text":"Clearly this c is exactly the solution that we\u0027re looking for. We are done."}],"ID":20911},{"Watched":false,"Name":"Exercise 10","Duration":"2m 50s","ChapterTopicVideoID":21568,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.320","Text":"In this exercise, we have 2 differentiable functions,"},{"Start":"00:04.320 ","End":"00:12.194","Text":"f and g and all of R. Suppose that we have the condition that for any point x,"},{"Start":"00:12.194 ","End":"00:13.950","Text":"f prime of x,"},{"Start":"00:13.950 ","End":"00:17.805","Text":"g of x is not equal to f of x, g prime of x."},{"Start":"00:17.805 ","End":"00:21.960","Text":"We have to show that between any 2 roots of f,"},{"Start":"00:21.960 ","End":"00:26.925","Text":"there exists at least 1 root of g. In case you forgot what route is,"},{"Start":"00:26.925 ","End":"00:29.050","Text":"here is a definition."},{"Start":"00:30.500 ","End":"00:35.910","Text":"Let\u0027s take 2 roots of f and let\u0027s call them a and b,"},{"Start":"00:35.910 ","End":"00:41.820","Text":"and assume that a is the smaller 1. f of a equals f of b equals 0."},{"Start":"00:41.820 ","End":"00:46.530","Text":"Now, this condition applies to a,"},{"Start":"00:46.530 ","End":"00:49.520","Text":"so f prime of a, g of a is not equal to f of a,"},{"Start":"00:49.520 ","End":"00:52.975","Text":"g prime of a. f of a is 0."},{"Start":"00:52.975 ","End":"00:56.360","Text":"This product is not equal to 0,"},{"Start":"00:56.360 ","End":"00:58.820","Text":"so neither of these is 0 and in particular,"},{"Start":"00:58.820 ","End":"01:00.815","Text":"g of a is not 0."},{"Start":"01:00.815 ","End":"01:02.720","Text":"Now instead of a,"},{"Start":"01:02.720 ","End":"01:07.760","Text":"we use b, we would get that g of b is not equal to 0."},{"Start":"01:07.760 ","End":"01:14.280","Text":"We have to show that g has a root in the open interval a,"},{"Start":"01:14.280 ","End":"01:19.545","Text":"b, meaning strictly between a and b. I will do it by contradiction."},{"Start":"01:19.545 ","End":"01:23.190","Text":"We\u0027ll suppose that g has no root in a,"},{"Start":"01:23.190 ","End":"01:26.980","Text":"b. g is not equal to 0,"},{"Start":"01:26.980 ","End":"01:29.900","Text":"not just in the open interval but on the closed interval,"},{"Start":"01:29.900 ","End":"01:34.700","Text":"because here we have g of a is not 0 and g of b is not 0."},{"Start":"01:34.700 ","End":"01:38.139","Text":"If g is not 0 in all of this interval,"},{"Start":"01:38.139 ","End":"01:44.920","Text":"we can define a function h as f over g. The denominator is not 0,"},{"Start":"01:44.920 ","End":"01:46.945","Text":"so this is well-defined."},{"Start":"01:46.945 ","End":"01:50.110","Text":"Since f and g are both differentiable,"},{"Start":"01:50.110 ","End":"01:52.675","Text":"then h is differentiable."},{"Start":"01:52.675 ","End":"01:57.670","Text":"Also notice that h of a and h of b are both"},{"Start":"01:57.670 ","End":"02:05.170","Text":"0 because it\u0027s true of f and f over g will also be 0 if f is 0."},{"Start":"02:05.170 ","End":"02:10.520","Text":"By Rolle\u0027s theorem, it follows from this and that h is differentiable,"},{"Start":"02:10.520 ","End":"02:16.625","Text":"that there is some c between a and b such that h prime of c equals 0."},{"Start":"02:16.625 ","End":"02:20.885","Text":"Now, h prime, just using the quotient rule"},{"Start":"02:20.885 ","End":"02:26.460","Text":"on f over g and substituting c we get this expression."},{"Start":"02:26.460 ","End":"02:28.939","Text":"If a fraction is 0,"},{"Start":"02:28.939 ","End":"02:31.910","Text":"then the numerator has to be 0."},{"Start":"02:31.910 ","End":"02:39.900","Text":"But this contradicts our condition that for no x, is this true?"},{"Start":"02:39.900 ","End":"02:42.470","Text":"This is contradiction, and"},{"Start":"02:42.470 ","End":"02:46.100","Text":"the contradiction came from this supposition that g has no root,"},{"Start":"02:46.100 ","End":"02:51.330","Text":"and therefore it does have a root between a and b. We\u0027re done."}],"ID":26236},{"Watched":false,"Name":"Exercise 11","Duration":"3m 22s","ChapterTopicVideoID":21569,"CourseChapterTopicPlaylistID":113412,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:05.070","Text":"In this exercise, we have a function defined on the positive numbers."},{"Start":"00:05.070 ","End":"00:09.809","Text":"It satisfies this identity that it takes a product"},{"Start":"00:09.809 ","End":"00:15.120","Text":"into a sum and f is differentiable at x equals 1."},{"Start":"00:15.120 ","End":"00:19.310","Text":"We have to show that f is differentiable on all of its domain."},{"Start":"00:19.310 ","End":"00:22.455","Text":"What\u0027s more that the derivative is given by the formula"},{"Start":"00:22.455 ","End":"00:26.940","Text":"f prime of x is the derivative of 1 divided by x."},{"Start":"00:26.940 ","End":"00:29.715","Text":"We\u0027re given this hint."},{"Start":"00:29.715 ","End":"00:36.330","Text":"Now, the hint would be easily provable if you\u0027ve studied group theory,"},{"Start":"00:36.330 ","End":"00:37.980","Text":"part of abstract algebra."},{"Start":"00:37.980 ","End":"00:42.020","Text":"If you have, you\u0027ll see that this hint comes out immediately,"},{"Start":"00:42.020 ","End":"00:45.920","Text":"basically, the positives form a group under multiplication"},{"Start":"00:45.920 ","End":"00:49.265","Text":"and the reals form a group under addition."},{"Start":"00:49.265 ","End":"00:51.935","Text":"This function preserves the group operation."},{"Start":"00:51.935 ","End":"00:54.905","Text":"I\u0027ll say no more in case you haven\u0027t studied group theory."},{"Start":"00:54.905 ","End":"00:58.940","Text":"Any case, we can write that f of 1 is f of 1 times 1."},{"Start":"00:58.940 ","End":"01:01.925","Text":"By this rule, it\u0027s f of 1 plus f of 1."},{"Start":"01:01.925 ","End":"01:05.180","Text":"Of course, f of 1 is 0 as the first part."},{"Start":"01:05.180 ","End":"01:08.150","Text":"I\u0027m doing it without group theory now."},{"Start":"01:08.150 ","End":"01:10.280","Text":"This second part, well,"},{"Start":"01:10.280 ","End":"01:15.250","Text":"we could take f of x and write x as y times x over y."},{"Start":"01:15.250 ","End":"01:17.880","Text":"Then this product becomes a sum."},{"Start":"01:17.880 ","End":"01:20.360","Text":"Then you just bring f of y over to this side"},{"Start":"01:20.360 ","End":"01:23.945","Text":"and we get the second part of the hint."},{"Start":"01:23.945 ","End":"01:28.865","Text":"This is going to really help us in showing that f is differentiable everywhere."},{"Start":"01:28.865 ","End":"01:32.320","Text":"The derivative is given by this limit."},{"Start":"01:32.320 ","End":"01:36.105","Text":"The 1 over h bring outside the quotient."},{"Start":"01:36.105 ","End":"01:42.430","Text":"Then this difference can be written as a quotient f of x plus h over x."},{"Start":"01:42.430 ","End":"01:44.900","Text":"Now, make a substitution."},{"Start":"01:44.900 ","End":"01:47.525","Text":"Let h equals kx,"},{"Start":"01:47.525 ","End":"01:52.415","Text":"think of x as fixed and h is the one that\u0027s varying, its going to 0."},{"Start":"01:52.415 ","End":"01:55.850","Text":"Also, h goes to 0 if and only if k goes to 0"},{"Start":"01:55.850 ","End":"01:58.925","Text":"because they\u0027re proportional to each other by a factor x,"},{"Start":"01:58.925 ","End":"02:01.165","Text":"which is not 0."},{"Start":"02:01.165 ","End":"02:06.034","Text":"F prime of x is the limit as k goes to 0,"},{"Start":"02:06.034 ","End":"02:09.260","Text":"1 over h is 1 over kx."},{"Start":"02:09.260 ","End":"02:13.080","Text":"Here instead of h we write Kx."},{"Start":"02:13.080 ","End":"02:15.635","Text":"Now you see why we did the substitution."},{"Start":"02:15.635 ","End":"02:24.465","Text":"The x cancels and what we\u0027re left with is here 1 plus k inside the brackets."},{"Start":"02:24.465 ","End":"02:29.230","Text":"This is over kx limit as k goes to 0."},{"Start":"02:29.230 ","End":"02:32.180","Text":"Now, this is equal to,"},{"Start":"02:32.180 ","End":"02:35.780","Text":"we can take the 1 over x outside is it doesn\u0027t depend on k."},{"Start":"02:35.780 ","End":"02:40.150","Text":"We can write f of 1 plus k,"},{"Start":"02:40.150 ","End":"02:45.154","Text":"as f of 1 plus k minus f of x1 because f of 1 is 0 from the hint."},{"Start":"02:45.154 ","End":"02:47.690","Text":"Now, you probably recognize this limit here."},{"Start":"02:47.690 ","End":"02:49.130","Text":"We\u0027re used to having it with h,"},{"Start":"02:49.130 ","End":"02:51.350","Text":"but it will work with k also."},{"Start":"02:51.350 ","End":"02:57.905","Text":"This is precisely the derivative of f at 1 and the x from here put underneath."},{"Start":"02:57.905 ","End":"02:59.860","Text":"That\u0027s what we had to show."},{"Start":"02:59.860 ","End":"03:02.120","Text":"So we are done."},{"Start":"03:02.120 ","End":"03:05.600","Text":"I\u0027ll just add a, by the way, f of x,"},{"Start":"03:05.600 ","End":"03:11.390","Text":"in fact, turns out to be the logarithm of x for some positive a,"},{"Start":"03:11.390 ","End":"03:13.070","Text":"which is not 1."},{"Start":"03:13.070 ","End":"03:17.030","Text":"That makes sense because the logarithm takes a product to a sum,"},{"Start":"03:17.030 ","End":"03:18.395","Text":"so it fits in."},{"Start":"03:18.395 ","End":"03:20.150","Text":"Anyway, that\u0027s aside."},{"Start":"03:20.150 ","End":"03:22.350","Text":"We are done."}],"ID":26237}],"Thumbnail":null,"ID":113412},{"Name":"Taylors Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"2m 59s","ChapterTopicVideoID":20172,"CourseChapterTopicPlaylistID":113421,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.810","Text":"In this exercise, f is a real-valued function on a, b. Implicitly,"},{"Start":"00:06.810 ","End":"00:10.230","Text":"we understand that it has a second derivative,"},{"Start":"00:10.230 ","End":"00:15.554","Text":"and f double prime of x is non-negative on all the interval."},{"Start":"00:15.554 ","End":"00:20.729","Text":"What we have to show is that given any x_0 in this interval,"},{"Start":"00:20.729 ","End":"00:23.040","Text":"that the following inequality holds."},{"Start":"00:23.040 ","End":"00:27.855","Text":"The geometric meaning of this is that the graph of f,"},{"Start":"00:27.855 ","End":"00:30.355","Text":"that\u0027s the left side of the inequality,"},{"Start":"00:30.355 ","End":"00:32.445","Text":"lies above the tangent line,"},{"Start":"00:32.445 ","End":"00:37.515","Text":"that\u0027s the right side of the inequality the tangent line where x equals x_0."},{"Start":"00:37.515 ","End":"00:42.700","Text":"We also say that f is concave up."},{"Start":"00:42.700 ","End":"00:48.950","Text":"Sometimes we say convex when the second derivative is non-negative."},{"Start":"00:48.950 ","End":"00:52.640","Text":"That was part a, and part b is to use part a"},{"Start":"00:52.640 ","End":"00:57.235","Text":"to prove a certain inequality as written here."},{"Start":"00:57.235 ","End":"01:03.210","Text":"Here\u0027s a sketch of what it means for the graph to be above the tangent line."},{"Start":"01:03.460 ","End":"01:06.760","Text":"We\u0027ll use Taylor\u0027s theorem."},{"Start":"01:06.760 ","End":"01:13.280","Text":"What it says is that there exists some point c between x_0 and x,"},{"Start":"01:13.650 ","End":"01:21.015","Text":"such that f of x is f of x_0 plus f prime of x_0 x minus"},{"Start":"01:21.015 ","End":"01:27.785","Text":"x_0 plus 1/2 factorial x minus x_0 squared and f double prime at this point c. Now,"},{"Start":"01:27.785 ","End":"01:33.395","Text":"this f double prime of c is going to be non-negative,"},{"Start":"01:33.395 ","End":"01:36.550","Text":"and because it\u0027s non-negative,"},{"Start":"01:36.550 ","End":"01:40.000","Text":"then if we drop this last term,"},{"Start":"01:40.000 ","End":"01:44.185","Text":"then we\u0027ll get that the left-hand side is bigger or equal to what\u0027s left."},{"Start":"01:44.185 ","End":"01:47.125","Text":"Now, on to part b."},{"Start":"01:47.125 ","End":"01:52.450","Text":"We\u0027ll apply part a to the function f of x equals"},{"Start":"01:52.450 ","End":"01:59.665","Text":"cosine x on the interval from 90 degrees-270 degrees."},{"Start":"01:59.665 ","End":"02:04.035","Text":"That\u0027s where the cosine is negative."},{"Start":"02:04.035 ","End":"02:07.790","Text":"The second derivative is minus cosine,"},{"Start":"02:07.790 ","End":"02:12.865","Text":"and on this interval that will be positive, well, non-negative anyway."},{"Start":"02:12.865 ","End":"02:19.640","Text":"We have this inequality from part a and applying it, we get,"},{"Start":"02:19.640 ","End":"02:25.555","Text":"in our case, f of x is cosine x and f prime is minus sine,"},{"Start":"02:25.555 ","End":"02:27.380","Text":"and we have this inequality."},{"Start":"02:27.380 ","End":"02:29.010","Text":"We can just relabel,"},{"Start":"02:29.010 ","End":"02:31.140","Text":"instead of x and x_0,"},{"Start":"02:31.140 ","End":"02:34.080","Text":"we can call them y and x."},{"Start":"02:34.080 ","End":"02:37.184","Text":"Just a renaming gives us this,"},{"Start":"02:37.184 ","End":"02:42.140","Text":"and just rearrange, bring the cosine x to the left,"},{"Start":"02:42.140 ","End":"02:45.445","Text":"and bring the y minus x in front of the sine."},{"Start":"02:45.445 ","End":"02:49.340","Text":"The minus of y minus x is x minus y."},{"Start":"02:49.340 ","End":"02:52.805","Text":"This equals this is what we had to show."},{"Start":"02:52.805 ","End":"02:56.405","Text":"Go back up. That\u0027s exactly what we have to show,"},{"Start":"02:56.405 ","End":"02:59.250","Text":"and so we are done."}],"ID":20981}],"Thumbnail":null,"ID":113421},{"Name":"Power Series, Taylor Series","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"2m 43s","ChapterTopicVideoID":20173,"CourseChapterTopicPlaylistID":113414,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"In this exercise, we have a function f which"},{"Start":"00:04.290 ","End":"00:08.040","Text":"is infinitely differentiable on an open interval a,"},{"Start":"00:08.040 ","End":"00:11.340","Text":"b and we have a particular point,"},{"Start":"00:11.340 ","End":"00:13.125","Text":"x naught in a, b."},{"Start":"00:13.125 ","End":"00:17.055","Text":"Now, suppose that there is some positive number M,"},{"Start":"00:17.055 ","End":"00:24.900","Text":"such that this inequality holds for all n and for all x in the interval."},{"Start":"00:24.900 ","End":"00:29.445","Text":"Inequality is that the nth derivative of f at the point"},{"Start":"00:29.445 ","End":"00:34.815","Text":"x in absolute value is less than or equal to M to the power of n,"},{"Start":"00:34.815 ","End":"00:37.045","Text":"the same n here and here."},{"Start":"00:37.045 ","End":"00:41.150","Text":"What we have to do is show that the Taylor series of f around"},{"Start":"00:41.150 ","End":"00:46.190","Text":"x_0 converges to f of x for all x in the interval."},{"Start":"00:46.190 ","End":"00:51.950","Text":"We can apply Taylor\u0027s theorem for the given n and x,"},{"Start":"00:51.950 ","End":"01:00.770","Text":"and we get that f of x is equal to the Taylor series up to the nth exponent or"},{"Start":"01:00.770 ","End":"01:04.745","Text":"the nth derivative plus the remainder or"},{"Start":"01:04.745 ","End":"01:11.685","Text":"error of order n. This holds for some c between x_0 and x."},{"Start":"01:11.685 ","End":"01:18.890","Text":"Basically, what we have to show is that E_ n of x turns to 0 as n goes to infinity."},{"Start":"01:18.890 ","End":"01:20.390","Text":"In fact, we\u0027ll show more than that,"},{"Start":"01:20.390 ","End":"01:24.985","Text":"that it turns to 0 uniformly in a way that doesn\u0027t depend on x."},{"Start":"01:24.985 ","End":"01:28.565","Text":"Let\u0027s estimate the error function E_n."},{"Start":"01:28.565 ","End":"01:32.930","Text":"The absolute value of E_n of x is equal to,"},{"Start":"01:32.930 ","End":"01:38.480","Text":"just putting what\u0027s here in absolute value is equal to this."},{"Start":"01:38.480 ","End":"01:43.115","Text":"This thing is less than M to the n plus 1,"},{"Start":"01:43.115 ","End":"01:47.270","Text":"the numerator from this inequality here,"},{"Start":"01:47.270 ","End":"01:55.250","Text":"and M to the n plus 1 times x minus x_0 to the n plus 1 absolute value here."},{"Start":"01:55.250 ","End":"01:59.420","Text":"This is equal to A to the n plus 1 If we just substitute or"},{"Start":"01:59.420 ","End":"02:04.340","Text":"define A equals M times absolute value of x minus x_0."},{"Start":"02:04.340 ","End":"02:07.280","Text":"We now have a bound for the absolute value of E_n"},{"Start":"02:07.280 ","End":"02:11.210","Text":"of x in a way that actually doesn\u0027t depend on x."},{"Start":"02:11.210 ","End":"02:19.115","Text":"Now, this turns to 0 by the ratio test if you just take the ratio of 2 consecutive terms,"},{"Start":"02:19.115 ","End":"02:23.300","Text":"you\u0027ll get capital A over n plus 1"},{"Start":"02:23.300 ","End":"02:27.410","Text":"or n plus 2 depending on which 2 consecutive terms you get,"},{"Start":"02:27.410 ","End":"02:30.245","Text":"and that goes to 0."},{"Start":"02:30.245 ","End":"02:35.780","Text":"E_n of x goes to 0 as n goes to infinity,"},{"Start":"02:35.780 ","End":"02:44.460","Text":"which means that the Taylor series converges to the function. We are done."}],"ID":20982},{"Watched":false,"Name":"Exercise 2","Duration":"8m 59s","ChapterTopicVideoID":20174,"CourseChapterTopicPlaylistID":113414,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.640","Text":"In this exercise, we\u0027re given a sequence a_n of non-negative real numbers,"},{"Start":"00:05.640 ","End":"00:11.595","Text":"and suppose that the sequence a_n to the 1 over n is a bounded sequence."},{"Start":"00:11.595 ","End":"00:15.990","Text":"Note that this makes sense because a_n is non-negative,"},{"Start":"00:15.990 ","End":"00:20.760","Text":"so we can take the nth root of a_ n. Now, for each n,"},{"Start":"00:20.760 ","End":"00:26.550","Text":"define A_n as the supremum,"},{"Start":"00:26.550 ","End":"00:30.600","Text":"the least upper bound of a_k to the 1 over"},{"Start":"00:30.600 ","End":"00:35.240","Text":"k for all the k\u0027s which are bigger or equal to n. Again,"},{"Start":"00:35.240 ","End":"00:39.230","Text":"it\u0027s basically the supremum of the tail of this sequence,"},{"Start":"00:39.230 ","End":"00:43.580","Text":"the tail starting from index n. Of course,"},{"Start":"00:43.580 ","End":"00:51.110","Text":"it\u0027s decreasing because the supremum of a set that keeps shrinking is decreasing,"},{"Start":"00:51.110 ","End":"00:56.775","Text":"and it\u0027s also bounded below because they\u0027re all non-negative."},{"Start":"00:56.775 ","End":"01:00.680","Text":"Decreasing and bounded below means that this sequence"},{"Start":"01:00.680 ","End":"01:04.310","Text":"converges and therefore it has a limit,"},{"Start":"01:04.310 ","End":"01:09.800","Text":"call it L. We have that A_n tends to L as n goes to infinity for some"},{"Start":"01:09.800 ","End":"01:16.880","Text":"non-negative L. Question a is to show that if L is less than 1,"},{"Start":"01:16.880 ","End":"01:20.120","Text":"then the infinite series converges,"},{"Start":"01:20.120 ","End":"01:23.655","Text":"and if L is bigger than 1,"},{"Start":"01:23.655 ","End":"01:27.090","Text":"then the series diverges."},{"Start":"01:27.090 ","End":"01:31.130","Text":"Then in part b, we\u0027ll use part a to"},{"Start":"01:31.130 ","End":"01:34.564","Text":"show that the radius of convergence of the power series,"},{"Start":"01:34.564 ","End":"01:42.095","Text":"the sum of a_n x to the n is 1 over L. I guess we\u0027re assuming that L is bigger than 0,"},{"Start":"01:42.095 ","End":"01:48.770","Text":"it actually is true if L is 0 on the convention that 1 over 0 is infinity."},{"Start":"01:48.770 ","End":"01:50.670","Text":"If L is 0, the radius is infinity."},{"Start":"01:50.670 ","End":"01:54.895","Text":"Anyway, we\u0027ll just be concerned with L strictly positive."},{"Start":"01:54.895 ","End":"01:57.510","Text":"Beginning with part a,"},{"Start":"01:57.510 ","End":"01:59.130","Text":"this has 2 sub-parts,"},{"Start":"01:59.130 ","End":"02:03.270","Text":"L less than 1 and L greater than 1."},{"Start":"02:03.270 ","End":"02:05.535","Text":"Let\u0027s start with L less than 1."},{"Start":"02:05.535 ","End":"02:12.030","Text":"Choose Epsilon so that L plus Epsilon comes out between L and 1."},{"Start":"02:12.030 ","End":"02:14.840","Text":"You just have to find any number between L and"},{"Start":"02:14.840 ","End":"02:17.980","Text":"1 and subtract L from it and we get Epsilon."},{"Start":"02:17.980 ","End":"02:20.920","Text":"Since the limit of A_n is L,"},{"Start":"02:20.920 ","End":"02:24.395","Text":"but the definition of a limit using Epsilon n,"},{"Start":"02:24.395 ","End":"02:29.255","Text":"there exists a natural number N such that,"},{"Start":"02:29.255 ","End":"02:32.865","Text":"whenever n is bigger or equal to N,"},{"Start":"02:32.865 ","End":"02:37.695","Text":"then A_n is less than L plus Epsilon."},{"Start":"02:37.695 ","End":"02:39.110","Text":"Normally with the limit,"},{"Start":"02:39.110 ","End":"02:40.820","Text":"we take a two-sided interval,"},{"Start":"02:40.820 ","End":"02:44.690","Text":"so say between L plus Epsilon and L minus Epsilon,"},{"Start":"02:44.690 ","End":"02:48.265","Text":"but we only care about this half of the inequality."},{"Start":"02:48.265 ","End":"02:51.470","Text":"Now, a_n to the 1 over n is less than or equal"},{"Start":"02:51.470 ","End":"02:56.240","Text":"to a_n because a_n to the 1 over n belongs to this set."},{"Start":"02:56.240 ","End":"02:58.565","Text":"It\u0027s the first member,"},{"Start":"02:58.565 ","End":"03:03.335","Text":"and any member of a set is less than or equal to the supremum of that set,"},{"Start":"03:03.335 ","End":"03:04.780","Text":"so we have this,"},{"Start":"03:04.780 ","End":"03:10.830","Text":"which means that a_n to the 1 over n is less than L plus Epsilon, which is less than 1,"},{"Start":"03:10.830 ","End":"03:16.580","Text":"and this holds for all n bigger or equal to N. Since I enter the 1"},{"Start":"03:16.580 ","End":"03:22.535","Text":"over n is less than this for all n from a certain point onwards,"},{"Start":"03:22.535 ","End":"03:26.550","Text":"then the limit superior of a_n to the 1 over"},{"Start":"03:26.550 ","End":"03:28.910","Text":"n is going to be less than necessarily but"},{"Start":"03:28.910 ","End":"03:31.640","Text":"certainly less than or equal to L plus Epsilon,"},{"Start":"03:31.640 ","End":"03:33.395","Text":"but it\u0027s still less than 1,"},{"Start":"03:33.395 ","End":"03:37.885","Text":"and by the Root Test or at least one of the versions of it,"},{"Start":"03:37.885 ","End":"03:42.075","Text":"this lim sup being less than 1,"},{"Start":"03:42.075 ","End":"03:45.240","Text":"guarantees convergence of the series."},{"Start":"03:45.240 ","End":"03:51.170","Text":"We\u0027ve done the first part where L is less than 1 and we\u0027ve shown that it converges."},{"Start":"03:51.170 ","End":"03:54.170","Text":"Now, let\u0027s take the case bigger than 1,"},{"Start":"03:54.170 ","End":"03:57.650","Text":"and hopefully we\u0027ll show that it diverges."},{"Start":"03:57.650 ","End":"03:59.555","Text":"If L is bigger than 1,"},{"Start":"03:59.555 ","End":"04:07.180","Text":"this time we\u0027ll choose the Epsilon so that L minus Epsilon lies between L and 1."},{"Start":"04:07.180 ","End":"04:11.580","Text":"Now, A_n tends to L then goes to infinity."},{"Start":"04:11.580 ","End":"04:17.645","Text":"There exists the N such that for all k bigger or equal to this N,"},{"Start":"04:17.645 ","End":"04:22.230","Text":"A_k is bigger than L minus Epsilon."},{"Start":"04:22.230 ","End":"04:26.655","Text":"I want to replace A_k by its definition,"},{"Start":"04:26.655 ","End":"04:29.565","Text":"to switch k and n in the index."},{"Start":"04:29.565 ","End":"04:33.005","Text":"This means that a_n to the 1 over n,"},{"Start":"04:33.005 ","End":"04:37.940","Text":"the supremum of this set is bigger than L minus Epsilon,"},{"Start":"04:37.940 ","End":"04:46.800","Text":"supremum on n bigger or equal to k. That means that for each k bigger or equal to N,"},{"Start":"04:46.800 ","End":"04:51.165","Text":"we can choose an index n_k bigger or equal to k,"},{"Start":"04:51.165 ","End":"04:52.830","Text":"that was just from this set,"},{"Start":"04:52.830 ","End":"04:58.945","Text":"such that a_nk to the 1 over n_k is bigger than L minus Epsilon."},{"Start":"04:58.945 ","End":"05:03.870","Text":"This is because the supremum is bigger than this."},{"Start":"05:03.870 ","End":"05:12.515","Text":"If all the n_k\u0027s satisfied that this is less than L minus Epsilon,"},{"Start":"05:12.515 ","End":"05:17.755","Text":"then the supremum would be less than or equal to L minus Epsilon,"},{"Start":"05:17.755 ","End":"05:19.300","Text":"which is not true."},{"Start":"05:19.300 ","End":"05:22.970","Text":"By contradiction, we can choose such an n_k,"},{"Start":"05:22.970 ","End":"05:31.485","Text":"I\u0027d like to take these indices n_k to get a sub-sequence of the a_n,"},{"Start":"05:31.485 ","End":"05:37.080","Text":"but for subsequence we need the indices to be increasing."},{"Start":"05:37.080 ","End":"05:40.175","Text":"Often we omit this step, but technically,"},{"Start":"05:40.175 ","End":"05:42.815","Text":"we should say that we can adjust"},{"Start":"05:42.815 ","End":"05:46.995","Text":"the sequence so that it\u0027s increasing just by thinning it out."},{"Start":"05:46.995 ","End":"05:51.800","Text":"We get a strictly increasing set of indices, n_k,"},{"Start":"05:51.800 ","End":"05:55.325","Text":"and then we will get a sub-sequence,"},{"Start":"05:55.325 ","End":"05:58.165","Text":"a_nk to the 1 over n_k."},{"Start":"05:58.165 ","End":"06:05.080","Text":"It\u0027s a sub-sequence where each of the members is bigger than L minus Epsilon,"},{"Start":"06:05.080 ","End":"06:06.590","Text":"and if that\u0027s so,"},{"Start":"06:06.590 ","End":"06:09.075","Text":"then the lim sup,"},{"Start":"06:09.075 ","End":"06:15.430","Text":"limit superior of a_n to the 1 over n is bigger or equal to L minus Epsilon,"},{"Start":"06:15.430 ","End":"06:17.845","Text":"actually would even write bigger than here."},{"Start":"06:17.845 ","End":"06:19.780","Text":"The reason for that is,"},{"Start":"06:19.780 ","End":"06:23.110","Text":"you can do a little contradiction here suppose that the lim sup"},{"Start":"06:23.110 ","End":"06:27.205","Text":"is less than L minus Epsilon."},{"Start":"06:27.205 ","End":"06:34.435","Text":"That means that only a finite number can be above L minus Epsilon."},{"Start":"06:34.435 ","End":"06:37.600","Text":"Most of the sequence all but a finite number will be"},{"Start":"06:37.600 ","End":"06:41.345","Text":"less than or equal to L minus Epsilon,"},{"Start":"06:41.345 ","End":"06:46.820","Text":"and that would contradict this because we have a whole sub-sequence bigger than,"},{"Start":"06:46.820 ","End":"06:48.670","Text":"it\u0027s not just a finite number,"},{"Start":"06:48.670 ","End":"06:52.200","Text":"and so by the root test,"},{"Start":"06:52.200 ","End":"06:55.680","Text":"when the lim sup is bigger than 1,"},{"Start":"06:55.680 ","End":"06:59.040","Text":"then the series diverges."},{"Start":"06:59.040 ","End":"07:05.700","Text":"That\u0027s part a, and now on to part b."},{"Start":"07:05.700 ","End":"07:09.315","Text":"Let\u0027s fix point x,"},{"Start":"07:09.315 ","End":"07:13.490","Text":"and we\u0027re going to try and reduce this to case a."},{"Start":"07:13.490 ","End":"07:20.730","Text":"Let a_n prime be a_n absolute value of x to the n,"},{"Start":"07:20.730 ","End":"07:28.130","Text":"and A_n prime will be the supremum of a_k prime to the 1 over k,"},{"Start":"07:28.130 ","End":"07:33.260","Text":"where k is bigger or equal to n. If we take the nth root of this,"},{"Start":"07:33.260 ","End":"07:35.870","Text":"we get that a_n prime to the 1 over"},{"Start":"07:35.870 ","End":"07:39.825","Text":"n is I enter the 1 over n times the absolute value of x,"},{"Start":"07:39.825 ","End":"07:42.750","Text":"just change n to k,"},{"Start":"07:42.750 ","End":"07:47.465","Text":"we get this and so when we take the supremum,"},{"Start":"07:47.465 ","End":"07:51.980","Text":"we also have it multiplied by absolute value of x."},{"Start":"07:51.980 ","End":"07:56.750","Text":"If you multiply the members of a set by a non-negative constant,"},{"Start":"07:56.750 ","End":"08:00.160","Text":"then the supremum is multiplied by that also."},{"Start":"08:00.160 ","End":"08:04.695","Text":"Now, A_n prime tends to L prime,"},{"Start":"08:04.695 ","End":"08:08.500","Text":"if we define L prime to be absolute value of x times L,"},{"Start":"08:08.500 ","End":"08:12.120","Text":"where A_n without the prime goes to L,"},{"Start":"08:12.120 ","End":"08:14.790","Text":"and we\u0027ll consider 2 cases."},{"Start":"08:14.790 ","End":"08:20.915","Text":"Case 1 will be where absolute value of x is less than 1 over L,"},{"Start":"08:20.915 ","End":"08:26.225","Text":"then we\u0027ll show that the series converges and in the other case, it\u0027ll diverge."},{"Start":"08:26.225 ","End":"08:28.550","Text":"Here we have L prime,"},{"Start":"08:28.550 ","End":"08:30.650","Text":"which is the absolute value of x times L,"},{"Start":"08:30.650 ","End":"08:33.515","Text":"is now going to be less than 1 from this."},{"Start":"08:33.515 ","End":"08:38.270","Text":"We can apply part a and get that the series,"},{"Start":"08:38.270 ","End":"08:40.310","Text":"the sum of a_n prime converges,"},{"Start":"08:40.310 ","End":"08:42.355","Text":"but this is just equal to this,"},{"Start":"08:42.355 ","End":"08:45.170","Text":"so this series converges."},{"Start":"08:45.170 ","End":"08:51.440","Text":"Similarly, if absolute value of x is bigger than 1 over L, then this diverges."},{"Start":"08:51.440 ","End":"08:54.905","Text":"Again, just doing a similar thing and using part a."},{"Start":"08:54.905 ","End":"08:59.820","Text":"That\u0027s part b, and we are done."}],"ID":20983}],"Thumbnail":null,"ID":113414},{"Name":"Areas of Surfaces of Revolution, Pappuss Theorems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"2m 20s","ChapterTopicVideoID":20191,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In this exercise, we consider"},{"Start":"00:01.890 ","End":"00:08.640","Text":"an equilateral triangle and its base is on the x axis as in the picture."},{"Start":"00:08.640 ","End":"00:12.450","Text":"The length of the side we\u0027re given is a."},{"Start":"00:12.450 ","End":"00:19.920","Text":"We have to use Pappus\u0027s theorem to evaluate the volume the solid of revolution,"},{"Start":"00:19.920 ","End":"00:24.570","Text":"we take this triangle and we revolve it around the line y"},{"Start":"00:24.570 ","End":"00:30.250","Text":"equals minus a. Pappus\u0027s theorem is more than 1."},{"Start":"00:30.250 ","End":"00:35.690","Text":"The 1 that we want involves solid of revolution and this is the theorem,"},{"Start":"00:35.690 ","End":"00:37.865","Text":"you can pause the clip and read it,"},{"Start":"00:37.865 ","End":"00:41.375","Text":"and this is the main formula that we\u0027ll be using."},{"Start":"00:41.375 ","End":"00:46.250","Text":"The main task initially is to find the centroid or"},{"Start":"00:46.250 ","End":"00:52.175","Text":"specifically the distance from the centroid to the axis of revolution."},{"Start":"00:52.175 ","End":"00:58.760","Text":"The centroid is where the 3 medians or the 3 heights intersect."},{"Start":"00:58.760 ","End":"01:03.245","Text":"It\u0027s the same thing for an equilateral triangle and the height of the triangle,"},{"Start":"01:03.245 ","End":"01:05.300","Text":"just by simple trigonometry,"},{"Start":"01:05.300 ","End":"01:11.555","Text":"it\u0027s a times tangent of 60 degrees over 2."},{"Start":"01:11.555 ","End":"01:19.780","Text":"The distance of the centroid to the base of the triangle is 1/3 of the height."},{"Start":"01:19.780 ","End":"01:21.820","Text":"It\u0027s well known, for example,"},{"Start":"01:21.820 ","End":"01:27.755","Text":"that the medians of a triangle intersect at 1/3 and 2/3 of the distance,"},{"Start":"01:27.755 ","End":"01:31.650","Text":"with the 1/3 part being the closer to the base."},{"Start":"01:31.650 ","End":"01:33.555","Text":"This is what we get."},{"Start":"01:33.555 ","End":"01:40.225","Text":"Then the distance of the centroid from the axis of revolution is what we got before,"},{"Start":"01:40.225 ","End":"01:44.220","Text":"plus a, that\u0027s this a here."},{"Start":"01:44.220 ","End":"01:47.130","Text":"We need the area of the triangle,"},{"Start":"01:47.130 ","End":"01:49.125","Text":"1/2 base times height,"},{"Start":"01:49.125 ","End":"01:53.400","Text":"and that comes out to be a squared root 3 over 4."},{"Start":"01:53.400 ","End":"01:57.140","Text":"Now we\u0027re ready to plug all the quantities into"},{"Start":"01:57.140 ","End":"02:03.290","Text":"Pappus\u0027s theorem and this is the essence of what the theorem is, this formula."},{"Start":"02:03.290 ","End":"02:09.185","Text":"We get 2 Pi and then Rho is this from before,"},{"Start":"02:09.185 ","End":"02:12.125","Text":"and a is the area is this."},{"Start":"02:12.125 ","End":"02:21.040","Text":"After simplification it comes out to be this and that\u0027s our answer. We\u0027re done."}],"ID":21027},{"Watched":false,"Name":"Exercise 2","Duration":"2m 21s","ChapterTopicVideoID":20192,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.765","Text":"In this exercise, we\u0027re asked to use Pappus\u0027s theorem to find the centroid of the region."},{"Start":"00:07.765 ","End":"00:09.810","Text":"It\u0027s the 1 in the picture."},{"Start":"00:09.810 ","End":"00:14.280","Text":"It\u0027s a 1/4 circle of radius 2 centered at the origin,"},{"Start":"00:14.280 ","End":"00:17.715","Text":"the bit that\u0027s in the first quadrant."},{"Start":"00:17.715 ","End":"00:22.910","Text":"What we\u0027re going to do is use the formula for the volume of a sphere."},{"Start":"00:22.910 ","End":"00:26.630","Text":"In this case it will be half a sphere because if we revolve"},{"Start":"00:26.630 ","End":"00:31.385","Text":"this about the x-axis or the y-axis,"},{"Start":"00:31.385 ","End":"00:34.850","Text":"then we get 1/2 a sphere of radius 2."},{"Start":"00:34.850 ","End":"00:37.835","Text":"Now our shape has a line of symmetry."},{"Start":"00:37.835 ","End":"00:41.990","Text":"The line y equals x is its symmetry so"},{"Start":"00:41.990 ","End":"00:46.550","Text":"the centroid will lie on that and so we\u0027ll call it a,"},{"Start":"00:46.550 ","End":"00:49.459","Text":"a and a is what we have to find."},{"Start":"00:49.459 ","End":"00:51.824","Text":"The area of the shape,"},{"Start":"00:51.824 ","End":"00:53.915","Text":"it\u0027s a 1/4 of a circle,"},{"Start":"00:53.915 ","End":"00:58.145","Text":"so it\u0027s a 1/4 of Pi r squared and the r is 2,"},{"Start":"00:58.145 ","End":"01:00.875","Text":"2 of course is the square root of 4."},{"Start":"01:00.875 ","End":"01:03.370","Text":"We just get Pi."},{"Start":"01:03.370 ","End":"01:07.810","Text":"The distance of the centroid from the axis is a."},{"Start":"01:07.810 ","End":"01:10.885","Text":"If the coordinates are a, a,"},{"Start":"01:10.885 ","End":"01:15.710","Text":"then the height is the same as the y of the point is a,"},{"Start":"01:15.710 ","End":"01:19.210","Text":"that will be row in the formula."},{"Start":"01:19.210 ","End":"01:22.850","Text":"This is the formula that we\u0027re going to use."},{"Start":"01:22.850 ","End":"01:24.650","Text":"This is Pappus\u0027s theorem."},{"Start":"01:24.650 ","End":"01:28.645","Text":"This part relates to volume of revolution."},{"Start":"01:28.645 ","End":"01:31.650","Text":"We can express the volume in 2 ways."},{"Start":"01:31.650 ","End":"01:34.175","Text":"1 way is just by direct computation."},{"Start":"01:34.175 ","End":"01:41.930","Text":"It\u0027s 1/2 the formula for the volume of a sphere of radius 2,"},{"Start":"01:41.930 ","End":"01:46.370","Text":"it\u0027s 4/3Pi r cubed and 1/2 of it, so we get this."},{"Start":"01:46.370 ","End":"01:52.565","Text":"On the other hand, Pappus\u0027s theorem tells us that the volume is 2Pi row a,"},{"Start":"01:52.565 ","End":"01:59.480","Text":"so we equate these 2 and after we substitute v and a,"},{"Start":"01:59.480 ","End":"02:01.875","Text":"and the Pi cancels,"},{"Start":"02:01.875 ","End":"02:07.185","Text":"what we get is that a is 16 over 6Pi."},{"Start":"02:07.185 ","End":"02:11.610","Text":"Once we have a, the centroid,"},{"Start":"02:11.610 ","End":"02:13.155","Text":"its coordinates are a,"},{"Start":"02:13.155 ","End":"02:18.165","Text":"a so the answer is 16 over 6Pi,"},{"Start":"02:18.165 ","End":"02:22.270","Text":"16 over 6Pi. We\u0027re done."}],"ID":21028},{"Watched":false,"Name":"Exercise 3","Duration":"1m 42s","ChapterTopicVideoID":20193,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.930","Text":"In this exercise, we inscribe a regular hexagon in the circle given by this equation,"},{"Start":"00:06.930 ","End":"00:08.760","Text":"so it\u0027s center is at 2,"},{"Start":"00:08.760 ","End":"00:12.030","Text":"0, and the radius is 1."},{"Start":"00:12.030 ","End":"00:16.650","Text":"Then we revolve the hexagon about the y-axis,"},{"Start":"00:16.650 ","End":"00:24.700","Text":"and we have to find the surface area and the volume for the solid generated."},{"Start":"00:25.420 ","End":"00:29.000","Text":"Now, the hexagon might not be like this,"},{"Start":"00:29.000 ","End":"00:30.815","Text":"it could be rotated,"},{"Start":"00:30.815 ","End":"00:33.139","Text":"but in any case by symmetry,"},{"Start":"00:33.139 ","End":"00:34.880","Text":"the center is going to be at 2,"},{"Start":"00:34.880 ","End":"00:38.025","Text":"0, center meaning the centroid."},{"Start":"00:38.025 ","End":"00:39.330","Text":"That\u0027s all we care about,"},{"Start":"00:39.330 ","End":"00:41.705","Text":"so it doesn\u0027t have to be like in the picture."},{"Start":"00:41.705 ","End":"00:45.790","Text":"We\u0027ll use both of Pappus\u0027s theorems, I\u0027ll remind you."},{"Start":"00:45.790 ","End":"00:48.784","Text":"Here they are, you can pause the video."},{"Start":"00:48.784 ","End":"00:50.390","Text":"These are the 2 formulas,"},{"Start":"00:50.390 ","End":"00:54.365","Text":"one gives us the volume in terms of the area,"},{"Start":"00:54.365 ","End":"00:59.000","Text":"and the other gives us the surface area in terms of the length."},{"Start":"00:59.000 ","End":"01:01.129","Text":"Each case we need Rho,"},{"Start":"01:01.129 ","End":"01:05.990","Text":"which is the distance from the centroid to the axis of revolution."},{"Start":"01:05.990 ","End":"01:08.925","Text":"In our case, the area,"},{"Start":"01:08.925 ","End":"01:11.745","Text":"just do some simple trigonometry,"},{"Start":"01:11.745 ","End":"01:15.240","Text":"you\u0027ll get that it\u0027s 3 root 3 over 2."},{"Start":"01:15.240 ","End":"01:17.410","Text":"The length is more obvious,"},{"Start":"01:17.410 ","End":"01:21.155","Text":"it\u0027s 6-sided and each side is of length 1,"},{"Start":"01:21.155 ","End":"01:25.510","Text":"and the distance from the center to the axis is 2."},{"Start":"01:25.510 ","End":"01:27.750","Text":"We\u0027ll use both of Pappus\u0027s theorems,"},{"Start":"01:27.750 ","End":"01:29.550","Text":"one will give us the volume,"},{"Start":"01:29.550 ","End":"01:31.620","Text":"one will give us the surface area."},{"Start":"01:31.620 ","End":"01:33.870","Text":"One is 2 Pi Rho A,"},{"Start":"01:33.870 ","End":"01:36.920","Text":"the other\u0027s 2 Pi Rho L. If you do the computations,"},{"Start":"01:36.920 ","End":"01:38.540","Text":"this is what we get for the volume,"},{"Start":"01:38.540 ","End":"01:43.110","Text":"this is what we get for the surface area, and we\u0027re done."}],"ID":21029},{"Watched":false,"Name":"Exercise 4","Duration":"1m 36s","ChapterTopicVideoID":20194,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.430","Text":"In this exercise, we have a circular disk and this is its equation."},{"Start":"00:05.430 ","End":"00:08.340","Text":"It\u0027s a solid disk including the interior."},{"Start":"00:08.340 ","End":"00:11.175","Text":"That\u0027s why there\u0027s the less than or equal to here."},{"Start":"00:11.175 ","End":"00:15.720","Text":"The center is at 4,0 clearly and the radius is 2,"},{"Start":"00:15.720 ","End":"00:17.775","Text":"which is the square root of 4."},{"Start":"00:17.775 ","End":"00:20.780","Text":"We revolve it about the line y equals x,"},{"Start":"00:20.780 ","End":"00:22.430","Text":"which is this line."},{"Start":"00:22.430 ","End":"00:26.255","Text":"We have to find the volume of the solid generated"},{"Start":"00:26.255 ","End":"00:31.310","Text":"just like a doughnut shape and we\u0027ll use Pappus\u0027s theorem."},{"Start":"00:31.310 ","End":"00:34.040","Text":"Pappus\u0027s theorem to remind you,"},{"Start":"00:34.040 ","End":"00:35.930","Text":"this is the formula we\u0027re going to use,"},{"Start":"00:35.930 ","End":"00:37.910","Text":"V equals 2Pi Rho A."},{"Start":"00:37.910 ","End":"00:40.880","Text":"You can pause the video and read this."},{"Start":"00:40.880 ","End":"00:44.569","Text":"Now this dotted line is just an auxiliary line."},{"Start":"00:44.569 ","End":"00:48.185","Text":"The centroid is at 4,0 by symmetry."},{"Start":"00:48.185 ","End":"00:51.635","Text":"We want to know the distance from the centroid to this line,"},{"Start":"00:51.635 ","End":"00:56.540","Text":"so what I did is dropped a perpendicular from the centroid to this line,"},{"Start":"00:56.540 ","End":"01:01.550","Text":"that\u0027s the dotted line and clearly this intersection is"},{"Start":"01:01.550 ","End":"01:07.370","Text":"2,2 and the distance here is twice the square root of 2."},{"Start":"01:07.370 ","End":"01:11.900","Text":"We can just divide the diagonal by square root of 2 and get this."},{"Start":"01:11.900 ","End":"01:14.465","Text":"That\u0027s the centroid and its distance,"},{"Start":"01:14.465 ","End":"01:16.745","Text":"the radius is 2 like we said."},{"Start":"01:16.745 ","End":"01:21.250","Text":"The area Pi r squared is 4Pi."},{"Start":"01:21.250 ","End":"01:23.030","Text":"Now we have the ingredients,"},{"Start":"01:23.030 ","End":"01:25.250","Text":"we have A and we have Rho,"},{"Start":"01:25.250 ","End":"01:29.540","Text":"so we can plug this into the formula from Pappus,"},{"Start":"01:29.540 ","End":"01:31.565","Text":"and after the computation,"},{"Start":"01:31.565 ","End":"01:37.150","Text":"we get 16 root 2Pi squared and we are done."}],"ID":21030},{"Watched":false,"Name":"Exercise 5","Duration":"2m 33s","ChapterTopicVideoID":20195,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.734","Text":"In this exercise we have a semicircular arc,"},{"Start":"00:03.734 ","End":"00:08.790","Text":"that\u0027s this and it\u0027s given by the equation of the circle."},{"Start":"00:08.790 ","End":"00:11.910","Text":"But we restrict y to be bigger or equal to 2,"},{"Start":"00:11.910 ","End":"00:14.969","Text":"so it gives us the upper half of this circle."},{"Start":"00:14.969 ","End":"00:18.540","Text":"Now we rotate this arc about this line,"},{"Start":"00:18.540 ","End":"00:21.735","Text":"which is the line y plus 2x equals 0"},{"Start":"00:21.735 ","End":"00:25.575","Text":"and we have to find the area of the surface generated."},{"Start":"00:25.575 ","End":"00:28.440","Text":"We\u0027re going to use Pappus\u0027s theorem of course."},{"Start":"00:28.440 ","End":"00:30.585","Text":"We\u0027ll do it in 2 steps."},{"Start":"00:30.585 ","End":"00:34.020","Text":"First, we\u0027ll revolve around this axis"},{"Start":"00:34.020 ","End":"00:38.070","Text":"and then find the centroid and once we have the centroid,"},{"Start":"00:38.070 ","End":"00:41.365","Text":"then we\u0027ll revolve around the requested axis."},{"Start":"00:41.365 ","End":"00:43.464","Text":"So it\u0027s 2 steps."},{"Start":"00:43.464 ","End":"00:48.530","Text":"Now the centroid is by symmetry somewhere where x equals 2."},{"Start":"00:48.530 ","End":"00:49.670","Text":"I don\u0027t know exactly."},{"Start":"00:49.670 ","End":"00:53.855","Text":"Say here that we\u0027ll call it 2, 2 plus Rho,"},{"Start":"00:53.855 ","End":"01:00.605","Text":"where Rho is the distance from the centroid to the axis y equals 2."},{"Start":"01:00.605 ","End":"01:03.995","Text":"Now by Pappus, and I\u0027ll remind you,"},{"Start":"01:03.995 ","End":"01:07.565","Text":"this is the second of 2 Pappus\u0027s theorems."},{"Start":"01:07.565 ","End":"01:10.750","Text":"We need this formula here basically."},{"Start":"01:10.750 ","End":"01:13.245","Text":"He has this formula copied."},{"Start":"01:13.245 ","End":"01:20.100","Text":"The area of the sphere 4 Pi r squared is by Pappus\u0027s theorem,"},{"Start":"01:20.100 ","End":"01:23.420","Text":"2 Pi Rho times the length of the curve."},{"Start":"01:23.420 ","End":"01:29.040","Text":"Now the length of a semicircle is just Pi r. The full circle is 2 Pi r,"},{"Start":"01:29.040 ","End":"01:31.965","Text":"so this is Pi r. So this is what we get."},{"Start":"01:31.965 ","End":"01:38.760","Text":"We cancel Pi 4 times 2 square root of 16 equals 4 Pi Rho,"},{"Start":"01:38.760 ","End":"01:41.835","Text":"which gives us that Rho is 4 over Pi,"},{"Start":"01:41.835 ","End":"01:44.385","Text":"and from this, the centroid is at 2,"},{"Start":"01:44.385 ","End":"01:46.995","Text":"2 plus 4 over Pi."},{"Start":"01:46.995 ","End":"01:54.575","Text":"We want the distance of the centroid from the axis y plus 2x equals 0."},{"Start":"01:54.575 ","End":"01:58.850","Text":"So we just use the formula for distance of a point from a line."},{"Start":"01:58.850 ","End":"02:03.290","Text":"I\u0027ll leave it to you to look the formula up and check this computation."},{"Start":"02:03.290 ","End":"02:07.350","Text":"But this is what we get for the distance of this point from"},{"Start":"02:07.350 ","End":"02:11.570","Text":"this line and then we use Pappus again,"},{"Start":"02:11.570 ","End":"02:15.830","Text":"this time with the other axis and the length is still the same."},{"Start":"02:15.830 ","End":"02:23.525","Text":"The 2 Pi is still the 2 Pi and then we\u0027re still using the formula that S is 2 Pi Rho L,"},{"Start":"02:23.525 ","End":"02:27.080","Text":"so the surface area is 2 Pi and this is"},{"Start":"02:27.080 ","End":"02:30.920","Text":"Rho and this is L and little bit of simplification,"},{"Start":"02:30.920 ","End":"02:34.320","Text":"and this is what we get and we are done."}],"ID":21031},{"Watched":false,"Name":"Exercise 6","Duration":"2m 57s","ChapterTopicVideoID":20196,"CourseChapterTopicPlaylistID":113417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.605","Text":"In this exercise, we have a curve which is a piece of a parabola."},{"Start":"00:04.605 ","End":"00:07.275","Text":"The parabola\u0027s y equals 1/2 of x squared plus 1."},{"Start":"00:07.275 ","End":"00:10.350","Text":"We just take the bit where x goes from 0-1,"},{"Start":"00:10.350 ","End":"00:12.760","Text":"it\u0027s solid green curve."},{"Start":"00:13.010 ","End":"00:15.630","Text":"The centroid of this curve,"},{"Start":"00:15.630 ","End":"00:18.675","Text":"we call it x bar, y bar."},{"Start":"00:18.675 ","End":"00:23.115","Text":"Then we want to find just x bar using Pappus\u0027s theorem."},{"Start":"00:23.115 ","End":"00:26.910","Text":"We need the x coordinate of the centroid."},{"Start":"00:26.910 ","End":"00:31.320","Text":"Well, let L be the length of the curve and S"},{"Start":"00:31.320 ","End":"00:36.360","Text":"the surface area that we get by revolving it around the y-axis."},{"Start":"00:36.360 ","End":"00:40.100","Text":"We use the formula for arc length,"},{"Start":"00:40.100 ","End":"00:43.250","Text":"which is this, general from a to b,"},{"Start":"00:43.250 ","End":"00:45.325","Text":"in this case, from 0-1."},{"Start":"00:45.325 ","End":"00:47.180","Text":"We\u0027ll compute L, we\u0027ll compute S,"},{"Start":"00:47.180 ","End":"00:48.950","Text":"and then we\u0027ll apply Pappus\u0027s theorem."},{"Start":"00:48.950 ","End":"00:53.125","Text":"Continuing with L, dy by dx is just x."},{"Start":"00:53.125 ","End":"00:55.170","Text":"This is 1/2 x squared plus 1/2."},{"Start":"00:55.170 ","End":"00:57.255","Text":"We differentiate it, you get just x."},{"Start":"00:57.255 ","End":"00:58.590","Text":"So we get this."},{"Start":"00:58.590 ","End":"01:02.240","Text":"We can find this integral using integration by parts."},{"Start":"01:02.240 ","End":"01:07.135","Text":"I\u0027ll leave it to you to check but what we get is this."},{"Start":"01:07.135 ","End":"01:09.840","Text":"Then we substitute 0 and 1,"},{"Start":"01:09.840 ","End":"01:13.710","Text":"and we get this expression for L. Next,"},{"Start":"01:13.710 ","End":"01:16.070","Text":"we need S. Again,"},{"Start":"01:16.070 ","End":"01:20.815","Text":"we have a formula for surface area of revolution."},{"Start":"01:20.815 ","End":"01:25.185","Text":"It\u0027s 2 Pi x square root of 1 plus dy by dx squared."},{"Start":"01:25.185 ","End":"01:28.770","Text":"Again, dy by dx is just x."},{"Start":"01:28.770 ","End":"01:30.630","Text":"This time, we get this integral,"},{"Start":"01:30.630 ","End":"01:33.135","Text":"Pi comes out in front of the integral,"},{"Start":"01:33.135 ","End":"01:41.390","Text":"and we can make a substitution that will let u be x squared plus 1 du be 2xdx."},{"Start":"01:41.390 ","End":"01:45.380","Text":"This just comes out to be the square root of udu."},{"Start":"01:45.380 ","End":"01:49.635","Text":"This u^1/2, so it\u0027s u ^1 and 1/2,"},{"Start":"01:49.635 ","End":"01:52.490","Text":"which is u root u divided by 3/2,"},{"Start":"01:52.490 ","End":"01:54.695","Text":"which is like multiplying by 2/3."},{"Start":"01:54.695 ","End":"01:57.740","Text":"Also, we substitute u instead of x,"},{"Start":"01:57.740 ","End":"02:00.175","Text":"so we get u goes from 1 to 2,"},{"Start":"02:00.175 ","End":"02:01.770","Text":"when x is 0 and 1,"},{"Start":"02:01.770 ","End":"02:03.645","Text":"u is 1 and 2,"},{"Start":"02:03.645 ","End":"02:06.180","Text":"so we get this and then substitute."},{"Start":"02:06.180 ","End":"02:08.030","Text":"This is what S is."},{"Start":"02:08.030 ","End":"02:10.955","Text":"Now that we have L and we have S,"},{"Start":"02:10.955 ","End":"02:14.330","Text":"then we\u0027ll apply Pappus\u0027s theorem,"},{"Start":"02:14.330 ","End":"02:15.470","Text":"which to remind you,"},{"Start":"02:15.470 ","End":"02:17.155","Text":"this is the second of 2."},{"Start":"02:17.155 ","End":"02:19.720","Text":"This is the formula we need."},{"Start":"02:19.720 ","End":"02:24.275","Text":"Pappus, row is just x bar,"},{"Start":"02:24.275 ","End":"02:27.740","Text":"that\u0027s the distance of the centroid from"},{"Start":"02:27.740 ","End":"02:31.685","Text":"the y-axis is just the x-coordinate of the centroid."},{"Start":"02:31.685 ","End":"02:33.860","Text":"Here, we have L and here,"},{"Start":"02:33.860 ","End":"02:39.600","Text":"we have S. We substitute the S here and the L here."},{"Start":"02:39.600 ","End":"02:41.040","Text":"Now stuff cancels."},{"Start":"02:41.040 ","End":"02:43.725","Text":"The Pi cancels with the Pi here,"},{"Start":"02:43.725 ","End":"02:47.115","Text":"and the 2 cancels with this 1/2 and this 1/2."},{"Start":"02:47.115 ","End":"02:53.565","Text":"What we end up with after we extract x bar is this expression."},{"Start":"02:53.565 ","End":"02:57.370","Text":"That\u0027s the answer we\u0027re looking for, and we\u0027re done."}],"ID":21032}],"Thumbnail":null,"ID":113417},{"Name":"Principal Normal, Curvature","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"4m 41s","ChapterTopicVideoID":20240,"CourseChapterTopicPlaylistID":113418,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/20240.jpeg","UploadDate":"2019-11-03T14:02:11.2500000","DurationForVideoObject":"PT4M41S","Description":null,"MetaTitle":"Exercise 1 - Principal Normal, Curvature: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Principal Normal, Curvature practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/calculus-i%2c-ii-and-iii/advanced-theory-exercises/principal-normal%2c-curvature/vid21033","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.600","Text":"In this exercise, we have a parametric curve,"},{"Start":"00:03.600 ","End":"00:09.270","Text":"r of t. T is the unit tangent vector."},{"Start":"00:09.270 ","End":"00:11.790","Text":"We denote by the letter s,"},{"Start":"00:11.790 ","End":"00:14.625","Text":"as usual, the arc length parameter."},{"Start":"00:14.625 ","End":"00:16.020","Text":"We\u0027re going to abbreviate,"},{"Start":"00:16.020 ","End":"00:17.520","Text":"everything is a function of t,"},{"Start":"00:17.520 ","End":"00:22.215","Text":"but we\u0027ll drop the parentheses t. Let\u0027s assume that"},{"Start":"00:22.215 ","End":"00:27.540","Text":"r has a 2nd derivative and that t has a first derivative."},{"Start":"00:27.540 ","End":"00:31.185","Text":"We have to show 4 things and they build on each other."},{"Start":"00:31.185 ","End":"00:33.180","Text":"Let\u0027s take them at a time."},{"Start":"00:33.180 ","End":"00:36.120","Text":"In a, we have to show,"},{"Start":"00:36.120 ","End":"00:39.960","Text":"I won\u0027t read it out, this equation written here, the identity."},{"Start":"00:39.960 ","End":"00:43.770","Text":"For a, we can first write r prime,"},{"Start":"00:43.770 ","End":"00:45.660","Text":"which is dr by dt,"},{"Start":"00:45.660 ","End":"00:48.525","Text":"and we can use the chain rule."},{"Start":"00:48.525 ","End":"00:52.320","Text":"We\u0027ll let r vary with t via s,"},{"Start":"00:52.320 ","End":"00:58.890","Text":"r is a function of s and s is a function of t. Dr by ds times ds by dt gives us dr by"},{"Start":"00:58.890 ","End":"01:06.365","Text":"dt and dr by ds is well-known to be the unit tangent vector."},{"Start":"01:06.365 ","End":"01:08.750","Text":"We were left with this equation,"},{"Start":"01:08.750 ","End":"01:13.840","Text":"and now we can differentiate this with respect to t again,"},{"Start":"01:13.840 ","End":"01:19.325","Text":"and then we get r double prime and we use the product rule for derivatives."},{"Start":"01:19.325 ","End":"01:22.370","Text":"First of all, we have ds by dt,"},{"Start":"01:22.370 ","End":"01:29.450","Text":"and then we differentiate this with respect to t. The derivative of this,"},{"Start":"01:29.450 ","End":"01:32.990","Text":"which is this times the t as is,"},{"Start":"01:32.990 ","End":"01:36.245","Text":"and this is exactly what we had to show here."},{"Start":"01:36.245 ","End":"01:38.300","Text":"That does part a."},{"Start":"01:38.300 ","End":"01:41.735","Text":"Now for part b, this is what we have to show."},{"Start":"01:41.735 ","End":"01:47.250","Text":"Notice that we have to here compute r double prime cross r prime."},{"Start":"01:47.250 ","End":"01:53.020","Text":"In part a, we have expressions for r prime and for r double prime."},{"Start":"01:53.020 ","End":"01:59.015","Text":"We can plug in these 2 into the left-hand side here to the cross product."},{"Start":"01:59.015 ","End":"02:00.890","Text":"Well, this is just what I said."},{"Start":"02:00.890 ","End":"02:02.330","Text":"Now the cross product,"},{"Start":"02:02.330 ","End":"02:07.020","Text":"so we plug in r double prime and r prime,"},{"Start":"02:07.020 ","End":"02:10.160","Text":"I colored them so you can see what belongs to what."},{"Start":"02:10.160 ","End":"02:13.000","Text":"Now, this is equal 2."},{"Start":"02:13.000 ","End":"02:16.780","Text":"We have from the first term ds by dt squared,"},{"Start":"02:16.780 ","End":"02:18.875","Text":"and then this crossed with this,"},{"Start":"02:18.875 ","End":"02:23.250","Text":"plus this, and then this crossed with this."},{"Start":"02:23.250 ","End":"02:27.245","Text":"A vector crossed with itself is always 0,"},{"Start":"02:27.245 ","End":"02:30.215","Text":"and so this term drops off,"},{"Start":"02:30.215 ","End":"02:32.360","Text":"and we\u0027re just left with this."},{"Start":"02:32.360 ","End":"02:35.735","Text":"That is what we had to show if you go back and look."},{"Start":"02:35.735 ","End":"02:38.360","Text":"That\u0027s part b. Part c,"},{"Start":"02:38.360 ","End":"02:40.915","Text":"this is what we have to show."},{"Start":"02:40.915 ","End":"02:46.160","Text":"For part c, we need to remember that the derivative of"},{"Start":"02:46.160 ","End":"02:52.235","Text":"the unit tangent vector is perpendicular to the unit tangent vector."},{"Start":"02:52.235 ","End":"02:55.470","Text":"The norm of the cross product of"},{"Start":"02:55.470 ","End":"02:59.930","Text":"2 perpendicular vectors is just the product of the norms."},{"Start":"02:59.930 ","End":"03:03.680","Text":"I mean, normally it would be times sine of the angle between them,"},{"Start":"03:03.680 ","End":"03:08.140","Text":"but if the angle between them is 90 degrees then that\u0027s sine is 1."},{"Start":"03:08.140 ","End":"03:13.114","Text":"Also, we know that T is a unit vector,"},{"Start":"03:13.114 ","End":"03:14.300","Text":"so its norm is 1,"},{"Start":"03:14.300 ","End":"03:16.760","Text":"so all we\u0027re left with is this."},{"Start":"03:16.760 ","End":"03:19.445","Text":"From what we did above in part b,"},{"Start":"03:19.445 ","End":"03:21.710","Text":"we got that this equation holds,"},{"Start":"03:21.710 ","End":"03:26.480","Text":"this identity, and so we can take the norm of both sides."},{"Start":"03:26.480 ","End":"03:29.555","Text":"This is a scalar, so it comes out."},{"Start":"03:29.555 ","End":"03:33.244","Text":"This we\u0027ve computed up here,"},{"Start":"03:33.244 ","End":"03:36.425","Text":"so all we get is this is equal to this,"},{"Start":"03:36.425 ","End":"03:42.225","Text":"and ds by dt is equal to the norm of r prime,"},{"Start":"03:42.225 ","End":"03:44.310","Text":"and we get this."},{"Start":"03:44.310 ","End":"03:50.570","Text":"Then just by dividing both sides by the norm of r prime squared,"},{"Start":"03:50.570 ","End":"03:52.955","Text":"this goes into the denominator here,"},{"Start":"03:52.955 ","End":"03:56.790","Text":"we get this equality, this identity."},{"Start":"03:56.790 ","End":"03:58.325","Text":"That\u0027s what we had to show."},{"Start":"03:58.325 ","End":"04:02.570","Text":"That\u0027s part c. Now on to part d,"},{"Start":"04:02.570 ","End":"04:05.450","Text":"to scroll back to show you what is part d,"},{"Start":"04:05.450 ","End":"04:09.185","Text":"You have to show the curvature Kappa is this expression."},{"Start":"04:09.185 ","End":"04:14.330","Text":"Now by definition, Kappa is this expression."},{"Start":"04:14.330 ","End":"04:16.070","Text":"Then from the above,"},{"Start":"04:16.070 ","End":"04:19.330","Text":"we can still see it here, we\u0027ve got this."},{"Start":"04:19.330 ","End":"04:28.084","Text":"Kappa is equal to this divided by r prime norm,"},{"Start":"04:28.084 ","End":"04:32.575","Text":"just means that we have to increase this 2 to a 3."},{"Start":"04:32.575 ","End":"04:35.040","Text":"This is the expression for Kappa,"},{"Start":"04:35.040 ","End":"04:37.515","Text":"and that\u0027s exactly what we had to show."},{"Start":"04:37.515 ","End":"04:41.890","Text":"That concludes part d and the whole exercise."}],"ID":21033},{"Watched":false,"Name":"Exercise 2","Duration":"4m 51s","ChapterTopicVideoID":20241,"CourseChapterTopicPlaylistID":113418,"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":"This exercise has several parts;"},{"Start":"00:03.330 ","End":"00:05.490","Text":"a, b, and c, and they\u0027re all independent."},{"Start":"00:05.490 ","End":"00:10.080","Text":"In each case, r of t is a 3-dimensional vector function on"},{"Start":"00:10.080 ","End":"00:16.770","Text":"the reals and in each case we have to find the unit tangent vector,"},{"Start":"00:16.770 ","End":"00:19.795","Text":"the principle normal and the curvature."},{"Start":"00:19.795 ","End":"00:22.770","Text":"Just a routine computational exercise,"},{"Start":"00:22.770 ","End":"00:25.925","Text":"let\u0027s start with a copied it."},{"Start":"00:25.925 ","End":"00:30.890","Text":"Then we get the unit tangent vector is given by"},{"Start":"00:30.890 ","End":"00:36.160","Text":"this expression and if we differentiate this, we get this,"},{"Start":"00:36.160 ","End":"00:39.650","Text":"I\u0027m not going to go into the little details and then we have to divide"},{"Start":"00:39.650 ","End":"00:43.595","Text":"by the norm of this using the standard formula,"},{"Start":"00:43.595 ","End":"00:47.360","Text":"the square root of x square plus y square plus z squared."},{"Start":"00:47.360 ","End":"00:50.525","Text":"Then a bit of simplification."},{"Start":"00:50.525 ","End":"00:53.150","Text":"The denominator comes out to be this,"},{"Start":"00:53.150 ","End":"00:55.580","Text":"and this is the numerator."},{"Start":"00:55.580 ","End":"00:57.725","Text":"I used 1 of the formulas,"},{"Start":"00:57.725 ","End":"01:00.260","Text":"sine squared plus cosine squared is 1."},{"Start":"01:00.260 ","End":"01:02.090","Text":"You should memorize this."},{"Start":"01:02.090 ","End":"01:06.755","Text":"Now let\u0027s move on to the standard unit normal."},{"Start":"01:06.755 ","End":"01:11.165","Text":"We just differentiate this and divide by its magnitude,"},{"Start":"01:11.165 ","End":"01:14.959","Text":"by its norm and just as before,"},{"Start":"01:14.959 ","End":"01:19.625","Text":"this is the derivative of what we had here,"},{"Start":"01:19.625 ","End":"01:22.355","Text":"differentiate it, we get this,"},{"Start":"01:22.355 ","End":"01:24.350","Text":"then take the norm of it."},{"Start":"01:24.350 ","End":"01:25.970","Text":"When you have a scalar in front,"},{"Start":"01:25.970 ","End":"01:30.845","Text":"you can leave that scalar as long as it\u0027s positive and just take the norm of what\u0027s left."},{"Start":"01:30.845 ","End":"01:32.735","Text":"It\u0027s the square root of this, plus this,"},{"Start":"01:32.735 ","End":"01:37.125","Text":"plus this squared and do the computation,"},{"Start":"01:37.125 ","End":"01:45.785","Text":"the denominator comes out to be 1 and so in the end we have the unit normal is this,"},{"Start":"01:45.785 ","End":"01:47.555","Text":"that\u0027s 2 out of 3."},{"Start":"01:47.555 ","End":"01:50.885","Text":"Then finally, we have to compute the curvature,"},{"Start":"01:50.885 ","End":"01:55.080","Text":"and this is the formula r prime,"},{"Start":"01:55.080 ","End":"01:58.925","Text":"the norm of is equal to what\u0027s here in blue,"},{"Start":"01:58.925 ","End":"02:05.300","Text":"square root of 2 and the norm of T prime is here,"},{"Start":"02:05.300 ","End":"02:09.050","Text":"is this or actually it\u0027s equal to this,"},{"Start":"02:09.050 ","End":"02:10.670","Text":"which is 1 ultimately."},{"Start":"02:10.670 ","End":"02:13.025","Text":"This is what I\u0027m copying this 1"},{"Start":"02:13.025 ","End":"02:17.305","Text":"here and so this is the answer for the curvature, okay, that\u0027s part a."},{"Start":"02:17.305 ","End":"02:20.295","Text":"Now part b, this is r,"},{"Start":"02:20.295 ","End":"02:23.090","Text":"next we need big T,"},{"Start":"02:23.090 ","End":"02:24.800","Text":"which is equal to this,"},{"Start":"02:24.800 ","End":"02:29.240","Text":"which is equal to the derivative of this divided by its norm."},{"Start":"02:29.240 ","End":"02:32.795","Text":"Then route a is 2 root 2,"},{"Start":"02:32.795 ","End":"02:35.065","Text":"so this is what we get."},{"Start":"02:35.065 ","End":"02:45.005","Text":"Next we want the normal which is T prime over the norm of T prime, which is this."},{"Start":"02:45.005 ","End":"02:47.570","Text":"Oh, and of course here once again,"},{"Start":"02:47.570 ","End":"02:51.170","Text":"I used the sine squared plus cosine squared is 1,"},{"Start":"02:51.170 ","End":"02:53.705","Text":"and we\u0027ll use it again here."},{"Start":"02:53.705 ","End":"03:01.025","Text":"1 over root 2 times 2 is root 2 and then what we have left is equal to 1."},{"Start":"03:01.025 ","End":"03:04.885","Text":"Just what\u0027s in black goes on to the next line."},{"Start":"03:04.885 ","End":"03:07.235","Text":"That\u0027s the normal."},{"Start":"03:07.235 ","End":"03:11.540","Text":"Lastly, the curvature using this formula, well,"},{"Start":"03:11.540 ","End":"03:14.884","Text":"I\u0027ve colored the relevant bits r prime,"},{"Start":"03:14.884 ","End":"03:16.400","Text":"is this the root 8,"},{"Start":"03:16.400 ","End":"03:21.050","Text":"and T prime is the root 2 here and so we get"},{"Start":"03:21.050 ","End":"03:26.090","Text":"root 2 over root 8 and root 8 is 2 root 2 so it\u0027s a half,"},{"Start":"03:26.090 ","End":"03:28.720","Text":"and that concludes part b."},{"Start":"03:28.720 ","End":"03:33.620","Text":"In part c, this is r of t and don\u0027t forget that we were"},{"Start":"03:33.620 ","End":"03:38.465","Text":"told that t is positive because later we\u0027re going to divide by t first the unit tangent."},{"Start":"03:38.465 ","End":"03:41.960","Text":"This is the formula and this is what it comes out"},{"Start":"03:41.960 ","End":"03:46.355","Text":"to and the denominator as usual will be using this formula,"},{"Start":"03:46.355 ","End":"03:49.420","Text":"so we end up with this."},{"Start":"03:49.420 ","End":"03:51.560","Text":"We can simplify."},{"Start":"03:51.560 ","End":"03:56.510","Text":"Take the t squared outside the square root."},{"Start":"03:56.510 ","End":"03:58.010","Text":"Remember that t is positive,"},{"Start":"03:58.010 ","End":"04:00.470","Text":"so we don\u0027t need an absolute value and then t cancels"},{"Start":"04:00.470 ","End":"04:03.395","Text":"with t. What we\u0027re left with is this."},{"Start":"04:03.395 ","End":"04:05.240","Text":"Next onto the normal."},{"Start":"04:05.240 ","End":"04:10.955","Text":"Normal is this formula and this is equal to what we have from here,"},{"Start":"04:10.955 ","End":"04:16.070","Text":"differentiated and then divided by the norm of"},{"Start":"04:16.070 ","End":"04:21.800","Text":"the numerator here and this simplifies to this,"},{"Start":"04:21.800 ","End":"04:25.750","Text":"and finally simplifies to this and then"},{"Start":"04:25.750 ","End":"04:32.360","Text":"finally the curvature is this over this and the bits I\u0027ve colored,"},{"Start":"04:32.360 ","End":"04:35.330","Text":"this is the norm of r prime."},{"Start":"04:35.330 ","End":"04:37.775","Text":"This is the norm of T prime,"},{"Start":"04:37.775 ","End":"04:42.015","Text":"so this is equal to 1 over root 5 over root 5t,"},{"Start":"04:42.015 ","End":"04:47.900","Text":"which is 1 over 5t because the root 5 goes with root 5 times anyway."},{"Start":"04:47.900 ","End":"04:52.500","Text":"That\u0027s it. That\u0027s the last part of part c, so we\u0027re done."}],"ID":21034},{"Watched":false,"Name":"Exercise 3","Duration":"3m 11s","ChapterTopicVideoID":20242,"CourseChapterTopicPlaylistID":113418,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we have the curve y equals natural log of x."},{"Start":"00:04.890 ","End":"00:07.830","Text":"This is not a parametric curve here."},{"Start":"00:07.830 ","End":"00:12.385","Text":"It\u0027s just the graph of a function y in terms of x."},{"Start":"00:12.385 ","End":"00:18.180","Text":"We have to find the point on the curve which has the greatest curvature."},{"Start":"00:18.180 ","End":"00:22.785","Text":"We start off by computing the curvature at a general point."},{"Start":"00:22.785 ","End":"00:25.470","Text":"Now when we have y as a function of x,"},{"Start":"00:25.470 ","End":"00:28.455","Text":"this is the formula we use."},{"Start":"00:28.455 ","End":"00:33.940","Text":"We\u0027re here, I should say f of x is natural log of x, of course."},{"Start":"00:34.130 ","End":"00:36.169","Text":"Let\u0027s do the computation."},{"Start":"00:36.169 ","End":"00:39.440","Text":"First of all, the derivatives f of x is natural log of x,"},{"Start":"00:39.440 ","End":"00:41.060","Text":"f prime of x is 1 over x,"},{"Start":"00:41.060 ","End":"00:43.310","Text":"f double-prime minus 1 over x squared."},{"Start":"00:43.310 ","End":"00:45.830","Text":"Now, plug it into here."},{"Start":"00:45.830 ","End":"00:50.915","Text":"We have, well, lets you look at it and see that I have made a mistake here."},{"Start":"00:50.915 ","End":"00:55.400","Text":"We get this and this simplifies to this."},{"Start":"00:55.400 ","End":"00:58.640","Text":"We just put this as the common denominator."},{"Start":"00:58.640 ","End":"01:06.345","Text":"The denominator here, the x squared becomes x cubed and x cubed goes up to the numerator."},{"Start":"01:06.345 ","End":"01:10.310","Text":"Here we have x cubed over x squared, which is the x."},{"Start":"01:10.310 ","End":"01:14.090","Text":"The rest of it is just x squared plus 1 that remains to the 3 over"},{"Start":"01:14.090 ","End":"01:18.665","Text":"2 Now that we have an expression for k of x to find its maximum,"},{"Start":"01:18.665 ","End":"01:22.430","Text":"we treat this as function that needs to be maximized."},{"Start":"01:22.430 ","End":"01:26.135","Text":"We differentiate and look for critical points."},{"Start":"01:26.135 ","End":"01:29.660","Text":"Prime of x is this."},{"Start":"01:29.660 ","End":"01:33.770","Text":"We use the quotient rule which I put here."},{"Start":"01:33.770 ","End":"01:36.305","Text":"Anyway, this is just computations."},{"Start":"01:36.305 ","End":"01:37.940","Text":"We get this."},{"Start":"01:37.940 ","End":"01:45.440","Text":"Then we simplify by dividing top and bottom by x squared plus 1 to the power of a 1/2."},{"Start":"01:45.440 ","End":"01:46.715","Text":"This disappears."},{"Start":"01:46.715 ","End":"01:49.025","Text":"This 3 over 2 just becomes 1,"},{"Start":"01:49.025 ","End":"01:51.800","Text":"and this 3 becomes 5 over 2 here."},{"Start":"01:51.800 ","End":"01:56.834","Text":"Then this x squared minus 3x squared minus 2x squared."},{"Start":"01:56.834 ","End":"02:01.235","Text":"The critical point is when k prime is 0."},{"Start":"02:01.235 ","End":"02:05.455","Text":"New page, so this is 0 when the numerator is 0."},{"Start":"02:05.455 ","End":"02:07.695","Text":"1 minus 2x squared is 0,"},{"Start":"02:07.695 ","End":"02:11.055","Text":"so x squared is a 1/2."},{"Start":"02:11.055 ","End":"02:13.130","Text":"We were given that x is positive,"},{"Start":"02:13.130 ","End":"02:14.480","Text":"so there\u0027s only 1 solution,"},{"Start":"02:14.480 ","End":"02:17.450","Text":"x equals 1 over the square root of 2."},{"Start":"02:17.450 ","End":"02:21.590","Text":"Question is, is this a maximum or a minimum?"},{"Start":"02:21.590 ","End":"02:25.910","Text":"Let\u0027s check the value of k prime to the left and to the right of this."},{"Start":"02:25.910 ","End":"02:28.280","Text":"In the interval to the left,"},{"Start":"02:28.280 ","End":"02:31.460","Text":"we have k prime is positive."},{"Start":"02:31.460 ","End":"02:35.360","Text":"Because if x is less than 1 over root 2,"},{"Start":"02:35.360 ","End":"02:41.620","Text":"then x squared is less than a 1/2 2x squared is less than 1,1 minus this is positive."},{"Start":"02:41.620 ","End":"02:44.060","Text":"When x is bigger than 1 over root 2,"},{"Start":"02:44.060 ","End":"02:47.480","Text":"similarly, k prime is negative."},{"Start":"02:47.480 ","End":"02:52.010","Text":"The function k goes from increasing to decreasing."},{"Start":"02:52.010 ","End":"02:55.460","Text":"That means that the point is a maximum."},{"Start":"02:55.460 ","End":"02:59.355","Text":"Plug the x into the function and we\u0027ve got"},{"Start":"02:59.355 ","End":"03:04.205","Text":"natural log of 1 over root 2 because we were asked for the actual point on the curve."},{"Start":"03:04.205 ","End":"03:06.545","Text":"The point on the curve, it\u0027s 1 over root 2,"},{"Start":"03:06.545 ","End":"03:12.100","Text":"natural log of 1 over root 2. Okay, we\u0027re done."}],"ID":21035}],"Thumbnail":null,"ID":113418},{"Name":"Change of Variables in a Triple Integral, Area of a Parametric Surface","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"4m 45s","ChapterTopicVideoID":20243,"CourseChapterTopicPlaylistID":113419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.960","Text":"In this exercise, we have a solid D and it"},{"Start":"00:03.960 ","End":"00:08.370","Text":"lies inside the cylinder x squared plus y squared equals 1."},{"Start":"00:08.370 ","End":"00:14.190","Text":"Below the cone, z equals square root of 4 times x squared plus y squared,"},{"Start":"00:14.190 ","End":"00:17.025","Text":"and above the plane z equals naught."},{"Start":"00:17.025 ","End":"00:27.030","Text":"Our task is to evaluate the triple integral of x squared dxdydz over the solid D,"},{"Start":"00:27.170 ","End":"00:31.080","Text":"and there\u0027s a picture here that might help imagine it."},{"Start":"00:31.080 ","End":"00:34.500","Text":"What I\u0027ve done is taken a cross section."},{"Start":"00:34.500 ","End":"00:36.225","Text":"This is the xy plane,"},{"Start":"00:36.225 ","End":"00:40.215","Text":"cut anywhere in the plane through the z-axis."},{"Start":"00:40.215 ","End":"00:45.995","Text":"This thing has a rotational symmetry and this is what it looks like."},{"Start":"00:45.995 ","End":"00:49.310","Text":"This is the cone,"},{"Start":"00:49.310 ","End":"00:56.839","Text":"so this part in this shading is below the cone and above the xy plane."},{"Start":"00:56.839 ","End":"01:02.180","Text":"If we project this solid onto the xy plane of from the side,"},{"Start":"01:02.180 ","End":"01:04.850","Text":"it just looks like a line from minus 1-1,"},{"Start":"01:04.850 ","End":"01:07.340","Text":"but from above it would look like a circle,"},{"Start":"01:07.340 ","End":"01:11.150","Text":"x squared plus y squared less than or equal to 1."},{"Start":"01:11.150 ","End":"01:14.539","Text":"Actually it\u0027s a disk not a circle because it contains the inside,"},{"Start":"01:14.539 ","End":"01:16.675","Text":"and in cylindrical coordinates,"},{"Start":"01:16.675 ","End":"01:21.860","Text":"this condition just says that r squared is less than or equal to 1,"},{"Start":"01:21.860 ","End":"01:25.295","Text":"which is the same as saying that r is less than or equal to 1,"},{"Start":"01:25.295 ","End":"01:28.325","Text":"because r is always non-negative."},{"Start":"01:28.325 ","End":"01:35.990","Text":"The solid D is bounded below by the plane z equals naught and above by"},{"Start":"01:35.990 ","End":"01:44.300","Text":"the cone z equals to r. This we got because x squared plus y squared is r squared,"},{"Start":"01:44.300 ","End":"01:46.310","Text":"and if we take the square root of 4r squared,"},{"Start":"01:46.310 ","End":"01:49.140","Text":"we get 2r, we don\u0027t need absolute value,"},{"Start":"01:49.140 ","End":"01:50.584","Text":"r is always positive."},{"Start":"01:50.584 ","End":"01:52.250","Text":"For example, you could plug in,"},{"Start":"01:52.250 ","End":"01:55.430","Text":"say y equals 0 and just look at the zx plane,"},{"Start":"01:55.430 ","End":"01:59.090","Text":"then we would get z equals twice absolute value of x."},{"Start":"01:59.090 ","End":"02:01.460","Text":"It looks like an absolute value with the point here."},{"Start":"02:01.460 ","End":"02:06.305","Text":"In polar, we can describe D as the set of all triples r Theta z,"},{"Start":"02:06.305 ","End":"02:09.200","Text":"where r is less than or equal to 1, like here,"},{"Start":"02:09.200 ","End":"02:12.755","Text":"and z goes between 0 and 2r."},{"Start":"02:12.755 ","End":"02:16.160","Text":"We also have to convert x squared to polar,"},{"Start":"02:16.160 ","End":"02:22.345","Text":"and in polar x squared is r cosine Theta squared because x is r cosine Theta."},{"Start":"02:22.345 ","End":"02:27.590","Text":"What we get is the triple integral in cartesian or rectangular"},{"Start":"02:27.590 ","End":"02:32.120","Text":"coordinates over d becomes a triple integral in polar."},{"Start":"02:32.120 ","End":"02:37.534","Text":"Now dxdydz is always rdzdrd Theta,"},{"Start":"02:37.534 ","End":"02:42.560","Text":"x squared we said is r cosine Theta squared or r"},{"Start":"02:42.560 ","End":"02:48.635","Text":"squared cosine squared Theta and the limits z goes from 0 to 2r."},{"Start":"02:48.635 ","End":"02:56.054","Text":"That\u0027s z, and then r goes from 0 to 1,"},{"Start":"02:56.054 ","End":"03:01.230","Text":"the radius and Theta full circle from 0 to 2Pi,"},{"Start":"03:01.230 ","End":"03:07.030","Text":"and the rest of the exercise is just computation."},{"Start":"03:07.030 ","End":"03:09.584","Text":"Move to a new page,"},{"Start":"03:09.584 ","End":"03:11.285","Text":"so let\u0027s do the computing."},{"Start":"03:11.285 ","End":"03:19.070","Text":"What we can do is take the cosine squared Theta outside the integrals for z and r,"},{"Start":"03:19.070 ","End":"03:21.170","Text":"because it only depends on Theta."},{"Start":"03:21.170 ","End":"03:25.805","Text":"What we have here is r-squared with this r is r cubed."},{"Start":"03:25.805 ","End":"03:28.430","Text":"So do the integral with respect to z first,"},{"Start":"03:28.430 ","End":"03:33.830","Text":"and that comes out to be r cubed c. I think,"},{"Start":"03:33.830 ","End":"03:38.600","Text":"I suppose I could have taken the r cubed in front of this integral dz, it doesn\u0027t matter."},{"Start":"03:38.600 ","End":"03:44.920","Text":"All cubes are constant as far as z goes to the integrals r cubed z between 0 and 2r."},{"Start":"03:44.920 ","End":"03:46.710","Text":"At 0, we get nothing,"},{"Start":"03:46.710 ","End":"03:49.680","Text":"at 2r, we get 2r^4."},{"Start":"03:49.680 ","End":"03:53.490","Text":"Because just studying z equals 2r to r times r cubed and"},{"Start":"03:53.490 ","End":"03:57.290","Text":"then this we take from 0 to 1, the integral."},{"Start":"03:57.290 ","End":"03:59.375","Text":"The integral will be 2/5,"},{"Start":"03:59.375 ","End":"04:02.170","Text":"r^5 between 0 and 1,"},{"Start":"04:02.170 ","End":"04:05.595","Text":"0 gives 0, the 1 gives 2/5."},{"Start":"04:05.595 ","End":"04:09.840","Text":"Now we can take the 1/5 out in front,"},{"Start":"04:09.840 ","End":"04:13.675","Text":"we are going to leave the 2 here because it\u0027s a trigonometric formula,"},{"Start":"04:13.675 ","End":"04:19.135","Text":"that 2 cosine squared Theta is 1 plus cosine 2 Theta."},{"Start":"04:19.135 ","End":"04:21.890","Text":"This is a straightforward integral."},{"Start":"04:21.890 ","End":"04:24.020","Text":"The integral of 1 is Theta,"},{"Start":"04:24.020 ","End":"04:25.310","Text":"for cosine 2 Theta,"},{"Start":"04:25.310 ","End":"04:26.450","Text":"we get sine 2 Theta,"},{"Start":"04:26.450 ","End":"04:28.520","Text":"but we have to divide by 2."},{"Start":"04:28.520 ","End":"04:33.770","Text":"Then the limits of integration at 0, everything is 0."},{"Start":"04:33.770 ","End":"04:37.040","Text":"At 2 Pi the sine is also 0,"},{"Start":"04:37.040 ","End":"04:42.800","Text":"but Theta isn\u0027t, so we just get 2 Pi and the fifth from here."},{"Start":"04:42.800 ","End":"04:45.930","Text":"This is the answer and we\u0027re done."}],"ID":21036},{"Watched":false,"Name":"Exercise 2","Duration":"3m 52s","ChapterTopicVideoID":20244,"CourseChapterTopicPlaylistID":113419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.770","Text":"In this exercise, we\u0027re given a triple iterated integral in rectangular coordinates."},{"Start":"00:07.770 ","End":"00:10.755","Text":"Let\u0027s see what is the solid,"},{"Start":"00:10.755 ","End":"00:15.190","Text":"the body that this volume represents."},{"Start":"00:15.260 ","End":"00:18.000","Text":"You could do it from left to right or right to left."},{"Start":"00:18.000 ","End":"00:19.110","Text":"If we do it from left to right,"},{"Start":"00:19.110 ","End":"00:25.020","Text":"x goes from minus 2 to 2 and then y goes from this to this,"},{"Start":"00:25.020 ","End":"00:29.460","Text":"which is a lower semicircle to an upper semicircle of radius 2."},{"Start":"00:29.460 ","End":"00:35.190","Text":"In other words, x and y cover the disk of radius 2 in"},{"Start":"00:35.190 ","End":"00:41.965","Text":"the xy-plane and then z goes from x squared plus y squared to 4."},{"Start":"00:41.965 ","End":"00:46.564","Text":"If we represent this as the xy-plane looking from the side,"},{"Start":"00:46.564 ","End":"00:48.325","Text":"and this is the z-axis,"},{"Start":"00:48.325 ","End":"00:51.710","Text":"then z equals x squared plus y squared is a paraboloid."},{"Start":"00:51.710 ","End":"00:54.530","Text":"Really, it\u0027s z equals r squared if we think already"},{"Start":"00:54.530 ","End":"00:59.850","Text":"cylindrical and z equals 4 is just plain or from the side,"},{"Start":"00:59.850 ","End":"01:02.925","Text":"it\u0027s just a line, z equals 4."},{"Start":"01:02.925 ","End":"01:04.880","Text":"This is the body."},{"Start":"01:04.880 ","End":"01:07.280","Text":"We rotate this around the z-axis,"},{"Start":"01:07.280 ","End":"01:13.720","Text":"you get the paraboloid and the shadow on the xy-plane is this."},{"Start":"01:13.720 ","End":"01:16.910","Text":"D is the paraboloid I described."},{"Start":"01:16.910 ","End":"01:22.250","Text":"I already hinted we\u0027re going to use cylindrical coordinates here so D is bounded"},{"Start":"01:22.250 ","End":"01:28.925","Text":"below by z equals x squared plus y squared or z equals r squared."},{"Start":"01:28.925 ","End":"01:34.740","Text":"You can actually think of this as r. The parabola above by z equals"},{"Start":"01:34.740 ","End":"01:37.890","Text":"4 and the projection on"},{"Start":"01:37.890 ","End":"01:42.440","Text":"the xy-plane is the disk x squared plus y squared less than or equal to 4,"},{"Start":"01:42.440 ","End":"01:46.105","Text":"or in polar r is less than or equal to 2."},{"Start":"01:46.105 ","End":"01:50.075","Text":"Now we can say what the integral is in polar."},{"Start":"01:50.075 ","End":"01:55.010","Text":"The body D can be described, like we said,"},{"Start":"01:55.010 ","End":"01:57.875","Text":"z goes from r squared to 4."},{"Start":"01:57.875 ","End":"02:00.605","Text":"These vertical cross-sections."},{"Start":"02:00.605 ","End":"02:05.530","Text":"The radius r goes from 0 to 2,"},{"Start":"02:05.530 ","End":"02:11.530","Text":"and the angle is the whole thing from 0 to 2 Pi for theta."},{"Start":"02:11.530 ","End":"02:22.170","Text":"Then x is r cosine theta and dz dy dx is always r dz dr d theta."},{"Start":"02:22.170 ","End":"02:23.940","Text":"Sometimes I say z, sometimes z,"},{"Start":"02:23.940 ","End":"02:26.150","Text":"depending whether you\u0027re in England or America."},{"Start":"02:26.150 ","End":"02:27.769","Text":"Now we just have to evaluate,"},{"Start":"02:27.769 ","End":"02:30.040","Text":"this is just computational."},{"Start":"02:30.040 ","End":"02:33.890","Text":"What we can do is the cosine theta can be"},{"Start":"02:33.890 ","End":"02:38.050","Text":"brought in front of the integral dz dr it doesn\u0027t depend on z,"},{"Start":"02:38.050 ","End":"02:41.475","Text":"or r. R, with this r makes r squared,"},{"Start":"02:41.475 ","End":"02:46.515","Text":"and I can pull that in front of the first integral because it doesn\u0027t depend on dz."},{"Start":"02:46.515 ","End":"02:52.150","Text":"We\u0027re just left with the 1 here as the integrant in the inner integral."},{"Start":"02:52.150 ","End":"02:57.000","Text":"The integral of 1dz would just be z and"},{"Start":"02:57.000 ","End":"03:01.850","Text":"a substitute 4 and r squared and subtract it we get 4 minus r squared here."},{"Start":"03:01.850 ","End":"03:04.050","Text":"That\u0027s this middle integral,"},{"Start":"03:04.050 ","End":"03:07.260","Text":"and then we still have what was left r squared."},{"Start":"03:07.260 ","End":"03:12.840","Text":"Anyway, multiply out you got 4 r squared minus r to the 4th,"},{"Start":"03:12.840 ","End":"03:17.790","Text":"4r squared gives us 4/3 r cubed minus r to the 4th,"},{"Start":"03:17.790 ","End":"03:21.565","Text":"gives us minus a 1/5 r to the 5th."},{"Start":"03:21.565 ","End":"03:25.430","Text":"Plug-in 0 and 2 and subtract and then we get"},{"Start":"03:25.430 ","End":"03:30.484","Text":"this and integral of cosine theta is sine theta."},{"Start":"03:30.484 ","End":"03:33.545","Text":"Subtract to this fraction,"},{"Start":"03:33.545 ","End":"03:36.350","Text":"and we get 64/15."},{"Start":"03:36.350 ","End":"03:38.360","Text":"You should check that."},{"Start":"03:38.360 ","End":"03:43.535","Text":"Integral of sine is minus cosine,"},{"Start":"03:43.535 ","End":"03:46.640","Text":"but at 0 and at 2Pi it\u0027s the same thing."},{"Start":"03:46.640 ","End":"03:47.990","Text":"The difference becomes 0,"},{"Start":"03:47.990 ","End":"03:52.860","Text":"so we\u0027re just left with 0 and we\u0027re done."}],"ID":21037},{"Watched":false,"Name":"Exercise 3","Duration":"5m 22s","ChapterTopicVideoID":20245,"CourseChapterTopicPlaylistID":113419,"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, there are 3 sub exercises."},{"Start":"00:04.110 ","End":"00:12.435","Text":"In each 1, we\u0027re given a description of a region in terms of rectangular coordinates,"},{"Start":"00:12.435 ","End":"00:18.930","Text":"and we have to convert that to a description in terms of spherical coordinates."},{"Start":"00:18.930 ","End":"00:21.030","Text":"Let\u0027s take them 1 at a time."},{"Start":"00:21.030 ","End":"00:25.514","Text":"In a, we have a region that\u0027s inside this sphere,"},{"Start":"00:25.514 ","End":"00:28.140","Text":"and outside this sphere."},{"Start":"00:28.140 ","End":"00:31.470","Text":"We\u0027ll start with the 2nd one."},{"Start":"00:31.470 ","End":"00:35.145","Text":"X squared plus y squared plus z squared,"},{"Start":"00:35.145 ","End":"00:39.120","Text":"bigger or equal to 1, because of the outside."},{"Start":"00:39.120 ","End":"00:42.270","Text":"This is Rho squared,"},{"Start":"00:42.270 ","End":"00:48.820","Text":"bigger or equal to 1, so Rho is bigger or equal to 1 since Rho is non-negative."},{"Start":"00:48.980 ","End":"00:52.505","Text":"For the 2nd sphere, we want the inside,"},{"Start":"00:52.505 ","End":"00:55.770","Text":"so it\u0027s less than or equal to."},{"Start":"00:56.510 ","End":"01:01.640","Text":"If you expand x squared plus y squared plus z squared minus 4 z,"},{"Start":"01:01.640 ","End":"01:04.045","Text":"which comes over to the other side."},{"Start":"01:04.045 ","End":"01:06.630","Text":"The plus 4 cancels with the plus 4,"},{"Start":"01:06.630 ","End":"01:08.745","Text":"so we\u0027re left with this inequality,"},{"Start":"01:08.745 ","End":"01:10.355","Text":"this is Rho squared,"},{"Start":"01:10.355 ","End":"01:12.815","Text":"z is Rho cosine Phi,"},{"Start":"01:12.815 ","End":"01:15.695","Text":"canceled by Rho, and we\u0027ve got this."},{"Start":"01:15.695 ","End":"01:18.260","Text":"Now compare this to this."},{"Start":"01:18.260 ","End":"01:19.610","Text":"We have 2 conditions on Rho,"},{"Start":"01:19.610 ","End":"01:22.535","Text":"a bigger or equal to, and the less than or equal to."},{"Start":"01:22.535 ","End":"01:31.325","Text":"They overlap when 4 cosine Phi is bigger or equal to 1,"},{"Start":"01:31.325 ","End":"01:32.750","Text":"or dividing by 4,"},{"Start":"01:32.750 ","End":"01:34.810","Text":"cosine Phi bigger equal to 1/4,"},{"Start":"01:34.810 ","End":"01:38.135","Text":"and it has to be also less than or equal to 1."},{"Start":"01:38.135 ","End":"01:41.315","Text":"As far as the range on Phi,"},{"Start":"01:41.315 ","End":"01:45.700","Text":"it\u0027s between 0 and that cosine of 1/4."},{"Start":"01:45.700 ","End":"01:47.955","Text":"For a given Phi,"},{"Start":"01:47.955 ","End":"01:54.445","Text":"Rho varies from 1-4 cosine Phi."},{"Start":"01:54.445 ","End":"01:56.930","Text":"We can describe the region in spherical."},{"Start":"01:56.930 ","End":"01:59.725","Text":"It\u0027s the set of all Rho, Theta, Phi,"},{"Start":"01:59.725 ","End":"02:01.380","Text":"no restriction on Theta,"},{"Start":"02:01.380 ","End":"02:03.720","Text":"meaning from 0-2 Pi."},{"Start":"02:03.720 ","End":"02:05.430","Text":"Phi, like we said,"},{"Start":"02:05.430 ","End":"02:06.900","Text":"is in this range,"},{"Start":"02:06.900 ","End":"02:09.330","Text":"and Rho, as we said,"},{"Start":"02:09.330 ","End":"02:12.990","Text":"from 1-4 cosine Phi."},{"Start":"02:12.990 ","End":"02:17.450","Text":"That\u0027s part a. Now, part b,"},{"Start":"02:17.450 ","End":"02:21.920","Text":"I just scrolled back so you can see what the question is."},{"Start":"02:21.920 ","End":"02:26.315","Text":"We\u0027re going to convert those 2 equations to spherical."},{"Start":"02:26.315 ","End":"02:29.285","Text":"The first one it said equals."},{"Start":"02:29.285 ","End":"02:32.660","Text":"You might wonder why I put less than or equal to here,"},{"Start":"02:32.660 ","End":"02:33.890","Text":"and not greater than or equal to."},{"Start":"02:33.890 ","End":"02:37.160","Text":"How do we know what\u0027s below the sphere,"},{"Start":"02:37.160 ","End":"02:38.750","Text":"and what\u0027s above the sphere?"},{"Start":"02:38.750 ","End":"02:41.120","Text":"We can just check by substituting a point."},{"Start":"02:41.120 ","End":"02:44.780","Text":"Let\u0027s say we took the center of this sphere,"},{"Start":"02:44.780 ","End":"02:47.590","Text":"which is where z equals 1/2,"},{"Start":"02:47.590 ","End":"02:52.585","Text":"and x and y are 0, then we get 1/4 less than or equal to 1.5."},{"Start":"02:52.585 ","End":"02:54.445","Text":"That\u0027s the right way around."},{"Start":"02:54.445 ","End":"02:56.555","Text":"Similarly for the cone,"},{"Start":"02:56.555 ","End":"02:59.690","Text":"above the cone, so bigger or equal to,"},{"Start":"02:59.690 ","End":"03:01.070","Text":"we can again check."},{"Start":"03:01.070 ","End":"03:05.725","Text":"For example, you could use this point, 0, 0, 1."},{"Start":"03:05.725 ","End":"03:08.010","Text":"Z is 1, x and y are 0,"},{"Start":"03:08.010 ","End":"03:10.425","Text":"and 1 is bigger or equal to 0."},{"Start":"03:10.425 ","End":"03:16.130","Text":"You can always figure out which side you want by just taking a sample point."},{"Start":"03:16.130 ","End":"03:19.510","Text":"Converting to spherical, this one"},{"Start":"03:19.510 ","End":"03:23.670","Text":"becomes Rho squared less than or equal to Rho cosine, Phi canceled by Rho,"},{"Start":"03:23.670 ","End":"03:25.245","Text":"and we\u0027ve got this."},{"Start":"03:25.245 ","End":"03:30.380","Text":"The other one, I\u0027ll let you take a look at the computations."},{"Start":"03:30.380 ","End":"03:35.615","Text":"Just need to remember that this expression is Rho sine Phi,"},{"Start":"03:35.615 ","End":"03:37.595","Text":"one of the conversion formulas."},{"Start":"03:37.595 ","End":"03:44.975","Text":"Anyway, we get this constraint and I want to take the intersection of these 2."},{"Start":"03:44.975 ","End":"03:51.770","Text":"We get the region defined by Theta is unrestricted from 0-2 Pi,"},{"Start":"03:51.770 ","End":"03:55.995","Text":"Phi is less than or equal to Pi over 4,"},{"Start":"03:55.995 ","End":"03:59.920","Text":"but bounded below by 0."},{"Start":"04:01.160 ","End":"04:05.055","Text":"Rho is bounded below by 0,"},{"Start":"04:05.055 ","End":"04:07.415","Text":"in spherical, that\u0027s the smallest it can be."},{"Start":"04:07.415 ","End":"04:09.950","Text":"That completes part b."},{"Start":"04:09.950 ","End":"04:14.380","Text":"Now on to part c. I\u0027ll let"},{"Start":"04:14.380 ","End":"04:18.865","Text":"you go back and look at what the equations were. I\u0027ll repeat them here."},{"Start":"04:18.865 ","End":"04:22.205","Text":"We were given 3 equations,"},{"Start":"04:22.205 ","End":"04:24.630","Text":"2 planes, z equals 1,"},{"Start":"04:24.630 ","End":"04:30.925","Text":"z equals 2, and a cone z equals square root of 3 times x squared plus y squared."},{"Start":"04:30.925 ","End":"04:34.290","Text":"We wanted the region enclosed by these,"},{"Start":"04:34.290 ","End":"04:38.080","Text":"so clearly it\u0027s above the cone,"},{"Start":"04:38.080 ","End":"04:41.285","Text":"below the plane z equals 2,"},{"Start":"04:41.285 ","End":"04:43.315","Text":"and above the plane z equals 1,"},{"Start":"04:43.315 ","End":"04:45.700","Text":"this is the direction of the inequalities."},{"Start":"04:45.700 ","End":"04:48.430","Text":"Now convert to spherical,"},{"Start":"04:48.430 ","End":"04:53.910","Text":"and this one becomes Rho bigger or equal to 1 over cosine Phi,"},{"Start":"04:53.910 ","End":"04:58.380","Text":"or secant Phi, this one Rho less than or equal to 2 secant Phi,"},{"Start":"04:58.380 ","End":"05:04.500","Text":"and this one becomes where Phi is less than or equal to Pi over 6."},{"Start":"05:04.500 ","End":"05:09.125","Text":"These restrictions taken together give us the region,"},{"Start":"05:09.125 ","End":"05:12.380","Text":"no restriction on Theta, Phi from here,"},{"Start":"05:12.380 ","End":"05:22.240","Text":"and Rho between secant Phi and 2 secant and Phi. That\u0027s it."}],"ID":21038},{"Watched":false,"Name":"Exercise 4","Duration":"4m 30s","ChapterTopicVideoID":20246,"CourseChapterTopicPlaylistID":113419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.300","Text":"In this exercise, D is the solid bounded above by the plane z equals"},{"Start":"00:06.300 ","End":"00:13.080","Text":"4 and below by the cone z equals the square root of x squared plus y squared."},{"Start":"00:13.080 ","End":"00:17.175","Text":"We have to evaluate this triple integral over"},{"Start":"00:17.175 ","End":"00:20.865","Text":"the solid and this is in rectangular coordinates."},{"Start":"00:20.865 ","End":"00:22.605","Text":"There\u0027s a picture here,"},{"Start":"00:22.605 ","End":"00:24.270","Text":"it\u0027s a view from the side,"},{"Start":"00:24.270 ","End":"00:28.080","Text":"this is the whole xy plane, the z-axis."},{"Start":"00:28.080 ","End":"00:31.080","Text":"If you rotate this figure about the z-axis,"},{"Start":"00:31.080 ","End":"00:33.555","Text":"you get the 3-dimensional body."},{"Start":"00:33.555 ","End":"00:35.745","Text":"We want this part here."},{"Start":"00:35.745 ","End":"00:39.770","Text":"Seems to me that we could use either cylindrical or spherical."},{"Start":"00:39.770 ","End":"00:44.625","Text":"I went for spherical because the integrand then becomes just Rho."},{"Start":"00:44.625 ","End":"00:47.190","Text":"Let\u0027s convert these equations."},{"Start":"00:47.190 ","End":"00:49.040","Text":"Well, actually, we\u0027ll turn them into inequalities."},{"Start":"00:49.040 ","End":"00:51.740","Text":"We want z less than or equal to 4 because we\u0027re"},{"Start":"00:51.740 ","End":"00:55.980","Text":"below the plane and bigger or equal to this."},{"Start":"00:55.980 ","End":"00:59.615","Text":"Let\u0027s take z less than or equal to 4 first."},{"Start":"00:59.615 ","End":"01:04.245","Text":"That comes out to be Rho cosine Phi less than 4,"},{"Start":"01:04.245 ","End":"01:07.700","Text":"so Rho is less than 4 secant."},{"Start":"01:07.700 ","End":"01:11.560","Text":"The other 1 bigger or equal to,"},{"Start":"01:11.560 ","End":"01:14.835","Text":"and z is Rho cosine Phi."},{"Start":"01:14.835 ","End":"01:18.830","Text":"The square root of x squared plus y squared is Rho sine Phi."},{"Start":"01:18.830 ","End":"01:23.360","Text":"We get that cosine Phi is bigger or equal to sine Phi."},{"Start":"01:23.360 ","End":"01:28.010","Text":"Then dividing both sides by the cosine and switching sides,"},{"Start":"01:28.010 ","End":"01:30.770","Text":"the tangent is less than or equal to 1."},{"Start":"01:30.770 ","End":"01:37.470","Text":"Phi is between 0 and 45 degrees or Pi over 4."},{"Start":"01:38.120 ","End":"01:42.150","Text":"Now, we need to convert the integral,"},{"Start":"01:42.150 ","End":"01:46.755","Text":"so dxdydz, this is the expression for it."},{"Start":"01:46.755 ","End":"01:50.685","Text":"The integrand is just Rho like I mentioned."},{"Start":"01:50.685 ","End":"01:54.440","Text":"If we substitute all that, what we get,"},{"Start":"01:54.440 ","End":"01:57.150","Text":"this Rho is here,"},{"Start":"01:57.150 ","End":"02:03.845","Text":"this is here, and then we just have to figure out the limits."},{"Start":"02:03.845 ","End":"02:05.795","Text":"Like we said here,"},{"Start":"02:05.795 ","End":"02:08.720","Text":"Rho is less than or equal to 4 secant Theta,"},{"Start":"02:08.720 ","End":"02:10.790","Text":"so down to 0."},{"Start":"02:10.790 ","End":"02:15.300","Text":"These are the limits for Rho and that\u0027s the d Rho,"},{"Start":"02:15.300 ","End":"02:22.379","Text":"then wrapped around that is Phi and that goes from 0 to Pi over 4."},{"Start":"02:22.379 ","End":"02:27.820","Text":"Theta\u0027s unrestricted, so it\u0027s 0 to 2Pi d Theta."},{"Start":"02:27.820 ","End":"02:30.395","Text":"Now we just have to evaluate that."},{"Start":"02:30.395 ","End":"02:34.730","Text":"Rho time Rho squared is Rho cubed and the sine Phi can"},{"Start":"02:34.730 ","End":"02:39.820","Text":"be pulled out in front of the integral with respect to Rho."},{"Start":"02:39.820 ","End":"02:43.809","Text":"Now, the integral of Rho cubed d Rho,"},{"Start":"02:43.809 ","End":"02:47.765","Text":"the indefinite integral is Rho to the 4th over 4."},{"Start":"02:47.765 ","End":"02:49.820","Text":"If we plug in 0, we get nothing,"},{"Start":"02:49.820 ","End":"02:53.640","Text":"and if we plug in 4 secant Theta, then we get, well,"},{"Start":"02:53.640 ","End":"02:57.900","Text":"it should be 4 to the 4th secant to the 4th Theta divided by 4,"},{"Start":"02:57.900 ","End":"03:01.910","Text":"and that\u0027s what knocks the 4 down to 3 in the exponent."},{"Start":"03:01.910 ","End":"03:05.600","Text":"They also separated the integral into 2 integrals."},{"Start":"03:05.600 ","End":"03:08.220","Text":"The d Theta is not related."},{"Start":"03:08.220 ","End":"03:10.260","Text":"Here we only have Phi and Rho,"},{"Start":"03:10.260 ","End":"03:12.995","Text":"so it\u0027s the product of these 2 integrals."},{"Start":"03:12.995 ","End":"03:21.835","Text":"The integral of d Theta is just 2Pi and the 4 cubed pull out in front."},{"Start":"03:21.835 ","End":"03:25.550","Text":"For the rest of it, we do it by substitution."},{"Start":"03:25.550 ","End":"03:32.910","Text":"We could say t equals cosine Phi and then dt is minus sine Phi d Phi."},{"Start":"03:32.910 ","End":"03:38.820","Text":"Here we\u0027ll write the secant to the 4th as cosine to the 4th in the denominator."},{"Start":"03:38.820 ","End":"03:47.010","Text":"After the substitution, we get the cosine to the 4th Phi becomes t to the 4th,"},{"Start":"03:47.010 ","End":"03:50.960","Text":"sine Phi d Phi becomes minus dt."},{"Start":"03:50.960 ","End":"03:52.880","Text":"I put the minus 1 here."},{"Start":"03:52.880 ","End":"03:54.710","Text":"4 cubed is 64,"},{"Start":"03:54.710 ","End":"03:57.040","Text":"times 2 is 128."},{"Start":"03:57.040 ","End":"04:03.860","Text":"Now, the integral of this is 1 over t cubed over 3."},{"Start":"04:03.860 ","End":"04:06.140","Text":"It\u0027s minus t to the minus 4,"},{"Start":"04:06.140 ","End":"04:11.030","Text":"increased the power by 1 and divide by the minus 3,"},{"Start":"04:11.030 ","End":"04:12.695","Text":"so we get this."},{"Start":"04:12.695 ","End":"04:16.715","Text":"Then we have to substitute the upper and lower limits,"},{"Start":"04:16.715 ","End":"04:18.520","Text":"and that gives us,"},{"Start":"04:18.520 ","End":"04:20.730","Text":"plug in 1, it\u0027s just 1,"},{"Start":"04:20.730 ","End":"04:22.545","Text":"plug in 1 over root 2,"},{"Start":"04:22.545 ","End":"04:26.375","Text":"it\u0027s root 2 to the power of 3, which is 2 root 2."},{"Start":"04:26.375 ","End":"04:31.080","Text":"Anyway, this is the answer. We are done."}],"ID":21039},{"Watched":false,"Name":"Exercise 5","Duration":"3m 53s","ChapterTopicVideoID":20247,"CourseChapterTopicPlaylistID":113419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.055","Text":"In this exercise, D is the solid enclosed by 2 spheres."},{"Start":"00:05.055 ","End":"00:09.060","Text":"There\u0027s the smallest sphere, this 1."},{"Start":"00:09.060 ","End":"00:10.965","Text":"This is its equation."},{"Start":"00:10.965 ","End":"00:13.050","Text":"There\u0027s the largest sphere,"},{"Start":"00:13.050 ","End":"00:16.590","Text":"which is this 1."},{"Start":"00:16.590 ","End":"00:22.830","Text":"What we have to do is to give a triple integral in"},{"Start":"00:22.830 ","End":"00:31.420","Text":"spherical coordinates that gives us the volume of this overlap of the 2 spheres."},{"Start":"00:31.520 ","End":"00:37.890","Text":"Let\u0027s convert these to spherical and we\u0027ll change them to inequalities."},{"Start":"00:37.890 ","End":"00:42.820","Text":"Here we\u0027ll have a less than or equal to."},{"Start":"00:43.250 ","End":"00:46.510","Text":"I guess here we\u0027ll have less than or equal to also."},{"Start":"00:46.510 ","End":"00:51.070","Text":"It\u0027s got to be contained in both of these spheres in the overlap."},{"Start":"00:51.070 ","End":"00:56.790","Text":"The first 1 is this and here I\u0027ll let you"},{"Start":"00:56.790 ","End":"01:02.575","Text":"follow the computations later and you can pause this."},{"Start":"01:02.575 ","End":"01:06.625","Text":"We get 2 Rho less than or equal to 2 cosine Phi."},{"Start":"01:06.625 ","End":"01:13.225","Text":"As for the other 1, it just comes out to Rho less than or equal to root 3."},{"Start":"01:13.225 ","End":"01:17.185","Text":"Not surprising because it\u0027s centered at the origin."},{"Start":"01:17.185 ","End":"01:19.960","Text":"If we compare the 2 Rhos,"},{"Start":"01:19.960 ","End":"01:22.030","Text":"equate them and solve,"},{"Start":"01:22.030 ","End":"01:24.385","Text":"we get Phi equals Pi over 6,"},{"Start":"01:24.385 ","End":"01:26.890","Text":"which is the Phi for this point."},{"Start":"01:26.890 ","End":"01:31.040","Text":"It\u0027s actually a whole circle from this point to this point."},{"Start":"01:31.040 ","End":"01:33.340","Text":"Anyway, this is the side view."},{"Start":"01:33.340 ","End":"01:37.340","Text":"To say that Rho is less than or equal to this and less than or equal to this"},{"Start":"01:37.340 ","End":"01:40.880","Text":"means that Rho is less than or equal to the minimum of these 2."},{"Start":"01:40.880 ","End":"01:43.760","Text":"We take whichever 1 is smaller."},{"Start":"01:43.760 ","End":"01:47.449","Text":"Now, in this angle here,"},{"Start":"01:47.449 ","End":"01:51.110","Text":"the large circle is the smaller 1."},{"Start":"01:51.110 ","End":"02:00.135","Text":"When we get outside of this angle in this part here and this part here, then take 2."},{"Start":"02:00.135 ","End":"02:05.990","Text":"Now when Phi is less than Pi over 6,"},{"Start":"02:05.990 ","End":"02:09.980","Text":"it\u0027s going to be in this part within this wedge."},{"Start":"02:09.980 ","End":"02:14.435","Text":"Then the smaller 1 of the 2 belongs to the large circle."},{"Start":"02:14.435 ","End":"02:20.540","Text":"Then beyond that were in the smaller circle here and which is the same as this."},{"Start":"02:20.540 ","End":"02:24.510","Text":"We break it up into 2 parts."},{"Start":"02:25.480 ","End":"02:29.495","Text":"I said, for Phi less than or equal to Pi over 6,"},{"Start":"02:29.495 ","End":"02:32.430","Text":"we take the root 3,"},{"Start":"02:32.430 ","End":"02:34.380","Text":"which is the large circle,"},{"Start":"02:34.380 ","End":"02:39.485","Text":"and for Phi bigger or equal to Pi over 6 in this part here or here."},{"Start":"02:39.485 ","End":"02:42.140","Text":"Then we take the other 1,"},{"Start":"02:42.140 ","End":"02:45.650","Text":"which is the 2 cosine Phi."},{"Start":"02:45.650 ","End":"02:47.870","Text":"Let\u0027s see what do we get?"},{"Start":"02:47.870 ","End":"02:50.510","Text":"We get that the volume is, well,"},{"Start":"02:50.510 ","End":"02:53.750","Text":"it\u0027s the same integrand,"},{"Start":"02:53.750 ","End":"02:55.640","Text":"It\u0027s just dx, dy,"},{"Start":"02:55.640 ","End":"02:59.250","Text":"dz in spherical, which is this."},{"Start":"02:59.250 ","End":"03:04.480","Text":"Then the limits are as follows."},{"Start":"03:04.820 ","End":"03:12.820","Text":"Take 2 in the 1 case where Phi is less than or equal to Pi over 6,"},{"Start":"03:12.820 ","End":"03:14.945","Text":"that\u0027s this part here."},{"Start":"03:14.945 ","End":"03:19.755","Text":"Rho goes from 0 to root 3."},{"Start":"03:19.755 ","End":"03:24.350","Text":"Then the other part where Phi is bigger or equal to"},{"Start":"03:24.350 ","End":"03:29.240","Text":"Pi over 6 means from Pi over 6 up to the maximum which is Pi over 2."},{"Start":"03:29.240 ","End":"03:36.315","Text":"Then the Rho goes from 0 to 2 cosine Phi,"},{"Start":"03:36.315 ","End":"03:40.770","Text":"that\u0027s the smaller circle."},{"Start":"03:40.770 ","End":"03:46.595","Text":"Theta is unrestricted in both cases, it\u0027s from 0 to 2Pi."},{"Start":"03:46.595 ","End":"03:49.729","Text":"We weren\u0027t asked to evaluate the integrals,"},{"Start":"03:49.729 ","End":"03:53.220","Text":"so basically we\u0027re done."}],"ID":21040}],"Thumbnail":null,"ID":113419}]

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1.1

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