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00:04:560Paolo Guiotto: Okay, good morning.
00:08:880Paolo Guiotto: So, is it now okay with the slides? You have free access? Okay.
00:16:930Paolo Guiotto: Okay, last time, we introduced the A definition of directional derivative.
00:28:470Paolo Guiotto: Which is written here, which you may say, basically, is a real variable derivative.
00:37:770Paolo Guiotto: We have seen some examples on how to compute it.
00:43:310Paolo Guiotto: And, a significant example was this one.
00:47:860Paolo Guiotto: Because it says that, basically, the concept of directional derivative is not a good concept of derivative. This because we may have that there exists a directional derivative along each direction, but still the function is not even continuous.
01:07:430Paolo Guiotto: Now, a well-known and remarkable factor from calculus in one variable is that when you are differentiable, you are continuous.
01:16:930Paolo Guiotto: So differentiability must be stronger, okay? So we should expect that this concept should imply continuity, which is not the case.
01:28:40Paolo Guiotto: So this is not the good definition of derivative. Nonetheless, there are some special directional derivatives that, as you will see, play an important role. So let's say some…
01:46:890Paolo Guiotto: Important, very… important,
01:57:680Paolo Guiotto: directional derivatives.
02:05:740Paolo Guiotto: Now, let me just remind that… I recall that.
02:16:40Paolo Guiotto: A directional derivative is, by definition, I will keep, the case of numerical functions.
02:27:430Paolo Guiotto: But the same holds for vector-valid function is a limiter for T going to zero of
02:36:250Paolo Guiotto: F of X plus DV.
02:40:890Paolo Guiotto: minus F of X divided by T.
02:46:100Paolo Guiotto: Now, important directional derivatives are those in direction of the vectors of the canonical basis of RD.
02:56:370Paolo Guiotto: So… let… V be a vector, one of the vector of the canonical basis, so we say EJ,
03:08:60Paolo Guiotto: Well, EJ is the vector with all components equal to zero, except for the jth one.
03:16:780Paolo Guiotto: Jth component, which is equal to 1.
03:21:130Paolo Guiotto: Now, let's see what happens if we plug this special vector into this definition.
03:27:470Paolo Guiotto: We have them.
03:30:330Paolo Guiotto: So, D… EJF at point X.
03:38:170Paolo Guiotto: is the limit.
03:40:790Paolo Guiotto: when t goes to 0, for the moment I just copy, F of X plus T EJ
03:49:340Paolo Guiotto: minus F of X divided by D.
03:54:750Paolo Guiotto: Now, the fact is that TEJ
03:59:650Paolo Guiotto: So, we know that, you know that we multiply components by component, okay? So, take vector EJ and multiply by t. Since all components are zero, except the jth component, which is 1, TJ is the vector 0, 0, etc, 0, and the jth component.
04:18:190Paolo Guiotto: We have tea.
04:20:769Paolo Guiotto: And then 0, 0, 0.
04:23:570Paolo Guiotto: So, if you do X vector X plus TEJ,
04:29:980Paolo Guiotto: Let's say that the vector X has components X1, X2, XD.
04:37:550Paolo Guiotto: So we have to do vector X1, X2, XD,
04:43:180Paolo Guiotto: plus TJ, which is the vector, 0, 0 everywhere, except at jth coordinate.
04:51:130Paolo Guiotto: So what happens when you do this sum?
04:53:750Paolo Guiotto: You see that none of the components change, except for the component XJ that you have here.
05:01:40Paolo Guiotto: Because that one will become, so X1, X2, etc, until XJ-1.
05:10:720Paolo Guiotto: Then you have… let's, write in red, XJ plus T,
05:16:780Paolo Guiotto: And the next one will be XJ… hmm… XJ plus 1…
05:24:00Paolo Guiotto: XJ plus 2, etc, until XD.
05:28:100Paolo Guiotto: So basically, you see that, that, vector
05:32:610Paolo Guiotto: has changed the component of X only in the J8 component, replacing with J… XJ plus T.
05:39:690Paolo Guiotto: So now, let's go back to that limit.
05:43:90Paolo Guiotto: So… The directional derivative along direction EJ of F, point X,
05:52:130Paolo Guiotto: is the limit for t going to zero off. Now, this here is the array where I have to evaluate the first F, no? F of X plus TEJ. So I write F
06:07:550Paolo Guiotto: I put all components, X1 until XJ-1, then there is the jth component, XJ plus T.
06:17:260Paolo Guiotto: and then XJ plus 1 until XD.
06:23:790Paolo Guiotto: Okay? What you see here is, this, no?
06:28:990Paolo Guiotto: F evaluated at X plus TJ is this one.
06:36:580Paolo Guiotto: Then I have minus F evaluated at point X.
06:41:640Paolo Guiotto: So I will do this, F of X1, etc, XJ-1, I still color in red the jade component, which is just XJ, and then I continue with black.
06:54:70Paolo Guiotto: XJ plus 1 until XD.
06:59:270Paolo Guiotto: And let's say that this is in blue.
07:03:30Paolo Guiotto: this box here.
07:05:380Paolo Guiotto: This is F of X.
07:07:510Paolo Guiotto: Then we have to divide everything by t.
07:14:970Paolo Guiotto: Now, for a second, Try to… well, let's say that we call our tea also in red.
07:21:830Paolo Guiotto: And we call it the fraction also in red.
07:27:80Paolo Guiotto: Now, for a second, let's forget of the black letters.
07:31:330Paolo Guiotto: the X1XG.
07:34:90Paolo Guiotto: minus 1, XG plus 1XT, you know? So this looks… like…
07:42:730Paolo Guiotto: If I would forget of the black coordinates, I would see something like, let's put a red also here.
07:51:460Paolo Guiotto: So, it put, like, as F evaluated at XJ plus T,
07:56:570Paolo Guiotto: minus F evaluated at XJ divided by T, and we take the limit when t goes to 0.
08:04:300Paolo Guiotto: If you look at this, this is… What is it?
08:13:930Paolo Guiotto: I'm sorry, I don't after you.
08:16:150Paolo Guiotto: Yeah, no, but… Why normal? It's the usual derivative, the first year calculus derivative. It is a,
08:29:890Paolo Guiotto: what you called F prime, no? F prime evaluated where? At point XJ.
08:37:490Paolo Guiotto: If F were a function of one single variable, that would be the derivative with respect to that variable. You see?
08:46:840Paolo Guiotto: Now, the derivative is what? f of x plus h minus f of x divided by h, you send H at to 0, and you get the derivative.
08:54:890Paolo Guiotto: But our F actually depends on all these other variables.
09:00:480Paolo Guiotto: Which are, however, here, constant. They are not interested by T.
09:07:50Paolo Guiotto: So, what is this thing?
09:09:200Paolo Guiotto: This is just like this one, so it is the ordinary derivative.
09:13:780Paolo Guiotto: of what? Of the function F,
09:17:950Paolo Guiotto: computed respect to the jth variable. So you forget the other variables, they are like parameters, constant, fixed, and you differentiate only respect to the jth variable.
09:32:20Paolo Guiotto: This quantity is what we call partial derivative of F with respect to the, jth variable.
09:41:640Paolo Guiotto: So, this quantity…
09:51:60Paolo Guiotto: is called… partial.
09:58:410Paolo Guiotto: derivative.
10:01:50Paolo Guiotto: of F… with respect to the Jade, variable.
10:11:590Paolo Guiotto: Well, it is, in fact, It is.
10:17:700Paolo Guiotto: An ordinary… When I say ordinary, it means, one… variable.
10:28:140Paolo Guiotto: first year calculus. Derivative.
10:32:870Paolo Guiotto: Of what? Of the function f
10:37:960Paolo Guiotto: as a function only of the jth variable, so where you freeze the other variable, X1, 2XJ minus 1,
10:47:840Paolo Guiotto: And you look at this as just function of XJ, XJ plus 1 to XD.
10:56:480Paolo Guiotto: So you look at this, As… oh…
11:01:980Paolo Guiotto: As a function where you are interested only in the… how it depends on the jade variable.
11:10:650Paolo Guiotto: So, for example.
11:16:570Paolo Guiotto: Well, let's do a stupid example so the calculations are not particularly complicated, but let's imagine that we have this function, F of two variables, to make simple. FXY, which is X cosine, sorry, sine…
11:34:70Paolo Guiotto: Y…
11:36:460Paolo Guiotto: And I want to compute the derivative with respect to the second direction, for example. Let's compute… let's compute…
11:49:480Paolo Guiotto: The derivative, respect to here, there are only two vectors in the basis, so there is 1, 0, and 0, 1. So I compute the derivative with respect to
11:59:280Paolo Guiotto: the direction 01 of this F at the generic point XY. Let's see what is the value.
12:08:710Paolo Guiotto: So, formally, we have to do limit. So, basically, what I'm doing here is this in a particular case. So, limit when T goes to 0 of F at point XY plus t vector 01.
12:25:700Paolo Guiotto: minus F point XY.
12:29:90Paolo Guiotto: divided by T.
12:31:120Paolo Guiotto: Now, T times 01 is… 0T. So we have limit…
12:38:490Paolo Guiotto: for T going to 0 of FXY plus 0T means F of X, Y plus T,
12:47:90Paolo Guiotto: minus F of XY,
12:49:710Paolo Guiotto: divided by t. Now, this calculation is for a generic F. Let's now plug the particular expression we have for F. So this means that we have to compute the limit for t going to 0 of… our F is… F of XY is X sine y.
13:07:680Paolo Guiotto: So here I have F of XY plus t, so it would be X times sine of Y plus T.
13:15:430Paolo Guiotto: Minus. F of XY is X sine y.
13:20:300Paolo Guiotto: divided by t.
13:22:420Paolo Guiotto: So, as you can see, we can factorize an X. It comes limit for T going to 0. X times sine
13:29:900Paolo Guiotto: of Y plus T minus sine Y divided by T.
13:36:970Paolo Guiotto: Now, if you want, you can consider this as a limit. When t goes to 0, the argument of sine goes to Y. Sorry, there is an error here, is sine of y. So sine of Y minus sine of y goes to 0, denominator goes to 0, 0 versus 0. But if you look carefully, that's the derivative of sine.
13:54:540Paolo Guiotto: You see? Sine of y plus t minus sine of Y divided T when t goes to 0. So this is the derivative of sine
14:02:700Paolo Guiotto: evaluated at this point, why?
14:05:460Paolo Guiotto: So, it is cosine of Y, and therefore we get X cos y.
14:12:90Paolo Guiotto: So, you see what we… we obtained the derivative with respect to direction 01 of F at point XY. Well, let's write the F. The F was
14:23:970Paolo Guiotto: X, sine Y.
14:28:60Paolo Guiotto: It is X cosine y. So what we did, it is like if we took this function, and we differentiate respect to the letter Y, considering X as a constant. If you would do the derivative of this function with respect to Y,
14:45:390Paolo Guiotto: Forgetting of X, let's say, considering X as a constant, as if X is 3. If you have the derivative with respect to Y of 3 sine y, you would say it is 3 cosine of Y.
14:58:300Paolo Guiotto: Now, what if you have X sine Y? X is a constant.
15:03:30Paolo Guiotto: Respect to Y, no? Imagine that we are taking the derivative only with respect to Y, considering X as a constant. We would get X times the derivative of this, which is cosine of Y. So that's what this directional derivative… I will introduce a notation.
15:21:320Paolo Guiotto: For this reason, Let's finish the example. So, 4.
15:27:990Paolo Guiotto: this.
15:30:220Paolo Guiotto: Reason… We… Right.
15:37:110Paolo Guiotto: So, we write this. The directional derivative in direction EJ of F at point X will be just denoted by definition as DJF
15:50:290Paolo Guiotto: Excellent.
15:51:640Paolo Guiotto: Or, this is a more old-fashioned notation. This is the notation originally introduced by Leibniz. He wrote this DF over DXJ.
16:05:280Paolo Guiotto: Evaluated at point X.
16:08:340Paolo Guiotto: So this is the ancient way to denote the derivatives.
16:12:380Paolo Guiotto: Like, because this reminds that the derivative is a sort of ratio, no?
16:18:40Paolo Guiotto: But we will, use this one.
16:22:330Paolo Guiotto: So, in other words, so…
16:26:370Paolo Guiotto: If I have to compute the derivative… well, let's say that, when… 4.
16:35:970Paolo Guiotto: functions, F.
16:39:410Paolo Guiotto: off.
16:40:660Paolo Guiotto: Two or three variables.
16:44:240Paolo Guiotto: So, you know that for two or three variables, we do not use indexes, so we do not say F of X1, X2. We prefer to write F as F of XY for functions of two variables, and for functions of three variables, we prefer to use the typical three letters, XYZ, no?
17:04:450Paolo Guiotto: So, this means that for functions of two variables, it means that we have possibly
17:11:390Paolo Guiotto: two partial derivatives. The derivatives DD1F is the derivative with respect to the first variable.
17:19:720Paolo Guiotto: OE is the first variable is X. So that's why we also use, for this specific case, the notation DXF, okay? That's the specific notation we use in this case, and the D2F is the derivative with respect to the second variable, which is the DYF.
17:39:170Paolo Guiotto: And similarly, down here, we have that D1F is DXF.
17:45:730Paolo Guiotto: D2F is DYF, and D3F is DZF.
17:54:550Paolo Guiotto: And how do you compute this? As we said, these partial derivatives are ordinary derivatives, so you differentiate, as you learned, for functions of one variable, considering all the other variables as constants. So, for example, when I have to compute
18:15:540Paolo Guiotto: The derivative with respect to X, let's say, of X e to y.
18:23:290Paolo Guiotto: Well, I don't need to use the formula
18:27:360Paolo Guiotto: reminding me that it is a directional derivative, because at the end, if I look this as a function of X,
18:35:960Paolo Guiotto: Okay, this is X times a constant. So what is the derivative? If you want, you have the constant, which is e to y, times the derivative with respect to X, of the variable. What is the derivative of X?
18:52:410Paolo Guiotto: 1. So this will come E to Y. So if I have to compute DY of sine
19:01:360Paolo Guiotto: of X plus Y squared.
19:06:50Paolo Guiotto: So I have to look at this as a function of Y. Yes, there is also X, but for this operation, X is a constant.
19:13:260Paolo Guiotto: So, if you look at Y, this is… derivative of sine is cosine of the argument, which is X plus Y squared.
19:22:440Paolo Guiotto: Then I have times the derivative of the argument with respect to Y. The argument is X plus Y squared. Now, what is the derivative of X with respect to Y?
19:35:490Paolo Guiotto: DX… DY of X is.
19:40:470Paolo Guiotto: is 0, because DX is a constant for Y. You are differentiating with respect to Y, so X is seen as a constant, so that's zero.
19:51:760Paolo Guiotto: While the derivative with respect to Y of Y squared is… to Y.
19:57:430Paolo Guiotto: And therefore, this comes to Y cosine of X plus Y squared.
20:05:40Paolo Guiotto: Derivative with respect to Z of XE to X squared plus ZY. I don't know.
20:16:610Paolo Guiotto: That's why this.
20:18:890Paolo Guiotto: So, remind, whatever is not depending on Z is a constant. So, for example, I could say it is derivative with respect to Z of X, e to x squared, which is a constant. This is a constant
20:35:660Paolo Guiotto: in Z.
20:37:540Paolo Guiotto: as a function of Z is independent of Z, therefore it's a constant. Then you have another part here that contains Z, so that's not a constant. So since it is a derivative of a product, the constant comes out
20:50:520Paolo Guiotto: so it is XZ to X squared, then I have to differentiate with respect to Z to ZY. Now, the derivative of the exponential is the exponential.
21:00:180Paolo Guiotto: So I will have e to ZY, exponential of the argument, which is ZY, times the derivative with respect to Z of the exponent, ZY. And what is this derivative?
21:17:510Paolo Guiotto: It is equal to… Why?
21:21:380Paolo Guiotto: Okay, so these are just ordinary derivatives, so they are nice, because we already know how to compute it. There is a little bit more of complexity, because there are several variables, but when you have to compute a partial derivatives, it is easy.
21:37:520Paolo Guiotto: You have just to compute the ordinary derivative.
21:41:740Paolo Guiotto: Of course, be careful.
21:45:90Paolo Guiotto: So let's write warning.
21:51:230Paolo Guiotto: You… R. A load.
22:01:360Paolo Guiotto: To use… ordinary.
22:09:10Paolo Guiotto: calculus.
22:10:530Paolo Guiotto: Rules.
22:13:970Paolo Guiotto: when… It is possible.
22:22:810Paolo Guiotto: But not when it is impossible. For example.
22:28:10Paolo Guiotto: Imagine I have this function, FXY,
22:32:510Paolo Guiotto: Of course, to create a problem, I will…
22:36:940Paolo Guiotto: takes, take X times Y.
22:40:660Paolo Guiotto: divided X squared plus Y squared.
22:43:940Paolo Guiotto: for X, And Y different from 0, 0.
22:49:350Paolo Guiotto: And for XY equal, we take 0.
22:53:780Paolo Guiotto: For XY, different here, equal.
22:57:740Paolo Guiotto: So, now we have a well-defined function for every point xy.
23:03:570Paolo Guiotto: And let's see, discuss.
23:11:750Paolo Guiotto: Existence, and… Value…
23:17:980Paolo Guiotto: of the derivatives with respect to X of F at 0, 0, and the same derivative with respect to Y of F at 0, 0.
23:31:490Paolo Guiotto: Okay.
23:32:900Paolo Guiotto: Now, we have a function.
23:35:290Paolo Guiotto: Let's see if what we learned works. Well, apparently we may say, okay, what is the DX of F at point XY?
23:45:960Paolo Guiotto: Well, here I have a first problem. The function is defined to
23:50:530Paolo Guiotto: through two different formulas. So, it depends where point XY is. So I may say that for XY different from 00,
24:00:140Paolo Guiotto: probably I would use the first line, so I have to do the derivative with respect to X of this guy, X times Y divided the X square… sorry.
24:12:570Paolo Guiotto: When you want the gum, it doesn't come. And when you don't want it, it's always there.
24:18:410Paolo Guiotto: Okay.
24:20:450Paolo Guiotto: Now, this is the derivative of a fraction, right? So how do you do the derivative? You do the square of the denominator. X square plus y squared squared.
24:33:220Paolo Guiotto: And then you do derivative of numerator. So what is the derivative with respect to X of the numerator?
24:40:510Paolo Guiotto: Y times denominator, which is X squared plus Y squared, minus numerator, X times Y times derivative with respect to X of denominator. It is…
24:53:570Paolo Guiotto: 2X. 2X. Do you see?
24:56:810Paolo Guiotto: So when you take the derivative with respect to X of this, no? It is 2X, the derivative of this is 0, no? So 2X. So that's the derivative, I can simplify, but…
25:10:240Paolo Guiotto: It is not particularly important. Okay, so let's keep there, because it's practically useless. You see that we cannot plug 00 inside this formula.
25:19:890Paolo Guiotto: So, but… this… is useless.
25:29:270Paolo Guiotto: 4… XY exactly equal to 0, 0.
25:36:150Paolo Guiotto: So what can be… what could be the conclusion from this?
25:40:270Paolo Guiotto: I would say that some of you would think that there is not partial derivative at 0, because there is not dysphoria.
25:50:100Paolo Guiotto: Do you see?
25:52:960Paolo Guiotto: I mean, I am asking, is there a partial derivative at 0, 0 of that function? We computed the partial derivative with respect to X, we obtained that quantity, but this calculation works for XY different from 00, because that's F for XY different from 0, 0.
26:11:240Paolo Guiotto: So, in particular, this means that we are never allowed to put X and Y equals 0 and 0 here. Because that formula definitely doesn't hold for XY equal to 0, 0.
26:24:660Paolo Guiotto: Now, the question is, does this mean that the derivative does not exist because the formula, that formula does not make sense?
26:34:820Paolo Guiotto: I would be sure that some of you would think that that's the correct answer. That's wrong.
26:40:320Paolo Guiotto: Okay? The unique thing I can say here is that that formula does not make any sense for X and Y both 0. So…
26:48:910Paolo Guiotto: How can I discuss that?
26:51:610Paolo Guiotto: We could say, well, at 0, the function is equal to 0, so the derivative is 0, perhaps. But that's a wrong argument, because it's just one single point.
27:01:990Paolo Guiotto: Where the function is zero, and you never compute the derivative of one single value. Otherwise, all derivatives would be zero.
27:11:170Paolo Guiotto: So how can we do? Well, actually, here, we can discuss the existence and value of the partial derivative by reminding that the partial derivative is a directional derivative, and we can compute it by the definition.
27:25:560Paolo Guiotto: to discuss
27:31:430Paolo Guiotto: Existence and value.
27:36:690Paolo Guiotto: off.
27:37:790Paolo Guiotto: the partial derivative of F at 0, 0, We remind them.
27:47:180Paolo Guiotto: So in this case, we cannot just solve the problem saying, let's compute the derivative with respect to X, and then put X equals 0, because this doesn't work.
27:57:460Paolo Guiotto: We can compute the derivative with respect to X,
28:00:840Paolo Guiotto: as an usual derivative, but unfortunately, that formula is valid for all points except the unique one who is requested here. So, we remind that the derivative with respect to X of F at 0, 0 is what? Is a directional derivative
28:19:230Paolo Guiotto: In which direction?
28:21:110Paolo Guiotto: Remind that the derivative with respect to the jth variable is the derivative with respect to the j… the vector of the canonical base. So this is the vector 1, 0, the first variable, first vector of the canonical base.
28:36:280Paolo Guiotto: So, this is what is the directional derivative. So, it is the limit for t going to 0 of F of… the point is 0, 0, plus T, the vector is 1, 0,
28:49:270Paolo Guiotto: minus F at 00, divided by T.
28:54:400Paolo Guiotto: So this is limit for T going to 0. This thing is 00 plus T10. T10 is T0. 00 plus T0 is T0, so we have to evaluate F at point T0.
29:08:980Paolo Guiotto: Minus F at point 00 divided by T. What is F at point T0? I have to look at the formula, and which one of the two should I take, the first or the second, for F of T0?
29:26:110Paolo Guiotto: What do you think?
29:28:40Paolo Guiotto: T0 is the first one.
29:31:310Paolo Guiotto: Yeah, but why… In other words, YT cannot be 0,
29:37:580Paolo Guiotto: Because I'm doing a limit, and T is different from 0. So my point, T0, is not 0, definitely, so it's different from 0, 0, and therefore I use the first line. What is F of T0? F of T0…
29:53:150Paolo Guiotto: since T0 is not 0, 0, is X times Y, so T times 0 divided T squared plus 0 square. So it is 0 divided T squared, it is 0.
30:07:770Paolo Guiotto: So that quantity is 0. What about F00? By definition, it is equal to 0.
30:14:820Paolo Guiotto: So another value here, I have 0. So my limit is the limit when t goes to 0 of numerator is 0 minus 0, 0 divided t, so that limit is 0. So as you can see, the derivative exists, and it is equal to 0.
30:31:140Paolo Guiotto: Okay? Something that I cannot see from this calculation.
30:36:710Paolo Guiotto: Okay, so…
30:39:00Paolo Guiotto: There exists the direct… the partial derivative of F at 0, which is, by definition, the directional derivative at point… at direction 1,0 of F, 0, and this is equal to 0.
30:53:480Paolo Guiotto: Same calculation.
30:56:970Paolo Guiotto: calculation for…
30:59:570Paolo Guiotto: the other partial derivative with respect to Y at 0, 0, you actually don't need to repeat the calculation. Why?
31:07:480Paolo Guiotto: Because if you look at F, it depends on Y as it depends on X. So you will get exactly the same calculation with the letter X replaced by Y. So at the end, you get the same value, 0.
31:20:150Paolo Guiotto: So, as you can see, these two partial derivatives actually exist.
31:24:350Paolo Guiotto: Even if, and this is not in contradiction, this thing, does not have any meaning at 0, 0. There is no contradiction, because this formula is valid for XY, different from 0, 0. It doesn't tell anything about what is the value at 0, 0.
31:43:870Paolo Guiotto: Something is valid only there, and they cannot tell anything about what happens here.
31:49:780Paolo Guiotto: You see?
31:51:40Paolo Guiotto: So, this formula is correct. We calculated with the… as if it is an ordinary derivative with respect to X, but unfortunately, it works only at… only. It works at every point except one point.
32:07:500Paolo Guiotto: Which is exactly the point where the problem asks to discuss the existence and value of the directional derivative, the . So does this mean that the derivative does not exist at 0, 0? No.
32:23:250Paolo Guiotto: It means I cannot use this formula at 0, 0.
32:26:920Paolo Guiotto: So, how can I discuss them?
32:29:160Paolo Guiotto: I have always to remind that I have a definition for this thing.
32:34:590Paolo Guiotto: So if I go back to the original definition, and I compute directly the directional derivative, which is the directional derivative.
32:42:420Paolo Guiotto: First variable, first vector of the canonical basis.
32:46:570Paolo Guiotto: Second variable, second vector, third variable, third vector, and so on. So here it is with respect to the direction is 1, 0. I wrote, I computed, and I discovered that it exists and it is equal to zero. Okay?
33:01:980Paolo Guiotto: Okay.
33:03:700Paolo Guiotto: So… We will,
33:09:990Paolo Guiotto: We will compute lots of partial derivatives, so we will, of course, return on this calculation. For a moment, we just…
33:20:300Paolo Guiotto: suspend here this discussion, because…
33:23:790Paolo Guiotto: Partial derivatives are special directional derivatives, and even if directional derivatives are not sufficient to ensure the differentiability, the continuity, sorry, we think that this concept is not a good concept of derivative.
33:41:140Paolo Guiotto: So, now, the goal is…
33:49:250Paolo Guiotto: We need… another… definition.
33:59:20Paolo Guiotto: Off.
34:00:200Paolo Guiotto: derivative.
34:03:00Paolo Guiotto: That must at least imply the continuity, must be stronger.
34:07:570Paolo Guiotto: Now, let's go back once again to the original definition, okay? We remind…
34:19:449Paolo Guiotto: That, huh?
34:21:650Paolo Guiotto: We cannot…
34:26:790Paolo Guiotto: computer, the limit… or H going to 0.
34:35:350Paolo Guiotto: Well, let's stay on the case of numerical function for a second.
34:41:100Paolo Guiotto: F of X plus H, minus F of X.
34:48:800Paolo Guiotto: divided by H.
34:52:409Paolo Guiotto: Or more in general.
35:02:380Paolo Guiotto: Or… Vector.
35:07:770Paolo Guiotto: valued.
35:10:460Paolo Guiotto: functions.
35:14:200Paolo Guiotto: So the same thing with the function f that now has vector values. So limit for H going to 0 of
35:24:230Paolo Guiotto: Let's use the capital letter F of X plus H.
35:29:960Paolo Guiotto: minus F… of AXA.
35:34:140Paolo Guiotto: divided by H.
35:36:640Paolo Guiotto: And we cannot do this because we cannot divide by vector. So it doesn't matter if the function is vector-valued or numerical-valued. In any case, we have to divide by age to write this formula, and we can do this operation, okay?
35:52:880Paolo Guiotto: Because,
35:58:130Paolo Guiotto: We… cannot, divide.
36:06:40Paolo Guiotto: by vector H.
36:09:180Paolo Guiotto: So, how can we, how can we circumvent this, this difficulty, this technical difficulty?
36:18:540Paolo Guiotto: To find out the solution, we go back to the classical case, the one-dimensional case.
36:26:330Paolo Guiotto: Now, to, to solve… this.
36:32:560Paolo Guiotto: Issue.
36:34:860Paolo Guiotto: We… go back, To… the… one… variable case.
36:49:10Paolo Guiotto: So we're reminded that,
36:51:520Paolo Guiotto: So here, F is a function of X. X is a real variable, so it is defined on some domain of the real line.
37:01:740Paolo Guiotto: Real valued, okay?
37:04:480Paolo Guiotto: This is first-year calculus.
37:06:640Paolo Guiotto: F prime of X is, by definition, the limit
37:11:890Paolo Guiotto: when H goes to zero of f of x plus h minus f of x divided by H, no? That's the classical definition.
37:24:390Paolo Guiotto: Which is the definition we cannot extend in general. We may notice that this is equivalent to say that if we carry F' on the same side of the limit, we have limit
37:38:250Paolo Guiotto: when H goes to zero off, the ratio f of x plus h minus F of X divided by h minus F prime x, all this equals 0.
37:50:560Paolo Guiotto: Huh?
37:51:710Paolo Guiotto: And since f prime of x is a constant.
37:55:220Paolo Guiotto: It does not depend on H, which is the variable here we are doing the limit. We can also carry inside the limit, because that constant will come out of the limit.
38:06:590Paolo Guiotto: But if we write this and we do the common denominator, this is equivalent, again, to limit
38:13:280Paolo Guiotto: for H going to zero of fraction.
38:17:280Paolo Guiotto: F of X plus H minus F of X minus f prime of X times h divided by H is 0.
38:29:850Paolo Guiotto: For the moment, also this, which is equivalent to the definition of
38:35:180Paolo Guiotto: derivative cannot be written with H vector, because there is always this…
38:43:800Paolo Guiotto: There is always this H at the denominator.
38:47:590Paolo Guiotto: However, this property is saying that when a limit is 0, and as you can see, both quantities are going to zero.
38:58:520Paolo Guiotto: It means that when a fraction is zero.
39:01:690Paolo Guiotto: It means that the numerator must be much smaller than the denominator. And I say much smaller because both are going to zero. In particular, this denominator is small. H is going to zero. So this one…
39:14:460Paolo Guiotto: nor the fraction goes to zero, this one must be much smaller than this one. There is a precise definition for that.
39:23:400Paolo Guiotto: We say that, and this is equivalent, that the numerator, which is FX last H minus FX.
39:33:640Paolo Guiotto: minus F prime X H is of smaller order of age, or, in other symbols, is little o of h.
39:45:600Paolo Guiotto: Because what does it mean that someone is little O of H? It just means that someone divided by age goes to zero, okay? By definition.
39:58:440Paolo Guiotto: a quantity, let's say G of H,
40:02:230Paolo Guiotto: is little O of H if and all if, when you compute the limit for h going to 0, G of H divided by H is 0.
40:14:740Paolo Guiotto: Because the idea behind this, and that's the fundamental point in limits, is that you do not compare quantities in the sense that you say one is boller than the other because it is less or equal.
40:28:300Paolo Guiotto: Or one is bigger than another because it is greater or equal. No. You compare quantities in the ratio. You say that one is smaller than another if ratio goes to zero.
40:39:910Paolo Guiotto: You say that one is bigger than another if ratio goes to infinity.
40:44:100Paolo Guiotto: You say they are the same, the word is asymptotic, if ratio goes to 1.
40:50:180Paolo Guiotto: So that's the key concept when we do limits. The comparison is done in ratio, and the meaning is not an exact meaning, but it's qualitative meaning. It means that to have this fraction going to zero, this quantity must be much smaller than this one.
41:06:810Paolo Guiotto: No? Otherwise, the function… the fraction will never go to zero.
41:11:770Paolo Guiotto: If they have the same size, no?
41:14:530Paolo Guiotto: This means this quantity should be much smaller than this one.
41:20:550Paolo Guiotto: And now this is the point where we can transform this relation into relation valid for our case, where F is a function of an array of variables. And even F is vector-valued, so…
41:37:840Paolo Guiotto: What if, if, we… Right… now.
41:47:670Paolo Guiotto: That, F… Let's say… let's now pass to the general vector value… vector variable function.
41:58:300Paolo Guiotto: We write the FEs.
42:00:190Paolo Guiotto: differentiable, at… X, if we can say that. I take just that expression, I box here.
42:13:390Paolo Guiotto: And I replace the new letters. Okay, let's see what happens. If…
42:19:940Paolo Guiotto: F, evaluated at X plus H, minus F.
42:27:790Paolo Guiotto: of X.
42:29:800Paolo Guiotto: Minus F prime.
42:32:530Paolo Guiotto: of X.
42:34:500Paolo Guiotto: Now, let's suspend what doesn't mean times, but we imagine that it will be some kind of product.
42:42:90Paolo Guiotto: Now, we will see that, apparently, it's impossible, because it seems that we are still having the problem of doing products or divisions between vectors, but that's not the case, as you will see.
42:54:670Paolo Guiotto: Equal little O, that now, as you will see, should be an array, but let's wait a second for this.
43:01:600Paolo Guiotto: So, as you can see, it is exactly the same property, but with some arrows, no?
43:06:900Paolo Guiotto: I'd say.
43:08:100Paolo Guiotto: Okay, now, the idea is that we can now give a precise meaning to this one, and this will define, in a unique way, the quantity F prime of X.
43:20:50Paolo Guiotto: Okay, let's see… First of all, what should be this object?
43:27:320Paolo Guiotto: So, we have several problems to solve here. First, problem.
43:35:460Paolo Guiotto: So, the first problem is what… Kinda.
43:41:910Paolo Guiotto: God, and…
43:46:710Paolo Guiotto: off… Well, let's say mathematical, of course, Object, which seems very rough.
43:55:370Paolo Guiotto: It's not just an object, but let's say, we should say mathematical entity, or to say that what kind of object is,
44:04:550Paolo Guiotto: All that. Should be…
44:10:740Paolo Guiotto: this, F prime… of X.
44:16:910Paolo Guiotto: This is the most important thing that we have to understand first.
44:20:800Paolo Guiotto: Because for ordinary functions, where I say ordinary functions, I mean first-year functions, or functions of real variable, real value, the numerical functions you are used. In this relation, this quantity here, the derivative.
44:40:820Paolo Guiotto: is what? What is the derivative?
44:44:260Paolo Guiotto: is a number, constant, if you want, okay? This is a number.
44:49:640Paolo Guiotto: everything here are numbers, okay, is a number. And this is an algebraic… Product.
44:56:970Paolo Guiotto: Okay?
44:58:430Paolo Guiotto: You do this times this, the algebra.
45:02:240Paolo Guiotto: Now, look at this. This expression is a little bit more complicated, because Let's start with this object.
45:12:400Paolo Guiotto: What kind of object is this? Well, imagine that here, F is a function of
45:19:350Paolo Guiotto: variable X, which is a vector.
45:22:580Paolo Guiotto: And it gives you a vector.
45:26:70Paolo Guiotto: So X is a vector of a certain domain that now belongs to a space RD.
45:33:310Paolo Guiotto: for example, R2, two variables, RT, three variables, and we'll take values in another vector space, which, in principle, has nothing to do with RD. Let's say RM.
45:47:340Paolo Guiotto: So, the most important, the first thing to understand is that this quantity, f of x.
45:54:910Paolo Guiotto: belongs here, so it's an image. So it is a vector of Rn.
46:02:990Paolo Guiotto: So this means that this guy belongs to RM.
46:07:80Paolo Guiotto: Well, this is a guy of the same nature, he's an F of something, so it will be in our M.
46:14:550Paolo Guiotto: So, if I stop here, the first part of this expression is the subtraction between two vectors of the same space, and this is well possible. I can do this algebraic thing.
46:29:740Paolo Guiotto: So, now I have minus another object here that necessarily must be of the same nature, otherwise this expression does not make any sense, no? So, since here I'm taking a vector of RM minus a vector of RM, minus what is this? This will be another vector of RM, otherwise this operation does not have any.
46:51:420Paolo Guiotto: Any… any sense. So, this means that also this one should be a vector in RM.
47:01:420Paolo Guiotto: Okay.
47:02:730Paolo Guiotto: But this H is what?
47:06:710Paolo Guiotto: a vector of… This belongs… is here, is inside the argument.
47:13:590Paolo Guiotto: This lives in the starting space, which is RD.
47:19:50Paolo Guiotto: So this is a vector of RD,
47:21:980Paolo Guiotto: As well as X is a vector of RD.
47:26:190Paolo Guiotto: And this makes sense, two vectors, their sum is in RD, which is the argument of F. Same here. So this H is a vector of RD.
47:36:510Paolo Guiotto: And now, what should be this F prime of X?
47:40:560Paolo Guiotto: Such that when you multiply this quantity by a vector of RD, you obtain a vector of RM.
47:48:710Paolo Guiotto: What does this?
47:51:250Paolo Guiotto: in the multiplication.
47:54:770Paolo Guiotto: And metrics, that's right.
47:57:660Paolo Guiotto: It's a matrix, you know, product line by column.
48:01:960Paolo Guiotto: This matrix must have the same number of columns of the number of components of this. This must have D columns.
48:10:100Paolo Guiotto: M must have the number of lines as the components of these vectors, so M lines.
48:16:530Paolo Guiotto: This is a D by M method. This should be, okay? So… So, we realized…
48:31:00Paolo Guiotto: That.
48:33:620Paolo Guiotto: in order.
48:38:370Paolo Guiotto: the product.
48:40:940Paolo Guiotto: I write the product, even if I don't know exactly what kind of product it should be, you know, but for the moment, the product…
48:49:770Paolo Guiotto: F prime of X.
48:54:180Paolo Guiotto: times H takes a vector of RD.
48:58:480Paolo Guiotto: And, yields, at the end, a vector of RM,
49:04:460Paolo Guiotto: Well, that F prime of X
49:13:00Paolo Guiotto: must be… We sighed. I repeat.
49:20:380Paolo Guiotto: Imagine how this is a vector, H, with D components.
49:25:870Paolo Guiotto: H1, H2, etc. When you do the product line by columns, what do you do? The first entry here times the first component. Plus.
49:35:700Paolo Guiotto: Second entry, second component, plus third entry, third component, and so on, no? So you must have the number of columns equal to the number of components of this age. So these metrics must have D columns.
49:50:970Paolo Guiotto: Okay?
49:53:840Paolo Guiotto: D… by M metrics.
49:59:420Paolo Guiotto: If you want, we can see this, no, in an ex… in an expanded way. F prime… X.
50:09:840Paolo Guiotto: H…
50:11:190Paolo Guiotto: is we have our vector H, which is the vector H1, H2, H3, etc, HD. Remind that H belongs to the
50:22:350Paolo Guiotto: the initial space at the… So here we have a matrix.
50:28:790Paolo Guiotto: And how this product works, I have numbers in line, so let's say A11, A12, I'm listing the elements of the first line. How many?
50:41:40Paolo Guiotto: You know that when you do the product, you will do A11 times H1, plus A12 times H2, A13 times H3, plus H14 times H4, etc, until you arrive here, and therefore you must have an A1D. This means that there are D columns here.
51:03:780Paolo Guiotto: Once you have done this, you will have the first component, so you will have A11H1 plus A12H2, etc.
51:17:550Paolo Guiotto: A1dhd.
51:21:460Paolo Guiotto: And now… You have a second component.
51:25:690Paolo Guiotto: How many components should you have? You remember that this vector at the end will have M component, because it is someone that belongs into our M. And so, it seems for each line you have a component, this means only that there are M lines. So you have a second line, A21,
51:43:350Paolo Guiotto: A22 A to D. A third line, a fourth line, an nth line, AM1, AM to… A.M.
51:55:510Paolo Guiotto: D. And so, the last line will be AM1H1 plus AM2H2, etc, AM.
52:06:640Paolo Guiotto: DHD. That's the unique possibility to do what we expect this operation does.
52:14:880Paolo Guiotto: So this says that,
52:17:510Paolo Guiotto: The object we are going to define, which is the derivative, is not what we can expect, a number, but it is a matrix.
52:27:90Paolo Guiotto: So you have to start keeping in mind this first idea. The derivative for us will be a matrix, not a number.
52:34:610Paolo Guiotto: So, as you can imagine, it becomes to be complicated, because you do not visualize geometrically what is metrics, okay? It's a table of numbers, these metrics.
52:46:00Paolo Guiotto: Okay, so this is the, let's say, the possible answer to the… I'm not yet given the definition, but to arrive to the definition, I could give you the definition immediately, but it would be, like, to hit you with a gun, no? So I'm trying to prepare you to the definition, so step by step.
53:05:20Paolo Guiotto: And so the first step was understanding of what is F' of X. Now, do we want to take a break?
53:16:100Paolo Guiotto: It's not yet finished, because the next will be understanding of what is this, okay?
53:22:20Paolo Guiotto: Okay, let's take a break. How much?
53:26:410Paolo Guiotto: 5, 10 minutes, 10? Okay, 10 minutes back.
53:50:600Paolo Guiotto: Okay, so, from this first discussion, we understood that
53:58:220Paolo Guiotto: The derivative of F, if F is a function from RD to RM, Okay? Must be a matrix.
54:09:100Paolo Guiotto: with the same number of columns of the dimension of the domain, D, and the same number of lines as the dimension of the codomain, M, okay?
54:22:160Paolo Guiotto: Now, the last, and the second problem is to understand what should be this little O, because we are writing an identity.
54:32:690Paolo Guiotto: Now, at the left, we have…
54:35:620Paolo Guiotto: a difference between three vectors of our m.
54:39:700Paolo Guiotto: This means that the right-hand side will be a vector of RN.
54:43:70Paolo Guiotto: So, first of all, to be correct, I should put an arrow here.
54:47:100Paolo Guiotto: And I should say that also this guy will belong to RM.
54:52:280Paolo Guiotto: And, and this is the most important problem. What does it mean to be little o? So, second.
55:01:350Paolo Guiotto: problems.
55:03:590Paolo Guiotto: So… walk… does… It means… Little O of H.
55:17:360Paolo Guiotto: That's not a trivial question, because if we go back a second to what we said above.
55:23:630Paolo Guiotto: It is written here, no?
55:26:860Paolo Guiotto: You are… this is the usual definition. We say that a quantity, G of H, is little o of h if the limit when H goes to 0 of g of h divided by H is 0.
55:40:110Paolo Guiotto: If we put arrows here, we return to the old problem, because H is an arrow, and we cannot divide by H.
55:48:510Paolo Guiotto: But if we think a second, Here, what is important
55:54:60Paolo Guiotto: is the fact that that quantity G of H should be smaller than H.
56:01:670Paolo Guiotto: So, if I replace H by a quantity, numerical quantity, related to age.
56:08:940Paolo Guiotto: that represents the size of H,
56:12:770Paolo Guiotto: This could be a good compromise. What is the quantity? Numerical quantity of age that gives the size of age. It is the norm of age. That's why we give this definition. This is now the first definition, the second will be the definition of derivative.
56:33:220Paolo Guiotto: So… We say… That.
56:41:190Paolo Guiotto: A function… well, let's keep uppercase letters, because these are arrows.
56:47:310Paolo Guiotto: G of H.
56:51:40Paolo Guiotto: is little o of H,
56:55:970Paolo Guiotto: If I do not write G of H divided H, I cannot do that, but I can write this limit when H goes to 0,
57:07:30Paolo Guiotto: of G of H… not divided by H, but divided by the norm of H. This is 0.
57:19:230Paolo Guiotto: Now, the idea here is that, of course, I'm no more dividing by a vector, this is a number, so this can be done.
57:28:300Paolo Guiotto: And, norm of H is the length of the vector, so it's representative of the size of H. So it is small exactly when H is small.
57:39:380Paolo Guiotto: No more, no less than this. So, it's natural definition, and that's what we take as definition.
57:45:910Paolo Guiotto: So, we have now all the ingredients to formalize the definition of what is the derivative.
57:51:620Paolo Guiotto: We… Now.
57:56:170Paolo Guiotto: Mold… to… define… Differentiability.
58:05:730Paolo Guiotto: fork… functions.
58:11:140Paolo Guiotto: of the general type we are considering here. So, functions of vector, variable, vector, value.
58:18:560Paolo Guiotto: So, definition.
58:22:790Paolo Guiotto: Let F, let's do everything with the kale.
58:29:860Paolo Guiotto: function of X defined on a domain D of RD, with values in RM.
58:40:450Paolo Guiotto: I use different letters because they are not necessarily the same, but it could be R2 to R2. Why not? R3 to RT. These type of transformations, same dimension to same dimension, are quite important.
58:53:410Paolo Guiotto: For example, this will be the change of variables for calculus of integration. Per calculus integral, in integration theory, they will be functions RD to RD.
59:04:270Paolo Guiotto: But in general, let's say that we have a function from RD to RM.
59:10:20Paolo Guiotto: We say that…
59:15:710Paolo Guiotto: F.
59:18:430Paolo Guiotto: is differentiable.
59:26:250Paolo Guiotto: At… point.
59:30:450Paolo Guiotto: X… in domain D.
59:34:440Paolo Guiotto: If… there exist
59:40:740Paolo Guiotto: a matrix, M by T. No, sorry. Yeah, the…
59:48:750Paolo Guiotto: Yes, first we usually write the number of lines and then the number of columns, okay? M by D.
59:54:490Paolo Guiotto: metrics.
00:00:50Paolo Guiotto: denoted.
00:04:130Paolo Guiotto: Well, maybe it's better if I write… let me just change the form, if there exists.
00:10:780Paolo Guiotto: F prime.
00:13:700Paolo Guiotto: of X.
00:16:680Paolo Guiotto: M by the matrix.
00:22:130Paolo Guiotto: Such that,
00:28:600Paolo Guiotto: So we write the same formula, just a copy, F of X plus H, minus F of X.
00:39:170Paolo Guiotto: minus F prime.
00:41:620Paolo Guiotto: of X times H, this times is the usual product matrix times vector, is a little o of h
00:56:90Paolo Guiotto: Or… equivalently.
01:05:600Paolo Guiotto: Since little o of h means that the quantity divided norm of H goes to 0 when H goes to 0, the quantity is this one, this is the G of H.
01:17:280Paolo Guiotto: So, limit.
01:19:900Paolo Guiotto: when H goes to zero.
01:23:450Paolo Guiotto: off.
01:24:580Paolo Guiotto: F of X… plus age.
01:30:360Paolo Guiotto: minus F of X.
01:34:200Paolo Guiotto: minus F prime of X times… H… divided by norm of age, is equal to 0.
01:47:530Paolo Guiotto: zero vector.
01:51:850Paolo Guiotto: Okay? It's a complex definition, and there is a very important difference with the definition of derivative you have seen last year.
02:00:950Paolo Guiotto: I will return in a moment here, but let me scroll back and, until,
02:09:90Paolo Guiotto: this, okay? So last year, you said a function f is differentiable at point x if there exists this limit
02:18:360Paolo Guiotto: and it is finite, and you call the value of the limit f prime of X. So this definition is, of course, simpler with respect to the definition we have seen here, because it's less complicated, but the principal difference is that you know how to compute F prime.
02:37:310Paolo Guiotto: Because it says F' is this limit.
02:40:670Paolo Guiotto: Okay? The biggest difference with the definition we have here is that it is much more involved, but there is no way to simplify this. Because, look, apart for details, it says if there exists a matrix, there exists
02:57:750Paolo Guiotto: Metrics. You don't see anywhere the formula for this metrics.
03:02:620Paolo Guiotto: It says, if you are able to find a matrix that you call f prime of x such that this is true.
03:09:240Paolo Guiotto: But it's not written…
03:11:450Paolo Guiotto: F prime of X equal. So, the first question from this definition is, yes, nice, but how do I compute this matrix?
03:19:880Paolo Guiotto: Because if I have to invent the matrix software to look, but how? How can I look at these metrics? It's not clear. It's a sort of implicit definition, it's not so explicit. However, we will see in a moment that this matrix has a precise form.
03:36:860Paolo Guiotto: This matrix deserve a special name. It is called the Jacobian Matrix.
03:44:400Paolo Guiotto: F prime over X.
03:46:830Paolo Guiotto: is called up.
03:50:370Paolo Guiotto: Jacobian… metrics.
03:59:200Paolo Guiotto: of F, At… point.
04:04:430Paolo Guiotto: X.
04:05:890Paolo Guiotto: This comes from Jack Corby, that was a mathematician.
04:09:860Paolo Guiotto: A couple centuries ago.
04:11:830Paolo Guiotto: Okay.
04:14:20Paolo Guiotto: Okay, now, first remark… This definition That's not.
04:26:570Paolo Guiotto: provide… And it… formula.
04:36:800Paolo Guiotto: for… the… Jacobian metrics.
04:45:00Paolo Guiotto: So, it seems to basically be a sort of,
04:49:690Paolo Guiotto: such difficult definition to be checked. How can I find a metric such that this happens? No?
04:56:670Paolo Guiotto: It's like, you see the difference? This says, you want to compute the derivative, compute that limit. You have an operational way to compute that quantity.
05:08:960Paolo Guiotto: But this is different now. This is saying you have to find a number, it should be from the case of the derivative, you should find a number such that this happens.
05:18:520Paolo Guiotto: Imagine that someone would define the derivative at first-year calculus in this way. The derivative is a number for which this happens.
05:27:720Paolo Guiotto: Say, yeah, thanks, but how do I compute that number?
05:31:420Paolo Guiotto: Nobody does this in first-year calculus, because it would be too complex, and mostly because we have a simpler definition, which is
05:42:160Paolo Guiotto: We say that the function is differentiable if that limit exists, and the value of the limit is the derivative. So you have also, at the same time, a definition and an operational way to compute the derivative.
05:53:730Paolo Guiotto: Unfortunately, this cannot be done for vector-valued case, because we don't know how to compute that limit. We don't know how to give a meaning to this, because of the problem of the definition by age. So we needed to turn around this problem.
06:09:770Paolo Guiotto: And we arrive at this, which is the best we can say. So the function is differentiable if this happens, so if there exists a matrix that we call f prime of x, the Jacobian matrix, for which this limit is 0. But as you can see, there is not a formula for F'.
06:27:760Paolo Guiotto: So a first question is, of course, how do we compute this Jacobian matrix? Unfortunately, there is a nice way that involves the partial derivatives. This comes after a proposition.
06:43:580Paolo Guiotto: A very important proposition. It can be proved the following fact.
06:48:630Paolo Guiotto: If the function f is differentiable, at point X,
06:58:240Paolo Guiotto: So if there exists these Jacobian matrix, then…
07:02:980Paolo Guiotto: So it says that this concept is much stronger than the directional derivatives. There exist all the directional derivatives of F at point X,
07:15:400Paolo Guiotto: for every vector V,
07:18:100Paolo Guiotto: The vector V is a vector that stays in the domain, so here is in RD.
07:25:00Paolo Guiotto: So all these derivatives exist. So this… see, if I stop here, is saying that differentiable is more than having directional derivatives, okay? But the nice thing is that it says.
07:39:550Paolo Guiotto: what is the relation between these quantities and the Jacobian matrix? It says that this is exactly F prime of X
07:48:740Paolo Guiotto: times V.
07:52:160Paolo Guiotto: And thanks to this formula, we will deduce the formula for the Jacobian matrix, as you will see. But let's first see this, which is a simple calculation, basically.
08:06:130Paolo Guiotto: So, we have to verify that there are the directional derivative, and that formula applies. So, the directional derivative, let's write the definition. Direction V, point X, is the limit when T goes to 0 of F…
08:25:340Paolo Guiotto: of X plus TV.
08:29:399Paolo Guiotto: minus F… of X… divided by T, right? That's the definition of directional derivative.
08:40:850Paolo Guiotto: Okay, if you look at this numerator, it looks like f of x plus h minus f of x, where h is this guy here.
08:51:390Paolo Guiotto: So that's what we want to do. We know something about f of x plus h minus f of x, which is written there. Now we want to plug that thing inside this limit here, somewhere.
09:03:260Paolo Guiotto: So what I do is to reconstruct the numerator of this fraction, imagine with H equal TV, okay? So, what I need here is, I have F of X plus H is this one. Imagine that I call for a moment this H.
09:20:939Paolo Guiotto: F of X plus H, minus F of X. To have that numerator here, I miss just the last term, minus F prime of X times H. So what I do, I make it to appear, so I write limit when t goes to 0,
09:37:340Paolo Guiotto: Fraction, it would be pretty long.
09:40:350Paolo Guiotto: F over X… plus TV.
09:46:250Paolo Guiotto: minus F of X.
09:50:479Paolo Guiotto: Up to this point, it's the same. Now, I want to create this term here.
09:57:440Paolo Guiotto: minus F prime X times H.
10:00:320Paolo Guiotto: Where age is the vector TV. So, I do not have… how can I do? I add and subtract. So, I first subtract F prime of X.
10:11:790Paolo Guiotto: applied… of course not, there is no H here, but my H is TV, applied to vector TV,
10:20:30Paolo Guiotto: So let's write with parentheses, because that vector is multiplied by the matrix F prime. So since that quantity was not there, I have… I subtracted, now I have to add back. So F prime…
10:34:830Paolo Guiotto: Except… TV.
10:39:430Paolo Guiotto: And as you can see now, it's the same, because I added and subtracted the same quantity. But now I break fraction here.
10:46:920Paolo Guiotto: So I do. Limit, when t goes to 0, of F of X… plus TV.
10:57:740Paolo Guiotto: minus F of X.
11:01:730Paolo Guiotto: minus F prime… of X, huh?
11:06:330Paolo Guiotto: applied to vector TV, divided by T.
11:10:450Paolo Guiotto: And then I have a blossom.
11:13:200Paolo Guiotto: F prime.
11:15:710Paolo Guiotto: of X applied to TV, divided by T. All this is into the limit.
11:23:930Paolo Guiotto: And so far, isn't identical.
11:26:940Paolo Guiotto: Now, let's first look at this second part.
11:31:650Paolo Guiotto: I have matrix F prime x, Multiply the line by colon, with the vector TV.
11:42:110Paolo Guiotto: Let's see what you remind about this story.
11:47:590Paolo Guiotto: Tell me, can I say that this is T times F prime of x applied to vector V?
12:00:150Paolo Guiotto: Yes, why? Of course, it is true. There are…
12:04:840Paolo Guiotto: several possible ways to explain this. But first is, do the calculation. You will see all the entries here are multiplied by the cost and t, so you will have that when you do the product line by column, all factors will be multiplied by T, you can put everything in front of.
12:23:380Paolo Guiotto: But a better way is that to remind that the operation multiplying metrics by vector as a function of the vector is… what kind of function?
12:35:770Paolo Guiotto: It's a special type of function.
12:38:770Paolo Guiotto: D functions, metrics.
12:41:700Paolo Guiotto: Times vector, this function, is what?
12:47:230Paolo Guiotto: is what is called A?
12:49:530Paolo Guiotto: linear function, okay? What is the linear function? The features of linear function is that M, if you multiply by a constant, a vector, this constant just flips to
13:02:670Paolo Guiotto: and also M of H plus W, no? It is equal to MH plus MW, right?
13:12:860Paolo Guiotto: This is linearity. Linear… function.
13:18:790Paolo Guiotto: So that's why the coefficient t can be written in front of this. If you want, you do the product with metrics line by column, and you realize that this is correct. Now, if you can do this.
13:33:810Paolo Guiotto: So when I have this fraction.
13:37:450Paolo Guiotto: So, when I have F prime, OVEX.
13:41:880Paolo Guiotto: Evaluated on TV, divided by T.
13:46:420Paolo Guiotto: So I flip T out here, so T times F prime of X,
13:52:490Paolo Guiotto: times V divided by T. As you can see, this T and T cancel, and what remains is independent of T.
14:02:00Paolo Guiotto: You see? So it's a constant in T.
14:05:970Paolo Guiotto: So, here, you can say that this is equal to F prime X, evaluated
14:14:460Paolo Guiotto: on vector V, multiply the line by columns on V.
14:19:120Paolo Guiotto: So, this is about the second fraction. Now, let's look at the first fraction. Let's see what we can expect.
14:26:650Paolo Guiotto: You're reminded that the origin of this operation was to recreate
14:32:360Paolo Guiotto: The numerator we have here in the definition.
14:36:280Paolo Guiotto: Oh, by definition, we know that this limit is zero. So when we take this numerator with a vector H, and we divide by the norm of H,
14:47:150Paolo Guiotto: and we send H to 0, this fraction goes to zero, okay? So let's write it down here.
14:54:650Paolo Guiotto: Now… Since… F.
15:00:260Paolo Guiotto: well, let's write in this way, exists the derivative of F at point x. That means F is differentiable at X. The limit
15:07:930Paolo Guiotto: When H goes to 0,
15:11:220Paolo Guiotto: of F… I'm just copying the definition. F of X plus H minus F of X
15:20:710Paolo Guiotto: minus F prime X times A chun?
15:26:270Paolo Guiotto: divided by the norm of H, this quantity goes to zero.
15:32:770Paolo Guiotto: So, we now claim… look at this… so, well, let me copy. So…
15:40:250Paolo Guiotto: I now copied this block here. So, the limit.
15:46:110Paolo Guiotto: Look carefully at the two expressions, they are not exactly the same, but they look very similar.
15:52:810Paolo Guiotto: F of X plus TV.
15:57:790Paolo Guiotto: Minus F of X.
16:01:670Paolo Guiotto: minus F prime.
16:04:320Paolo Guiotto: of X, applied to TV, divided by T,
16:10:270Paolo Guiotto: I pretend to compute the second limit by using the first one.
16:15:700Paolo Guiotto: What is, what do you have to see here?
16:19:140Paolo Guiotto: Call for a second TV equal age.
16:23:510Paolo Guiotto: Call this age.
16:25:670Paolo Guiotto: So you see that you have an H also here.
16:28:590Paolo Guiotto: So that numerator is F of X plus H, minus f of x, minus F prime of x, this is an S prime, sorry, I forgot.
16:38:260Paolo Guiotto: This is F prime, huh? F prime of X times H. Exactly the quantity we have here.
16:44:900Paolo Guiotto: divided by TT is not normal of age. But what is norm of age?
16:50:960Paolo Guiotto: Norm of H, if H is TV, this is norm of TV.
16:56:630Paolo Guiotto: And what is this?
17:00:470Paolo Guiotto: Not exactly, almost.
17:03:210Paolo Guiotto: Absolute value, that's right. Absolute value of T times norm of V.
17:09:110Paolo Guiotto: Okay, I don't see the absolute value, I don't see the norm of V.
17:14:450Paolo Guiotto: But since I don't see I can't always create, because I'm God, no? I can create from nothing, everything. So I can say that, okay, let's push out this, 1 over T, and let's replace by modules of t. Of course, I cannot do that in this way, I have to multiply and divide.
17:31:20Paolo Guiotto: T is different from 0 because we are going to 0, so t is not 0.
17:36:600Paolo Guiotto: Now, to multiply, to make here, to appear the norm of V, it's a little bit more complicated, because this could be 0.
17:45:230Paolo Guiotto: Who says that vector V is not 0. Okay, let's take for a second V different from 0, because V equals 0 is trivial, in fact.
17:55:630Paolo Guiotto: And if V is different from 0,
17:59:680Paolo Guiotto: We can now multiply and divide by the norm of V, because it's a number different from 0. And now we have recreated here the norm of age.
18:13:170Paolo Guiotto: And now we can say, all this block here, Goes to zero.
18:21:200Paolo Guiotto: And what about what remains? What remains is… so we have limit for T going to 0,
18:30:160Paolo Guiotto: off. Let's call this quantity as 0T, okay? Something that goes to zero.
18:35:690Paolo Guiotto: Then, what remains is the absolute value of T divided T times norm of V.
18:42:90Paolo Guiotto: Well, absolute value of T divided T is T divided, absolute value of t is the sine.
18:48:170Paolo Guiotto: of T, which of course has not a limit when t goes to zero. But who cares? It is multiplied by someone who is going to zero. So this goes to zero. This quantity is just bounded.
19:01:480Paolo Guiotto: An absolute value is constant, even, okay?
19:05:370Paolo Guiotto: And therefore, the limit will be zero.
19:08:740Paolo Guiotto: Okay? So now we can conclude…
19:12:660Paolo Guiotto: So, block 1, this, we learned, goes to 0. Block 2 is constantly equal to that, the limit is the quantity radius there.
19:22:180Paolo Guiotto: So we get the data, so the conclusion.
19:30:690Paolo Guiotto: So we started from the calculation of the directional derivative…
19:36:160Paolo Guiotto: The directional derivative of F, okay? So the directional derivative.
19:42:130Paolo Guiotto: of F in direction V at point X. Conclusion says that, blah blah blah, it is equal to this red quantity, F prime of X times V.
19:56:690Paolo Guiotto: And this also… If V is different from 0,
20:04:620Paolo Guiotto: Right? Because we added this restriction. What if V is equal to 0?
20:08:900Paolo Guiotto: Well, you reminded what is the directional derivative in direction 0?
20:14:180Paolo Guiotto: It is equal to…
20:20:300Paolo Guiotto: The directional derivative in direction 0 of whatever is
20:26:760Paolo Guiotto: Well, think about it, it is the limit when t goes to zero.
20:33:460Paolo Guiotto: What'd you say?
20:38:660Paolo Guiotto: Yeah, but the value… Zero, that's right, because you see that it is equal to X plus T0.
20:47:630Paolo Guiotto: minus F of X, divided by T.
20:52:770Paolo Guiotto: T0 is 0.
20:55:410Paolo Guiotto: X plus 0 is X. F of X minus F of X is 0, so that's a 0.
21:01:430Paolo Guiotto: So the left-hand side is zero.
21:04:170Paolo Guiotto: And the right-hand side?
21:11:400Paolo Guiotto: What if I multiply by zero?
21:13:700Paolo Guiotto: It's a linear stuff, no? You don't need to remind how this operation works. They sell always 0 into 0, okay? So this is 0, and so it means that the identity holds also for V equals 0, so now I can say for every V.
21:30:50Paolo Guiotto: So, actually, for every V.
21:33:750Paolo Guiotto: in RD. And that finishes the proof.
21:38:250Paolo Guiotto: Okay, now, from this identity, we find out the entries of F', because… Now… from…
21:50:990Paolo Guiotto: The identity, directional derivative of F.
21:55:760Paolo Guiotto: at point X equal F prime.
21:59:810Paolo Guiotto: X.
22:01:360Paolo Guiotto: times V, we determine… D… entries.
22:12:410Paolo Guiotto: of the Jacobian metrics.
22:21:420Paolo Guiotto: Let's see why. F prime of X…
22:24:620Paolo Guiotto: is, we said, a name by D matrix.
22:28:970Paolo Guiotto: M lines, D columns. So let's say that it is something like A11 etc, A1D.
22:38:550Paolo Guiotto: And then we have AM1AMD.
22:45:90Paolo Guiotto: Now, do you know what I have to do to extract the generic term here, AIJ, so J…
22:54:110Paolo Guiotto: Line… no, sorry, height line, and Jade column.
22:59:440Paolo Guiotto: This one. I want this one.
23:06:440Paolo Guiotto: Yeah, that's the point. We have to do the multiplication, we have to put a vector V
23:14:740Paolo Guiotto: that… notice, if you multiply by V equal, let's take D,
23:20:20Paolo Guiotto: I'm sure that I will do a mess with the indexes. I try… at worst, I have the IJ inversed. Okay, let's apply this with the EJ. What we obtain is F prime
23:37:340Paolo Guiotto: X, huh?
23:38:950Paolo Guiotto: multiplied by EJ, it means that you are multiplying by this vector, which is a vector made of zeros everywhere, except at the jth position.
23:50:660Paolo Guiotto: So now imagine that you multiply your metrics with the A's.
23:54:450Paolo Guiotto: A11, A1D, etc. As you can see, when you do line by column, there is only one of these A that is saved. It is that one which is in position J, okay? So this will be the A1J, A2J, A3J, AMJ.
24:14:780Paolo Guiotto: Jay.
24:16:500Paolo Guiotto: Okay? Because if you do this product, you get A11 times 0, 0. A12 times 0, 0, etc. A1J times 1, A1J, then A1J plus 1 times 0, 0. So when you do this product, what remains is exactly
24:35:270Paolo Guiotto: the jth column, A1J, A1J, A2J, etc, A, M, J. So when I multiply this matrix by the jth vector of the canonical basis, I get the J column of the matrix.
24:53:620Paolo Guiotto: So if you want the J column of the matrix.
24:56:870Paolo Guiotto: Now, using this identity, so the J column, A1J.
25:04:840Paolo Guiotto: A, M, J. This is… F prime.
25:09:840Paolo Guiotto: X.
25:11:30Paolo Guiotto: multiplied by EJ, which is the derivative, the direction derivative in the j direction of F.
25:19:820Paolo Guiotto: of X.
25:22:770Paolo Guiotto: Okay, but now I want this element, the element in line I.
25:28:960Paolo Guiotto: So this is my… AIJ.
25:33:820Paolo Guiotto: How do I extract this one from this vector? I simply multiply it by the scalar product with now EI.
25:42:20Paolo Guiotto: So take this and multiply in the scalar product with EI.
25:49:920Paolo Guiotto: So what happens? You are doing, A1J, A, M, J…
25:59:150Paolo Guiotto: scalar product with 0, 0, 1,
26:03:740Paolo Guiotto: 000. This is in the iH position. So when you do A1J times 00, plus A2J times 00, etc, this will give you Aij, the element we want.
26:17:960Paolo Guiotto: So this is equal, then, to the product of this, DEJ, F… Excellent.
26:27:460Paolo Guiotto: scalar with the EI vector.
26:31:670Paolo Guiotto: And what is this?
26:33:210Paolo Guiotto: Now, assume that our F, which is vector-valued.
26:38:580Paolo Guiotto: This is an array with them components. These components are functions, no? Because F is a function. So these functions are F1, F2, etc, FM.
26:53:260Paolo Guiotto: So these are the components
26:59:590Paolo Guiotto: of the vector F.
27:02:710Paolo Guiotto: arrow. We have still 3 minutes, so we… We can't finish.
27:09:330Paolo Guiotto: Okay, so what happens when we do the scalar product with the vector EI?
27:14:420Paolo Guiotto: The same thing, you have your vector, F1, FM…
27:19:520Paolo Guiotto: You do the product with the vector EI, so 0, 1, 0, in the height.
27:28:710Paolo Guiotto: position, and you get the ith component. So this is FI.
27:34:760Paolo Guiotto: So this means that you are doing here the J… the directional derivative with respect to the EJ factor of the canonical basis of the i-th component of F.
27:51:680Paolo Guiotto: So, that's… we have the formula.
27:54:630Paolo Guiotto: AI…
27:57:370Paolo Guiotto: The element, line I, column J, of the Jacobian matrix, is the derivative. Now, that's a directional derivative in one of the vectors of the canonical basis, so it's a partial derivative.
28:11:240Paolo Guiotto: So it is the partial derivative with respect to the jth variable of what? Of the ith component of F.
28:20:270Paolo Guiotto: Okay?
28:22:40Paolo Guiotto: So that's the formula we get.
28:24:480Paolo Guiotto: So, in other words, Conclusion.
28:31:960Paolo Guiotto: So, suppose that our F… Which is a function.
28:37:300Paolo Guiotto: of vector X. It is vector-valued.
28:41:600Paolo Guiotto: It goes from RD to RM.
28:45:800Paolo Guiotto: So this means that each F of X is an array, okay, with M components. So we call these components F1 of X, these are functions of X, F2 of X, etc, FM .
29:04:570Paolo Guiotto: Okay?
29:06:240Paolo Guiotto: Then, the Jacobian matrix, F prime.
29:11:580Paolo Guiotto: of X.
29:13:690Paolo Guiotto: is the matrix where the element at line i, column J, is the partial derivative with respect to the jth variable of the it component. What does it mean?
29:25:810Paolo Guiotto: Let's take the first line. In the first line, we have A11, A12, A13, A14, okay?
29:34:110Paolo Guiotto: So, in the first line, I is 1.
29:38:800Paolo Guiotto: F1 is the first component of F.
29:43:840Paolo Guiotto: So what do you find in the first line? You find the derivative with respect to the first variable of the first component.
29:51:600Paolo Guiotto: The derivative with respect to the second variable still of the first component.
29:56:820Paolo Guiotto: The derivative of the first component with respect to the third variable until you finish the variables, the derivative of the first component with respect to the last variable, the fth variable. Second line, now I is 2, so it means that now you work on the second component of capital F.
30:15:830Paolo Guiotto: So, you compute all the partial derivatives of this second component, so you will have here.
30:22:00Paolo Guiotto: D1 of F2.
30:25:210Paolo Guiotto: So, second component, first partial derivative of the second company, or partial derivative with respect to the first variable. Then you have the partial derivative
30:35:150Paolo Guiotto: with respect to the second variable, still of the second component. Then you have the partial derivative with respect to the third variable of the second component, until you finish all the variables.
30:47:150Paolo Guiotto: Next line, you work on the third component until you finish all the components. Last line will be the set of the array of the derivatives of the last component with respect to all the possible variables.
31:05:520Paolo Guiotto: Okay.
31:08:910Paolo Guiotto: Nice,
31:11:460Paolo Guiotto: Now, we should, we should, do you have, do 2 minutes? We do a, just a quick example, just to fix ideas. Example.
31:22:60Paolo Guiotto: So let's take a… you see complicated, but normally we have a low number of variables.
31:28:600Paolo Guiotto: Imagine that we have a function f of xy with two components. Say that they are X times Y and X plus Y, to make easier.
31:39:270Paolo Guiotto: So this is F1, this is F2. These are the two components.
31:44:510Paolo Guiotto: So this means that the Jacobian matrix of this
31:49:680Paolo Guiotto: is made by? First of all, you see that we start from a vector with two components, so this belongs to R2,
31:59:920Paolo Guiotto: And we end into a vector of R2.
32:03:270Paolo Guiotto: So this means that B is 2, and them is 2. Jacobian matrix will be a 2x2 matrix, okay?
32:10:310Paolo Guiotto: Now, in the first line, we have, take the first component, F1, and compute all possible derivatives. There are only two, because there are two variables. So we have the derivative with respect to X of F1, derivative with respect to Y of F1.
32:25:230Paolo Guiotto: Second line, you now take the second component of F, F2, and you again compute the two derivatives, F, DXF2, DYF2.
32:35:940Paolo Guiotto: Now, specifically, here we have derivative DXF1 is… Why? The YF1?
32:49:530Paolo Guiotto: X.
32:51:380Paolo Guiotto: DXF2.
32:55:300Paolo Guiotto: DXF2, the derivative with respect to X of F2, One…
33:02:160Paolo Guiotto: Derivative with respect to Y of F2,
33:05:80Paolo Guiotto: One. And that's the Jokomian matrix.
33:08:410Paolo Guiotto: Okay? So, as you can see, it's not an impossible, an impossible task. Okay, have a nice weekend. I do not leave you any new exercise, because we need still to work a bit on this. So, see you on, Tuesday, okay?