AI Assistant
Transcript
00:19:420Paolo Guiotto: Okay, good morning.
00:21:930Paolo Guiotto: We start with an exercise, exercise 18, 5, 4… AB2 positive parameters.
00:34:380Paolo Guiotto: A different from B.
00:36:700Paolo Guiotto: And we define the function FAB of X,
00:41:210Paolo Guiotto: has e to minus a absolute value of X minus E to minus the absolute value of X, always divided by X.
00:52:100Paolo Guiotto: So, there are two questions. First, is FAB in L1 realigned?
01:02:260Paolo Guiotto: So, in this case, if yes, compute the Fourier transform of these functions.
01:13:110Paolo Guiotto: So… What can be said about the integrability for this function?
01:20:120Paolo Guiotto: This function is well-defined, but for x equals 0. For x different from 0, we could say that the function is continuous in the real line, except 0.
01:33:500Paolo Guiotto: So basically, the problem is at zero at plus minus infinity.
01:38:120Paolo Guiotto: At zero, we may observe that, at…
01:44:780Paolo Guiotto: X equals 0. So when X goes to 0, we see that the numerator goes to 0, but also numerator, because both exponentials, they go to E to 0, so they go to 1.
01:58:650Paolo Guiotto: And, recalling that, for example, e to t is 1 plus t plus little O of T, we may write FABX
02:12:850Paolo Guiotto: as 1 minus a absolute value of X plus little o. Well, A is a coefficient. We never write this in the little o, and we can just write little O of X.
02:26:380Paolo Guiotto: minus 1 minus B, absolute value of X, plus, again, little O of X.
02:32:960Paolo Guiotto: all this divided by X.
02:36:260Paolo Guiotto: So, we can cancel the one.
02:39:810Paolo Guiotto: we can put together the two absolute values, so we have B minus A,
02:45:330Paolo Guiotto: models of x plus little O of X, all this divided by X.
02:52:300Paolo Guiotto: From this, it is clear that the function cannot be extended in a continuous way at zero, but there is finite limit from left to right, so there is no problem with the integrability.
03:07:380Paolo Guiotto: So this, when X goes to 0, from the right, it goes to B minus A, from the left, it goes to…
03:17:380Paolo Guiotto: minus B minus A, so in general, these are two different values, but since they are finite.
03:25:200Paolo Guiotto: we can say that, in any case, F, A, B, is integral.
03:32:780Paolo Guiotto: at… Zero.
03:35:780Paolo Guiotto: At infinity, we do not have any problem, because, we can say at,
03:43:660Paolo Guiotto: plus minus infinity. You take models of F, A, B, X,
03:51:90Paolo Guiotto: is less or equal, you can throw away the minus, because, of course, when X goes to plus-minus infinity, the exponential will go to zero fast enough to kill
04:01:470Paolo Guiotto: Well, not actually… you don't need to kill, because that 1 over X is helping going to 0. So you could say that for at least X greater than 1,
04:13:320Paolo Guiotto: greater than 1.
04:15:700Paolo Guiotto: You can even throw away that 1 over X and say that this is bounded… this, plus.
04:24:200Paolo Guiotto: these.
04:26:350Paolo Guiotto: So, they are L1 functions,
04:30:740Paolo Guiotto: Integral at plus infinity at minus infinity.
04:34:660Paolo Guiotto: Okay? Of course, this boundary is not correct when X is closer to 0, because I cannot throw away the X, but in any case, this is sufficient to say that F, A, B,
04:49:300Paolo Guiotto: is in L1 real eye.
04:53:700Paolo Guiotto: So, question two, we can compute the Fourier transform.
04:57:590Paolo Guiotto: Now…
04:59:80Paolo Guiotto: Of course, there is always the definition we should try to compute the integral, which is not the idea here.
05:05:290Paolo Guiotto: Because we know something about this, no? There is this function, which is the exponential function, for which we already know the Fourier transform, no? We remind that the Fourier transform of this type of function, evaluated at point X, is, A…
05:24:780Paolo Guiotto: I do not mind if it is A or 2A, I will check now.
05:29:950Paolo Guiotto: We did this a couple of times below.
05:34:810Paolo Guiotto: exponential, so deflated form is two-way, okay?
05:40:310Paolo Guiotto: So this is just for precision. So we know this, and of course, for B, we have the formula. So how can we use this to compute the freelance form of this function?
05:51:240Paolo Guiotto: You'll notice that if I multiply by X, I notice that… We… Notice… that.
06:01:870Paolo Guiotto: XFAB, X, is equal to the difference between the two exponentials.
06:13:300Paolo Guiotto: So I can say that the Fourier form of the left-hand side will be the same of the Fourier form of the right-hand side, so…
06:20:320Paolo Guiotto: This is a variable, FAB.
06:24:680Paolo Guiotto: valuable.
06:26:580Paolo Guiotto: Evaluated XC. Now, the fluid transform is a linear operation, no, because it is an integral, so…
06:34:900Paolo Guiotto: the Fourier transform of the difference, right? Some difference is the sum difference of the Fourier transforms, so I have 2A over A squared plus C squared minus 2B over B squared plus C squared.
06:51:880Paolo Guiotto: And now, what do I know here? Here, I should remind of a remarkable fact that the Fourier transform of XF is related to the derivative of the Fourier transform.
07:10:370Paolo Guiotto: Now, this is the convolution.
07:12:990Paolo Guiotto: We have seen yesterday the formula.
07:17:750Paolo Guiotto: Which is about here. Now, if a function is L1 and XF is in L1, then there is this formula. So if I add this minus i, which means let's multiply both sides by minus i.
07:32:780Paolo Guiotto: I can write this, so minus i, let's put the minus i as a factor here. And this one.
07:43:260Paolo Guiotto: is, clearly, we are in the conditions of the theorem, because we already observed that FAB is in L1, no? And X FAB is exactly, in our case, is E2 minus A modulus X minus E2 minus B modulus X, which is, of course, in L1.
08:05:550Paolo Guiotto: So we are in condition to apply that fact that says this is equal to the derivative respect to C of the Fourier transform of function f.
08:18:100Paolo Guiotto: So, at the end, I have this,
08:21:20Paolo Guiotto: So, I get that derivative with respect to C of the Fourier transform, sorry, of FAB at point C is minus i times this 2A over A squared plus C squared.
08:39:809Paolo Guiotto: For the moment, it's not convenient to do any calculation here, to simplify.
08:47:490Paolo Guiotto: So, the function I want, the Fourier transform of F,
08:52:90Paolo Guiotto: is a function whose derivative is this one, so I have to compute a primitive of this function. So this means that my function
09:03:380Paolo Guiotto: F, A, B, hat.
09:06:50Paolo Guiotto: C will be a primitive of this.
09:10:160Paolo Guiotto: in the variable C, plus a certain suitable constant that we have to determine.
09:17:120Paolo Guiotto: Okay, so let's do the calculation. D minus i is a factor, so I'll put here. I have to do the primitive of 2A divided the A squared plus C squared.
09:30:350Paolo Guiotto: in variable C.
09:32:710Paolo Guiotto: minus the primitive of 2B divided B squared plus C squared, invaluable to see.
09:42:00Paolo Guiotto: And then, I have to add a constant to the final, resolved.
09:48:680Paolo Guiotto: Now, what are these two primitives? They are similar. So, what is primitive of 2A divided A squared plus C squared C?
09:58:650Paolo Guiotto: This looks like…
10:06:500Paolo Guiotto: If you put A equal 1, you read that 2 divided 1 plus C squared, the primitive is
10:13:520Paolo Guiotto: Well, you can carry the two outside,
10:17:470Paolo Guiotto: it would be the arctangent of C. Now, what if we have an A? So, we can rework this, we factorize A squared, so we get primitive of 1 plus C over A squared.
10:35:800Paolo Guiotto: So we simplify an A here, 1. It is better if we give the A to the variable.
10:46:670Paolo Guiotto: So in such a way that you see that this is the 2 octangent of C divided A.
10:55:970Paolo Guiotto: And similarly, the other term would be 2x tangent of C divided B. So we have that the Fourier transform of our function
11:06:970Paolo Guiotto: is,
11:11:420Paolo Guiotto: is, minus i, there is a common factor of 2 times arc tangent.
11:20:620Paolo Guiotto: of C divided A minus R tangent of C divided B. This is for the calculation of the primitive, then we have to add a constant.
11:32:380Paolo Guiotto: Now, basically, the calculation of the fluid transform is over, apart for that constant. How do we determine this constant?
11:42:80Paolo Guiotto: How can we determine well, normally.
11:50:50Paolo Guiotto: Possibility number one, you know a value of the Fourier transformer for some specific value of X.
11:59:780Paolo Guiotto: Or possibility number two, it's about the same.
12:03:900Paolo Guiotto: But it's not exactly a value, it is what happens when C goes to some limit, no? For example, plus infinity. Here we know something, because we're reminded that the Fourier transform of an L1 function
12:23:780Paolo Guiotto: Goes to what? When the variable goes to plus-minus infinity?
12:28:600Paolo Guiotto: It goes to… Remind… Zero. This is the Riemann Lebeg… Lemma.
12:39:460Paolo Guiotto: And so, when we go to infinity, you see that arctangent here, A is… A and B are supposed to be both positive. So, for example, C goes to plus infinity, C over A goes to plus infinity, arctangent goes to P over 2.
12:56:390Paolo Guiotto: So this goes to P over 2, but also this one goes to P over 2, so parentheses goes to 0.
13:02:480Paolo Guiotto: So it means that that formula, when we send C to plus infinity, goes to the constant C, when C goes to plus, or the same minus infinity.
13:16:20Paolo Guiotto: And since this must be equal to 0, we get that the value of the constant is 0, and therefore we have the final conclusion. The Free a transformer of F is equal to minus 2i R tangent
13:34:570Paolo Guiotto: C over A minus R tangent of C over B.
13:45:890Paolo Guiotto: And this is the frequent form for this function.
13:51:120Paolo Guiotto: Okay, now, the, main, the main cause of today is the inversion formula.
14:09:970Paolo Guiotto: Inversion formula is that formula we derived informally
14:14:930Paolo Guiotto: introducing the Fourier transform. It is this remarkable identity. f of x equals 1 over 2 pi, double hat f evaluated at minus X. So let's just copy, and let's start talking about this.
14:31:60Paolo Guiotto: So the inversion… formula.
14:39:20Paolo Guiotto: is this relation, f of x equal 1 over 2 pi, the double hat of F evaluated.
14:49:630Paolo Guiotto: 0. Now, the question, the main question is, under which assumption is this formula true? Now, it turns out that,
14:59:990Paolo Guiotto: This formula is true under, let's say, the natural assumptions
15:06:800Paolo Guiotto: That you need to write the formula itself. What are the national assumptions? You need to compute first the hat of F, so this demands F in L1,
15:20:50Paolo Guiotto: And then, since we have seen that it's not necessarily true that the transform of F is an L1 function, to do the second operation, we need again that also F at being L1.
15:32:360Paolo Guiotto: Well, it turns out that if both F and the Fourier transform of F are L1 functions, now, you know that we have seen, for example, a sufficient condition
15:45:920Paolo Guiotto: It's not an if and only if. It's just a sufficient condition to ensure that FAT is in a 1. That condition, it's written right here, comes from the relations between
15:58:540Paolo Guiotto: regularity of the function, and decay at infinity of Foulet transform.
16:04:500Paolo Guiotto: And this says that, basically, if f is regular enough, the Fourier transform is integral, okay? So, the K is sufficiently fast at infinity. But this is not, of course, an if and only if in general.
16:18:610Paolo Guiotto: So let's say, if F, F at are in L1, Then… the inversion… formula.
16:33:710Paolo Guiotto: Also… The unique care is in the meaning of the formula.
16:40:140Paolo Guiotto: Because there is, so, for, let's say, for which X this formula is correct. Because you have at the left the function f.
16:52:940Paolo Guiotto: The function f is just an L1 function, so in principle, it's an irregular function. It's not necessarily a continuous function.
17:01:760Paolo Guiotto: At right, we have this constant, 1 over 2 pi, then we have the double hat of F, which is an hat, no? It is the hat of F hat. But as every hat, as every Fourier transform, it is continuous.
17:16:790Paolo Guiotto: So the right-hand side, this function double X here, is a continuous function. It's clear that if I change the X into minus X, I cannot change the continuity. It is still continuous. So the right-hand side is a continuous function.
17:36:610Paolo Guiotto: So, this might… may… may be a little bit disturbing, because what does it mean that this, which is an L1, is equal to this, which is a continuous function?
17:48:650Paolo Guiotto: Well, it makes sense once we, say that these forms are not necessary for every act, but for almost every act, that's the…
17:59:160Paolo Guiotto: meaning of this, almost every X in R.
18:04:810Paolo Guiotto: Now, let's see… The proof of this.
18:10:770Paolo Guiotto: Now, if we start, or if we try to, let's say, verify this identity, starting from this quantity, computing the double hat of F, evaluating at minus X, and trying to show that it is equal to F , let's see what happens.
18:30:570Paolo Guiotto: If you write it in double at.
18:33:250Paolo Guiotto: Well, let's first write the double add evaluated at point X, then we will switch to minus, minus X.
18:42:270Paolo Guiotto: So now this is an head, okay?
18:45:500Paolo Guiotto: So, it will be an integral on the real line of what? This is the hat of F hat, you see?
18:55:210Paolo Guiotto: So the function you have to put here, it is F at.
18:59:180Paolo Guiotto: Let's use its natural variable with the letter X C. So here, I have D exponential E minus C,
19:08:200Paolo Guiotto: times… be careful, because the integration variable is the Xi, is the variable of the function you are transforming, which is the F at, okay? So it means that the variable for the Fourier transform here is this X.
19:26:340Paolo Guiotto: So if you want, the order should be X times XC, no, if you think about formula, but X times C or C times X is the same thing.
19:37:130Paolo Guiotto: Okay.
19:39:180Paolo Guiotto: Now, let's plug into this formula now what is F hat. Now, this is integral on r of integral on R, and this is now the Fourier transform
19:52:500Paolo Guiotto: Since X is already used here for this variable, I will not use for the integration variable the letter X, but, for example, letter Y. F of Y, E minus I. Now, the variable is X, so CY, DY,
20:10:580Paolo Guiotto: This quantity here is the, F at of C.
20:17:960Paolo Guiotto: times the character, which is E minus IXXC, or TX.
20:25:650Paolo Guiotto: Now, this is integrated into SIP.
20:28:800Paolo Guiotto: Okay, so this is the double hat of F evaluated at X. So, the double hat of F evaluated at minus X will be…
20:39:490Paolo Guiotto: The double F of F, evaluated at minus X will be exactly this formula, where I have to change X into minus X.
20:49:580Paolo Guiotto: You see that minus X is ER.
20:54:640Paolo Guiotto: So, if I change X with minus X in that formula, I get this integral in R of integral.
21:03:770Paolo Guiotto: of FYE minus IC.
21:08:500Paolo Guiotto: Y, this is in the variable Y, E, I, C, X, Pixie.
21:16:910Paolo Guiotto: This is literally the formula. Now, the formula makes sense because under our assumptions, we have F is in L1, so this integral makes sense. It is the tropelier transform of F, so well defined.
21:38:70Paolo Guiotto: Because… F is in a 1.
21:44:410Paolo Guiotto: And then, this is, again, the function f of c. So the second integral makes sense as well, because this is supposed to be L1, okay? We are supposing that this is L1. So what is written?
22:00:340Paolo Guiotto: Does not require any further, assumption.
22:06:430Paolo Guiotto: Now, what would you do at this point? Since we have a double integral, let's flip into a unique integral. So, this would be an integral on R2 of F of Y,
22:19:760Paolo Guiotto: then you see, I could put together the two exponentials. So, for example, I get E2 minus Xi… well, let's say E2X…
22:33:370Paolo Guiotto: Well, it is better if we keep this notation, doesn't matter whether plus, minus, And then, yeah, the DXCDY.
22:43:320Paolo Guiotto: The problem is that this second integral does not make sense.
22:48:320Paolo Guiotto: So, you are not allowed to do this, because The problem is that, this… lost.
23:01:110Paolo Guiotto: Integral.
23:03:350Paolo Guiotto: does.
23:04:790Paolo Guiotto: Not… Meg.
23:08:760Paolo Guiotto: And it… sensor.
23:12:950Paolo Guiotto: Because if you look as an integral on R2 of a certain function of these two variables, C and Y, to make sense, you need that this function be an L1 in the two variables. So.
23:27:150Paolo Guiotto: Because if you look at L1 means that you need the integral on R2 of the absolute value of this stuff here, F of YE minus iC times X minus Y,
23:44:360Paolo Guiotto: D, C, DY, this must be filed.
23:48:930Paolo Guiotto: But what is this?
23:51:260Paolo Guiotto: This is the integral on R2. I split the modulus into modulus FY times the modulus of the exponential.
24:03:290Paolo Guiotto: And again, as we have seen many times, this is a modulus 1 number, because it is E2i some real. C is real, X is real, Y is real, everything is real, except for I, so this is E2I theta, no?
24:20:30Paolo Guiotto: with theta real, so it's a modulus 1 number, and this means that you have integral in R2 of modulus f of y
24:28:450Paolo Guiotto: If it is an integral only in Y, it's perfectly okay, but it's an integral only also in the other variable, C.
24:36:600Paolo Guiotto: And that, as you can see, there is no dependence. The constant function in C cannot be integral, unless the function is degenerate is equal to 0. So this is equal to plus infinity
24:50:30Paolo Guiotto: Unless… F is equal to 0, so unless everything is trivial.
24:58:500Paolo Guiotto: So… This… means that we can proceed like that. Now, we will do now a trick here.
25:09:590Paolo Guiotto: to make this passage correct. So, we will add something that depends on C, which is the missing variable, which is good enough to be integral. And this something is a Gaussian, okay? So, we are going to modify the integral, and we won't have any more this quantity, because the new integral won't be this one, would be a modification of this one.
25:33:600Paolo Guiotto: So, what I'm doing is the following. Let's… modify…
25:41:900Paolo Guiotto: The double at F at minus X as follows.
25:48:210Paolo Guiotto: So the double at is this double nested integral that you see there. Now, what I put here is an extra factor, E minus… we will use a parameter that will disappear at the end.
26:06:160Paolo Guiotto: So this, I put here as epsilon, C squared.
26:13:440Paolo Guiotto: In such a way that when epsilon goes to zero, this will go to 1, so it will disappear.
26:21:890Paolo Guiotto: So, let's write this.
26:25:400Paolo Guiotto: Integral.
26:27:70Paolo Guiotto: on R… integral on R, f of y, F of Y, Well, actually…
26:41:270Paolo Guiotto: It is better for scaling reasons for computations and other reasons we put divided by 2.
26:48:390Paolo Guiotto: So we have, E2 minus I, sorry.
26:54:610Paolo Guiotto: Yes. See, yes, this is CY.
26:59:300Paolo Guiotto: E minus… well, let's put him below this.
27:03:170Paolo Guiotto: E minus epsilon C squared over 2, which is, for this integral, a constant.
27:12:230Paolo Guiotto: EY, then we have the second integration times E to iCXC C.
27:22:420Paolo Guiotto: Now, this is no more the previous integral, because the previous integral is with that, if you want, with epsilon equals zero, the previous integral. So let's call this i epsilon
27:35:910Paolo Guiotto: integral epsilon. Since after the integrations, this is a quantity that depends on X, so let's call it I epsilon…
27:47:820Paolo Guiotto: Well, let's keep still writing X or minus X is the same.
27:54:260Paolo Guiotto: Let's write i epsilon of minus X. Now, the big claim is this.
28:03:140Paolo Guiotto: is this, that when I send epsilon to zero, it will happen that this i epsilon of minus X, well, as I may expect by looking at this expression.
28:16:280Paolo Guiotto: It should go to the case when epsilon is 0. When epsilon is 0, that exponential is 1, so hopefully this will go to the double hat of F evaluated at minus X.
28:29:330Paolo Guiotto: Well, so we restore back the integral we started from. But what will happen, and this will be something new.
28:38:860Paolo Guiotto: that you don't see in this formula for the moment. This will go also to 2 pi f of x.
28:47:220Paolo Guiotto: In some suitable sense. It won't be the same sense for the two cases, okay? But in any case, we will be able, at the end, to conclude that from this, we get the identity of these two things, which is the inversion formula.
29:02:530Paolo Guiotto: Okay? So this is the strategy. So, when epsilon, of course, is sent to zero pot.
29:10:520Paolo Guiotto: Now, let's see the first claim, which is this one, which is a bit easier, a bit more immediate, because
29:23:40Paolo Guiotto: If you look, so let's say, let's call this 1.
29:28:860Paolo Guiotto: If you look at the expression I epsilon minus X, as you can see, this term is independent of the integration variable, it's there, but you could put out here.
29:40:880Paolo Guiotto: And so I can write my integral in this way. Integral on R, when you carry outside this exponential, what is inside is D hat of F, so F hat, evaluated at point C, times P minus epsilon
29:59:950Paolo Guiotto: C squared over 2… E-I-C-X.
30:06:950Paolo Guiotto: Yes, so… You have this.
30:10:220Paolo Guiotto: Now, I want to pass to the limit.
30:13:580Paolo Guiotto: And, so this is the classical problem. You see, you have an integral.
30:19:490Paolo Guiotto: Now, this function, sorry, this is an integral in letter C, not letter X.
30:27:750Paolo Guiotto: I can look at this as a function, well, let's give a different letter, let's call it capital F. That depends on the integration variable, but there is this X that here is playing the role of a parameter.
30:47:90Paolo Guiotto: It is not a divider with respect to we want to do the limit, but we want to do the limit with respect to epsilon.
30:54:800Paolo Guiotto: So we want to see what happens when epsilon goes to zero. If you want to take epsilon equal 1 over n, and this becomes, instead of writing F epsilon, you write FN, okay?
31:06:970Paolo Guiotto: So this is a classical problem where you want to pass to the limit into a sequence of objects into an integral. So the idea is we apply the dominated convergent theorem.
31:22:160Paolo Guiotto: we apply…
31:27:600Paolo Guiotto: dominated convergence. So, to apply dominated convergence, we need to know, first, what happens when we send this epsilon to zero, point-wise. Well, this is easy, because when you send epsilon to zero.
31:43:320Paolo Guiotto: 0 plus, of this quantity, F hetera C, E minus epsilon, C squared over 2,
31:51:890Paolo Guiotto: E, I, C, X, you see that… well, there is no B. You see that epsilon is only in this middle term, all the other terms are constants in epsilon.
32:03:900Paolo Guiotto: And when epsilon goes to 0, this quantity goes to 1. So the limit is definitely F at C,
32:10:280Paolo Guiotto: E, I, C, X, and this is… sorry.
32:16:110Paolo Guiotto: This is a… for every C, for every X, let's say almost every C, almost every X.
32:27:870Paolo Guiotto: And second, we need to dominate this thing uniformly in epsilon, right?
32:34:270Paolo Guiotto: But what is the domination here? With what? A function that should be integrable in the variable… integration variable, which is the letter XC. So now, if I take the absolute value of all this expression.
32:54:870Paolo Guiotto: As you can see, we have that this is less or equal modulus F at Z.
33:00:490Paolo Guiotto: Then we have the exponential, which is positive, E minus epsilon C squared over 2. And then there is this E to iCX, which is a modulus 1 number that we don't need to carry around.
33:14:770Paolo Guiotto: And, look at this. This contains epsilon, but it is, since our epsilon is positive, this is a negative exponential, so it would be bounded by 1. So I can say that this is less or equal than modulus F at C, which is a good function, G of C,
33:34:850Paolo Guiotto: is a good integral, dominant, because we assume that F hat is in L1, so this is an L1 function.
33:44:870Paolo Guiotto: So we are in condition to apply dominated convergence, and we can pass to the limit when epsilon goes to zero, computing the limit inside. So we have that. So, limit…
33:58:770Paolo Guiotto: for epsilon going to zero positive of this EI epsilon minus X is the integral of the limit
34:08:420Paolo Guiotto: when epsilon goes to zero positive of what is inside, but we said what is inside has limit this F hat CEICX, so what I see is integral on R of F hat C.
34:26:10Paolo Guiotto: E, I, C, X, and that's exactly C.
34:32:230Paolo Guiotto: And that's exactly the Fourier transform of F hat, so the double hat of F, but evaluated at minus X. Now, this X can be, has never, been touched, so can be whatever, so for every X in R.
34:52:190Paolo Guiotto: And so this is the first limit. We proved that, then, this I epsilon minus X goes… now we can say how? Pointwise, to this double F head of minus X.
35:08:400Paolo Guiotto: Okay, now let's come to the second part.
35:11:750Paolo Guiotto: Which is this one.
35:14:690Paolo Guiotto: Let's see that this i epsilon minus X goes to 2 pi f of x in some sense. But to see that this happens, we have to work on this formula. So, let's take the formula EI epsilon.
35:36:750Paolo Guiotto: I just copy, AI epsilon minus X is integral on R, integral on R, so we have the F at C,
35:46:550Paolo Guiotto: E minus ICY, then we added this E minus epsilon C squared over 2. This is in the most integral, VY. Then we have e to icy.
36:02:90Paolo Guiotto: Now, the first factor is that we, here, we can write this as a double integral, so we can apply the Fabini theorem, because this is an integral on R2 of F at C,
36:17:730Paolo Guiotto: E minus iC, well, let's factorize these two, Y minus X, E2 minus epsilon C squared over 2, dxi 3Y,
36:32:50Paolo Guiotto: It's because now the function that you see here, this is in L1R2.
36:40:470Paolo Guiotto: Differently from the previous calculation that was basically redemption and equals zero.
36:46:710Paolo Guiotto: So this, because…
36:51:700Paolo Guiotto: If you take the integral of the absolute value of that stuff here.
36:57:940Paolo Guiotto: what remains is integral on F2 of what? Of course, modulus F at X,
37:06:840Paolo Guiotto: the modulus of the unitary exponential will be 1 again, and then we have E minus epsilon C squared over 2, and this is integrated index C,
37:24:240Paolo Guiotto: Sorry.
37:25:720Paolo Guiotto: I perhaps didn't.
37:28:630Paolo Guiotto: Possibly.
37:41:420Paolo Guiotto: Yeah, no, but… yes, so this should… sorry, yes, you're right. Thank you. This is F of Y.
37:52:120Paolo Guiotto: Right? So this is… Because, yes, I just, copied the… in a… with,
38:04:10Paolo Guiotto: wrong thing. So, this is F of Y.
38:08:950Paolo Guiotto: So this is F of Y.
38:13:470Paolo Guiotto: Okay, so now it works. So now, this is integral, because, as you can see, it's a function of two variables. C and Y is actually a function of C times a function of Y, so you can split the two integrals into integral of modules f of y.
38:33:130Paolo Guiotto: in dy times integral of E minus epsilon C squared over 2 digxi. But in any case, this thing is fine.
38:43:230Paolo Guiotto: So, we can say that this passage now makes sense.
38:49:90Paolo Guiotto: And what we do is now we flip the two integrations. So let's now write this formula with integration first in C, and then in Y. So the initial formula is you first integrate in Y and then in C.
39:05:520Paolo Guiotto: Now, we use the Fubility theorem to flip this into integration
39:11:760Paolo Guiotto: Again, it's a double integration, but same thing, so we have F of Y,
39:18:930Paolo Guiotto: E minus IXCY minus X, E minus epsilon C squared over 2, and then this is first in C, and then in Y.
39:32:580Paolo Guiotto: Since the second integration is in Y, and the first is in C, now you see that this quantity is a constant for, for this integral, which is in C. So we can carry outside of this
39:49:310Paolo Guiotto: So we write integral on R, f of y.
39:53:360Paolo Guiotto: And the other part remaining remains because we have integration in C, you should see here, and this is the variable. So integration in R of E minus IC, Y minus X,
40:08:240Paolo Guiotto: E minus epsilon C squared over 2, X D1.
40:15:240Paolo Guiotto: But this guy here is a Fourier transform. We have to recognize it is a Fourier transform, because if you write this way, integral of E minus epsilon C squared over 2 times E minus iCT
40:34:70Paolo Guiotto: in the variable C, You see, this is a Fourier integral.
40:42:210Paolo Guiotto: a function.
40:43:810Paolo Guiotto: of C, no, function phi of C, P minus iCT. Now, C is not the variable of the Fourier transform. The variable of the Fourier transform is this number T. So this is the Fourier transform of phi evaluated at T.
41:04:600Paolo Guiotto: when you have this integral in the variable C.
41:09:800Paolo Guiotto: So this means that this is the Fourier transform of this, E minus epsilon, the square over 2,
41:19:330Paolo Guiotto: evaluated at point T.
41:23:900Paolo Guiotto: Now, this is the Gaussian distribution.
41:26:980Paolo Guiotto: If you want to carry to the standard Gaussian writing, you should put the sigma square here, so put this epsilon in the denominator of 2, so this becomes a 1 over epsilon here, and this is our sigma square.
41:47:140Paolo Guiotto: Now, the written form for the Gaussian, E minus square, divided to sigma squared.
41:54:990Paolo Guiotto: The Fourier transform of this is root of 2 pi sigma squared E minus, there is no mean, so minus 1 half sigma square t squared. That's the Fourier transform.
42:08:280Paolo Guiotto: So we apply this, and we get root of… 2 pi…
42:17:380Paolo Guiotto: Now, sigma square is 1 over epsilon, so let's write 1 over epsilon.
42:24:270Paolo Guiotto: E minus 1F, sigma squared is 1 over epsilon d squared.
42:33:70Paolo Guiotto: So, mmm… let's say…
42:40:530Paolo Guiotto: Well, root of 2 pi over epsilon, E minus… let's put 2 epsilon down here, T squared.
42:50:80Paolo Guiotto: Now, our tea was… this, Y minus X.
42:55:790Paolo Guiotto: So, it means that… The integral, let's, rewrite here. So, integral.
43:04:430Paolo Guiotto: of E minus C squared over epsilon, C squared over 2. E minus iCY minus X,
43:16:420Paolo Guiotto: DXC is this thing, root of 2 pi.
43:21:620Paolo Guiotto: over epsilon, E minus Y minus X squared over 2 epsilon.
43:32:320Paolo Guiotto: Now, we plug this into this formula, so this part here is what we have computed down here.
43:44:00Paolo Guiotto: is this one.
43:46:30Paolo Guiotto: So, let's, let's, re… rewrite, let's…
43:52:760Paolo Guiotto: clean up something. So we started from this, I epsilon minus X.
43:57:950Paolo Guiotto: And let's see what we obtained.
44:00:230Paolo Guiotto: Conclusion.
44:02:710Paolo Guiotto: We obtain that I epsilon minus X is equal to…
44:08:760Paolo Guiotto: So, after a few steps, it was integral on r of f of y times that box, so integral on R.
44:18:00Paolo Guiotto: F of Y, times the box, which is this, E minus Y minus X squared over 2 epsilon.
44:32:820Paolo Guiotto: Well, let's write this way. Let's write divided by root of 2 pi epsilon.
44:38:980Paolo Guiotto: So, since I want this 2 pi at the denominator, I need to multiply and divide by 2 pi, so this creates a factor of 2 pi here.
44:48:260Paolo Guiotto: D, this is an integral in DY.
44:52:20Paolo Guiotto: So, take this 2 pi outside, and you get 2 pi.
44:57:280Paolo Guiotto: integral in R of FY, E minus Y minus X squared divided 2 epsilon.
45:06:750Paolo Guiotto: Over root of 2 pi epsilon.
45:13:600Paolo Guiotto: Now, if you look carefully to this, well, since Y minus X squared is also X minus Y squared, so let's write in that way, in such a way that we recognize better what is this.
45:26:320Paolo Guiotto: You see that this is F of Y times a function of X minus y. You see that?
45:32:390Paolo Guiotto: F of Y times G of X minus y, and that's a convolution.
45:38:260Paolo Guiotto: So this is 2 pi, deconvolution between F And this function, delta epsilon, What, that type, Ceylon?
45:50:680Paolo Guiotto: is the function, say, of t equal, E minus T squared over 2 epsilon over the root of 2P epsilon.
46:02:680Paolo Guiotto: And that's exactly, The Goshen unit.
46:15:740Paolo Guiotto: approximate unit.
46:19:130Paolo Guiotto: This is quite important because we know that when we send epsilon to zero.
46:26:270Paolo Guiotto: We mentioned this, we have not proved that.
46:29:590Paolo Guiotto: Since, we have that,
46:33:90Paolo Guiotto: F star delta epsilon, when you have a function F, which is, as in this case, in L1, this goes in L1 norm to the function F itself when epsilon goes to 0.
46:48:460Paolo Guiotto: Because that motion basically concentrates more and more around each of point X, so the average becomes the value of the function at point X.
47:00:820Paolo Guiotto: The fact is that the convergence is in L1. So this means that, I… our I epsilon
47:10:330Paolo Guiotto: minus X, which is 2 pi, this convolution, F star delta epsilon, this quantity goes, in L1 sense, to 2 pi F.
47:24:240Paolo Guiotto: So that's the second, part of this puzzle. So we now know that
47:30:870Paolo Guiotto: In this schema, the i epsilon minus X goes to 2 pi f in L1 sense.
47:39:200Paolo Guiotto: And now we can combine everything to draw the conclusion.
47:48:620Paolo Guiotto: So, this… I repeat here the picture. This proves… That.
47:57:810Paolo Guiotto: our I epsilon minus X, We know it goes pointwise to F.
48:05:670Paolo Guiotto: OVAX.
48:08:880Paolo Guiotto: And at the same time, it goes L1 to 2 pi F of X.
48:18:690Paolo Guiotto: Sorry, it goes to… this goes to the double at… at minus X, so this was the first
48:27:920Paolo Guiotto: part, no? The one is…
48:32:390Paolo Guiotto: Yes, I epsilon goes to minus X, goes pointwise to the double h of f at minus X. The second part is,
48:41:350Paolo Guiotto: I epsilon minus X goes to 2 pi f of x, but in L1 sense.
48:46:860Paolo Guiotto: Now, how we can combine the two.
48:49:890Paolo Guiotto: Since it goes in a one sense, it does not go pointwise, otherwise the confusion would be immediate, but…
48:56:950Paolo Guiotto: If we extract a subsequence, what does it mean? What is the subsequence here? There is no sequence, apparently. But here, we are taking epsilon going to zero. So you may think, take epsilon equal 1 over n. So now you create a sequence of epsilon, you have a sequence of these things, i epsilon.
49:17:770Paolo Guiotto: So that sequence goes in L1 to 2 pi F. A subsequence will go almost everywhere. So we can say that if we look to a suitable sequence of epsilon, we will have that this limit is also point-wise almost everywhere.
49:34:630Paolo Guiotto: And, point-wise, almost everywhere.
49:39:280Paolo Guiotto: Let's specify how long… Sweet about… Sequence… off.
49:52:240Paolo Guiotto: Epsilon. We don't care, because at the end, this epsilon disappears. So, once we know that this goes also pointwise, this goes already pointwise for every X, this for every X, this for almost every X, so in particular, these two must be the same almost every X.
50:11:350Paolo Guiotto: And this is the conclusion.
50:14:350Paolo Guiotto: So… F double hat at minus X will be equal to 2 pi f of x.
50:24:610Paolo Guiotto: almost every accent, and the finishes the proof.
50:29:820Paolo Guiotto: Okay, so let's see, because this is, we may say, the most important,
50:36:140Paolo Guiotto: result, we, we see on this, part, of course, on Fourier transform.
50:43:270Paolo Guiotto: Let's see a number of applications of this.
50:48:680Paolo Guiotto: So, as usual, we continue till the end, non-stop, right?
50:54:440Paolo Guiotto: So, the example one… is the Fourier transform of Hoshi… distribution.
51:11:630Paolo Guiotto: Well, the Cauchy distribution is this.
51:15:160Paolo Guiotto: let's say FAX is this function, 1 over A squared plus X squared. We write A squared just to say that it is positive, okay? And for convenience, we take A positive, even if A negative doesn't change this formula, but let's assume that A is positive.
51:35:180Paolo Guiotto: So we don't have to specify later. Now, this function is clearly in L1.
51:42:600Paolo Guiotto: R…
51:44:170Paolo Guiotto: If you want, you could compute the Fourier transform directly, but in this case, you should use methods coming from,
51:54:210Paolo Guiotto: polymorphic function theory, so functions of complex variable methods, integrations for functions with complex hardware that I don't know if you have never seen. But in any case, we don't need to do this calculation now.
52:10:310Paolo Guiotto: Because, as you may observe, this looks like what?
52:14:640Paolo Guiotto: This looks like the Fourier transformer.
52:19:20Paolo Guiotto: Of the exponential is what we used here, right?
52:23:750Paolo Guiotto: So that's the key fact that we are now going to use. So let's remind of this.
52:31:170Paolo Guiotto: So our goal is now to compute the Fourier transform of this.
52:35:680Paolo Guiotto: So, remind, recall, Beth.
52:41:920Paolo Guiotto: we have this remarkable fact. The Fourier transform of an exponential, E minus A, the modulus, evaluated at C, is 2A divided A squared plus C squared, right?
52:57:290Paolo Guiotto: Now, changing letter, so writing this in letter X, so this means that if you, switch this
53:06:270Paolo Guiotto: into the letter X. This means that this evaluated at X is X square here.
53:14:200Paolo Guiotto: The letters are letters. You understand that I can call the variable of the written form as I like. This letter X is just an agreement.
53:22:790Paolo Guiotto: But in particular, this says that if you divide by 2A, so you leave 1, and you put 1 over 2A here, you are allowed to do, because 2A is a constant, you can carry in and out from the Fourier transform because it is linear.
53:37:580Paolo Guiotto: So, let's clean up this. We have that the Fourier transform of 1 over 2A, E minus A, the absolute value of the variable, evaluated at point x, is exactly the function we want to compute the Fourier transform.
53:54:850Paolo Guiotto: So if we now take the Foulet transform to both sides, what happens is that at left, I have the double hat of the 1 over 2A E minus A, the absolute value.
54:12:460Paolo Guiotto: Okay?
54:14:60Paolo Guiotto: Now, let's give the name Xi to the variable of this Fourier transform, and this will be the hat of 1 over A squared plus E squared, evaluated XC, which is the quantity I want to compute, you see?
54:31:150Paolo Guiotto: This is what we are looking for.
54:34:680Paolo Guiotto: And what happens at left? That's a double head of something.
54:40:510Paolo Guiotto: And so the point is that this double F seems to be going to be more or less 2 pi F. So let's see this. Can I say exactly that inversion formula applies to this function since F has been used, so let's call it this function G.
55:01:190Paolo Guiotto: So G of X is the function 1 over 2AE minus A modulus of X.
55:10:130Paolo Guiotto: Now, is the function G an L1 function? Yes, we know.
55:16:940Paolo Guiotto: Is its Fourier transform an L1 function? Yes, we can say, because what is the Fourier transform of this? Is 2A divided A squared plus C squared, right? Which is an L1.
55:30:690Paolo Guiotto: So, G is a function for which G and G-hat are both in L1. So, for G, inversion formula holds.
55:40:300Paolo Guiotto: So… inversion formula.
55:47:480Paolo Guiotto: applies.
55:50:710Paolo Guiotto: And what we have is, let's write for G. Invention formula says that, literally G of X is 1 over 2 pi, the double hat of G, evaluated at minus X.
56:06:700Paolo Guiotto: Or if you want, equivalently, you can write that the double hat of G, evaluated at minus X is 2 pi g of x.
56:16:570Paolo Guiotto: Of course, almost every X, right?
56:21:280Paolo Guiotto: Or, equivalently, again, if I want to look at this formula as a formula for the double hat, so called minus Xx, this becomes double hat of G at point X is 2 pi G at minus X, right?
56:40:630Paolo Guiotto: So, this is what happens because of the inversion formula to G. The double transform of G is just 2 pi G evaluated at minus X.
56:52:320Paolo Guiotto: So, returning back here, the double head of this 1 over 2ae minus A modulus is 2 pi, that function evaluated at minus X.
57:05:250Paolo Guiotto: So… the double head of 1 over 2A, E minus A, the absolute value.
57:15:680Paolo Guiotto: evaluated at, well, this is evaluated… I know that this is disturbing you, but you have to… to…
57:25:520Paolo Guiotto: to adapt to this. So, evaluated at X, so let's call the variable at C, because this is because we are computing the Fourier transform of this guy, so we will use the lesser C.
57:38:960Paolo Guiotto: Now, this is, because of this formula here, is 2 pi, the function itself, so 1 over 2a, E minus A, the absolute value, evaluated at minus the variable. So, this case, the variable is C, I will write minus C.
57:58:520Paolo Guiotto: Now, the two councils.
58:02:740Paolo Guiotto: So, what remains is pi over A, B minus A. Now, absolute value of minus C is the same of absolute value of C.
58:12:420Paolo Guiotto: So the double add of that function is this, but the double add of that function was also the Fourier transform of the Cauchy distribution.
58:22:100Paolo Guiotto: So we got the doctor.
58:24:180Paolo Guiotto: the Fourier transform of 1 over A squared plus square
58:31:50Paolo Guiotto: Evaluated at C is pi over A, E minus A modulus of C.
58:38:660Paolo Guiotto: Now, for which city?
58:40:680Paolo Guiotto: Now, we should say, for almost every X, in principle, because this comes from the application of the inversion formula. So, at a certain point, there is an almost every. But I can actually say that it's an every C. Why? Because we know that this function is difficult to form of an L1 function, so it will be continuous.
59:00:840Paolo Guiotto: And the right-hand side is continuous.
59:03:850Paolo Guiotto: So if two continuous functions coincide almost everywhere, they must coincide everywhere.
59:09:550Paolo Guiotto: It's the usual evidence. If F is equal to G almost everywhere, and they are both continuous, they must be the same, because you take the difference, F minus G equal to 0 almost everywhere, it cannot be different from 0 at the point is continuous, because it would be different from 0 in average, so in the positive measure set.
59:29:170Paolo Guiotto: So now, I can say that because of continuity, this is actually forever exit. It's a detail, it's not particularly important, but…
59:38:510Paolo Guiotto: Just to understand how it works.
59:42:980Paolo Guiotto: Okay.
59:44:580Paolo Guiotto: Let's see, this was an old exam.
59:48:860Paolo Guiotto: These are the exercise example, 19… 1, 3…
59:55:440Paolo Guiotto: So, the goal of this exercise is to compute the Fourier transform of this function.
00:03:420Paolo Guiotto: 1 over 1 plus X squared squared.
00:08:900Paolo Guiotto: Now, there are, steps,
00:16:100Paolo Guiotto: So… It says, compute the Fourier transform of XF of X, basically.
00:29:770Paolo Guiotto: And there isn't hinter.
00:33:880Paolo Guiotto: XF of X is the derivative.
00:38:90Paolo Guiotto: of something. Because basically, to compute this, the trick is to recognize that this is the derivative of something, this is more or less. So, number two, determine the Fourier transform of N.
00:54:320Paolo Guiotto: Number 3… compute.
00:59:940Paolo Guiotto: these two integrals, integrals from 0 to plus infinity, or 1 over 1 plus X squared squared.
01:07:720Paolo Guiotto: So you could compute this integral, this one, with the elementary method, because it's an integral operational function, so you have an velvet to do that. But it would be a little bit longer.
01:20:670Paolo Guiotto: The second one you cannot do with elementary methods, so integral 0 plus infinity sine x divided 1 plus X squared squared.
01:31:220Paolo Guiotto: So we will see how to compute these integrals by using the Fourier transform of this F.
01:38:550Paolo Guiotto: So, let's see the solution.
01:42:20Paolo Guiotto: So, the first asks to compute the Fourier's form of XF of X. Let's see what is XF of X.
01:48:980Paolo Guiotto: XF of X is X divided 1 plus X squared squared.
01:57:550Paolo Guiotto: Now, there is a need to recognize that this is a derivative, basically.
02:02:690Paolo Guiotto: This could be the derivative with respect to X of Y.
02:07:270Paolo Guiotto: More or less, then we will, put the D.
02:11:700Paolo Guiotto: the correct numbers. But this should come computing the derivative of…
02:21:980Paolo Guiotto: Exactly, 1 over 1 plus X squared. More or less, maybe there is some number to fix it. In fact, if we do the derivative of 1 over 1 plus X squared.
02:33:870Paolo Guiotto: This is minus the fraction, denominator becomes the square, 1 plus X squared squared, and then numerator is the derivative of the denominator. I'm using the formula d, X of 1 over something, G is minus G prime over G squared.
02:54:770Paolo Guiotto: So, this is 2X.
02:57:990Paolo Guiotto: So, as you can see, I just need a minus 2 to have a correct formula. I do not have. I create. This can be done because the derivative is linear. You can always multiply, divide.
03:10:150Paolo Guiotto: And, you are okay. So, this means that XF of X is apart for minus 1 half is the derivative.
03:18:650Paolo Guiotto: So this means that…
03:22:300Paolo Guiotto: I have to compute the Fourier transform, so first, I should be sure that this calculation can be done. Also, because I'm going to use properties that demand something. So, first of all, is it possible to compute the Fourier transform of X, F of X? We need to know if this is an L1 function.
03:42:60Paolo Guiotto: And this can be easily seen. Notice that things
03:48:680Paolo Guiotto: XF of X is equal to X divided 1 plus X squared squared. So clearly, this is a continuous function, there is no problem anywhere, so it's a continuous function.
04:02:510Paolo Guiotto: in the interior line, and therefore, for integrability, you just need to check what happens at plus minus infinity. At plus-minus infinity, it's the same, numerator is X, denominator is X to power 4,
04:19:100Paolo Guiotto: No? So, at the end, I have 1 over X cubed, which is integral, at… plus minus infinity.
04:32:40Paolo Guiotto: So, I concluded that XF of X is an L1R function.
04:40:950Paolo Guiotto: So the Fourier transform makes sense.
04:43:960Paolo Guiotto: And therefore, I can write also that the free transform of XF of X
04:50:760Paolo Guiotto: Here, you see, it's not a good idea to put the letters there, because then you have another letter, which is the variable of the Fugger transform, and you would do a mess. So let's, use another symbol that hides the variable, like, the sharp, the ballot, what you want.
05:10:70Paolo Guiotto: So this is, following the hint, is the Fourier transform of, what? We say the minus 1 half
05:22:300Paolo Guiotto: derivative with respect to X of,
05:30:680Paolo Guiotto: So I did a mess, and… sorry. Did this minus 2…
05:39:140Paolo Guiotto: I'm doing things a little bit faster. There is no minus 2 here.
05:46:380Paolo Guiotto: Because think about, you do the derivative of this 1 over, and you get it is minus 2X tie over, no? I'm looking at this formula. So it means that if this is the term X divided 1 plus X squared squared.
06:03:230Paolo Guiotto: I have to divide by minus 2. There is only this one. This is correct, but the minus 2 there.
06:08:750Paolo Guiotto: was a mistake. So minus 1 half derivative with respect to X of 1 over 1 plus D squared.
06:18:290Paolo Guiotto: Now, the minus 1 half goes outside bilinearity. Then we have the fluidence form of a derivative.
06:25:780Paolo Guiotto: Right?
06:28:110Paolo Guiotto: And, you know that there is a formula, Fourier transform of derivative, what we call the F prime, derivative with respect to each variable, is equal to IC, the Fourier transform of the function F, provided
06:44:210Paolo Guiotto: F, F prime are both in L1.
06:48:300Paolo Guiotto: Which is the case… well, I'm doing a mess here for you, especially because DF here is not the F. I'm concerned, so let's use another letter here. Let's call it G. G prime is this, provided GG prime are in L1, okay?
07:06:560Paolo Guiotto: Now, I'm using this formula with this G. You see? This is the G we are considering here, because there is the probability of that pump.
07:17:60Paolo Guiotto: And that function is in a 1.1 plus X squared is in a 1, and it's derivative, but the derivative is, just this X at the back, so here you can check that it is in a 1.
07:28:720Paolo Guiotto: So, we are, safe here. We can use the formula, and this says that it is ik times the Fourier transform of 1 over 1 plus D squared.
07:42:00Paolo Guiotto: evaluated at sea.
07:44:290Paolo Guiotto: Right?
07:45:460Paolo Guiotto: That's the Cauchy distribution, we just come to computer, so we get…
07:51:700Paolo Guiotto: minus i half, which is a constant, C,
07:56:640Paolo Guiotto: Then we have the Fourier transform of 1 over 1 plus this square. We have here the formula with A equal 1. This is pi e to minus modulus
08:09:390Paolo Guiotto: So it is pi e to minus modulus of C.
08:15:840Paolo Guiotto: So let's, so we computed the Fourier transform of X, F of X.
08:24:990Paolo Guiotto: And this, evaluated at point C, turns out to be minus i half C. Well, let's put the pi here.
08:33:279Paolo Guiotto: CE minus models of C.
08:37:250Paolo Guiotto: Okay?
08:41:370Paolo Guiotto: So now we, pass to second point.
08:46:229Paolo Guiotto: The first point was, compute this Fourier transform. Second, determine the Fourier transform of F.
08:52:740Paolo Guiotto: No.
08:53:710Paolo Guiotto: Of course, use this calculation to determine the Fourier transform of that. Let's see how. Well, first.
09:00:479Paolo Guiotto: Can we compute the Fourier transform of F? Yes, because F, which is this function, 1 over 1 plus X squared squared.
09:10:560Paolo Guiotto: is clearly an L1R function.
09:14:529Paolo Guiotto: It is continuous, it goes to zero fast as 1 over X to power 4.
09:21:200Paolo Guiotto: F is continuous.
09:25:500Paolo Guiotto: And at plus minus infinity, well, you might have a problem. It is asymptotic to 1 over x to power 4, so it is integral
09:38:960Paolo Guiotto: At plus minus infinity.
09:42:960Paolo Guiotto: So this function is in L1, and its fully transparent makes sense.
09:47:850Paolo Guiotto: Now, of course, the point is, how can I use this to compute the Fourier transform of F?
09:56:60Paolo Guiotto: Now, here you should remind of one of the properties we have seen yesterday. This is related to the derivative of the Fourier transform.
10:04:930Paolo Guiotto: And precisely, we recall that Recall that, huh?
10:13:60Paolo Guiotto: Here, we apply to F, finally, so if I have F and XF are both in L1, then it turns out that Fourier transform of F is differentiable, so there exists the derivative with respect to C of F at C.
10:32:610Paolo Guiotto: And this is D hat of minus i.
10:37:250Paolo Guiotto: XF, no, the Fourier's form of XF multiplied by minus i, evaluated at point C.
10:47:580Paolo Guiotto: So now is this formula we are going to use.
10:50:780Paolo Guiotto: Because we have, basically, the right-hand side, apart for minus i, which is just a factor, so I know this is so…
11:00:340Paolo Guiotto: In our case, DCF hat c will be minus i, you can also see this, you carry outside of the Fourier transform by linearity, D hat of XF of X.
11:17:620Paolo Guiotto: We have this.
11:19:350Paolo Guiotto: This is, given by that formula. So I have a minus i times minus i pi over 2, C…
11:31:510Paolo Guiotto: Oh, it is C? Yeah. CE minus modulus of C.
11:38:380Paolo Guiotto: So, minus, minus, plus, I, I, minus, minus… by Hoff C.
11:46:430Paolo Guiotto: E minus modulus of C.
11:49:420Paolo Guiotto: So what we have is not yet the fluid transform of F, but it is its derivative, dxc F hat. C,
11:58:930Paolo Guiotto: Is minus pi half C, E minus modulus of C.
12:06:730Paolo Guiotto: If we want to go back to F, we have to integrate this relation. It's saying the root of F hat is this, F hat is a primitive.
12:16:850Paolo Guiotto: So, F hat of C… Will be a primitive.
12:24:830Paolo Guiotto: of, this minus pi half primitive CE minus modulus of C,
12:33:620Paolo Guiotto: plus constant. Now, how do we compute the primitive? Well, since there is that modulus that changed the exponential, let's do the calculation for T positive, C negative.
12:47:100Paolo Guiotto: For Xi positive, I have minus pi half primitive of CE minus X, Dixie plus constant.
12:59:240Paolo Guiotto: Right?
13:01:300Paolo Guiotto: Now, this is, minus pi half, I can do by parts.
13:06:270Paolo Guiotto: So I use this as a derivative. Perhaps it is better to give the minus to the exponential in such a way that it is the derivative of E minus C. So let's put the minus here.
13:20:510Paolo Guiotto: So, I give this to E, and so it becomes E… no, sorry, I'm saying that it's not a… it's not a definite integral, it's…
13:32:290Paolo Guiotto: So it is XC E minus X minus primitive. Now, derivative moves on factor of XC, so derivative is 1, e minusc.
13:44:940Paolo Guiotto: plus the constant.
13:48:550Paolo Guiotto: by half.
13:50:670Paolo Guiotto: CE minus C minus… now, that minus can be given to the exponential in such a way that's the derivative of E minus C. So, primitive derivative is the function E minus C.
14:06:990Paolo Guiotto: Last. Honestly.
14:09:690Paolo Guiotto: Boxy negative.
14:11:710Paolo Guiotto: Similar calculation. We have minus by half primitive of Xi. Now it is E2 minus modulus, modulus is minus, so it is E2C, dxi.
14:27:870Paolo Guiotto: It's done.
14:31:40Paolo Guiotto: So it is minus pi half.
14:33:480Paolo Guiotto: Again, by parts, so E t is the derivative of itself, so we have CE to C.
14:40:600Paolo Guiotto: minus, we move the derivative on factor of C, we get 1 primitive of e toxi.
14:47:160Paolo Guiotto: plus constant.
14:50:560Paolo Guiotto: minus by half… C, E to C minus E took C plus constant.
15:00:400Paolo Guiotto: So, we now, put together these two. This is the expression of F at C for C positive.
15:09:670Paolo Guiotto: This is the expression of F at CFOXC negative.
15:14:550Paolo Guiotto: Now, there is this constant, and there is something that should be fixed. So, first of all, of course, determine the value of the constants. How do we determine? Well, for example, we could look at the value at infinity, because the hat of F
15:29:620Paolo Guiotto: We know that goes to zero when C goes to plus minus infinity.
15:35:980Paolo Guiotto: Because of Riemann LeBague lemon.
15:39:440Paolo Guiotto: You can see that in both cases, the first line, when C goes to pass C3, that's for C positive.
15:46:380Paolo Guiotto: The two exponentials kill the C, and they go to zero as well, they are both zero, and therefore the constant must be 0. And the same is here, because here is what's going to minus infinity to get the same. So, the unique possibility is that C equals 0.
16:08:250Paolo Guiotto: And so we have this formula, FXC is equal to…
16:13:800Paolo Guiotto: Well, here, you should have a little bit of care, because the first formula is FoxC positive, the second one Foxy negative. Let's add this equals 0.
16:23:560Paolo Guiotto: We must be sure that it is a continuous function, because it remains it's Fourier transform, so it must be continuous. And since we have this double definition, we should also check that this is fine at zero.
16:35:600Paolo Guiotto: For the first, we have pi half, then we have… well, we can factorize E minus C, so let's do that. E minus C times C plus 1.
16:51:430Paolo Guiotto: And for the second is pi half… It is minus by half.
17:00:330Paolo Guiotto: Minus by half.
17:05:40Paolo Guiotto: E. Took C times C minus 1.
17:11:470Paolo Guiotto: So when C is 0 in the first line, you get pi half. When C is 0 in the second line, you get pi half. So it is continuous. Well, it's not there.
17:22:450Paolo Guiotto: And we can write, maybe, in a unique formula, so pi half E2C, E2 minus C is E2 modulus C, proxy positive, negative.
17:34:980Paolo Guiotto: Then, you'll see that we have C plus 1, C minus 1,
17:41:40Paolo Guiotto: So, maybe it is better if we give the minus to the… to this, so we write as 1 minus C.
17:49:790Paolo Guiotto: And this has 1 plus X, because they become both 1 plus or minus X.
17:56:800Paolo Guiotto: And that's the modulus of Xia.
17:59:610Paolo Guiotto: Proxy, positive, and negative. So, it doesn't matter, it's just to have a unique formula. No, we need to, to… we have still to the third question.
18:10:50Paolo Guiotto: Which is the question concerning the integrals, okay? So…
18:14:690Paolo Guiotto: This, ends the calculation of the Fourier transform of our function, F at X equal pi half
18:25:180Paolo Guiotto: E minus modulus C1 plus modulus C.
18:31:970Paolo Guiotto: Oxy Korea.
18:35:220Paolo Guiotto: Now, we have question 3 that concerns the integrals.
18:40:120Paolo Guiotto: We have to compute number 1,
18:42:730Paolo Guiotto: It was required that the integral from 0 plus infinity of that quantity, 1 over 0.
19:00:670Paolo Guiotto: Okay, so the trick is, try to understand the,
19:06:270Paolo Guiotto: Well, there are a number of tricks to compute these things easily, but now what we know is the Fourier transform. What is the F at of C?
19:18:260Paolo Guiotto: Well, the effort of C is this integral, is the integral on R of our function, F, so 1 over 1 plus
19:26:680Paolo Guiotto: X squared squared e to minus iCXDX.
19:33:900Paolo Guiotto: So, you see that I get more or less perfectly disciplined when I do well.
19:48:760Paolo Guiotto: I spent some reason.
19:50:660Paolo Guiotto: No, it's not, because that's… I have to make this factor equal 1. This happens when…
19:59:910Paolo Guiotto: I think the volumes.
20:04:180Paolo Guiotto: Because models needs algae garbage models inside your quantitative model.
20:10:40Paolo Guiotto: Special values of C, which value makes that part equal 1?
20:16:280Paolo Guiotto: 0. So, if you put the X equals 0, you get that F at zero is the… in fact, this is, in general, is the integral of the function.
20:28:780Paolo Guiotto: You see?
20:30:370Paolo Guiotto: So, this means that it is integral on r of 1 over 1 plus X squared squared.
20:38:30Paolo Guiotto: Which is not exactly our integral, but since the function is symmetric, No? Is, even.
20:47:90Paolo Guiotto: This integral will be 2 times the other one, so this will be two integrals from 0 plus infinity, what we have to compute.
20:59:180Paolo Guiotto: So, in other words, the integral we are looking for
21:03:960Paolo Guiotto: is half of the value of the Fourier transform at zero.
21:12:350Paolo Guiotto: Since we have the Fourier transformer, It is this thing.
21:17:120Paolo Guiotto: At 0, we see that when x is 0, we get pi half e to 0. 1, parenthesis is 1 plus 0, 1, so it is pi half. This is pi…
21:29:60Paolo Guiotto: Half total is pi over 4, and that's the value of this integral.
21:36:960Paolo Guiotto: The second integral is, integral 0 plus infinity of sine x divided 1 plus X squared square DX.
21:53:270Paolo Guiotto: Now, this one is again somehow related to the Fourier transform, because remember that this guy here is cosine of minus CX plus I sine
22:08:740Paolo Guiotto: minus CX, since of the minus, cos of minus CX is cos of CX.
22:16:820Paolo Guiotto: And, i sine, becomes a minus sign. So, it is clear that there is some kind of, relation between… it is like if you have, you see, no? It is… it's like if you have, integral of cost CX divided,
22:36:410Paolo Guiotto: 1 plus X squared up, squirrel, and signs, yes. Okay, so, this could be a way to…
22:44:390Paolo Guiotto: carry the Fourier transformer to this integral.
22:48:800Paolo Guiotto: or… Which I don't know if it is convenient here.
22:58:590Paolo Guiotto: Because the sine integral is zero, no? If we use that here.
23:03:860Paolo Guiotto: you could say that I have the Fourier transform FXC, right? Which is integral on R of 1 over 1 plus X squared, and now squared. I put cos CX
23:18:320Paolo Guiotto: minus i sine CX.
23:22:380Paolo Guiotto: But here, you understand that if you split this into two sub… two integrals.
23:28:800Paolo Guiotto: One is the integral on r of cosine CX divided 1 plus X squared squared.
23:38:200Paolo Guiotto: EX, then you have minus i integral on r of the sine
23:44:220Paolo Guiotto: CX divided 1 plus X squared squared.
23:49:10Paolo Guiotto: So you may think, well, it is done, takes SQL1, okay?
23:54:130Paolo Guiotto: But unfortunately, this part here is zero.
23:58:200Paolo Guiotto: You can see, because the Fourier transform we computed, unless it is wrong, is real, you see?
24:05:40Paolo Guiotto: It's not, as not imaginary parts, it's real numbers. So, if this is the expression of the Fourier transform, that cannot be imaginary part, it must be zero.
24:14:740Paolo Guiotto: However, even if you download the transform and you look at this particular, the function you have here is odd, because when you change X into minus X, this does not change, but this changes R. And when you integrate a NOT function on a symmetrical development perspective, the RTS.
24:34:930Paolo Guiotto: Okay, this is the dollar. So that integral is zero, so we cannot get that way, which is not the case of this one, because we are not neglecting one asymmetry.
24:46:180Paolo Guiotto: Either that's all.
24:47:640Paolo Guiotto: So we should understand how to make visa.
24:51:30Paolo Guiotto: to recreate the other.
24:55:410Paolo Guiotto: So, another trick could be, okay, why don't we write this sign by using Euler formulas? Sine is e to iX, right, minus… e to minus iX divided by 2i.
25:14:550Paolo Guiotto: Hmm.
25:20:490Paolo Guiotto: So let's see if we can get something from this. So, let's give a name to our integral to do not carry all this stuff around. So this is, 1 over 2i.
25:32:720Paolo Guiotto: integral, 0 to plus infinity, e to iX1 over
25:39:700Paolo Guiotto: So, 1 plus X squared squared, Minus integrals 0 plus infinity.
25:49:490Paolo Guiotto: minus IX… over 1 plus X squared squared dx.
26:19:850Paolo Guiotto: Okay.
26:27:140Paolo Guiotto: if… here.
26:32:420Paolo Guiotto: to change variable, y equals minus X. When x goes from 0 plus infinity, this goes from minus infinity to 0, becomes the integral minus infinity to 0 of e to iY, say, over 1 plus
26:47:940Paolo Guiotto: X squared or Y square are the same thing, because this is a square. And this is DY. So, as you can see, it's the same integral.
26:59:250Paolo Guiotto: But now completed from minus 15 to 0, so it is 1 over 2i.
27:04:270Paolo Guiotto: integral from minus infinity plus infinity E to IX divided 1… sorry, I used the same letter X for both integrals.
27:15:670Paolo Guiotto: But there is something which is disturbing.
27:22:290Paolo Guiotto: Because this is the Fourier transform, right?
27:26:740Paolo Guiotto: More or less, no?
27:28:930Paolo Guiotto: If you want to see the Fourier integral, you have to put the minus here, you can put the minus 1 here.
27:37:720Paolo Guiotto: But the problem is that this patterned one of twoi, that's… there is something wrong here.
27:47:400Paolo Guiotto: Because this looks like 1 over 2i, then this is the Fourier transform of F evaluated at minus 1.
27:58:260Paolo Guiotto: But then, then, this is not possible, because this would say that I is imaginary, so what is wrong here?
28:12:880Paolo Guiotto: What is the address?
28:15:700Paolo Guiotto: That's the formula for sign, isn't it?
28:19:440Paolo Guiotto: this one.
28:21:300Paolo Guiotto: So, if we plug this… We can split into two integrals, and it seems correct.
28:29:840Paolo Guiotto: I changed variable in the right integral, y equals minus X. Y goes from minus infinity to zero, e to IY, 1 over 1 plus Y squared, dy.
28:59:70Paolo Guiotto: Is that correct, the change of variable?
29:02:730Paolo Guiotto: Yes, because I give the X, DY, no.
29:09:280Paolo Guiotto: Yeah, here is where there is the error.
29:14:70Paolo Guiotto: I guess that this… the error is that this remains minus, because if I do the change of variables.
29:20:180Paolo Guiotto: 0 remains 0. So formally, if I want to use this minus and give the X in such a way that minus DX becomes DY,
29:32:330Paolo Guiotto: I will have here integral from 0 to minus infinity of MIY over 1 plus Y squared.
29:42:270Paolo Guiotto: squared DY, and that's minus this integral, so… That's not correct.
29:53:20Paolo Guiotto: Hmm.
29:57:950Paolo Guiotto: So now…
30:03:630Paolo Guiotto: And now… can sign out this.
30:15:710Paolo Guiotto: Let me check. It's 1 minute we still have.
30:36:400Paolo Guiotto: So we should work a bit on this, because I don't see now…
30:47:110Paolo Guiotto: Okay, so I don't want to waste time. I will, I will, write the solution, later.
30:54:880Paolo Guiotto: okay, let's just use a minute to say… Mmm…
31:10:280Paolo Guiotto: Okay, actually, time is… would be over…
31:14:340Paolo Guiotto: So, okay, we could, it's, we… since time is over, we stop here. Let me just,
31:22:50Paolo Guiotto: Leave you with some of the problems, so do… Exercises, 19, 3… 1… the number 2…
31:36:440Paolo Guiotto: the number 3… well, actually… These, problems,
31:43:70Paolo Guiotto: concern the calculation of the Fourier original. We have not yet… so it's the inverse of the Fourier transform, so perhaps it's better.
31:54:140Paolo Guiotto: Let me see if you can do… okay, so the number one… Forget 2 and 3.
32:10:470Paolo Guiotto: Maybe you can do the number 7.
32:18:260Paolo Guiotto: you can try, we see tomorrow, example, this one, so I will do…
32:26:290Paolo Guiotto: Maybe the number one…
32:30:50Paolo Guiotto: Tomorrow, I will do this, number one. Okay, why don't we do all these three?
32:36:410Paolo Guiotto: Okay, so do these three problems for tomorrow.
32:40:610Paolo Guiotto: I will think about how to handle this issue, and I will write the solution later.
32:48:50Paolo Guiotto: Okay.
32:51:210Paolo Guiotto: But it's doing that.
32:57:790Paolo Guiotto: Okay, let's stop here. That's today.