AI Assistant
Transcript
00:00:00Paolo Guiotto: Time to start.
00:03:120Paolo Guiotto: Could you confirm that you received me well?
00:08:189Parvin Yavari: Yes, we hear you, and there's not a.
00:10:340Paolo Guiotto: Okay.
00:11:490Paolo Guiotto: Thank you.
00:14:910Paolo Guiotto: Okay, so let's start… And, we start finishing the exercise. We,
00:24:80Paolo Guiotto: We started yesterday, so let me quickly review the exercise.
00:28:650Paolo Guiotto: We, were computing the integral on a certain domain D of the function X times Y.
00:36:940Paolo Guiotto: The main D is defined by these conditions, X times Y between 1 and 3, and Y between X and 3X. And there is a hint that says, use the change of variable UV equal XY,
00:54:390Paolo Guiotto: Y over X. So, U equal X times Y, and V equal Y over X.
01:00:260Paolo Guiotto: We did this, so we described the domain in the Cartagian plane. We noticed that domain is closed and bounded. It's the red region you see here in the figure.
01:15:480Paolo Guiotto: And then we describe the domain D in the new coordinates, UV,
01:24:980Paolo Guiotto: And we found that in the new coordinates, the domain becomes U between 1 and 3, and V between 1 and 3, so this is the image of the set D through the map that defines the change of variable, which is the map you see here.
01:44:480Paolo Guiotto: Now, to compute the integral, we apply the change of variables formula.
01:51:180Paolo Guiotto: In this case, differently from the cases where we used the polar, spherical, and so on coordinates, where in those cases we had already this map.
02:04:230Paolo Guiotto: So the, starting coordinates function of the new coordinates
02:10:550Paolo Guiotto: In this exercise, we have rather this format. New coordinates function of initial coordinates. So, you see that the map is defined this way. You have these new coordinates are functions of these old coordinates, X and Y. So, it is a lot important to have clear what we have, if it is the map
02:35:100Paolo Guiotto: phi or the phi minus 1, because you have to put the appropriate functions here and there.
02:42:520Paolo Guiotto: So, we have to transform our integral only into an integral on the pair UY, so we say that this becomes an integral. We have here the formula.
02:56:510Paolo Guiotto: The change of variable formula.
02:59:710Paolo Guiotto: So this is the domain in terms of U and V, so this is the phi of D. Then we have the function. The function was just X times Y, the starting function.
03:13:750Paolo Guiotto: Perhaps it's breathing here, yeah. That's the function. And since in the new variables X times Y is just U, this is what happens to the function when we rewrite the function in the new variables.
03:27:660Paolo Guiotto: Now, the quantity that remains to compute is the modulus of the determinant of the phi minus 1 prime.
03:35:470Paolo Guiotto: So what we did, was to compute the phi minus 1,
03:40:910Paolo Guiotto: And this is the phi-1. As you can see here, this is the map that gives XY, which are the initial coordinates, functions of UV, which are the new coordinates. So if you think to the case of polar coordinates, it's exactly the kind of
03:58:690Paolo Guiotto: map now, because here, in polar coordinates, you have U equal raw and V equal theta.
04:04:250Paolo Guiotto: So, I told you to finish the exercise, so let's see what you have done. I want to finish also to show you something about the way to compute this determinant. So, let's now do directly, since we have phi minus 1, phi minus 1
04:20:720Paolo Guiotto: prime, is a function of UV, is the Jacobian matrix, so it's the Jacobian matrix.
04:31:40Paolo Guiotto: of phi minus 1 in variables UV. So this is what? Phi minus 1 is the map you see here. Well, let me just copy down here. Phi-1 of UV is a map that maps R2 into R2.
04:50:570Paolo Guiotto: And the first component is root of U over V, the second component is root of UV.
04:58:740Paolo Guiotto: So these are the two components.
05:00:930Paolo Guiotto: And, so the Jacobian matrix, since phi minus 1, as well as for phi maps are 2 to a 2, this is a 2x2 matrix.
05:10:850Paolo Guiotto: Well, now, the derivatives are respect to U and respect to V.
05:15:580Paolo Guiotto: So we have the first component is this one. We have to do derivative with respect to U and derivative with respect to V. I would prefer to write in terms of powers, because it's much better for the calculation. So I write to U to 1 half
05:32:480Paolo Guiotto: V to minus 1 half, and here U1 half V1 half. This is the same, but written in terms of powers. So when we do this, the derivative with respect to U is 1 half u to 1 half minus 1, so minus 1 half.
05:50:710Paolo Guiotto: Then there is V to minus 1 half, which is just a constant for U.
05:55:320Paolo Guiotto: Derivative of this first component with respect to V lets me move a little bit this.
06:01:780Paolo Guiotto: so…
06:04:200Paolo Guiotto: derivative of the first component, still this one with respect to V, we have U to 1 half, then the derivative with respect to V is minus 1 half times V to minus 1 half minus 1 minus 3 half.
06:19:730Paolo Guiotto: Now, the second component is this one, and we have to do the same, the two derivatives with respect to U and V.
06:26:170Paolo Guiotto: So the derivative with respect to U is 1 half U2 minus 1 half times V to 1 half.
06:33:860Paolo Guiotto: And the derivative with respect to V is U2, 1 half times 1 half
06:40:140Paolo Guiotto: V to 1 half minus 1, so minus 1 half.
06:43:930Paolo Guiotto: So that's the Jacobian matrix of phi minus 1. Now we have to do the determinant of this matrix, so determinant of phi minus 1 prime.
06:55:710Paolo Guiotto: as function of UV, and then we have to take the absolute value, which is the quantity we have into the integral. You have the modulus of the determinant of the few minus 1 prime.
07:08:660Paolo Guiotto: Now, the phi minus 1 prime is written there, so we have modulus of
07:13:520Paolo Guiotto: So we have this times this, so 1 fourth…
07:18:490Paolo Guiotto: U minus 1 half times U1 half gives U to 0, 1, and V minus 1 half V minus 1 half gives V to minus 1. Then we have minus this.
07:31:570Paolo Guiotto: Minus. There is a minus here, the minus in front of this, so this will yield a plus here. 1 half, 1 half, 1 fourth, again, then U1 half times U minus 1 half, again, U to 0, 1.
07:50:210Paolo Guiotto: And V minus 3F plus V times V, 1 half is minus 3F plus 1 half is, again, V minus 1.
07:58:650Paolo Guiotto: So, at the end, we have modulus, 2 times this, so 1 half V minus 1.
08:05:720Paolo Guiotto: Now, remind that this V
08:08:610Paolo Guiotto: This V, this V is Y over X, and remind that in our domain, or if you want, you can see also here, V is greater than 1. So, in the domain, V is positive.
08:24:710Paolo Guiotto: And so we have 1FV-1, or if you prefer, 1 over 2V, because V is between 1 and 3, so the absolute value is positive. So this is the quantity you have to put into the integral. So we can say that the integral on the of F, X, Y
08:44:790Paolo Guiotto: DX, DY… Well, the function was X times Y, so let's put the right function.
08:53:200Paolo Guiotto: becomes the integral on the new domain, which is U between 1 and 3, and V between 1 and 3.
09:02:320Paolo Guiotto: Of, this was you.
09:08:180Paolo Guiotto: U times… now there is the modulus of determinant, which is 1 over 2V.
09:13:930Paolo Guiotto: DUDV. And this is the new integral to be computed, which you can see it's pretty easy, because we can carry this outside, so we have 1 half.
09:24:850Paolo Guiotto: And then we apply the reduction formula for the parameterization. There is no problem, because you can say, for example, let's start integrating in U, and then we do the second integration in V. U will be between 1 and 3, when V will be between 1 and 3. The function is U over V.
09:42:810Paolo Guiotto: Because I put the one half outside.
09:45:460Paolo Guiotto: So now this goes outside, because it's a constant for this integration, and therefore we have 1 half
09:53:960Paolo Guiotto: You also may notice here that this integral in U does not contain anymore the V variable, so I can put outside. So at the end, I get the product between the integrals between 1 and 3 of 1 over V.
10:08:270Paolo Guiotto: times the integral between 1 and 3 of UDU, which are two very simple integrals.
10:15:850Paolo Guiotto: So here we have log of absolute value of V between 1 and 3 times U squared over 2 between 1 and 3.
10:26:470Paolo Guiotto: So the conclusion is 1 half log of 3 minus log of 1, which is 0, so log of 3.
10:36:140Paolo Guiotto: times, here we have 1 half times U square, when U is 3 is 9, minus U square, when U is 1 is 1. So, 8 half, which is 4, divided 2 is 2, so the end is
10:53:600Paolo Guiotto: 2 logarithms of 3. And this is the final result.
10:59:430Paolo Guiotto: Have you obtained something similar or not?
11:06:810Paolo Guiotto: You can… you can just put… post the message in the chat, just to tell me if you obtained this.
11:15:910Paolo Guiotto: Meanwhile, that you respond to this question, I will do this remark, which is…
11:25:200Paolo Guiotto: An alternative and sometimes useful way
11:29:580Paolo Guiotto: to compute this determinant here, okay?
11:34:220Paolo Guiotto: So, this is an alternative.
11:41:500Paolo Guiotto: calculation… off.
11:44:710Paolo Guiotto: modulus determinant phi minus 1 prime. But in this case, maybe it could have been helpful.
11:54:100Paolo Guiotto: Now.
11:55:110Paolo Guiotto: to compute this determinant, in principle, I need first to compute phi minus 1, then to compute its Jacobian matrix that you see here, and then to do its determinant. So I have to do these three steps.
12:13:740Paolo Guiotto: So, the first step was here.
12:16:690Paolo Guiotto: So, taking your change of variable and explicit XY function of UV,
12:23:840Paolo Guiotto: That's something we have done. Then we have done the calculations of the Jacobian matrix, and then finally we computed the term. So, it seems to be a very straightforward way, but with long calculations. We may notice here the following remarkable fact.
12:43:740Paolo Guiotto: Now, we start noticing that
12:51:660Paolo Guiotto: Since phi and phi-1 are each the inverse matrix of the other, so we can write something like this, phi composed with phi-1,
13:03:330Paolo Guiotto: Well, let's to, to write, this is more general than this example, so let's use generic notations. So let's say that this is the variable Y for phi minus 1.
13:16:160Paolo Guiotto: Well, since this composition is the composition between a function and its inverse, the result is the identity. So, we will get Y, that's for every Y.
13:29:420Paolo Guiotto: So, in particular, when you compute the derivative.
13:36:80Paolo Guiotto: you have this. If you do the derivative of the right-hand side.
13:42:220Paolo Guiotto: So, the Jacobian metrics of this map
13:45:570Paolo Guiotto: This map is the identity, so let's call identity. Identity of Y is the function that gives Y, that associates simply Y to Y, to itself. Now, the identity has a Jacobian matrix.
14:05:430Paolo Guiotto: that we can easily determine, because if identity of Y is Y, it means that the components of the identity are the components of the vector Y. So let's say Y1
14:17:320Paolo Guiotto: Y, YM, where this is the dimension of the space you are considering.
14:25:60Paolo Guiotto: Now, the Jacobian matrix of this identity, so the derivative of identity, is a matrix which is a NEM
14:35:530Paolo Guiotto: by M matrix, where here you have the gradients of the component. So I have to take the gradient of the first component, which is Y1, the gradient of the second component, which is Y2, etc, the gradient of the last component, which is YM.
14:52:160Paolo Guiotto: Now, remind that these gradients are arrays of partial derivatives, so there will be the derivative with respect to the first variable, derivative with respect to the second variable, etc, until we finish the variables. So this means that…
15:09:250Paolo Guiotto: If we look at the lines of this matrix, so what is the gradient of this variable, Y1?
15:16:980Paolo Guiotto: Are you able to write what are the components of this gradient here?
15:23:840Paolo Guiotto: I see that nobody is answering me, that's not a good thing.
15:35:570Paolo Guiotto: But do you hear me? Yes or no?
15:38:820Parvin Yavari: Yes, Professor, you're hearing you. I guess at the end, we get a diagonal matrix, because at each term, we will have the partial derivative of that specific term.
15:52:220Paolo Guiotto: Yeah, exactly. So, in fact, if we do the derivatives with respect to Y1, Y2, which are the variables, YM,
16:02:10Paolo Guiotto: Here, the derivative of Y1 with respect to Y1 is 1. The derivative of Y1 with respect to Y2 is 0, and so on for all the other components.
16:13:920Paolo Guiotto: For the second line, you have to do the derivatives of Y2 with respect to all the variables. So when you differentiate with respect to Y1, you get 0, respect to Y2, you get 1, and all the other components are 0.
16:26:510Paolo Guiotto: So, in other words, when you do all these calculations, you will discover that this is a diagonal matrix with ones on the diagonal, so it's the identity matrix. So, IM, which is the identity…
16:42:710Paolo Guiotto: metrics, of RN.
16:49:380Paolo Guiotto: Okay, so this is the right-hand side of this identity. Now, what is the left-hand side of the identity? It is this composition. And what is the derivative of this composition, phi composed by phi minus 1 of Y?
17:07:960Paolo Guiotto: Now, if these are one variable functions, this is a composition, a chain, so we have the chain rule.
17:19:349Paolo Guiotto: And we would write, it is the derivative of phi, so phi prime, evaluated at the phi minus 1 of Y,
17:27:310Paolo Guiotto: times the derivative of the inner function, which is phi minus 1, so phi minus 1 prime y.
17:35:260Paolo Guiotto: Well, the same formula holds for, vector K, for the vector Ks, but be careful that now.
17:44:220Paolo Guiotto: These two objects are matrices.
17:47:760Paolo Guiotto: And these are the Jacobian metrics.
17:54:80Paolo Guiotto: of, in this case, phi.
17:58:70Paolo Guiotto: Here, phi, and this is phi minus 1. So these are matrices, and since this map, so phi and phi minus 1 goes from RM to RM, this is a name, sorry, an M.
18:16:410Paolo Guiotto: by N metrics, and also this is an M by M matrix.
18:21:390Paolo Guiotto: So the product of these two makes sense.
18:24:890Paolo Guiotto: Okay, so this means that we have this remarkable identity, that
18:30:500Paolo Guiotto: phi prime, so the Jacobian matrix of phi evaluated at phi minus 1Y, Times phi minus 1 prime.
18:41:300Paolo Guiotto: of Y.
18:43:60Paolo Guiotto: It is equal to the identity of RM.
18:47:730Paolo Guiotto: Because if this is the identity, by doing the derivative, at both sides, we get the same.
18:54:240Paolo Guiotto: At right-hand side, we obtain the identity, as we checked. At left-hand side, we get this one by the chain rule. So, the two derivatives must be the same, because they come from an identity, and we get this.
19:08:50Paolo Guiotto: So, in particular, as you can see, this guy is the guy we are interested in, because this is the matrix that appears in the change of variable formula, you see?
19:20:400Paolo Guiotto: The change of variable formula here contains the determinant of the phi minus 1 prime.
19:25:920Paolo Guiotto: This relation says that… This matrix, the phi… oh, sorry, the phi minus 1
19:34:340Paolo Guiotto: prime, the Jacobian matrix of phi minus 1,
19:38:410Paolo Guiotto: It is what? Well, I cannot write 1 over, because these are matrices, but when I multiply by the inverse matrix of this times the identity, I get that this is D phi prime at phi minus 1Y,
19:55:880Paolo Guiotto: inverse metrics. So, the relation between the two is that this Ease. D.
20:05:380Paolo Guiotto: Inverse… metrics… off… this.
20:14:430Paolo Guiotto: So this is the relation between these two mattresses. And therefore, if there is this relation, the determinant of the matrix phi minus 1 prime, which is the quantity.
20:27:140Paolo Guiotto: we are interested in, because of the change of variable formula, will be the determinant of this matrix. But you know what is the relation between the determinant of a matrix and determinant of its inverse?
20:43:550Paolo Guiotto: The relation is that the determinant is the reciprocal, so this is 1 over the determinant of phi prime evaluated at phi minus 1Y.
20:55:370Paolo Guiotto: So now, this remarkable formula…
20:58:470Paolo Guiotto: I'm doing once because this is general, okay? It's not depending on that specific change of variable. Test that you could compute this quantity, that is, the quantity that appears in the change of variable formula, by computing the reciprocal of this quantity.
21:18:550Paolo Guiotto: Let's see the practical example in the previous case. So, for example… in the… previews… exercise. We have.
21:33:790Paolo Guiotto: So the direct map fee… Fee, XY,
21:40:70Paolo Guiotto: is UV equals phi XY. That was defined as X times Y and Y over X, right? So this is the direct map.
21:50:510Paolo Guiotto: So to compute the determinant of the inverse, the Jacobian matrix of the inverse, we could invert this relation, then do the Jacobian matrix, then do the determinant, and it is what we have done above. But now I want to do the same calculation by choosing the right-hand side of this identity. So what I do is…
22:09:550Paolo Guiotto: determine… compute first the Jacobian matrix of phi. Now, the Jacobian matrix is in X and Y, so it will be still a 2x2 matrix, but now the derivatives are with respect to the variables of phi, which are X and Y.
22:24:510Paolo Guiotto: This is the first component, so we have to do the derivatives of this with respect to X and with respect to Y.
22:31:80Paolo Guiotto: When we do the derivative with respect to X, we get Y. When we do the derivative with respect to Y, we get X.
22:38:70Paolo Guiotto: And the second component is Y over X. When we do the derivative with respect to X, we get, this is X to minus 1, so minus Y over X squared.
22:49:210Paolo Guiotto: And when we do this derivative with respect to Y, we get 1 over X.
22:53:770Paolo Guiotto: So, as you can see, it is, in this particular case, it is, let's say, simpler than this one, which is a little bit dangerous because of these exponents, it's very easy to do a mistake. While here.
23:08:700Paolo Guiotto: it seems to be much easier. Now, let's compute the determinant of E prime.
23:16:50Paolo Guiotto: at point XY,
23:17:970Paolo Guiotto: then we will have to compute what does it mean, this thing, okay? Let's take a second here. So the determinant of C prime is the determinant of the matrix you see here, which is Y over X minus X times minus Y over X squared.
23:37:960Paolo Guiotto: As you can see, we have a simplification, this with this, this we come plus.
23:43:410Paolo Guiotto: So at the end is 2 times Y over X.
23:48:00Paolo Guiotto: And now, what is the determinant of phi prime evaluated… sorry, phi prime?
23:55:580Paolo Guiotto: Evaluated at phi-1 UV.
24:00:100Paolo Guiotto: Well, this is, this is the point with the X and the Y that corresponds to U and V. And you remind that in the position, we have that U is just X times Y, and V is Y over X. So if you see this formula, this Y over X is just V, so I get that this is 2V.
24:21:810Paolo Guiotto: in, variable COB.
24:23:980Paolo Guiotto: So, finally, I used this identity, and they said the determinant of phi minus 1 prime in the variables UV, determinant of phi minus 1 prime in the variables UV, is 1 over the determinant of phi prime.
24:43:810Paolo Guiotto: of phi minus 1 UVE.
24:47:210Paolo Guiotto: But the quantity that we have done here has been computed, it is equal to 2V1 over 2V. And look what we obtained.
25:01:280Paolo Guiotto: It is exactly the same, of course.
25:06:90Paolo Guiotto: Okay?
25:08:90Paolo Guiotto: So, now, it's just, say, this is just a possibility to… an alternative, possibility to compute this determinant. It is… there is not a rule, it's not always better to doing this way or the other way. It depends on the kind of calculations we have to do.
25:30:900Paolo Guiotto: Okay, so let's see if we have understood anything. Let's repeat the calculation with the exercise number 2. So we are here integral on Y squared.
25:41:240Paolo Guiotto: DX, the Y… Domain D, is the set of points X, Y,
25:52:90Paolo Guiotto: with X times Y between 1 and 2.
25:57:900Paolo Guiotto: And the second condition is XY squared between 1 and 2.
26:04:960Paolo Guiotto: This is the domain D.
26:07:190Paolo Guiotto: And it is said that, use… The change of variable.
26:15:540Paolo Guiotto: that says U equal X times Y and V equal X times Y squared. This is the…
26:24:280Paolo Guiotto: Hint, in this case.
26:28:710Paolo Guiotto: So let's try to calculate this without doing figures.
26:33:410Paolo Guiotto: Okay, so, the change of variable is given in the form I have new coordinates, UV, function of all the coordinates, XY. So, again, going back to,
26:48:60Paolo Guiotto: Sorry, I would prefer to put this in the new, page, so… just one second.
26:56:490Paolo Guiotto: How it works, the copy and paste, I never remind this thing.
27:07:250Paolo Guiotto: What's that? No.
27:09:30Paolo Guiotto: This is just moving… where is it? Copy and paste, insert, maybe. No.
27:17:20Paolo Guiotto: What stupid software… Okay,
27:25:700Paolo Guiotto: Well, let me… I do not remind what is the… The problem with the finger.
27:38:950Paolo Guiotto: Oh, shh…
27:42:230Paolo Guiotto: I canceled very well. Okay, so let's just copy, again. Number 2. So we have integral on D of, Y squared dxty y.
27:57:60Paolo Guiotto: So the domain D is a set of points, XY,
28:01:850Paolo Guiotto: with, XY between 1 and 2, and,
28:07:370Paolo Guiotto: Also XY square between 1 and 2.
28:13:20Paolo Guiotto: So it says to use this change variable, u equal X times Y, and V equal XY square, okay?
28:22:520Paolo Guiotto: So…
28:25:30Paolo Guiotto: So, as you can see, the change of variable is in the form of new variables, function of the old one. So, this is the case, we have the map phi naught, phi minus 1.
28:37:130Paolo Guiotto: So the change of variable will be, in general, integral on D, Y squared dx, dy, equal the integral on the image through the map fee of the domain D.
28:49:930Paolo Guiotto: The function is, so this is F, here we have F of phi minus 1,
28:57:990Paolo Guiotto: UV. Then we have the modulus of the determinant of phi minus 1 prime.
29:04:860Paolo Guiotto: UV… in DUDV.
29:10:130Paolo Guiotto: Okay?
29:12:780Paolo Guiotto: Now, let's see the ingredients. So, first, let's determine what is the condition, what are the conditions for UNV. So, I have that, UV
29:29:810Paolo Guiotto: belongs to phi of D.
29:34:820Paolo Guiotto: So, if and only if UV is phi of XY, so this means X times Y and X times Y squared.
29:47:980Paolo Guiotto: 4…
29:49:860Paolo Guiotto: XY that are in D. But XY in D means, if we look at the conditions, means X times Y between 1 and 2, and X times Y squared between 1 and 2 as well.
30:09:00Paolo Guiotto: And these are exactly U and V. So we read if and only if U is between 1 and 2, and V is between 1 and 2.
30:22:440Paolo Guiotto: So what are the U and V that are in phi of D, those for which these two conditions are verified.
30:30:890Paolo Guiotto: Second, what is the function in the new coordinates? So, here, I should compute this map, because I need to express what is Y as function of UV.
30:45:500Paolo Guiotto: So, as you can see, we have that U, and this will be useful also to determine phi minus 1. U is X times Y, V is XY squared.
30:57:160Paolo Guiotto: So, if you want to extract U and V, what could you do? For example, you could divide the first by the second, so you see that V divided by U is XY squared divided XY.
31:13:530Paolo Guiotto: which yields, of course, X is simplified, Y.
31:18:230Paolo Guiotto: disappear, and so we have Y equal V over U.
31:23:900Paolo Guiotto: Then we will also have an X.
31:27:320Paolo Guiotto: X can be determined, for example, by this first relation, is U over Y, but Y is V over U, so this will be U square over V. So these are the conditions. So if you want UV,
31:47:40Paolo Guiotto: is phi of XY And this is this relation. If and only if XY is phi minus 1.
32:00:960Paolo Guiotto: of UV. And this is these relations. So let's write better, if and all if X is equal to U squared divided V,
32:13:720Paolo Guiotto: And Y is V divided U. So, this is the map V minus 1.
32:23:580Paolo Guiotto: So, in particular, this means that…
32:26:140Paolo Guiotto: this function, the function is Y squared, we have to put in place of y this quantity, so it will be V of U squared.
32:36:970Paolo Guiotto: So we can say that integral on D of Y squared, D, X, DY, is transformed into.
32:46:960Paolo Guiotto: So we say that the domain phi of D is given by these conditions, so U between 1 and 2, and V between 1 and 2 as well.
33:01:630Paolo Guiotto: The function is Y squared, Y is V over U, so it will be V over U squared.
33:07:840Paolo Guiotto: And then now we have the modulus of the determinant of the phi minus 1 prime UV.
33:16:390Paolo Guiotto: Dio Divi.
33:18:920Paolo Guiotto: Now, to compute this model's determinant of the Jacobian matrix of phi minus 1, we have phi minus 1, we could compute this, the Jacobian matrix of this map.
33:31:520Paolo Guiotto: Or, in alternative, you could take this, which is the direct map, and use the trick we have seen in the previous remark. Let's do that way.
33:41:780Paolo Guiotto: We noticed that.
33:48:740Paolo Guiotto: the phi prime of XY, phi is a function of XY, so phi prime will be a function of XY,
33:56:420Paolo Guiotto: We use a Jacobian matrix, in this case a 2x2 matrix, of the direct map fee, which is this one. This is the map fee.
34:05:200Paolo Guiotto: So the two components are X times Y and X times Y squared. I have to do the derivatives of the two components, so if you want the gradient of X times Y, this is the first line, this is the gradient of X times Y squared, respectively, respect to the derivatives, which are derivatives with respect to X and derivative with respect to Y.
34:28:500Paolo Guiotto: So we get here, derivative with respect to X of XY is Y. Derivative with respect to Y of XY is X. Second line, derivative with respect to X of XY squared is Y squared. Derivative with respect to Y of X times Y squared is 2XY.
34:49:90Paolo Guiotto: So, the determinant…
34:52:449Paolo Guiotto: of the phi prime is equal to, as you can see, 2XY squared minus XY squared, so at the end, XY squared. So this is the determinant of phi prime, okay? Now.
35:10:390Paolo Guiotto: Seen, sir.
35:12:490Paolo Guiotto: The determinant of the inverse, which is the quantity we are really interested in, so the determinant of the
35:20:490Paolo Guiotto: Jacobian matrix of the inverse in variables UV, it is equal to 1 over the determinant of the direct map, so the Jacobian of the direct map phi prime.
35:32:690Paolo Guiotto: at phi minus 1 of UV,
35:36:910Paolo Guiotto: And this determinant is written here, it is XY squared. That, in coordinates UV, incidentally, it is exactly equal to V, as you can see.
35:49:760Paolo Guiotto: So this is V.
35:52:10Paolo Guiotto: So this means that we have 1 over V,
35:55:910Paolo Guiotto: And therefore, the modulus of the determinant of V minus 1 prime UV
36:03:390Paolo Guiotto: will be the modulus of 1 over V.
36:08:640Paolo Guiotto: But remind that on our domain, in particular, V is between 1 and 2. So, since V is greater or equal than 1, it is positive, this means that we get 1 over V.
36:24:20Paolo Guiotto: And so we have now all the ingredients to write how this change of variable formula looks like. So let's put a star here, and we continue down here.
36:37:210Paolo Guiotto: So we have an integral on U between 1 and 2.
36:42:320Paolo Guiotto: V between 1 and 2.
36:45:760Paolo Guiotto: It was V over U square.
36:52:610Paolo Guiotto: times the modulus of the determinant, blah blah blah, is 1 over V, DU, DV.
36:59:990Paolo Guiotto: And now we can simplify. We have integral u from 1 to 2, and V from 1 to 2 as well.
37:09:630Paolo Guiotto: Then we have V square over V is V divided U squared du dv.
37:16:550Paolo Guiotto: Now, this is very easy. We can apply the reduction formula.
37:21:850Paolo Guiotto: and split this into two integrations. It's different which one is the first. We could decide to start with U and finish with V, so the function is V divided U squared. The range for U is 1 to 2 when V is from 1 to 2.
37:38:920Paolo Guiotto: Now, this V is a constant, I can carry outside and put a 1 here.
37:45:240Paolo Guiotto: And now, also, I noticed that this integral is independent of the variable B, so it's a constant for that integration, so everything boils down to this, integral from 1 to 2 of V dV.
37:58:430Paolo Guiotto: times integral from 1 to 2 of 1 over u squared du.
38:05:490Paolo Guiotto: And now it's easy, because this is V square over 2 between 1 and 2 times, this is, what is it? Minus 1 over U, between 1 and 2.
38:19:860Paolo Guiotto: So, the conclusion is… When V is 2, we get 4, so 1 half times 4 minus…
38:30:430Paolo Guiotto: When V is 1, we get 1, 1 times
38:34:160Paolo Guiotto: Here, when U is 2, we get minus 1 half.
38:38:250Paolo Guiotto: Minus… when U is 1, we get minus 1, so plus 1.
38:43:850Paolo Guiotto: So this is another 1 half, this is 3, so 3 over 4, and that's the final result.
38:57:50Paolo Guiotto: Do you have any comment?
39:03:370Parvin Yavari: No, thank you, it was clear.
39:05:470Parvin Yavari: We're at least.
39:09:740Paolo Guiotto: Any other comments?
39:15:280Paolo Guiotto: So, there are.
39:16:950MEHMET EREN ASLAN: Yeah?
39:22:820Paolo Guiotto: No questions?
39:25:210Paolo Guiotto: Okay, there are, do the exercise, 5, 7, 6… Number 3…
39:37:10Paolo Guiotto: Also do the 577, which is… Hmm… Nothing special.
39:46:600Paolo Guiotto: It's, standard, the same nature of what we have seen.
39:50:580Paolo Guiotto: Maybe I will show you just, just for the sake of,
39:56:290Paolo Guiotto: doing an exercise, the 578, which is a little bit more technical. The point is that…
40:05:660Paolo Guiotto: there are a little bit more complicated… I don't know, let's see. So it says, we have this function, F of XY,
40:19:440Paolo Guiotto: It is defined as X to exponent 3 halves divided root of y minus X.
40:27:640Paolo Guiotto: times e to minus X times Y to 3 half.
40:35:730Paolo Guiotto: where XY belongs to this set, D, which is
40:41:940Paolo Guiotto: made of these points. X, Y, in, up to… Well…
40:49:880Paolo Guiotto: 0, less or equal than X, less or equal than Y.
40:55:200Paolo Guiotto: Notice that in particular, Since these exponents 3AF,
41:01:860Paolo Guiotto: are like the cube of the root, and the coordinates XY are positive, basically everything here is defined.
41:12:500Paolo Guiotto: Now, let's, use… the question is, use… Change of variables.
41:24:410Paolo Guiotto: Again, given in the form of new variables.
41:27:880Paolo Guiotto: function of all the variables, so X times Y, X over Y.
41:34:850Paolo Guiotto: to compute.
41:38:730Paolo Guiotto: the integral only of the function f.
41:50:280Paolo Guiotto: Okay.
41:55:260Paolo Guiotto: So this is the change of variable, this is the map phi in the notations of the change of variable formula. So the change of variable formula says that integral on… of a function f on the
42:08:750Paolo Guiotto: becomes the integral on the image through the map C of the domain D of F composed with phi-1 UV,
42:18:90Paolo Guiotto: The modulus of the determinant of phi minus 1 prime UV, Diyo. DV.
42:28:900Paolo Guiotto: So this is the formula. We have to understand what is phi of D, so what is the range for the new coordinates UV?
42:37:880Paolo Guiotto: which is perhaps here less evident, but let's see. What is the function? And then, to compute this. So let's start from the domain. Let's look at this.
42:54:920Paolo Guiotto: So we have that, we have to understand what is the condition for UV. UV belongs to phi of D, if and only if.
43:05:990Paolo Guiotto: UV is phi of some point, XY, that belongs to D.
43:15:780Paolo Guiotto: Now, this means that this is the value of phi is X times Y, X over Y.
43:27:210Paolo Guiotto: Where XY belongs to D, and this means 0 less or equal than X less or equal than Y.
43:37:580Paolo Guiotto: Now, we have to understand what does it mean, this in terms of UV. So, U is X times Y,
43:46:110Paolo Guiotto: And V is X over Y.
43:49:890Paolo Guiotto: Okay.
43:51:990Paolo Guiotto: what can we say? We have to… there is a unique condition. This is the condition, so we have to understand what this condition implies in terms of U and V.
44:04:620Paolo Guiotto: So what could you notice here?
44:09:630Paolo Guiotto: Do you notice anything?
44:25:10Paolo Guiotto: Violence.
44:27:330Paolo Guiotto: Divide by Y. Yes, the Y is positive, so we can say that this is the same of divided by Y. This X over Y is less or equal than 1. So this means…
44:40:370Paolo Guiotto: Since the new variable V is X over Y, this means V between 0 and 1.
44:47:750Paolo Guiotto: And, what about you?
44:53:650Paolo Guiotto: Is there any condition for U… U is the product X times Y.
44:59:550Paolo Guiotto: When asked, why verify this condition here? What could you say about U?
45:07:80Edip Deniz Angi: Can we multiply by Y?
45:12:240Paolo Guiotto: Yeah, but if I multiply by Y, I get, multiplying by Y, 0 less or equal X times Y less or equal Y squared.
45:22:850Paolo Guiotto: So this would mean U between 0 and Y squared, but so this says what?
45:39:140Paolo Guiotto: we would need to express Y in function of UV. How can we get Y function of UV?
45:51:130Edip Deniz Angi: U… U divided by V?
45:54:300Paolo Guiotto: Okay, if you divide by V, well, here you notice that in this domain, everything is positive. Well, when it is zero, these are just spatial cases, but for most of the cases, the coordinates are positive. So we can divide. If we divide, we get…
46:11:820Paolo Guiotto: from this, we get U divided V is equal to XY divided
46:17:890Paolo Guiotto: X over Y, so this X disappear. What remains is exactly Y square, no? So this is…
46:27:500Paolo Guiotto: U over V. But you see what happens? What we get is U is between 0 and U over V. This is the condition we obtain.
46:41:740Paolo Guiotto: What about this condition?
46:48:260Paolo Guiotto: be careful, because I know that V is between 0 and 1, so the number U over V is always true, is always greater than this one, so always…
47:00:310Paolo Guiotto: when V is between less than… is less than 1, between 0 and 1, because U over V will be greater than
47:07:720Paolo Guiotto: 1 over V, 1 over V will be greater or equal 1, so U over V will be always greater or equal U. So this is a fake condition. It's a condition that says it's always verified, and this means that only one condition remains, which is this one. So we can say that…
47:29:220Paolo Guiotto: So X… UV is in phi of D, if and only if V is between 0 and 1, and U is between 0 and whatever. So these are the conditions for U and V, okay? So this is now
47:47:40Paolo Guiotto: the, set phi of D is made by pairs UV, where U is positive and V is between 0 and 1.
47:57:390Paolo Guiotto: Okay, so now we can start to, do the change of variables. So integral on D of F in the XDY becomes integral on… the set is u greater or equal than zero.
48:11:870Paolo Guiotto: V between 0 and 1.
48:15:410Paolo Guiotto: This will be, at the end, an integral in UV.
48:19:340Paolo Guiotto: We have to write the function f in the new variables. We have to… in principle, I should replace to X, where is it, the inverse map.
48:34:570Paolo Guiotto: We… we computed somewhere…
48:41:60Paolo Guiotto: Because we have the Y squared here, I do not… we have not yet done this, okay. But it doesn't matter. We have to look at the function fxy.
48:50:220Paolo Guiotto: which is, what is, x to 3 halves, divided the root of Y minus X, Y minus X,
49:02:550Paolo Guiotto: E to minus XY to 3 half. This is the function, right?
49:10:540Paolo Guiotto: And we have to look at this in variables UV. Remind that the change of variable is… U is equal to X times Y, and V is equal to X of Y.
49:24:180Paolo Guiotto: So, you see that, for example, this X times Y is U.
49:30:580Paolo Guiotto: Okay, so what could we do? For example, if you factorize the Y here, so this is X to 3F, that I can see as X times X to 1 half.
49:48:190Paolo Guiotto: divided. I factorize a Y, and I take out of the root, so Y to 1 half the root of 1 minus X over Y, and this X over Y is now V.
50:03:280Paolo Guiotto: So this is E2 minus U to 3 half.
50:08:500Paolo Guiotto: Right?
50:10:310Paolo Guiotto: So you see that there is another X over Y here, so this is, equal to X times the root of X over Y, which is, again, V.
50:26:720Paolo Guiotto: divided root of 1 minus V, which is down there.
50:34:340Paolo Guiotto: And, exponential of minus U to 3 half.
50:40:790Paolo Guiotto: We are almost.
50:43:230Paolo Guiotto: So this is X root of V divided root of 1 minus V, E to minus U to 3F.
50:54:750Paolo Guiotto: So we needed to see what is X in function of UV.
50:59:790Paolo Guiotto: And this, can be done…
51:05:330Paolo Guiotto: If we… if we multiply the two equations from these two, we have that U times V, it is X times X, X squared, and Y times 1 over Y is 1. So X…
51:21:70Paolo Guiotto: square is UV, so this means that X is either plus or minus the root of UV.
51:27:920Paolo Guiotto: But, remind that in our domain, X,
51:33:260Paolo Guiotto: is positive, so, I have to take the positive root, because X is positive. So, finally, we have equal to… this becomes,
51:47:660Paolo Guiotto: root of UV.
51:51:490Paolo Guiotto: times root of V divided the root of 1 minus V, E minus E to… U to 3 halves.
52:02:110Paolo Guiotto: We can now do some simplification. For example, put the 2 root of V together, and we get V over root of 1 minus VE.
52:17:320Paolo Guiotto: For this other, I cannot do any simplification, so root of UE minus U to 3 half.
52:26:460Paolo Guiotto: And this is the function F, invaluable UV, so it is F of phi minus 1 UV.
52:36:40Paolo Guiotto: So, in other words, instead of computing phi minus 1 and then, composing with F, what I did is, I, is this. I took the function f, and I tried to recognize in the function
52:54:650Paolo Guiotto: the two new variables, U and V, that are defined by this formula. If you don't trust or don't like this, you could compute first V minus 1. So, in alternative, let's do this. So, from UV equal XYX over Y,
53:14:740Paolo Guiotto: I compute the inverse map first, which is something that I could need to compute the Jacobian of phi minus 1.
53:25:520Paolo Guiotto: So, to get the inverse, I repeat what I've done somewhere here. If you multiply the two equations, you get X squared equal UV, and if you divide the two equations, you get Y squared equal
53:44:910Paolo Guiotto: So the first divided by the second, U over V, right? So at the end, we have, since both X and Y are positive.
53:54:340Paolo Guiotto: X equal the root of UV, and Y is the root of U over V. Now, this is the inverse, because this function gives phi minus the coordinates XY function of UV, while this one gives UV function of XY.
54:14:580Paolo Guiotto: So if you now want to compute the F of phi minus 1 UV,
54:20:700Paolo Guiotto: you have to compute F, where instead of X, you put root of UV, instead of Y, you put root of U divided V.
54:31:500Paolo Guiotto: Now, you take your function that is X to 3 half, so you have to take this root of UV, it's better if we write in terms of power, because it's X to 3 half.
54:46:710Paolo Guiotto: So X is UV to 1 half, 1 half.
54:52:280Paolo Guiotto: divided by the root of Y minus X,
54:56:700Paolo Guiotto: So why is U to 1 half divided V to 1 half minus X, which is U to 1 half V to 1 half.
55:10:440Paolo Guiotto: times exponential of minus X times Y, which is U, to 3 halves.
55:19:570Paolo Guiotto: So you see that the reason why this exercise has a plus is because the calculations are…
55:27:260Paolo Guiotto: A little bit, a little bit, tedious.
55:33:470Paolo Guiotto: So… but doing step-by-step, we… we can, we can survive. So, we have U to 1 half to 3 half makes U to 3 fourths. Then we have V to 1 half to 3 half makes V the same, V to 3 fourths.
55:53:590Paolo Guiotto: divided by… I can do the common denominator here, put this V to 1 half outside of the root. This big root is another exponent
56:05:230Paolo Guiotto: one half.
56:07:480Paolo Guiotto: So, when I put outside, I have 1 over V to 1 fourth times U to 1 half minus U to 1 half, and when we do the common denominator, this is V…
56:23:00Paolo Guiotto: to 1… to 1 half.
56:27:660Paolo Guiotto: and E minus U cubed, U to 3 halves.
56:33:340Paolo Guiotto: Okay, now this, 1 over V goes upstairs times V to 1 fourth.
56:41:730Paolo Guiotto: And V to 3 fourth times V to 1 fourth, these two together makes just V. So we have V.
56:50:870Paolo Guiotto: Then I can factorize this U to 1 half, so let's write U3 fourth over U to 1 fourth, because U to 1 half, then there is another 1 half here.
57:04:170Paolo Guiotto: And what remains into the root is 1 minus V, E minus U cubed U to 3F,
57:14:370Paolo Guiotto: Sorry, this is a 3F exponent.
57:20:10Paolo Guiotto: Okay, we are almost done, because now I put upstairs this, this becomes 3 fourths minus 1 fourth, so this is… this exponent is 2 fourths, so which is 1 half. So we have V, U to 1 half divided by…
57:37:160Paolo Guiotto: 1 minus V to 1 half
57:41:310Paolo Guiotto: E2 minus U2 3 halves. I don't see any further simplification. Let's check that we get the same
57:50:940Paolo Guiotto: result obtained above. That's the function. You see V times root of U, and it is V times root of u here, divided by the root of 1 minus V, which is here.
58:06:30Paolo Guiotto: times the exponential u minus 2 3 halves that you have here. So you see that we obtained the same function, okay?
58:15:530Paolo Guiotto: So now this is the function f in the new coordinates. What we need still to compute is the modulus of the determinant
58:23:750Paolo Guiotto: of the phi minus 1 prime in coordinates UV,
58:29:570Paolo Guiotto: We have everything here, because we just computed the inverse map of phi, so if you want, we can proceed directly.
58:38:550Paolo Guiotto: Or we can proceed inversely by doing the Jacobian matrix of phi, and then restoring the lattice UV.
58:50:90Paolo Guiotto: Since, I don't… I don't want to do lots of calculation with those exponents, I use this one of them, the modulus of the tenant of phi prime, phi minus 1 UVE.
59:05:990Paolo Guiotto: Now, let's compute the direct Jacobian.
59:09:10Paolo Guiotto: the determinant of phi prime as function of XY. This is the determinant of the matrix.
59:18:140Paolo Guiotto: Now, phi is written here, this is phi.
59:23:460Paolo Guiotto: So, as you can see, the two components are 1 and 2, and we have to put the two gradients there. So, gradient of X times Y
59:33:460Paolo Guiotto: gradient of X over Y.
59:38:320Paolo Guiotto: So let's finish this exercise, then we do the break. So this comes, the determinant of… derivative with respect to X is Y, derivative with respect to Y is X.
59:49:850Paolo Guiotto: In the second line, we have derivative with respect to X is 1 over Y. Derivative with respect to Y is minus X over Y squared.
59:59:00Paolo Guiotto: So that determinant is minus…
00:03:540Paolo Guiotto: So we get the X over Y from this, then we have minus, this time this, another X over Y, so minus 2X over Y.
00:15:00Paolo Guiotto: So this is the determinant of phi prime.
00:17:950Paolo Guiotto: Therefore, the absolute value of the determinant of phi prime at phi minus 1 UV
00:27:670Paolo Guiotto: It is equal to the modulus of minus 2.
00:31:420Paolo Guiotto: X over Y, let's keep for a second the X and Y, I don't need to put X and Y, because X over Y is already the variable V, no?
00:42:580Paolo Guiotto: So we just replace V. So it is minus 2V in absolute value, so 2 absolute value of V, but this is positive, so 2V.
00:56:20Paolo Guiotto: So finally, we can say that the integral on the… of our function is the integral. So we say that for U, we have a positive for V between 0 and 1, if I'm not wrong.
01:13:540Paolo Guiotto: Yes. Now, we have to put the function and the modulus of the determinant of phi minus 1.
01:21:300Paolo Guiotto: The function is written here, so let's copy.
01:26:360Paolo Guiotto: VU to 1 half over 1 minus V to 1 half
01:33:810Paolo Guiotto: E to minus U to 3 halves, this is the function.
01:38:230Paolo Guiotto: times the modulus of determinant, it is not this one, but it is the reciprocal, it is this one, so 1 over 2V.
01:47:930Paolo Guiotto: So we have a 1 over 2V here.
01:52:170Paolo Guiotto: DUDV. And finally, this is the integral we have to compute.
01:58:890Paolo Guiotto: So we can simplify this fee with this fee.
02:04:220Paolo Guiotto: And therefore, we now apply the reduction formula.
02:09:670Paolo Guiotto: we may decide to integrate. Here, it's rather indifferent, because as you can see, the function is splitted into a product, so… and also the domain is simple, so I can say, let's start with V, so I have to integrate this.
02:27:100Paolo Guiotto: 1 minus V to exponent 1 half.
02:30:430Paolo Guiotto: Then all the other thing is independent of V, so I put outside, this is the factor 1 half. Then I have U to 1 half E minus U to 3 half.
02:41:760Paolo Guiotto: And this is in DV, this will be in the U.
02:45:770Paolo Guiotto: And for you, I have only this, so 0 to plus infinity.
02:52:340Paolo Guiotto: Now, again, you see that this innermost integral.
02:57:860Paolo Guiotto: does not contain anymore U, so it's a constant for the second integration, so this means that we split everything into a product.
03:06:300Paolo Guiotto: Integral 0 to plus infinity of u exponent 1 half E minus U to 3 half.
03:14:390Paolo Guiotto: DU times integral 0 to 1 of 1 over 1 minus V3, not 3F, 1 half DV.
03:28:440Paolo Guiotto: Now, these two are, not particularly complicated, because we have integrals 0 to 1 of 1 over 1 minus V2,
03:39:240Paolo Guiotto: 1 half in the V.
03:42:630Paolo Guiotto: This, more or less, comes from doing the derivative with respect to V of 1 minus V.
03:50:200Paolo Guiotto: to minus 1 half plus one, so to one and a half. In fact, we obtain 1 half
03:55:940Paolo Guiotto: 1 minus V to 1 half minus 1, so minus 1 half, and this is exactly this. Then there is the derivative of the argument with respect to V, which is minus 1. So to have this, I need a minus 1 half.
04:11:630Paolo Guiotto: So I create this minus 1 half here by multiplying.
04:18:360Paolo Guiotto: by minus 1 half, also I have to divide, so it means I multiply outside by minus 2.
04:28:70Paolo Guiotto: So now this is, the derivative of this.
04:33:980Paolo Guiotto: So the integral will be minus 2, evaluation of 1 minus V to exponent, 1 half between V equals 0, V equals 1.
04:46:690Paolo Guiotto: At V equals 1, I get 0. At V equals 0, I get to 1 to 1 half, so the difference is minus 1 to 1 half, which is minus 1, time minus 2, final result, plus 2.
05:03:570Paolo Guiotto: About the other integral, we have integral 0 to plus infinity, due to one path.
05:13:410Paolo Guiotto: E to minus U to 3 half, huh?
05:19:140Paolo Guiotto: Is that right? Yeah.
05:21:30Paolo Guiotto: Well, also here, you should recognize that there is a derivative.
05:25:680Paolo Guiotto: and the derivative of this guy, because if I do the derivative with respect to U of E minus U to 3 half.
05:33:640Paolo Guiotto: I get E minus U to 3 alpha, then there is the derivative of the exponent, which is minus…
05:43:30Paolo Guiotto: 3 halves U to 1 half.
05:47:930Paolo Guiotto: So, if I put a minus 3 half here, I have a derivative, so I need to have a minus 3 half here, that means to have a minus 2 thirds out here.
06:00:120Paolo Guiotto: So this becomes minus 2 thirds, and then we have the evaluation of E minus U to 3 half.
06:09:250Paolo Guiotto: Between u equals 0 and u equals plus infinity.
06:14:350Paolo Guiotto: at plus infinity. You have U at minus infinity, so, you get 0. Minus. At U equals 0, you get e to 0, 1. So, finally, we have… that parenthesis is minus 1 times minus 2 thirds equal plus 2 third.
06:33:640Paolo Guiotto: And therefore, final conclusion, the integral on D of that function f
06:40:550Paolo Guiotto: It is equal to 1 half that we have here.
06:48:100Paolo Guiotto: times the integral of the U1 half E minus U to 3 half is… 2 thirds…
06:56:960Paolo Guiotto: Times the other integral is equal to plus 2.
07:04:230Paolo Guiotto: So, say, 2 thirds.
07:10:550Paolo Guiotto: Okay.
07:12:350Paolo Guiotto: Okay, to be clear, this is much beyond the level required, so I would suggest you to do the 579, just,
07:24:870Paolo Guiotto: As a training, a mental training, not because this is the level that you should acquire.
07:33:340Paolo Guiotto: with these calculations, okay? But the difference between this exercise and the previous one is that we have a lot of long and tedious calculations. Conceptually, there is no difference.
07:47:640Paolo Guiotto: Okay, let's take a break.
07:50:780Paolo Guiotto: Let's do, yeah, it's about 1.
07:55:910Paolo Guiotto: 140, so we… we still would have 20 minutes. However, let's do a break.
08:04:300Paolo Guiotto: 5 minutes, huh?
08:06:260Paolo Guiotto: Of Breck, okay?
08:09:650Parvin Yavari: Thank you.
08:25:319Paolo Guiotto: Okay, do you hear me?
08:28:490Parvin Yavari: Yes, perhaps.
08:30:220Paolo Guiotto: Okay.
08:33:840Paolo Guiotto: Market.
08:40:640Paolo Guiotto: Okay.
08:42:00Paolo Guiotto: So, in this final part, 20 minutes we still have.
08:47:970Paolo Guiotto: We will talk about a nice, interesting formula, which is called the green… wins formula.
09:03:80Paolo Guiotto: I want to do the proof. I will just limit to the statement.
09:08:359Paolo Guiotto: Which is a formula that allows to write the circulation of a plane
09:17:109Paolo Guiotto: a planar vector field in terms of a double integral. Let's see how it works. So, let's imagine that we have a vector field F,
09:32:69Paolo Guiotto: a planar vector field, so a vector field function of two variables, XY, with the two components, little f…
09:42:630Paolo Guiotto: XY, and little g, XY.
09:47:439Paolo Guiotto: be a vector field.
09:56:580Paolo Guiotto: on a domain D contained into R2.
10:05:300Paolo Guiotto: Now, let's help also the intuition with a certain figure. So let's imagine that this is the domain D.
10:13:870Paolo Guiotto: Now, you remind that, an important,
10:20:80Paolo Guiotto: An important, property of, fields.
10:26:340Paolo Guiotto: is that they are conservative if and only if all the circulations are equal to zero. So, we recall that…
10:38:210Paolo Guiotto: We recall… that,
10:43:260Paolo Guiotto: This, factor holds, F.
10:47:450Paolo Guiotto: is conservative.
10:53:350Paolo Guiotto: So…
10:54:510Paolo Guiotto: F is the gradient of some function f, which is the potential of the field. If, and only if this condition holds, all circulations of F are zero for every gamma.
11:12:970Paolo Guiotto: For every gamma, CO quit.
11:17:610Paolo Guiotto: contained in D.
11:19:880Paolo Guiotto: Okay, so, the property is that whatever is the, the circuit we take into D, let's say gamma is the circuit, the circulation of the vector field along this, along this,
11:35:340Paolo Guiotto: this circuit is zero. Now, the green formula provides an interesting formula to compute the circulation of a vector field of this type.
11:51:900Paolo Guiotto: Provided, let's say, the circuit is like, more or less, what you see in the figure. That is, the circuit is the boundary of a set.
12:06:820Paolo Guiotto: Well, the field inside is a good field, so the two components of the field are regular functions, so continuous with continuous derivatives. And it says the following factor. So, theorem…
12:25:190Paolo Guiotto: So, let's… Effort… V.
12:31:550Paolo Guiotto: C1… Vector.
12:36:970Paolo Guiotto: Field.
12:39:150Paolo Guiotto: So this means that the two components, little f and little g, continuous.
12:49:520Paolo Guiotto: with all the derivatives, DX, F, DYF, DXG, DYG, Continuous.
13:05:400Paolo Guiotto: Then… Let also gamma.
13:10:840Paolo Guiotto: contained, when I say, on… on D, okay? That's gamma B, circuit contained in D, B.
13:23:900Paolo Guiotto: Circuit.
13:28:590Paolo Guiotto: Such that…
13:34:250Paolo Guiotto: So we need that, number one, we need that, now I, I, I, I will not write this condition exactly in the precise meaning. I will give a sort of,
13:49:410Paolo Guiotto: figurative idea, so I say that gamma… is D.
13:55:880Paolo Guiotto: Boundary.
14:01:400Paolo Guiotto: of a set omega contained into D.
14:07:390Paolo Guiotto: Mmm… Gamma is the boundary of set omega contained in D. And number two,
14:19:220Paolo Guiotto: I will show you what exactly means, what is behind this condition. And number two, there is another very qualitative, condition that, intuitively, it's easy, but it's not practically… it's not easy to formalize. That the circuit gamma
14:40:720Paolo Guiotto: is, counter… clock.
14:50:110Paolo Guiotto: wise, oriented.
14:57:530Paolo Guiotto: With respect to points… off.
15:05:270Paolo Guiotto: Omega.
15:06:600Paolo Guiotto: So what… what this condition means? So, let's repeat the figure. This is the domain D.
15:13:70Paolo Guiotto: Okay? Well, we have a circuit.
15:16:580Paolo Guiotto: That must be the boundary of a region, which is this in blue I'm coloring, and this is the omega set.
15:30:170Paolo Guiotto: Now, saying this, This is the condition one.
15:35:430Paolo Guiotto: We in particular say that, for example, this… it's a sort of qualitative description, there cannot be, holes of the into omega, because imagine that
15:51:850Paolo Guiotto: D were this one. Suppose that D is this black region.
15:56:870Paolo Guiotto: And it is just with that hole, okay?
16:02:580Paolo Guiotto: So that hole is not in D, so this is decomplementary, this is decomplementary.
16:09:330Paolo Guiotto: So you see a sort of plane region with a hole inside. If I take this circuit here.
16:17:750Paolo Guiotto: This is a good circuit because it is the boundary of something of an omega entirely contained into the domain, so this omega is contained into D.
16:31:100Paolo Guiotto: But this is different from this other secret. Imagine that you have a secret that turns around the, the hole. Now, that red circuit is the boundary of this thing, as you can see.
16:45:730Paolo Guiotto: That includes also the whole, because if you have to think to that red line as the boundary of something, this would be the omega. But this omega would not be contained into D, because the hole that you have there is made of points which are not in D. So this omega cannot be contained in D, otherwise…
17:06:670Paolo Guiotto: Otherwise, otherwise… Say, the whole… would be… contained in D, and…
17:24:310Paolo Guiotto: While, let's say, while… oh.
17:29:40Paolo Guiotto: while…
17:33:240Paolo Guiotto: It is contended into the complementary.
17:36:80Paolo Guiotto: So, let's say the first one is an example of a circuit verifying condition 1. So, this gamma…
17:44:540Paolo Guiotto: this gamma… verifies, condition…
17:51:810Paolo Guiotto: 1. It is the boundary of a region which is contained in D. And this second circuit does not verify
18:07:570Paolo Guiotto: 1. Because it is a circuit, yes, it is contained in D, yes, but it is not the boundary of something which is contained in D. It is a boundary, yes, but the boundary of a domain which is not contained in D.
18:22:900Paolo Guiotto: So this second secret does not verify the first condition, so it's a kind of secret we cannot consider for this theorem.
18:31:330Paolo Guiotto: So, for this theorem, we should consider only secrets, which are contained in D, and in such a way they are boundary of regions anteriorly contained in D. Another way to say, it means that there cannot be holes of D
18:48:150Paolo Guiotto: Inside the, the circuit, the circuit, gamma.
18:54:110Paolo Guiotto: This is, intuitively what it means.
18:57:610Paolo Guiotto: About the second condition, the second condition says that this gamma should be a circuit that turns around in a counterclockwise
19:09:970Paolo Guiotto: way around the points of omega, that omega, for which gamma is the boundary.
19:18:670Paolo Guiotto: So you have to imagine that you are around here.
19:23:130Paolo Guiotto: And you see the circuit turning around in a counterclockwise way, so you would see, imagine that there is a car moving along this circuit, you would see the car moving in that way, like the Formula 1 circuit, most of the Formula 1 circuits are in counterclockwise orientation, like that.
19:41:670Paolo Guiotto: Okay, so this is, this is, verifies… condition…
19:51:280Paolo Guiotto: 1, and also condition 2. But, for example, if I take… this is, again, the domain D, and this is my circuit, it is boundary of something, but now it is oriented in that way.
20:05:580Paolo Guiotto: So,
20:08:340Paolo Guiotto: In this case, I have that my secret is still the boundary of a region entirely contained into the domain, but if you put a point here, you would see the cars moving around the boundary, moving in a clockwise way. So this verifies
20:28:110Paolo Guiotto: Verifies… One, but… not.
20:36:80Paolo Guiotto: 2.
20:37:100Paolo Guiotto: You could have also this case, for example, you have your domain, the circuit.
20:43:00Paolo Guiotto: and maybe a hole, so let's say that there is a hole here, so this black region is not in D, so D is this one. And now you have your secret, which is properly oriented in a counterclockwise way.
20:57:930Paolo Guiotto: But in this case, it's the boundary of something not entirely contained in D. Because of this part, the bad part is the hole, which is outside of D. So this one, verifies…
21:14:860Paolo Guiotto: to… but… not.
21:18:410Paolo Guiotto: 1. Okay?
21:20:370Paolo Guiotto: So the unique circuits, in other words, that are good are those of this type. So those that are boundaries of regions entirely contained into D, and such that if we are inside the circuit, we see the circuit turning in a counterclockwise
21:38:710Paolo Guiotto: Way around the domain, around the point.
21:42:80Paolo Guiotto: Now, if this is the case, then we have this remarkable formula. The circulation of the field F around this secret gamma is this double integral, is the integral on the domain omega.
21:58:530Paolo Guiotto: for which gamma is the boundary of this function, the Y… the derivative with respect to Y of the first component, minus the derivative with respect to X of the second component in the XDY.
22:13:930Paolo Guiotto: This is the green formula.
22:22:120Paolo Guiotto: Now, we don't prove it.
22:25:340Paolo Guiotto: Because, now it's, it's late for this class to start proving this, and, it's not particularly difficult. There is the proof in notes, if you are curious.
22:40:910Paolo Guiotto: But let's say that we accepted just a statement. I want to do a couple of remarks on this formula. Remark number one.
22:51:330Paolo Guiotto: You notice that,
22:53:870Paolo Guiotto: This quantity here that you have in this integral should be something familiar to you, because if the field F has the two components, F and G,
23:07:180Paolo Guiotto: Well, these two derivatives, D, Y, F,
23:13:730Paolo Guiotto: and DXG are what? This is the… the YF is the derivative of the first component with respect to the second variable, while this is the derivative of the second component with respect to the first variable.
23:31:470Paolo Guiotto: Now, you remind that this is connected to some check about the field F, huh?
23:36:70Paolo Guiotto: F is irrotational.
23:39:970Paolo Guiotto: Ear rotational.
23:42:470Paolo Guiotto: If, and only if DYF is identical, equal to DXG. So, in other words, if and only if that difference, DYF minus DXG, is equal to 0, is identical equal to zero.
24:00:260Paolo Guiotto: But if that difference is equal to zero, well, we know that this is a necessary condition to be conservative, so F conservative implies,
24:12:390Paolo Guiotto: F, irrotational.
24:14:510Paolo Guiotto: That means that that difference is zero, and as you can see, this is current with what we already know. If that difference is zero, the integral on omega would be zero, and therefore the circulation would be zero. But we already know, and we proved by other means, that when F is conservative, the circulations are always equal to zero, okay?
24:36:400Paolo Guiotto: Now, what is interesting in this is the opposite, that means that, Io.
24:45:880Paolo Guiotto: The field is irrotational, that means this is zero.
24:50:500Paolo Guiotto: Then, by looking at this formula, we would have that… so… So, if… F… Ease.
25:02:740Paolo Guiotto: Irrotational.
25:04:760Paolo Guiotto: We would have that DYF.
25:07:790Paolo Guiotto: minus DXG would be identically zero. And so, in particular, we would have that the circulation along a secret gamma with the previous conditions, so gamma.
25:21:360Paolo Guiotto: such that, gamma is the boundary of an omega, omega interior contended in D, gamma, counter…
25:36:400Paolo Guiotto: clock.
25:38:880Paolo Guiotto: wise.
25:40:400Paolo Guiotto: oriented. So, for a gamel of that type.
25:45:580Paolo Guiotto: We would have that. The circulation would be, thanks to the green formula.
25:51:240Paolo Guiotto: would be the integral on that domain omega of the YF minus DXGE, but that would be identically equal to 0, so this would be 0.
26:01:770Paolo Guiotto: So we would have that if F is irrotational, the circulation of F along this particular circuit would be always equal to 0, which looks to be very close to say that the field is conservative. You remind that
26:21:740Paolo Guiotto: Conservative is if and only if all the circulations are zero.
26:28:140Paolo Guiotto: And this says that if F is irrational, then we have all circulations equals 0 when the circulations are taken on sequence, which are boundaries of domains that are entirely contained into the domain D.
26:45:50Paolo Guiotto: So, when this becomes a sufficient condition, we have this corollary.
26:52:550Paolo Guiotto: So, if the unique possibility is that if domain D is such that
27:04:900Paolo Guiotto: For every gamma circuit.
27:11:660Paolo Guiotto: which is, contained in D. Then…
27:16:980Paolo Guiotto: Gamma is the boundary of something.
27:21:810Paolo Guiotto: contained in India.
27:24:90Paolo Guiotto: then we would have that the, the… all the circulations along gamma would be equal to zero for every gamma, and therefore F would be conservative.
27:40:450Paolo Guiotto: Sorry.
27:42:130Paolo Guiotto: I have to put the…
27:44:900Paolo Guiotto: I missed an assumption. So, let…
27:49:140Paolo Guiotto: F, B, which is the fundamental assumption, irrotational.
27:58:710Paolo Guiotto: It says, if D is a domain.
28:01:960Paolo Guiotto: such that for every gamma secret, then gamma is the boundary of some domain entirely contained in D, then all circulation would be equal to 0, and therefore the field will be conservative.
28:16:710Paolo Guiotto: This is a condition on the domain, so in particular, it means that, for example, if domain D is made like this, suppose that domain D
28:27:670Paolo Guiotto: equal R2 minus a point, typically the origin, so everything except the origin. This condition here is not verified. You cannot say that if it is irrotational, it is conservative, because if you take a circuit that turns around the origin, like this one, even if it is counterclockwise oriented.
28:51:880Paolo Guiotto: This would be the boundary of something, this one, that must contain also the bad point, the zero.
28:59:900Paolo Guiotto: So, in this case, it would not be true that for every circuit gamma, gamma is the boundary of something contained into D, so the theorem wouldn't apply. In this case, theorem…
29:14:170Paolo Guiotto: Thus… Not… Old.
29:19:150Paolo Guiotto: for this type of domain. But if, for example, domain D is, for example, R2,
29:27:460Paolo Guiotto: So, in F2 means that domain is everything, so you can see that whatever is the circuit.
29:34:240Paolo Guiotto: The circuit will be always a boundary of something which isn't really contended into D, because D is everything. So that omega, whatever it is, would be contained in D, because D is everything. So, for example, I would have this, that if F is irrotational.
29:54:360Paolo Guiotto: on, on R2, it is also conservative.
30:07:190Paolo Guiotto: So, in other words, the green formula provides a partial, vice versa.
30:16:850Paolo Guiotto: to the fact that, in general, we know a conservative field is irrotational, but not the vice versa. The vice versa is false. We have seen several examples of this issue, okay? That we cannot say that just checking that the field is irrotational is enough to say that it is conservative.
30:36:570Paolo Guiotto: But through the green formula, we can see that if the domain needs at least four plane fields, so 4 fields on R2, if the domain has this property.
30:48:50Paolo Guiotto: The property is that every circuit is the boundary of a domain interiorly contained into D,
30:57:330Paolo Guiotto: Then, automatically, the circulations are zero, so the field is conservative. So at the end, irrotational implies conservative.
31:05:490Paolo Guiotto: I've done these two examples to show you that on the first example, domain is everything except the origin, so there is a hole, or several holes, more in general.
31:17:70Paolo Guiotto: The theorem does not apply, because in this case, there are circulations, like this one. There are circuits, which are boundary of domains, the disk here, the blue disk, but these domains are not contained into the domain D. So, in this case, I cannot apply this corollary.
31:37:700Paolo Guiotto: But in other cases, like, for example, the domain D is R2, so we have a field on R2. It is true that whatever is the circuit we take, it is a boundary of something contained in the domain, because the domain is everything, so that condition is obviously verified.
31:58:210Paolo Guiotto: And therefore, in that case, for example, this corollary applies, and it says that
32:04:410Paolo Guiotto: If you have any rotational field, that field is automatically conservative. So this provides a sufficient condition in order a rotational field be also conservative.
32:18:830Paolo Guiotto: Okay, let's stop here for today.
32:24:340Paolo Guiotto: So, today is Wednesday, we have class on Friday, right? So we will meet on Friday morning regularly in class.
32:36:60Paolo Guiotto: I will do next time, spend a few minutes to show you some
32:43:340Paolo Guiotto: application, some further application of the green formula, and do some of the exercises. You could, in any case, try to do by yourself, the exercise
32:57:390Paolo Guiotto: 5, 7, 10, which is an exercise on the application of the green formula. So, let's see what you do. So, there are circulations to compute, and you have to apply the green formula.
33:12:170Paolo Guiotto: in a suitable way, okay? So, you… you… you have everything written, so there are four circulations to compute. However, try to do this, and then we see, we see the solution on Friday.
33:29:650Parvin Yavari: Just one quick question, Professor. For Chapter 5, we need… we kind of skipped Barry Center, center of mass. That part, will not.
33:40:780Paolo Guiotto: I want to do, it's just a reading. Well, the definitions, yes, maybe the definitions are important. I have a few definitions. What is the body center, the center of mass?
33:53:490Paolo Guiotto: Yes, because maybe I could give this to compute the bicenter of the domain as an exercise, but that's just the… if you, if you, if you, if you read, it's just a couple of definitions, nothing…
34:09:180Parvin Yavari: Okay.
34:09:920Paolo Guiotto: So, you can read the section 5-5, because it's just two definitions and some application. Well, I… we don't need to do… there is purpose theorem, which is a formula for the volume of rotation solid, but that… that's not needed, because we do directly, it's not important.
34:29:230Paolo Guiotto: This part, okay?
34:31:270Parvin Yavari: Okay, thank you, Professor.
34:33:220Paolo Guiotto: Okay, see you on Friday. Have a nice day.
34:36:120Gordafarid Akbarzadeh: Goodbye. See you.
34:37:610Paolo Guiotto: Bye, bye.