AI Assistant
Transcript
00:20:440Paolo Guiotto: va.
00:23:180Paolo Guiotto: But then, no.
00:29:40Paolo Guiotto: Vatican.
00:31:770Paolo Guiotto: Okay, good morning.
00:40:780Paolo Guiotto: Good afternoon.
00:42:100Paolo Guiotto: Bye.
00:44:810Paolo Guiotto: I'm sorry, but today I'm still learning the schedules. I went… I came here this time today at 8.30.
00:53:900Paolo Guiotto: thinking that I had class here, but I had class in another building, so I am a little bit late because of that.
01:00:950Paolo Guiotto: Okay, so I asked you to do a couple of exercises. Exercise… 181… I hope you try them.
01:13:510Paolo Guiotto: these two.
01:15:270Paolo Guiotto: In any case, when I will give you exercises to do, maybe the class after, or…
01:22:940Paolo Guiotto: in a couple of times, I will return on these exercises, I will do solutions in class, but probably not of all the exercises. In any case, I will publish solutions on Moodle, okay? So there will be a file that I will publish somewhere today.
01:40:260Paolo Guiotto: A unique file where I will add, from time to time, solutions of the assignment I give you to do.
01:47:480Paolo Guiotto: Okay, so let's see this one. By using the definition of norm, approve the parallelogram identity, which is the following, norm of X plus Y.
02:00:50Paolo Guiotto: square, plus… Norm of X minus Y.
02:06:60Paolo Guiotto: Squared equal to… that multiplies norm of X squared plus norm of y.
02:15:240Paolo Guiotto: square.
02:16:860Paolo Guiotto: You know why this is called the parallelogram identity?
02:21:810Paolo Guiotto: The idea is that if you use the agreement to represent vectors with arrows, Like this.
02:30:290Paolo Guiotto: Now, X plus Y is… the geometrical sum is this thing. I told you last time that you can represent in this way.
02:40:660Paolo Guiotto: This is the vector X plus Y.
02:46:700Paolo Guiotto: And, X minus Y, well.
02:49:330Audio shared by Paolo Guiotto: Minus?
02:49:730Paolo Guiotto: Y is just Y,
02:53:60Paolo Guiotto: reversed the… so this is minus Y. So X minus Y is like X plus minus Y, so it is exactly this other, say, diagonal.
03:06:720Paolo Guiotto: So this is X… Minus Y, which is, by the way, this one.
03:13:290Paolo Guiotto: It's the same of this.
03:15:380Paolo Guiotto: So this is X… minus Y.
03:19:230Paolo Guiotto: So, as you can see, this figure here…
03:24:270Paolo Guiotto: is a parallelogram where X plus Y and X minus 1 are the two diagonals, X and Y are the two sides. So that formula says that the sum of the squares of the diagonals is two times the sum of the squares of the sides. That is called parallelogram identity.
03:43:560Paolo Guiotto: This is just for the geometrical meaning of this name. However, to prove the identity, so solution, Well, we…
04:17:190Paolo Guiotto: the sum of XJYJ, which is also known as the scalar product of vector X with vector Y. This, with other notation, would be denoted by X
04:30:190Paolo Guiotto: dot Y.
04:32:580Paolo Guiotto: And it is called this color… product.
04:39:370Paolo Guiotto: off… X.
04:42:540Paolo Guiotto: And… Why?
04:45:690Paolo Guiotto: Now, we don't need any particular, information about this. Now, take this identity and rewrite with… for X minus Y.
04:56:950Paolo Guiotto: So basically, you replace Y with the minus Y. What happens? We get norm of X squared.
05:04:890Paolo Guiotto: plus norm of minus Y squared.
05:08:540Paolo Guiotto: plus 2 sum XJ. Now, when I have vector minus Y, it has components minus YJ, so I have times minus YJ.
05:19:540Paolo Guiotto: Now, this, remains norm of X squared, plus the norm of minus Y is…
05:29:890Paolo Guiotto: the norm of Y, you know? The length,
05:32:950Paolo Guiotto: of the two vectors are the same. And here, we can take out that minus from the sum, and we write minus sum XJYJ.
05:44:110Paolo Guiotto: Now, take these two identities, 1 and 2, and SUM.
05:50:310Paolo Guiotto: what you get is the conclusion, because if you do the sum, you get norm X plus Y,
05:58:140Paolo Guiotto: squared plus norm of X minus Y squared equal… you see that we have norm of X squared, norm of X squared, this gives 2 norm of X
06:10:380Paolo Guiotto: square.
06:11:910Paolo Guiotto: Again, we have norm of Y. A norm of Y, so 2 norm of Y.
06:17:760Paolo Guiotto: squared, and then we have a plus the sum of XJYJ minus the sum XJYZ is 0. And that's the conclusion.
06:28:770Paolo Guiotto: Okay, the second exercise, 1A2, Asks you to verify
06:37:540Paolo Guiotto: This is also something which is, Let's say, useful to know.
06:45:120Paolo Guiotto: Maybe not now, but in the future.
06:48:150Paolo Guiotto: In more advanced courses, you will encounter these kind of ideas. So, the fact that the length of a vector is not defined in a unique way can be defined in many ways, okay? You may change the way to compute lengths.
07:05:530Paolo Guiotto: For example, here we have two proposals.
07:09:90Paolo Guiotto: One says that to define the length of vector X, we use this subindex 1, as the sum of the modulus of the coordinates.
07:21:900Paolo Guiotto: J goes from 1 to M, if M is there, or D, sorry.
07:27:300Paolo Guiotto: Use the letter D.
07:29:580Paolo Guiotto: If the vector X is in RD.
07:33:970Paolo Guiotto: For your information, this is called also Manhattan…
07:40:620Paolo Guiotto: Norma.
07:42:200Paolo Guiotto: You will see why in a second. And a second definition is the following. The infinity norm, which is the maximum of the modulus of DXJ when J range from 1 to D, so the components of the vector.
08:00:70Paolo Guiotto: These are two different ways to compute lengths. If you want, why that is called Manhattan… Manhattan Norm, because imagine you are in the plane.
08:13:30Paolo Guiotto: And you represent the town of Manhattan. As typical American cities, they have this orthogonal system of streets, so they maybe are like that, no?
08:25:260Paolo Guiotto: In Manhattan, there is only the Broadway that cuts in diagonal.
08:29:390Paolo Guiotto: So, suppose that you are here, in this point, say, the origin, and you want to go, I don't know, like, there.
08:39:360Paolo Guiotto: Of course, you could measure distances by taking the geometrical distance, but if this must be the distance that you needed to do to work from yellow point to green point, that distance is going to be not the distance you have to do.
08:54:250Paolo Guiotto: But instead, you will have to do something like this. You go stride this way, and then you go stride that way, or a combination of other things. The result is always the same.
09:05:720Paolo Guiotto: Now, in this way, you see that you are measuring if this is the point of coordinates, X1, X2, this means that this is X1 and this is X2.
09:17:60Paolo Guiotto: So that, the length of this green path would be X1 plus X2.
09:23:360Paolo Guiotto: Now, if the point is at left, X1 is negative here, so you cannot say X1 plus X2, because X1 is not the distance, it is just the abscessa, so you take the absolute value.
09:34:310Paolo Guiotto: Okay? And you get that formula. So, in general, if the point has D components, you sum the absolute values of the components, and you get a definition.
09:45:680Paolo Guiotto: Instead, this one takes the maximum of the distances, so it is like if you use this as distance to the audience, this side, okay? This is another choice.
09:57:190Paolo Guiotto: Now… Apart for this interpretation, you have to check that these are… Check.
10:07:220Paolo Guiotto: that, they… Verify… the… Same.
10:20:330Paolo Guiotto: properties.
10:23:220Paolo Guiotto: of the… Euclidean… Nor.
10:30:440Paolo Guiotto: So, for example, I will do for the case of one norm. So, if you have not yet done, please try to do, following the same kind of work you have to do for the infinity norm.
10:42:470Paolo Guiotto: So… Let's do… for… the one… Norma.
10:55:490Paolo Guiotto: So we have to check the four properties are positivity.
11:03:770Paolo Guiotto: Then we have a vanishing.
11:09:110Paolo Guiotto: Then we have homogeneity.
11:14:150Paolo Guiotto: And finally, the triangular inequality.
11:22:160Paolo Guiotto: Okay? So positivity means that we have to verify that the norm of vector X is always greater, well, then 0 for every X in RD.
11:34:760Paolo Guiotto: Okay? Now, this comes immediate… immediately from the definition, because what is this norm is the sum of the absolute values of the coordinates. So, since this is,
11:47:840Paolo Guiotto: Well…
11:50:150Paolo Guiotto: That's right. Since, the norm of X.
11:55:400Paolo Guiotto: The one norm is this quantity, sum modulus XJ, this is clearly positive, because it is the sum of positive quantities.
12:04:950Paolo Guiotto: Well, let's do down here.
12:08:20Paolo Guiotto: Okay, so vanishing.
12:13:510Paolo Guiotto: So here we have to check that norm of X.
12:17:460Paolo Guiotto: in this one norm is 0, if and only if, that's the point, vector X is vector 0. Now, start from this. What does it mean that norm of X
12:29:910Paolo Guiotto: is 0. Here, this zero is not the vector 0, it's the number 0. Norm is a number, numerical value. So this means that this sum,
12:43:00Paolo Guiotto: J goes 1 to D of absolute values of XJ is 0.
12:48:980Paolo Guiotto: Once again, we see, we have seen Argent like that.
12:51:980Paolo Guiotto: This is a sum of positive quantities.
12:55:460Paolo Guiotto: So, to be zero, it must be that all the quantities are equal to zero, otherwise that sum would be strictly positive. So this is that modulus XJ equals 0 for every J,
13:08:820Paolo Guiotto: And since modulus is 0 if and not if XJ is 0, this means XJ equals 0 for every j. This means that your vector, X, is the vector with all components equal 0, and that's what we call 0.
13:22:550Paolo Guiotto: So this is the vanishing.
13:25:90Paolo Guiotto: Now, let's go to the homogeneity.
13:30:360Paolo Guiotto: We have to check that the norm of lambda X The one normal.
13:36:770Paolo Guiotto: Here, lambda is a scalar, X is a vector, E is equal to modulus of lambda times the norm of X.
13:43:890Paolo Guiotto: for every lambda scholar, so real.
13:47:540Paolo Guiotto: And for every X vector.
13:50:600Paolo Guiotto: RD.
13:53:110Paolo Guiotto: Okay. Again, we start from the definition. These are straightforward arguments. Do you see there is no idea behind this?
14:01:720Paolo Guiotto: And this was also for the Euclidean norm. By the way, up to this point, the Euclidean norm becomes difficult for the triangular inequality. This is a general fact. Normally, what is difficult is the triangular inequality.
14:17:340Paolo Guiotto: And sometimes it could be difficult even the vanishing. Even the vanishing, as you can see, demands a little bit of discussion, okay?
14:25:330Paolo Guiotto: Now, here we just apply the definition. This is sum over J. What are the components of lambda X?
14:31:200Paolo Guiotto: We know that lambda times X means that we multiply by lambda all the components, so this will be absolute value of lambda XJ.
14:40:50Paolo Guiotto: Now, we have to use something here. Modulus of a product, we know that this is the product of the moduluses, so we have this.
14:48:440Paolo Guiotto: Now, you see that there is a constant factor that multiplies all the XJ. So use distributivity, and you put outside the… you factorize the lambda, you get this modulus of lambda times sum of j modulus XJ. But the quantity that you see here is exactly norm of X in,
15:07:440Paolo Guiotto: One norm of X, by definition, and that's the conclusion.
15:12:880Paolo Guiotto: And finally, the triangular inequality.
15:16:260Paolo Guiotto: Which, in this case, is actually much easier than the case of the Euclidean norm. It's easy, yeah. Because we start, so the triangular inequality is this.
15:32:360Paolo Guiotto: We have to prove that the one norm of the sum is less or equal than the sum of the one norms.
15:43:500Paolo Guiotto: Okay, so this is the goal, to be clear, is what we have to prove.
15:48:690Paolo Guiotto: It is not what we know, okay? So we start…
15:53:280Paolo Guiotto: Let's say in a naive way, let's start computing this.
15:58:70Paolo Guiotto: Okay, this is, so the norm of X plus Y
16:04:660Paolo Guiotto: One norm. So the one norm takes the sum of the absolute values of the components of the vector.
16:12:740Paolo Guiotto: Our vector is X plus Y.
16:15:400Paolo Guiotto: how the sum works.
16:17:720Paolo Guiotto: you sum components by component. So the jth component of X plus Y is the sum of the jth components of X plus that one of Y. So it is XJ plus YJ.
16:29:840Paolo Guiotto: Okay, so, now what can we do? The modules of the sum is not definitely the sum of the models, so we cannot use something like this.
16:38:430Paolo Guiotto: We know that the modulus of the sum, let's say, for any two numbers, also the modulus fulfilled the triangular inequality, so this is going to be less than modulus of A plus modulus of B.
16:50:580Paolo Guiotto: So, we can say that if we apply this to these quantities, I will have that modulus of XJ plus YJ
16:58:180Paolo Guiotto: will be less or equal than modulus XJ plus modulus YJ.
17:03:780Paolo Guiotto: Here I have the sum.
17:05:480Paolo Guiotto: So if I now sum all these inequalities, it remains an inequality. Because if I sum numbers, modulus XJ plus YJ,
17:16:310Paolo Guiotto: Since each of these numbers is less than the right-hand side there, when I sum the right-hand sides, so modulus XJ plus modulus yj.
17:27:10Paolo Guiotto: I get something bigger.
17:29:400Paolo Guiotto: Okay? Each of the numbers at the left is less or equal than each of the numbers at light, so the sum will be less or equal than the sun.
17:36:820Paolo Guiotto: So I got this.
17:38:140Paolo Guiotto: Now, this is more or less the conclusion, because at left, you have the one norm, okay? So this is, in other words, X plus Y.
17:49:580Paolo Guiotto: one norm of X plus Y. Less or equal than. Now, we have sum of modulus XJ plus some modulus YJ. I can split this into sum modulus XJ plus sum of modulus YJ. And you recognize that each of these two is a norm. This is the norm of X,
18:08:210Paolo Guiotto: In one sense, and this is the norm of Y still in Y sense.
18:13:680Paolo Guiotto: And so, plus the 2. You have the lesser equal, you are done.
18:18:820Paolo Guiotto: Okay, so you do the same, if you have not yet done, for the, infinity norm.
18:27:960Paolo Guiotto: Okay.
18:29:630Paolo Guiotto: These, were just the two exercises I left you to do. Now we return back to, the… no, we had class… no class yesterday, right? We had class on Wednesday.
18:41:450Paolo Guiotto: So we return to what we…
18:44:960Paolo Guiotto: What was the conclusion of Wednesday's class?
18:49:970Paolo Guiotto: So we concluded the class. Let me go back here.
18:54:140Paolo Guiotto: With this definition of limit, okay?
18:57:300Paolo Guiotto: So we introduced a norm of vectors, a way to measure length of vectors, because we needed to define the operation of limit. At this point, it's just for limits of sequences, but soon we will extend it to the limits of functions, which is the important case here.
19:17:310Paolo Guiotto: So, let's just restart from this point.
19:23:630Paolo Guiotto: And, just, rewrite the definition. So, we introduced the…
19:34:10Paolo Guiotto: the… definition.
19:37:790Paolo Guiotto: of limit…
19:43:930Paolo Guiotto: limit of… a sequence… of… vectors…
19:54:120Paolo Guiotto: So we say that that is sequence XN of vectors in RD.
20:01:320Paolo Guiotto: There exists the limit of this sequence when n goes to infinity.
20:07:610Paolo Guiotto: And, of course, this is, intuition, if we are thinking to R2, these are points, R2 moving somewhere. If they go anywhere, that anywhere will be an object of the same nature. It will be a vector of R2, not a vector of R1 or R4. So it will be an L that belongs to RD.
20:27:860Paolo Guiotto: If and all if, and formally, we have this property. For every epsilon positive, there exists an initial index capital N, such that the distance between Xn and L
20:38:200Paolo Guiotto: gets smaller than epsilon for every n larger than capital n. That's the formal definition.
20:45:600Paolo Guiotto: Here we can do, before we start launching into examples, we do an important remark.
20:55:850Paolo Guiotto: Which is the following.
20:58:220Paolo Guiotto: If we notice carefully this definition.
21:02:430Paolo Guiotto: Go for a moment, these quantities, let's call them delta N.
21:09:990Paolo Guiotto: N, because that depends on n. Here, n is an important variable. These quantities are what kind of object? Vectors or numbers?
21:24:10Paolo Guiotto: It's a norm, it's a number, okay? So, the sequence delta N, if I call it delta N norm of XN minus L,
21:36:670Paolo Guiotto: This is a sequence of numbers, so the sequence delta N, is a sequence
21:45:850Paolo Guiotto: Of, actually, not just any number, because they are positive, Positive, Numbers.
21:57:770Paolo Guiotto: Now, if we look at that property, which is written in red, a box here with yellow.
22:05:790Paolo Guiotto: And we just rewrite with delta N, we see this. So, the sequence limit…
22:14:620Paolo Guiotto: In n of Xn is equal to F,
22:19:780Paolo Guiotto: If not if, for every epsilon positive, there exists an initial index n such that delta N, which is that difference, the distance.
22:30:350Paolo Guiotto: is less or equal than epsilon. Remind also that it is positive, so it's not written above, because it's, how to say, it's implicit. A norm is positive.
22:41:770Paolo Guiotto: Okay, so what I add here…
22:44:90Paolo Guiotto: For every n larger or equal than capital N.
22:48:430Paolo Guiotto: Now, if you look at this condition, since delta N is positive, this means also that the modulus of delta N is less than epsilon for every n larger than capital N.
23:00:690Paolo Guiotto: But now, forget of our story of vectors and limits of vectors, let's look at just this thing.
23:09:160Paolo Guiotto: What this thing says…
23:11:170Paolo Guiotto: For every epsilon, there exists initial index such that the absolute value of delta n less than epsilon for every n larger than.
23:19:140Paolo Guiotto: Well, we can see this as the distance between number delta n and 0 is less than epsilon.
23:27:220Paolo Guiotto: And written like that, it says that…
23:33:340Paolo Guiotto: Says nothing.
23:36:520Paolo Guiotto: Of course, it says something. What, what?
23:43:510Paolo Guiotto: the sequence delta N,
23:53:800Paolo Guiotto: So we are doing… which course are we doing? Calculus.
23:57:630Paolo Guiotto: No? Analysis, which is based on… Limits, right?
24:05:40Paolo Guiotto: If you don't have a limit, you don't have a derivative, you don't have an integral, you don't have a fucking nothing, okay? So, limits is the key word that you must have in mind.
24:14:220Paolo Guiotto: What is saying this?
24:17:970Paolo Guiotto: That delta N?
24:21:10Paolo Guiotto: No, delta N is not the limit. Who is the limit?
24:26:180Paolo Guiotto: For every epsilon positive, the distance between delta N and 0 is getting small, provided n is large enough. This is saying that delta N is going to zero. Oh, that's equivalent.
24:38:680Paolo Guiotto: So, delta N…
24:41:480Paolo Guiotto: the limit in n of delta N, which is not the same as the other limit, because the other limit is the limit of vectors, this is the limit of numbers, is 0.
24:51:730Paolo Guiotto: So, in fact, since now let's restore the notation.
24:56:720Paolo Guiotto: So this means limit in N of norm Xn minus L
25:04:750Paolo Guiotto: equals zero. So, we have… this remark says that the limit
25:10:350Paolo Guiotto: In n of the sequence Xn of arrays, vectors is L, if and only if.
25:19:390Paolo Guiotto: LRD, let's put this, because…
25:22:450Paolo Guiotto: Today, we will also introduce the infinite limit, we will see. If and only if the limit in n of norm of XN
25:32:120Paolo Guiotto: minus L… is zero.
25:36:110Paolo Guiotto: So…
25:38:240Paolo Guiotto: what is the interest in this remark is that if you want to check that the limit of the sequence of vectors, Xn is L,
25:48:60Paolo Guiotto: You just have to do… this is a practical instruction, you have to do compute the norm of XN minus L. That's… it's now a number, because norm is a number. And then do the limiting N of that quantity.
26:04:300Paolo Guiotto: That quantity is a numerical stuff, so it's something that you have learned last year.
26:09:160Paolo Guiotto: So, in principle, this is how to reduce this problem to a problem that you already know how to discuss.
26:15:730Paolo Guiotto: You see? Now I will be showing examples, in practice, how it works. Okay, so let's see examples.
26:26:950Paolo Guiotto: example, this is a one-star example. So, one star means very easy, calculation's easy, so we don't have to focus on technical difficulties, but focus on the, on…
26:42:680Paolo Guiotto: on the ideas, no? So the question is, I show that… That,
26:50:790Paolo Guiotto: the limit in n of the sequence 1 over n, n plus 1 over n is 01.
27:04:620Paolo Guiotto: Of course, here we are in R2, you see? There are vectors with the two coordinates. We can have even an idea of what is going on. In general, it's hard to do figures here, because
27:19:360Paolo Guiotto: It's difficult to do… if the dimension is 3, it's very difficult if the dimension is 4 is impossible, okay? Dimension 2 is relatively easy.
27:29:910Paolo Guiotto: So, here we are in R2.
27:34:80Paolo Guiotto: We have our sequence of arrays, which is this one. Of course, the sequence will start from index 1, index 0. That's not…
27:42:590Paolo Guiotto: make any sense. So, for example, this means that X1 is the array 1 over 1, and the other is 1 half, no, sorry, 2.
27:56:440Paolo Guiotto: Right?
27:57:530Paolo Guiotto: X2 is… so let's see, what is it? Abshysa 1, ordinate 2, so let's say that this is,
28:06:620Paolo Guiotto: 1, and this is 2, our point, X1, is here.
28:13:120Paolo Guiotto: Okay? You see?
28:15:740Paolo Guiotto: Now, X2 is what? It is 1 half, and then we have 3 halves.
28:23:330Paolo Guiotto: So, it missed the app, she says, at middle here, and the ordinate is 3 half, so let's say that this is 1.
28:31:690Paolo Guiotto: 3 half is about here, so let's say this is the point X2.
28:39:740Paolo Guiotto: Then we have X3. Of course, I cannot write down all points. There is the formula there. But let's see the first few examples, just to
28:49:440Paolo Guiotto: Have an idea. 1 3rd, and then we have, 4 thirds here.
28:54:650Paolo Guiotto: Now, one-third is a bit, so this is… this was one half, one-third is about here, well, this is…
29:01:110Paolo Guiotto: More or less here.
29:02:820Paolo Guiotto: So you see that the Abshesa is moving to the left.
29:07:430Paolo Guiotto: And the ordinate 4 thirds is a little bit less than 3 halves.
29:11:810Paolo Guiotto: So we are, let's say, about, yeah, still above 1.
29:16:30Paolo Guiotto: But let's say that X3 is here.
29:19:670Paolo Guiotto: So, as you can see, these are points moving, and I have infinitely many points, and it says they go to 0, 1.
29:27:350Paolo Guiotto: So, abhisha 0, ordinate 1. Who is this? This is the boy.
29:32:720Paolo Guiotto: So, of course, the figure has some… a certain degree of approximation, but it's like if we move… imagine that we see a lamp switching on at each second, we see this lamp moving and approaching that green point. So this is what he's saying.
29:50:670Paolo Guiotto: Okay, now, to verify this, So, solution.
29:57:160Paolo Guiotto: To verify this, I just apply what I learned here. I compute the norm of XN minus the candidate limit, L, which is given here. This is L, so…
30:09:100Paolo Guiotto: And then I check if the distance goes to zero. So let's start first computing the… well, let's compute first the distance, XN minus L.
30:19:360Paolo Guiotto: the difference. Xn minus L is… so the vector Xn is 1 over N. N plus 1 over n, this is… these are the two components, minus L, which is 0, 1.
30:34:40Paolo Guiotto: So, you know that we do the sum difference components by component. So, 1 over n minus 0, n plus 1 over n minus 1. These are the two components. If you want, maybe it is convenient to split this as 1 plus 1 over n, you can cancel the ones.
30:55:340Paolo Guiotto: Here… this with this. And so what remains is the vector 1 over n, 1 over n.
31:04:780Paolo Guiotto: But you say that this was a scalar. No, this is a vector. I'm doing difference of vectors. The scalar is the norm of this quantity, okay? So now, we computed the norm of XN minus L,
31:19:320Paolo Guiotto: So it is the norm of the vector 1 over n, 1 over n.
31:25:190Paolo Guiotto: And now we use the Euclidean stuff, no? So this is the root of.
31:30:160Paolo Guiotto: square of the first component, so 1 over n power 2, so it means 1 over n squared, plus square of the second component, another 1 over n squared.
31:41:70Paolo Guiotto: So it gives 2 over n squared. If we want, we can slightly simplify this thing, saying that it is root of 2 divided the root of n squared, which is…
31:55:320Paolo Guiotto: Now, in this case, we can write N, because we know that N
31:59:390Paolo Guiotto: is a natural greater than positive, right. Okay, so now we have that the norm of XN minus L is root of 2 divided by N.
32:11:780Paolo Guiotto: And what do we do with this?
32:13:690Paolo Guiotto: Well, what we do is now verify if the limit in n of this thing, which is a sequence of numbers, is no more a sequence of vectors. So, let's compute the limit in N.
32:25:260Paolo Guiotto: Of this sequence of norms.
32:27:910Paolo Guiotto: Xn minus L…
32:30:370Paolo Guiotto: We have to do the limit in n, of course, when n goes to plus infinity, blah blah blah, root of 2 divided to n, which is clearly equal to
32:39:790Paolo Guiotto: 0. So, as you can see, this is verified, and therefore, the other conclusion holds. So, this means that we can conclude that there exists the limit
32:53:910Paolo Guiotto: Inan.
32:55:70Paolo Guiotto: going to plus infinity of our sequence 1 over n.
32:59:290Paolo Guiotto: N plus 1 over n, and that limit is exactly vector 0, 1, as plate.
33:08:360Paolo Guiotto: Okay?
33:10:320Paolo Guiotto: Do you have any questions?
33:15:430Paolo Guiotto: So, you see, it's… this check is apparently simple. Compute the norm of the difference, and then do the limit.
33:30:230Paolo Guiotto: Now, what is the problem?
33:37:90Paolo Guiotto: Where is the problem with this procedure? You say, there is no problem, of course, yes, there is no problem, but there is a problem.
33:46:470Paolo Guiotto: Yes, it may happen, but that's not exactly what they mean.
33:50:660Paolo Guiotto: This procedure To be applied, demands that you already know what the limit.
34:00:640Paolo Guiotto: So normally, if you think about whenever you add the two compute limits in the first year, maybe you could have
34:07:860Paolo Guiotto: Mmm.
34:09:210Paolo Guiotto: you could think that these are stupid and nonsense problems. Why should I compute thousands of limits? What is the interest? Of course, you are learning a technique, limits are important for many applications, but the key point is that you always compute a limit of something that you don't know if it has the limit and what is the limit.
34:29:710Paolo Guiotto: And you never check the limit, knowing already the answer.
34:34:280Paolo Guiotto: Let's say that that would be too easy. In the real world, you never know if the limit exists, and even if the limit exists, what is the value of the limit? So the problem with this check is that you need to have the candidate L, otherwise what vector put… you can put there.
34:53:170Paolo Guiotto: Now, here, from this example, for example, we can learn something interesting, because look at what happens.
35:00:640Paolo Guiotto: If you focus on this simple example.
35:05:250Paolo Guiotto: You see that if you look at the first coordinate, this is the sequence 1 over n, that goes to
35:12:450Paolo Guiotto: Zero.
35:13:900Paolo Guiotto: So, in fact, you see that the first coordinate goes exactly to the first coordinate of the limit.
35:19:570Paolo Guiotto: And the second coordinate of the vector is n plus 1 over n. If you remember, we can write this as 1 plus 1 over n, where you see that this goes to
35:32:240Paolo Guiotto: 1… one of them goes to zero, but 1 plus goes to 1. And this is exactly the second company.
35:40:710Paolo Guiotto: So, as you can see, what is happening here is that the first component
35:45:830Paolo Guiotto: Go to the first component of the limit.
35:49:120Paolo Guiotto: The second component goes to the second component of the limit. So this, in some sense, could suggest that maybe we could compute the limit of vectors by doing limit component by component. Is that correct or not? Now, let's see this.
36:05:970Paolo Guiotto: So, this is a remarker, important remark.
36:13:620Paolo Guiotto: So… Let's try to formalize. Now, this can become a little bit complicated, because… let's take a sequence XN.
36:25:960Paolo Guiotto: of vectors.
36:27:640Paolo Guiotto: Each Xn is a vector. It means that it will have a certain number of D components.
36:34:740Paolo Guiotto: Now, how can we write the components? Because components, they deserve an index. Now, you write X1, X2, XD. But there is already an index there, XN. So it means that XN, as the component, they should be XN1, XN2, XND.
36:51:310Paolo Guiotto: You see? So you have to accept that we need the two indexes, in this case. So these components will be XN1, XN2, XND. So these are the components
37:08:720Paolo Guiotto: of vector Xn.
37:12:310Paolo Guiotto: So the two indexes have different, entirely different meaning, because this second one says in which position of the array you are, no?
37:21:780Paolo Guiotto: While the first one tells you in which position of the sequence you are.
37:26:720Paolo Guiotto: So, X1 million one is the first component of the 1 millionth vector of the sequence, okay? So they are different. The second index ranges from 1 to D, the first one ranges from 0 to plus infinity, so they are entirely different. You cannot confuse the two.
37:45:750Paolo Guiotto: But now that we have this, huh?
37:49:270Paolo Guiotto: And the limit, L,
37:51:840Paolo Guiotto: has itself components that we can call… here, there is not a problem with the double index, L1, LD.
38:00:420Paolo Guiotto: Now, can we say that…
38:08:290Paolo Guiotto: that, the… Vector Xn.
38:13:810Paolo Guiotto: I now use this notation with the arrow, that means,
38:18:550Paolo Guiotto: This is a short link, or writing limit in n of,
38:24:00Paolo Guiotto: Xn equal L, or it is equivalent, an equivalent notation.
38:30:100Paolo Guiotto: Can I say that, XN goes to L if and only if…
38:36:650Paolo Guiotto: So how can we write that property that was suggested by the example?
38:41:360Paolo Guiotto: The first com- the first component, the sequence of the first components, the RX and 1,
38:49:870Paolo Guiotto: So when I go with n to infinity, this is a sequence of numbers, because they are components, goes to the first component of L.
38:59:530Paolo Guiotto: This sequence of second components goes to the second component of L, and so on. The last component, D component, the sequence of D components goes to the D components of L.
39:13:680Paolo Guiotto: Is that true?
39:15:640Paolo Guiotto: The answer is yes, provided everything is fine. So, XN goes to limit L in RD if and only if the first component goes to limit L1 in R, limit L2 in R, limit LD in R.
39:32:260Paolo Guiotto: Okay?
39:36:350Paolo Guiotto: Well, I don't want to prove the proof, it's just brighter definitions.
39:42:810Paolo Guiotto: probably it is an exercise, I don't know, don't remember what I wrote here, but let's use this, okay?
39:51:170Paolo Guiotto: So, example.
39:54:110Paolo Guiotto: Here, I'm taking the exercise 183.
39:59:450Paolo Guiotto: And, number one, So, here we have one star. This means very technically, technically, very simple.
40:07:810Paolo Guiotto: and focus on the basics, the definitions, the first property we have seen. The first example is, take this sequence XN equal E to minus N comma 1.
40:22:650Paolo Guiotto: So we are in R2.
40:27:70Paolo Guiotto: Okay? So, it will converge if it converges to something of R2. So it won't converge to a number, this thing. It converts to a vector with two components, if it converges. Now.
40:41:280Paolo Guiotto: We don't have it here, so the exercise… the problem is, is there a limit for this XN, and what is the limit?
40:50:480Paolo Guiotto: So you see that here, I do not have the limit. It's different from previous exercise, where you had just to check that that vector 01 is the limit there. Here, we don't know what is the limit, even if the limit exists.
41:05:630Paolo Guiotto: So, the question is a little bit more open, so we have to think a little bit more. But as we understood here, we could attack this problem by looking what happens to the components. So, what are the components? Yeah, I could say…
41:22:40Paolo Guiotto: If you want DXN1, the sequence of the first component is this one.
41:28:840Paolo Guiotto: is the sequence E2 minus n.
41:32:310Paolo Guiotto: What happens to this sequence of numbers when n goes to plus infinity?
41:38:00Paolo Guiotto: It goes to… to 1, Y1?
41:44:710Paolo Guiotto: We put 0 and 1.
41:47:230Paolo Guiotto: numbers, if… Yeah, but let's think. N is going to plus infinity. Minus n is going to… Minus infinity.
41:58:220Paolo Guiotto: It exponential at minus infinity is… Is what?
42:06:780Paolo Guiotto: 2?
42:09:50Paolo Guiotto: Okay, so let's look at this way. This is also 1 over e to n. Is it correct?
42:16:360Paolo Guiotto: Okay. E to n goes to E2N, the denominator there. This goes to…
42:24:460Paolo Guiotto: When n goes to plus infinity, That's a lot of material.
42:30:300Paolo Guiotto: And now, you have to review the exponential, because I'm a bit confused. This goes to plus infinity, and therefore this fraction goes to zero, so the limit is 0.
42:39:840Paolo Guiotto: The second… the sequence of second components is constantly equal to 1.
42:46:230Paolo Guiotto: So, it's a constant sequence, it will go to the constant value 1. So, as you can see, we have that there is no third component, because there are only two.
42:55:470Paolo Guiotto: We are exactly in this case, no? The sequence of the first components converges to SAML1 finite. The sequence of the second components converges to SAML2 finite. Therefore, we can say that…
43:08:830Paolo Guiotto: The sequence of the vectors XN will converge to the vector made up of these two components, 0, 1.
43:18:370Paolo Guiotto: Okay? So this is nice, because it is saying that at least that level of sequences of vectors, basically we can do the limit component by component. We will use this terminology, component-wise, okay?
43:33:880Paolo Guiotto: But be careful, because the story is a bit more complicated, and when we will go to limits of functions would be entirely different, but we will return later.
43:43:480Paolo Guiotto: Okay, so this was the example.
43:47:590Paolo Guiotto: For example, the number 3… Okay?
43:54:260Paolo Guiotto: Let's do still this one. Here we have this sequence. Xn is the sequence made of vector 1 over n, 1 over n squared, and third component, sine of 1 over n.
44:12:180Paolo Guiotto: So here you see there are three components, so we are in R3.
44:19:420Paolo Guiotto: I don't do a figure, I don't need… it's too complicated, because I have three axes, the figure could be a little bit complicated, but…
44:30:220Paolo Guiotto: Here we can say that if you look at the first components, the sequence of first components is 1 over N. That clearly goes to 0. Second components.
44:42:650Paolo Guiotto: the, the sequence is 1 over n squared that clearly goes to 0. The third component is…
44:51:470Paolo Guiotto: sine of 1 over n. And this goes to…
44:55:920Paolo Guiotto: 0, because the argument goes to zero, sine is continuous, so sine of 0, which is 0. So, we are exactly in this situation. Each of the components has a finite limit, and therefore we conclude that the full vector, XN
45:14:610Paolo Guiotto: Goes to the vector equal to zero, which all the components are zero.
45:21:60Paolo Guiotto: Okay.
45:28:560Paolo Guiotto: Yes, just before we do the break,
45:34:410Paolo Guiotto: Let's return a bit here.
45:37:140Paolo Guiotto: So this says array Xn goes to limit L in RD, even though if this happens.
45:43:680Paolo Guiotto: So, in particular, how do I interpret the case when I have the first component goes to limit L1 finite, the second component has not a limit?
45:54:890Paolo Guiotto: Can I say that that vector will have a limit?
45:58:650Paolo Guiotto: No, because A is if and only if.
46:01:810Paolo Guiotto: To have a limit, you must have limits for each of the components. If one of the components, just one, has not a limit, there is no limit for the vector. You see?
46:12:340Paolo Guiotto: So, for instance, if you do… this is not there, but, example, if you take, xn equal
46:22:430Paolo Guiotto: 1 over n, and I don't know, minus 1 to N,
46:28:680Paolo Guiotto: Okay, here you see that the first component, XN1, is 1 over n, and this is good because it goes to 0. But the second component, XN2, is the sequence minus 1 to N. So that sequence minus 1 plus 1 minus 1 plus 1 forever. And this has not a limit.
46:47:650Paolo Guiotto: So, since that is an if and only if, I can conclude that the sequence XN Has. No.
46:58:80Paolo Guiotto: limit.
47:01:00Paolo Guiotto: in… R2 in this case.
47:07:460Paolo Guiotto: Yeah.
47:10:290Paolo Guiotto: Does that ensure tobacco?
47:13:870Paolo Guiotto: It's just what we were saying. It's an if and don't live, okay?
47:19:70Paolo Guiotto: It has… be careful, because at this point, it seems clear, but later we will, in the next… after the break, we will introduce the infinite limiter, and this will change a bit the interpretation, but let's say that this
47:36:570Paolo Guiotto: remind this, XN has a finite limit vector, so a limit L in RD
47:45:560Paolo Guiotto: If and of if this happens.
47:47:820Paolo Guiotto: So, if this does not happen, because one of the components has troubles, this means that there is not a finite limit, but maybe there is some other limit. For example, for the infinite limit, it may happen, it is not an if and all if.
48:01:900Paolo Guiotto: But we will have to see what is infinite limit, because infinite limit, as you expect, even in R, infinity plus infinity minus infinity are not numbers. So the definition is different, it is a sort of other story.
48:14:880Paolo Guiotto: Okay, but we still consider it as a limit operation. It is still defined. Even here, there is a limit for xn going to infinity, but we have a little bit to think about what is the infinity of, for example, the plane or the space. Okay? Because the idea we have with plus infinity minus infinity is a bit different here.
48:35:100Paolo Guiotto: Okay, I would say we can take 5 minutes break, and then we continue, okay?
48:58:390Paolo Guiotto: Okay, please take your seats.
49:10:170Paolo Guiotto: Selected.
49:23:590Paolo Guiotto: So, this is an important definition.
49:28:490Paolo Guiotto: We want to define what does it mean that a sequence of vectors, Xn, goes to infinity.
49:36:300Paolo Guiotto: Now, here, the problem is that the…
49:39:290Paolo Guiotto: Of course, you are used with plus infinity and minus infinity in the real line.
49:45:450Paolo Guiotto: Because in the real line, there is a left and right, so you know that, down here, let's say roughly like that, there is plus infinity, and at the left, there is minus infinity.
49:56:810Paolo Guiotto: So, basically, on the rear line, we have two infinities.
50:00:580Paolo Guiotto: But even if you think in the case, the simplest case of a dimension greater than 1, the plane, there is not a plus infinity. You may still continue to think that this is minus, plus infinity and this is minus infinity. But what is this?
50:16:920Paolo Guiotto: Or what is this? Or what is this? Maybe you can go there, No?
50:22:860Paolo Guiotto: Infinity, in some sense, means that the quantity becomes big. That's the idea, no?
50:30:250Paolo Guiotto: Now, the idea we use here actually simplifies a bit the story between plus infinity minus infinity, because at the end, it does not distinguish between the two. So we say that… we will talk about the unique infinity, in fact.
50:47:230Paolo Guiotto: We say that, huh?
50:53:740Paolo Guiotto: a sequence XN.
50:56:660Paolo Guiotto: of vectors of, let's return to general discussion, RD.
51:03:550Paolo Guiotto: Haas.
51:05:270Paolo Guiotto: infinite…
51:10:230Paolo Guiotto: limit.
51:13:60Paolo Guiotto: So we write limit when n goes to infinity of Xn. So we don't write plus infinity minus infinity, because this
51:23:130Paolo Guiotto: simply do not make sense. We just write infinity, and perhaps to remind that this is something like an infinity in the space of D, let's put a D here to remind that it's the infinity of D.
51:37:750Paolo Guiotto: Of course, it's not the array with all the entries equal to plus infinity or minus infinity, it's not that. It's not an array, it's not a point of RD.
51:46:860Paolo Guiotto: The definition is simple at the end. If the quantity norm of XN Goes to plus infinity.
51:58:120Paolo Guiotto: What does it mean that the norm of Xn goes to plus infinity? Remind that norm of a vector represents the length of the vector. So, you said that the sequence goes to infinity if the distance to the origin goes to plus infinity. So, if the limit
52:15:680Paolo Guiotto: in n of this is plus infinity.
52:19:440Paolo Guiotto: Be careful in the reading of this condition, because since this is a sequence of norms, these are numbers. So you are talking about the sequence of positive numbers that goes to plus infinity. It makes sense.
52:36:60Paolo Guiotto: It's not an error, because at the right, I should write an array. There is no array here, no? The array is inside here. I take the norm, it becomes a Scala, okay? And therefore, the definition is simple. You compute the norm of Xn and check if it goes to plus infinity.
52:55:650Paolo Guiotto: Or not. So, it is immediately operational. For example.
53:02:230Paolo Guiotto: Suppose I have the sequence 1 over N and N.
53:08:480Paolo Guiotto: So what do you think about this sequence?
53:11:660Paolo Guiotto: The first component goes to zero, so it's good.
53:16:100Paolo Guiotto: But the second goes to plus infinity, so we cannot say that each of the components has a finite limit, so we cannot say, or better, we can say that this sequence has not a limit in RD, in R2.
53:28:320Paolo Guiotto: Okay? If you want.
53:30:910Paolo Guiotto: We can say this. Since,
53:34:590Paolo Guiotto: 1 over N, the first component, XN1 is 1 over N, goes to 0.
53:40:590Paolo Guiotto: But… The sequence of second components, N, goes to plus infinity, We cannot say we… Cannot Say…
53:56:890Paolo Guiotto: Better, we can say, let's turn it in positive. We can say… We can.
54:05:350Paolo Guiotto: Say… that, there is no limit.
54:13:750Paolo Guiotto: of sequence Xena. Of course, this is the sequence Xena.
54:19:560Paolo Guiotto: in R2.
54:23:150Paolo Guiotto: Okay?
54:24:310Paolo Guiotto: Because this limit exists even only if coordinates have a finite limit.
54:29:690Paolo Guiotto: And the limits of the code, it makes an array, which is the limit array.
54:35:360Paolo Guiotto: However.
54:40:220Paolo Guiotto: Since, if we compute the norm of XN, This is the root of.
54:48:200Paolo Guiotto: Square of the first component, so 1 over n squared, plus square of the second, n squared.
54:56:190Paolo Guiotto: As you can see, this quantity, when you do the limit, it has a limit, because this goes to zero. But this goes to plus infinity. So, there is no indeterminate form here. Under the root, the quantity goes to plus infinity, the root will go to plus infinity as well.
55:14:340Paolo Guiotto: So you see?
55:15:820Paolo Guiotto: And therefore, we can say that XN goes to infinity
55:22:140Paolo Guiotto: In the plane. There is no contradiction with the previous statement, because the previous statement says there is no limit vector in R2. Infinity 2 is not a vector of R2.
55:33:360Paolo Guiotto: is even not a vector, it's an agreement. The agreement is a definition that the norm goes to plus infinity.
55:39:710Paolo Guiotto: Okay?
55:41:480Paolo Guiotto: Here, for example, you can have a little intuition, because if you look at these points, the points are 1 over NN. So, for example, X1 is the vector 1, 1.
55:55:690Paolo Guiotto: X2 is, 1 half 2.
56:00:590Paolo Guiotto: X3… is 1 3rd, 3, and so on.
56:06:930Paolo Guiotto: So you see that the first point is 11, so let's say that it is more or less here.
56:14:50Paolo Guiotto: 1, 1. Then the second is 1 half 2, so we have this, 1 half, and 2 here. So it is about here.
56:24:860Paolo Guiotto: The third point is 1 3rd 3. So, 3 is here, 1 3rd here, so let's say that this point is about here.
56:35:320Paolo Guiotto: Imagine that you join this with the line, and you see that this point is moving
56:42:410Paolo Guiotto: Upward is getting close to the y-axis and escaping, no? Upward in the plane.
56:51:160Paolo Guiotto: So in that way, it's going to infinity. Let me show you another interesting example.
56:57:540Paolo Guiotto: Because from this example, you may see, if you look at what happened.
57:03:910Paolo Guiotto: Okay, I cannot say that there is a finite limit, because one of the components, the second one, has not a finite limit. But since it has an infinite limit, the limit of the array will be infinite, infinite, no? I will go to infinite in the plane.
57:20:630Paolo Guiotto: So, probably, let's say… adapting this, we could say that Xn goes to infinity.
57:32:310Paolo Guiotto: If and only if at least one of the coordinates go to plus, minus infinity. That's wrong. Look at this example.
57:42:240Paolo Guiotto: So, the component-wise limit does not work for infinity, because this is the message. Look at this example.
57:50:10Paolo Guiotto: Take us back to our XN, this. N cosine n… N sine n.
58:00:730Paolo Guiotto: So where is this vector? It's not particularly complicated to understand where is it when n is fixed, because it's like…
58:11:160Paolo Guiotto: N times cos… N, sign, N.
58:19:840Paolo Guiotto: So, apart for n, the vector cosa and sine n belongs on a circle.
58:24:540Paolo Guiotto: centrade in the origins, radius 1. With angle n, well, we do not have a formula for sine and cosine of naturals.
58:32:930Paolo Guiotto: But, it doesn't matter, so we can have an idea. For n equal 1, you have X1 is cos 1 sine 1, 1 is between 0 and pi of, so let's say that X1 is about… if this is the unitary circle.
58:51:340Paolo Guiotto: So, let's say, to understand this is 1. Let's say that X1 Could be about here.
58:59:280Paolo Guiotto: X2 belongs to the circle of radius 2.
59:03:900Paolo Guiotto: Because it's 2 times a unitary vector, a vector of the circle of radius 1. You see that factor N in front. When you take 2, so X1 is…
59:15:900Paolo Guiotto: The vector cos 1 sine 1.
59:19:940Paolo Guiotto: So it belongs to the circle, centered at the origin radius 1. X2 is 2 cos 2 Sine 2.
59:29:500Paolo Guiotto: So, cos 2 sine 2 still belongs to the unitary circle.
59:34:370Paolo Guiotto: Two is, between pi over 2 and pi, you know?
59:40:670Paolo Guiotto: So, let's say that, more or less, we are talking about an angle of this type. So, the point cos 2 sine 2 could be here. When I multiply by 2, it means that I'm now moving on the circle of radius 2.
59:57:760Paolo Guiotto: So let's say that this is 2, so my second point could be actually here.
00:03:100Paolo Guiotto: X2.
00:05:180Paolo Guiotto: Now, X3 is where? X3 is 3 times cos 3, Sign 3.
00:13:750Paolo Guiotto: Now, cos 3, sine 3, is still on the unitary circle.
00:17:950Paolo Guiotto: 3 is a little bit less than 3.14, also, it's less than pi, and between pi, half, and pi. So we are still in the second quarter, so let's say that maybe we are very close to pi, so like, like here. So the black point is cost 3 signed 3, okay?
00:34:800Paolo Guiotto: Now, I multiply by 3, so I make it 3 times longer, so I have to go on the circle, this time with the radius 3, something like this.
00:46:750Paolo Guiotto: And my point will be about here. That's X3.
00:52:00Paolo Guiotto: Then I have X4, which is 4 times cos 4 sine 4, and you can continue like that.
00:57:150Paolo Guiotto: So, let's say that, roughly, these points will be turning continuously around the origin and getting far and far, so turning like a spiral, no? But going…
01:09:870Paolo Guiotto: Far away from the origin.
01:12:560Paolo Guiotto: And in fact, I show you that now this vector is going to infinity.
01:18:60Paolo Guiotto: Now… Notice that…
01:26:180Paolo Guiotto: If I compute the norm of XN, It is the root of…
01:32:920Paolo Guiotto: Square of the first component, which is n cos n.
01:37:760Paolo Guiotto: plus square of the second company, which is n sine n.
01:43:770Paolo Guiotto: This is, once you factorize N square.
01:47:920Paolo Guiotto: You get cos square n plus sine square n.
01:55:60Paolo Guiotto: Because square plus sine squared is well known. It is equal to 1,
01:59:960Paolo Guiotto: And therefore, we remain with root of n squared, which is n, because n is positive. So this quantity goes clearly to plus infinity. So this says, without doubt, that this XN
02:16:70Paolo Guiotto: Goes to infinity 2.
02:19:450Paolo Guiotto: But if you look at components, none of the two components has a limit.
02:25:200Paolo Guiotto: Notice that…
02:28:560Paolo Guiotto: Well, that's not an easy thing to show, if I take the first component, XN1, is N, sine… so, sorry, cosine.
02:42:360Paolo Guiotto: Cost N.
02:44:230Paolo Guiotto: Well… It's not easy, because the sequence cosine is not easy.
02:49:460Paolo Guiotto: The behavior of this sequence is very complicated.
02:52:920Paolo Guiotto: And,
02:54:880Paolo Guiotto: However, perhaps you, you, you can imagine that, probably you, you, you know this even, that this sequence, cosine of natural, okay.
03:06:390Paolo Guiotto: Not cosine on anything that goes to infinity, no, cosine of natural.
03:12:290Paolo Guiotto: Now, this sequence is very complicated, it obstulates between minus 1 and 1, it has not a limit, and actually, you can extract sub-sequences from this and go everywhere you would like, you know, you… if you want, there is a sequence of integers such that cosine of n goes to 0, for example.
03:29:470Paolo Guiotto: Well, it's not clear what happens when you multiply by n, because there is a competition between these two, no? This is going to be big times this, which is obstulating. But you understand that if this can take values close to 1, this is close to N. So…
03:47:560Paolo Guiotto: For those values, you are going to plus infinity.
03:50:400Paolo Guiotto: When this is going to be close to minus 1, this is going minus n, approximately, so it goes to minus infinity.
03:57:630Paolo Guiotto: And so, if this sequence has two… This kind of behavior.
04:02:430Paolo Guiotto: Big close to plus infinity, big close to minus infinity, definitely it has not a limit. So there is no limit for this thing.
04:11:160Paolo Guiotto: When n goes to plus infinity.
04:14:300Paolo Guiotto: And similarly, for the second component, which is n sine… And the same happens.
04:22:420Paolo Guiotto: So, in this case, even both components have no limit.
04:27:00Paolo Guiotto: Okay?
04:28:340Paolo Guiotto: So you see that this is different from what happens with sequences with finite limits. When the limit is finite, you can say, finite as vector, you can say, each of the companies must have a finite limit, and that's an if and only if.
04:44:880Paolo Guiotto: So it is, let's say, easy. I want to know if it is a finite limit, I check the limit of the components. If they come finite, I can conclude that the full vector has a limit, and the limit is made by the array of the limits.
05:00:10Paolo Guiotto: But with the infinite limit, there is not such an if and only if.
05:04:690Paolo Guiotto: However, it's not such a big problem, because with the infinite limit, what… if you suspect that the limit is infinity, you just compute the norm of their vector, and that's become a sequence of positive numbers.
05:21:420Paolo Guiotto: If it goes to plus infinity, it is clear, okay?
05:25:630Paolo Guiotto: So… It's not a big problem that you cannot do the check component by component.
05:33:210Paolo Guiotto: I want to repeat this point, because maybe it's not clear.
05:38:690Paolo Guiotto: If I have to check a finite limiter, and I want to use… Basically the definition.
05:47:740Paolo Guiotto: You see, check that this norm goes to zero. There is a little problem, that I needed to have the candidate L.
05:55:180Paolo Guiotto: But to have the candidate L,
05:57:840Paolo Guiotto: because of this property, I should compute each of the components. And it turns out that if each of the components has a finite limit, I already know that that's a limit for the vector. So I don't need to do that check, it's stupid. It's like if I check two times that the limit exists, you understand this?
06:16:730Paolo Guiotto: So, if you would like to use this, you would need to compute L, but how would you compute L? You would compute L by doing the component-wise limits, so by each of the components, and once you have done that, if you have a finite array, you already automatically know that that's true. You don't need to verify.
06:38:100Paolo Guiotto: while in here…
06:40:470Paolo Guiotto: If I have to verify that infinity is the limit, well, you see that you don't need any limit, you know? You just have to compute the norm of XL, so you have all the ingredients you need, the sequence, you compute the norm, and try to see if you're able to compute the limit of this quantity.
06:59:220Paolo Guiotto: If that limit is plus infinity, you can conclude that the limit is infinite.
07:03:360Paolo Guiotto: And here, you don't have any check for the components.
07:07:370Paolo Guiotto: Okay? So, it's clear that if one of the components goes to infinity, you can conclude, because you may notice
07:17:340Paolo Guiotto: Let's, tell this as a remark.
07:22:40Paolo Guiotto: So, let's say that, better than the remark, let's say warning.
07:32:940Paolo Guiotto: So, it is clear… That, if… One… of the… components…
07:52:240Paolo Guiotto: of… Xena… ghosts.
07:58:590Paolo Guiotto: 2 plus minus infinity, since we are talking about components, they are numbers.
08:03:940Paolo Guiotto: Okay? Then the sequence of the array will go to infinity, infinity in the space.
08:13:50Paolo Guiotto: And that's why, because if you think about, I'm saying, take your XN,
08:18:240Paolo Guiotto: which will have components X and 1, accent to… XN3, and so on, XND.
08:26:689Paolo Guiotto: And suppose that, for some reason, you see immediately that one of the components, so for example, XN2,
08:35:420Paolo Guiotto: Go to plus infinity.
08:38:660Paolo Guiotto: Can you conclude that the array goes to infinity? Of course, yes, because the norm of XN
08:46:189Paolo Guiotto: You know that the norm of XN is the root.
08:51:250Paolo Guiotto: of the sum of the squares of the components. So I should do X and J square and sum over J.
09:00:710Paolo Guiotto: Now, if you take this sum, you forget of all the components, and you save only the second one. Clearly, all this sum is larger than XN2 square.
09:13:729Paolo Guiotto: So since the sum is larger than that number, the root will be larger, because the root respects the order. So this will be larger than XN2 square, which is the absolute value of XN2.
09:26:960Paolo Guiotto: So if this guy pushes to plus infinity.
09:31:470Paolo Guiotto: By what is called the two policemen, or the…
09:39:140Paolo Guiotto: Huh?
09:40:740Paolo Guiotto: Right, the squeeze, thank you, the squeeze theorem. This guy who is pushing up to plus infinity will force the other guy to go to plus infinity as well. So, from this, you get norm of XN goes to plus infinity, and that's the conclusion, you see?
09:58:550Paolo Guiotto: So if you have just one single component that goes to infinity, plus, minus, doesn't matter, actually, you don't even know that it goes to plus infinity or minus infinity, because, as you can see, the condition is modulus of the component that goes to plus infinity. But however, this is a detail.
10:14:210Paolo Guiotto: If you see that one of the components clearly goes to infinity, you can say it goes to infinity, the vector. There is no question.
10:22:700Paolo Guiotto: But!
10:25:610Paolo Guiotto: As this example shows, And this you should impress in your mind.
10:31:610Paolo Guiotto: In this example, you go to infinity, but none of the coordinates goes anywhere, actually.
10:39:200Paolo Guiotto: They do not have a limit.
10:41:450Paolo Guiotto: So you cannot say that this thing is an if-and-all leaf.
10:46:410Paolo Guiotto: I go to infinity in space if and or if at least one of the components, coordinates, as you like, goes to infinity as a sequence of numbers.
10:57:870Paolo Guiotto: No, that's false. And the example is written just above here. Okay, so… Let's, say that. But… this…
11:13:630Paolo Guiotto: You know, that the evil hides in details, and that's a classical example, but this is… not.
11:23:240Paolo Guiotto: N… If and only if, so a double arrow implication.
11:29:960Paolo Guiotto: The example… The… Example… that we come to see, XN.
11:41:130Paolo Guiotto: equal N cos n.
11:44:780Paolo Guiotto: N. Sine N… shows… that.
11:52:60Paolo Guiotto: We can have… we can well have that Xn goes to infinity, But…
11:59:510Paolo Guiotto: None of the components has a limit.
12:17:940Paolo Guiotto: Okay?
12:19:360Paolo Guiotto: So, you have to be careful. If limit is finite, I can say, yes, limit is component by component. If limit is infinite, no, limit is not component by component. But, if I see that one of the components, or more, of course, goes to infinity.
12:38:570Paolo Guiotto: Then I can conclude, yes, it is true that the vector goes to infinity. It's… the proof is here.
12:46:210Paolo Guiotto: But that's not necessarily true, okay? So if I don't see, I cannot exclude that it goes to infinity. So you have to be careful.
12:59:420Paolo Guiotto: Of course, here you have an exercise, that exercise we were saying, I would suggest you try to do, and maybe on…
13:07:880Paolo Guiotto: We see on… Monday? No. Tuesday.
13:12:310Paolo Guiotto: Tuesday. I have to learn the schedules.
13:15:730Paolo Guiotto: So, let's… I leave you, for the moment, the exercises that are on this order of ideas do.
13:24:280Paolo Guiotto: 1.83… Number 4… Number 5… Yeah, and that's… the number 5, are, like, statements, no?
13:40:240Paolo Guiotto: Where you have to understand if it's true or false.
13:44:480Paolo Guiotto: If it is true, you have to prove. So, write… try to write a proof.
13:51:800Paolo Guiotto: If it is false, you just need to show a counterexample. So an example that shows you that that property is non-true, okay?
14:01:850Paolo Guiotto: Let's see what you do, and on Tuesday, we will…
14:07:530Paolo Guiotto: give a loop to some of these problems. Okay.
14:11:610Paolo Guiotto: Let's say that now we have defined the, operation of limit for sequences.
14:21:750Paolo Guiotto: What is the main interest in analysis is to have the operation of limits for functions. So, let's start talking about this.
14:33:460Paolo Guiotto: Limit.
14:35:300Paolo Guiotto: off.
14:36:790Paolo Guiotto: function.
14:43:20Paolo Guiotto: Now… Let's spend the first minute to say what kind of function are we going to consider here.
14:51:760Paolo Guiotto: So, in this course.
14:58:790Paolo Guiotto: We consider…
15:04:480Paolo Guiotto: functions.
15:08:470Paolo Guiotto: Well, normally, we use the usual letters F, but here, to, let's say, I don't know if it makes easier, the story, but, let's try to use, for general functions, we will use uppercase letters.
15:25:850Paolo Guiotto: And then we use lowercase letters for certain types of functions.
15:30:930Paolo Guiotto: Let me explain why. Well, let's consider… we consider, several… types…
15:43:30Paolo Guiotto: of functions. So let me, show you some examples, then we will unify in a unique definition.
15:50:690Paolo Guiotto: So, imagine that I have a function of, to begin, two variables, X and Y, something like XY squared, you know, it doesn't matter the formula, okay?
16:02:560Paolo Guiotto: What kind of function is this?
16:05:270Paolo Guiotto: You see that this function takes this object, which is a vector.
16:10:190Paolo Guiotto: And it gives you what? So this is an object of R2,
16:14:590Paolo Guiotto: And this is an object of R.
16:17:340Paolo Guiotto: So, formally, this function f is defined on R2, and takes values in R.
16:26:970Paolo Guiotto: So, well, we could call this a numerical function, because the values taken by F are numbers.
16:34:110Paolo Guiotto: Do not confuse values.
16:36:650Paolo Guiotto: with variables. The values are the F of, no? So, F of 1 minus 5 is 1 times minus 5 squared, so it is 25. This quantity here is the value taken by F at point 1 minus 5, or on the vector 1 minus 5.
16:56:890Paolo Guiotto: So, for example, I can have function of three variables, XYZ, X plus Y sine Z, I don't know.
17:07:290Paolo Guiotto: I'm writing random expressions. Now, here you see that we have the object in argument of F is a vector of R3, and the value taken is a number. So again, we have a function which is similar to the previous one, because we start
17:26:240Paolo Guiotto: From a vector space, and we end into Set of real numbers.
17:31:210Paolo Guiotto: So, a general paradigm for this will be, in general.
17:40:520Paolo Guiotto: we will have a function F,
17:44:420Paolo Guiotto: of a vector, which will have a certain number of components. When we do arguments in general, we will not use XYZ, because there are only 24 letters, and we may have 1 million of variables.
17:58:980Paolo Guiotto: We need a more efficient way, and that's why we use indexes. But, if we are talking about a function of two variables, it's better if we use XY instead of X1, X2.
18:10:350Paolo Guiotto: or 3 variables, XYZ instead of X1, X2, X3, okay?
18:15:380Paolo Guiotto: In general, we use this notation, or this one, f of x, if we want to have a more compact notation.
18:23:180Paolo Guiotto: And this F is, you know, functions are not necessarily defined everywhere, so they will have a domain, domain that is a subset of the space RD, and they take value in R. These kind of functions are… F is a function.
18:43:580Paolo Guiotto: Sorry, numerical.
18:50:290Paolo Guiotto: function.
18:52:680Paolo Guiotto: off… vector… variable.
18:57:70Paolo Guiotto: So the variable, the argument of F, is a vector.
19:01:80Paolo Guiotto: The value is a number, so that's why we call it a numerical function.
19:06:320Paolo Guiotto: These functions are very important because an optimization problem involves functions of this type.
19:13:250Paolo Guiotto: You want to maximize, minimize something. To find the minimum, maximum, you need that these values are numbers, not vectors.
19:21:850Paolo Guiotto: But, but, but it is important to have in mind also a little bit more general types of functions. For example, and here I will change slightly the letter.
19:32:770Paolo Guiotto: So now take this, I use the uppercase F with an array above, take this F of XY,
19:41:670Paolo Guiotto: equal, I don't know, X plus Y times, comma XY.
19:48:20Paolo Guiotto: Now, look at this thing.
19:50:240Paolo Guiotto: This takes… you see, the argument is a vector that belongs, in this case, to R2,
19:57:800Paolo Guiotto: And the value is not a number. The F of XY is what? Is itself another vector. And in this case, you have a vector of R2.
20:10:100Paolo Guiotto: So in this case, you would have a function f that is defined on R2 with values in R2.
20:18:60Paolo Guiotto: You know, functions can be defined on every set with values on any other set, so there is no rule about that. It depends on what you have to do with the function. A function like that could be a transformation of the plane, so something that moves points from one part to another.
20:34:980Paolo Guiotto: How do you represent a displacement, a plane displacement?
20:39:670Paolo Guiotto: A displacement is a rule that says, you know where this point goes? It goes here.
20:46:230Paolo Guiotto: So, if you want to use a function to describe a movement in plane, you need an object like that, a function that transforms a point into another point, or a vector into another vector.
21:00:570Paolo Guiotto: But you could use also functions to do other, more general and complicated things.
21:06:420Paolo Guiotto: But before we enter in this, look, this function is difficult to visualize.
21:12:680Paolo Guiotto: Because, normally, How do we represent a function? We represent functions with graphs.
21:22:370Paolo Guiotto: When we do a graph, we put somewhere the domain, and then we have another vertical axis for the codomain. Imagine that now we have to do a figure here. The domain has two dimensions, so the domain is something like this, R2.
21:36:370Paolo Guiotto: We take points to compute the F there. But then, if we want to see the codomain, we need to put that at two axis, because the codomain is not R. I is R2, so we need another axis, and maybe another axis again, R2. So, it's something like 4 dimensions.
21:54:460Paolo Guiotto: And we cannot draw this.
21:57:110Paolo Guiotto: So, it's impossible to have an idea with… at least with a traditional representation of a graph of this type of functions. So, forget it.
22:09:860Paolo Guiotto: Okay? So…
22:13:40Paolo Guiotto: And we can have even more complicated objects. I could have a function that takes a vector with two components, and it transforms into a vector with three components.
22:24:620Paolo Guiotto: I don't know.
22:27:490Paolo Guiotto: Why not?
22:30:370Paolo Guiotto: In physics, there's plenty of this kind of objects, engineering.
22:34:690Paolo Guiotto: Okay? So, what is the general paradigma? In general.
22:48:370Paolo Guiotto: we… consider… functions.
22:56:390Paolo Guiotto: let's say F. I put the array here because the values… to remind me that the values of this thing are arrays, not numbers.
23:06:620Paolo Guiotto: Okay, so this makes maybe easier to remind me that this object is an object with values which are vectors, is vector-valued, and the domain is itself a vector. So I could say f of x, where X is a vector.
23:25:830Paolo Guiotto: let's say, of a space RD, and the image is not necessarily a vector of the same size. As you can see here, I start from R2, and I end into R3.
23:40:80Paolo Guiotto: So, in general, I have to expect that this quantity will belong to RM, let's say, with the M not necessarily equal to D.
23:49:810Paolo Guiotto: So, the picture will be F equal F of X,
23:55:220Paolo Guiotto: will be a function defined on some domain, as every function there will be a domain, a subset of RD with values in RM.
24:06:310Paolo Guiotto: This is the most general kind of function we consider in this course.
24:14:400Paolo Guiotto: we will not treat, we… let's say that we will not work too much with this kind of generality, but…
24:22:290Paolo Guiotto: For example, when we will discuss integration, just to make you an example.
24:29:850Paolo Guiotto: If I do integration in two variables, and I want to do a change of variables.
24:35:220Paolo Guiotto: There are now two integration variables. When you do a change of variables in the integral, you say, let's introduce a new variable that simplifies my calculation, right? So you are integrating in variable X, and you set Y equal a certain function of X. That's the change of variable.
24:51:650Paolo Guiotto: Now, here we will have to compute integrals in two variables, let's say X and Y, and we would want to change it into other two variables, because probably the change of variable will change two old variables into two new variables. So my change of variable will be represented by a function that takes XY and gives me ZW.
25:14:150Paolo Guiotto: are the two new variables. So it's a function R2 to R2, or for a triple integrals.
25:20:80Paolo Guiotto: So, for the integral in 3 variables, I will have a transformation, 3 variables into 3 variables. So, function R3 to R3.
25:28:240Paolo Guiotto: So it is important that we have the keys to handle also these kind of functions. So that's why we will see, when we will do the differential calculus, you will see the derivative for a function of this type.
25:42:310Paolo Guiotto: which is an object, a relatively complicated object, which is a matrix, okay? The derivative will be a matrix here. So last year was a scholar, here it will be a matrix of numbers. However, it will be, it will be easy, everybody.
25:57:170Paolo Guiotto: learn this. At this point, maybe it may appear a bit complex, because you don't have an intuition, not… neither me, it's not your fault. It's that none of us has an intuition about this.
26:12:630Paolo Guiotto: So, it is better, however, to keep on simple things. So, think that this is a function that takes vectors into vectors. We have a way to measure distances from vectors in every space, so we have a norm, Euclidean norm here, we have an Euclidean norm here.
26:29:540Paolo Guiotto: And that's the important thing to know. Don't be, don't try to, visualize these spaces, because that's simple, simply impossible.
26:40:910Paolo Guiotto: Okay, after this premise, Let's return here. The goal is to introduce the operation of limit.
26:50:720Paolo Guiotto: Now, what we want to do is, for such F, so we will give the definition for this general.
26:58:560Paolo Guiotto: Of course, then the calculation will be done in a deeper function of two variables, okay? Numerical, so much easier. But what we see works for the general.
27:09:170Paolo Guiotto: Okay, so, we want… the goal is… we want… 2.
27:17:140Paolo Guiotto: define…
27:19:110Paolo Guiotto: So, what is the limit? So, let's write it, even if we don't know what is exactly now. So, we want to do the limit.
27:26:610Paolo Guiotto: of a function f, A vector variable, vector valued.
27:32:450Paolo Guiotto: When the variable goes somewhere.
27:34:890Paolo Guiotto: So now, the variable is a vector, X, that will go, probably, to some vector, to some point. I don't know what is the notation, let me keep the notation in the same…
27:50:580Paolo Guiotto: I call the P.
27:54:220Paolo Guiotto: That stands for point.
27:57:110Paolo Guiotto: In some book, you find X0, but I don't want to put a zero here, because it makes confusion with indexes that will go… so I just write the point B.
28:06:620Paolo Guiotto: So, if you eliminate the arrows, this is the limit you have seen last year, no? So, what is the idea behind this, that this limit is equal to L, that will be, of course, a vector here.
28:19:100Paolo Guiotto: The idea is that you want to give the meaning, a precise meaning, and even operational rules to this. You want to understand what happens to this quantity, f of x, when x gets close to P.
28:34:150Paolo Guiotto: Okay?
28:35:510Paolo Guiotto: I want to say that this is going to L.
28:38:860Paolo Guiotto: Now, the fact that XPL are arrays is irrelevant, because the idea is that if this is close to this, and I can say that because of the norm, the norm tells me when I am close to, then this is close to this. Now.
28:55:630Paolo Guiotto: I don't know what kind of definition you have seen last year for ordinary limit, but I will try to refresh a definition.
29:03:650Paolo Guiotto: So, let's refresh.
29:08:470Paolo Guiotto: Because, there are different approaches to this problem. The…
29:15:230Paolo Guiotto: Well, let's say a possible definition of limit.
29:18:540Paolo Guiotto: Possible.
29:20:770Paolo Guiotto: definition.
29:22:610Paolo Guiotto: of limit… four functions F, of a real variable.
29:31:410Paolo Guiotto: X is real.
29:34:900Paolo Guiotto: So, definition… So let's say this is a sort of calculus…
29:41:210Paolo Guiotto: One definition. We have a function defined on some domain contained in the real line, real value.
29:49:80Paolo Guiotto: Now, we need an important definition, which is the definition of accumulation point.
29:58:170Paolo Guiotto: Those are points where we can compute limit. I don't want to enter here, we will return, but it's not important. Let's write it, just that P must be an accumulation point for the domain D.
30:12:250Paolo Guiotto: Okay, so I will return on this, but I want to focus on the definition. How do we say that the limit of f of x when x goes to P is L? So, we say that…
30:25:160Paolo Guiotto: the limit…
30:27:30Paolo Guiotto: When x goes to P of F of X is L, here l is just a value, R, If…
30:37:620Paolo Guiotto: what happens is, well, we want to say that no matter how we get close to point P, the value F of X gets close to the limit L.
30:47:850Paolo Guiotto: A way to say this is using sequences. It says that for every sequence XN, Contained in the domain.
30:57:170Paolo Guiotto: Different from the limit point, you know that in limits, we never consider what happens exactly at the limit point, because the function normally is not even defined in the limit point.
31:09:750Paolo Guiotto: Or, if it is defined, the value could have nothing to do with the limit, but such that Xn goes to P,
31:17:440Paolo Guiotto: then what happens? That if you go to P, the function of F at point xn will drive you to limit L.
31:27:50Paolo Guiotto: So this is the way to say, no matter how you get to point P, provided that you don't touch the point P,
31:35:710Paolo Guiotto: And you are in the domain D, otherwise you cannot compute F, okay?
31:40:450Paolo Guiotto: then the function will go always to the limit L, whatever is the way you use. That's the forever sequence, okay?
31:50:250Paolo Guiotto: Now, what we do is that we can take this definition, put arrows, and we have the definition of limit for the vector functions. So…
31:59:950Paolo Guiotto: definition.
32:03:500Paolo Guiotto: Calculus 2.
32:06:320Paolo Guiotto: So, letter F… With the arrays, so function of an array.
32:13:660Paolo Guiotto: A vector, vector valued, so will be defined on a subset now of RD with values in RM.
32:22:810Paolo Guiotto: and pick a point, P,
32:25:740Paolo Guiotto: There must be an accumulation point for the domain. We have not yet introduced, I will do next time, but it's just to close the point to have a definition here.
32:37:850Paolo Guiotto: So we say, That.
32:42:570Paolo Guiotto: Exist the limit when X… goes to P.
32:50:900Paolo Guiotto: of F of X.
32:53:900Paolo Guiotto: And this limit is L.
33:01:390Paolo Guiotto: Is it clear? If…
33:03:830Paolo Guiotto: And the property is the same, just the same, but just with the arrow. So, the same philosophy.
33:09:890Paolo Guiotto: And the unique fact that changes is that now I have sequence of vectors that are moving, but we define what is the limit, right? So, if whatever is the sequence, XN,
33:22:540Paolo Guiotto: in the domain D,
33:24:850Paolo Guiotto: different from the limit point P, so they don't… again, they do not have to touch the limit point, because maybe there, the function usually is not even defined.
33:35:880Paolo Guiotto: But such that XN goes to point P,
33:40:880Paolo Guiotto: Then, the values F at vector Xn
33:45:630Paolo Guiotto: always go to the limit L.
33:49:510Paolo Guiotto: Now, this definition, the nice of this definition.
33:53:130Paolo Guiotto: what I like a lot of this definition is that with a unique definition, we can make sense to all possible cases. So, the cases when L is finite.
34:03:610Paolo Guiotto: But also when L is infinity, in this case, sorry, I did an error here.
34:09:910Paolo Guiotto: Just here, there is an arrow. Why?
34:13:580Paolo Guiotto: Because look at the function F, takes values in RM.
34:17:880Paolo Guiotto: So, the limit must be in the set in codomain of F, because it's a limit of these objects. These objects are vectors of RM, so it must be in RM. It's like if R2 to R3, I'm saying that the limit of vector in the space is a vector in the plane. It's something that does not make sense. So, the error is that I wrote here D, but it should be M.
34:45:970Paolo Guiotto: Okay, apart from this, now you see that the philosophy is the same. So now we have a definition, and the point is, okay, we have a definition, but how do we compute the limit in practice?
34:57:240Paolo Guiotto: Well, that's the same you had with the Calculus 1. You never computed the limit with the definition of limit, because that would be crazy.
35:05:830Paolo Guiotto: Okay? So we have to learn a bit of techniques of calculus, and that's what we do the next week, okay? So we stop here for today.