[{"Name":"Introduction to Fourier Transform","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Absolute Integrability","Duration":"8m 19s","ChapterTopicVideoID":28198,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"Before we get into the Fourier transform,"},{"Start":"00:03.420 ","End":"00:06.569","Text":"there\u0027s a topic that may not have been covered,"},{"Start":"00:06.569 ","End":"00:08.220","Text":"want to make sure you know about it."},{"Start":"00:08.220 ","End":"00:13.420","Text":"It\u0027s called Absolute Integrability and the space G(R)."},{"Start":"00:13.420 ","End":"00:16.185","Text":"Let\u0027s start with the definition."},{"Start":"00:16.185 ","End":"00:19.005","Text":"Let D be domain in R,"},{"Start":"00:19.005 ","End":"00:20.820","Text":"typically this is an interval,"},{"Start":"00:20.820 ","End":"00:22.755","Text":"could be the whole real line."},{"Start":"00:22.755 ","End":"00:25.830","Text":"The function which could be complex or"},{"Start":"00:25.830 ","End":"00:31.800","Text":"real-valued and with domain D is called absolutely integrable."},{"Start":"00:31.800 ","End":"00:35.415","Text":"If the integral on D,"},{"Start":"00:35.415 ","End":"00:39.090","Text":"the absolute value of f is finite."},{"Start":"00:39.090 ","End":"00:41.315","Text":"In a moment, we\u0027ll give some examples."},{"Start":"00:41.315 ","End":"00:44.660","Text":"There\u0027s another concept that goes together with this,"},{"Start":"00:44.660 ","End":"00:46.820","Text":"which is piecewise continuity,"},{"Start":"00:46.820 ","End":"00:49.690","Text":"but I believe you\u0027ve studied that already."},{"Start":"00:49.690 ","End":"00:53.675","Text":"Let\u0027s get into the definition of the space."},{"Start":"00:53.675 ","End":"00:58.550","Text":"Sometimes it\u0027s called L^1_PC(R),"},{"Start":"00:58.550 ","End":"01:03.050","Text":"and it\u0027s often called G(R)."},{"Start":"01:03.050 ","End":"01:04.864","Text":"This is a vector space,"},{"Start":"01:04.864 ","End":"01:06.620","Text":"which is actually a normed space,"},{"Start":"01:06.620 ","End":"01:11.660","Text":"although we may not need the norm of piecewise continuous functions,"},{"Start":"01:11.660 ","End":"01:14.310","Text":"which are also absolutely integrable,"},{"Start":"01:14.310 ","End":"01:16.490","Text":"meaning what we said here."},{"Start":"01:16.490 ","End":"01:21.605","Text":"But d here is all the real line minus infinity to infinity."},{"Start":"01:21.605 ","End":"01:23.530","Text":"Again, this space,"},{"Start":"01:23.530 ","End":"01:24.970","Text":"which is also called this,"},{"Start":"01:24.970 ","End":"01:27.520","Text":"consists of functions that have 2 properties that"},{"Start":"01:27.520 ","End":"01:30.985","Text":"piecewise continuous and they are absolutely integrable."},{"Start":"01:30.985 ","End":"01:33.955","Text":"By the way, the norm like I said we may not use it,"},{"Start":"01:33.955 ","End":"01:38.455","Text":"is defined as a norm with a 1 here and the norm of f"},{"Start":"01:38.455 ","End":"01:43.255","Text":"is the integral from minus infinity to infinity of absolute value of f(x) dx."},{"Start":"01:43.255 ","End":"01:48.220","Text":"We know that this is finite because it has to be absolutely integrable."},{"Start":"01:48.220 ","End":"01:50.230","Text":"Should have said the norm on this space."},{"Start":"01:50.230 ","End":"01:52.090","Text":"But I think that\u0027s understood."},{"Start":"01:52.090 ","End":"01:53.920","Text":"That\u0027s it for definitions."},{"Start":"01:53.920 ","End":"01:55.900","Text":"Now let\u0027s do some examples."},{"Start":"01:55.900 ","End":"01:57.970","Text":"Here\u0027s an example exercise,"},{"Start":"01:57.970 ","End":"01:59.800","Text":"which is really 4 in 1."},{"Start":"01:59.800 ","End":"02:02.135","Text":"For each of these functions,"},{"Start":"02:02.135 ","End":"02:05.240","Text":"well, they\u0027re all called f, but they\u0027re separate."},{"Start":"02:05.240 ","End":"02:13.435","Text":"We have to check if the f here is a member of this space G(R) or L^1_PC(R)."},{"Start":"02:13.435 ","End":"02:16.280","Text":"By the way, the PC stands for piecewise continuous,"},{"Start":"02:16.280 ","End":"02:18.155","Text":"the 1 I won\u0027t get into that here."},{"Start":"02:18.155 ","End":"02:19.640","Text":"L is for the bag."},{"Start":"02:19.640 ","End":"02:20.750","Text":"These are the 4 functions,"},{"Start":"02:20.750 ","End":"02:22.190","Text":"I\u0027ll take 1 at a time."},{"Start":"02:22.190 ","End":"02:26.195","Text":"The first one, 1 over 1 plus x^2."},{"Start":"02:26.195 ","End":"02:28.070","Text":"This is positive."},{"Start":"02:28.070 ","End":"02:30.875","Text":"Note that, and it\u0027s continuous."},{"Start":"02:30.875 ","End":"02:35.180","Text":"Since it\u0027s continuous, in particular, it\u0027s piecewise continuous."},{"Start":"02:35.180 ","End":"02:37.055","Text":"That\u0027s one of the properties."},{"Start":"02:37.055 ","End":"02:41.140","Text":"Now we want the other one, absolute integrability."},{"Start":"02:41.140 ","End":"02:44.060","Text":"Let\u0027s take the integral from minus infinity to infinity of"},{"Start":"02:44.060 ","End":"02:47.660","Text":"the absolute value of f. Now because this is positive,"},{"Start":"02:47.660 ","End":"02:49.985","Text":"we can throw out the absolute value."},{"Start":"02:49.985 ","End":"02:54.350","Text":"The integral of 1 over 1 plus x^2 is known to be arctangent of x."},{"Start":"02:54.350 ","End":"02:57.085","Text":"We want this from minus infinity to infinity,"},{"Start":"02:57.085 ","End":"03:01.055","Text":"and the arctangent, I remind you it looks like this."},{"Start":"03:01.055 ","End":"03:03.215","Text":"When x goes to infinity,"},{"Start":"03:03.215 ","End":"03:05.870","Text":"this thing goes to Pi over 2."},{"Start":"03:05.870 ","End":"03:09.560","Text":"Basically because when y goes to Pi over 2,"},{"Start":"03:09.560 ","End":"03:11.975","Text":"tangent y goes to infinity."},{"Start":"03:11.975 ","End":"03:16.265","Text":"Similarly here, minus Pi over 2 and minus infinity."},{"Start":"03:16.265 ","End":"03:20.525","Text":"What we get is Pi over 2 minus minus Pi over 2,"},{"Start":"03:20.525 ","End":"03:25.950","Text":"and that is equal to Pi which is certainly finite."},{"Start":"03:25.950 ","End":"03:30.485","Text":"So f is absolutely integrable and it\u0027s piecewise continuous."},{"Start":"03:30.485 ","End":"03:36.500","Text":"Therefore, it belongs to the space G(R) or L^1_PC(R)."},{"Start":"03:36.500 ","End":"03:39.650","Text":"Now we come to the second question,"},{"Start":"03:39.650 ","End":"03:41.270","Text":"which is this function."},{"Start":"03:41.270 ","End":"03:42.739","Text":"We need to check 2 things,"},{"Start":"03:42.739 ","End":"03:46.685","Text":"piecewise continuous and absolutely integrable."},{"Start":"03:46.685 ","End":"03:50.435","Text":"This is continuous, so piecewise continuous."},{"Start":"03:50.435 ","End":"03:54.980","Text":"That\u0027s easy. Note that f is an odd function."},{"Start":"03:54.980 ","End":"03:58.250","Text":"Absolute value of f is an even function."},{"Start":"03:58.250 ","End":"04:01.910","Text":"If we take the integral for minus infinity to infinity,"},{"Start":"04:01.910 ","End":"04:06.025","Text":"you can replace it by twice the integral from 0 to infinity."},{"Start":"04:06.025 ","End":"04:11.480","Text":"Now, x over 1 plus x^2 is positive when x is positive."},{"Start":"04:11.480 ","End":"04:16.295","Text":"So we can throw out the absolute value and put the 2 here."},{"Start":"04:16.295 ","End":"04:20.660","Text":"Because then the numerator is the derivative of the denominator."},{"Start":"04:20.660 ","End":"04:26.810","Text":"When this happens, then the integral is the natural logarithm of the denominator."},{"Start":"04:26.810 ","End":"04:31.310","Text":"Yeah, natural log of 1 plus x^2 from 0 to infinity."},{"Start":"04:31.310 ","End":"04:35.270","Text":"We can symbolically say it\u0027s 1 plus infinity^2."},{"Start":"04:35.270 ","End":"04:37.355","Text":"It\u0027s clearly infinity."},{"Start":"04:37.355 ","End":"04:39.395","Text":"This is symbolic, like I said."},{"Start":"04:39.395 ","End":"04:42.200","Text":"Because this integral is not finite,"},{"Start":"04:42.200 ","End":"04:45.485","Text":"then f is not absolutely integrable."},{"Start":"04:45.485 ","End":"04:51.920","Text":"In conclusion, we can say that f is not in the space G(R)."},{"Start":"04:51.920 ","End":"04:53.940","Text":"Also written like this."},{"Start":"04:53.940 ","End":"04:55.530","Text":"That\u0027s part 2."},{"Start":"04:55.530 ","End":"04:58.605","Text":"Let\u0027s move on to part 3."},{"Start":"04:58.605 ","End":"05:01.725","Text":"This is the function f(x)."},{"Start":"05:01.725 ","End":"05:03.795","Text":"It\u0027s 1 over x^2,"},{"Start":"05:03.795 ","End":"05:06.300","Text":"but only when x is bigger than 1."},{"Start":"05:06.300 ","End":"05:09.600","Text":"Elsewhere, it\u0027s just 0."},{"Start":"05:09.600 ","End":"05:12.575","Text":"It looks like there\u0027s a jump discontinuity here."},{"Start":"05:12.575 ","End":"05:16.490","Text":"Now, f is continuous on each of these pieces"},{"Start":"05:16.490 ","End":"05:20.570","Text":"from minus infinity to 1 and from 1 to infinity."},{"Start":"05:20.570 ","End":"05:27.074","Text":"Just a question of what happens here and for piecewise continuity,"},{"Start":"05:27.074 ","End":"05:31.610","Text":"for the limit of the function to exist on the right and on the left,"},{"Start":"05:31.610 ","End":"05:34.175","Text":"they don\u0027t have to be equal, but they both have to exist."},{"Start":"05:34.175 ","End":"05:38.399","Text":"Indeed they do. When x goes to 1 from the left, it\u0027s 0."},{"Start":"05:38.399 ","End":"05:40.530","Text":"When x goes to 1 from the right,"},{"Start":"05:40.530 ","End":"05:44.475","Text":"we get to 1 because it\u0027s 1 over 1^2."},{"Start":"05:44.475 ","End":"05:47.180","Text":"F is piecewise continuous."},{"Start":"05:47.180 ","End":"05:49.280","Text":"Now, what about the absolute integrability?"},{"Start":"05:49.280 ","End":"05:50.900","Text":"We will check this integral."},{"Start":"05:50.900 ","End":"05:54.920","Text":"This comes out to be the integral from 1 to infinity because yeah,"},{"Start":"05:54.920 ","End":"05:57.265","Text":"we just need the part where it\u0027s not 0,"},{"Start":"05:57.265 ","End":"05:59.850","Text":"1 over x^2 dx positive,"},{"Start":"05:59.850 ","End":"06:01.835","Text":"so we don\u0027t need the absolute value even,"},{"Start":"06:01.835 ","End":"06:05.015","Text":"and this integral is minus 1 over x."},{"Start":"06:05.015 ","End":"06:07.310","Text":"We substitute 1 and infinity."},{"Start":"06:07.310 ","End":"06:11.390","Text":"Symbolically, we get minus 1 over infinity."},{"Start":"06:11.390 ","End":"06:14.310","Text":"We have 0 plus 1 over 1."},{"Start":"06:14.310 ","End":"06:18.030","Text":"That gives us one which is finite."},{"Start":"06:18.030 ","End":"06:20.680","Text":"F is absolutely integrable."},{"Start":"06:20.680 ","End":"06:25.265","Text":"We already said it\u0027s piecewise continuous so both conditions are met."},{"Start":"06:25.265 ","End":"06:28.490","Text":"F is in this space,"},{"Start":"06:28.490 ","End":"06:31.040","Text":"and that was part 3."},{"Start":"06:31.040 ","End":"06:34.625","Text":"Now we come to part 4 the last part."},{"Start":"06:34.625 ","End":"06:39.395","Text":"This was the function and show you a graph of it."},{"Start":"06:39.395 ","End":"06:43.955","Text":"It\u0027s 1 over twice the square root of x here,"},{"Start":"06:43.955 ","End":"06:46.895","Text":"from 0-1 and 0 elsewhere."},{"Start":"06:46.895 ","End":"06:51.008","Text":"For a change, we\u0027ll first check the absolute integrability. Let\u0027s see."},{"Start":"06:51.008 ","End":"06:55.415","Text":"This integral of absolute value of f minus infinity to infinity."},{"Start":"06:55.415 ","End":"06:59.090","Text":"We only have to take the part from 0-1 because outside of"},{"Start":"06:59.090 ","End":"07:04.090","Text":"that it\u0027s 0 and here it\u0027s 1 over twice the square root of x."},{"Start":"07:04.090 ","End":"07:09.545","Text":"We know the integral of 1 over twice root x is just root x."},{"Start":"07:09.545 ","End":"07:12.365","Text":"We have to evaluate this from 0-1."},{"Start":"07:12.365 ","End":"07:14.915","Text":"It\u0027s root 1 minus root 0,"},{"Start":"07:14.915 ","End":"07:18.245","Text":"which is 1 and it\u0027s finite."},{"Start":"07:18.245 ","End":"07:20.840","Text":"That means that f is absolutely integrable."},{"Start":"07:20.840 ","End":"07:24.620","Text":"The other part is the piecewise continuous."},{"Start":"07:24.620 ","End":"07:29.855","Text":"Now, it is continuous on each of the 3 pieces minus infinity to 0,"},{"Start":"07:29.855 ","End":"07:33.690","Text":"from 0-1, and from 1 to infinity."},{"Start":"07:33.690 ","End":"07:36.740","Text":"What happens at the point 0 and 1? We have to check it."},{"Start":"07:36.740 ","End":"07:39.245","Text":"It has left and right-handed limits."},{"Start":"07:39.245 ","End":"07:41.930","Text":"Well, at x=1 we\u0027re okay."},{"Start":"07:41.930 ","End":"07:46.040","Text":"Because on this side it\u0027s 0 and from this side it\u0027s 1."},{"Start":"07:46.040 ","End":"07:48.485","Text":"But what about x=0?"},{"Start":"07:48.485 ","End":"07:52.790","Text":"Well, limit from the left is okay, that\u0027s 0,"},{"Start":"07:52.790 ","End":"07:55.850","Text":"but the limit from the right is infinity,"},{"Start":"07:55.850 ","End":"07:58.430","Text":"which is not considered to be a limit,"},{"Start":"07:58.430 ","End":"08:02.120","Text":"not for the purposes of piecewise continuity."},{"Start":"08:02.120 ","End":"08:04.745","Text":"That means because of this,"},{"Start":"08:04.745 ","End":"08:08.435","Text":"there\u0027s no right limit and hence it\u0027s not piecewise continuous."},{"Start":"08:08.435 ","End":"08:10.250","Text":"Because it failed on this,"},{"Start":"08:10.250 ","End":"08:12.830","Text":"f does not belong to R space,"},{"Start":"08:12.830 ","End":"08:15.440","Text":"also known as G(R)."},{"Start":"08:15.440 ","End":"08:19.920","Text":"That concludes part 4 and we are done."}],"ID":29427},{"Watched":false,"Name":"Fourier Transform - Definition","Duration":"5m 51s","ChapterTopicVideoID":28199,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.970","Text":"Now a new topic, The Fourier Transform."},{"Start":"00:02.970 ","End":"00:06.600","Text":"In this clip we\u0027ll give a definition as well as an example."},{"Start":"00:06.600 ","End":"00:10.995","Text":"Let f be a function defined on all the reals,"},{"Start":"00:10.995 ","End":"00:14.070","Text":"and it can be complex-valued or real-valued."},{"Start":"00:14.070 ","End":"00:17.880","Text":"It has to be piecewise continuous and absolutely integrable."},{"Start":"00:17.880 ","End":"00:21.345","Text":"In case you\u0027ve forgotten what absolutely integrable means,"},{"Start":"00:21.345 ","End":"00:24.060","Text":"it means that the integral from minus infinity to"},{"Start":"00:24.060 ","End":"00:28.020","Text":"infinity of the absolute value is finite."},{"Start":"00:28.020 ","End":"00:31.785","Text":"We can also say that f belongs to g(r)."},{"Start":"00:31.785 ","End":"00:36.179","Text":"That\u0027s what it means to be in piecewise continuous and absolutely integrable."},{"Start":"00:36.179 ","End":"00:38.415","Text":"Now we\u0027re going to define another function,"},{"Start":"00:38.415 ","End":"00:42.825","Text":"usually denoted as f with a carrot like a hat on it."},{"Start":"00:42.825 ","End":"00:48.635","Text":"We define it as f^ of a different variable, Omega,"},{"Start":"00:48.635 ","End":"00:55.460","Text":"is 1/2Pi, the integral of f(x) e_minus i Omega x dx."},{"Start":"00:55.460 ","End":"00:59.040","Text":"This new function f^ is called the Fourier"},{"Start":"00:59.040 ","End":"01:03.155","Text":"transform of f. Some people I\u0027ve seen in the Wikipedia,"},{"Start":"01:03.155 ","End":"01:05.405","Text":"they used the Greek letter XI,"},{"Start":"01:05.405 ","End":"01:08.870","Text":"instead of the Greek letter Omega, which is here."},{"Start":"01:08.870 ","End":"01:12.470","Text":"In physics, if x represents time,"},{"Start":"01:12.470 ","End":"01:16.055","Text":"then Omega represents frequency in the inverse units."},{"Start":"01:16.055 ","End":"01:19.055","Text":"So that if x is measured in seconds,"},{"Start":"01:19.055 ","End":"01:21.410","Text":"then Omega is measured in hertz,"},{"Start":"01:21.410 ","End":"01:23.435","Text":"which means per second."},{"Start":"01:23.435 ","End":"01:25.085","Text":"Let\u0027s do an example."},{"Start":"01:25.085 ","End":"01:31.355","Text":"Find the Fourier transforms of the function f(x)=e to the minus absolute value of x."},{"Start":"01:31.355 ","End":"01:38.750","Text":"Let\u0027s first show that f really is piecewise continuous and absolutely integrable."},{"Start":"01:38.750 ","End":"01:40.280","Text":"Well, it\u0027s continuous,"},{"Start":"01:40.280 ","End":"01:43.674","Text":"so piecewise continuous and absolutely integrable."},{"Start":"01:43.674 ","End":"01:45.200","Text":"Let\u0027s do the computation."},{"Start":"01:45.200 ","End":"01:51.755","Text":"The integral of absolute value of f. Because this is an even function,"},{"Start":"01:51.755 ","End":"01:55.009","Text":"many of you replace x by minus x, makes no difference."},{"Start":"01:55.009 ","End":"01:58.685","Text":"Then we can take twice the integral from 0 to infinity."},{"Start":"01:58.685 ","End":"02:03.205","Text":"When x is positive, it\u0027s e_minus x."},{"Start":"02:03.205 ","End":"02:08.355","Text":"This is equal to, integral of e_minus x is minus e_minus x."},{"Start":"02:08.355 ","End":"02:11.915","Text":"We get minus 2e_minus x from 0 to infinity."},{"Start":"02:11.915 ","End":"02:15.500","Text":"Symbolically, can plug in infinity,"},{"Start":"02:15.500 ","End":"02:19.220","Text":"and 0, because a minus infinity really means"},{"Start":"02:19.220 ","End":"02:23.390","Text":"e_minus x as x goes to infinity, but that is 0."},{"Start":"02:23.390 ","End":"02:26.000","Text":"Here we just have minus 2 times minus 1,"},{"Start":"02:26.000 ","End":"02:29.210","Text":"which is 2, which is certainly finite."},{"Start":"02:29.210 ","End":"02:33.140","Text":"Now that we\u0027ve confirmed that f really is a valid function,"},{"Start":"02:33.140 ","End":"02:36.815","Text":"meets the conditions, then we can do the computation."},{"Start":"02:36.815 ","End":"02:40.610","Text":"Let\u0027s start computing f^ of Omega."},{"Start":"02:40.610 ","End":"02:42.710","Text":"By the definition, this is what it is."},{"Start":"02:42.710 ","End":"02:47.075","Text":"In our case, f(x) is e to the minus absolute value of x."},{"Start":"02:47.075 ","End":"02:49.370","Text":"Now we can break this up into 2 integrals,"},{"Start":"02:49.370 ","End":"02:50.990","Text":"from minus infinity to 0,"},{"Start":"02:50.990 ","End":"02:52.639","Text":"and from 0 to infinity."},{"Start":"02:52.639 ","End":"02:57.980","Text":"We do that because absolute value of x changes definition between positive and negative,"},{"Start":"02:57.980 ","End":"03:02.045","Text":"from minus infinity to 0 and 0 to infinity."},{"Start":"03:02.045 ","End":"03:05.510","Text":"On this part absolute value of x is minus x,"},{"Start":"03:05.510 ","End":"03:08.000","Text":"so this becomes just plus x."},{"Start":"03:08.000 ","End":"03:09.860","Text":"Here, absolute value of x is x,"},{"Start":"03:09.860 ","End":"03:12.155","Text":"so it\u0027s e_minus x."},{"Start":"03:12.155 ","End":"03:15.454","Text":"Now we can just use the rules of exponents."},{"Start":"03:15.454 ","End":"03:17.510","Text":"We add the exponents."},{"Start":"03:17.510 ","End":"03:20.735","Text":"Here also, the product means that we add the exponents."},{"Start":"03:20.735 ","End":"03:23.315","Text":"We can take x out the brackets."},{"Start":"03:23.315 ","End":"03:30.205","Text":"The integral of e_x times something is e_x times something over that something."},{"Start":"03:30.205 ","End":"03:35.635","Text":"Similarly here, e_x times this over this."},{"Start":"03:35.635 ","End":"03:37.518","Text":"In each case you have 1/2Pi,"},{"Start":"03:37.518 ","End":"03:40.930","Text":"and we have to take the integral here from minus infinity to 0,"},{"Start":"03:40.930 ","End":"03:42.850","Text":"here from 0 to infinity."},{"Start":"03:42.850 ","End":"03:46.120","Text":"We substitute 0 and minus infinity."},{"Start":"03:46.120 ","End":"03:49.495","Text":"Or strictly speaking, you have to take a limit."},{"Start":"03:49.495 ","End":"03:52.165","Text":"When we plug in 0, it\u0027s e_0, which is 1."},{"Start":"03:52.165 ","End":"03:57.700","Text":"Minus infinity really means the limit as x goes to minus infinity this."},{"Start":"03:57.700 ","End":"04:02.380","Text":"Similarly here, 0, we just get 1."},{"Start":"04:02.380 ","End":"04:04.675","Text":"But infinity means the limit."},{"Start":"04:04.675 ","End":"04:06.955","Text":"Let\u0027s compute these 2 limits."},{"Start":"04:06.955 ","End":"04:10.580","Text":"First, this limit. We\u0027ll show that this is 0."},{"Start":"04:10.580 ","End":"04:12.410","Text":"If the limit of the absolute value is 0,"},{"Start":"04:12.410 ","End":"04:15.890","Text":"then the limit of the thing itself is 0. This is equal too."},{"Start":"04:15.890 ","End":"04:20.585","Text":"We can break it up into the product e_x, e_minus i Omega x."},{"Start":"04:20.585 ","End":"04:22.865","Text":"Each of the x is positive real,"},{"Start":"04:22.865 ","End":"04:25.490","Text":"so we can take it out of the absolute value."},{"Start":"04:25.490 ","End":"04:32.180","Text":"The absolute value of e to the something imaginary i times something real is always 1."},{"Start":"04:32.180 ","End":"04:33.845","Text":"It\u0027s on the unit circle."},{"Start":"04:33.845 ","End":"04:37.460","Text":"What we get is just e_x,"},{"Start":"04:37.460 ","End":"04:39.365","Text":"as x goes to minus infinity."},{"Start":"04:39.365 ","End":"04:41.270","Text":"This is 0, you can say symbolically,"},{"Start":"04:41.270 ","End":"04:42.725","Text":"it\u0027s e to the minus infinity,"},{"Start":"04:42.725 ","End":"04:45.665","Text":"and e to the minus infinity is 0."},{"Start":"04:45.665 ","End":"04:48.320","Text":"As for the other one, same thing."},{"Start":"04:48.320 ","End":"04:50.090","Text":"We can break it up into 2 pieces,"},{"Start":"04:50.090 ","End":"04:53.624","Text":"e_minus x, and e_i something."},{"Start":"04:53.624 ","End":"04:56.075","Text":"This is 1 in absolute value."},{"Start":"04:56.075 ","End":"05:01.775","Text":"We get e_minus x as x goes to plus infinity."},{"Start":"05:01.775 ","End":"05:04.040","Text":"But it\u0027s still e to the minus infinity."},{"Start":"05:04.040 ","End":"05:06.859","Text":"Unlike here, it\u0027s equal to 0."},{"Start":"05:06.859 ","End":"05:09.380","Text":"Now we can go and substitute back here,"},{"Start":"05:09.380 ","End":"05:11.920","Text":"where we have these limits we put 0."},{"Start":"05:11.920 ","End":"05:13.380","Text":"Now we can combine them,"},{"Start":"05:13.380 ","End":"05:17.194","Text":"1/2Pi times 1/1 minus Omega."},{"Start":"05:17.194 ","End":"05:19.940","Text":"Here, we can get rid of the minus, here, here, and here."},{"Start":"05:19.940 ","End":"05:22.470","Text":"It\u0027s 1/1 plus i Omega,"},{"Start":"05:22.470 ","End":"05:28.430","Text":"cross multiply 1 plus i Omega plus 1 minus i Omega, is 2."},{"Start":"05:28.430 ","End":"05:34.205","Text":"The product of these using difference of squares is 1^2 plus i^2 Omega squared,"},{"Start":"05:34.205 ","End":"05:39.680","Text":"which means minus i^2 Omega squared,"},{"Start":"05:39.680 ","End":"05:41.135","Text":"which is plus Omega squared."},{"Start":"05:41.135 ","End":"05:42.829","Text":"Then the 2 with the 2 cancels,"},{"Start":"05:42.829 ","End":"05:43.940","Text":"put the Pi in here,"},{"Start":"05:43.940 ","End":"05:45.770","Text":"and we can say that the answer is this."},{"Start":"05:45.770 ","End":"05:49.385","Text":"This is what the Fourier transform of f is."},{"Start":"05:49.385 ","End":"05:51.900","Text":"We are done."}],"ID":29428},{"Watched":false,"Name":"Fourier Transform – Properties","Duration":"4m 35s","ChapterTopicVideoID":28200,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.470","Text":"This clip is about some of the properties of the Fourier transform."},{"Start":"00:04.470 ","End":"00:06.150","Text":"First, a couple of notes."},{"Start":"00:06.150 ","End":"00:11.820","Text":"Assume all the functions are in G(R) which is also known as this,"},{"Start":"00:11.820 ","End":"00:15.930","Text":"say functions of x as opposed to functions of Omega."},{"Start":"00:15.930 ","End":"00:19.095","Text":"Other note is that the Fourier transform,"},{"Start":"00:19.095 ","End":"00:24.765","Text":"you can view it as an operator that takes a function to its Fourier transform."},{"Start":"00:24.765 ","End":"00:27.585","Text":"We could say f(f) is f^."},{"Start":"00:27.585 ","End":"00:31.740","Text":"Also, a reminder what the Fourier transform is, we have f,"},{"Start":"00:31.740 ","End":"00:35.310","Text":"then f^ the function of Omega which is defined thus,"},{"Start":"00:35.310 ","End":"00:36.780","Text":"and here\u0027s the Omega."},{"Start":"00:36.780 ","End":"00:40.165","Text":"First property is linearity."},{"Start":"00:40.165 ","End":"00:45.260","Text":"If we take the Fourier transform of Alpha f plus Beta g,"},{"Start":"00:45.260 ","End":"00:51.346","Text":"I should have said the Alpha and Beta are scalars and f and g are functions,"},{"Start":"00:51.346 ","End":"00:53.480","Text":"then we just take Alpha times"},{"Start":"00:53.480 ","End":"00:57.485","Text":"the Fourier transform of f plus Beta times the Fourier transform of g."},{"Start":"00:57.485 ","End":"01:00.710","Text":"That\u0027s one of the reasons I showed you the notation with the"},{"Start":"01:00.710 ","End":"01:04.760","Text":"F. We don\u0027t have to use this notation."},{"Start":"01:04.760 ","End":"01:09.470","Text":"We could say that if h is Alpha f plus Beta g,"},{"Start":"01:09.470 ","End":"01:13.700","Text":"then h^ is Alpha f^ plus Beta g^."},{"Start":"01:13.700 ","End":"01:15.200","Text":"We could also, I suppose,"},{"Start":"01:15.200 ","End":"01:16.440","Text":"write it this way,"},{"Start":"01:16.440 ","End":"01:19.725","Text":"that Alpha f plus Beta g all hat."},{"Start":"01:19.725 ","End":"01:21.630","Text":"But it looks funny that way."},{"Start":"01:21.630 ","End":"01:24.619","Text":"Anyway, let\u0027s get onto the next property."},{"Start":"01:24.619 ","End":"01:26.630","Text":"F^ is continuous."},{"Start":"01:26.630 ","End":"01:29.705","Text":"Whenever f is in this space,"},{"Start":"01:29.705 ","End":"01:33.219","Text":"then the transform Fourier of f is continuous."},{"Start":"01:33.219 ","End":"01:36.045","Text":"Another property due to Riemann-Lebesgue,"},{"Start":"01:36.045 ","End":"01:43.055","Text":"the limit as Omega goes to infinity or minus infinity of f^ of Omega is 0."},{"Start":"01:43.055 ","End":"01:46.310","Text":"Next property, if f is real,"},{"Start":"01:46.310 ","End":"01:48.785","Text":"then the following property holds."},{"Start":"01:48.785 ","End":"01:53.630","Text":"If you take the transform of f and then take its conjugate,"},{"Start":"01:53.630 ","End":"01:57.380","Text":"it\u0027s like taking the transform of minus Omega."},{"Start":"01:57.380 ","End":"02:00.410","Text":"This is fairly easy to prove. Let\u0027s do it."},{"Start":"02:00.410 ","End":"02:03.005","Text":"The first thing is to use the definition."},{"Start":"02:03.005 ","End":"02:07.730","Text":"The Fourier transform of f is given by this formula."},{"Start":"02:07.730 ","End":"02:10.835","Text":"Just conjugate this and conjugate this."},{"Start":"02:10.835 ","End":"02:14.020","Text":"Now, the 1 over 2 Pi is real,"},{"Start":"02:14.020 ","End":"02:17.090","Text":"and the conjugate will go inside the integral,"},{"Start":"02:17.090 ","End":"02:18.770","Text":"but also f is real,"},{"Start":"02:18.770 ","End":"02:21.275","Text":"so we only have to conjugate this part."},{"Start":"02:21.275 ","End":"02:27.435","Text":"The conjugate of e^minus i Omega x is just e^plus i Omega x."},{"Start":"02:27.435 ","End":"02:31.535","Text":"We can write Omega as minus minus Omega."},{"Start":"02:31.535 ","End":"02:33.605","Text":"We could have it like this."},{"Start":"02:33.605 ","End":"02:39.535","Text":"This corresponds exactly to the Fourier transform of minus Omega."},{"Start":"02:39.535 ","End":"02:42.545","Text":"That\u0027s what it is and that\u0027s what we had to prove."},{"Start":"02:42.545 ","End":"02:44.600","Text":"That\u0027s Property 4."},{"Start":"02:44.600 ","End":"02:46.850","Text":"Now Property 5."},{"Start":"02:46.850 ","End":"02:49.535","Text":"If f is an even function,"},{"Start":"02:49.535 ","End":"02:53.420","Text":"then the transform of f is given by this formula,"},{"Start":"02:53.420 ","End":"02:57.185","Text":"1 over Pi integral from 0 to infinity of f(x),"},{"Start":"02:57.185 ","End":"02:59.195","Text":"the exponent we have cosine."},{"Start":"02:59.195 ","End":"03:00.815","Text":"That\u0027s also easy to prove."},{"Start":"03:00.815 ","End":"03:04.420","Text":"The definition of f hat is like this."},{"Start":"03:04.420 ","End":"03:11.935","Text":"E^minus i Omega x can be written as cosine of Omega x minus i sine Omega x."},{"Start":"03:11.935 ","End":"03:16.395","Text":"Then we can substitute that here."},{"Start":"03:16.395 ","End":"03:18.774","Text":"You get that this equals the following."},{"Start":"03:18.774 ","End":"03:23.185","Text":"Just see we break it up the e^minus Omega x into this bit and this bit."},{"Start":"03:23.185 ","End":"03:26.530","Text":"Now because f is even and cosine is even,"},{"Start":"03:26.530 ","End":"03:28.735","Text":"f times cosine is even,"},{"Start":"03:28.735 ","End":"03:33.550","Text":"and also f is even and sine is odd so F times sine is odd."},{"Start":"03:33.550 ","End":"03:36.190","Text":"We know what happens when we integrate"},{"Start":"03:36.190 ","End":"03:38.695","Text":"on a symmetric interval and even or an odd function."},{"Start":"03:38.695 ","End":"03:39.985","Text":"In the case of even,"},{"Start":"03:39.985 ","End":"03:43.665","Text":"it\u0027s just double the integral from 0 to infinity."},{"Start":"03:43.665 ","End":"03:46.245","Text":"The 2 cancels with the 2 and it\u0027s 1 over Pi."},{"Start":"03:46.245 ","End":"03:47.958","Text":"In case of an odd function,"},{"Start":"03:47.958 ","End":"03:50.050","Text":"the integral is 0 on a symmetric interval,"},{"Start":"03:50.050 ","End":"03:52.300","Text":"so this part just drops out."},{"Start":"03:52.300 ","End":"03:55.200","Text":"That\u0027s the proof of Part 5."},{"Start":"03:55.200 ","End":"03:57.280","Text":"Then Part 6 is similar,"},{"Start":"03:57.280 ","End":"04:01.250","Text":"but involves an odd function. Here it is."},{"Start":"04:01.370 ","End":"04:04.234","Text":"The proof is very similar."},{"Start":"04:04.234 ","End":"04:08.350","Text":"I\u0027ll just note that it\u0027s got a minus i over Pi here."},{"Start":"04:08.350 ","End":"04:11.810","Text":"Again, we start the same way as we did before and make"},{"Start":"04:11.810 ","End":"04:14.840","Text":"the same substitution, only this time,"},{"Start":"04:14.840 ","End":"04:19.060","Text":"this part is the odd function and this part is the even function,"},{"Start":"04:19.060 ","End":"04:20.660","Text":"so this drops out,"},{"Start":"04:20.660 ","End":"04:24.560","Text":"and this we can double make it from 0 to infinity."},{"Start":"04:24.560 ","End":"04:28.370","Text":"The 2 cancels with the 2 and we have minus i 1 over Pi,"},{"Start":"04:28.370 ","End":"04:31.055","Text":"the integral from 0 to infinity of this."},{"Start":"04:31.055 ","End":"04:35.430","Text":"That\u0027s the last property. We are done."}],"ID":29429},{"Watched":false,"Name":"Fourier Transform – Alternatve Definitions and More Properties","Duration":"5m 24s","ChapterTopicVideoID":28201,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.440","Text":"In this video, we\u0027ll talk about some alternate definitions of"},{"Start":"00:04.440 ","End":"00:09.825","Text":"the Fourier transform and also some more properties of the transform."},{"Start":"00:09.825 ","End":"00:13.440","Text":"Now, our definition is Definition 1."},{"Start":"00:13.440 ","End":"00:15.810","Text":"I probably should highlight it."},{"Start":"00:15.810 ","End":"00:18.990","Text":"That\u0027s the one we\u0027ll be working with in this course."},{"Start":"00:18.990 ","End":"00:22.980","Text":"But there are others and there are only two others,"},{"Start":"00:22.980 ","End":"00:26.715","Text":"these are the three most popular ones, I would say."},{"Start":"00:26.715 ","End":"00:30.690","Text":"Definition 2 is the same as Definition 1,"},{"Start":"00:30.690 ","End":"00:34.545","Text":"except it doesn\u0027t have the 1/2Pi here."},{"Start":"00:34.545 ","End":"00:37.320","Text":"Definition 3, which some use,"},{"Start":"00:37.320 ","End":"00:39.990","Text":"and there\u0027s reasons for this but I won\u0027t go into it,"},{"Start":"00:39.990 ","End":"00:44.670","Text":"has a 1 over square root 2Pi in front."},{"Start":"00:44.670 ","End":"00:49.175","Text":"Only differ by this constant that comes in front of the integral."},{"Start":"00:49.175 ","End":"00:53.840","Text":"As some of the properties we learned about holds for all the definitions,"},{"Start":"00:53.840 ","End":"00:55.940","Text":"but some of them depend on the definitions."},{"Start":"00:55.940 ","End":"01:00.245","Text":"Let\u0027s start with properties that don\u0027t depend on the definition."},{"Start":"01:00.245 ","End":"01:03.110","Text":"Here are some of the ones we\u0027ve talked about already."},{"Start":"01:03.110 ","End":"01:08.615","Text":"Linearity and complex conjugation this one is called."},{"Start":"01:08.615 ","End":"01:10.400","Text":"These two we have discussed already,"},{"Start":"01:10.400 ","End":"01:12.050","Text":"now there\u0027s some more."},{"Start":"01:12.050 ","End":"01:16.010","Text":"These are more that are not effected by which three definitions."},{"Start":"01:16.010 ","End":"01:19.640","Text":"There is one called scaling and shifting can actually get"},{"Start":"01:19.640 ","End":"01:23.225","Text":"formula for scaling separate and shifting separately but here it\u0027s all combined."},{"Start":"01:23.225 ","End":"01:26.585","Text":"A is the scaling factor and b is the shifting amount."},{"Start":"01:26.585 ","End":"01:31.655","Text":"Then we have these formulas for cosine or sine modulation."},{"Start":"01:31.655 ","End":"01:36.485","Text":"If you take the function and multiply it by a cosine Omega Naught x."},{"Start":"01:36.485 ","End":"01:41.120","Text":"Then what happens is in the frequency domain,"},{"Start":"01:41.120 ","End":"01:43.250","Text":"instead of f^ of Omega,"},{"Start":"01:43.250 ","End":"01:49.250","Text":"we get the average of Omega minus Omega Naught and Omega plus Omega Naught,"},{"Start":"01:49.250 ","End":"01:52.460","Text":"the average of the function."},{"Start":"01:52.460 ","End":"01:55.670","Text":"If we put a sine here instead of a cosine,"},{"Start":"01:55.670 ","End":"01:58.190","Text":"we get something similar but different."},{"Start":"01:58.190 ","End":"02:01.550","Text":"Instead of the plus, we have a minus and we have an extra i here."},{"Start":"02:01.550 ","End":"02:05.210","Text":"Then there\u0027s a formula for differentiation."},{"Start":"02:05.210 ","End":"02:10.355","Text":"If you differentiate a function n times and take the Fourier transform,"},{"Start":"02:10.355 ","End":"02:13.340","Text":"it\u0027s like taking the Fourier transform of"},{"Start":"02:13.340 ","End":"02:18.965","Text":"the original function and then multiplying it by i Omega^n."},{"Start":"02:18.965 ","End":"02:22.145","Text":"Should\u0027ve said most these are for reference,"},{"Start":"02:22.145 ","End":"02:24.080","Text":"just so you have an idea."},{"Start":"02:24.080 ","End":"02:28.015","Text":"I mean, there\u0027ll be there when you need them and you may not need them."},{"Start":"02:28.015 ","End":"02:30.280","Text":"Yeah, differentiation, next,"},{"Start":"02:30.280 ","End":"02:33.020","Text":"multiplication by x to the n. If you take your function,"},{"Start":"02:33.020 ","End":"02:34.430","Text":"multiply it back to the n,"},{"Start":"02:34.430 ","End":"02:36.740","Text":"what happens to the Fourier transform?"},{"Start":"02:36.740 ","End":"02:41.930","Text":"Well, it gets differentiated and times and also multiplied by i to"},{"Start":"02:41.930 ","End":"02:47.820","Text":"the n. Now some formulas that are affected by which of the three definitions?"},{"Start":"02:48.650 ","End":"02:52.120","Text":"Let\u0027s say they\u0027re for reference."},{"Start":"02:52.580 ","End":"02:55.690","Text":"You know what, I\u0027ll skip these two for a moment."},{"Start":"02:55.690 ","End":"03:00.879","Text":"You don\u0027t know what convolution is perhaps inverse transform is actually the formula"},{"Start":"03:00.879 ","End":"03:06.324","Text":"that if you have the Fourier transform and you want to get back to the original function,"},{"Start":"03:06.324 ","End":"03:10.015","Text":"the formula is integral from minus infinity to infinity,"},{"Start":"03:10.015 ","End":"03:13.929","Text":"e to the i Omega x times the function."},{"Start":"03:13.929 ","End":"03:16.450","Text":"It doesn\u0027t exactly take you back."},{"Start":"03:16.450 ","End":"03:18.580","Text":"If it\u0027s a continuous point of f,"},{"Start":"03:18.580 ","End":"03:19.660","Text":"it will take you back."},{"Start":"03:19.660 ","End":"03:24.640","Text":"But if not, it\u0027ll take you to the average of the value on the left and on the right."},{"Start":"03:24.640 ","End":"03:27.130","Text":"In other words, if there was a jump in the original"},{"Start":"03:27.130 ","End":"03:30.250","Text":"f. Let me do a transform and inverse transform,"},{"Start":"03:30.250 ","End":"03:31.865","Text":"we come to the average,"},{"Start":"03:31.865 ","End":"03:33.895","Text":"the midpoint of the jump."},{"Start":"03:33.895 ","End":"03:35.960","Text":"It\u0027s always the same with the other two,"},{"Start":"03:35.960 ","End":"03:39.305","Text":"except that some constants at the beginning."},{"Start":"03:39.305 ","End":"03:44.705","Text":"There\u0027s a formula called Plancherel formula or theorem."},{"Start":"03:44.705 ","End":"03:47.480","Text":"This is really the inner product in"},{"Start":"03:47.480 ","End":"03:53.200","Text":"L2 space and what Plancherel says that it preserves inner product."},{"Start":"03:53.200 ","End":"03:56.920","Text":"Well, preserved inner product and this Definition 3,"},{"Start":"03:56.920 ","End":"04:01.450","Text":"otherwise it\u0027s multiplied by a constant or divided by a constant."},{"Start":"04:01.450 ","End":"04:04.290","Text":"But on the left and on the right,"},{"Start":"04:04.290 ","End":"04:05.855","Text":"well, let\u0027s look at it here."},{"Start":"04:05.855 ","End":"04:08.270","Text":"This is the inner product of f and g,"},{"Start":"04:08.270 ","End":"04:10.955","Text":"and this is the inner product of f^ and g^."},{"Start":"04:10.955 ","End":"04:12.920","Text":"The last one here is duality."},{"Start":"04:12.920 ","End":"04:17.270","Text":"If you apply the transform and then you apply the transform"},{"Start":"04:17.270 ","End":"04:22.985","Text":"again to the Fourier transform already and you get almost back to the original."},{"Start":"04:22.985 ","End":"04:26.390","Text":"Well, in this case, you very closely get back to the original ethics,"},{"Start":"04:26.390 ","End":"04:28.610","Text":"that is f of minus x and here,"},{"Start":"04:28.610 ","End":"04:32.765","Text":"there\u0027s an extra constant 2Pi here, 1/2Pi."},{"Start":"04:32.765 ","End":"04:35.585","Text":"Yeah, the ones with the convolution,"},{"Start":"04:35.585 ","End":"04:37.580","Text":"convolution of two functions,"},{"Start":"04:37.580 ","End":"04:46.495","Text":"f and g is defined as f asterisk g. What it does to x is it takes the integral of f(x),"},{"Start":"04:46.495 ","End":"04:49.040","Text":"g(x) minus ( t )dt."},{"Start":"04:49.040 ","End":"04:52.179","Text":"Where t goes from minus infinity to infinity,"},{"Start":"04:52.179 ","End":"04:55.825","Text":"doesn\u0027t matter where you take the minus t here or here."},{"Start":"04:55.825 ","End":"04:57.585","Text":"It\u0027s symmetric."},{"Start":"04:57.585 ","End":"05:01.935","Text":"F star g is g star f. With that definition,"},{"Start":"05:01.935 ","End":"05:06.860","Text":"these two rules say that a convolution of two functions gets"},{"Start":"05:06.860 ","End":"05:12.365","Text":"taken by the Fourier transform to the product up to a constant."},{"Start":"05:12.365 ","End":"05:16.220","Text":"Here it\u0027s exactly, here\u0027s a different constant."},{"Start":"05:16.220 ","End":"05:19.520","Text":"Just like the convolution goes to a product to product goes"},{"Start":"05:19.520 ","End":"05:22.640","Text":"to a convolution up to a constant."},{"Start":"05:22.640 ","End":"05:25.500","Text":"I think we\u0027re done with this clip."}],"ID":29430},{"Watched":false,"Name":"Excerise 1","Duration":"2m 25s","ChapterTopicVideoID":28202,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.815","Text":"In this exercise, we\u0027re going to compute the Fourier transform of this function."},{"Start":"00:04.815 ","End":"00:08.400","Text":"This is called the characteristic function or an indicator"},{"Start":"00:08.400 ","End":"00:12.600","Text":"function of the interval minus 1,1."},{"Start":"00:12.600 ","End":"00:18.285","Text":"This is defined as 1 if x belongs to this set and 0 otherwise."},{"Start":"00:18.285 ","End":"00:22.500","Text":"In general, Chi of a subset is"},{"Start":"00:22.500 ","End":"00:27.090","Text":"the function which is 1 if x belongs to the subset and 0 otherwise."},{"Start":"00:27.090 ","End":"00:29.570","Text":"In this case, it\u0027s also called the"},{"Start":"00:29.570 ","End":"00:34.115","Text":"rectangular window function or a rectangular window function."},{"Start":"00:34.115 ","End":"00:38.600","Text":"First a reminder of what the Fourier transform of a function f is."},{"Start":"00:38.600 ","End":"00:40.420","Text":"It\u0027s the following formula."},{"Start":"00:40.420 ","End":"00:41.820","Text":"Here instead of f,"},{"Start":"00:41.820 ","End":"00:43.395","Text":"we have this function,"},{"Start":"00:43.395 ","End":"00:45.720","Text":"Chi sub minus 1,1,"},{"Start":"00:45.720 ","End":"00:47.595","Text":"and the Fourier transform,"},{"Start":"00:47.595 ","End":"00:53.405","Text":"you\u0027ll use the notation of this curly f. This is the formula here."},{"Start":"00:53.405 ","End":"01:00.285","Text":"This function is just 1 but only from minus 1,1 and 0 otherwise so we get this integral."},{"Start":"01:00.285 ","End":"01:09.000","Text":"Integral of e^minus i Omega x is e^minus i Omega x over minus i Omega and the 1/2pi."},{"Start":"01:09.000 ","End":"01:12.695","Text":"This will only work if Omega is not 0."},{"Start":"01:12.695 ","End":"01:20.270","Text":"We can drop the minus and swap the upper and lower limits of substitution."},{"Start":"01:20.270 ","End":"01:23.930","Text":"Then plugging in here, minus 1,"},{"Start":"01:23.930 ","End":"01:25.415","Text":"we get e^i Omega,"},{"Start":"01:25.415 ","End":"01:28.880","Text":"minus e^minus i Omega and rearrange also."},{"Start":"01:28.880 ","End":"01:33.905","Text":"The Omega can go here and the 2 can go here and we get this."},{"Start":"01:33.905 ","End":"01:39.905","Text":"The reason for rearranging it like this is because this is one of the formulas for sine,"},{"Start":"01:39.905 ","End":"01:43.890","Text":"this case sine of Omega over the pi Omega."},{"Start":"01:43.890 ","End":"01:46.020","Text":"For Omega equals 0,"},{"Start":"01:46.020 ","End":"01:49.580","Text":"we could just use the continuity that the Fourier transform is always"},{"Start":"01:49.580 ","End":"01:53.120","Text":"continuous and take the limit as omega goes to 0,"},{"Start":"01:53.120 ","End":"01:55.860","Text":"write the 1 over pi separately and this is"},{"Start":"01:55.860 ","End":"01:58.985","Text":"a well-known limit sine Omega over omega tends to 1,"},{"Start":"01:58.985 ","End":"02:01.025","Text":"so the answer is 1 over pi."},{"Start":"02:01.025 ","End":"02:06.590","Text":"Or we could do it by just plugging in instead of x,"},{"Start":"02:06.590 ","End":"02:08.420","Text":"we could put in 0,"},{"Start":"02:08.420 ","End":"02:12.905","Text":"and then we\u0027ll get this formula with a 0 here."},{"Start":"02:12.905 ","End":"02:15.770","Text":"This comes out to be 1 over 2pi, the integral of 1."},{"Start":"02:15.770 ","End":"02:17.150","Text":"This integral is 2,"},{"Start":"02:17.150 ","End":"02:22.280","Text":"so 2 over 2 pi is 1 over pi, same thing."},{"Start":"02:22.280 ","End":"02:26.100","Text":"That\u0027s the end of this exercise."}],"ID":29431},{"Watched":false,"Name":"Exercise 2","Duration":"2m 33s","ChapterTopicVideoID":28203,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.745","Text":"This exercise, we\u0027re going to compute the Fourier transform of this function."},{"Start":"00:05.745 ","End":"00:12.120","Text":"It\u0027s 1 minus absolute value of x when absolute value of x is less than 1,"},{"Start":"00:12.120 ","End":"00:16.905","Text":"in other words, from minus 1 to 1 and 0 otherwise."},{"Start":"00:16.905 ","End":"00:22.935","Text":"Here\u0027s the formula for the Fourier transform of a function f. We call it f hat."},{"Start":"00:22.935 ","End":"00:26.040","Text":"In our case, f is an even function."},{"Start":"00:26.040 ","End":"00:28.830","Text":"There\u0027s a formula for the Fourier transform of"},{"Start":"00:28.830 ","End":"00:32.010","Text":"an even function where instead of the e^ minus I Omega x,"},{"Start":"00:32.010 ","End":"00:37.965","Text":"we put cosine Omega x and we take the integral from 0 to infinity,"},{"Start":"00:37.965 ","End":"00:40.860","Text":"and 1 over Pi instead of 1 over 2Pi."},{"Start":"00:40.860 ","End":"00:45.480","Text":"This is the formula we have and in our case,"},{"Start":"00:45.480 ","End":"00:47.940","Text":"it\u0027s 0 from 1 to infinity."},{"Start":"00:47.940 ","End":"00:55.740","Text":"We just need to take from 0 to 1 and f(x) here is 1 minus x."},{"Start":"00:55.740 ","End":"00:58.220","Text":"We\u0027ll do it as an integration by parts,"},{"Start":"00:58.220 ","End":"00:59.960","Text":"the formula you should be familiar with,"},{"Start":"00:59.960 ","End":"01:01.855","Text":"just as a reminder."},{"Start":"01:01.855 ","End":"01:06.020","Text":"What we get is 1 minus x times the integral of this,"},{"Start":"01:06.020 ","End":"01:12.325","Text":"which is sine Omega x over Omega minus derivative of this is minus 1 times this."},{"Start":"01:12.325 ","End":"01:17.860","Text":"Now, this part comes out to be 0 because sine of Omega times 0 is 0,"},{"Start":"01:17.860 ","End":"01:19.100","Text":"that takes care of the zeros,"},{"Start":"01:19.100 ","End":"01:22.330","Text":"as for the 1, here 1 minus x becomes 0."},{"Start":"01:22.330 ","End":"01:26.120","Text":"In either case, you get 00 minus 0 is 0,"},{"Start":"01:26.120 ","End":"01:30.605","Text":"we just have this integral which comes out to be"},{"Start":"01:30.605 ","End":"01:36.810","Text":"minus cosine Omega x over Pi Omega squared,"},{"Start":"01:36.810 ","End":"01:39.570","Text":"these two minuses cancel."},{"Start":"01:39.570 ","End":"01:43.340","Text":"Takeaway the minus, put 0 here and 1 here."},{"Start":"01:43.340 ","End":"01:46.160","Text":"When x is 0, we get cosine 0 is 1."},{"Start":"01:46.160 ","End":"01:47.360","Text":"When x is 1,"},{"Start":"01:47.360 ","End":"01:51.080","Text":"we get minus cosine Omega and the Pi Omega squared."},{"Start":"01:51.080 ","End":"01:54.080","Text":"This doesn\u0027t work for Omega equals 0."},{"Start":"01:54.080 ","End":"01:56.360","Text":"If Omega is 0,"},{"Start":"01:56.360 ","End":"02:00.845","Text":"we can return to this point here and let Omega equals 0,"},{"Start":"02:00.845 ","End":"02:02.960","Text":"so cosine is 1."},{"Start":"02:02.960 ","End":"02:08.070","Text":"We just have the 1 minus x. Integral from 0 to 1,"},{"Start":"02:08.070 ","End":"02:09.840","Text":"1 minus x dx,"},{"Start":"02:09.840 ","End":"02:12.315","Text":"which is a simple computation."},{"Start":"02:12.315 ","End":"02:14.085","Text":"Comes out to be 1 minus a 1/2,"},{"Start":"02:14.085 ","End":"02:17.745","Text":"which is the 1/2 times 1 over Pi, 1 over 2Pi."},{"Start":"02:17.745 ","End":"02:21.684","Text":"Or you could take the limit of this as Omega goes to 0."},{"Start":"02:21.684 ","End":"02:24.740","Text":"You might happen to know that 1 minus cosine Omega over"},{"Start":"02:24.740 ","End":"02:28.070","Text":"Omega squared turns to a 1/2 as Omega goes to 0."},{"Start":"02:28.070 ","End":"02:30.170","Text":"Again, we get 1 over 2Pi."},{"Start":"02:30.170 ","End":"02:33.480","Text":"That concludes this exercise."}],"ID":29432},{"Watched":false,"Name":"Exercise 3","Duration":"2m 25s","ChapterTopicVideoID":28204,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.090","Text":"In this exercise, we\u0027re going to compute the Fourier transform of this function,"},{"Start":"00:06.090 ","End":"00:12.765","Text":"which is e^-x on the positives and 0 on the negatives."},{"Start":"00:12.765 ","End":"00:15.705","Text":"It has a jump discontinuity here."},{"Start":"00:15.705 ","End":"00:22.470","Text":"Certainly piecewise, continuous, and it\u0027s also absolutely integrable."},{"Start":"00:22.470 ","End":"00:24.510","Text":"Check these things from time to time we supposed to do"},{"Start":"00:24.510 ","End":"00:27.825","Text":"a mental check that we\u0027re in the right space of functions."},{"Start":"00:27.825 ","End":"00:30.570","Text":"The Fourier transform has this formula and we\u0027ll"},{"Start":"00:30.570 ","End":"00:36.900","Text":"substitute this function for our f(x) and we\u0027ll get the integral."},{"Start":"00:36.900 ","End":"00:39.150","Text":"We only need it from 0 to infinity,"},{"Start":"00:39.150 ","End":"00:42.600","Text":"where the function is e^-x."},{"Start":"00:42.600 ","End":"00:46.765","Text":"Now, do some algebra of exponents."},{"Start":"00:46.765 ","End":"00:49.765","Text":"We can combine these and get this."},{"Start":"00:49.765 ","End":"00:58.585","Text":"The integral of e to the minus something x is itself divided by the coefficient of x,"},{"Start":"00:58.585 ","End":"01:01.615","Text":"which is minus 1 plus i Omega here."},{"Start":"01:01.615 ","End":"01:05.665","Text":"We need to plug in infinity and 0 and subtract."},{"Start":"01:05.665 ","End":"01:10.930","Text":"Simplify by taking this part in front that doesn\u0027t contain x."},{"Start":"01:10.930 ","End":"01:14.890","Text":"We just need this part from 0 to infinity."},{"Start":"01:14.890 ","End":"01:17.770","Text":"Formally, we would get the following,"},{"Start":"01:17.770 ","End":"01:21.310","Text":"except we\u0027re not really supposed to substitute infinity."},{"Start":"01:21.310 ","End":"01:23.290","Text":"Let\u0027s spell it out,"},{"Start":"01:23.290 ","End":"01:25.510","Text":"a claim that this is equal to 0."},{"Start":"01:25.510 ","End":"01:27.920","Text":"To show something tends to 0,"},{"Start":"01:27.920 ","End":"01:30.600","Text":"we can show its absolute value tends to 0."},{"Start":"01:30.600 ","End":"01:34.429","Text":"We will compute the limit as x goes to infinity of this absolute value."},{"Start":"01:34.429 ","End":"01:39.880","Text":"Then this can be split up into the absolute value of e^-x,"},{"Start":"01:39.880 ","End":"01:42.995","Text":"which is e^-x to the positive,"},{"Start":"01:42.995 ","End":"01:49.105","Text":"times the absolute value of e^-x i Omega."},{"Start":"01:49.105 ","End":"01:51.900","Text":"Now, this is equal to 1,"},{"Start":"01:51.900 ","End":"01:55.830","Text":"because Omega is a real parameter,"},{"Start":"01:55.830 ","End":"02:01.820","Text":"and e^-i, something real is 1 in absolute value."},{"Start":"02:01.820 ","End":"02:03.395","Text":"It\u0027s on the unit disk."},{"Start":"02:03.395 ","End":"02:06.155","Text":"We know that e^-x goes to 0."},{"Start":"02:06.155 ","End":"02:08.735","Text":"Something tends to 0 times something,"},{"Start":"02:08.735 ","End":"02:10.940","Text":"which is 1, still tends to 0,"},{"Start":"02:10.940 ","End":"02:12.320","Text":"the limit is 0."},{"Start":"02:12.320 ","End":"02:15.275","Text":"Instead of this, we can put 0."},{"Start":"02:15.275 ","End":"02:20.195","Text":"That just leaves us cancel the minus here with the minus here,"},{"Start":"02:20.195 ","End":"02:25.260","Text":"and we get this and this is the answer, and the end of the clip."}],"ID":29433},{"Watched":false,"Name":"Exercise 4","Duration":"3m 50s","ChapterTopicVideoID":28205,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.135","Text":"In this exercise, we\u0027re going to compute the Fourier transform of this function."},{"Start":"00:06.135 ","End":"00:12.930","Text":"Probably easiest to understand from the diagram from minus 1 to 1, it\u0027s 1."},{"Start":"00:12.930 ","End":"00:15.630","Text":"Here, it\u0027s 2, here it\u0027s 2,"},{"Start":"00:15.630 ","End":"00:17.625","Text":"and everywhere else it\u0027s 0,"},{"Start":"00:17.625 ","End":"00:21.015","Text":"reminder of what the Fourier transform is."},{"Start":"00:21.015 ","End":"00:22.860","Text":"But in our case,"},{"Start":"00:22.860 ","End":"00:25.440","Text":"f is an even function."},{"Start":"00:25.440 ","End":"00:27.915","Text":"It\u0027s easy to see from the picture."},{"Start":"00:27.915 ","End":"00:29.990","Text":"We use the other formula,"},{"Start":"00:29.990 ","End":"00:32.360","Text":"especially for an even function."},{"Start":"00:32.360 ","End":"00:34.190","Text":"This is the formula."},{"Start":"00:34.190 ","End":"00:37.010","Text":"It\u0027s an integral from 0 to infinity,"},{"Start":"00:37.010 ","End":"00:38.645","Text":"so the 3 pieces,"},{"Start":"00:38.645 ","End":"00:39.815","Text":"0 to 1,"},{"Start":"00:39.815 ","End":"00:41.015","Text":"1 to 2,"},{"Start":"00:41.015 ","End":"00:44.700","Text":"and 2 to infinity but the last part is all 0s."},{"Start":"00:44.700 ","End":"00:46.335","Text":"We\u0027re just going to have to integrals;"},{"Start":"00:46.335 ","End":"00:51.555","Text":"the part from 0 to 1 where the function is 1 and the part from 1 to 2,"},{"Start":"00:51.555 ","End":"00:53.395","Text":"where the function is 2."},{"Start":"00:53.395 ","End":"00:58.400","Text":"Here we are integral from 0 to 1 of 1 times the cosine integral from"},{"Start":"00:58.400 ","End":"01:03.960","Text":"1 to 2 of 2 times the cosine and 1 over Pi in front."},{"Start":"01:03.960 ","End":"01:06.829","Text":"Integral of cosine is sine."},{"Start":"01:06.829 ","End":"01:09.305","Text":"We have to divide by the derivative."},{"Start":"01:09.305 ","End":"01:12.920","Text":"Again, the derivative of cosine is sine."},{"Start":"01:12.920 ","End":"01:19.445","Text":"Here we have a 2 in front and here we have from 0 to 1 and here from 1 to 2."},{"Start":"01:19.445 ","End":"01:23.350","Text":"Let\u0027s see, combine the Pi with the Omega."},{"Start":"01:23.350 ","End":"01:28.695","Text":"From here we get sine 1 Omega minus sine 0 Omega."},{"Start":"01:28.695 ","End":"01:34.745","Text":"Here we get twice sine 2 Omega minus sine 1 Omega."},{"Start":"01:34.745 ","End":"01:40.760","Text":"Just rearrange this, put it all over Pi Omega and collect like terms."},{"Start":"01:40.760 ","End":"01:46.895","Text":"We get 2 sine 2 Omega minus sine Omega over Pi Omega."},{"Start":"01:46.895 ","End":"01:52.280","Text":"Notice that when we got to here from here,"},{"Start":"01:52.280 ","End":"01:54.410","Text":"we assume that Omega is not 0,"},{"Start":"01:54.410 ","End":"01:56.720","Text":"otherwise, this wouldn\u0027t make sense."},{"Start":"01:56.720 ","End":"02:02.570","Text":"We really have to also evaluate f hat at 0."},{"Start":"02:02.570 ","End":"02:04.813","Text":"There\u0027s actually 2 ways of doing this,"},{"Start":"02:04.813 ","End":"02:07.265","Text":"1 way is to use continuity."},{"Start":"02:07.265 ","End":"02:09.840","Text":"I\u0027ll do that 1 first."},{"Start":"02:10.760 ","End":"02:17.505","Text":"We can take this and let Omega 10 to 0 take the limit."},{"Start":"02:17.505 ","End":"02:19.035","Text":"Sine 2 Omega,"},{"Start":"02:19.035 ","End":"02:22.370","Text":"there\u0027s a trigonometric formula and identity that"},{"Start":"02:22.370 ","End":"02:26.540","Text":"says that this is 2 sine Omega cosine Omega."},{"Start":"02:26.540 ","End":"02:27.980","Text":"Now that we have this,"},{"Start":"02:27.980 ","End":"02:29.060","Text":"we can break it up."},{"Start":"02:29.060 ","End":"02:31.330","Text":"The sine Omega comes out."},{"Start":"02:31.330 ","End":"02:38.660","Text":"In fact, we can take the sine Omega here and here and put that with the Omega."},{"Start":"02:38.660 ","End":"02:40.160","Text":"Probably see where I\u0027m going."},{"Start":"02:40.160 ","End":"02:42.320","Text":"This is a famous limit."},{"Start":"02:42.320 ","End":"02:48.840","Text":"What\u0027s left is 2 times 2 is 4 cosine Omega minus 1 over Pi."},{"Start":"02:49.450 ","End":"02:53.795","Text":"We said this goes to 1."},{"Start":"02:53.795 ","End":"02:57.350","Text":"This goes to 0,"},{"Start":"02:57.350 ","End":"02:58.725","Text":"cosine goes to 1,"},{"Start":"02:58.725 ","End":"03:01.515","Text":"4 times 1 minus 1 is 3."},{"Start":"03:01.515 ","End":"03:04.270","Text":"We get 3 over Pi,"},{"Start":"03:04.270 ","End":"03:07.055","Text":"that\u0027s f^ at 0."},{"Start":"03:07.055 ","End":"03:12.230","Text":"The other way is just to go back to the step before we"},{"Start":"03:12.230 ","End":"03:18.630","Text":"divided by Omega and put x equals 0 here."},{"Start":"03:18.630 ","End":"03:22.695","Text":"We\u0027ve got f^ of 0."},{"Start":"03:22.695 ","End":"03:25.140","Text":"Here, x is 0,"},{"Start":"03:25.140 ","End":"03:26.670","Text":"the cosine is 1,"},{"Start":"03:26.670 ","End":"03:31.635","Text":"and here the cosine is 1 so it\u0027s just 1 and 2 dx,"},{"Start":"03:31.635 ","End":"03:33.180","Text":"dx here from 0 to 1,"},{"Start":"03:33.180 ","End":"03:36.200","Text":"here from 1 to 2, very basic integrals."},{"Start":"03:36.200 ","End":"03:37.310","Text":"This integral is 1,"},{"Start":"03:37.310 ","End":"03:38.705","Text":"this integral is 2,"},{"Start":"03:38.705 ","End":"03:42.305","Text":"1 plus 2 is 3 so it\u0027s 3 over Pi."},{"Start":"03:42.305 ","End":"03:46.870","Text":"That\u0027s exactly the same as what we got here, which it should be."},{"Start":"03:46.870 ","End":"03:50.680","Text":"That concludes this exercise."}],"ID":29434},{"Watched":false,"Name":"Exercise 5","Duration":"4m 4s","ChapterTopicVideoID":28206,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.685","Text":"In this exercise, we\u0027ll compute the Fourier transform of this function."},{"Start":"00:05.685 ","End":"00:12.570","Text":"It\u0027s e^ minus ax for positive x and e^bx for negative x."},{"Start":"00:12.570 ","End":"00:16.559","Text":"A and b are positive real numbers."},{"Start":"00:16.559 ","End":"00:19.185","Text":"The graph of it looks something like this,"},{"Start":"00:19.185 ","End":"00:20.430","Text":"starts out at one,"},{"Start":"00:20.430 ","End":"00:23.385","Text":"and tends asymptotically to 0 on both sides."},{"Start":"00:23.385 ","End":"00:30.600","Text":"It\u0027s easy to show that this is absolutely integrable and it\u0027s also piecewise continuous."},{"Start":"00:30.600 ","End":"00:35.400","Text":"This is the formula for the Fourier transform of f,"},{"Start":"00:35.400 ","End":"00:37.860","Text":"takes it to a function f hat,"},{"Start":"00:37.860 ","End":"00:39.605","Text":"I like to call it."},{"Start":"00:39.605 ","End":"00:43.945","Text":"In our case, we\u0027ll break it up into 2 integrals."},{"Start":"00:43.945 ","End":"00:45.320","Text":"Because of the way it\u0027s defined,"},{"Start":"00:45.320 ","End":"00:50.600","Text":"it\u0027s natural to break it up to 0 and from 0,"},{"Start":"00:50.600 ","End":"00:54.245","Text":"because it\u0027s defined up to 0 as e^ bx and the other one,"},{"Start":"00:54.245 ","End":"00:57.455","Text":"e^ minus ax from 0 to infinity."},{"Start":"00:57.455 ","End":"01:04.625","Text":"We have this sum and then combine the exponents like so."},{"Start":"01:04.625 ","End":"01:06.410","Text":"Getting used to doing this,"},{"Start":"01:06.410 ","End":"01:07.760","Text":"in the many exercises,"},{"Start":"01:07.760 ","End":"01:11.825","Text":"we do this kind of exponent computation."},{"Start":"01:11.825 ","End":"01:16.820","Text":"Now we can take the integral of each."},{"Start":"01:16.820 ","End":"01:24.425","Text":"So for this one we get e^ b minus I Omega x divided by the b minus I Omega."},{"Start":"01:24.425 ","End":"01:26.300","Text":"Remember the coefficient times x,"},{"Start":"01:26.300 ","End":"01:28.445","Text":"we just divide by that coefficient."},{"Start":"01:28.445 ","End":"01:30.445","Text":"These are the limits."},{"Start":"01:30.445 ","End":"01:32.745","Text":"Here, also,"},{"Start":"01:32.745 ","End":"01:37.310","Text":"e^minus a plus I omega x divided by,"},{"Start":"01:37.310 ","End":"01:40.099","Text":"the minus just put it in front."},{"Start":"01:40.099 ","End":"01:42.560","Text":"Now let\u0027s substitute."},{"Start":"01:42.560 ","End":"01:48.245","Text":"What I\u0027m going to do is just as if infinity and minus infinity were a number."},{"Start":"01:48.245 ","End":"01:50.150","Text":"So if x is 0,"},{"Start":"01:50.150 ","End":"01:52.130","Text":"we have e^0 is 1."},{"Start":"01:52.130 ","End":"01:56.690","Text":"Here we put in minus infinity. We subtract."},{"Start":"01:56.690 ","End":"02:01.130","Text":"Here, put x equals infinity."},{"Start":"02:01.130 ","End":"02:03.740","Text":"When x is 0, it\u0027s minus 1."},{"Start":"02:03.740 ","End":"02:06.490","Text":"Now, this is not quite right. It\u0027s symbolic."},{"Start":"02:06.490 ","End":"02:08.560","Text":"So really we should treat it as a limit."},{"Start":"02:08.560 ","End":"02:12.970","Text":"In fact, we show that each of these is 0,"},{"Start":"02:12.970 ","End":"02:20.245","Text":"meaning that the limit here as x goes to minus infinity and put x back and here,"},{"Start":"02:20.245 ","End":"02:25.300","Text":"x goes to infinity again with the x replacing the infinity."},{"Start":"02:25.300 ","End":"02:27.805","Text":"If we put absolute values,"},{"Start":"02:27.805 ","End":"02:35.290","Text":"that will work because if the absolute value goes to 0 then the thing itself goes to 0,"},{"Start":"02:35.290 ","End":"02:37.945","Text":"and it\u0027s easier to compute with absolute value."},{"Start":"02:37.945 ","End":"02:40.030","Text":"So this limit is equal to,"},{"Start":"02:40.030 ","End":"02:46.355","Text":"we can break this up as a product e^bx and e^minus I Omega x."},{"Start":"02:46.355 ","End":"02:51.415","Text":"Now e^ bx is positive because bx is real."},{"Start":"02:51.415 ","End":"02:54.595","Text":"So it doesn\u0027t even need the absolute value when we pulled it out."},{"Start":"02:54.595 ","End":"02:58.975","Text":"This is e to the power of something purely imaginary."},{"Start":"02:58.975 ","End":"03:04.520","Text":"That means it\u0027s on the unit circle and the absolute value is 1,"},{"Start":"03:04.520 ","End":"03:08.470","Text":"and because b is positive and x goes to minus infinity,"},{"Start":"03:08.470 ","End":"03:11.034","Text":"bx goes to minus infinity,"},{"Start":"03:11.034 ","End":"03:14.800","Text":"and e to the power of something that goes to minus infinity is 0."},{"Start":"03:14.800 ","End":"03:16.885","Text":"0 times 1 is 0."},{"Start":"03:16.885 ","End":"03:25.270","Text":"Similarly here, this is what we have and when x goes to infinity and a is positive,"},{"Start":"03:25.270 ","End":"03:27.585","Text":"then minus ax goes to minus infinity."},{"Start":"03:27.585 ","End":"03:30.375","Text":"Again we get this going to 0,"},{"Start":"03:30.375 ","End":"03:32.580","Text":"this thing being 1 constantly,"},{"Start":"03:32.580 ","End":"03:34.215","Text":"so this goes to 0."},{"Start":"03:34.215 ","End":"03:37.925","Text":"We can replace this by 0 and this by 0."},{"Start":"03:37.925 ","End":"03:43.670","Text":"From here we just get the 1 over the b minus i Omega."},{"Start":"03:43.670 ","End":"03:46.820","Text":"From here the, minus,"},{"Start":"03:46.820 ","End":"03:51.715","Text":"minus is plus 1 over a plus I Omega."},{"Start":"03:51.715 ","End":"03:54.890","Text":"I guess it changed the order here because I wanted to put the"},{"Start":"03:54.890 ","End":"03:57.800","Text":"a before the b, no particular reason."},{"Start":"03:57.800 ","End":"03:59.450","Text":"The 1 over 2Pi here,"},{"Start":"03:59.450 ","End":"04:01.940","Text":"and this is the answer."},{"Start":"04:01.940 ","End":"04:05.160","Text":"That concludes this exercise."}],"ID":29435},{"Watched":false,"Name":"Exercise 6","Duration":"2m 16s","ChapterTopicVideoID":28191,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.940","Text":"In this exercise, we\u0027re going to compute the Fourier transform of f (x),"},{"Start":"00:05.940 ","End":"00:09.250","Text":"which is the following step function,"},{"Start":"00:09.250 ","End":"00:13.560","Text":"a window function, and it\u0027s defined as follows."},{"Start":"00:13.560 ","End":"00:15.450","Text":"The picture is here."},{"Start":"00:15.450 ","End":"00:19.830","Text":"Reminder of the formula for a Fourier transform is this."},{"Start":"00:19.830 ","End":"00:24.420","Text":"We just have to substitute F for what it is in our case,"},{"Start":"00:24.420 ","End":"00:31.710","Text":"which will give us the integral instead of f Just 1 from 0-1."},{"Start":"00:31.710 ","End":"00:34.605","Text":"Outside of that, F is 0."},{"Start":"00:34.605 ","End":"00:38.835","Text":"We can use Euler\u0027s formula for expanding this."},{"Start":"00:38.835 ","End":"00:45.755","Text":"This becomes cosine of Omega x minus I sine of Omega x."},{"Start":"00:45.755 ","End":"00:50.910","Text":"The integral of this is sine Omega x over Omega."},{"Start":"00:50.910 ","End":"00:56.435","Text":"For this, the integral of minus sine x is cosine x,"},{"Start":"00:56.435 ","End":"00:59.225","Text":"this case cosine Omega x over Omega."},{"Start":"00:59.225 ","End":"01:04.445","Text":"We can substitute when x is 0, it just gives 0."},{"Start":"01:04.445 ","End":"01:08.240","Text":"When x is 1, we have sine Omega over Omega."},{"Start":"01:08.240 ","End":"01:11.240","Text":"Similarly, this expression,"},{"Start":"01:11.240 ","End":"01:16.590","Text":"we have cosine of 1 Omega minus cosine 0 Omega,"},{"Start":"01:16.590 ","End":"01:18.805","Text":"cosine 0 is 1."},{"Start":"01:18.805 ","End":"01:22.580","Text":"So this is the expression we get."},{"Start":"01:22.580 ","End":"01:26.540","Text":"That\u0027s our transformed function,"},{"Start":"01:26.540 ","End":"01:31.835","Text":"except that this formula doesn\u0027t work when Omega is 0."},{"Start":"01:31.835 ","End":"01:33.230","Text":"We have 2 choices."},{"Start":"01:33.230 ","End":"01:37.865","Text":"We can either find the limit of this as Omega goes to 0,"},{"Start":"01:37.865 ","End":"01:39.860","Text":"which might be a bit tricky,"},{"Start":"01:39.860 ","End":"01:42.065","Text":"probably use L\u0027Hopital\u0027s formula."},{"Start":"01:42.065 ","End":"01:48.765","Text":"The other choice is to go back here and let Omega equal 0."},{"Start":"01:48.765 ","End":"01:53.205","Text":"So Fourier transform of f at 0,"},{"Start":"01:53.205 ","End":"01:58.475","Text":"1 over 2 Pi the integral of 1 e to the minus I Omega 0,"},{"Start":"01:58.475 ","End":"02:00.630","Text":"all these just becomes 1."},{"Start":"02:00.630 ","End":"02:05.310","Text":"We have the integral of 1 from 0-1, that\u0027s 1."},{"Start":"02:05.310 ","End":"02:08.655","Text":"The answer is 1 over 2 Pi."},{"Start":"02:08.655 ","End":"02:14.055","Text":"You\u0027d get the same thing if you took the limit as omega goes to 0."},{"Start":"02:14.055 ","End":"02:17.330","Text":"That concludes this exercise."}],"ID":29436},{"Watched":false,"Name":"Exercise 7","Duration":"1m 42s","ChapterTopicVideoID":28192,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.635","Text":"In this exercise, we\u0027ll compute the Fourier transform of this function f(x),"},{"Start":"00:05.635 ","End":"00:09.620","Text":"which is described as follows and in the diagram."},{"Start":"00:09.620 ","End":"00:14.500","Text":"Remember the formula for the Fourier transform of a function f,"},{"Start":"00:14.500 ","End":"00:18.745","Text":"which gives us a function f hat. Let\u0027s start."},{"Start":"00:18.745 ","End":"00:23.620","Text":"In our case, the function f is either 1, 2, or 0."},{"Start":"00:23.620 ","End":"00:25.870","Text":"We don\u0027t have to take the 0 part."},{"Start":"00:25.870 ","End":"00:31.485","Text":"It\u0027s 1 between 0 and 1 and it\u0027s 2 between 1 and 2."},{"Start":"00:31.485 ","End":"00:34.330","Text":"We get this sum of integrals."},{"Start":"00:34.330 ","End":"00:42.625","Text":"The integral of e^minus i Omega x is itself divided by minus i Omega."},{"Start":"00:42.625 ","End":"00:46.530","Text":"Here also, this over minus i Omega."},{"Start":"00:46.530 ","End":"00:48.800","Text":"It brings that part to the front."},{"Start":"00:48.800 ","End":"00:51.620","Text":"Just have to substitute plug in 0,"},{"Start":"00:51.620 ","End":"00:53.820","Text":"we get 1, plug in 1,"},{"Start":"00:53.820 ","End":"00:56.450","Text":"it\u0027s e^minus Omega, and so on and so on."},{"Start":"00:56.450 ","End":"01:02.165","Text":"Then we just have to collect terms simplify and we get this expression."},{"Start":"01:02.165 ","End":"01:05.840","Text":"Note that this doesn\u0027t work for Omega equals 0."},{"Start":"01:05.840 ","End":"01:07.850","Text":"In fact, already at this stage,"},{"Start":"01:07.850 ","End":"01:10.490","Text":"we couldn\u0027t make the passage from here to here,"},{"Start":"01:10.490 ","End":"01:11.850","Text":"if Omega is 0."},{"Start":"01:11.850 ","End":"01:13.880","Text":"We better compute that separately."},{"Start":"01:13.880 ","End":"01:20.015","Text":"F hat(0), just put 0 here instead of Omega."},{"Start":"01:20.015 ","End":"01:21.775","Text":"This is what we get."},{"Start":"01:21.775 ","End":"01:24.930","Text":"Now, e^0 is 1."},{"Start":"01:24.930 ","End":"01:29.920","Text":"This thing simplifies integral of 1 plus the integral of 2."},{"Start":"01:29.920 ","End":"01:31.535","Text":"These are the limits."},{"Start":"01:31.535 ","End":"01:33.780","Text":"This comes out to be 1,"},{"Start":"01:33.780 ","End":"01:35.805","Text":"and this comes out to be 2."},{"Start":"01:35.805 ","End":"01:39.690","Text":"Altogether 1 plus 2 is 3 over 2Pi."},{"Start":"01:39.690 ","End":"01:43.060","Text":"That concludes this exercise."}],"ID":29437},{"Watched":false,"Name":"Exercise 8","Duration":"1m 19s","ChapterTopicVideoID":28193,"CourseChapterTopicPlaylistID":277672,"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\u0027ll compute the Fourier transform of f(x),"},{"Start":"00:04.320 ","End":"00:08.190","Text":"which is e^x when x is between 0 and 1,"},{"Start":"00:08.190 ","End":"00:12.060","Text":"there it is, and it\u0027s 0 everywhere else."},{"Start":"00:12.060 ","End":"00:16.005","Text":"Here\u0027s a reminder of the formula that converts"},{"Start":"00:16.005 ","End":"00:20.055","Text":"function f to its transform f hat, f caret."},{"Start":"00:20.055 ","End":"00:22.125","Text":"Let\u0027s compute it in our case,"},{"Start":"00:22.125 ","End":"00:24.570","Text":"we just have to take the integral from 0-1."},{"Start":"00:24.570 ","End":"00:27.480","Text":"We only want the non-zero part and that\u0027s from 0-1,"},{"Start":"00:27.480 ","End":"00:29.700","Text":"which gives us here 0 and 1."},{"Start":"00:29.700 ","End":"00:32.985","Text":"Use the law of exponents to combine these 2."},{"Start":"00:32.985 ","End":"00:37.230","Text":"We get e^x times 1 minus i Omega."},{"Start":"00:37.230 ","End":"00:38.990","Text":"That\u0027s a straightforward integral."},{"Start":"00:38.990 ","End":"00:43.745","Text":"We just take it as it is and divide by 1 minus i Omega."},{"Start":"00:43.745 ","End":"00:48.060","Text":"Then we have to substitute 1 and 0 and subtract."},{"Start":"00:48.060 ","End":"00:51.785","Text":"What we get, if x is 0,"},{"Start":"00:51.785 ","End":"00:54.485","Text":"then we get just here,"},{"Start":"00:54.485 ","End":"01:00.480","Text":"e^0 which is 1 and if x is 1 here we have e^1 minus i Omega,"},{"Start":"01:00.480 ","End":"01:07.010","Text":"so it\u0027s e^1 minus i Omega minus 1 over the same denominator and the 1 over 2Pi stays."},{"Start":"01:07.010 ","End":"01:08.750","Text":"That\u0027s really it."},{"Start":"01:08.750 ","End":"01:13.220","Text":"Just remark there\u0027s no place where denominator could be 0 because"},{"Start":"01:13.220 ","End":"01:19.520","Text":"Omega is real so 1 minus omega can\u0027t be 0. We\u0027re done."}],"ID":29438},{"Watched":false,"Name":"Exercise 9","Duration":"2m 54s","ChapterTopicVideoID":28194,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"In this exercise, we\u0027ll compute the Fourier transform of f(x),"},{"Start":"00:04.710 ","End":"00:08.250","Text":"which is equal to each of the 2ix,"},{"Start":"00:08.250 ","End":"00:11.550","Text":"but only one x is between minus 1 and 1 and 0."},{"Start":"00:11.550 ","End":"00:16.170","Text":"Otherwise, I didn\u0027t bring a sketch of this because it\u0027s complex valued,"},{"Start":"00:16.170 ","End":"00:18.345","Text":"won\u0027t fit into a 2D graph."},{"Start":"00:18.345 ","End":"00:22.125","Text":"Here\u0027s a reminder of the formula for the Fourier transform."},{"Start":"00:22.125 ","End":"00:24.750","Text":"Familiar. In our case,"},{"Start":"00:24.750 ","End":"00:30.405","Text":"what we have is the integral just from minus 1 to 1 of this function,"},{"Start":"00:30.405 ","End":"00:33.195","Text":"e to the 2, ix times the rest of it,"},{"Start":"00:33.195 ","End":"00:35.305","Text":"using the law of exponents."},{"Start":"00:35.305 ","End":"00:40.345","Text":"This combines to each of the i to minus Omega x, dx."},{"Start":"00:40.345 ","End":"00:43.010","Text":"We\u0027ll use the formula due to Euler e to"},{"Start":"00:43.010 ","End":"00:47.080","Text":"the i Theta equals cosine theta plus I sine Theta."},{"Start":"00:47.080 ","End":"00:57.155","Text":"That will give us the following cosine of 2 minus Omega x and here sign to minus Omega x."},{"Start":"00:57.155 ","End":"01:03.620","Text":"Now, the cosine of 2 minus Omega x is an even function in x."},{"Start":"01:03.620 ","End":"01:06.020","Text":"If you put minus x instead of x,"},{"Start":"01:06.020 ","End":"01:08.450","Text":"it\u0027s the same and this is a symmetric interval."},{"Start":"01:08.450 ","End":"01:15.785","Text":"In this case, we can simplify this to twice the integral from 0 to 1."},{"Start":"01:15.785 ","End":"01:21.530","Text":"Here we have an odd function on a symmetric interval."},{"Start":"01:21.530 ","End":"01:24.290","Text":"In that case the integral is 0,"},{"Start":"01:24.290 ","End":"01:26.300","Text":"so we don\u0027t have to add anything else."},{"Start":"01:26.300 ","End":"01:28.565","Text":"We just have this to compute."},{"Start":"01:28.565 ","End":"01:32.690","Text":"Integral of cosine is sine,"},{"Start":"01:32.690 ","End":"01:34.850","Text":"but we have to divide by the inner derivative,"},{"Start":"01:34.850 ","End":"01:36.470","Text":"the 2 minus Omega."},{"Start":"01:36.470 ","End":"01:39.050","Text":"Note that this won\u0027t work for Omega equals 2."},{"Start":"01:39.050 ","End":"01:41.645","Text":"We\u0027ll do that case at the end."},{"Start":"01:41.645 ","End":"01:47.795","Text":"Substitute x=1, and we get sine of 2 minus Omega."},{"Start":"01:47.795 ","End":"01:49.450","Text":"When x is 0, it\u0027s just 0,"},{"Start":"01:49.450 ","End":"01:51.005","Text":"so this is the answer."},{"Start":"01:51.005 ","End":"01:55.580","Text":"But for Omega not equal to 2 and for Omega equals 2,"},{"Start":"01:55.580 ","End":"01:57.530","Text":"this 2 possible ways of doing it."},{"Start":"01:57.530 ","End":"02:02.685","Text":"One is to use the fact that f hat is always continuous."},{"Start":"02:02.685 ","End":"02:06.590","Text":"Then we can just take the limit of this as Omega goes to 2."},{"Start":"02:06.590 ","End":"02:08.225","Text":"Let\u0027s do that method first."},{"Start":"02:08.225 ","End":"02:13.955","Text":"When Omega goes to 2 we can replace 2 minus Omega by theta."},{"Start":"02:13.955 ","End":"02:16.130","Text":"Then theta goes to 0."},{"Start":"02:16.130 ","End":"02:18.530","Text":"We have the limit of sin theta over theta,"},{"Start":"02:18.530 ","End":"02:21.230","Text":"which is famous limit and that\u0027s equal to 1."},{"Start":"02:21.230 ","End":"02:23.850","Text":"But we also have the 1 over Pi here."},{"Start":"02:23.850 ","End":"02:25.490","Text":"The answer is 1 over Pi."},{"Start":"02:25.490 ","End":"02:30.160","Text":"The other way of doing it is just to return to this integral."},{"Start":"02:30.160 ","End":"02:35.160","Text":"Let Omega equals 2 here and cosine of 2 minus 2,"},{"Start":"02:35.160 ","End":"02:37.665","Text":"it\u0027s cosine is 0, that\u0027s 1."},{"Start":"02:37.665 ","End":"02:44.415","Text":"We have the integral of 1dx from 0 to 1 that\u0027s equal to 1."},{"Start":"02:44.415 ","End":"02:48.075","Text":"Then 1 over Pi is the answer."},{"Start":"02:48.075 ","End":"02:50.390","Text":"Luckily it\u0027s the same answer as here,"},{"Start":"02:50.390 ","End":"02:51.815","Text":"otherwise we\u0027d have a problem."},{"Start":"02:51.815 ","End":"02:54.660","Text":"That\u0027s it."}],"ID":29439},{"Watched":false,"Name":"Exercise 10","Duration":"3m 31s","ChapterTopicVideoID":28195,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.730","Text":"In this exercise, we\u0027re going to compute"},{"Start":"00:02.730 ","End":"00:07.580","Text":"the Fourier transform of f(x) which is equal to sin(x),"},{"Start":"00:07.580 ","End":"00:12.770","Text":"but just when x is between minus 1 and 1 and 0 otherwise."},{"Start":"00:12.770 ","End":"00:14.685","Text":"This is what it looks like."},{"Start":"00:14.685 ","End":"00:19.440","Text":"It\u0027s clearly piecewise continuous and absolutely integrable."},{"Start":"00:19.440 ","End":"00:21.885","Text":"It\u0027s in the space g(r),"},{"Start":"00:21.885 ","End":"00:27.260","Text":"we called it and reminder of what the Fourier transform is."},{"Start":"00:27.260 ","End":"00:33.140","Text":"But we\u0027re not going to be using this formula because this is an odd function."},{"Start":"00:33.140 ","End":"00:36.455","Text":"There\u0027s an easier formula for an odd function."},{"Start":"00:36.455 ","End":"00:39.100","Text":"We had the following formula."},{"Start":"00:39.100 ","End":"00:43.505","Text":"I guess we\u0027re not going to use this one in this exercise."},{"Start":"00:43.505 ","End":"00:48.640","Text":"Minus 1 over Pi and the integral of f(x) sine Omega x."},{"Start":"00:48.640 ","End":"00:54.910","Text":"Now f(x), we just have to take it between 0 and 1 here,"},{"Start":"00:54.910 ","End":"00:59.040","Text":"because after 1, it\u0027s equal to 0."},{"Start":"00:59.040 ","End":"01:05.430","Text":"We get integral from 0-1 and f(x) is sin(x) here."},{"Start":"01:05.430 ","End":"01:08.160","Text":"Now we have sine something, sine something."},{"Start":"01:08.160 ","End":"01:12.050","Text":"The trigonometric identity for sine Alpha, sine Beta,"},{"Start":"01:12.050 ","End":"01:17.150","Text":"and we\u0027ll use this identity to develop this."},{"Start":"01:17.150 ","End":"01:23.659","Text":"We get cosine of this minus this,"},{"Start":"01:23.659 ","End":"01:26.975","Text":"minus cosine of this plus this."},{"Start":"01:26.975 ","End":"01:31.228","Text":"We can break it up into,"},{"Start":"01:31.228 ","End":"01:34.225","Text":"and the integral of cosine is sine."},{"Start":"01:34.225 ","End":"01:38.630","Text":"This is really x times 1 minus Omega,"},{"Start":"01:38.630 ","End":"01:41.240","Text":"so we have to divide by 1 minus Omega."},{"Start":"01:41.240 ","End":"01:44.450","Text":"Similarly here, the integral of cosine is sine,"},{"Start":"01:44.450 ","End":"01:47.155","Text":"but we have to divide by 1 plus Omega."},{"Start":"01:47.155 ","End":"01:50.070","Text":"I don\u0027t know whether there\u0027s a blank line here."},{"Start":"01:50.070 ","End":"01:54.320","Text":"This works only if Omega is not equal to plus or minus 1."},{"Start":"01:54.320 ","End":"01:56.120","Text":"This doesn\u0027t work when Omega is minus 1,"},{"Start":"01:56.120 ","End":"01:57.920","Text":"this doesn\u0027t work when Omega is 1."},{"Start":"01:57.920 ","End":"02:00.350","Text":"Take care of this later."},{"Start":"02:00.350 ","End":"02:03.800","Text":"Now we just have to substitute in both of these."},{"Start":"02:03.800 ","End":"02:06.110","Text":"If we plug in x=0,"},{"Start":"02:06.110 ","End":"02:11.000","Text":"we get 0, so we just need to put x=1 here and here."},{"Start":"02:11.000 ","End":"02:13.730","Text":"We get sin(1) minus Omega here,"},{"Start":"02:13.730 ","End":"02:14.940","Text":"sin(1) plus Omega,"},{"Start":"02:14.940 ","End":"02:16.920","Text":"and the rest of it is the same."},{"Start":"02:16.920 ","End":"02:20.210","Text":"That\u0027s our Fourier transform."},{"Start":"02:20.210 ","End":"02:24.455","Text":"But we have to take care now of the special cases."},{"Start":"02:24.455 ","End":"02:31.700","Text":"What we\u0027ll do is use the continuity property of the transform and then we\u0027ll also"},{"Start":"02:31.700 ","End":"02:34.715","Text":"use the formula that we use a lot that the limit"},{"Start":"02:34.715 ","End":"02:38.500","Text":"as Theta goes to 0 of sine Theta over Theta is 1."},{"Start":"02:38.500 ","End":"02:40.960","Text":"If Omega is 1,"},{"Start":"02:40.960 ","End":"02:47.870","Text":"then what we get is that the limit as Omega goes to 1 is like Theta goes to 0,"},{"Start":"02:47.870 ","End":"02:50.110","Text":"or Theta is 1 minus Omega."},{"Start":"02:50.110 ","End":"02:53.130","Text":"This tends to 1 and this one,"},{"Start":"02:53.130 ","End":"02:56.260","Text":"when Omega is 1 is just sin(2) over 2."},{"Start":"02:56.260 ","End":"02:59.435","Text":"On the other hand, if we take the minus 1,"},{"Start":"02:59.435 ","End":"03:02.160","Text":"then 1 plus Omega is 0."},{"Start":"03:02.160 ","End":"03:04.530","Text":"When Omega tends to minus 1, 1 plus Omega,"},{"Start":"03:04.530 ","End":"03:08.415","Text":"which is equal to Theta goes to 0."},{"Start":"03:08.415 ","End":"03:12.285","Text":"We get sine Theta over Theta as Theta tends to 0,"},{"Start":"03:12.285 ","End":"03:15.030","Text":"that comes out to be 1, that\u0027s this one."},{"Start":"03:15.030 ","End":"03:17.100","Text":"Here if Omega is minus 1,"},{"Start":"03:17.100 ","End":"03:18.720","Text":"1 minus minus 1 is 2,"},{"Start":"03:18.720 ","End":"03:20.850","Text":"so that\u0027s sin(2) over 2."},{"Start":"03:20.850 ","End":"03:26.030","Text":"We have the formula where f hat of Omega is this except for"},{"Start":"03:26.030 ","End":"03:32.040","Text":"plus or minus 1 where we\u0027ve computed them separately as this and this and we are done."}],"ID":29440},{"Watched":false,"Name":"Exercise 11","Duration":"2m 17s","ChapterTopicVideoID":28196,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.680","Text":"Here we have to compute the Fourier transform of this function f(x),"},{"Start":"00:04.680 ","End":"00:09.930","Text":"which is equal to x when the absolute value of x is less than a."},{"Start":"00:09.930 ","End":"00:16.050","Text":"Which means that x is between a and minus a and 0 otherwise,"},{"Start":"00:16.050 ","End":"00:18.480","Text":"a is some positive parameter."},{"Start":"00:18.480 ","End":"00:23.020","Text":"Now, usual formula for the Fourier transform is this"},{"Start":"00:23.020 ","End":"00:27.450","Text":"but we have a special formula for odd and even functions and in our case,"},{"Start":"00:27.450 ","End":"00:30.990","Text":"this function is clearly an odd function so we use"},{"Start":"00:30.990 ","End":"00:36.775","Text":"this formula which we studied in the properties of the Fourier transform."},{"Start":"00:36.775 ","End":"00:40.040","Text":"Substitute f(x) equals x,"},{"Start":"00:40.040 ","End":"00:42.335","Text":"but only up to a."},{"Start":"00:42.335 ","End":"00:44.930","Text":"We don\u0027t need to go down to minus a because we\u0027re starting at 0,"},{"Start":"00:44.930 ","End":"00:46.370","Text":"so from 0 to a."},{"Start":"00:46.370 ","End":"00:47.840","Text":"This is the part,"},{"Start":"00:47.840 ","End":"00:52.295","Text":"we\u0027re taking the integral of times sine Omega x."},{"Start":"00:52.295 ","End":"00:54.500","Text":"This will do as an integration by parts."},{"Start":"00:54.500 ","End":"00:56.630","Text":"This is a reminder of the formula."},{"Start":"00:56.630 ","End":"00:59.495","Text":"We get x times the integral of this,"},{"Start":"00:59.495 ","End":"01:06.360","Text":"which is this minus derivative of this times again the integral of this."},{"Start":"01:06.360 ","End":"01:09.735","Text":"This will only work if Omega is non-zero."},{"Start":"01:09.735 ","End":"01:13.085","Text":"Continuing. The minus cancels with the minus."},{"Start":"01:13.085 ","End":"01:15.920","Text":"The integral of cosine is sine,"},{"Start":"01:15.920 ","End":"01:19.130","Text":"but we need to divide again by Omega."},{"Start":"01:19.130 ","End":"01:25.190","Text":"So we get sine Omega x over Omega squared and from 0 to a."},{"Start":"01:25.190 ","End":"01:28.130","Text":"This part, we substitute a,"},{"Start":"01:28.130 ","End":"01:31.700","Text":"we get minus a cosine Omega over Omega and we plug in 0,"},{"Start":"01:31.700 ","End":"01:34.025","Text":"it\u0027s 0 because when x is 0 at all 0."},{"Start":"01:34.025 ","End":"01:38.075","Text":"We still have to substitute x equals a here,"},{"Start":"01:38.075 ","End":"01:40.100","Text":"when x is 0, it comes out 0."},{"Start":"01:40.100 ","End":"01:42.125","Text":"So just x equals a,"},{"Start":"01:42.125 ","End":"01:46.160","Text":"which gives us sine Omega a over Omega squared here."},{"Start":"01:46.160 ","End":"01:48.290","Text":"Now tidy up a bit."},{"Start":"01:48.290 ","End":"01:52.910","Text":"Just bring it all over a common denominator of Omega squared here."},{"Start":"01:52.910 ","End":"01:56.960","Text":"This is the answer except that it\u0027s only for Omega non-zero."},{"Start":"01:56.960 ","End":"01:58.970","Text":"For Omega equals 0."},{"Start":"01:58.970 ","End":"02:01.640","Text":"If we go back, we have the integral."},{"Start":"02:01.640 ","End":"02:05.915","Text":"Here it is. Just put x equals 0,"},{"Start":"02:05.915 ","End":"02:07.460","Text":"sine of 0 is 0."},{"Start":"02:07.460 ","End":"02:09.665","Text":"So it\u0027s just the integral of 0,"},{"Start":"02:09.665 ","End":"02:13.705","Text":"which is 0 times minus I over pi is still 0."},{"Start":"02:13.705 ","End":"02:18.030","Text":"That completes the answer. We are done."}],"ID":29441},{"Watched":false,"Name":"Exercise 12","Duration":"1m 5s","ChapterTopicVideoID":28197,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.520","Text":"In this exercise, we\u0027re asked to prove that there isn\u0027t any function f in"},{"Start":"00:05.520 ","End":"00:12.585","Text":"the space G(R) such that the Fourier transform of f of Omega is the following."},{"Start":"00:12.585 ","End":"00:15.150","Text":"I\u0027ve drawn a picture of it,"},{"Start":"00:15.150 ","End":"00:20.295","Text":"This is the graph of f hat and this is the Omega axis."},{"Start":"00:20.295 ","End":"00:22.785","Text":"Turns out that the answer is no."},{"Start":"00:22.785 ","End":"00:28.365","Text":"Because one of the properties of the Fourier transform is that if f is in this space,"},{"Start":"00:28.365 ","End":"00:33.690","Text":"meaning it\u0027s absolutely integrable and piecewise continuous,"},{"Start":"00:33.690 ","End":"00:37.110","Text":"then it has to be continuous for all Omega."},{"Start":"00:37.110 ","End":"00:40.035","Text":"But this function isn\u0027t continuous."},{"Start":"00:40.035 ","End":"00:42.030","Text":"Easy to see from the picture."},{"Start":"00:42.030 ","End":"00:46.535","Text":"But yes, if you check at Omega equals plus or minus 1/2,"},{"Start":"00:46.535 ","End":"00:49.175","Text":"if we approach, say, 1/2 from the right,"},{"Start":"00:49.175 ","End":"00:51.635","Text":"it\u0027s 0, but 1/2 on the left,"},{"Start":"00:51.635 ","End":"00:53.555","Text":"we would get 1 minus 1/2,"},{"Start":"00:53.555 ","End":"00:58.380","Text":"which is 1/2, there\u0027s definitely a jump discontinuity here."},{"Start":"00:58.380 ","End":"01:03.080","Text":"Because of that, it can\u0027t be the Fourier transform of anything."},{"Start":"01:03.080 ","End":"01:05.580","Text":"That\u0027s all, we\u0027re done."}],"ID":29442},{"Watched":false,"Name":"Fourier Transform, Useful Properties and Examples","Duration":"9m 37s","ChapterTopicVideoID":29368,"CourseChapterTopicPlaylistID":277672,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.990","Text":"In this clip, we\u0027ll see some useful rules and formulas for the Fourier transform."},{"Start":"00:06.990 ","End":"00:13.950","Text":"I especially chose the ones that will be most useful in the rest of this chapter,"},{"Start":"00:13.950 ","End":"00:18.615","Text":"and especially for the section on convolution theorem."},{"Start":"00:18.615 ","End":"00:24.065","Text":"We\u0027ll start with a particular Fourier transform which turns out to be very useful,"},{"Start":"00:24.065 ","End":"00:25.250","Text":"it\u0027s of a Gaussian."},{"Start":"00:25.250 ","End":"00:28.880","Text":"The Gaussian is a bell curve, usually,"},{"Start":"00:28.880 ","End":"00:32.510","Text":"e^minus x^2, but it could be more general."},{"Start":"00:32.510 ","End":"00:39.440","Text":"The Fourier transform of e^minus x^2 is given as follows."},{"Start":"00:39.440 ","End":"00:42.335","Text":"This is also a bell curve, a Gaussian."},{"Start":"00:42.335 ","End":"00:51.055","Text":"It\u0027s just stretched by a factor of 2 horizontally because it\u0027s only got over 2^2."},{"Start":"00:51.055 ","End":"00:55.910","Text":"It\u0027s compressed by a factor of 2 root Pi vertically."},{"Start":"00:55.910 ","End":"00:59.495","Text":"Anyway, Fourier transform of a bell curve is another bell curve."},{"Start":"00:59.495 ","End":"01:03.990","Text":"Here\u0027s the first of 4 rules that I\u0027ll bring here."},{"Start":"01:03.990 ","End":"01:06.485","Text":"This one is called scaling and shifting,"},{"Start":"01:06.485 ","End":"01:08.690","Text":"which is basically what happens when you take"},{"Start":"01:08.690 ","End":"01:12.920","Text":"the Fourier transform of f, and instead ax,"},{"Start":"01:12.920 ","End":"01:14.540","Text":"put ax plus b,"},{"Start":"01:14.540 ","End":"01:19.230","Text":"a and b are real constants and we assume a is not 0."},{"Start":"01:19.230 ","End":"01:23.280","Text":"What we do is we modify f-hat."},{"Start":"01:23.280 ","End":"01:25.410","Text":"Instead of the argument Omega,"},{"Start":"01:25.410 ","End":"01:31.564","Text":"we replace it by Omega over a and we multiply it by this."},{"Start":"01:31.564 ","End":"01:35.010","Text":"This is commonly used when b is 0,"},{"Start":"01:35.010 ","End":"01:38.915","Text":"so I made a special rule for the case b=0."},{"Start":"01:38.915 ","End":"01:45.400","Text":"B is 0 so this thing drops out because it\u0027s e^0 and we\u0027re left with this."},{"Start":"01:45.400 ","End":"01:52.070","Text":"Sometimes it\u0027s convenient to have a on the denominator so we just invert this."},{"Start":"01:52.070 ","End":"02:00.100","Text":"As an example, let\u0027s find the Fourier transform of e^minus 4x^2."},{"Start":"02:00.100 ","End":"02:04.410","Text":"Of course, we will use this formula here."},{"Start":"02:05.870 ","End":"02:11.175","Text":"Let\u0027s let g(x) be e^minus x^2."},{"Start":"02:11.175 ","End":"02:17.010","Text":"The Fourier transform of Omega is from the above this."},{"Start":"02:17.010 ","End":"02:19.825","Text":"We get that fx,"},{"Start":"02:19.825 ","End":"02:22.385","Text":"which is e^minus,"},{"Start":"02:22.385 ","End":"02:27.200","Text":"you can write 4x^2 is 2x^2 is g of 2x."},{"Start":"02:27.200 ","End":"02:32.540","Text":"We can use this formula with x replaced by 2x using this."},{"Start":"02:32.540 ","End":"02:37.545","Text":"I\u0027ll copy this but replace f by g so we get this formula."},{"Start":"02:37.545 ","End":"02:39.990","Text":"In our case, a is equal to 2,"},{"Start":"02:39.990 ","End":"02:42.651","Text":"so it becomes g of 2x."},{"Start":"02:42.651 ","End":"02:45.540","Text":"This is what it is."},{"Start":"02:45.540 ","End":"02:52.725","Text":"Just replacing a by 2 here and here and then g-hat of Omega over 2 is,"},{"Start":"02:52.725 ","End":"02:54.720","Text":"this is g-hat of Omega,"},{"Start":"02:54.720 ","End":"02:59.915","Text":"so we replace here Omega by Omega over 2, which is this."},{"Start":"02:59.915 ","End":"03:04.580","Text":"Now we can collect the constant 2 with 2 gives 4,"},{"Start":"03:04.580 ","End":"03:07.280","Text":"Omega over 2^2 is Omega squared over 4,"},{"Start":"03:07.280 ","End":"03:11.270","Text":"and the 4 with the 4 give 16 so we get e^minus Omega"},{"Start":"03:11.270 ","End":"03:15.500","Text":"squared over 16 here and 1/4 root Pi in front,"},{"Start":"03:15.500 ","End":"03:18.065","Text":"and that is the answer."},{"Start":"03:18.065 ","End":"03:21.500","Text":"Now we come to the second rule, differentiation."},{"Start":"03:21.500 ","End":"03:28.415","Text":"This tells us how to figure out the Fourier transform of the nth derivative of f,"},{"Start":"03:28.415 ","End":"03:33.410","Text":"assuming we know the Fourier transform of f. What we do is"},{"Start":"03:33.410 ","End":"03:39.230","Text":"multiply the result by i Omega^n and best to give an example,"},{"Start":"03:39.230 ","End":"03:46.610","Text":"Fourier transform of 1 over x^2 plus 1 is a 1/2e to the minus absolute value of Omega."},{"Start":"03:46.610 ","End":"03:51.350","Text":"We have to prove that the Fourier transform of this is this."},{"Start":"03:51.350 ","End":"03:53.180","Text":"Where\u0027s the derivative?"},{"Start":"03:53.180 ","End":"03:58.025","Text":"Well, let\u0027s just try differentiating this first and see what we get."},{"Start":"03:58.025 ","End":"04:01.290","Text":"We get minus 2x over x^2 plus 1^2."},{"Start":"04:01.290 ","End":"04:07.535","Text":"Very similar to this just a factor of minus 2 will take care of the minus 2 shortly."},{"Start":"04:07.535 ","End":"04:14.040","Text":"Let\u0027s just apply this rule to this with n=1."},{"Start":"04:14.040 ","End":"04:19.370","Text":"We get the Fourier transform of the derivative is i Omega^n,"},{"Start":"04:19.370 ","End":"04:23.555","Text":"which is i Omega^1 times the Fourier transform of this."},{"Start":"04:23.555 ","End":"04:25.265","Text":"Now, this we\u0027re given."},{"Start":"04:25.265 ","End":"04:28.025","Text":"This is equal to, where is it?"},{"Start":"04:28.025 ","End":"04:33.935","Text":"This. Here, we know that the derivative is from here,"},{"Start":"04:33.935 ","End":"04:36.964","Text":"minus 2x over x^2 plus 1^2."},{"Start":"04:36.964 ","End":"04:39.260","Text":"Now by linearity of the transform,"},{"Start":"04:39.260 ","End":"04:43.850","Text":"we can take the 2 and the minus outside in fact bring them over to the right-hand side,"},{"Start":"04:43.850 ","End":"04:46.625","Text":"put the 2 here and the minus."},{"Start":"04:46.625 ","End":"04:55.845","Text":"We get i Omega^1 is just i Omega and then over 4 with a minus and that\u0027s the answer."},{"Start":"04:55.845 ","End":"04:58.925","Text":"Now, the third rule called duality,"},{"Start":"04:58.925 ","End":"05:03.185","Text":"which deals with what happens when we take the transform of a transform."},{"Start":"05:03.185 ","End":"05:06.770","Text":"It turns out that we don\u0027t exactly get back to the original,"},{"Start":"05:06.770 ","End":"05:08.810","Text":"but we get close to it, that f(x),"},{"Start":"05:08.810 ","End":"05:11.350","Text":"we have a minus x and we have a 1 over 2Pi,"},{"Start":"05:11.350 ","End":"05:15.390","Text":"this will help us to find an inverse transform."},{"Start":"05:15.390 ","End":"05:17.330","Text":"When we have a function of x,"},{"Start":"05:17.330 ","End":"05:19.475","Text":"the transform is defined like this."},{"Start":"05:19.475 ","End":"05:24.229","Text":"We use the letter Omega here and we get a function of Omega because the integral is dx."},{"Start":"05:24.229 ","End":"05:27.050","Text":"Now if we switch the x and the Omega,"},{"Start":"05:27.050 ","End":"05:33.680","Text":"then we get that the transform of a function g of Omega is using the same formula,"},{"Start":"05:33.680 ","End":"05:36.680","Text":"but with Omega replaced by x and vice versa."},{"Start":"05:36.680 ","End":"05:38.810","Text":"This time we have to d Omega,"},{"Start":"05:38.810 ","End":"05:42.985","Text":"so we get back to a function of x."},{"Start":"05:42.985 ","End":"05:44.660","Text":"We start with a function of x,"},{"Start":"05:44.660 ","End":"05:45.950","Text":"we get a function of Omega,"},{"Start":"05:45.950 ","End":"05:47.300","Text":"start with a function of Omega,"},{"Start":"05:47.300 ","End":"05:48.965","Text":"we get a function of x."},{"Start":"05:48.965 ","End":"05:52.165","Text":"As an example, we\u0027re asked to find the function"},{"Start":"05:52.165 ","End":"05:56.735","Text":"f(x) such that its transform is e^minus Omega squared."},{"Start":"05:56.735 ","End":"06:00.410","Text":"Instead of thinking about an inverse transform,"},{"Start":"06:00.410 ","End":"06:03.290","Text":"we\u0027ll take again a transform because"},{"Start":"06:03.290 ","End":"06:06.470","Text":"the transform of a transform gets you closely back to"},{"Start":"06:06.470 ","End":"06:12.995","Text":"the original and recall the formula from the beginning of the clip, we\u0027ll use this."},{"Start":"06:12.995 ","End":"06:15.535","Text":"Start by copying this rule,"},{"Start":"06:15.535 ","End":"06:21.649","Text":"and now replace this f of f(x) by e^minus Omega squared,"},{"Start":"06:21.649 ","End":"06:24.590","Text":"so this part e^minus Omega squared."},{"Start":"06:24.590 ","End":"06:26.420","Text":"Now applying this formula,"},{"Start":"06:26.420 ","End":"06:28.775","Text":"but reversing x and Omega,"},{"Start":"06:28.775 ","End":"06:33.410","Text":"we get that this is equal to the following,"},{"Start":"06:33.410 ","End":"06:37.510","Text":"like what we had with Omega, but x instead."},{"Start":"06:37.510 ","End":"06:39.470","Text":"Now do 2 things,"},{"Start":"06:39.470 ","End":"06:44.845","Text":"switch sides and multiply both sides by 2Pi."},{"Start":"06:44.845 ","End":"06:49.370","Text":"We get that f of minus x is 2Pi over 2 root Pi,"},{"Start":"06:49.370 ","End":"06:53.750","Text":"which is just root Pi e^minus x^2 over 4."},{"Start":"06:53.750 ","End":"06:56.960","Text":"There\u0027s one more step to go because we want f(x),"},{"Start":"06:56.960 ","End":"06:59.830","Text":"so replace x by minus x."},{"Start":"06:59.830 ","End":"07:02.570","Text":"Here we have minus x^2."},{"Start":"07:02.570 ","End":"07:04.220","Text":"Well, it doesn\u0027t matter if it\u0027s squared,"},{"Start":"07:04.220 ","End":"07:06.020","Text":"you can throw out the minus."},{"Start":"07:06.020 ","End":"07:08.315","Text":"This is the answer,"},{"Start":"07:08.315 ","End":"07:11.390","Text":"and that\u0027s a third example."},{"Start":"07:11.390 ","End":"07:13.565","Text":"Now, the fourth and last rule,"},{"Start":"07:13.565 ","End":"07:16.195","Text":"which is called Plancherel\u0027s Theorem,"},{"Start":"07:16.195 ","End":"07:17.930","Text":"this is what it says."},{"Start":"07:17.930 ","End":"07:20.000","Text":"There are certain conditions attached to f,"},{"Start":"07:20.000 ","End":"07:22.595","Text":"but let\u0027s not worry about the technical stuff."},{"Start":"07:22.595 ","End":"07:24.020","Text":"This is what the rule says,"},{"Start":"07:24.020 ","End":"07:29.630","Text":"that the integral of the function squared in absolute value is the"},{"Start":"07:29.630 ","End":"07:36.605","Text":"same as the integral of the transformed function squared but with a factor of 2Pi here."},{"Start":"07:36.605 ","End":"07:42.180","Text":"By the way, there is a generalized Plancherel\u0027s Theorem which has 2 functions,"},{"Start":"07:42.180 ","End":"07:46.385","Text":"f and g and this is a special case when g=f."},{"Start":"07:46.385 ","End":"07:49.390","Text":"But we won\u0027t be using this now."},{"Start":"07:49.390 ","End":"07:56.005","Text":"The example is a theoretical one to prove that there\u0027s no function f(x)."},{"Start":"07:56.005 ","End":"07:58.700","Text":"Let\u0027s not worry about the technical."},{"Start":"07:58.700 ","End":"08:02.914","Text":"It just means that all the integrals exist and we can take the Fourier transform."},{"Start":"08:02.914 ","End":"08:09.680","Text":"There\u0027s no function f such that the transform of f is the following function of Omega."},{"Start":"08:09.680 ","End":"08:14.365","Text":"I\u0027m going to prove this by contradiction by supposing that such an f(x) exists."},{"Start":"08:14.365 ","End":"08:16.720","Text":"Applying the Plancherel\u0027s Theorem,"},{"Start":"08:16.720 ","End":"08:18.470","Text":"we get the following."},{"Start":"08:18.470 ","End":"08:20.120","Text":"Well, I wrote it in reverse."},{"Start":"08:20.120 ","End":"08:23.869","Text":"I mean take the 2Pi over here and reverse sides."},{"Start":"08:23.869 ","End":"08:28.085","Text":"Now we know that this integral converges."},{"Start":"08:28.085 ","End":"08:31.090","Text":"In other words, it\u0027s less than infinity,"},{"Start":"08:31.090 ","End":"08:34.680","Text":"supposedly this integral should converge 2."},{"Start":"08:34.680 ","End":"08:38.690","Text":"Note that f-hat is an even function."},{"Start":"08:38.690 ","End":"08:42.515","Text":"Because if you replace Omega minus Omega is the same thing."},{"Start":"08:42.515 ","End":"08:49.210","Text":"We can make this integral double the integral from 0 to infinity."},{"Start":"08:49.210 ","End":"08:52.715","Text":"Now, from 0 to infinity,"},{"Start":"08:52.715 ","End":"08:53.930","Text":"Omega is positive,"},{"Start":"08:53.930 ","End":"08:59.540","Text":"so we can throw out the absolute value get the integral from 0 to infinity."},{"Start":"08:59.540 ","End":"09:01.235","Text":"I forgot a 2 here."},{"Start":"09:01.235 ","End":"09:02.990","Text":"It doesn\u0027t really matter."},{"Start":"09:02.990 ","End":"09:06.980","Text":"We get twice the integral of 1 over 1 plus Omega."},{"Start":"09:06.980 ","End":"09:11.585","Text":"That\u0027s from squaring this and remembering that Omega is positive,"},{"Start":"09:11.585 ","End":"09:15.795","Text":"integral of 1 over 1 plus Omega is natural log of 1 plus Omega,"},{"Start":"09:15.795 ","End":"09:18.780","Text":"evaluated from 0 to infinity."},{"Start":"09:18.780 ","End":"09:21.420","Text":"At 0, it\u0027s just 0."},{"Start":"09:21.420 ","End":"09:23.340","Text":"At infinity, it\u0027s infinity."},{"Start":"09:23.340 ","End":"09:26.720","Text":"We get twice infinity, which is infinity."},{"Start":"09:26.720 ","End":"09:32.075","Text":"That\u0027s a contradiction because it\u0027s equal to something that\u0027s less than infinity."},{"Start":"09:32.075 ","End":"09:37.320","Text":"That concludes this example and this clip."}],"ID":30983}],"Thumbnail":null,"ID":277672},{"Name":"The Convolution Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Convolution Theorem","Duration":"4m 20s","ChapterTopicVideoID":29371,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29371.jpeg","UploadDate":"2022-06-28T20:01:09.8730000","DurationForVideoObject":"PT4M20S","Description":null,"MetaTitle":"The Convolution Theorem: Video + Workbook | Proprep","MetaDescription":"Fourier Transform - The Convolution Theorem. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/calculus-i%2c-ii-and-iii/fourier-transform/the-convolution-theorem/vid30984","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.234","Text":"Now, we come to the convolution theorem."},{"Start":"00:03.234 ","End":"00:07.495","Text":"But first, let\u0027s review some definitions."},{"Start":"00:07.495 ","End":"00:14.405","Text":"We had the space written like this, L^1 _PC(R)."},{"Start":"00:14.405 ","End":"00:18.025","Text":"The L^1 means it\u0027s the L^1 norm,"},{"Start":"00:18.025 ","End":"00:21.910","Text":"PC piecewise continuous and it\u0027s over the reals,"},{"Start":"00:21.910 ","End":"00:25.735","Text":"and it\u0027s also called G(R) or just plain G,"},{"Start":"00:25.735 ","End":"00:31.240","Text":"is the space of functions that are piecewise continuous."},{"Start":"00:31.240 ","End":"00:35.530","Text":"They could be complex valued or real valued"},{"Start":"00:35.530 ","End":"00:41.040","Text":"and they\u0027re absolutely integrable on R. And to remind you,"},{"Start":"00:41.040 ","End":"00:45.065","Text":"that means that the integral,"},{"Start":"00:45.065 ","End":"00:49.744","Text":"which is actually the L^1 norm of f is finite,"},{"Start":"00:49.744 ","End":"00:51.695","Text":"meaning the integral converges."},{"Start":"00:51.695 ","End":"00:55.670","Text":"And it can be shown that if we take f and g from"},{"Start":"00:55.670 ","End":"01:00.740","Text":"this space G(R) then the following integral,"},{"Start":"01:00.740 ","End":"01:05.060","Text":"which is basically the convolution integral converges."},{"Start":"01:05.060 ","End":"01:09.275","Text":"So the convolution exists if we\u0027re working in this space."},{"Start":"01:09.275 ","End":"01:11.764","Text":"And now the convolution theorem,"},{"Start":"01:11.764 ","End":"01:15.190","Text":"if we have f and g in the space G,"},{"Start":"01:15.190 ","End":"01:18.815","Text":"then we take the convolution of f and g,"},{"Start":"01:18.815 ","End":"01:22.685","Text":"and then take the Fourier transform of that,"},{"Start":"01:22.685 ","End":"01:28.340","Text":"what we get is 2 Pi times the Fourier transform of f times"},{"Start":"01:28.340 ","End":"01:34.445","Text":"the Fourier transform of g. You have to remember that this is extra 2 Pi here."},{"Start":"01:34.445 ","End":"01:38.600","Text":"If it wasn\u0027t for that, we could say the Fourier of the convolution is"},{"Start":"01:38.600 ","End":"01:43.835","Text":"the product of the transforms but needs a 2 Pi."},{"Start":"01:43.835 ","End":"01:47.750","Text":"It turns out that it not only works in the Fourier transform,"},{"Start":"01:47.750 ","End":"01:51.380","Text":"but also for Laplace transform and several other transforms,"},{"Start":"01:51.380 ","End":"01:54.260","Text":"but we just need it for the Fourier transform."},{"Start":"01:54.260 ","End":"01:55.880","Text":"And if you want to put this in words,"},{"Start":"01:55.880 ","End":"01:58.130","Text":"you could say the transform of"},{"Start":"01:58.130 ","End":"02:02.855","Text":"the convolution is 2 Pi times the product of the transforms."},{"Start":"02:02.855 ","End":"02:04.340","Text":"Now let\u0027s prove it,"},{"Start":"02:04.340 ","End":"02:06.245","Text":"it\u0027s not a difficult proof,"},{"Start":"02:06.245 ","End":"02:08.495","Text":"just messing around with integrals."},{"Start":"02:08.495 ","End":"02:12.140","Text":"We first apply the Fourier transform to"},{"Start":"02:12.140 ","End":"02:18.635","Text":"the convolution and we use the formula for the Fourier Transform."},{"Start":"02:18.635 ","End":"02:24.850","Text":"It takes a function of x to a function of Omega and uses this formula,"},{"Start":"02:24.850 ","End":"02:29.240","Text":"we multiply it by e to the minus i Omega x,"},{"Start":"02:29.240 ","End":"02:32.720","Text":"and we take the integral dx and we get a function of Omega,"},{"Start":"02:32.720 ","End":"02:35.340","Text":"and we have to divide by 2 Pi."},{"Start":"02:35.530 ","End":"02:42.115","Text":"Now we apply the definition of the convolution for this and this is what we get."},{"Start":"02:42.115 ","End":"02:44.795","Text":"Just rearrange it a bit."},{"Start":"02:44.795 ","End":"02:48.200","Text":"Let\u0027s make it a double integral dtdx."},{"Start":"02:48.200 ","End":"02:51.230","Text":"Put the e to the minus i Omega x inside,"},{"Start":"02:51.230 ","End":"02:54.935","Text":"and we\u0027ll make a change of variables."},{"Start":"02:54.935 ","End":"03:01.730","Text":"Let x minus t equal u and then du equals dx."},{"Start":"03:01.730 ","End":"03:05.630","Text":"And if x goes from minus infinity to infinity,"},{"Start":"03:05.630 ","End":"03:11.930","Text":"then so does u for any particular t. And since x is u plus t,"},{"Start":"03:11.930 ","End":"03:15.740","Text":"we replace this x here by u plus t,"},{"Start":"03:15.740 ","End":"03:18.725","Text":"and here the x minus t is u."},{"Start":"03:18.725 ","End":"03:23.360","Text":"Now we can multiply out the exponent,"},{"Start":"03:23.360 ","End":"03:28.690","Text":"get this e to the minus i Omega t e to the minus i Omega u."},{"Start":"03:28.690 ","End":"03:30.540","Text":"Now everything is separated,"},{"Start":"03:30.540 ","End":"03:33.755","Text":"we can get this as the product of two integrals."},{"Start":"03:33.755 ","End":"03:37.565","Text":"We have the integral of f(t) to the minus Omega tdt."},{"Start":"03:37.565 ","End":"03:42.505","Text":"Similarly with u only it\u0027s g instead of f. Now,"},{"Start":"03:42.505 ","End":"03:44.240","Text":"just playing around with the constants,"},{"Start":"03:44.240 ","End":"03:46.234","Text":"we can put a 2 Pi in front,"},{"Start":"03:46.234 ","End":"03:48.200","Text":"keep the 1/2 Pi here,"},{"Start":"03:48.200 ","End":"03:50.785","Text":"and put another 1/2 Pi here."},{"Start":"03:50.785 ","End":"03:56.120","Text":"I\u0027ve colored them so you can see that each of these is the Fourier transform,"},{"Start":"03:56.120 ","End":"04:03.830","Text":"this one for f and this one for g. What we get is 2 Pi times the Fourier transform of f,"},{"Start":"04:03.830 ","End":"04:06.335","Text":"which we can write as f with a hat on,"},{"Start":"04:06.335 ","End":"04:08.940","Text":"and this G with a hat on."},{"Start":"04:08.940 ","End":"04:10.640","Text":"This is what we had to show."},{"Start":"04:10.640 ","End":"04:13.910","Text":"We had to show that the Fourier transform of"},{"Start":"04:13.910 ","End":"04:18.485","Text":"the convolution is 2 Pi times the product of the transform,"},{"Start":"04:18.485 ","End":"04:20.640","Text":"and we are done."}],"ID":30984},{"Watched":false,"Name":"Definition of Convolution","Duration":"3m 20s","ChapterTopicVideoID":29370,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.479","Text":"In this clip, we\u0027ll define the convolution"},{"Start":"00:03.479 ","End":"00:06.674","Text":"of two functions in case you don\u0027t know already."},{"Start":"00:06.674 ","End":"00:10.589","Text":"This will be used in the convolution theorem,"},{"Start":"00:10.589 ","End":"00:15.100","Text":"which is useful in the Fourier Transform."},{"Start":"00:15.140 ","End":"00:17.790","Text":"We start with two functions,"},{"Start":"00:17.790 ","End":"00:19.785","Text":"f and g, which are real-valued,"},{"Start":"00:19.785 ","End":"00:22.890","Text":"but it could also be complex-valued on"},{"Start":"00:22.890 ","End":"00:27.930","Text":"the whole real line and we are going to combine them to get a third function."},{"Start":"00:27.930 ","End":"00:31.020","Text":"Not going to be plus or minus or times."},{"Start":"00:31.020 ","End":"00:34.365","Text":"It\u0027s going to be a new operation called convolution."},{"Start":"00:34.365 ","End":"00:36.960","Text":"We denote it f*g,"},{"Start":"00:36.960 ","End":"00:38.940","Text":"though there are other notations."},{"Start":"00:38.940 ","End":"00:44.330","Text":"Some people use the tensor product like a circle with an x in it."},{"Start":"00:44.330 ","End":"00:47.690","Text":"We use the asterisk and it\u0027s defined as follows."},{"Start":"00:47.690 ","End":"00:49.850","Text":"The convolution of f and g is"},{"Start":"00:49.850 ","End":"00:54.350","Text":"a new function and its value at x is the following integral which"},{"Start":"00:54.350 ","End":"01:01.640","Text":"depends on f and g. This is a function of x because this involves x and t,"},{"Start":"01:01.640 ","End":"01:03.620","Text":"but the integral is dt."},{"Start":"01:03.620 ","End":"01:07.535","Text":"The definition will work as long as the integral converges."},{"Start":"01:07.535 ","End":"01:11.750","Text":"That\u0027s basically it except of course that you\u0027d like an example."},{"Start":"01:11.750 ","End":"01:15.140","Text":"Let\u0027s give an example and then end the clip."},{"Start":"01:15.140 ","End":"01:17.930","Text":"The example exercise would be,"},{"Start":"01:17.930 ","End":"01:22.070","Text":"compute the convolution of f and g,"},{"Start":"01:22.070 ","End":"01:25.895","Text":"where f is this and g is this,"},{"Start":"01:25.895 ","End":"01:27.680","Text":"f is piecewise-defined,"},{"Start":"01:27.680 ","End":"01:32.705","Text":"it\u0027s 1 between -1 and 1 and 0 outside of that."},{"Start":"01:32.705 ","End":"01:35.465","Text":"For the solution, we\u0027ll work off the definition."},{"Start":"01:35.465 ","End":"01:42.680","Text":"We\u0027ll just copy this here and then we want to replace f and g with what they are here."},{"Start":"01:42.680 ","End":"01:48.340","Text":"Now notice that f is 0 outside the interval from -1 to 1,"},{"Start":"01:48.340 ","End":"01:50.490","Text":"and inside it\u0027s =1."},{"Start":"01:50.490 ","End":"01:56.330","Text":"What we can do is just throw f(t) out or replace it by 1 and change the limits of"},{"Start":"01:56.330 ","End":"02:02.775","Text":"integration from -1 to 1 and then what we get is the integral."},{"Start":"02:02.775 ","End":"02:08.000","Text":"The f part is taken care of and we need g(x)-t. Looking at g(x),"},{"Start":"02:08.000 ","End":"02:12.865","Text":"we just replace x by x-t and this is the integral we need now."},{"Start":"02:12.865 ","End":"02:15.605","Text":"We\u0027ll do this using a substitution."},{"Start":"02:15.605 ","End":"02:19.700","Text":"We let x-t be a new variable,"},{"Start":"02:19.700 ","End":"02:23.940","Text":"u, u=x-t, so du=-dt."},{"Start":"02:24.010 ","End":"02:27.580","Text":"These are limits for t,"},{"Start":"02:27.580 ","End":"02:32.720","Text":"so we have to do x minus this to get u."},{"Start":"02:32.720 ","End":"02:36.770","Text":"Here we have x-1, and here we have x-(-1),"},{"Start":"02:36.770 ","End":"02:44.215","Text":"which is x+1 and x-t=u and dt=-du."},{"Start":"02:44.215 ","End":"02:47.270","Text":"But what we can do is get rid of this minus and switch"},{"Start":"02:47.270 ","End":"02:52.060","Text":"the order of the upper and lower limits of integration like so."},{"Start":"02:52.060 ","End":"03:01.110","Text":"Now the integral of 1 over u^2+1 dun is a well-known integral, it\u0027s the arctangent."},{"Start":"03:01.110 ","End":"03:08.210","Text":"We get the arctangent of u and we need to evaluate between x-1 and x+1,"},{"Start":"03:08.210 ","End":"03:11.880","Text":"so we get arctangent(x+1)-arctangent(x-1)."},{"Start":"03:13.850 ","End":"03:21.210","Text":"That\u0027s the answer for the convolution of f and g and that concludes this clip."}],"ID":30985},{"Watched":false,"Name":"Exercise 1","Duration":"4m 15s","ChapterTopicVideoID":29372,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.425","Text":"In this exercise, we\u0027re asked to compute the convolution of this function with itself."},{"Start":"00:07.425 ","End":"00:13.710","Text":"Now, this function is the characteristic function on the interval minus 1 to 1."},{"Start":"00:13.710 ","End":"00:18.090","Text":"Characteristic function means that its value for"},{"Start":"00:18.090 ","End":"00:24.215","Text":"a given x is either 1 if x is in the interval and 0 otherwise."},{"Start":"00:24.215 ","End":"00:25.955","Text":"In fact in general,"},{"Start":"00:25.955 ","End":"00:32.030","Text":"the chi function for a subset of R is defined as"},{"Start":"00:32.030 ","End":"00:36.440","Text":"chi_a is 1 if x belongs to the set a"},{"Start":"00:36.440 ","End":"00:41.360","Text":"and 0 otherwise and there are other notations for it."},{"Start":"00:41.360 ","End":"00:45.350","Text":"More modern to use the one notation and"},{"Start":"00:45.350 ","End":"00:49.960","Text":"then it\u0027s called an indicator function rather than a characteristic function."},{"Start":"00:49.960 ","End":"00:51.960","Text":"There\u0027s a hint to divide into cases."},{"Start":"00:51.960 ","End":"00:54.090","Text":"By the formula for convolution,"},{"Start":"00:54.090 ","End":"00:55.620","Text":"this is what we have."},{"Start":"00:55.620 ","End":"01:02.490","Text":"We have 2 chi functions multiplied by each other but they\u0027re in different places."},{"Start":"01:02.490 ","End":"01:07.945","Text":"Let\u0027s first of all take the case where these two don\u0027t overlap."},{"Start":"01:07.945 ","End":"01:12.730","Text":"This is a function of t and the other one is also a function of t for a given x."},{"Start":"01:12.730 ","End":"01:15.490","Text":"We think of x as a parameter and we fix it for now."},{"Start":"01:15.490 ","End":"01:18.310","Text":"But let\u0027s say x is here far to the right,"},{"Start":"01:18.310 ","End":"01:22.360","Text":"then this function won\u0027t overlap with this function,"},{"Start":"01:22.360 ","End":"01:24.175","Text":"and if you multiply them,"},{"Start":"01:24.175 ","End":"01:28.254","Text":"we get 0, because this is 1 here and 0 otherwise."},{"Start":"01:28.254 ","End":"01:31.540","Text":"This is 1 here. There\u0027s no place where they\u0027re both non-zero,"},{"Start":"01:31.540 ","End":"01:33.460","Text":"so the product is 0."},{"Start":"01:33.460 ","End":"01:37.330","Text":"Case 1 is where the pictures like this,"},{"Start":"01:37.330 ","End":"01:40.795","Text":"which means x minus 1 is to the right of 1,"},{"Start":"01:40.795 ","End":"01:43.420","Text":"1 is less than x minus 1."},{"Start":"01:43.420 ","End":"01:46.075","Text":"That gives us x bigger than 2."},{"Start":"01:46.075 ","End":"01:50.955","Text":"In this case, this integral is simply 0"},{"Start":"01:50.955 ","End":"01:57.005","Text":"so the convolution at this point x is 0 for x satisfying this."},{"Start":"01:57.005 ","End":"01:59.605","Text":"Now let\u0027s go to the next case."},{"Start":"01:59.605 ","End":"02:02.360","Text":"In this case, they do have an overlap."},{"Start":"02:02.360 ","End":"02:05.575","Text":"We move this blue one further to the left."},{"Start":"02:05.575 ","End":"02:08.160","Text":"This is determined by,"},{"Start":"02:08.160 ","End":"02:14.875","Text":"as an inequality that x minus 1 falls between minus 1 and 1."},{"Start":"02:14.875 ","End":"02:19.800","Text":"That gives us that x is between 0 and 2."},{"Start":"02:19.800 ","End":"02:22.565","Text":"In this case, we get an overlap."},{"Start":"02:22.565 ","End":"02:26.400","Text":"In this part they overlap between x"},{"Start":"02:26.400 ","End":"02:30.390","Text":"minus 1 and 1 and the product here is 1 and 0 elsewhere,"},{"Start":"02:30.390 ","End":"02:39.420","Text":"so the integral simply becomes the integral from x minus 1 to 1(1) which is 2 minus x,"},{"Start":"02:39.420 ","End":"02:43.490","Text":"because it\u0027s 1 minus x minus 1 is 2 minus x."},{"Start":"02:43.490 ","End":"02:45.515","Text":"Now we\u0027ll come to Case 3,"},{"Start":"02:45.515 ","End":"02:50.750","Text":"where we\u0027re going to move this blue rectangles further to the left and this gives us"},{"Start":"02:50.750 ","End":"02:57.535","Text":"the inequality that x plus 1 is between minus 1 and 1."},{"Start":"02:57.535 ","End":"03:07.940","Text":"That is that x is between minus 2 and 0 and the overlap is here between minus 1 and x"},{"Start":"03:07.940 ","End":"03:16.605","Text":"plus 1 and so the integral is the integral of 1 from minus 1 to x plus 1 which comes out"},{"Start":"03:16.605 ","End":"03:20.900","Text":"to be x plus 2 or 2 plus x and the last case is"},{"Start":"03:20.900 ","End":"03:25.415","Text":"when we move this rectangle further to the left so that they don\u0027t overlap at all,"},{"Start":"03:25.415 ","End":"03:33.255","Text":"like so, and this is determined by x plus 1 less than minus 1"},{"Start":"03:33.255 ","End":"03:37.590","Text":"or x is less than minus 2 and"},{"Start":"03:37.590 ","End":"03:41.975","Text":"in this case the integral again is 0 because there\u0027s no overlap."},{"Start":"03:41.975 ","End":"03:44.725","Text":"All we have to do now is summarize."},{"Start":"03:44.725 ","End":"03:49.460","Text":"We\u0027ll just take the two cases that are not 0 and everything else"},{"Start":"03:49.460 ","End":"03:54.105","Text":"is lumped under the else or otherwise case is 0."},{"Start":"03:54.105 ","End":"03:57.149","Text":"We had that here, for example,"},{"Start":"03:57.149 ","End":"04:03.885","Text":"x between minus 2 and 0 is in the interval minus 2,0 and there is 2 plus x."},{"Start":"04:03.885 ","End":"04:11.640","Text":"Previously we had x between 0 and 2 and it was 2 minus x and everything else was 0."},{"Start":"04:11.640 ","End":"04:15.310","Text":"This is the answer and we\u0027re done."}],"ID":30986},{"Watched":false,"Name":"Exercise 2","Duration":"4m 23s","ChapterTopicVideoID":29373,"CourseChapterTopicPlaylistID":294465,"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 asked to compute the convolution of f with itself,"},{"Start":"00:05.640 ","End":"00:10.800","Text":"where f is the function that\u0027s e to the minus x"},{"Start":"00:10.800 ","End":"00:16.860","Text":"when x is positive and 0 when x is negative and this is a picture of it."},{"Start":"00:16.860 ","End":"00:20.835","Text":"We\u0027re given a hint to divide into 2 cases,"},{"Start":"00:20.835 ","End":"00:23.340","Text":"x positive and x negative."},{"Start":"00:23.340 ","End":"00:25.440","Text":"By the definition of the convolution,"},{"Start":"00:25.440 ","End":"00:27.945","Text":"this is the integral we get."},{"Start":"00:27.945 ","End":"00:33.900","Text":"Note that f(t) is 0 when t is"},{"Start":"00:33.900 ","End":"00:39.850","Text":"negative so we can just make this integral from 0 to infinity,"},{"Start":"00:39.850 ","End":"00:42.495","Text":"and when t is positive,"},{"Start":"00:42.495 ","End":"00:49.710","Text":"then f(t) is e to the minus t. We replace this by e to the minus t and this by 0."},{"Start":"00:49.750 ","End":"00:56.059","Text":"For each x, we get an integral in t. We look at x as a parameter."},{"Start":"00:56.059 ","End":"00:58.160","Text":"This is a function of t. Take the integral,"},{"Start":"00:58.160 ","End":"01:00.005","Text":"we get a number,"},{"Start":"01:00.005 ","End":"01:03.170","Text":"and it turns out that there\u0027s 2 main cases,"},{"Start":"01:03.170 ","End":"01:04.835","Text":"x positive, x negative,"},{"Start":"01:04.835 ","End":"01:06.755","Text":"just like in the hint."},{"Start":"01:06.755 ","End":"01:11.450","Text":"If x is positive and that\u0027ll be our case 1,"},{"Start":"01:11.450 ","End":"01:15.125","Text":"then what we have is like following."},{"Start":"01:15.125 ","End":"01:20.120","Text":"The horizontal axis is t and the red function is"},{"Start":"01:20.120 ","End":"01:27.155","Text":"just like e to the minus x on each e to the minus t and 0 when t is negative."},{"Start":"01:27.155 ","End":"01:30.080","Text":"The other function f(x) minus t,"},{"Start":"01:30.080 ","End":"01:34.280","Text":"it is a shift and a reflection."},{"Start":"01:34.280 ","End":"01:37.660","Text":"We shift to where t is equal to x."},{"Start":"01:37.660 ","End":"01:41.095","Text":"It goes the other way like this was a mirror."},{"Start":"01:41.095 ","End":"01:46.279","Text":"These are the 2 functions we have and we multiply them and take the integral."},{"Start":"01:46.279 ","End":"01:51.710","Text":"Now it\u0027s going to be 0 whenever t is less than 0 or t is greater than x,"},{"Start":"01:51.710 ","End":"01:55.615","Text":"we only care about the interval between 0 and x."},{"Start":"01:55.615 ","End":"02:00.480","Text":"Let me write this. When t is less than x,"},{"Start":"02:00.480 ","End":"02:07.430","Text":"then x minus t is positive and really we\u0027re only taking positive t. Yeah,"},{"Start":"02:07.430 ","End":"02:11.180","Text":"because we\u0027re taking the integral from 0 to infinity."},{"Start":"02:11.180 ","End":"02:18.890","Text":"Just repeating the first step is where we replace f(t) by e to the minus t and took"},{"Start":"02:18.890 ","End":"02:27.064","Text":"the integral from 0 to infinity and now we can replace the upper infinity by x."},{"Start":"02:27.064 ","End":"02:29.780","Text":"Because when we get beyond x,"},{"Start":"02:29.780 ","End":"02:32.260","Text":"the blue one is 0,"},{"Start":"02:32.260 ","End":"02:36.185","Text":"so the product is 0 so we get e to the minus t,"},{"Start":"02:36.185 ","End":"02:42.694","Text":"e to the minus x minus t minus x minus t makes it t minus x."},{"Start":"02:42.694 ","End":"02:45.860","Text":"Then, using rules of exponents,"},{"Start":"02:45.860 ","End":"02:48.965","Text":"what we get here is just e to the minus x,"},{"Start":"02:48.965 ","End":"02:50.905","Text":"from 0 to x,"},{"Start":"02:50.905 ","End":"02:53.930","Text":"the integral is dt so e to the minus x is a"},{"Start":"02:53.930 ","End":"02:57.530","Text":"constant as far as t goes and pull it out of the integral."},{"Start":"02:57.530 ","End":"03:03.585","Text":"This integral is just the length of the interval because it\u0027s 1 for the integrand."},{"Start":"03:03.585 ","End":"03:08.245","Text":"It\u0027s x minus 0 is x times e to the minus x comes out x,"},{"Start":"03:08.245 ","End":"03:09.925","Text":"e to the minus x,"},{"Start":"03:09.925 ","End":"03:12.710","Text":"and that\u0027s case 1."},{"Start":"03:12.950 ","End":"03:16.750","Text":"So now let\u0027s move to case 2,"},{"Start":"03:16.750 ","End":"03:19.480","Text":"which is where x is negative,"},{"Start":"03:19.480 ","End":"03:22.165","Text":"and if x is negative,"},{"Start":"03:22.165 ","End":"03:28.560","Text":"then x minus t is negative because t is positive,"},{"Start":"03:28.560 ","End":"03:30.330","Text":"goes from 0 to infinity."},{"Start":"03:30.330 ","End":"03:39.710","Text":"And so f(x) minus t is 0 because f is 0 for negative argument."},{"Start":"03:39.710 ","End":"03:49.955","Text":"This is the picture of this one is f(t) and the other one is f(x) minus t and this is x."},{"Start":"03:49.955 ","End":"03:54.650","Text":"But actually, we only take the part from 0 to infinity."},{"Start":"03:54.650 ","End":"03:58.025","Text":"So obviously, the product is 0 because one of the functions is"},{"Start":"03:58.025 ","End":"04:05.530","Text":"0. f convolution f(x) is 0 in case 2."},{"Start":"04:05.530 ","End":"04:10.910","Text":"When we summarize, we get that the convolution of f with"},{"Start":"04:10.910 ","End":"04:16.430","Text":"f at the point x is either x e to the minus x,"},{"Start":"04:16.430 ","End":"04:19.040","Text":"that\u0027s our case 1 when x is positive,"},{"Start":"04:19.040 ","End":"04:21.619","Text":"and 0 when x is negative,"},{"Start":"04:21.619 ","End":"04:24.810","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":30987},{"Watched":false,"Name":"Exercise 3","Duration":"4m 6s","ChapterTopicVideoID":29374,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.870","Text":"In this exercise, we have to find a function f in our space G,"},{"Start":"00:06.870 ","End":"00:08.370","Text":"and to remind you,"},{"Start":"00:08.370 ","End":"00:09.780","Text":"this is a space of piecewise"},{"Start":"00:09.780 ","End":"00:15.540","Text":"continuous and absolutely integrable functions on the whole real line,"},{"Start":"00:15.540 ","End":"00:18.330","Text":"such that f hat,"},{"Start":"00:18.330 ","End":"00:26.085","Text":"meaning the Fourier transform of f of Omega is sine Omega over Omega^2."},{"Start":"00:26.085 ","End":"00:30.885","Text":"Now, squared means this times itself,"},{"Start":"00:30.885 ","End":"00:32.180","Text":"and that reminds us,"},{"Start":"00:32.180 ","End":"00:36.515","Text":"or it should remind us of convolution theorem, the plan is,"},{"Start":"00:36.515 ","End":"00:44.585","Text":"find a function g such that the Fourier transform of g is sine Omega over Omega,"},{"Start":"00:44.585 ","End":"00:50.930","Text":"and then apply the convolution theorem to g convolved with itself."},{"Start":"00:50.930 ","End":"00:56.630","Text":"F of the convolution g star g will be,"},{"Start":"00:56.630 ","End":"01:00.200","Text":"don\u0027t forget the 2Pi Fourier transform of g,"},{"Start":"01:00.200 ","End":"01:04.095","Text":"this is g with a hat on it, times itself."},{"Start":"01:04.095 ","End":"01:06.405","Text":"This is equal to 2Pi,"},{"Start":"01:06.405 ","End":"01:12.915","Text":"and this is going to be sine Omega over Omega times itself, meaning squared."},{"Start":"01:12.915 ","End":"01:16.435","Text":"Now, we just have a matter of 2Pi,"},{"Start":"01:16.435 ","End":"01:18.710","Text":"since Fourier transform is linear,"},{"Start":"01:18.710 ","End":"01:21.185","Text":"if we put a 1/2Pi here,"},{"Start":"01:21.185 ","End":"01:23.690","Text":"we\u0027ll get here exactly what we want."},{"Start":"01:23.690 ","End":"01:27.725","Text":"Our function little f is going to be this."},{"Start":"01:27.725 ","End":"01:29.690","Text":"Now, what remains is to find"},{"Start":"01:29.690 ","End":"01:36.290","Text":"g. I want to refer you to a previous exercise, you know what?"},{"Start":"01:36.290 ","End":"01:38.510","Text":"I\u0027ll take you there."},{"Start":"01:38.510 ","End":"01:41.810","Text":"We had an exercise to compute the Fourier transform of"},{"Start":"01:41.810 ","End":"01:45.625","Text":"the characteristic function on the interval minus 1,1,"},{"Start":"01:45.625 ","End":"01:48.365","Text":"and at the end of this exercise,"},{"Start":"01:48.365 ","End":"01:49.850","Text":"we got the answer,"},{"Start":"01:49.850 ","End":"01:52.835","Text":"Sine Omega over Pi Omega,"},{"Start":"01:52.835 ","End":"02:00.065","Text":"returning, we showed that the Fourier transform of this is sine Omega over Pi Omega."},{"Start":"02:00.065 ","End":"02:02.645","Text":"Again, very close to what we want,"},{"Start":"02:02.645 ","End":"02:04.795","Text":"just an extra Pi here."},{"Start":"02:04.795 ","End":"02:11.535","Text":"We\u0027ll take g to be Pi times this characteristic function,"},{"Start":"02:11.535 ","End":"02:15.440","Text":"and then the Fourier transform of g will be this without the Pi here,"},{"Start":"02:15.440 ","End":"02:17.210","Text":"which is exactly what we want."},{"Start":"02:17.210 ","End":"02:19.490","Text":"Just to summarize the important parts,"},{"Start":"02:19.490 ","End":"02:29.730","Text":"we found the function g to equal this and the function f that we want is equal to this."},{"Start":"02:29.730 ","End":"02:37.435","Text":"That gives us that f is 1/2Pi g star g,"},{"Start":"02:37.435 ","End":"02:40.295","Text":"which is equal to 1/2Pi,"},{"Start":"02:40.295 ","End":"02:46.385","Text":"and then g star g is the convolution of this with itself,"},{"Start":"02:46.385 ","End":"02:48.365","Text":"so we get Pi^2,"},{"Start":"02:48.365 ","End":"02:54.750","Text":"and then this convolved with itself and let\u0027s just tidy up the fractions,"},{"Start":"02:54.750 ","End":"03:00.000","Text":"and so we have that f is Pi/2, that\u0027s this."},{"Start":"03:00.000 ","End":"03:08.715","Text":"The convolution of Chi with Chi on the interval minus 1,1,"},{"Start":"03:08.715 ","End":"03:12.666","Text":"but we\u0027ve done this exercise already."},{"Start":"03:12.666 ","End":"03:14.500","Text":"In the previous exercise,"},{"Start":"03:14.500 ","End":"03:18.100","Text":"we showed that the convolution of this Chi minus"},{"Start":"03:18.100 ","End":"03:23.215","Text":"1,1 with itself is the following definition."},{"Start":"03:23.215 ","End":"03:27.130","Text":"I\u0027ll remind you, this was"},{"Start":"03:27.130 ","End":"03:32.380","Text":"the exercise that we did compute the convolution and I\u0027ll just scroll down,"},{"Start":"03:32.380 ","End":"03:37.105","Text":"you can look at it later or find the clip for this exercise,"},{"Start":"03:37.105 ","End":"03:40.324","Text":"and in the end, this is what we got,"},{"Start":"03:40.324 ","End":"03:42.795","Text":"which is what is here."},{"Start":"03:42.795 ","End":"03:46.280","Text":"All that remains now is the matter of the Pi/2,"},{"Start":"03:46.280 ","End":"03:51.200","Text":"so if we multiply this by Pi/2,"},{"Start":"03:51.200 ","End":"03:57.315","Text":"just multiply each of these cases by Pi/2 and we get the following,"},{"Start":"03:57.315 ","End":"04:00.486","Text":"get of 2 plus x Pi/2 times 2 plus x."},{"Start":"04:00.486 ","End":"04:01.580","Text":"Everything is the same,"},{"Start":"04:01.580 ","End":"04:03.305","Text":"except with the Pi/2 here and here,"},{"Start":"04:03.305 ","End":"04:04.745","Text":"and this is our answer,"},{"Start":"04:04.745 ","End":"04:07.050","Text":"and we are done."}],"ID":30988},{"Watched":false,"Name":"Exercise 4","Duration":"4m 56s","ChapterTopicVideoID":29375,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"In this exercise, we\u0027ll let E be"},{"Start":"00:03.090 ","End":"00:07.020","Text":"the subspace of G consisting of those functions which have"},{"Start":"00:07.020 ","End":"00:16.229","Text":"a second derivative in the space G. Our task is to find a function g such that for all f,"},{"Start":"00:16.229 ","End":"00:19.035","Text":"the following equation is satisfied."},{"Start":"00:19.035 ","End":"00:22.425","Text":"This equation is called an integral equation,"},{"Start":"00:22.425 ","End":"00:25.620","Text":"and the reason we require that f\u0027\u0027 exists"},{"Start":"00:25.620 ","End":"00:29.490","Text":"is that f \u0027\u0027 appears here as part of the condition."},{"Start":"00:29.490 ","End":"00:32.880","Text":"The space G is good for performing convolutions."},{"Start":"00:32.880 ","End":"00:36.930","Text":"If you take any two functions in G, the convolution exists."},{"Start":"00:36.930 ","End":"00:41.270","Text":"Now note, if we look at f minus f\u0027\u0027 of the function,"},{"Start":"00:41.270 ","End":"00:45.680","Text":"we have some function of t times another function of x minus"},{"Start":"00:45.680 ","End":"00:51.740","Text":"t. This should remind us of convolution and we can write this,"},{"Start":"00:51.740 ","End":"00:56.375","Text":"the left-hand side of the convolution of f minus f\u0027\u0027 with"},{"Start":"00:56.375 ","End":"01:01.500","Text":"g. The convolution operation is linear,"},{"Start":"01:01.500 ","End":"01:03.950","Text":"so if we take a difference here,"},{"Start":"01:03.950 ","End":"01:07.565","Text":"it becomes the difference of the convolutions."},{"Start":"01:07.565 ","End":"01:12.520","Text":"Now, we\u0027ll apply the Fourier transform to this whole equation."},{"Start":"01:12.520 ","End":"01:19.280","Text":"The Fourier transform of a convolution is 2 Pi times the product,"},{"Start":"01:19.280 ","End":"01:24.650","Text":"and similarly here, 2 Pi times the product of the transforms."},{"Start":"01:24.650 ","End":"01:31.430","Text":"Now, we\u0027re going to use a rule to figure out the transform of the second derivative."},{"Start":"01:31.430 ","End":"01:34.190","Text":"The rule we need is this one,"},{"Start":"01:34.190 ","End":"01:38.645","Text":"which gives us in general the Fourier transform of an nth derivative,"},{"Start":"01:38.645 ","End":"01:43.970","Text":"is the transform times i Omega to the power of n and in our case,"},{"Start":"01:43.970 ","End":"01:45.665","Text":"n equals 2,"},{"Start":"01:45.665 ","End":"01:49.340","Text":"and so this becomes this."},{"Start":"01:49.340 ","End":"01:51.860","Text":"Now i^2 is minus 1."},{"Start":"01:51.860 ","End":"01:54.290","Text":"The minus will cancel with the minus."},{"Start":"01:54.290 ","End":"01:57.140","Text":"We can cancel the 2 here, here,"},{"Start":"01:57.140 ","End":"02:02.045","Text":"and here, and what we get is the following."},{"Start":"02:02.045 ","End":"02:06.410","Text":"I\u0027ve highlighted f hat of Omega because we\u0027re going to take it out"},{"Start":"02:06.410 ","End":"02:10.910","Text":"the brackets and get that Pi times"},{"Start":"02:10.910 ","End":"02:15.095","Text":"g hat plus Pi Omega squared"},{"Start":"02:15.095 ","End":"02:21.410","Text":"g hat minus 1 times f hat of Omega equals 0."},{"Start":"02:21.410 ","End":"02:27.825","Text":"Now, this is true for all f and so this has to be equal to 0."},{"Start":"02:27.825 ","End":"02:31.250","Text":"For example you could let f be this thing"},{"Start":"02:31.250 ","End":"02:35.695","Text":"itself and then you get something squared is 0, so it\u0027s 0."},{"Start":"02:35.695 ","End":"02:40.640","Text":"We get that this is equal to 0 and bring the one over to the other side,"},{"Start":"02:40.640 ","End":"02:42.740","Text":"and this is what we have now."},{"Start":"02:42.740 ","End":"02:51.215","Text":"We can isolate g hat of Omega because it\u0027s this times Pi plus Pi Omega squared,"},{"Start":"02:51.215 ","End":"02:54.490","Text":"or Pi times 1 plus Omega squared,"},{"Start":"02:54.490 ","End":"02:56.558","Text":"so dividing both sides,"},{"Start":"02:56.558 ","End":"02:59.870","Text":"don\u0027t know why there\u0027s a blank line here, never mind,"},{"Start":"02:59.870 ","End":"03:05.240","Text":"so g prime is 1 over Pi times 1 plus Omega squared."},{"Start":"03:05.240 ","End":"03:09.920","Text":"Now we want to get from g hat back to g. We look in"},{"Start":"03:09.920 ","End":"03:15.830","Text":"the table of Fourier transforms and we note the following rule,"},{"Start":"03:15.830 ","End":"03:20.785","Text":"which is pretty much what we have if a is equal to 1,"},{"Start":"03:20.785 ","End":"03:24.485","Text":"and so the inverse transform of what we have is"},{"Start":"03:24.485 ","End":"03:28.295","Text":"e to the minus a absolute value of x with a equals 1,"},{"Start":"03:28.295 ","End":"03:32.345","Text":"so e to the minus absolute value of x."},{"Start":"03:32.345 ","End":"03:36.125","Text":"That\u0027s the answer and we are done."},{"Start":"03:36.125 ","End":"03:41.900","Text":"But don\u0027t go yet, I think I\u0027ve cheated you by just pulling this formula out of a table."},{"Start":"03:41.900 ","End":"03:43.925","Text":"I\u0027ll prove it."},{"Start":"03:43.925 ","End":"03:47.390","Text":"I looked back and in the introduction section there was"},{"Start":"03:47.390 ","End":"03:53.090","Text":"an exercise to show that if f(x) is this function,"},{"Start":"03:53.090 ","End":"03:57.365","Text":"then the Fourier transform of f is this function."},{"Start":"03:57.365 ","End":"04:00.260","Text":"If we take b equals a,"},{"Start":"04:00.260 ","End":"04:01.955","Text":"first of all note,"},{"Start":"04:01.955 ","End":"04:06.965","Text":"here I can replace x by absolute value of x because it\u0027s positive."},{"Start":"04:06.965 ","End":"04:16.090","Text":"Here, I can replace x by minus absolute value of x because x is negative."},{"Start":"04:16.090 ","End":"04:22.450","Text":"So this is e to the minus b absolute value of x, but b is a."},{"Start":"04:22.450 ","End":"04:27.210","Text":"So we get e to the minus a absolute value of x,"},{"Start":"04:27.210 ","End":"04:30.655","Text":"and then from this formula replacing b by a,"},{"Start":"04:30.655 ","End":"04:32.395","Text":"we get this expression."},{"Start":"04:32.395 ","End":"04:36.850","Text":"Then a simple addition in complex numbers,"},{"Start":"04:36.850 ","End":"04:38.440","Text":"we get cross-multiplying,"},{"Start":"04:38.440 ","End":"04:40.065","Text":"a minus i Omega,"},{"Start":"04:40.065 ","End":"04:41.740","Text":"and here, a plus i Omega,"},{"Start":"04:41.740 ","End":"04:44.434","Text":"the plus minus i Omega cancels, we get 2a."},{"Start":"04:44.434 ","End":"04:47.470","Text":"This times this, it\u0027s the product of conjugates,"},{"Start":"04:47.470 ","End":"04:49.495","Text":"so a^2 plus Omega squared,"},{"Start":"04:49.495 ","End":"04:51.280","Text":"and then the 2 cancels with the 2,"},{"Start":"04:51.280 ","End":"04:52.660","Text":"the Pi goes here,"},{"Start":"04:52.660 ","End":"04:55.160","Text":"and this is the answer and this is what we wanted."},{"Start":"04:55.160 ","End":"04:57.420","Text":"Now, we\u0027re really done."}],"ID":30989},{"Watched":false,"Name":"Exercise 5","Duration":"2m 34s","ChapterTopicVideoID":29376,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.615","Text":"In this exercise, we\u0027re given two functions: f(x) is 1 over x squared plus 4,"},{"Start":"00:06.615 ","End":"00:09.750","Text":"and g(x) is 1 over x squared plus 1."},{"Start":"00:09.750 ","End":"00:14.054","Text":"Our task is to compute the convolution of these two functions,"},{"Start":"00:14.054 ","End":"00:21.000","Text":"f*g. We\u0027re given a useful formula from the table of Fourier transforms,"},{"Start":"00:21.000 ","End":"00:24.645","Text":"but the Fourier transform of 1 over x squared plus a squared"},{"Start":"00:24.645 ","End":"00:29.085","Text":"is 1 over 2^minus a absolute value of Omega."},{"Start":"00:29.085 ","End":"00:33.029","Text":"You could try to compute it directly using the formula of a convolution,"},{"Start":"00:33.029 ","End":"00:35.610","Text":"but it comes up pretty difficult which is why we need"},{"Start":"00:35.610 ","End":"00:39.585","Text":"another approach and we\u0027re going to use the convolution theorem."},{"Start":"00:39.585 ","End":"00:44.660","Text":"Instead of computing f*g will compute first the Fourier transform,"},{"Start":"00:44.660 ","End":"00:48.140","Text":"and later we\u0027ll come back from the transform."},{"Start":"00:48.140 ","End":"00:53.511","Text":"Fourier transform of a convolution is 2 Pi times the product of the transforms,"},{"Start":"00:53.511 ","End":"01:02.225","Text":"and this will give us 2Pi times the transform of f which is 1 over x squared plus 2^2."},{"Start":"01:02.225 ","End":"01:05.435","Text":"I\u0027m writing it like this because we\u0027re going to use this formula."},{"Start":"01:05.435 ","End":"01:07.700","Text":"The other one, 1 over x squared plus 1,"},{"Start":"01:07.700 ","End":"01:10.280","Text":"is 1 over x squared plus 1 squared."},{"Start":"01:10.280 ","End":"01:11.630","Text":"From this formula,"},{"Start":"01:11.630 ","End":"01:14.695","Text":"one\u0027s with a=2 and one\u0027s with a=1,"},{"Start":"01:14.695 ","End":"01:17.670","Text":"we\u0027ll get the following and we can simplify this."},{"Start":"01:17.670 ","End":"01:22.380","Text":"We have 2Pi times 1/4 times 1/2."},{"Start":"01:22.380 ","End":"01:25.430","Text":"The half with the 2 goes so it\u0027s Pi over 4,"},{"Start":"01:25.430 ","End":"01:33.120","Text":"and then using the rules of exponents we get here e^minus 3 absolute value of Omega."},{"Start":"01:33.120 ","End":"01:35.159","Text":"Together with the Pi over 4 like I said,"},{"Start":"01:35.159 ","End":"01:38.330","Text":"let\u0027s see if we can manipulate this a bit to get it in"},{"Start":"01:38.330 ","End":"01:42.950","Text":"the form 1/2a^minus a absolute value of Omega."},{"Start":"01:42.950 ","End":"01:45.150","Text":"Well, a is got to be 3,"},{"Start":"01:45.150 ","End":"01:47.655","Text":"so we want here 1/6."},{"Start":"01:47.655 ","End":"01:49.685","Text":"If I multiply and divide,"},{"Start":"01:49.685 ","End":"01:53.960","Text":"you can take here the 2a which is 6,"},{"Start":"01:53.960 ","End":"01:58.060","Text":"put a 6 here to compensate, and then simplify."},{"Start":"01:58.060 ","End":"02:03.420","Text":"What we get is 6Pi over 4 is 3Pi over 2,"},{"Start":"02:03.420 ","End":"02:11.720","Text":"and this part it becomes f(1) over x squared plus 3^2."},{"Start":"02:11.720 ","End":"02:16.650","Text":"Since the Fourier transform of this is equal to 3Pi over"},{"Start":"02:16.650 ","End":"02:21.900","Text":"2 the transform of this and the transform is linear with respect to constants,"},{"Start":"02:21.900 ","End":"02:26.970","Text":"we can just take this function multiply it by 3Pi over 2."},{"Start":"02:26.970 ","End":"02:31.185","Text":"F*g is 3 Pi over 2 times 1 over x^2 plus 9."},{"Start":"02:31.185 ","End":"02:34.450","Text":"That\u0027s the answer and we\u0027re done."}],"ID":30990},{"Watched":false,"Name":"Exercise 6 Part 1","Duration":"4m 36s","ChapterTopicVideoID":29377,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"This exercise has 2 parts."},{"Start":"00:02.790 ","End":"00:09.105","Text":"In part a, we\u0027re going to compute the Fourier transform of this function of x,"},{"Start":"00:09.105 ","End":"00:13.110","Text":"and part b, we\u0027ll read when we come to it."},{"Start":"00:13.110 ","End":"00:17.790","Text":"Part a, we\u0027re going to do it directly without using table of transforms or anything,"},{"Start":"00:17.790 ","End":"00:19.514","Text":"just from the formula."},{"Start":"00:19.514 ","End":"00:22.800","Text":"The Fourier transform of this function is 1/2Pi,"},{"Start":"00:22.800 ","End":"00:27.855","Text":"the integral of the function times e to the minus i Omega x dx."},{"Start":"00:27.855 ","End":"00:29.970","Text":"This gives us a function of Omega."},{"Start":"00:29.970 ","End":"00:36.070","Text":"E to the minus i Omega x is cosine Omega x minus i sine Omega x,"},{"Start":"00:36.070 ","End":"00:40.025","Text":"and then we can break this up into 2 integrals."},{"Start":"00:40.025 ","End":"00:41.510","Text":"The first one with the cosine,"},{"Start":"00:41.510 ","End":"00:43.190","Text":"the second one with the sine."},{"Start":"00:43.190 ","End":"00:48.145","Text":"This one is an even function because this is even and this is even,"},{"Start":"00:48.145 ","End":"00:50.885","Text":"and the other part is an odd function because an"},{"Start":"00:50.885 ","End":"00:53.705","Text":"even times an odd function is an odd function,"},{"Start":"00:53.705 ","End":"00:58.400","Text":"so the integral of an odd function over a symmetric interval is 0."},{"Start":"00:58.400 ","End":"01:00.355","Text":"This drops off."},{"Start":"01:00.355 ","End":"01:04.190","Text":"Integral of an even function on a symmetric interval is"},{"Start":"01:04.190 ","End":"01:09.335","Text":"twice the integral of just the positive part of the interval."},{"Start":"01:09.335 ","End":"01:14.480","Text":"We have the integral from 0 to infinity and the 2 cancels with the 2 here,"},{"Start":"01:14.480 ","End":"01:16.945","Text":"so it\u0027s just 1/Pi."},{"Start":"01:16.945 ","End":"01:22.505","Text":"From the cosine, we can go back to exponential form using this formula,"},{"Start":"01:22.505 ","End":"01:24.590","Text":"and then we can rearrange a bit,"},{"Start":"01:24.590 ","End":"01:26.180","Text":"put the 2 in front,"},{"Start":"01:26.180 ","End":"01:31.085","Text":"also multiply out the e to the minus x with the 2 exponents here,"},{"Start":"01:31.085 ","End":"01:32.540","Text":"and we get this."},{"Start":"01:32.540 ","End":"01:35.030","Text":"In this integral, we\u0027ll do by parts."},{"Start":"01:35.030 ","End":"01:38.705","Text":"This is the formula for the integration by parts."},{"Start":"01:38.705 ","End":"01:43.050","Text":"We\u0027ll take this part as f in the formula and this is g\u0027,"},{"Start":"01:43.050 ","End":"01:45.020","Text":"so what we need is f,"},{"Start":"01:45.020 ","End":"01:50.855","Text":"g. This is f and g is the integral of this, minus,"},{"Start":"01:50.855 ","End":"01:54.680","Text":"and then here we have f\u0027 with g. f\u0027 is 1,"},{"Start":"01:54.680 ","End":"01:59.455","Text":"so it just disappears and g again is this from here."},{"Start":"01:59.455 ","End":"02:02.525","Text":"Now to evaluate this at the upper limit,"},{"Start":"02:02.525 ","End":"02:05.029","Text":"it\u0027s a limit as x goes to infinity."},{"Start":"02:05.029 ","End":"02:06.860","Text":"When x goes to infinity,"},{"Start":"02:06.860 ","End":"02:11.315","Text":"this is e to the minus x times e to the i something,"},{"Start":"02:11.315 ","End":"02:16.160","Text":"e to the power of an imaginary number is on the unit circle, so bounded."},{"Start":"02:16.160 ","End":"02:19.190","Text":"We just really look at e to the minus x,"},{"Start":"02:19.190 ","End":"02:22.460","Text":"which goes to 0 when x goes to infinity,"},{"Start":"02:22.460 ","End":"02:25.145","Text":"and similarly, here also e to the minus x."},{"Start":"02:25.145 ","End":"02:28.220","Text":"Not only does this go to 0 as well as this,"},{"Start":"02:28.220 ","End":"02:33.300","Text":"but this is an exponential decay to 0,"},{"Start":"02:33.300 ","End":"02:37.740","Text":"which is much quicker than 1 plus x goes to infinity,"},{"Start":"02:37.740 ","End":"02:42.060","Text":"this wins, so even the product goes to 0."},{"Start":"02:42.060 ","End":"02:45.595","Text":"Now, plug in x=0,"},{"Start":"02:45.595 ","End":"02:48.950","Text":"but we need to take it with a minus because it\u0027s the lower limit,"},{"Start":"02:48.950 ","End":"02:51.830","Text":"this is 1, and this also is 1,"},{"Start":"02:51.830 ","End":"02:54.655","Text":"but we need minus and minus."},{"Start":"02:54.655 ","End":"02:59.030","Text":"The other integral, just similar to the integral"},{"Start":"02:59.030 ","End":"03:02.510","Text":"we just had, I\u0027m going to do the same thing,"},{"Start":"03:02.510 ","End":"03:06.575","Text":"but this time we\u0027ll have the constant in the denominator squared,"},{"Start":"03:06.575 ","End":"03:11.060","Text":"minus 1 plus i Omega^2 and minus 1 minus i Omega^2,"},{"Start":"03:11.060 ","End":"03:15.090","Text":"and again from 0 to infinity."},{"Start":"03:15.410 ","End":"03:19.760","Text":"Again, when x goes to infinity,"},{"Start":"03:19.760 ","End":"03:20.930","Text":"this goes to 0,"},{"Start":"03:20.930 ","End":"03:22.400","Text":"this goes to 0,"},{"Start":"03:22.400 ","End":"03:26.450","Text":"we don\u0027t even have the 1 plus x part so it\u0027s more"},{"Start":"03:26.450 ","End":"03:30.945","Text":"straightforward and just similar to what we did here."},{"Start":"03:30.945 ","End":"03:36.980","Text":"The infinity part is the 0 and then minus this and minus this."},{"Start":"03:36.980 ","End":"03:42.380","Text":"This minus will cancel this minus and this minus,"},{"Start":"03:42.380 ","End":"03:46.820","Text":"and if we put a common denominator here and here,"},{"Start":"03:46.820 ","End":"03:50.560","Text":"common denominator, this is what we get the straightforward."},{"Start":"03:50.560 ","End":"03:53.700","Text":"Then here the 2 with the 2 cancels,"},{"Start":"03:53.700 ","End":"03:56.805","Text":"we get 1/Pi 1 plus Omega^2."},{"Start":"03:56.805 ","End":"04:01.265","Text":"I got the details here."},{"Start":"04:01.265 ","End":"04:04.430","Text":"It comes out to 2 minus 2 Omega^2,"},{"Start":"04:04.430 ","End":"04:06.500","Text":"and then we can cancel the 2 again."},{"Start":"04:06.500 ","End":"04:08.675","Text":"If we take a common denominator,"},{"Start":"04:08.675 ","End":"04:14.020","Text":"here we have 1 minus Omega^2 over the 1 plus Omega^2 squared."},{"Start":"04:14.020 ","End":"04:16.520","Text":"Here, we have to compensate."},{"Start":"04:16.520 ","End":"04:19.340","Text":"We have to multiply top and bottom by 1 plus Omega^2"},{"Start":"04:19.340 ","End":"04:23.880","Text":"again and so we get 1 plus Omega ^2 from the 1."},{"Start":"04:23.880 ","End":"04:27.765","Text":"Now the Omega^2 minus Omega^2 cancels,"},{"Start":"04:27.765 ","End":"04:30.570","Text":"which brings us to the answer,"},{"Start":"04:30.570 ","End":"04:34.310","Text":"2 over Pi 1 plus 0mega^2 squared."},{"Start":"04:34.310 ","End":"04:37.320","Text":"That concludes part a."}],"ID":30991},{"Watched":false,"Name":"Exercise 6 Part 2","Duration":"2m 15s","ChapterTopicVideoID":29378,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.655","Text":"Now we come to part b."},{"Start":"00:02.655 ","End":"00:07.125","Text":"In part a, we computed the Fourier transform of this function,"},{"Start":"00:07.125 ","End":"00:09.375","Text":"and this was the result we got."},{"Start":"00:09.375 ","End":"00:14.445","Text":"In part b, we\u0027re going to use this to solve the following integral equation."},{"Start":"00:14.445 ","End":"00:18.780","Text":"We need to solve it for f. Here again is that integral equation."},{"Start":"00:18.780 ","End":"00:20.415","Text":"Now, quite often,"},{"Start":"00:20.415 ","End":"00:22.920","Text":"this integral equations have a convolution in them,"},{"Start":"00:22.920 ","End":"00:26.460","Text":"and you can see this by the x minus t and the t here,"},{"Start":"00:26.460 ","End":"00:29.460","Text":"so it\u0027s the convolution of these 2 functions."},{"Start":"00:29.460 ","End":"00:33.260","Text":"It\u0027s the e to the minus absolute value of x function"},{"Start":"00:33.260 ","End":"00:38.585","Text":"convolved with the f of x function, and this as is."},{"Start":"00:38.585 ","End":"00:42.470","Text":"Now, let\u0027s take the Fourier transform of both sides."},{"Start":"00:42.470 ","End":"00:43.730","Text":"For the left-hand side,"},{"Start":"00:43.730 ","End":"00:47.510","Text":"we\u0027ll use the convolution theorem and get"},{"Start":"00:47.510 ","End":"00:53.420","Text":"the transform of a convolution is 2pi times the product of the transform,"},{"Start":"00:53.420 ","End":"00:56.165","Text":"so 2pi transform of this,"},{"Start":"00:56.165 ","End":"00:58.850","Text":"times the transform of this."},{"Start":"00:58.850 ","End":"01:02.780","Text":"Now, we\u0027ve seen this before in Fourier transforms,"},{"Start":"01:02.780 ","End":"01:04.475","Text":"but in case not,"},{"Start":"01:04.475 ","End":"01:07.805","Text":"this is the entry in the table that we\u0027re going to use,"},{"Start":"01:07.805 ","End":"01:10.250","Text":"in our case, a equals 1,"},{"Start":"01:10.250 ","End":"01:13.480","Text":"so we get from here with a equals 1,"},{"Start":"01:13.480 ","End":"01:16.485","Text":"1 over pi times 1 plus Omega squared."},{"Start":"01:16.485 ","End":"01:20.025","Text":"The transform of f will be f hat."},{"Start":"01:20.025 ","End":"01:25.260","Text":"Here, this is the result of part a, that\u0027s this."},{"Start":"01:25.260 ","End":"01:31.695","Text":"This will cancel with the 2 here,"},{"Start":"01:31.695 ","End":"01:37.270","Text":"and also this 2 will cancel with this 2 and this will be a 1,"},{"Start":"01:37.270 ","End":"01:39.095","Text":"whoops, I forgot to cancel the pi."},{"Start":"01:39.095 ","End":"01:45.995","Text":"Yeah, what we\u0027re left with is pi times f hat of Omega is 1 over 1 plus Omega squared."},{"Start":"01:45.995 ","End":"01:49.090","Text":"Now just divide both sides by pi,"},{"Start":"01:49.090 ","End":"01:53.940","Text":"and we get that f hat of Omega is 1 over pi 1 plus Omega squared."},{"Start":"01:53.940 ","End":"01:55.800","Text":"This should look familiar."},{"Start":"01:55.800 ","End":"02:00.080","Text":"We just use this a moment ago here,"},{"Start":"02:00.080 ","End":"02:02.435","Text":"and we can also use it backwards."},{"Start":"02:02.435 ","End":"02:05.180","Text":"If the transform of f is this,"},{"Start":"02:05.180 ","End":"02:10.955","Text":"then f is e to the minus a absolute value of x with a equals 1."},{"Start":"02:10.955 ","End":"02:15.750","Text":"It\u0027s this, and that\u0027s the answer and we\u0027re done."}],"ID":30992},{"Watched":false,"Name":"Exercise 7 Part 1","Duration":"3m 48s","ChapterTopicVideoID":29379,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:03.150","Text":"In this exercise of 2 parts,"},{"Start":"00:03.150 ","End":"00:06.795","Text":"let\u0027s just do Part a in this clip and Part b in the next."},{"Start":"00:06.795 ","End":"00:11.250","Text":"We have to compute the following convolution of this function with itself,"},{"Start":"00:11.250 ","End":"00:16.460","Text":"and this function which is the characteristic function of the interval 0,"},{"Start":"00:16.460 ","End":"00:19.650","Text":"1 is defined at the point x to be 1 or"},{"Start":"00:19.650 ","End":"00:23.654","Text":"0 depending on whether x belongs to the interval or doesn\u0027t."},{"Start":"00:23.654 ","End":"00:25.380","Text":"Yeah, In Part b later."},{"Start":"00:25.380 ","End":"00:31.305","Text":"The convolution of Psi with itself is just using the definition"},{"Start":"00:31.305 ","End":"00:37.910","Text":"of the convolution of this function of t and the same function of x minus t,"},{"Start":"00:37.910 ","End":"00:41.195","Text":"the integral from minus infinity to infinity of the product."},{"Start":"00:41.195 ","End":"00:44.075","Text":"We can simplify this expression."},{"Start":"00:44.075 ","End":"00:51.815","Text":"Psi 0, 1 of x minus t is 1 if and only if x minus t is in the interval 0,"},{"Start":"00:51.815 ","End":"00:55.235","Text":"1, meaning 0 less than or equal to x minus t less than or equal to 1."},{"Start":"00:55.235 ","End":"00:56.555","Text":"This is equivalent."},{"Start":"00:56.555 ","End":"00:59.420","Text":"If you multiply by minus 1,"},{"Start":"00:59.420 ","End":"01:03.770","Text":"we get that minus 1 less than or equal to t minus x less than or equal to 0."},{"Start":"01:03.770 ","End":"01:09.230","Text":"Add x everywhere and we get x minus 1 less than or equal"},{"Start":"01:09.230 ","End":"01:14.945","Text":"to t less than or equal to x which means that t is in the interval from x minus 1 to x,"},{"Start":"01:14.945 ","End":"01:20.530","Text":"and so this is the characteristic function on the interval x minus 1x"},{"Start":"01:20.530 ","End":"01:26.957","Text":"(t) and we can replace that here so we get the following."},{"Start":"01:26.957 ","End":"01:30.140","Text":"We have 2 expressions, Psi (t),"},{"Start":"01:30.140 ","End":"01:31.880","Text":"one on the interval 0, 1,"},{"Start":"01:31.880 ","End":"01:34.985","Text":"and one on the interval x minus 1-x."},{"Start":"01:34.985 ","End":"01:38.360","Text":"What we\u0027ll do is split up into cases."},{"Start":"01:38.360 ","End":"01:41.360","Text":"In the first case we\u0027ll assume that"},{"Start":"01:41.360 ","End":"01:45.200","Text":"this characteristic function is completely to the right of this,"},{"Start":"01:45.200 ","End":"01:49.000","Text":"meaning the interval x minus 1x is to the right of 0, 1,"},{"Start":"01:49.000 ","End":"01:55.485","Text":"and that means that 1 is less than x minus 1 which means that x is bigger than 2."},{"Start":"01:55.485 ","End":"01:58.550","Text":"In this case the product of these 2 functions is"},{"Start":"01:58.550 ","End":"02:03.040","Text":"0 because at every point one or the other is 0,"},{"Start":"02:03.040 ","End":"02:11.120","Text":"and so we get that the integral that we want is the integral of 0 which is 0."},{"Start":"02:11.120 ","End":"02:16.420","Text":"The next case we\u0027ll move this to the left until there\u0027s some overlap,"},{"Start":"02:16.420 ","End":"02:21.275","Text":"and that will be the case where x minus 1"},{"Start":"02:21.275 ","End":"02:26.615","Text":"is between 0 and 1 which means that x is between 1 and 2."},{"Start":"02:26.615 ","End":"02:30.050","Text":"In this case there is an overlap here,"},{"Start":"02:30.050 ","End":"02:36.780","Text":"and we get the integral of 1 times 1 on"},{"Start":"02:36.780 ","End":"02:44.735","Text":"the interval from x minus 1-1 which comes out to be 1 minus x minus 1 which is 2 minus x."},{"Start":"02:44.735 ","End":"02:46.355","Text":"That\u0027s case 2."},{"Start":"02:46.355 ","End":"02:50.015","Text":"Case 3 will be when we move this x minus 1x interval"},{"Start":"02:50.015 ","End":"02:54.750","Text":"further to the left until it overlaps on the other side,"},{"Start":"02:54.750 ","End":"02:59.470","Text":"and in this case the overlap is from 0-x."},{"Start":"02:59.470 ","End":"03:03.576","Text":"Case 3 x is between 0 and 1;"},{"Start":"03:03.576 ","End":"03:07.940","Text":"the integral of 1 times 1 from 0-x,"},{"Start":"03:07.940 ","End":"03:10.355","Text":"and this comes out to be just x."},{"Start":"03:10.355 ","End":"03:16.460","Text":"The fourth case is when this one is completely to the left of this one."},{"Start":"03:16.460 ","End":"03:19.460","Text":"Again, the integral\u0027s going to be 0,"},{"Start":"03:19.460 ","End":"03:24.910","Text":"so this is the case where x is less than 0 and the integral comes out to be 0."},{"Start":"03:24.910 ","End":"03:28.620","Text":"That\u0027s 4 cases and let\u0027s summarize"},{"Start":"03:28.620 ","End":"03:32.525","Text":"them and we really only care about the cases where it\u0027s non-zero,"},{"Start":"03:32.525 ","End":"03:40.590","Text":"so we had from case 3 that the t=x on this interval and this was case 2."},{"Start":"03:40.590 ","End":"03:45.584","Text":"X is between 1 and 2 and it\u0027s 2 minus x and 0 in other cases,"},{"Start":"03:45.584 ","End":"03:48.660","Text":"and that concludes Part a."}],"ID":30993},{"Watched":false,"Name":"Exercise 7 Part 2","Duration":"5m 4s","ChapterTopicVideoID":29380,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.005","Text":"We just did part a."},{"Start":"00:02.005 ","End":"00:04.695","Text":"Now we come to part b where we have to prove that this"},{"Start":"00:04.695 ","End":"00:08.120","Text":"improper integral is equal to Pi over 3."},{"Start":"00:08.120 ","End":"00:11.584","Text":"We\u0027re given a hint to use Plancherel\u0027s theorem."},{"Start":"00:11.584 ","End":"00:16.910","Text":"This with the function f being X convolution"},{"Start":"00:16.910 ","End":"00:23.160","Text":"with X. X on the interval 0,1 we evaluated this in part a."},{"Start":"00:23.160 ","End":"00:28.050","Text":"This is equal to the piecewise defined function, this one."},{"Start":"00:28.050 ","End":"00:35.010","Text":"Before we apply Plancherel\u0027s theorem where we\u0027ll need the Fourier transform of this,"},{"Start":"00:35.010 ","End":"00:38.790","Text":"let\u0027s first compute the Fourier transform of just X itself."},{"Start":"00:38.790 ","End":"00:43.470","Text":"We\u0027ll just drop the 0,1 for notation, it becomes cumbersome."},{"Start":"00:43.470 ","End":"00:49.960","Text":"X hat using the definition of the Fourier transform is this integral."},{"Start":"00:49.960 ","End":"00:56.615","Text":"The function X is zero outside the interval from 0 to1."},{"Start":"00:56.615 ","End":"01:02.915","Text":"It\u0027s one inside so this drops off on the limits change to 0,1."},{"Start":"01:02.915 ","End":"01:11.075","Text":"This integral is e to the minus i Omega x divided by minus i Omega from 0 to 1."},{"Start":"01:11.075 ","End":"01:17.205","Text":"Get rid of this minus and reverse the 0,1, like so."},{"Start":"01:17.205 ","End":"01:21.150","Text":"Then this is equal to when X is 0,"},{"Start":"01:21.150 ","End":"01:22.815","Text":"this is equal to 1."},{"Start":"01:22.815 ","End":"01:24.315","Text":"When X is 1,"},{"Start":"01:24.315 ","End":"01:27.135","Text":"this is e to the minus i Omega."},{"Start":"01:27.135 ","End":"01:29.235","Text":"This is what we have."},{"Start":"01:29.235 ","End":"01:36.170","Text":"We can write e to the minus i Omega in terms of sine and cosine as follows."},{"Start":"01:36.170 ","End":"01:39.395","Text":"Now that we have the Fourier transform of X,"},{"Start":"01:39.395 ","End":"01:45.755","Text":"we\u0027re ready to apply Plancherel\u0027s theorem on X convolution X,"},{"Start":"01:45.755 ","End":"01:49.430","Text":"reminder of what the Plancherel\u0027s theorem is."},{"Start":"01:49.430 ","End":"01:54.515","Text":"What we get in our case is the following."},{"Start":"01:54.515 ","End":"01:56.795","Text":"This is meant to be a hat."},{"Start":"01:56.795 ","End":"01:59.455","Text":"It doesn\u0027t look like a hat, but it is."},{"Start":"01:59.455 ","End":"02:02.720","Text":"For this, we\u0027ll use the convolution theorem."},{"Start":"02:02.720 ","End":"02:09.125","Text":"The transform of a convolution is 2Pi times the product of the transforms."},{"Start":"02:09.125 ","End":"02:11.610","Text":"We get 2PiXhatXhat."},{"Start":"02:13.810 ","End":"02:20.805","Text":"We have 2Pi^2 and here are 2Pi, so together 2Pi^3."},{"Start":"02:20.805 ","End":"02:23.840","Text":"Then we have X,X and will this squared,"},{"Start":"02:23.840 ","End":"02:27.760","Text":"so it\u0027s the absolute value of X^4."},{"Start":"02:27.760 ","End":"02:30.675","Text":"We know what X hat is."},{"Start":"02:30.675 ","End":"02:32.295","Text":"It\u0027s equal to this."},{"Start":"02:32.295 ","End":"02:34.545","Text":"Let\u0027s substitute that here."},{"Start":"02:34.545 ","End":"02:37.150","Text":"This is what we get."},{"Start":"02:37.610 ","End":"02:45.705","Text":"The 2Pi^3 with 1/2Pi^4 gives us 1/2Pi."},{"Start":"02:45.705 ","End":"02:48.020","Text":"Then we\u0027ll write this numerator."},{"Start":"02:48.020 ","End":"02:55.900","Text":"Just group it as 1 minus cosine Omega plus i sine Omega, real and imaginary."},{"Start":"02:56.080 ","End":"03:02.960","Text":"The absolute value of a plus ib^2 is a^2 plus b^2."},{"Start":"03:02.960 ","End":"03:04.445","Text":"We can apply that here,"},{"Start":"03:04.445 ","End":"03:07.580","Text":"but still we need an x plus squared."},{"Start":"03:07.580 ","End":"03:10.300","Text":"Also i^4 is 1."},{"Start":"03:10.300 ","End":"03:11.985","Text":"We get the following."},{"Start":"03:11.985 ","End":"03:15.040","Text":"Now we need to square this."},{"Start":"03:15.320 ","End":"03:19.730","Text":"Just using regular algebra, this is what we get."},{"Start":"03:19.730 ","End":"03:22.310","Text":"Cosine squared plus sine squared is 1,"},{"Start":"03:22.310 ","End":"03:28.180","Text":"so this becomes 2 minus 2 cosine Omega, but still squared."},{"Start":"03:28.180 ","End":"03:30.345","Text":"We can take 2 out,"},{"Start":"03:30.345 ","End":"03:33.315","Text":"but it\u0027s 2^2 and 2^2 over 2 is 2."},{"Start":"03:33.315 ","End":"03:37.330","Text":"This becomes 2 over Pi times this integral."},{"Start":"03:37.330 ","End":"03:41.950","Text":"This equals this and we can write it the other way round."},{"Start":"03:41.950 ","End":"03:46.190","Text":"What we had to evaluate was essentially this,"},{"Start":"03:46.190 ","End":"03:47.900","Text":"but it was given in terms of x."},{"Start":"03:47.900 ","End":"03:51.515","Text":"But x is just a dummy variable here in this integral,"},{"Start":"03:51.515 ","End":"03:54.020","Text":"then we can replace Omega by x."},{"Start":"03:54.020 ","End":"04:00.735","Text":"Now a reminder of what convolution with X was from before, it\u0027s this."},{"Start":"04:00.735 ","End":"04:02.870","Text":"If we take the integral of this,"},{"Start":"04:02.870 ","End":"04:09.240","Text":"we need either x^2 or 2 minus x^2 or 0^2 depending on where X is."},{"Start":"04:09.240 ","End":"04:11.160","Text":"Of course we don\u0027t need the 0 part."},{"Start":"04:11.160 ","End":"04:15.075","Text":"But from 0-1, we have x^2."},{"Start":"04:15.075 ","End":"04:16.380","Text":"From 1 to 2,"},{"Start":"04:16.380 ","End":"04:18.990","Text":"we have 2 minus x^2."},{"Start":"04:18.990 ","End":"04:21.570","Text":"The integral of x^2 is x^3 over 3."},{"Start":"04:21.570 ","End":"04:24.560","Text":"The integral of 2 minus x^2 is 2 minus x^3 over 3,"},{"Start":"04:24.560 ","End":"04:27.650","Text":"but with a minus here because of the integer derivative."},{"Start":"04:27.650 ","End":"04:36.420","Text":"Then this integral is equal to 1^3 over 3 minus 0, which is 1/3."},{"Start":"04:36.420 ","End":"04:41.550","Text":"Here, when x is 2, we get 0."},{"Start":"04:41.550 ","End":"04:43.155","Text":"When x is 1,"},{"Start":"04:43.155 ","End":"04:46.080","Text":"then we get here,"},{"Start":"04:46.080 ","End":"04:48.590","Text":"1^3 over 3 with a minus."},{"Start":"04:48.590 ","End":"04:50.900","Text":"But we also subtract it because it\u0027s a lower limit,"},{"Start":"04:50.900 ","End":"04:53.005","Text":"so it becomes a plus here."},{"Start":"04:53.005 ","End":"04:58.485","Text":"This is just 2/3 times a 1/2Pi, which is 1/3Pi."},{"Start":"04:58.485 ","End":"05:00.020","Text":"This is what we had to show,"},{"Start":"05:00.020 ","End":"05:02.135","Text":"go back and look at the original question."},{"Start":"05:02.135 ","End":"05:05.640","Text":"That proves it. We are done."}],"ID":30994},{"Watched":false,"Name":"Exercise 8","Duration":"3m 23s","ChapterTopicVideoID":29381,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.480","Text":"In this exercise, we\u0027re going to compute the convolution of this function with itself."},{"Start":"00:06.480 ","End":"00:09.885","Text":"This is kai on the interval 1-2."},{"Start":"00:09.885 ","End":"00:14.355","Text":"Applying the definition of convolution of 2 functions,"},{"Start":"00:14.355 ","End":"00:16.140","Text":"we have the integral,"},{"Start":"00:16.140 ","End":"00:17.760","Text":"the first function of t,"},{"Start":"00:17.760 ","End":"00:19.980","Text":"the second function of x minus t,"},{"Start":"00:19.980 ","End":"00:23.220","Text":"and the integral from minus infinity to infinity dt,"},{"Start":"00:23.220 ","End":"00:25.170","Text":"which leaves us with a function of x."},{"Start":"00:25.170 ","End":"00:29.550","Text":"We can simplify the second function here, the second part,"},{"Start":"00:29.550 ","End":"00:34.710","Text":"Kai 1,2 of x minus t. Let\u0027s see when it\u0027s equal to 1."},{"Start":"00:34.710 ","End":"00:40.635","Text":"That\u0027s if and only if x minus t is between 1 and 2,"},{"Start":"00:40.635 ","End":"00:42.840","Text":"because that\u0027s the 1,2 here,"},{"Start":"00:42.840 ","End":"00:45.915","Text":"and then multiply it by minus 1."},{"Start":"00:45.915 ","End":"00:49.930","Text":"We get that t minus x is between minus 2 and minus 1."},{"Start":"00:49.930 ","End":"00:56.105","Text":"Now add x and we get that t is between x minus 2 and x minus 1."},{"Start":"00:56.105 ","End":"01:00.470","Text":"Which means that essentially this is the same as"},{"Start":"01:00.470 ","End":"01:06.410","Text":"the characteristic function of x minus 2x minus 1 and we can substitute"},{"Start":"01:06.410 ","End":"01:11.675","Text":"that here and get that what we have to evaluate is the following integral"},{"Start":"01:11.675 ","End":"01:17.555","Text":"for all x and we\u0027ll split this up into different cases of x."},{"Start":"01:17.555 ","End":"01:25.485","Text":"The first case we will take is where the blue one is totally to the right of the red one."},{"Start":"01:25.485 ","End":"01:30.195","Text":"This happens when 2 is less than x minus 2,"},{"Start":"01:30.195 ","End":"01:32.940","Text":"meaning that x is bigger than 4,"},{"Start":"01:32.940 ","End":"01:34.645","Text":"and in this case,"},{"Start":"01:34.645 ","End":"01:38.620","Text":"the integral just comes out to be zero because the product of these 2 functions"},{"Start":"01:38.620 ","End":"01:42.660","Text":"is 0 at every point 1 or the other is 0."},{"Start":"01:42.660 ","End":"01:45.435","Text":"Now we come to case 2,"},{"Start":"01:45.435 ","End":"01:49.830","Text":"where we move the blue one further to the left so there\u0027s some overlap"},{"Start":"01:49.830 ","End":"01:56.090","Text":"here and that means that x minus 2 is between 1 and 2,"},{"Start":"01:56.090 ","End":"01:58.795","Text":"which is the same as saying that x is between 3 and 4,"},{"Start":"01:58.795 ","End":"02:02.110","Text":"and in this case,"},{"Start":"02:02.110 ","End":"02:08.595","Text":"we get the integral from x minus 2 to 2 of 1,"},{"Start":"02:08.595 ","End":"02:12.835","Text":"which is 2 minus x minus 2 which is 4 minus x."},{"Start":"02:12.835 ","End":"02:14.335","Text":"That\u0027s case 2."},{"Start":"02:14.335 ","End":"02:16.480","Text":"Now let\u0027s get to case 3."},{"Start":"02:16.480 ","End":"02:20.140","Text":"This is where we move the blue one still further to the left,"},{"Start":"02:20.140 ","End":"02:23.390","Text":"but there\u0027s still overlap on the other side."},{"Start":"02:23.390 ","End":"02:28.665","Text":"This time, x minus 1 is between 1 and 2,"},{"Start":"02:28.665 ","End":"02:32.184","Text":"which means that x is between 2 and 3,"},{"Start":"02:32.184 ","End":"02:40.560","Text":"and the overlap is from 1 to x minus 1 so we get the integral from 1 to x minus 1 of 1,"},{"Start":"02:40.560 ","End":"02:45.735","Text":"which comes out to be x minus 2 and there\u0027s 1 more case to go."},{"Start":"02:45.735 ","End":"02:48.545","Text":"Here again, there\u0027s no overlap."},{"Start":"02:48.545 ","End":"02:52.210","Text":"This is when x minus 1 is less than 1,"},{"Start":"02:52.210 ","End":"02:55.395","Text":"which means that x is less than 2,"},{"Start":"02:55.395 ","End":"02:59.900","Text":"and in this case, the integral also comes out to be 0."},{"Start":"02:59.900 ","End":"03:01.160","Text":"We have 4 cases."},{"Start":"03:01.160 ","End":"03:03.185","Text":"Now we need to summarize."},{"Start":"03:03.185 ","End":"03:07.725","Text":"We had in case 3 that when x is between 2 and 3,"},{"Start":"03:07.725 ","End":"03:09.565","Text":"it\u0027s x minus 2."},{"Start":"03:09.565 ","End":"03:11.120","Text":"That\u0027s still up here."},{"Start":"03:11.120 ","End":"03:16.090","Text":"In case 2, we had 4 minus x for x between 3 and 4."},{"Start":"03:16.090 ","End":"03:17.300","Text":"In the other 2 cases,"},{"Start":"03:17.300 ","End":"03:20.795","Text":"the first and the fourth combined with just the 0 case."},{"Start":"03:20.795 ","End":"03:23.970","Text":"This is the answer and we\u0027re done."}],"ID":30995},{"Watched":false,"Name":"Exercise 9","Duration":"3m 36s","ChapterTopicVideoID":29382,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.910","Text":"In this exercise, we\u0027re going to compute the convolution of this function with itself,"},{"Start":"00:05.910 ","End":"00:10.425","Text":"where this is the characteristic function on the interval 0,2,"},{"Start":"00:10.425 ","End":"00:12.375","Text":"this is the Greek letter Chi,"},{"Start":"00:12.375 ","End":"00:17.010","Text":"and it\u0027s equal to 1 if x is in the interval and 0 otherwise."},{"Start":"00:17.010 ","End":"00:19.680","Text":"Using the definition of convolution,"},{"Start":"00:19.680 ","End":"00:22.650","Text":"we take the integral from minus infinity to infinity."},{"Start":"00:22.650 ","End":"00:23.895","Text":"In the first function,"},{"Start":"00:23.895 ","End":"00:27.135","Text":"we replace the argument by t, and in the second one,"},{"Start":"00:27.135 ","End":"00:28.935","Text":"by x minus t,"},{"Start":"00:28.935 ","End":"00:32.665","Text":"then take the integral dt and we get a function of x."},{"Start":"00:32.665 ","End":"00:34.970","Text":"This x minus t here is not convenient."},{"Start":"00:34.970 ","End":"00:40.010","Text":"What we can do is we can figure out an alternative way to express this."},{"Start":"00:40.010 ","End":"00:44.750","Text":"This function at x minus t is 1, well,"},{"Start":"00:44.750 ","End":"00:48.950","Text":"you can follow this line of inequality if and only if t is"},{"Start":"00:48.950 ","End":"00:50.990","Text":"in the interval from x minus 2 to"},{"Start":"00:50.990 ","End":"00:54.320","Text":"x so that this is the same as"},{"Start":"00:54.320 ","End":"00:58.700","Text":"the characteristic function on the interval from x minus 2 to x."},{"Start":"00:58.700 ","End":"01:02.830","Text":"Next, we\u0027re going to split up into different cases for x."},{"Start":"01:02.830 ","End":"01:06.349","Text":"But before that, we can rewrite this as follows,"},{"Start":"01:06.349 ","End":"01:09.530","Text":"just replacing this by what it\u0027s equal to."},{"Start":"01:09.530 ","End":"01:14.660","Text":"The first case will be when x is to the right enough"},{"Start":"01:14.660 ","End":"01:20.479","Text":"that this Chi function doesn\u0027t overlap with this Chi function."},{"Start":"01:20.479 ","End":"01:24.670","Text":"For that to happen, we want x minus 2 to be bigger than 2."},{"Start":"01:24.670 ","End":"01:26.790","Text":"If x minus 2 is bigger than 2,"},{"Start":"01:26.790 ","End":"01:29.745","Text":"it means that x is bigger than 4."},{"Start":"01:29.745 ","End":"01:31.310","Text":"In this case, obviously,"},{"Start":"01:31.310 ","End":"01:34.250","Text":"the product is equal to 0,"},{"Start":"01:34.250 ","End":"01:41.030","Text":"so the integral is equal to 0 and so the convolution of our function is 0."},{"Start":"01:41.030 ","End":"01:46.085","Text":"In the next case, we\u0027re going to move this function further to the left, so it overlaps."},{"Start":"01:46.085 ","End":"01:50.820","Text":"We get something like this where this overlaps partly"},{"Start":"01:50.820 ","End":"01:57.035","Text":"and this situation will apply when x minus 2 is between 0 and 2,"},{"Start":"01:57.035 ","End":"01:59.390","Text":"which means that x is between 2 and 4."},{"Start":"01:59.390 ","End":"02:06.185","Text":"In this case, we get an overlap between x minus 2 and 2,"},{"Start":"02:06.185 ","End":"02:09.965","Text":"so we get the integral from x minus 2-2 of 1dt,"},{"Start":"02:09.965 ","End":"02:12.800","Text":"which comes out to be 2 minus x minus 2,"},{"Start":"02:12.800 ","End":"02:14.695","Text":"which is 4 minus x."},{"Start":"02:14.695 ","End":"02:16.920","Text":"So that\u0027s our Case 2."},{"Start":"02:16.920 ","End":"02:19.365","Text":"Now, in Case 3,"},{"Start":"02:19.365 ","End":"02:22.720","Text":"we move the blue ones still further to the left,"},{"Start":"02:22.720 ","End":"02:30.430","Text":"so it overlaps like this and this is the case when x is between 0 and 2."},{"Start":"02:30.430 ","End":"02:35.250","Text":"This time, the overlap is from 0 to x and the"},{"Start":"02:35.250 ","End":"02:40.660","Text":"integral comes out to be integral from 0 to x of 1, which is just x."},{"Start":"02:40.660 ","End":"02:47.697","Text":"We have 1 more case to go when this is still further to the left, like this."},{"Start":"02:47.697 ","End":"02:53.320","Text":"This is similar to the first case where the integral is going to be 0,"},{"Start":"02:53.320 ","End":"02:55.030","Text":"the convolution will be 0."},{"Start":"02:55.030 ","End":"03:01.930","Text":"This is the case when x is less than 0 and the convolution comes out to be 0."},{"Start":"03:01.930 ","End":"03:07.535","Text":"That\u0027s all 4 cases and now we just have to summarize them in a table."},{"Start":"03:07.535 ","End":"03:11.785","Text":"So this was Case 3."},{"Start":"03:11.785 ","End":"03:15.450","Text":"That\u0027s where it\u0027s equal to x from 0-2."},{"Start":"03:15.450 ","End":"03:21.220","Text":"This is Case 2 and this is both Cases 1 and 4,"},{"Start":"03:21.220 ","End":"03:23.795","Text":"which is the zero case."},{"Start":"03:23.795 ","End":"03:26.510","Text":"You\u0027re curious what this function looks like,"},{"Start":"03:26.510 ","End":"03:28.955","Text":"this is what it is."},{"Start":"03:28.955 ","End":"03:34.040","Text":"It\u0027s mostly 0 except for this triangle here."},{"Start":"03:34.040 ","End":"03:37.050","Text":"That concludes this exercise."}],"ID":30996},{"Watched":false,"Name":"Exercise 10","Duration":"3m ","ChapterTopicVideoID":29383,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In this exercise, we\u0027re going to compute"},{"Start":"00:02.550 ","End":"00:06.585","Text":"the convolution of 2 characteristic chi functions,"},{"Start":"00:06.585 ","End":"00:08.490","Text":"one on the interval 0, 1,"},{"Start":"00:08.490 ","End":"00:11.220","Text":"and the other on the interval 1, 2."},{"Start":"00:11.220 ","End":"00:16.875","Text":"Just a reminder, the chi function on an interval or on any set for that matter is"},{"Start":"00:16.875 ","End":"00:22.890","Text":"equal to 1 if x belongs to the set and 0, otherwise."},{"Start":"00:22.890 ","End":"00:29.265","Text":"We\u0027ll use the definition of the convolution that the convenience will reverse the order."},{"Start":"00:29.265 ","End":"00:34.800","Text":"We get the first function of t and the second one of x minus"},{"Start":"00:34.800 ","End":"00:40.760","Text":"t. This we\u0027ve done in a previous exercise, we\u0027ve simplified this."},{"Start":"00:40.760 ","End":"00:42.830","Text":"Here\u0027s the calculations."},{"Start":"00:42.830 ","End":"00:50.240","Text":"It comes out to be the characteristic function on the interval from x minus 1 to x."},{"Start":"00:50.240 ","End":"00:51.875","Text":"We can replace it,"},{"Start":"00:51.875 ","End":"00:55.070","Text":"and we\u0027ve got this expression."},{"Start":"00:55.070 ","End":"00:59.074","Text":"This is what we have to compute for different values of x."},{"Start":"00:59.074 ","End":"01:01.565","Text":"There are 4 basic cases."},{"Start":"01:01.565 ","End":"01:03.125","Text":"In the first case,"},{"Start":"01:03.125 ","End":"01:06.575","Text":"the blue one will be completely to the right of the red one."},{"Start":"01:06.575 ","End":"01:10.405","Text":"That\u0027s when x minus 1 is to the right of 2,"},{"Start":"01:10.405 ","End":"01:13.725","Text":"so we get x bigger than 3,"},{"Start":"01:13.725 ","End":"01:17.640","Text":"and in this case, the product is obviously 0."},{"Start":"01:17.800 ","End":"01:21.065","Text":"The answer to the convolution,"},{"Start":"01:21.065 ","End":"01:23.825","Text":"which is this, is 0,"},{"Start":"01:23.825 ","End":"01:25.435","Text":"that\u0027s for Case 1."},{"Start":"01:25.435 ","End":"01:26.955","Text":"In Case 2,"},{"Start":"01:26.955 ","End":"01:30.975","Text":"we\u0027ll move the blue one to the left,"},{"Start":"01:30.975 ","End":"01:36.590","Text":"so it overlaps partially with the interval 1, 2, like so."},{"Start":"01:36.590 ","End":"01:40.100","Text":"This is the case when x minus 1 is between 1 and 2,"},{"Start":"01:40.100 ","End":"01:42.110","Text":"or x is between 2 and 3."},{"Start":"01:42.110 ","End":"01:48.300","Text":"In this case, the overlap is between x minus 1 and 2 as in the picture."},{"Start":"01:48.300 ","End":"01:51.090","Text":"What we get is 2 minus x minus 1,"},{"Start":"01:51.090 ","End":"01:54.488","Text":"which is 3 minus x, that\u0027s Case 2."},{"Start":"01:54.488 ","End":"01:58.870","Text":"And in Case 3, we\u0027re going to move this still further to the left,"},{"Start":"01:58.870 ","End":"02:02.480","Text":"like so, but there\u0027s still some overlap here."},{"Start":"02:02.480 ","End":"02:06.605","Text":"From the picture, this means that x is between 1 and 2,"},{"Start":"02:06.605 ","End":"02:11.215","Text":"and we get the integral from 1 to x of 1,"},{"Start":"02:11.215 ","End":"02:14.190","Text":"which is just x minus 1."},{"Start":"02:14.190 ","End":"02:16.214","Text":"That\u0027s the third case."},{"Start":"02:16.214 ","End":"02:18.950","Text":"Now the last case is when we move"},{"Start":"02:18.950 ","End":"02:21.710","Text":"the blue one still further to the left so there is no overlap."},{"Start":"02:21.710 ","End":"02:24.020","Text":"It\u0027s a bit like the first case."},{"Start":"02:24.020 ","End":"02:26.450","Text":"This is when x is less than 1,"},{"Start":"02:26.450 ","End":"02:28.670","Text":"and just like in Case 1,"},{"Start":"02:28.670 ","End":"02:32.600","Text":"the integral, which is the convolution, is 0."},{"Start":"02:32.600 ","End":"02:35.285","Text":"What we have to do now is summarize."},{"Start":"02:35.285 ","End":"02:37.735","Text":"This is Case 3,"},{"Start":"02:37.735 ","End":"02:39.810","Text":"this is Case 2,"},{"Start":"02:39.810 ","End":"02:44.255","Text":"and this is Cases 1 and 4 combined where the convolution is 0."},{"Start":"02:44.255 ","End":"02:46.760","Text":"Just to complete the clip,"},{"Start":"02:46.760 ","End":"02:50.620","Text":"I\u0027ll show you a picture of what this convolution looks like."},{"Start":"02:50.620 ","End":"02:53.730","Text":"It\u0027s mostly 0, and then from 1-2,"},{"Start":"02:53.730 ","End":"02:56.425","Text":"it climbs to 1 and then down to 0,"},{"Start":"02:56.425 ","End":"03:00.550","Text":"again continues being 0. That\u0027s it."}],"ID":30997},{"Watched":false,"Name":"Exercise 11 Part 1","Duration":"2m 58s","ChapterTopicVideoID":29384,"CourseChapterTopicPlaylistID":294465,"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\u0027re going to compute the following convolution in 2 ways."},{"Start":"00:04.890 ","End":"00:08.430","Text":"By the way, this is the function"},{"Start":"00:08.430 ","End":"00:12.930","Text":"like f(x) equals e to the minus x^2 is not a numerical computation."},{"Start":"00:12.930 ","End":"00:14.685","Text":"It\u0027s convolution of functions."},{"Start":"00:14.685 ","End":"00:16.500","Text":"We\u0027re going to do it in 2 ways."},{"Start":"00:16.500 ","End":"00:20.594","Text":"First way using definition of convolution and secondly,"},{"Start":"00:20.594 ","End":"00:27.060","Text":"using the convolution theorem and also some properties of the Fourier transform."},{"Start":"00:27.060 ","End":"00:32.010","Text":"For part a, we\u0027ll be using a well-known integral result that the integral of e"},{"Start":"00:32.010 ","End":"00:37.380","Text":"to the minus x^2 from minus infinity to infinity is square root of Pi."},{"Start":"00:37.380 ","End":"00:41.225","Text":"Let\u0027s start. Using the definition of convolution,"},{"Start":"00:41.225 ","End":"00:44.600","Text":"we have here the integral from minus infinity to infinity."},{"Start":"00:44.600 ","End":"00:49.310","Text":"Here, we replace x by t and here we replace x by x minus"},{"Start":"00:49.310 ","End":"00:54.510","Text":"t. Squaring this e to the minus x^2 doesn\u0027t depend on t,"},{"Start":"00:54.510 ","End":"00:57.260","Text":"so we can pull it in front of the integral."},{"Start":"00:57.260 ","End":"01:02.420","Text":"What we\u0027re left with is 2xt minus t^2 minus t^2,"},{"Start":"01:02.420 ","End":"01:04.505","Text":"which is minus 2t ^2."},{"Start":"01:04.505 ","End":"01:08.015","Text":"Into a bit of algebra on this exponent,"},{"Start":"01:08.015 ","End":"01:11.480","Text":"we can take 2 out if we can take minus 2 out,"},{"Start":"01:11.480 ","End":"01:14.210","Text":"so we can reverse the order to complete the square,"},{"Start":"01:14.210 ","End":"01:21.020","Text":"we add a quarter t^2 and then to compensate since we have minus twice a quarter,"},{"Start":"01:21.020 ","End":"01:24.235","Text":"we have to add a half of the x^2."},{"Start":"01:24.235 ","End":"01:28.880","Text":"Now, this part is t minus a half x^2,"},{"Start":"01:28.880 ","End":"01:31.020","Text":"and the 2 we can write as root 2^2,"},{"Start":"01:31.020 ","End":"01:37.370","Text":"so we can combine the root t with t minus half x and get all that squared with a minus 2."},{"Start":"01:37.370 ","End":"01:41.105","Text":"This half x^2 in the exponent,"},{"Start":"01:41.105 ","End":"01:44.255","Text":"we can combine it with e to the minus x^2."},{"Start":"01:44.255 ","End":"01:49.510","Text":"What we\u0027re left with is e to the minus x^2 over 2."},{"Start":"01:49.510 ","End":"01:55.100","Text":"The exponent here from this multiplying the route 2 by the bracket,"},{"Start":"01:55.100 ","End":"02:02.060","Text":"we get root 2t minus 1 over root 2x and that\u0027s this squared."},{"Start":"02:02.060 ","End":"02:05.090","Text":"This point, we make a substitution."},{"Start":"02:05.090 ","End":"02:08.600","Text":"If this was just x^2 would be okay with the integral."},{"Start":"02:08.600 ","End":"02:10.160","Text":"But if it\u0027s not x, it\u0027s this,"},{"Start":"02:10.160 ","End":"02:15.230","Text":"so let u equal root 2t minus x over root t and then"},{"Start":"02:15.230 ","End":"02:21.060","Text":"du is root 2dt and dt is du divided by root 2."},{"Start":"02:21.060 ","End":"02:28.710","Text":"We can put 1 over root 2 in the beginning and then that becomes du and this becomes u."},{"Start":"02:28.710 ","End":"02:33.260","Text":"This is what we have after the substitution and this integral, we know what it is."},{"Start":"02:33.260 ","End":"02:36.020","Text":"I mean, it got it written with you now instead of x,"},{"Start":"02:36.020 ","End":"02:37.325","Text":"but that doesn\u0027t matter."},{"Start":"02:37.325 ","End":"02:40.645","Text":"This integral is square root of Pi."},{"Start":"02:40.645 ","End":"02:43.680","Text":"This is our answer but let\u0027s just tidy it up,"},{"Start":"02:43.680 ","End":"02:49.355","Text":"combine the 1 over root 2 with the root Pi to get root Pi over 2,"},{"Start":"02:49.355 ","End":"02:55.490","Text":"and recall that the left-hand side was what we were looking for this convolution,"},{"Start":"02:55.490 ","End":"02:59.640","Text":"so this is our answer and we are done."}],"ID":30998},{"Watched":false,"Name":"Exercise 11 Part 2","Duration":"3m 49s","ChapterTopicVideoID":29385,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.780","Text":"Now we come to part b of this exercise."},{"Start":"00:03.780 ","End":"00:06.270","Text":"We just did part a and we got the answer so we\u0027ll"},{"Start":"00:06.270 ","End":"00:09.389","Text":"check if doing it a different way gives the same result."},{"Start":"00:09.389 ","End":"00:12.300","Text":"It turns out that b is not so straightforward,"},{"Start":"00:12.300 ","End":"00:15.000","Text":"or at least it involves the Fourier transform because"},{"Start":"00:15.000 ","End":"00:17.760","Text":"of the convolution theorem and we\u0027ll need to look up"},{"Start":"00:17.760 ","End":"00:23.660","Text":"some Fourier transforms in a table and also some of the properties of Fourier transform."},{"Start":"00:23.660 ","End":"00:25.145","Text":"Anyway, let\u0027s get started."},{"Start":"00:25.145 ","End":"00:29.210","Text":"For convenience, we\u0027ll let h(x) be the answer."},{"Start":"00:29.210 ","End":"00:33.815","Text":"We\u0027re looking for the convolution of these two functions or this function with itself."},{"Start":"00:33.815 ","End":"00:36.635","Text":"Now we\u0027ll apply the convolution theorem,"},{"Start":"00:36.635 ","End":"00:40.661","Text":"which says that if we apply Fourier transform of a convolution,"},{"Start":"00:40.661 ","End":"00:44.180","Text":"we get 2Pi times the product of the transforms."},{"Start":"00:44.180 ","End":"00:48.150","Text":"The first step is to compute Fourier transform of e to the minus x squared."},{"Start":"00:48.150 ","End":"00:49.440","Text":"Well, we won\u0027t compute it,"},{"Start":"00:49.440 ","End":"00:52.130","Text":"we\u0027ll look it up in a table of Fourier transforms,"},{"Start":"00:52.130 ","End":"00:54.575","Text":"or we might have done it in a previous exercise."},{"Start":"00:54.575 ","End":"00:59.150","Text":"Then the event, the result is this as a function of Omega."},{"Start":"00:59.150 ","End":"01:00.500","Text":"If we just substitute that here,"},{"Start":"01:00.500 ","End":"01:03.395","Text":"we get 2Pi times this times itself,"},{"Start":"01:03.395 ","End":"01:08.210","Text":"which simplifies the 2/2,"},{"Start":"01:08.210 ","End":"01:09.825","Text":"2 is 1/2,"},{"Start":"01:09.825 ","End":"01:13.898","Text":"and the Pi with the root Pi, root Pi cancels."},{"Start":"01:13.898 ","End":"01:18.440","Text":"This times itself is e to the power of twice the exponents."},{"Start":"01:18.440 ","End":"01:20.720","Text":"This is what we get."},{"Start":"01:20.720 ","End":"01:27.065","Text":"What we have to do now to find h is to somehow figure out the inverse transform of this."},{"Start":"01:27.065 ","End":"01:30.575","Text":"Now we know the straightforward transform,"},{"Start":"01:30.575 ","End":"01:36.695","Text":"or at least we know the transform of something similar as we did here."},{"Start":"01:36.695 ","End":"01:42.545","Text":"What we can do is use the duality property of the transform."},{"Start":"01:42.545 ","End":"01:50.980","Text":"We take the transform of the transform of a function f. The result is 1 over 2Pi f(-x)."},{"Start":"01:50.980 ","End":"01:53.795","Text":"When we take the Fourier transform of a function of x,"},{"Start":"01:53.795 ","End":"01:54.980","Text":"we get a function of Omega."},{"Start":"01:54.980 ","End":"01:57.095","Text":"When we take the Fourier transform again,"},{"Start":"01:57.095 ","End":"02:00.165","Text":"we go back from Omega to x."},{"Start":"02:00.165 ","End":"02:03.680","Text":"Let f=h, and what we have is this,"},{"Start":"02:03.680 ","End":"02:09.330","Text":"but with h instead of f. Then we have f of this."},{"Start":"02:09.330 ","End":"02:10.620","Text":"But we can take the 1/2 out,"},{"Start":"02:10.620 ","End":"02:13.190","Text":"so it\u0027s 1/2 Fourier transform of this."},{"Start":"02:13.190 ","End":"02:15.910","Text":"Now this is not e to the minus Omega squared,"},{"Start":"02:15.910 ","End":"02:18.535","Text":"t to the minus Omega squared over 2."},{"Start":"02:18.535 ","End":"02:23.105","Text":"Just bringing the 2 over to the other side and the Pi over to this side."},{"Start":"02:23.105 ","End":"02:27.605","Text":"We can write this as something squared Omega over root 2 squared."},{"Start":"02:27.605 ","End":"02:29.915","Text":"It\u0027s still not Omega squared."},{"Start":"02:29.915 ","End":"02:33.860","Text":"We know that the Fourier transform of e to the minus x squared is this."},{"Start":"02:33.860 ","End":"02:34.910","Text":"Or in our case,"},{"Start":"02:34.910 ","End":"02:36.590","Text":"if we swap x and Omega,"},{"Start":"02:36.590 ","End":"02:38.300","Text":"we could write it this way."},{"Start":"02:38.300 ","End":"02:41.990","Text":"What we\u0027ll do, because this is Omega over root 2 is to apply"},{"Start":"02:41.990 ","End":"02:46.214","Text":"the scaling property that if we replace the function,"},{"Start":"02:46.214 ","End":"02:49.085","Text":"let\u0027s say f(x/a),"},{"Start":"02:49.085 ","End":"02:52.490","Text":"then the result is multiplied by absolute value"},{"Start":"02:52.490 ","End":"02:56.075","Text":"of a and we\u0027ll also stick in a inside the argument."},{"Start":"02:56.075 ","End":"03:00.890","Text":"If we apply this to e to the minus x squared with a being the square root of 2,"},{"Start":"03:00.890 ","End":"03:03.905","Text":"what we get is like this,"},{"Start":"03:03.905 ","End":"03:07.010","Text":"but we have an extra root 2 here, which is a,"},{"Start":"03:07.010 ","End":"03:10.310","Text":"and we also instead of x, put root 2x."},{"Start":"03:10.310 ","End":"03:15.740","Text":"Simplifying root 2x squared is 2x squared over 4 is x squared over 2."},{"Start":"03:15.740 ","End":"03:18.721","Text":"Then root 2 over 2 is 1 over root 2,"},{"Start":"03:18.721 ","End":"03:20.390","Text":"Pi over root Pi is root Pi,"},{"Start":"03:20.390 ","End":"03:22.605","Text":"so we get square root of Pi/2,"},{"Start":"03:22.605 ","End":"03:24.500","Text":"e to the minus x squared over 2."},{"Start":"03:24.500 ","End":"03:27.680","Text":"It\u0027s still not h, it\u0027s h(-x)."},{"Start":"03:27.680 ","End":"03:30.460","Text":"To get h(x), we replace x by minus x."},{"Start":"03:30.460 ","End":"03:34.370","Text":"Of course, that doesn\u0027t make any difference because it\u0027s squared."},{"Start":"03:34.370 ","End":"03:38.680","Text":"We get our answer that h(x) is this."},{"Start":"03:38.680 ","End":"03:40.490","Text":"h(x) is what we were looking for,"},{"Start":"03:40.490 ","End":"03:41.735","Text":"so this is our answer."},{"Start":"03:41.735 ","End":"03:44.115","Text":"If you compare it to the result of part a,"},{"Start":"03:44.115 ","End":"03:45.905","Text":"you\u0027ll see it\u0027s the same as what we got."},{"Start":"03:45.905 ","End":"03:49.980","Text":"Anyway, that concludes part b and this exercise."}],"ID":30999},{"Watched":false,"Name":"Exercise 12","Duration":"6m 17s","ChapterTopicVideoID":29386,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In this exercise, we are asked to solve"},{"Start":"00:02.550 ","End":"00:05.910","Text":"the following integral equation and it should jump"},{"Start":"00:05.910 ","End":"00:09.398","Text":"at you right away that the left-hand side;"},{"Start":"00:09.398 ","End":"00:15.840","Text":"the integral, is just the convolution of function f with itself."},{"Start":"00:15.840 ","End":"00:21.075","Text":"We can write it as f convolution f(x) equals"},{"Start":"00:21.075 ","End":"00:28.115","Text":"this function and let\u0027s make the exponents minus something squared."},{"Start":"00:28.115 ","End":"00:29.710","Text":"You\u0027ll see in a moment why."},{"Start":"00:29.710 ","End":"00:32.900","Text":"We take the convolution of both sides and on"},{"Start":"00:32.900 ","End":"00:37.106","Text":"the left we apply the convolution theorem, and get,"},{"Start":"00:37.106 ","End":"00:42.665","Text":"its 2Pi transform of f times transform of f. Here,"},{"Start":"00:42.665 ","End":"00:46.535","Text":"the Fourier transform of the right-hand side."},{"Start":"00:46.535 ","End":"00:52.280","Text":"We write it like this in the form ax plus b because we\u0027re going to use a formula."},{"Start":"00:52.280 ","End":"00:54.350","Text":"The first formula will be"},{"Start":"00:54.350 ","End":"00:58.580","Text":"the Fourier transform of e^ minus x^2 that should be well-known,"},{"Start":"00:58.580 ","End":"01:01.099","Text":"and it\u0027s equal to the following."},{"Start":"01:01.099 ","End":"01:04.185","Text":"But here we have e^minus something else squared,"},{"Start":"01:04.185 ","End":"01:06.390","Text":"you could think of it as ax plus b,"},{"Start":"01:06.390 ","End":"01:11.570","Text":"and then we can apply the scaling and shifting property."},{"Start":"01:11.570 ","End":"01:14.360","Text":"If we take f of ax plus b,"},{"Start":"01:14.360 ","End":"01:16.385","Text":"we get the following."},{"Start":"01:16.385 ","End":"01:22.160","Text":"In our case, a and b are both square root of 3/2 here and here."},{"Start":"01:22.160 ","End":"01:24.890","Text":"We\u0027ll work on this right-hand side and try to figure"},{"Start":"01:24.890 ","End":"01:27.830","Text":"out the Fourier transform of this function."},{"Start":"01:27.830 ","End":"01:31.880","Text":"What we have is instead of this function,"},{"Start":"01:31.880 ","End":"01:37.069","Text":"we replace Omega by Omega over a."},{"Start":"01:37.069 ","End":"01:39.870","Text":"A is root 3 over 2,"},{"Start":"01:39.870 ","End":"01:42.690","Text":"so 1 over a is root 2/ over 3."},{"Start":"01:42.690 ","End":"01:47.610","Text":"Similarly here, 1 over absolute a is also root 2 over 3."},{"Start":"01:47.610 ","End":"01:52.400","Text":"Here we have b over a which is 1 because a equals b,"},{"Start":"01:52.400 ","End":"01:55.985","Text":"but I wrote it in gray just so you can see."},{"Start":"01:55.985 ","End":"01:58.115","Text":"Bringing this down here,"},{"Start":"01:58.115 ","End":"02:03.630","Text":"we have that 2Pif-hat times f-hat is this,"},{"Start":"02:03.630 ","End":"02:06.330","Text":"just slightly simplified again."},{"Start":"02:06.330 ","End":"02:07.925","Text":"If we square this,"},{"Start":"02:07.925 ","End":"02:14.045","Text":"we get 2/3 of Omega squared and 2/3 over 4 is 1/6,"},{"Start":"02:14.045 ","End":"02:16.280","Text":"so this is what we have."},{"Start":"02:16.280 ","End":"02:19.424","Text":"Let\u0027s bring the 2Pi over to the right,"},{"Start":"02:19.424 ","End":"02:24.690","Text":"and then we just get f-hat of Omega squared 1 over 2Pi times this"},{"Start":"02:24.690 ","End":"02:30.935","Text":"lot and just a slight simplification of the numerical part."},{"Start":"02:30.935 ","End":"02:38.552","Text":"Now, take the square root of both sides just to extract f-hat of Omega."},{"Start":"02:38.552 ","End":"02:41.239","Text":"Just take the square root of the denominator,"},{"Start":"02:41.239 ","End":"02:42.860","Text":"divide this exponent by 2,"},{"Start":"02:42.860 ","End":"02:45.025","Text":"and divide this by 2."},{"Start":"02:45.025 ","End":"02:49.220","Text":"Recall that f-hat is the Fourier transform of"},{"Start":"02:49.220 ","End":"02:53.495","Text":"f. We would like to do an inverse transform to get back to f,"},{"Start":"02:53.495 ","End":"02:54.770","Text":"but we don\u0027t have to go backwards."},{"Start":"02:54.770 ","End":"02:57.395","Text":"We can go forwards because of duality."},{"Start":"02:57.395 ","End":"02:59.870","Text":"If we take the Fourier transform of both sides,"},{"Start":"02:59.870 ","End":"03:01.100","Text":"we get the Fourier transform."},{"Start":"03:01.100 ","End":"03:05.940","Text":"The Fourier transform of f is equal to this"},{"Start":"03:05.940 ","End":"03:08.660","Text":"where we just apply the Fourier transform to this part"},{"Start":"03:08.660 ","End":"03:11.630","Text":"because this is the constant and the Fourier transform is linear."},{"Start":"03:11.630 ","End":"03:12.950","Text":"We\u0027re going to apply 2 rules."},{"Start":"03:12.950 ","End":"03:17.540","Text":"One is the duality that we know that Fourier transform of Fourier"},{"Start":"03:17.540 ","End":"03:26.010","Text":"transform of f(x) is f of minus x and then times 1 over 2Pi."},{"Start":"03:26.010 ","End":"03:32.240","Text":"On the right, we can use the formula that multiplying a function by e^i some constant"},{"Start":"03:32.240 ","End":"03:39.240","Text":"times x is a shift to the right of constant c. What we do is,"},{"Start":"03:39.240 ","End":"03:41.450","Text":"it\u0027s not great notation,"},{"Start":"03:41.450 ","End":"03:47.780","Text":"but take the e^i Omega over 2 as e^icx only."},{"Start":"03:47.780 ","End":"03:49.470","Text":"We\u0027re not working with x,"},{"Start":"03:49.470 ","End":"03:51.400","Text":"we\u0027re working with Omega."},{"Start":"03:51.400 ","End":"03:56.705","Text":"The Fourier transform will take us back from Omega to x like a duality."},{"Start":"03:56.705 ","End":"03:58.824","Text":"C is a half,"},{"Start":"03:58.824 ","End":"04:04.390","Text":"so we take this as a function of x and then replace x by x minus 1/2."},{"Start":"04:04.390 ","End":"04:06.750","Text":"That\u0027s what is meant by this."},{"Start":"04:06.750 ","End":"04:08.870","Text":"Plug Fourier transform, get a function of x,"},{"Start":"04:08.870 ","End":"04:10.790","Text":"replace x by x minus 0.5."},{"Start":"04:10.790 ","End":"04:13.430","Text":"Continuing, that\u0027s due to operations."},{"Start":"04:13.430 ","End":"04:15.905","Text":"We\u0027ll bring the 2Pi over to the right,"},{"Start":"04:15.905 ","End":"04:21.215","Text":"and also we\u0027ll replace x by minus x it\u0027s only one place, x it\u0027s here."},{"Start":"04:21.215 ","End":"04:25.130","Text":"Instead of x write minus x."},{"Start":"04:25.130 ","End":"04:31.550","Text":"Now, we\u0027re going to compute the Fourier transform of e^minus Omega squared over 12."},{"Start":"04:31.550 ","End":"04:36.226","Text":"We can write this as e^minus something squared,"},{"Start":"04:36.226 ","End":"04:39.460","Text":"that something is 1 over root 12x."},{"Start":"04:39.460 ","End":"04:44.906","Text":"Once again we can apply the formula for e^minus x^2."},{"Start":"04:44.906 ","End":"04:48.074","Text":"Only we don\u0027t have x we have x over root 12,"},{"Start":"04:48.074 ","End":"04:56.625","Text":"so we can use this rule where a is root 12 and then we get root 12."},{"Start":"04:56.625 ","End":"05:00.590","Text":"Then the Fourier transform of e^minus x^2 only"},{"Start":"05:00.590 ","End":"05:05.974","Text":"replace Omega by root 12 Omega because we said a is root 12."},{"Start":"05:05.974 ","End":"05:08.980","Text":"Replacing Omega by root 12 Omega,"},{"Start":"05:08.980 ","End":"05:10.930","Text":"root 12^2 is 12,"},{"Start":"05:10.930 ","End":"05:12.860","Text":"12 over 4 is 3,"},{"Start":"05:12.860 ","End":"05:17.510","Text":"so we get here e^minus 3 Omega squared."},{"Start":"05:17.510 ","End":"05:19.745","Text":"Another simplification."},{"Start":"05:19.745 ","End":"05:25.385","Text":"Now that we figured out the Fourier transform of to the minus x^2 over 12 which is this,"},{"Start":"05:25.385 ","End":"05:29.140","Text":"we can put this in this formula here."},{"Start":"05:29.140 ","End":"05:36.025","Text":"First, let\u0027s repeat this because it\u0027s going to scroll off and now replace this by this."},{"Start":"05:36.025 ","End":"05:39.725","Text":"Well, we\u0027ve changed Omega to x."},{"Start":"05:39.725 ","End":"05:43.625","Text":"The transform of something of x is something of Omega and vice versa."},{"Start":"05:43.625 ","End":"05:47.710","Text":"Anyway, it should be minus 3x^2."},{"Start":"05:47.710 ","End":"05:51.935","Text":"This means replace x by minus x minus a half."},{"Start":"05:51.935 ","End":"05:55.396","Text":"What we get is the following,"},{"Start":"05:55.396 ","End":"05:59.690","Text":"just bring the constant out front and minus 3 times this squared."},{"Start":"05:59.690 ","End":"06:02.570","Text":"That\u0027s it, just one more step to go."},{"Start":"06:02.570 ","End":"06:04.314","Text":"Tidy up these constants,"},{"Start":"06:04.314 ","End":"06:05.644","Text":"I\u0027ll let you check."},{"Start":"06:05.644 ","End":"06:09.035","Text":"This comes out to be the fourth root of 6 over Pi."},{"Start":"06:09.035 ","End":"06:12.230","Text":"Here we can forget the minus and minus because it\u0027s squared,"},{"Start":"06:12.230 ","End":"06:18.520","Text":"so we have x plus 1/2 squared and this is the answer and we\u0027re done."}],"ID":31000},{"Watched":false,"Name":"Exercise 13","Duration":"1m 58s","ChapterTopicVideoID":29387,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:05.100","Text":"In this exercise, we\u0027re given f and this L^1 space and that it\u0027s"},{"Start":"00:05.100 ","End":"00:09.915","Text":"continuous and it satisfies the following integral equation."},{"Start":"00:09.915 ","End":"00:14.130","Text":"We have to prove that f is identically 0."},{"Start":"00:14.130 ","End":"00:15.915","Text":"Now when you look at this,"},{"Start":"00:15.915 ","End":"00:22.830","Text":"you think we could complete the square if we had also an x^2 here. That\u0027s what we\u0027ll do."},{"Start":"00:22.830 ","End":"00:27.600","Text":"We\u0027ll take this and multiply both sides by e to"},{"Start":"00:27.600 ","End":"00:34.065","Text":"the minus x^2 by cross-multiplying 0 by some degrees in 0."},{"Start":"00:34.065 ","End":"00:35.415","Text":"Now that we have this,"},{"Start":"00:35.415 ","End":"00:39.920","Text":"we can write it as minus x minus y^2."},{"Start":"00:39.920 ","End":"00:46.930","Text":"Now you can see there\u0027s a convolution of f(x) with the function e to the minus x^2."},{"Start":"00:46.930 ","End":"00:49.940","Text":"This will be replaced by y and this by x minus y. Yeah,"},{"Start":"00:49.940 ","End":"00:52.595","Text":"usually we have t, but this time we have y."},{"Start":"00:52.595 ","End":"00:59.660","Text":"Now, we apply the Fourier transform to both sides and use the convolution theorem."},{"Start":"00:59.660 ","End":"01:03.710","Text":"F of a convolution is going to be the product of the transforms,"},{"Start":"01:03.710 ","End":"01:05.075","Text":"but times 2 Pi,"},{"Start":"01:05.075 ","End":"01:09.305","Text":"so 2Pi transform of f is f hat."},{"Start":"01:09.305 ","End":"01:13.175","Text":"The transform of e to the minus x^2 is well known is this."},{"Start":"01:13.175 ","End":"01:16.730","Text":"You bring 2Pi to the other side, it\u0027s not 0."},{"Start":"01:16.730 ","End":"01:23.215","Text":"Also, this is not 0 and this is not 0."},{"Start":"01:23.215 ","End":"01:26.720","Text":"Really all we\u0027re left with is f-hat of Omega,"},{"Start":"01:26.720 ","End":"01:28.955","Text":"and that\u0027s identically 0."},{"Start":"01:28.955 ","End":"01:31.580","Text":"Now if the Fourier transform of f is 0,"},{"Start":"01:31.580 ","End":"01:34.160","Text":"doesn\u0027t mean that f is 0."},{"Start":"01:34.160 ","End":"01:36.440","Text":"Well, in general no,"},{"Start":"01:36.440 ","End":"01:38.480","Text":"but if f is continuous,"},{"Start":"01:38.480 ","End":"01:39.820","Text":"then it will be true."},{"Start":"01:39.820 ","End":"01:44.270","Text":"For example, if you changed f at a finite number of places,"},{"Start":"01:44.270 ","End":"01:46.625","Text":"from 0-1 or whatever,"},{"Start":"01:46.625 ","End":"01:49.325","Text":"then the Fourier transform wouldn\u0027t change."},{"Start":"01:49.325 ","End":"01:50.690","Text":"But when it\u0027s continuous,"},{"Start":"01:50.690 ","End":"01:53.690","Text":"we don\u0027t have any possibility to mess around with it,"},{"Start":"01:53.690 ","End":"01:56.638","Text":"it has to be 0 identically."},{"Start":"01:56.638 ","End":"01:59.520","Text":"That concludes this exercise."}],"ID":31001},{"Watched":false,"Name":"Exercise 14","Duration":"3m 6s","ChapterTopicVideoID":29388,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"In this exercise,"},{"Start":"00:01.830 ","End":"00:05.250","Text":"we\u0027re going to prove two properties of the convolution."},{"Start":"00:05.250 ","End":"00:11.415","Text":"In part A, we\u0027ll show that the convolution of 2 even functions is even."},{"Start":"00:11.415 ","End":"00:16.110","Text":"In part B, we\u0027ll show that the convolution operation is commutative,"},{"Start":"00:16.110 ","End":"00:23.670","Text":"f* g is the same as g* f. I will start with a and let f and g be even functions."},{"Start":"00:23.670 ","End":"00:26.880","Text":"We want to show that if h is the convolution,"},{"Start":"00:26.880 ","End":"00:28.710","Text":"then h is even."},{"Start":"00:28.710 ","End":"00:33.870","Text":"We\u0027ll start with a definition of h from the definition of convolution,"},{"Start":"00:33.870 ","End":"00:35.560","Text":"which is this,"},{"Start":"00:35.560 ","End":"00:44.440","Text":"and we\u0027ll make a substitution we\u0027ll replace t by minus tau so that dt is minus d tau."},{"Start":"00:44.440 ","End":"00:48.739","Text":"Note also that we have to change the upper and lower limits."},{"Start":"00:48.739 ","End":"00:51.590","Text":"If t goes from minus infinity to infinity,"},{"Start":"00:51.590 ","End":"00:54.560","Text":"tau goes from infinity to minus infinity."},{"Start":"00:54.560 ","End":"00:59.270","Text":"But there\u0027s a minus here from the minus d tau and"},{"Start":"00:59.270 ","End":"01:04.880","Text":"also these are upside down so we can throw out the minus and put these the right way up."},{"Start":"01:04.880 ","End":"01:06.935","Text":"F and g are both even."},{"Start":"01:06.935 ","End":"01:08.645","Text":"That\u0027s the given."},{"Start":"01:08.645 ","End":"01:16.075","Text":"We can replace minus tau by tau and x plus t by minus x minus tau."},{"Start":"01:16.075 ","End":"01:19.080","Text":"I\u0027m going to do an unnecessary step and replace"},{"Start":"01:19.080 ","End":"01:22.895","Text":"tau backed by t because It\u0027s a dummy variable."},{"Start":"01:22.895 ","End":"01:27.020","Text":"But I want you to see more clearly that this is a convolution."},{"Start":"01:27.020 ","End":"01:29.600","Text":"If you think of minus x as the variable,"},{"Start":"01:29.600 ","End":"01:32.525","Text":"this is the convolution of minus x,"},{"Start":"01:32.525 ","End":"01:40.790","Text":"meaning h of minus x is what you get here if you replace x by minus x, which is this."},{"Start":"01:40.790 ","End":"01:44.300","Text":"This comes out to be h of minus x,"},{"Start":"01:44.300 ","End":"01:46.545","Text":"and that\u0027s part A."},{"Start":"01:46.545 ","End":"01:48.390","Text":"On to part B,"},{"Start":"01:48.390 ","End":"01:52.640","Text":"we need to show that if you change the order, it doesn\u0027t matter."},{"Start":"01:52.640 ","End":"01:54.755","Text":"It\u0027s a bit similar to part A."},{"Start":"01:54.755 ","End":"01:56.195","Text":"What we\u0027re going to do,"},{"Start":"01:56.195 ","End":"02:01.000","Text":"is start with a definition and again, make a substitution."},{"Start":"02:01.000 ","End":"02:03.670","Text":"This time, a different 1."},{"Start":"02:03.670 ","End":"02:07.585","Text":"We\u0027re going to let tau equals x minus t and then"},{"Start":"02:07.585 ","End":"02:12.340","Text":"d tau will equal minus dt because of the minus here."},{"Start":"02:12.340 ","End":"02:17.360","Text":"Once again, we will get the minus here and the limits will be"},{"Start":"02:17.360 ","End":"02:24.085","Text":"reversed because if t is minus infinity then the tau is plus infinity and vice versa."},{"Start":"02:24.085 ","End":"02:27.550","Text":"Note that just as x minus t is tau,"},{"Start":"02:27.550 ","End":"02:33.410","Text":"we can swap around and get that t is x minus tau."},{"Start":"02:33.410 ","End":"02:37.480","Text":"We get the following integral and like above,"},{"Start":"02:37.480 ","End":"02:41.420","Text":"we can throw out the minus here and reverse the order here."},{"Start":"02:41.420 ","End":"02:45.260","Text":"Also, we can change the order of these two."},{"Start":"02:45.260 ","End":"02:46.790","Text":"Now, if you look at it,"},{"Start":"02:46.790 ","End":"02:51.830","Text":"this is exactly the definition of g convolution f. Well,"},{"Start":"02:51.830 ","End":"02:55.220","Text":"if you are used to it with t, let\u0027s replace tau by t,"},{"Start":"02:55.220 ","End":"02:56.750","Text":"just a dummy variable."},{"Start":"02:56.750 ","End":"03:02.555","Text":"Now we can see that this is g convolution f at the point x."},{"Start":"03:02.555 ","End":"03:06.840","Text":"That concludes part b and this exercise."}],"ID":31002},{"Watched":false,"Name":"Exercise 15","Duration":"5m 52s","ChapterTopicVideoID":29389,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.660","Text":"In this exercise, we\u0027re going to find a function"},{"Start":"00:03.660 ","End":"00:07.860","Text":"f(x) whose Fourier transform is f hat of Omega,"},{"Start":"00:07.860 ","End":"00:10.335","Text":"which is this expression."},{"Start":"00:10.335 ","End":"00:14.970","Text":"We can write this square as something times itself."},{"Start":"00:14.970 ","End":"00:18.165","Text":"The idea is to use the convolution theorem."},{"Start":"00:18.165 ","End":"00:26.295","Text":"Now, if we had a function f whose transform is 1/1 plus Omega squared would be good."},{"Start":"00:26.295 ","End":"00:27.700","Text":"We have something close,"},{"Start":"00:27.700 ","End":"00:31.905","Text":"we know that the Fourier transform of e to the minus absolute value of x."},{"Start":"00:31.905 ","End":"00:34.800","Text":"It\u0027s like this, but with an extra Pi here."},{"Start":"00:34.800 ","End":"00:40.895","Text":"We can fix that if we put a Pi in here because of linearity,"},{"Start":"00:40.895 ","End":"00:42.920","Text":"now we have the function Pi e to"},{"Start":"00:42.920 ","End":"00:47.395","Text":"the minus absolute value of x its Fourier transform is this."},{"Start":"00:47.395 ","End":"00:51.275","Text":"We can apply the convolution theorem and say"},{"Start":"00:51.275 ","End":"00:55.459","Text":"that if we take the convolution of this with itself,"},{"Start":"00:55.459 ","End":"01:01.520","Text":"Fourier transform will be 2 Pi times the 1/1 plus Omega squared times itself,"},{"Start":"01:01.520 ","End":"01:03.290","Text":"dividing by the 2 Pi,"},{"Start":"01:03.290 ","End":"01:06.965","Text":"one of the Pi\u0027s cancels and the 2 goes onto here."},{"Start":"01:06.965 ","End":"01:13.390","Text":"We have that Fourier transform of this is exactly what we want."},{"Start":"01:13.390 ","End":"01:18.925","Text":"The function we\u0027re looking for is what\u0027s in here."},{"Start":"01:18.925 ","End":"01:24.515","Text":"That by the definition of the convolution is the following integral."},{"Start":"01:24.515 ","End":"01:28.490","Text":"Once replacing x by t and once replacing x by x"},{"Start":"01:28.490 ","End":"01:30.980","Text":"minus t. But we can\u0027t say that this is"},{"Start":"01:30.980 ","End":"01:33.500","Text":"the answer because this is in the form of an integral."},{"Start":"01:33.500 ","End":"01:36.485","Text":"You want to get it more specific, more explicit."},{"Start":"01:36.485 ","End":"01:40.955","Text":"Now, there are two special points where one of these absolute values changes sign."},{"Start":"01:40.955 ","End":"01:44.100","Text":"One is when t is 0,"},{"Start":"01:44.100 ","End":"01:46.665","Text":"and 1 is where t is equal to x."},{"Start":"01:46.665 ","End":"01:51.605","Text":"0 and x are special and we want to break this integral up into 3."},{"Start":"01:51.605 ","End":"01:53.240","Text":"But there\u0027s 2 cases,"},{"Start":"01:53.240 ","End":"01:56.480","Text":"0 could be less than x or x could be less than 0."},{"Start":"01:56.480 ","End":"02:01.130","Text":"Let\u0027s, first of all, take the case where x is less than 0,"},{"Start":"02:01.130 ","End":"02:06.830","Text":"and then the integral can be split up as minus infinity to x,"},{"Start":"02:06.830 ","End":"02:08.705","Text":"and then x to 0,"},{"Start":"02:08.705 ","End":"02:10.610","Text":"and then 0 to infinity."},{"Start":"02:10.610 ","End":"02:12.715","Text":"Inside each of these,"},{"Start":"02:12.715 ","End":"02:16.754","Text":"the t and the x minus t won\u0027t change sign."},{"Start":"02:16.754 ","End":"02:24.665","Text":"We get that our f(x) is the sum of 3 integrals according to this."},{"Start":"02:24.665 ","End":"02:27.110","Text":"Let\u0027s see what happens inside each one."},{"Start":"02:27.110 ","End":"02:29.755","Text":"From minus infinity to x,"},{"Start":"02:29.755 ","End":"02:31.520","Text":"if t is in this interval,"},{"Start":"02:31.520 ","End":"02:34.475","Text":"it\u0027s less than 0 and it\u0027s less than x."},{"Start":"02:34.475 ","End":"02:38.185","Text":"We can take up the absolute values here and here,"},{"Start":"02:38.185 ","End":"02:41.780","Text":"and say, this is just e^t."},{"Start":"02:41.780 ","End":"02:43.530","Text":"T is less than 0,"},{"Start":"02:43.530 ","End":"02:46.640","Text":"so absolute value of t is minus t. Here,"},{"Start":"02:46.640 ","End":"02:47.840","Text":"t is less than x,"},{"Start":"02:47.840 ","End":"02:49.430","Text":"x minus t is positive,"},{"Start":"02:49.430 ","End":"02:51.605","Text":"so we just drop the absolute value."},{"Start":"02:51.605 ","End":"02:53.420","Text":"That\u0027s the first one. Now,"},{"Start":"02:53.420 ","End":"02:55.145","Text":"in the second one,"},{"Start":"02:55.145 ","End":"02:58.475","Text":"t is between x and 0."},{"Start":"02:58.475 ","End":"03:01.340","Text":"It\u0027s still less than 0, it\u0027s e^t."},{"Start":"03:01.340 ","End":"03:03.830","Text":"But here the x minus t changes sign,"},{"Start":"03:03.830 ","End":"03:06.115","Text":"so we don\u0027t have the minus."},{"Start":"03:06.115 ","End":"03:08.220","Text":"Then the last one,"},{"Start":"03:08.220 ","End":"03:10.457","Text":"t is already bigger than 0,"},{"Start":"03:10.457 ","End":"03:15.710","Text":"so absolute value of t is t. This is e^-t and this"},{"Start":"03:15.710 ","End":"03:21.080","Text":"again is e to the x minus t. We\u0027ve gotten rid of the absolute values."},{"Start":"03:21.080 ","End":"03:23.240","Text":"Now, all of these integrals are dt."},{"Start":"03:23.240 ","End":"03:29.375","Text":"We can bring the parts with just x and without t in front of the integral."},{"Start":"03:29.375 ","End":"03:31.190","Text":"For the first one,"},{"Start":"03:31.190 ","End":"03:34.580","Text":"we can take e^-x in front of the integral,"},{"Start":"03:34.580 ","End":"03:39.155","Text":"and then we have e^t and another e^t, so it\u0027s e^2t."},{"Start":"03:39.155 ","End":"03:43.025","Text":"In this case, we can bring the e^x in front,"},{"Start":"03:43.025 ","End":"03:46.130","Text":"and we have e to the t to the minus t which is 1,"},{"Start":"03:46.130 ","End":"03:47.660","Text":"and then the last one,"},{"Start":"03:47.660 ","End":"03:49.565","Text":"we can bring e^x in front."},{"Start":"03:49.565 ","End":"03:53.225","Text":"Then we have e^-t, e^-t, e^-2t."},{"Start":"03:53.225 ","End":"03:56.645","Text":"We have 3 straightforward integrals."},{"Start":"03:56.645 ","End":"04:01.685","Text":"The integral of e^2t from minus infinity to x"},{"Start":"04:01.685 ","End":"04:07.550","Text":"is e^2x minus e to the minus twice infinity,"},{"Start":"04:07.550 ","End":"04:10.520","Text":"which is 0, so it\u0027s just e^2x."},{"Start":"04:10.520 ","End":"04:13.835","Text":"Here., we have just the length of the interval,"},{"Start":"04:13.835 ","End":"04:16.580","Text":"which is 0 minus x."},{"Start":"04:16.580 ","End":"04:18.320","Text":"Then the last one,"},{"Start":"04:18.320 ","End":"04:22.560","Text":"this integral is minus 0.5 e^-2t."},{"Start":"04:22.560 ","End":"04:25.335","Text":"If you plug in infinity, it\u0027s 0."},{"Start":"04:25.335 ","End":"04:27.135","Text":"If you plug in 0,"},{"Start":"04:27.135 ","End":"04:29.235","Text":"this becomes 1,"},{"Start":"04:29.235 ","End":"04:30.930","Text":"we just have the minus 0.5,"},{"Start":"04:30.930 ","End":"04:33.540","Text":"but it\u0027s subtracted, so it\u0027s plus 0.5."},{"Start":"04:33.540 ","End":"04:35.250","Text":"This is what we have now,"},{"Start":"04:35.250 ","End":"04:40.555","Text":"and then just collecting e^-x, e^2x, e^x."},{"Start":"04:40.555 ","End":"04:45.190","Text":"Here, just x e^x with a minus."},{"Start":"04:45.190 ","End":"04:49.280","Text":"Here 0.5 with Pi over 2 is Pi over 4 e^x."},{"Start":"04:49.280 ","End":"04:53.765","Text":"Of course, these 2 combine to give Pi over 2 e^x."},{"Start":"04:53.765 ","End":"04:55.715","Text":"After we factorize it,"},{"Start":"04:55.715 ","End":"05:02.405","Text":"take the Pi over 2 out and then we have e^x minus xe^x."},{"Start":"05:02.405 ","End":"05:05.719","Text":"Anyway, this is what we get after factorization,"},{"Start":"05:05.719 ","End":"05:08.165","Text":"and after combining these 2, very good."},{"Start":"05:08.165 ","End":"05:10.985","Text":"Now we found what f(x) is explicitly,"},{"Start":"05:10.985 ","End":"05:13.925","Text":"but remember, this is for x less than 0."},{"Start":"05:13.925 ","End":"05:18.400","Text":"Now, do we have to do this whole work again for x bigger than 0?"},{"Start":"05:18.400 ","End":"05:24.075","Text":"Well, no. Recall that f(x) is this convolution."},{"Start":"05:24.075 ","End":"05:25.925","Text":"This is an even function,"},{"Start":"05:25.925 ","End":"05:28.490","Text":"and this is an even function."},{"Start":"05:28.490 ","End":"05:34.400","Text":"We know that an even function is also an even function,"},{"Start":"05:34.400 ","End":"05:36.305","Text":"that was the previous exercise."},{"Start":"05:36.305 ","End":"05:42.320","Text":"Mirroring this, just replacing x by minus x for when x is bigger than 0,"},{"Start":"05:42.320 ","End":"05:49.395","Text":"so this just copied and here x minus x. X is minus x, making it plus."},{"Start":"05:49.395 ","End":"05:50.885","Text":"This is the answer,"},{"Start":"05:50.885 ","End":"05:53.100","Text":"and we are done."}],"ID":31003},{"Watched":false,"Name":"Exercise 16","Duration":"4m 18s","ChapterTopicVideoID":29369,"CourseChapterTopicPlaylistID":294465,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:04.920","Text":"In this exercise, we define f(x) to equal the"},{"Start":"00:04.920 ","End":"00:09.480","Text":"following double integral which may look frightening,"},{"Start":"00:09.480 ","End":"00:10.785","Text":"but it\u0027s not so bad."},{"Start":"00:10.785 ","End":"00:16.530","Text":"We have to find an explicit expression for f(x) without integrals."},{"Start":"00:16.530 ","End":"00:20.000","Text":"Here\u0027s a reminder of what the Chi function,"},{"Start":"00:20.000 ","End":"00:22.460","Text":"the characteristic function on an interval is."},{"Start":"00:22.460 ","End":"00:26.710","Text":"It\u0027s sometimes called the window function."},{"Start":"00:26.710 ","End":"00:31.580","Text":"We can break it up like this because we see that here is"},{"Start":"00:31.580 ","End":"00:36.635","Text":"an Omega minus t and here\u0027s a t. Looks very much like a convolution."},{"Start":"00:36.635 ","End":"00:40.895","Text":"Which it is, so we can write that bit as Chi of"},{"Start":"00:40.895 ","End":"00:46.535","Text":"this interval convolution with itself that the point Omega."},{"Start":"00:46.535 ","End":"00:49.490","Text":"Now, this looks like a Fourier transform except"},{"Start":"00:49.490 ","End":"00:52.564","Text":"that there should be a minus here. No problem."},{"Start":"00:52.564 ","End":"00:59.300","Text":"Make a substitution that Omega is minus u and then d omega is minus du."},{"Start":"00:59.300 ","End":"01:07.640","Text":"We get to the Omega here we have minus u and to the Omega here we have minus u."},{"Start":"01:07.640 ","End":"01:12.095","Text":"Now, because this is an even function and this is an even function,"},{"Start":"01:12.095 ","End":"01:14.040","Text":"the convolution is also even."},{"Start":"01:14.040 ","End":"01:15.815","Text":"We can drop this minus,"},{"Start":"01:15.815 ","End":"01:20.390","Text":"but it\u0027s still not Fourier transform because there\u0027s something missing."},{"Start":"01:20.390 ","End":"01:22.390","Text":"There\u0027s the 1 over 2 Pi."},{"Start":"01:22.390 ","End":"01:26.240","Text":"Put 1/2 Pi here and compensate by putting"},{"Start":"01:26.240 ","End":"01:30.425","Text":"2Pi here and here notice the minus is dropped out."},{"Start":"01:30.425 ","End":"01:35.700","Text":"Now, it really does look like the Fourier transform of this."},{"Start":"01:35.840 ","End":"01:41.175","Text":"We have the Fourier transform of this."},{"Start":"01:41.175 ","End":"01:45.410","Text":"But the Fourier transform of this will be a function of x."},{"Start":"01:45.410 ","End":"01:48.275","Text":"We can tell because of the x here."},{"Start":"01:48.275 ","End":"01:52.960","Text":"We can take the 2Pi in front and also a function of u(x),"},{"Start":"01:52.960 ","End":"01:55.265","Text":"it\u0027s just a function of x."},{"Start":"01:55.265 ","End":"01:59.120","Text":"Now, applying the convolution theorem to this convolution,"},{"Start":"01:59.120 ","End":"02:01.370","Text":"we get the 2Pi from here,"},{"Start":"02:01.370 ","End":"02:03.740","Text":"the 2Pi from the theorem,"},{"Start":"02:03.740 ","End":"02:09.500","Text":"Fourier transform of the first one of Omega then times itself."},{"Start":"02:09.500 ","End":"02:12.740","Text":"Just put a box around this to say this is what we\u0027re going to"},{"Start":"02:12.740 ","End":"02:15.680","Text":"evaluate the side and then return here."},{"Start":"02:15.680 ","End":"02:18.905","Text":"We want to compute this Fourier transform,"},{"Start":"02:18.905 ","End":"02:26.415","Text":"this Fourier transform is 1 over 2Pi the integral of this thing e^minus i Omega xdx."},{"Start":"02:26.415 ","End":"02:29.280","Text":"My definition of Fourier transform."},{"Start":"02:29.280 ","End":"02:34.275","Text":"This window function is 1 between minus Pi over 2 and Pi over 2."},{"Start":"02:34.275 ","End":"02:36.110","Text":"We can just truncate this integral,"},{"Start":"02:36.110 ","End":"02:40.805","Text":"make it from minus Pi over 2 to Pi over 2 and this is just 1."},{"Start":"02:40.805 ","End":"02:43.965","Text":"We have the integral of e^ minus Omega x."},{"Start":"02:43.965 ","End":"02:48.795","Text":"The integral of this is e to the minus i Omega x divided by minus"},{"Start":"02:48.795 ","End":"02:53.715","Text":"i Omega and substitute the 2 limits of integration."},{"Start":"02:53.715 ","End":"02:58.025","Text":"We can get rid of the minus if we reverse the order here."},{"Start":"02:58.025 ","End":"03:02.150","Text":"Also, I\u0027d like to put the 2 in the denominator here and bring"},{"Start":"03:02.150 ","End":"03:06.030","Text":"the Omega to here so we get the following."},{"Start":"03:06.030 ","End":"03:09.800","Text":"You\u0027ll see why I did this in a moment."},{"Start":"03:09.800 ","End":"03:12.905","Text":"It\u0027s because we have the following formula."},{"Start":"03:12.905 ","End":"03:17.690","Text":"Sine Theta, e^i Theta minus e to the minus i Theta over 2i."},{"Start":"03:17.690 ","End":"03:20.230","Text":"This is what we have here,"},{"Start":"03:20.230 ","End":"03:25.665","Text":"with Theta being Pi Omega over 2."},{"Start":"03:25.665 ","End":"03:28.950","Text":"This part is Sine Pi Omega over 2 and then we"},{"Start":"03:28.950 ","End":"03:32.505","Text":"have the 1 over Pi Omega put in the denominator."},{"Start":"03:32.505 ","End":"03:36.125","Text":"That gives us the Fourier transform of this bit."},{"Start":"03:36.125 ","End":"03:38.585","Text":"Now, returning to what we had,"},{"Start":"03:38.585 ","End":"03:43.220","Text":"we had that f of Omega was 2Pi times 2Pi times this Fourier transform times this."},{"Start":"03:43.220 ","End":"03:45.550","Text":"We\u0027ve just evaluated this here."},{"Start":"03:45.550 ","End":"03:49.515","Text":"There\u0027s another one of them here and there\u0027s also the 2Pi 2Pi."},{"Start":"03:49.515 ","End":"03:52.715","Text":"What we get is the following,"},{"Start":"03:52.715 ","End":"03:56.630","Text":"which simplifies the Pi and the Pi cancel with Pi and Pi,"},{"Start":"03:56.630 ","End":"03:58.160","Text":"2 times 2 is 4."},{"Start":"03:58.160 ","End":"03:59.930","Text":"This is what we get."},{"Start":"03:59.930 ","End":"04:02.060","Text":"We found the function x only we have"},{"Start":"04:02.060 ","End":"04:07.310","Text":"the argument Omega and we really wanted to find f(x) so just to replace Omega with x,"},{"Start":"04:07.310 ","End":"04:09.740","Text":"it doesn\u0027t matter the functions are the same."},{"Start":"04:09.740 ","End":"04:13.700","Text":"We have that f(x) is 4 sine^2 Omega x over"},{"Start":"04:13.700 ","End":"04:18.660","Text":"2 divided by x^2 and that\u0027s the answer and we\u0027re done."}],"ID":31004}],"Thumbnail":null,"ID":294465}]

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1.1

1