AI Assistant
Transcript
00:03:320Paolo Guiotto: Okay, so… Last time, we… Huh.
00:09:680Paolo Guiotto: We, finished the… The class introducing the definition of What is a submercial?
00:19:880Paolo Guiotto: He's why not?
00:22:600Paolo Guiotto: this is fine by… functions.
00:30:400Paolo Guiotto: timing or not?
00:33:940Paolo Guiotto: Okay.
00:36:290Paolo Guiotto: M functions, G1, GM, which are the constraint For this problem.
00:42:610Paolo Guiotto: Now, when we have a unique constraint, so a unique function.
00:47:130Paolo Guiotto: We said that submersion is when the gradient is different from zero on there.
00:54:130Paolo Guiotto: Now, we could equivalently say this, saying that the rank of this one single vector is 1.
01:02:220Paolo Guiotto: Gradient different from 0 means that the vector is different from 0, the gradient vector, so the vector generates a space of dimension 1. A single vector can generate a space of dimension 1 when it's different from 0, a space of dimension 0 when it is 0, so…
01:22:300Paolo Guiotto: Now, the general condition is when we add… M constraint.
01:27:680Paolo Guiotto: is… we say that G1GM means a submersion if the rank of that matrix is M, exactly the number of constraints, okay?
01:38:370Paolo Guiotto: So this, this extends this particular cave. Now…
01:44:470Paolo Guiotto: How do we check that this rank?
01:48:880Paolo Guiotto: This rank is M.
01:52:140Paolo Guiotto: Well, let's, do remark I had here.
01:57:330Paolo Guiotto: rank of these metrics, gradient G1.
02:02:940Paolo Guiotto: Radiant GM.
02:06:260Paolo Guiotto: is equal to a month.
02:09:110Paolo Guiotto: if and only if these n vectors are linearly independent, If and only if,
02:17:840Paolo Guiotto: Well, or, let's say in another way, the rank is dimension… dimension of… Factor… space… generated… by… lines.
02:38:540Paolo Guiotto: In this case, lungs are these, ambassadors.
02:43:250Paolo Guiotto: R.
02:45:930Paolo Guiotto: columns.
02:51:270Paolo Guiotto: is… M.
02:54:850Paolo Guiotto: So…
02:55:890Paolo Guiotto: If and all if, these vectors are linearly independent. If and all if, what? Now, let's look at this metric here.
03:03:580Paolo Guiotto: The metrics made of these gradients, G1, GM.
03:09:410Paolo Guiotto: Remind that each gradient is an array of partial derivatives, so there is derivative with respect to the first variable of G1, derivative with respect to the second variable of G1, etc.
03:20:360Paolo Guiotto: Derivative with respect to the last variable, let's say that D is the number of variables, M is the number of constructs.
03:28:210Paolo Guiotto: So we have, the same here for G2.
03:36:260Paolo Guiotto: No, no, no, there is a mistake here, G1.
03:40:710Paolo Guiotto: G2, and so on. The same for GM.
03:51:810Paolo Guiotto: Now, the rank is M if the dimension of the vector space generated by lines or by columns, yeah, it's important, this second factor, is M.
04:03:520Paolo Guiotto: So, if we look at columns, here we have D columns.
04:08:800Paolo Guiotto: And M lines, huh?
04:12:600Paolo Guiotto: Now, to be M, this rank, a first remark is that the number D, the number of variables, must be at least equal to M.
04:23:250Paolo Guiotto: Because if the number D is less than M, We noticed that, huh?
04:36:360Paolo Guiotto: Necessarily.
04:42:330Paolo Guiotto: This D must be at least equal to M.
04:46:570Paolo Guiotto: In other words, we cannot have,
04:49:320Paolo Guiotto: The number of constraints which is larger than the number of variables.
04:55:390Paolo Guiotto: So, for example, there are two variables. I can have, at most, two constraints. I cannot have three constraints for two variables.
05:04:440Paolo Guiotto: As, I have 3 variables, I cannot have 4 constraints.
05:09:800Paolo Guiotto: Okay, the number of variables must be greater than the number of questions. Why? Because if the country happens, if T is less than M,
05:21:570Paolo Guiotto: then if you look at columns, there are D columns, which is less than M, so there are
05:27:330Paolo Guiotto: D column vectors, a number of vectors which is inferior to the number M. These D vectors cannot generate a space which is of dimension M, because they are less than M.
05:40:630Paolo Guiotto: So to generate a space of dimension 1, you need at least one vector.
05:44:430Paolo Guiotto: Right? To generate a space of dimension 2, you need two linearly independent vectors, but you need at least two vectors. You cannot do a space of dimension 2 with one vector.
05:55:570Paolo Guiotto: If you want to do a space of dimension 3, you need 3 vectors, 3 linearly independent vectors. But if you have only 2, you can generate a plane, not a space, you see? So, if you want to generate a space of dimension M, which is the condition we need here.
06:12:80Paolo Guiotto: We must have at least M lines and at least M columns. We have M lines, but we have D columns, so if B is less than M, we cannot generate a space of dimension M with D columns, you understand?
06:27:650Paolo Guiotto: If this is less than M, then necessarily rank
06:31:620Paolo Guiotto: would be… the rank is always less than the number of lines and number of columns, so would be less liquid than D, which is less than M, so…
06:40:80Paolo Guiotto: So, it… cannot be… equal to N.
06:47:620Paolo Guiotto: So, first remark is that the number of lines, so the number of columns, sorry, so the number of variables must be at least equal to M.
06:57:520Paolo Guiotto: Now, this means that, in general, this could be a square matrix if the two numbers are the same, or a rectangular matrix where B is bigger than N.
07:07:550Paolo Guiotto: Now, to have rank M, in this case, you need that there are at least M columns which are linearly independent. So let's say that you have column 1, column 2, column 3, etc, until column M. So let's say that these are M columns.
07:25:270Paolo Guiotto: If these are linearly independent.
07:30:220Paolo Guiotto: The submetrics that you have here, this is a submetrix with M lines, always, and then columns, so it's a square matrix.
07:41:140Paolo Guiotto: That matrix will have linearly independent vectors, so it's a rank matrix, and this means that its determinant will be different from zero.
07:51:860Paolo Guiotto: So, if determinant is different from 0 for this matrix, you are sure that the rank of the big matrix is M.
08:00:360Paolo Guiotto: But in general, if you form a matrix with M columns out of D, And,
08:09:180Paolo Guiotto: The determinant of this submetrix is different from zero, this tells that the big matrix has around 10.
08:15:760Paolo Guiotto: So the rank is M as soon as at least one M by M submetrix has determinant different from 0.
08:26:620Paolo Guiotto: So…
08:30:760Paolo Guiotto: We have… That, huh?
08:35:370Paolo Guiotto: That has nothing to do with the gradient, so… okay, it's a fact from mathemesis. So, we have that,
08:42:640Paolo Guiotto: rank, of this matrix, so the matrix made by the gradients G1, GM, is Emma.
08:54:750Paolo Guiotto: If, and only if, at least One.
09:01:910Paolo Guiotto: M by M.
09:04:690Paolo Guiotto: submetrics.
09:10:890Paolo Guiotto: Pause.
09:13:280Paolo Guiotto: determinant.
09:14:730Paolo Guiotto: Different from zero.
09:17:480Paolo Guiotto: Okay?
09:18:890Paolo Guiotto: Once the matrix of this one.
09:24:150Paolo Guiotto: And in particular, so when the rank 1 PM, Equivalently…
09:37:70Paolo Guiotto: Because this is what we will prove. If you think to the case M equals 1,
09:42:120Paolo Guiotto: We never check when the gradient is different from zero. We looked at points where gradient is zero, where this condition fails. If we are able to prove that points where gradient of G1 is 0 are not in D, it means that on B, gradient of G1 is not…
10:00:600Paolo Guiotto: And here we do the same. Instead of checking that the rank is M,
10:05:150Paolo Guiotto: we check that the rank cannot be less than M, so that is rank of the matrix.
10:12:380Paolo Guiotto: Gradient G1. Gradient GM.
10:17:60Paolo Guiotto: is less than M,
10:20:160Paolo Guiotto: cannot be bigger than M, because there are exactly M lines, so at most it's M.
10:25:730Paolo Guiotto: It is not M. If and only if, what should happen? So, to be M, at least one M by M submetrix must have the terminal different from 0.
10:37:330Paolo Guiotto: And so, if it must be less than M, this must be false. And this means that all
10:43:820Paolo Guiotto: M by M. Symmetrices.
10:51:670Paolo Guiotto: have.
10:53:10Paolo Guiotto: determinant.
10:54:970Paolo Guiotto: equals zero.
10:57:450Paolo Guiotto: And this means that we will have a subsystem of equations to determine these points. Let's do an example.
11:05:480Paolo Guiotto: For example, And this is a typical case, huh?
11:14:200Paolo Guiotto: D is 3, so we have 3 variables, X, Y, Z, R, variables.
11:25:590Paolo Guiotto: And M is 2. So, G1, G2, constraints.
11:39:700Paolo Guiotto: So we have two constraints on, on, tree variables.
11:45:900Paolo Guiotto: So this means that that matrix has two lines, so the matrix made of the two gradients, G1, G2,
11:53:900Paolo Guiotto: is the metrics, So DX, G1, DY, G1DZ.
12:02:480Paolo Guiotto: G1, this is the gradient of G1, and similarly, DXG2, DY, G2…
12:10:300Paolo Guiotto: DZ. G2, that's the gradient of G2.
12:13:820Paolo Guiotto: As you can see, this is a 2x3 matrix.
12:20:590Paolo Guiotto: Okay?
12:21:880Paolo Guiotto: The biggest possible value for the rank is to… cannot be 3.
12:28:180Paolo Guiotto: Okay, because if you look at lines, you have only two lines, two vectors, they can generate at most a space of dimension 2. They cannot generate a space of dimension 3.
12:38:490Paolo Guiotto: You may say, but the three columns may generate a space of dimension 3. That's indeed the false thing to the components. These are plane vectors. You see, these are vectors with two components. So they are in the Caucasian plane.
12:52:140Paolo Guiotto: Yes, if you are in the plane, at most you generate a space of dimension 2, not a space of dimension 3.
12:58:540Paolo Guiotto: demand tool.
12:59:860Paolo Guiotto: the plane, okay?
13:01:780Paolo Guiotto: So, in any case, the rank of this matters can be at most 2.
13:07:90Paolo Guiotto: So, which is exactly the number of constraints. So, when the rank is known to, Rank of these metrics.
13:17:40Paolo Guiotto: G1, greater than G2 is less than 2.
13:20:990Paolo Guiotto: If and all if, as we said.
13:24:120Paolo Guiotto: There is not a 2x2 submetrix with the terminal
13:29:110Paolo Guiotto: different from zero. So, all 2x2 submetrices of this have determinants equal to zero. How many matrices can we do in that way?
13:38:830Paolo Guiotto: This is one matrix.
13:41:660Paolo Guiotto: This is a second matrix.
13:45:360Paolo Guiotto: And you remind that this condition, we already met, something like this. And this is a third matrix.
13:51:270Paolo Guiotto: So this means that we have a system with the three determinants. Determinant of the blue… I don't copy the entries, but the blue matrix equals 0.
14:01:780Paolo Guiotto: Then you have determinant of the yellow matrix.
14:06:940Paolo Guiotto: equals zero.
14:08:820Paolo Guiotto: And finally, determinant of the red.
14:12:220Paolo Guiotto: Matrix, sir?
14:14:160Paolo Guiotto: equals zero.
14:15:360Paolo Guiotto: All these mattresses are 2x2, okay? So, 2x2, 2x2… 2 by 2.
14:23:740Paolo Guiotto: you get a system with 3 equations, with the three unknowns, X, Y, Z.
14:29:50Paolo Guiotto: Solutions of this system are points where the rank is not 2, so are points where
14:36:350Paolo Guiotto: the map G1G2 is not submerged.
14:39:590Paolo Guiotto: those solutions.
14:42:90Paolo Guiotto: Oh.
14:43:30Paolo Guiotto: If any, of course, huh?
14:46:760Paolo Guiotto: off.
14:48:610Paolo Guiotto: this.
14:50:290Paolo Guiotto: System.
14:54:970Paolo Guiotto: points.
14:57:440Paolo Guiotto: Wow.
15:00:20Paolo Guiotto: the map, G1, G2… is not.
15:07:00Paolo Guiotto: Submersion.
15:11:480Paolo Guiotto: Okay, so let's see an example, okay?
15:15:740Paolo Guiotto: Let's switch on the map. But maybe for this example, I will stay… Hey, junk.
15:23:920Paolo Guiotto: So, I want to take a… A part of an exercise.
15:28:830Paolo Guiotto: So this is, exercise, 2… 9… 17, huh?
15:38:260Paolo Guiotto: The domain D is described as a set of points X, Y, Z in R3.
15:47:510Paolo Guiotto: Such that we have two equations. The first one is Z squared equal X squared plus Y squared plus 1.
15:57:390Paolo Guiotto: Together with the second one, which is Z equal to X squared plus Y squared.
16:04:350Paolo Guiotto: You'll see there are two equations that… two conditions on points XYZ.
16:09:520Paolo Guiotto: So let's discuss the first question of the problem. Xu.
16:16:990Paolo Guiotto: that D is non-empty.
16:20:530Paolo Guiotto: And, B.
16:23:880Paolo Guiotto: Ease.
16:26:170Paolo Guiotto: It's zero.
16:28:410Paolo Guiotto: Set.
16:30:260Paolo Guiotto: Fall.
16:31:880Paolo Guiotto: Stop motion.
16:38:300Paolo Guiotto: on… Let's say.
16:41:390Paolo Guiotto: So, well, the first question is, let's just find the point. Since, yeah, we have two conditions, it's better, perhaps, to do not
16:49:190Paolo Guiotto: exaggerate with the zeros in the coordinate. For example, if I put X and Y both 0, I get the first equation is Z equal 1, the second is Z equals 0, so there is no solution.
17:01:430Paolo Guiotto: So, for example, here I could say…
17:04:329Paolo Guiotto: Why don't we take a point like,
17:08:450Paolo Guiotto: I cannot even take a Z equals 0, otherwise I put force, X and Y being equal to 0, so let's say…
17:14:770Paolo Guiotto: Let's keep calling X0Z. Belongs to D if and only if we have a system, Z square equal X squared plus 1.
17:25:930Paolo Guiotto: and Z equal 2X.
17:29:480Paolo Guiotto: square, right?
17:32:70Paolo Guiotto: So we got this.
17:34:40Paolo Guiotto: So we can use, this equation and plug into the first one.
17:39:770Paolo Guiotto: to get that Z squared must be equal… X squared is Z divided 2.
17:47:330Paolo Guiotto: Z divided 2 plus 1.
17:51:280Paolo Guiotto: So this is a second-degree equation.
17:54:30Paolo Guiotto: to Z squared minus Z minus 2 equals 0.
17:59:760Paolo Guiotto: So we have solutions here. Z, let's see, 1, 2, equal, plus 1, plus minus the root of…
18:09:720Paolo Guiotto: 1 square 1 minus 4… So, times 416 plus 16.
18:18:710Paolo Guiotto: divided by 4. So something like 1 plus minus root of 17 divided 4. So we have these values for Z, but it's not yet over, because we have to find X. So we have Z equal 1 plus minus root of 17 divided by 4.
18:37:340Paolo Guiotto: And since I have X squared equals Z divided 2,
18:41:110Paolo Guiotto: This means, 1 plus minus root of 17 divided, 4 divided by 2, so divided by 8.
18:49:670Paolo Guiotto: Now, you notice that with the minus, I do not have solutions here, okay? So, Z equal…
18:55:940Paolo Guiotto: 1 minus root of 17 divided 2, and X squared equal 1 minus root of 17 divided 8,
19:06:430Paolo Guiotto: Here, there are no solutions, no? Because that one is negative.
19:11:380Paolo Guiotto: I cannot be X squared. So here we are… we are with the reals.
19:15:770Paolo Guiotto: Or the alternative is Z equal 1 plus root of 17 divided 8, now this is positive, and X squared equals this number, whatever it is.
19:26:650Paolo Guiotto: It's a positive number, so we have solutions here. So X plus minus the root of this.
19:32:880Paolo Guiotto: So it means that points like plus, minus the root of 1 plus the root of 17 divided 8
19:40:880Paolo Guiotto: The Y was 0.
19:43:460Paolo Guiotto: And the Z is 1 plus root of 17 divided 8.
19:48:50Paolo Guiotto: These are points in the Saudi East.
19:51:240Paolo Guiotto: non-empty.
19:52:700Paolo Guiotto: However, this has nothing to do with the…
19:55:360Paolo Guiotto: with what we are going to check now. The second part, which is the Interesting path to understand how
20:02:300Paolo Guiotto: these, these tests works, okay?
20:05:900Paolo Guiotto: Let's first identify the constraints. So, we can say that D is the set where G1 of XYZ is 0.
20:17:920Paolo Guiotto: together with the G2XYZ equals 0.
20:24:200Paolo Guiotto: Well… what is G1?
20:28:580Paolo Guiotto: If you look at the two equations, the first one is Z squared equals X squared plus Y squared minus 1. I can carry a Z square on the other side. I have X squared plus Y squared minus Z squared plus 1, that's G1.
20:43:130Paolo Guiotto: about G2…
20:45:810Paolo Guiotto: This is Z equal 2x squared plus Y squared. I carry Z on the other side, I get 2X squared plus Y squared minus Z. These are the two constraints.
20:58:600Paolo Guiotto: Now, clearly, these are… Differentiable, because they are polynomials.
21:05:320Paolo Guiotto: Okay, now… We have to check.
21:12:490Paolo Guiotto: We… the goal is, Huh.
21:19:340Paolo Guiotto: Check.
21:22:930Paolo Guiotto: If… G1. G2… is a submersion.
21:35:600Paolo Guiotto: on Domain D.
21:37:600Paolo Guiotto: Okay.
21:39:600Paolo Guiotto: So, this means if and only if… Iran.
21:44:560Paolo Guiotto: of the matrix, gradient G1, gradient G2 is equal to the number of constants 2, on…
21:54:520Paolo Guiotto: So, at every point of time.
21:57:40Paolo Guiotto: Now, instead of checking this, we check where this rank is known to, is necessarily less than 2, because it cannot be…
22:06:60Paolo Guiotto: Now.
22:08:600Paolo Guiotto: Rank of this matrix, gradient G1, Gradient G2.
22:15:50Paolo Guiotto: Which is, by the way, the matrix. Let's write down these gradients. So G1 is written over there. So we have…
22:22:270Paolo Guiotto: 2X, derivative with respect to X, 2Y, and minus 2Z. That's the gradient of G1.
22:30:540Paolo Guiotto: Then gradient of G2, we have 4X.
22:33:760Paolo Guiotto: 2Y and minus 1.
22:37:220Paolo Guiotto: This rank must be less than
22:39:940Paolo Guiotto: No must be. Now, the rank is less than 2, it's not 2.
22:44:80Paolo Guiotto: If and only if… All 2x2 sub-determinants of this matrix are zero.
22:50:840Paolo Guiotto: So we have a determinant of this first matrix here.
22:57:330Paolo Guiotto: So we are.
22:58:580Paolo Guiotto: 2X… to, 4X… to Y, this equals 0.
23:06:760Paolo Guiotto: Second condition, we take the determinant of this.
23:11:500Paolo Guiotto: determinant… of 2Y minus 2Z, 2Y minus 1 equals 0.
23:21:470Paolo Guiotto: And last, we have the determinant of this.
23:26:650Paolo Guiotto: So, determinant of… 2X.
23:32:40Paolo Guiotto: minus 2Z… 4X minus 1.
23:37:840Paolo Guiotto: sequence.
23:40:400Paolo Guiotto: These three conditions must be verified together, okay?
23:45:70Paolo Guiotto: So, they are a system. Let's write down what they are. So, the first determination is 4XY.
23:55:170Paolo Guiotto: minus 8XY equals 0.
24:00:450Paolo Guiotto: The second is, minus 2Y plus… 4YZ equals 0.
24:09:80Paolo Guiotto: The third one is minus 2X.
24:12:430Paolo Guiotto: plus 8XZ equals 0.
24:17:100Paolo Guiotto: Now, let's simplify and clean up. Of course, the first one is minus 4XY, so XY.
24:23:500Paolo Guiotto: equals 0. The second one we can divide by 2, factorize the Y, so we get Y that multiplies 2Z minus 1.
24:33:150Paolo Guiotto: equals 0, and the last one is similar, we factorize X, we divide by 2, so 4 said.
24:40:880Paolo Guiotto: Minus 1.
24:43:900Paolo Guiotto: equals… Now, these three questions are basically the same.
24:48:470Paolo Guiotto: So… Which is, of course, the first one, because it's directly
24:53:410Paolo Guiotto: to zeros, this yields two possibilities.
24:56:660Paolo Guiotto: One is X equals 0, the other one is Y equals 0.
25:03:10Paolo Guiotto: Plugging X equals 0 into the second equation, nothing changed, so it means Y times 2z minus 1 equals 0. Into the third equation, we get 0 equals 0, so we can cancel this one.
25:15:630Paolo Guiotto: Doing the same for the second case, we get the second equation 0 plus 0, the third one is X times 4Z minus 1 equals
25:25:960Paolo Guiotto: Okay, now, continuing here, we have X equals 0. Looking at the second equation, we have an alternative. Y equals 0, or…
25:34:700Paolo Guiotto: Again, X equals 0, and 2Z minus 1 equals 0, so Z equals 1,
25:41:140Paolo Guiotto: Oh, here we have Y equals 0,
25:45:290Paolo Guiotto: Again, either X equals 0, or…
25:49:30Paolo Guiotto: Y equals 0, and Z equal 14.
25:54:110Paolo Guiotto: Now, we read the dissolutions. The first one are 00Z.
26:00:530Paolo Guiotto: It said, real, no condition.
26:03:680Paolo Guiotto: The second one is zero. There is no information on Y, so it's generic.
26:09:380Paolo Guiotto: one half.
26:10:770Paolo Guiotto: Again, white area.
26:13:510Paolo Guiotto: They are different. The third one are the same of the first one.
26:18:280Paolo Guiotto: default ones are X, now condition, 0, 1 fourth, with X.
26:28:270Paolo Guiotto: Now, what are these?
26:30:360Paolo Guiotto: These are points where the rank is less than 2. So these are points where G1 is not a submercial.
26:38:160Paolo Guiotto: Now, we have to understand if these points are in D, because if they are not in D, we can say that only
26:44:460Paolo Guiotto: that rank is 2. You see the point or not?
26:48:60Paolo Guiotto: Okay?
26:49:380Paolo Guiotto: So, now… 0.000Z belongs D… If and only if, here, you have two equations.
27:04:330Paolo Guiotto: Okay, so you plug zero.
27:07:540Paolo Guiotto: I do not remember what is the point.
27:09:970Paolo Guiotto: 0, 0, so for me, it's better if I look here at the equations.
27:15:30Paolo Guiotto: So, in the first equation, 000Z yields Z squared equals 1. In the second equation, yields Z equals 0.
27:24:320Paolo Guiotto: So what is the conclusion?
27:26:570Paolo Guiotto: impossible. So if it is impossible, it means that none of these points are in Okay?
27:34:670Paolo Guiotto: So, no solution.
27:40:90Paolo Guiotto: Second point, 0Y1 half.
27:44:10Paolo Guiotto: This belongs to thee, if and only if…
27:47:10Paolo Guiotto: I take the equations, they are z squared equal X squared plus Y squared plus 1, so Z square is 1 fourth equal X squared, so 0 plus Y squared plus 1. Do I need to write the second equation?
28:01:900Paolo Guiotto: No, because you see, Y squared plus 1 is greater than 1, cannot be 1 fourth.
28:07:410Paolo Guiotto: So, doesn't matter here, impossible.
28:12:470Paolo Guiotto: And about the third, X01 fourth.
28:17:740Paolo Guiotto: This belongs to the if and only if, again.
28:21:430Paolo Guiotto: First equation is Z squared equal X squared plus Y squared plus 1. So, Z is 1 over 4 squared is 1 over 16 equal X squared plus 1.
28:34:370Paolo Guiotto: Again, impulse.
28:37:570Paolo Guiotto: So what does it mean?
28:39:690Paolo Guiotto: It means that none of these points where the rank is less than 2 belongs to D. So.
28:45:150Paolo Guiotto: It means that… that rank of this…
28:49:630Paolo Guiotto: metrics, G1G2, gradient G1G2, is equal to 2 on…
28:57:420Paolo Guiotto: And in particular, this means that there's G1, G2, Ease.
29:04:360Paolo Guiotto: Submission.
29:09:670Paolo Guiotto: on… Okay?
29:15:330Paolo Guiotto: We have to repeat a number of times this check, and we will… this will become more or less transparent.
29:22:280Paolo Guiotto: Okay, now we know what is a submission, so…
29:28:30Paolo Guiotto: what is the, the point here? The point is that we want to know what is the condition for minimum-maximum points similar to the
29:38:740Paolo Guiotto: Lagrange theorem, no? This was the case when we have a single constraint represented by the function G.
29:47:400Paolo Guiotto: So, the theorem says.
29:49:160Paolo Guiotto: F is the target function, the function to be minimized or maximized. G is the function that defines the domain.
29:56:810Paolo Guiotto: the equation G equals 0.
29:58:780Paolo Guiotto: If G is a submerged on that domain.
30:01:830Paolo Guiotto: At any minimum maximum point, we must have gradient F proportional to gradient G. This was the message. Now, what is the version of this theorem? Well, we have several constraints. That's what we are going to see now. So, let's say general
30:23:980Paolo Guiotto: Lagrange.
30:29:510Paolo Guiotto: Multipliers, because as you will see, there are several multipliers.
30:34:390Paolo Guiotto: Motley.
30:36:850Paolo Guiotto: Fly ass.
30:41:580Paolo Guiotto: So, let's say. The problem is… Searched… Searching…
30:56:30Paolo Guiotto: Or… Minimum.
30:58:810Paolo Guiotto: or maximum.
31:00:850Paolo Guiotto: of a function f, let's say, of back to valuable X, on a domain which is defined by a certain number of
31:08:870Paolo Guiotto: constraints, G1 of X equals 0,
31:12:160Paolo Guiotto: G2 of X equals 0, GM of X.
31:16:450Paolo Guiotto: Equals zero.
31:19:830Paolo Guiotto: Now, the theorem says, So, legs… First of all, F… G1, G2, GM, B… differentiable, functions.
31:38:100Paolo Guiotto: We cannot talk of.
31:39:860Paolo Guiotto: Anything here, if these functions are not differentiable.
31:44:800Paolo Guiotto: And, suppose that.
31:54:400Paolo Guiotto: The key assumption is that the map, G1, GM, is a submersion.
32:06:40Paolo Guiotto: on… Domain D, which is the domain defined by these equations, G1 of X.
32:12:970Paolo Guiotto: equals 0, G2 of X equals 0, etc, GM… of X, sir.
32:23:220Paolo Guiotto: Then…
32:28:460Paolo Guiotto: if, well, that's right in red.
32:32:220Paolo Guiotto: If… So, underline, if, if, if…
32:37:530Paolo Guiotto: Okay, so this does not prove anything about extra if extra… Is a minimum.
32:45:520Paolo Guiotto: or a maximum.
32:48:220Paolo Guiotto: 4F.
32:50:360Paolo Guiotto: on D.
32:51:780Paolo Guiotto: point that must be in D, let's specify.
32:56:660Paolo Guiotto: What happens? It happens that… the gradient of F at that point
33:04:850Paolo Guiotto: Now, when we have a unique constraint, we say it is proportional to the constraint. In other words, we may say it's a linear combination of the unit gradient of the constraint.
33:16:920Paolo Guiotto: here is that the message is that that gradient is a linear combination of the gradients of the constraints at that point. So lambda 1, gradient G1, plus lambda 2 gradient G2 at that point, and so on.
33:35:610Paolo Guiotto: plus… lambda M… Gradient GM.
33:42:580Paolo Guiotto: That's volume.
33:46:240Paolo Guiotto: So, then there exists… Lambda 1, lambda 2, lambda M,
33:52:980Paolo Guiotto: These are real numbers, they are cool.
33:55:900Paolo Guiotto: Lagrange multipliers, no?
33:58:980Paolo Guiotto: Now, as you have now understood, we will never basically check that this happens.
34:06:60Paolo Guiotto: Okay, because this would mean to look for points X star where this condition is verified, so the unknown is the point.
34:14:780Paolo Guiotto: With its number of coordinates, plus these coefficients.
34:19:400Paolo Guiotto: So this is a very huge system of equations.
34:22:330Paolo Guiotto: But we don't need to involve these lambdas in the search.
34:27:300Paolo Guiotto: Let's say, let's put aside this remark.
34:34:360Paolo Guiotto: Condition gradient F is linear combination of the gradients of G1 GM.
34:47:699Paolo Guiotto: Can be seen, again, as a linear algebra property.
34:53:449Paolo Guiotto: Imagine that I have these vectors, gradient G1, gradient G2, gradient GM, and I add now gradient of F.
35:02:840Paolo Guiotto: Since gradient of F is a linear combination of the others, it means that when I put together these vectors, how many they are.
35:12:920Paolo Guiotto: They are.
35:17:260Paolo Guiotto: M plus 1, exactly.
35:19:450Paolo Guiotto: Because there are the gradients of the G1GM, they are M vectors, plus the last one, gradient F. They are M plus 1 vector.
35:27:810Paolo Guiotto: So… since the G is a submersion, these vectors are linearly independent, because you remind, submersion
35:40:380Paolo Guiotto: implies that the rank of this matrix, gradient G1, gradient GM, is M.
35:48:790Paolo Guiotto: So they generate a space of dimension M, and if you say M vectors generate a space of dimension M, it means that they are linear.
35:59:690Paolo Guiotto: But, if I added that last one, so they now becomes N plus 1, they did not generate a space of dimension n plus 1, because the last one is a linear combination of the other.
36:10:170Paolo Guiotto: So this condition means that the rank of the matrix made of gradient F plus all the gradients G1,
36:20:950Paolo Guiotto: Until GM.
36:25:580Paolo Guiotto: cannot be M plus 1, because if it is M plus 1, it would mean that these vectors are linearly independent, and this is not what this relation is saying. So this is less than m plus 1.
36:40:70Paolo Guiotto: And, how do we verify this? Well, it's similar to what we have seen before for the, for the,
36:49:150Paolo Guiotto: rank equal M of this matrix. We say that we actually won't check that the rank is M, but rather we will check where rank is less than M. So when rank is less than the maximum, if and if all MIM subdeterminants are 0.
37:07:350Paolo Guiotto: Okay?
37:08:660Paolo Guiotto: here is the same. When the rank is not 10 plus 1,
37:13:190Paolo Guiotto: If and only if all M plus 1 times n plus 1 subdeterminants are 0.
37:19:480Paolo Guiotto: All.
37:21:110Paolo Guiotto: M plus 1.
37:23:570Paolo Guiotto: times M plus 1.
37:29:150Paolo Guiotto: sub-determinants.
37:32:650Paolo Guiotto: I'll see you.
37:34:380Paolo Guiotto: So let's see in the practical example, of, so.
37:40:700Paolo Guiotto: For example?
37:44:970Paolo Guiotto: well, let's see a couple of standard examples, but please do not memorize these things, okay? Remind the general rule, which is much easier, because it is very easy to
37:55:680Paolo Guiotto: confuse one condition with another and write the wrong condition, okay? For example, I have to find the maximum of,
38:05:700Paolo Guiotto: This is what we already found, by the way, yeah? A maximum of a function of three variables on a domain which is defined by a unique constraint.
38:15:720Paolo Guiotto: We already discussed this problem.
38:19:450Paolo Guiotto: Okay, the condition rank… of the matrix, gradient F, gradient G,
38:26:550Paolo Guiotto: In this case, the matrix has only two lines, because there is the gradient of F and the gradient of the unit constant. This trunk is known to…
38:34:940Paolo Guiotto: If and only if all 2x2.
38:39:720Paolo Guiotto: sub-determinants, Are equal to zero.
38:44:830Paolo Guiotto: This is what we have seen until yesterday. So, if we go back a second to yesterday's class, let's just retake one exercise
38:55:310Paolo Guiotto: We have done, for example… yeah, yeah.
38:59:330Paolo Guiotto: We have this, no?
39:01:20Paolo Guiotto: we had to find the points at maximum distance, so we have to minimize, maximize that function, X squared plus Y squared plus X squared, that's F, and there is a unique constraint, you see? So the equation is, the G here was,
39:16:680Paolo Guiotto: Where is it?
39:19:10Paolo Guiotto: the G was a bit and above.
39:22:300Paolo Guiotto: However, is the… Eric, yeah, yeah.
39:26:600Paolo Guiotto: Z squared minus XY minus 1, okay?
39:29:790Paolo Guiotto: Now, what have we done? Well, apart for the existence.
39:34:750Paolo Guiotto: We probed that, here, F is differentiable, so if, you look here, if XYZ is a minimum maximum point for F, then gradient F must be proportional to gradient G. That means the rank of this method is less than 2. Now, how do we check that this rank is less than 2?
39:52:870Paolo Guiotto: We check by imposing equal to zero all the 2x2 subdeterminants, because if one of them is different from zero, the rank becomes 2, okay? And here you have this system, all 2x2 sub-determinants equal 0, and that's exactly what we have written here.
40:11:80Paolo Guiotto: No? In this case, M plus 1 is 2, because there is one constraint, M is 1, m plus 1 is 2. All 2x2 subdeterminates are 0, okay? So we get, in this case, in this case, this matrix is…
40:28:510Paolo Guiotto: There is the gradient of F, so DXF, DYF,
40:32:430Paolo Guiotto: the ZF, then we have the gradient of G, the XG, the YG,
40:39:470Paolo Guiotto: DZG, so you see, it's a 2x3 matrix, and to have that the rank is less than 2, you impose that all 2x2 sub-determinants are 0. There are 3 2x2 sub-determinants, so determinant of… well, let's use colors. These, the blue.
40:59:390Paolo Guiotto: equals zero… determinant of… Let's say the green equals zero.
41:09:550Paolo Guiotto: And we have a third one, which is a determinant of the, say, red.
41:17:440Paolo Guiotto: Which is made by the first and third column equal zero.
41:21:310Paolo Guiotto: And we got this.
41:22:640Paolo Guiotto: Bye.
41:23:870Paolo Guiotto: Now, what if I have, now, which is the other phase with three variables.
41:30:690Paolo Guiotto: So, let's say, for example, K is, one, huh?
41:36:470Paolo Guiotto: Case 2.
41:38:970Paolo Guiotto: I have, huh?
41:40:390Paolo Guiotto: To determine minimum, or maximum, of an F, of 3 variables.
41:47:980Paolo Guiotto: On a domain defined now by two constraints. So, equation G1XYZ equals 0,
41:55:350Paolo Guiotto: Plus, there is a second condition, G2XYZ equals 0.
42:01:950Paolo Guiotto: Okay? Now, what happens to the Lagrange theorem?
42:05:850Paolo Guiotto: Actor.
42:07:600Paolo Guiotto: Minimum.
42:09:760Paolo Guiotto: Maximum.
42:11:360Paolo Guiotto: points… We have… this is by Lagrange.
42:19:680Paolo Guiotto: The gradient F must be a linear combination of the gradients of G12, so formally, gradient F is lambda 1, gradient G1 plus lambda 2, gradient G2, but we won't check this condition.
42:34:740Paolo Guiotto: Instead, we will check that the rank of the metrics made of these three vectors, gradient F, gradient G1, gradient G2, cannot be maximum. The maximum is here…
42:48:250Paolo Guiotto: 3. So it must be less than 3.
42:52:40Paolo Guiotto: Okay?
42:53:930Paolo Guiotto: And this means, if and only if all…
42:58:390Paolo Guiotto: M plus… this is the M plus 1, so the 3 by 3 sub-determinants.
43:07:20Paolo Guiotto: Zero.
43:08:250Paolo Guiotto: How many 3x3 sub-determinants are there?
43:13:440Paolo Guiotto: Do you see?
43:14:630Paolo Guiotto: Without writing the metrics.
43:18:550Paolo Guiotto: There is just one, because this is a 3x3 matrix. Look.
43:23:640Paolo Guiotto: If we have this, this is… the gradient of F is DXF,
43:29:190Paolo Guiotto: DYFDZF. That's the gradient of F. Then we have the gradient of G1. DXG1, DX, DYG1, DZ, G1.
43:41:10Paolo Guiotto: Then we have the gradient of G2. So DXG2, DYG2, DZ, G2.
43:48:740Paolo Guiotto: And as you can see, this is a 3x3 matrix.
43:52:260Paolo Guiotto: So there is just a unique determinant here.
43:55:340Paolo Guiotto: And it is the determinant of this matrix.
43:58:340Paolo Guiotto: So that condition means determinant of this equals 0. So there is not a system for this case, but a unique equation.
44:08:610Paolo Guiotto: Okay?
44:11:450Paolo Guiotto: Okay, let's see at work this, huh?
44:15:980Paolo Guiotto: By completing the solution of the exercise, we started.
44:20:840Paolo Guiotto: Here.
44:21:970Paolo Guiotto: at the end of this slide, no? So, I will continue now.
44:28:270Paolo Guiotto: the solution of Exercise 2917, we solved the first question of this.
44:35:870Paolo Guiotto: I prefer, if we do this, such a way that we
44:39:560Paolo Guiotto: I've seen a complete argument instead of doing
44:42:790Paolo Guiotto: the brack now. We should do in 2 minutes the BRAC, but…
44:46:710Paolo Guiotto: I will take this problem and complete this.
44:50:100Paolo Guiotto: So, the exercise is this one. We have this domain, show that these don't empty, and this is the zero set of submission. Go ahead and check that.
44:58:410Paolo Guiotto: Okay, so I will… do you prefer I continue down here?
45:02:860Paolo Guiotto: Maybe it's better.
45:04:190Paolo Guiotto: Because…
45:05:270Paolo Guiotto: So, let's continue here. Of course, you have to remind that what you see here is in the next slide.
45:12:560Paolo Guiotto: So, there is a question, too, which says, the usual question is, the compact…
45:22:560Paolo Guiotto: So this has nothing to do with the stuff we are watching, but it will…
45:27:80Paolo Guiotto: it will have to do with the next question. So let's see…
45:31:540Paolo Guiotto: So, D, first of all, is defined by two equations, G1 equals 0, G2 equals 0. G1, G2 are polynomials, so certainly they are continuous function.
45:43:690Paolo Guiotto: And therefore, D is, closed, and that's the standard part of the story.
45:50:660Paolo Guiotto: Now, is D bounded?
45:56:430Paolo Guiotto: And this is, in general, there is not a…
46:00:920Paolo Guiotto: There are a few general tests, but Yeah, hardly.
46:05:440Paolo Guiotto: usable here. So we have to understand if, from conditions that characterize D, we get the coordinates of points are bounded.
46:17:10Paolo Guiotto: So we have that, these two conditions also to be indeed.
46:21:550Paolo Guiotto: Z square equal X squared plus Y squared plus 1.
46:25:800Paolo Guiotto: and Z equal 2… X squared plus Y squared plus… Nothing.
46:34:40Paolo Guiotto: Things like that.
46:37:370Paolo Guiotto: Okay, be careful, because a typical error would be to rewrite this first one, for example, as, I don't know, X squared plus Y squared minus Z squared, maybe equal to minus 1,
46:51:870Paolo Guiotto: then you deduce that X, Y, and Z are bounded by 1 for a set of reason, because the sum of squares is always bounded by 1.
47:01:130Paolo Guiotto: Okay, so what do you do by this?
47:04:240Paolo Guiotto: By the way, this would be suspected, because if I put Y equals that, I have that Y squared minus Z squared 0.
47:13:830Paolo Guiotto: So I would get X equal 1… well, let's see what happens. So, let's see what happens if we explore this type of point.
47:23:790Paolo Guiotto: X, Y, and Y.
47:27:200Paolo Guiotto: Let's see.
47:28:860Paolo Guiotto: Maybe we discovered that the set is unbounded or not. Let's see what happens.
47:34:490Paolo Guiotto: So, if and only if, from the first equation, I prefer to write this, Y squared equal X squared plus Y squared…
47:45:770Paolo Guiotto: Last one, and here I'm in.
47:47:850Paolo Guiotto: A raid me.
47:49:50Paolo Guiotto: out, because X squared plus 1 is never.
47:52:670Paolo Guiotto: Zero. So these are not points.
47:55:490Paolo Guiotto: Yes, X squared equals minus 1, by the way.
47:58:190Paolo Guiotto: So, no, no, they are not interesting. Okay, so let's focus on these two conditions.
48:04:810Paolo Guiotto: And let's see what, we… Do you have an idea?
48:25:500Paolo Guiotto: I know, you put the… But this seemed to be some ideas could be an idea.
48:37:890Paolo Guiotto: Okay, I want to follow your idea, even if it's not… the idea I would use…
48:44:770Paolo Guiotto: Let's see what happens. So, you say, you will use Z equal to 2X squared plus Y squared.
48:52:800Paolo Guiotto: We plug into the first line, so this becomes 2x squared.
48:56:820Paolo Guiotto: plus Y squared, squared.
48:59:730Paolo Guiotto: equal X squared plus Y squared plus 1, right?
49:12:920Paolo Guiotto: So you say, use, this…
49:16:860Paolo Guiotto: Yes, that's also what I would do, because… but, however, this is another way. So let's see from this one what we get.
49:26:540Paolo Guiotto: Now, the interesting fact is that the second line is now an equation that has reduced a bit the complexity, because we have only X and Y, and no more XYZ. That's a good fact.
49:39:900Paolo Guiotto: And moreover, the intuition suggests what?
49:44:910Paolo Guiotto: Well, you see that at the left-hand side.
49:47:970Paolo Guiotto: Unfortunately, we do not have X squared plus Y square. Otherwise, thinking in polar coordinates, we would have rot of power 4…
49:56:190Paolo Guiotto: Equal raw square plus 1. This means that the raw cannot be huge.
50:01:540Paolo Guiotto: Because, the left-hand side is much bigger than the right-hand side, so the rock cannot be…
50:07:640Paolo Guiotto: unbounded. Raw, I mean X squared plus Y squared. Now, the point is that can we use this idea or not, even if we have the two?
50:18:780Paolo Guiotto: Yes, we can do, for example, doing this little trick, huh?
50:23:260Paolo Guiotto: That says just take 2X squared plus Y squared, and notice that it is clearly greater than X squared plus Y squared, no?
50:32:120Paolo Guiotto: So I could say that if this holds, since this, when I square, will be greater than this.
50:40:860Paolo Guiotto: I will get necessarily that X squared plus Y squared squared less or equal than X squared plus Y squared plus 1?
50:54:890Paolo Guiotto: So now, I have that. If I call raw square the quantity X squared plus Y squared, be careful, because this raw is not the norm of XYZ. It's only X squared plus Y squared.
51:08:100Paolo Guiotto: This says that the raw square is less than raw square… no, raw power 4.
51:15:440Paolo Guiotto: Or if you want, let's use another letter to avoid the confusion. Let's call you this.
51:22:440Paolo Guiotto: So this is saying that, huh?
51:25:270Paolo Guiotto: U square is less than U plus 1, right?
51:30:290Paolo Guiotto: Now, this is an inequality in U,
51:33:410Paolo Guiotto: It says that U squared minus u minus 1 is less or equal than zero.
51:40:370Paolo Guiotto: Now, if you want, you can solve, but this is a parabola.
51:44:160Paolo Guiotto: with this concavity, so you know that it is negative between two values. It doesn't matter you determine the two values, because it says there will be a U1 and a U2, and your U must be between…
52:01:60Paolo Guiotto: these two, so it says U less than, greater than.
52:05:270Paolo Guiotto: forget of this one, but this is important. U less than U2, because this says that U is X squared, so X squared plus Y squared is less than a value U2.
52:19:760Paolo Guiotto: that you want, you can determine is the right solution of this. It is 1 plus, what is it? Root of 5 divided 2. It's this one.
52:30:370Paolo Guiotto: If you want.
52:31:390Paolo Guiotto: Now, this is very good, because we've obtained a bound for X and Y.
52:37:190Paolo Guiotto: So now it's a question to get the bound also for Z.
52:40:570Paolo Guiotto: I can go back to the system and use one of the equations. I would say the first one rather… well, actually, also the second one, but since I have X squared plus Y square, it's better to use the first one.
52:53:190Paolo Guiotto: So… by… First.
52:58:270Paolo Guiotto: equation.
53:00:370Paolo Guiotto: Z squared is equal to X squared plus Y squared plus 1, but X squared plus Y squared, this guy, is less than 1 plus root of 5 over 2, whatever it is, I get that also Z squared is bounded. It's bounded by
53:17:80Paolo Guiotto: This 1 plus root of 5 divided 2 plus 1.
53:23:70Paolo Guiotto: So now we have…
53:24:530Paolo Guiotto: Z squared is bounded. We already knew that both X squared and Y squared are bounded, and therefore, X squared plus Y squared plus Z squared is bounded. If you want, you can't compute that.
53:38:330Paolo Guiotto: Constantive.
53:39:330Paolo Guiotto: Doesn't matter.
53:40:800Paolo Guiotto: So this, means that this set is bounded.
53:44:510Paolo Guiotto: So D is bound later.
53:48:470Paolo Guiotto: And therefore, V is compact.
53:54:780Paolo Guiotto: Okay, so now we have been able to discuss this. And now there is the final point, which is…
54:03:300Paolo Guiotto: determine points. There is a typo here.
54:07:320Paolo Guiotto: points… Off… D at… minimum… Maximum.
54:17:360Paolo Guiotto: Here is written, quote, We mean that, we… have… to… Determine.
54:30:850Paolo Guiotto: The quote is the third variable. So, the minimum And the maximum…
54:37:100Paolo Guiotto: of what… on D, of the function that gives the third coordinate of Z.
54:43:10Paolo Guiotto: This is our F, X, Y, Z for this problem.
54:51:00Paolo Guiotto: Okay, existence.
54:56:460Paolo Guiotto: This is easy now, because D is… Compare.
55:05:20Paolo Guiotto: F is continuous, so we can apply Baissance theorem. There exists Minimum.
55:15:90Paolo Guiotto: and maximum.
55:16:970Paolo Guiotto: of F, only.
55:19:90Paolo Guiotto: So now we know they exist, we determine.
55:23:560Paolo Guiotto: So here is where we use Lagrange.
55:27:190Paolo Guiotto: We apply…
55:31:730Paolo Guiotto: La grandeur.
55:34:50Paolo Guiotto: Theorem?
55:36:160Paolo Guiotto: So, clearly, F, G1, G2, R.
55:42:60Paolo Guiotto: Differentiable.
55:43:950Paolo Guiotto: This is evident, they are polynomials, blah blah blah.
55:47:940Paolo Guiotto: We already checked that G1, G2 is a submersion on D.
55:56:00Paolo Guiotto: ease it.
55:57:560Paolo Guiotto: Submission.
56:02:280Paolo Guiotto: on D.
56:04:130Paolo Guiotto: So this was the first question we discussed above.
56:09:630Paolo Guiotto: So… if, X, YZ Indeed.
56:18:880Paolo Guiotto: Ease.
56:20:540Paolo Guiotto: Annie.
56:22:20Paolo Guiotto: Minimum.
56:23:60Paolo Guiotto: Or maximum.
56:25:840Paolo Guiotto: point.
56:27:500Paolo Guiotto: 4.
56:28:440Paolo Guiotto: F… So, Lagrange says that, We have…
56:37:120Paolo Guiotto: Once we write first the plane space, let's say gradient F is a linear combination of… here we have two gradients, so gradient G1…
56:46:230Paolo Guiotto: plus gradient G2.
56:49:380Paolo Guiotto: Or, equivalently, this means that the rank
56:53:910Paolo Guiotto: of the matrix made of gradient F, gradient G1, and gradient G2… This rank cannot be 3.
57:04:380Paolo Guiotto: Because if it is 3, it means that the three vectors are linearly independent, so you cannot have that gradient have its combination of the other two. So it must be less than 3, sorry.
57:17:520Paolo Guiotto: Now, this is the rank of the matrix. Let's write the matrix. The gradient of F
57:23:260Paolo Guiotto: F is Z, so derivative with respect to X is 0, respect to Y is 0, respect to Z is 1, 0, 0, 1.
57:32:810Paolo Guiotto: About G1, G2, we already computed the gradients, so let's go back and copy.
57:39:520Paolo Guiotto: So they were 2X to Y minus 2Z, and 4X to Y minus 1, I want to remind.
57:47:340Paolo Guiotto: I already forgot.
57:51:970Paolo Guiotto: Minus 2Z. And the other one is…
57:57:710Paolo Guiotto: was 4, right? 4X… to Y… N… Minus…
58:05:10Paolo Guiotto: Now, the rank of this must be less than 3.
58:07:930Paolo Guiotto: Now, this would mean all 3x3 subdeterminants equal to zero, but there is just one 3x3 subdeterminant.
58:16:830Paolo Guiotto: determinant itself. So the condition is determinant of this matrix, equal to zero.
58:24:140Paolo Guiotto: Now, to compute that determinant, we choose the first line.
58:28:140Paolo Guiotto: So we have 0 times that determinant, good case, minus 0 times another.
58:33:210Paolo Guiotto: plus one… times determinant of this box, 2X.
58:40:770Paolo Guiotto: to Y… 4XA… to Y.
58:46:160Paolo Guiotto: Now… It's really fundamental that you know how to complete the demo here, okay?
58:53:570Paolo Guiotto: So, I, I, I, I, here, I applied the, the.
58:57:840Paolo Guiotto: What is the name of the formula? Do not remind me.
59:01:930Paolo Guiotto: Okay, at the end, you see that we have only this thing to compute, so 4XY,
59:08:900Paolo Guiotto: minus 8, XY equals 0.
59:11:960Paolo Guiotto: So, at the end, all this boils down into this Incredible.
59:19:60Paolo Guiotto: 5.
59:20:390Paolo Guiotto: equation, X times Y equals 0, okay? So don't be scared by, oh, they rank less than all, maybe it's an infinite problem. At the end, it's a stupid problem. So this means, be careful in the interpretation. You have an alternative. X equals 0 or Y equals 0.
59:40:20Paolo Guiotto: X equals 0 means what?
59:46:380Paolo Guiotto: Zero?
59:48:450Paolo Guiotto: Why is that? So there is no information on the other coordinates, this means they can be whatever.
59:53:430Paolo Guiotto: So YZ can be any number.
59:56:520Paolo Guiotto: And here we get the same, X0Z.
00:01:300Paolo Guiotto: X, Z, E, R. Now… What are these points?
00:07:100Paolo Guiotto: These are points where the gradient of F is combination of the other two.
00:13:140Paolo Guiotto: But they are not necessarily on deed.
00:16:930Paolo Guiotto: So we must check if they are indeed. So now…
00:21:410Paolo Guiotto: 0.0YZ belongs to D, and here you have to go back to the equations, plug this into the equation.
00:31:500Paolo Guiotto: The two equations are Z squared equal X squared plus Y squared plus 1, so Z squared equal X squared 0 plus Y squared plus 1, and the second is Z equal 2.
00:46:00Paolo Guiotto: X squared 0 plus Y squared, so we got this.
00:52:970Paolo Guiotto: So we can,
00:54:790Paolo Guiotto: plug Z equals Y squared, we have already done this kind of calculation. So we have Z squared plus Z plus 1,
01:03:890Paolo Guiotto: We solve for Z, we get the values Z equal… 1…
01:10:30Paolo Guiotto: plus, minus… what is it? Root of 5.
01:14:570Paolo Guiotto: Divided 2.
01:17:170Paolo Guiotto: And now, Y squared must be 1 plus minus root of 5 divided 2, the minus…
01:27:550Paolo Guiotto: does not… Here, any solution here, so if and only if…
01:34:910Paolo Guiotto: we have… so, the unique case is Z equals 1 plus root of 5,
01:40:740Paolo Guiotto: divided 2, and Y squared is 1 plus root of 5 divided 2, so it means Y equal plus minus this number. 1 plus root of 5
01:51:900Paolo Guiotto: Divided, sweet.
01:54:670Paolo Guiotto: So, from this, we get these points.
01:58:410Paolo Guiotto: Zero.
01:59:760Paolo Guiotto: The Y is a plus-minus root of 1 plus root of 5 divided 2, and this add is 1 plus root of 5 divided
02:11:180Paolo Guiotto: Two points.
02:14:140Paolo Guiotto: What are these?
02:15:750Paolo Guiotto: A minimum, maximum points.
02:18:770Paolo Guiotto: You don't know yet.
02:20:70Paolo Guiotto: But, minimum-maximum are among these points.
02:24:200Paolo Guiotto: We have another point to check, the point X.
02:28:170Paolo Guiotto: 0Z.
02:29:860Paolo Guiotto: This belongs to… B… It's not the same. No, it's not exactly the same, but similar.
02:37:00Paolo Guiotto: So Z square equal X squared plus Y squared plus 1, so X squared plus 1.
02:42:470Paolo Guiotto: And the second equation, Z equals 2X squared, plus Y.
02:49:440Paolo Guiotto: square. So, this.
02:51:710Paolo Guiotto: Again, we get more or less the same, so…
02:56:110Paolo Guiotto: This means that the Z square is… Z half.
03:01:380Paolo Guiotto: We have already solved this, right?
03:03:700Paolo Guiotto: It wasn't the… When we checked the… Something, yeah.
03:10:150Paolo Guiotto: I remind of this, but they do not remind well of me.
03:13:730Paolo Guiotto: It wasn't here, right?
03:15:830Paolo Guiotto: You see, point X zeroZ. So we already found this point. So let's just copy, plus, minus.
03:22:980Paolo Guiotto: Root of 1 plus root of 17 divided.
03:28:560Paolo Guiotto: So we find… Peace.
03:32:410Paolo Guiotto: plus minus, root of 1 plus root of 17 divided by 8, 0, and the other is 1 plus root of 17 divided
03:44:440Paolo Guiotto: Okay, so now these are points of D where the gradient of F is proportional, a combination of the two gradients.
03:54:270Paolo Guiotto: Okay?
03:55:540Paolo Guiotto: And that's all for the Lagrange theorem.
03:58:20Paolo Guiotto: Now, to say which one is minimum-maximum, as usual, we need to just compute the function. The function is the third coordinate, so if I compute F here, I get value 1 plus root of 5 divided 2. If I do the same here, I get 1 plus root of 17.
04:16:140Paolo Guiotto: divided by A.
04:17:840Paolo Guiotto: Which one is bigger?
04:20:290Paolo Guiotto: So root of 5 is about 4, 4 plus 1, 5, 5 halves, so 2.5 something.
04:26:510Paolo Guiotto: The other one is root of 17,
04:29:350Paolo Guiotto: which is, about, root of 16, 4. 4 plus 1, 5 over 8, so that's less than 1, the other is bigger than 1. So these are mean points.
04:41:820Paolo Guiotto: And these are max points.
04:46:940Paolo Guiotto: Okay, so we can now deserve a bracket.
04:53:500Paolo Guiotto: Let's put in pause this.
05:01:20Paolo Guiotto: What is this?
05:03:330Paolo Guiotto: It's black, but…
05:07:610Paolo Guiotto: Perhaps something's wrong.
05:12:90Paolo Guiotto: Okay, let's do another exercise.
05:17:110Paolo Guiotto: This is exercise, 2.9.20.
05:24:680Paolo Guiotto: there is the domain, which is this set of points XYZ in R3.
05:40:110Paolo Guiotto: Where?
05:41:630Paolo Guiotto: X squared plus Y squared… minus Z squared equals 0.
05:49:850Paolo Guiotto: And X squared minus Y squared equals 1.
05:54:830Paolo Guiotto: No.
05:55:670Paolo Guiotto: This must be clear. This, notation means that the two conditions must be together verified.
06:05:30Paolo Guiotto: So it's… formally, it's an intersection.
06:08:60Paolo Guiotto: Those points will verify the first, And the second, no?
06:14:300Paolo Guiotto: So the questions are, number one, as the usual, G is non-empty, And,
06:23:470Paolo Guiotto: this is non-empty, and D… Jeez.
06:28:930Paolo Guiotto: Zero… Set.
06:32:90Paolo Guiotto: Of. A. Submers.
06:36:160Paolo Guiotto: Number 2 is the compactor.
06:43:570Paolo Guiotto: Number 3…
06:48:100Paolo Guiotto: determine… Beautiful.
06:52:440Paolo Guiotto: Annie?
06:54:560Paolo Guiotto: points.
06:56:180Paolo Guiotto: That's… Minimum… or maximum… distance.
07:02:900Paolo Guiotto: to the audience.
07:05:970Paolo Guiotto: points of…
07:10:870Paolo Guiotto: So… Solution.
07:14:330Paolo Guiotto: Well, as you… I don't know if you have seen, but I published, some,
07:22:50Paolo Guiotto: examples of, the final exam. Have you seen, or not?
07:29:10Paolo Guiotto: Yeah, you don't have just to look now at this, because we may be scared. I am able to do one exercise, perhaps, out of five, but I mean, now you will see the arguments will
07:43:170Paolo Guiotto: Much shorter, because we have done, let's say, critical mass.
07:50:550Paolo Guiotto: the Corsa.
07:51:990Paolo Guiotto: However, this was just to say that sometimes you will see this as a classical exam.
07:58:60Paolo Guiotto: like… problem.
08:01:310Paolo Guiotto: So, question one, D is non-empty, we just need to find a point.
08:05:610Paolo Guiotto: So, for example, I wouldn't take X equals Y because the second question is not verified.
08:12:50Paolo Guiotto: But perhaps, Y equals Z.
08:16:910Paolo Guiotto: Now, because this force is X equals 0. So let's put Y equals 0. Let's say a point X0Z belongs to D,
08:25:920Paolo Guiotto: I… you see, if I… if I use the two coordinates equals zero, since I have two equations, there will be two equations for one remaining unknown.
08:34:370Paolo Guiotto: This seems to be… have, you know? Because two equations for one unknown.
08:40:420Paolo Guiotto: In this case, I have two unknowns, two equations, it may work. I don't know, it's not…
08:45:600Paolo Guiotto: For granted, but let's say…
08:48:310Paolo Guiotto: Let's try. So this is X squared plus 0 minus Z squared equals 0, and the second equation becomes X squared
08:58:560Paolo Guiotto: equal 1.
09:01:819Paolo Guiotto: So it can work, because this says X equal plus minus 1.
09:07:890Paolo Guiotto: Then from X squared equal 1, we get z square equal X squared, so equal 1. This means Z equals plus minus 1.
09:17:399Paolo Guiotto: So basically, this says that plus, minus 1, 0, plus, minus 1,
09:22:380Paolo Guiotto: With all possible combination of signs, so these actually are four points, and they belong short.
09:29:740Paolo Guiotto: Okay, now let's come to the important part.
09:33:710Paolo Guiotto: D is the zero set of submersion, so let's first write D as… there are two conditions, so there will be one first condition, G1 equals 0, G2 equals 0, second condition.
09:47:180Paolo Guiotto: G1 is, for example, it's already written, X squared plus Y squared minus Z squared, and for G2…
09:58:590Paolo Guiotto: We can just write X squared minus Y squared minus 1.
10:03:720Paolo Guiotto: So, X squared minus Y squared minus Y.
10:09:530Paolo Guiotto: Now, clearly, these two functions, G1.
10:16:820Paolo Guiotto: and G2 are differentiable.
10:23:150Paolo Guiotto: They are polynomials, so I don't need even the test.
10:27:540Paolo Guiotto: differentiability tester.
10:29:840Paolo Guiotto: So, The map, G1, G2.
10:38:990Paolo Guiotto: Jeez.
10:41:460Paolo Guiotto: Submersion.
10:46:170Paolo Guiotto: on D.
10:47:730Paolo Guiotto: If and only if the definition says rank.
10:51:430Paolo Guiotto: of the matrix, made of the two gradients, G1, G2,
10:56:570Paolo Guiotto: is maximum, is equal to the number of constraints. It's 2.
11:01:580Paolo Guiotto: Now, instead of checking this, we check when rank is less than true.
11:07:870Paolo Guiotto: Rank?
11:09:00Paolo Guiotto: of this matrix, greater than G1, greater than G2. Let's compute the matrix. It is the matrix rank.
11:17:440Paolo Guiotto: gradient G1 is 2X2Y minus 2Z.
11:23:580Paolo Guiotto: Gradient G2 is 2X minus 2Y and 0.
11:29:610Paolo Guiotto: This is less than 2.
11:33:20Paolo Guiotto: If and only if…
11:35:850Paolo Guiotto: So it's not maximum. There cannot be a 2x2 submetrix with determinant different from 0, so we get a system, determinant of this, 2X, 2Y, 2X,
11:50:380Paolo Guiotto: minus 2Y, this equals 0. 10.
11:54:690Paolo Guiotto: Together with determinant 2X minus 2Z, 2X0 equals 0.
12:03:510Paolo Guiotto: Together with determinant of… Yeah, we have to take 1st and 3rd.
12:10:220Paolo Guiotto: 2X, no.
12:12:330Paolo Guiotto: Second, 2Y minus 2Z.
12:17:110Paolo Guiotto: minus 2Y and 0, this equals 0.
12:21:840Paolo Guiotto: We got a system. Now, seems a complicated system, but when we do the calculations.
12:28:740Paolo Guiotto: So the first one is minus 4XY,
12:32:740Paolo Guiotto: minus another 4XY, so minus 8XY equals 0.
12:38:410Paolo Guiotto: You can already cancel this minus 8.
12:41:570Paolo Guiotto: Then, for the second line, 2X times 0, 0 minus, minus or plus.
12:48:210Paolo Guiotto: 4XZ equals 0, and we can cancel the 4, and similarly here for the last one, so basically YZ equals
12:59:10Paolo Guiotto: So at the end, we got a very simple system. XY equals 0,
13:04:450Paolo Guiotto: XZ equals 0, YZ equals 0. So let's see…
13:10:590Paolo Guiotto: What happens? Let's choose an equation, for example, the first. Here, it's completely indifferent.
13:16:20Paolo Guiotto: So we get two cases, X equals 0, or… Y equals 0.
13:22:670Paolo Guiotto: In the first case, the second equation becomes 0 equals 0, so we can discard it, the third one remains YZ equals 0.
13:29:900Paolo Guiotto: In the second alternative, the second equation remains, XZ equals 0, the third one becomes 0 equals 0, so we can cancel.
13:37:670Paolo Guiotto: We can continue here, saying that either X equals 0, Y equals 0, R.
13:45:460Paolo Guiotto: X equals 0, Z equals 0.
13:49:270Paolo Guiotto: In the second case, we have Y equals 0, and X equals 0, or Why? What's zero.
13:56:420Paolo Guiotto: and Z equals 0.
14:00:220Paolo Guiotto: There is this case which is redundant, so here we produce .000Z.
14:08:260Paolo Guiotto: Is that any real?
14:11:30Paolo Guiotto: Here, we have .080.
14:14:970Paolo Guiotto: with Y and Real.
14:17:520Paolo Guiotto: And here we have points X00, 0.
14:21:100Paolo Guiotto: Redux and REA.
14:24:40Paolo Guiotto: Now, what are these points?
14:26:40Paolo Guiotto: These are points where the rank of that matrix is not maximum, it's not true. But we have to understand if the rank is 2 on D, so now it… we have to check
14:37:940Paolo Guiotto: If those points are indie or not.
14:41:720Paolo Guiotto: Now.
14:43:810Paolo Guiotto: 0.000 set. Belongs to D… You take the characteristic equations, which are written In the text them.
14:55:950Paolo Guiotto: And, you write the system for this pointer, so considering that you have X and Y equals 0,
15:04:940Paolo Guiotto: This means the first equation becomes minus Z squared equals 0, second equation is 0 equals 1, so… Impossible.
15:15:960Paolo Guiotto: So, these points are points where the function is not the submersion, but they are not in the…
15:23:510Paolo Guiotto: And similarly, for 0Y0, This is in the if and only if. In this case, we get…
15:33:890Paolo Guiotto: So Y squared equals 0 from the first equation.
15:39:220Paolo Guiotto: And, from the second equation, minus Y squared equals 1.
15:44:380Paolo Guiotto: Which, again, is impossible.
15:49:200Paolo Guiotto: Third point, X equals 0. X zero zero. Belongs to D, if, and only if.
15:56:480Paolo Guiotto: When we plug this into the two equations, we get X squared equal to 0,
16:02:780Paolo Guiotto: And from the second one, we get X squared equal to 1.
16:09:870Paolo Guiotto: That one is impossible.
16:13:760Paolo Guiotto: Okay?
16:14:650Paolo Guiotto: So at the end, we can say that, yes, there are points where the rank of that matrix is non-2, but those points are not in D, so only the rank must be 2.
16:25:190Paolo Guiotto: So this is the conclusion.
16:27:990Paolo Guiotto: This map, G1, G2.
16:31:690Paolo Guiotto: ease.
16:33:50Paolo Guiotto: Submission.
16:38:370Paolo Guiotto: on.
16:40:750Paolo Guiotto: Question number 2.
16:43:670Paolo Guiotto: About the compactness of the domain.
16:47:130Paolo Guiotto: So, as usual, we have a standard part of the story. This is a set defined by number of equations, and these are continuous, we already noticed this.
17:00:540Paolo Guiotto: small that they are the French ball.
17:02:850Paolo Guiotto: And therefore, from this, it follows that it is closed.
17:10:220Paolo Guiotto: So now the question, is it bounded?
17:19:230Paolo Guiotto: And, here we have to give a look to the conditions. So we have…
17:23:980Paolo Guiotto: Well, let's say that point XYZ belongs.
17:30:630Paolo Guiotto: B… If and only if… The two conditions are X squared plus
17:38:550Paolo Guiotto: Y squared minus Z squared is equal to 0.
17:43:790Paolo Guiotto: And X squared minus Y squared is equal to…
17:53:350Paolo Guiotto: So… What do you think here?
17:58:150Paolo Guiotto: So the alternatives are either we are able to get that the coordinates are bounded, or we just need to prove that there are points with a bitterly large coordinates set.
18:10:140Paolo Guiotto: So, mmm…
18:20:280Paolo Guiotto: You think, right? It is?
18:22:400Paolo Guiotto: unbounded. Why?
18:48:890Paolo Guiotto: Yeah, but I've not understood exactly what is the point you want to propose.
18:54:330Paolo Guiotto: In any case, let's see if I understood the idea. So, you could say that, for example.
19:02:60Paolo Guiotto: If I look at the second equation, I have X squared equal… A 1 plus white square.
19:08:70Paolo Guiotto: No?
19:09:490Paolo Guiotto: Let's plug this into the first equation. Let's see what happens. We get 1 plus Y squared plus another Y squared, so 2Y squared, minus Z squared equals 0.
19:24:190Paolo Guiotto: So this means that, for example.
19:27:60Paolo Guiotto: Yeah, you see that I am writing something X function of Y.
19:31:900Paolo Guiotto: So when Y is big, X is bigger. And yeah, I can do the same for the second equation. So you see, this is Z squared equal 1 plus 2Y squared.
19:44:210Paolo Guiotto: I don't know if it is… Okay.
19:47:210Paolo Guiotto: So this means that, we could say X equals plus minus the root of 1 plus Y squared, which is always defined, and same for Z plus minus the root of 1 plus 2Y squared.
20:00:450Paolo Guiotto: which is always defined. So my point is plus-minus root of 1 plus Y squared
20:07:990Paolo Guiotto: Y plus minus the root of 1 plus 2Y squared.
20:15:10Paolo Guiotto: Doesn't matter, the look.
20:16:980Paolo Guiotto: of this point, but to understand that these points can be taken for every Y real.
20:23:220Paolo Guiotto: And the three coordinates are unbounded, because when you go to the infinity, this point goes to the
20:29:360Paolo Guiotto: So this is, bounded.
20:33:140Paolo Guiotto: So… the… let's say the conclusion would be that the He is closed.
20:43:710Paolo Guiotto: And, unbounded.
20:48:40Paolo Guiotto: Okay, now we can move to the third part.
20:52:50Paolo Guiotto: Which is determine points of this domain at minimum, maximum distance. This is pretty standard, because we have to minimize, maximize the distance, or better, the square of the distance. We…
21:09:210Paolo Guiotto: But… So… Determined.
21:13:290Paolo Guiotto: a minimum… And the maximum… of this function, X squared plus Y squared plus Z squared.
21:21:960Paolo Guiotto: is not the distance, it is the square of the distance. Is there anything outside?
21:27:260Paolo Guiotto: No.
21:28:320Paolo Guiotto: on domain.
21:30:70Paolo Guiotto: Now, the main D is not compacted, and even if the function FE is continuous, we cannot…
21:36:790Paolo Guiotto: Bison's theorem. However, since, this domain D is closed and unbounded.
21:49:100Paolo Guiotto: F is continuous.
21:51:210Paolo Guiotto: And, moreover, Clearly, the limit of F at infinity.
21:58:300Paolo Guiotto: Of the space, since this is the square of the norm, this is plus infinity. We can say that, for sure, there won't be the maximum.
22:07:490Paolo Guiotto: of this F on D, so there won't be points of the maximum distance, but there will be the minimum.
22:16:610Paolo Guiotto: of F on B.
22:18:610Paolo Guiotto: So these are points that we can determine now.
22:23:410Paolo Guiotto: To determine minimum Points.
22:28:630Paolo Guiotto: We… apply.
22:32:790Paolo Guiotto: Lagrange Theorem.
22:36:280Paolo Guiotto: So… We already noticed that,
22:40:440Paolo Guiotto: So F, all the functions involved, G1D2 are…
22:46:730Paolo Guiotto: Differentiable, that's evident, they are polynomials.
22:50:590Paolo Guiotto: G1, G2 is a submission.
22:58:580Paolo Guiotto: Bye.
23:00:00Paolo Guiotto: Wang.
23:01:770Paolo Guiotto: And, so, if XYZ In V is… a minimum point.
23:11:720Paolo Guiotto: We talk about only minimums here.
23:13:940Paolo Guiotto: Point.
23:15:290Paolo Guiotto: 4F… Necessarily, the gradient F at that point is a linear combination of the two.
23:24:150Paolo Guiotto: of the two constants.
23:28:670Paolo Guiotto: Or, equivalently. Now, I'm always writing this property, even if we never use that property, really.
23:36:660Paolo Guiotto: Because in applications, this property is important, okay? So, it's not now, but in the future, if you never have to work with optimizations, with constraint optimizations, this could be important to know.
23:52:440Paolo Guiotto: this fact.
23:54:590Paolo Guiotto: Okay, so the rank of the metrics made of gradient F, the gradients of G1 and G2,
24:02:920Paolo Guiotto: This cannot be 3, so it must be less than 3.
24:07:820Paolo Guiotto: Now, this is the rank of the matrix.
24:12:240Paolo Guiotto: The gradient of F is 2x2Y, 2Z.
24:17:950Paolo Guiotto: The gradient G1 is… 2X, 2Y minus 2Z.
24:25:480Paolo Guiotto: And the gradient of G2 is 2X minus 2Y is 0.
24:32:380Paolo Guiotto: This rent must be less than 3.
24:36:730Paolo Guiotto: So this means all 3x3 sub-determinants must be 0, but in this case, there is only one 3x3 subdeterminant, which is the determinant of the matrix itself, of… determinant of this
24:50:200Paolo Guiotto: Equal to zero.
24:52:310Paolo Guiotto: To compute the determinant, we, of course, try to profit up the
24:57:100Paolo Guiotto: Line or column weapons is zero.
25:00:120Paolo Guiotto: So now the formula is 2X, so the sine are plus, minus, plus, so 2X times determinant of this guy here.
25:10:870Paolo Guiotto: So, 2Y… to Zadda.
25:14:520Paolo Guiotto: 2Y minus 2Z.
25:18:290Paolo Guiotto: Then I have, sign here is minus, doesn't matter if there is the minus, so yeah, but…
25:24:520Paolo Guiotto: Minus, so if you want to put in red the design, this sign.
25:28:640Paolo Guiotto: These are the signs for the expansion.
25:31:900Paolo Guiotto: So, I have, plus the first, minus the second. Now, the second is minus 2Y.
25:39:490Paolo Guiotto: And there is the determinant of the matrix made by this and this. So, 2X… shoes at the…
25:49:970Paolo Guiotto: 2X minus 2Z.
25:53:710Paolo Guiotto: And then I have, of course, the plus zero.
25:57:710Paolo Guiotto: Who cares?
26:00:610Paolo Guiotto: Of something, I don't write. All these must be equal to zero, okay?
26:05:710Paolo Guiotto: Now, let's do the calculations. We have…
26:09:140Paolo Guiotto: 2X times… that determinant is minus 4YZ minus another 4YZ, so minus 8YZ.
26:21:630Paolo Guiotto: Here we have minus, minus plus 2Y.
26:24:470Paolo Guiotto: And that is, minus 4XZ minus another 4 minus 8.
26:30:930Paolo Guiotto: X.
26:32:40Paolo Guiotto: Zed.
26:32:950Paolo Guiotto: So, all this mess to get… what?
26:36:680Paolo Guiotto: to get XYZ equals 0, you see?
26:39:830Paolo Guiotto: You can cancel the 2… D minus 8…
26:44:690Paolo Guiotto: So what remains is, X,
26:47:580Paolo Guiotto: times Y times Z equals 0, which yields three alternatives. X equals 0, or Y equals 0, or Z equals 0.
26:59:230Paolo Guiotto: Now we have to think in terms of points, because we are looking for points, XYZ. The first one are points with X equals 0. This means 0YZ.
27:10:150Paolo Guiotto: These ones are points with Y equals 0, so they would be something like X0Z, and these ones are points with Z equal, so XYZ.
27:23:630Paolo Guiotto: Okay, so now we have found what? We have found points where the rank is less than 3.
27:31:550Paolo Guiotto: We have now to check which among them are indeed. So, we start…
27:37:480Paolo Guiotto: Checking 1.0YZ that belongs to D.
27:43:220Paolo Guiotto: Yeah, I'm always cool.
27:44:960Paolo Guiotto: Equations, so this is,
27:48:200Paolo Guiotto: the first equation, X squared plus Y squared minus Z squared, so…
27:52:930Paolo Guiotto: Y squared minus Z squared minus equal to 0, then I have X squared minus Y squared equal to 1.
28:03:130Paolo Guiotto: As you can see, the second equation is impossible, so it means that none of this is on D.
28:10:650Paolo Guiotto: Let's try with the X0Z.
28:16:00Paolo Guiotto: So this means that, it…
28:18:350Paolo Guiotto: happens. It well happens that you've got points that are not
28:22:500Paolo Guiotto: In North Domain, so you don't get that point.
28:26:140Paolo Guiotto: So now we have X squared plus Y squared minus Z squared equals 0.
28:35:970Paolo Guiotto: And the second is X squared minus Y squared, so X squared equals 1.
28:41:200Paolo Guiotto: So this means that we get X square equals 1, z square equals Y squared equals 1, so these are the four points, plus, minus 1, 0, plus, minus 1.
28:53:570Paolo Guiotto: Be careful, 4 points.
28:56:160Paolo Guiotto: So this means all combinations of sine.
28:59:830Paolo Guiotto: Okay?
29:00:800Paolo Guiotto: Because they are independent, no? X squared equals 1 means that X can be both plus or minus 1. X squared equals 1 means the same.
29:09:60Paolo Guiotto: third point is the point, XYZ0, right?
29:14:520Paolo Guiotto: XYZ0, we again plug into the equations, we get…
29:20:840Paolo Guiotto: So, X squared plus Y squared… minus z square equals 0.
29:26:430Paolo Guiotto: And the second equation is X squared minus Y squared equals 1.
29:32:820Paolo Guiotto: About this?
29:35:540Paolo Guiotto: Huh?
29:36:970Paolo Guiotto: Impossible, because the first one forces X and YB both equal 0. You plug into the second, you get 0 equals 1. So this says X equals 0, Y equals 0, but then 0 equals 1, so that's impossible.
29:54:630Paolo Guiotto: So, what is the conclusion? Now, we found these four points, they are what?
30:00:900Paolo Guiotto: They are Lagrange points, so points where the gradient path is combination of the two gradients, and they are on the…
30:09:230Paolo Guiotto: We know that minimum is among them. Now we need to compute the function, but as you can see, F at these four points yields the same value.
30:21:510Paolo Guiotto: The function is X squared plus Y squared plus X squared, so it is 1 plus 1, 2.
30:26:980Paolo Guiotto: So this means that all of them take the same value, and since the minimum is among them, they are all minimum points.
30:34:790Paolo Guiotto: So, conclusion… Points plus minus 1, 0 plus minus 1.
30:42:780Paolo Guiotto: R… Minimum.
30:46:960Paolo Guiotto: points.
30:48:510Paolo Guiotto: And that's it for the exercise.
30:52:40Paolo Guiotto: Okay, now, do exercise, Well, we have done the 17…
31:02:540Paolo Guiotto: The 20, there is a 21.
31:06:510Paolo Guiotto: So, what is 2921?
31:11:170Paolo Guiotto: The 22 is with one constraint.
31:15:650Paolo Guiotto: the 25… Maybe the 27th?
31:23:870Paolo Guiotto: the 28th… Okay, so all these are exercises similar to this one.
31:30:310Paolo Guiotto: Good, let's stop here. Have a nice weekend, and see you next week.