AI Assistant
Transcript
00:01:780Paolo Guiotto: Yeah, I know everyone got sick, oh, I know, I know.
00:06:840Paolo Guiotto: Okay.
00:08:760Paolo Guiotto: I have no time to waste now.
00:14:210Paolo Guiotto: Okay, so for today, I won't, to…
00:22:620Paolo Guiotto: Do some, to start with the… some exercise… Of those, I left you.
00:32:119Paolo Guiotto: Then today we'll finish with exercises. I will not do all the exercises, 1, 8, 10, but I will do…
00:43:20Paolo Guiotto: Sam, and the… The remaining, I will publish the solutions today afternoon.
00:51:140Paolo Guiotto: So, let's take, for example, I do not remind exactly what we
00:55:670Paolo Guiotto: did previously, but I found that there are interesting and not easy exercises here.
01:05:580Paolo Guiotto: So, for example, let's start giving a look
01:09:730Paolo Guiotto: To number 3, which is somehow easy, but it demands, A bit of,
01:20:980Paolo Guiotto: It's… it's not so straightforward.
01:23:790Paolo Guiotto: So, we have limit at infinity X squared, times Y squared plus X squared plus Y squared minus X times Y.
01:35:220Paolo Guiotto: So this is our function, FXY, clearly well-defined on the interior Cartesian plane, R2.
01:44:750Paolo Guiotto: So, we have to compute the limit. We can easily see that, for example, if you look at the section X0 along the x-axis, this is X squared, so it goes to plus infinity when x goes to plus minus infinity, which is the condition.
02:01:710Paolo Guiotto: For which point X0 goes to infinity in plane.
02:07:670Paolo Guiotto: So we may say that if a limit exists, it is equal to plus infinity at this point.
02:13:810Paolo Guiotto: Now, the question is, should we try to prove that the limit exists, and it's equal to plus infinity, or should we try to look for sections along which we have a different limit?
02:27:150Paolo Guiotto: Now, best thing to do is to use polar coordinates. So, in polar coordinates, f of rock cosine theta.
02:36:110Paolo Guiotto: Raw sine theta.
02:38:380Paolo Guiotto: we see that X squared, Y squared is a rho power 4,
02:43:310Paolo Guiotto: Each X is raw sine rho cos, X and Y is raw square, rho squared.
02:49:70Paolo Guiotto: 2 times 4 power 4, then we have cop…
02:51:990Paolo Guiotto: square theta sine square theta. I know, someone else who is attending, right. Another one.
03:00:310Paolo Guiotto: Huh?
03:02:240Paolo Guiotto: All of you are sick, right? Is that the answer? Okay.
03:07:110Paolo Guiotto: I see.
03:08:610Paolo Guiotto: There should be some pandemic here, because there's no other explanation. X squared plus Y squared is raw square minus XY is raw square cos theta.
03:20:100Paolo Guiotto: sine theta. You know what happens if you do like that? I will suspend recording.
03:26:210Paolo Guiotto: Okay, I will suspend recording.
03:30:720Paolo Guiotto: Because that's not the way to do.
03:33:210Paolo Guiotto: Okay, so what do we see here? Rap over 4… cos square theta.
03:38:680Paolo Guiotto: sine squared theta.
03:41:950Paolo Guiotto: Plus raw square 1 minus cos theta.
03:46:430Paolo Guiotto: sine theta.
03:49:440Paolo Guiotto: And wherever you arrive even later, even online, huh?
03:53:610Paolo Guiotto: So…
03:57:590Paolo Guiotto: Looking at this, we know that rho is going to plus infinity. It looks like if it is raw square, rho power 4, the leading term.
04:05:890Paolo Guiotto: But be careful, because cos theta or sine theta could be zero. So that term could be vanishing, so…
04:13:860Paolo Guiotto: It will not go to infinity.
04:16:790Paolo Guiotto: So we should not consider too much that term, because for certain theta, it is zero, and in fact, along the axis, this is X squared, Y squared. So when one of the two coordinates is zero, the term is 0.
04:29:890Paolo Guiotto: But remind what is the goal? Since I,
04:33:880Paolo Guiotto: I want to see if this thing goes to plus infinity. Well, what can I say about this is just that this is positive.
04:42:210Paolo Guiotto: And nothing more, because I said it could be 0. What about this? We have a raw square times this parenthesis, where if we write this as 1 half 2, this becomes raw square 1 minus 1 half sine 2 theta.
05:00:850Paolo Guiotto: And in this way, I can see that this parenthesis is away from zero, because that minus 1 half can be at most 1 half. So this is clearly always greater than 1 half, 1 minus 1 half.
05:16:740Paolo Guiotto: And so this says that my F is greater than… I throw away this part here, because it's useless, basically, 0, plus 1 half raw square. So now, I am in good condition, because this says that F is bounded below by a function of raw.
05:34:940Paolo Guiotto: which is going to plus infinity when raw goes to plus infinity. So this is,
05:41:530Paolo Guiotto: Sufficient to conclude that there exists the limit of
05:46:310Paolo Guiotto: F when point XY goes to infinity in a plane, and it is equal to plus infinity.
05:56:980Paolo Guiotto: Let's give a look to number 4, which is a bit tricky.
06:01:950Paolo Guiotto: It is a bit tricky because we have here the limit for XYZ.
06:11:400Paolo Guiotto: going to the infinity of F, sorry, of…
06:16:790Paolo Guiotto: This is the function X power 4 plus Y power 4 plus Z power 4 minus XYZ.
06:27:180Paolo Guiotto: Yes.
06:28:760Paolo Guiotto: Now, again, this is our F, it is well defined, it is defined on domain D is the full space R3.
06:40:820Paolo Guiotto: Okay?
06:42:300Paolo Guiotto: Clearly, I can see one of the sections, for example, along the axis, X00 gives X to power 4, that goes to plus infinity when x goes to minus plus infinity, so when the point X00 goes to infinity in space.
07:02:510Paolo Guiotto: These are equivalent.
07:04:740Paolo Guiotto: So this again says that if we have a limit, this limit can be only plus infinity.
07:10:940Paolo Guiotto: Now, the natural, change of variable here would be the use of spherical coordinates. But there is a problem here, because if I replace spherical coordinates, I won't do that, because it's a mess, basically. I will see that the first three terms are not X squared plus Y squared plus
07:30:700Paolo Guiotto: Z squared, so I have a raw square. They are a power 4, so that I could expect they are something like… so I could say, intuitively, there is not the rational reason behind this for the moment, that this could look like X squared plus Y squared plus Z square, maybe.
07:50:600Paolo Guiotto: So they…
07:52:10Paolo Guiotto: yield a contribution like raw power 4. And in fact, if you do the change of variables, you will see that there is a rho power 4 factorizing there. While in the other factor, there is a raw cube. So this is like a cubic factor, this is, like, fourth power.
08:10:680Paolo Guiotto: But the problem is that, unfortunately, when I do the change of variable, here I get some coefficient that depends on the angles, theta and phi, and I have to remove this. So I want to do something like this is greater than or equal than some constant.
08:27:320Paolo Guiotto: And that's not exactly easy to do. So let's see how we can do that.
08:31:900Paolo Guiotto: Now, in ADR, there are many ways. Here, I'm not showing a standard way. There is no standard solution. You have to forget that.
08:41:280Paolo Guiotto: There are attempts, ways, ideas of what we can do. I'm showing you some techniques. For example.
08:48:130Paolo Guiotto: If I understand that X power 4 plus Y4 plus Z4 is like that x squared plus Y squared plus n squared squared, I would like to try a relation… to look for a relation between these two quantities.
09:02:330Paolo Guiotto: So let's take the second one and develop the square, and let's see what happens. We notice that
09:08:700Paolo Guiotto: X squared plus Y squared plus Z squared
09:14:440Paolo Guiotto: squared is X power 4 plus Y power 4 plus Z power 4, which is the term I have.
09:22:940Paolo Guiotto: But then there are the double products, so something like 2x square y squared, plus 2X square Z squared plus 2Y square Z squared. So there is a little difference.
09:37:870Paolo Guiotto: Now, here I see that there is an easy relation, because all these terms are
09:43:400Paolo Guiotto: What would you say here? R?
09:46:200Paolo Guiotto: Positive, okay. So if they are positive, it means that the left-hand side is at this thing here.
09:55:980Paolo Guiotto: Is this plus positive.
09:59:00Paolo Guiotto: So what is the relation between the yellow and the purple box?
10:07:70Paolo Guiotto: The yellow is… is bigger. So I get this relation, and this is unfortunately the useless.
10:14:200Paolo Guiotto: relation. I get this, that X squared plus Y squared plus Z squared
10:22:340Paolo Guiotto: is greater or equal than X power 4 plus Y power 4 plus Z power 4. Let me explain you why this is basically useless.
10:35:920Paolo Guiotto: Unfortunately.
10:39:770Paolo Guiotto: Because I could say from this that my F, which is FXYZ, equals that term, X power 4, let's say, let's do in red.
10:50:340Paolo Guiotto: X power 4 plus Y power 4 plus Z power 4, and then there is the minus XYZ,
11:00:590Paolo Guiotto: Well, if I replace the red term with that square.
11:04:880Paolo Guiotto: my function gets bigger, so I have an upper bound, not a lower bound. And if the goal is to prove that the limit is plus infinity, the upper bound is basically useless. I would have this X square… sorry.
11:19:580Paolo Guiotto: Let's write in red, X squared plus Y squared plus Z squared squared minus…
11:28:190Paolo Guiotto: XYZ. Now, this would be nice, because here, if I would replace the spherical coordinates, this would become equal to the quantity X squared plus Y squared plus Z squared is raw square, right? So, with the spherical
11:45:910Paolo Guiotto: coordinates.
11:48:280Paolo Guiotto: This becomes raw square to power 2. Well, let's emphasize this,
11:54:390Paolo Guiotto: raw square to power 2, then I have minus… when I replace the XYZ with raw something, raw something else, and so on, I have a raw cube here, then if you want to write the terms, this is a cos phi… no, it's a sine phi.
12:13:850Paolo Guiotto: So, cube, let's write in black. Sine phi cos theta, that's for X. For Y, I have some sine phi sine theta, so another sine phi, and for Z, I have cos phi.
12:28:900Paolo Guiotto: Now, it doesn't matter what is this coefficient, because
12:32:720Paolo Guiotto: It's something that varies between minus 1 and 1. So I could say that, since I am doing an upper bound, at this point, it's not… has not sensitive to put a lower bound, because I'm considering something which is bigger, okay? So I should continue in that direction. I will have something like rho power 4,
12:51:900Paolo Guiotto: minus, at most, this quantity, so this is actually becomes plus raw cube, because since I am bounding above, I have…
13:03:790Paolo Guiotto: Cool.
13:04:710Paolo Guiotto: This is a fake minus, because this could be a minus, so it could be a plus, actually, no? So, the biggest quantity I have here is raw cube times minus this factor that at most could be 1. So, at most I have this.
13:20:90Paolo Guiotto: So I'm saying that my F is bounded above
13:25:290Paolo Guiotto: By this function, which is called because it goes to class infinity, but useless, because an upper bound of this type does not have anything.
13:33:800Paolo Guiotto: It says…
13:34:940Paolo Guiotto: F is less than someone who is going to infinity, but F would be 0, for example, constantly 0. So it doesn't mean that F has limit to infinity. Be careful, because…
13:46:990Paolo Guiotto: He's a popper.
13:48:610Paolo Guiotto: In half of them, we do that. We say that from this, it follows that have had to do finally clean.
13:55:370Paolo Guiotto: Okay? So this makes the difference in the quality of your convention.
14:02:60Paolo Guiotto: Because if you do not understand what is the difference, in this case, between an upper bound and a lower bound, it means that you are trying just to repeat something that you have not really understood.
14:12:320Paolo Guiotto: Okay? So, this is useless.
14:16:380Paolo Guiotto: I'm not saying you are stupid, I'm saying you have to understand what we are doing, okay? That's most important than doing exact calculations, okay? So, this is useless.
14:30:830Paolo Guiotto: To prove… that the limit of F is equal to plus infinity. You'll see why, you know?
14:40:50Paolo Guiotto: I'm saying I'm smaller than this guy who is going to plus infinity, but I could be zero.
14:45:770Paolo Guiotto: And so I want to go to infinity, okay? So, we need… what we need is… we need…
14:54:420Paolo Guiotto: a lower bound.
14:59:690Paolo Guiotto: So, something like F, X, Y.
15:03:220Paolo Guiotto: Z greater or equal than some G of rh that goes to plus infinity when rho goes to plus infinity. That's what we need.
15:13:500Paolo Guiotto: So we have to go back.
15:15:200Paolo Guiotto: Because where did we obtain this upper bound? This comes from this point. And this comes from this inequality.
15:23:990Paolo Guiotto: So the point is that this inequality is useless. Now you understand why I wrote useless, okay, for this problem. Okay, it's interesting, but it's not useful. So let's go back here, where we have the identity.
15:39:470Paolo Guiotto: And so what I should try to do is not to observe that that quantity is greater or equal than zero. That leads to this bound, useless, but I should try to do something like, this is less or equal than what? Let's see what should be the goal.
15:57:70Paolo Guiotto: Okay, let me rewrite the identity. So the identity is, I take X squared plus Y squared plus Z squared is equal to X power 4 plus Y power 4 plus Z power 4, then we have the double products, 2X squared, Y squared.
16:15:790Paolo Guiotto: plus 2X squared Z squared plus 2Y square Z squared.
16:23:960Paolo Guiotto: Okay?
16:25:190Paolo Guiotto: Now, suppose that I am able to say that this is less or equal
16:30:890Paolo Guiotto: then this guy, or a constant. Constant times this guy, X power 4 plus Y power 4 plus Z power 4.
16:41:170Paolo Guiotto: If I'm able to do that, so I will have that this will be less or equal than, as if you want constant plus 1, my quantity X power 4 plus Y power 4 plus Z power 4.
16:55:750Paolo Guiotto: And from this, we would get exactly the opposite of this, because we would get X4 plus Y4 plus Z4 greater or equal than, apart for a factor, so from this, we would get
17:10:89Paolo Guiotto: X power 4 plus Y power 4 plus Z power 4, greater or equal than 1 over C plus 1, that quantity. X squared plus Y squared plus Z square. And this would be good, because at this point, since I have in my F,
17:28:660Paolo Guiotto: enough, I have this term. In this… this… this way, I will be able to lower bound this by…
17:35:970Paolo Guiotto: something like this, and this would be good. So that's the general strategy. So we want to try to bound this.
17:43:220Paolo Guiotto: Now, the point is, how can we bound that?
17:45:840Paolo Guiotto: Well, we can bound by using an elementary inequality that we already met here.
17:51:280Paolo Guiotto: So, look at this temp, these are temps, 2 something times something.
17:55:780Paolo Guiotto: Do you remind me of this?
17:57:680Paolo Guiotto: 2AB is less or equal
18:05:480Paolo Guiotto: A square plus B squared. That's a very stupid inequality, apparently, apparently stupid inequality. This comes from saying that B minus A is greater or equal than zero, so that's why I'm saying it's stupid, because it's a trivial fact.
18:20:580Paolo Guiotto: But it yields this inequality that it's not so stupid, because if you plug this into that expression.
18:29:560Paolo Guiotto: So now you can see that if you take a 2X square, Y squared, one of these double products, so this is my A and this is my B, I will have that this is less or equal than the square of A, so the square of X squared is X power 4,
18:46:490Paolo Guiotto: plus the square of Y squared, so Y power 4.
18:50:240Paolo Guiotto: And similarly, when I have 2X square Z squared, I have… this is less or equal X power 4 plus Z power 4. And the same, of course, for the term 2Y square Z squared. This will be less or equal than Y power 4 plus Z power 4.
19:08:640Paolo Guiotto: Now, you sum up these three inequalities, because you have these three terms here, you see.
19:15:970Paolo Guiotto: So, summing up these, you get that…
19:19:760Paolo Guiotto: the sum 2X squared plus Y squared, sorry, 2X squared Y squared plus 2X square Z squared plus 2X
19:33:450Paolo Guiotto: Y squared, Z squared, is less or equal than… when you sum up, you get X squared plus X squared, 2… sorry, X power 4 plus X power 4, 2 times this, plus you get 2 times Y power 4 plus 2 times Z power 4. So that's exactly what we get.
19:52:990Paolo Guiotto: X power 4 plus Y power 4 plus Z power 4. And now this is the key.
19:59:890Paolo Guiotto: Step, because… Returning back to F…
20:04:690Paolo Guiotto: So, our F is X power 4 plus Y power
20:10:250Paolo Guiotto: 4 plus Z power 4… no, sorry, let me first, no, no, I forgot one step, so let's return back to this identity.
20:22:660Paolo Guiotto: Uns… So… this one. So I have the square of this sum, so…
20:30:130Paolo Guiotto: X squared plus Y squared plus Z squared, so here in square, is equal to X power 4 plus Y power 4 plus Z power 4, and then we have the sum of the double products, which is less or equal than this, so I can say plus…
20:47:100Paolo Guiotto: 2X power 4
20:50:110Paolo Guiotto: plus Y power 4 plus Z power 4. So at the end, I have 3 times X power 4 plus Y power 4 plus Z power 4, which is what I wanted. That's precisely this inequality, you see.
21:07:890Paolo Guiotto: The C plus 1 is 3.
21:11:710Paolo Guiotto: Okay, so this says that, let's emphasize this, that this quantity, X power 4 plus Y power 4 plus Z power 4 is greater than or equal than 1 third
21:24:890Paolo Guiotto: the square of X squared plus Y squared plus Z squared.
21:31:610Paolo Guiotto: Now we can go back to F, huh?
21:35:370Paolo Guiotto: So, F, which is X… well, let's use it called RAT.
21:41:90Paolo Guiotto: X power 4 plus Y power 4 plus Z power 4.
21:45:380Paolo Guiotto: Then we have the minus XYZ term, huh?
21:50:130Paolo Guiotto: Now, since that X power 4 plus Y power 4 plus Z power 4 is greater or equal than 1 third that square, I can say that if I replace here, I get something which is smaller.
22:01:700Paolo Guiotto: And that's the right way. So it is 1 3rd…
22:06:810Paolo Guiotto: X square, Y… that's what's it, right?
22:11:150Paolo Guiotto: 1 3rd… X squared plus Y squared plus Z squared squared.
22:20:30Paolo Guiotto: and then minus XYZ.
22:24:120Paolo Guiotto: And so here, if you replace now the spherical coordinates, you get The right information, spherical coordinates.
22:33:940Paolo Guiotto: You get 1 third, this becomes raw square.
22:38:930Paolo Guiotto: raised power 2, minus raw cube, and then we have all the cosine sine, no? This was a sine phi cos theta, that's for X,
22:50:470Paolo Guiotto: Then we have for Y sine phi sine theta, so sine phi squared here, sine theta, and then cos phi here.
23:01:10Paolo Guiotto: So, here we have… 1 third, that's the coefficient, raw to power 4.
23:09:380Paolo Guiotto: minus… rock cuba, and all this bunch of sine and cosine. Sine squared, phi, cos theta, sign… Tita cos.
23:23:240Paolo Guiotto: And that's the right way, because now we can say that
23:27:390Paolo Guiotto: At worst, this quantity here, what we subtract from rho power 4, is minus rho cubed, because that product, at most is 1. It cannot be 1, as one can easily see, but it doesn't matter here, we don't need to be such refined. So we can say that this is larger than
23:46:960Paolo Guiotto: 1 3rd rho power 4 minus rho cubed, that's the function g or rho we were looking for. So this goes to plus infinity when rho goes to plus infinity, and this allows us to conclude that, so there exists the limit.
24:03:840Paolo Guiotto: when point XYZ goes to infinity.
24:10:320Paolo Guiotto: in plane of function f, and this is equal to plus infinity, and this ends the exercise.
24:18:910Paolo Guiotto: Okay? So, you see, it's not, this one was, definitely not, not easy.
24:28:560Paolo Guiotto: I want to show you also number 5, perhaps.
24:33:890Paolo Guiotto: Which is still a bit tricky, in a different sense, not because of For example, Okay, let's see.
24:48:690Paolo Guiotto: Okay, number 5, we have this limit.
24:52:380Paolo Guiotto: still in three variables, X, Y, Z, that go to… Infinity for you.
25:04:580Paolo Guiotto: This is number 5. So we have X squared plus Y squared
25:10:440Paolo Guiotto: plus Z power 4 minus XZ.
25:19:520Paolo Guiotto: Okay, so, also here, this is our F.
25:23:420Paolo Guiotto: of 3 variables, XYZ,
25:27:560Paolo Guiotto: And I can see that F of X 00, for example, this is equal to X squared, so it goes to plus infinity when modulus of X goes to plus infinity.
25:40:970Paolo Guiotto: A very common error is to conclude here, the limit exists and it is equal to plus infinity.
25:47:570Paolo Guiotto: Lots of people say that.
25:50:340Paolo Guiotto: Okay?
25:51:710Paolo Guiotto: So be careful. At this stage, you cannot conclude anything. The unique thing you can say, if there is a limit, this limit is plus infinity. That's the unique thing.
26:02:430Paolo Guiotto: Okay, now we have to understand if it is worth to spend time trying to disprove this value for the limit, or trying to prove that that's the limit.
26:13:330Paolo Guiotto: And to do that, we should use, again, polar coordinates, or spherical coordinates here.
26:19:650Paolo Guiotto: As you can see, we do not have exactly the quantity which is raw square. Raw square is X squared plus Y squared plus Z squared.
26:32:430Paolo Guiotto: No?
26:33:400Paolo Guiotto: So this is interesting because
26:36:630Paolo Guiotto: We know, the point XYZ goes to infinity in space, if and only if,
26:45:860Paolo Guiotto: raw goes to plus infinity, and about theta and phi, there is no particular information, so we know only that. The other two angles, phi is between 0 and
26:56:680Paolo Guiotto: Bye.
26:57:770Paolo Guiotto: And theta between 0 and 2 pi.
27:02:880Paolo Guiotto: But we don't know anything else. We don't… we cannot say they go to something, okay? Think always to the point, turning around the origin, but getting away from the origin.
27:14:640Paolo Guiotto: And, so, the advantage is that you don't have any information on these other two coordinates, but
27:21:620Paolo Guiotto: Somehow the limit reduces to a one variable limit.
27:26:600Paolo Guiotto: So, there is an advantage, and something, of course, that you have to pay for that advantage.
27:33:80Paolo Guiotto: So that's why it would be reasonable to change variable to have something expressed in terms of raw, for which you can easily discuss the limit value.
27:42:500Paolo Guiotto: But unfortunately, we do not have X squared plus Y squared plus Z squared.
27:47:700Paolo Guiotto: We have X squared plus Z squared plus Y squared plus Z power 4.
27:51:870Paolo Guiotto: So, let's say that a naive approach could say, okay, we do not have, we create. What does it mean? I do not have a Z square, I add and subtract Z squared. So, let's do that. I can notice that my F, which is X squared plus Y squared plus Z power 4 minus XZ,
28:11:540Paolo Guiotto: is equal… can be also written as a…
28:14:980Paolo Guiotto: So, let's emphasize this. X squared plus Y squared plus Z squared… I'm not saying that Z4 is Z squared. I'm just adding. So, since I added, I have to subtract, so I give to Z power 4.
28:31:680Paolo Guiotto: minus Z squared, and then you have minus XZ.
28:36:640Paolo Guiotto: This would make easy that part, but it remains still the last term inside.
28:42:540Paolo Guiotto: Now, you could here decide, of course, to change variable.
28:48:600Paolo Guiotto: But here, you would have some trouble, because this term…
28:54:50Paolo Guiotto: Now, if you imagine, put Z equal the raw cost fee.
29:00:610Paolo Guiotto: Now, that term is not a fourth power of rho. Clearly, Z power 4 is raw power 4. But there is a cosine phi to power 4.
29:12:950Paolo Guiotto: Which means that cosine phi is 0, and this can be… take phi equals pi half, this is 0. So this means that that's not really fourth-order term. So you have to be careful, because it's not that term which leads you to infinity.
29:29:560Paolo Guiotto: But reasonably, it will be the first red term. And here, you have to be careful, because this is the second degree term, it goes in competition with this one within the second degree.
29:41:510Paolo Guiotto: imagine you put X equal raw times whatever, but raw times whatever discounts across square, times a bunch of sine into sine.
29:50:460Paolo Guiotto: Okay? So there is a sort of competition between the software and the software.
29:55:810Paolo Guiotto: Now, what I do is that I…
29:58:170Paolo Guiotto: Take the B-spot, yeah, with the right mindset set, right?
30:02:510Paolo Guiotto: And then I will show you that this one can be discarded in some way.
30:07:400Paolo Guiotto: Okay? Because the first two are… the first and the last, so this part here, is sufficient. So let's give a look to that part.
30:17:600Paolo Guiotto: So on.
30:18:900Paolo Guiotto: let's call it F tilde.
30:21:390Paolo Guiotto: this, X squared plus Y squared plus Z squared minus XZ. And here, we do the change of variables, so we use spherical
30:32:900Paolo Guiotto: coordinates.
30:35:190Paolo Guiotto: This becomes, of course, raw square here, minus raw square. For X, we have a sine P, cosine theta, and for Z, we have a cos p.
30:51:260Paolo Guiotto: Now… We factorize the raw square, we get 1 minus, cos theta.
30:59:780Paolo Guiotto: Well, if I write this way, sine phi, cost P…
31:06:470Paolo Guiotto: Apparently, I cannot say that this is a raw square, because that parenthesis could be zero.
31:12:960Paolo Guiotto: If that product cost sine cost is 1, in principle, this could be 0, so it could destroy raw square.
31:21:930Paolo Guiotto: But that's not possible. We know that that cannot be 1, because to be 1, everything must be plus 1, minus 1, but we cannot have sine phi cos phi at the same moment, plus 1, minus 1, no? If one is 1, the other is 0.
31:37:220Paolo Guiotto: But this is not sufficient. So here we can use it, we are lucky, we can use the duplication formula, so put a 2 here, and this becomes a sine of 2 phi.
31:49:470Paolo Guiotto: Because this is sufficient to say that we actually have a raw square 1 minus 1 half this product, cos theta, sine 2 phi.
32:00:280Paolo Guiotto: this quantity It's definitely between minus 1 and 1.
32:05:870Paolo Guiotto: So, it's larger than minus 1, it's smaller than 1. Multiply by 1 half, it means that this factor here is between minus 1 half and plus 1 half.
32:19:490Paolo Guiotto: So the worst can happen. Why I'm thinking about the worst? Because if I… I'm trying to bet on plus infinity limit, I need a lower bound, not an upper bound. So I want something, like, greater or equal.
32:33:650Paolo Guiotto: So I have to say, what can be that parenthesis in the worst case?
32:38:280Paolo Guiotto: It can be 1 minus the worst I can subtract, which is 1 half. So, it can be, at worst, 1 minus 1 half, which is 1 half.
32:47:570Paolo Guiotto: Which is good, because this alone goes to plus infinity.
32:51:980Paolo Guiotto: But this is not the lower bound for F, it is the lower bound for a part of F, so it remains this
33:00:20Paolo Guiotto: ping here.
33:01:240Paolo Guiotto: What can be said about that?
33:03:430Paolo Guiotto: Now, since I have a good bound for this, I know that this part here is greater than 1 half raw square, I don't need a bound for that, really. I need just a lower bound, like it's greater than a constant.
33:16:980Paolo Guiotto: Okay? Now, you understand that this is, reasonable, apparently… I do think it's important.
33:25:280Paolo Guiotto: In terms of that square, so that's… That's an island.
33:30:650Paolo Guiotto: Politics, going to infinity does not mean that regardless.
33:36:140Paolo Guiotto: So we have to be a little more careful here.
33:39:140Paolo Guiotto: And the argument is, however, this.
33:42:340Paolo Guiotto: Let's, define H of Z, that function.
33:47:10Paolo Guiotto: Z power 4 minus Z squared.
33:51:00Paolo Guiotto: Now, what kind of function is this one? It is an even function, that's not important here, but however, let's use it. So it's a symmetric respect to the…
34:02:560Paolo Guiotto: the ordinates axis, so this is Z and values H of Z.
34:07:940Paolo Guiotto: Do not use Y, which is used for other things.
34:11:690Paolo Guiotto: So, in 0 is 0.
34:14:780Paolo Guiotto: at plus infinity is plus infinity, because Z4 dominates, so it's like that. At minus infinity is like that, because it is even. So whatever you can imagine, this function will… will have a minimum, cannot be negatively unbounded.
34:32:550Paolo Guiotto: In fact, we can do a simple check, compute the derivative, this is 4Z cubed minus 2Z, that means Z times 2z times 2Z squared minus 1.
34:46:989Paolo Guiotto: So, since it is, even, we can say, let's look at Z positives, only the half of the axis. So, for Z positive.
34:57:110Paolo Guiotto: So, we have that H prime is positive, this means H is increasing.
35:05:30Paolo Guiotto: If and only if… now it's this parenthesis, because Z is positive, 2Z squared minus 1 must be positive. So this means Z squared greater than 1 half, and this means, since here Z is supposed to be positive, Z is greater than 1 over root of 2.
35:25:540Paolo Guiotto: So it means that, we have a point, 1 over root of 2.
35:30:170Paolo Guiotto: After 1 over root of 2 arrives, the function increases. At left, decreases. It means that that's a minimum point. In fact, the function has a plot like that.
35:40:720Paolo Guiotto: And at the left, we have the symmetric picture, so this will be minus 1 over root of 2…
35:47:700Paolo Guiotto: So, these are minimum points for this age.
35:53:130Paolo Guiotto: So it means that the minimum value of H is the value that function h takes at 1 over root of 2.
36:01:260Paolo Guiotto: So, minimum.
36:03:910Paolo Guiotto: for Z in 0 plus infinity of H of Z,
36:11:280Paolo Guiotto: That, because the function is symmetric, H.
36:16:880Paolo Guiotto: even.
36:18:340Paolo Guiotto: Otherwise, you could discuss the negative part similarly.
36:22:290Paolo Guiotto: turns out to be the minimum of values H of Z for Z in the interior line.
36:29:720Paolo Guiotto: it is equal to H at point 1 over root of 2. You can also…
36:34:610Paolo Guiotto: See, what is it? Because the function h is Z power 4 minus Z squared, so 1 over root of 2 to power 4 is
36:42:910Paolo Guiotto: 1 over 4… Minus 1 over 2, so it is equal to minus 1 over 4.
36:50:620Paolo Guiotto: So this means that our quantity, Z power 4 minus Z squared, is always greater or equal than minus 1 fourth, whatever is Z real.
37:02:500Paolo Guiotto: Okay? So now we have this bound
37:06:490Paolo Guiotto: for this part Z4 minus Z squared. We have this bound for the other part, we put together, we get the bound for F.
37:16:330Paolo Guiotto: So we have that F… of X, Y, Zed, huh?
37:22:560Paolo Guiotto: is equal, so let me just rewrite. We wrote in this way, X squared plus Y squared plus Z squared minus XZ, we reorganized it in this way, plus Z power 4 minus Z squared.
37:36:350Paolo Guiotto: We know that this part here, we proved it is greater or equal than 1 half raw square.
37:42:890Paolo Guiotto: This guy here is greater than or equal than minus 1 fourth.
37:46:900Paolo Guiotto: So everything will be greater or equal than 1 half raw square minus 1 fourth, and that's our desired G of raw that goes to plus infinity when rho that goes to plus infinity.
38:04:360Paolo Guiotto: So we are now in the conditions to conclude that there exists a limit for XYZ going to infinity.
38:14:900Paolo Guiotto: of this F, and the limit is equal to plus infinity.
38:21:610Paolo Guiotto: Okay.
38:23:30Paolo Guiotto: Good.
38:24:490Paolo Guiotto: So, I'd say that,
38:27:290Paolo Guiotto: I will publish solution of the remaining exercises about this.
38:33:780Paolo Guiotto: So, if you have, not been able to… I don't know if you tried to do these exercises, there is, I've done number 6,
38:46:350Paolo Guiotto: Number 7, it's much easier.
38:49:690Paolo Guiotto: You should understand.
38:52:360Paolo Guiotto: No, I've done number 5. Okay, you can do, for example, if you have not yet done, do number 6, which is similar to number 5, so this last one. So, a similar idea, you have to think about…
39:08:340Paolo Guiotto: then you will see the solution I will publish this afternoon, maybe this afternoon, late afternoon, or tomorrow morning. By the way.
39:15:900Paolo Guiotto: Tomorrow… tomorrow I cannot do class, so, there is no class tomorrow, so we will meet on Friday.
39:23:810Paolo Guiotto: Okay?
39:24:960Paolo Guiotto: Do you want to take a shot back?
39:27:370Paolo Guiotto: Okay, 5 minutes.
39:37:820Paolo Guiotto: Okay, as I said,
39:40:620Paolo Guiotto: For… for the moment, we stop here, we're doing exercises on calculus of limits. We will do, again, soon, because,
39:51:590Paolo Guiotto: When we enter into differential calculus, we will have to compute limits, as you will see.
39:57:60Paolo Guiotto: So we will return on this kind of problems later.
40:00:930Paolo Guiotto: Now,
40:04:650Paolo Guiotto: So we have to finish this part, of course, so we have to do, basically, a couple of things. First of all, to talk about continuity.
40:19:800Paolo Guiotto: One of the…
40:21:540Paolo Guiotto: First, applications of the definition of limit is to the definition of what is a continuous function.
40:29:530Paolo Guiotto: So, the definition is exactly the same. You have seen for one variable functions. So, we have a function F defined on a domain D, which is a subset of some space, say, RD,
40:45:220Paolo Guiotto: With values in… oh, okay, so we are talking about general functions, so let's take… the general case is a vector-valued function.
40:54:220Paolo Guiotto: into our M, say.
40:57:730Paolo Guiotto: So take a point, X, X star.
41:06:500Paolo Guiotto: In the domain,
41:08:550Paolo Guiotto: So the idea is that we want to say that the function f is continuous at point x star if the limit of the function when x goes to X star is the value of the function.
41:19:10Paolo Guiotto: So, let's leave some… something here, some space, but that's where we want to arrive, a limit of, to say this, limit when X goes to X star.
41:30:680Paolo Guiotto: of, F, of X, is equal to F evaluated at point X star.
41:40:780Paolo Guiotto: Now, to write this, we need…
41:43:80Paolo Guiotto: First, that we can evaluate F at point X star, so that's why you see X star must be a point in the domain. So this means that continuity is meaningless outside of the domain of the function.
41:56:170Paolo Guiotto: Where the function is not defined, it doesn't make any sense to say the function is continuous or is not continuous. This is not a novelty, no? You say function 1 over x is not defined at x equals 0.
42:11:740Paolo Guiotto: So, you say it is discontinuous. No, it's not defined, so discontinuity does not make any sense, you see?
42:20:50Paolo Guiotto: And second, since we need to compute the limit of F, we need that… that point external is not any point, but it must be also an accumulation point for the domain.
42:33:290Paolo Guiotto: Now, here you may wonder, is it not sufficient to say that X in the domain, X star in the domain, is automatically an accumulation point for the domain. The answer is no. Let me show you with a figure, because if domain I color in red the domain.
42:52:550Paolo Guiotto: Domain is made of this set here. So this is domain, huh?
42:57:380Paolo Guiotto: Plus a single point somewhere here.
43:01:430Paolo Guiotto: So if this is the domain, so this is in the domain here.
43:07:560Paolo Guiotto: That point is not an accumulation point for the domain, because you remember, accumulation point means there exists a sequence of points of the domain different from the point that goes to that point, and there is no sequence that does this.
43:23:110Paolo Guiotto: Okay? So, that's why being in a domain is not automatically implying that you are an accumulation point of the domain, that's why you have to put both.
43:33:240Paolo Guiotto: In most of the cases, let's say that once you are in the domain, you are also an accumulation point, unless your point is an isolated point of the domain. But, okay.
43:43:930Paolo Guiotto: So, we say that,
43:51:320Paolo Guiotto: F…
43:56:900Paolo Guiotto: is continuous.
44:00:110Paolo Guiotto: At… point.
44:02:930Paolo Guiotto: X star… if this condition holds. So, if the limit
44:10:690Paolo Guiotto: for X going to X star of the function f is equal, to the value of the function at point x star. So we may say, in other words, that if we already know that the function is continuous, computing its limit is easy.
44:26:340Paolo Guiotto: Because you just evaluate the function at the point, and you have the value of the limit. You don't need to do any kind of calculation.
44:33:800Paolo Guiotto: If you think about to all the previous examples, we have seen of calculus of limits. Do not look at cases when we computed the limit at infinity. Infinity is not the point of the domain. But when we computed, for example, many limits at zero.
44:50:470Paolo Guiotto: What
44:55:600Paolo Guiotto: So we had the, for example, let's take this one. This was a limiting 0. The function you have there is not defined at zero, so definitely cannot be continuous, okay? Because continuity makes sense only at points of the domain. If your point is not in the domain, it means that the function cannot be defined continuous.
45:15:210Paolo Guiotto: So this means that you cannot, of course, just replace X equals 0, y equals 0, and compute the limit. Here, it is clear, because if you do that, you see that you get 0 divided 0, which is not defined.
45:27:420Paolo Guiotto: So, in general, when you compute a limit, an interesting limit is more or less always where the function is not defined.
45:37:130Paolo Guiotto: You do that to see what happens to the function, to be not defined, because that point maybe is a point where the definition, loses some… some meaning. For example, you have a division by zero, a logarithm defined at zero, and so on.
45:56:120Paolo Guiotto: But if you already know that the function is continuous there, you can, you could, in principle, compute the limit by just doing the evaluation. This will never happen, computing the limit, but let's say this just for your information.
46:10:260Paolo Guiotto: Okay, so what, what is important to know about, discontinuity? As you will see, the continuity is an important assumption in many results, so,
46:25:800Paolo Guiotto: It is important to understand how to check continuity, So, how… can… We… Sure.
46:39:30Paolo Guiotto: Or let's say, quickly.
46:44:940Paolo Guiotto: Check continuity.
46:48:150Paolo Guiotto: Of course,
46:50:440Paolo Guiotto: you understand that you cannot verify continuity with the definition, that is, computing the limit at every point of the domain, because the domain will be made of infinitely many points, and in general, this operation won't be easy. So we need some criteria that makes this check easy.
47:10:210Paolo Guiotto: So, now I will give you some informations without, basically any proof, because they are more or less, acceptable. You may expect that this is… these properties are true, and so we don't need to justify
47:25:840Paolo Guiotto: In a particular way. So the first remark is the… the following, so facts.
47:35:650Paolo Guiotto: So number one, a function F, sorry, a vector-valued function F, is continuous.
47:49:390Paolo Guiotto: At some point. If and only if… ALL.
47:54:970Paolo Guiotto: Eats.
47:56:880Paolo Guiotto: components. Now let's, components.
48:04:190Paolo Guiotto: are continuous.
48:07:830Paolo Guiotto: What are we saying here?
48:09:900Paolo Guiotto: So, this is a… this is a vector-valued function. So, when you look at this F,
48:17:110Paolo Guiotto: This is a F array vector of X vector.
48:23:210Paolo Guiotto: So it means that this guy here is not a number, it's a vector, no? It's a vector of, we say, the RM, right, in this notation.
48:33:670Paolo Guiotto: So it means that it will have M components.
48:37:590Paolo Guiotto: So normally, we can write this as an array with M components.
48:44:270Paolo Guiotto: And each of these components, of course, since this is a function.
48:49:430Paolo Guiotto: So this depends on X to X. Also, the confidence will depend on X, obviously. So there will be functions
48:59:560Paolo Guiotto: Components are numbers, so numerical functions, so let's say F1…
49:04:630Paolo Guiotto: of vector X, F2 of vector X, and so on, FM of vector X.
49:13:70Paolo Guiotto: So these are the components.
49:17:710Paolo Guiotto: components.
49:22:400Paolo Guiotto: off F.
49:24:970Paolo Guiotto: No? So…
49:26:360Paolo Guiotto: Just an example, no? Imagine that you have a F array of XY, so function of two variables, so an array made of two numbers.
49:38:550Paolo Guiotto: This equal to, I don't know, X plus Y, X times YX divided Y. So you see, this is a function that takes a vector XY in
49:51:480Paolo Guiotto: R2… And it gives you a vector, F of XY, that belongs… Ehh…
50:03:960Paolo Guiotto: They have all my admiration, as you may imagine.
50:08:490Paolo Guiotto: And it yields you a vector with three components, so this guy belongs to R3.
50:14:500Paolo Guiotto: Now, let me explain another reason, that's… I know it's personal, so you… you would say, who cares about that?
50:22:50Paolo Guiotto: But imagine that all of you
50:24:40Paolo Guiotto: attend classes online, so I would be here alone, talking to Dunabdi, and moreover, you would be there in Zoom, maybe with any image. You see, I see a letter A here, so probably you are… you are playing at PlayStation or something like that, or with your mobile, and not Antenemaker. So imagine how can I
50:48:80Paolo Guiotto: Do the class in this way.
50:50:260Paolo Guiotto: Talking to… to nobody.
50:52:370Paolo Guiotto: And I need your eyes. I need your response, because I can understand… I can have an immediate feedback. Maybe this part is not so easy, so I should repeat, because I see that you are not completely convinced on what you are doing. Or maybe something is clear. You see, your feedback for me.
51:13:80Paolo Guiotto: And for any of my colleagues, it's important. That's also why you should be here.
51:18:670Paolo Guiotto: Because the class can be better. Otherwise, I do not have any feedback, and therefore, I don't know what I'm doing, actually. Maybe what I'm doing is clear to me, but not to you.
51:31:650Paolo Guiotto: You know, I don't want to read your justifications, okay?
51:37:150Paolo Guiotto: You may have all the nice motivations, but…
51:42:890Paolo Guiotto: Okay, so in this case, as you can see, we have a function, from R2 to R3. So these are the three components, F1, F2, F3, and as you can see, these are three functions which takes numerical values, no? So F1 of XY,
52:02:250Paolo Guiotto: is X plus Y.
52:05:430Paolo Guiotto: F2 of XY is X times Y.
52:10:730Paolo Guiotto: F3 of XY is X over Y.
52:16:880Paolo Guiotto: So, the… these are…
52:21:950Paolo Guiotto: We are not recording.
52:25:420Paolo Guiotto: No, is it… it is important.
52:28:200Paolo Guiotto: Okay. So, these are the three components of this example. So, the components, you have to imagine that this F of X is an array with M components. Each of these components is actually F function, so the components are, so…
52:47:160Paolo Guiotto: we normally write F… if we write something like this, F equals F1, F2, etc, FM…
52:57:740Paolo Guiotto: Of course, we mean that each of them, so if F is a function defined on a domain D of RD with values in RM,
53:09:490Paolo Guiotto: The domain of definition is, of course, the set of vectors for which the F of X makes sense, but all the components should make sense, otherwise the array will not make sense. So, this is equivalent to say that we have functions, F1, F2, etc. F, M defined on domain D,
53:32:180Paolo Guiotto: contained off on RD with values in R.
53:37:30Paolo Guiotto: Okay, so these are the components. Now, this property is saying that to check continuity is sufficient to check the continuity of the components, which are still functions of an array, but they are numerical functions, so it's a sort of reduction of the…
53:56:50Paolo Guiotto: Oh, the second point… so,
53:59:840Paolo Guiotto: In particular, important class of functions that we use often is the class of polynomials, as you have seen in lots of examples, we… here, we most of the time computed the limits with polynomials, so sum of products of powers of the variables, okay?
54:22:880Paolo Guiotto: So, the fact is that a polynomial So, polynomials…
54:36:40Paolo Guiotto: off… vector variable, R, continuous, functions.
54:46:30Paolo Guiotto: on… RD.
54:49:70Paolo Guiotto: So, a polynomial means that a polynomial is a function that we normally denote with letter P of an array X defined on the full RD, if there are no restrictions.
55:02:680Paolo Guiotto: real valued, and normally how a polynomial is made is P of X,
55:11:300Paolo Guiotto: is a sum of something like a coefficient, then we have X1 to some exponent K1, X2 to some exponent k2, and so on, XD to some exponent KD.
55:27:660Paolo Guiotto: And then we do sum of terms like this, a finite sum.
55:32:240Paolo Guiotto: Find it.
55:34:380Paolo Guiotto: sum, no? So, for example.
55:37:530Paolo Guiotto: all functions we have seen in most of the examples of this type. A polynomial in true variable could be, I don't know, right, randomly, X power 3 plus XY squared minus 4XY plus 3. That's a polynomial.
55:56:220Paolo Guiotto: Okay? A polynomial in three variables, XYZ, can be XYZ plus Z power 4, for example, okay?
56:07:990Paolo Guiotto: Now, this kind of functions are well-defined for every point, every vector, and they are continuous
56:17:200Paolo Guiotto: As you may expect, this, you know, it's a basic factor from functions of one variable. Polynomials are, always continuous.
56:26:70Paolo Guiotto: The third important fact for us is that
56:31:100Paolo Guiotto: Now, imagine we have a function F, which is continuous, F, function, numerical function of vector variables. So this is defined on D, contained in RD, with values in R.
56:49:960Paolo Guiotto: And suppose that we have a second function, let's call it G,
56:54:930Paolo Guiotto: Now, what we want to do is… let's leave some space here. What we want to do is G of F of X, so a composition. I don't want to discuss the composition in general, because it becomes a mess with indexes, but let's treat
57:12:10Paolo Guiotto: Simple and most, frequent cases.
57:15:730Paolo Guiotto: So imagine that, now, if I want to do g of f of x, f of x is a number, so I just need that G be a function of real variables. So, to simplify things, let's call the variable of GY, and assume that this is a function R to R.
57:32:430Paolo Guiotto: This such that the function F and G are both continuous.
57:42:870Paolo Guiotto: Then, decomposition, G of F of X is continuous.
57:49:980Paolo Guiotto: Where?
57:51:790Paolo Guiotto: Define.
57:54:620Paolo Guiotto: Actually, we could state a general fact that says when you compose continuous functions, you always get a continuous function, provided it is defined.
58:05:340Paolo Guiotto: So this is an important fact that we use with functions of one real variable. That is, the composition of continuous functions is continuous where defined, okay?
58:18:20Paolo Guiotto: Here, I'm a bit specifying in a particular case, because, for example, I want to say something like this. Suppose that I have my function, and now
58:29:210Paolo Guiotto: let's call it H for a second, XY be the function, I don't know, logarithm of 1 plus X plus Y.
58:39:70Paolo Guiotto: Okay.
58:41:790Paolo Guiotto: Now, imagine that you have to discuss where this function is defined and continuous.
58:47:320Paolo Guiotto: determine… Where… H is defined And… continuous.
59:00:10Paolo Guiotto: Now, for the domain of definition.
59:05:790Paolo Guiotto: The domain is always the set of points, XY, for which your function makes sense.
59:13:640Paolo Guiotto: Now, what do we need to… to have this thing defined?
59:18:180Paolo Guiotto: We have the argument of log is 1 plus X plus Y. It's a polynomial, okay? As you can see, this is a polynomial.
59:31:620Paolo Guiotto: And this is always defined. There is no restriction. Then we have to compute the log of that quantity. Log is not always defined. It demands that the argument be strictly positive. So we put the condition 1 plus X plus Y positive.
59:48:200Paolo Guiotto: And that's the domain.
59:49:960Paolo Guiotto: In this case, we may have an idea of what is this domain, because it is a subset of R2, so we can do some figure here, because 1 plus X plus Y positive.
00:03:90Paolo Guiotto: is equivalent to say you, for example, could write a Y greater than minus 1 minus X.
00:10:500Paolo Guiotto: Now, what is this?
00:12:650Paolo Guiotto: It is the plane region of points XY,
00:17:750Paolo Guiotto: for which the Y is greater than minus 1 minus X.
00:21:630Paolo Guiotto: How can I plot that? Well, I can plot, which is not in the region, the set where y is equal to minus 1 minus X. It's not in the region. But let's plot this, because it is easy, that's a straight line, no?
00:36:140Paolo Guiotto: Slope is minus 1, so it is about… this is point minus 1, it is about something like this. This red dashed line
00:47:760Paolo Guiotto: is the line Y equal minus 1 minus X.
00:53:120Paolo Guiotto: So what does it mean, Y equal minus 1 minus X? It means that if you pick a point here, DY is equal to 1 minus X. So if this is DX, this is DY. DY is equal.
01:08:760Paolo Guiotto: So, if you want that Y being greater, you must have that the Y is bigger than that. So, you are above the red line. So, the domain is this.
01:20:10Paolo Guiotto: part of the plane.
01:22:240Paolo Guiotto: It's an aft plane, actually.
01:25:560Paolo Guiotto: So in green, you see what is the domain, above that line.
01:30:930Paolo Guiotto: Okay.
01:32:600Paolo Guiotto: Well, it is not particularly important to be able to do this kind of figure, but if possible, you must be able to do, at least in simple cases, and you will see that it's not so easy, okay? Because you have always to think in terms of a plane region.
01:49:380Paolo Guiotto: So, for example, if I say, let me just, I tell you, are you able to plot the region where X is positive? What is this?
01:59:280Paolo Guiotto: But people might think it's 0 to plus infinity. But in plain R2,
02:05:180Paolo Guiotto: Points where X is strictly positive, where points where x is 0 is the y-axis. This is…
02:12:390Paolo Guiotto: the set of points where x is 0.
02:14:840Paolo Guiotto: The set of points where x is positive are those at the right of this, so in this case, the domain would be enough plane like that.
02:23:790Paolo Guiotto: So the set where X is positive is not an interval.
02:27:550Paolo Guiotto: is a half plane, okay? You have to be careful, because we are always working in some vector space, so the condition is on all the coordinates. So if one of the coordinates has no condition, it means it is free to vary in its range.
02:43:900Paolo Guiotto: Okay, so we will return on these kind of operations when we do exercises on this.
02:50:230Paolo Guiotto: And of course, it won't be always possible to draw a figure, because it could be impossible to have an idea on how the domain is made. But normally, when we do exercises, just to practice, we keep… we try to keep things easy, so you should be always more or less able to plot the figure.
03:10:890Paolo Guiotto: If required, you should be able. If it is not required, at least that's probably it's too difficult, so forget it. Okay, now we know that that's the domain.
03:21:340Paolo Guiotto: What the… so the question was, determine where H is defined and continuous. So we know that that def is defined in that domain. Now, what about the continuity? As you can see, our H… so let's discuss continuity.
03:42:200Paolo Guiotto: Our age… Can be… so, which is log of,
03:48:160Paolo Guiotto: Again, look at the, at the arguments at the examiner, not at the specific calculations, which are clearly very easy here. I can see this as a G of F of X as a composition.
04:03:180Paolo Guiotto: It is a G of F of XY,
04:08:760Paolo Guiotto: Where XY can be taken this one, the polynomial.
04:14:230Paolo Guiotto: And G is this one, is the logarithm.
04:17:770Paolo Guiotto: So, you see, we are exactly in this, in this situation. We have a function defined, the domain is in R2, real value. A second function, if both are continuous, the composition is continuous, where defined.
04:32:620Paolo Guiotto: Now, here we can say that the set of points where this composition is defined is the domain of the function H.
04:41:580Paolo Guiotto: Okay? Because the composition is defined when the argument of log is positive, which is the domain. So we can conclude that since
04:51:240Paolo Guiotto: Both.
04:53:530Paolo Guiotto: F… equal 1 plus X plus Y.
04:59:650Paolo Guiotto: and G. Now, since Y is used for the coordinate, I will use a different letter. Let's say G of T equals log of t, okay? Since these are…
05:12:140Paolo Guiotto: continuous.
05:14:210Paolo Guiotto: functions.
05:17:580Paolo Guiotto: Well… defined.
05:23:110Paolo Guiotto: Then, direct composition, which is our H, so F followed by G, which is our H, is continuous, Where?
05:34:830Paolo Guiotto: defined, huh?
05:37:340Paolo Guiotto: So… H is continuous in
05:41:510Paolo Guiotto: its natural domain Z, okay? So it is continuous where it is defined. And that's it for this question.
05:48:860Paolo Guiotto: As we will see continuity, you know, where the function is continued.
05:53:220Paolo Guiotto: is important, for main applications, so… Okay, so let me… to act about,
06:06:830Paolo Guiotto: Okay, well, let's see, yes, something that it's,
06:11:750Paolo Guiotto: Also useful to know. Second example.
06:17:390Paolo Guiotto: Now, consider this function, which is the norm as a function, huh? So, if you take the norm.
06:25:490Paolo Guiotto: as a function.
06:27:450Paolo Guiotto: So I will write it with this symbol. I put dot to say that this is not the norm of a specific vector, but the norm as a function. So the function that takes a vector of RD and gives you a number. Actually, it is a positive number, but we write R. So the function that takes X,
06:46:100Paolo Guiotto: and it's the norm of X.
06:51:240Paolo Guiotto: Okay, so we have a function defined on our B numerical function. Now, it turns out that this function, NORMA,
06:58:900Paolo Guiotto: is continuous, on… Pardon.
07:06:590Paolo Guiotto: Let's see why.
07:09:390Paolo Guiotto: Because what is the norm of X?
07:13:210Paolo Guiotto: Well, if you look at the definition, this is the Euclidean norm, this is the square root of X1 squared plus X2 squared plus, etc, XD squared.
07:27:780Paolo Guiotto: So we may look at this as a composition, you see?
07:31:670Paolo Guiotto: Because, we do, G.
07:36:150Paolo Guiotto: of F of X, huh?
07:39:920Paolo Guiotto: Where?
07:43:410Paolo Guiotto: Let's say, what do we do first on X? The first thing we calculate is this quantity here.
07:51:620Paolo Guiotto: And this quantity, as a function of X, is what kind of function?
07:58:320Paolo Guiotto: Numerical, yes, but it's a specific type of function.
08:02:380Paolo Guiotto: He's the son of… Powers is a polynomial, okay? So where? F of X is the function
08:11:120Paolo Guiotto: X1 squared plus X2 squared plus, etc, XD squared, that's a polynomial.
08:21:90Paolo Guiotto: So a function which is continuous everywhere.
08:24:460Paolo Guiotto: And the second one, G of, you can use the letter, let's write T, is the square root of T. That's the second function.
08:33:979Paolo Guiotto: This is, continuous.
08:38:69Paolo Guiotto: 4… positive or zero. It's continuous everywhere.
08:44:10Paolo Guiotto: Where it is defined.
08:45:950Paolo Guiotto: Now, we concluded that the composition F followed by G, which is the norm as a function.
08:54:870Paolo Guiotto: is continuous.
08:57:170Paolo Guiotto: Where… Defined. And where the norm is defined.
09:02:930Paolo Guiotto: It's always defined, because we do the root of a positive number, so… So, on,
09:13:779Paolo Guiotto: So again, we are using some of the facts we have seen, no? Polynomials are a large class of continuous functions, and whenever we do some either non-polynomial function, like the root.
09:27:300Paolo Guiotto: Or maybe I do the exponential of sine, cosine, no? So, for example,
09:35:540Paolo Guiotto: If I give you, I don't know, this function, F of XY equal, I don't know, first component sine X plus Y, second component E2X, I don't know, cosine Y, for example.
09:55:150Paolo Guiotto: This is a function F.
09:58:60Paolo Guiotto: clearly is a function defined on domain V. We can see here that there is no restriction for the coordinates, because we can always do X plus Y, and then sine of X plus Y, as well as into X cosine of Y, so domain is everything.
10:13:120Paolo Guiotto: And it is valued in R2.
10:17:220Paolo Guiotto: Okay?
10:18:520Paolo Guiotto: Now, discuss, determine where… F… is continuous.
10:30:120Paolo Guiotto: So, how can we do this discussion? So, we can use some of the facts we mentioned above. The first fact says that if you want to check the continuity for a vector-valued function, it's equivalent to check the continuity for the components.
10:47:130Paolo Guiotto: Okay, so let's take each of the components.
10:51:10Paolo Guiotto: So, Latin… F1, the first component, is, this.
10:58:610Paolo Guiotto: And F2 is this, so F1 is sine of X plus Y.
11:04:620Paolo Guiotto: and the F2… is, XY.
11:10:240Paolo Guiotto: E2X costs Y.
11:13:820Paolo Guiotto: Now, as you can see, so let's discuss,
11:19:360Paolo Guiotto: I will not ask you to do this kind of discussion every time, but this is just to make you confident that continuity is more or less always verified whenever you compose good functions. Let's discuss
11:36:800Paolo Guiotto: continuity.
11:38:860Paolo Guiotto: of F1, and… F2.
11:44:40Paolo Guiotto: Okay, so F1.
11:47:310Paolo Guiotto: F1 can be seen as a composition. It is a sign.
11:53:960Paolo Guiotto: of a polynomial, PXY.
11:58:960Paolo Guiotto: Where this is, the polynomial X plus Y.
12:06:40Paolo Guiotto: It's a polynomial.
12:08:380Paolo Guiotto: And, so the composition, no, so we know that this is a continuous function wherever, no? So in R2.
12:17:340Paolo Guiotto: sine is a continuous function. In the real line, the composition is always defined, we get that F1 is continuous web defined, which is in F2.
12:30:900Paolo Guiotto: about F2… which is E2X times cos y.
12:39:800Paolo Guiotto: Well, it is clear that this is a, we know, a continuous function of X, this is a continuous function of Y,
12:46:910Paolo Guiotto: And that product will be a continuous function of X and Y.
12:51:500Paolo Guiotto: So, this will be a continuous function on R2.
12:55:540Paolo Guiotto: So since the two components are continuous on R2, the vector is continuous on at 2.
13:08:500Paolo Guiotto: So… Unless explicitly stated that you don't need to… Verify.
13:16:760Paolo Guiotto: Anything in particular. Remark?
13:25:660Paolo Guiotto: which is just something we already said, but not in this form. And since this kind of functions is important, for example, in algebra, it was to be mentioned. So, linear
13:41:710Paolo Guiotto: linear… functions.
13:46:760Paolo Guiotto: R… Continuos.
13:52:800Paolo Guiotto: So what does it mean, linear function?
13:56:150Paolo Guiotto: A linear function is a function defined, then I will put the, all the…
14:03:670Paolo Guiotto: formal things. So you have the metrics.
14:09:820Paolo Guiotto: Let's say that we have M by D metrics, so we have M lines by D column, so D
14:19:10Paolo Guiotto: M… So this is A11, A12, etc.
14:24:430Paolo Guiotto: A1D.
14:26:950Paolo Guiotto: Then here we have a second line, A21, A22, etc, A2B.
14:33:190Paolo Guiotto: And so on. The last line is M, AM1, AM2, AM…
14:41:690Paolo Guiotto: Now, we can do the product line by column, and we have an operation matrix times vector. So, if we have a vector with exactly the same number of columns, so this will have D components, X1, XD,
14:59:520Paolo Guiotto: This is the, the definition of the operation. We have an array, which is made like that, A11X1 plus A12X2, and so on, plus A1DXD.
15:17:750Paolo Guiotto: And the second line, similarly, and the last line, AM1, X1, AM2.
15:25:330Paolo Guiotto: X2, and so on.
15:28:880Paolo Guiotto: AMD.
15:30:460Paolo Guiotto: XD.
15:33:650Paolo Guiotto: So, in this way, we have a function, F, that takes, a vector of RD,
15:42:550Paolo Guiotto: and gives you a vector of Rn.
15:47:760Paolo Guiotto: Okay? There is a compact way to write this. If you call A.
15:53:990Paolo Guiotto: M by D matrix.
15:59:330Paolo Guiotto: This can be written in this way. F of X, is simply A.
16:05:680Paolo Guiotto: times X. We do not write the dot, because it is clear if A is a matrix, X is a vector. A times X means exactly this thing.
16:16:400Paolo Guiotto: this operation.
16:18:520Paolo Guiotto: Now, this function, F, is a continuous function. Why? Because if we look at components, these are the components of the function.
16:28:380Paolo Guiotto: These components are what kind of functions?
16:31:790Paolo Guiotto: polynomials. They are first-degree polynomials. So they are continuous, and therefore the F is continuous.
16:39:630Paolo Guiotto: So…
16:44:220Paolo Guiotto: So F… ease.
16:48:100Paolo Guiotto: Continuos.
16:49:760Paolo Guiotto: on.
16:51:140Paolo Guiotto: V… Because… all.
16:58:160Paolo Guiotto: weeks.
16:59:490Paolo Guiotto: components.
17:05:590Paolo Guiotto: which are, if you want, FJX1, XD.
17:12:580Paolo Guiotto: So the J component is the J line of this array, so we will have AJ1X1 plus AJ2X2, and so on.
17:28:370Paolo Guiotto: A J, D, X, D.
17:32:390Paolo Guiotto: These… these components are… polynomials.
17:41:410Paolo Guiotto: for J equal 1, 2, 3, 2, M.
17:46:540Paolo Guiotto: There are M components. So this is an important class, it's not, of course, the most general class of functions, but these functions are important because in linear algebra, these are the kind of functions you use, and they are continuous functions.
18:02:500Paolo Guiotto: Okay.
18:03:800Paolo Guiotto: So, in principle, we should have still a few minutes.
18:09:160Paolo Guiotto: But, since we should now start with
18:18:60Paolo Guiotto: other topics, I… I prefer to stop here.
18:22:870Paolo Guiotto: Does it?
18:23:970Paolo Guiotto: We would just, introduce,
18:27:350Paolo Guiotto: few definitions without any comments. So let's stop here for today, and a reminder, we will meet on Friday, okay?
18:37:450Paolo Guiotto: Have a nice day.