AI Assistant
Transcript
00:10:410Paolo Guiotto: Okay.
00:11:410Paolo Guiotto: We start with some of the exercises I left you.
00:18:900Paolo Guiotto: That's… So, exercise…
00:35:740Paolo Guiotto: Exercise 1815.
00:39:220Paolo Guiotto: We did the first two. Now, let's do number 3.
00:46:730Paolo Guiotto: We still have to say if the set is open, closed, bounded, compact. The set D is the set of points
00:57:340Paolo Guiotto: XY… in our tool… Such that modulus of X plus modulus of Y is strictly less than 1.
01:11:550Paolo Guiotto: Okay, as you can see, this set is defined to a strict inequality.
01:17:380Paolo Guiotto: involving this function, so let, say G of XY.
01:23:900Paolo Guiotto: be the function modulus of X plus modulus of Y minus 1. So, our domain D is the set of points where G is less than 0.
01:37:30Paolo Guiotto: Now, this function is not a polynomial, because multiples of X is not X.
01:42:330Paolo Guiotto: However, modulus of X is continuous, modulus of Y is continuous, since… this function G is continuous.
01:54:150Paolo Guiotto: on R2… We conclude that this set D is open because of the strict inequality.
02:06:100Paolo Guiotto: Now, is it closed? To be closed, should be both open and closed.
02:13:590Paolo Guiotto: So, I can say that D is closed.
02:19:60Paolo Guiotto: implies that D is, both… Open.
02:27:320Paolo Guiotto: and close them.
02:29:760Paolo Guiotto: And we say that this is possible if and only if D is either empty or
02:37:480Paolo Guiotto: The full set up, too.
02:40:320Paolo Guiotto: Is that the case? No, because as you can see, D is non-empty. For example, 0.00 belongs to D.
02:48:740Paolo Guiotto: But… 00 is a point that belongs to V,
02:55:680Paolo Guiotto: You plug this into the condition, modulus of 0 plus modulus of 0 is 0, strictly less than 1.
03:03:320Paolo Guiotto: So, this says that D is non-empty.
03:08:720Paolo Guiotto: And, D is, is not R2, because there are points of R2 which are not there. For example, 1-1.
03:18:110Paolo Guiotto: is not in D, so this says that D cannot be R2.
03:25:210Paolo Guiotto: So the conclusion is that, D… Cannot… be closed.
03:36:300Paolo Guiotto: So it won't be compact, because to be compact, it must be closed and bounded.
03:54:460Paolo Guiotto: But it is bounded, in this case.
03:59:170Paolo Guiotto: Let's see this… the… Ease.
04:02:860Paolo Guiotto: bounded.
04:05:290Paolo Guiotto: Because bounded means that the norm of vector of this set is bounded by a constant. Here, models of X plus models of Y is not the Euclidean norm, but, however, from this constraint.
04:22:130Paolo Guiotto: Yeah?
04:23:350Paolo Guiotto: We can deduce something, indeed.
04:28:100Paolo Guiotto: I told you there is not a general way to prove if a set is bounded.
04:35:270Paolo Guiotto: But the goal is clear. You have to get a bound for the coordinates, or a bound for the normal. Now, from this, you can see that…
04:45:110Paolo Guiotto: Now, since this sum is less than 1,
04:48:700Paolo Guiotto: and we are summing positive numbers, this is important, I can deduce that each of them must be less than 1.
04:59:150Paolo Guiotto: Because if one of them is larger than 1, the sum would be larger than 1.
05:06:810Paolo Guiotto: If, one… Off.
05:11:720Paolo Guiotto: modulus X is larger than 1, or modulus Y is larger than 1, then clearly, it's not true that if the sum of two numbers is less than 1, the two numbers must be less than 1. This can be false.
05:26:820Paolo Guiotto: 100 minus 100 is 0, so I can say 100 plus minus 100 is 0, which is less than 1.
05:35:600Paolo Guiotto: But the one number is 100, so you see?
05:39:20Paolo Guiotto: This is because they are positive. Since modulus of X and modulus of Y are, in this case, positive numbers, then you have that modulus X plus modulus y would be greater or equal than, for example, modules of X that is greater than 1, but this should be less than 1, and this is impossible, okay? So this is the…
06:02:180Paolo Guiotto: argument behind this.
06:04:110Paolo Guiotto: Now, once we have this, we have a bound for the coordinates.
06:07:920Paolo Guiotto: So this should be… it is sufficient. If you want, you can say that if you want to compute the norm, the norm of XY is…
06:17:540Paolo Guiotto: the root of X squared plus Y squared, but since you have this bound here.
06:25:470Paolo Guiotto: modulus of X less than 1, modulus of Y less than 1, so it means X between minus 1 and 1, and Y between minus 1 and 1. When you do X squared plus Y squared, both these numbers are less or equal than 1.
06:39:230Paolo Guiotto: So this is no larger than root of 2, and this is the constant m that holds for every point XY of the set D. So this explains precisely why this set is bounded.
06:57:810Paolo Guiotto: Okay, now let's try to… To see… well, let's do the number 5 first, which is easy.
07:06:810Paolo Guiotto: We have this domain, it is made now points XYZ in R3.
07:13:450Paolo Guiotto: Such that We have this condition, X squared, plus 2Y square…
07:21:290Paolo Guiotto: plus 3Z to power 4, less or equal than 4.
07:35:80Paolo Guiotto: Okay, so as you can see here.
07:38:600Paolo Guiotto: We can say that our domain D
07:41:190Paolo Guiotto: is defined to an inequality like G less or equal 0, where…
07:48:340Paolo Guiotto: the function G of XYZ is, of course.
07:53:660Paolo Guiotto: X squared plus 2Y squared plus 3Z to power 4 minus 4.
08:02:630Paolo Guiotto: It's a polynomial, and so it is continuous. G is… Continos on R3.
08:10:930Paolo Guiotto: And we know that this is a standard fact. Whenever a set is defined by a large or weak inequality, it is a closed set. So, this is the first conclusion. D is closed.
08:26:570Paolo Guiotto: Now, to say if it is also open, as you know, if it is… if it is open, it should be open and closed, then it should be equal to either empty or R3. But in this case, as you can see.
08:42:409Paolo Guiotto: For example.000 belongs to D, so this says that D is non-empty.
08:49:950Paolo Guiotto: And if you want a point which is not indie, take 111, for example, 111.
08:56:350Paolo Guiotto: Because when you put XYZ equal 111, you get 1 square, 1 plus 2Y square, Y squared is 1, so 2, 1 plus 2 is 3, plus 3Z power 4, which is, again, 1, so plus 3, 1 plus 2, 2 plus 3 is
09:12:750Paolo Guiotto: 6, let's request for false. So this is not in the…
09:17:380Paolo Guiotto: So this means that D is not the full space artery, and so this means that D cannot B.
09:29:880Paolo Guiotto: Open.
09:33:00Paolo Guiotto: Street That's funny.
09:37:20Paolo Guiotto: What?
09:39:770Paolo Guiotto: is strictly missing.
09:46:680Paolo Guiotto: Okay.
09:47:980Paolo Guiotto: So… However, it doesn't change this.
09:53:190Paolo Guiotto: No, it changed. Okay, we can just flip the two words, so it is open.
10:00:980Paolo Guiotto: The same conclusion happens here, so it cannot be closed.
10:07:990Paolo Guiotto: Thank you.
10:10:800Paolo Guiotto: Okay, so, it cannot be… compactor?
10:18:620Paolo Guiotto: Because to be compact, you must be both closed and bounded. So if one of two is false, you cannot be compact. It's not…
10:28:580Paolo Guiotto: Compacten?
10:32:760Paolo Guiotto: What about bounded?
10:35:340Paolo Guiotto: Here, you see there is something that looks like, the norm.
10:42:590Paolo Guiotto: For example, here we could use the argument we have seen before, is, D… bounded?
10:52:830Paolo Guiotto: So I noticed that XYZ in D,
10:57:240Paolo Guiotto: implies that this quantity X squared plus 2Y squared plus 3Z power 4 is less than 4. Less, frictly less.
11:10:600Paolo Guiotto: Now, you see that this is some… Off.
11:16:50Paolo Guiotto: positive numbers.
11:22:40Paolo Guiotto: So, since the sum is the sum of positive number, which is less than 4,
11:27:990Paolo Guiotto: it must be that each of these three terms must be less than 4. Otherwise, if one is bigger than 4, the sum will be bigger than 4. So I get that
11:38:830Paolo Guiotto: All these three conditions holds together. X squared must be less or equal to 4, 2Y squared must be less or equal than 4, and the 3
11:48:290Paolo Guiotto: Z power of 4, less or equal than 4, from which we get that modulus of X must be less or equal than 2, modulus of Y must be less or equal, Y squared less or equal than 2, so Y in modulus less or equal than root of 2,
12:05:430Paolo Guiotto: And, Z modulus of Z, less or equal than the fourth root of 4 thirds.
12:13:00Paolo Guiotto: Motava di ques.
12:16:950Paolo Guiotto: What changed? If it is… if it is strict, it is also less or equal, right? Here, I need just a bound, the less or equal is not important, okay? So, from this, I get that X squared plus Y squared plus Z squared
12:35:450Paolo Guiotto: will be less or equal than 4 plus 2 plus the square of that, so the root of 4 thirds, whatever it is, it doesn't matter, because I just need a constant. This is the constant M that bounds the square
12:53:890Paolo Guiotto: Actually, this is the square of the normal, so if you want the normal, we just take the root, both sides.
13:02:80Paolo Guiotto: So we have this thing, bounce the norm of XYZ,
13:09:20Paolo Guiotto: And this happens for every point X, Y, Z in… D, so D is bounded.
13:23:410Paolo Guiotto: And this finishes the question.
13:26:890Paolo Guiotto: Is there any question, comment?
13:31:850Paolo Guiotto: Okay, let's see another one. For example, the number 8.
13:38:30Paolo Guiotto: So these last are a little bit more complicated.
13:42:60Paolo Guiotto: Not particularly. So in this case, we have the domain D of points XYZ, in R3… Such that…
13:53:710Paolo Guiotto: XY is Z plus 1, and…
13:59:510Paolo Guiotto: X squared plus Y squared is less or equal than 1.
14:06:670Paolo Guiotto: So now, in this case, the set is defined as…
14:11:10Paolo Guiotto: Well, here we have a mixed situation. There is an equality and a large inequality, but for equalities and large inequalities, it's the same.
14:19:340Paolo Guiotto: So we can say that this domain is defined as, say, G1XYZ equal 0, and G2XY.
14:32:100Paolo Guiotto: Z less or equal 0, where these G1, G2 are the constraints.
14:37:730Paolo Guiotto: And, precisely they are… G1 is, for example, X times Y minus Z minus 1, and G2 is equal to, of XYZ, it is X squared plus Y squared minus 1.
14:54:660Paolo Guiotto: It doesn't matter if the function does not depend on Z here, okay?
15:01:410Paolo Guiotto: It's a function of three variables, it can be independent of one or two of them, okay? Or maybe of all or three of them, it's a constant in that case.
15:12:740Paolo Guiotto: Now, clearly, these two functions are continuous functions in R2, in R3, sorry.
15:18:980Paolo Guiotto: And we know that when the set is defined by…
15:22:910Paolo Guiotto: Large inequalities and equalities, that's a closed set, okay? So, this is a standard fact, the ease.
15:32:630Paolo Guiotto: Closed.
15:36:170Paolo Guiotto: Again, it cannot be open, otherwise it should be equal to empty on the full space, so we have to exclude that.
15:44:540Paolo Guiotto: So, let's, see… let's find a point which is in that,
15:52:820Paolo Guiotto: in that set. For example, saying here it's a bit more complicated, because we have to fulfill two conditions.
16:00:350Paolo Guiotto: So, if I take Y equals 0, and Z equals 0.
16:06:360Paolo Guiotto: So take a point like X00. X00 belongs to D if and only if the two conditions boil down to what? The first one becomes Z… no, sorry, that's wrong.
16:22:350Paolo Guiotto: Let's put the… we have to still keep a degree here of freedom. Let's take a point like X00Z, and let's see when it is in D. It is in D when we plug into these two conditions, even though if the first one becomes 0 equals Z plus 1,
16:41:980Paolo Guiotto: And the second one becomes X squared plus 0 squared, less or equal than 1.
16:48:40Paolo Guiotto: So this means that Z must be equal to minus 1,
16:52:700Paolo Guiotto: And X squared and the modulus of X is less or equal than 1. So any X less or equal in absolute value than 1 is good. So, for example, I don't know, 1, 0, minus 1 belongs to D. Also, I don't know, 0, 0 minus 1,
17:11:970Paolo Guiotto: belongs to D, and so on. I need just one point. So this says that D is non-empty.
17:19:200Paolo Guiotto: On the other hand, D is not the full space RD, because, for example, if you take point 000,
17:30:990Paolo Guiotto: belongs to D, if and only if.
17:34:420Paolo Guiotto: Now, a point to be, indeed, must fulfill the two constraints, not just one.
17:40:180Paolo Guiotto: In fact, you see that for the first constraint, since the three coordinates are 0, at left you have 0, at right you have 0 plus 1, so 1. So this becomes to 0 equal 1, which is impossible, of course, so I can stop even here.
17:55:710Paolo Guiotto: It doesn't matter if the second condition is verified. The second condition is 0 square plus 0 squared, so 0 less or equal than 1.
18:04:170Paolo Guiotto: So, but in any case, since the two conditions must be verified together, and one of them is false, this means that the system of two is not verified. So this means that particular point 000
18:21:750Paolo Guiotto: is not in D, and so this means that D cannot be the full space at 3.
18:28:930Paolo Guiotto: So, in particular, the conclusion is that D is not.
18:36:320Paolo Guiotto: Open.
18:39:250Paolo Guiotto: Okay?
18:40:630Paolo Guiotto: Now…
18:41:810Paolo Guiotto: Since it is closed, let's see if it is also bounded. It's true that if this is true, it will be also compact.
18:50:740Paolo Guiotto: And, so, D bounded.
18:55:590Paolo Guiotto: So, let's see what does it mean to be in D. XYZ is in D. We rewrite once again. This condition means to fulfill this system. X times Y equals Z plus 1, and X squared plus Y squared less or equal than 1.
19:14:410Paolo Guiotto: Okay, now, in this case, I told you there is not a standard approach, but we can look at the conditions and the constraints, and we can try to see if data has something.
19:28:720Paolo Guiotto: And one of these two tells something. Which one?
19:33:340Paolo Guiotto: Tell something about the information bound.
19:37:80Paolo Guiotto: Yes, the second one says that…
19:43:540Paolo Guiotto: Yeah, that's the usual argument. Notice that this is not the norm of XYZ, okay?
19:50:160Paolo Guiotto: Because that is X… the square of the norm would be X squared plus Y squared plus n squared. However, this says that, from this, we gain that the modules of X is less or equal to 1, and modulus of Y is also less or equal to 1. So at least 2 out of 3 coordinates are bounded.
20:07:880Paolo Guiotto: Now, can we use this to get something about Z?
20:13:590Paolo Guiotto: Yes, of course, I can use the first equation, because the two must be verified together. So, by…
20:21:560Paolo Guiotto: The… First… equations.
20:27:310Paolo Guiotto: We have that we isolate Z, that we can do, fortunately, here. It is XY minus 1.
20:34:500Paolo Guiotto: Now, what can be said about this? Well, notice I have to get a bound for the coordinate. Bound means not Z less or equal, but the absolute value. So Z is between this and that, okay? So, what happens if I take the absolute value of X? I have the absolute value of XY minus 1.
20:53:140Paolo Guiotto: Now, I can use the triangular inequality. This is less or equal than the modulus of X times Y, plus the modulus of minus 1.
21:02:440Paolo Guiotto: which is, of course, equal 1, and then here I have modulus X, modulus times modulus y plus 1. I know this, now I use
21:13:760Paolo Guiotto: And this is less or equal than 1 times 1 plus 1, so at the end, it is 2. So the conclusion is that I gained also modulus of Z is less or equal than 2.
21:26:500Paolo Guiotto: So, as you can see in this situation, for example, one of the two constraints gives a bound for two coordinates, not for the three coordinates.
21:37:380Paolo Guiotto: So this means that if you stop here.
21:40:650Paolo Guiotto: at this constraint, this constraint is not bounded. It's bounded in X and Y, but it's free in Z. You know what kind of set is this one?
21:51:410Paolo Guiotto: the set of points XYZ for which that condition is verified.
21:56:800Paolo Guiotto: What ca…
22:01:130Paolo Guiotto: No, I'm asking what kind of set is this one?
22:05:340Paolo Guiotto: The set described by just this condition.
22:12:520Paolo Guiotto: I'm asking something different. I'm asking, what is this?
22:16:740Paolo Guiotto: It's an elementary object.
22:19:880Paolo Guiotto: It's not like… it is a cylinder, because if you look in the space, XYZ,
22:26:260Paolo Guiotto: In plane XY, of course, this is a disk centered in the origin radius 1, so we have these points here.
22:34:950Paolo Guiotto: But since DZ is free, so when I look at points X, Y, Z that verify this thing, it means that X and Y must be in the disk, and Z is free. So this means an interior infinite vertical cylinder with rotation axis DZ axis.
22:53:240Paolo Guiotto: So this is an unbounded set.
22:56:50Paolo Guiotto: Alone.
22:57:440Paolo Guiotto: But together with this other condition.
23:01:880Paolo Guiotto: Together with this one, which is… you can see also this one is an unbounded, even if it is not an easy set to imagine.
23:10:660Paolo Guiotto: But the two are both unbounded, but together, they become a bounded set, okay?
23:18:360Paolo Guiotto: So, the conclusion is that now I have that modulus of X modulus of Y is less than 1, and since by the other condition I get also modulus of Z is less than 2, so I can say that the norm of XY
23:33:590Paolo Guiotto: Z, which is the square root of X squared plus Y squared plus Z squared. It is less or equal than root of… for X and Y, I have the bound 1,
23:47:580Paolo Guiotto: So 1 plus 1, when I do the square, it remains 1. And for Z, I have, 2 squares, so 4.
23:55:130Paolo Guiotto: Doesn't matter, it comes root of 6. And this holds for every point XYZ of D.
24:04:20Paolo Guiotto: So now we have the conclusion.
24:06:320Paolo Guiotto: D is bounded.
24:11:530Paolo Guiotto: And therefore, there is also… Compact.
24:19:60Paolo Guiotto: which is just a name we give to closed and bounded sets. Why are they so important? Because of the Viasas theorem, we have seen the statement last time.
24:36:50Paolo Guiotto: And the Weissance theorem, it is an important result concerning a fundamental problem, a fundamental applied problem.
24:45:120Paolo Guiotto: searching for minimum, maximum of a certain function. Of course, it does not provide a method, it does not tell you how to find these points, but it tells you something apparently very weak, you know, that they exist.
24:59:260Paolo Guiotto: Okay? Even if you don't know how to find them.
25:03:610Paolo Guiotto: So this is why these kind of sets, compact sets.
25:08:290Paolo Guiotto: Are so important, and we need to understand how to verify if they are of this nature.
25:19:110Paolo Guiotto: Okay, so I would say that you can finish… I will publish soon, if you have not yet done, so I will publish solution, finish…
25:29:990Paolo Guiotto: Exercise, 1-815, huh?
25:37:290Paolo Guiotto: Okay, so let's return on devices here, and so…
25:42:710Paolo Guiotto: Let's say that, compact… Set, s… Important.
25:57:160Paolo Guiotto: Because… off.
26:02:770Paolo Guiotto: Off via Slas.
26:10:620Paolo Guiotto: DLM.
26:12:880Paolo Guiotto: Let me just shortly remind. So, if we have a function F, which is continuous on a domain D,
26:22:130Paolo Guiotto: F, just to remind, is a function of vector variable, but real-valued. Otherwise, minimum, maximum, they do not have any sense, okay?
26:35:880Paolo Guiotto: Okay, so, such that D is compact, Ben… that exist then.
26:45:530Paolo Guiotto: minimum of F on D, and maximum of F on D.
26:52:610Paolo Guiotto: These are the minimum-maximum values, so…
26:56:190Paolo Guiotto: the minimum, maximum points are, so there exist equivalently, there exist two points, Xmin, And that's Maxer.
27:06:20Paolo Guiotto: It does not tell you, for example, if they are unique, it does not tell you how many of these points there can be. There could be infinitely many minimums, maximums. It does not tell you any kind of information of this type.
27:18:790Paolo Guiotto: Okay? It's just pure existing, very weak result, apparently. It says that there is these two points such that
27:28:90Paolo Guiotto: the values of F are between the value achieved at Xmin
27:35:280Paolo Guiotto: And the value achieved at X max.
27:40:210Paolo Guiotto: So these are the minimum-maximum points.
27:43:160Paolo Guiotto: For every action.
27:45:100Paolo Guiotto: Only, okay?
27:47:510Paolo Guiotto: Now, a question, an important question, because not always, unfortunately, we are in this context, so we have to maximize, minimize something on a domain which is not compact. What can be said in that case? So, a natural question is, what
28:07:720Paolo Guiotto: if… the… is not.
28:12:390Paolo Guiotto: Compact.
28:15:910Paolo Guiotto: Well, in general, it cannot be… we cannot say anything about this, but this is not because we are in… with the function of several variables, this already happens with functions of real variable, okay? So, in general.
28:36:450Paolo Guiotto: Nothing… can… Beat.
28:43:790Paolo Guiotto: side. I think you have seen this kind of examples, but just to refresh the ideas, for example, this…
28:54:620Paolo Guiotto: is, independent, off.
29:00:950Paolo Guiotto: RD, the fact that we are working on RD. It happens in R1. For example, So let's say.
29:10:730Paolo Guiotto: Examples,
29:12:580Paolo Guiotto: So our, D is a subset of R, and we have a traditional numerical function f of x.
29:22:240Paolo Guiotto: Now, you know that the Weiss theorem in that case, becomes if F is continuous on a closed and bounded interval, which is actually a particular case of this, because a closed and bounded interval is a closed and bounded sector, okay?
29:39:510Paolo Guiotto: Then, the function has both minimum and maximum. But what if we are not on a closed and bounded interval? So, it could happen basically everything. So, for example, take F, defined on interval 01 without the endpoints.
29:56:620Paolo Guiotto: Real valued, and just take f of x equal X.
30:00:190Paolo Guiotto: So this function is a straight function.
30:03:630Paolo Guiotto: which is defined… of course, you can say, it is defined in the interior line. Okay, but who cares? Now, your domain is interval 0, 1, and you want to know, is this func… has this function a minimum-maximum on that interval? If you look at the graph of this, since 0 is not in the domain, you don't have that point on the graph. Then you have a straight line that does like this, and also at 1,
30:27:490Paolo Guiotto: You don't have the value.
30:30:290Paolo Guiotto: Now, the key point is that the natural minimum-maximum for this function should be at X equals 0 and at X equals 1, but since X equals 0 and 1 are not in domain, it means that in domain, there is not any minimum-maximum point. You can see, because if this were the minimum.
30:48:700Paolo Guiotto: So if this is the minimum value, well, it is plenty of points where the function takes a smaller value, so that cannot be a minimum, okay?
30:57:780Paolo Guiotto: And the same for the, for the maximum. So this shows that, in general, if you remove this condition, closed and bounded, the TRM is no more… is no longer true.
31:09:930Paolo Guiotto: However, there is an important case
31:14:190Paolo Guiotto: that we can introduce with a graphical intuitive idea. So imagine that we have a function defined… let's start with the case of dimension 1.
31:26:120Paolo Guiotto: On the full real line. So the real line is definitely an unbounded set, so we cannot expect that the Beyster's theorem applies, no? For a function, even if the function is continuous on the real line.
31:39:320Paolo Guiotto: But imagine that, you know, that the function does this at plus minus infinity, so it goes to plus infinity.
31:48:710Paolo Guiotto: So how can you imagine a continuous function such that at plus minus infinity, it goes to plus infinity.
31:56:590Paolo Guiotto: Well, whatever is the line you draw here.
32:00:20Paolo Guiotto: You will see that sooner or later, you will have a minimum point, somewhere.
32:05:440Paolo Guiotto: It's like if you have a mountain and there will be a valet somewhere, no? Between two mountains.
32:12:740Paolo Guiotto: And, in fact, this factor comes… turns out to be a general result, or position, that we can now write directly in the version for the D-dimensional case. So, assume that we have a function f continuous on domain D,
32:30:900Paolo Guiotto: So, as above, the function f is a function of vector variable defined on the real value.
32:39:600Paolo Guiotto: with the domain D, Which is assumed to be… Closed them.
32:48:250Paolo Guiotto: But unbounded.
32:50:530Paolo Guiotto: Closed and bounded is compact, so biases theorem. Here we have closed and unbounded.
33:02:980Paolo Guiotto: Unbounded means just not bounded, okay? We say that this practically means that there exists at least a sequence of points that goes to the infinity, okay? So you see this sort of complementary case to the previous one.
33:19:340Paolo Guiotto: Now, even if we know this, for example, R in dimension 1 is closed but unbounded, okay? The full space is closed, so in RD, RD is closed because it's complementary.
33:32:420Paolo Guiotto: is empty, which is open, so RD is closed, but unbounded. So this theorem applies in particular, to the case when the domain is the full space.
33:43:560Paolo Guiotto: So, in particular, let's write here, in particular, the… equal RD.
33:52:970Paolo Guiotto: verifies these conditions. Now, these two conditions alone are not sufficient to ensure the existence of anything, but if we add this factor, that the function blows up at infinity, and this can be said in this way, suppose…
34:14:420Paolo Guiotto: Moreover, That… the limit… when you send X to the infinity of this space of the function f.
34:30:350Paolo Guiotto: Is equal to plus infinity.
34:33:340Paolo Guiotto: Now, you may imagine why I insisted a little bit on limits with value equal to plus infinity. That's because of this fact. Now.
34:42:50Paolo Guiotto: It is exactly this situation, where now we have, on the horizontal axis, of course, this is not the case, because RD will be a multidimensional space, but imagine that the flat line, the x-axis, is RD,
34:58:180Paolo Guiotto: So, here now we have, vectors here.
35:01:890Paolo Guiotto: So, X, that belongs to here, and we know that when we send our X to the infinity, that in this case means far away, so far from the origin, okay? Not to the right, there is not right and left, if the domain has the dimension greater than 1.
35:20:310Paolo Guiotto: And you have that the limit of F is plus infinity, so the F is big at infinity. Well, in this case.
35:30:270Paolo Guiotto: it happens that, of course, F cannot have a maximum.
35:34:640Paolo Guiotto: Because F cannot have a val… there cannot be a point where F at that point is bigger than all the other value of F. This is because the values of F are unbounded, no? So you cannot put a bound for the values of F.
35:49:890Paolo Guiotto: But there is a minimum in this case. Then, There exists a point Xmin in D.
36:00:590Paolo Guiotto: such that this point is a global minimum point. So the value of F at that point is less or equal than any other value of F.
36:11:260Paolo Guiotto: For every acts in… D.
36:15:330Paolo Guiotto: So this point, Xmin, is, global.
36:24:700Paolo Guiotto: Mean… point.
36:27:870Paolo Guiotto: for F.
36:32:500Paolo Guiotto: Okay, I will not, prove this, even if the proof is nice.
36:39:550Paolo Guiotto: But it is, still based on the Weissa's theorem.
36:43:470Paolo Guiotto: So, let's say that the idea behind this, I can do a justification with a figure, it's not a proof, but it gives an idea why this fact is true. Imagine the function is this one, no? It's blowing at infinity, it's blowing up at infinity, and it is continuous.
37:02:790Paolo Guiotto: So, graphically, we see that there is a minimum point, okay, but we have to justify. Well, the idea is that, since the function is going to plus infinity.
37:12:630Paolo Guiotto: at infinity, you can discard. Oh.
37:17:400Paolo Guiotto: when you have to search for the minimum, you can forget of what happens far away from the origin, because the function is going to be big, so there won't be the minimum at right of that… this bar, or at left of this one. So, here, there are no…
37:37:550Paolo Guiotto: I'm sorry.
37:41:120Paolo Guiotto: not mean.
37:43:350Paolo Guiotto: points here.
37:45:600Paolo Guiotto: and norming points also here. So, in other words, you can prove that you can restrict your search to a region which turns out to be closed and bounded. And that is where you apply VISTAS, so this yellow region is closed bound.
38:06:970Paolo Guiotto: bounded.
38:09:130Paolo Guiotto: So, compactor.
38:12:550Paolo Guiotto: So here is where you apply Weiser, and Weister tells you that there is a minimum point.
38:18:750Paolo Guiotto: There is also a maximum, of course, but that maximum is useless because it won't be a maximum for F, because F is exposing at infinity, okay? This, is granted…
38:33:70Paolo Guiotto: is ensured.
38:36:830Paolo Guiotto: Bye.
38:38:360Paolo Guiotto: Vias TLM.
38:41:310Paolo Guiotto: So, this is basically the idea behind the proof, no? You show that you can restrict your search to a suitable, compact set, and there you apply the vast theory. So, in any case, you don't have, again, any information about how to determine that point.
39:00:740Paolo Guiotto: It says just that there is a minimum point.
39:04:240Paolo Guiotto: And of course, you will have a dual statement, which is the dual statement.
39:11:820Paolo Guiotto: If…
39:13:330Paolo Guiotto: The limit for x going to infinity of the function f is now equal to minus infinity, so this is the case when you have that this
39:27:490Paolo Guiotto: profile falls down at infinity. So, in this case, what you will have is the existence of the maximum, okay? So, in this case, there will be
39:40:660Paolo Guiotto: global.
39:43:470Paolo Guiotto: Maximum.
39:45:630Paolo Guiotto: for F.
39:48:310Paolo Guiotto: on the… So, this is what we need, we will use later.
39:54:620Paolo Guiotto: But of course, it is, let's say, it is not sufficient to start searching for minimum-maximum, because actually these theorems are just existence theorems. They do not tell anything about how to determine these points. And that's where we have to enter in the next part of the course, which is the part relative to the French calculus.
40:18:610Paolo Guiotto: Because differential calculus will provide a method to determine these points.
40:23:660Paolo Guiotto: But as we will see, it will work a bit differently from what you are used to.
40:29:700Paolo Guiotto: Okay, I would say that it is a good point to take a break, you want?
40:37:370Paolo Guiotto: Okay, let's take, 5 minutes, but if you want, we can do… since it is,
40:45:550Paolo Guiotto: Well, actually, we started at 45,
40:48:560Paolo Guiotto: No, 5 minutes, Dai, okay?
40:52:170Paolo Guiotto: Let's put in pause here.
40:58:680Paolo Guiotto: Okay, so, there is still a little part, but
41:03:430Paolo Guiotto: I think that you can try to do this, read by your own. The section 1.7 is just a half page, so you do this, read…
41:16:710Paolo Guiotto: section…
41:20:250Paolo Guiotto: 1.7. It talks about what are connected sets, which is a concept that roughly means a set made of one single piece, okay?
41:32:720Paolo Guiotto: Okay, and
41:38:980Paolo Guiotto: you can try, also, that would be a nice theoretical exercise to do the exercise 1816. This is a theoretical exercise, so it is classified by 3 stars, because it involves a little bit abstract reasoning.
41:58:90Paolo Guiotto: A similar exercise is the next one, the 17, which is based on the concept introduced here.
42:05:740Paolo Guiotto: Okay, now we enter in the next part, which is, let's say the…
42:11:290Paolo Guiotto: We could say the principal part of this course, which is differential calculus.
42:21:20Paolo Guiotto: Differential calculus, as you know, is what has to deal with the concept of derivative, okay? So let's refresh to… well, let's say that the goal is, first of all.
42:34:630Paolo Guiotto: To introduce, introduce… A definition of derivative.
42:49:810Paolo Guiotto: for a function of the kind we are considering. So, function, let's take the question from the more general point of view. So, function F
43:02:50Paolo Guiotto: of vector variable, vector valued, so this is defined on some domain D of RD with values into some RM.
43:15:680Paolo Guiotto: So, we want to define, what should be F prime?
43:20:970Paolo Guiotto: of X.
43:22:930Paolo Guiotto: Now, since we have a definition for this, when all dimensions are 1, We recall that.
43:30:800Paolo Guiotto: we recall… that.
43:37:580Paolo Guiotto: For a function, now I will use a traditional lowercase letters, F equals f of x, defined on D, which is a subset of the real line, real valued.
43:51:410Paolo Guiotto: We say that, that… the function f is differentiable.
44:02:490Paolo Guiotto: at point X, if there existed this quantity.
44:08:810Paolo Guiotto: There exists the limit when H goes to 0 of this ratio, FX plus H minus FX divided by H, and this limit is finite.
44:23:80Paolo Guiotto: This happens, we call the limit the derivative of F at point x, by definition, okay?
44:32:280Paolo Guiotto: Now, as you know, this quantity has many important applications.
44:37:700Paolo Guiotto: not only to mathematics, but in physics, it is used to define quantities like velocity, acceleration, so mechanical quantities. It is used in many, many problems. In economics, it is used to define rates. In demography, the same type of
45:01:320Paolo Guiotto: quantities, so it is used basically everywhere. So, that's why it's so important. For the moment right here, we do not speculate on what is the interest we will see later, so let's just focus on the problem of defining something similar for the type of functions we have here.
45:20:340Paolo Guiotto: So what happens if we just take this definition and we transpose literally to our case?
45:27:930Paolo Guiotto: So, trying, to… use this… as a definition, off… F prime.
45:47:490Paolo Guiotto: of action.
45:49:740Paolo Guiotto: we see that something, non-trivial happens. So we say, what should be this? Well, now we have the limit… well, before we write what is H, let's write the fraction.
46:02:460Paolo Guiotto: We expect that the numerator will be something like F of X plus H. As you understand, since H is summed to X, it will be a vector of the same nature of X.
46:16:210Paolo Guiotto: So, we expect that down here, there will be a limit when that vector h goes to the vector 0, okay, of f of x plus h minus f of x.
46:26:810Paolo Guiotto: divided by H.
46:30:960Paolo Guiotto: Now, the problem here is that it doesn't matter if the function f is vector-valued.
46:36:700Paolo Guiotto: Even if it is numerical valued, but the variable is a vector.
46:42:870Paolo Guiotto: This quantity cannot be defined. Why? Because the problem is that we do not have a division by vectors. So we have what? We have sum between vectors, difference. We have product, but by scalars, not by vectors.
46:59:640Paolo Guiotto: So the operation of product between two vectors is not defined, and since the product between two vectors is not defined, you don't have even the ratio, which should be the inverse operation, no? The division is the opposite of the multiplication.
47:13:980Paolo Guiotto: But since we do not have a multiplication, this quantity is… cannot be defined in this way.
47:20:260Paolo Guiotto: But let's say this even if the function f Ease.
47:31:90Paolo Guiotto: numerical.
47:34:910Paolo Guiotto: valued.
47:38:330Paolo Guiotto: So, I mean, we have F equal F of vector variable.
47:45:500Paolo Guiotto: So, basically, the kind of functions we consider it
47:48:670Paolo Guiotto: until now, no? In all the examples, we always considered functions of several variables, but numerical functions. We computed limited for numerical functions, and we did exercises so far for numerical functions. So, this is a function defined on the domain D of RD.
48:05:880Paolo Guiotto: But REAL valued them.
48:09:70Paolo Guiotto: D-limit…
48:15:610Paolo Guiotto: limit, when age… Goes to zero.
48:20:920Paolo Guiotto: of F of X plus H.
48:25:760Paolo Guiotto: minus F of X.
48:28:260Paolo Guiotto: divided by H, Cannot be defined.
48:39:770Paolo Guiotto: This because if you look at this, the numerator is what? What kind of object is. Well, you know that F
48:47:540Paolo Guiotto: It's numbers, so F of this, whatever it is, is a number. Minus F of this is a number, so numerator is a number, so we have a scalar.
48:59:40Paolo Guiotto: But, denominator is a vector.
49:01:820Paolo Guiotto: In RD, as well as here, if you look at this expression, here, the values of F arrow, capital F, are vectors. So that is a vector of what? Where the values of capital F are.
49:18:190Paolo Guiotto: What is it? The capital F.
49:20:800Paolo Guiotto: takes values in RM. So that's an array with M component. So this is in RM, this is in RM, that's the difference between two vectors of RM, so it is a vector of RM. So the numerator is a vector of RM, and the denominator is a vector of RD. And we do not have such operation.
49:40:790Paolo Guiotto: Okay, we don't know what does it mean to do the division between this and this, and we don't even know what does it mean to divide a number by a vector.
49:49:780Paolo Guiotto: We know what does it mean to multiply number by vector, but we don't know what does it mean dividing by a vector. That's the problem. And this problem, as you can see.
50:00:490Paolo Guiotto: As soon as the dimension D is greater than 1,
50:05:550Paolo Guiotto: So in dimension 2, 3, 4, etc, this will be always a vector.
50:11:320Paolo Guiotto: The only case you can define in this way is D equals 1, which is the case you studied last year.
50:18:50Paolo Guiotto: So this means that some work will be required to have a definition of derivative that cannot be defined in this way, okay?
50:32:890Paolo Guiotto: So, the problem is that, because… Because… We… do… not.
50:44:140Paolo Guiotto: have… a division.
50:50:530Paolo Guiotto: Bye.
50:52:430Paolo Guiotto: a vector.
50:55:180Paolo Guiotto: So this operation simply does not exist.
51:00:00Paolo Guiotto: Okay, now, so we can close here, go home, because it's over. No, of course, mathematician tried to find a way to solve this puzzle.
51:11:960Paolo Guiotto: So the first solution is a sort of cheap solution. Now, we could say, with the modern language, a fake solution. That is, we try to transform this limit into a limit where the denominator is a number, so that way that we can do that limit.
51:30:140Paolo Guiotto: Let's see, this heals to the concept of the first definition of derivative, which is the directional
51:45:10Paolo Guiotto: derivative.
51:48:930Paolo Guiotto: That, as you will see, it won't be the answer to our problem, it won't be the right concept of derivative.
51:57:440Paolo Guiotto: But it is important because special directional derivatives are very, very, very important quantities in this subject. So, let's have a function f.
52:10:340Paolo Guiotto: We give the definition for the general case, so function f is a function of vector variable, vector-valued.
52:19:930Paolo Guiotto: So, it will be defined on a domain D, which is in a sum RD, with values in sum RM. There is no intuition, you cannot visualize it. It's not even me. I can have an idea of what is this kind of thing. It's just a function that transforms vectors into vectors, okay?
52:38:210Paolo Guiotto: Now, what is the idea? Wealth.
52:43:650Paolo Guiotto: imagine that in this formula, instead of… the idea is that vector H is going to 0, so it's a small vector, okay? Now, you could say, what if I replace this vector H by something like TV, where T is a scalar.
53:02:340Paolo Guiotto: that goes to zero. And V is fixed. So, in this way, I am basically taking increments which are proportional to a fixed vector V,
53:13:140Paolo Guiotto: So you imagine that since V is fixed, then T is moving to zero, to make the vector TV going to zero. Dividing by TV, or just dividing by T,
53:26:180Paolo Guiotto: should give the idea that we are trying to measure how much the increment of F with respect to the increment of the variable. So the idea is the following. 4.
53:40:930Paolo Guiotto: V in RD. Be careful, this is a vector of the same… of the same space
53:49:220Paolo Guiotto: Because it belongs to the domain, this thing, huh?
53:52:460Paolo Guiotto: We define, huh?
53:59:10Paolo Guiotto: this quantity that we denote by this symbol, this sort of fancy D, D.
54:07:690Paolo Guiotto: V of F at point X.
54:12:950Paolo Guiotto: what is, by definition, the limit for T going to zero off. We do the incremental ratio, so F of X plus the vector h becomes TV,
54:28:30Paolo Guiotto: So V is fixed here, and we are taking the limit not in V, but in this scalar, T. T is a scalar, it's a number.
54:37:330Paolo Guiotto: minus F of X, huh?
54:41:310Paolo Guiotto: And to make this fraction well-defined, I do not divide by TD, otherwise I'm still having the same problem, but just by T.
54:50:420Paolo Guiotto: Now I'm dividing by a scar, and this makes sense, because the operation… you can see this as a multiplication of the scalar, if you want formally. It is 1 over T times this, which is a vector, F of X plus
55:06:220Paolo Guiotto: Tv minus F of X.
55:10:490Paolo Guiotto: What you see here in parentheses, this is a vector that belongs to the range of F.
55:17:570Paolo Guiotto: Where are vectors F of something? R in RM, so this is a vector of RM,
55:24:260Paolo Guiotto: while 1 over T is a scalar, is a number.
55:28:570Paolo Guiotto: And this operation, scalar times vector, is defined, okay? So this definition makes sense. So, for V in RD, we define the… this quantity that we call directional derivative.
55:49:560Paolo Guiotto: derivative.
55:52:370Paolo Guiotto: of the function F, at point X.
55:59:330Paolo Guiotto: Along… the direction V.
56:04:20Paolo Guiotto: Because it enters also the direction. You change direction, you change this quantity. Of course, we define this quantity provided that quantity has a finite limit.
56:16:160Paolo Guiotto: Now, what kind of object is this one?
56:18:900Paolo Guiotto: Since this is the product between this color and this vector.
56:24:110Paolo Guiotto: this will be a vector of the same nature of this one, so it is a vector of RM. So this…
56:34:410Paolo Guiotto: in RM.
56:36:410Paolo Guiotto: So we say that the function f has a directional derivative at this point x along direction V, and we call this, with this symbol, this directional derivative, if this limit that you see here is fine.
56:52:860Paolo Guiotto: Okay? So, in particular, Let's specialize this for the case of numerical function.
57:01:420Paolo Guiotto: In particular, If the function f
57:08:180Paolo Guiotto: is a function of vector variable, but numerical, so defined on the in RD.
57:15:540Paolo Guiotto: Real valued.
57:19:180Paolo Guiotto: the directional derivative along direction V, which is still a vector. This is in the domain of F at point X.
57:28:910Paolo Guiotto: is the limit when t goes to 0, so there is no arrow here, F of X plus TV.
57:38:540Paolo Guiotto: Of course, there are the arrows in the argument, because the argument is a vector, but the values of this F are numbers, okay? So this numerator here is a number divided by a number, so this quantity is a number.
57:54:770Paolo Guiotto: So we define the directional derivative as this.
58:00:00Paolo Guiotto: Okay, now we have a definition, and let's explore this definition. First of all, by computing some directional derivative. So here we have the example 212, which is a one-star example, means very easy, no technicalities, just a simple calculation. So it asks to compute the directional derivative along direction 11,
58:24:200Paolo Guiotto: Most of these examples, I would say all the examples will be for numerical valued function.
58:31:110Paolo Guiotto: But to do not complicate too much the story, but the concept makes sense for vector-valued functions.
58:37:860Paolo Guiotto: So, FX at point 0, 0.
58:42:690Paolo Guiotto: Where… the function F, XY is X cosine Y.
58:53:110Paolo Guiotto: Let's see what is this quantity. So, the solution consists just in applying the formula, okay? So, D11F00 means the directional derivative of F at 0.00 along direction 11. So, this is…
59:12:730Paolo Guiotto: our V… This is the vector V. In above notations, this is point X.
59:20:510Paolo Guiotto: Okay?
59:23:160Paolo Guiotto: So we have to do the limit.
59:26:790Paolo Guiotto: or T going to zero. Off. Let's, since it is the first time.
59:31:430Paolo Guiotto: the best thing to learn something is not to try to memorize for 3 quarters of hours that formula, but practicing, no? So I have to do F of
59:43:270Paolo Guiotto: Let's specialize, well, let's write for the first time the definition. X plus TV minus F of X.
59:53:100Paolo Guiotto: divided by T. Okay, now we specialize, we put inside the ingredients, this is F at point X is 00, plus T, V is 11,
00:06:640Paolo Guiotto: You see? Minus F at 0.00 divided by T.
00:11:960Paolo Guiotto: Okay?
00:13:270Paolo Guiotto: So you see, this is X.
00:15:310Paolo Guiotto: This is V.
00:17:550Paolo Guiotto: Now, we have to do some algebra there, so we have a limit for t going to 0. You'll see I'm not yet replacing the function F. For the moment, this is indifferent to F. So 00 plus T11 is what?
00:38:690Paolo Guiotto: What is T11?
00:41:520Paolo Guiotto: So that's TT.
00:43:700Paolo Guiotto: This is TT, okay? You multiply all the components by… now this is a multiplication, scala by vector. How it works? Go back to class number 1.
00:56:520Paolo Guiotto: So… You were worried that we are going slowly, but for the moment, we are going slowly because
01:04:140Paolo Guiotto: you have to be, you must get confident with these operations, because they are new, you are not used to compute with vectors and so on. As soon as this becomes easy, we will go a little bit faster, because we don't have every time to remind, what does it mean this operation or that, how do we do this calculation, because it becomes natural, okay? But it takes a little bit of time.
01:28:90Paolo Guiotto: So we have to do slowly. So that's YTT. And 00 plus TT is…
01:33:910Paolo Guiotto: TT. So that's the argument of F minus F00.
01:38:570Paolo Guiotto: divided by t. Now it's time to replace what is our F. Remind that F of X… Y is X cos Y, no? F of X.
01:50:50Paolo Guiotto: Y is X cosine Y. So, FTT will be T cos T.
01:58:40Paolo Guiotto: Minus F00 will be 0 cos 0, divided T, 0 cos 0 is 0. So, at the end, we have the limit when t goes to 0 of T cos t.
02:10:710Paolo Guiotto: divided by t. Now, we can't… before we compute the limit, we can simplify something. What remains is the limit of cos T when t goes to 0, so cos 0,
02:21:460Paolo Guiotto: equal to 1. So we concluded that the derivative of this F at 0, direction 1, is equal to 1. That's the value.
02:35:750Paolo Guiotto: Okay, and that's… The answer to this question.
02:42:480Paolo Guiotto: Okay, before we launch into the next example, let me… Point out to you that…
02:55:630Paolo Guiotto: we have an exercise here, do exercise 291, which is an exercise similar to this one. Let me do, maybe…
03:06:750Paolo Guiotto: a couple of examples. So, for example, number two, compute… the directional derivative, in direction 2-2.
03:23:330Paolo Guiotto: of F at 0.10.
03:26:630Paolo Guiotto: Where the function F, XY, is equal to arctangent
03:33:560Paolo Guiotto: octangent to X plus Y.
03:38:440Paolo Guiotto: So it's just to apply a definition.
03:41:140Paolo Guiotto: So, let's say that we have a D22…
03:45:640Paolo Guiotto: F, 1, 0. I recommend you to do these exercises, even… there is nothing behind them, but it's just to learn the definition without the need of memorizing, which is completely useless here. So you do the limit at T goes to 0 of F, the point is 1, 0.
04:03:210Paolo Guiotto: plus T, the vector, the direction is 2, 2.
04:08:170Paolo Guiotto: minus F, point is 1, 0, divided by t. So that's the limit we have to compute according to the definition. Now, we do the algebra, limit when t goes to zero. Off. Denominator, downstairs you have t.
04:25:910Paolo Guiotto: So, in the first argument, we have 10 plus T22, which is 2T, 2T,
04:35:250Paolo Guiotto: So this becomes F at point 1 plus 2t.
04:40:600Paolo Guiotto: 0 plus 2T, so this.
04:43:250Paolo Guiotto: minus F10.
04:46:170Paolo Guiotto: And now it's time to put the function, which is arctangent of the sum of the two coordinates. So this is limit for t going to 0 of the arctangent
05:00:130Paolo Guiotto: of 1 plus 2t plus 2T, you see? This is X, this is Y.
05:09:290Paolo Guiotto: The function is FXY is arctangent of X plus Y, right?
05:14:410Paolo Guiotto: minus F10. It is arctangent of 1 plus 0 arctangent of 1.
05:22:840Paolo Guiotto: divided by… T.
05:25:740Paolo Guiotto: So at the end, this is the limit when t goes to 0.
05:30:290Paolo Guiotto: of arc tangent… of 1 plus 40,
05:36:400Paolo Guiotto: Minus arc tangent of 1. If we want, we know that arc tangent of 1 is Pi over 4…
05:44:650Paolo Guiotto: divided by T.
05:46:940Paolo Guiotto: Now, what is the, say, the good thing with this definition of directional derivative? That you see here concretely? What do you see? That we have to compute a limit in one variable.
06:00:510Paolo Guiotto: something that you are supposed to know how to do, okay? However, you have a number of tools to compute this kind of limits. So basically, the advantage with this definition of derivative
06:14:30Paolo Guiotto: is that it reduces the calculation to something we already know how to deal. Limits in one real variable.
06:22:10Paolo Guiotto: Okay? It can be complicated, can be easy, but it's something that we already know. So, since it is a limit, we treat it as a limit. What happens when t goes to 0? Dominator goes to zero.
06:32:700Paolo Guiotto: The argument of the R tangent goes to 1, so this goes to the ar tangent of 1, which is pi over 4, the numerator goes to 0, so we have the 0 over 0 in determinate form.
06:44:860Paolo Guiotto: So what can we do now?
06:47:290Paolo Guiotto: I know, I know, I know. It's like to go in a church and do a robbery in a church, like, something like this.
06:57:190Paolo Guiotto: So, it's too… it's too predictable.
07:01:270Paolo Guiotto: You think that the orbital rule will save you forever. That's completely false, okay?
07:08:720Paolo Guiotto: Allo. So, but in this case, it works, so let's use it. So the derivative of the denominator is 1, derivative of numerator, the minus P over 4, goes away, so we have the derivative of the arctangent, which is 1 over 1 plus the square of the argument, which is 1 plus 4t.
07:26:660Paolo Guiotto: square, then we have the derivative of the argument, which is 4. That's the derivative.
07:33:90Paolo Guiotto: Do you agree?
07:34:750Paolo Guiotto: So when T goes to zero, now we do not have any
07:38:310Paolo Guiotto: any further problem, because we have, this limit, 4 divided 1 plus, 1 plus 40,
07:47:360Paolo Guiotto: square, so when t goes to 0, this is easy, because that denominator goes to 2, so 4 over 2 is 2, and that's the value of the derivative.
08:00:260Paolo Guiotto: Well, let's see something a little bit more spicy, let's say.
08:06:100Paolo Guiotto: For example, the number 4,
08:11:390Paolo Guiotto: you have to compute the derivative at minus… along direction minus 1, 1 of F at 0.00, where F?
08:19:930Paolo Guiotto: XY… is defined in this way. Well, there is a double definition, basically, because it is said that
08:27:450Paolo Guiotto: It is, XY.
08:31:720Paolo Guiotto: divided, X square… plus Y fourth, when point XY is different from 0, 0.
08:46:180Paolo Guiotto: And when point XY is 0, the value is 0. So when point XY is equal to 0.
08:54:410Paolo Guiotto: Okay? It doesn't matter if there is this double definition. It is a function, no?
09:00:319Paolo Guiotto: Of course, you see that you cannot use the first line to define a function also at 0, 0, because it… you get 0 divided 0. However.
09:09:830Paolo Guiotto: So the answer is that we have to compute the limit, so let's say, wheat.
09:17:850Paolo Guiotto: But… to compute…
09:25:109Paolo Guiotto: the directional derivative minus 1, 1 of F at 0, 0, which is, by definition, the limit
09:33:140Paolo Guiotto: for T going to 0 of F. The point is 0, 0.
09:38:790Paolo Guiotto: plus T, the direction is minus 1, 1.
09:43:680Paolo Guiotto: minus F at 00.
09:46:760Paolo Guiotto: divided by T.
09:49:950Paolo Guiotto: I'm pretty sure that once you have done 3 times this, you don't need to remind the definition.
09:56:300Paolo Guiotto: So this is the limit when t goes to zero. So now this will be minus TT plus 0, so minus TT. So F of minus TT
10:07:930Paolo Guiotto: minus F00 divided by T.
10:12:670Paolo Guiotto: Now let's plug the function f.
10:15:180Paolo Guiotto: Be careful here, because we have the double definition. So it is clear that F00 is…
10:22:40Paolo Guiotto: 0. So, now, the point is, what is this F minus TT?
10:27:10Paolo Guiotto: Should I use the first line or the second line of that definition?
10:37:420Paolo Guiotto: Yeah, the first line, but why?
10:41:250Paolo Guiotto: Because remember that when T goes to zero, t is different from zero. In a limit, we never consider the limit point. So, since T is different from zero, so T going to zero.
10:55:850Paolo Guiotto: In particular, It means that T is different from 0.
11:03:870Paolo Guiotto: Okay?
11:06:490Paolo Guiotto: So, the point minus TTE cannot be the .00, because to be 00
11:14:380Paolo Guiotto: it should be t equals 0, so that's different from 0, and that's why the definition of F I have to take there is the first line, okay? So, this means that F of minus T
11:28:190Paolo Guiotto: is… the function is X times Y, so minus T times T divided by X squared plus Y power 4, so T squared
11:39:810Paolo Guiotto: Well, minus T squared plus T power 4. So, minus T squared over T squared plus T power 4. We can simplify a bit. This is minus T squared divided T squared times 1 plus T squared.
11:58:600Paolo Guiotto: This disappeared, it means 1, so minus 1 over 1 plus t squared. This is the value of the function at point minus TT, and that's the quantity we have to plug here.
12:12:840Paolo Guiotto: So we have.
12:15:230Paolo Guiotto: Equal. Limit.
12:17:340Paolo Guiotto: when T goes to zero, off.
12:21:200Paolo Guiotto: F minus TT, which is minus 1 over 1 plus t squared.
12:27:720Paolo Guiotto: minus F00, which is 0, divided by P.
12:33:330Paolo Guiotto: So, at the end, we have a limit when t goes to 0 of minus 1 over T times 1 plus t squared.
12:45:700Paolo Guiotto: Right?
12:46:910Paolo Guiotto: What do you think about this limit?
12:54:70Paolo Guiotto: You see that denominator goes to…
12:57:200Paolo Guiotto: 0. The numerator is constant equal to 1, so this is not necessarily an indeterminate form. However.
13:13:360Paolo Guiotto: Now, it comes that it depends.
13:15:780Paolo Guiotto: Because if T goes to 0 positive.
13:20:690Paolo Guiotto: means that t goes to 0 with positive values. Then, this quantity goes to zero, this T times 1 plus t squared, yes, it goes to 0, but you see that with the sign of this, because the other factor is positive, so it is 0 plus. So 1 over…
13:39:730Paolo Guiotto: So minus 1 over T times 1 plus t squared, we go to… that's 1 over 0 plus is plus infinity with the minus… Minus infinity.
13:51:710Paolo Guiotto: Whereas, if t goes to 0, negative, now this quantity, t times 1 plus t squared, goes to 0, negative, and this means that minus 1 over t times 1 plus t squared goes to plus infinity. And what is the conclusion?
14:11:510Paolo Guiotto: There is no limit, that's right.
14:14:190Paolo Guiotto: So this means there is no limit. So if there is no limit, what does it mean? That this was, by definition, the directional derivative.
14:22:700Paolo Guiotto: no limit, no directional derivative. So, the conclusion is, Conclusion.
14:31:400Paolo Guiotto: There exists not the, proposed directional derivative that was at minus 1, 1.
14:39:490Paolo Guiotto: Is that clear?
14:45:880Paolo Guiotto: Okay. Now…
14:48:540Paolo Guiotto: let's say, so far, so good, but in fact, it is not so good, because now I show you an example.
14:55:490Paolo Guiotto: That is, is going to destroy the idea that this is a good definition of derivative.
15:03:20Paolo Guiotto: Okay.
15:05:130Paolo Guiotto: So this example is a… Example 2, 1.
15:11:10Paolo Guiotto: Three.
15:12:300Paolo Guiotto: And we may say, we may title this, Why… the, directional… derivative.
15:23:670Paolo Guiotto: is not… Gout.
15:28:560Paolo Guiotto: definition.
15:30:730Paolo Guiotto: off.
15:32:240Paolo Guiotto: derivative.
15:35:930Paolo Guiotto: Well, let's see what happens here. So here, we have this function, so let…
15:40:760Paolo Guiotto: FXY be defined, it's, actually, it's an old friend. We have already met this function as an example of a function not having a limited zero. Perhaps you remind, however, I will refresh to you this story.
16:04:500Paolo Guiotto: Now, take this function.
16:07:150Paolo Guiotto: Now, we already proved, huh?
16:17:210Paolo Guiotto: Roved it.
16:19:380Paolo Guiotto: that, huh?
16:22:40Paolo Guiotto: There is no limit for this function.
16:27:390Paolo Guiotto: When point XY.
16:30:780Paolo Guiotto: Goes to 0, 0.
16:34:170Paolo Guiotto: Okay?
16:35:410Paolo Guiotto: Well, this… we just go back to one of the first examples, so…
16:41:850Paolo Guiotto: Let me find out. This is accumulation points, these are the first limits.
16:49:480Paolo Guiotto: Yeah. Well, it is changed with X and Y. Here is…
16:53:700Paolo Guiotto: XY squared divided X squared plus Y fourth. Now we have, with different letters, if we want, we just flip X with Y. But it… you see, you recognize that it is the same function, apart for the letters, okay? If you want the same function, okay, let's do the same function.
17:11:240Paolo Guiotto: So this will be different from that one. So we put the square here, square power 4 here, like that. That's now the same function.
17:21:200Paolo Guiotto: Now here, in this class, we proved that there is no limit.
17:28:270Paolo Guiotto: Okay, for this function at 0, 0.
17:33:210Paolo Guiotto: Now, what we prove now is the following.
17:37:280Paolo Guiotto: Let's… Check.
17:42:470Paolo Guiotto: That this function Has all possible directional derivatives, so there exist
17:51:280Paolo Guiotto: The directional derivative along direction V of this function at 0.00, this happens for every V
17:59:610Paolo Guiotto: for every vector V, which is, in this case, in R2. Okay?
18:06:220Paolo Guiotto: So this function, we have to check, but for a second, accept this, and what is weird in this example?
18:14:690Paolo Guiotto: It is like if this function is having all possible directional derivative, but it does not the limit, so in particular, since it is… this limit does not exist, we cannot say that this function is continuous, because we cannot say that the limit is the value of the function.
18:34:80Paolo Guiotto: That's all.
18:36:980Paolo Guiotto: F… will be… a function.
18:46:870Paolo Guiotto: not continuous.
18:49:680Paolo Guiotto: At… 0, 0.
18:53:850Paolo Guiotto: And let's say… Because of one.
19:02:850Paolo Guiotto: One is actually more than discontinuous, because there is not even the limit.
19:08:180Paolo Guiotto: Okay? What does it mean to be continuous? That there exists a limit, and the limit is the value of the function. But in this case, there exists there is no limit for F, so it cannot be continuous.
19:21:680Paolo Guiotto: But… So F will be a function, not continuous 0, but… having… all… the… directional… derivative.
19:41:510Paolo Guiotto: at 0, 0.
19:46:700Paolo Guiotto: You see, this is strange why.
19:50:560Paolo Guiotto: Because if you go back to the derivative for function of one real variable, one basic remarkable feature of the derivative is that once you have the derivative, the function is automatically continuous.
20:05:470Paolo Guiotto: So… for… functions… F… of X, with the X, real.
20:17:880Paolo Guiotto: the existence of F' of X implies F continues at point X. So the differentiability is stronger than continuity, because differentiability means there is a tangent, so the function must be smooth.
20:36:160Paolo Guiotto: The function is discontinuous, means there is a jump, something like this, okay? So if there is a jump, that cannot be a derivative.
20:44:960Paolo Guiotto: So, this is important because it says that differentiability is expected to be a property which is stronger than continuity, while here, we have an example that says we can have all possible directional derivatives, and still the function is not continuous.
21:02:700Paolo Guiotto: So, the differentiability, according to this definition, directional derivatives, is not sufficient to ensure continuity.
21:10:440Paolo Guiotto: And this is what is not so nice, because we expect that differentiability means more than continuity, okay? So that's why this example basically is saying that this concept of differentiability is not a good concept.
21:26:410Paolo Guiotto: So let's do the calculation for the directional derivative at a generic vector V. So let's say that vector V, has…
21:35:440Paolo Guiotto: Components alpha, beta.
21:38:180Paolo Guiotto: So here, alpha-beta are generic.
21:41:410Paolo Guiotto: is a generic vector in R2.
21:44:500Paolo Guiotto: So we may notice that, well, there is a trivial directional derivative. If V is the vector 0, well, you can see that whatever is the function, whatever is the point, the directional derivative along direction 0 is trivial, because what is this?
22:05:750Paolo Guiotto: It is the limit when t goes to 0 of F of X plus Tv.
22:12:340Paolo Guiotto: minus F of X.
22:15:490Paolo Guiotto: divided by T.
22:17:390Paolo Guiotto: This is an argument that works for whatever is the function, okay? So if the vector V is 0, it means that you are doing T times 0. What is T times 0?
22:29:910Paolo Guiotto: is 0. And so what is the f of x plus dv?
22:35:590Paolo Guiotto: F of X, so that's the limit when t goes to 0 of f of x minus F of X divided by t. But that numerator will be 0, so you are doing the limit of 0 divided, the limit is…
22:51:340Paolo Guiotto: Zero. So, this is a general fact, okay? Whatever… Ease.
23:01:730Paolo Guiotto: F, the directional derivative in direction 0 of F at whatever is the point, is just 0 for every X.
23:12:630Paolo Guiotto: Okay, so there is nothing to compute. If the vector V is different from 0, we have still 1 minute.
23:20:940Paolo Guiotto: Let's see if we can finish. If V is different from 0,
23:25:70Paolo Guiotto: So the directional derivative along direction V of F at 0, 0 is the limit when t goes to 0 of F of 00 plus Tv. The vector V here is vector alpha-beta.
23:45:20Paolo Guiotto: These are the two components, minus F00.
23:49:850Paolo Guiotto: divided by T.
23:51:700Paolo Guiotto: Now, T alpha beta is the vector Yeah, plenty of things.
23:59:250Paolo Guiotto: T alpha T beta, plus 0, 0, it remains T alpha T beta. Okay, F00 is what? If you go back to the definition of this F, the value at 0 is 0. So this is definitely equal to 0.
24:13:800Paolo Guiotto: Therefore, we have the limit for t going to zero of f of t alpha T beta divided by t. Now, what is FT alpha t beta?
24:26:690Paolo Guiotto: So what branch of this definition should I take? The first line or the second?
24:33:80Paolo Guiotto: Remind that I take the second only when F is evaluated at .00.
24:38:450Paolo Guiotto: Now, you see that here, this point is different from 0, 0, because to be 0, you must have that T alpha and T beta must be equal to 0.
24:49:180Paolo Guiotto: Since t is going to zero, it is different from zero. This happens only if alpha and beta are 0, which is not the case, because the vector of V is assumed to be different from zero.
24:59:570Paolo Guiotto: So, I have to take the first line, so we have limit when t goes to 0.
25:04:290Paolo Guiotto: So it was X, which is T alpha, times Y squared, T beta squared, divided X squared, T alpha squared, plus T beta power 4.
25:19:480Paolo Guiotto: So what we get, doing, the algebra…
25:24:450Paolo Guiotto: is T cubed alpha beta squared divided the T squared that we can factorize. It is alpha squared plus t squared beta power 4.
25:37:580Paolo Guiotto: So, as you can see, we have a T squared, a cube that simplifies this T-square.
25:45:350Paolo Guiotto: Now, what is the limit when t goes to 0?
25:48:930Paolo Guiotto: Here, there is, a little k, because,
25:52:960Paolo Guiotto: you would say it is equal, I'm sure that you will say that it is equal to…
25:59:140Paolo Guiotto: Oh, sorry, I forgot to divide by T, you see? There is this guy here. This is just F. So there is a divided by T.
26:09:480Paolo Guiotto: So, divided by T,
26:11:620Paolo Guiotto: everything, so this makes this a T cube, sorry. So this disappears. So, let me do one line more. Limit when T goes to 0, alpha beta squared divided alpha squared plus t squared beta power 4.
26:28:330Paolo Guiotto: What would you see? What would you say about this limit?
26:33:470Paolo Guiotto: I'm sure that you would say this is going to zero, so it is alpha beta square divided alpha squared.
26:40:280Paolo Guiotto: Almost, so we can say that this is alpha beta square.
26:45:300Paolo Guiotto: Divided the alpha squared, so beta squared divided alpha.
26:50:750Paolo Guiotto: Provided alpha is different from zero. Otherwise, it is not defined.
26:55:710Paolo Guiotto: But what if alpha is zero?
26:58:430Paolo Guiotto: Well, if alpha is 0, there's no problem, because in the case alpha equals 0, this becomes 0, that becomes 0, so you have 0 divided t beta fourth, and beta is different from 0, otherwise the vector alpha beta would be 0, 0. So this, that's 0 divided this, the limit will be 0.
27:19:50Paolo Guiotto: So for our equals 0, it becomes this. So as you can see at the end, we computed the directional derivative of F at 0.00, whatever is the vector V. So this calculation takes a generic V,
27:35:110Paolo Guiotto: So we separated the case V equals 0 from the case V different from 0, where the calculation is a bit more complicated. But the point is that at the end, we arrive to compute whatever is the vector V, the directional derivative.
27:50:980Paolo Guiotto: So, we just have this situation. We have all directional derivatives, but
27:58:470Paolo Guiotto: This function has not even the limit, so it cannot be continuous.
28:01:980Paolo Guiotto: So that's why this definition of derivative is really insufficient for us.
28:07:680Paolo Guiotto: Okay, guys, let's stop here.
28:10:440Paolo Guiotto: I inform you that tomorrow I can't do classes, so next class will be on Friday morning, okay?
28:19:680Paolo Guiotto: Okay…