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00:00:660Paolo Guiotto: Let's start recording.
00:19:930Paolo Guiotto: So, let's start with the exercise 9.
00:27:660Paolo Guiotto: give the definition of the Fourier transformer.
00:31:110Paolo Guiotto: I think that that one's right.
00:33:580Paolo Guiotto: Here's the definition, end of convolution product, F star G for FG in L1 prove that the transform of the convolution is equal to the product of the Fourier transform, and this is the calculation I don't do here. Now.
00:51:850Paolo Guiotto: It says, number one, compute the Fourier transform of E minus A modulus. We are ready.
01:00:20Paolo Guiotto: West.
01:03:770Paolo Guiotto: That's a sort of.
01:05:760Paolo Guiotto: The question one, should I repeat? Because it is just what we have done in class, no? Calculation of the Fourier transform of E minus A, the modulus you can do directly, and then use this to deduce the Fourier transform of the Cauchy distribution.
01:24:540Paolo Guiotto: That was by using the inversion formula, because once you have that this is a two-way over A squared plus
01:32:900Paolo Guiotto: C-square,
01:37:710Paolo Guiotto: you have that, so from this, it follows that 1 over 2A E minus A, the modulus hat of this is,
01:49:660Paolo Guiotto: 1 over A squared plus C squared. So this implies that the hat of 1 over A squared plus
02:00:130Paolo Guiotto: The square of the variable, let's use the letter X for this, is the double head of 1 over 2A, E minus A absolute value.
02:13:550Paolo Guiotto: At X, now you use the inversion formula that says the double that is 2 pi times inversion formula.
02:25:270Paolo Guiotto: that you should justify, it applies here, because both the function and its politics form are in L1
02:31:870Paolo Guiotto: And this is, 1 over 2A, E minus A, evaluated at minus X. So this simplifies this, and finally, you get P over AE minus A modulus of X, and that's…
02:47:410Paolo Guiotto: the answer.
02:50:10Paolo Guiotto: Now, we used the previous factor, this…
02:53:760Paolo Guiotto: The question is to… to compute the…
02:56:880Paolo Guiotto: Fourier transform of, this, 1 over…
03:02:230Paolo Guiotto: A squared plus X squared squared.
03:07:270Paolo Guiotto: For a positive difference. Now, use previous facts, you should combine what you… please, the exercises.
03:16:750Paolo Guiotto: Once you, combine,
03:19:650Paolo Guiotto: something you have seen above. So here, you may know… you should notice that this 1 over A squared plus X squared squared is 1 over A squared plus X squared times 1.
03:34:710Paolo Guiotto: over A squared plus X squared, and these two are Fourier transforms.
03:40:130Paolo Guiotto: So, this is the hat of pi over AE minus A, the modulus.
03:47:510Paolo Guiotto: And this is the hat of the same thing, part of an AE minus AD modulus. So probably the exercise asks to do this, the hat of the convolution.
04:01:320Paolo Guiotto: So we can put this factor outside, so pi squared over A squared, E minus A modulus star E minus A
04:12:560Paolo Guiotto: Come on, the Wilson.
04:15:840Paolo Guiotto: And, at point, at point… Excellent.
04:27:680Paolo Guiotto: Yeah, because we have to compute the Fourier transform of this function, so we should say that then…
04:38:10Paolo Guiotto: the hat of 1 over A squared plus variable squared squared.
04:45:140Paolo Guiotto: at point C, let's say, is pi square of A squared, the pattern of this pattern.
04:55:620Paolo Guiotto: E minus A modulus, star, E minus A modulus.
05:02:240Paolo Guiotto: evaluated at point C.
05:07:610Paolo Guiotto: Now, inversion formula here applies because the convolution between two L1 functions is an L1 function, so this guy is an L1, and since we know that the hat of that function
05:22:740Paolo Guiotto: is apart for a factor. This one, which is in L1,
05:28:460Paolo Guiotto: We know that both E minus A modulus star, E minus A modulus this belongs to L1, as convolution…
05:41:850Paolo Guiotto: of L1 functions.
05:46:740Paolo Guiotto: And it's at…
05:50:350Paolo Guiotto: E minus A modulus star E minus A modulus. Also, this one, which is a part for the factor
06:00:870Paolo Guiotto: A squared over P squared, so it is A squared over P squared, 1 over A squared plus X squared squared.
06:10:330Paolo Guiotto: So this is clearly an L1 function.
06:14:220Paolo Guiotto: So, since we have the function and its head are in L1, inversion formula applies.
06:21:130Paolo Guiotto: here, and we get pi square of A squared, then E minus A modulus star E minus A modulus. This convolution evaluated, there is probably
06:37:10Paolo Guiotto: The double hat is 2 pi, there is a 2 pi, right? This… times evaluated at minus X.
06:47:280Paolo Guiotto: You see this?
06:51:380Paolo Guiotto: So, sorry, at minus C, the letter of the transform was C at minus C.
06:59:250Paolo Guiotto: So, we have to compute this convolution.
07:04:350Paolo Guiotto: Now, E minus A modulus star, E minus A modulus.
07:11:710Paolo Guiotto: But let's say at point X, then we will flip the argument into minus C,
07:18:380Paolo Guiotto: This is the integral, R.
07:23:190Paolo Guiotto: We already computed this, I do not remind. E minus A, let's say eta.
07:29:130Paolo Guiotto: E minus A modulus C minus eta.
07:34:270Paolo Guiotto: the eta.
07:40:500Paolo Guiotto: Now, what is this? So… We split the integral into minus infinity to 0, E minus a eta, here.
07:55:780Paolo Guiotto: E minus A modulus C minus eta.
08:00:590Paolo Guiotto: the eta plus integral 0 plus infinity, it becomes EA eta.
08:07:00Paolo Guiotto: E minus A modulus C minus eta.
08:17:490Paolo Guiotto: Okay… Now, perhaps it is better to split.
08:24:920Paolo Guiotto: Yeah.
08:27:410Paolo Guiotto: So, we split, the calculation for…
08:35:370Paolo Guiotto: Let's see, C positive, C negative.
08:38:780Paolo Guiotto: So, if C is positive, this can be. So, this first integral…
08:46:60Paolo Guiotto: since eta is negative, C minus eta is positive, so the model is itself, we have integral minus infinity to 0e minus a eta.
09:00:120Paolo Guiotto: E minus A, C minus eta, eta.
09:06:330Paolo Guiotto: About this one… This will be split into integrals from 0 to C.
09:12:900Paolo Guiotto: EA eta. Here, C minus eta is positive, because eta is smaller than C, so we still have E minus A, C minus eta, eta.
09:27:220Paolo Guiotto: plus the integral from C2 plus infinity, EA eta, in this case now changes, and this becomes E to AC
09:37:10Paolo Guiotto: Minus eta, d eta.
09:42:460Paolo Guiotto: So, about this one, we have E minus…
09:51:20Paolo Guiotto: That there's something wrong here.
10:08:380Paolo Guiotto: There's something wrong here, because A minus A eta simplifies the E plus A eta, and then we have an integral of a constant of minus infinity zero, that's not correct.
10:19:950Paolo Guiotto: So where… where is the error? Yes, the error is because I forgot that…
10:26:310Paolo Guiotto: Since I am integrating from minus infinity to zero, that's a plus a meter, not a minus.
10:33:20Paolo Guiotto: So that's a plus.
10:35:420Paolo Guiotto: So we have the first integral here is E minus Xi. Then we have integral minus infinity to zero e to a eta.
10:49:140Paolo Guiotto: beta…
10:51:270Paolo Guiotto: Okay, about to the other side, we have plus e to minus AXC, then we have integral 0 to c of…
11:01:310Paolo Guiotto: E2A, eta.
11:05:450Paolo Guiotto: And this last is,
11:10:480Paolo Guiotto: No, that is… I did wrong at the… I did the opposite. This is a minus, right.
11:20:500Paolo Guiotto: So this is a minus because, is, eta is positive here, so a minus here…
11:27:370Paolo Guiotto: So we have a minus also here.
11:31:330Paolo Guiotto: Dan, there is something wrong here.
11:34:840Paolo Guiotto: No, it minus immediately 0 is correct, the E to 2A eta is this one, that probably is wrong, because from 0 to Xi, the A eta cancels, and then it remains 1.
11:48:920Paolo Guiotto: And the last one is integral from C to plus infinity.
11:54:90Paolo Guiotto: E2 minus 2A eta, then there is an E2A XE down, out here.
12:00:690Paolo Guiotto: Okay, up for you.
12:02:860Paolo Guiotto: Now, it is correct. So we have E minus Xi, then we have e to 2A eta divided 2A, evaluated from minus infinity to 0.
12:16:780Paolo Guiotto: Now, this evaluation at 0, the exponential is 1. At minus infinity, the exponential is 0. Remind that H is positive.
12:27:410Paolo Guiotto: So we have an exponent going to minus infinity, so this evaluation is 1 over 2A, and that's okay. Then here we have E minus E minus Xe times X, so CE minus XC.
12:44:540Paolo Guiotto: From this term, yeah?
12:47:520Paolo Guiotto: And for the last one, we have E2AXE, and again… the evaluation E minus 2A.
12:57:600Paolo Guiotto: eta divided by minus 2A, this time from C to plus infinity. At plus infinity, you get 0.
13:07:240Paolo Guiotto: for the same reason, it is positive. At X, you get minus. With the other minus here, it becomes plus.
13:16:650Paolo Guiotto: E22A minus 2A, C divided 2A.
13:24:540Paolo Guiotto: So, in conclusion, we get 1 over 2A E minus AC plus C, E minus AC.
13:35:630Paolo Guiotto: And then we have, there is an 8C here. E minus 8C divided to A.
13:48:100Paolo Guiotto: Yes?
13:52:10Paolo Guiotto: So, it should be… E minus AX, that multiplies 1 over 8.
14:01:40Paolo Guiotto: plus X seat.
14:03:890Paolo Guiotto: This Foxy positive.
14:07:860Paolo Guiotto: Now, Foxy negative, we could repeat, but…
14:15:160Paolo Guiotto: Since we are doing the convolution between, this is,
14:20:510Paolo Guiotto: particular situation where you have F star, but that we have F star F.
14:27:170Paolo Guiotto: But I think it's not important. The key point is that the function is…
14:33:190Paolo Guiotto: symmetric, no? It's even. When you change X into minus X, nothing changes. So…
14:40:610Paolo Guiotto: Is it true that the function has some symmetry? So what happens when I do this at minus X? This would be integral f of y. Let's use this since we have F star F, but probably it is with F star G and FG even, something like this.
14:59:560Paolo Guiotto: So we have minus X, minus Y, BY.
15:05:690Paolo Guiotto: Okay.
15:07:630Paolo Guiotto: Now, what could we say? So, since the function is symmetric, this is F at minus X plus Y, right?
15:15:860Paolo Guiotto: So it is equal to F of X plus Y.
15:19:170Paolo Guiotto: The function is even. Okay.
15:21:730Paolo Guiotto: F even, so F of minus something equal to F of something.
15:29:350Paolo Guiotto: So this becomes the integral of FY
15:33:410Paolo Guiotto: But I can also write this as F minus Y,
15:37:470Paolo Guiotto: F of X plus Y, DY, yes. Now, we change variable, we set Z equal minus Y. This becomes integral f of z. F of X plus Y becomes minus Z.
15:56:330Paolo Guiotto: And then DY becomes DZ, because here we are doing,
16:01:290Paolo Guiotto: the change of variable, keeping the domain, no? Otherwise, I should put the endpoints and invert the endpoints. At the end, you have, in this case, you have demodruus, and this becomes exactly S star F,
16:14:380Paolo Guiotto: at X. So we proved that the convolution is, symmetric, so F, star F is, even.
16:26:60Paolo Guiotto: And therefore, since we compute the convolution FOXC positive, FOXC negative is the same value.
16:35:310Paolo Guiotto: By using the… for X negative, we use the value at minus X, no? So, in a unit world, you can say that E minus A modulus
16:46:650Paolo Guiotto: star E minus A modulus.
16:50:240Paolo Guiotto: at point, let's say, C, it is equal to that thing. We symmetrize this by replacing C with the absolute value of C, so E minus
17:01:750Paolo Guiotto: A absolute value of C, 1 over A plus absolute value of C.
17:08:430Paolo Guiotto: Okay, so this is the convolution.
17:14:310Paolo Guiotto: So, going back here… We have the Fourier transform of the quantity we want.
17:22:770Paolo Guiotto: 1 over A squared plus square squared is 2 pi cubed divided A squared. That convolution at minus C, since it is semantic, is the same at C. So we have this conclusion.
17:38:40Paolo Guiotto: Conclusion…
17:40:300Paolo Guiotto: The F of 1 over A squared plus D squared squared at point C is 2 pi cubed over A squared, E minus A absolute value of C,
17:56:900Paolo Guiotto: times 1 over A plus absolute value.
18:00:810Paolo Guiotto: Obviously.
18:03:610Paolo Guiotto: And this is, the answer. I don't know if…
18:07:670Paolo Guiotto: There might be errors somewhere, but… Should be this one.
18:12:730Paolo Guiotto: Did you watch something like that?
18:18:690Paolo Guiotto: I see only one answering me, but…
18:26:00Paolo Guiotto: Next one was… Number?
18:33:690Paolo Guiotto: Number 10.
18:35:880Paolo Guiotto: So, exercise 10 is an exercise on hild spaces. We have H is L2 minus 11.
18:46:760Paolo Guiotto: with the usual, N2 product.
18:50:90Paolo Guiotto: VI case, so integral from minus 1, 1 of F times G.
18:56:130Paolo Guiotto: Now, it says define as…
19:00:110Paolo Guiotto: as the set of F in H, such that integral minus 1,0F equal integral 01F.
19:11:550Paolo Guiotto: the X, okay.
19:16:520Paolo Guiotto: Number one, S… is well.
19:21:380Paolo Guiotto: defined… and… closed subspace.
19:32:50Paolo Guiotto: All aged.
19:34:40Paolo Guiotto: Number two, determining the orthogonal projection on S of F for a genetic F in H.
19:44:370Paolo Guiotto: In particular, Compute the projection of this function.
19:53:60Paolo Guiotto: So, I remember, as far as I remember, when this was given.
20:00:120Paolo Guiotto: Even if it is simple, I don't know what you have talked about, even if it is simple, lots of people,
20:07:960Paolo Guiotto: Get, confused about this condition.
20:11:920Paolo Guiotto: Because, if you look, it's not the integral from minus 1, so…
20:17:610Paolo Guiotto: what should we do? Well, it was… the idea was simple. We, well, let's say, first of all, if you want to say that this is well-defined, we should check that since there are integrals, and these are integrals of F, F is in L2, but that means that the integral of F is rather F is in L1, so this requires a little
20:41:880Paolo Guiotto: justification, okay? That's why there is this well-defined here. So, however, I noticed that since,
20:52:770Paolo Guiotto: F in L2, this is a general fact that, comes from Cauchifold's inequality. This implies,
21:03:810Paolo Guiotto: that F is in a 1 minus 1,
21:09:790Paolo Guiotto: And so, in particular, that F is in L1 on the two subintervals, 01 and, minus 1, 0, so the two integrals make sense.
21:24:960Paolo Guiotto: Okay, this could be a justification, so…
21:30:910Paolo Guiotto: In other words, I don't need… how it's written S makes sense. I don't need any other condition. So S is… well…
21:40:230Paolo Guiotto: defined.
21:41:660Paolo Guiotto: A better justification that would help you to find out what is S would be the following alternative.
21:52:790Paolo Guiotto: I could notice that integral minus 1,0 of f equal integral 01 of s, right, could be written as…
22:03:220Paolo Guiotto: I could try to put under a unique integral from minus 1 to 1, because that looks like, you know, a product of… so what should I do? Well, I could write the integer minus 1 to 1 of F, that is multiplied by, for example, 1 when X is negative.
22:22:360Paolo Guiotto: So, if you want to carry everything on the other side, you get this minus 1 is 0,
22:28:210Paolo Guiotto: Let's do it.
22:29:670Paolo Guiotto: Lowly.
22:31:00Paolo Guiotto: this equals 0, right? This is the condition. So I could represent this as integral from minus 1 to 1 of F,
22:39:520Paolo Guiotto: Multiplied by the indicator of interval minus 1, 0, right?
22:44:970Paolo Guiotto: And this one, I could say, plus integral 0, or sorry, minus 1, 1, again, over, times minus the indicator of 0, 1.
22:54:590Paolo Guiotto: It is the same thing. So in this way, I could see that this condition
23:01:320Paolo Guiotto: This condition is equivalent to this one, minus 1, 1 of F times indicator of minus 1, 0.
23:11:20Paolo Guiotto: Minus indicator of 0, 1, This equals to 0.
23:18:420Paolo Guiotto: And this is an orthogonality condition, because clearly this guy here belongs to L2,
23:26:350Paolo Guiotto: No? And therefore, what is written here is written F, scalar. That, that stuff here, let's call it, I don't know.
23:39:630Paolo Guiotto: U0.
23:41:530Paolo Guiotto: U0, let's say it's a specific factor, equals 0. So, in other words, S is the set of elements of H, such that F, color, U0, equals 0. In this way, you don't have to justify anything, because
23:59:500Paolo Guiotto: That condition, the scalpad, is always defined.
24:03:840Paolo Guiotto: And you see that this means also that
24:06:840Paolo Guiotto: Since you are perpendicular to U0, this is the orthogonal of U0, no? This is the orthogonal of the space made… of the set made by the single U0, but if you want, this is the same of the
24:23:810Paolo Guiotto: space generated by U0, by the span of U0.
24:28:210Paolo Guiotto: Okay.
24:30:270Paolo Guiotto: Now, this is important because maybe S is not finally dimensional. However, you can easily prove that it is… what is required here is a closed subspace. Well, if you want, since,
24:43:790Paolo Guiotto: Since S is the orthogonal of some U, where U is the span.
24:50:170Paolo Guiotto: of U0. It's a general fact that whatever is that U, that S is a linear subspace, and it is also closed.
25:00:900Paolo Guiotto: is, general.
25:04:410Paolo Guiotto: true.
25:07:520Paolo Guiotto: factor.
25:09:620Paolo Guiotto: That's… S is a closed subspace.
25:17:740Paolo Guiotto: off. Age.
25:20:190Paolo Guiotto: Okay, so now we can move to question two, that is to determine the projection For a genetic vector.
25:28:590Paolo Guiotto: F on this S. Now, since this S is your terminal of U, and it is better to determine the projection on U, because this is finally dimensional, I use another remarkable fact, since
25:47:550Paolo Guiotto: S is…
25:56:270Paolo Guiotto: closed up.
26:01:40Paolo Guiotto: And… It is the orthogonal of U. We know that the projection
26:07:750Paolo Guiotto: on S is F minus the projection on U of F. It's this one, no? So let's determine this one.
26:19:170Paolo Guiotto: And, the projection on U of F, since U is a space generated by
26:27:70Paolo Guiotto: a single vector U0, no? This is just, if you want, is something like lambda U0, or a certain lambda, that must verify the orthogonality conditions, such that
26:43:380Paolo Guiotto: Where… Lambda is determined.
26:50:430Paolo Guiotto: Bye.
26:52:220Paolo Guiotto: There is a unique condition here, because there is a unique vector, U0. So, by the orthogonality condition, which is F minus PUF scalar, the unique generator, U0 equals 0.
27:07:490Paolo Guiotto: This means, F minus lambda F U0.
27:14:350Paolo Guiotto: Sorry, lambda U0.
27:17:410Paolo Guiotto: Equals zero.
27:21:160Paolo Guiotto: So you get easily that F scalar U0.
27:25:490Paolo Guiotto: minus lambda U0, Scala U0, this must be equal to 0. So lambda… is equal to FU0,
27:40:670Paolo Guiotto: norm of U0 squared.
27:44:840Paolo Guiotto: So, whatever is this number, we can compute, but we can write that, so pi…
27:53:590Paolo Guiotto: UF will be, that number, F stalar U0, divided by
28:01:50Paolo Guiotto: Norm of U0 square times U0.
28:05:170Paolo Guiotto: And, the projection.
28:09:960Paolo Guiotto: on S of F will be F minus this thing.
28:14:690Paolo Guiotto: F scalar U0 divided by norm of U0, square U0. This is the formula for… then, if we want, since U0 is that specific vector, we can compute this.
28:30:220Paolo Guiotto: In particular, the projection on S.
28:37:60Paolo Guiotto: of X, well, we noticed that F of X equal X is clearly a function of the space, L2 minus 1, there's no question in this.
28:49:60Paolo Guiotto: And so this is X minus… we have to compute the
28:53:950Paolo Guiotto: well, sharp, let's call it. U0 divided by the norm of U0 square times U0, which is that indicator.
29:05:880Paolo Guiotto: So, the sharp…
29:08:900Paolo Guiotto: Scalar U0 is the integral from minus 1 1 of the function, which is X times that U0, which is the indicator of minus 1, 0, minus the indicator of 01.
29:24:150Paolo Guiotto: Yes. So, in other words, this is the integral from minus 1,0 of X minus the integral from 0 to 1 of X DX.
29:36:710Paolo Guiotto: The, the, the two values… are, the function is,
29:46:750Paolo Guiotto: is symmetric, no? Is odd, this one.
29:50:720Paolo Guiotto: So the integral… is opposite, so they are a tuna.
29:57:920Paolo Guiotto: Or if you want, you compute X squared… X squared over 2 between minus 1, 0, minus X squared over 2 between 0, 1.
30:10:390Paolo Guiotto: So… you get zero, right?
30:15:180Paolo Guiotto: So this is, no.
30:20:490Paolo Guiotto: This is, is that correct?
30:23:980Paolo Guiotto: I'm missing in a… Glass of water.
30:29:380Paolo Guiotto: Yes, 1 half… no, no, no, zero minus… that's not zero.
30:35:660Paolo Guiotto: That's strange. So, the value at 0 is 0 minus the value at minus 1, which is 1 half, so they get minus 1 half, minus another one half is minus 1. This is minus 1. Sorry.
30:47:670Paolo Guiotto: Now, the norm of U0 square…
30:53:190Paolo Guiotto: Is the integral from minus 1 to 1 of that indicator of minus 1, 0.
31:01:90Paolo Guiotto: Minus indicator of 01 squared.
31:06:990Paolo Guiotto: Now, when you do the square, you get indicator of minus 10 squared is still the indicator, because the value is 0, 1, right?
31:16:660Paolo Guiotto: So when you do the square of 0, you get 0. When you do the square of 1, you get 1.
31:21:270Paolo Guiotto: Plus the other indicator, 01 square, which is, again, equal to the indicator itself.
31:28:420Paolo Guiotto: Minus the double product, but the double product would be zero everywhere, except at most of 0. So when you do the integral, you get zero. So, in fact, they are two orthogonal vectors. So you get this, and this is…
31:42:490Paolo Guiotto: Equal to the indicator of minus 11.
31:45:860Paolo Guiotto: So it is equal to 1, and so this integral is equal to 2.
31:52:430Paolo Guiotto: So, the projection of S of X is equal to X minus… this color product is minus 1, divided, the square of the norm is 2, and the U0 is an indicator minus 1 is 0, minus indicator 0 of what?
32:13:270Paolo Guiotto: And so we get, this, plus one half this. This is the autonominal projection.
32:22:10Paolo Guiotto: Okay.
32:23:420Paolo Guiotto: Okay.
32:25:220Paolo Guiotto: So the next one is…
32:32:150Paolo Guiotto: the 11…
32:38:340Paolo Guiotto: So, here we have this function G of C equal sine C minus C cosine C, divided by c cubed.
33:00:540Paolo Guiotto: So, there are three questions. Number one, is G in L1?
33:06:360Paolo Guiotto: L1R, of course, is G in LT?
33:12:50Paolo Guiotto: Justify carefully. Number two… discuss the existence of fully original 4G, existence of… 48 original.
33:26:340Paolo Guiotto: Or, G.
33:28:50Paolo Guiotto: Now, unfortunately, this year, we had to cut the… just the last part, which is on the Fourier transform in L2.
33:37:730Paolo Guiotto: So, in some exercises of the past year exams, you may find something concerning the
33:45:710Paolo Guiotto: inversion in L2, not in L1. You just skip this thing, because you needed, yeah, the… this transformer.
33:56:930Paolo Guiotto: Then number 3 says, show that CG over C…
34:06:760Paolo Guiotto: It gives this hint. This is the derivative with respect to C of something.
34:13:570Paolo Guiotto: Then, use this to determine the Fourier original of G. Justify carefully the properties you use to answer. So, use this…
34:27:630Paolo Guiotto: to determine… Original.
34:34:860Paolo Guiotto: of…
34:37:699Paolo Guiotto: Now, when you say justifying carefully, it means just you have to… to show me that you are… you master what you apply, no? You justify, I'm using this because of this fact, etc, no?
34:53:500Paolo Guiotto: You do not… I do not ask you to write the statement in detail, but at least if you are applying something non-trivial, you should justify this.
35:07:100Paolo Guiotto: Okay, solution of 1. Is G in L1? What can be said? First of all, this G is defined everywhere apart 0, and when it is defined, it is a continuous function in the real line, except 0. This is sufficient to say that it is a measurable function.
35:25:250Paolo Guiotto: Now, for integrity, we have to discuss the behavior of G at 0 at plus, minus infinity.
35:32:90Paolo Guiotto: So, for… Integrability… We determine… The… behavioral…
35:46:230Paolo Guiotto: of, actually.
35:48:750Paolo Guiotto: maybe I never stressed too much on this, but we should go check the absolute value of G, because we have to check absolute equality probability.
35:58:630Paolo Guiotto: Okay?
36:00:300Paolo Guiotto: at… C equals 0 and plus minus something.
36:05:810Paolo Guiotto: Now, since these are different, because at plus-minus infinity, let's start with that.
36:11:670Paolo Guiotto: at C equal plus minus infinity, so when C goes to plus minus infinity.
36:17:390Paolo Guiotto: I take modulus of G. What can be said? Well, when C goes to infinity, I have this sine, cosine, they do not have any particular behavior, they just obstinate between minus 1 and 1, so I cannot say it is like to have 1, or it is like to have Xi, which is not correct, because X is going to infinity, so…
36:35:690Paolo Guiotto: sine C is bounded, C is not bounded, so you cannot say that. That is correct in 0, but it's not correct at infinity.
36:43:740Paolo Guiotto: So what can be said here? However, I could notice that downstairs I have XC cube.
36:49:920Paolo Guiotto: which is going to infinity, so 1 of X cubed is going good at infinity as an integral function. Numerator is not clear, but at worst, I have this
37:01:800Paolo Guiotto: C, so, like, the first power. So the good idea, this would be to bound this, I say, bound model sine C
37:10:770Paolo Guiotto: minus C costs…
37:12:800Paolo Guiotto: C divided modulus C to power 3, then I just throw away everything here in the rough way, so modulus of sine plus modulus of X equals C,
37:27:780Paolo Guiotto: Then, this is less than 1. This is modulus of T times a factor less than 1. Everything is less than…
37:38:30Paolo Guiotto: modulus C plus 1 over modulus C cubed.
37:43:00Paolo Guiotto: C cubed. Now, if this week, which is bigger, is integral, the other one which is smaller, will be integral. Now, this one, I can say, which is not 100% correct, this right thing, but it is formal, if I'm saying right things, at plus-minus infinite, this goes like one of a modulus C squared.
38:02:750Paolo Guiotto: which is integrable at plus minus infinity. So I have that. Models of G is dominated by someone who is integral, because it is asymptotic to someone else, which is integral at plus-minus infinity. I deduce that G is…
38:19:800Paolo Guiotto: in L1 at plus minus infinity. This means that it is integral on any interval that contains plus infinity, and for the moment, X does not take 0.
38:34:110Paolo Guiotto: At X equals 0, the situation is a bit different, because at X equals 0, now the behavior of sine and cosine matters, because the numerator is going to zero, the denominator is going to zero, so I have to check how
38:48:940Paolo Guiotto: badly is going to zero, that thing. The denominator with respect to the numerator, because if this
38:56:310Paolo Guiotto: you know, if it is going to explode, I could have some problems. So I need to understand how the numerator is going to zero. Here, I can use the asymptotic formulas that say sine C equals C plus little O of C, the Maclaurin expansion, no?
39:15:900Paolo Guiotto: This one won't be sufficient, so let's prepare for the other one. C minus C cubed over 6 plus O of C cubed.
39:26:800Paolo Guiotto: For cosine, I have 1 minus c squared over 2 plus O of C squared.
39:34:410Paolo Guiotto: This should be sufficient.
39:37:690Paolo Guiotto: So, when I look at G of C,
39:40:750Paolo Guiotto: This is equal over XC cubed downstairs. Numerator is sine. You see that if I put C plus little o of c, then I have minus C cosine, which is 1 minus C squared over 2 plus O of C squared.
40:00:980Paolo Guiotto: the… what happens here is that the two C cancel here, with this one.
40:07:120Paolo Guiotto: And then you're in with a little O of C,
40:10:740Paolo Guiotto: plus minus something that is going to be c cubed plus a little O of C cubed, which is negligible. So, if I do that, I would remain just with O of C over C cubed, which is not sufficient, because numerator is going to 0 faster than C, but we don't know how fast.
40:30:710Paolo Guiotto: And we compare with denominator, which goes to zero as C cubed, so we don't know yet what happens. That's why we need to use the, the third order approximation for sine.
40:44:620Paolo Guiotto: And this one will solve the puzzle, so we have C times 1 minus C squared over 2 plus little O of C squared.
40:55:910Paolo Guiotto: So, divided by C cubed. Now, if we do repeat the calculation, this disappears again, of course.
41:03:210Paolo Guiotto: But now we have minus C cubed over 6 here, plus O of C cubed.
41:09:570Paolo Guiotto: Then we have a plus C cubed over 2 plus CO of C squared, you know that these little O are like powers. So when you multiply by C, by the variable you can carry inside, that's a little O of X cubed.
41:26:330Paolo Guiotto: divided by Xi cubed.
41:29:860Paolo Guiotto: So, basically, the point is that these two together, they are…
41:36:540Paolo Guiotto: 3 over 6 minus 1 over 6 makes 2 over 6, which is,
41:42:700Paolo Guiotto: 2 over 6 is 1 over 3. So, it is, at the end, 1 third of C cubed
41:49:600Paolo Guiotto: plus little O of C cubed divided by XC cubed. So even, we discover that this, when C goes to 0,
41:59:80Paolo Guiotto: Numerator is 1 3rd C cubed, terminator is c cubed with this goes to 1 third. It has a finite limit, so this function could be considered as a continuous function, because we extend by adding the value 1 3rd at X equals 0, so…
42:16:930Paolo Guiotto: we can… Consider… G. Continuos.
42:24:500Paolo Guiotto: at C equals 0. So this means that there is no problem with integrability at zero.
42:32:340Paolo Guiotto: So, gee… modules G is integral.
42:37:760Paolo Guiotto: At C equals 0, it is integral at plus minus infinity, it is integral on the interior line, so G is in L1.
42:47:350Paolo Guiotto: Now, what changed for L2?
42:51:360Paolo Guiotto: nothing, because G is in L2 if and only if integral of G squared is finite, integral of r is finite.
43:01:790Paolo Guiotto: It is measurable, we already know, etc. So basically, if and only if G squared is in L1, no?
43:09:930Paolo Guiotto: But what changed to G-square?
43:12:570Paolo Guiotto: At 0, nothing, because G goes to 1 third, the G squared will go to 1 over 9, so it's still continuous. At plus infinity, minus infinity, we have seen that G
43:23:530Paolo Guiotto: is dominated by someone who is, like, 1 over XC squared. When you do the square, it will go even better, because it will go to 0 as 1 over XC to power 4.
43:34:30Paolo Guiotto: So we can say that, so, models of G at C equals 0 square, is continuous.
43:48:800Paolo Guiotto: let's say it's not correct, but we can do like that. A modulus, G squared is dominated by something like constant over C to the power of 4 at C equal plus minus infinity. So, also, G squared
44:08:770Paolo Guiotto: is integral.
44:11:190Paolo Guiotto: on time.
44:13:180Paolo Guiotto: And therefore, G is in L2.
44:16:720Paolo Guiotto: So we don't need to repeat, of course, if we have already done the algo.
44:22:480Paolo Guiotto: Now, question 2.
44:26:300Paolo Guiotto: This is about the free original. Free original means a function, So, F… Pourier Original.
44:36:270Paolo Guiotto: of G if F is an L1 function such that F at is equal to G.
44:47:130Paolo Guiotto: That's simply what is this.
44:49:150Paolo Guiotto: Now, we know that, huh?
44:56:420Paolo Guiotto: my… the… inversion.
45:01:430Paolo Guiotto: formula.
45:06:50Paolo Guiotto: If both G and G-hat are in L1, 10 such original exist.
45:17:170Paolo Guiotto: than D.
45:19:740Paolo Guiotto: 48.
45:21:80Paolo Guiotto: Originally.
45:22:760Paolo Guiotto: exists.
45:25:660Paolo Guiotto: And it is equal, we have also formula, F of X,
45:31:290Paolo Guiotto: is the inverse Fourier transform is 1 over 2 pi, the hat of G evaluated at minus X.
45:39:120Paolo Guiotto: So, what should be checked is this part of the story, that E hat of G is in L1.
45:47:530Paolo Guiotto: We already know that G is in L1, that's 4.1. We need to know if also the height of G is in L1.
45:55:990Paolo Guiotto: Now, to check this, either we compute the F of G, and then we see what is it. This would be, in fact, the…
46:05:520Paolo Guiotto: I don't know if it will be the goal, really, let's see later. Because at the end, we will compute the… since we will compute in next point, DF, we will have automatically the hat of G. No matter how do we compute that F,
46:21:00Paolo Guiotto: either directly by computing the Fourier transform or G, or indirectly by discovering that G is the hat of someone, so we are the F. At the… at the end, we will have this function, G hat, so we will be able to say directly.
46:36:280Paolo Guiotto: But let's see if we can say without computing that, and there is a condition for that, this…
46:45:400Paolo Guiotto: is true.
46:48:490Paolo Guiotto: If G, G prime, and G2nd, these three are in L1.
46:54:740Paolo Guiotto: This is just a secretion condition, it's not an if and all if, but sometimes it's a cheap test, because you can do directly on the function, okay? Now, we already know that this is true, because we checked above, so let's see what happens for G prime of C.
47:13:130Paolo Guiotto: It's a bit tedious calculation, but remind that the scope is to check if this is in L1, so we don't have to do… we have to do just the derivative. Now, the function G was…
47:28:220Paolo Guiotto: Denominator is Xi cubed, so we do the square of this becomes Xi to power 6,
47:35:180Paolo Guiotto: Then we have derivative of numerator, well…
47:39:170Paolo Guiotto: So, numerator is sine, so cos C minus, the derivative of C sine C. So, this is sine…
47:53:50Paolo Guiotto: No, no, no, sorry. It's a C cosine on the right, so, koski?
48:01:660Paolo Guiotto: minus X… the derivative, of course, is minus X, so there is a plus here, sine C. This is the derivative of numerator times the denominator, which is XC cubed.
48:15:280Paolo Guiotto: minus the numerator, which is sine of C.
48:20:290Paolo Guiotto: minus C cosine C.
48:23:100Paolo Guiotto: times the derivative of C cubed, which is 3x squared.
48:30:90Paolo Guiotto: We can simplify a bit this.
48:37:560Paolo Guiotto: Don't think anything more than this, we can…
48:40:930Paolo Guiotto: Factorize the Xi square, so this remains XC, and this becomes X power 4.
48:47:880Paolo Guiotto: So, after these cancellations, we have XC squared sine C, if I'm not wrong.
48:58:650Paolo Guiotto: minus 3… sine C minus C cos C.
49:07:790Paolo Guiotto: divided by Xi to power 4. That's the derivative, no?
49:13:50Paolo Guiotto: This is continuous on R minus 0, so now for integrity, we have to do the same checks as above.
49:21:630Paolo Guiotto: 4.
49:23:260Paolo Guiotto: integrability.
49:26:240Paolo Guiotto: We have… at zero.
49:44:370Paolo Guiotto: Is it good or bad?
49:46:630Paolo Guiotto: That's… we have… have we studied c-sign, the numerator there?
49:55:40Paolo Guiotto: C, sine C minus X cosine X, right?
50:00:610Paolo Guiotto: That was Xi Cube, huh?
50:11:750Paolo Guiotto: Yes, of course. So this was, as you reminded, c cube of 3, right? Plus little O of C cubed.
50:24:120Paolo Guiotto: So this, means that since this can be seen as C plus little O of C,
50:30:390Paolo Guiotto: With this tree, you see that the G of C
50:34:660Paolo Guiotto: Sounds like c cube plus O of C cube.
50:40:210Paolo Guiotto: minus 3 times that, which is C cubed plus O of C cubed.
50:47:670Paolo Guiotto: So the 2xi cubed cancel, divided by Xi power 4, and we remain with the x c cube divided by Xi power 4, which is not sufficient at 0. So this means that we have to extend a bit the expansions.
51:05:930Paolo Guiotto: So it's going to be dramatic, but not particularly
51:11:210Paolo Guiotto: hard, so we have a C squared, so the sign here is C minus C cubed over 6 plus Oxy cube.
51:22:380Paolo Guiotto: Let's try to go quickly. Minus 3. Now, we have 4 sine C cubed knock C. Sorry, C.
51:31:620Paolo Guiotto: minus c cubed over 6. Well, this is not sufficient, because, it's going to simplify, so I need the next term.
51:41:210Paolo Guiotto: C to power 5 divided by whatever, what is 5 factorial?
51:46:680Paolo Guiotto: Right, let's, let's fight, probably will, will, will not be… Needed this one. See?
51:54:120Paolo Guiotto: Then we have the C cosine, so that is a minus. Minus C, the cosine is 1 minus C squared over 2, and c fourth of a 4 factorial.
52:08:60Paolo Guiotto: Oxy 4. Let's see what remains after the cancellations of all this divided by Xi to power 4.
52:15:710Paolo Guiotto: So, from the previous calculation, we have the XC cube that comes from this, that will be killed by the… well, the X is canceled already here. Then there will be the XC cube that comes from combining these two.
52:30:420Paolo Guiotto: No? This is the term that produces the cube of a 3 times 3. This will be canceled… all these terms are canceled by
52:40:230Paolo Guiotto: by this one.
52:41:890Paolo Guiotto: So what is survive now is A minus C to power 5 over 6,
52:49:320Paolo Guiotto: plus O of C to power 5, minus 3. Here, we have C to power 5 over 5 factorial, plus the little o of c power 5. But the good term comes…
53:03:850Paolo Guiotto: No.
53:05:750Paolo Guiotto: I… I thought that it was C to power 4. No, we have this, minus C to power 4 over 4 factorial… no, sorry, it's power of 5, because of 4 times 1.
53:20:480Paolo Guiotto: plus O of C power phi. At the end of the day, what we have here
53:28:740Paolo Guiotto: You see that the numerator, they won't cancel this terms, so it's like a constant C to power 5 plus little o of c power 5 divided C power 4.
53:41:520Paolo Guiotto: So, definitely goes to 0 when X goes to 0.
53:46:140Paolo Guiotto: So again, we can consider G prime can… be considered continuous, at… C equals 0.
53:58:540Paolo Guiotto: At C equal, plus, minus something is much easier.
54:03:130Paolo Guiotto: at… C equal plus minus infinity. We have modulus G of X,
54:15:420Paolo Guiotto: I want you,
54:17:890Paolo Guiotto: So, to be honest, since at the next point we will compute the… for the original, this check becomes, let's say…
54:27:740Paolo Guiotto: superfluous, because once we have the solution, who cares to verify if the ad has a greater form? Once I will compute the fully original, no? You see? But here, the scope of the exercise was just to make you working on integrability, okay?
54:47:560Paolo Guiotto: So, this, this was the… the… the… the scope of this point.
54:52:840Paolo Guiotto: At plus minus infinity, the G prime…
54:56:920Paolo Guiotto: Oh, this is G prime, sorry, yeah?
55:00:100Paolo Guiotto: the G prime, which is here. Now, do as before, throw away sine, cosine. What you have here is less or equal than. Downstairs, we have the XC to power 4. Upstairs, we have C squared sine C, so C squared, okay?
55:17:370Paolo Guiotto: minus 3, so plus 3 sine C1, plus C cos C cos C1. So, at the end, we have, like, C squared.
55:27:620Paolo Guiotto: plus, 3, design is 1, plus models of C.
55:34:80Paolo Guiotto: So… Hmm, there is something wrong here.
55:46:470Paolo Guiotto: No, it is correct, okay? So, what is the behavior of this at plus minus infinity? Well, numerator, the dominating factor is C squared, so C squared over Xi to power 4, that gives 1 of XC squared, which is integral.
56:05:200Paolo Guiotto: At plus minus infinity.
56:09:00Paolo Guiotto: So this means that G prime is in L1.
56:14:410Paolo Guiotto: Now, we can understand that the same kind of checks are for G-Secum.
56:20:530Paolo Guiotto: Same checks,
56:26:870Paolo Guiotto: for Gisecondo.
56:30:970Paolo Guiotto: Now, once I… but who ensures? Well, actually, you see that what happens to the derivative is that it's going better, not worse.
56:40:480Paolo Guiotto: And, so it shouldn't… it shouldn't be… we accept that, these kind of checks, okay? Let's pass to question 3, which is,
56:50:430Paolo Guiotto: the inversion problem, so determining the inverse. Now, it says, the starting point, it says, take XG of C, and observe that it is a derivative of something. So let's do that, and let's see what is it.
57:12:700Paolo Guiotto: Now, if I look CG of C, remind that, G was, sine…
57:20:180Paolo Guiotto: C minus C cos C, if I'm not wrong, divided at C cubed.
57:27:570Paolo Guiotto: So when I multiply, I get this.
57:30:760Paolo Guiotto: Now, it says this is a derivative.
57:35:160Paolo Guiotto: Well, do you see what kind of derivative?
57:39:40Paolo Guiotto: Could be.
57:40:750Paolo Guiotto: Don't… don't… I'm not asking you to tell exactly the answer, but…
57:48:520Paolo Guiotto: So, this is a fraction. You see that denominator is a square, so this should come from doing the derivative of a ratio.
57:56:70Paolo Guiotto: So let's say, if I do the derivative, do not use F and G, but phi over C.
58:02:390Paolo Guiotto: The formula says it is a fraction. You do C squared, then you have phi prime times C minus phi times C prime. So we have to understand if we can recognize these two guys. So one could be, of course, this one.
58:19:720Paolo Guiotto: And, so the numerator should be the derivative
58:26:280Paolo Guiotto: Times, well, this factor, you see that the factor appears here, no?
58:32:780Paolo Guiotto: So if my factor is C, this other guy should be the derivative of the numerator.
58:39:230Paolo Guiotto: So this should come from sine. So now, the idea could be, what if I do the derivative of sine C over Xi? Let's see. This is C squared
58:52:410Paolo Guiotto: Now, derivative of numerator cosc times denominator sine c times derivative of denominator 1, which is, apart for the sine, exactly this guy.
59:07:800Paolo Guiotto: So, I can say that CG of C is the derivative respect to C of minus sine
59:16:830Paolo Guiotto: C over C.
59:22:130Paolo Guiotto: I see. Now, this is an exercise on L2, huh?
59:27:180Paolo Guiotto: Let's see why. Because now, the point is, the goal is we should compute the Fourier transform of G, or…
59:35:260Paolo Guiotto: which is basically the same. Determine the function f, which is the fully original.
59:41:590Paolo Guiotto: Okay, G.
59:43:970Paolo Guiotto: Now…
59:48:320Paolo Guiotto: Yeah, the point is that here…
59:52:800Paolo Guiotto: There is a point that you cannot do with the L1 transform, because the idea would be now to take the hat of both sides, no?
00:02:180Paolo Guiotto: But the problem is that the L1 transform cannot be computed here, because if you look at this function here.
00:10:70Paolo Guiotto: This is not L1.
00:12:300Paolo Guiotto: If we go back, here…
00:15:580Paolo Guiotto: I'm sorry to have given you this exercise. However, the L2 transform works with the same rules, so that's why it works. This is G, no?
00:26:140Paolo Guiotto: If you do… you multiply by Xi, look at this bound, this bound becomes 1 of XC, which is not integral.
00:32:950Paolo Guiotto: So you have a problem at infinity, you do not have a problem at zero, because you multiply by something that goes to 0. So the problem is that this function is not in a1, so I wouldn't be allowed to compute the Fourier transform here.
00:49:620Paolo Guiotto: But this function would be in L2 instead, and that… so I could compute the so-called L2 transform, we have not seen.
00:59:00Paolo Guiotto: So, let's say that we couldn't finish… we know that there is a pre-original, no, but we don't… we can't do the calculation in this way.
01:13:780Paolo Guiotto: Let me think if we can…
01:21:460Paolo Guiotto: I should think about if there is another way. However, let's do informally how… is how, if, we could do. So this would be the… this, huh?
01:33:890Paolo Guiotto: So the hat of this…
01:44:730Paolo Guiotto: Do I really need to do this?
01:52:630Paolo Guiotto: No, maybe.
01:57:480Paolo Guiotto: No, maybe we… no, maybe the idea… we could… let's see if we can do without doing this transport.
02:05:160Paolo Guiotto: We know that this is a transform, right?
02:09:730Paolo Guiotto: This is the transform of direct, more or less. So I remind that the indicator of minus AA
02:17:810Paolo Guiotto: has a Fourier transform exactly to… what is it? 2A sign?
02:27:420Paolo Guiotto: XC over XC, right? To be this.
02:32:310Paolo Guiotto: So if I put the A equal 1, the indicator of minus 1, C is 2 sign…
02:41:640Paolo Guiotto: C over C.
02:43:870Paolo Guiotto: So, in particular, this sine C of X is the hat of this, and this is correct. We don't…
02:51:210Paolo Guiotto: So, this is the derivative with respect to t, well, there is also this minus, let's put also the minus, of the hat of minus 1 half the indicator minus 1, 1.
03:05:530Paolo Guiotto: C.
03:12:920Paolo Guiotto: And now…
03:20:130Paolo Guiotto: This one is the derivative of the Fourier transformer, so it is the Fourier transformer.
03:27:200Paolo Guiotto: You know that today, if, If a function, say…
03:34:810Paolo Guiotto: Let's do not use F. We know that H and XH are both in L1, then we have that there exists the derivative with respect to C of the transform of H.
03:49:400Paolo Guiotto: And this is the hat of minus i, the variable H.
03:56:940Paolo Guiotto: And this is the case, this is our H. H is in L1, and when you multiply by X, nothing changes, because it's just a function which is 0 outside interval minus 1, 1. So, if H is 1 half minus 1 half indicator of minus 1, this belongs to L1, but also
04:20:270Paolo Guiotto: XH belongs to L1.
04:25:10Paolo Guiotto: So this is the transform of minus i Sharpen.
04:33:450Paolo Guiotto: H, which is that function, so let's put 1 half here, indicator minus 1, 1.
04:42:930Paolo Guiotto: Okay.
04:45:710Paolo Guiotto: So this is the transform of the ascendant. I'm trying to… to do the exercise,
04:54:410Paolo Guiotto: Without computing the hat here, but here it's where you would need to do the hat.
05:01:140Paolo Guiotto: To have the double head there.
05:09:820Paolo Guiotto: I suspect that we cannot do without the L2 transform.
05:18:660Paolo Guiotto: No, we can't do, because at this point, I have a C, G of C,
05:24:770Paolo Guiotto: is the height of that function, I have… The variable.
05:31:690Paolo Guiotto: times, H, where H… Well, times the indicator, not H.
05:39:500Paolo Guiotto: Indicator of minus 11.
05:44:70Paolo Guiotto: Now, if I could take the hat once more, but I cannot do with the L1 stuff, because this is not in L1,
05:58:610Paolo Guiotto: But it is in L2.
06:01:740Paolo Guiotto: I know, I'm sorry, I've not checked this. I thought it was an L1…
06:07:540Paolo Guiotto: problem. So, suppose that you can do
06:10:640Paolo Guiotto: Even if the function is L2. So we would have the double hat of i half variable indicator minus 11. This is just to show you the conclusion.
06:22:680Paolo Guiotto: This should be equal to the integrator of this sharp G sharp. Let's put at point X at point X here.
06:32:260Paolo Guiotto: But now, this is the double hat. The double hat is 2 pi the original function, so I have X indicator minus 1, 1,
06:45:430Paolo Guiotto: Excellent.
06:48:500Paolo Guiotto: And about this one, huh?
06:51:570Paolo Guiotto: again, if you multiply by everything by minus i in order to have minus i here, so you have minus i sharp G.
07:01:960Paolo Guiotto: sharp detransform of this, evaluated at X, is… well, we can simplify here. When we multiply by minus i on this, we get II minus 1 times minus plus.
07:16:860Paolo Guiotto: Pi X indicator minus 1, 1,
07:24:240Paolo Guiotto: Yes, this is at the minus axon.
07:27:720Paolo Guiotto: Because of the inversion.
07:30:480Paolo Guiotto: At the left here, this is the head of this, which is the derivative with respect to X of the h of g.
07:39:240Paolo Guiotto: So we get this, that the derivative of the f of g is pi x. Now, the indicator is symmetric, so indicator minus 1, 1 of X.
07:50:870Paolo Guiotto: is the same.
07:52:310Paolo Guiotto: And then we can integrate this to get the function g-hat.
07:58:310Paolo Guiotto: Because this is equal to pi x, when x is between minus 1 and 1.
08:06:80Paolo Guiotto: And 0, when X is less than minus 1, X is greater than 1.
08:12:980Paolo Guiotto: So, if you do G hat of X,
08:26:130Paolo Guiotto: Now, to integrate this would be better to split X positive, X negative.
08:41:69Paolo Guiotto: So, for X… No. Mmm…
08:57:500Paolo Guiotto: Because this… this integral is,
09:03:410Paolo Guiotto: Or Outlook, how can we say?
09:07:290Paolo Guiotto: minus G hat 0, this would be the integral from 0 to X of the derivative of this thing, so pi
09:16:50Paolo Guiotto: by Y indicator of… minus 11Y.
09:23:899Paolo Guiotto: DY…
09:29:370Paolo Guiotto: So… The problem is, when X is,
09:36:359Paolo Guiotto: for x less than 1, this becomes… the indicator is 1, is pi integral from 0 to X of Y, which is Y square of 2.
09:48:69Paolo Guiotto: with the evaluation will become X squared over 2.
09:52:120Paolo Guiotto: When X is greater or equal than 1,
09:56:700Paolo Guiotto: This becomes just the integral between 0, 1, so it will be 0, integral between 0 and 1 of Y, so again, X squared over 2 evaluated between 0, 1, so it gives 1 half
10:10:370Paolo Guiotto: So this means that, and for X negative, I think that nothing changed.
10:18:330Paolo Guiotto: So, since this value is undetermined, and remind that this is a Fourier transformer, so,
10:26:80Paolo Guiotto: G hat X will be equal to… Bye.
10:36:280Paolo Guiotto: can I say phi X squared over 2 indicator of minus 1, 1,
10:42:950Paolo Guiotto: plus pi half minus the value at 0. But since this value cannot be different from zero, otherwise this function
10:53:680Paolo Guiotto: is not integral. This means that this must be equal to 0, and that's the function G at
11:03:550Paolo Guiotto: of X is pi x squared over 2 indicator minus 11 Excellent.
11:12:660Paolo Guiotto: There is a little, for example, this factor confirms that this is not the little sum of N1…
11:25:50Paolo Guiotto: Function, because this length is not continuous.
11:29:620Paolo Guiotto: So that's an L2 Fourier transform. So, in any case… I… I…
11:40:440Paolo Guiotto: I apologize for this, that, we, we…
11:44:660Paolo Guiotto: we had this, problem with the L2 fluid transform. We have not seen this yet, so…
11:52:590Paolo Guiotto: Is there any other exercises left?
11:58:50Paolo Guiotto: I think there are other two.
12:04:230Paolo Guiotto: Okay, 25… So, have you stopped here? With this, what have you done?
12:12:620Paolo Guiotto: That's so good, Peter.
12:15:620Paolo Guiotto: thought of that would be taken.
12:18:640Paolo Guiotto: Yeah, because you realize that So… Actually, I hate to jump like this.
12:31:790Paolo Guiotto: So you realized that there was a problem with L1?
12:35:470Paolo Guiotto: Okay, I'm sorry for that.
12:39:870Paolo Guiotto: In any case. So, this one says you have f of t equal integral on R1 minus cosine CX,
12:49:530Paolo Guiotto: divided the X square, X squared plus 1.
12:54:890Paolo Guiotto: Yeah, so… Question one, I mean, domain… off.
13:03:760Paolo Guiotto: this function, capital F, number two, domain D prime contained into D, where…
13:16:810Paolo Guiotto: there exists the derivative respect to C of this F.
13:21:40Paolo Guiotto: Number 3, domain D second, containing D prime, where there exists the second derivative of F.
13:32:500Paolo Guiotto: and computer. And number 4… use Foullier transformation to… compute the… The second delivery of F.
13:47:810Paolo Guiotto: and then use this to determine F.
13:54:350Paolo Guiotto: Okay, so, we have a function in the integral, depending on parameter C,
14:02:200Paolo Guiotto: So we treat this as an integral depending on parameter. The domain of definition of the function is the set of C, reals.
14:11:50Paolo Guiotto: Such that the integral makes sense. So the function F, as function of X,
14:18:210Paolo Guiotto: for the parameter X is integral in the real life.
14:23:740Paolo Guiotto: So this is the domain. Now, we have to check what happens to this function. So the function is, as a function of X,
14:34:150Paolo Guiotto: for every value of C fixed is definitely a continuous function in the real line, except point 0, where we have this problem due to the factor X squared.
14:47:550Paolo Guiotto: We may notice that for C equals 0, spatial case, since the numerator is constant equal to 0, not plus 0 is 1,
14:56:450Paolo Guiotto: F of,
14:58:800Paolo Guiotto: for value of parameter is identically equal to 0, so definitely it will belong to L1, okay? So this… we may say that X equals 0 belongs to the domain, for sure.
15:14:930Paolo Guiotto: So, for C different from zero.
15:21:110Paolo Guiotto: we have… to discuss… integrability at. So we have a problem at 0 and plus minus infinity.
15:38:810Paolo Guiotto: at zero, So, the function fxc is 1 minus cos CX over X squared 1 plus X squared.
15:54:630Paolo Guiotto: So bear in mind that, cos T is 1 minus T squared over 2 plus O of T squared.
16:03:610Paolo Guiotto: So, 1 minus cost is T squared over 2.
16:08:190Paolo Guiotto: So, basically, we have, CX squared over 2 plus a correction of the X. So, since we care for the variable X here, X is just a plan that we write, just
16:20:970Paolo Guiotto: Oxi squared, X squared, and down we have X squared times 1 plus X squared. So, as you can see, when X goes to 0, this goes to 1, so the denominator is like X squared. Denominator is like a C squared, X squared over 2,
16:39:620Paolo Guiotto: So, when you go to 0, it is like X square over 2. X squared in the denominator, X squared in the denominator, this is 1. So, in particular, this means that you have a finite limit, the function can be considered as continuous.
16:55:720Paolo Guiotto: at 0, so F… is, integral.
17:02:900Paolo Guiotto: at X equals 0 for every value of forensic C, real.
17:09:920Paolo Guiotto: at x equal plus minus infinity, since, again, the numerator has not a particular behavior, we just bound. We have modulus FXC, we throw away that 1 minus cos, which is bounded by 1,
17:29:330Paolo Guiotto: And we remain with X times X squared plus X squared plus 1, so which is asymptotic to 1 over X to power 4, which is integral at plus minus infinity.
17:43:830Paolo Guiotto: So, the conclusion is that domain D
17:47:950Paolo Guiotto: which is the set of values C for which the function F, as function of X is L1, is the entire real line.
17:57:780Paolo Guiotto: domain B is R.
18:02:40Paolo Guiotto: Question 2. Now we have to discuss the differentiability, and of course, we will compute the differentiability… the derivative by carrying the derivative inside. So…
18:14:570Paolo Guiotto: Applying…
18:19:380Paolo Guiotto: differentiability under… Integral sine.
18:27:320Paolo Guiotto: We have that derivative with respect to C of F,
18:33:750Paolo Guiotto: is equal to the integral on R of the derivative with respect to C of the little f x c.
18:43:530Paolo Guiotto: provided.
18:48:970Paolo Guiotto: So there are two conditions. Number one, there exists the partial derivative with respect to C of FXC, but that's evident because we have to do the derivative with respect to C, 1 minus cos CX divided
19:05:370Paolo Guiotto: X squared times 1 plus X squared. Now, here, X is just… constant.
19:14:320Paolo Guiotto: for this derivative, so we have 1 over X squared times 1 plus X squared.
19:20:940Paolo Guiotto: Times derivative of the numerator is the derivative 1, 0, minus cosine is sine.
19:27:250Paolo Guiotto: CX, times the derivative of the argument respective C, so this provides another X.
19:35:190Paolo Guiotto: And so at the end, we have, we can cancel this with this, we have a sign…
19:41:580Paolo Guiotto: CX over X times 1 plus X squared.
19:46:890Paolo Guiotto: Now, this calculation, of course, can be done for every X.
19:52:400Paolo Guiotto: Except for X equal to 0, so we can say almost every X. That is the condition for the
20:00:700Paolo Guiotto: integral variable… the variable of the integral, and this can be done for every value C real. There is no particular issue.
20:13:00Paolo Guiotto: Second, we need a bound for the derivative, no? The bound that must be independent of the parameter.
20:23:90Paolo Guiotto: Now, since this is the value of
20:26:870Paolo Guiotto: The derivative, we got the sine…
20:30:30Paolo Guiotto: CX divided the X, 1 plus X squared.
20:36:200Paolo Guiotto: Now, a temptation could be say that this is less than 1 for sine, then we leave the modules of X 1 plus X squared.
20:45:740Paolo Guiotto: which is independent of the parameter, which is integral at plus infinity, minus infinity, but it is not integral at zero. So this thing is not, strictly speaking, an L1 function in real time, so we have to do better.
21:01:440Paolo Guiotto: Now, the better is to remind that the modulus of sine t is controlled by T.
21:10:70Paolo Guiotto: So, I can say that this is less or equal than modulus CX, so modulus C modulus X, divided by modulus X1 plus X squared. This counts as this one, so eliminates the problem at the origin, but still keeps that C inside.
21:29:150Paolo Guiotto: So now, if I want to have a bound uniform in C, I need to bound C. So I can say that this is controlled by, I don't know, R divided 1 plus X squared, so I will call this function, say, GRX.
21:46:900Paolo Guiotto: which is now an L1 function in the real liner.
21:50:990Paolo Guiotto: And this holds, of course, again, almost every X in R, but not for all C, but for every C,
22:00:710Paolo Guiotto: in the interval minus RR with this R that can be fixed there, but can be taken a bit really large. So the theorem can be applied on this set.
22:12:760Paolo Guiotto: the differentiability. So, if you replace here the interval minus RR, it's still valid. So you have that on that interval, you have the differentiability and the control of the derivative. So it means that
22:29:520Paolo Guiotto: there exists the derivative with respect to C of capital FC, and it is equal to the integral of the derivative, so in particular, that function, sine CX divided X1 plus
22:43:460Paolo Guiotto: X squared… This for every C in the interval minus RR.
22:51:790Paolo Guiotto: But now, since this R is arbitrary, since… Jeez.
23:01:570Paolo Guiotto: And… positive.
23:05:580Paolo Guiotto: Real?
23:07:440Paolo Guiotto: This means that, in fact, that derivative holds for every C, no? Because imagine, this is the real line. You pick XC wherever you want, you pick an error which is big enough, you apply the theorem on that interval, minus RR,
23:23:840Paolo Guiotto: And you have that at York C, the function is differentiable. So the conclusion is that there exists the derivative with respect to X of F at every point C of the real line.
23:36:310Paolo Guiotto: And therefore, the domain for the derivative that set D prime, which is in principle a subset of the domain of the function, is, again, the domain of the function R.
23:50:810Paolo Guiotto: Question 3, we have to repeat once more this to compute the second derivative.
23:57:830Paolo Guiotto: Again,
24:01:400Paolo Guiotto: applying,
24:06:210Paolo Guiotto: differentiation beyond the… Integral sine…
24:14:310Paolo Guiotto: We will have that. The second derivative with respect to X of the function f
24:19:830Paolo Guiotto: will be the integral on R, Okay, off.
24:26:330Paolo Guiotto: D derivative with respect to C of the previous derivative, no?
24:31:170Paolo Guiotto: the derivative respect to T of F, X, C.
24:37:950Paolo Guiotto: Now, provide.
24:43:30Paolo Guiotto: Number 1.
24:45:940Paolo Guiotto: There exists this derivative.
24:49:500Paolo Guiotto: Of the derivative we computed above.
24:54:720Paolo Guiotto: This is the derivative with respect to P of, now we have sine…
24:59:100Paolo Guiotto: CX divided by X1 plus X squared.
25:06:170Paolo Guiotto: Apart from the usual problem at X equals 0, where we cannot even write a function, we have 4x different from 0. This is cosine CX
25:19:520Paolo Guiotto: times the derivative of the argument with respect to X, it is X, divided by X1 plus X squared.
25:29:140Paolo Guiotto: So at the end, we have this cos CX divided 1 plus C squared, and this for every valued C where you can write this thing, so…
25:40:160Paolo Guiotto: Second, we need the uniform control of this derivative.
25:50:50Paolo Guiotto: the uniform control in the Bible C. Here, it is much easier than the previous case, because
25:57:530Paolo Guiotto: We just, bound that cosine by 1.
26:03:110Paolo Guiotto: And this now holds whatever is the X.
26:06:920Paolo Guiotto: With this, we choose an L1 function in the real liner.
26:11:360Paolo Guiotto: And this holds for every value C.
26:14:500Paolo Guiotto: And of course, again, almost every X.
26:19:290Paolo Guiotto: So, the theorem applies, and we get that…
26:25:500Paolo Guiotto: the second derivative with respect to C of F exists.
26:29:400Paolo Guiotto: And it is equal to the integral on R of… this quantity, cosine CX,
26:37:70Paolo Guiotto: over 1 plus X squared dx. This, for every real
26:46:580Paolo Guiotto: Now we have the question number 4, which is,
26:50:00Paolo Guiotto: compute the second derivative using the Fourier transform, and then determine the function f.
26:58:410Paolo Guiotto: Wait, for example… could say that, I don't know, cosine theta is EI theta.
27:10:40Paolo Guiotto: plus E minus i theta divided by 2.
27:17:50Paolo Guiotto: Or we could say cos Xia, or cos theta, let's take it this notation.
27:24:810Paolo Guiotto: is the real part of E2Y theta.
27:30:590Paolo Guiotto: I think, yeah.
27:32:450Paolo Guiotto: It's basically… indifferent.
27:37:730Paolo Guiotto: Okay.
27:38:900Paolo Guiotto: So if we take this one, for example, this says that that's the second derivative.
27:46:320Paolo Guiotto: with respect to C of that capital F, is the integral on R of the real part.
27:54:70Paolo Guiotto: of 1 over 1 plus X squared E.
28:00:140Paolo Guiotto: I see X.
28:02:850Paolo Guiotto: Now, real, real, CC is the real part of a sum, it's the sum of the real parts. We can switch with the integral.
28:11:900Paolo Guiotto: We can say a real part of this, and this is the Fourier transform, more or less. It is the Fourier transform of 1 over 1 plus the square.
28:21:520Paolo Guiotto: but evaluated at minus X.
28:27:90Paolo Guiotto: So, since we know that the, that is the full transform of the Cauchy distribution, so it is pi over… this is A is 1, so E minus A1 modulus of minus C.
28:43:990Paolo Guiotto: Now, this quantity is real, so we can suppress the real and say pi minus modulus of C.
28:52:60Paolo Guiotto: So that's the second derivative of this, and now we have to go back to F by doing a two-step integration. So since this is, let's say, F second of C,
29:09:920Paolo Guiotto: What is F prime?
29:15:740Paolo Guiotto: Well, you know that if you differentiate E to minus, we need to do…
29:21:250Paolo Guiotto: e to minus modulus C, the derivative of this is E to minus modulus c, then there is the derivative of this,
29:29:830Paolo Guiotto: Argument, which is, minus the modulus sign, minus sign, or if you want sine, of minus C.
29:42:20Paolo Guiotto: No, I, I needed to do separate…
29:45:350Paolo Guiotto: Maybe there is a way, but I don't. So, F second of X is equal to pi E minus C for XC positive.
29:56:450Paolo Guiotto: pi into C for C negative.
30:00:910Paolo Guiotto: So, F prime of C will be…
30:04:240Paolo Guiotto: Foxy positive, minus pi E minus, C, plus a constant.
30:11:900Paolo Guiotto: And the other case would be pi e toxi plus constant.
30:17:220Paolo Guiotto: In principle, I should say the two constants are different.
30:25:760Paolo Guiotto: Now… How do we determine the constants?
30:34:660Paolo Guiotto: Well, we know that this function
30:41:20Paolo Guiotto: has a derivative, so it must be continuous. You notice that this Foxy positive, this Foxy negative.
30:49:160Paolo Guiotto: So, the derivative from the right is…
30:54:950Paolo Guiotto: K minus pi, the derivative from the left.
31:04:970Paolo Guiotto: Well, this is another K.
31:08:530Paolo Guiotto: K tilde class pi.
31:12:280Paolo Guiotto: The two values must be the same.
31:16:20Paolo Guiotto: So… The difference between these two is 2 pi.
31:25:910Paolo Guiotto: But this is not yet enough.
31:35:730Paolo Guiotto: What is F'?
31:38:750Paolo Guiotto: F prime is here.
31:43:490Paolo Guiotto: Well, we… we may use something here.
31:48:510Paolo Guiotto: That is, if I put X equals 0,
31:52:380Paolo Guiotto: It's more… a little bit better. If you put X equals 0, you see that sine is 0.
31:58:460Paolo Guiotto: So the integral is integral of zero, you get 0. So it's better if you use the condition. So, since,
32:05:900Paolo Guiotto: F prime of 0 equal integral on r of sine 0x divided X1 plus X squared.
32:18:100Paolo Guiotto: That's, you know…
32:20:50Paolo Guiotto: So, we must have that F prime of C has be equal… when you put X equals 0,
32:30:810Paolo Guiotto: In these formulas, you get, in the first line, you get K minus P that must be 0, so K must be equal to pi.
32:39:260Paolo Guiotto: So it is minus pi, E minus C, plus by…
32:46:760Paolo Guiotto: And downstairs, you get the opposite pi, E to C minus pi. That's for C positive, and that's for C negative.
32:57:140Paolo Guiotto: Now we can go to F.
32:59:860Paolo Guiotto: So F of C will be, this is the pi e to minus c, because the derivative of this is this one.
33:08:840Paolo Guiotto: plus pi c plus constant.
33:14:410Paolo Guiotto: And the second line is pi. It took C minus pike C.
33:19:600Paolo Guiotto: plus constant, that's FOC C positive, FOC C negative.
33:31:240Paolo Guiotto: And maybe, also, here we have some particular value.
33:36:30Paolo Guiotto: The function F, you see that when C is set, we set to 0.
33:40:690Paolo Guiotto: 1 minus cos, cos of 0 is 1, the numerator is 0. So again, we already noticed that F was 0. And since,
33:51:550Paolo Guiotto: F of 0 must be equal to 0. We get f of x equal…
33:58:120Paolo Guiotto: So, pi E minus C plus pi X. When you put this equal to 0, you get Bye!
34:09:910Paolo Guiotto: Pi, and so this must be minus pi to get 0.
34:15:330Paolo Guiotto: Second line is pi e tox c minus pi c.
34:20:950Paolo Guiotto: So when X is easy, we get pi again, minus pi here.
34:26:120Paolo Guiotto: So we can, unify this Fox C positive, this Fox C negative.
34:31:900Paolo Guiotto: So, the pi exponential… no, this is plus.
34:37:470Paolo Guiotto: The exponential is just e to minus modulus of C, right?
34:43:530Paolo Guiotto: Then you see that we have piC positive minus piCOC C negative. This is pi modules of C.
34:53:340Paolo Guiotto: And then we have, minus 5.
34:56:510Paolo Guiotto: And that's the function f of x.
35:00:420Paolo Guiotto: The value of the function.
35:03:930Paolo Guiotto: If there are no mistakes.
35:07:280Paolo Guiotto: Okay, there is one more, but time is over. I don't know if the room is occupied. Which one is the last one?
35:17:310Paolo Guiotto: Sure, please.
35:18:780Paolo Guiotto: 38… No, this is 35.
35:26:310Paolo Guiotto: This one is on the sequence, do you want to do, or…
35:30:870Paolo Guiotto: Or in any case, I will publish the solution.
35:33:720Paolo Guiotto: So, if you want, we do. I don't know if the room is occupied, but we have the room until 12.30 for sure.
35:43:800Paolo Guiotto: If it is enough that you check the solution, we stop here. If you want to see, you have to tell me, I just…
35:51:30Paolo Guiotto: Quickly continue.
35:54:650Paolo Guiotto: Okay, very good.
35:59:150Paolo Guiotto: Okay, so, I'm sorry for that exercise. Of course, it won't, happen in, the…
36:08:970Paolo Guiotto: In the exam, you will have only at one Booleans 4.
36:18:160Paolo Guiotto: Okay.
36:20:450Paolo Guiotto: Let's stop here.