AI Assistant
Transcript
00:15:00Paolo Guiotto: I'm not sharing with you.
00:17:980Paolo Guiotto: Right?
00:21:870Paolo Guiotto: Okay, so to the, start doing, some exercise, the novel.
00:36:670Paolo Guiotto: We have seen this formula yesterday.
00:43:660Paolo Guiotto: One second.
00:46:610Paolo Guiotto: So… The green formula, What is going on So, green formula.
01:03:840Paolo Guiotto: Says that, we have a vector field, F.
01:10:840Paolo Guiotto: defined on the domain D of R2, R2…
01:17:440Paolo Guiotto: The components of the vector field, Say, FG…
01:22:490Paolo Guiotto: So here, of course, F and G are both functions of two variables, X and Y.
01:29:140Paolo Guiotto: G function of X and Y.
01:33:210Paolo Guiotto: defined on the domain V, where the field is defined, but they are real bad.
01:38:680Paolo Guiotto: Because they are the components of the filler.
01:41:580Paolo Guiotto: We assume that,
01:44:670Paolo Guiotto: F, G, together with their derivatives, so let's write shorty gradient.
01:52:290Paolo Guiotto: F gradient G, all these things are continuous on the domain.
01:59:590Paolo Guiotto: So… We have a feature, which is the main V of…
02:08:970Paolo Guiotto: Where the field, and therefore the components, are defined.
02:12:770Paolo Guiotto: Then we have also a… circuit, gamma, contained in the circuit.
02:22:30Paolo Guiotto: And here we have to… To be a little bit intuitive, because the formal definitions are… more complicated, but…
02:31:790Paolo Guiotto: They are not necessary for our purposes. So, gamma is a secret in the… Such that,
02:41:10Paolo Guiotto: We have these two conditions that must be verified.
02:44:790Paolo Guiotto: Gamma is the boundary of something which is anteriorly contained in it. So, whatever surrounds this gamma
02:53:810Paolo Guiotto: is competitive indeed, and it is this, so… Gamma.
03:01:180Paolo Guiotto: Jeez.
03:03:510Paolo Guiotto: boundary.
03:08:180Paolo Guiotto: Off.
03:09:70Paolo Guiotto: region, omega, entirely contained.
03:12:590Paolo Guiotto: And I told you.
03:14:530Paolo Guiotto: why this is not always true, because it depends on how this is made. If our domain has a whole.
03:24:820Paolo Guiotto: For example, like this.
03:26:880Paolo Guiotto: So this is a hula. I hope you understand that this is this thing.
03:31:450Paolo Guiotto: Or if you want, let's do it in another way.
03:34:350Paolo Guiotto: colors. So let's imagine that this is what a color with Maybe.
03:41:900Paolo Guiotto: It's color, like, yellow.
03:47:830Paolo Guiotto: this one.
03:50:10Paolo Guiotto: So this is… this is D. What you see in yellow is D.
03:56:50Paolo Guiotto: We'll take half an hour to color all this.
04:06:780Paolo Guiotto: Well, let's do this.
04:10:830Paolo Guiotto: D is this.
04:13:580Paolo Guiotto: Okay.
04:14:830Paolo Guiotto: That's the way.
04:15:860Paolo Guiotto: Let me see. That there is a hole. The hole means that the white thing is not…
04:21:850Paolo Guiotto: So, three results you see here.
04:25:200Paolo Guiotto: Ben… This gamma would be a circuit until it contained in B, But…
04:32:350Paolo Guiotto: And it is the boundary of something. The something is this region. But as you can see, that region, omega.
04:40:520Paolo Guiotto: Contains a part which is not in me.
04:43:880Paolo Guiotto: So this omega is not contained in the…
04:48:290Paolo Guiotto: Okay, because there are points of this omega, the sponsor?
04:53:300Paolo Guiotto: Which are not.
04:55:530Paolo Guiotto: So… boundary of… Until they contain it indeed. So, let's say… That this is true.
05:10:610Paolo Guiotto: And this is… Yes.
05:19:550Paolo Guiotto: Yeah, yeah, yeah, that's… that's a right remark. Instead, while this could be decays.
05:28:860Paolo Guiotto: But this, this is,
05:34:580Paolo Guiotto: This is a bit more complicated, so let's say that this is our, set with the roller.
05:42:880Paolo Guiotto: I don't want to enter in such a… implications, because,
05:47:670Paolo Guiotto: You see, we are already using a definition which is not extremely clear, what does mean.
05:54:660Paolo Guiotto: So, how do you… do you practically verify that this is true? But suppose that, for example, I have this circuit…
06:10:840Paolo Guiotto: No?
06:11:780Paolo Guiotto: For example, this liquid is the boundary of a region, this one.
06:17:880Paolo Guiotto: Which is now contained into this.
06:21:160Paolo Guiotto: So the key point is, basically, if you want, roughly speaking, this circuit shouldn't turn around the holes of the domain. Because if you turn around the hole, it means that the region delimited by the circuit contains the hole.
06:37:260Paolo Guiotto: Okay?
06:38:430Paolo Guiotto: So, imagine the holes like lakes, so you cannot turn around the lake.
06:45:370Paolo Guiotto: Okay? You have to be… The region, the limited, must be always ground.
06:52:980Paolo Guiotto: However, intuitively, it is clear, because if you draw a figure, you can always distinguish which is, which is verified the
07:02:910Paolo Guiotto: condition. Formally, it's a bit more complicated to tell us what does it mean, so I don't want to enter in this, because it's
07:10:230Paolo Guiotto: It demands a lot of sophisticated mathematics, which is much beyond what we need.
07:17:330Paolo Guiotto: The second point is that this circuit is, counter… clock.
07:28:520Paolo Guiotto: wise.
07:30:330Paolo Guiotto: oriented.
07:32:870Paolo Guiotto: What does it mean? Also, this is not easy. It is easy to… everyone understand what does it mean, but how to precise… how to precisely formalize this condition is not easy.
07:46:120Paolo Guiotto: So it means that it, it must turn around
07:51:150Paolo Guiotto: points of omega in such a way that they see these points moving in a counterclockwise way. So if this is the case for this omega, this would be the orientation for this.
08:06:550Paolo Guiotto: So you see, that apparently, for the second example, Otherwise, again.
08:14:390Paolo Guiotto: No.
08:15:490Paolo Guiotto: Point, set up.
08:17:370Paolo Guiotto: inside of the…
08:35:429Paolo Guiotto: So…
08:37:210Paolo Guiotto: Well, this is not particularly important, because if it is clockwise oriented, it changes the sign of everything we are saying. So this is just for the formula that has the right sign.
08:50:950Paolo Guiotto: Well, then with these conditions, we have this, the circulation of F.
08:56:430Paolo Guiotto: Along this gamma.
08:59:440Paolo Guiotto: Can be transformed into this double integral, which is an integral on the region delimited by that comma.
09:07:660Paolo Guiotto: So, what we call Domiga…
09:09:770Paolo Guiotto: And we have this, this expression here. We have to take the derivative with respect to Y of the first component of the field, minus the derivative with respect to X of the second component of the field.
09:21:330Paolo Guiotto: And this is a double.
09:24:900Paolo Guiotto: Now, last… yesterday, we, not yesterday, Wednesday.
09:30:860Paolo Guiotto: Yes, WebPass Wednesday.
09:33:490Paolo Guiotto: We have seen that this, yields a sufficient condition
09:40:240Paolo Guiotto: to ensure that any rotational field, is conservative. Because, basically, this is a condition actually on the domain. If the domain B is such that whenever you take a circuit.
09:55:570Paolo Guiotto: This secret is the boundary of a region entirely containing the int omega. Now, it doesn't matter if it is clockwise or counterclockwise, because if it is clockwise oriented for you, this is clockwise, right?
10:10:830Paolo Guiotto: Is that…
10:11:750Paolo Guiotto: So it's a counterclockwise only. So this is clockwise oriented. You change the orientation, you change the sign to the circulation. If it is zero, it remains zero. So that's the point.
10:22:240Paolo Guiotto: So, in this case, if it is irrotational, the circulation is zero, because that condition is exactly the condition that means that it will be rotational. The two crossed derivatives are the same. So, the difference is zero.
10:40:170Paolo Guiotto: So, this provides a sufficient condition. Here, here is an example. So, for example, if you take as domain of the field in here playing R2,
10:50:190Paolo Guiotto: Whatever is the circuit, it will be the boundary of something that is contained in the domain, because the domain is everything.
10:57:740Paolo Guiotto: While this is not true, if the domain, for example, has an hole, because, for example, here, you see, when you take a secret that turns around that hole, here is the origin, you cannot say that whatever is delimited by the secret is contained in the domain, because in this case, that would be also the zero, which
11:17:860Paolo Guiotto: So you cannot apply that theorem to this domain.
11:22:170Paolo Guiotto: to a domain with holes, you cannot apply, but you can apply with domain without holes. So in this case, it says that, for example, on R2, a new rotational field is automatically conserved.
11:36:00Paolo Guiotto: So there is a potential.
11:38:990Paolo Guiotto: Okay, so now I want to show you some application of this. I asked…
11:43:380Paolo Guiotto: do some of the exercises 5, 7, 10, so let's see some of them.
11:51:430Paolo Guiotto: Let's see how it works, this formula. So, apply Green's formula to compute this. So, let's take number one. We have the circulation of a gamma of this field.
12:03:610Paolo Guiotto: XY… Good EX.
12:07:170Paolo Guiotto: plus 2Y.
12:09:200Paolo Guiotto: Where gamma is the boundary of this, minus 1, 1.
12:18:300Paolo Guiotto: square, so… It is the square.
12:22:800Paolo Guiotto: We had the Caucasian player in R2.
12:26:520Paolo Guiotto: the… that box is the interval minus 1 to 1 for X, minus 1 to 1 for Y.
12:33:660Paolo Guiotto: So it is the Cartesian product of the two intervals, and this is a set of points where the two coordinates are both between minus 1 and 1, so this is the square. So this is, how to say, the omega set.
12:47:700Paolo Guiotto: It's, this square.
12:50:330Paolo Guiotto: And we want to do the circulation of that field along the boundary of this.
12:58:740Paolo Guiotto: To a longer days back.
13:00:530Paolo Guiotto: Now, it is not written here, but since we apply a green formula, we will consider the path as counterclockwise oriented, so this way.
13:13:420Paolo Guiotto: So now the idea is that we apply the, we apply greens.
13:25:610Paolo Guiotto: Well, that's what we thought.
13:30:340Paolo Guiotto: formula.
13:33:580Paolo Guiotto: So this says that the circulation of a field F is equal to the integral on that omega of, let's rewrite the general formula, the YF minus DXG.
13:48:00Paolo Guiotto: So here… F… Isa… XY…
13:56:700Paolo Guiotto: And the second component is 3X.
13:59:780Paolo Guiotto: plus 2Y.
14:01:750Paolo Guiotto: So these are F and G, okay? Omega is the region delimited by that secret, so it's the square minus 1, 1,
14:12:730Paolo Guiotto: square, okay? So at the end, we have to compute the integral.
14:17:350Paolo Guiotto: on the set minus 1 1 square of DY of the F component, which is XY.
14:26:450Paolo Guiotto: minus DX of D.
14:29:750Paolo Guiotto: Y component, the second component, 3X plus 2Y.
14:34:600Paolo Guiotto: India, Iwa.
14:36:950Paolo Guiotto: So this is the literal application of the form.
14:39:610Paolo Guiotto: Now, the dy of this is X minus, the DX of this is 3.
14:45:650Paolo Guiotto: So this is what we have.
14:47:370Paolo Guiotto: compute. So we have to compute the integral from minus 1, 1 square of X minus 3 in the x divided.
14:57:390Paolo Guiotto: Well, for example, I could say this is the integral for minus 1, 1 squared, over X, DXDY,
15:05:980Paolo Guiotto: Minus 3, the integral of 1 EXDY, on this domain.
15:13:610Paolo Guiotto: We know that when we integrate one, we can interpret this, in the case of double integrals, as the area.
15:21:300Paolo Guiotto: of the domain minus 1, one square. So the area of this square
15:26:390Paolo Guiotto: It is a square of side 2, so it is 4.
15:30:640Paolo Guiotto: This value is 4.
15:33:30Paolo Guiotto: While this one is easily equal to zero, because here we can apply the reduction formula.
15:40:10Paolo Guiotto: Since X is between minus 1, 1, and Y the same, we can say, okay, let's integrate first in X, so the function is X. We will integrate between minus 1 and 1, then we integrate in Y again…
15:52:370Paolo Guiotto: 1, and 1.
15:53:900Paolo Guiotto: But the innermost integral, this one, is the integral of an odd function with respect to a symmetric integral. If you want, you can compute that you get 0. So this integration yields 0. So minus 3 times 4 minus 12, this is the value of the circulation.
16:12:450Paolo Guiotto: Okay?
16:13:480Paolo Guiotto: If I wanted the circulation for the clockwise oriented, I would have to change sound, so it would be plus 12.
16:24:30Paolo Guiotto: So, for example, number… 3…
16:32:360Paolo Guiotto: It says you have to compute the D… The…
16:43:950Paolo Guiotto: Okay, the circulation of this, X squared plus Y… XY.
16:54:50Paolo Guiotto: along… well, gamma, it says gamma of P,
16:59:730Paolo Guiotto: This time, it gives the parameterization of gamma directly. It's 1 plus cost the… Sign E.
17:10:180Paolo Guiotto: 20 is… Between 0 and 2 pi.
17:14:940Paolo Guiotto: Since I have the parameterization of gamma, I could proceed even in two ways.
17:20:970Paolo Guiotto: So, direct way…
17:26:260Paolo Guiotto: I use the definition of, line integral, so you remind that when you have gamma…
17:32:470Paolo Guiotto: the line integral, whatever is, this is a circulation, because if you see T equals 0, t equals 1 is the same point, so in fact, this is a circulation.
17:43:310Paolo Guiotto: But, well, it doesn't matter what is, so this is the integral from, if, let's say, the general formula is when gamma goes from AB to…
17:54:370Paolo Guiotto: Who are… M… So this will be integral from A to B of F of gamma, of T,
18:03:750Paolo Guiotto: Scala product with Gamma Prime T.
18:07:430Paolo Guiotto: This is the general definition. So this means that I should compute for this case, the integral from 0 to 2 pi.
18:16:530Paolo Guiotto: Now, F is this one.
18:19:290Paolo Guiotto: Okay?
18:20:350Paolo Guiotto: So you should replace, to the component of the letters X and Y, what?
18:27:930Paolo Guiotto: your gamma T is the vector which is made by those components, 1 plus cos and psi. So I should say that
18:36:810Paolo Guiotto: So this is going to be… seems to be a long calculation, but let's see…
18:41:900Paolo Guiotto: To show the differences between the two ways.
18:45:680Paolo Guiotto: So we have X squared, so 1 plus cos T… is DX.
18:51:720Paolo Guiotto: square.
18:53:770Paolo Guiotto: Is that clear, what I'm doing?
18:57:290Paolo Guiotto: So I'm doing two steps in one, I'm saying I'm computing this F of gamma of t, no? Gamma of t is this.
19:05:280Paolo Guiotto: So this is the point, so this is the X, and this is the Y.
19:08:770Paolo Guiotto: I have now to plug this into the components of the field. So, the first component is X squared plus Y, where X is 1 plus cos T and Y is sine T, so I will have 1 plus cos T squared plus sine T.
19:22:690Paolo Guiotto: You see, that's X squared plus Y. The second component is X times Y, 1 plus cos D,
19:30:440Paolo Guiotto: is DX times sine T is the y.
19:34:270Paolo Guiotto: Now, all this, this thing is F evaluated at gamma of t.
19:42:610Paolo Guiotto: Then I have to do this color product with gamma prime.
19:49:10Paolo Guiotto: Now, gamma prime means that you have to do the derivative with respect to t of this, that is the derivative component by component, you get minus sine t.
19:59:50Paolo Guiotto: costly.
20:01:240Paolo Guiotto: All this must be integrated in DT.
20:04:160Paolo Guiotto: So now we do the calculations, so we got integral 0, 2 pi, 1 plus cosi square…
20:14:870Paolo Guiotto: I don't know what to simplify here. Let's say 1 plus cos squared T plus 2 cos T,
20:21:270Paolo Guiotto: This is 1 plus cos squared.
20:23:990Paolo Guiotto: Plus sine T… These… times minus sine t. Well, let's not use the dot, otherwise…
20:35:100Paolo Guiotto: confused there. It's not the scalar product, that's an algebraic problem.
20:39:510Paolo Guiotto: Plus, second component, 1 plus, Maybe you're right, but… do that product, so scientists… Plus.
20:51:460Paolo Guiotto: scientific steel.
20:54:320Paolo Guiotto: Times another cost.
20:59:580Paolo Guiotto: So, at the end, it will be elementary, because there are integrals of sine, cosine, something like this, but it will be a little bit longer. So, we have, minus sine t…
21:11:520Paolo Guiotto: Then we have Cost Square… Times minus sign, so minus sine t cos square.
21:24:480Paolo Guiotto: then you have 2 cos T minus 70, so minus 2 cos… Sign team…
21:32:970Paolo Guiotto: minus sine square T, and we finished with all this. It is here.
21:41:20Paolo Guiotto: Then we have the second product, which is a little of scientific.
21:46:780Paolo Guiotto: cross T… And then, sign T… Cause…
21:56:980Paolo Guiotto: We can't simplify.
22:07:50Paolo Guiotto: There is this one that this… Go away with this.
22:11:660Paolo Guiotto: And we have bees, with this… And nothing else, right?
22:17:450Paolo Guiotto: Okay, so now we have minus negative 0 to pi over sine t. This is…
22:25:970Paolo Guiotto: Everyone should know what is the value of this. This is zero.
22:32:00Paolo Guiotto: Then… We have another minus the integral 0 to pi of sine squared t, this is non-zero.
22:45:680Paolo Guiotto: And that's it, right?
22:48:970Paolo Guiotto: No, no, no, no, I forgot there is also minus integral here.
22:53:540Paolo Guiotto: Bye.
22:54:820Paolo Guiotto: of cos T sine P.
22:57:690Paolo Guiotto: But also, this one is the… it is equal to zero.
23:01:370Paolo Guiotto: It's better to restore the tube.
23:03:370Paolo Guiotto: This is sine of 2P.
23:05:620Paolo Guiotto: So, this is an epiliotic function, even, all the function, and when you integrate on a period or a multiple of…
23:16:490Paolo Guiotto: So at the end, we have this integral to compute, minus integral 02 pi of sine…
23:21:500Paolo Guiotto: square t, and this can be computed by a by parts.
23:27:360Paolo Guiotto: So, for example, integral of sine squared t…
23:32:490Paolo Guiotto: Can be written as integral 0, 2 pi sine sine.
23:40:500Paolo Guiotto: This sign is the derivative of minus cosine.
23:45:90Paolo Guiotto: So this becomes minus sign.
23:47:690Paolo Guiotto: T, trustee… To be evaluated between 0 and 2 pi.
23:53:470Paolo Guiotto: Then we move the derivative on the other factor.
23:58:490Paolo Guiotto: derivative of sine is cosine, so we get minus cos… We have the… This evaluation yields 0, no?
24:07:560Paolo Guiotto: And this one, for minus plus, we use the identity. This is 1 minus sine squared.
24:17:630Paolo Guiotto: So we get that the integral of 1 is 2 pi minus the integral 0 2 pi of sine squared, which is the initial
24:24:730Paolo Guiotto: to grow.
24:26:280Paolo Guiotto: So we carry at left, it gives to the integral equal to pi, so the integral is good.
24:32:570Paolo Guiotto: Integr 2 pi of sine squared, Is equal to pi.
24:39:140Paolo Guiotto: So finally, we can conclude, because here, we can say that this is equal to… Minus.
24:47:840Paolo Guiotto: Bye.
24:49:360Paolo Guiotto: So, this is the value of the circulation, no?
24:53:880Paolo Guiotto: So, the situation… on gamma of this F is equal to… minus 5.
25:03:70Paolo Guiotto: Okay?
25:07:50Paolo Guiotto: Because there is a minus here.
25:10:260Paolo Guiotto: Now, with… green… Green's formula.
25:17:910Paolo Guiotto: Let's see what is it… this better world, I don't know
25:23:150Paolo Guiotto: So, in this case, we say that… so let's go back to the circulation.
25:30:820Paolo Guiotto: The difficulty here might be to identify what is the domain, huh?
25:36:70Paolo Guiotto: Of which this secret is the boundary.
25:40:270Paolo Guiotto: However, let's see. So this is the integral on, secret gamma of F.
25:48:360Paolo Guiotto: Which is, the field… X square.
25:53:880Paolo Guiotto: That's why.
25:56:570Paolo Guiotto: XY… So this will… turns out to be the integral on omega of DY of X squared plus Y.
26:07:970Paolo Guiotto: Minus DX of, with X, white sums.
26:15:230Paolo Guiotto: So here, X, Y.
26:21:880Paolo Guiotto: Now, this dy of X squared plus Y is 1 minus…
26:29:840Paolo Guiotto: I got something to say, something wrong.
26:32:530Paolo Guiotto: What is the mistake?
26:36:610Paolo Guiotto: I did a mess.
26:42:810Paolo Guiotto: Seems to be correct. The feed is X squared plus YX times Y.
26:47:520Paolo Guiotto: So I do dy of the first component minus DX of the second.
26:51:850Paolo Guiotto: So I get that the Y of the first component is 1, right? Minus DX of the R, Y.
26:59:600Paolo Guiotto: There is no mistake, but…
27:01:390Paolo Guiotto: I was thinking that it was 1, so the first would have been 0, and so on.
27:06:520Paolo Guiotto: So we get that we have to do the integral on omega of 1 minus y dxty y. Now, the problem is, what is this omega? Because here we do not have a domain omega, but it's a boundary.
27:19:170Paolo Guiotto: The boundary, let's see if you understand what is it, gamma of P.
27:25:930Paolo Guiotto: Is 1 plus, 1 plus cost D… Sign T… I could split this.
27:37:50Paolo Guiotto: 2… So this is 1 sine T.
27:44:150Paolo Guiotto: laugh.
27:45:480Paolo Guiotto: Posity… It's fine.
27:49:940Paolo Guiotto: Right?
27:52:240Paolo Guiotto: Now, if I have only this one…
27:56:350Paolo Guiotto: This point describes a unitary circle.
28:00:70Paolo Guiotto: Right?
28:02:90Paolo Guiotto: So I have a point here, which is moving around a military circle.
28:08:950Paolo Guiotto: But, I add a point which is moving on this curve. What is this? This is abshisha 1 ordinate sine T.
28:18:90Paolo Guiotto: So it means that in the plane XY, abscissa is always 1, and the ordinate is sine T, so this is XY.
28:30:340Paolo Guiotto: When P ranges from 0 to 2 pi. So this is basically, sort of.
28:37:460Paolo Guiotto: Graph of sine, because, You have,
28:44:530Paolo Guiotto: The point is, how do we…
28:47:730Paolo Guiotto: Okay, however, let's see. 1 plus 7T, so for T equals 0, it is,
28:53:660Paolo Guiotto: 0, so the point is, here. No, is here.
28:59:140Paolo Guiotto: then,
29:06:00Paolo Guiotto: Yeah, you are right.
29:07:940Paolo Guiotto: Right, right?
29:18:860Paolo Guiotto: What?
29:31:590Paolo Guiotto: No, no.
29:33:150Paolo Guiotto: Yes, you are right.
29:35:990Paolo Guiotto: No, no, you are right.
29:38:270Paolo Guiotto: So it's not this, sorry.
29:42:350Paolo Guiotto: It is one zero.
29:44:840Paolo Guiotto: I'm stupid.
29:47:600Paolo Guiotto: 10 plus this much easier, because it means 10 is this point, cost is empty turns around the origin, so this is just the circle translated by the point 1, 0, so it's a circle centered at 1,0 and radius 1, okay?
30:07:590Paolo Guiotto: So the omega is what is inside this.
30:12:180Paolo Guiotto: So, how can we describe? We can describe the set of points where X minus 1 squared plus Y squared is less or equal than 1. That's the omega.
30:25:820Paolo Guiotto: is the disk.
30:27:10Paolo Guiotto: Centered at 1, 0, and radius 1.
30:32:550Paolo Guiotto: Okay, so we have to compute the integral on this set, x minus 1 squared plus Y squared, less or equal than 1 of 1 minus y
30:44:950Paolo Guiotto: DX, DY.
30:47:350Paolo Guiotto: Now, we could split this into the integral of 1, dx, dy.
30:52:900Paolo Guiotto: Minus the integral of Y, DXDY, on the same domain of 4.
30:58:630Paolo Guiotto: But this is the area of Omega.
31:04:660Paolo Guiotto: So it's the area of the disk of radius 1, so it is pi times 1 square, so it is pi.
31:13:770Paolo Guiotto: Now, the problem is, what is this?
31:17:460Paolo Guiotto: But…
31:29:80Paolo Guiotto: What is this, integral? Do you see?
31:33:480Paolo Guiotto: integral of Y, DX, DY… in X minus 1 square plus Y squared less or equal 1.
31:45:950Paolo Guiotto: Now… we can also prove this in general. This function we are integrating…
31:52:330Paolo Guiotto: is, odd, in the sense that if you change Y with minus Y, the function changes sine. Y, the remainder is symmetric.
32:00:290Paolo Guiotto: So the integral will be 0. However, you don't see, we do in the ordinary way. We use polar coordinates, but not centrade at 0, 0, because this wouldn't,
32:11:850Paolo Guiotto: simplify that condition. We use polar coordinate, just translated in such a way that 1,0 billion. So we say something like X equals 1 plus 4 cosine theta.
32:22:770Paolo Guiotto: And Y equal raw styling people.
32:25:640Paolo Guiotto: You see?
32:26:850Paolo Guiotto: In this way, X minus 1 squared plus Y squared will become raw square.
32:33:250Paolo Guiotto: Since I do a translation, if you think to the Jacobian matrix, when you do the derivative, that one disappears.
32:40:540Paolo Guiotto: So you don't have to do any new calculation. The change of variable will be raw times the raw
32:46:270Paolo Guiotto: Pita?
32:47:650Paolo Guiotto: The function is Y, which is a Rosa and theta.
32:52:150Paolo Guiotto: The domain becomes raw square less or equal than 1.
32:56:450Paolo Guiotto: Now, condition on theta means theta between 0 and 2 pi.
33:01:950Paolo Guiotto: So we have to, integrate. Now, raw square less than 1 means raw between 0 and 1.
33:09:690Paolo Guiotto: Theta between 0 and 2 pi each.
33:13:180Paolo Guiotto: The function is raw square sine theta.
33:16:890Paolo Guiotto: B-roll.
33:19:410Paolo Guiotto: We do first integration in theta, so we apply the traction formula. It is integral 0 to pi of sine theta.
33:27:920Paolo Guiotto: theta, then we have a raw square in 0 that will be integrated between 0 and 1. But this integral is equal to 0, so you get that this is 0 as advertised.
33:40:890Paolo Guiotto: So, this means that by applying the green formula, We got this, apparently, there is some error.
33:52:180Paolo Guiotto: So we… we have that the circulation along gamma of F Is equal to… so…
34:02:50Paolo Guiotto: This was the integral. We transformed it into this double integral, which is this one. We split this into this, which is pi, and this, which is 0, so the total is pi.
34:16:130Paolo Guiotto: But you may raise your little finger and say, no, but we got minus pi in the first half.
34:22:840Paolo Guiotto: So there should be an error.
34:25:290Paolo Guiotto: So now the question is, is really an error or not?
34:29:550Paolo Guiotto: And how can you explain this?
34:36:520Paolo Guiotto: Yeah, this one, we should, verify.
34:40:530Paolo Guiotto: We should verify.
34:45:650Paolo Guiotto: What is strange is, actually, This is, this should be counterclockwise.
34:53:110Paolo Guiotto: Because, If we… you look at this, no?
34:57:770Paolo Guiotto: Cosine T, sine t turns in this way.
35:01:910Paolo Guiotto: Right?
35:03:190Paolo Guiotto: No?
35:04:800Paolo Guiotto: So, this means that gamma is already counterclockwise oriented.
35:11:160Paolo Guiotto: So now the results should be the same, because in the second case, we assumed that gamma was counterclockwise oriented, as it is, in fact, by this parameterization. So the value of this integral that we get from the green formula
35:27:470Paolo Guiotto: should be the same of the value we get by computing directly the integral by using the parameterization of gamma, what we have seen here.
35:37:450Paolo Guiotto: So, there is an error at this point. Where is it?
35:41:380Paolo Guiotto: The suspect could be that the first part is wrong for some sign here, with all this mess of calculations. So, now…
35:51:380Paolo Guiotto: Let's say this.
35:53:290Paolo Guiotto: There he is.
35:56:790Paolo Guiotto: an error.
36:01:70Paolo Guiotto: I… I'd say… I'd say… I'd say in the first part, because the second part is very…
36:07:890Paolo Guiotto: The calculations have… really… Trivial, no, no…
36:13:540Paolo Guiotto: particular difficulty, so I would say that there is an error.
36:17:510Paolo Guiotto: in… Fast stuff.
36:22:410Paolo Guiotto: wave.
36:23:380Paolo Guiotto: Which is very likely, because there are several calculations, and one single calculation yields an error, can yield an error.
36:33:510Paolo Guiotto: I don't know what is it?
36:37:720Paolo Guiotto: So, the, this, the, the chain… the substitutions, I think, are correct.
36:45:80Paolo Guiotto: Because that's an error, a sine error, so we have to verify where signs comes out. So the derivative is correct, cos minus sine cosine.
36:56:810Paolo Guiotto: We did the products, so with this minus…
37:01:960Paolo Guiotto: Minus, seems to be correct here.
37:06:250Paolo Guiotto: This is all plus.
37:08:200Paolo Guiotto: Then we simplify the minus 2 plus 1, and this yields a minus. However, this integral is 0, so who cares?
37:21:400Paolo Guiotto: So… Seems that the minus comes only from this one, right?
37:28:640Paolo Guiotto: So we have this, this is the responsible, and this will be definitely a negative number, because it is minus
37:36:120Paolo Guiotto: Something, so…
37:38:800Paolo Guiotto: This calculation cannot yield anything different, because you see that the value is correct, it's positive, it's the integral of positive function.
37:47:770Paolo Guiotto: So I don't see it, where is it, where is the arrow?
37:51:640Paolo Guiotto: Okay, well… You look for this.
37:56:400Paolo Guiotto: And maybe next time, tell me, where is the arrow?
38:03:850Paolo Guiotto: Indeed.
38:05:20Paolo Guiotto: Until it feels warm.
38:08:260Paolo Guiotto: However, I don't want to exaggerate to insist, I want to just show a nice,
38:16:120Paolo Guiotto: In consequence of the green formula, which is the area formula.
38:25:360Paolo Guiotto: So, we may, we may tell this story in this way.
38:31:660Paolo Guiotto: Imagine you have a lake.
38:40:290Paolo Guiotto: And you want to compute the area of the surface of the lake. Well.
38:45:120Paolo Guiotto: The green formula says that if you take a car, and you…
38:48:720Paolo Guiotto: Or a bicycle, if you are an ecologist.
38:51:540Paolo Guiotto: Or you go by woke, for example.
38:54:130Paolo Guiotto: And you go around the boundary of, the edge of the…
38:59:200Paolo Guiotto: Of the lake, so this means that
39:02:270Paolo Guiotto: you, at each time, know your position and your velocity. So, in other words, you described perfectly this,
39:11:950Paolo Guiotto: This circuit that turns around the lake in a counterclockwise way.
39:19:540Paolo Guiotto: Then, you have this.
39:22:40Paolo Guiotto: If this is, this will be an X of T and a Y of T.
39:29:430Paolo Guiotto: So, in this case, you have that if you do the circulation along this gamma of this particular
39:38:500Paolo Guiotto: field, so what is the green formula?
39:44:820Paolo Guiotto: DY, yeah?
39:46:580Paolo Guiotto: So, the field is 0… no, sorry, is Y0.
39:54:600Paolo Guiotto: You get that this is, according to the green formula, it is the integral on the omega, which is the
40:01:410Paolo Guiotto: lake… Of, say,
40:07:480Paolo Guiotto: DY of the first component, which is Y, minus the DX of the second component, which is 0. So you get that this is the integral on omega of 1,
40:19:220Paolo Guiotto: DXDY, which is the area of…
40:24:970Paolo Guiotto: So, by computing the circulation of this, you could have the value of the area. Now, the circulation depends only at the end by this gamma.
40:35:810Paolo Guiotto: So, in other words, this says that if you know…
40:39:40Paolo Guiotto: Exactly, time by time, what is your position and your velocity, so the complete trajectory.
40:46:440Paolo Guiotto: Of motion of the… this,
40:50:790Paolo Guiotto: this turning around the lake, you can determine what is the surface, the area of the lake, which is a nice result. Normally, since this would be, formally, this would be the integral from, say, A to B, where this is the range of the parameter for gamma.
41:10:710Paolo Guiotto: of, so now we should say F of gamma of T
41:17:820Paolo Guiotto: Times gamma prime of t. This is the general formula.
41:23:40Paolo Guiotto: Now, F of gamma, you see, F is the field of XY equal Y0.
41:30:560Paolo Guiotto: So when you compute F evaluated at point gamma of P,
41:35:750Paolo Guiotto: you have to evaluate F at point X of T, Y of T, but according to the definition of F, this is Y of T
41:46:870Paolo Guiotto: Zero.
41:48:480Paolo Guiotto: while the vector, velocity vector, is X prime, T, Y prime T.
41:56:940Paolo Guiotto: So when you do the calculation there, you have to do the scalar product, between YT, See it all?
42:05:610Paolo Guiotto: This is the back door.
42:07:450Paolo Guiotto: F… evaluated at point gamma of P.
42:12:460Paolo Guiotto: Scalar product with the Gamma Prime.
42:16:60Paolo Guiotto: which is here the vector X prime, Y prime.
42:22:440Paolo Guiotto: This is gonna rhyme.
42:26:580Paolo Guiotto: And then you reintegrate.
42:28:270Paolo Guiotto: Well, this color product, because of that zero, simplifies to this integral from A to B of Y of T…
42:37:170Paolo Guiotto: X prime. T.
42:39:720Paolo Guiotto: VT.
42:42:430Paolo Guiotto: Okay, so we got this formula, the so-called area formula.
42:47:440Paolo Guiotto: area.
42:49:820Paolo Guiotto: of, omega.
42:52:370Paolo Guiotto: is equal to the integral from A to B.
42:55:820Paolo Guiotto: of why.
42:58:730Paolo Guiotto: X prime.
43:00:820Paolo Guiotto: DTE.
43:01:900Paolo Guiotto: So this, it's important because,
43:09:520Paolo Guiotto: the position XY as function of T, so they do not depend, apparently, on the shape of the domain omega, and this is the one that
43:19:200Paolo Guiotto: Well, there is an informal way to write this integral.
43:24:150Paolo Guiotto: Since this is like the derivative of X, normally this integral is written in this way, integral of Y dx.
43:36:120Paolo Guiotto: This is just an informal notation, okay?
43:39:650Paolo Guiotto: to say this, okay? Do not confuse this as an integration in X, because this is not integral in the variable X.
43:47:460Paolo Guiotto: See, the integral invariables.
43:49:340Paolo Guiotto: But in any case, this is called the area formula.
43:57:690Paolo Guiotto: So, for example, here we have the exercise 5711. Use the area formula to compute the area.
44:08:260Paolo Guiotto: compute…
44:13:420Paolo Guiotto: Here, yeah.
44:16:60Paolo Guiotto: Yeah, that's off.
44:19:140Paolo Guiotto: playing.
44:21:340Paolo Guiotto: regions… the limited…
44:29:710Paolo Guiotto: Bye.
44:30:920Paolo Guiotto: And here we have several examples.
44:33:330Paolo Guiotto: Let's see that the area formula works in a simple example. The simple example is this number 1,
44:41:920Paolo Guiotto: Well, actually, it is not the number 1. Let's do the number 0, which is not there.
44:47:620Paolo Guiotto: The number zero is this. We take a region for which everyone knows the area, and the region is a circle, for example, a circle centered in the origin, and the radius R.
45:00:790Paolo Guiotto: So, we give the parameterization gamma of t equals R cos T… Sign T…
45:10:480Paolo Guiotto: when T is between 0 and 2 pi. So we expect that the area formula yields
45:17:840Paolo Guiotto: 2, no, pi r squared, okay?
45:21:120Paolo Guiotto: Let's see that this is the case.
45:25:90Paolo Guiotto: So, according to the area formula, the area?
45:30:100Paolo Guiotto: of the region omega, which is delimited by this particular segment. I know that everyone knows that the area is pi r squared, but let's verify that this formula here, this one, produces the same value. So this is the integral from 0 to pi.
45:48:290Paolo Guiotto: This is the integration ranger for the parameter T.
45:53:440Paolo Guiotto: You see the formula saying Y of T, X prime of t. So let's write, once again, in general, YTX prime.
46:02:140Paolo Guiotto: DTE… Well, this is X of T, and this is Y of T.
46:08:240Paolo Guiotto: So we got integral 0, 2 pi.
46:11:170Paolo Guiotto: Y of T is R sine t.
46:15:110Paolo Guiotto: times X prime of t, this is X of t, this is minus R, scientists… P.T.
46:25:920Paolo Guiotto: So at the end, we get a minus R squared, integral 02 pi, of sign… square T.
46:36:310Paolo Guiotto: And we get the wrong value again, huh?
46:44:490Paolo Guiotto: Yes. Because we computed above sine square, integral of sine square, we got, what? Pi.
46:51:830Paolo Guiotto: No, no, bye.
46:53:750Paolo Guiotto: That's good.
46:54:950Paolo Guiotto: So the formula is, is, is correct, modular assigner.
46:59:870Paolo Guiotto: So we get minus pi as square, let's see, because this depends, again, by the orientation, so let me, one second, verify.
47:10:260Paolo Guiotto: So, this is the formula.
47:12:960Paolo Guiotto: That we get for this field, assuming that gamma is counterclockwise.
47:19:720Paolo Guiotto: So we get that this is the integral of omega of 1, and this is the area. So gamma must be counterclockwise.
47:27:520Paolo Guiotto: we say that this, making explicit replacement of the components, is this one, so YX prime, I see YX prime.
47:38:930Paolo Guiotto: So this is correct.
47:40:560Paolo Guiotto: Why this is wrong.
47:49:750Paolo Guiotto: It seems the same error we have above.
47:53:430Paolo Guiotto: We get something wrong with sine, but this is, this is counterclockwise oriented.
48:00:630Paolo Guiotto: Definitely.
48:11:180Paolo Guiotto: We should be getting… It's… I don't want that yet.
48:19:30Paolo Guiotto: I don't want that there is an error in…
48:30:440Paolo Guiotto: Yay.
48:33:770Paolo Guiotto: as well.
48:35:220Paolo Guiotto: Why not?
48:39:180Paolo Guiotto: Okay, I will have to verify. Apparently, there is a…
48:43:420Paolo Guiotto: a systemic error that seems that I wrote the formula with the wrong sign, but this is the formula. I have in notes, and as fast as I know, this is correct.
48:55:600Paolo Guiotto: So I don't know now why…
48:58:00Paolo Guiotto: The area formula is the same, huh?
49:03:570Paolo Guiotto: Yes, here it's… no, minus a YDX, so it's the same, it's correct.
49:10:830Paolo Guiotto: Yeah, I also did the example.
49:29:350Paolo Guiotto: Apparently, looking at the example, there is even an error in the example, which is not… So nice.
49:36:640Paolo Guiotto: Okay, I will verify. It should come with glass. I'm sure that this is correct, but… Stranger, so…
49:48:440Paolo Guiotto: I don't see what is the error. However, let's take it back, so I will think about, now, 10 minutes.
49:55:450Paolo Guiotto: We'll start later.
50:04:820Paolo Guiotto: Okay, I… I guess that there is a,
50:10:360Paolo Guiotto: A systemic error. That means,
50:14:670Paolo Guiotto: That the formula, the green formula, is wrong.
50:19:620Paolo Guiotto: Because looking at the proof, it seems that there is something wrong with the sign, yeah?
50:25:710Paolo Guiotto: So, I… I suspect that the right formula is with this or the exchange, so you have,
50:35:60Paolo Guiotto: the DX of the second component minus the DY of the first component.
50:41:500Paolo Guiotto: This is the… The strong enough.
50:44:380Paolo Guiotto: Well…
50:45:420Paolo Guiotto: about these, remarks on irritational fields, nothing changed, because that quantity is zero. So, also here, there would be the XG minus YF.
50:59:20Paolo Guiotto: the XG minus DYF, but nothing changed in the argument we have seen here.
51:06:290Paolo Guiotto: About the exercise we have seen before, this would explain why we got the wrong sign.
51:15:940Paolo Guiotto: If the order is, opposite, so if we have, Yikes, so… here, of, of XY.
51:27:230Paolo Guiotto: minus DY of X squared plus Y, we would have here Y minus 1.
51:45:870Paolo Guiotto: Also, ChatGPT says. So, it is, it is, it is correct.
51:52:00Paolo Guiotto: So, here we have,
52:00:770Paolo Guiotto: be careful with ChatGPT, because it is not always correct.
52:08:810Paolo Guiotto: Well… It works when something is known.
52:13:520Paolo Guiotto: Work when it is unknown.
52:15:180Paolo Guiotto: So, at the end, this would be minus pi, and the area formula… with this,
52:23:340Paolo Guiotto: agreement, it is… so, it is just the opposite. So, here we have,
52:29:880Paolo Guiotto: DX0 minus the Y. Y, so it is minus this.
52:37:940Paolo Guiotto: And so the correct area formula with the minus sign, yeah.
52:45:420Paolo Guiotto: And, here, of course, we would get this minus, minus, now plus, and finally plus.
52:52:530Paolo Guiotto: by our square.
52:55:880Paolo Guiotto: So, for example, let's do the number 2.
52:59:490Paolo Guiotto: Kia gamma.
53:01:950Paolo Guiotto: of T is, this.
53:06:230Paolo Guiotto: Sign T… Las.
53:09:390Paolo Guiotto: signing…
53:14:260Paolo Guiotto: sine squared T… then minus plus E… Minus sine… Costy.
53:27:200Paolo Guiotto: PE025.
53:33:320Paolo Guiotto: Now, to apply the area formula, we… transform this, area.
53:42:410Paolo Guiotto: Of the region, delimited by, by… by gamma.
53:48:750Paolo Guiotto: As the circulation We say the now minus, the correct version.
53:55:40Paolo Guiotto: the circulation on gamma of Y0.
53:58:680Paolo Guiotto: Now, the point is that, We should, verify… That this gamma…
54:06:40Paolo Guiotto: is counterclockwise oriented, so how can we do that here? Well, it's not a big problem, because if it is clockwise oriented, we don't get the area, but the opposite.
54:18:950Paolo Guiotto: So, if we get the negative value, it means that the
54:21:920Paolo Guiotto: the path is clockwise oriented, so the area would be minus that value, so… because the paths can be only either clockwise or counterclockwise oriented. If one gives the area the area, the other one gives minus the area.
54:35:920Paolo Guiotto: When you change… The way you…
54:40:430Paolo Guiotto: But you change the sign of the situation.
54:43:650Paolo Guiotto: So, we don't care too much to verify this, so let's compute this integral. So this is minus integral 0 to 2 pi. We said that this is Y, these are X of t, this is Y of T. So the formula is Y of P, X prime T,
55:04:950Paolo Guiotto: So this means minus integral 0 to 2 pi, y of t is…
55:09:380Paolo Guiotto: minus cos T, minus sine T.
55:13:550Paolo Guiotto: Costy.
55:15:460Paolo Guiotto: times this is YT. Now, we have to do X prime, which is a derivative of sine cos T.
55:23:790Paolo Guiotto: plus 2 sine T… Fosky.
55:28:290Paolo Guiotto: And this is X prime.
55:31:630Paolo Guiotto: Okay?
55:32:630Paolo Guiotto: Now, we, do the multiplication, so maybe…
55:36:600Paolo Guiotto: We put a minus here, so it becomes plus integral 025.
55:41:110Paolo Guiotto: So we have cost, cost, cost, square.
55:44:270Paolo Guiotto: P plus, to 2.
55:48:360Paolo Guiotto: sine T… costs… square T.
55:53:950Paolo Guiotto: Then we have a sine tica squared piece, so another one, so next 3.
55:58:100Paolo Guiotto: plus, sine P, 2 sine… Square T… cos square t.
56:08:690Paolo Guiotto: Okay… So… I would put together these two, but leave alone this one, because this is the river.
56:21:20Paolo Guiotto: If you look carefully… This is the derivative with respect to t of something about cosine T.
56:28:240Paolo Guiotto: Cuba.
56:29:660Paolo Guiotto: If you do the derivative, you get 3 cos… Squio.
56:34:500Paolo Guiotto: Then the duty, of course, is minus sine t.
56:38:80Paolo Guiotto: So if we put the minus, yeah, we have the derivative.
56:41:970Paolo Guiotto: So we have, integral 0 to pi. Let's factorize the cos squared t.
56:48:650Paolo Guiotto: times 1 plus 2 sine squared t.
56:59:160Paolo Guiotto: Vt, then we have,
57:02:770Paolo Guiotto: minus… I put a minus 2 times here, minus, minus. So the minus 3, etc, is the derivative.
57:11:870Paolo Guiotto: So I will have the evaluation of cost T.
57:16:530Paolo Guiotto: Cuba… Between 0 and 2 pi.
57:23:140Paolo Guiotto: I could have said that it is equal to 0, because the function is odd, and we are on a period. However, if you do the calculation, cos of 2 pi is 1, cos of 0 is 1, the difference is 0, so this is it.
57:36:550Paolo Guiotto: So now we have the integral of, sine… I would say that integral 0 to pi of cos square, we already computed for sine squared, it is the same value.
57:47:680Paolo Guiotto: This is equal to pi.
57:50:560Paolo Guiotto: Then we have to do a final integration to integral 0 2 pi.
57:56:780Paolo Guiotto: of the costs… We are tea… Sine squared T.
58:05:540Paolo Guiotto: Which we can do by farther, so take this one only.
58:11:450Paolo Guiotto: So we write as integer 0 to pi, for example, sine squared t times cos T. This is more or less the derivative.
58:21:390Paolo Guiotto: costly, yeah.
58:24:440Paolo Guiotto: This is the derivative of,
58:28:600Paolo Guiotto: of, should come from Sainti, Cuba.
58:33:860Paolo Guiotto: This is 3 sine T squared.
58:37:480Paolo Guiotto: Times the deluity of sine is cosine.
58:40:840Paolo Guiotto: So we put a tree here, 1 third out here.
58:45:460Paolo Guiotto: And this is now the derivative of sine t cubed. So we integrate by parts. This is sine T…
58:54:80Paolo Guiotto: Cuba.
58:55:800Paolo Guiotto: times cos T to be evaluated between 0 and 2 pi.
59:00:620Paolo Guiotto: Minus the integer 02 pi, now the derivative Moves from Sainti, Cuba.
59:09:10Paolo Guiotto: 2 cos T, which is minus sign, so another plus time T to power 4.
59:17:730Paolo Guiotto: Back to… However, So this evaluation yields zero.
59:23:580Paolo Guiotto: Because the sign is zeroed between points, so we have one-third… Integru pi.
59:31:390Paolo Guiotto: of sine T.
59:34:440Paolo Guiotto: Super powerful.
59:42:830Paolo Guiotto: Okay, we could split into sine square, sine square…
59:50:380Paolo Guiotto: Pine square, sine squared… This is 1 minus cos square.
59:59:170Paolo Guiotto: So this is 1 third…
00:02:690Paolo Guiotto: the integral from 0 to pi of sine squared t, we already know this value.
00:08:720Paolo Guiotto: It is equal to pi.
00:11:290Paolo Guiotto: Minus the integral 0 2 pi of sine squared…
00:16:50Paolo Guiotto: P cos squared t, which is exactly the initial integral, if I'm not wrong.
00:22:220Paolo Guiotto: We started from this, no?
00:25:620Paolo Guiotto: So we get that, so if we call i the integral, i, it is equal to 1 third…
00:34:720Paolo Guiotto: pi minus I… So, pi over 3 minus i over 3.
00:43:170Paolo Guiotto: You carry on the other side, together with this, it is 1 plus 1 3rd is 4 thirds.
00:49:10Paolo Guiotto: Aye?
00:50:80Paolo Guiotto: equal pi over 3, so finally, i is equal… 3 over 4, this… So, pi over 4.
01:01:590Paolo Guiotto: So this is the value of that integral, so returning back to that formula.
01:09:270Paolo Guiotto: So we got that area.
01:11:390Paolo Guiotto: is equal to… you see, is equal to 5 plus 2 times this, which is I, So, area…
01:25:120Paolo Guiotto: Omega is equal to pi.
01:28:340Paolo Guiotto: plus 2 times i, which is pi over 4.
01:33:180Paolo Guiotto: So, pi over 2… So, 3 halves.
01:36:660Paolo Guiotto: Bye.
01:38:20Paolo Guiotto: And this is the value.
01:41:400Paolo Guiotto: Okay, so, nothing special, do the other exercises.
01:46:950Paolo Guiotto: 3, 4, 5.
01:52:100Paolo Guiotto: 2… 7… 5, 7, 11… Numbers… 5.
02:02:260Paolo Guiotto: Okay. Let's say that this,
02:06:400Paolo Guiotto: With this, we closed, basically, the part on… I told you… the next,
02:13:600Paolo Guiotto: Chapter is on surface integration, but Oh, sir.
02:20:160Paolo Guiotto: so, we just, sure.
02:27:860Paolo Guiotto: notes to Chapter 7, and we…
02:31:170Paolo Guiotto: Start an entirely new topic that has connection with the differential cycles we have seen in the initial part.
02:38:910Paolo Guiotto: This topic is called the polymorphic,
02:45:520Paolo Guiotto: functions.
02:49:20Paolo Guiotto: Well, what is this strange world?
02:52:730Paolo Guiotto: Now… For several reasons that you will see when you study the
03:00:650Paolo Guiotto: Mainly, information theory and things like that.
03:03:890Paolo Guiotto: It is convenient to deal with functions of complex variables.
03:09:520Paolo Guiotto: mostly for, let's say, computational reasons.
03:15:600Paolo Guiotto: In the sense that, in many cases, it is,
03:22:60Paolo Guiotto: Better to use complex numbers, because we have an algebraic structure which is, which is,
03:32:960Paolo Guiotto: helpful in certain calculations that Happens, when you study signals and, similar.
03:42:260Paolo Guiotto: Things, huh?
03:43:470Paolo Guiotto: So, here we are not interested in the reasons why you need this type of calculator.
03:51:30Paolo Guiotto: There is, also an interest in, in, having, practical, Confidence with the…
03:59:750Paolo Guiotto: This kind of, objects.
04:02:490Paolo Guiotto: From the formal point of view, every of you knows what is a complex number, no? So let's just refresh quickly the few things we need.
04:13:480Paolo Guiotto: And then we define with that our goals.
04:16:40Paolo Guiotto: So the set of complex numbers is a set that,
04:21:460Paolo Guiotto: It's a set of numbers of the form in algebraic form. A plus ID, well, AD2, A, and B are both real numbers.
04:31:670Paolo Guiotto: So, for example, objects like 3 plus I2, that, equivalently we often write 3 plus 2i.
04:43:350Paolo Guiotto: or, I don't know, 5 plus I minus 7, that actually we write as a 5 minus 7i.
04:54:290Paolo Guiotto: Or numbers like 0 plus I, that actually is written as I, and so on.
05:02:720Paolo Guiotto: Now, this letter I aims to represent a number that does not exist in real numbers, so I is called the imaginary unit.
05:21:470Paolo Guiotto: And,
05:23:80Paolo Guiotto: Its main feature that everyone must know is that I square, so I times i, is equal to minus 1. So it arises as a root of a negative number.
05:35:840Paolo Guiotto: And actually, this is the reason why these numbers were invented.
05:40:220Paolo Guiotto: more than 5… about 5 cent is our goal.
05:45:710Paolo Guiotto: Also, at that time, there were not a real reason, but, say, a convenience reason to introduce these numbers, because
05:54:720Paolo Guiotto: When mathematicians tried to solve third-degree equations, so equations like,
06:02:50Paolo Guiotto: X cubed plus 3X minus 1 equals 0, these kind of equations.
06:10:00Paolo Guiotto: They found that They always have a real solution, at least one real solution.
06:17:510Paolo Guiotto: But there was a particular mathematician who, for the first time, found a formula for that solution, and what was strange is that to write that formula, he needed to write root of negative numbers, even if the final result was a real number.
06:32:950Paolo Guiotto: And, so, this was a little bit strange, so they, they…
06:39:570Paolo Guiotto: At that time, they started to think that if they enlarge the concept of number, they would have been able to write these formulas, so that's how complex number…
06:54:570Paolo Guiotto: We're… we're born, Five, centuries ago.
07:01:190Paolo Guiotto: Now, nowadays, we consider numbers as any other, even if, normally, apparently, we do not find this kind of numbers in real life.
07:12:880Paolo Guiotto: but for certain problems, it might be convenient to use these numbers. For example…
07:23:900Paolo Guiotto: We know, everyone knows that a way to represent a number of this type, since we need the two real numbers, is to a point in the Cartagian plane.
07:39:980Paolo Guiotto: So we may identify the number A plus IB with the point AB in the Cartagen plane.
07:48:620Paolo Guiotto: Now, the Cartesian plane is not a numerical sector.
07:52:220Paolo Guiotto: We know that there is an ordering, for example, you cannot say, Boeing.
07:56:990Paolo Guiotto: less than another point, so… but there is an algebraic structure. For example, we can add points, and this is the structure of vector space.
08:07:890Paolo Guiotto: But it is not evident in operation of, as a multiplication of points.
08:13:740Paolo Guiotto: So the complex number makes this possible in a certain sense. We can introduce an operation like an algebraic structure on the Cartesian plane.
08:22:970Paolo Guiotto: So there are many good reasons for which we introduce these numbers, and So,
08:29:10Paolo Guiotto: We do not spend any further time in justifying the existential.
08:41:189Paolo Guiotto: I'm thinking, well, you know that, on, on C… Sweet.
08:51:479Paolo Guiotto: the sum… It's probably good.
08:56:580Paolo Guiotto: With their, with their inverse operation.
09:02:40Paolo Guiotto: We'd…
09:10:170Paolo Guiotto: operations.
09:12:00Paolo Guiotto: Reach out.
09:13:890Paolo Guiotto: difference, and…
09:16:630Paolo Guiotto: division between numbers. So we know that the way this formula works is the following. Whenever you have to sum two numbers, so number A plus IB, and number C plus ID,
09:32:430Paolo Guiotto: Well, actually, this is the sum of the vector space, so…
09:37:640Paolo Guiotto: We do the sum of the real,
09:40:220Paolo Guiotto: fats, so we go A plus C plus I, B plus C, or if you want, we'll treat this as
09:48:100Paolo Guiotto: a formal operation where we factorize the eye.
09:51:939Paolo Guiotto: As you are… Tuesday.
09:55:840Paolo Guiotto: High school with, ordinary algebra.
10:01:450Paolo Guiotto: And we have also a multiplication. This one, it's a bit… important, because, it's not particularly natural.
10:11:700Paolo Guiotto: operation.
10:12:890Paolo Guiotto: It comes, again, formally by developing this product.
10:16:760Paolo Guiotto: When you develop this product, you will discover that there are a number of options, so something like AC plus,
10:24:60Paolo Guiotto: I, A, B… plus IBC, plus… then you have this one, I square, B…
10:33:650Paolo Guiotto: Now, since we know that this number I arises as the root of minus 1, so I squared is minus 1, we could rearrange this as AC minus P.
10:45:910Paolo Guiotto: plus IAV.
10:49:330Paolo Guiotto: plus BC.
10:51:100Paolo Guiotto: And this is actually how the product is defined.
10:55:20Paolo Guiotto: So the idea is that the product works as the algebraic,
11:00:540Paolo Guiotto: Formal manipulation would suggest doing it.
11:04:390Paolo Guiotto: Using the… property… We arranged everything as,
11:10:280Paolo Guiotto: separating what is multiplied by i and what is not multiplied by I.
11:17:70Paolo Guiotto: But if you want to have an interpretation of this product, this is a little bit more complicated, and actually.
11:23:460Paolo Guiotto: It demands the introduction of the trigonometric notation we will see in a moment.
11:31:680Paolo Guiotto: So then you can prove that a division is possible, so you can divide by numbers different from zero, so you can do something like A plus ID, you can do the subtraction, A plus ID minus C plus ID, this is simple. You work as…
11:49:280Paolo Guiotto: to space, so you do A minus C plus IP minus D. This is easy.
11:55:770Paolo Guiotto: for the division, A plus IB divided C plus ID. The nice thing is that, here, you can do a division.
12:06:520Paolo Guiotto: Notice that this is,
12:09:00Paolo Guiotto: This is non-trivial, because identifying numbers with points, it is like if we are doing a division between vectors.
12:16:590Paolo Guiotto: We said a lot of time ago that we do not have that separation, really, in the Cartesian plane.
12:24:370Paolo Guiotto: So, in general, you cannot divide vectors by vectors, but in complex numbers, we may say that thinking points as complex numbers, for this particular case, we could do the division between two.
12:37:460Paolo Guiotto: And this division works in this way, so basically the idea is that you multiply the denominator by
12:46:420Paolo Guiotto: So we multiply this fraction, by, so, a unit, so something like, C minus ID, C minus IDE.
12:57:400Paolo Guiotto: Well, this number deserves a special name, and we will return in a moment, please. When we do the product down here, we discover that this comes C squared plus D squared.
13:07:550Paolo Guiotto: So it's real and different from zero. It can be zero only when both C and B are zero, which is
13:15:740Paolo Guiotto: number 0, so this is different from 0 if C plus ID
13:21:990Paolo Guiotto: From the zero of complex numbers.
13:24:450Paolo Guiotto: which is the number with both C and D equal to 0.
13:27:600Paolo Guiotto: And then, in the numerator, we get something which is a product of these two.
13:33:350Paolo Guiotto: So it is something like, A, C…
13:36:510Paolo Guiotto: now it comes, plus BD.
13:40:180Paolo Guiotto: Then plus I… B, C… minus AB.
13:48:740Paolo Guiotto: So, at the end, you can…
13:50:640Paolo Guiotto: split this into the number AC plus BD divided C squared plus D squared plus I,
13:59:730Paolo Guiotto: BC minus AD divided C squared plus D squared. I'm not writing this formula because we have to remind of this. We have just to remind that it is possible to do these operations.
14:11:880Paolo Guiotto: So we have an algebra, a quite rich algebra, that contains all the same operations we have with real numbers, some with the same property. So the sum is commutative, so you can
14:28:110Paolo Guiotto: split the order between the two elements of the sum and nothing changes. It's associative. There is the zero, which is the number 0 plus 90.
14:37:970Paolo Guiotto: And there is the opposite, and the same for the product. The product is commutative, is associative. In this case… in the case of the product, there is the unit, so the zero of complex numbers, I will write only one 0C.
14:53:880Paolo Guiotto: Otherwise, we will write always zero, is the number 0 plus I0?
15:00:220Paolo Guiotto: And the unit number for the complex numbers is the number 1 plus I0.
15:05:260Paolo Guiotto: We continue writing 0 and 1, because we may think that these are the 0 and 1 of the real numbers.
15:14:610Paolo Guiotto: We have a division between complex numbers, so we have a structure which is 99% the same of real numbers, but there is a 1% of difference, and the 1% is that there cannot be an order, for example.
15:28:500Paolo Guiotto: remark,
15:34:30Paolo Guiotto: It is… That's possible.
15:40:210Paolo Guiotto: 2.
15:41:170Paolo Guiotto: Define, huh?
15:44:320Paolo Guiotto: What is called a total order.
15:53:100Paolo Guiotto: on C. A total order means that whenever you take two complex numbers, you can say if they are not equal.
16:01:500Paolo Guiotto: which one is bigger and which one is smaller. This happens with the real numbers. If you take two real numbers, you can always say, one, if they are not equal, one is bigger, the other is smaller. This is impossible with the complex numbers, because
16:17:20Paolo Guiotto: The idea is, this one.
16:21:120Paolo Guiotto: Since, number i is different from number 0.
16:27:510Paolo Guiotto: Because the number 0 is 0 plus I0, formally, this is 0 plus I0, while this is 0 plus I1, so they are not the same.
16:39:450Paolo Guiotto: So, you wonder if, if, yves… There is.
16:49:600Paolo Guiotto: In a total order.
16:55:60Paolo Guiotto: Order.
16:57:190Paolo Guiotto: on C…
16:59:230Paolo Guiotto: then we should say that either I is greater than zero, since it is different, or I would be less than 0.
17:10:420Paolo Guiotto: But both of these properties cannot be verified, because take the first one.
17:15:40Paolo Guiotto: And if you multiply both sides by i, since you are multiplying by a positive number, this inequality should be preserved. So you would have that i times i should be greater than 0 times i, but 0 times i is 0, and I times i is minus 1.
17:32:80Paolo Guiotto: So you would get minus 1 greater than 0.
17:36:220Paolo Guiotto: And the same would happen here, because we… if we still multiply by i, in this case, assuming that i is a negative number.
17:45:60Paolo Guiotto: We divide by a negative number, we would invert the order of the inequality, so we would have that i times i would be greater than 0 times i, so we boils down to
17:55:730Paolo Guiotto: the same problem.
17:57:520Paolo Guiotto: So there is not an order. But apart from the order, we have an algebraic structure which is very similar to the
18:05:960Paolo Guiotto: structure of R.
18:08:900Paolo Guiotto: Now, this is quite important because, jumping a little bit,
18:17:230Paolo Guiotto: In the future, in the next future.
18:20:810Paolo Guiotto: If you remind, when we considered a function of two variables, we have seen this
18:30:140Paolo Guiotto: Let's see, when? About here.
18:33:320Paolo Guiotto: Not yet here.
18:39:420Paolo Guiotto: So when we consider the function of even two real variables, numerical function, when we define the derivative.
18:48:490Paolo Guiotto: We have immediately a problem, because if you want to import the classical definition of derivative, we should have a division between vectors, and this is impossible. Or, at least, even if it's numerical, we should be able to divide by a vector.
19:04:140Paolo Guiotto: If now we consider a function where the variable is a complex number.
19:08:660Paolo Guiotto: So, formally, all… now I've invited Patricia, but imagine for a second that X is a complex number, H is a complex number, F of X, F of X plus H are a complex number, and so also H is a complex number. This formula makes sense.
19:24:560Paolo Guiotto: So, this is interesting because for functions of complex variable.
19:29:940Paolo Guiotto: The condition of derivative can be given by this limit, and this makes this thing interesting.
19:36:900Paolo Guiotto: So, our goal is, in fact, to introduce and discuss Let's come to the goal.
19:47:740Paolo Guiotto: So our… goal… Ease.
19:53:680Paolo Guiotto: So… Introduce.
20:02:790Paolo Guiotto: Discuss.
20:06:290Paolo Guiotto: differentiability.
20:16:370Paolo Guiotto: for… functions.
20:21:730Paolo Guiotto: function of… normally we use letter Z for complex numbers, so which is defined on a domain, D, that will be now a subset of the complex plane with values in C. So, functions, complex values of complex variables.
20:38:870Paolo Guiotto: So something like F prime of Z equal to the limit when H goes to 0 of FZ plus H minus Fz divided by
20:53:20Paolo Guiotto: H. Well, now everything here is complex. So this number is complex, the numerator of this fraction is complex, and also this denominator is complex. There is no issue in defining this thing, because we have a division between complex numbers, so whatever they are, we know that this is possible.
21:12:580Paolo Guiotto: And we don't know what happens when we do this, but let's say that since this, in principle, is possible, we want to explore a bit this definition.
21:23:810Paolo Guiotto: And this will, this will,
21:27:310Paolo Guiotto: reveal a number of surprises that makes this definition extremely particular and different from the… what we might expect from the definition of derivative we have seen previously. So.
21:42:680Paolo Guiotto: It's the same definition, but this will have a certain number of important consequences.
21:49:590Paolo Guiotto: Before we can take this, the first step is to
21:54:970Paolo Guiotto: Take confidence a bit better with…
21:58:20Paolo Guiotto: functions of complex variables, because you have never seen this, you have seen just the algebra of complex numbers, but functions of complex variables is something new.
22:09:40Paolo Guiotto: So, we will start talking about, we… we'll… Start.
22:18:650Paolo Guiotto: Talking.
22:22:30Paolo Guiotto: about, functions.
22:28:580Paolo Guiotto: Off.
22:30:140Paolo Guiotto: Complex.
22:34:290Paolo Guiotto: valuable.
22:36:670Paolo Guiotto: So… Let's start there.
22:39:430Paolo Guiotto: introducing,
22:41:890Paolo Guiotto: first examples of functions. Well, a function, you know, it will be always the same concept. It's a rule to assign it to a number, another number, okay? So, a function f of the complex Bible Z will be defined on some domain B,
22:58:840Paolo Guiotto: which is now a subset of complex numbers with values in C. So you associate to a complex number Z,
23:07:220Paolo Guiotto: In C, another complex number that we call F of Z that will be still
23:12:740Paolo Guiotto: in C. Now, let's start… Taking confidence with the example of these functions.
23:27:660Paolo Guiotto: Okay, so a 3DL example is the simplest possible function is a constant. So, a function f of z constantly equals to some number, C, complex number.
23:41:910Paolo Guiotto: Is a constant function.
23:49:960Paolo Guiotto: is a, say, stupid example of function. Stupid… it's an important type of function, also constants deserve the name of functions. So if you want to have a known constant function, you could take, for example, F of Z equals Z itself.
24:05:250Paolo Guiotto: This is a function.
24:06:800Paolo Guiotto: Or a little… more in general, well, let's say a different example, I could take 3Z plus 1.
24:14:270Paolo Guiotto: So this is a first-degree polynomial, no? So the…
24:18:420Paolo Guiotto: it belongs to a more generalized class, something like, A… well, let's say that… let's use… for a second, let's use letter A. AZ plus B.
24:32:30Paolo Guiotto: But notice that these numbers, A and B, can be complex, not only something like 3Z plus 1, I could say, for example, IZ plus,
24:46:440Paolo Guiotto: to Y. Why not? This is a function of this type, no? This is my A, and this is my B.
24:53:710Paolo Guiotto: Well, since,
24:55:880Paolo Guiotto: Let me use the above letter A and B for the real and imaginary part of the number we did.
25:01:670Paolo Guiotto: We will use different letters here.
25:04:200Paolo Guiotto: But in any case, this is the example of a first… Decoon.
25:12:420Paolo Guiotto: polynomial.
25:17:390Paolo Guiotto: we can describe second-degree polynomials, I don't know, something like Z squared, or Z squared plus 1.
25:27:370Paolo Guiotto: plus… plus I, Y naught, z squared plus 1, or IZ square minus 3.
25:35:570Paolo Guiotto: Or more in general, AZ square plus BZ plus C. This is a second-degree polynomial.
25:45:690Paolo Guiotto: So notice that to define these functions, I just need the algebra.
25:49:810Paolo Guiotto: Because,
25:54:680Paolo Guiotto: Probably not.
25:59:710Paolo Guiotto: Yeah, the algebra.
26:01:310Paolo Guiotto: these things.
26:03:170Paolo Guiotto: So you see that, for example, if I think that Z is a vector, so is an XY, I cannot do Z squared. I've not defined what is XY squared for vectors, so I do not have…
26:16:250Paolo Guiotto: This type of functions.
26:18:410Paolo Guiotto: But for a complex number, it seems to be easier. I can take just the variable and get a square.
26:24:260Paolo Guiotto: So, in general, I will have an extension of this class will be a generic polynomial.
26:31:360Paolo Guiotto: Okay, a polynomial of a certain degree, say, n, so it will be… now we have to use indexes to say this, so something like ANZ to the N plus AN minus 1, z to n minus 1, etc, plus…
26:49:250Paolo Guiotto: We have A1Z21 plus A0.
26:54:420Paolo Guiotto: Where these numbers, A0, A1.
27:01:420Paolo Guiotto: etc. AN can be any complex numbers. So this is an ant… 10th degree.
27:12:900Paolo Guiotto: polynomial.
27:16:790Paolo Guiotto: Now, to have a really nth degree, I need that nth degree coefficient at nth degree.
27:24:420Paolo Guiotto: Fujito, provided the… Yan.
27:28:350Paolo Guiotto: went from Syria.
27:30:890Paolo Guiotto: So, in this way, we have a certain class of functions, polynomials.
27:36:340Paolo Guiotto: Now, let's see some function which is not of polynomial type, some simple, easy function.
27:42:750Paolo Guiotto: For example, I could define a function in this way. You take your Z,
27:48:550Paolo Guiotto: let's use letters like Z for the variables, for the complex variable, and XY for the real variables. So in such a way that representing number Z as X plus IY, where X and Y are real.
28:05:720Paolo Guiotto: Well, a function could do the following. Take Zadda, which is the number X plus Y, and give
28:12:470Paolo Guiotto: give to this value X. Now, X is what is formerly what we call the real part of Z.
28:18:580Paolo Guiotto: So the function is the real part of
28:21:150Paolo Guiotto: the complex number Z. So if you want to do F of i is the real part of i, what is the real part of i?
28:29:830Paolo Guiotto: Zero, okay?
28:31:500Paolo Guiotto: And, similarly, we could take the imaginary part of Z.
28:37:370Paolo Guiotto: Remind that the imaginary part is always real.
28:40:960Paolo Guiotto: Even if it is called imaginary. It is the coefficient of I, okay?
28:46:170Paolo Guiotto: So this is another example. These are not polynomials, even if they are somehow first-degree things in the variable, but the variable is not Z. You see, it's not F of Z equals Z.
28:59:690Paolo Guiotto: it is equal to real part of Z, so…
29:02:770Paolo Guiotto: That's not a polynomial function, in fact, even if it is a first-degree function in some sense.
29:09:10Paolo Guiotto: Another important function could be the conjugate.
29:13:560Paolo Guiotto: So the conjugate, means that if we want to see the function in action on the complex number X plus IY,
29:23:70Paolo Guiotto: This does this. It gives the value to X plus AY gives the value X minus.
29:29:660Paolo Guiotto: I, wow, this is the candidate.
29:32:60Paolo Guiotto: Okay? So if you want to compute what is F of 3 plus I2, it is equal to 3 minus i2. This is…
29:42:980Paolo Guiotto: Okay, so it's a clear, load that gives to any complex number a value.
29:51:00Paolo Guiotto: A final example I want to give you is this one.
29:55:210Paolo Guiotto: the models…
29:56:560Paolo Guiotto: of the complex number Z, you know that the models is defined by taking the root of the real part of Z squared.
30:07:750Paolo Guiotto: plus the imaginary part of that, Square.
30:12:590Paolo Guiotto: So, well, let's write in with the algebraic notation. This means that F of X plus IY is the root of X squared plus Y squared.
30:25:570Paolo Guiotto: No? So, if I want to see what is of X of i, f of i, what is it?
30:31:540Paolo Guiotto: This is F of 0 plus I1, right? So X is 0, Y is 1. This gives the root of 0 squared plus 1 squared, so root of 1 equal to 1. So the value that this function assigns to i is 1.
30:49:250Paolo Guiotto: What is the value of, 2 minus i? This is,
30:54:560Paolo Guiotto: as if you want F of 2 plus i minus 1.
30:59:580Paolo Guiotto: So, it means root of 2 squared plus minus 1 squared.
31:05:280Paolo Guiotto: So it is a root of 5.
31:07:770Paolo Guiotto: This is the value. Why not? Okay? It doesn't matter if the…
31:12:310Paolo Guiotto: the image, in this case, is a real number. We always think that this is a real, but we see always reals as part of complex numbers. Reels are those complex numbers, A plus AB,
31:29:660Paolo Guiotto: where the imaginary part, the B, is equal to 0. Those are reals, okay?
31:35:990Paolo Guiotto: So we always think reals as a subset of complex numbers.
31:42:280Paolo Guiotto: Okay, we'll stop here for today. Next time, we will explore more complex functions, okay?
31:48:590Paolo Guiotto: Don't get pregnant.
31:50:940Paolo Guiotto: What?