AI Assistant
Transcript
00:00:10Paolo Guiotto: Peace.
00:02:270Paolo Guiotto: Okay, let's start.
00:04:740Paolo Guiotto: Today, we will come to an end to this first part, of course.
00:11:30Paolo Guiotto: We will basically start with some applications of the inversion formula. So let me recall this factor, important factor, that inversion
00:24:160Paolo Guiotto: formula.
00:28:590Paolo Guiotto: I'm sorry, sir.
00:39:530Paolo Guiotto: It says that if we have an F, which is… It's awful.
00:48:490Paolo Guiotto: He's in Aloha.
00:50:950Paolo Guiotto: with also the Fourier transform of ethanol 1,
00:55:400Paolo Guiotto: Okay? Then we have this remarkable formula that says a double transform of F, evaluated at point x, is 2 pi
01:10:780Paolo Guiotto: F… of… minus tracks while I write in this way, or,
01:21:390Paolo Guiotto: Well, let's say, let's write it this way. It doesn't matter, oh, if you want, we can write F at minus X equal to pi F.
01:31:140Paolo Guiotto: But normally, if I want to say F equal, I will use this formula with f of x equal 1 over 2 pi d double theta F.
01:41:530Paolo Guiotto: Now, a first remark about consequences is the following, which is, the so-called uniqueness.
01:55:80Paolo Guiotto: Ugh.
01:56:70Paolo Guiotto: Fourier transform. This says that if you have two functions, F and G, such that the Fourier transforms coincide.
02:06:260Paolo Guiotto: So, if F and G are L1 functions, such that…
02:11:650Paolo Guiotto: This, of course, is identically verified. Then, necessarily, the two functions must be the same almost everywhere.
02:24:00Paolo Guiotto: And this is just a very simple, proof, because if you have that the two transforms coincide, you carry
02:33:530Paolo Guiotto: the two transforms on the left-hand side, for example, you have F hat minus G hat is identically zero. You know that the transform operation… operator is a linear operation, so this means that the Fourier transform of F minus G is 0.
02:50:900Paolo Guiotto: Now, let's look at the F minus G.
02:53:360Paolo Guiotto: Since F and G are in L1,
02:58:190Paolo Guiotto: their difference, F minus G, will be in L1.
03:02:550Paolo Guiotto: And, because the default transform of F minus G is 0, And since,
03:09:750Paolo Guiotto: the transform of this function is 0, and 0 is clearly an L1 function, we have that, both F minus G and the hat of F minus G are L1. So inversion formula applies to F minus G.
03:26:890Paolo Guiotto: inversion.
03:29:930Paolo Guiotto: formula.
03:33:530Paolo Guiotto: for… F minus G, and this says that the function f minus G
03:40:830Paolo Guiotto: at pointer X is 1 over 2 pi, the double hat of F minus G, at point minus X.
03:52:280Paolo Guiotto: But this thing inside, the hat of F minus G is 0, okay? So this is identically zero. So we are doing 1 over 2 pi d hat of 0, evaluated at minus X, and the hat of 0 is, of course, equal to 0.
04:14:00Paolo Guiotto: Now, of course, you remind that the inversion formulas, this identity, is an identity almost everywhere, because on one side, we have F, which is an L1 function, so not necessarily always defined. On the right-hand side, we have a hat. A double hat is a hat.
04:33:700Paolo Guiotto: So,
04:36:570Paolo Guiotto: In this case, we have a continuous function. So this is an almost everywhere identity, and therefore, this is an almost everywhere identity, from which we get the function.
04:49:230Paolo Guiotto: Now, this little fact is important because we may solve a number of problems. Let me illustrate this on a couple of… on three different situations. One is, for example, we do exercise
05:06:900Paolo Guiotto: 19… 3… 7.
05:11:800Paolo Guiotto: which is, let for A positive, let FAX be the Cauchy distribution, 1 over A squared plus X squared.
05:25:40Paolo Guiotto: So, use… If we are transformer to computer.
05:33:60Paolo Guiotto: the convolution between FA and FB
05:38:40Paolo Guiotto: For A and D pockets.
05:43:820Paolo Guiotto: Now, this convolution, in principle, could be computed by hand, because it is an integral operational function, so in principle we could do that, but it's a long calculation. Let's see how to use the Fourier transform.
05:57:180Paolo Guiotto: Do a, in this case, a smarter calculation.
06:01:20Paolo Guiotto: So we take P, F, A, and F. First of all, is the convolution well-defined? Convolution is never a trivial operation, it demands something. Well, we know that convolution is defined when one function is in L1 and the other is in LP, with NEP, and the product is in LP. In this case, these functions, F, A, and FB,
06:23:320Paolo Guiotto: are both L1 function in realign.
06:28:300Paolo Guiotto: So the convolution is well-defined.
06:31:600Paolo Guiotto: And we know also that it is an L1 function. This is the young URM.
06:44:830Paolo Guiotto: So, now, since it is in L1, we can compute its Fourier transform, and we have that FA convolution FP theta, evaluated at point C. We now know that there is a remarkable property of the Fourier transform with respect to the convolution.
07:03:690Paolo Guiotto: It transforms a convolution product into an algebraic product.
07:07:610Paolo Guiotto: So this becomes FA hat C times FB hat.
07:13:690Paolo Guiotto: Si.
07:15:400Paolo Guiotto: Now, we determined, last time, I guess.
07:20:660Paolo Guiotto: the Fourier transform of the… or perhaps the
07:27:690Paolo Guiotto: Yes, the… of the Cauchy distribution, and we discovered that, the…
07:33:660Paolo Guiotto: We get transform of this is P over A, E minus A modus C, okay? So let's use this formula.
07:41:10Paolo Guiotto: Cool. So, yeah, this is a P over A, E minus A modulus of C. This is similarly P over B, E minus B modulus of C.
07:53:900Paolo Guiotto: So, as you can see, the product is very easy, because it is P squared over AB times E minus, we can sum the exponents, and this means that we have this.
08:08:270Paolo Guiotto: And this is the Fourier transform of the convolution.
08:12:110Paolo Guiotto: Now, what is the point? That, of course, I want this quantity here in the… under the hat.
08:19:710Paolo Guiotto: And the point is that we should recognize at the right that there is basically a fuller form.
08:25:710Paolo Guiotto: In fact, just we come to see that the Fourier transform of a Cauchy distribution, the hat of 1 over, let's say, C squared plus sharp square c is pi over C, E minus C modulus.
08:45:160Paolo Guiotto: So this function, apart for a constant factor that we can always arrange, is basically a function of this type, with C equal A plus B. So, if we reproduce P over C in front, we have F gets us 4. Of course, we can do that.
09:02:370Paolo Guiotto: We have B over AB. We need to have B over A plus B. We have the pi, but we do not have A plus B, so we multiply, divide.
09:12:910Paolo Guiotto: And then we have E minus A plus B modulus of C.
09:18:840Paolo Guiotto: Now, here we recognize that this is the hat of 1 over A plus B squared plus variable square.
09:31:630Paolo Guiotto: And since, this in front of, this function is just a coefficient, we can carry bin linearity into the Fourier transform, and we get hat of, let's say, A plus B over AB,
09:48:690Paolo Guiotto: buy… Then we have, one of,
09:53:50Paolo Guiotto: A plus B square plus sharp square C.
09:58:800Paolo Guiotto: Now, this is the calculation of the Fourier transform of convolution between FA and FB.
10:08:450Paolo Guiotto: at point C.
10:09:930Paolo Guiotto: Now, here is where you use uniqueness, because we have these two functions, the convolution and the right-hand side under the head, they have the same Fourier transform, of course, for every point C. They must coincide almost everywhere, so we have that from this FA star.
10:29:770Paolo Guiotto: FB, this is by uniqueness.
10:36:570Paolo Guiotto: at points X must be this, pi A plus B.
10:41:800Paolo Guiotto: over AB, 1 over A plus B squared plus X squared.
10:50:40Paolo Guiotto: And that's the solution.
10:55:570Paolo Guiotto: Okay?
10:58:440Paolo Guiotto: Let's see another example of this kind of idea. So, when I know that, I have to compute the Fourier transform. I know that the Fourier transform is the Fourier transform of something known, my function is that one, okay?
11:11:570Paolo Guiotto: So this can be used to solve, for example, certain particular types of equations, like this one, exercise…
11:20:930Paolo Guiotto: 19.38. This is called a convolution
11:31:470Paolo Guiotto: equation.
11:34:150Paolo Guiotto: There are certain models where we can have this type of equations. Here, the problem is solver.
11:42:780Paolo Guiotto: the equation… integral on r of f of x minus Y, E minus absolute value of Y, DY equal…
11:55:90Paolo Guiotto: E minus 2X.
11:57:990Paolo Guiotto: 4X, Ria.
12:01:260Paolo Guiotto: Well, the unknown… is supposed to be F in L1R.
12:10:600Paolo Guiotto: Okay, so this is an equation where you have the unknown, which is the function f in that,
12:18:400Paolo Guiotto: in that identity.
12:20:270Paolo Guiotto: Of course, here we recognize that that integral is a convolution integral, so integral on r of f of x minus y
12:30:710Paolo Guiotto: G of Y, DY is, by definition, what is the convolution between F and G at point X.
12:38:860Paolo Guiotto: So, here… So the integral on R of f of x minus y minus modulus of Y
12:51:180Paolo Guiotto: is the convolution between F and E minus G modulus.
12:56:750Paolo Guiotto: at Bontex.
12:59:710Paolo Guiotto: This convolution is well-defined, because we are looking for NF which is an L1 function.
13:07:360Paolo Guiotto: So, exponential is, well, a good L1 function, so this is L1 because it is an L1. This is L1 by assumption.
13:22:260Paolo Guiotto: So, the convolution is well-defined.
13:30:430Paolo Guiotto: is… Well… defined.
13:37:280Paolo Guiotto: Now, what is the equation? So, the equation…
13:43:280Paolo Guiotto: is… can be written in this, let's say, more, let's say, elegant, perhaps former. The convolution between F and this exponential at point X is the function E minus 2 modulus of X.
14:01:980Paolo Guiotto: Now, to solve this equation, we could use the Fourt transform, because this operation will transform the convolution product into an algebraic product.
14:12:70Paolo Guiotto: This will make possible to explicit
14:15:420Paolo Guiotto: Not F, but it's Fourier transformer. So, at the end, we have the Fourier transform of F, and if we are lucky, we recognize who is F. Let's see what happens. So, applying…
14:32:710Paolo Guiotto: For a transform, we have that the transform of F star E minus modulus at point C will be the transform of E minus 2 modulus at point C.
14:47:530Paolo Guiotto: Now, this is the written form of an exponential.
14:51:600Paolo Guiotto: We remind that the transform of E minus A modulus is, if I'm not wrong, 2A at point C, 2A divided A squared plus C squared.
15:05:590Paolo Guiotto: So when we do that, A is 2, we get 4,
15:09:350Paolo Guiotto: divided the 4 plus Z squared.
15:14:500Paolo Guiotto: This is the right-hand side of the equation. On the left-hand side, we have that V splits into F hat C times the hat of the exponential, which is the same, it is 2, now it is with A equal 1, 1 plus C squared.
15:32:210Paolo Guiotto: Okay?
15:42:530Paolo Guiotto: There is something wrong here.
15:54:810Paolo Guiotto: There's something wrong with the exercise, because you see, now, what happens? I can explicit the F header.
16:03:290Paolo Guiotto: I suspect there is a typo in the assignment, let's see, probably there is something missing.
16:09:670Paolo Guiotto: So I have 4 divided 2, and then I have 1 plus C squared divided to 4 plus C squared.
16:18:480Paolo Guiotto: But this problem is impossible.
16:21:110Paolo Guiotto: Why?
16:29:540Paolo Guiotto: This can… that cannot be possible, because the right-hand side cannot be a Fourier transform. And why?
16:39:290Paolo Guiotto: No, we know that the Fourier transform is not necessarily L1, but what do we know about the Fourier transform? It is continuous, always, and that's continuous.
16:49:700Paolo Guiotto: And it goes to 0 at infinity, and that's… goes to 2 at infinity.
16:55:410Paolo Guiotto: So this, goes to, 2 at… as C goes to plus-minus infinity.
17:02:430Paolo Guiotto: So… There cannot be an F in L1 such that… This star.
17:14:980Paolo Guiotto: is verified.
17:20:380Paolo Guiotto: Okay, so we could say that this equation has not solution, as in one function,
17:31:600Paolo Guiotto: Let's see if we modify… let's modify a bit the problem.
17:35:960Paolo Guiotto: Let's… well, at least let's take another pro, ironically.
17:42:820Paolo Guiotto: So let me wait one second, I have to look on the… because here we do not have…
17:50:510Paolo Guiotto: I want to take an example from the exercises of, for the exam.
18:01:00Paolo Guiotto: Well, let's see where we have some of these.
18:08:130Paolo Guiotto: Okay, so let me take… let's take… what is it?
18:17:860Paolo Guiotto: I want to take a… Okay, we have exercise, 14… Exam… Set.
18:31:170Paolo Guiotto: So, here we have,
18:33:210Paolo Guiotto: Well, there is the first question. Let FG in L1. What is the convolution of F and G? So what should… you should answer. The convolution of F and G is this formula, okay?
18:45:960Paolo Guiotto: What important property of free transform holds in connection with convolution? It's just one property, the head of the convolution.
18:53:750Paolo Guiotto: It's the product of the ad.
18:55:550Paolo Guiotto: And write a precise statement, and provide a proof of it, of this property. Now, you know that it's just a straightforward calculation. You write the head of the convolution, and after a few steps, you arrive to show that it is the property. So, this is
19:11:710Paolo Guiotto: the first part of the question. The second part is consider the… equation… Lambda FX… Las.
19:27:260Paolo Guiotto: integral on R of f of y 1 over X minus y squared divided, Equal.
19:41:910Paolo Guiotto: 1 over… 1 plus X squared plus 1 over 4 plus X squared for X in i.
19:54:50Paolo Guiotto: Where lambda is just a parameterial parameter.
19:59:310Paolo Guiotto: So it says, question one, which is actually the question two, letter F, In L1, L1R.
20:10:710Paolo Guiotto: B… a solution.
20:15:550Paolo Guiotto: determine… the hat of F,
20:23:980Paolo Guiotto: And, determined, or… Which… Lambda T.
20:34:930Paolo Guiotto: equation.
20:38:850Paolo Guiotto: has.
20:40:910Paolo Guiotto: a unique… solution.
20:47:770Paolo Guiotto: Number 2…
20:49:770Paolo Guiotto: well, here it says solves for lambda equal to pi. That probably is a typo because, well, the story is that there are several formulas for the Fourier transform with different, factors here and there, and so this was probably with the…
21:11:190Paolo Guiotto: I'm not sure that it is this wonder value, but, let's see.
21:19:450Paolo Guiotto: So, look at the equation, what we see, we have lambda F plus the reason integral, and in that integral, you should recognize that it is a convolution-type integral. You see, f of y times a function of X minus y. So, the equation
21:38:70Paolo Guiotto: is… we can say lambda F plus F star what? You are doing the convolution with the 1 over 1 plus square, you see?
21:48:770Paolo Guiotto: So it is, 1 over 1 plus square.
21:53:110Paolo Guiotto: Evaluated at x equal, so lambda f of x equal 1 over 1 plus X squared.
22:02:80Paolo Guiotto: plus 1 over 4 plus X scorer.
22:06:610Paolo Guiotto: So we noticed that, since, F is in L1.
22:13:410Paolo Guiotto: And 1 over 1 plus D squared is an L1 function. As well, their convolution is well-defined.
22:22:910Paolo Guiotto: F star 1 over 1 plus square.
22:28:170Paolo Guiotto: is well-defined.
22:32:650Paolo Guiotto: and not only well-defined, it is also L1, because it is the convolution of two L1 functions, and… L1.
22:43:820Paolo Guiotto: So this authorizes to compute the Fourier transform of the left-hand side, because left-hand side is lambda F, F is in L1, plus convolution f start the other function, this guy is in L1. So the left-hand side is in L1, and also the right-hand side is in L1. So, computing…
23:06:50Paolo Guiotto: the… Create transformer… off… Booth… sides… of the… equation… We get.
23:24:260Paolo Guiotto: So, lambda F hat C
23:29:20Paolo Guiotto: plus. Now, the convolution, the hat of the convolution is the product of the hats, so the convolu- the, transform of F, then we have the transform of 1 over 1 plus square.
23:43:250Paolo Guiotto: Evaluated XC, then we have the transform of 1 over 1 plus square.
23:50:680Paolo Guiotto: plus the convol… the transform of 1 over 4 plus square. All these are Cauchy distributions.
24:00:540Paolo Guiotto: So we remind that the transform of 1 over A squared plus D squared
24:07:300Paolo Guiotto: at point C is pi over AE minus A D modulus of C.
24:15:140Paolo Guiotto: So, plugging this into this formula, we get…
24:19:500Paolo Guiotto: Well, you see that we can, first of all, at left, factorize an F hat, so we write F at X,
24:26:470Paolo Guiotto: times.
24:27:760Paolo Guiotto: lambda minus, you know, plus pi over a.
24:34:760Paolo Guiotto: Well, A is actually 1, so just pi.
24:41:170Paolo Guiotto: No, probably the lambda wasn't, no, perhaps the correct lambda is 2 pi, really. So, let's… let's keep the 2 pi of the exercise.
24:52:970Paolo Guiotto: E minus the modulus.
24:58:910Paolo Guiotto: You see, this is the left-hand side.
25:01:610Paolo Guiotto: equal… the right-hand side is, the distribution… the transform of the fescal distribution with A equal 1, so we have pi E minus modulus C.
25:13:600Paolo Guiotto: The second one, this is 2 squared, so it's with A equal 2, so we have pi half E minus 2 modules C.
25:25:630Paolo Guiotto: Should be something like this.
25:28:780Paolo Guiotto: So… We can… Now, say that… If we can divide…
25:39:720Paolo Guiotto: So, unless the parenthesis, the round parenthesis is 0,
25:43:760Paolo Guiotto: But, this means what? You see that we have this function is a function of T that will be, at most, a couple of values for which this can be 0 because of the modulus, or none. For example, if lambda is positive, this quantity is never equal to zero, so we can divide.
26:03:870Paolo Guiotto: So, however, he says that F at C is equal to…
26:10:750Paolo Guiotto: Let's see if we can reorganize anything here, so let's factorize P over 2.
26:17:540Paolo Guiotto: we can factor a E minus modulus C also, so we have, 2 plus E minus modulus C.
26:29:550Paolo Guiotto: divided by this thing, lambda plus pi E minus modulus of C.
26:39:850Paolo Guiotto: We have this.
26:44:740Paolo Guiotto: Now… We have a, the question asks to determine for which lambda equation has a unique solution.
26:53:50Paolo Guiotto: How we should understand from this formula which values of lambda are good and which values of lambda are not good.
27:02:110Paolo Guiotto: Now, you see that the unique, strange thing that may happen here, let me see, is denominator is 0.
27:15:370Paolo Guiotto: So we probably have a problem, because in one of this, you know, it functions somewhere in order to fill.
27:23:290Paolo Guiotto: And the function must be honest.
27:25:820Paolo Guiotto: So, this happens when?
27:28:940Paolo Guiotto: Well, we may suspect that if lambda is positive, this will never be, okay? So probably the problem is when, even with lambda, Brazil, this is never effective.
27:41:830Paolo Guiotto: So when translating negative, this could be 0, because imagine that you put the minus 1 here, if this value is plus 1,
27:51:760Paolo Guiotto: So we probably will have to discuss when that denominator can be zero.
27:59:890Paolo Guiotto: So what… what is the argument? Now, lambda plus pi E minus modulus C, it can be 0 if and only if, well, we have E minus modulus of C equal… equal minus lambda over pi.
28:19:280Paolo Guiotto: So we could say, never…
28:23:120Paolo Guiotto: If lambda is great… is greater or equal than zero, because the left-hand side is positive, the right-hand side would be negative or zero, and that's not possible.
28:38:200Paolo Guiotto: While, if lambda is negative.
28:43:80Paolo Guiotto: We can continue. This happens, if and only if… now, if lambda is negative, that right-hand side is positive.
28:51:60Paolo Guiotto: So we have, we can take log sub minus modulus C equal… log… of minus lambda over pi.
29:02:810Paolo Guiotto: So, this means, modulus of C equal minus log of minus lambda over pi.
29:12:260Paolo Guiotto: Okay, but this is not necessarily, this is not necessarily a solution, because the left-hand side is positive, the right-hand side we don't know if it's positive or negative. So, it is clear that if the right-hand side is negative, this is impossible.
29:30:440Paolo Guiotto: If it is positive, this is possible. If this is positive, both zero. So this equation has a solution.
29:38:960Paolo Guiotto: We, which… pause.
29:44:860Paolo Guiotto: solution, if and only if this quantity minus log of minus lambda… lambda over pi is greater or equal than zero. This happens if log…
29:59:440Paolo Guiotto: of minus lambda over pi is less or equal than 0. This happens taking exponentials. If minus lambda over pi is less or equal than e to 0, 1, if and only if lambda over pi is greater than minus 1,
30:17:940Paolo Guiotto: If, no, if lambda, is greater.
30:23:870Paolo Guiotto: Or equal, then, minus pi.
30:27:450Paolo Guiotto: So let me…
30:31:910Paolo Guiotto: Yeah, so if lambda is… so the conclusion is, if lambda is positive, that quantity is never zero. If lambda is negative, that quantity
30:43:770Paolo Guiotto: can be 0 if and only if lambda is greater or equal than minus pi. Now, for this lambda.
30:54:800Paolo Guiotto: 4… Lambda negative.
30:58:770Paolo Guiotto: And, greater or equal than minus pi.
31:03:650Paolo Guiotto: We have that the denominator vanishes at point… denominator… All.
31:12:640Paolo Guiotto: F at X C equals… Let me just copy.
31:18:140Paolo Guiotto: by half, huh?
31:21:960Paolo Guiotto: E minus… Models C.
31:25:800Paolo Guiotto: one.
31:32:50Paolo Guiotto: the two class E minus modulus C.
31:40:460Paolo Guiotto: Divided by lambda plus… by… P minus modular… it's quite complicated, this problem.
31:52:760Paolo Guiotto: So, this probably was an old exam.
31:59:720Paolo Guiotto: We simplified things in last years. So it is better we discovered now this.
32:04:690Paolo Guiotto: Okay, so denominator of can vanish.
32:10:120Paolo Guiotto: Well, at better, vanishes.
32:17:910Paolo Guiotto: at, we say, at C where minus module C equal minus…
32:24:420Paolo Guiotto: a log of minus lambda over pi at modulus C equals to minus log of minus lambda over pi. It doesn't matter where, because the point is that if this goes to zero, then F at
32:42:30Paolo Guiotto: See?
32:43:330Paolo Guiotto: is not continuous.
32:46:280Paolo Guiotto: at this point, at modules C equal this.
32:51:850Paolo Guiotto: And in particular, it cannot be affiliate transform.
32:56:10Paolo Guiotto: So… There is not.
32:59:370Paolo Guiotto: L1, F in L1, such that F at X is equal to this thing.
33:08:470Paolo Guiotto: Because this thing is not continuous.
33:16:940Paolo Guiotto: Yeah.
33:19:180Paolo Guiotto: I want to notice that if lambda is zero.
33:23:870Paolo Guiotto: Because we said if lambda is greater or equal than zero, this never vanishes. But if lambda is zero, something else happened.
33:31:900Paolo Guiotto: If lambda is equal to zero.
33:36:580Paolo Guiotto: you see that this formula reduces to… then F at X should be equal to…
33:43:920Paolo Guiotto: pi half E minus modulus C.
33:48:610Paolo Guiotto: 2 plus, E minus mod was C.
33:53:480Paolo Guiotto: divided by lambda 0, this is pi E minus mod risk c.
33:59:40Paolo Guiotto: Now, this cancels all this, so we should have 1 half times 2 plus E minus modulus C,
34:08:290Paolo Guiotto: Which is a continuous function, but…
34:12:30Paolo Guiotto: Yeah, exactly. It goes to 1 when C goes to plus-minus infinity.
34:19:989Paolo Guiotto: So, also, so in this case, we have that the hat is continuous, but it does not go to zero.
34:28:110Paolo Guiotto: While… so… no solution.
34:40:20Paolo Guiotto: while for lambda positive, this goes to zero. Now, it asks to solve the case lambda equal to pi. So we have, in this case, FXC must be equal to
34:58:870Paolo Guiotto: So, we have to put the lambda equal to pi, so this means pi half E minus modulus c times 2 plus E minus modulus C,
35:11:180Paolo Guiotto: Divided by lambda, which is 2 pi.
35:15:850Paolo Guiotto: plus pi E minus modules C.
35:21:210Paolo Guiotto: So, this case… Simplifies a bit, because we can,
35:27:460Paolo Guiotto: Right, pi half E minus modulus C times 2 plus E minus modulus C.
35:36:650Paolo Guiotto: divided, if we factor the pi, we have 2 plus E minus modus C.
35:44:00Paolo Guiotto: So we have that this simplifies this, and also this cancels this.
35:49:240Paolo Guiotto: So at the end, the F at should be 1 half E minus modulus of X.
35:57:200Paolo Guiotto: which we recognize, it is a Fourier transform.
36:01:550Paolo Guiotto: This comes from the Cauchy distribution that we mentioned above, so the formula is this one, when A is 1.
36:09:740Paolo Guiotto: If A is each one, you get pi E minus the modulus. So, we do not have pi, we add a pi, like that.
36:19:130Paolo Guiotto: And so, now, this is the hat of 1 over 1 plus the square.
36:26:760Paolo Guiotto: In such a way that everything becomes 1 over 2 pi, 1 over 1 plus square.
36:35:230Paolo Guiotto: At point C.
36:37:150Paolo Guiotto: So, the F hat of X must be the hat of that thing, but since the latest form is unique, uniqueness.
36:49:890Paolo Guiotto: we get f of x equal 1 over 2 pi
36:56:220Paolo Guiotto: times 1 plus X squared, and that's the solution.
37:00:300Paolo Guiotto: of the equation.
37:01:940Paolo Guiotto: But this exercise was definitely a bit complicated.
37:07:950Paolo Guiotto: There should be something easier next.
37:15:450Paolo Guiotto: Let's show another remarkable application. Here, I won't be extremely formal.
37:23:780Paolo Guiotto: Which is, the application
37:33:980Paolo Guiotto: Application…
37:39:640Paolo Guiotto: to… PDs… I'll just show you an example, and in particular, to the heat equation.
37:58:60Paolo Guiotto: Now, the heat equation is a PD, I am sure you have seen already in this.
38:03:650Paolo Guiotto: So, it's a PDE where the unknown is a function u of time and space.
38:11:160Paolo Guiotto: Here we consider space, in, R.
38:16:250Paolo Guiotto: And, the question is this one, the derivative of the solution U respects
38:23:350Paolo Guiotto: 2T is a coefficient, positive coefficient, so let's say sigma squared. Let's write over 2 for computational reasons. We could write just sigma squared, but this will simplify the calculations later.
38:38:10Paolo Guiotto: So I prefer to write sigma square half the second derivative with respect to X of the function u.
38:45:610Paolo Guiotto: This function U represents the temperature on an infinite, say, an infinite line at point X.
38:57:410Paolo Guiotto: time t. Normally, this problem consists in solving the equation with an initial condition. So, at time t equals 0, you know that the temperature at any point is given by a function p of x.
39:14:330Paolo Guiotto: So this is, no, no, no.
39:20:250Paolo Guiotto: Now…
39:21:190Paolo Guiotto: The Foulette transformer is an interesting tool for special types of PDEs, particularly those with polynomial coefficients, because Fourier transform transforms derivatives into multiplications by powers, and this
39:41:100Paolo Guiotto: somehow… Helps in transforming this differential equation into a sort of algebraic equation.
39:49:360Paolo Guiotto: So let's see how…
39:51:160Paolo Guiotto: Now, the idea here is to take the Fourier transform of the solution with respect to the X variable, not with respect to the t variable. U is a function of TX. There is a Fourier transform of two variables, of three variables. We have not seen the definition, but it's the same. But here, we just use the transform we have seen so far.
40:13:380Paolo Guiotto: So we define this function, ZT,
40:16:380Paolo Guiotto: C as D. Fourier transform, we may say, of UT sharp.
40:26:790Paolo Guiotto: function of X. So at the end, you will see a function of P, of T and Xi.
40:34:980Paolo Guiotto: So we assume implicitly that this can be done, so we assume that UT be a function, an L1 function, in the variable X.
40:47:210Paolo Guiotto: And the idea will be to, as we have seen in this case of integral equations, to take the Fourier transform to both sides, and see what happens. So we want to compute the Fourier transform of the derivative and the Fourier transform of the right-hand side.
41:04:760Paolo Guiotto: Now, we will see that these two are a bit different, because, if,
41:10:900Paolo Guiotto: If we compute the Fourier transform of the derivative with respect to T,
41:16:920Paolo Guiotto: So, we may notice that the hat of DTU sharp, so we are doing the Fourier transform with respect to the X variable of this. Formally, this is the integral on R,
41:32:260Paolo Guiotto: of your function, which is in this case DTUTX,
41:37:650Paolo Guiotto: E minus I, C, X, EX.
41:42:340Paolo Guiotto: Now, as you can see, the variable t is not the integration variable, and the derivative with respect to t is the derivative with respect to the parameter. So we may expect that under good assumptions, this derivative will come out.
41:55:560Paolo Guiotto: Of course, since I don't know yet which are the right assumptions, and yeah, we are not doing a formal argument, so I assume that for a moment this operation can be done.
42:06:210Paolo Guiotto: At the end of all the story, I will arrive to a formula for you, and on that formula, I can show that doing the right assumptions, they are very weak in particular, in this case, this operation can be performed.
42:20:360Paolo Guiotto: So let's assume that we can do that. So the derivative goes outside of the integral, and we have this. Now, here, this is the Fourier transform of U
42:32:110Paolo Guiotto: T, with respect to the variable X, no? So this is the quantity we defined above the head of UT sharp.
42:41:520Paolo Guiotto: at point C.
42:43:370Paolo Guiotto: So that's the function that we called P ,
42:48:800Paolo Guiotto: So, in other words, we have this, that the hat of the derivative with respect to t is the derivative with respect to t of the hat.
43:00:240Paolo Guiotto: And this is, let's say, reasonable, because these two operations, the derivative in T and the Fourier transform in X, they work on different variables. So you may expect that you can switch one with the other, because
43:14:100Paolo Guiotto: between the two, there is an integral, so that's the point. So now we have this second ingredient, the hat of the derivative respect to T of UT sharp at point C is the derivative of
43:30:230Paolo Guiotto: be… I remind you that VTXC is the Fourier transform of UT sharp.
43:41:540Paolo Guiotto: Now, what happens when I take a Fourier transform to the right-hand side?
43:47:890Paolo Guiotto: So, when I do the Fourier transform of the sigma squared over 2, derivative with respect to X 2 times of QT sharper.
43:57:640Paolo Guiotto: Now, the factor sigma square half is just a multiplicative factor. Then I have to do this, the Fourier transform of the derivative with respect to the variable of which I do the Fourier transform.
44:14:170Paolo Guiotto: So this is exactly as the formula of the derivative, Fourier transform of the derivative. We may remind that the Fourier transform of F prime
44:24:510Paolo Guiotto: Where F prime is the derivative of F with respect to its variable x, so this is the DXF.
44:32:140Paolo Guiotto: So let's write with this notation that reminds us of partial derivatives. This is a multiplication by factor iC the Fourier transform of F, right?
44:44:380Paolo Guiotto: If I iterate this, so I take the second derivative with respect to X of F at point C, well, this is what? The derivative of the derivative.
44:59:170Paolo Guiotto: I apply once the formula for the derivative of… the fluid transform of the derivative, I get the factor ik out, and then I have dxf hat C.
45:11:850Paolo Guiotto: I apply once again, I have a new I-axis, so this becomes IC square.
45:18:360Paolo Guiotto: F at C. And this is the formula.
45:21:980Paolo Guiotto: iC square is I2 is minus 1, T squared is square, so minus C squared, F at C.
45:31:620Paolo Guiotto: This is what happens if I have just a function of X. And the same happens here, because C is just a parameter for this transport, no? It's not touched by the integration, so I may expect that when I will do the hat of this.
45:52:790Paolo Guiotto: I will get minus C squared, the head of UT sharp.
45:59:950Paolo Guiotto: at point C, and this is, again, my V, so minus C squared VTC.
46:08:990Paolo Guiotto: And this is, let's say, the sack on the…
46:11:840Paolo Guiotto: formula, so D hat of DXX were better, sigma squared over 2, dxx UT.
46:19:560Paolo Guiotto: sharp, I do the hat of all this, evaluated at C, is equal to sigma square over 2, the coefficient, times minus C squared, so minus sigma squared over C squared, and then we have VTX.
46:38:650Paolo Guiotto: So once we have done this,
46:41:110Paolo Guiotto: we can apply all these calculations to the equation. So… The equation DTU, TX,
46:53:660Paolo Guiotto: equals sigma square over 2, dxx.
46:58:10Paolo Guiotto: UTX, huh?
47:01:610Paolo Guiotto: will imply that the hat of DTU D sharp, huh?
47:08:970Paolo Guiotto: at point C, is sigma squared half, we just computed, minus sigma squared half C squared.
47:17:530Paolo Guiotto: V, VX, huh? VTXE.
47:21:440Paolo Guiotto: But the left-hand side, we say that the derivative with respect to t, yeah, we have the formula, is the derivative with respect to T of P. So this is DT.
47:32:250Paolo Guiotto: B, T, C.
47:35:170Paolo Guiotto: So we get that applying the Fourier transform to both sides, provided this transform can be computed.
47:44:490Paolo Guiotto: we get that the function V, so B of Txi solves this equation DTV6C…
47:59:410Paolo Guiotto: Equal minus sigma squared half C squared V tick C.
48:08:110Paolo Guiotto: Which is still a differential equation, so apparently we have not…
48:13:600Paolo Guiotto: done such a big simplification. But in fact, this is a very simple equation.
48:21:190Paolo Guiotto: You see, the equation for U is a partial differential equation. So you have derivatives in T, derivatives in X.
48:28:570Paolo Guiotto: But this one is an ordinary differential equation, because you can consider c as a constant, and you look at the variable T, where there is the derivative, this is an equation which is a first-order linear equation.
48:43:230Paolo Guiotto: So this is a PDE.
48:45:740Paolo Guiotto: And this is what is called an ODE, ordinary differential equation. And moreover, this is a very simple equation, because it is like when you have Y prime t equal this for the letter T is a constant. It's like CY of T.
49:04:780Paolo Guiotto: You know, to solve this equation. Now, what is the function such that y prime is C times y is the exponential?
49:12:920Paolo Guiotto: with the coefficient C. So this says that it is YT is E to CT, maybe times a constant K, because all these functions verify this type of equation. So we can do the same for this, reminding now that C is this number.
49:32:700Paolo Guiotto: So it depends on x, but not on T, so for T is a constant. So we will get that
49:38:380Paolo Guiotto: from this V of tick C, Is equal to, constant.
49:45:800Paolo Guiotto: Now, that constant is a constant in time.
49:50:780Paolo Guiotto: But, there is also Xi parameters, so you may expect that this constant will be a constant in C. It's constant in T, but it's a function of C. E minus this coefficient, sigma squared, well, better, let's write that. E minus 1 half sigma square C squared.
50:10:980Paolo Guiotto: And remind that this is the head of our UT shell. So what we have here, we have almost found
50:21:50Paolo Guiotto: the Fourier transform of the solution, the Fourier transform with respect to the X value, not respect to the pair DX. Now, the point is, what is this coefficient? How can I determine the coefficient?
50:35:270Paolo Guiotto: As you can see, here we have not yet used, in any place, the initial condition.
50:41:360Paolo Guiotto: And that's where what we are going to do now, because since we know that
50:49:780Paolo Guiotto: By the initial condition.
50:56:520Paolo Guiotto: Sorry, I forgot here to put the letter T.
51:02:400Paolo Guiotto: That is here, right?
51:05:120Paolo Guiotto: So, by the initial condition.
51:09:490Paolo Guiotto: I know that U at 0x is equal to this data phi of X, which is supposed to be known.
51:18:690Paolo Guiotto: So, how can we use this now to, determine this coefficient K?
51:26:80Paolo Guiotto: Let's put t equals 0, let's see what happens. So, if I put t equals 0, I notice that V of 0x c becomes exactly here. When you set T equals 0, the exponential is e to 0, so 1, becomes exactly that coefficient KC.
51:46:300Paolo Guiotto: But on the other side, this is the head of U0 sharp.
51:53:280Paolo Guiotto: And U0 sharp is phi.
51:56:180Paolo Guiotto: So this is the hat of phi evaluated at Z.
52:01:240Paolo Guiotto: So, what is K of C?
52:05:440Paolo Guiotto: From this, we get that K of X.
52:09:10Paolo Guiotto: which is V0C, which is the hat of U0 sharp evaluated at Xi, but U0 is the initial data, phi, evaluated at C. We did use that that coefficient, KXC,
52:27:330Paolo Guiotto: is nothing but the head of phi of C.
52:32:630Paolo Guiotto: So now we have a conclusion, because we have found that VTICE
52:38:890Paolo Guiotto: which is the hat of UT sharp.
52:44:730Paolo Guiotto: Evaluated at point C. This is equal to KC, which is fiat C,
52:52:420Paolo Guiotto: times exponential minus 1 half sigma squared C squared.
53:00:210Paolo Guiotto: So, we have not yet found the solution, U, but we have found explicitly, we may say, the Fourier's form is starting to get spiral of the solution.
53:11:760Paolo Guiotto: Because, this depends on this part, which is, known as an explicit function, then it's the sigma squared coefficient that we are in front of.
53:20:860Paolo Guiotto: the equation, and this is if we get a form of initial data. So, since also the initial data is supposed to be known, we can say, partly that
53:32:910Paolo Guiotto: This is an explicit formula for the Fourier transform of the solution.
53:39:890Paolo Guiotto: Now, we want to go back to the solution.
53:43:460Paolo Guiotto: So the idea is that now we should recognize that at the right-hand side, there is a Fourier transform.
53:50:140Paolo Guiotto: In fact, we have something which is visible, there is this hat here.
53:55:420Paolo Guiotto: And then there is something else here.
53:58:750Paolo Guiotto: But now we should remind something important, that the Fullerton's form of the Gaussian, E minus square divided 2 sigma squared.
54:10:500Paolo Guiotto: C, square… C squared. No. Good night.
54:16:10Paolo Guiotto: So, I normally prefer, because for me it's easier to remind in the scaled version, so this formula here, this is E minus 1 half sigma squared C squared.
54:27:750Paolo Guiotto: So in this way, you see that this guy here is the head of something. So we have that. The hat of U, T-sharp.
54:38:800Paolo Guiotto: evaluated at X is the product between the hat of phi and the hat of this Gaussian distribution.
54:48:360Paolo Guiotto: E minus square divid… divided by…
54:53:460Paolo Guiotto: 2 sigma squared over root of 2 pi sigma squared.
55:00:410Paolo Guiotto: Si.
55:02:560Paolo Guiotto: Now, this is the product of… algebraic product of two Fourier transforms, and we know the rule that this comes from the Fourier transform of the convolution.
55:13:660Paolo Guiotto: So this is also the Fourier transform of T star D ocean kernel.
55:21:730Paolo Guiotto: E minus R squared over 2 sigma squared, divide the root of 2 pi.
55:26:570Paolo Guiotto: sigma squared.
55:28:390Paolo Guiotto: evaluated at sea.
55:30:480Paolo Guiotto: And now we have the conclusion at hand, because…
55:35:680Paolo Guiotto: We are saying that if we're a transform of UTE,
55:41:120Paolo Guiotto: Okay, so for a fixed value.
55:44:50Paolo Guiotto: Yes, I… I know, I'm… I apologize, because I'm still forgetting this letter T. You see that there is the letter T, yeah? So, I apologize. There is a T here.
56:00:550Paolo Guiotto: So there is,
56:03:60Paolo Guiotto: Okay, this formula that I wrote here is correct, but it is not exactly the same. So if I want to see that coefficient, I have to give the T to sigma squared, put the t with sigma square.
56:19:970Paolo Guiotto: It's basically, you change sigma squared with sigma squared t, so you put a T here.
56:26:150Paolo Guiotto: a T down here in the root, and you get a T here, T sigma square axis square.
56:32:540Paolo Guiotto: So, sorry for that.
56:34:680Paolo Guiotto: This is, 2 sigma squared t, there is a sigma squared T down here.
56:40:670Paolo Guiotto: Okay, and so we have that. Create transform of UT as transformer variable X is the transform of this convolution.
56:52:660Paolo Guiotto: But this means that, now by uniqueness of the Fourier transform.
57:01:340Paolo Guiotto: that you have, finally, this beautiful formula. UTX is the convolution between phi and E minus sharp squared 2 sigma squared t divided by root of 2 pi
57:18:950Paolo Guiotto: sigma squared T.
57:21:120Paolo Guiotto: Excellent.
57:22:570Paolo Guiotto: Or, in integral terms, this is integral on R, phi of Y,
57:30:30Paolo Guiotto: E minus X minus Y squared over 2 sigma squared t over root of 2 pi.
57:41:310Paolo Guiotto: sigma squared T, DY.
57:45:450Paolo Guiotto: So this is the formula.
57:48:980Paolo Guiotto: Where the formula for the solution of the heat equation comes from.
57:55:540Paolo Guiotto: Now…
57:59:450Paolo Guiotto: what have we done? Basically, we've done a calculation. So, we assumed, and this is, in general, the… the strategy, no?
58:08:190Paolo Guiotto: If you don't know what are the right conditions, you take your equation, you apply outside the Fourier transform. Now, here, for this case, we use the Fourier transform only in the next viable. We couldn't do predictive because,
58:24:940Paolo Guiotto: The physics of this problem is asymmetric in time. So, once you fix the initial condition, you can go on in the future and not in the past. We cannot have T negative. Yeah, and that's because if you put T negative.
58:41:340Paolo Guiotto: everything here gets wrong, because you mentioned T is negative here, this becomes a positive exponential, cannot be identical, so all this argument crashes. And also, this final formula.
58:56:250Paolo Guiotto: It's a consensus negative, because we have a group of negatives, which are strong as numbers. This explanation comes as a positive explanations of the story changes in theory.
59:06:680Paolo Guiotto: So this problem has, is well-defined only for T-positive, so T positive…
59:16:210Paolo Guiotto: and the X real. That's why, basically, if you want to use the Fourier transform.
59:24:550Paolo Guiotto: There are other equations for which, for example, you can have two positive, for example, for the wave equation R.
59:32:390Paolo Guiotto: Secondly, and it's similar, this one, but with another derivative. That way, you could use different letters forming both variables.
59:42:850Paolo Guiotto: But in this case, we have… we can do only this. In any case, the general idea is we take the Fourier transform of the solution, and we apply the Fourier transform to both sides. We use a
59:58:410Paolo Guiotto: We assume that certain steps can be done, for example, this one, no?
00:04:620Paolo Guiotto: And at the end of all this, we get,
00:09:660Paolo Guiotto: the goal is to… to solve for the Fouliatens form of U, no? So at the end, we obtained a formula for the Fouliaten's form of U.
00:23:300Paolo Guiotto: Now, in this case, we recognize that this is F in the transform, we use uniqueness, and we arrive to transform that.
00:30:560Paolo Guiotto: So, let's say that we're not clear… how's clear at this point.
00:36:320Paolo Guiotto: So now, what could we prove the real company?
00:39:670Paolo Guiotto: Now, once you have this, you could prove that if the function tree is in 1, this function is well-defined X, because it is a convolution between this and the notion there, and so there's no problem. You can, doing by hand, the derivatives, respectively, respect to X two times.
00:58:260Paolo Guiotto: And just to find that you can differentiate down the length of a sign, but let's say, the motion.
01:03:940Paolo Guiotto: is fast enough at infinity that helps you to control how the derivations on the integral sign, because if you want to complete the derivative perspective here, you have to carry the derivative in the inside here. If you want to complete derivative with respect to X, you have the one to differentiate this one, no?
01:22:990Paolo Guiotto: But in any case, you can show, with a little bit of…
01:27:330Paolo Guiotto: work, that this function has derivatives with respect to t, with respect to X, 1 times 2 times. Actually, it turns out that this has derivative with respect to X with any overlays in the same feeling function. It's much more regular than what is needed.
01:44:750Paolo Guiotto: And by computing these derivatives, so dt of u and dxx of u, and plugging into the density, you can verify that it is real true that dt of this U is equal to sigma squared to the DXX of this U.
02:01:490Paolo Guiotto: I don't want to solve this one.
02:05:620Paolo Guiotto: So, this is how usually… It was, huh?
02:10:550Paolo Guiotto: with this.
02:12:870Paolo Guiotto: Okay.
02:14:120Paolo Guiotto: Now, for example, here.
02:18:660Paolo Guiotto: you are able to solve for you once you arrive at this point, because you can recognize that this is a transform, and then reduce unit of multiple to the act of something, so U equal to something.
02:36:390Paolo Guiotto: Now, the problem, more general is…
02:39:860Paolo Guiotto: Can I always invert this matrix form? So, suppose that you don't see that this is n-hat of something directly, so what can you say about the fact that a function is, in general, n-hat of something? This is the inversion problem.
02:57:620Paolo Guiotto: So, inversion…
03:02:830Paolo Guiotto: problem.
03:04:230Paolo Guiotto: So, the inversion problem consists in… so, the direct problem is you have an F, you compute its Fourier transform, you have an F at…
03:13:870Paolo Guiotto: Now, the inverse problem is, imagine that you have a function of, let's say, let XC, to think that it should be a free transform. We wonder if this function is the eth of something, okay? So, problem is…
03:29:30Paolo Guiotto: Given… a function G, let's say function of C, See, Riah?
03:38:860Paolo Guiotto: Detail me.
03:40:540Paolo Guiotto: an F, let's say function of X, Axial.
03:46:940Paolo Guiotto: Such that,
03:48:930Paolo Guiotto: the transform of F be our function G.
03:53:870Paolo Guiotto: The function f is called the Fourier original of G.
03:59:770Paolo Guiotto: F is older.
04:04:70Paolo Guiotto: 48.
04:08:370Paolo Guiotto: Original.
04:12:250Paolo Guiotto: off.
04:14:150Paolo Guiotto: Jeez.
04:16:480Paolo Guiotto: Now, an actual question is, under which assumptions this F exists, So, question one.
04:24:180Paolo Guiotto: I'm the… Which… assumptions… Such.
04:35:610Paolo Guiotto: and F exists.
04:38:450Paolo Guiotto: And question 2… How can be computed?
04:44:840Paolo Guiotto: How… F… 10.
04:48:370Paolo Guiotto: be determined?
04:51:350Paolo Guiotto: by… G?
04:56:600Paolo Guiotto: Of course, we may say that if G is yet of someone, necessarily, we must have that G is continuous, it's bounded, it goes to zero at infinity, so these are, first,
05:13:60Paolo Guiotto: Remarks, if, G is the head of F, at least for F in L1.
05:23:830Paolo Guiotto: Then, necessarily, G must be continuous, and G at plus minus infinity should be equal to zero.
05:35:470Paolo Guiotto: Now, these two conditions alone are not sufficient. The example is not easy to be shown, I can do with the tools we have here.
05:45:60Paolo Guiotto: But, however, it can be proved It can… B.
05:52:790Paolo Guiotto: Proved, huh?
05:56:440Paolo Guiotto: that, There are functions.
06:03:230Paolo Guiotto: Well, let's say that these, conditions… are… not.
06:12:750Paolo Guiotto: sufficient.
06:15:460Paolo Guiotto: to… determine… To… But to have… the existence of F.
06:27:100Paolo Guiotto: Fourier, original.
06:30:70Paolo Guiotto: object, okay? So even if you have a continuous function that goes to zero at infinity, it is not necessarily true that this is the Fourier form of something, okay?
06:40:970Paolo Guiotto: An example could be constructed with a function with these features, which is… which is not the free transform of anything. There is no F for which F at is your G.
06:52:750Paolo Guiotto: Something that we can say, relatively cheap, is that If… the function G, is also in L1, huh?
07:07:750Paolo Guiotto: Then, we can notice this. Then… if, G,
07:15:670Paolo Guiotto: is the cut of F with the F in L1, huh?
07:21:910Paolo Guiotto: we would be in this situation. F is in L1, and its hat is in L1, because we are assuming that G is in L1. So, we…
07:34:570Paolo Guiotto: have.
07:36:700Paolo Guiotto: both F and FF in L1. So, inversion formula would apply.
07:46:580Paolo Guiotto: And we would get that by doing D hat, we would have that G hat would be the double hat of F. But we know that when we do the double hat, we go back to F, basically.
07:59:480Paolo Guiotto: Let's write precisely. Now, since I'm writing a G hat, I will use the letter X, not the letter X, because X here is supposed to be the variable of G, okay? And since this is going to be F, let's say that we use the letter X for F, we use the letter F
08:17:569Paolo Guiotto: the letter X for the variable here. So, precisely, what happens is that G hat of X would be the double hat
08:26:939Paolo Guiotto: of F at point X.
08:30:930Paolo Guiotto: Now, by the inversion formula, this is 2 pi, F evaluated at minus X.
08:39:850Paolo Guiotto: And from this, we would get the formula for F. So…
08:45:670Paolo Guiotto: replacing X with minus X, I would, have that f of x is 1 over 2 pi.
08:53:220Paolo Guiotto: the hat of G evaluated at minus X.
08:58:750Paolo Guiotto: So, we can say that these cells
09:02:600Paolo Guiotto: If we assume that our G is in L1, and it is the Fourier transform of some F in L1,
09:13:609Paolo Guiotto: Then, that F is this one.
09:16:840Paolo Guiotto: So it gives you the way to compute that.
09:20:140Paolo Guiotto: So basically, as you can see, apart from the fact that 1 over to 5, apart from the fact that the value is not X, but minus X,
09:28:510Paolo Guiotto: F is latent form of G. So, the inverse of this later transform you see is written form. That's what we say. Now, this is… this says… this… logically, this proves what? This proves that if you have an F1 function, which is…
09:45:649Paolo Guiotto: get the form of an L1 function, F, then that F is this one, no?
09:51:510Paolo Guiotto: But, let's say that also the vice versa holds, and the same argument proves. So, this proves that
10:06:390Paolo Guiotto: If G is equal to F, h, Weed, then.
10:14:340Paolo Guiotto: G and F, both in L1.
10:18:700Paolo Guiotto: Then? F.
10:21:840Paolo Guiotto: is determined by these formulas. 1 over 2 pi, G hat minus X.
10:31:590Paolo Guiotto: Vice versa.
10:33:400Paolo Guiotto: Let's say…
10:39:20Paolo Guiotto: So, this does not yet respond completely to the question. Now, the question is, given G, why the reason F is such that G is at F? Because this is just for me, suppose that G is at F. So, you know that there is this F. This just says, then F must be this one.
10:59:250Paolo Guiotto: Okay?
11:00:370Paolo Guiotto: This is not proving the existence of F. Instead, if F exists, it is this one, but it does not tell you that it exists. However, he says also that if F exists, it must be different over G.
11:13:990Paolo Guiotto: Okay, so this one is the unique possibility. Now you can say this, vice versa. If G is in L1,
11:24:480Paolo Guiotto: And, also, So you can compute its transform. And also, the hat of G is in L1,
11:34:320Paolo Guiotto: Now, we know that this is not automatic, no? If you have an L1 function, the hat is not necessarily L1. We have the case of rectangle, no? D hat is sine X over X, which is not L1, no? So…
11:49:820Paolo Guiotto: We have to put that condition, because it's not automatically verified, but once we know this, setting…
11:59:120Paolo Guiotto: the candidate, this one, F of X.
12:05:310Paolo Guiotto: you put, by definition, 1 over 2 pi g-hat evaluated at minus X. Now, this thing makes sense, no? Because, you know, G is not 1, this one is well-defined, you get even a continuous function, this defines something, right?
12:20:520Paolo Guiotto: And you are assuming that also the height of 2 is in a 1, so this is in advance. If you transform this.
12:29:760Paolo Guiotto: And you can do that. The hat of F will be 1 over 2 pi, well, G hat of minus the variable.
12:40:500Paolo Guiotto: the at of all these. Let's say that for the at, we use the letter X for this section.
12:46:550Paolo Guiotto: And now, let's see what is this.
12:48:860Paolo Guiotto: Now, the coefficient 1 over 2 pi is a multiplicative factor we put in front. 1 over 2 pi. Then we have D hat of D hat of G evaluated at minus the variable.
13:01:270Paolo Guiotto: Appoint seal.
13:03:460Paolo Guiotto: There is… there are general properties for the Fourier transform when you translate the argument, or when you multiply by a constant. We can see directly here, because this would be the integral. Now, be careful, because the variable
13:18:420Paolo Guiotto: of is Xi, no? So we say the integral of this function, so G hat at minus X,
13:27:370Paolo Guiotto: E minus iCX, this is integral in X. Okay, you see? This is, by definition, what is the Fourier transform of the function G hat minus X.
13:41:60Paolo Guiotto: But, you change variable here, you put Y equal minus X, you get… this becomes the integral in R of G hat Y,
13:50:580Paolo Guiotto: This becomes C to ICY, and DX becomes just the Y in the absolute value.
13:58:430Paolo Guiotto: So this is, N hat.
14:01:200Paolo Guiotto: It is the hat of G hat, as you can see, evaluated at
14:12:490Paolo Guiotto: Minus X, exactly. So, what happens here, you see that the minus that you have inside has gone outside.
14:21:240Paolo Guiotto: So, this means that if I continue here.
14:26:520Paolo Guiotto: I have 1 over 2 pi, then there is the double hat of G, evaluated at minus Z.
14:34:530Paolo Guiotto: But, because of the inversion formula, what is the double head of, something?
14:42:130Paolo Guiotto: So this is 1 over 2 pi, inversion
14:47:140Paolo Guiotto: formula, it is G, no, the two ads are canceled, at minus the variable, which is minus X, times 2 pi, I'm sorry, 2 pi G is minus minus X. So, at the end of all this story, you get exactly G of Xi.
15:06:890Paolo Guiotto: So what shows this is that
15:09:740Paolo Guiotto: If G is in L1, and G hat is also in L1, then defining this thing, f of x equals 1 over 2 by G hat of minus X, which is the unique possible Fourier original, you check that this is actually a Fourier original.
15:27:690Paolo Guiotto: Because the Fourier transform of F turns out to be, G of C.
15:35:910Paolo Guiotto: So, the conclusion is that… F is a Fourier.
15:42:700Paolo Guiotto: Original of… G.
15:46:810Paolo Guiotto: So, the conclusion is that…
15:55:840Paolo Guiotto: So if G is in L1,
15:59:900Paolo Guiotto: with the G hat is in a 1.
16:03:370Paolo Guiotto: there exists a unique F in L1, such that…
16:09:570Paolo Guiotto: D, hat of F is G.
16:14:750Paolo Guiotto: We have…
16:18:110Paolo Guiotto: that this F is this function, f of x is 1 over 2 pi g hat at minus X.
16:28:310Paolo Guiotto: So basically, the inverse of the Fourier transformer what goes back, no?
16:35:10Paolo Guiotto: The Fourier transform takes an F and gives you the hat of F. Now, imagine you take a G and you want to find the F such that F hat is G. What is DF? It's basically a gain, apart for the exchange of sine.
16:55:90Paolo Guiotto: Now, let's see some example of users.
16:59:320Paolo Guiotto: Of this, again, from the exercises.
17:06:250Paolo Guiotto: What is it?
17:09:720Paolo Guiotto: Okay, so…
17:17:790Paolo Guiotto: exercise… 1932.
17:28:560Paolo Guiotto: We have, this function, G, A, B, of C, It's defined as 1 over…
17:38:440Paolo Guiotto: C squared plus A square.
17:42:200Paolo Guiotto: times C squared plus B squared.
17:45:860Paolo Guiotto: Here, C is real, and A and B are positive values.
17:54:180Paolo Guiotto: with the A different from B.
18:03:200Paolo Guiotto: Now, question one, show… That G. A B.
18:10:100Paolo Guiotto: has… Fourier Original.
18:15:750Paolo Guiotto: in L1.
18:20:760Paolo Guiotto: and compute it.
18:28:680Paolo Guiotto: Question 2.
18:31:510Paolo Guiotto: show that.
18:35:430Paolo Guiotto: also, C, G, A, B.
18:39:900Paolo Guiotto: C?
18:41:460Paolo Guiotto: As for original.
18:51:470Paolo Guiotto: and find it in terms of the Fourier original of G.
18:55:780Paolo Guiotto: And then… Express.
19:01:900Paolo Guiotto: this… in terms… of Fourier origin out.
19:11:130Paolo Guiotto: All.
19:12:190Paolo Guiotto: Gee, but let's see what happens.
19:16:170Paolo Guiotto: Okay, so… Do not focus on the specific example.
19:25:240Paolo Guiotto: So let's, see.
19:27:170Paolo Guiotto: Now, first of all, show that F has a fully original. Of course, if I…
19:33:80Paolo Guiotto: If I can say, well, this function is the free original because the freedom form of this function is that G,
19:40:180Paolo Guiotto: It's a perfectly fine answer, okay? But sometimes you should distinguish the fact that you are able to say there is a free original from the fact that you can explicitly determine the free original.
19:54:670Paolo Guiotto: Because these are two different stories.
19:56:990Paolo Guiotto: If you want to say that there is a free original V cells, just check that G is in L1 and the transform of G is in L1. Then you are sure that there is free original, and that's… Now, to check that G hat is in L1, useful G hat, basically you have that h, you see?
20:14:110Paolo Guiotto: Now, if you are able to compute G-hat, this means that you are already at your
20:20:20Paolo Guiotto: But this is not necessarily required, because to check that the transform is in L1 can be done with, you know, with the regularity test. If the function, the derivative and the second derivative, i in L1,
20:35:280Paolo Guiotto: Then we have that free transform is in your mind. This was a consequence of the…
20:42:190Paolo Guiotto: relation between Fourier transform and derivative. So, we don't need to do… we can do directly on G all these checks without computing the Fourier transform, okay?
20:52:440Paolo Guiotto: So… If a… So we know that.
21:01:460Paolo Guiotto: that.
21:03:280Paolo Guiotto: If, G… well, let's just write G and not GAV.
21:08:710Paolo Guiotto: He's in L1.
21:10:570Paolo Guiotto: Well, this is clear, no?
21:13:310Paolo Guiotto: It's a continuous function.
21:15:730Paolo Guiotto: those plus A squared plus B squared never equal to 0, the denominator, so no problem at find it at infinity, it's like 1 over XC to power 4, no?
21:25:940Paolo Guiotto: True.
21:29:00Paolo Guiotto: Because G is continuous in the real line.
21:34:120Paolo Guiotto: G of C at plus, minus infinity. Do you see the asymptotic behavior?
21:43:680Paolo Guiotto: So, C squared plus A squared A is a constant, goes like C squared. The other goes like C squared. The product will go as C to power 4. So, everything goes as 1 over C to power 4, which is integral
22:00:860Paolo Guiotto: at… plus min… plus minus… Infinity.
22:10:970Paolo Guiotto: And… G hat is in L1. Now, how do we know this?
22:17:990Paolo Guiotto: Either we compute G-hat and we check if it is in L1, that's a possibility, but this means that we need to be able to compute that full term. We will do later.
22:28:890Paolo Guiotto: Or, we could use that, this remark, because…
22:36:290Paolo Guiotto: We know that if we have a function such that function G with its first and second derivative are both in L1, then the hat of G is in L1.
22:50:370Paolo Guiotto: We know that G is in L1 are already checked, no?
23:01:270Paolo Guiotto: About G prime, we do 4G prime, you understand easily that 4G second is similar. 4G prime, well, we have to compute a derivative here, which is a reminder, we don't need to do heavy calculations for nothing, we just need to know if the derivative is integral, okay?
23:20:190Paolo Guiotto: So, for example, that's 1 over something. I would do the derivative in the form minus. Denominator is the square, so I will have C squared plus A square, square. C squared plus B squared squared.
23:33:940Paolo Guiotto: That numerator is the derivative of denominator, that product.
23:39:100Paolo Guiotto: Now, what is the product? You will have, if you want, the derivative is 2xi times the other factor, 2C times C squared plus B squared, and minus 2C times C squared plus A squared. You don't need now to simplify.
23:59:60Paolo Guiotto: Because, clearly, this is a continuous function in the real line.
24:03:450Paolo Guiotto: There are no issues with the denominator. And if you look at the behavior at plus, minus infinity, well, let's take the absolute value.
24:14:550Paolo Guiotto: So, it is, so the miners disappear, you have, perhaps a 2,
24:21:540Paolo Guiotto: Well, no, the C cubed term actually cancels, no? You see, there is 2C cubed minus 2C cubed. So, at the end, this term, will be something first order, no?
24:35:450Paolo Guiotto: It's like C, you know? You have 2CB square minus 2CA2, so it's about constant,
24:43:900Paolo Guiotto: constant modulus of Xi divided downstairs, we have a lot of power, because Xi to power 4 squared, so C to power A.
24:53:50Paolo Guiotto: So it's about, constant divided models of Xi to power 7, which is integral At plus minus infinity.
25:04:930Paolo Guiotto: And same check for the second.
25:07:960Paolo Guiotto: Same for…
25:10:410Paolo Guiotto: G second, I don't see… you see that it is even… it is even better, the behavior, than the behavior of G. G was going to one of… as 1 of XC4, this goes as 1 of XC to 7, the next one would be even faster, at infinity.
25:26:550Paolo Guiotto: So, since we have the seasonal one.
25:29:800Paolo Guiotto: and jihat is also in L1, we can use that general result we just come to see to say that there exists an F in L1,
25:40:390Paolo Guiotto: Such that f hat is equal to G. And we know that f of x is 1 over 2 pi the hat of g at minus X.
25:52:790Paolo Guiotto: Now, the problem is, we should compute the D hat of G,
25:58:600Paolo Guiotto: Or we should, we should, now, we should…
26:04:520Paolo Guiotto: Or we should see that this G is the Earth of someone.
26:09:800Paolo Guiotto: How can we do here? Now, you recognize that these denominators are suspect, notice that the denominator of the Cauchy distribution.
26:17:490Paolo Guiotto: So I may wonder, how can I get this from the Cauchy distribution?
26:22:760Paolo Guiotto: So that seems the common denominator of something, right? So let's see what you have by doing this. 1 of XC squared plus A squared times X c squared plus C squared. What if I try to do 1 over X c squared.
26:39:400Paolo Guiotto: plus A squared plus 1 over, plus or minus, let's say, XC squared plus E squared. I want to get that fraction. You see that if I sum, I get the same denominator, then I have to do the sum of the two denominators in denominator. So I do not cancel the X squared, you see?
26:59:630Paolo Guiotto: If I put the plus, I will get something like 2C squared plus I squared plus B squared. It's not what I want. So, with the minus here, it happens that if you do the common denominator, you get
27:14:150Paolo Guiotto: C squared plus A square times C squared plus B squared.
27:19:760Paolo Guiotto: And then with the minus, I have C squared plus B squared, minus… C squared plus A square.
27:30:590Paolo Guiotto: So this cancels these two guys, and what remains is B squared minus A squared, which is just a constant.
27:39:580Paolo Guiotto: over C squared plus A squared, C squared plus B squared. So basically, apart from the coefficient B squared minus A squared, that's my function G of C.
27:53:700Paolo Guiotto: So I see that if I want now to extract G, I have that G of C is this coefficient, 1 over B squared minus A squared.
28:06:630Paolo Guiotto: You remind that in the beginning of the problem, A was different from B, so that number is not zero.
28:14:580Paolo Guiotto: So, I'm safe with this.
28:17:130Paolo Guiotto: because… A is different from B. Then we have one other, C squared plus A squared.
28:28:730Paolo Guiotto: minus… 1 over C squared plus B squared.
28:34:170Paolo Guiotto: I know that these two are Fourier transforms, right?
28:38:20Paolo Guiotto: These are the Cauchy distributions, no?
28:41:120Paolo Guiotto: So, not… not exactly. So, what is the… the point? They come from this one, no? E minus modulus, let's say C modulus the variable. This thing has returned 2C divided C squared plus C squared, right?
28:59:490Paolo Guiotto: Food
29:00:560Paolo Guiotto: So I need to have the 2A, the 2B. I can do, because I can put a 2A here, and I divide by 2A, a 2B here, I divide by 2.
29:13:100Paolo Guiotto: So we get… this is one of a B squared minus A squared times
29:18:920Paolo Guiotto: 1 over 2A, then we have the hat of E minus A, the modulus.
29:25:460Paolo Guiotto: C.
29:26:630Paolo Guiotto: And here we have minus 1 over 2B, the head of the minus B, the modulus.
29:36:80Paolo Guiotto: Okay, now we can see how this can be made a unique hat, because we can still carry in and out the coefficients, we can put together the sums, so we have a unique big hat, 1 over B squared minus A squared.
29:52:360Paolo Guiotto: times 1 over 2A, E minus A modulus.
29:57:680Paolo Guiotto: minus 1 over 2B, E minus B demobulus.
30:03:980Paolo Guiotto: all this evaluated at C. That function is our G , so here you have written in front the Fourier original.
30:14:590Paolo Guiotto: Okay, so if you have done this without doing this previous discussion, it's correct. It's perfectly fine, because you are solving at once the existence and the exact value of the solution.
30:30:140Paolo Guiotto: So that's perfectly fine. So from this, we get… we don't have even… even to use this formula, just to… just… just to be clear, no?
30:38:910Paolo Guiotto: We don't need to compute the GF, because we already read who is F, so F of X is 1 over B squared minus A squared times 1 over 2A, E minus A modulus of X minus 1 over 2B.
30:58:370Paolo Guiotto: E minus B modulus of X.
31:03:460Paolo Guiotto: And that's the answer to this question.
31:08:590Paolo Guiotto: Okay, now time is over, and there is a second question, which is, well, let's see, just to start. There is a question about XC. What about XCG of C?
31:23:760Paolo Guiotto: Now… This is L1.
31:27:240Paolo Guiotto: And for the same reasons as before, also the hat of this, CG,
31:33:770Paolo Guiotto: will be in L1. Actually, what is the hat of CG?
31:39:340Paolo Guiotto: So, variable times G.
31:49:810Paolo Guiotto: Yeah, that's the derivative. Now, we remind that minus I sharp G.
31:55:700Paolo Guiotto: sharp C, that's the derivative of the Fourier transform of G.
32:09:940Paolo Guiotto: Okay.
32:11:230Paolo Guiotto: So…
32:13:620Paolo Guiotto: I already have the free transform of G, which is basically this function here. You compute the derivative.
32:23:200Paolo Guiotto: Well, actually, this will be in the variable X, not variable C.
32:28:810Paolo Guiotto: And you have that this will be in L1 easily, okay?
32:33:490Paolo Guiotto: Okay, so, since these two are verified, there is a fully original.
32:40:530Paolo Guiotto: And, the Fourier original is… in this case, perhaps it is convenient to use the formula, because there exists F such that F hat is G, that F is, we say, the 1 over 2 pi d hat
33:00:800Paolo Guiotto: Well, not G, it's, XCG of C.
33:06:30Paolo Guiotto: So, is 1 over 2 pi the hat of CG of C,
33:12:790Paolo Guiotto: for better the variable, G, the variable, evaluated at minus X, but this, basically, we have computed, it is the derivative. So you put the minus i here, a minus i here, and this is the derivative with respect to X of the function F you found.
33:31:800Paolo Guiotto: Well, sorry, I'm doing a… I'm running here, but let's call it H. There is an H such that H hat is
33:40:270Paolo Guiotto: this function, that age… That age is this.
33:46:320Paolo Guiotto: which is the derivative with respect to X of F, where f is this function here. So if you want to do the calculations, you have this.
33:57:930Paolo Guiotto: Okay, so, two exercises, 19… 3… 3…
34:10:940Paolo Guiotto: Well, 3-4 is a little bit heavy of calculations.
34:15:550Paolo Guiotto: 3-5… These are all problems with the inversion.
34:22:630Paolo Guiotto: And… 9.
34:26:590Paolo Guiotto: Okay, that's it. So, as I told you, I… I will give a last homework that you should do along the weekend, okay? And I will also give, by tonight, maybe tomorrow morning, the XM simulation.
34:44:400Paolo Guiotto: That means I will give you one or a combination of exercises from previous, exams.
34:52:240Paolo Guiotto: And I will be in class on Tuesday morning to do the solution of these exercises, okay?
34:59:850Paolo Guiotto: If you want to do the exam simulation, try to do, I would say, maybe on Monday, and possibly on Monday you have classes.
35:11:290Paolo Guiotto: In the morning.
35:12:730Paolo Guiotto: So take some time in the afternoon, put yourself in exam conditions, so 2 hours, 30 minutes, no interruption, no one that bothers you, okay? No material, apart for formulas.
35:28:430Paolo Guiotto: And try to see if you can work out the solution in these conditions, okay? I do exercises here and there. Just simulate the exam.
35:38:680Paolo Guiotto: Good.
35:43:530Paolo Guiotto: Let's place this.