AI Assistant
Transcript
00:03:300Paolo Guiotto: Okay, good morning.
00:08:670Paolo Guiotto: So… Yesterday, we, good morning.
00:16:149Paolo Guiotto: It's 12.30.
00:19:430Paolo Guiotto: Yesterday, we, finished, our class.
00:24:130Paolo Guiotto: with this extension of the Lagrangian multiplier theorem, extension to the case of functions of generic number of variables.
00:35:210Audio shared by Paolo Guiotto: So…
00:36:260Paolo Guiotto: The problem is always the same. We want to find the minimum or the maximum of a certain function f on a domain D, which is defined here through an equation.
00:49:480Paolo Guiotto: of type G equals 0.
00:51:760Paolo Guiotto: Now, this theorem says that if the function F we want to minimize, maximize, and the constraint G are good functions, so differentiable functions, and G is a submersion on the optimization domain.
01:08:800Paolo Guiotto: So this means that gradient of G is never zero at every point of the set where G is 0.
01:17:630Paolo Guiotto: Then, at any minimum-maximum point for F on D, we have this condition. Gradient F is proportional to gradient G.
01:27:120Paolo Guiotto: Now, in practice, how do we check that gradient F is proportional to gradient G? We say that this means that two vectors are linearly dependent.
01:37:530Paolo Guiotto: And, well, if the two vectors are two by two vectors, so they have two components, and this is the case when the dimension is two.
01:49:260Paolo Guiotto: That's only that case.
01:52:230Paolo Guiotto: An easy test is the following. We take the matrix made of the two vectors, the two gradients. This is a square matrix, two by two matrix.
02:03:170Paolo Guiotto: And the two vectors are linearly independent if and only if determinant of the matrix is different from zero. The two vectors are linearly dependent if and only if the determinant of the matrix is zero.
02:16:940Paolo Guiotto: Now, if D is not 2, for example, if we have functions of 3 variables, so D is 3, this is no more a square matrix. It's, for example, a 2 by 3 matrix.
02:30:990Paolo Guiotto: Be careful, because lots of you will do this error. They will write determinant equals zero.
02:37:430Paolo Guiotto: But this is a nonsense. There is no determinant for a matrix which is not a square matrix. So, what is the condition for this case?
02:48:190Paolo Guiotto: Now, there is a determinant condition, but it's a little bit more involved. In fact, we want to say that these two vectors are linearly dependent. Now.
03:02:920Paolo Guiotto: in a way, equivalent way, to say this, gradient F.
03:07:770Paolo Guiotto: And gradient G.
03:11:80Paolo Guiotto: R.
03:12:320Paolo Guiotto: linearly. Dependent.
03:15:270Paolo Guiotto: Well, let's say, if and only if the dimension of the vector space generated by these two vectors is, at most.
03:25:580Paolo Guiotto: think about while I write.
03:29:120Paolo Guiotto: D.
03:30:490Paolo Guiotto: dimension of… vector space generated.
03:39:350Paolo Guiotto: by these two.
03:41:480Paolo Guiotto: Gradient F, gradient G.
03:43:850Paolo Guiotto: is, at most, Is it most?
03:48:870Paolo Guiotto: It's mostly done.
03:53:20Paolo Guiotto: One, that's right.
03:55:650Paolo Guiotto: So, think about, you have two vectors, you have two fingers, okay?
04:00:700Paolo Guiotto: What can you do with two fingers?
04:02:830Paolo Guiotto: Either you can have one-dimensional vector space, the two vectors are proportional.
04:08:860Paolo Guiotto: or two-dimensional vector space. But you cannot have three-dimensional vector space with the two You'll see?
04:16:829Paolo Guiotto: So, at most, you can have a dimension 2.
04:20:640Paolo Guiotto: But this is when they are linearly independent. If they are linearly independent, they cannot be a dimension 2. It is at most 1.
04:29:70Paolo Guiotto: Okay, now, there is a quantity that I'm sure you heard.
04:34:640Paolo Guiotto: Some, sometimes in linear algebra, which is the rank of a matrix.
04:41:780Paolo Guiotto: Now, this, this rank.
04:46:380Paolo Guiotto: This dimension of the vector space is also called the rank of the metrics, Gradient F, gradient G.
04:59:890Paolo Guiotto: So when we say rank is 1, we mean the dimension of the vector space generated by lines, and it can be proved that it is the same also by columns. So the rank… there is a rank for lines, a rank for columns.
05:16:280Paolo Guiotto: Well, this number is the same, so the rank is always the dimension of the vector space generated either by lines or by columns of the matrix.
05:26:110Paolo Guiotto: Notice that here there is a difference, because we have two lines, for example, and D columns. So if D is 3, we have 3 columns, so 3 vectors.
05:36:820Paolo Guiotto: Now, this rank must be not 2, must be less than 2, which is the maximum possible rank.
05:44:690Paolo Guiotto: Now, when this happens.
05:46:790Paolo Guiotto: Well, and this I don't know if you… if you never heard of this, this happens, the rank is not maximum, if and only if all 2x2 sub-determinants of this matrix are zero.
06:02:890Paolo Guiotto: all.
06:04:70Paolo Guiotto: 2 by 2.
06:06:510Paolo Guiotto: Sub. Determinants.
06:09:920Paolo Guiotto: of these metrics, gradient F, gradient G, R… equal to zero.
06:18:640Paolo Guiotto: Okay? Now, for example.
06:21:440Paolo Guiotto: This is, let's say, D is equal to 3, so we have functions of 3 variables, XYZ, and G is a function of XYZ as well.
06:36:120Paolo Guiotto: The metrics made of the two vectors. This is a common example that we will have in many applications, but it's not a unique case, as you will see.
06:46:680Paolo Guiotto: is now a matrix with two lines and three columns. The three columns are… corresponds to the three partial derivatives with respect to the three variables of the two functions. This is for F, and this is for G.
07:02:560Paolo Guiotto: As you can see, we have a 2x3 metric, so determinant of this equals 0 is a nonsense.
07:10:600Paolo Guiotto: But what this condition says is that all 2x2 sub-determinants… what are these? Well, in this matrix, you have 3 submetrises, which are 2x2. So, for example, we have this one.
07:26:740Paolo Guiotto: is a 2x2 submetrix. We have… this one is another 2x2 submetrix, and there is a third one which is made of, for example, this factor plus this one, okay? So let's say this plus this.
07:46:330Paolo Guiotto: So you can always have three 2x2 submetrixes in a matrix like that. Now, these three sub-determinants must be equal to zero, okay? So, the condition rank
08:00:850Paolo Guiotto: of this gradient F, gradient G,
08:04:870Paolo Guiotto: Is less than 2, so this is the condition that means the two are linearly dependent.
08:11:790Paolo Guiotto: If and only if we get a system of three equations. So, determinant… do not memorize this, we will do in practice. So, determinant of the blue matrix, I don't want to write this equal to zero.
08:28:130Paolo Guiotto: Determinant of the yellow matrix Equal to zero.
08:36:860Paolo Guiotto: And, at the same time, determinant of the green matrix
08:42:559Paolo Guiotto: equal to zero. So, you see, you get 3 equations That must be verified together.
08:51:400Paolo Guiotto: Because if only one of these three determinants is different from zero, it means that, for example, if the blue determinant is different from zero, it means that that blue matrix, this one, is invertible. But this is invertible if these two columns are linearly independent. So it means that in these three vectors, there are two, at least.
09:15:280Paolo Guiotto: which are linearly independent. So these three vectors, these three columns, generate a space of dimension 2.
09:24:160Paolo Guiotto: So the rank of this matrix is 2, because the rank is the dimension of the space generated either by lines or by columns. It's the same. So if you know that there are two columns that are linearly independent, it means that also the two lines are linearly independent, okay? So you cannot have that one of them or two of them are different from zero. Otherwise.
09:46:690Paolo Guiotto: The rank of this matrix would be equal to 2.
09:49:920Paolo Guiotto: So they must be all equal to zero, and this is a system, okay, of three equations.
09:57:320Paolo Guiotto: So now, the best thing to do is to see this condition at work, So, for example…
10:07:70Paolo Guiotto: Let's do… let's see, simple example to begin. Exercise… 2916, number 1.
10:18:350Paolo Guiotto: So, we have to find the minimum Maximum… minimum, maximum of… this function.
10:29:220Paolo Guiotto: As you will see, very easy. It's a linear function, X minus 2Y plus 2Z.
10:35:150Paolo Guiotto: on the domain D, which is defined by this equation, X squared plus Y squared plus Z squared equal 9.
10:44:300Paolo Guiotto: Okay?
10:45:680Paolo Guiotto: So let's see all the ingredients at work here.
10:48:620Paolo Guiotto: Here, we have that function F, XYZ, is the function we want to optimize, is X minus 2Y…
11:01:360Paolo Guiotto: plus 2Z, then.
11:03:580Paolo Guiotto: And domain D is the set where G is 0.
11:07:800Paolo Guiotto: And, for example, you can say that G…
11:11:620Paolo Guiotto: is, you just rewrite this equation equals 0, so X squared plus Y squared plus Z squared minus 9.
11:21:480Paolo Guiotto: Okay, so the problem is determining, searching for minimum-maximum of this F on domain G equals 0.
11:31:530Paolo Guiotto: So we are in this canonical situation.
11:34:590Paolo Guiotto: Now, as said many times, this theorem is like Fermat theorem.
11:41:410Paolo Guiotto: So the condition, gradient F proportional to gradient G does not necessarily test that there are minimum-maximum points. You have to justify differently this. So we have always to, in this kind of problem, to justify existence
12:00:360Paolo Guiotto: This follows normally from a standard argument, so the function F is a continuous function, clearly. Domain B is closed and bounded. It is closed because it is defined by an equation that involves a continuous function.
12:20:660Paolo Guiotto: Because G is continuous.
12:23:750Paolo Guiotto: and bounded.
12:26:10Paolo Guiotto: Well, this is evident, because the constraint is norm equal to 3, no? Because that's the square of the norm.
12:35:110Paolo Guiotto: So, the existence of minimum and maximum, minimum and maximum for F on D is ensured by…
12:50:130Paolo Guiotto: via SaaS.
12:52:20Paolo Guiotto: pure.
12:53:350Paolo Guiotto: So now we know these points exist, and let's determine that.
12:58:90Paolo Guiotto: So, if, X, Y, Z Indeed.
13:05:930Paolo Guiotto: Well, before, let's write, to determine Mean max points.
13:18:180Paolo Guiotto: We… Wish… to… apply, Lagrange, multiplier.
13:33:120Paolo Guiotto: to Europe.
13:35:40Paolo Guiotto: So now, to apply that theorem, there is a technical check to do before we can apply it, and the technical check is this one. We need to verify that the function G, the constraint, is a submersion on that domain. In other words, we have to verify
13:51:250Paolo Guiotto: that gradient of G is always different from 0 on the domain, okay?
13:56:250Paolo Guiotto: So… First, though.
14:01:350Paolo Guiotto: G… is a submersion.
14:09:280Paolo Guiotto: on D, where D is the domain where G is equal to 0. Let's see why. Well, how do we do this check? We take G, we have to verify that gradient is different from 0, so let's start computing the gradient.
14:23:750Paolo Guiotto: So we have the partial derivatives with respect to X of G. These are very simple here, so it's just 2X.
14:34:170Paolo Guiotto: The partial derivative with respect to Y is 2Y,
14:39:60Paolo Guiotto: And the partial derivative with respect to Z is 2Z.
14:42:440Paolo Guiotto: So clearly, these are continuous functions, and this says that this function G is
14:48:570Paolo Guiotto: Differentiable, which is a basic requirement that you need to
14:55:780Paolo Guiotto: have all these results, differentiable functions, so it's a formal check, but it should be always, done. Then.
15:04:20Paolo Guiotto: Gradient, instead of checking if gradient is different from 0, let's check where gradient is 0. Gradient of G is zero, if and only if the three partial derivatives must be equal to 0. So we get 2X equals 0, 2Y equals 0, 2Z equals 0. So this means
15:24:20Paolo Guiotto: that point XYZ is the point 000.
15:29:690Paolo Guiotto: So there is a point where, things are not good, because this pointer… at this point, the gradient is zero.
15:39:540Paolo Guiotto: However, this point is not in the domain, because the domain is the set of points where X squared plus Y squared plus Z squared is 9, and 4.000, this quantity is 0. So I can say that this is not in D,
15:57:310Paolo Guiotto: And in particular, this will mean that the gradient of G will be always different from 0 on D.
16:04:700Paolo Guiotto: So, this says that G is a submersion
16:13:890Paolo Guiotto: on D. So we now can use this theorem. So the theorem now says So… if… XYZ.
16:27:180Paolo Guiotto: Indeed.
16:28:770Paolo Guiotto: Is. Any.
16:31:280Paolo Guiotto: Mean max point.
16:35:630Paolo Guiotto: And we know they exist, so we know that we can say, let's take XYZ at any minimum maximum point. Then, at this point, we must have gradient F proportional to gradient G at point
16:51:730Paolo Guiotto: XYZ. In general, gradient F won't be proportional to the gradient of G. But at minimum point, this must happen.
17:01:30Paolo Guiotto: And we can now say that, equivalently, the rank of the matrix made of the two gradients, gradient F, gradient G,
17:10:200Paolo Guiotto: cannot be 2, because that rank means dimension of the vector space generated by the lines, or the columns of these matrix, in particular by these two vectors. So this cannot be 2.
17:21:530Paolo Guiotto: Now, this means that the rank of
17:24:619Paolo Guiotto: Gradient F is what? F is just a linear function, X minus 2Y plus 2Z, so the gradient is 1 minus 2, 2. You see?
17:35:970Paolo Guiotto: Okay, so this means that the rank of this matrix
17:40:750Paolo Guiotto: And about G, we have 2X, 2Y, 2Z.
17:45:710Paolo Guiotto: This must be less than 2.
17:48:260Paolo Guiotto: Now we come to the system of determinants. This happens if and only if the system made by the three 2x2 sub-determinants of this matrix, which are determinant of
18:02:750Paolo Guiotto: 1 minus 2, 2X, 2Y, Equals zero.
18:09:480Paolo Guiotto: determinant of minus 22Y2Z equals 0.
18:17:810Paolo Guiotto: And there is another one, determinant of 1, 2, 2x2Z equals 0.
18:29:300Paolo Guiotto: do not be scared from the appearance of this, because it's the understandable problem. We are ready to
18:37:350Paolo Guiotto: get used, for example, solving for gradient F equals 0 into systems of three equations in three unknowns. That's another system, once we write these determinants.
18:47:650Paolo Guiotto: Okay, so now let's proceed in computing the determinants we have. For the first line, 1 times 2Y, 2Y, minus… minus 2 times 2X, so…
18:59:530Paolo Guiotto: plus 4X.
19:01:950Paolo Guiotto: equals 0. That's the first equation. The second one is minus 4Z minus 4Y equals 0. The third one is 2Z minus 4X equals 0.
19:15:170Paolo Guiotto: Now, let's simplify, reduce as much as possible. We have… so we, clearly we simplify dividing by 2, so X plus,
19:24:920Paolo Guiotto: 2X plus Y equals 0, then the second line is Y plus Z equals 0, and the third line is
19:35:560Paolo Guiotto: Z minus 2X equals 0.
19:39:990Paolo Guiotto: So now what happens here? Let's solve it's a linear system, so it is, simple.
19:47:860Paolo Guiotto: system, but as you will see, there are lots of solutions, because, for example, I could say, from the third line, I get Z equal to X.
19:57:300Paolo Guiotto: From the second line, I get Y equals minus Z, which is 2X minus 2X.
20:04:690Paolo Guiotto: Well, from the first line, if you put Y equals minus 2X in the first line, you get 0 equals 0. You see?
20:12:690Paolo Guiotto: So the first line is just equal to the second line, so it's a redundant equation.
20:19:680Paolo Guiotto: So, here you end with these two equations with three unknowns. So, the number of unknowns is higher than the number of equations, and what else can you do here? Nothing more than this, because this says that solutions of these systems are points with X, whatever.
20:36:990Paolo Guiotto: Y is minus 2X, and Z is 2X. This, for the moment, whatever is X in R.
20:45:270Paolo Guiotto: Now, are these all the possible minimum-maximum points? Not yet, because what these points are.
20:55:290Paolo Guiotto: Okay? Be careful. These points are these solutions of this equation.
21:04:830Paolo Guiotto: So, we just wrote, what are the solutions of that equation? That equation is equivalent to this thing. So, the solution of the equation gradient F proportional to gradient G.
21:19:380Paolo Guiotto: Why these are not all the solutions we are interested in.
21:23:520Paolo Guiotto: Because we need for… to look for solutions that are in D. So, in general, it is not true that this point is in D. So, we have to verify now which of these belong to D. Now…
21:40:730Paolo Guiotto: Point X minus X, minus 2X, 2X, belongs to the if and only if
21:48:220Paolo Guiotto: Remind that it is the… this sphere.
21:52:80Paolo Guiotto: the surface of the sphere, X squared plus Y squared plus Z squared equals 9. So you just plug there, you get X squared plus Y squared, which is minus 2x squared, plus Z squared, which is 2X squared, equal 9.
22:07:390Paolo Guiotto: So, calculating this, you get, 9X square equals 9,
22:13:630Paolo Guiotto: And from this, X squared equals 1.
22:18:40Paolo Guiotto: So, X equal plus or minus 1.
22:22:680Paolo Guiotto: So this means that only, actually, two points of this are in D. So we get points… so when X is plus 1 is 1, minus 2, 2, and when X is minus 1, we get minus 1 plus 2 minus 2.
22:41:390Paolo Guiotto: Now, what are these points?
22:44:160Paolo Guiotto: These points are points for which gradient F is proportional to gradient G, and they are indeed… so they are exactly the solutions we are looking for.
22:55:740Paolo Guiotto: Now…
22:57:270Paolo Guiotto: we… what is the argument now? The argument is, so, if XYZ is indeed, is any minimum maximum points, then it must verify this condition. But the points, indeed, that verify this condition are only these two. So…
23:13:550Paolo Guiotto: Minimum, maximum points are among them. So this has restricted the search to just 2 points.
23:20:430Paolo Guiotto: So… Mean… And max points.
23:27:310Paolo Guiotto: R…
23:29:270Paolo Guiotto: Well, of course, yeah, we have two points, so probably one will be the minimum, the other will be the maximum. R, one…
23:37:820Paolo Guiotto: off.
23:39:230Paolo Guiotto: 1 minus 2, 2…
23:42:350Paolo Guiotto: and minus 1, 2, minus 2. And how to decide? It's just two points. We look at the values of the function. F at 1 minus 2, 2,
23:52:00Paolo Guiotto: Is the function
23:58:450Paolo Guiotto: Aye.
24:00:300Paolo Guiotto: Okay, so remind that I brought the solution of this exercise in yesterday's live.
24:06:930Paolo Guiotto: So, you won't find in today's slide. However, the function is this one, X minus 2Y plus 2Z. So, 1 minus 2Y, which is minus 2, plus 2Z, which is 2.
24:23:70Paolo Guiotto: So this, yields, what is it? 9.
24:27:990Paolo Guiotto: And F at minus 1 plus 2, minus 2 is X minus 2Y, plus 2Z.
24:39:680Paolo Guiotto: And this is minus 9. So now you understand that this is the maximum point
24:48:460Paolo Guiotto: And this is the minimum point.
24:54:200Paolo Guiotto: Okay?
24:56:260Paolo Guiotto: Do you have any question?
25:02:30Paolo Guiotto: Okay, so let's, change page, and let's see, another example.
25:11:670Paolo Guiotto: Okay, now, I do one of these exercises,
25:16:160Paolo Guiotto: exercise 2, 9, 18. This is, let's say, an exam…
25:24:620Paolo Guiotto: Style exercise with the more or less
25:28:250Paolo Guiotto: same difficulty. So it is written like this. Let d be the domain of points XYZ In R3,
25:41:460Paolo Guiotto: Which is the 18, yeah.
25:45:430Paolo Guiotto: such that Z squared is X times Y plus 1.
25:53:860Paolo Guiotto: So there are three questions. Number one, D is non-empty, show that D is non-empty.
26:01:850Paolo Guiotto: And… D is D.
26:08:100Paolo Guiotto: Zero.
26:10:450Paolo Guiotto: with zero set, off… a submersion
26:20:230Paolo Guiotto: on… be itself.
26:25:970Paolo Guiotto: Number two… Let's write the questions, then we comment.
26:30:650Paolo Guiotto: is… D… Well, actually, here, it's affirmative, let's put it in interrogative 4. Is the compactor… Question 3.
26:48:420Paolo Guiotto: determine… points.
26:53:220Paolo Guiotto: of D at… minimum, Maximum… distance… to the origin.
27:05:860Paolo Guiotto: So, 20.06.
27:09:450Paolo Guiotto: Okay.
27:15:190Paolo Guiotto: So, first of all, D is non-empty. We just need to show that there is one point, no? So let's look for a point.
27:24:730Paolo Guiotto: This thing is not a particularly interesting question, but it might be interesting, especially for the question 2 sometimes.
27:34:700Paolo Guiotto: Because…
27:36:970Paolo Guiotto: Of course, you can… you just need to exhibit one point to respond to this question, huh? So, for example, I could notice that if I take both X and Y0,
27:46:990Paolo Guiotto: that product X times Y is 0, so it remains Z squared equal 1. So, if I take Z equal plus minus 1, this is okay. So these points are in D, and so D is non-empty.
28:02:700Paolo Guiotto: And that's all for this first part of the question.
28:07:80Paolo Guiotto: But as we will return on this later, a better analysis of this will be helpful for the question too. We will return.
28:16:860Paolo Guiotto: Now, what does it mean this second part of the question? Show that D is the zero set of a submersion on D itself? What we have to check is that D is G equals 0 for some G,
28:36:290Paolo Guiotto: Which… is A.
28:40:840Paolo Guiotto: Submersion.
28:45:580Paolo Guiotto: on that domain, on G equals 0. So, and where is G? Well, of course, is that you write this equation as something equals 0. So, G is, for example, Z squared minus XY minus 1. You see?
29:02:540Paolo Guiotto: Because G equals 0 means that equation is verified.
29:05:890Paolo Guiotto: So the domain can be seen as G of XYZ equals 0, you carry everything on one side, and what is on that side is your function G.
29:17:940Paolo Guiotto: Now, we have to check that this G here
29:22:670Paolo Guiotto: Is a submersion on that domain.
29:26:260Paolo Guiotto: So, to do that, we start computing the partial derivatives of G,
29:35:790Paolo Guiotto: DDX is minus Y, DDY is minus X, DDZ is 2Z.
29:42:890Paolo Guiotto: So clearly, these are continuous functions, and therefore, the function G is differentiable.
29:52:800Paolo Guiotto: So, now, to discuss,
29:56:920Paolo Guiotto: whether it… is it a submersion or not, we, discuss where gradient of G is equal to 0.
30:04:980Paolo Guiotto: Now, gradient of G at point XYZ Is equal to zero.
30:12:430Paolo Guiotto: if and only if, what should happen? Should happen that minus Y is 0.
30:19:520Paolo Guiotto: minus X is 0, and 2Z is 0. So, in other words, the point XYZ must be .000. So, there is a point where gradient is 0.
30:35:380Paolo Guiotto: Now, is that point on the set where G is 0, which is our domain, Is that in DE?
30:47:250Paolo Guiotto: Although we check, we just plug this point into the condition. Now, this happens if and only if and only if Z squared, 0 square, is equal to X times Y zero times 0 plus 1.
31:04:90Paolo Guiotto: And this is false.
31:06:60Paolo Guiotto: So we can conclude that gradient of G is different from 0 on D, so this means on every point of the domain D, so this means that G is a submersion
31:27:490Paolo Guiotto: on D, which is the set where G is 0.
31:33:250Paolo Guiotto: Okay, so we now know that the domain is the zero set of a function, which is as a measure on the domain itself.
31:41:250Paolo Guiotto: Let's now move to question 2. Is decompact? The compact means closed and bounded.
31:47:420Paolo Guiotto: And there is no relation in general between these two properties, so we have discussed separately.
31:53:110Paolo Guiotto: So, first of all, we can say that since D is G equals 0, and G is a continuous function.
32:01:890Paolo Guiotto: So, domain is defined by an equation that involves a continuous function. Domain D is closed. That's standard, we don't need to justify.
32:14:00Paolo Guiotto: Now, the question is, is D also bounded there?
32:18:730Paolo Guiotto: So this means, roughly our coordinates of points of D bounded, the three coordinates, we should get… we should get that from the condition we see there.
32:31:430Paolo Guiotto: Now, you have to understand if that condition could be verified or not with big coordinates, for example.
32:41:180Paolo Guiotto: So, what do you think?
32:46:550Paolo Guiotto: Where would you put your hand that I can cut?
32:52:120Paolo Guiotto: No?
32:55:630Paolo Guiotto: And you're scared that… Now, don't tell me bounded or unbounded?
33:04:770Paolo Guiotto: I'm bounded.
33:06:170Paolo Guiotto: Why?
33:09:330Paolo Guiotto: No, let's, let's listen.
33:13:780Paolo Guiotto: No, but if you… if you want to prove that it is unbounded, you have to show me that there are points with big coordinates that verify that one.
33:30:630Paolo Guiotto: What? No. I wanted to show that there are points that verify that condition with big coordinates.
33:47:110Paolo Guiotto: Do you have any idea?
33:50:160Paolo Guiotto: So that's why I say that doing this first part could be useful, because I could notice that
33:56:580Paolo Guiotto: You don't need to put both coordinates equals zero, you just need one coordinate equals 0. So notice that point, for example, with the X equals 0, YZ belongs to domain D, if and only if, let's take a point of this type.
34:11:530Paolo Guiotto: Now, we need Z squared equals 0 times Y plus 1. So, this is, again, 1. So, if and only if Z is plus minus 1.
34:22:449Paolo Guiotto: But there is no condition on why.
34:25:00Paolo Guiotto: So this means that these points, 0, y plus minus 1, are, indeed, whatever is the Y, so Y real. And so this means that there is no bound for the Y coordinate.
34:39:80Paolo Guiotto: Okay? By the way, this means that if you look at these points in the space, X, YZ,
34:46:150Paolo Guiotto: We cannot draw that domain, huh?
34:49:360Paolo Guiotto: To understand what is it, but these points are points with X equals 0, so we are in the plane ZY.
34:58:30Paolo Guiotto: The Z is plus 1 minus 1, so they… they can… they can be, for example, Z equals 1 is here, Z equal minus 1 is here, and the Y is arbitrary, so this is a straight line like that.
35:12:870Paolo Guiotto: in the plane… in the plane, YZ. So these are two lines like this.
35:19:240Paolo Guiotto: to infinite lines. So the domain contains these two lines, and many other things.
35:25:560Paolo Guiotto: Okay, so this is sufficient to say that D is… Unbounded.
35:34:220Paolo Guiotto: And therefore, in particular, the… is not…
35:39:260Paolo Guiotto: compact. So, the answer is negative to the question.
35:45:630Paolo Guiotto: Okay, let's now move to question 3.
35:48:650Paolo Guiotto: Because this seems to be new, but actually it consists just in
35:53:90Paolo Guiotto: setting the problem in the right way. Determine points of the at minimum maximum distance to the origin.
36:00:270Paolo Guiotto: Well, what is the distance?
36:02:360Paolo Guiotto: If I have a point in space.
36:08:530Paolo Guiotto: say, with coordinates, X, Y, Z,
36:12:140Paolo Guiotto: The distance to the origin is what?
36:16:350Paolo Guiotto: The norm, okay? So, distance.
36:19:870Paolo Guiotto: Between XYZ.
36:23:160Paolo Guiotto: And the origin is…
36:25:920Paolo Guiotto: the norm of XYZ. So, formally, it's the root of X squared plus Y squared plus Z squared.
36:37:660Paolo Guiotto: So what I have to find, the problem consists in determining points of D at minimum, maximum distance to the origin. The distance is that function root of X squared plus Y squared plus X squared, so I have to find points where this quantity is maximum or minimum points of D.
36:56:930Paolo Guiotto: So, the problem… consists… in search… searching… for… Minimum.
37:14:560Paolo Guiotto: Maximum.
37:16:50Paolo Guiotto: of functions root of X squared plus Y squared plus Z squared on domain D,
37:25:290Paolo Guiotto: which is our domain, so the domain where Z squared is equal X times Y plus 1.
37:32:780Paolo Guiotto: So, as you can see, this is the classical optimization problem. So, minimum Maximum.
37:40:490Paolo Guiotto: of a function f, in this case of three variables.
37:44:620Paolo Guiotto: on a domain which is defined by a constraint, GXYZ equals 0.
37:51:70Paolo Guiotto: Okay, so to solve this problem, we will use the Lagrange multiplier method, but before, we have to show existence, etc.
38:02:270Paolo Guiotto: But before we enter into this, in this case, I want to show you a little trick that can be used in many other circumstances similar to this one.
38:12:590Paolo Guiotto: Now, if we go a little bit, farther,
38:18:190Paolo Guiotto: you imagine that at a certain moment, we have to compute the derivatives the gradient of F, no? The gradient of F with that kind of function means to carry around that root, because if you…
38:31:700Paolo Guiotto: For example, dxf here is the derivative of the root. So it is 1 over 2, the root of X squared plus Y squared plus Z squared, and then we have the derivative with respect to X of the argument, which is 2X. So this is the derivative. So you see that there is this…
38:50:670Paolo Guiotto: This route that basically bothers, because we have to write a lot.
38:56:140Paolo Guiotto: So there is a little trick that we can do here, and it is the following. Notice that…
39:07:520Paolo Guiotto: The root of X squared plus Y squared plus Z squared is minimum, or maximum.
39:17:60Paolo Guiotto: Exactly, if and only if what is the argument of the root is minimum or maximum, because the root is just an increasing function of its argument.
39:28:540Paolo Guiotto: So when the X squared plus Y squared plus Z squared is maximum, then the root will be maximum. When X squared plus Y squared plus Z squared will be minimum, the root will be minimum.
39:40:480Paolo Guiotto: So, if and only if X squared plus Y squared plus Z square is minimum, or maximum.
39:49:850Paolo Guiotto: So the general idea is that I have a function of X squared plus Y squared plus Z squared, which is an increasing function.
39:58:110Paolo Guiotto: So, this increasing function will take maximum value when the argument is maximum, the minimum value when the argument is minimum.
40:06:430Paolo Guiotto: And working with this function instead of this one is much better, because the gradient is just 2X, 2Y to Z, instead of having X divided the root and things like that. So…
40:22:460Paolo Guiotto: minimum.
40:25:250Paolo Guiotto: Maximum.
40:27:680Paolo Guiotto: points.
40:29:600Paolo Guiotto: of the router, X squared plus Y squared plus Z squared, are the same off…
40:41:370Paolo Guiotto: minimum maximum points of X squared plus Y squared plus Z squared.
40:47:740Paolo Guiotto: I'm not saying minimum, maximum values, because they will be different.
40:52:640Paolo Guiotto: You have one quantity and the root of that quantity, so if the quantity is 4, the root is 2, they cannot be the same.
41:00:330Paolo Guiotto: Except for very particular case, when the value is 0, when the value is 1, but in general, the values are different.
41:07:480Paolo Guiotto: But where the function, the two functions, attain minimums or maximums, they are the same points. So, instead of solving with this F,
41:19:150Paolo Guiotto: So… we, solve… for this. Minimum.
41:29:70Paolo Guiotto: maximum of X squared plus Y squared plus Z squared on the domain, which is Z squared equal XY plus 1.
41:43:140Paolo Guiotto: So this will be our F.
41:46:310Paolo Guiotto: Which is much better for calculations. Okay, now it's time to, to solve the problem. So we have first to discuss existence.
42:00:690Paolo Guiotto: Let me stress once more, I know that perhaps if you have not yet grasped this, you won't, catch right now, but let's repeat once more. Why don't I don't study, and why I want,
42:18:190Paolo Guiotto: I will penalize you if you have this kind of problem at the exam, if you just start applying the Lagrange or the Fermat theorem, without any discussion about the existing… because suppose that you solve this system without knowing that these points are minimum-maximum. So what can you say?
42:37:700Paolo Guiotto: So you would say, okay, I've done all the calculation. For example, this example here.
42:42:870Paolo Guiotto: You don't know there exists minimum, maximum, and you do look at the second part only, you know? You say, okay, if XYZ is a point of minimum, maximum, then gradient F is proportional gradient G, okay, blah blah blah. At the end, I arrive to these two points, I evaluate function.
43:01:500Paolo Guiotto: Now, I see that at the first point, I have value 9, and second, minus 9.
43:07:210Paolo Guiotto: What can I draw from this?
43:09:860Paolo Guiotto: Well, these are like points, potentially, with the horizontal tangent, but not minimum nor maximum.
43:16:620Paolo Guiotto: So if you don't know that they are minimum-maximum, you, at this stage, can say, okay, if minimum-maximum exist is one of these two, and if it is one of these two, the first one is maximum, the second is minimum. But you don't know yet, they exist.
43:32:110Paolo Guiotto: And this argument does not show they exist.
43:35:300Paolo Guiotto: It just says, if they exist, they must be these ones.
43:40:280Paolo Guiotto: But it does not prove the existence, you see? Okay? So this is the problem, that we need always to justify the existence. Now here, we have a little problem, because in point 2, we discovered the domain is not compact.
43:59:40Paolo Guiotto: So we cannot use Weissler's theorem in a standard way, but what do we know about D? D is closed
44:10:20Paolo Guiotto: and unbounded.
44:17:500Paolo Guiotto: If we know anything about this, F is continuous.
44:21:60Paolo Guiotto: Yes F is X squared plus Y squared plus N squared.
44:27:100Paolo Guiotto: You know that if the domain is unbounded, we have a result of this type, that if the function goes to infinity, at infinity, with plus or minus, we can have one of the two, minimum and not maximum, or maximum and not minimum. Here.
44:43:510Paolo Guiotto: The additional, and since, the limit…
44:50:850Paolo Guiotto: when XYZ goes to infinity 3 of F,
44:57:10Paolo Guiotto: is what? Well, I don't need to compute this limit, I hope, because this is the limit, for XYZ going to infinity 3.
45:09:50Paolo Guiotto: of the quantity
45:12:80Paolo Guiotto: X squared plus Y squared plus z squared, which is the square of the norm of vector XYZ. So, it is by definition that when XYZ goes to infinity, that quantity will go to plus infinity. There is nothing to do here.
45:27:660Paolo Guiotto: So, now, if I have… since I have this, I can say that for sure, there won't be maximum, because the function is unbounded. There is not
45:38:790Paolo Guiotto: maximum, for F on that domain D, but there will be a minimum.
45:46:320Paolo Guiotto: for F on D.
45:49:370Paolo Guiotto: So this means that, returning to the problem, the problem consists in searching for points at minimum, maximum distance to the origin.
45:59:460Paolo Guiotto: Since there is no maximum distance, it means that there are no points at maximum distance. And this is perfectly fine, because we understood that
46:09:620Paolo Guiotto: There are points, like these ones, which are arbitrarily far from the origin. So there cannot be a point which is the point at maximum distance to the origin, okay? Because there are points with arbitrarily large distance to the origin.
46:26:770Paolo Guiotto: So, no maximum points, but there will be points at minimum distance to the orbit. So now, to determine them.
46:34:330Paolo Guiotto: So let's finish before we take the break. Now, it is here where we, of course, apply the Lagrange theorem.
46:43:100Paolo Guiotto: So let's… Apply… Lagrange…
46:53:200Paolo Guiotto: URM.
46:55:190Paolo Guiotto: Now, first of all, we don't need to check that… so the assumptions of the theorem are, we need to have two functions
47:06:560Paolo Guiotto: F and G, which are differentiable, and such that G be a submersion on the set G equals 0. This was already found in point 1, okay? So we already know that G is differentiable, and it is a submersion on the domain D, so this, of course, can be used.
47:25:210Paolo Guiotto: So, in… 1… we… Proved.
47:32:710Paolo Guiotto: that G is differentiable.
47:37:70Paolo Guiotto: And that… It is… also… a submersion.
47:46:940Paolo Guiotto: on Domain T.
47:49:830Paolo Guiotto: about F, F is differentiable.
47:52:960Paolo Guiotto: This is clear.
47:54:680Paolo Guiotto: Yeah, early.
47:56:830Paolo Guiotto: F is differentiable.
48:00:560Paolo Guiotto: you know, the gradient of F is…
48:04:650Paolo Guiotto: vector 2X to Y to Z. Clearly, it is a continuous function.
48:14:550Paolo Guiotto: So… If, here we are talking about only minimums, if X, Y, Z, indeed.
48:27:440Paolo Guiotto: Is a minimum.
48:30:400Paolo Guiotto: point… 4.
48:33:30Paolo Guiotto: Fo?
48:34:620Paolo Guiotto: At that point, we must have gradient F is proportional to gradient G.
48:41:950Paolo Guiotto: which is not the condition we're going to solve. We treat this equivalently, saying that the rank of the matrix made of gradient F, gradient G,
48:53:250Paolo Guiotto: which is the dimension of the vector space generated by lines or by columns, it is the same of this matrix cannot be 2.
49:03:290Paolo Guiotto: Okay, because this would mean they are linearly independent.
49:07:10Paolo Guiotto: while this condition says they are linearly dependent. If you don't see this, it means gradient F minus lambda gradient G equals 0, right? This means that there is a non-trivial linear combination of the two vectors that produces
49:23:370Paolo Guiotto: the vector is zero, okay? So this means linearly dependent.
49:27:180Paolo Guiotto: So, proportional is a special case of linear dependence. Now, this means that the rank of this matrix is what? The gradient F is 2x.
49:38:70Paolo Guiotto: 2Y, 2Z. The gradient G, if I'm not wrong, was minus Y, minus X, 2Z.
49:48:980Paolo Guiotto: Now, this rank must be less than 2,
49:52:560Paolo Guiotto: Now, here we have the system of the three 2x2 sub-determinants.
49:59:360Paolo Guiotto: So, determinant of metrics.
50:02:620Paolo Guiotto: 2X, 2Y minus Y minus X, this must be equal to 0.
50:10:830Paolo Guiotto: Determinant of… 2X2Z minus Y2Z. Also, this must be equal to zero, together with the first one.
50:24:60Paolo Guiotto: And finally, there is also a third one.
50:26:870Paolo Guiotto: which is the first and third column, 2X.
50:30:680Paolo Guiotto: No, okay, I missed the second and third. So 2Y2Z minus X, 2Z, this one equals 0.
50:44:100Paolo Guiotto: Okay? So it's a system, okay? Be careful. Do not split the solution into, I solve equation 1, then I solve equation 2, then I solve equation 3, and I put together a solution, because this is wrong.
50:57:430Paolo Guiotto: What you're doing in that case, you are doing the union of solutions, while here you are doing the intersection. You have to take the three… the points that verify the three questions together.
51:09:60Paolo Guiotto: So, now, let's write the determinants. We have, for the first line, we have minus 2X squared plus 2Y squared equals 0.
51:20:790Paolo Guiotto: For the second, we have 4XZ plus 2YZ equals 0.
51:32:00Paolo Guiotto: And the last one is 4YZ.
51:35:430Paolo Guiotto: plus 2XZ equals 0. Okay, now let's clean up.
51:41:460Paolo Guiotto: simplifying as much as possible. So, for example, we eliminate these two.
51:48:590Paolo Guiotto: We have Y squared minus X squared equals 0, but we can factor as Y minus X times Y plus X.
51:58:920Paolo Guiotto: Always remind that since we are solving equations like something equals zero, it is convenient to try to factorize as much as possible.
52:08:510Paolo Guiotto: I can also do the same with the second, so I put a 2 here, I factor a Z, that multiplies 2X plus Y equals 0, and similarly for the third one.
52:22:980Paolo Guiotto: So again, Z times, 2Y plus X equals 1.
52:29:980Paolo Guiotto: Okay, now we have to enter in the solution of this system, and the strategy is always the same. We should choose one of the equations.
52:40:500Paolo Guiotto: Be careful, because here, a very frequent error is, for example, you take this one equals zero, then this one equals zero. No. You have to choose a line and say, when this is… when the product, for example, I choose the first line, when this product is zero.
52:59:80Paolo Guiotto: If, not if one of the two factors is zero, and so I get two subsystems. Number one, Y minus X equals 0, so let's say Y equal X.
53:12:240Paolo Guiotto: And system 2 is Y plus X equals 0, so Y equals minus X. Then, I plug this into the other two equations. What happens? So, when Y is equal to X, the second is Z times 3x equals 0. So, ZX equals 0.
53:33:00Paolo Guiotto: But also, the last one is ZX equals 0. So it's redundant, you just eliminate. When Y is equal to minus X, this becomes, again, ZX equals 0, and the last one is the same.
53:48:310Paolo Guiotto: Okay, that's good, because we are now reduced to two sub-cases, so we are…
53:54:860Paolo Guiotto: Y equals X, Z equals 0, or…
53:59:330Paolo Guiotto: Y equal X, and X equals 0.
54:03:800Paolo Guiotto: Or, and the second part, same thing. Y equals minus X and Z equals 0, or Y equals minus X and X equals 0.
54:17:300Paolo Guiotto: Now, here we have the solution, because here we have points, so there is no condition on X. Y must be equal to X, so these are points XX. For Z, we have 0.
54:28:80Paolo Guiotto: X is a bittery, because you don't see any other restrictions, so X real.
54:33:680Paolo Guiotto: And similarly here, X is 0, Is that X? Yes.
54:38:760Paolo Guiotto: Y is equal to X, so it will be 0. There's no condition on Z, so you will have Z3 here.
54:46:560Paolo Guiotto: Here, we have Z again equals 0, but Y equals minus X, so there will be X minus X 0.
54:54:490Paolo Guiotto: And these are different from the first ones, no? For example, 110 and 1 minus 1, 0, you see?
55:03:110Paolo Guiotto: They… you get the same point only when X is 0, for one value of X, but for all values of X, they are different.
55:10:440Paolo Guiotto: And the last one is, again, 00Z, Z in R. This one was already found, so we can…
55:19:940Paolo Guiotto: Okay. Now, what these points are… well, these are just points that verify this condition. We have not yet imposed this one.
55:30:680Paolo Guiotto: These are just XYZ for which gradient F is proportional to gradient G. So, to finish, we have to check which one of these are, indeed.
55:40:690Paolo Guiotto: Now.XX0… belongs to D, if and only if.
55:49:180Paolo Guiotto: D, what is the condition? Z squared, so 0 square equal X times Y, so X squared plus 1.
55:59:20Paolo Guiotto: What do you think?
56:03:870Paolo Guiotto: impossible. So what does it mean? That all these points we found here are solution of gradient F proportional to gradient G, but none of them is in the domain D, so we don't care about these points.
56:18:90Paolo Guiotto: Okay, let's check for 000Z, then?
56:23:780Paolo Guiotto: They are, indeed, if and only if.
56:27:930Paolo Guiotto: Z squared equal X times Y, which is 0, plus 1. So this yields Z squared equal 1, Z equal plus minus 1. So here we have two points, 0, 0 plus minus 1.
56:44:940Paolo Guiotto: And we save for future use. And finally, X minus X
56:52:50Paolo Guiotto: 0. This belongs to D if and all if, again, 0 square, so 0, is equal to X times Y, which is now minus X squared plus 1. So this is not impossible, because this X squared equals 1.
57:09:170Paolo Guiotto: So, X equal plus or minus 1.
57:14:340Paolo Guiotto: This yields 2 points.
57:16:750Paolo Guiotto: Because when X is 1, we get the 1 minus 1, 0.
57:21:770Paolo Guiotto: And the other case is minus 1, 0.
57:26:310Paolo Guiotto: So now, what is the conclusion?
57:29:500Paolo Guiotto: These are points of D that verify the condition gradient F proportional to gradient G, okay? So, minimum and maximum points, actually, only minimum points, because there is no maximum, are among them.
57:47:650Paolo Guiotto: So to decide which one are minimum points, we just evaluate the function f of 00 plus minus 1. Now, the function is X squared plus Y squared plus n squared, it is 1.
58:01:940Paolo Guiotto: F of 1 minus 1, 0 is the same of F minus 1, 0. This is 1 squared plus minus 1 squared, so 1, 1, 2, plus 0.
58:15:630Paolo Guiotto: So now you see which one, is the minimum, are the minimums? These are minimum points, so 0, 0, plus minus 1,
58:27:790Paolo Guiotto: are minimum points, so returning back to the question, the question was determine points of D. At minimum distance to the origin, maximum distance, we already proved that there are no points, so R.
58:41:640Paolo Guiotto: points.
58:43:500Paolo Guiotto: of D at minimum, distance, to the audience.
58:51:630Paolo Guiotto: And descends the exercise.
58:53:810Paolo Guiotto: Just one second.
58:56:20Paolo Guiotto: And what about these points, minus 1, 1, 0, minus 1, 1, 0?
59:02:890Paolo Guiotto: We don't know.
59:05:140Paolo Guiotto: Okay? So, think about… imagine that you have not discussed the existence part.
59:14:20Paolo Guiotto: So, imagine that you start from this.
59:17:420Paolo Guiotto: So you don't know. Minimum, maximum exists, you don't know anything. You proceed this way, at the end, you have these points, you evaluate the function, you get this. You would be led to think that the first one are minimums, and the second one are maximums.
59:34:340Paolo Guiotto: Which is wrong, because the second ones are not maximums, there are no maximum points here. So that's why you need always to know if minimum, maximum exists. Because otherwise, just the application of Lagrange theorem, like Fermat theorem, does not tell you.
59:52:290Paolo Guiotto: What they are.
59:54:660Paolo Guiotto: Okay?
59:57:230Paolo Guiotto: Good, let's take a break.
00:03:490Paolo Guiotto: is on.
00:05:170Paolo Guiotto: Okay.
00:09:10Paolo Guiotto: Do exercise, to 921.
00:16:20Paolo Guiotto: No, 20… the 19th, so… do.
00:23:70Paolo Guiotto: 2919.
00:25:690Paolo Guiotto: It's a bit heavy in terms of calculations.
00:29:940Paolo Guiotto: But you can survive.
00:32:990Paolo Guiotto: So we do the 2-9… 22…
00:39:190Paolo Guiotto: There are many other problems that I'm skipping, because they have domains with two constraints, which are pretty similar, but we needed to extend a bit the Lagrange theorem, so let's do one more example on this.
00:55:770Paolo Guiotto: So here, the domain D is the set of points X, Y, Z,
01:02:280Paolo Guiotto: Such that we have X squared plus Y squared minus Z squared equals 1.
01:12:600Paolo Guiotto: There are the usual three questions, show that D is non-empty.
01:18:20Paolo Guiotto: And D is the zero set.
01:21:670Paolo Guiotto: with the… G… submersion…
01:31:850Paolo Guiotto: on the itself.
01:35:300Paolo Guiotto: Question 2 is the compact.
01:42:900Paolo Guiotto: Question 3.
01:46:340Paolo Guiotto: Basically, it's the same problem.
01:49:730Paolo Guiotto: points… of the… at… Minimum.
01:56:850Paolo Guiotto: Maximum.
01:58:900Paolo Guiotto: these stones.
02:01:20Paolo Guiotto: to the origin. So, basically, I just changed the name. The same problem.
02:10:210Paolo Guiotto: So, question one.
02:14:300Paolo Guiotto: as I said above, did not empty, sufficient to show that one point is indeed, but
02:21:710Paolo Guiotto: Let's take also this, as an opportunity to understand
02:30:840Paolo Guiotto: Maybe do not… here, we have 3 coordinates, the temptation would be, let's put two of them
02:36:470Paolo Guiotto: equal to 0, so we remain, we do a unique equation in unique and known. That could be the idea. For example, I put Y and Z equal 0. I see that this means having points like X00, this belongs to D if and only if the equation reduces to X square equals 1, if and only if X equals plus minus 1.
02:59:380Paolo Guiotto: I've done, so now I know that there are two points, at least, in the…
03:05:640Paolo Guiotto: But here, I could be a little bit more careful, because you see that that Y square minus Z square can be made 0.
03:13:460Paolo Guiotto: just with Y equals Z. So, for example, XYY belongs to D if and only if I get exactly the same condition, you see?
03:25:800Paolo Guiotto: And therefore, I get that points plus minus 1Y, Y are all in D, and now I do not have any condition on Y for every Y in R.
03:39:260Paolo Guiotto: And these points,
03:41:380Paolo Guiotto: can be taken with arbitrary large coordinates for second and third coordinates, so this means that domain is not bounded. We have yet the information on this. So, but let's re… let's return on this later. Now.
03:57:230Paolo Guiotto: Domain D can be written as G equals 0,
04:01:930Paolo Guiotto: with, of course, G equal to, just put everything on one side, X squared plus Y squared minus Z squared minus 1.
04:14:410Paolo Guiotto: Now, we have that the three derivatives of G are 2X, 2Y… And,
04:28:80Paolo Guiotto: minus 2Z. Clearly, they are continuous functions, so G is differentiable.
04:39:560Paolo Guiotto: And, so to check…
04:46:500Paolo Guiotto: If, G is a submersion.
04:56:220Paolo Guiotto: on the domain D,
04:59:770Paolo Guiotto: We look for points where gradient of G is 0. Well, actually, we should prove that gradient is non-zero, but it is easier to prove where gradient is zero. That's why I always set the problem this way.
05:13:480Paolo Guiotto: We look… at points… Where?
05:22:770Paolo Guiotto: gradient of G is equal to 0.
05:25:820Paolo Guiotto: Because this means that we take the three components, 2X, to why…
05:31:680Paolo Guiotto: minus 2Z, and we impose that they are all equal to zero.
05:36:590Paolo Guiotto: And this yields only the point XYZ equal to 000.
05:45:400Paolo Guiotto: Now, is this point indeed?
05:48:670Paolo Guiotto: D is defined by this equation, X squared plus Y squared minus Z squared equal 1. If you put X, Y, and Z0, you get 0 equals 1, which is not verified, so not in the
06:02:900Paolo Guiotto: So this means that gradient of G is different from 0.
06:09:640Paolo Guiotto: on D, and this is exactly, by definition, what
06:13:960Paolo Guiotto: means to be a submersion on the cheese, a submersion
06:23:210Paolo Guiotto: this is just a name to this property, okay? We could say.
06:27:130Paolo Guiotto: in equivalent way, there are no stationary points of G on D, okay? Because stationary points are points where gradient is zero, which is the same thing.
06:38:810Paolo Guiotto: We call this in this way because there are many other reasons that we don't touch here.
06:46:410Paolo Guiotto: Okay, so this is about question one. Question two, is the set compact?
06:52:120Paolo Guiotto: D is the set defined by the equation G equals 0. Clearly, G is continuous, we said it is differentiable, so D is closed.
07:04:220Paolo Guiotto: And that's a standard story. Now, since we have done this work here, we can say that since
07:13:250Paolo Guiotto: points plus minus 1, Y, Y, R in D, for every Y in R,
07:22:700Paolo Guiotto: Well, necessarily, you see that you cannot say that the coordinates are bounded.
07:27:650Paolo Guiotto: Because there are these points, with arbitrarily big Y and Z. So, domain D is unbounded.
07:44:350Paolo Guiotto: And in particular, the… is not.
07:50:530Paolo Guiotto: compact.
07:54:290Paolo Guiotto: Okay.
07:55:680Paolo Guiotto: Be careful, let's say, warning.
08:00:380Paolo Guiotto: You know what,
08:03:30Paolo Guiotto: I usually see in an exam here that people would say, you have X squared, Y squared, minus X squared. Of course.
08:13:800Paolo Guiotto: you have not realized this thing, okay? You look at the constraint, and you say that from this I get X squared, Y squared, and Z squared must be less or equal than 1.
08:27:790Paolo Guiotto: So you are using this argument. The argument that works with plus, if I have the sum of three positive numbers, which is 1, each of them must be less than 1, because if they are positive, if one of them is larger than 1, the sum will be larger than 1. There is no question.
08:47:700Paolo Guiotto: But here, we have not the sum of three positive numbers, because there is a minus, and this means that there is the difference
08:55:40Paolo Guiotto: I can have difference of positive number, which is zero, and the number are big. 1 million minus 1 million is 0.
09:02:560Paolo Guiotto: But the number is big, you see?
09:05:240Paolo Guiotto: So that minus tells you that you cannot say this. This is completely wrong. It's a very bad error.
09:14:170Paolo Guiotto: And be careful, because this would change the story in this problem. For example, if you say that this holds, you are induced to say that the set is compact, and then the next part will be wrong.
09:27:550Paolo Guiotto: Okay? So, you have to be careful on these things, and this normally happens because people do not think on what they are doing, they just repeat what they have seen.
09:41:979Paolo Guiotto: I know that there are lots of repetitions here, we are doing… it seems we are repeating the same time, same things every time, but all this is an argument. I have to write the details to explain step-by-step why this follows on that, and so on.
10:00:380Paolo Guiotto: Even if it is a little bit repetitive, it doesn't matter. You have always to be forced on what you are doing.
10:07:450Paolo Guiotto: Okay? It's not just a standard application of this or that rule.
10:12:310Paolo Guiotto: So, in fact, here.
10:14:670Paolo Guiotto: Domain D is not compact, so we cannot say anything immediately about existence of minimum-maximum. However.
10:21:520Paolo Guiotto: Since, so, let's say the problem was… it's exactly the same, so I won't repeat the same discussion. The problem is, still to determine minimum-maximum points and minimum-maximum distance to the origin, from this. So, we have…
10:41:600Paolo Guiotto: We have… to solve… or… minimum… maximum…
10:51:900Paolo Guiotto: of X squared plus Y squared plus Z squared. I already explained why we consider this instead of the root, which would be the distance, on the main D.
11:05:780Paolo Guiotto: So now, let's call F this function.
11:08:940Paolo Guiotto: Domain D is not compact, so I cannot say immediately, even if F is continuous, that the conclusion, the existence, follows from the Weiss theorem. Domain D is, however, closed
11:26:660Paolo Guiotto: and unbounded.
11:30:280Paolo Guiotto: And, we also have trivially here that limit
11:36:950Paolo Guiotto: when point XYZ goes to the infinity of this particular function is plus infinity.
11:46:360Paolo Guiotto: So, we are in the right conditions to say that there is not maximum
11:53:460Paolo Guiotto: for this function F, so there won't be points at maximum distance to the origin.
11:59:660Paolo Guiotto: But there will be minimum. We are exactly in the same Case of the previous example.
12:07:810Paolo Guiotto: Now, to… Determine minimum.
12:13:90Paolo Guiotto: points.
12:15:90Paolo Guiotto: we apply… Lagrange Theorem.
12:22:330Paolo Guiotto: So we already checked point 1, that G is a submersion.
12:27:970Paolo Guiotto: So we… I'll read it.
12:34:180Paolo Guiotto: Check.
12:37:900Paolo Guiotto: That… G is… differentiable.
12:44:620Paolo Guiotto: And… It is… a submersion, on the…
12:55:840Paolo Guiotto: Then, also, F is clearly differentiable.
13:00:920Paolo Guiotto: also… S is differentiable.
13:07:470Paolo Guiotto: Well, actually, all polynomials are always differentiable, so… because the partial derivatives are polynomials, and they are continuous, so you don't have… if the function is a polynomial, you can skip that kind of
13:21:300Paolo Guiotto: discussion.
13:24:670Paolo Guiotto: So… if, XYZ indeed, is a minimum point. We can talk only about this point.
13:39:240Paolo Guiotto: for F…
13:41:710Paolo Guiotto: We must have, at that point, that gradient of F at point XYZ is proportional to the gradient of G at point XYZ.
13:56:60Paolo Guiotto: Which is equivalent to say that the rank of the matrix generated by the two gradients
14:04:710Paolo Guiotto: Is less than 2.
14:08:370Paolo Guiotto: Now, this means that the rank of the matrix
14:13:920Paolo Guiotto: For a gradient of F, we ask the usual 2X, 2Y, 2Z.
14:19:710Paolo Guiotto: For the gradient of G, I think 2X to Y minus 2Z.
14:27:530Paolo Guiotto: This must be less than 2.
14:30:580Paolo Guiotto: Now, this happens if and only if all the 2x2 sub-determinants are zero.
14:38:620Paolo Guiotto: Now, the first one is determinant of 2x… 2Y, 2X, 2Y.
14:48:180Paolo Guiotto: Equals zero. By the way, I don't need to compute this determiner, because you see that the two vectors are the same, so the terminal will come zero, because these two vectors are… whatever X and Y are, are linearly dependent, are the same vector, just one vector. So this is just a throwaway condition.
15:06:640Paolo Guiotto: So we have determinant of 2x2Z, 2X minus 2Z, then.
15:14:660Paolo Guiotto: equal 0. And the third one is determinant of 2Y
15:21:600Paolo Guiotto: 2Z2Y minus 2Z. This is not the same case of the first line, huh? Because you see that,
15:31:650Paolo Guiotto: These vectors are not the same. That minus change the second component.
15:37:280Paolo Guiotto: Okay, so, well, you can use what you know about determinant, what you are supposed to know. For example, determinant is a linear function of lines and columns.
15:48:290Paolo Guiotto: So, if you have something that multiplies a line or a column, a factor, you could carry outside. For example, this 2 comes out, also this 2 comes out, so it's 2 square. We can cancel because it is a constant different from zero, so this means that we can simplify all these two.
16:07:760Paolo Guiotto: So, this means that the terminal of the first is minus XZ, then there is another minus XZ, so minus 2XZ.
16:16:740Paolo Guiotto: equals 0. And similarly here, minus 2YZ equals 0.
16:23:510Paolo Guiotto: So again, we can cancel this minus 2, so at the end, we have X times Z equals 0,
16:29:360Paolo Guiotto: and Y times Z equals 0. The two equations are perfectly the same, so we have the alternative, X equals 0, or Z equals 0, first equation.
16:41:360Paolo Guiotto: Well, second equation, YZ equals 0 here, and second equation here is 0 equals 0.
16:49:740Paolo Guiotto: So, in this side, what are the solutions?
16:54:80Paolo Guiotto: XYZ0. You see, there are no conditions on XY. On Z, there is zero. So this means X and Y are arbitrary, real numbers.
17:04:860Paolo Guiotto: About this one, we have two sub-cases, X equals 0, Y equals 0, or
17:12:320Paolo Guiotto: X equals 0, and Z equals 0.
17:16:650Paolo Guiotto: The first one produces 000Z, with Z real.
17:23:650Paolo Guiotto: The second one produces 0Y0 with Y real.
17:30:630Paolo Guiotto: You see that apart for special cases, that equals 0, y equals 0, these solutions are different from the other ones, okay?
17:40:910Paolo Guiotto: No. Maybe? No, no, no. The second one are already contained there, no? Because they… they are special case, so I can actually throw away this, but the first one, no. They are different.
17:52:780Paolo Guiotto: Okay, now what are these points? These are just points for which this condition is verified, because this point comes out from writing this condition. So we still have to verify this one. Now, we impose
18:10:110Paolo Guiotto: Now.00Z is indeed…
18:16:350Paolo Guiotto: The main thing was X squared plus Y squared minus Z squared equal 1. So this becomes minus Z squared equal 1,
18:28:170Paolo Guiotto: So?
18:30:350Paolo Guiotto: Impossible.
18:33:740Paolo Guiotto: So, no solutions. And the other one is XYZ belongs to D if and only if X squared plus Y squared equal 1.
18:47:310Paolo Guiotto: which is well possible, and there's plenty of solutions. We could say that, for example, these are solutions on the unit circle, so, like, cosine theta sine theta
19:00:390Paolo Guiotto: 0, if we want to describe, with theta in 0 to pi. So there are infinitely many points of this nature.
19:09:80Paolo Guiotto: Did that, indeed.
19:11:470Paolo Guiotto: Okay, now…
19:14:90Paolo Guiotto: To see… so the unique possibilities for minimums are now these points here, which are infinitely many, but they depend on this single parameter.
19:24:650Paolo Guiotto: Now, we have that the value of F
19:27:340Paolo Guiotto: at this point, we actually, as you will say at the end, don't need to write this thing. So at points like XYZ0, this is X squared plus Y squared. So if X squared plus Y squared is 1, at all these points, the value of the function is 1.
19:46:780Paolo Guiotto: So what is the conclusion?
19:48:910Paolo Guiotto: We saved it.
19:50:510Paolo Guiotto: There exists the minimum, not the maximum.
19:54:30Paolo Guiotto: the minimum, of F on D must be one of these points.
20:01:70Paolo Guiotto: since at these points, the value of the function is always equal to 1, it means that all these points are minimums, you see? Because the function takes the same values. So…
20:12:510Paolo Guiotto: all these… R… Minimum.
20:20:50Paolo Guiotto: Points, huh?
20:21:950Paolo Guiotto: So, and this means that the minimum point means points at minimum distance to the origin.
20:28:620Paolo Guiotto: That is… points.
20:33:990Paolo Guiotto: Off.
20:35:670Paolo Guiotto: D at minimum, distance.
20:41:990Paolo Guiotto: to the orange.
20:45:250Paolo Guiotto: And that's it for this exercise.
20:52:480Paolo Guiotto: Okay.
20:55:270Paolo Guiotto: Now, we have, few minutes.
21:02:470Paolo Guiotto: So, what is the next step? Now, let's say that, we are somehow, say, able, So… Now…
21:14:40Paolo Guiotto: We… Nope.
21:19:890Paolo Guiotto: So… solve… A problem, where you have to determine minimum, maximum.
21:28:710Paolo Guiotto: of a function f, let's return to general notation X, on a domain which is defined by a unique equation, G of X equals 0.
21:42:430Paolo Guiotto: The next step is to extend this to the case when the domain is defined by several equations, several conditions. It may happen, because why not? No, the domain is defined… you must verify this, and that, and that, and that, okay?
21:58:580Paolo Guiotto: D. Extension…
22:06:750Paolo Guiotto: of this… problem.
22:10:650Paolo Guiotto: Is searching for minimum, maximum, still of function F.
22:18:970Paolo Guiotto: on a domain which is defined by a certain number of equations. So let's say there will be G1 of X
22:26:820Paolo Guiotto: equal to 0, G2 of X,
22:30:550Paolo Guiotto: equal to 0, and so on. Let's say, until a certain number M of conditions, G… no, the arrow here is wrong.
22:42:430Paolo Guiotto: GM of X equal to 0.
22:47:780Paolo Guiotto: Okay, so this is the… A more general form of a constrained optimization problem we consider here.
22:58:990Paolo Guiotto: Now, what happens? It happens that, of course, there will be a…
23:03:920Paolo Guiotto: a version of this theorem adapted to these cases. The version will tell, I will not state the version now.
23:13:840Paolo Guiotto: Because we will now focus on the concept of submersion, what is a submersion here. But however, it will work in the similar way, so if we have a function f and certain number of constraints, G1, GM, differentiable, with the constraint.
23:30:970Paolo Guiotto: Being a submersion, so we have to understand what does it mean to be a submersion, when now we have several functions, and the condition is not that the single gradients are different from zero, as you will see. However.
23:46:360Paolo Guiotto: Once we have understood what does it mean, a submersion, the conclusion will be similar. If X star is any video of maximum points for F on D, then gradient of F is not proportional to one gradient.
23:59:120Paolo Guiotto: But the… this condition will be modified, saying that gradient of F is a linear combination of the gradients of the constraints.
24:08:50Paolo Guiotto: This is more or less the plan. Now, the first step is to understand what is a submersion for this case.
24:17:160Paolo Guiotto: So… We… Need… to define… What?
24:30:590Paolo Guiotto: Thus… Admin… that… well, actually, this will be a property that involves all the constraints, so G1,
24:50:260Paolo Guiotto: G… sorry, there is no array here. G1.
24:54:430Paolo Guiotto: GM… is a… Sugmation.
25:04:440Paolo Guiotto: Now, if we have one single constraint, so if M is 1, So there is one equation.
25:15:430Paolo Guiotto: So, G1 is a submersion.
25:22:900Paolo Guiotto: if and if the gradient of G1 is different from 0 on the domain D.
25:31:680Paolo Guiotto: Well, we wanted this be a submersion.
25:35:730Paolo Guiotto: Now, gradient, of G1 different from 0 means
25:42:430Paolo Guiotto: in terms of… you remind that gradient is formally a matrix, a one-line matrix, the Jacobian matrix. It means that one of the entries of this matrix, so we have D1, G1, D2, G1, etc.
26:01:970Paolo Guiotto: DG1, D is the number of variables. At least one of the coordinates of this vector must be different from zero. One of the entries of this vector is different from zero.
26:15:490Paolo Guiotto: So, equivalently, this vector is non-zero. Equivalently, the rank of this matrix is 1.
26:26:50Paolo Guiotto: Because this matrix, remind, the rank is the dimension of the vector space generated by line columns of a matrix. If I say that the rank of that matrix is 1, it means that that single vector generates a space of dimension 1.
26:44:420Paolo Guiotto: Now…
26:45:640Paolo Guiotto: What can be for a unique vector? You can generate a space of dimension 1, but also a space of dimension 0, and that happens when the vector is 0. If the vector is non-zero, you generate a line, a straight line, a space of dimension 1.
27:02:110Paolo Guiotto: So the condition can be seen in this way. If M is 1, saying that G1 is a submersion means that the rank of the matrix made by the gradient of G1 is 1.
27:12:740Paolo Guiotto: It's a bit exaggerated, because it's much easier to look at the gradient different from zero. But this is important because it provides the base for the definition.
27:24:300Paolo Guiotto: That we give now.
27:26:650Paolo Guiotto: We say that,
27:32:140Paolo Guiotto: So G1.
27:37:220Paolo Guiotto: GM.
27:38:510Paolo Guiotto: So the bunch of these functions is a submission
27:50:520Paolo Guiotto: on the yield?
27:53:150Paolo Guiotto: If we take the matrix made of the gradients of these constraints, G1, G2, so gradient G1, gradient G2, gradient GM,
28:08:550Paolo Guiotto: Now, the condition is the rank of this matrix must be equal to the number of constraints, so must be equal to M for every point X of the domain D.
28:24:670Paolo Guiotto: So, if M is 1, this just reduces to the previous case, because you have that these metrics are the unique line, and you are saying the rank of
28:35:210Paolo Guiotto: That unit line, gradient G1, is 1.
28:38:850Paolo Guiotto: Now, this means that the vector is non-zero.
28:43:780Paolo Guiotto: If M is 2, 3, or in general, any number of constraints, this condition is a bit different. Now, we have to understand what does it mean that this rank is exactly equal to M, but this will be the starting point of the next class, because now.
29:01:480Paolo Guiotto: time is over, okay? So let's, let's keep this, and please, when we restart on Friday.
29:10:340Paolo Guiotto: take just 5 minutes to review this, because I will start from this condition, trying to understand what does it mean, how do we check that the rank of these metrics is N.
29:22:520Paolo Guiotto: It would be a good idea
29:24:870Paolo Guiotto: If you could just spend at home 10 minutes reviewing what is the rank of a matrix, okay? We already used this, but it's better if you refresh.
29:39:230Paolo Guiotto: Okay, let's stop.