AI Assistant
Transcript
00:05:280Paolo Guiotto: Okay, so last time… We spent, a month of,
00:12:740Paolo Guiotto: The last part of the class.
00:16:40Paolo Guiotto: Reviewing these, definitions of, What is a positive, Negative definite metrics.
00:27:210Paolo Guiotto: Besides that, so let's quickly review, because we finished a bit,
00:36:420Paolo Guiotto: Fast. So, a positive definite matrix is a matrix for which we have this property. MV, scalar V is greater or equal than zero for every vector V.
00:49:90Paolo Guiotto: If you want to say strictly positive, this means that MD scalari is strictly positive.
00:55:410Paolo Guiotto: for all vectors except the unique one for which this cannot be, that is vector V equals 0. And similarly for negative definite.
01:05:370Paolo Guiotto: Now, this is going to be important because, for example, we have this result, I will return later on this, that says that the function is convex.
01:17:500Paolo Guiotto: If the second derivative, which is the Asian matrix, is positive definite on the concave, if it is negative definite.
01:26:600Paolo Guiotto: So, a question arises, how do we check if a matrix is positive definite or negative definite?
01:34:170Paolo Guiotto: Now, we noticed that we are interested in checking this for a particular type of mattresses, which are the Asian mattresses, but under our circumstances, are symmetric mattresses.
01:45:520Paolo Guiotto: Now, this makes this check a bit easier because of this important fact. The matrix is diagonalizable.
01:54:100Paolo Guiotto: So this means that my change of base can be reduced to a diagonal form. And in fact, positivity, negativity depends on the size of the eigenvalues, so the elements that you have on the diagonal.
02:07:70Paolo Guiotto: Precisely, we have seen that it is positive definite if and only if the eigen values are positive.
02:14:710Paolo Guiotto: Weak sense, negative definite, if they are negative, and strictly positive, they are strictly positive, but all eigenvalues, strictly positive, strictly negative, metrics, strictly negative, if the eigenvalues are strictly negative.
02:30:920Paolo Guiotto: Now, this is good because, in principle, we have a recipe to determine the eigenvalues, which is to solve the characteristic equation, which is a polynomial equation.
02:40:680Paolo Guiotto: That polynomial equation is easy for 2x2 matrices, because it is a secondary equation. But since it does the same degree of the size of the matrix, so for a D by D matrix, this is a D degree equation, so it could be difficult, you know?
02:59:140Paolo Guiotto: However, there is, and this was the final part.
03:03:400Paolo Guiotto: An important remark, that the fact that they are all positive, or all negative.
03:09:710Paolo Guiotto: can be seen also, in this way. So, if we consider the diagonal matrix with elements lambda 1, lambda B on the diagonal.
03:19:630Paolo Guiotto: And we take the square submetrises, one by one, two by two, three by three, etc, taking just the first D elements, line by column.
03:31:700Paolo Guiotto: Now we notice that if we compute the determinants of these matrices, they are the products, the consecutive products of the eigenvalues. So lambda 1 for the first, lambda 1, lambda 2 for the second, lambda 1, lambda 2 for lambda 3 for the third, and so on.
03:46:980Paolo Guiotto: But in the case that all are positive, these numbers are all positive.
03:52:440Paolo Guiotto: And actually, vice versa, perhaps I… I have not said this, because if these numbers are all positive, if this determinants are all positive, look, the first one says that lambda 1 is positive, right?
04:07:180Paolo Guiotto: The second one says that lambda 1, lambda 2 is positive, but since lambda 1 is already positive, it must be also lambda 2 positive.
04:15:410Paolo Guiotto: So, lambda 1, lambda 2 must be positive. Now you pass the third one. You know that lambda 1, lambda 2, lambda 3 is positive.
04:23:530Paolo Guiotto: And you already know that lambda 1 and lambda 2 are both positive. Necessarily, lambda 3 must be positive, otherwise the problem would be negative, you see? So with this mechanism, we have that these determinants are all positive, if and only if all the lambdas are positive.
04:42:20Paolo Guiotto: And the other column is the case when the lambdas are all negative.
04:46:750Paolo Guiotto: So if they are all negative, where lambda 1 is negative, then lambda 1, lambda 2 becomes positive. Product of negative, no?
04:54:410Paolo Guiotto: Lambda 1, lambda 2, lambda 3 is now negative, yeah? And so on.
04:59:680Paolo Guiotto: So, you see that if the lambdas are negative, the size of these determinants are minus plus, minus plus, and so on. So, they are alternating sides, starting with the minus.
05:12:710Paolo Guiotto: Vice versa. If you have the design are this way, minus plus, minus plus, from the first line, you need that lambda 1 must be negative.
05:23:580Paolo Guiotto: From the second line, you have lambda 1 times lambda 2 positive, okay? But you know that lambda 1 is negative, so lambda 2 must be negative, otherwise the product cannot be positive, you see? So, you get lambda 1, lambda 2 negative.
05:39:610Paolo Guiotto: Then you put this into the third one, lambda 1, lambda 2, lambda 3 is negative.
05:44:980Paolo Guiotto: You already know that the first two are negative, so lambda 1 times lambda 2 will be positive.
05:51:740Paolo Guiotto: times lambda 3 would be negative only if lambda 3 is negative.
05:56:580Paolo Guiotto: So you see that this sequence of signs of the determinants is exactly minus plus, minus plus, minus plus, and so on, if and only if all the lambdas are negative.
06:08:410Paolo Guiotto: Okay?
06:09:670Paolo Guiotto: So, here we can see that
06:13:540Paolo Guiotto: The sequence of these determinants for the diagonal matrix are
06:19:380Paolo Guiotto: alternating with the minus, initially, exactly if and only if we are in one of these two cases, so if and only if the diagonal matrix, at least, is positive or negative.
06:35:90Paolo Guiotto: Now, why this is the same for the metric sand?
06:38:230Paolo Guiotto: So, that's why I say that the conclusion that I did not justify. I want to convince you that this is the case.
06:45:460Paolo Guiotto: Because what happens is that if we take our metric exam.
06:51:660Paolo Guiotto: The math exam is made up of elements M11, M12, and so on.
06:57:700Paolo Guiotto: So let's say that we take the generic,
07:00:720Paolo Guiotto: submetics with K lines and K columns. So here we stop at the M1K, and this will be MK1.
07:10:740Paolo Guiotto: And this will be MKK.
07:13:740Paolo Guiotto: So this is a square matrix. So let's compute the determinant of this matrix, MK.
07:22:160Paolo Guiotto: Now, we keep in mind that the matrix is symmetric and diagonalizable, so there is this fact that T minus 1 times N times T is the diagonal matrix.
07:35:390Paolo Guiotto: Now… you can easily see it, trust me, that if you take the math exam, K,
07:42:740Paolo Guiotto: And you take the submetrics K by K of T, the first little k lines, first k column, so this is the DK.
07:51:90Paolo Guiotto: minus 1K, you get exactly the decay, so the submetrix, so the matrix D is this one, lambda 1, lambda 2, etc. So we have a certain point, lambda K, and we stop at lambda.
08:08:250Paolo Guiotto: D. Zeros, zeros here. So the matrix dec is this one.
08:14:490Paolo Guiotto: So this is the MK, and this is the DK.
08:18:270Paolo Guiotto: So there is this reaction.
08:20:350Paolo Guiotto: Now, when you plug determinant to both sides, determinant of DK minus 1, MK, PKE.
08:29:770Paolo Guiotto: must be equal to the determinant of the matrix dec, and that one is the product of the first K eigenvalues. So this is lambda 1 times lambda 2 times etc, lambda K.
08:44:380Paolo Guiotto: But now, I don't know if you're reminded of this. If you do not remind, or you don't know, you trust me, because there's no other way.
08:52:880Paolo Guiotto: There is a remarkable property of the Telinanta.
08:56:330Paolo Guiotto: If you have a determinant of square matrices, the dominant of the product is
09:05:300Paolo Guiotto: is the product of determinants, so you can…
09:08:710Paolo Guiotto: write this as determinant TK minus 1, determinant MK and determinant decay.
09:19:230Paolo Guiotto: Anger!
09:20:670Paolo Guiotto: just because the terminant of the product is the product of the tenant, you can recombine these two guys together and say that this is also the tenant of the product that comes from dec-1 times dec.
09:35:30Paolo Guiotto: finds determinant of MK.
09:38:560Paolo Guiotto: What is the determinant of inverse of the matrix times the matrix? Now, this matrix is what?
09:45:920Paolo Guiotto: When you multiply by a matrix by each inverse, you get the
09:49:960Paolo Guiotto: Identity metrics, which is the metrics?
09:52:950Paolo Guiotto: K by K with 1s on the diagonal, and that determinant will be equal to…
09:58:740Paolo Guiotto: 1, because it's just the product on the diagonal. So at the end, you get this formula that the determiner
10:06:520Paolo Guiotto: of the submetrics K by K submetrics built on M, taking the first k lines and k columns, is equal exactly to that product, the lambda 1, lambda k.
10:20:580Paolo Guiotto: So you have this, that, since we already discussed the sign of this product, we have that the matrix is positive definite, when the lambdas are positive, that means those products are all positive, okay?
10:37:20Paolo Guiotto: And so this means that also this must be positive, because they are the same.
10:41:680Paolo Guiotto: So, we have this conclusion. Conclusion.
10:47:50Paolo Guiotto: The metric sum is positive definite if and only if
10:52:30Paolo Guiotto: the determinants of the matrices MK are all positive.
10:58:20Paolo Guiotto: So, let's say, since the next one, R plus minus, I write this sign of this number.
11:07:520Paolo Guiotto: is equal… to 1 for every K.
11:11:630Paolo Guiotto: So this means they are code positive.
11:14:330Paolo Guiotto: While the matrix is negative, even only if the lambdas are negative, even only if those products are alternating signs, starting with the minus, so I have sine.
11:27:280Paolo Guiotto: of determinants.
11:29:10Paolo Guiotto: of MK is now not 1, it is minus plus minus plus, so we write this as minus 1 to K plus 1 for every K.
11:40:980Paolo Guiotto: And that's a test to decide
11:45:890Paolo Guiotto: If this is positive, definite, negative definite, no?
11:50:690Paolo Guiotto: So, let's do an example.
11:56:860Paolo Guiotto: So, letter F for XY…
12:00:100Paolo Guiotto: be this function, X squared plus Y squared, I don't know, minus XY. For XY, in October.
12:10:680Paolo Guiotto: So, compute.
12:15:130Paolo Guiotto: the Asian metrics.
12:17:910Paolo Guiotto: And… determine… It's… sign.
12:26:950Paolo Guiotto: Remember, the sign of the matrix does not mean that… it means exactly this. Positive definite or negative definite, okay?
12:36:530Paolo Guiotto: So let's see here…
12:39:510Paolo Guiotto: To compute the Asian matrix, we have to compute the second partial delivities, so we have to start with the first partial delivities. We have…
12:48:510Paolo Guiotto: for X, derivative with respect to X is 2X minus Y. Derivative with respect to Y is 2Y minus X. They are both continuous, so I did use F, differentiable
13:03:860Paolo Guiotto: on Archer.
13:05:950Paolo Guiotto: So the first derivative exists, basically.
13:09:190Paolo Guiotto: It exists, the first derivative.
13:11:930Paolo Guiotto: as a function from R to… in this case, notice it's not a function R to 2R, because the gradient is a vector. This is a function R22r.
13:21:690Paolo Guiotto: Now… Let's conclude the second partial derivatives.
13:25:660Paolo Guiotto: I, D… Second derivative with respect to AXA,
13:31:610Paolo Guiotto: is, you have to differentiate that thing with respect to that, so you get what?
13:37:150Paolo Guiotto: So, then you have the mixed derivative with respect to XY,
13:41:570Paolo Guiotto: Now, which one is this one? This should be… that one that you take, the second line, the Y alpha, you differentiate in X, so you see that you get minus 1, and this is exactly the same that you would get by doing this. Firstly, differentiating in X and then in Y.
13:59:460Paolo Guiotto: That's because they are continuous, so we can flip the order of these.
14:04:560Paolo Guiotto: And finally, the second derivative with respect to Y is… is against you.
14:12:540Paolo Guiotto: The one small, you have that. These are continuous functions, so, the gradient, if you want the literal, the gradient, that means, differentiable.
14:23:440Paolo Guiotto: is, no, two tiles differentiable. Twice… Differentiable.
14:34:710Paolo Guiotto: Underdeation metrics, Peace.
14:38:660Paolo Guiotto: The matrix is a 2x2 matrix with these entries. So.
14:44:390Paolo Guiotto: minus 1, minus 1, 2. Now we have a matrix. It's constant, it's simple.
14:51:900Paolo Guiotto: To understand if it is positive, negative, let's compute the sub-determinants. So we have, in this case, just two sub-determinants. One is 2,
15:01:680Paolo Guiotto: And the other is the determinant of the matrices. Okay, so let's say that in that notation, just to show once, M1 is made of one single entry, M2 is 2x2 that in this case coincides with the full matrix.
15:18:400Paolo Guiotto: So the determinant of M1 is equal to 2,
15:23:840Paolo Guiotto: The determinant of M2 is 2 times 2, 4 minus… minus 1 times minus 1, so it is minus 1, 3.
15:33:870Paolo Guiotto: You see that they are both positive, Right?
15:38:450Paolo Guiotto: So we are in this case.
15:41:670Paolo Guiotto: All determinants are positive, the matrix is positive.
15:46:220Paolo Guiotto: So this N, which is the second derivative, the ashomatics, is positive definite for every point XY, it's independent on XY, so there is no problem, for every XY in R2.
16:01:760Paolo Guiotto: Okay.
16:05:300Paolo Guiotto: We will repeat this check several times. I wanted just to give you a simple example to show you simply, without any technical complications. So you see there are… the derivatives are easy. At the end, we get a constant metric, so we don't have to discuss for which points were positive, definite, negative definite.
16:25:300Paolo Guiotto: Okay.
16:26:270Paolo Guiotto: So, in any case, you see what is the algorithm. You compute the second derivative, and then you compute the determinants of the submatrices. If you have a 3x3 matrix, because F depends on 3 variables.
16:40:70Paolo Guiotto: Then you will have 3 sub-determinants.
16:42:870Paolo Guiotto: If you have a 10 by 10 metrics, you will have to compute 10 sub the tenets, and so on.
16:49:00Paolo Guiotto: Okay, let me change the slide, yeah.
16:54:500Paolo Guiotto: Okay, so let's say that now we know how to check positivity, and let's see how to use. I told you yesterday that this is important because we, we introduced this problem
17:10:60Paolo Guiotto: With the idea of, extending this remarkable factor.
17:17:790Paolo Guiotto: That, to check… what is it?
17:23:690Paolo Guiotto: To check… okay, to check concavity convexibling, we should look at the second derivative and say it's positive or negative.
17:31:950Paolo Guiotto: We discussed what is the second derivative, we discovered it is a matrix.
17:37:290Paolo Guiotto: Now, to say, what does it mean, positive negative? The right concept is this one, of positive definite, or negative definite negative.
17:47:590Paolo Guiotto: Then we have seen how to check if it is positive or negative with the test that we discussed again, just a couple of minutes ago, okay? Now let's return on functions.
18:01:20Paolo Guiotto: So, first of all, mmm…
18:04:490Paolo Guiotto: Let's say this, that, the,
18:08:840Paolo Guiotto: The check-off positivity, or negativity, of the Asian matrix is not only important to understand that the function is concave or convex, which is a problem we will not consider too much here. I just told you, because
18:23:380Paolo Guiotto: Because, remind this to understand the full story.
18:29:920Paolo Guiotto: everything originated from this problem. When can we say that a stationary point is actually a minimum or a maximum? In general, only knowing that it is a point where gradient is zero is not sufficient to conclude.
18:46:110Paolo Guiotto: But if you have some more information, like it is concrete convex, which is an important case in many applications, I mentioned in economics, for example.
18:58:400Paolo Guiotto: that becomes a sufficient condition. Now, that's because of this theorem we have seen the sharp proof. Now, the function is concave, for example, at the stationary point, you have necessarily a maximum.
19:13:990Paolo Guiotto: Right?
19:15:520Paolo Guiotto: So, convex are important for this kind of review.
19:20:870Paolo Guiotto: However, what if the function is not concave convex?
19:24:980Paolo Guiotto: Again, points where gradient F equals 0 are not necessarily minimum or maximum. We have already seen this in several cases, examples.
19:37:490Paolo Guiotto: what can be said anything on these points? We have seen that, sometimes we can say some of them are minimum, some of them are maximum, but there are many others for which grade it is zero that we don't know what they are, no?
19:50:850Paolo Guiotto: Now, the action metrics can give some kind of information about the nature of these points.
19:58:480Paolo Guiotto: Even out of concavity convexity.
20:01:630Paolo Guiotto: And this is, so deep… sighing… off… Asian… metrics?
20:16:950Paolo Guiotto: is… Use a pool.
20:22:620Paolo Guiotto: So… Guts.
20:27:60Paolo Guiotto: Song.
20:29:640Paolo Guiotto: Father… information…
20:37:620Paolo Guiotto: D… Not sure.
20:42:150Paolo Guiotto: of a stationary point.
20:46:380Paolo Guiotto: So a point where gradient of F, at that point, say, X star is equal to vector 0.
20:56:80Paolo Guiotto: This because of,
20:59:30Paolo Guiotto: a consequence of the so-called Taylor formula. You have seen Taylor formula for functions of one variable.
21:07:520Paolo Guiotto: So, let's remind…
21:14:840Paolo Guiotto: the… Thank you, Lord.
21:20:310Paolo Guiotto: formula.
21:22:640Paolo Guiotto: I will limit to the Taylor formula, add to the second derivative, okay?
21:29:20Paolo Guiotto: You may imagine that we could define also, in this case, the third derivative, fourth derivative, but it's going to be a very complicated object, okay? Until the second derivative, we can handle, then the third derivative is complex. However, for what we have to do in this course, we will just move to this.
21:45:890Paolo Guiotto: first two derivatives. So we have that if the TRM says that, if the function f
21:54:60Paolo Guiotto: which is now a function of real variable, f of x, so X is in a certain domain of the real life, real value.
22:04:390Paolo Guiotto: Let's say F is, twice, differentiable, At a certain point, X star.
22:16:690Paolo Guiotto: So this means there exists the first derivative at point X star, there exists also the second derivative at point X star.
22:27:230Paolo Guiotto: Then, we have this formula, which has many important applications.
22:32:800Paolo Guiotto: Because it can be used for numerical calculation, it says it can compute approximately F by using a polynomial. That says it is equal to F to X star.
22:43:580Paolo Guiotto: plus F prime, X star.
22:46:630Paolo Guiotto: times X minus X star. I know you have C with X0, but I don't want to use X0 here, because we have already indexes around, so let's use X star. Then you have it the next time you remind what is…
23:05:370Paolo Guiotto: Do you mind what is this next one?
23:10:510Paolo Guiotto: You do not remind me this.
23:12:520Paolo Guiotto: I would kill you.
23:15:410Paolo Guiotto: Second derivative of F at point X star, divided by 2 factorial, which is 2, X minus X star squared.
23:24:580Paolo Guiotto: That's the second.
23:26:60Paolo Guiotto: And then there is a correction, which is a little O of X minus X star squared. That's…
23:32:170Paolo Guiotto: A term which is negligible with respect to all these other terms.
23:37:440Paolo Guiotto: Okay, now, is there a shape for this formula in the case of a function of the vector variable?
23:44:630Paolo Guiotto: And the answer is yes.
23:48:810Paolo Guiotto: 4… A function f, now function of vector variable.
23:55:620Paolo Guiotto: still real value, so the final domain be Offer the real value, no?
24:03:830Paolo Guiotto: We… Vote.
24:08:510Paolo Guiotto: Following…
24:13:650Paolo Guiotto: extension.
24:17:850Paolo Guiotto: It's not important we prove. It's not particularly complicated. The proof is based on this formula, by the way.
24:25:650Paolo Guiotto: But the nice thing to do is to understand how it should be written, this formula when x is now an array, no? So we have this theorem. We suppose that function f is twice
24:43:80Paolo Guiotto: differentiable.
24:44:850Paolo Guiotto: at point… X star.
24:47:980Paolo Guiotto: So this, this point is now a vector, so this means that we have the first derivative, which is the gradient.
24:56:160Paolo Guiotto: at point X star, and the second derivative, which is now the Asian matrix, that we denoted by this symbol.
25:07:740Paolo Guiotto: The forgotten arrow here.
25:12:100Paolo Guiotto: Then… Now, look at this formula. It says, F of X, it will be F of X, no?
25:20:220Paolo Guiotto: Only X is an array here. Equal. The value of the function at the center of this expunction, so this is F of X star.
25:30:830Paolo Guiotto: Which is now redacted, though.
25:33:430Paolo Guiotto: the point, not the value. The value is still a number.
25:37:90Paolo Guiotto: That's it.
25:38:360Paolo Guiotto: I have to write here, well, this… we met this problem when we defined the derivative. We have to replace a frame with gradient, and in that case, this is a vector, so that product is a product between a matrix and vector, so we know what that means. Gradient F at point X star.
25:57:260Paolo Guiotto: times line by column, X minus X star.
26:04:930Paolo Guiotto: plus. And now, what is this? F second divided by 2? Well, divided by 2 means, let me write 1 part.
26:14:860Paolo Guiotto: Then, there is F second, that is now the second derivative, the addition matrix. So I will have this second F at point at star.
26:26:290Paolo Guiotto: And now, what is X minus X tau squared?
26:30:330Paolo Guiotto: Here, it is the algebraic square, no? The square of the number.
26:35:40Paolo Guiotto: But for us, X minus X star is a vector, so we cannot do product, vector times another vector. We don't have that operation.
26:43:900Paolo Guiotto: But what we can do is broader matrix times vector, so…
26:48:60Paolo Guiotto: what means, I can always write here this, let's say, in a bit,
26:55:780Paolo Guiotto: exaggerated way as X minus X star times itself, right?
27:01:300Paolo Guiotto: is not wrong.
27:03:620Paolo Guiotto: But now, the interpretation of this becomes, you take the Asian matrix, you apply to vector X minus X star.
27:13:380Paolo Guiotto: What is this guy?
27:15:440Paolo Guiotto: Once you have done this, this is a matrix.
27:21:130Paolo Guiotto: And this is a backdoor.
27:23:300Paolo Guiotto: So when you do matrix times vector, you get
27:28:590Paolo Guiotto: What kind of object you get?
27:31:670Paolo Guiotto: you get the vector. This is a square matrix, a D by D matrix.
27:36:780Paolo Guiotto: that you multiply line by column with a vector, which has the same number of components of D, because D is the number of the variables, so it is a B-dimensional vector. So what you get here is a D-dimensional vector, vector with D components.
27:53:240Paolo Guiotto: And then here you have another vector with D component that you have somehow to multiply to get a scala.
28:00:480Paolo Guiotto: So what kind of operation can you imagine does this?
28:04:820Paolo Guiotto: Huh?
28:06:170Paolo Guiotto: This color product, dot product.
28:08:710Paolo Guiotto: Okay, this guy is the scholar.
28:13:70Paolo Guiotto: product.
28:15:860Paolo Guiotto: And finally, of course, there is a little O,
28:19:760Paolo Guiotto: We already met the problem of little o. What does mean little o minus X star for vectors? It means little o of norm of X minus X star.
28:31:780Paolo Guiotto: So, since we have square, we square out this norm, and we are done.
28:37:410Paolo Guiotto: Now, this formula here is called the Taylor formula for our F, okay?
28:52:170Paolo Guiotto: Now, this formula is important for the test we are going to see. I will not give a proof, because it's a little bit…
28:59:900Paolo Guiotto: longer… the idea is, relatively, quick.
29:05:690Paolo Guiotto: From this formula.
29:15:220Paolo Guiotto: We… Gat, huh?
29:18:480Paolo Guiotto: Bethesda.
29:21:320Paolo Guiotto: for… Local, Minimums, or maximums.
29:34:100Paolo Guiotto: So… What is the idea?
29:38:80Paolo Guiotto: Well, let's… since I'm not going to do any proof here on this factor, let's do an informal argument that will help to have a conclusion, and then we write a statement.
29:52:710Paolo Guiotto: So, suppose that, then…
29:58:290Paolo Guiotto: So, it's a sort of a way… how do a theorem arises?
30:03:240Paolo Guiotto: For example, suppose that I know that X star is a stationary point, is stationary point.
30:13:570Paolo Guiotto: 4… F. So, gradient F at point X star is 0.
30:24:460Paolo Guiotto: Now, I want to discuss if this point is a minimum or a maximum. We know that it is not necessarily a minimum or a maximum. Now, the usual example of points where tangent is the reason that function is included. This is for functions
30:37:570Paolo Guiotto: of one real variable. For functions of several variables, we have something that you don't see for functions of one variable. It is the case of saddle points, so where you have a maximum along a direction, a minimum along another.
30:52:180Paolo Guiotto: At that point, the derivative is zero, the gradient is zero, but that point is… cannot be a minimum, because only direction you have a maximum. There cannot be a maximum, because only direction you have a minimum. A saddle, okay? We're now going to give a precise definition to what is a saddle.
31:08:430Paolo Guiotto: Okay, so let's suppose that we know that gradient F is 0.
31:14:980Paolo Guiotto: I look at this formula, and let's see what happens to this formula if gradient F is 0. So if you plug 0 in that formula, you see that it is here, the gradient term. If that factor is 0, you are multiplying back to 0 times another vector, so what you get?
31:36:390Paolo Guiotto: I have a back to zero, line by column time, you get zero. So, internal formula.
31:48:10Paolo Guiotto: We get, huh?
31:51:140Paolo Guiotto: let me just write in this way, I carry this value on the left-hand side, okay? We get f of x
32:01:680Paolo Guiotto: minus F of X star.
32:06:90Paolo Guiotto: is equal to… so you see, I carried this first term to the left-hand side. The second block is 0.
32:14:970Paolo Guiotto: So, I don't have anything, and then it remains that one facet, one calf, The Asian metrics, huh?
32:23:700Paolo Guiotto: at point X star, times X minus X star.
32:31:850Paolo Guiotto: Scholar with X minus X star.
32:37:330Paolo Guiotto: Plus a little correction.
32:44:590Paolo Guiotto: Now, this little o is something that when X is close to a star.
32:50:380Paolo Guiotto: So this quantity is approximately zero. This number is negligible respect to each one. That's the meaning of the literal.
32:58:950Paolo Guiotto: So I can say that 1X is close to X0,
33:03:310Paolo Guiotto: If X is close to X… sorry, X star, not X0,
33:09:720Paolo Guiotto: I can't forget of that reminder.
33:13:260Paolo Guiotto: And I can see only this, 1 half the Asian, F, at point X star.
33:20:150Paolo Guiotto: times X minus X star.
33:23:890Paolo Guiotto: Scarlet product with X.
33:26:580Paolo Guiotto: minus X star.
33:30:380Paolo Guiotto: So I've got this, let me copy the formula in red.
33:33:880Paolo Guiotto: the increment, or the difference, between F and,
33:39:210Paolo Guiotto: F at point X star is approximately 1 half
33:45:370Paolo Guiotto: the gradient… sorry, the Asian matrix of F at point X star times X minus X star.
33:55:890Paolo Guiotto: Scala with X star, X minus X star.
34:01:20Paolo Guiotto: Okay? So this is what comes out after we assume that gradient of F at point X star is 0. That is, X star is a stationary point for F.
34:13:10Paolo Guiotto: So, now…
34:14:489Paolo Guiotto: remind that our interest is what can be said about that point. Is that a minimum? Is that a maximum?
34:22:520Paolo Guiotto: Now, to be a minimum, Next star is minimum.
34:30:429Paolo Guiotto: If and not if the value of F at point X is larger than F at point X star, right?
34:38:989Paolo Guiotto: But this is equivalent of saying that the difference between F of X and FX star must be… Possibly.
34:48:130Paolo Guiotto: So, and similarly, maximum if this is less or equal, and this is negative.
34:55:310Paolo Guiotto: Now, look at that formula. That formula says exactly how the difference between these two values is made, and the difference is positive or negative if and or if the right-hand side is positive or negative. You see? So, this if and only if
35:12:860Paolo Guiotto: One Earth, the Asian, at point X star.
35:19:70Paolo Guiotto: times X minus X star, It's got a X minus X star.
35:28:330Paolo Guiotto: Is, for example, positive or negative in the other case.
35:34:320Paolo Guiotto: Does it remind anything to you, this?
35:39:220Paolo Guiotto: call this, reaction, you call it N.
35:43:540Paolo Guiotto: call this vector U.
35:46:140Paolo Guiotto: This is the same vector U.
35:49:10Paolo Guiotto: So this exactly means… well, forget of 1 half, because one half won't change the size of this thing. It is M, U, scholar U, greater or equal than zero, or less or equal than zero.
36:04:390Paolo Guiotto: So, what does it mean? M, positive definite or negative definite?
36:09:480Paolo Guiotto: M positive, or M negative definite.
36:14:960Paolo Guiotto: So this is not a proof. It looks like it is a proof, but there are many steps in which we are a bit vague. For example, where I eliminated the little O, I should be precise. But let's say that this gives this idea. If I am in a stationary point.
36:32:960Paolo Guiotto: That stationary point seems to be a minimum if and only if M is positive definite, a maximum if and only if M is definite negative.
36:44:330Paolo Guiotto: Now, this would be nice, because to understand the nature of that point, I compute variation, and I see if variation is positive or negative. I can say, I don't know if it is global, because it works only locally, but it will say something about that point.
36:59:630Paolo Guiotto: And in fact, this becomes a true theorem in its former proposition.
37:05:60Paolo Guiotto: Salt.
37:06:270Paolo Guiotto: Lat… F… equal F of X, defined on domain D of RD, Real value, though.
37:17:120Paolo Guiotto: B… twice… but we say we introduced the notation C2, right? Am I wrong?
37:24:700Paolo Guiotto: C2 is, the partial derivatives continuous until the second one. C2 only.
37:32:300Paolo Guiotto: So in this way, we have everything differentiable, second derivative, etc. Then… If X star Peace.
37:45:290Paolo Guiotto: Of course, indeed.
37:47:90Paolo Guiotto: It's a stationary bond.
37:51:980Paolo Guiotto: That fun?
37:53:230Paolo Guiotto: So, gradient of F at point X star equals 0.
38:01:740Paolo Guiotto: In the case when the Asian matrix is positive definite, We have that.
38:08:410Paolo Guiotto: This is a minimum.
38:14:90Paolo Guiotto: Let's say like that. And…
38:19:570Paolo Guiotto: the ancient matrix at FX star. Now, to have this conclusion, you cannot have a…
38:26:650Paolo Guiotto: just weakly positive deficit. You mean… you need to have stronger positive, so strictly positive deficit.
38:33:970Paolo Guiotto: Then, X star, is… Local, Minimum.
38:44:10Paolo Guiotto: 4F.
38:46:30Paolo Guiotto: While, if the dual statement, the Asian metrics, is strictly negative definite.
38:57:320Paolo Guiotto: Next stop.
38:58:760Paolo Guiotto: Is a local maximum.
39:05:910Paolo Guiotto: Okay?
39:07:900Paolo Guiotto: So this is a sufficient condition that it's a one way, because for the other way, we cannot ensure that it is strictly positive.
39:16:980Paolo Guiotto: The other way, which is not interesting for us, because it says that if you have a local minimum, the Asian is not necessarily strictly positive, but it must be positive in large sense, okay?
39:32:810Paolo Guiotto: So you… it's not an if-and-off. You cannot say that it is a minimum if and all if, the action is simply positive, because you can have a minimum with the action metrics equal to zero. I will show an example in a moment. So it doesn't work as a current impression. However, this is still part of this. If this will…
39:52:790Paolo Guiotto: It is a sufficient condition that says, if you know the character sign of the asion matrix, you know the nature of the point, which is normally what we are interested in, because we study a function, we determine stationary points, and we know if they are minimums or maximums, no?
40:12:230Paolo Guiotto: This, however, remains a local test. It says what happens to that point just in a neighborhood of that point. It doesn't tell you if it is a global meeting.
40:22:500Paolo Guiotto: So, global minimum, global maximum must be still determined with the meet of the FC. This, let's say, that provides some information for those points that we don't know if they are global, local minimums, you know, what they are. This gives a little condition for which we can understand the nature of that point.
40:40:830Paolo Guiotto: Okay, let's do remarks and examples. The first remark… It's actually a warming.
40:54:100Paolo Guiotto: Huge annual.
40:57:800Paolo Guiotto: at minimum, And maximum, for the same reason, point.
41:05:120Paolo Guiotto: Deion is only large positive or large negative.
41:12:610Paolo Guiotto: But it is not necessarily strictly positive-negative.
41:17:70Paolo Guiotto: But… Asian strictly positive, or strictly negative, might be… False.
41:29:630Paolo Guiotto: I'm sure that some of you is wondering, but why? You said that the Asian is positive at minimum. No, I said if the Asian is positive, you have a minimum.
41:41:480Paolo Guiotto: Of course, the point must be a stationary point of the bike is a sport.
41:45:500Paolo Guiotto: But what here is saying is that it is not true that you have always the notation that is positive at a minimum. It could be zero. And this is the example. Take this function, FXY, I take X power 4 plus Y power 4.
42:03:650Paolo Guiotto: It should be clear, even just looking at this function, that 00 is a global mean, right?
42:11:410Paolo Guiotto: No? You see that this is greater or equal than zero, which is the value at 0, 0. So, we… without doing any calculation, we know that 0, 0 is a…
42:24:210Paolo Guiotto: global.
42:27:730Paolo Guiotto: Minimum.
42:29:690Paolo Guiotto: But… If it is global, it is also local. It is the vice versa, which is non-true, okay?
42:36:480Paolo Guiotto: But what happens if we complete the H here? So let's do the job. DXF is 4X cubed.
42:44:520Paolo Guiotto: DYF is 4Y cubed. They are both continuous, so F is differentiable.
42:55:410Paolo Guiotto: Fuck.
42:56:580Paolo Guiotto: Done.
42:58:150Paolo Guiotto: Let's go on the second partial derivatives. VXX is equal to 12X squared
43:05:830Paolo Guiotto: D of F, the second XY of F is 0, as well as the second Y axon of X.
43:16:960Paolo Guiotto: And this second YY of F is, again, 12Y squared.
43:23:210Paolo Guiotto: So, all of them are continuous. So, gradient F is differentiable.
43:32:530Paolo Guiotto: And so we have this second derivative.
43:35:980Paolo Guiotto: The second derivative is, in this case, the matrix made by DXX F?
43:43:910Paolo Guiotto: The second, this is XYX. The second YX…
43:50:560Paolo Guiotto: But it's the same at the end.
43:54:50Paolo Guiotto: YX.
43:55:740Paolo Guiotto: And, these are from the YWOW.
44:01:310Paolo Guiotto: So this is to have X squared 00, to have Y squared.
44:09:60Paolo Guiotto: At 0, what happens?
44:12:60Paolo Guiotto: The Asian matrix at 00 is just All entries equal to zero.
44:19:390Paolo Guiotto: It's a degeneric case.
44:21:730Paolo Guiotto: It is not stakely positive.
44:25:170Paolo Guiotto: Because if you compute the determinants, these are 0 and 0.
44:29:800Paolo Guiotto: Okay?
44:31:140Paolo Guiotto: So, it's a… it's a degenerate case.
44:34:510Paolo Guiotto: So, it's not positive.
44:37:350Paolo Guiotto: But, but nonetheless.
44:46:460Paolo Guiotto: 0, 0.
44:48:130Paolo Guiotto: is a global meeting.
44:52:910Paolo Guiotto: So this is also important, because if you do not arrive to classify
44:58:560Paolo Guiotto: this point, by getting that the Asia matrix is positive definite or negative definite, you cannot conclude anything.
45:08:520Paolo Guiotto: we will see in examples. In Lucia, this is a little example, no? Suppose that,
45:15:790Paolo Guiotto: In this example, I am blind, so I don't see that 00 is clearly a minimum.
45:23:880Paolo Guiotto: So, suppose that I do not recognize, and I… I just… I just proceed as a donkey, okay? So what the donkey does, if…
45:36:780Paolo Guiotto: I… I'm saying I, I'm not saying you. Okay, I proceed.
45:43:140Paolo Guiotto: Oz.
45:44:270Paolo Guiotto: That's Barney Key.
45:47:490Paolo Guiotto: So what should I do? Now, I have a function, I want to determine minimum, maximum, the donkey does this.
45:53:30Paolo Guiotto: I start computing gradient F and looking for points where gradient F is 0. So that's easy, because I computed gradient F, I got 4x cubed, 4Y cubed, so this means vector 4X cubed, 4Y cubed.
46:09:290Paolo Guiotto: equal back to 0, and that's possible if and only if XY are both equal to 0, so I get .00. I think that the donkey does this, okay?
46:20:240Paolo Guiotto: But then, the donkey computed the second derivative, and discovers that at 00 is equal to this.
46:30:580Paolo Guiotto: So…
46:32:00Paolo Guiotto: Probably, if the donkey applies the test we have seen here with the sub-determinants. Well, actually, it was in the previous slide, down here.
46:46:170Paolo Guiotto: the test that we just finished to discuss this morning, okay?
46:53:430Paolo Guiotto: The donkey would compute the determinants here.
46:59:390Paolo Guiotto: So, I have that determinant of the submetrics one by one is zero. Determinant of matrix 000 is 0.
47:10:210Paolo Guiotto: So, the donkey would come through that M, so…
47:14:920Paolo Guiotto: radiation is not positive definite, so the 0.00 is not.
47:22:590Paolo Guiotto: A minimum.
47:24:820Paolo Guiotto: And of course, since it is not also negative definite, 00 is not a maximum.
47:33:560Paolo Guiotto: Which is wrong, of course.
47:36:100Paolo Guiotto: Well, the point is that this final step is wrong. This is the implication that you cannot do.
47:44:10Paolo Guiotto: Let's say that the donkey up to that point, has done the right thing, no? Computed the gradient, found the stationary point.
47:53:280Paolo Guiotto: discussed the design of the Asian metrics. The point is that the conclusion is wrong. You cannot say that since
48:00:820Paolo Guiotto: The Asian matrix is not positive definite, strictly positive definite, to be precise. That point cannot be a minimum, and since it is not also strictly negative definite, that point is not a maximum.
48:13:810Paolo Guiotto: That's wrong, because you can be a minimum, exactly this example shows, having the asymetics which is not centrally positive.
48:23:30Paolo Guiotto: That's why, as I told you in the statement, the right implication on the other direction, if you are the minimum, is not variation matrix's problem.
48:33:920Paolo Guiotto: But they're slightly weaker with that. The Hat is weaker than zero, so it allows also the case Asian matrix equals zero, which is exactly the case we have here.
48:45:290Paolo Guiotto: Okay, so I would say that it's time to take a break, and then we continue with some exercise.
48:56:640Paolo Guiotto: Okay, now I want to show you with, some example, how we can use this, this, this material we have seen.
49:06:840Paolo Guiotto: This is the example, 5… 9…
49:13:780Paolo Guiotto: It says you have this function, F of XM,
49:22:200Paolo Guiotto: F of X equal… X over, Y equal X times X power 4 plus Y squared.
49:31:310Paolo Guiotto: minus Y squared X cubed up.
49:36:00Paolo Guiotto: Seems that we are going to have a lot of calculations. So, the problem is, fine… And what?
49:45:700Paolo Guiotto: classify…
49:52:190Paolo Guiotto: the… Stationary points… off.
50:01:320Paolo Guiotto: And then discuss existence of local, global.
50:05:760Paolo Guiotto: X-ray muscle, let's say 1, 2…
50:09:900Paolo Guiotto: local, it is included in the classified. What does mean classify? Let's say, mean, discuss mean max of F,
50:24:400Paolo Guiotto: on us, too.
50:29:300Paolo Guiotto: But there is also number 3F of R3.
50:35:440Paolo Guiotto: Okay, so what does it mean, classify? Classify means, are you able to say the stationary points are minimums, maximums, and particularly local minimum, local maximums, okay? Because in general, a stationary point, if it is an extrema, is not necessarily global, okay?
50:55:100Paolo Guiotto: While it is global, it is also local. So, local, global. Are you able to say that? So, let's, first of all, start determining these points. So, here, we start computing the partial derivatives, DX, F,
51:08:670Paolo Guiotto: is now, doing… well, it's better to write X power 5 XY squared minus X cubed Y squared. So we have the real deal with respect to X is 5X power 4 plus Y squared minus 3X squared y squared.
51:28:830Paolo Guiotto: derivative with respect to Y is, 0 plus, 2XY.
51:38:480Paolo Guiotto: minus, 2X cubed Y.
51:43:750Paolo Guiotto: So, as you can see, they are both continuous functions on the full plane R2, so this means that differentiability test
51:57:910Paolo Guiotto: The function f is differentiable on… Too.
52:04:770Paolo Guiotto: Okay?
52:06:290Paolo Guiotto: Now, stationary points is that this, we don't think we've got mediation for stationary points.
52:13:130Paolo Guiotto: We have to just work here. XY is… a stationary point.
52:21:390Paolo Guiotto: Whoa.
52:23:430Paolo Guiotto: If not, if gradient over, at point XY is equal to 0.
52:30:730Paolo Guiotto: That means we have to impose these two derivatives equal to zero, we get system.
52:36:460Paolo Guiotto: 5X.
52:38:170Paolo Guiotto: power 4 plus Y squared.
52:41:510Paolo Guiotto: minus 3X square, Y squared, there is no possibility
52:46:550Paolo Guiotto: To simplify this, at least seems. While in the second, we could factorize a 2, an X, a Y, so we have 2XY times 1 minus X squared.
53:00:790Paolo Guiotto: Which is much better, because as you can see, this is a product, no? So you write this as 1 minus X, 1 plus X, and you have all the alternatives. So I'd say, definitely, let's work on this.
53:14:570Paolo Guiotto: equation.
53:15:700Paolo Guiotto: So we have a lot of alternatives.
53:18:560Paolo Guiotto: But first, we simplify this, too.
53:21:410Paolo Guiotto: We have X equals 0. This is the first alternative.
53:25:760Paolo Guiotto: Hold on.
53:26:980Paolo Guiotto: Second alternative, Y equals 0. R.
53:30:970Paolo Guiotto: Third alternative X equals 1, or 4th alternative X equals minus 1. You see that?
53:38:480Paolo Guiotto: Now you plug all this into the first equation.
53:41:670Paolo Guiotto: If I put X equals 0, the first equation becomes Y squared equals 0. So that's nice, because this means I have a unit point 0, 0.
53:51:960Paolo Guiotto: If I put Y equals 0, I get 5x power over 4 equals 0, so again, the same point, zero there.
54:00:40Paolo Guiotto: If an output X equals 1 in the first line, I get 5.
54:06:160Paolo Guiotto: 5 plus Y squared minus 3Y square equals 0.
54:13:290Paolo Guiotto: Which is, not, immediate, so let's, let's continue. And X equals minus 1 is the same, because it was X squared X4. So it is, again, 4, 5, so basically minus 2Y square equals 0.
54:27:660Paolo Guiotto: So these two are similar, because you see the same equation for Y.
54:32:290Paolo Guiotto: So, here we are, buddy.
54:34:770Paolo Guiotto: X equals 1, and so 5 minus 2Y squared equals 0, so Y squared equals 5 alpha.
54:43:690Paolo Guiotto: And this is X equals minus 1Y square equals 5F.
54:49:890Paolo Guiotto: And so, finally, this, is two points.
54:54:420Paolo Guiotto: 1 and plus minus root of the 5 half.
54:59:780Paolo Guiotto: This is minus 1 plus minus a root of 5 off.
55:05:480Paolo Guiotto: So the conclusion is…
55:10:580Paolo Guiotto: D.
55:11:620Paolo Guiotto: stationary boils… R.
55:16:150Paolo Guiotto: 0, zero.
55:18:640Paolo Guiotto: 1 plus minus root of 5 half, and minus 1 plus minus, so I can just write plus minus here. All possibilities, all combinations of signs, okay? So these are 4 points, not 2.
55:33:580Paolo Guiotto: Okay?
55:36:770Paolo Guiotto: combinations.
55:43:160Paolo Guiotto: Offside.
55:48:680Paolo Guiotto: So this is, about,
55:52:190Paolo Guiotto: No, this is not yet about question one, because now we see what does mean classified.
55:58:460Paolo Guiotto: For the moment, we know they are the stationary point. We don't know their nature, if they are minimums, maximums.
56:05:190Paolo Guiotto: In any case, what we're going to use is this criteria here that won't tell you that they are global or local. It will tell you if something happens, they are local, only local. It won't tell you they are global.
56:23:870Paolo Guiotto: So, to classify these points.
56:36:150Paolo Guiotto: So, let's… Nope.
56:41:870Paolo Guiotto: the Asian metrics.
56:46:330Paolo Guiotto: To compute this, we have to do the secondary. The second reality with respect to X. What is this?
56:53:330Paolo Guiotto: We have to take derivative with respect to X and differentiate again with respect to X, so we get 20x cubed.
57:04:70Paolo Guiotto: when we differentiate Y square 0, minus 6XY squared.
57:10:710Paolo Guiotto: Right.
57:12:590Paolo Guiotto: minus 6… Exit vicepaw.
57:17:380Paolo Guiotto: Then we have the Nixon derivative, XY. Since these functions are polynomials, you understand that these derivatives are the same, with respect to
57:26:980Paolo Guiotto: Switch in the order, so take the one you prefer and differentiate with respect to the other, blah blah blah. For example, I can take the first and differentiate with respect to Y, or the second and differentiate with respect to X.
57:39:780Paolo Guiotto: you get the same result. So if I do with the first line, I get 2Y minus 6x2Y,
57:48:160Paolo Guiotto: So, 2Y minus 6X squared Y.
57:52:980Paolo Guiotto: And then there is the second derivative with respect to Y.
57:59:620Paolo Guiotto: So this is 2X minus 2X cubed.
58:07:680Paolo Guiotto: Okay, so these are the data. From this, you see that they are continuous. So, again, I'm using the differentiability test.
58:17:190Paolo Guiotto: apply the knobs to F, but to the derivative of F, to the gradient. So gradient of F is differentiable
58:25:540Paolo Guiotto: So, there existed the second derivative that we call, Asian metrics. Reminder, it's not just a formality, because the Jacobian matrix, this is the Jacobian matrix, the Jacobian matrix of the gradient.
58:43:500Paolo Guiotto: the Jacobian matrix can be written without the function is differentiable, because it needs only the partial index, okay? And since you need differentiability to all these general parts.
58:54:950Paolo Guiotto: Differentiability is everywhere, so here is implicit because F is 2, but you see…
59:02:310Paolo Guiotto: Yeah, twice differentiable. You need all of these properties. You should always, in principle, be sure that you can apply the theorems you are using.
59:11:990Paolo Guiotto: So, the patient matrix at generic point XY, if you want to see these metrics, is this is the first entry, 20X cubed minus 6,
59:23:70Paolo Guiotto: Y squared, then on the diagonal, on the two…
59:28:420Paolo Guiotto: opposite of the diagonal, 2Y minus 6x2Y, and 2Y minus 6x2Y, yeah?
59:38:650Paolo Guiotto: And this is 2X minus 2X cubed.
59:43:340Paolo Guiotto: So now we have to look at this, what is this in this, four, five bottles?
59:49:300Paolo Guiotto: You immediately see that at 0, the matrix degenerates.
59:53:890Paolo Guiotto: Because the Asian matrix at 00 is… All entries equal zero.
00:02:220Paolo Guiotto: So, from this, we cannot conclude anything, because
00:06:240Paolo Guiotto: The unique way to use this test
00:10:600Paolo Guiotto: Is to have a street sign here.
00:14:950Paolo Guiotto: Okay? So if it is not strictly positive, it is not strictly negative, you can do this. Then you can do anything with that test.
00:24:250Paolo Guiotto: Okay, so here, we cannot… use… V… But let's call it Asian… Best.
00:40:60Paolo Guiotto: So, how can we do… okay, we cannot do radiation testing. Is there any other way?
00:46:460Paolo Guiotto: Now, the other way is to do by hand, okay? There is always that way. In any case, in any situation of real life, you can always do by hand. How do we do by hands to understand what is .00?
01:01:680Paolo Guiotto: But an idea could be, let's try to see if we section this function, and we understand anything about the behavior. For example, what if I have F of X 0? Let's see what is this.
01:15:250Paolo Guiotto: But let's say that… In this case…
01:24:780Paolo Guiotto: to classify.
01:28:300Paolo Guiotto: point.
01:29:680Paolo Guiotto: Zero. Zero.
01:32:60Paolo Guiotto: Sweet.
01:33:260Paolo Guiotto: so… With the prehistorically needed.
01:41:290Paolo Guiotto: by facts.
01:43:10Paolo Guiotto: So, this means that there is no rule. Whatever is correct is good, okay? The unique requirement is that it must be correct.
01:52:10Paolo Guiotto: So if I do F of X0, what is this?
01:56:430Paolo Guiotto: Well, I am lucky if I do have a vaccine, because if you do have a vaccine.
02:01:950Paolo Guiotto: You see what happens to the function when you put Y equals 0. The second and the third black term are 0. It remains X to the power of 5.
02:12:00Paolo Guiotto: So if you see this, What would you conclude about 00?
02:24:330Paolo Guiotto: Look, how is the function X to power 5? If you have to plot X to power 5, this is the x-axis.
02:32:180Paolo Guiotto: Of course, for X, around 0, so 0 is 0. X to power 5 is like…
02:43:180Paolo Guiotto: Yes, no maximum, no minimum, but to plot, it's like a cubic, it's like that.
02:49:310Paolo Guiotto: Okay, so this is the function X to power 5. Now, this is the section of F. It means that, if you want to have an idea, we are here in the Cartesian plane XY,
03:03:90Paolo Guiotto: So imagine that you move the point along the x-axis, and the value… the value F of XY… now, I should plot a surface, but if I restrict to points X0, what I see is that red line in space, so I see something like this.
03:21:370Paolo Guiotto: goes down. Of course, the red line is in the vertical plane XZ, let's say. Let's call this tensed axis.
03:30:20Paolo Guiotto: Okay, it's not,
03:32:450Paolo Guiotto: So, we can say that this function cannot have any minimum… the 00 is, is, is not…
03:42:100Paolo Guiotto: is not… It's Anita.
03:46:790Paolo Guiotto: Mean?
03:48:140Paolo Guiotto: Local meal.
03:50:520Paolo Guiotto: And we know local.
03:54:40Paolo Guiotto: Max.
03:55:650Paolo Guiotto: So, none of the two.
03:57:570Paolo Guiotto: So at the end, we classified. We said that it is a point which is not a local extreme.
04:05:150Paolo Guiotto: Okay, now let's compute deviation on the other points.
04:10:90Paolo Guiotto: So, before we launch in computing, I, I, I don't know if…
04:15:80Paolo Guiotto: We need, unfortunately, to write four different matrices.
04:19:420Paolo Guiotto: So, I will do one of them, and then you do the other three. Because, you see, there are the cubes, there is Y, so these still are… they change when you change sign. So, let's do 4.1 root of 5R. Let's see what is it.
04:36:310Paolo Guiotto: the Asian matrix of F at 0.1
04:40:820Paolo Guiotto: Root of 5 halves, we have to do a little bit of calculations. So, X is 1,
04:47:310Paolo Guiotto: So, yeah, 20, 1 cubed 20.
04:52:560Paolo Guiotto: minus…
04:54:410Paolo Guiotto: 6XY squared. So X is 1, Y is root of 5 half, Y squared would be 5 halves, so it is minus 6 times 1 times 5,
05:06:380Paolo Guiotto: That's the first entry.
05:07:990Paolo Guiotto: The second one is 2Y. Y is root of 5 half. You see that it multiplies everything? So maybe it's better if we write root of 5 half that multiplies
05:20:390Paolo Guiotto: 2 minus 6x squared. X is 1, so 2 minus 6 minus 4.
05:26:720Paolo Guiotto: Fortunately, the matrix is semantic, so I have the same entry here, minus 4 root of 5 half.
05:33:450Paolo Guiotto: And down here, I have 2X minus 2X cubed.
05:38:120Paolo Guiotto: which is X is 1, 2, minus 2, 0. So this is… Now, the fact that you see an entry equals 0 is not… is not at all a bad thing, okay?
05:49:220Paolo Guiotto: Because, well, let's, rework this. So this is, 30 divided 215, 20 minus 15, 5.
05:59:360Paolo Guiotto: But here, we cannot do anything better than this, I think. 4 root of 5 alpha minus 4 root of 5 alpha, and 0 here.
06:09:730Paolo Guiotto: Now, this is definitely not to degenerate.
06:13:460Paolo Guiotto: And so let's see if we can determine the sign. To determine the sign, we have to compute these two determinants. The determinant of the 1x1 submetics, which is this one, so the matrix made by the first line column, which is 5, it is positive. And the determinant of the second, it is the full matrix.
06:33:720Paolo Guiotto: So this one is 5 times 0, 0, minus… then there is the product of these two, which is plus, so 16, 5 out, it doesn't matter, it's negative.
06:45:320Paolo Guiotto: So here we are the case. First, the diminutivity is positive, second determinant is negative.
06:50:690Paolo Guiotto: Which is not one of the… it's an alternating sign, but with the wrong order, no? You should start from minus plus.
06:58:90Paolo Guiotto: So, what happens in this case? Because we are, we have a, we are… you see that,
07:05:310Paolo Guiotto: It is not positive, it is not a negative death.
07:09:570Paolo Guiotto: Because to be positive, definitely, all sub-determinants must are not disclosed in the other slide.
07:17:440Paolo Guiotto: Sub-determinants must be equal to… must be positive. And to be negative, all subdeterminants must be alternating sides with first sign minus, and that's the opposite.
07:29:610Paolo Guiotto: Now, in this case, we can say that,
07:35:260Paolo Guiotto: It… it is, it is respected, you know, because,
07:39:500Paolo Guiotto: It is like, if one, if you take the eigenvalues, this means that the first eigenvalue, lambda 1, is positive, and the product lambda 1, lambda 2 is negative, so it means that this guy must be negative.
07:55:530Paolo Guiotto: So we have an eigenvalue positive, an eigenvalue negative, so this should be something like this. On a direction, the derivative is positive, so… and on the other direction, the derivative is negative. So it should be a saddle point.
08:10:690Paolo Guiotto: But to… to make this, clear, so we need,
08:21:740Paolo Guiotto: Let's say, quite decent, because…
08:29:529Paolo Guiotto: So, it is like if you have the second derivative along one direction is positive. When the second derivative is positive, it means that the function is convex, no? You have a minimum on that direction.
08:43:279Paolo Guiotto: But there is another direction for which the second derivative is negative.
08:47:529Paolo Guiotto: So this means that on this direction, you would see the function with the concavity pointing downward.
08:53:490Paolo Guiotto: So you have a direction on which the function has a minimum, a direction on which a function has a maximum. This is what is a saddle point.
09:02:950Paolo Guiotto: And this happens whenever you have,
09:07:770Paolo Guiotto: So let's put a definition. I forgot to put this as definition. If,
09:13:920Paolo Guiotto: Determine if, the determinants, of the Asia metrics.
09:21:550Paolo Guiotto: at some point, X star.
09:27:60Paolo Guiotto: of, let's say, the K by K submetuses.
09:32:00Paolo Guiotto: Are different from zero.
09:34:670Paolo Guiotto: bots.
09:37:149Paolo Guiotto: So, do you understand this notation?
09:43:330Paolo Guiotto: So this is the… K by K sub… metrics, huh?
09:53:470Paolo Guiotto: of the Asian magnetic zone.
10:00:90Paolo Guiotto: But, the Asian matrix is not… Positive.
10:06:620Paolo Guiotto: And it is not negative.
10:09:650Paolo Guiotto: So, all other cases, basically, because positive definite means all these determinants are positive. Negative definite means these determinants have alternative sides, starting with minus.
10:22:560Paolo Guiotto: You know, other possibilities, so for example, all these determinants, negative.
10:28:120Paolo Guiotto: Or you have an alternating sequence with not minus, plus, minus plus, or you could have any other kind of thing. So, for example, suppose I have minus, plus, minus, minus plus, I see this sequence of size, you see that it starts nice, but at this point, I should have a plus.
10:47:550Paolo Guiotto: So in this case, I am still in this condition. This point is a saddle point.
10:53:830Paolo Guiotto: Next thought.
10:55:960Paolo Guiotto: is older.
10:59:860Paolo Guiotto: saddle…
11:04:320Paolo Guiotto: Point.
11:06:220Paolo Guiotto: It could be proved that along certain directions, there is a maximum, along certain other there is meaning.
11:12:360Paolo Guiotto: Okay?
11:13:400Paolo Guiotto: So, we can say that this poem,
11:25:840Paolo Guiotto: Yeah, I'm not negative, thank you.
11:28:530Paolo Guiotto: So, in our case, 0.1 root of 5 half… is, sadly.
11:40:370Paolo Guiotto: Let me see if one of these is, like.
11:46:250Paolo Guiotto: No, I… the same happens to all these points, so… So, same… Conclusion.
11:57:40Paolo Guiotto: for… Oh, father.
12:01:910Paolo Guiotto: stationary bonds.
12:04:620Paolo Guiotto: So, check…
12:06:550Paolo Guiotto: you have to compute deviation metrics on the other points and see what happens. Let's do an exercise where we see something.
12:17:130Paolo Guiotto: So here, at the end of the chapter.
12:21:630Paolo Guiotto: Now, you could, review the exercise 296,
12:27:120Paolo Guiotto: Where there is now… you understand what does mean classify. Classify means this. So, do exercise 296, classifying it.
12:44:180Paolo Guiotto: stationary points.
12:48:550Paolo Guiotto: Now, I want to do another exercise, which is a quite comprehensive exercise, which is two 9… 13.
12:59:770Paolo Guiotto: And you do the… due to 9-14, I will publish the solution.
13:11:460Paolo Guiotto: Okay, let's see the 2, 9, 14 here, 13, here we have a function f is equal to…
13:18:140Paolo Guiotto: X squared times Y squared minus X minus 1 squared.
13:26:200Paolo Guiotto: for XY in R2.
13:31:260Paolo Guiotto: Question 1.
13:33:460Paolo Guiotto: There exists a limit, of F at infinity.
13:39:160Paolo Guiotto: Question 2.
13:43:140Paolo Guiotto: If he has computed, okay. Question 2, determine and classify stationary points.
13:58:440Paolo Guiotto: Personally, minimum, maximum.
14:03:760Paolo Guiotto: of F on R3.
14:11:320Paolo Guiotto: Then there is a second problem. Basically, it's a second exercise. Question 4 is,
14:18:20Paolo Guiotto: Show that, there exists the mean and max,
14:25:170Paolo Guiotto: For this F on D, where D is now a subdomain of F2, is domain of points XY. These are two exercises in one,
14:35:220Paolo Guiotto: Y also equal than zero, X between 0 and Y plus 1.
14:45:260Paolo Guiotto: And then, what is F4D?
14:49:680Paolo Guiotto: So it's a sort of comprehensive exercise with several questions.
14:54:520Paolo Guiotto: Number one.
14:57:230Paolo Guiotto: So what about the limit?
15:00:820Paolo Guiotto: So, always, when you see some minus, probably you have some suspects that you don't see the typical quantities X squared, Y squared.
15:10:540Paolo Guiotto: So, let's see what should happen here.
15:14:580Paolo Guiotto: You don't see anything?
15:23:10Paolo Guiotto: I don't… What do you want to do?
15:29:360Paolo Guiotto: Bye.
15:30:740Paolo Guiotto: Yes, I see. Okay.
15:33:810Paolo Guiotto: But, I mean, we have to discuss the limit.
15:42:680Paolo Guiotto: Yup.
15:44:50Paolo Guiotto: Yeah, but do you have… do you see anything here directly? Should we try to go?
15:49:240Paolo Guiotto: Boom.
15:51:230Paolo Guiotto: Proving that the limit is something, or, you don't see, okay.
15:58:320Paolo Guiotto: Let's see, for example, let's look at sections. I have F of 0Y. I start with this because of factor X squared that kills everything, so you see that this is constantly going to 0, so it goes to 01.0.
16:18:860Paolo Guiotto: 0, Y goes to the infinity of space, means Y goes to plus-minus infinity.
16:26:30Paolo Guiotto: So this function won't go to infinity, first of all.
16:31:130Paolo Guiotto: Now, if there is a limit, it must be zero.
16:34:410Paolo Guiotto: So, do you think that this function is going to zero at infinity?
16:40:820Paolo Guiotto: It does.
16:43:770Paolo Guiotto: Okay, I don't know what intuition do you have, but let's take FX0. Let's kill the other coordinate. What we get is X square…
16:54:840Paolo Guiotto: times minus X minus 1 squared.
16:58:560Paolo Guiotto: So, in other words, minus X squared
17:01:420Paolo Guiotto: X minus 1 square. Now, when I send point X0 to infinity, what happens?
17:11:870Paolo Guiotto: This quantity has a limit, yes or no.
17:16:80Paolo Guiotto: X0 goes to infinity. This silence… I don't like this silence. You have to talk here, okay? We're doing practice. We have to do together. So X0 goes to infinity means X goes where?
17:30:180Paolo Guiotto: plus minus infinity. What happens when x goes to plus minus infinity, that quantity, X squared, times X minus 1 squared?
17:38:480Paolo Guiotto: It goes to minus infinity, so what is the conclusion?
17:43:360Paolo Guiotto: You see? Your intuition.
17:48:440Paolo Guiotto: your intuition was already killed. Okay, there is no limit at infinity.
17:53:890Paolo Guiotto: Okay, so for the moment, we answered the first question, so we don't need to do anything more than this, because we found sections along which we have two different limits. This says the limit does not exist.
18:08:150Paolo Guiotto: Exactly these things, because you see that if you put X equals 0, you kill everything, you have 0. Then XY going to infinite means B coordinates, this quantity will always be equal to zero. With B coordinates. Starts to be strange, you know?
18:28:90Paolo Guiotto: For example, other things that you should notice here, for example, there are other interesting sections. For example, if you take the section Y equal X minus 1, you move around parentheses, also this one, you get the function equal to 0. However, what we have done is sufficient to conclude the question to
18:46:40Paolo Guiotto: Stationary points.
18:48:720Paolo Guiotto: So let's start computing the partial derivatives. DXF is equal to… Now…
18:58:90Paolo Guiotto: Okay, so 2X times the parentheses…
19:03:420Paolo Guiotto: Y squared minus X minus 1 square.
19:08:110Paolo Guiotto: Plus X squared, I'm differentiating as a product, okay?
19:12:550Paolo Guiotto: then I have to differentiate the parentheses, Y squared is 0 minus 2X minus 1.
19:19:630Paolo Guiotto: This is the derivative with respect to X. Derivative with respect to Y is easier, because I have X squared, which is a constant, times… the parenthesis contains Y if they are Y squared minus X minus 1 square. When I differentiate with respect to Y, I get to Y.
19:36:290Paolo Guiotto: You see that these are continuous, so the function f differentiability, Best.
19:44:810Paolo Guiotto: F is differentiable on R2.
19:51:280Paolo Guiotto: Okay. Now, XY is stationary point, for that.
20:00:520Paolo Guiotto: if NW is gradient F, equals zero.
20:05:310Paolo Guiotto: at point XY, of course. This is a system
20:09:960Paolo Guiotto: Can we simplify anything here? So, when I put the X equals 0, I can cancel the 2, I can factorize an X, so I can write X times Y squared minus X minus 1 squared.
20:25:810Paolo Guiotto: Then, since I'm factorizing an X here, I get minus X times X minus 1.
20:33:850Paolo Guiotto: Perhaps we can simplify a bit.
20:37:320Paolo Guiotto: more, but the second equation seems to be better, because the second is just 2x squared y equals 0, so I will start to work on this one.
20:49:410Paolo Guiotto: This is the alternative. Either X is zero form.
20:56:300Paolo Guiotto: Why is it?
20:57:520Paolo Guiotto: When x is 0, I plug into the first line, I get 0.0.
21:02:70Paolo Guiotto: So, what does it mean? This…
21:06:20Paolo Guiotto: All points near Y are solutions for every Y in R. All these are stationary points.
21:13:20Paolo Guiotto: Y equals 0. When Y equals 0, I get X.
21:16:910Paolo Guiotto: times, well, perhaps I can factor…
21:22:160Paolo Guiotto: A minus put outside a factor X minus 1,
21:26:380Paolo Guiotto: And then it remains X minus 1 plus X, so 2X minus 1, if I'm not wrong.
21:36:780Paolo Guiotto: I've done the algebra into that parenthesis.
21:40:20Paolo Guiotto: So this is 3 sub-cases. Y equals 0, X equals 0.
21:47:720Paolo Guiotto: Y equals zero?
21:50:450Paolo Guiotto: X equals 1, or Y equals 0.
21:54:760Paolo Guiotto: X equals 1 power.
21:57:230Paolo Guiotto: From the first day of .00, that I can consider already there, so…
22:02:490Paolo Guiotto: That new point. Here I have one zero, which is new, and here I have 1 half 0, which is, again.
22:09:690Paolo Guiotto: Okay, so we have these bonds. These are infinitely many.
22:14:290Paolo Guiotto: These are just two other points. We have to classify So, we used the ashen, We tried.
22:26:430Paolo Guiotto: So… Apply.
22:30:10Paolo Guiotto: We'll try to apply.
22:35:870Paolo Guiotto: patient… faster.
22:39:450Paolo Guiotto: So we need to compute this second a little bit now.
22:43:780Paolo Guiotto: Okay, so… huh.
22:47:280Paolo Guiotto: What a nice expression. So, DXF…
22:55:70Paolo Guiotto: Sooner or later, we have tools.
23:00:770Paolo Guiotto: So, the second derivative with respect to X,
23:04:100Paolo Guiotto: I don't know what is convenient to do. For the moment, I just… Don't simplify this.
23:10:80Paolo Guiotto: Because I don't see any particular simplification, so let's see the calculation. So I consider still this as a problem. So, 2X times this. So, differentiating with respect to X, I have derivative of 2x, which is 2, times the parenthesis, Y squared minus X minus 1.
23:28:140Paolo Guiotto: Screw it up.
23:30:740Paolo Guiotto: plus 2X times the derivative of the parenthesis, which is minus 2X minus 1.
23:41:740Paolo Guiotto: Then, we have this subtle product, minus 2, That multiplies.
23:51:320Paolo Guiotto: I have X squared by an X minus 1. Differentiating X squared, you have 2X times X minus 1,
23:58:520Paolo Guiotto: plus X squared times the deliberative of X minus 1, which is 1, and that's it.
24:05:760Paolo Guiotto: Then I have the mixed derivative, dxy second F. This is better if I differentiate from this.
24:13:580Paolo Guiotto: No? So differentiate with this with respect to X, okay? Because this is the Y, so I get 4XY,
24:24:440Paolo Guiotto: And the second derivative with respect to Y, Is, 2X square.
24:35:10Paolo Guiotto: Now, let's see the… well, we conclude that all these are continuous, so, gradient F is differentiable.
24:45:50Paolo Guiotto: So… F is twice… differentiable.
24:51:790Paolo Guiotto: Because F is differentiable, and the gradient is differentiable. F and it's further.
24:56:740Paolo Guiotto: Now, let's compute the Asian metrics. I don't want to put on this mess, but let's see on this point. So, let's compute for… because there is the zero here that we'll add, perhaps.
25:09:860Paolo Guiotto: So, for X equals 0, the asia matrix at .0Y is,
25:16:600Paolo Guiotto: When we put this X equal to 0,
25:21:70Paolo Guiotto: So we see that this first entry is a 2Y square minus 1, right?
25:29:920Paolo Guiotto: Battle this is zero.
25:31:990Paolo Guiotto: 4D mixed term is 0.
25:34:590Paolo Guiotto: And unfortunately for this one, it's zero. So, this means that, as you can see, the determinant
25:41:670Paolo Guiotto: Of this, sub, let's say…
25:47:490Paolo Guiotto: I don't know, we should use some notation. The one-dimensional matrix is just 2Y square minus 1,
25:56:470Paolo Guiotto: But when we take the 2x2 submatics, which is the matrix itself, this is zero.
26:04:480Paolo Guiotto: So, definitely this is, this session matrix is not positive, and it is not negative.
26:13:700Paolo Guiotto: So I can throw away the tester.
26:17:300Paolo Guiotto: And so… And we must solve by some direct discussion.
26:23:590Paolo Guiotto: by… cancer.
26:27:590Paolo Guiotto: Which is never easy.
26:31:590Paolo Guiotto: So, let's… let's, try to understand. The point is .0Y, so this is the domain of XY. These points are here.0Y.
26:45:890Paolo Guiotto: What is the value of the function at this point? Let's remind that the function fxy is X squared times y squared minus X minus 1 squared, right?
26:57:490Paolo Guiotto: So, first of all, when I evaluate F0, why I get the zero value.
27:03:970Paolo Guiotto: So the function is, along this axis is constantly equal to zero.
27:13:350Paolo Guiotto: Okay.
27:19:910Paolo Guiotto: Yeah. Now, how can we understand the…
27:24:00Paolo Guiotto: Well, there could be… okay, so we could do the following, discussion, for example. I'm saying…
27:31:840Paolo Guiotto: There is not a general strategy here, so…
27:35:420Paolo Guiotto: For example, since here the function can be factorized, as you suggested, X squared times 2Y,
27:42:290Paolo Guiotto: minus X minus 1, so Y minus X plus 1 times Y plus X minus 1.
27:51:170Paolo Guiotto: We could even study the sign that could tell something, because the function is value 0 at this point. So if function is positive, I can say that that value is meaning. If function is negative, it's maximum, something like this. So let's discuss the sign. F of XY, let's see where it is positive.
28:10:600Paolo Guiotto: That means greater than f of 0.01, you see? So, well, it is a mean.
28:18:570Paolo Guiotto: Now, this is positive, so definitely, if and not infant.
28:23:130Paolo Guiotto: Y minus X plus 1 times Y plus X minus 1 is positive.
28:32:970Paolo Guiotto: So, either both positive, both negative.
28:37:460Paolo Guiotto: Now, be very careful, because there are several complications here. The sine, the study of the sine, the fact that this is an equality in two variables. So, we have either X minus Y plus 1 is positive, and
28:52:460Paolo Guiotto: X plus… Y plus X minus 1 is positive.
28:56:670Paolo Guiotto: Or…
28:58:00Paolo Guiotto: Y minus X plus 1 is negative, and at the same type, Y plus X minus 1 is negative.
29:08:520Paolo Guiotto: What is this?
29:10:980Paolo Guiotto: Well, Y minus X plus 1 is positive if and believe Y is greater or equal than X minus 1.
29:21:540Paolo Guiotto: This can be easily drilled, no? Because this is,
29:25:370Paolo Guiotto: A line that passes through this point, 45 degrees. Look, so it is above.
29:32:00Paolo Guiotto: Above that line.
29:34:270Paolo Guiotto: And the other condition is Y plus X minus 1 greater or equal than zero, if and all if, similarly, Y is greater or equal.
29:46:750Paolo Guiotto: then minus X plus 1, which is another straight line, and it is exactly this one.
29:54:860Paolo Guiotto: So, if we must be above the two, it means that we are here.
30:04:90Paolo Guiotto: So, what is this green region? It's where these two are verified, the first two.
30:10:760Paolo Guiotto: It's the region where this is correct.
30:13:780Paolo Guiotto: And this is part of the region where F is possible. So, we can say that in that region, F is plus. So here, plus sign here, here, here, here.
30:28:830Paolo Guiotto: We have not yet finished with F positive, because we have also the second line, but that would be similar, because Y minus X plus 1 less or equal than 0 means Y
30:40:190Paolo Guiotto: Below… or equal X minus 1.
30:44:680Paolo Guiotto: Which is this one, this is Y equal X minus 1.
30:49:20Paolo Guiotto: And the other one is Y plus X minus 1 less or equal than zero, even if Y is below minus X plus 1, so below both. So it means that also here we have F plus.
31:07:370Paolo Guiotto: You see?
31:08:840Paolo Guiotto: So again, also here, positive sign. And therefore, it means that in everywhere, in the remainder, F is negative.
31:19:410Paolo Guiotto: Okay?
31:21:430Paolo Guiotto: So this is the sine of F.
31:25:100Paolo Guiotto: Okay, we knew that.
31:27:520Paolo Guiotto: at the… on the y-axis, function F is what, you reminder?
31:35:80Paolo Guiotto: You must be rich dawn, otherwise…
31:38:830Paolo Guiotto: Here, F of 0y is constantly equal to 0 at every point here.
31:44:910Paolo Guiotto: So think about, I take a little neighborhood here, what I see, that the function is zero, at that point, it is positive, around that point it is a…
31:55:720Paolo Guiotto: a minimum.
31:57:160Paolo Guiotto: And the same year.
31:59:480Paolo Guiotto: So this argument works for every point on the y-axis, except one
32:04:900Paolo Guiotto: Disgraceful point, which is this one.
32:07:960Paolo Guiotto: For which, whatever is the neighborhood that you take, you have both positive and negative function… negative values. So that point won't be anything.
32:19:30Paolo Guiotto: So it means all points, except that Y
32:23:350Paolo Guiotto: But that Y, which is Y equals minus 1, are local minimums. That part is not a local minimum and made a local maximum.
32:35:560Paolo Guiotto: Yeah, because, in fact, you see that along the y-axis, sorry, along the… yeah, you see that,
32:43:120Paolo Guiotto: If you can imagine this function, this function would be somewhere here above, somewhere below, so it sounds like a saddle.
32:51:610Paolo Guiotto: However, we can say that the .0Y
32:55:760Paolo Guiotto: for Y different from minus 1, Ease… Local numero.
33:03:970Paolo Guiotto: 0.0 minus 1.
33:06:930Paolo Guiotto: is… Not local. Minimum.
33:14:40Paolo Guiotto: local maximum is… is normal 2. Okay, so this was for D.01. Let's see now what happens for D.10.
33:24:270Paolo Guiotto: So again, that's computation at 0.10.
33:32:610Paolo Guiotto: So now it is Y to be 0, so the entities are here.
33:38:360Paolo Guiotto: So, Y is… you see that Y is here. For example, this means that the two elements out of the diagonal are zero.
33:46:110Paolo Guiotto: And what about for the soluble thing? So, for Y equals 0, that one is 2. Okay, so what we can write immediately is that this is 0, this is 0, this is 2. What about the first entry?
34:03:290Paolo Guiotto: So we have Y is 0, so… and X is 1, so these terms, X minus 1, all equal to 0. So basically, you see that the first is 0, the second is 0, then we have minus 2 times 1, so we get minus 2.
34:19:680Paolo Guiotto: So here we have minus 2.
34:22:70Paolo Guiotto: It's a diagonal, so automatically here, what you see are the eigenvalues.
34:26:810Paolo Guiotto: The matrix is already diagonal, and one is negative, the other is positive.
34:33:240Paolo Guiotto: So that if you have… if you see the eigenber values, they must be all positive, all negative.
34:37:820Paolo Guiotto: If one is negative, the other is positive. This is a saddle. And in fact, two complete determinants, the first one is minus, and the second one
34:49:470Paolo Guiotto: the two determinants are minus minus, you see? The first one is minus 2, the second is minus 4, so we are not in the case, okay? So…
35:01:190Paolo Guiotto: This is a saddle.
35:06:470Paolo Guiotto: And, now it's amazing, but time is over.
35:10:280Paolo Guiotto: This point, what is this?
35:14:330Paolo Guiotto: 1 half zero.
35:16:590Paolo Guiotto: Well, which is pretty much the same, because the Y is 0, 1 after 0. The Y is 0, this means that these two elements are 0.
35:24:570Paolo Guiotto: The X is, 1 half, so the element down here is 2x squared, so 1 half square is 1 fourth times 2 is 1 half, so here we have 1 half.
35:37:490Paolo Guiotto: About the first one, Here we have to do some calculation, unfortunately.
35:45:100Paolo Guiotto: So Y is 0?
35:47:230Paolo Guiotto: So we get 2 times X minus 1, X minus 1 half, minus 1 minus 1 half square, 1 fourth, so times minus 1 fourth.
35:57:50Paolo Guiotto: plus 2X. X is 1 half, so 2, 1 half.
36:02:440Paolo Guiotto: Minus 2 times 1 half minus 1 is minus 1 half.
36:08:650Paolo Guiotto: minus 2 times 2, 1 half times minus 1 half plus 1 fourth. So what is all this mass? It is minus 1 half… this is a plus one.
36:25:430Paolo Guiotto: This is, minus 1 half plus 1.
36:29:950Paolo Guiotto: This is one…
36:31:810Paolo Guiotto: Minus 1 half plus 140 is minus 1 fourth. Minus 1 fourth times minus 2 is plus 1 half. Minus 1 half is plus 1. I hope I've done the calculations correctly. So now you see this? What is?
36:50:890Paolo Guiotto: Last plus, so it is, yes, so it is… 1 of 0 is…
37:00:650Paolo Guiotto: If the deviation is positive, that means that we have A.
37:04:370Paolo Guiotto: Nuh-
37:05:740Paolo Guiotto: Meaning, a local… Minimum.
37:13:460Paolo Guiotto: Can be a global minimum.
37:16:980Paolo Guiotto: We don't know yet, huh? You're saying that we don't know?
37:21:790Paolo Guiotto: We know.
37:23:630Paolo Guiotto: Look.
37:25:460Paolo Guiotto: The function goes down to minus infinity, there is no minimum for this function, so this point will remain only a local minimum. It won't be a local… a global minimum.
37:35:610Paolo Guiotto: Okay.
37:37:410Paolo Guiotto: So, since time is, is over.
37:41:330Paolo Guiotto: let's say we completed the question 2, now you can easily respond to question 3, it remains the maximum. For the minimum, there is no minimum. For the maximum, think about it, and then do problem 4. I will write the solution here, or I will publish, okay?
37:59:970Paolo Guiotto: Online. Okay.
38:02:510Paolo Guiotto: You'll… Finish.
38:06:210Paolo Guiotto: Have a nice weekend, and…