Transcript
00:35:130Paolo Guiotto: Okay, good morning.
00:39:100Paolo Guiotto: Okay, let's start with an exercise.
00:43:750Paolo Guiotto: of checking differentiability, just to see one more example. 292, actually, I would defy, slightly the…
00:55:200Paolo Guiotto: exercise, I take the number 4, let's say, modified.
01:04:959Paolo Guiotto: So we have this function, FXY, Which is equal to… X squared, Y… Minos…
01:23:220Paolo Guiotto: Okay, so let's modify in this way.
01:33:830Paolo Guiotto: No, sorry. X, way, X squared, Y square, minus YX squared… Minus, Y cubed.
01:49:770Paolo Guiotto: divided X by X squared plus Y square.
01:57:410Paolo Guiotto: Okay.
01:58:640Paolo Guiotto: This for XY different from 00.
02:03:790Paolo Guiotto: And 0 for XY.
02:07:860Paolo Guiotto: Equal 000.
02:11:520Paolo Guiotto: There are three questions. Number one, is F… Continos at… Zero, zero?
02:22:810Paolo Guiotto: Number two, computer… There exists derivatives partial derivatives of F at 0, 0.
02:31:370Paolo Guiotto: And, if, in affirmative case, compute them.
02:36:560Paolo Guiotto: is… number 3, is F differentiable?
02:41:380Paolo Guiotto: at 00.
02:45:530Paolo Guiotto: Okay.
02:46:800Paolo Guiotto: So, let's see the… First up.
02:51:240Paolo Guiotto: Born.
02:52:610Paolo Guiotto: To be continuous, since here, the point where we have to check everything is exactly where the definition changed, so there is no other thing that checking with the definition. F.
03:04:950Paolo Guiotto: Continos.
03:06:430Paolo Guiotto: at… 0, 0.
03:10:340Paolo Guiotto: If and only if…
03:12:560Paolo Guiotto: I hope you have seen my message this morning. Today, we will do 60 minutes non-stop, and then we finish the class for today. The limit for XY going to 0,
03:28:160Paolo Guiotto: of F, X, Y, This is equal to F00.
03:33:650Paolo Guiotto: Which is equal to zero, so basically we have to verify if that limit is zero or not.
03:39:570Paolo Guiotto: Now, for the limit, of course, since XY is going to 0, but it is different from 0, 0, FXY will be the first line of this formula.
03:50:510Paolo Guiotto: And, well, let's directly plug the polar coordinates, let's see what happens. So, if I take FXY…
03:57:490Paolo Guiotto: And I replace X by raw cosine theta, and Y by raw sine theta, I have.
04:07:300Paolo Guiotto: The denominator is X squared plus Y squared, which is raw square.
04:12:900Paolo Guiotto: Numerator is X squared y squared, so raw power 4 cos square theta sine square theta.
04:24:200Paolo Guiotto: Then I have, minus… Alright, Cuba.
04:30:710Paolo Guiotto: Yx, we are so, sine theta… Cos square theta.
04:40:560Paolo Guiotto: And the other one is Y cubed, so, minus raw cube sine… qubit theta.
04:50:170Paolo Guiotto: So, as you can see, we can simplify. We have a raw square here.
04:56:950Paolo Guiotto: Multiply by this factor, cos square theta sine squared theta.
05:04:100Paolo Guiotto: And we have a minus, raw.
05:10:330Paolo Guiotto: That is multiplied by sine theta, which one we can factorize.
05:17:190Paolo Guiotto: sine theta.
05:20:230Paolo Guiotto: Then we have cos squared theta plus sine
05:25:270Paolo Guiotto: Square theta, so which is actually equal to 1.
05:30:970Paolo Guiotto: And, so you see that when XY goes to 0, this means for raw goes to 0,
05:37:520Paolo Guiotto: These coefficients for raw square and raw are bounded functions of theta, so we can easily handle this and say that it goes to zero.
05:47:990Paolo Guiotto: And, how. So… the distance between FXY and 0, which is,
05:55:450Paolo Guiotto: The candidate to be the limit, is the absolute value of raw square cos square theta sine square theta minus raw sine theta.
06:10:120Paolo Guiotto: Now, models of sum differences less frequent sum of the models is, so I have raw square, then I have models of cos squared theta sine squared theta.
06:22:970Paolo Guiotto: Plus.
06:24:470Paolo Guiotto: I do not put minus, okay? Because it's a fake minus, there is also the sign here. So, absolute value of sine theta.
06:34:890Paolo Guiotto: Okay, this is less or equal 1, this is less or equal 1, so I have less or equal rho square plus rho that goes to 0 when rho goes to 0, which is equivalent
06:48:160Paolo Guiotto: off… This.
06:51:520Paolo Guiotto: XY goes to 0C.
06:54:370Paolo Guiotto: So the conclusion is that There exists the limit.
06:59:980Paolo Guiotto: for XY going to 0, 0.
07:04:250Paolo Guiotto: of this F, and this is equal to 0, which is the value at 0, 0. So, F is continuous.
07:12:490Paolo Guiotto: That's zero-ceto.
07:19:450Paolo Guiotto: And that's the first step.
07:21:430Paolo Guiotto: point.
07:22:810Paolo Guiotto: Second, the problem asks for the partial derivatives at 0, 0 of F.
07:29:980Paolo Guiotto: Now, since that point is the bad point where the diffusion changed, I cannot just complete the partial derivative of the first line, and then put X and Y equal 0, because
07:40:420Paolo Guiotto: the first line is valid for XY different from 0, 0. So, the unique road is to go back to the definition. DXF at 0, 0 is a directional derivative.
07:53:70Paolo Guiotto: direction derivative along the first vector of the canonical basis, 10 of F at 0, 0.
08:02:500Paolo Guiotto: Which is a limit.
08:04:320Paolo Guiotto: or T going to zero, of F… 0, 0 plus P.
08:11:70Paolo Guiotto: 1, 0… minus F00 divided by T.
08:17:840Paolo Guiotto: all this, the argument of F is 00 plus T0, so T0, so this is F at T0, minus F00.
08:29:100Paolo Guiotto: F00 is 0, what about FT0? T is going to 0, but it is different from 0, so we have to look at the first line. When you see that you put the Y equals 0, the numerator is 0.
08:44:330Paolo Guiotto: denominator remains X squared, so this would be 0 divided T squared, so in any case, 0.
08:52:140Paolo Guiotto: And the limit is limit for t going to 0 of 0 divided t, which is, of course, equals 0.
09:02:530Paolo Guiotto: And the similar calculation, unfortunately, the function is not, specular with respect to X and Y, so we have to…
09:10:70Paolo Guiotto: compute DYF at 0, 0 is the same. It is now the direction relative, the direction is 01.
09:18:890Paolo Guiotto: F00.
09:22:430Paolo Guiotto: So, we have the limit.
09:25:950Paolo Guiotto: when t goes to zero, alpha…
09:29:240Paolo Guiotto: Now, let's do these calculations without writing. So we have 00 plus T01, now it will be 0T, so this will be F0T.
09:41:210Paolo Guiotto: minus F00.
09:43:820Paolo Guiotto: divided by T.
09:46:440Paolo Guiotto: F00 is 0. What about F0T? We have, of course, to look at the first line. When x is 0, you see that the numerator remains minus Y squared, y cubed.
09:59:200Paolo Guiotto: the luminator is Y squared, so the total is…
10:03:830Paolo Guiotto: minus T cubed divided by T squared.
10:09:510Paolo Guiotto: So we have a limit for t going to 0.
10:13:610Paolo Guiotto: of fraction.
10:15:460Paolo Guiotto: That, this T… this square T cubed minus this over… this square is minus T.
10:22:340Paolo Guiotto: So, minus T divided T, That's… It's too big, this fraction.
10:28:30Paolo Guiotto: This is the limit when t goes to 0 of minus 1, which is minus 1.
10:35:270Paolo Guiotto: So the conclusion is that the two partial derivatives, they both exist, and one is equal to 0, the other one is equal to minus 1, okay?
10:44:60Paolo Guiotto: Now, question 3.
10:45:990Paolo Guiotto: Question 3 concerns the differentiability.
10:50:680Paolo Guiotto: So…
10:51:920Paolo Guiotto: I'm not going to use, yet the test of the differentiability test we have, mentioned at the end of, class… yesterday, class, where we do today.
11:02:760Paolo Guiotto: In a moment. So, for the moment, I used the definition.
11:07:720Paolo Guiotto: So, according…
11:13:430Paolo Guiotto: to the… definition… F is differentiable at 0, 0, 0,
11:24:110Paolo Guiotto: If and only if we have this limit.
11:28:570Paolo Guiotto: when H goes to zero H factor of F0 plus H.
11:36:480Paolo Guiotto: minus F0.
11:39:110Paolo Guiotto: minus the Jacobian matrix, which is the gradient, in this case, F at 00, applied to H. All this divided by the norm of H,
11:49:180Paolo Guiotto: This limit must be equal to zero.
11:53:330Paolo Guiotto: Now, let's see what is this limit.
11:56:410Paolo Guiotto: So, let's remind that F00 is, 0. This value is 0.
12:04:480Paolo Guiotto: That's… the gradient is the vector with component 0, 1.
12:09:120Paolo Guiotto: applied to the vector H, it is better if we introduce coordinates for this UV, say.
12:15:800Paolo Guiotto: So, line by column.
12:19:800Paolo Guiotto: 01 times UV is equal to…
12:23:120Paolo Guiotto: Sorry, it is minus 1.
12:26:220Paolo Guiotto: So it is equal to… Minus V. You have a minus V here.
12:31:670Paolo Guiotto: So, this is the limit.
12:35:270Paolo Guiotto: In coordinates. UV.
12:38:40Paolo Guiotto: Going to 0, 0, off.
12:42:280Paolo Guiotto: Well, downstairs we have the norm of H, which is the root of u squared plus B squared.
12:48:540Paolo Guiotto: At the numerator, we have F, this is F of H, the argument, so F of UV.
12:57:290Paolo Guiotto: minus F of 0.
12:59:310Paolo Guiotto: which is 0. Minus, minus that product, which is minus B.
13:07:80Paolo Guiotto: Now, FUV is… since UV is not 0, we have to look at the first line of this thing. So, X squared, Y squared.
13:19:80Paolo Guiotto: So that would be in UV, U square, V square.
13:23:550Paolo Guiotto: So the denominator is V squared plus V square, then what is it, this?
13:31:60Paolo Guiotto: minus B, U square minus the cube.
13:38:770Paolo Guiotto: So, minus, VU square… Minus V cubed.
13:45:220Paolo Guiotto: Okay.
13:46:600Paolo Guiotto: equals zero.
13:49:420Paolo Guiotto: If and only if the limit when UV goes to 0, 0, that's now…
13:57:530Paolo Guiotto: write a bit better this. Maybe you better write this as a power plus V squared equal to power 1 half.
14:07:880Paolo Guiotto: So this minus, minus V is a plus V.
14:13:410Paolo Guiotto: And we have to do the… the sum.
14:16:960Paolo Guiotto: of this, this plus B. So let's do the common denominator.
14:21:630Paolo Guiotto: underneath is your square plus V square R.
14:24:570Paolo Guiotto: Numerator is u squared v square minus V U squared minus V cubed, then I have plus V times U squared plus V squared.
14:34:670Paolo Guiotto: As you can see, This cancels this, and this cancels this.
14:40:780Paolo Guiotto: So, at the end, we have… That this must be zero.
14:46:460Paolo Guiotto: Of course, this has to be probed, yeah? This is an if and only if F is differentiable at 0, 0, if and only if this limit is 0. Let's continue. So, let's simplify. This is the limit
15:00:570Paolo Guiotto: for UV going to 0, 0.
15:05:390Paolo Guiotto: of U squared V squared divided by… there is u squared plus V squared to power 1 plus 1 half Pia.
15:17:290Paolo Guiotto: If this is equal to zero. Now, the question is, is that limit equal to zero?
15:23:260Paolo Guiotto: So this is a new function, let's call GUV.
15:28:680Paolo Guiotto: Let's see what is it in polar coordinates.
15:31:920Paolo Guiotto: G-U-V… With the U equal raw cosine theta, B equal rho sine theta.
15:42:980Paolo Guiotto: is raw power 4… Cos square theta sine… Square theta, divided by…
15:52:980Paolo Guiotto: U squared plus V squared is raw square to explain 3 times.
15:59:210Paolo Guiotto: So, at the end, we have rho power 4 divided… this is… Wrong?
16:07:530Paolo Guiotto: Dude… Then we have cos squared theta sine squared theta.
16:15:350Paolo Guiotto: We can simplify, this disappears, and it remains a raw.
16:20:140Paolo Guiotto: From this, we see that when raw goes to zero, that product will go to zero. The coefficient in theta is bounded, so we can reasonably see that this is going to zero. And in fact.
16:31:180Paolo Guiotto: We say, absolute value of GUV minus zero is exactly this quantity. There is no need of modulus, because it is positive.
16:42:520Paolo Guiotto: Cos square theta sine square theta.
16:47:50Paolo Guiotto: Now, all this coefficient is less or equal 1, so this is less or equal raw that goes to 0 when raw goes to 0, and this is exactly
16:58:770Paolo Guiotto: UV going to 0, 0.
17:03:920Paolo Guiotto: And this shows that, so there exists the limit
17:08:810Paolo Guiotto: when UV goes to 0, 0,
17:13:869Paolo Guiotto: of GUV, and that limit is 0, which is exactly what,
17:22:10Paolo Guiotto: what, what must be verified to have F differential. So the conclusion is that, yes, F is differentiable.
17:30:840Paolo Guiotto: Conclusion.
17:35:260Paolo Guiotto: F is differentiable.
17:39:930Paolo Guiotto: at 0, 0.
17:44:680Paolo Guiotto: If you have not yet done, I warmly suggest you to do the exercise 292.
17:50:280Paolo Guiotto: Because it is not particularly important for the…
17:57:10Paolo Guiotto: For the course, but it is important because it makes you to learn well the definition.
18:02:470Paolo Guiotto: how to check, there are details inside, so it is better if you train a bit on this. Now, yesterday we mentioned a test that should be
18:18:270Paolo Guiotto: a bit better for, checking differentiability. That says…
18:23:590Paolo Guiotto: If you have the existence of partial derivatives, which alone is not sufficient.
18:28:750Paolo Guiotto: But you know also that they are continuous, this is the plus that you need. With this extra assumption, then the function turns out to be differentiable. And also, this is nice because it gives this conclusion
18:41:960Paolo Guiotto: on a whole domain, no? If the continuity is on a domain D, you have differentiability automatically at all points of the domain D.
18:51:420Paolo Guiotto: Okay?
18:52:570Paolo Guiotto: So let's see an example of using this.
18:56:350Paolo Guiotto: With the exerciser, 293.
19:04:570Paolo Guiotto: So, here it says, show…
19:09:810Paolo Guiotto: Well, oh, Boveda, let's put things in a more open form. Discuss… differentiability, of F.
19:21:710Paolo Guiotto: XY defined as X times the square root of X squared plus Y squared.
19:30:270Paolo Guiotto: on… its natural domain.
19:34:100Paolo Guiotto: which is the set of points where this thing is defined. There is no restriction, because the unit problem could be the root that requires the argument positive, or zero.
19:45:610Paolo Guiotto: And this is the case for the quantity X squared plus Y squared 0. Domain D is the full plane R2.
19:55:600Paolo Guiotto: So this is the question. Let's see the answer.
20:02:80Paolo Guiotto: Now.
20:03:80Paolo Guiotto: Let's see if we can apply the differentiability. That's also because we cannot think to check for every point, X, Y, doing all this machinery of producing the limit, etc, okay?
20:16:640Paolo Guiotto: So, as you will see, it will be a little bit faster, much more faster than the other method.
20:23:340Paolo Guiotto: You have always to remember that this is not any characterization, that is not an if-and-all if.
20:30:70Paolo Guiotto: Okay? So, it might be that the function is differentiable, but that condition is not verified.
20:37:60Paolo Guiotto: Okay, that won't be the case for us, because 99% of cases, these conditions apply, but remind also that there is the 1% of things that go wrong.
20:49:220Paolo Guiotto: So, what do we do? We compute, we start computing the partial derivatives, and let's see what happens. Here…
20:57:650Paolo Guiotto: We can easily compute the partial derivatives, DX, F, X, Y. We treat this as a function of X, so forget Y. Y is a constant.
21:08:390Paolo Guiotto: So, as a function of X, I see a product, no? So I start differentiating the factor X, so I get 1 times the root, X squared plus Y squared, plus X. Now I differentiate the root.
21:23:560Paolo Guiotto: The derivative of the root is 1 over 2 the root, X squared plus Y squared, then I have to differentiate the argument.
21:32:240Paolo Guiotto: Respect to X, so Y is a constant, the derivative with respect to X is 2X.
21:37:510Paolo Guiotto: So at the end, I get root of X squared plus Y squared, the two cancel.
21:48:140Paolo Guiotto: plus X divided the root of X squared plus Y squared, or if you prefer, we can do the common denominator. It is the root of X squared plus Y squared. Then when you do the root times the root, you get X squared plus Y squared.
22:07:250Paolo Guiotto: plus X.
22:09:520Paolo Guiotto: Okay? Yeah.
22:12:790Paolo Guiotto: Should be what?
22:14:250Paolo Guiotto: Where?
22:17:510Paolo Guiotto: This one. Okay, thank you. So, at the end, this is, too expert, okay?
22:25:660Paolo Guiotto: Good. Now, When this is correct.
22:35:140Paolo Guiotto: Huh?
22:39:70Paolo Guiotto: No, I'm asking, for which XY is this correct?
22:45:720Paolo Guiotto: for every XY, No. You see that you cannot put X and Y equal 0.
22:52:940Paolo Guiotto: So this is for every point XY, different from point 00.
23:00:90Paolo Guiotto: Okay? Otherwise, it's a nonsense what is written for 0. So this means that this formula
23:07:120Paolo Guiotto: works for the partial derivative, except at one single point, okay?
23:13:80Paolo Guiotto: And I'd say that this will be helpful, because it, it, help
23:18:270Paolo Guiotto: It helps to discuss the differentiability for all points except one point, where we have to discuss a little bit more in detail. Same calculation for the DYF,
23:32:510Paolo Guiotto: this is even easier, because Y is only in the root, so X is a factor.
23:39:720Paolo Guiotto: So you see that when I have to do the dy of X times root of X squared plus Y squared, that X is a coefficient, so it comes out. X times the derivative, I have to the root at the denominator.
23:56:670Paolo Guiotto: And then I have the derivative of the argument, this time with respect to Y, so it is 2Y. So this simplifies, and I get XY divided by a root of X squared plus Y squared. Again, for which points this… is this correct?
24:14:600Paolo Guiotto: for every point XY, except the unique one for which we cannot write that thing, which is 0, 0.
24:24:860Paolo Guiotto: Okay, so let's see if we can apply… if we can apply differentiability tests at least for, for every point except 00.
24:36:70Paolo Guiotto: What can we say? Well, if you look at the expressions of the partial derivatives, this is the partial derivative with respect to X. This is the partial derivative with respect to Y.
24:48:800Paolo Guiotto: as functions of XY, These are compositions, fractions of continuous functions.
24:55:910Paolo Guiotto: So, they are continuous where they are defined, and this means except .00. So, from this, I see that DXFDYF are continuous function on the full plane, are 2, except 1.00.
25:16:160Paolo Guiotto: So, the differentiability test…
25:22:580Paolo Guiotto: Says. If you know that on some domain.
25:27:570Paolo Guiotto: the partial derivatives are continuous. On that domain, the function is differentiable. So, I know now that the function F, F,
25:37:600Paolo Guiotto: is differentiable, on… R2 minus 0.
25:46:470Paolo Guiotto: So, I discussed almost all points. All points except one.
25:51:740Paolo Guiotto: Okay? Without doing it… particularly complicated calculations. Now.
25:58:630Paolo Guiotto: What about… the question is, what about…
26:05:340Paolo Guiotto: 0, 0.
26:07:170Paolo Guiotto: Because at 0, I cannot use this, because…
26:12:30Paolo Guiotto: I do not have the derivatives there.
26:14:740Paolo Guiotto: For the moment. So let's start computing the derivatives at 0, 0.
26:19:470Paolo Guiotto: What is DXF at 00?
26:24:280Paolo Guiotto: Now…
26:25:480Paolo Guiotto: If I look at the expression of the DX, I see that I cannot plug in that formula X and Y equals 0.
26:34:180Paolo Guiotto: Does this mean that the derivative does not exist?
26:39:230Paolo Guiotto: No, this happened already in many other examples. So this means that if you want to see if this exists, you have to go back to the definition.
26:47:790Paolo Guiotto: So this is the directional derivative, direction 10F.00. So this is the limit for T going to 0 of
26:58:500Paolo Guiotto: F00 plus T10, so it is FT0, minus F00 divided by T.
27:07:410Paolo Guiotto: Now, let's take our function. Function FXY,
27:12:190Paolo Guiotto: is X root of X squared plus Y squared.
27:18:160Paolo Guiotto: So let's plug T0.
27:20:770Paolo Guiotto: If we put T0, we get t times root of T squared.
27:26:240Paolo Guiotto: T times root of T squared.
27:29:280Paolo Guiotto: F00 is 0, So, we remain with the limit.
27:34:340Paolo Guiotto: for T going to 0, of T, root of T squared, divided by T.
27:41:920Paolo Guiotto: the two Ts simplify, and
27:44:960Paolo Guiotto: This is absolute value of T, so I have to compute the limit.
27:49:350Paolo Guiotto: for T going to zero of absolute value of T, which is equal to 0.
27:54:700Paolo Guiotto: So the partial derivative with respect to X at 0, 0 exists, and it is equal to 0.
28:00:810Paolo Guiotto: What about the derivative with respect to Y at 0, 0?
28:04:840Paolo Guiotto: I gained.
28:06:120Paolo Guiotto: I cannot use that formula for the derivative with respect to Y, because it does not make any sense at 0, 0.
28:15:240Paolo Guiotto: So, does this mean that the derivative does not exist? No.
28:20:540Paolo Guiotto: It means that I have to use the definition.
28:23:710Paolo Guiotto: So this will be the derivative, direction 01 of F at 0, 0.
28:29:170Paolo Guiotto: So this is the limit.
28:31:380Paolo Guiotto: For T going to 0, all.
28:34:260Paolo Guiotto: F00 plus T01, which is the vector 0P.
28:39:210Paolo Guiotto: So F0T minus F00 divided by T.
28:44:780Paolo Guiotto: The F is above, so F0T, you see that when you set x equals 0, the function is 0, so this is 0, this is 0, numerator is 0, divided by t is 0, so the limit is 0.
28:57:450Paolo Guiotto: So this shows that the derivatives exist at 0, 0.
29:02:190Paolo Guiotto: Now, do I know that the function is differentiable at 00?
29:07:140Paolo Guiotto: Yes or no?
29:09:850Paolo Guiotto: No, because these are only the partial derivatives.
29:13:580Paolo Guiotto: So what are now the possibilities?
29:16:230Paolo Guiotto: Either I use the definition, as we did in the previous exercise, Okay?
29:22:710Paolo Guiotto: So now I should check that.
29:25:300Paolo Guiotto: This thing holds.
29:26:770Paolo Guiotto: Limit when H goes to 0, F0 plus H minus F0 minus gradient, blah blah blah, divided by norm of H equals 0. This is the definition.
29:36:520Paolo Guiotto: Or, or, and this is what I show you now, I could use the differentiability test
29:43:380Paolo Guiotto: How? I show that the partial derivatives are continuous at 0, okay?
29:50:370Paolo Guiotto: So, if they are continuous also at 0, it means that they are now continuous on R2, not only on R2 minus 00, but they are continuous also at 00s, okay? So…
30:03:180Paolo Guiotto: The question is, how… We… check.
30:11:20Paolo Guiotto: Differentiability… Now… Well, we have two alternatives.
30:30:190Paolo Guiotto: One… with definition.
30:37:790Paolo Guiotto: Okay, so… Checking… Limit… for H going to 0.
30:48:670Paolo Guiotto: F of… we are checking differentiability at 0, so it would be F0 plus H minus F0 minus the gradient F at 0. We have the gradient because we have the components, the two partial derivatives, times H, divided by norm of H, and we should check that this is equal to 0.
31:08:930Paolo Guiotto: Or… Number two… with the… D… differentiability Faster.
31:22:130Paolo Guiotto: That is… Checking.
31:31:390Paolo Guiotto: that the partial derivatives, DXF, DYF, continuous.
31:39:370Paolo Guiotto: at… 0, 0.
31:42:490Paolo Guiotto: Okay?
31:45:430Paolo Guiotto: So this means that I have to prove it's not an easier way, as you will see. It's not…
31:50:780Paolo Guiotto: much better than doing with the definition. That is, I have to prove that the limit
31:56:690Paolo Guiotto: for XY, going to 0, 0.
32:01:80Paolo Guiotto: of the partial derivative with respect to X of F at point XY coincide with the value of the partial derivative with respect to X at 0, which we computed one minute ago, and this was equal to
32:16:480Paolo Guiotto: Zero.
32:21:270Paolo Guiotto: How bold.
32:25:440Paolo Guiotto: Okay?
32:26:430Paolo Guiotto: So this is a new problem, which is, again, a limit, no? So now you understand why I insisted a lot with limit at zero, because they are important.
32:35:230Paolo Guiotto: So this is the limit for XY going to 0, 0. What is the expression for the partial derivative I have to use here?
32:44:690Paolo Guiotto: Well, I have the derivative at 0, 0, which is 0, and the derivative at point XY, different from 0, which is this formula
32:54:50Paolo Guiotto: Now I have to use this one, because this is the formula for the derivative when point XY is not 0, which is exactly the case here. This point is not 0, 0.
33:05:630Paolo Guiotto: Okay, so the formula is 2X squared plus Y squared divided by… I do not remember…
33:15:440Paolo Guiotto: root of X squared plus Y square.
33:20:630Paolo Guiotto: So this is the limit we have to compute.
33:23:860Paolo Guiotto: Which is not difficult, because you look at polar coordinates, this…
33:31:320Paolo Guiotto: root of X squared plus Y squared.
33:34:920Paolo Guiotto: in polar, coordinates.
33:38:940Paolo Guiotto: this becomes.
33:40:610Paolo Guiotto: The denominator is the root of X squared plus Y squared, it's the norm, so it's raw.
33:46:170Paolo Guiotto: Numerator, you get raw squared times 2 cos…
33:51:60Paolo Guiotto: Square theta plus sine squared theta.
33:56:40Paolo Guiotto: You see that this,
33:58:140Paolo Guiotto: raw simplifies a bit, so we remain with raw times this coefficient 2 cos squared theta plus
34:07:980Paolo Guiotto: Sine square theta, which is, in any case, bound.
34:11:989Paolo Guiotto: So now we can easily conclude, no? If you want, you can put an absolute value here, minus 0, which is…
34:18:900Paolo Guiotto: itself, because the quantity is positive. You get this, this is less or equal. Throw away everything here, this is by 2, this is by 1, 3 rows, so this goes to 0 when rho goes to 0. And therefore, this means that
34:35:250Paolo Guiotto: That limit here, let's put a star here.
34:39:560Paolo Guiotto: That limit exists, and it is equal to zero.
34:44:840Paolo Guiotto: And, fortunately, that zero is the value of the partial derivative of F at 0, 0. So it means that the derivative is also continuous at 0, 0.
34:56:690Paolo Guiotto: Same check for the Y, you do by… you can do.
35:02:130Paolo Guiotto: So, same.
35:04:880Paolo Guiotto: Check.
35:08:30Paolo Guiotto: 4.
35:09:410Paolo Guiotto: DYF, so the continuity at 0, 0. You have a limit, which is more or less the same.
35:16:440Paolo Guiotto: And, you will easily discover that the limit is, again, the value of the partial derivative.
35:23:810Paolo Guiotto: And so the conclusion is that we now know that the partial derivatives with respect to X and respect to Y, they were already continuous on R2 minus 0. They are also continuous at 0, so they are continuous on the full plane R2.
35:41:270Paolo Guiotto: So you apply to this domain, the differentiability test. This says that F is differentiable
35:49:160Paolo Guiotto: on R2. So, in particular, also at 00.
35:54:550Paolo Guiotto: Okay?
35:55:730Paolo Guiotto: So, you have here an example of application of different accessibility.
36:01:130Paolo Guiotto: faster.
36:02:450Paolo Guiotto: Of course, these are a little bit tricky problems, because the scope is to test you on this concept, the definition of differentiable, the application of differentiability tests. Normally, in normal use, we have much easier functions, okay, than these ones.
36:19:720Paolo Guiotto: Okay, good. Now… As anticipated yesterday, one of the main…
36:31:160Paolo Guiotto: As I described since the beginning of this course, one of the main reasons we need to develop differential calculus is because we want to discuss the problem of determining maximum-minimum of a function. So.
36:46:130Paolo Guiotto: Let's introduce a few definitions. Something has been already introduced.
36:53:970Paolo Guiotto: So, let's put this.
36:57:690Paolo Guiotto: let… So, F be a function of vector variable X,
37:05:460Paolo Guiotto: defined on a domain D of RD.
37:08:850Paolo Guiotto: real valued. It is clear that since I have to take the maximum value of F, F must be a number, okay?
37:19:300Paolo Guiotto: Because I want to say F of that point is greater than F of any other point, no?
37:25:520Paolo Guiotto: So we recall that.
37:30:170Paolo Guiotto: recall… That.
37:35:330Paolo Guiotto: a pointer, say, X max.
37:41:110Paolo Guiotto: Indeed.
37:42:760Paolo Guiotto: Ease.
37:43:950Paolo Guiotto: global, I prefer to use this, or if you want, absolute, Maximum.
37:54:460Paolo Guiotto: boil.
37:57:750Paolo Guiotto: 4.
37:59:310Paolo Guiotto: F on D.
38:02:10Paolo Guiotto: If the value of the function at that point
38:06:600Paolo Guiotto: is larger than any other value of the function. Sorry, here they are vectors, sorry.
38:13:180Paolo Guiotto: The points are vectors, but the values of the functions are numbers. Otherwise, you cannot write a greater or equal, no? You cannot say a vector is greater or equal than another vector. There is not such relation between vectors, but of course there is among numbers. So, for every X of domain D.
38:31:800Paolo Guiotto: And similarly, a minimum point is defined.
38:35:420Paolo Guiotto: And… Similarly.
38:42:610Paolo Guiotto: I mean… Well, better, global minimum.
38:48:340Paolo Guiotto: a mobile… Minimum.
38:52:790Paolo Guiotto: bonds.
38:54:210Paolo Guiotto: R.
38:55:540Paolo Guiotto: defined.
38:58:720Paolo Guiotto: Now, you know that, we have seen
39:03:950Paolo Guiotto: A very weak result, that says.
39:07:540Paolo Guiotto: It is the Beiss theorem. We have a continuous function on a good domain, closed and bounded, that function has definitely a maximum and a minimum on that domain.
39:19:350Paolo Guiotto: But unfortunately, that theorem does not tell you any method to determine these points. And in real-world applications, you want, of course, that it's very important
39:29:370Paolo Guiotto: It's crucial, it's the unique, say, point, to determine the minimum, maximum, the optimal configurations of your system.
39:37:440Paolo Guiotto: So you need a method to find them.
39:39:800Paolo Guiotto: And that method in one variable calculus, is provided by differential calculus, okay?
39:47:50Paolo Guiotto: Now, usually, in dimension, one, so if,
39:54:230Paolo Guiotto: Let's say that the problem is,
40:00:670Paolo Guiotto: how… Karen.
40:03:860Paolo Guiotto: Sweet.
40:05:70Paolo Guiotto: detail,
40:09:900Paolo Guiotto: these… points.
40:13:930Paolo Guiotto: Of course, provided
40:19:110Paolo Guiotto: Say… exists.
40:25:990Paolo Guiotto: Now, in dimension 1, so if,
40:29:710Paolo Guiotto: We say the RD, right? If D is 1,
40:33:770Paolo Guiotto: So, we are dealing with functions of real variable.
40:38:490Paolo Guiotto: X is a real number.
40:41:670Paolo Guiotto: What we do is more or less the following algorithm.
40:45:720Paolo Guiotto: We do… Number one, we compute F prime of X.
40:53:140Paolo Guiotto: Provided the function is differentiable.
40:55:910Paolo Guiotto: Number two, we study.
40:59:910Paolo Guiotto: Where derivative is positive or negative.
41:04:540Paolo Guiotto: Because this, on interval, says that the function is increasing or decreasing.
41:09:770Paolo Guiotto: And then, once you know that, on an interval, for example, the function is increasing.
41:15:820Paolo Guiotto: be up to a certain point, X star.
41:19:900Paolo Guiotto: and decreasing after, and there is continuous, you are a maximum at that point.
41:26:340Paolo Guiotto: No? At least on the income lab.
41:29:130Paolo Guiotto: Now, here we have a problem, initial problem with this,
41:33:190Paolo Guiotto: strategy, because we could compute here, I mean, with the function of vector variable. We could compute the derivative, now we know what is it, the derivative, and in this case, it is a vector, it is the gradient vector, so in our
41:48:850Paolo Guiotto: case, so this is dimension 1, let's say that dimension greater than 1
41:53:580Paolo Guiotto: Here we have the gradient of F at point X, which is an array… it is an array of RD. It has D components, the same number of the variables of the components of X.
42:06:210Paolo Guiotto: And so, this can be computed, but what does it mean, gradient of F greater or equal than zero?
42:13:470Paolo Guiotto: So maybe this zero should be a vector.
42:16:170Paolo Guiotto: You may think, all components are positive, but unfortunately, there is not such a property, okay? So this is something that it's like the division by vector. It's something that it is not defined, it cannot be defined in a reasonable way.
42:31:950Paolo Guiotto: The meaning is that we get contradictions if we try to define this ordering between vectors. So it is not clear what should mean
42:40:310Paolo Guiotto: the vector, the gradient is positive, so we do not have that.
42:45:170Paolo Guiotto: And so we cannot rely on monotonicity, for example, that even it's meaningless, because the fact that the function is increasing means that when you increase the variable, you increase the value, no?
43:00:360Paolo Guiotto: Now, think to the case we have with functions of several variables. What does it mean to increase the variable? I have a vector as a variable. I should say, a vector is greater than another. Again, the same problem. I do not have an ordering. I cannot say a vector is bigger than another. There is not such operation.
43:19:290Paolo Guiotto: So, here we are in trouble. It means that we cannot expect that we will compute the derivative, and then we will study the sine to determine whether function is increasing, decreasing. These things simply do not make sense. So, what can be done in this case? Well, in dimension 1,
43:37:940Paolo Guiotto: this story actually starts from an important remark. First, historically.
43:44:860Paolo Guiotto: The first remark that made it interesting to introduce differential calculus was made by Fermat a couple of centuries before differential calculus was even defined. So, he had not yet the definition of derivative, but he noticed
44:03:990Paolo Guiotto: a geometrical fact, that when you have a minimum or a maximum, the tangent is horizontal. No?
44:12:230Paolo Guiotto: So tangent horizontal means that the angular coefficient of the tangent line is zero.
44:18:70Paolo Guiotto: And since the angular coefficient is the derivative, these points are points where f prime of x equals 0.
44:26:790Paolo Guiotto: In fact, Ferma discovered that solving this equation, you have… you can find minimum-maximum points for F.
44:36:950Paolo Guiotto: Now, you have to be clear, because the condition F prime of x equals 0 alone does not distinguish between minimum and maximum, as you can see. It is verified by both.
44:49:510Paolo Guiotto: It does not tell you if you have a global minimum, a global maximum, because, for example, this point here is definitely not the global maximum point, because the function is higher here. But, however, here you have the tangent, which is a horizontal.
45:06:780Paolo Guiotto: So this is rather a maximum point for a part of the graph of F. So you may say, for this part, that looks as a maximum, but not for the complete graph, okay?
45:20:200Paolo Guiotto: And moreover, there could be points where the tangent is original, but they are not minimum-maximum. So, like, points like this one.
45:29:360Paolo Guiotto: Now, in this point, you have… the tangent is horizontal, but yet this is not a minimum, and neither a maximum, okay? So the equation F prime of x equals 0 actually
45:42:620Paolo Guiotto: Does not determine exactly minimum, maximum points.
45:47:430Paolo Guiotto: However, former mathematician of Fermat times, and even nowadays.
45:54:110Paolo Guiotto: Having an equation to identify possible solutions is better than having nothing. Look, Weiss just tells you there exists a minimum, there exists a maximum, yes, but how can I find that?
46:06:470Paolo Guiotto: You don't have any recipe.
46:08:360Paolo Guiotto: This one would say, okay, you have to look among this solution of this equation. This restricts a lot of the search, okay? Maybe you can find the solution of this, there are 3 solutions, you have only 3 possibilities.
46:22:10Paolo Guiotto: then you know that if there is a maximum, it is one of these three points. If there is a minimum, it's one of them.
46:28:320Paolo Guiotto: So you restrict a lot the search to a lower number of points. So that's why this condition that last year you have never considered, because you have much more powerful tools, like the sign of the derivative, here will become important, okay?
46:46:940Paolo Guiotto: And in this case, for us, this condition will become gradient F,
46:51:210Paolo Guiotto: equal 0. This equation is going to play a central role here, okay? Because this, in principle, can be solved. It is an equation
47:02:820Paolo Guiotto: you look for points X where the gradient is 0, okay?
47:07:780Paolo Guiotto: Okay, now, what we have to do is to introduce all these, to… to put all these, facts on our setup. We need first,
47:20:800Paolo Guiotto: As we said, that condition, derivative equals zero, is not a condition that is verified necessarily only by global minimum-maximum, but it should be… could be verified by local minimum-maximum.
47:35:160Paolo Guiotto: So we need first to introduce this concept, which is a sort of refinement of this one, of global minimum. So let's put into a definition. What is a local minimum, local maximum? So since here we have done the case of the maximum, we will do the case of the maximum.
47:53:10Paolo Guiotto: for local points. So, let F… be a function of X.
48:01:340Paolo Guiotto: defined on domain D, of RD, real value.
48:08:00Paolo Guiotto: We say that.
48:12:340Paolo Guiotto: a point we will still call Xmin, okay?
48:21:920Paolo Guiotto: Sorry, we are talking about the maximum, so let's call it X max.
48:26:450Paolo Guiotto: in domain D. Of course, minimum, maximum points.
48:30:830Paolo Guiotto: Doesn't matter if they are local or global, they must be in the domain. Why? Because you have to compute the value of the function at those points to say, there, the value is bigger than any other value. So you need to compute the function. That's why they must be in the domain.
48:47:140Paolo Guiotto: So we said that this point is a local Or sometimes relative.
48:56:680Paolo Guiotto: But I prefer to use local, okay?
48:59:610Paolo Guiotto: local, maximum point.
49:05:560Paolo Guiotto: 4.
49:07:190Paolo Guiotto: F on D.
49:09:860Paolo Guiotto: Well, what is the condition, for the local maximum? Well, let's do a figure. Now, imagine, of course, the figure, you have to… you need to have a lot of imagination, because I'm now saying this is RD.
49:24:270Paolo Guiotto: Let's say that domain that you look… it seems like an interval here, but it is not an interval, it is a subset of RD. Now, if these two, RD is… R2 is the plane, is a plane, okay? So assume that this is the point, this is the domain, this is the point X max.
49:43:590Paolo Guiotto: Now, we want to define this condition, huh? That point is not, maybe, a global maximum point.
49:52:760Paolo Guiotto: But locally, locally means around, when we are close to that point, so if we forget everything outside of a small
50:05:200Paolo Guiotto: Part of the domain around that point.
50:09:830Paolo Guiotto: If you look only at what is between the two lines, you can say that the value here is greater than all the other values when you take the X only here, not from the domain.
50:21:530Paolo Guiotto: on the full domain is not true, because there are points with a higher F, no? But if you just restrict your attention to the part of domain, it becomes true. How do we say that?
50:34:80Paolo Guiotto: Well…
50:35:120Paolo Guiotto: This, if you think, this is what in the real line, this is an interval, and the intervals are what we call bowls, okay?
50:42:880Paolo Guiotto: In dimension 1, an interval is a set of points whose distance from point X max is less than something.
50:50:750Paolo Guiotto: If I go in the plane, points where the distance is less than something is a disk. If I go in the space, points where the distance to X max is less than something is a ball, is a sphere, and so on. So, the right concept is the concept of ball.
51:09:90Paolo Guiotto: Okay, so we say that this is a local maximum point for F on D if… Their existence.
51:18:290Paolo Guiotto: So there exists Ebola.
51:20:930Paolo Guiotto: centered at that point X max, with some radius r. So, actually, this means there exists a radius R positive, such that
51:32:790Paolo Guiotto: When I compute the value of F at any point x.
51:38:180Paolo Guiotto: So the ball is the red stuff here.
51:41:890Paolo Guiotto: X Maxa.
51:45:180Paolo Guiotto: R.
51:46:760Paolo Guiotto: When I take F of X for X in that ball, not necessarily in all the domain, so the ball X max
51:56:140Paolo Guiotto: Radius R.
51:58:320Paolo Guiotto: Now, since in general, I want to know if this ball is contained in the domain, I put this additional condition in the section with D.
52:07:90Paolo Guiotto: In order that I am taking points in the ball that are also in D, this because I have to compute the value of the function, this is less or equal than the value of F at point X max.
52:20:890Paolo Guiotto: So, if you, let's say, cancel for a second this, that would be the definition of global max, huh? F of X less than FX max for every vertex in D.
52:31:830Paolo Guiotto: Okay? But even now say, this is maybe not true, but it's true, when I take point X in D,
52:38:570Paolo Guiotto: and close to point X max, so in what we call neighborhood of X max, well, this means that that point is a maximum, but only for a limited part of the domain, okay? And similar definition for the minimum.
52:54:240Paolo Guiotto: Similarly.
53:02:790Paolo Guiotto: if, X, well, let's say, Xmin…
53:10:950Paolo Guiotto: is a local.
53:13:880Paolo Guiotto: Minimum.
53:15:360Paolo Guiotto: point.
53:19:00Paolo Guiotto: If… The same, condition, there exists a ball centered at point X min, there is an arrow here.
53:28:100Paolo Guiotto: Ex Mina?
53:30:90Paolo Guiotto: with some radius r, where now the value of the function
53:37:60Paolo Guiotto: The values of the function at point x are larger than the value at point x mean.
53:43:500Paolo Guiotto: For WeChat, for all access, That belongs to that ball, centered at that point, Xmin.
53:54:20Paolo Guiotto: I'm missing lots of arrow.
53:57:230Paolo Guiotto: and that are in D, for which I can compute the function M.
54:02:710Paolo Guiotto: Okay, so these are, these are, let's say, local, relative, minimum, maximum points. A clear remark is that, remark…
54:17:50Paolo Guiotto: Of course, global points are local, but not vice versa.
54:22:660Paolo Guiotto: So, Annie… global.
54:29:460Paolo Guiotto: Minimum, or maximum, is also… Local, Minimum. Maximum.
54:42:350Paolo Guiotto: But, of course, not vice versa.
54:45:680Paolo Guiotto: But… not.
54:48:120Paolo Guiotto: vice versa.
54:52:10Paolo Guiotto: you don't need a specific example. Look at the figure, no? This one is a local minimum, but maybe the local maximum, the global maximum is this one, no?
55:02:940Paolo Guiotto: So you can do just a figure, like this, so you have your function.
55:08:390Paolo Guiotto: I may say this is a global maximum point.
55:13:890Paolo Guiotto: The global maximum point is this one, is in the domain. This one is a local Maximum… point.
55:23:110Paolo Guiotto: Maybe there are… there's another one, perhaps this one, local maximum point, also this one, this, and I don't know. So this one, why not? No? Also this one.
55:33:870Paolo Guiotto: It has a value which is higher than values around that point, but it is not the highest possible value, which is attained here.
55:42:930Paolo Guiotto: Okay?
55:45:330Paolo Guiotto: Now, the, the fundamental fact is, the, what we were saying before, this theorem, which is called also Fermat theorem.
56:02:110Paolo Guiotto: that says, suppose that F… be defined on, well, F of X,
56:10:490Paolo Guiotto: The final domain D, contained in RD, real-valued.
56:20:240Paolo Guiotto: And the F, the differentiable.
56:23:860Paolo Guiotto: on.
56:26:240Paolo Guiotto: Okay, so we take a function, a differentiable function.
56:31:270Paolo Guiotto: Then, it happens that if you have a minimum, maximum point.
56:38:860Paolo Guiotto: Which is anteriorly contained in this, so it belongs into the interior of this.
56:44:810Paolo Guiotto: Then.
56:48:920Paolo Guiotto: If… X Max.
56:54:410Paolo Guiotto: Or… X meeting.
56:57:380Paolo Guiotto: It holds for both minimum-maximum points.
57:00:970Paolo Guiotto: Ease.
57:02:610Paolo Guiotto: local… Maximum.
57:07:230Paolo Guiotto: Or minimum, the other case.
57:11:590Paolo Guiotto: Such that this point, X max.
57:16:490Paolo Guiotto: belongs to the interior of D, so it is in D, but it is Y containing in D.
57:24:300Paolo Guiotto: This is important, because as you will see, but this is not newer, it happens also for functions of one variable, this factor.
57:32:520Paolo Guiotto: that the point must be in the interior, the same X for the minimum, of course.
57:42:10Paolo Guiotto: Then, necessarily, the gradient of F at that point, X max.
57:49:550Paolo Guiotto: or Xmin must be equal to 0.
58:08:810Paolo Guiotto: So, it says, if you want to look for, local minimum points, well.
58:17:330Paolo Guiotto: If they are in the interior, they must solve this equation. Why this condition on the interior? I… I… time is almost over, so I… I will do the proof next time of this, but let's do some remark on this statement. Remarks.
58:33:750Paolo Guiotto: So, number one, Why is this condition? Why…
58:39:810Paolo Guiotto: Is the condition that… let's write only for the maximum, it's the same for the minimum.
58:44:990Paolo Guiotto: X maxi in the interior of the so important.
58:52:680Paolo Guiotto: Well, this is, not…
58:56:30Paolo Guiotto: Because we are in dimension higher than 1. It's something that happens also in dimension 1.
59:03:520Paolo Guiotto: this… is because…
59:10:430Paolo Guiotto: If the minimum maximum point, so let's say for the maximum, is not in the interior of B, We
59:20:390Paolo Guiotto: might… However… that the gradient of F at that point, Could be different.
59:31:220Paolo Guiotto: from zero.
59:34:510Paolo Guiotto: Well, it's not a surprise, I told you. Let's be in dimension 1.
59:39:90Paolo Guiotto: So, the gradient becomes the derivative, okay?
59:43:150Paolo Guiotto: So in that… this is R1.
59:45:790Paolo Guiotto: take a function like F of X equal X.
59:50:590Paolo Guiotto: for X in 01, in the interval 0, 1, okay? So what is the plot of this function? It is this one.
00:00:220Paolo Guiotto: This is the value at 1.
00:04:50Paolo Guiotto: So…
00:05:620Paolo Guiotto: You see that there is a global minimum and a global maximum. The minimum is here, at 0, and the maximum is here, at 1.
00:14:360Paolo Guiotto: So, X equals 0 is global.
00:21:260Paolo Guiotto: Minimum.
00:22:630Paolo Guiotto: For this function, because F of 0 is 0, which is less or equal than f of x, which is X, for every X between 0, 1. Do you agree?
00:36:350Paolo Guiotto: And the same, X equals 1 is A global maximum.
00:43:480Paolo Guiotto: Because if you look at the value, f of 1 is 1, which is greater than F , which is X,
00:52:200Paolo Guiotto: For every X, between 0 and 1.
00:55:960Paolo Guiotto: Well, I don't need to write this. Look at the figure now, where the function takes the biggest possible value at x equals 1, where it takes the smallest possible value at x equals 0.
01:07:170Paolo Guiotto: But… What happened to the derivative?
01:11:920Paolo Guiotto: The derivative is constantly equal to 1, so it is never equal to 0.
01:18:160Paolo Guiotto: So what goes wrong here?
01:20:420Paolo Guiotto: It goes wrong that these two points, 0 and 1, the domain is this one, let me do with this color. The domain is the interval 0, 1, that's the domain
01:32:700Paolo Guiotto: the two points, the minimum and the maximum, are not in the interior. They are the end points of this interval, so they are not contained with the one-dimensional boat, which is an interval, by the way, no? So, here…
01:48:600Paolo Guiotto: These two points, 0 and 1, they are both not in the interior of D.
01:56:470Paolo Guiotto: If they were in the interior, this wouldn't be possible. So, if they are not in the interior, you cannot say that the derivative is zero. But, when they are in the interior, there is a theorem. It is the Fermat theorem that says, necessarily, the derivative here, the gradient, is equal to zero.
02:14:510Paolo Guiotto: So this is the first remark.
02:20:80Paolo Guiotto: Well, the second remark, I, I repeat in words. We said, global points, which are normally the goal of a maximization problem.
02:31:540Paolo Guiotto: Are, in particular, local points.
02:35:140Paolo Guiotto: So, if you look for a global minimum, global maximum, and you know that for some reason it must be in the interior, then also for the global minimum, global maximum, this will solve the equation gradient f equals 0.
02:50:20Paolo Guiotto: Okay?
02:51:440Paolo Guiotto: However, exactly as we say graphically here, the condition gradient f equals 0, derivative equals 0, does not determine minimum, maximum points on the points where derivative is 0, and they are not
03:07:90Paolo Guiotto: minimum and neither maximum, you see? So the condition alone is not sufficient to distinguish between minimum and maximum, and it does not tell you for sure that you have found a minimum or a maximum. Could be something else.
03:21:920Paolo Guiotto: So what is the interest? Again, the interest in this condition is that it reduces the search of minimum-maximum points. It is an equation. You know that the solutions that you are looking for are among the solutions of this equation. So if you look… if you solve this equation, if you are able to solve this equation.
03:40:760Paolo Guiotto: You have lots of candidates to be the solution, and there cannot be anyone else than these points. You see, that's why that condition is so important. We give just the name and definition.
03:54:880Paolo Guiotto: A point.
03:58:230Paolo Guiotto: Where?
04:01:10Paolo Guiotto: gradient F is equal to 0, is called… stationary…
04:14:940Paolo Guiotto: point.
04:18:210Paolo Guiotto: Okay?
04:19:290Paolo Guiotto: Now, there is an exercise that I would recommend you to do.
04:24:30Paolo Guiotto: At the end…
04:27:960Paolo Guiotto: maybe I do one second example, the really simple, just to fix you the idea, then I need you to do this exercise, which is the 296. It's simple, it's just a calculation. We will do, we will see on Friday, right?
04:44:150Paolo Guiotto: So, do exercise, 2.
04:50:460Paolo Guiotto: 0.9.6… Well, let me do the number 2, just to show you. The exercise is determine… D… Stationary points.
05:04:510Paolo Guiotto: off.
05:05:580Paolo Guiotto: Here we have F, X, Y,
05:08:850Paolo Guiotto: equal, I do this one, which is a…
05:11:670Paolo Guiotto: break it faster. X squared plus Y squared plus XY.
05:19:450Paolo Guiotto: So let's see the solution.
05:23:790Paolo Guiotto: So, what are stationary points? We are… we just introduced XY
05:32:260Paolo Guiotto: Is a stationary point.
05:36:450Paolo Guiotto: if…
05:37:780Paolo Guiotto: the gradient of F at point XY is equal to 0. So, what is the gradient? That's a vector whose components are the partial derivative with respect to X and the partial derivative with respect to Y. So let's compute these two.
05:52:860Paolo Guiotto: We have DXF is… do you see what is it?
05:59:520Paolo Guiotto: to Axon?
06:01:50Paolo Guiotto: Plus Y. Right.
06:02:950Paolo Guiotto: DY is… to Y plus X. Now, as you can see, these functions are…
06:13:10Paolo Guiotto: Polynomials, so in particular, they are…
06:16:110Paolo Guiotto: Continuous wear on the full plane R2. So this says that differentiability test.
06:24:620Paolo Guiotto: The function f is differentiable on R2.
06:31:460Paolo Guiotto: Okay? So you see, the check is very fast in concrete cases.
06:36:350Paolo Guiotto: Then, the gradient F is now the vector, first component, 2X plus Y, Second component, 2Y plus X.
06:45:780Paolo Guiotto: Now, if we want to look for stationary points, we have to look for points XY, for which this is the vector 0, 0.
06:53:130Paolo Guiotto: And what does it mean, this? We have a system, because we need that together, the first component be zero, we get 2X plus Y equals 0, and the second component, 2Y plus X, be also 0. Together! Not one is 0 and the other is different from zero. That's why we have a system.
07:11:660Paolo Guiotto: Because each component is equal to zero, means that together you have a list of equations.
07:18:190Paolo Guiotto: Now, this is a simple 2x2 system that can be easily solved. It's a linear system, and as you can easily see, the solution is the unique point XY equals 0, 0.
07:29:900Paolo Guiotto: So, 0, 0.
07:32:390Paolo Guiotto: is… the… unique, stationary. Point.
07:40:560Paolo Guiotto: for… this.
07:43:770Paolo Guiotto: F.
07:44:800Paolo Guiotto: Now, this ends the exercise. The remaining are similar. You have an F, and you have to solve the system gradient F equals 0. So, try to do, and on Friday, we will see the solution.
07:59:710Paolo Guiotto: Have a nice day.