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00:00:00Paolo Guiotto: It's never gonna help me just…
00:14:870Paolo Guiotto: Okay.
00:20:180Paolo Guiotto: Nope.
00:28:680Paolo Guiotto: So that's even the truth.
00:38:260Paolo Guiotto: So… Okay. That's… I start… With the first step.
00:49:350Paolo Guiotto: but there are two sets. The first one starts with the exercise 36.
00:59:940Paolo Guiotto: So here we have a domain, D,
01:03:320Paolo Guiotto: Made of points XYZ in Ap tree.
01:07:840Paolo Guiotto: Such that…
01:09:580Paolo Guiotto: X squared plus Y squared equals Z squared, and Y squared plus Z minus 2 squared equal 1.
01:20:750Paolo Guiotto: Let's see questions one by one. Number one, D is non-empty.
01:27:430Paolo Guiotto: And, it is zero.
01:34:670Paolo Guiotto: Set of a submersion.
01:44:400Paolo Guiotto: So, about the first one, we just need to exhibit one point.
01:49:620Paolo Guiotto: So since there are two conditions, we maybe could impose just one. Let's look for a point with Z equal to…
02:00:640Paolo Guiotto: So a point of type XY2.
02:04:910Paolo Guiotto: This belongs to D if and only if the two conditions must be verified together. So we get X squared plus Y squared equals Z squared, which is… this is not Z, it is 2R.
02:19:670Paolo Guiotto: the number 2. So this is 4.
02:22:910Paolo Guiotto: And Y squared is equal to 1.
02:27:760Paolo Guiotto: So from the second, we have that Y is equal plus minus 1. From the first, when we plug a Y square equal 1, here we get X square equal 3, so X equal plus minus the root of 3.
02:43:770Paolo Guiotto: So we… we found, basically 4 points, plus, minus root of 3, Plus, minus 1, 2…
02:53:430Paolo Guiotto: If you want all sign.
02:58:760Paolo Guiotto: are indie.
03:02:440Paolo Guiotto: Now, the question is, of course, the important part of the question is this one.
03:10:110Paolo Guiotto: T is naturally defined as, by a pair of equations, G1 equals 0, G2 equals 0, where…
03:22:10Paolo Guiotto: the two functions, G1 and G2.
03:25:580Paolo Guiotto: could be… So for G1, X squared plus Y squared minus Z squared, and for G2,
03:35:620Paolo Guiotto: It is Y squared… Y square plus…
03:40:830Paolo Guiotto: Z minus 2 square minus 1. 0, 0. Equation of this is domain.
03:48:130Paolo Guiotto: So… instead of checking whether this is a submersion, we say that G1, G2, G1, G2.
04:01:80Paolo Guiotto: is… not.
04:05:180Paolo Guiotto: a submersion
04:11:540Paolo Guiotto: at point XYZ.
04:15:650Paolo Guiotto: If and of if, do you remind the condition?
04:19:700Paolo Guiotto: Greetings being zero.
04:24:410Paolo Guiotto: What?
04:27:320Paolo Guiotto: Okay, answer one, the exam is when? When the gradient? The exam is when?
04:35:500Paolo Guiotto: Fantastic.
04:36:950Paolo Guiotto: 2016, so it's about 10 days from now.
04:41:130Paolo Guiotto: So, 2 answers out of 2 are wrong. Let's hear some other answer.
04:48:40Paolo Guiotto: Comfort.
04:48:940Paolo Guiotto: radium of G1 and G2, it should be… And also, we read the top F, G1 and G2, and then we, take, no.
04:59:610Paolo Guiotto: She's… she's close, but not yet complete rank.
05:04:600Paolo Guiotto: So the rank of the matrix made of the two gradients, G1, G2,
05:10:570Paolo Guiotto: This rank must be less than 2. That's the condition.
05:15:730Paolo Guiotto: There is no determinant here. This is a 2x3 matrix. You see two lines, there are 3 columns. If we write the matrix.
05:24:260Paolo Guiotto: We got that.
05:26:110Paolo Guiotto: rank of… gradient G1 is 2X, 2Y minus 2Z, gradient of G2 is 0, 2Y, and 2Z minus 2. So, you see, there is no determinant here.
05:43:700Paolo Guiotto: So, the condition determinant… yeah, there will be determinants later, but it's not determinant of a matrix for this case. So, rank less than 2, and this is possible if and only if…
05:58:740Paolo Guiotto: All the 2x2.
06:00:940Paolo Guiotto: Submatrices have determinant equal zero, so we get determinant of this one, 2X, 2Y, 0, 2Y, this must be equal to 0.
06:14:690Paolo Guiotto: Together with the determinant of 2x minus 2Z 0, 2Z minus 2 equal 0, together with this determinant. Determinant of
06:28:780Paolo Guiotto: 2Y minus 2Z2Y2 times Z minus 2. This equals 0.
06:38:680Paolo Guiotto: We can do a little simplification here or later. Determinant is linear.
06:45:90Paolo Guiotto: function of lines and columns. So if all the entries of a line are multiplied by the same factor, like 2 here, we can carry outside the variety 2 here.
06:55:570Paolo Guiotto: And this is simplified because we can divide. And similarly, this two goes away. Also, these two here, here, all the two disappear. Oh, if you do not realize this, you will see later.
07:06:910Paolo Guiotto: Now, if we compute this determinant, we get XY equals 0. For the second one, we get X times Z minus 2 equals 0. For the third one, we get Y times Z minus 2,
07:25:440Paolo Guiotto: Minus, minus, plus YZ equals 0.
07:29:980Paolo Guiotto: So, let's clean up. We have XY equals 0,
07:36:80Paolo Guiotto: X times Z minus 2 equals 0. So we have, Y… that multiplies…
07:46:410Paolo Guiotto: 2Z minus 2, so this makes a Z minus 1, once I factorize the 2, equals 0.
07:53:700Paolo Guiotto: So these are basically the same type of equations, it is indifferent which one we choose, we take the first one. So this says that either X is equal to 0, or Y is equal to 0.
08:08:490Paolo Guiotto: For X equals 0, the second one becomes 0 equals 0, is redundant.
08:13:820Paolo Guiotto: And the third one is Y times Z minus 1 equals 0. For Y equals 0, it is the third one to be redundant, and the second one is X times Z minus 2 equals 0.
08:27:500Paolo Guiotto: So now we split this into subsystems. X equals 0, Y equals 0, or…
08:35:340Paolo Guiotto: X equals 0, and Z minus 1 equals 0, Z equals 1. Or, here we have Y equals 0 and X equals 0, or again, Y equals 0, and Z minus 2 equals 0, Z equals 2.
08:52:50Paolo Guiotto: So these are points of type 0, 0, 0, and Z.
08:57:70Paolo Guiotto: Zed is undetermined, so for the moment is real.
09:01:170Paolo Guiotto: These are points of time, 0Y1, with Y real.
09:07:900Paolo Guiotto: These ones are the same points of the first type, and the last ones are points where X is undetermined, 0 and 2, X real.
09:18:930Paolo Guiotto: Now, what are these points?
09:21:950Paolo Guiotto: These are points where the rank of that matrix is less than 2.
09:27:200Paolo Guiotto: they are not yet bad points, because we have to verify whether they are on D or not, because if they are not on D, we can say that on D, the rank cannot be less than 2, okay? So let's analyze these points. 000Z belongs to D,
09:47:210Paolo Guiotto: Now we have to plug this into the equations, which are X squared plus Y squared equals Z squared, so 0 squared plus 0 squared equals Z squared, that's the first equation. The second is Y squared, so 0 squared plus Z minus 2 squared equals 1.
10:06:680Paolo Guiotto: Well, you see that this is impossible, because the first one yields Z equals 0. When you plug into the second one, you get minus 2 squared, so 4 equals 1, that's impossible.
10:20:70Paolo Guiotto: So, none of these points are in there, but they are bad points, okay?
10:26:150Paolo Guiotto: But not in D, so who cares for this problem?
10:29:250Paolo Guiotto: So now 0Y1 is indeed. Same check. We plug into the first equation. 0 squared plus Y squared equal 1 squared.
10:39:420Paolo Guiotto: And the second one is Y squared plus Z is 1, so 1 minus 2 squared equal 1, okay?
10:49:740Paolo Guiotto: If I'm not wrong, yeah. So, the first one says Y squared equal 1, the second one, since this is 1, Y squared equals 0, which is impossible.
11:03:960Paolo Guiotto: So, also, these points are not in D.
11:06:490Paolo Guiotto: Third point. X02.
11:10:40Paolo Guiotto: X02. This is in the if and only if.
11:15:850Paolo Guiotto: X squared plus Y squared equals Z squared… sorry, that's a 2,
11:25:140Paolo Guiotto: 2, so this means, 2 squared 4.
11:29:610Paolo Guiotto: And the second one is 0 squared plus Z minus 2, so 2 minus 2.
11:35:940Paolo Guiotto: Square equal 1.
11:37:970Paolo Guiotto: Well, this says that X squared equal 4,
11:42:780Paolo Guiotto: But the second one, this is 0, 0 equals 1, and that's impossible.
11:49:80Paolo Guiotto: So at the end, we have seen that none of these points belongs to D, so the conclusion is that the rank of the matrix, gradient G1, gradient G2,
12:01:10Paolo Guiotto: is not less than 2, it is equal to 2, it cannot be 3, because there are, at maximum, 2 lines. Rank is true on D. That is…
12:14:600Paolo Guiotto: G2… is… at Submersion.
12:21:570Paolo Guiotto: on D.
12:24:570Paolo Guiotto: And it's just question one.
12:26:910Paolo Guiotto: Question 2. So be careful when you apply these things. You have to, to use the appropriate,
12:34:680Paolo Guiotto: tools, huh?
12:37:150Paolo Guiotto: And do not write nonsensus. Is the compactor…
12:45:160Paolo Guiotto: So, compact… D… is compactor.
12:52:380Paolo Guiotto: If and only if these two things closed, and bound.
13:01:780Paolo Guiotto: There's no relation between these two properties, so you have to discuss separately. Well, the first one is standard, the D is…
13:10:510Paolo Guiotto: Close the… Because…
13:17:900Paolo Guiotto: D is described by two equations, G1 equals 0, G2 equals 0, and clearly the functions G1, G2, are both continuous.
13:29:250Paolo Guiotto: The polynomial, so this is a standard fact.
13:33:560Paolo Guiotto: Now, the point is, is this set bounded? So there is no general, idea here…
13:43:890Paolo Guiotto: is debounded.
13:45:940Paolo Guiotto: We have to understand if it is bounded, then prove that it is bounded, or if we think that it is unbounded, to disprove this means to find points with big, arbitrarily big knob. Now, if we look at the two equations, X, Y, Z belongs to this.
14:04:400Paolo Guiotto: Even only if…
14:06:520Paolo Guiotto: The first one is X squared plus Y squared equal Z squared. The second one is Y squared plus Z minus 2 squared equal 1.
14:17:210Paolo Guiotto: Is there any one of them that says something about the boundlessness?
14:23:310Paolo Guiotto: Yeah.
14:26:310Paolo Guiotto: is this one. So let's start with this one. This says Y squared less or equal 1, and also Z minus 2 squared less or equal 1. Sum of positive things equal 1, but equally constant, they must be both less than a constant. And from this one, we have models of Y less or equal than 1,
14:45:10Paolo Guiotto: And from this one, modulus Z-2 is less or equal than 1, so if we want about Z, this means that Z minus 2 is between minus 1 and 1, so the Z is between, at most 3, at least 1.
15:03:130Paolo Guiotto: So we have bound for Z, bound for Y. So now we hope that we can get also the bound for X, and we have to look at this condition.
15:12:830Paolo Guiotto: Now, if we look at this condition, we can say that X squared is Z squared minus Y squared, so in particular, it is less or equal than
15:31:170Paolo Guiotto: You have to look at the simple. The simple is… I know that Z is bounded. I'm subtracting positive to Z, so it is less than Z squared. Z square is less than 9, because Z is less than 3.
15:45:430Paolo Guiotto: So, I get, finally, that X squared is less than 9, so models of X is less than 3. So, finally, we have
15:54:610Paolo Guiotto: X, Y, and Z bounded, so… D is bounded.
16:01:960Paolo Guiotto: And this ends question 2. Question 3…
16:07:290Paolo Guiotto: It says, find points… points of… D… at minimum, Maximum.
16:18:720Paolo Guiotto: distance… to the origin.
16:23:680Paolo Guiotto: Now, the distance to the origin distance to…
16:31:530Paolo Guiotto: The origin, is, the norm of XYZ.
16:39:390Paolo Guiotto: So, which is the root of X squared plus Y squared plus Z squared.
16:45:470Paolo Guiotto: We are 7 times this argument, so we have to minimize, maximize that quantity.
16:52:00Paolo Guiotto: But since this quantity is minimum-maximum, when the argument of the root is minimum-maximum, So, we… have…
17:04:599Paolo Guiotto: To determine the minimum Maximum.
17:10:880Paolo Guiotto: of this X squared plus Y squared plus Z squared.
17:15:990Paolo Guiotto: on the domain ID.
17:18:650Paolo Guiotto: Now, of course, we are going to use the Lagrange theorem to determine these points, but Lagrange gives only a necessary condition, not sufficient, so we have to establish existence before. So we can say that the function FXYZ
17:35:490Paolo Guiotto: BX squared plus Y squared plus Z squared. It is a continuous function. D is compactor.
17:47:880Paolo Guiotto: So there is an important theorem, Bayastras theorem.
17:52:700Paolo Guiotto: that says that there exist both minimum and maximum of F only. So now we know they exist, and we determine
18:04:110Paolo Guiotto: If possible, by using the Lagrange theorem to determine these… points… we apply… the Lagrange theorem.
18:22:40Paolo Guiotto: Okay, so, we already know that we are ready Checked,
18:35:360Paolo Guiotto: That, huh?
18:38:140Paolo Guiotto: the map for G1, G2, is a submersion.
18:46:940Paolo Guiotto: on the domain D, which is the zero set of these two maps, huh?
18:52:360Paolo Guiotto: Which is the, let's say, technical condition we need to apply this year.
18:58:380Paolo Guiotto: Moreover, F is clearly… differentiable.
19:06:90Paolo Guiotto: Which is the other technical points. So… If XYZ Indeed.
19:17:20Paolo Guiotto: is any.
19:19:370Paolo Guiotto: Minimal.
19:21:220Paolo Guiotto: or maximum.
19:22:980Paolo Guiotto: point… for F… We must have that the gradient of F, at that point, XYZ,
19:35:760Paolo Guiotto: is a linear combination of the gradients of the two constraints. So there will be a lambda 1, gradient G1 plus lambda 2 gradient G2.
19:48:210Paolo Guiotto: Or, equivalently, the matrix made of the gradient of F, the gradient of G1, and the gradient of G2,
19:58:390Paolo Guiotto: These metrics have… Rank less than 3, not maximum, because rank 3
20:07:410Paolo Guiotto: Means that the three vectors are linearly independent, so the gradient cannot be a linear combination of the other two.
20:15:470Paolo Guiotto: Now, this happens, this is now a 3x3 matrix.
20:20:130Paolo Guiotto: So, if we want a rank of…
20:23:440Paolo Guiotto: We, the gradient F is 2x2Y2Z. The gradient G is 2x2Y minus 2Z.
20:34:280Paolo Guiotto: And the green energy 2 is 0, 2Y, and 2Z minus 2.
20:42:380Paolo Guiotto: Now, the rank of this… this metric is less than 3. This is a 3x3 metric, sir.
20:48:600Paolo Guiotto: So, in principle, the rank is less than something if all the something-by-something submatrices have determinant equals 0. Here, there is only one 3 by 3 submatrix, which is the matrix itself. So here, yes, it is a unique determinant, okay?
21:05:860Paolo Guiotto: Determinant of this
21:09:90Paolo Guiotto: Equals zero. And now we have to write this. So it's convenient to use, of course, the zero.
21:16:100Paolo Guiotto: So we have 2X times this determinant.
21:21:130Paolo Guiotto: So determinant of 2Y minus 2Z2Y, 2 times Z minus 2.
21:30:70Paolo Guiotto: Then, minus 2X, these are the determinants, this plus this.
21:37:770Paolo Guiotto: And this yields, so determinant to Y2Z… to why… 2Z minus 2.
21:49:240Paolo Guiotto: And then there is plus 0 times another determinant. All this equals 0.
21:55:520Paolo Guiotto: So this is… As you can see, there is a factor 2X. We should factorize.
22:02:980Paolo Guiotto: 2X times… then we have,
22:07:970Paolo Guiotto: Oh, mad. Shit. So, we have… And now what happens?
22:18:340Paolo Guiotto: No, no, we should… I think that we should leave the building. So, we have to interrupt class because of fire alarm, so it's a simulation, so I will pause the recording.
22:36:560Paolo Guiotto: Okay, we can restart.
22:42:610Paolo Guiotto: But we were computing this, determinant, 2X, so we have, What is 4?
22:52:920Paolo Guiotto: 4Y times Z minus 2 minus plus 4YZ.
23:01:320Paolo Guiotto: Well, this is the first determinant minus. The second one is 4Y times Z minus 2.
23:12:820Paolo Guiotto: minus 4YZ.
23:17:580Paolo Guiotto: All this equal zero.
23:21:180Paolo Guiotto: As you can see, for example, this… term here…
23:25:940Paolo Guiotto: Simplifies with this one, because of the minus.
23:29:440Paolo Guiotto: But the other two.
23:31:340Paolo Guiotto: They do not simplify, they add, so we have, so 2X. In the parentheses, we have 8YZ, or this equals 0, so at the end, XYZ equals 0.
23:46:670Paolo Guiotto: This yields three possibilities. So, X equals 0, or Y equals 0, or Z equals 0.
23:55:460Paolo Guiotto: In terms of points, X equals 0 means that the first coordinate is 0, but the other two are free.
24:03:460Paolo Guiotto: This one means that we have points like X0Z, and this one is XYZ.
24:10:380Paolo Guiotto: Now, what are these points? These points are points where the rank is less than 3, so where the gradient is the combination of the other two gradients. The gradient F is the combination of the other two gradients.
24:22:700Paolo Guiotto: But they must be Andy, otherwise they are not of interest. So, now, the first point, 0YZ,
24:35:640Paolo Guiotto: belongs to D, if and only if, and we have to plug into the system, so…
24:42:370Paolo Guiotto: The first equation is X squared plus Y squared equals Z squared, so we get Y squared equals Z squared. The second equation is Y squared plus Z minus 2 squared equals 1.
24:55:610Paolo Guiotto: So this is non-trivial, at least at first look. So we could, for example, plug this into this one. So, keep the first equation, Y squared equals Z square, and the second equation becomes Z square plus… well, we could do the square…
25:15:460Paolo Guiotto: So Z squared minus 4Z plus 4 equal 1.
25:22:780Paolo Guiotto: This means that the second one is the second degree equation in Z.
25:29:00Paolo Guiotto: 2Z squared minus 4Z.
25:33:170Paolo Guiotto: plus 3 equals 0, then Y squared equals Z squared.
25:39:730Paolo Guiotto: So, from the first one, we get Z equal minus plus 4 plus minus the root of delta is minus 4 squared 16,
25:51:320Paolo Guiotto: Minus 4 times 2, 8 times 324, so it is negative.
25:58:690Paolo Guiotto: What is going on?
26:03:210Paolo Guiotto: It is a little problem.
26:06:940Paolo Guiotto: Maybe I gave two…
26:17:790Paolo Guiotto: Okay, let's see if we can do…
26:24:120Paolo Guiotto: Okay, so you see that this is negative, so it means that our coordinates are a real value, so no solution.
26:35:410Paolo Guiotto: So, the second point is X0Z. X…
26:40:140Paolo Guiotto: 0Z belongs to D if and only if we get X squared plus Y squared is 0 equals Z squared, and the second equation is Y squared 0 plus Z minus 2 squared equals 1.
26:56:800Paolo Guiotto: This one yields Z minus 2 equal plus minus 1.
27:02:740Paolo Guiotto: So, Z equal 3 plus 1… no, 2 plus 1, 3, 2 minus 1, 1, there are two values.
27:13:20Paolo Guiotto: So it means that we have a K is Z equal 3, and X square equals Z squared equals 9, so it means that X is equal plus minus 3.
27:25:880Paolo Guiotto: Or, the second case is Z equals 1,
27:29:940Paolo Guiotto: In this case, X squared equals Z squared is equal to 1, so X equals plus minus 1.
27:37:260Paolo Guiotto: So at the end, we have this point, so plus, minus 3.
27:41:00Paolo Guiotto: Zero. Three.
27:43:420Paolo Guiotto: And here, plus, minus 1, 0, 1.
27:49:470Paolo Guiotto: two points each. And the last are points XYZ0, right?
27:59:910Paolo Guiotto: Now, XYZ0 belongs to D if and only if the first equation is X squared plus Y squared equals Z squared, which is equal to 0.
28:09:150Paolo Guiotto: And the second one is Y squared plus 0 minus 2 minus 2 minus 2 square equal 1.
28:17:580Paolo Guiotto: Now, from the first, we get that that's possible if and only if X and Y are both equal to zero. But when you plug into the second, this becomes impossible.
28:32:60Paolo Guiotto: So, no solutions.
28:34:400Paolo Guiotto: Okay, so this means that the unique points where the gradient F is linear combination of the other two gradients are these four points.
28:43:230Paolo Guiotto: So the mean-max are among them. And if we look at the value of F, Since…
28:50:770Paolo Guiotto: F at plus minus 303. The function is the sum of the squares. This is equal to 18, while F at plus minus 101
29:03:80Paolo Guiotto: This is equal to 2.
29:05:140Paolo Guiotto: So we did use that. Plus, minus 3, 0.
29:11:290Paolo Guiotto: Points.
29:14:160Paolo Guiotto: of the… At… these are at max.
29:19:610Paolo Guiotto: Distance.
29:22:180Paolo Guiotto: to the origin.
29:25:230Paolo Guiotto: While the other, plus minus 101, points of D, at minimum, instance, to the arch.
29:38:770Paolo Guiotto: And this… Finishes the problem.
29:42:210Paolo Guiotto: Do you have any questions?
29:47:310Paolo Guiotto: No question.
29:50:00Paolo Guiotto: Okay, let's see the exercise number 37.
29:55:310Paolo Guiotto: This is an exercise, on, integral, multiple integral. It says by using The change of variable.
30:10:600Paolo Guiotto: U equal knowledge, sorry.
30:14:280Paolo Guiotto: X equal U square.
30:18:220Paolo Guiotto: X equals U squared, and Y equals V squared, compute.
30:27:350Paolo Guiotto: this integral. Integral on 0 plus infinity.
30:32:410Paolo Guiotto: square, so it's a double integral, this is the first quarter. E to minus X plus Y divided, the root of
30:43:290Paolo Guiotto: X squared Y plus… XY square.
30:49:900Paolo Guiotto: The XDY.
30:53:550Paolo Guiotto: Okay.
30:56:220Paolo Guiotto: So… so we are in the context of,
31:01:750Paolo Guiotto: integration, so here the function f we want to integrate is E minus X plus Y divided the root of X squared plus
31:14:430Paolo Guiotto: Y plus XY squared.
31:18:150Paolo Guiotto: Now, this function is well-defined. I don't want to enter into exactly the domain, but let's see that this is well-defined and continuous with respect to what is our domain when both coordinates are positive.
31:36:620Paolo Guiotto: And the domain D, integration domain, 0 plus infinity, is in any case not compact, because it is unbounded.
31:49:380Paolo Guiotto: So I cannot conclude anything immediately about the existence of the integral. However, the function is also positive, so, integrability
32:01:360Paolo Guiotto: off.
32:02:600Paolo Guiotto: F coincides with integrity of modulus of F, so we say everything for modulus of F, it's the same as saying everything for F. Now, we have that the integral 0 plus infinity of the function
32:18:00Paolo Guiotto: E minus X plus Y divided the root of X squared Y plus X.
32:26:90Paolo Guiotto: Y squared. It says, use this change of variable. Now, the change of variable is not one of the standard change of variables we know. It says X equals u squared, V… sorry, Y equals V squared.
32:41:790Paolo Guiotto: So in this way, I'm saying XY, which is the initial variable, is a function of a new variable.
32:50:290Paolo Guiotto: So this is… I'm giving XY as a function of new variable UV.
32:59:870Paolo Guiotto: So this is, let's say, the standard form. You may remind that when we have integral on the… of a function f in variables, say, XY, the XDY,
33:10:650Paolo Guiotto: This becomes the integral in the new variables UV, so let's say DUDV will be at the end, of the function F, where I have to put in place of XY exactly this, phi
33:23:730Paolo Guiotto: UV.
33:28:70Paolo Guiotto: Well, actually, with the notation we used in the theory, this was the phi minus 1, because it's the function that normally we use phi when we say UV is…
33:40:610Paolo Guiotto: the new variables are functions of the old. This is given in the contrary, the old variable has functioned in the new, so this is the… literally the phi minus 1, which is exactly what we have here, determinant of phi minus 1 prime.
33:56:720Paolo Guiotto: And the domain D is what should be the image through the direct map, the fee, we don't have, of the domain D. So this is the formula we have to use.
34:09:760Paolo Guiotto: Well, at the end, it is easier than what it seems, because, so let's say that this becomes… will become an integral in DUDV.
34:19:969Paolo Guiotto: Of what? Let's write the function. The function is E2 minus X plus Y. Now, since X is u squared, and Y is V squared, this will be u squared plus V squared.
34:32:449Paolo Guiotto: divided the root of… we have X square Y plus XY squared. How can we write this?
34:42:409Paolo Guiotto: Well, we could say that if x is u squared, so X squared would be U to power 4,
34:52:620Paolo Guiotto: Y is V square.
34:55:909Paolo Guiotto: Plus, this one is X, which is your square.
35:01:130Paolo Guiotto: Y squared is V to power 4.
35:05:690Paolo Guiotto: Now, there is the models of this determinant that we compute. So, determinant of the Jacobian matrix of phi minus 1. Since we already have the phi minus 1, it's written here, no? XY is U squared, B squared, this is the phi minus 1.
35:24:920Paolo Guiotto: So we have a… this is a map R2 to R2, where this is the first component, this is the second component. And here we have to compute the Jacobian matrix. It will be a 2x2 matrix. The derivatives are with respect to the variables u and v.
35:40:400Paolo Guiotto: So the first component is u squared, derivative with respect to U is 2U, derivative with respect to V is 0.
35:48:270Paolo Guiotto: The second component here is V squared, derivative with respect to V is 0, derivative with respect to V is 2V, so that's the Jacobier matrix. We have to take the modulus of this determinant, which is exactly 4 UV, absolute value.
36:02:240Paolo Guiotto: So, the quantity we have to put here is for Nuvi.
36:06:00Paolo Guiotto: Now, what about the D domain? Well, the domain is easy, fortunately, because XY belongs to D,
36:16:760Paolo Guiotto: If and if the domain D is 0 plus infinity square, so this means X is positive.
36:26:00Paolo Guiotto: And Y is positive.
36:28:660Paolo Guiotto: But these are, respectively, U squared and V square.
36:34:20Paolo Guiotto: Okay So…
36:37:290Paolo Guiotto: Now, here we have to be careful, because this map must be a big section. So, X equal U square means U equal what?
36:48:770Paolo Guiotto: equal, in principle, two values. If we want a unique value, we take the positive root plus root of X, and V equal plus root of y. So this becomes the integral for u and v positive.
37:04:30Paolo Guiotto: So, also here, U.
37:06:240Paolo Guiotto: and V positive.
37:08:520Paolo Guiotto: Of this quantity.
37:10:370Paolo Guiotto: Now, let's simplify.
37:13:240Paolo Guiotto: Since U and V are positive, we can throw away the modules, and we remain with the integral on u's positive, V positive.
37:22:780Paolo Guiotto: E minus U squared plus V square.
37:27:500Paolo Guiotto: Divided by the root.
37:31:190Paolo Guiotto: Well, you see that we could factorize u squared if down here we factorize u squared, V squared, what remains is U squared plus V square, right?
37:45:370Paolo Guiotto: So when we take the root of all this, root U square V squared would be the modulus of UV.
37:53:460Paolo Guiotto: Then we have the root of U squared plus V square.
37:59:40Paolo Guiotto: And then…
38:03:880Paolo Guiotto: There is.
38:05:470Paolo Guiotto: Can we integrate this thing? Well, let's see.
38:08:430Paolo Guiotto: for UVs.
38:10:90Paolo Guiotto: EU, DV.
38:13:850Paolo Guiotto: Okay, so now this cancels this, so we get 4 integrals on u positive, V positive, of this function. E minus u squared plus V squared divided the root of u squared plus V squared, the U divid.
38:34:530Paolo Guiotto: Do we see exactly how to compute this?
38:40:380Paolo Guiotto: comments? Questions? Yeah.
38:43:820Paolo Guiotto: Can you explain again why you can't be plus and minus… Yeah, because this map must be a big section, so you have… since if I put plus mism, there are two, no?
38:57:580Paolo Guiotto: I have to decide which one to take, so I will take the positive in such a way that it becomes a one-to-one correspondence. To each XY, there is a UV, and to each UV, there is an unique XY. Okay, now it becomes an objection.
39:11:320Paolo Guiotto: Otherwise, it means not a rejection, so minus 1 cannot be rejected.
39:16:660Paolo Guiotto: Yes, about computing this, we could use polar coordinates, so U equals rock cosine theta.
39:25:70Paolo Guiotto: V equal rhos sine theta.
39:27:680Paolo Guiotto: So this is a standard. Notice that UVR in the Cartesian plane in this range.
39:36:610Paolo Guiotto: So it means that the integral becomes 4 raw, any positive.
39:42:410Paolo Guiotto: For theta, angles are between 0 and pi over 2.
39:49:950Paolo Guiotto: Then, the function is E minus raw squared divided the root of u squared plus V squared is raw. There is the change of variable, which is raw, d raw, d theta.
40:00:660Paolo Guiotto: And that's it.
40:02:460Paolo Guiotto: So we see that this counts us, and we get 4 integral.
40:07:830Paolo Guiotto: We can now apply the reduction formula.
40:11:640Paolo Guiotto: Say, for example, we integrate lasting raw, we carry, and then in theta, 0 to pi over 2.
40:18:410Paolo Guiotto: Then we have E minus rho squared d theta, then 0. This goes out because it is independent of theta, and when we integrate from 0 pi half theta, we get pi half, so 4 times pi half. Then we have the integral from 0 to plus infinity of E minus rho squared zero.
40:39:360Paolo Guiotto: Which is not easy to be computed, but that's a Gaussian integral.
40:43:960Paolo Guiotto: So we remind that this particular formula, integral from minus infinity plus infinity, this is the standard Gaussian integral. This divided by 2D raw, this is root of 2 pi.
41:00:550Paolo Guiotto: This is the so-called standard Gaussian integral. If we want to have that, we need to put the root of 2, we do a little change of variable to have this. So, instead of having raw square, we want rho square over 2, so it means that I call R,
41:19:870Paolo Guiotto: raw divided by root of 2. I take this change of variable. So this becomes pi, the integral. Now, when raw goes from 0 to plus infinity, raw over root of 2 goes from 0 to plus infinity.
41:35:230Paolo Guiotto: E minus raw square… no, sorry, it's the contrary. It's,
41:41:700Paolo Guiotto: I want that raw be equal… I want… what I want is raw square becomes r squared divided 2. So the changing variable is R equal root of 2 raw.
41:54:340Paolo Guiotto: So, however, nothing changed for the range, because when raw goes from 0 to plus infinity, root of 2 times rho goes from 0 to plus infinity. Raw square becomes R squared divided 2. The D raw becomes dr divided root of 2.
42:14:930Paolo Guiotto: Yes, and so now you see that we have one half of the Gaussian integral.
42:21:50Paolo Guiotto: So, this is… so there is a 2 also here. I forgot this comes from this. So, 2 pi over root of 2
42:33:700Paolo Guiotto: The integral from 0 to plus infinity, E minus R squared divided 2DR,
42:40:430Paolo Guiotto: Now, if I want to put from minus infinity to plus infinity, I have to add a factor of 1 half, yeah.
42:46:990Paolo Guiotto: And this is now the Gaussian integral. All this is equal to root of 2 pi, so the total is pi over root of 2 times root of 2 pi.
42:58:920Paolo Guiotto: And this means, pi to 1 plus 1 half 3F.
43:03:520Paolo Guiotto: This is the value of the integral.
43:09:680Paolo Guiotto: Okay, I think there is no other question about this. Do you have any question?
43:15:130Paolo Guiotto: Where does that formula come from? What? We did in class.
43:22:850Paolo Guiotto: This is one of the most important integrals you use in probability. This will be always around, so it's better if you,
43:34:540Paolo Guiotto: remember.
43:36:260Paolo Guiotto: Any other questions?
43:40:610Paolo Guiotto: We can, switch to the exercise 38.
43:45:610Paolo Guiotto: Well, as you have seen from the simulations, the message is we have five exercises, of which one
43:54:410Paolo Guiotto: The last one usually has a more theoretical flavor, let's say.
44:01:230Paolo Guiotto: And since there are,
44:03:490Paolo Guiotto: more topics, I cannot include all the topics in an exam. So, there will be some topic which is not present in the exam.
44:13:140Paolo Guiotto: And this will rotate, okay? So you cannot expect that the exam will be an exercise on constrained optimization, an integral, than this one, which is an exercise on oligomorphic function, a differential equation.
44:27:270Paolo Guiotto: And it's a radical question on continuity or open setup. It's not that. In fact, that's why I gave you two examples to tell you, look, it will be viable. So you have… this is what you have, more or less, to expect. You have also the previous exams to have an idea.
44:46:650Paolo Guiotto: Okay, so this says, the argument here is, C differentiability. You have a function, f of x plus IY equal UXY.
45:00:240Paolo Guiotto: plus I… VXY.
45:04:500Paolo Guiotto: Well, question one, I don't want to respond to PR. It says, which properties of U and V characterize C differentiability of F at points Z equal, what is…
45:18:310Paolo Guiotto: of point X, let me see if I can…
45:26:790Paolo Guiotto: It's too much.
45:28:570Paolo Guiotto: Three wires are too much.
45:30:930Paolo Guiotto: Let's see.
45:35:420Paolo Guiotto: Because I have a problem with the battery of this one.
45:40:900Paolo Guiotto: So let's see. Unless you have this FDY for me, but… In that case, I…
45:55:160Paolo Guiotto: Awesome.
45:56:310Paolo Guiotto: No.
46:03:400Paolo Guiotto: It's too much, it's too exciting, but just isn't it?
46:07:350Paolo Guiotto: This should be… is supposed to be the innovative boom, huh? It's so innovative that there is… you see a lot… it seems like we are at the Stone Age with this case, look, that's the innovative boom.
46:21:150Paolo Guiotto: It's math.
46:25:470Paolo Guiotto: And these, these monitors, we have these wheels, it's simply… Okay.
46:34:860Paolo Guiotto: Same like this, but I think we're gonna have to…
46:37:740Paolo Guiotto: Okay, so the first question says, which properties of E characterize C differentiability? So, here you are supposed to respond
46:46:560Paolo Guiotto: the function f is C differentiable if and only if the U and V must be differentiable, in real sense, and they must verify the Cauchier-Riemann equations, and you write the Cauchier-Riemann equations, okay?
47:02:410Paolo Guiotto: If you forget, for example, of differentiability, it's a big problem. So, let's say that we start from question 2.
47:11:490Paolo Guiotto: It says, you have UXY, equal AX squared plus BXY plus CY squared.
47:23:290Paolo Guiotto: and VXY.
47:26:240Paolo Guiotto: equal expense what?
47:29:590Paolo Guiotto: Where A, B, C, real constant. A, B, C, are real.
47:34:970Paolo Guiotto: constants.
47:37:160Paolo Guiotto: So, the question is determine… ABC… Such a way that, such… that F is… C differentiable.
47:52:320Paolo Guiotto: on scene, so on every point of the conscious plane. So here we apply what you…
47:58:820Paolo Guiotto: should have, replied in question one. So, we say that, to be… C differentiable.
48:13:360Paolo Guiotto: UNV must… B… are differentiable.
48:21:610Paolo Guiotto: And this is clear, because they are polynomials.
48:25:200Paolo Guiotto: True.
48:27:540Paolo Guiotto: U and V, polynomials.
48:36:240Paolo Guiotto: And… they… must… Also, verify… the Kosheri Man… equations.
48:53:140Paolo Guiotto: Or conditions, as you prefer. So they are DXU equals DYV, And DYU is minus DXV.
49:04:130Paolo Guiotto: So let's see what happens if we write these two equations. DXU is…
49:09:340Paolo Guiotto: the U is there, it's 2AX, plus BY,
49:16:90Paolo Guiotto: This must be equal to DYV, which is X
49:22:90Paolo Guiotto: Attention, this will be for every point, since we want these conditions false at every point.
49:28:160Paolo Guiotto: At every Z means at every point XY.
49:33:540Paolo Guiotto: of R2.
49:35:540Paolo Guiotto: And for the second condition, the YU is BX.
49:42:380Paolo Guiotto: Plus… 2C?
49:45:630Paolo Guiotto: Why?
49:47:190Paolo Guiotto: This should be minus… DXV… DXV is Y.
49:53:540Paolo Guiotto: Again, for every XY.
49:55:540Paolo Guiotto: Now, these are identities between polynomials.
49:59:700Paolo Guiotto: Precisely, we have that the polynomial 2A minus 1X plus BY, this one must be identically 0, and the second line reads BX plus 2C plus 1Y must be identically equal to 0.
50:20:140Paolo Guiotto: Now, this is possible if and if all the coefficients of these two polynomials are zero. So we get…
50:28:970Paolo Guiotto: this. 2A minus 1 must be equal 0. B must be equal 0, which is the same that you have here, and the 2C plus 1 must be equal 0.
50:40:810Paolo Guiotto: So this means that… B is equal to 0, A is equal to 1 half
50:47:280Paolo Guiotto: And C is equal to minus 1 half.
50:50:160Paolo Guiotto: So, at the end, U, X, Y, Must have this, shape.
50:58:110Paolo Guiotto: So, A is 1 half, 1 half X squared.
51:02:700Paolo Guiotto: plus B is 0, so there is no XY, plus C, which is minus 1 half Y squared.
51:10:380Paolo Guiotto: and V, X, Y,
51:12:450Paolo Guiotto: is equal to X times Y. Only for these couples, this pair of U and V, you get, combining U plus AV, you get a C-differentiable function.
51:24:180Paolo Guiotto: Number 3, question 3… which values of A and BC?
51:33:140Paolo Guiotto: ABC… Now, for the values ABC of 2, well, let's write that to the question.
51:42:920Paolo Guiotto: determine… F of Z as… a function.
51:51:290Paolo Guiotto: of that. What is it?
51:53:990Paolo Guiotto: Well, F of Z, if Z is X plus IY,
52:02:720Paolo Guiotto: is equal to, U plus IV, so 1 half X squared minus
52:09:600Paolo Guiotto: Y squared plus IV, which is XY.
52:15:150Paolo Guiotto: Now, we want to see what is this… to write this thing in terms of Z, no?
52:22:200Paolo Guiotto: Well, you should recognize that, if you take, factorize the 1 half, this is X squared minus Y squared plus i to XY, so this sounds like the square.
52:35:190Paolo Guiotto: The fact, if we have an X plus IY,
52:38:380Paolo Guiotto: square, this is X squared plus IY squared, but I square is minus 1, so minus Y squared.
52:46:80Paolo Guiotto: plus the double product, I emphasize the i2XY. So this guy here is nothing but Z squared. So the function we are looking for is Z square over 2.
52:58:940Paolo Guiotto: And that's it.
53:00:780Paolo Guiotto: Do you have any questions?
53:07:210Paolo Guiotto: No question.
53:09:680Paolo Guiotto: Okay, so you see also that there are exercises which are shorter, exercises which are longer… we cannot put everything on the same size, because
53:19:910Paolo Guiotto: This is practically impossible. Exercise 39.
53:24:880Paolo Guiotto: So here we have a differential equation, Y prime equals 1 over Y minus log of T.
53:34:810Paolo Guiotto: So the first question is, determine Domain… the… domain… Of local existence and uniqueness.
53:50:750Paolo Guiotto: So, we have to identify the right-hand side of the equation, so the equation…
53:57:980Paolo Guiotto: equation is Y prime equal FTY.
54:03:710Paolo Guiotto: Well, this function is 1 over Y minus logarithm of t.
54:10:380Paolo Guiotto: So the domain is the domain of this function, where this function is defined and possibly continuous with derivative with respect to Y continuous. First of all, let's determine the domain.
54:22:900Paolo Guiotto: This is the set of TYs for which that function makes sense. As you see, there is log that demands the argument of log positive, so this is T positive.
54:35:20Paolo Guiotto: Then, we have also a,
54:38:920Paolo Guiotto: denominator, Y minus log of t, that must be different from 0.
54:49:890Paolo Guiotto: Sorry, sorry.
54:58:960Paolo Guiotto: Okay, so this means, again.
55:05:180Paolo Guiotto: So this means that T is positive, and Y is different of log of T.
55:15:270Paolo Guiotto: It's a huge event.
55:19:230Paolo Guiotto: Okay, so let's, see what does it mean. This, in plane TY. We have T positive, so the right half plane.
55:30:880Paolo Guiotto: And Y must be different of log of t. Log of t is a line.
55:36:990Paolo Guiotto: It's the plot of the logarithm.
55:39:870Paolo Guiotto: So this red line is not in the domain. Domain is everything else, okay? So domain is this part plus this part below and above the log.
55:50:410Paolo Guiotto: Now, clearly, The function f is a good function where it is defined.
56:00:280Paolo Guiotto: So it is continuous on the domain D, and the YF is also continuous in the domain B, because, as you can easily see, the YF is
56:12:970Paolo Guiotto: What is… this is 1 over Y, so it is minus the square of denominator, Y minus log of t.
56:22:760Paolo Guiotto: What is the derivative with respect to Y of that denominator?
56:34:30Paolo Guiotto: Are you following me? So I'm computing the derivative with respect to Y of that F, so I'm using the function
56:41:880Paolo Guiotto: Derivative of 1 over something is minus the derivative over the square. So I wrote the square downstairs, now I have to put the derivative of Y minus log T with respect to Y in the numerator. What is the…
56:56:110Paolo Guiotto: It is just one.
56:58:430Paolo Guiotto: So, as you can see, this is a continuous function domain. So, the domain of local existence uniqueness is that one.
57:05:860Paolo Guiotto: Question 2.
57:08:560Paolo Guiotto: stationary solutions.
57:14:860Paolo Guiotto: And then, regions where solutions, regions, of the… Wow.
57:25:820Paolo Guiotto: solutions… Increasing or decreasing.
57:32:200Paolo Guiotto: Now, stationary solution… a stationary solution is a constant solution. Y constant equal to C is a solution.
57:40:450Paolo Guiotto: We just plug into the equation, we get 0Y prime, equal 1 over Y, which is the constant minus log of t.
57:51:360Paolo Guiotto: So, what do you think about this?
57:54:620Paolo Guiotto: It cannot be possible. 1 over something cannot be zero.
57:58:270Paolo Guiotto: Impossible. So, no constant solutions.
58:07:540Paolo Guiotto: constant or stationary.
58:13:920Paolo Guiotto: solutions.
58:16:370Paolo Guiotto: Now, about the monotonicity, Y is increasing if and only if Y prime is greater or equal than 0. Y prime is 1 over Y
58:27:900Paolo Guiotto: minus logarithm of t, and this quantity can be positive if and if the denominator, Y minus log T, that cannot be 0.
58:37:990Paolo Guiotto: must be positive. So, Y above the logarithm of T?
58:44:490Paolo Guiotto: So it means that, going back to the figure, when we are above the red line, the solution is growing. When we are below, the solution is going down.
58:55:620Paolo Guiotto: Okay.
58:57:690Paolo Guiotto: Question 3, well, actually, it says, now, let's why…
59:05:320Paolo Guiotto: from Alpha, beta… 2.
59:08:730Paolo Guiotto: The solution of the Cauchy problem.
59:14:650Paolo Guiotto: with initial condition Y2 equals 0.
59:18:110Paolo Guiotto: Let's see what is it, this condition, the passage condition.
59:22:130Paolo Guiotto: Well, we have to put Y of 2 equals 0, it means the value is 0, so we are on the t-axis.
59:28:630Paolo Guiotto: at time t equal to, the value is 0. Now, what is time 2? We have to know that this timer, because it matters if we are at right or at left of that point, this is 1, because it is the log of t, no?
59:42:200Paolo Guiotto: This is Y equal log of T.
59:45:600Paolo Guiotto: So, time 2 will be about here, and the solution
59:50:740Paolo Guiotto: passes through that pod. So, by looking at this picture, we know that we are in the region where the solution is decreasing, and probably the solution will stay decreasing all time, because to change naturally, we should cross the red line, which is impossible.
00:06:720Paolo Guiotto: So we already expect something about this solution. So the question 3 says, So that Y is monotone.
00:18:980Paolo Guiotto: It's not specified as increasing, decreasing, but we understood that it will be decreasing.
00:25:10Paolo Guiotto: And what is the limit… limit when P goes to alpha, the first time of the solution?
00:33:380Paolo Guiotto: of the derivative.
00:35:810Paolo Guiotto: Okay, so, we said, why is incre… sorry, decreasing?
00:43:940Paolo Guiotto: Now, to give a formal argument.
00:49:220Paolo Guiotto: We should just prove that the solution stays all times in the region where solutions are decreasing.
00:56:90Paolo Guiotto: So, below the red line.
00:58:410Paolo Guiotto: These are the causal.
01:02:910Paolo Guiotto: Y of T is always less than log of t for all times t.
01:10:20Paolo Guiotto: Well, this is… this, in principle, should be proved. So, if you say this in an exam, well, yes, but why? So, you will receive this.
01:21:110Paolo Guiotto: Okay?
01:22:40Paolo Guiotto: You don't explain why.
01:24:150Paolo Guiotto: Okay, so let's see why.
01:26:780Paolo Guiotto: If false…
01:33:10Paolo Guiotto: Well, remind that it is true at least at one time.
01:39:920Paolo Guiotto: at P equal to Y of 2, which is 0, is less than log of 2, which is positive. So at time t equal 2 is true. If false, it means that there exists another time.
01:56:90Paolo Guiotto: t hat, such that Y of T hat is greater or equal than log of T hat.
02:04:420Paolo Guiotto: Now, equal cannot be.
02:07:370Paolo Guiotto: But… YT hat equals the log would mean that the solution crosses the logarithm at some point.
02:18:840Paolo Guiotto: So, there is a point T hat where this happens. Y of T hat is exactly the value of the logarithm at t hat. So the solution is on the red line, and this would mean that it is outside of the domain, and this is not possible.
02:34:200Paolo Guiotto: for a solution. This would imply that the point T-hat Y at T hat would not be in D, and this is impossible.
02:49:540Paolo Guiotto: And… if… The other possibility is that y at t hat is greater than log of T hat.
02:59:620Paolo Guiotto: Well, but since that time, T equals 2 is below.
03:05:750Paolo Guiotto: since… Y at time 2 is less than log of 2, by continuity.
03:15:910Paolo Guiotto: So you have a time where you have below. You have a time where you are above, you have a time where you are on.
03:21:700Paolo Guiotto: There would be another time, maybe T double hat, where Y at T double hat would be again equal to the log.
03:30:560Paolo Guiotto: But this would fall back to the previous contradiction. So it can never be above, it means it is always below, and therefore it is decreasing, and this is
03:41:370Paolo Guiotto: Complete the arc.
03:43:00Paolo Guiotto: Now, the limit. What about the limit? So…
03:48:440Paolo Guiotto: The idea is that the solution is, if we go back in the past, because we are going to the initial time alpha.
03:56:770Paolo Guiotto: So, we should bet that the solution will crash on this red line.
04:02:310Paolo Guiotto: Because if it doesn't crash, it means that it ends, it ends sometime before the crash, so the situation should be the following. Suppose that the solution, this is time 2, does this. It ends here.
04:19:490Paolo Guiotto: So these points here is the point with abscissa alpha and the limit value, let's call it Y of alpha, okay?
04:28:780Paolo Guiotto: So, the key point is to say that this point must be here, on the red line, cannot be inside. It cannot be inside because, you know, the general idea is the solution can never die inside the domain, and that the formal argument is the argument of the compact set.
04:46:320Paolo Guiotto: So we would have a box like this, but now we write precisely, that contains the solution forever in the past.
04:53:160Paolo Guiotto: So the solution cannot leave the box. And this is in contradiction with the properties of solution. So, the claim is… claim is that the point, alpha, Y alpha.
05:09:530Paolo Guiotto: Belongs to the line where Y is equal to the log of t.
05:15:760Paolo Guiotto: If this happens, we will see that the limit of the solution can be computed.
05:21:220Paolo Guiotto: Why?
05:23:300Paolo Guiotto: If false…
05:28:260Paolo Guiotto: we take this compactor, let's write the compactor, K, is for times, we take from alpha to time 2.
05:38:520Paolo Guiotto: From… for the ordinates, we take from 0 to the limit value, which is Y alpha.
05:45:530Paolo Guiotto: Now, if false, so… Why alpha is below the log of alpha.
05:55:850Paolo Guiotto: then this box K would be compact and contained in the domain.
06:01:440Paolo Guiotto: and solution.
06:03:200Paolo Guiotto: And… this solution.
06:07:530Paolo Guiotto: Why?
06:08:930Paolo Guiotto: Never.
06:10:740Paolo Guiotto: leaves.
06:13:400Paolo Guiotto: Okay… in the past.
06:17:940Paolo Guiotto: Which is impossible.
06:22:630Paolo Guiotto: So, this means that, the final value
06:27:430Paolo Guiotto: Y alpha must be equal to log of alpha. So, it means that the solution is here. And now, what happens to the limit of the derivative? Limit when t goes to alpha of Y prime t.
06:42:590Paolo Guiotto: We take the equation, this is the limit when t goes to alpha, of 1 over Y of T minus log of t.
06:52:530Paolo Guiotto: When T goes to alpha, what happens? This goes to Y of alpha.
06:56:780Paolo Guiotto: This goes to log of alpha.
06:59:380Paolo Guiotto: So, since we know that Y of alpha is equal to log of alpha, this quantity must go to 0.
07:09:630Paolo Guiotto: Zero, how?
07:14:880Paolo Guiotto: you come… to this point, from which side? Below. So, until solution is alive, we say above.
07:27:910Paolo Guiotto: It is here. Solution is always below the logarithm, so Y of T minus log of t is negative. So, in particular, this is a zero.
07:38:30Paolo Guiotto: Minds.
07:39:610Paolo Guiotto: So, 1 over 0 minus is…
07:42:820Paolo Guiotto: negative infinity. That's the limit of the solution. So what we see is the solution approaching that point with the vertical slope.
07:52:180Paolo Guiotto: There is another question.
07:54:700Paolo Guiotto: Question 4.
07:56:200Paolo Guiotto: concavity, concavity… and VITA.
08:07:290Paolo Guiotto: Plus… less than plus infinity, or equal to plus infinity.
08:12:770Paolo Guiotto: Well, let's see the concavity. Well, the concavity we need Y second?
08:17:00Paolo Guiotto: So we have Y prime equal 1 over Y minus log of t. Now we compute the derivative attention respect to t, so this is Y second of t.
08:28:939Paolo Guiotto: We have to do the derivative with respect to t of 1 over Y minus log of t.
08:34:819Paolo Guiotto: We do this derivative, reminding that Y is a function of T. So, first of all, it's a fraction, so… and a special type 1 over, so we do minus, denominator is the square of that denominator.
08:49:380Paolo Guiotto: And then here I have to put the derivative with respect to T of this guy here.
08:56:250Paolo Guiotto: What is the derivative of this, Y minus log of t.
09:07:979Paolo Guiotto: We're still informed.
09:10:220Paolo Guiotto: Y1?
09:12:40Paolo Guiotto: Y1, I'm not doing the derivative with respect to Y, but with respect to T, Y is Y of T, so this is Y prime.
09:20:470Paolo Guiotto: minus 1 over T.
09:24:149Paolo Guiotto: That's the deliverable.
09:25:930Paolo Guiotto: Now, what do we know about Y prime? We know something. Now, remind that the goal is to determine the sign of this thing. We know that Y is decreasing, so Y' is negative.
09:38:240Paolo Guiotto: Then I have minus a positive, so the numerator is negative. With the minus in front, it becomes… and this is a square, so this guy here is positive, so this becomes positive.
09:52:450Paolo Guiotto: Okay? So, it means that the second derivative is positive, it means that Y has the concavity pointing upward, or it is convex.
10:05:600Paolo Guiotto: So now we can have a… we can also start drawing the plot of the solution.
10:11:870Paolo Guiotto: So, we said there is nothing here.
10:14:640Paolo Guiotto: Here we have the… logarithm line.
10:19:120Paolo Guiotto: And our initial condition about here, time 2. So we already said that the solution is decreasing, so going back in the past, it is increasing, and it crunched
10:31:500Paolo Guiotto: here on the barrier, with the vertical slope. It is convex, so it's like that.
10:39:390Paolo Guiotto: decreasing.
10:40:940Paolo Guiotto: So, what can be expected about beta?
10:48:490Paolo Guiotto: So the question is beta finite, or beta equal plus infinity?
10:56:870Paolo Guiotto: Yes, but we have to justify it, right? So, betifying it, how it can be? Not like this, because in this case, the solution would die inside, so there should be something like that, but that's in contradiction with the concavity, so that's the argument we have to put. So, the claim is…
11:16:970Paolo Guiotto: Claim is beta equal plus infinity.
11:22:530Paolo Guiotto: Why? So, we have to argue by contradiction. If beta is less than plus infinity.
11:30:480Paolo Guiotto: Then, we have two possibilities. Dan.
11:34:980Paolo Guiotto: Either the solution at beta is finite, so in particular here is not too big negative, so it's greater than minus infinity, or the solution at beta is equal to minus infinity, because it is decreasing. It cannot go to plus infinity, for example.
11:54:370Paolo Guiotto: But in this case, we have the argument of combat sets that says the solution will be trapped in this box forever in the future. So in this case.
12:05:690Paolo Guiotto: The solution, the point TYT, would be in the box from… this is time 2 to time beta.
12:14:520Paolo Guiotto: which is, since beta is finite, this is closed and bounded, times 0 to the final value, y beta, which is, again, finite, because we are assuming that this value is finite. This is a compact
12:30:910Paolo Guiotto: content, indeed.
12:32:870Paolo Guiotto: And this would be for every T, for every T positive after initial time 2.
12:40:910Paolo Guiotto: Okay? So the solution never leaves the box in the future, and that's impossible.
12:49:270Paolo Guiotto: So, now, why this cannot be possible?
12:53:330Paolo Guiotto: We have to exclude it. Intuitively, you could say the function is convexed, to go down to minus infinity, it should change the voncavity.
13:01:350Paolo Guiotto: I can accept this, even if it is not perfectly,
13:07:270Paolo Guiotto: solid argument. Well, a solid argument could be the following. We use the concavity because of the convexity.
13:15:960Paolo Guiotto: Convexity means that the function is above each of its tangents. So, for example, you take the tangent at 2, it's a straight line like that, and the function is always above that line.
13:28:740Paolo Guiotto: So, since the tangent cannot go to minus infinity in a finite time, it's a straight line, so, let's see. If y of beta were equal to minus infinity, then
13:42:100Paolo Guiotto: by convexity, Y of T would be larger than the tangent at…
13:53:550Paolo Guiotto: t equal 2. What is the tangent? It is Y of 2 plus Y prime of 2 times T minus 2. Whatever it is, it doesn't matter.
14:03:370Paolo Guiotto: No? So, in particular, if you take t equal betta, you would have that Y of beta would be greater than Y of 2 plus Y prime at 2 times beta minus 2. If beta is finite, as here.
14:19:370Paolo Guiotto: is finite. This number is greater than minus infinity, and that's a contradiction. This is in contradiction with this.
14:31:440Paolo Guiotto: Okay? So they cannot stay together. So it means that the solution, if beta is finite, we could have two possibilities. Either the solution ends at some finite value, first case, or it goes down to minus infinity, second case. In the first case, we are in contradiction with the argument of convert sets.
14:51:310Paolo Guiotto: In the second case.
14:53:10Paolo Guiotto: the solution is in contradiction with the convexity. So none of these possibilities can be, and therefore there is a unique case left, which is this one, okay? So beta equal…
15:05:900Paolo Guiotto: plus infinity.
15:07:410Paolo Guiotto: And the plot is now clear. It is this one.
15:12:60Paolo Guiotto: So, again, It's got the negative part here.
15:16:350Paolo Guiotto: The red line of the logarithm.
15:20:100Paolo Guiotto: This is the plane TY, and the solution starts at time t equal 2 from 0, so it is convex, decreasing, so it will be something like this.
15:30:880Paolo Guiotto: Okay, so this is the qualitative plot of the solution.
15:35:530Paolo Guiotto: Do you have any questions?
15:39:310Paolo Guiotto: Yeah.
15:44:470Paolo Guiotto: The argument, well, here we are saying, for example, why… I don't know, we say that beta should be plus infinity.
16:00:280Paolo Guiotto: If beta is fine, we have two possibilities. The solution is going down, no?
16:06:710Paolo Guiotto: So let's redo this figure.
16:10:960Paolo Guiotto: The solution is going down. What could happen to stop here at some time beta? Either it stops here at some finite point, or it goes down at minus infinity. So these are the two possibilities, number one and number two.
16:28:490Paolo Guiotto: Now, we want to say that none of this can be
16:31:380Paolo Guiotto: The first one… well, the first one, the idea is that the solution would die inside the domain, and this is not compatible with that property, which is called the property of compact sets.
16:42:500Paolo Guiotto: That is, whatever is the compact you take in the domain, the solution must live in future and in past this compat. In this case, you would have this
16:53:670Paolo Guiotto: rectangle here, the green rectangle, so we are in the first case, so in this case, that would contain the solution forever in the future.
17:05:530Paolo Guiotto: The solution cannot live, because…
17:08:660Paolo Guiotto: There are all future times we arrive up to beta, which is supposed to be the last time of life for the solutions, and all values for the solutions.
17:18:510Paolo Guiotto: So it means the solution is inside that box forever in the future. Of course, it leaves the box in the past, but the general theorem says that it must live both in future and past, and this does not live in future. So this is in contradiction with a general fact.
17:38:70Paolo Guiotto: Okay, there is a general theorem that says whatever is the box, it must leave the box, provided it is compact and inside the domain. Okay? So this is the contradiction here.
17:49:320Paolo Guiotto: The second one is a different contradiction, because in the second case, we are saying, okay, this is the case where the solution should have a vertical asylum, should do something like this, no?
18:02:940Paolo Guiotto: At time beta. This, intuitively, the argument could be… this is in contradiction with the fact that the solution is convex. You see that here you should change the concavity. The solution should point down.
18:19:180Paolo Guiotto: Well, this could be accepted as a qualitative, very qualitative description. A more formal argument is the following. By convexity, the function… I know the convexity says that the graph
18:33:190Paolo Guiotto: is always above each of the tangents I take. For example, the tangent at time t equal 2, which is this one.
18:41:970Paolo Guiotto: And from this inequality, when you take t equal betta, you get this one. It doesn't matter what are the values. What it says is that here, this quantity is finite, because Y of 2 is finite, it is 0, Y prime is 2 is whatever is finite, beta is finite, because we are assuming that beta is finite.
19:03:350Paolo Guiotto: Okay?
19:05:80Paolo Guiotto: So the right-hand side is finite, and this one, instead, is going to minus infinity, it's minus infinity. So how can be minus infinity greater than finite number? That's a contradiction.
19:16:780Paolo Guiotto: Condition is not possible, okay? So, since now, we say that if beta is finite, we have two possibilities, and none of them
19:27:920Paolo Guiotto: Let's say. And both of them, it's a contradiction, it means that the initial assumption must be wrong.
19:36:90Paolo Guiotto: So, the conclusion is that this is true.
19:40:270Paolo Guiotto: Okay?
19:41:700Paolo Guiotto: We are still using logical arguments.
19:48:120Paolo Guiotto: Okay, so now we have the exercise 40.
19:52:800Paolo Guiotto: That's a theoretical question.
19:55:510Paolo Guiotto: So, that is the first part, which is a definition. What does it mean that a status is open?
20:01:300Paolo Guiotto: So you should provide the definition. But please, when it is asked for a definition, you have to provide the precise definition, not by words.
20:10:870Paolo Guiotto: So, if they… you have to say, for example, if it is open, if it is, through that, for every point, there is a ball which is entirely contained in the set. You have to write down this, not even by word, by symbols, if possible, okay?
20:25:960Paolo Guiotto: Then says, let F be a continuous function on RD, so let's take this second part. F is a function that goes from, here, RD to RM.
20:39:960Paolo Guiotto: Continos.
20:41:650Paolo Guiotto: on… RD, so at every point of work.
20:46:680Paolo Guiotto: To prove the following property, that, prove that.
20:54:380Paolo Guiotto: If S is contained into RM, so which is the arrival space, is open.
21:03:750Paolo Guiotto: Then, also the anti-image… sorry, capital F… the anti-image of the set S is… is open.
21:16:230Paolo Guiotto: So this is in RM, and this is a counter image, so it will be in the domain, so this will be in RD.
21:26:880Paolo Guiotto: There is, well, it's recalled that, for your convenience, that the anti-image of a set is the set of points X in RD, such that F of X belongs to the set S. This is what is…
21:45:580Paolo Guiotto: And there is a hint, suppose, that for some open S, F minus 1 of S.
21:51:550Paolo Guiotto: Is not open.
21:57:540Paolo Guiotto: I don't know if, if this hint is really… of,
22:11:90Paolo Guiotto: Okay, let's, let's, let's see, because I'm sure that… okay, let's see what the hint would suggest.
22:21:360Paolo Guiotto: If… F… minus 1 of S.
22:28:850Paolo Guiotto: is not… Open.
22:33:200Paolo Guiotto: And let's continue this. So, the starting point's just this one.
22:38:740Paolo Guiotto: I'm pretty sure that some of you would say that F minus 1 of S is closed, right?
22:45:500Paolo Guiotto: You say no, but… letting the 30% of you that will say this. Okay, so…
22:53:430Paolo Guiotto: If it is not open, what does it mean? Okay, it means that…
22:59:790Paolo Guiotto: So, I don't know what kind… if a figure could be of any add, but let's say we have an F, which is moving from here to here, so our S is here. Well, since it is open, let's use the typical dashed notation for open sets.
23:19:540Paolo Guiotto: And so we are taking the anti-image of this.
23:24:50Paolo Guiotto: Well, it is clear that if it is empty, it can be, no, because there are no X for which f of x belongs to S. If it is empty, if the empty image would be empty, it would be open, okay? So let's assume that it is non-empty.
23:39:830Paolo Guiotto: Because it is saying… suppose that it is not open. So we have this anti-image, F-1 of S, is some set here.
23:48:730Paolo Guiotto: And, suppose that it is not open.
23:52:290Paolo Guiotto: So what could be side?
23:55:530Paolo Guiotto: It must be closed? Why, why close now?
24:01:540Paolo Guiotto: Why close? I see, that's the same trap I was saying before. I think there are two possibilities, if I'm not… I don't remember it wrongly. F, if it's, not closed, if it's… But why close? What, what? I think when it's continuous…
24:19:880Paolo Guiotto: But, let's stop, please. Why are you talking about closed?
24:25:650Paolo Guiotto: Because you are doing what I said just one minute ago, sets are either open or closed, so if it is not open, it is closed. Right. That's wrong.
24:35:870Paolo Guiotto: Okay. It doesn't depend on if F is continuous or not? No. Continuity will come later. No, we have to think about what does it mean if F minus 1 is not open.
24:48:300Paolo Guiotto: So, what is open means that whatever is the point you pick here, there is a little ball
24:56:380Paolo Guiotto: Until we contain it there.
24:59:70Paolo Guiotto: Now, what does it mean that this does not happen?
25:02:290Paolo Guiotto: It means that there is one bastard point for which.
25:07:600Paolo Guiotto: Let's see later. For which, what happens?
25:12:530Paolo Guiotto: So there exists a bad point here.
25:15:780Paolo Guiotto: So, this means that there exists a point, X, let's call it X star, in, F-1.
25:27:420Paolo Guiotto: of S.
25:29:430Paolo Guiotto: Such that there is no ball centered at that point anteriorly contended into the set.
25:36:980Paolo Guiotto: So if there is no ball anteriorly containing into the set, it means that for every bowl, what happens?
25:46:340Paolo Guiotto: The main difficulty in doing this kind of exercises
25:49:930Paolo Guiotto: is a logical argument, no? What does it mean that there is not a single ball entirely contained? It means that whatever is the ball you take, the ball is not contained, no? This is not entirely contained into that set. But what does it mean this?
26:07:270Paolo Guiotto: It means that…
26:09:330Paolo Guiotto: If you have to do a figure, this is your set. So imagine you take a ball, it means that, again, it doesn't mean that it isn't a complementary.
26:19:590Paolo Guiotto: It is not contained until it means that it contains something of the complementary. So there is some point here, which is in the complementary and not in the set. So, we have that, so, in other words, this implies that for every ball, B, X, star.
26:39:600Paolo Guiotto: R… There exists a point, in the bowl, Well, let's say…
26:50:590Paolo Guiotto: XR, let's call it XR, in that ball, B. Excellent.
26:58:750Paolo Guiotto: Stop.
27:01:260Paolo Guiotto: Such that this point is not… the point is the red point here in the figure, is not in the set F-1 of S. But if it is not in the set, the set is this one, is the set of points that F sends into S.
27:16:590Paolo Guiotto: So this means that F of that point, XR, is not in S.
27:27:470Paolo Guiotto: You see? Because if it is in the anti-image, F should send the point into S.
27:35:470Paolo Guiotto: Okay, now we know that for every ball, there exists a point.
27:40:800Paolo Guiotto: In that ball, that F sends out of S.
27:55:850Paolo Guiotto: that for every ball, means not one single ball, means whatever is the radius. So normally, when you have these things where you have different balls, you have a point. In each ball, you find a point, you see, you have to imagine dynamically, this thing. Take a variable radius, R equal 1 over n.
28:14:150Paolo Guiotto: So this means that there is a point that now we baptize XN in the bowl centered at point X star.
28:21:950Paolo Guiotto: radius 1 over n. So since the radius is going to 0, this sequence is moving, and it's going to the point X star, because the distance is going to 0, such that the values of this… the F on these points are not in S.
28:41:760Paolo Guiotto: But let's see what we can draw from this. From this, we have that the sequence XN
28:48:50Paolo Guiotto: goes to the point X star.
28:51:620Paolo Guiotto: Because the distance between X and And the point, X star.
28:57:940Paolo Guiotto: is bounded by 1 over n that goes to 0. So that, by definition of limits, we go there.
29:10:800Paolo Guiotto: Have we used the continuity?
29:13:250Paolo Guiotto: Not yet. Maybe it's time to use. We have a sequence of points moving in the domain of F that goes to this point X star. What the images will do. F of XN
29:28:20Paolo Guiotto: We do what?
29:33:830Paolo Guiotto: What is continuity? If Xn is moving to X star, F of Xn will go to…
29:39:700Paolo Guiotto: To my mother, which is… We go where?
29:45:170Paolo Guiotto: we go to F of X star.
29:50:260Paolo Guiotto: Right?
29:51:840Paolo Guiotto: By continuity.
29:53:990Paolo Guiotto: And now… Okay, so let's see the field.
29:58:890Paolo Guiotto: So this is, the anti-major, F minus 1 of S.
30:04:150Paolo Guiotto: We know, by contradiction, we said that there is a point, and of course, in this figure, this point, X star cannot be here, because if point would be this one, we would have a ball.
30:15:190Paolo Guiotto: We have to imagine, probably your intuition was correct, that this point could be on the boundary. So let's imagine that the point X star is here.
30:24:350Paolo Guiotto: And at this point, it says, is approached by points, these red points are DXN, that move to him, but they are not in the anti-image, so they are outside.
30:39:240Paolo Guiotto: However, the function does this. When you map everything to this mapF, you now go into the setS,
30:47:990Paolo Guiotto: Because this point, X star, is in S.
30:53:30Paolo Guiotto: sorry, the image of, X star.
30:57:270Paolo Guiotto: Why?
30:59:810Paolo Guiotto: Where have we taken the point X star into the anti-image of S. So when we map to, to… with F, it will be inside S. Because what is the anti-image? It's the set of points that F maps into S. So, my point X star will be mapped into some point which is in S.
31:22:220Paolo Guiotto: And what do we know about this?
31:24:90Paolo Guiotto: S is… Noticeable.
31:31:970Paolo Guiotto: So, you have to be concentrated, because we are proving that when X is open, the anti-image is open, okay? So, you must have these things in… like stone, iron in your…
31:45:190Paolo Guiotto: brain, okay? So S is open, okay, S is open. And then what happens if S is open? That…
31:55:440Paolo Guiotto: I have a point into an open set, so…
32:01:710Paolo Guiotto: Huh?
32:05:250Paolo Guiotto: I don't.
32:07:980Paolo Guiotto: Yeah, so there is a little bulb entirely contained, but now there is something wrong here, because…
32:14:440Paolo Guiotto: What is doing this red sequence when I map with F? This. They are going to F of X star, so maybe they are outside, but sooner or later, they will be in the green ball, because they have to go to the point. So these points here are the points F of XN.
32:33:760Paolo Guiotto: So, I don't know, maybe some of them they will be outside of F, but they definitely must enter into the green bowl because of this.
32:42:690Paolo Guiotto: Okay? So this means that this must be into the ball centered at F of X star.
32:50:830Paolo Guiotto: radius, I don't know, R?
32:55:30Paolo Guiotto: this bowl, which is contained in S, for N large.
33:02:610Paolo Guiotto: But this is impossible, because if you are in S, it would mean that also the point XN would be in the anti-image
33:12:550Paolo Guiotto: Because by definition, this is the set of points that are sent into S for N large.
33:19:580Paolo Guiotto: And this is impossible, because we built this sequence as if they are not
33:26:70Paolo Guiotto: In the anti-image. And that's a contradiction.
33:35:480Paolo Guiotto: So the contra… we got the contradiction, it means that F minus 1 of S cannot be
33:42:60Paolo Guiotto: not open. It must be open, okay?
33:47:840Paolo Guiotto: Okay.
33:52:900Paolo Guiotto: Well, actually, it's 1218, huh?
33:58:150Paolo Guiotto: So… There is another,
34:03:10Paolo Guiotto: I can do just one,
34:06:580Paolo Guiotto: One more exercise. In the next one, we have… in any case, I will publish the solution, okay?
34:12:950Paolo Guiotto: We have, again, an exercise on constrained optimization, so I wouldn't say that it's very different from this one.
34:20:680Paolo Guiotto: There is an exercise on, fields, this diamond, which is in the same family of,
34:28:180Paolo Guiotto: the exercise we have seen with the complex functions. It's not… we have an integral with an area.
34:36:830Paolo Guiotto: Maybe the Exercise 44, On the complex function.
34:42:990Paolo Guiotto: I don't know.
34:44:720Paolo Guiotto: But it's this exercise?
34:47:600Paolo Guiotto: Which is a bit different.
34:50:540Paolo Guiotto: from the previous one, this says you have this function, F of Z, it is defined in this way.
34:58:490Paolo Guiotto: It is a Z conjugate square of a Z, or Z different from 0.
35:05:910Paolo Guiotto: And 0 for Z equals 0.
35:10:690Paolo Guiotto: Hello, it says, question one, U and V, so are the real and the imaginary part of F, so F is U plus AV.
35:22:260Paolo Guiotto: verify the Kosher-Riemann equations.
35:31:420Paolo Guiotto: And the question 2… It tasks easily.
35:37:370Paolo Guiotto: FC differentiable.
35:40:460Paolo Guiotto: at z equals 0.
35:48:840Paolo Guiotto: You will understand at the end why this… what is the reason of this exercise, what is the message behind this.
35:56:490Paolo Guiotto: So, since we do not have U and V here, we have F of Z, we have first to compute these two functions. How can we do? Well, F of Z is F of Z if Z is equal to X plus IY,
36:13:240Paolo Guiotto: is F of X plus a y.
36:16:540Paolo Guiotto: And it is equal to, according to that definition, 0 if X plus IY Is equal to zero.
36:24:60Paolo Guiotto: If X plus IY is different from 0, I have to do Z conjugate, square, and divide by Z, okay? Now, Z conjugate is X minus IY. I do the square, and I divide by X, by Z, X plus IY.
36:42:890Paolo Guiotto: This is. Okay, this here you don't see U plus IV, because it's a fraction, so we have to calculate this fraction and to put the fraction in such a way that we can read what is the real part of F and what is the imaginary part of F.
36:59:70Paolo Guiotto: So how can we do? Well, of course, we develop the numerator. The square is X squared, then we have IY squared, so minus Y squared, and then the double product is minus i2XY. This is the square of the numerator, and then we have to divide by X plus IY. Now.
37:18:610Paolo Guiotto: Do you remind how… How you can compute the ratio between two complex numbers.
37:25:740Paolo Guiotto: To put the number under an algebraic form.
37:31:770Paolo Guiotto: There is a little algebraic trick to do the division. Exactly. We multiply and divide by the conjugate of the denominator.
37:42:330Paolo Guiotto: Because this case, this product, X plus IY times X minus Y is a Z times Z conjugate, is the modulus of Z squared. So this becomes… the denominator now becomes real, and X squared plus Y squared.
37:58:280Paolo Guiotto: We do the numerator, so with a little bit of patience, we have,
38:04:560Paolo Guiotto: So, X times X squared minus Y squared. Then, for the real part, I have also to do the product of this with this. So, minus, minus plus II minus 2XY squared. Then, I have the imagined.
38:27:640Paolo Guiotto: No, I have to multiply and divide by the conjugate of this.
38:32:660Paolo Guiotto: When we have a minus, then i is with the power of 2 minus? Yeah, but…
38:43:690Paolo Guiotto: whatever is plus minus, A plus B squared is A squared.
38:49:430Paolo Guiotto: plus B squared, plus the double product, right?
38:53:850Paolo Guiotto: Now, my B is minus IY, the minus square is plus.
38:59:280Paolo Guiotto: The I-squared is minus, and the Y squared is Y squared. That's why you see minus Y squared.
39:06:230Paolo Guiotto: And the double product is 2 times X times minus IY, so minus i2XY.
39:16:370Paolo Guiotto: You're not convinced.
39:19:260Paolo Guiotto: Minus 12, okay, but minus Y in power…
39:24:400Paolo Guiotto: You are not convinced about this one?
39:27:650Paolo Guiotto: B minus… Well, X squared. Then do minus IY squared.
39:33:930Paolo Guiotto: plus, minus i squared. Minus square is plus. I square is minus…
39:40:540Paolo Guiotto: So you have minus. Now, Y squared is Y squared, that's why you see Y squared.
39:45:800Paolo Guiotto: Okay.
39:47:500Paolo Guiotto: Okay, so now, with the eye…
39:53:360Paolo Guiotto: So maybe we put the minus… what?
39:57:520Paolo Guiotto: Yeah, let's put the minus instead of plus.
40:01:520Paolo Guiotto: Yeah.
40:02:720Paolo Guiotto: So we have minus i, so we get 2XY times X, so X squared.
40:11:410Paolo Guiotto: And then we have minus IY times…
40:28:750Paolo Guiotto: X cuba… XY squared minus XY squared minus 2XY squared is minus 3XY squared.
40:38:470Paolo Guiotto: divided by X squared plus Y squared.
40:42:400Paolo Guiotto: plus I… well, actually, let's write minus, maybe.
40:48:420Paolo Guiotto: Yeah, it's better if we've got the minus here.
40:52:240Paolo Guiotto: And then we have 3X squared Y minus Y cubed.
40:58:260Paolo Guiotto: Well, maybe we give the plus… let's do this.
41:02:880Paolo Guiotto: We give the plus here to the coefficient, so it comes Y cubed minus 3X squared Y divided X squared plus Y squared. Now you see who are the U and the P. So this guy is the real part, UXY,
41:21:370Paolo Guiotto: And this guy is the imaginary part of VXY.
41:26:530Paolo Guiotto: So, UXY is finally.
41:29:380Paolo Guiotto: equal to…
41:30:850Paolo Guiotto: This thing, X cubed minus 3XY squared divided by X squared plus Y squared, these for XY different from 0, 0.
41:43:270Paolo Guiotto: Y, from X… for XY equals 00, We get zero.
41:50:940Paolo Guiotto: And similarly for V, VXY is… EY Cuba, minus 3XY squared, right?
42:04:820Paolo Guiotto: Oh, X square Y.
42:08:340Paolo Guiotto: divided by X squared plus Y squared, if XY is different from 0, 0.
42:16:630Paolo Guiotto: and 0 if XY is equal to 0, 0.
42:22:450Paolo Guiotto: Okay, now, we should verify that this fulfilled the Cauchier-Riemann equation. So, at .00…
42:33:600Paolo Guiotto: Well, here it's at 00, I forgot to… At… 0, 0.
42:41:460Paolo Guiotto: So, to verify the Kosheriemann equation, we needed to compute the partial derivatives, right?
42:46:950Paolo Guiotto: So, we have to compute the…
42:49:350Paolo Guiotto: DXU at 0, DYU at 0, and the same for DXV and DYV.
43:01:150Paolo Guiotto: Now, of course, 00 is a bad point, because there is the change of definition, so we cannot just take the derivative and evaluate, because this will not work. So we have to proceed back with the definition. So DXU at 00 is what?
43:16:180Paolo Guiotto: Well, it is the directional derivative, so it's the limit when t goes to 0 of u of 00 plus T10,
43:27:820Paolo Guiotto: minus U, 00… divided by T.
43:32:680Paolo Guiotto: Well, this is UT0.
43:38:970Paolo Guiotto: Well, you see that if you put, UT0, you get what?
43:47:60Paolo Guiotto: So this is the limit.
43:49:930Paolo Guiotto: when T goes to zero. Well, this is the value at 0, it is 0. UT0, you have to be careful, because it is this one with the Y equals zero and X equals T, so it is T cubed minus 0, so T cubed divided T squared, so T.
44:07:300Paolo Guiotto: So you get T minus 0 divided t, easily the limit is 1.
44:13:920Paolo Guiotto: Now, if… let's do first this derivative, the DYV at 00, to see if it is the same value, because the Kosheriemann equation says that these two should be the same, so this is what we should verify.
44:32:480Paolo Guiotto: Now this, again, we have to use the, direct formula. So this is the derivative with respect to Y, so it is V of 0T minus V00, 0.
44:43:630Paolo Guiotto: divided by T.
44:45:540Paolo Guiotto: This is zero.
44:47:390Paolo Guiotto: V0T is what? When X is 0, you see that we get numerator is T cubed, Yes.
44:56:270Paolo Guiotto: It is T cubed 0 divided T squared, so it's again equal to T.
45:03:70Paolo Guiotto: And so we have T minus 0 divided T, so the limit comes 1.
45:07:490Paolo Guiotto: So the first question remaining question, at least, is verified, and you can check that it is verified also, the second one.
45:15:70Paolo Guiotto: Now, the question, too, Similarly.
45:22:530Paolo Guiotto: we verify it.
45:26:230Paolo Guiotto: that, DY.
45:28:490Paolo Guiotto: U at 00 is minus DXB at 00.
45:33:980Paolo Guiotto: Do the check.
45:35:960Paolo Guiotto: The question tool… is the following. Is that…
45:42:660Paolo Guiotto: C differentiable at 0, 0. So now we should know that U and V verify the Kosheriemann equation. Can we conclude that F is differentiable at 0, 0?
45:54:130Paolo Guiotto: Well, you should know that the answer is no, we cannot conclude, okay? So we should check. And, so what are the possibilities?
46:03:00Paolo Guiotto: Either we, we discuss the differentiability of these two functions at 0, 0, this is a way, or…
46:12:920Paolo Guiotto: We discuss directly the derivative on F by… It's completely gone.
46:23:680Paolo Guiotto: By using the fact that we know the formula, maybe we can, say if F is C differentiable, okay?
46:32:730Paolo Guiotto: So, two… Check.
46:37:680Paolo Guiotto: Differentiability, we… Should.
46:45:330Paolo Guiotto: Either… discussed.
46:51:550Paolo Guiotto: We are differentiability.
46:54:990Paolo Guiotto: of U and V.
46:57:800Paolo Guiotto: Or, it's, let's say, an alternative, check directly.
47:07:610Paolo Guiotto: on Africa.
47:10:940Paolo Guiotto: Way there.
47:15:800Paolo Guiotto: F is differentiable or not.
47:23:560Paolo Guiotto: I don't know which is better, because the differentiability
47:27:510Paolo Guiotto: It's a neat to have you.
47:33:590Paolo Guiotto: Well, let's say I would try directly. Let's take this second way. So, the differentiability is at 0, so we have to check that the limit… we have to see if this limit exists. Now we are in CR. F of 0 plus H minus F of 0 divided by H.
47:53:760Paolo Guiotto: Now, this… I take this one because F of 0 is 0. F of H is given. This is the limit when H goes to 0.
48:02:550Paolo Guiotto: Zero, and nothing else.
48:05:290Paolo Guiotto: of F of H, which is H conjugate squared divided by 8, this is F of H, all this divided by H, which is this one.
48:16:960Paolo Guiotto: So at the end, I have to compute, to discuss the limit when H goes to 0 of H conjugate squared divided H squared.
48:26:920Paolo Guiotto: It's a bit suspect, because,
48:31:10Paolo Guiotto: In fact, you can see, if H goes to 0, and it is real, so the number H is something like X plus I0, so we move our H along the x-axis, so like this.
48:46:770Paolo Guiotto: Now, H and H conjugate are the same.
48:50:820Paolo Guiotto: Right? So, we get that the limit in this case is the limit of one.
48:58:00Paolo Guiotto: Because here, H, H conjugate is still X plus I0. So the limit would be 1.
49:05:460Paolo Guiotto: If we move on the y-axis, this means that we have numbers 0 plus IY.
49:11:460Paolo Guiotto: Okay?
49:12:940Paolo Guiotto: In this case, you see that if H is 0 plus IY, the conjugate is not the same, it's minus IY.
49:23:30Paolo Guiotto: Well, actually, in this case, because there is the square, so we get the same at the end,
49:34:620Paolo Guiotto: So, no, one second. So, H over H, H conjugate over H, in this case would be minus 1 squared, so it… in any case, we get 1. So.
49:47:370Paolo Guiotto: This is not a good choice, mmm…
49:53:680Paolo Guiotto: Okay… And what if we go to mobile along that distance?
49:59:940Paolo Guiotto: So this is something like X plus IX.
50:10:600Paolo Guiotto: If we do this, let's say H is X plus IX,
50:16:270Paolo Guiotto: And we do H bar over H is X plus IX, and so minus, sorry, divided by X…
50:28:790Paolo Guiotto: plus IVX, so… Now… if we…
50:34:850Paolo Guiotto: Do the division, so multiplying by…
50:38:680Paolo Guiotto: the conjugate, this becomes X minus IX squared divided X squared plus X squared.
50:48:400Paolo Guiotto: So this is… what?
50:51:570Paolo Guiotto: X squared minus X squared minus i2X squared.
50:59:680Paolo Guiotto: Yes?
51:00:780Paolo Guiotto: Square of the first, plus square of this. You see? Minus becomes 1, I becomes minus, X becomes X squared, and double product. Okay, this cancels, and downstairs we have 2X squared.
51:16:470Paolo Guiotto: So we get that this is minus I, if I'm not wrong.
51:22:450Paolo Guiotto: And so when you do the square… the ratio of the square.
51:26:350Paolo Guiotto: which is H, conjugate square divided H squared. This is like doing minus i square, so now we get minus 1.
51:37:390Paolo Guiotto: You see, if we move along this direction, the limiter, so the limiter…
51:44:100Paolo Guiotto: of H bar square over H squared, which is the limit of F of H minus F0 divided H, the limit we should compute to compute the derivative, in this case, comes minus 1.
51:59:650Paolo Guiotto: But that's not possible, because on one side we got 1, on the other side we get minus 1, so this means that there is no derivative of this F.
52:08:610Paolo Guiotto: at zero, zero.
52:10:860Paolo Guiotto: So, the scope of this example was to show you that
52:16:370Paolo Guiotto: It could be possible that U and V verify the Cauchiermann equations, but still the function is not C-differentiable, okay? So the Cauchyriemann equations are not sufficient.
52:28:640Paolo Guiotto: Okay, I'd say that we can stop here, if you agree.
52:34:290Paolo Guiotto: Yes, you can ask, let me just stop.