AI Assistant
Transcript
00:02:780Paolo Guiotto: Okay, so welcome to this intimate Christmas lecture.
00:09:180Paolo Guiotto: So… what is Okay.
00:17:230Paolo Guiotto: Today, we are going to introduce a very important characterization of see differentiability.
00:28:790Paolo Guiotto: That, somehow will explain some of the… apparently…
00:35:500Paolo Guiotto: surprising facts we have seen last time. So let me just quickly review what we have seen in last class.
00:43:600Paolo Guiotto: We introduced the definition of what does it mean derivative of a function of complex variable, which is the same definition we have for the derivative of a function of real variable.
00:57:220Paolo Guiotto: We have seen with a few examples that apparently it works in the same way, so we have the same derivatives for powers,
01:07:880Paolo Guiotto: We have the same rules of calculus.
01:10:780Paolo Guiotto: We have also seen that, we mentioned this fact that power cities are…
01:17:900Paolo Guiotto: always differentiable in the disk of convergence, and you can differentiate by doing the derivative, let's say, term by term. So, like, if you could make derivative of an infinite sum equals sum of derivatives.
01:32:700Paolo Guiotto: Here, you have to be careful, because the general rule says that the derivative of the sum is the sum of the derivatives. When the two functions have a derivative, this formula extends to any finite sum of functions, so if I have 1 million of functions.
01:48:620Paolo Guiotto: the sum of 1 million of functions, the derivative of this sum would be always the sum of the derivatives. The difference here is that we have infinitely many terms.
01:58:930Paolo Guiotto: And here, it's a bit more…
02:01:440Paolo Guiotto: complicated to prove that the derivative of the sum is equal to the sum of the derivatives in general, but for this particular case, it works. And from this, we get some nice formula, as the derivative of the exponential is the exponential function.
02:18:620Paolo Guiotto: The same formulas we have already seen for real versions of these functions. So, derivative of hyperbolic cosine is hyperbolic sine, derivative of hyperbolic sine is hyperbolic cosine. You can check easily that derivative of sine is cosine, derivative of cosine is minus sine.
02:37:970Paolo Guiotto: Okay, but at a certain point, we experienced that with some apparently innocent functions, like this one, f of z equal real part of Z,
02:52:330Paolo Guiotto: I say innocent because, normally the idea of being not differentiable is associated with something should happen to the function.
03:02:710Paolo Guiotto: there is some irregularity, but if I look at how this function is defined, it takes the number X plus IY, and it associates X. So…
03:14:50Paolo Guiotto: It seems to be an innocent function, but however, this function turns out to be not differentiable at any point.
03:21:970Paolo Guiotto: And, I hope you tried with these other examples, they are very similar. So, also these functions, imaginary part of Z, or the modulus of Z squared, where you see the function is X squared plus Y squared. Y shouldn't be differential?
03:40:320Paolo Guiotto: or Z conjugate. These functions are not differentiable. Now, this morning, we are going to understand why this happens, and actually, we have a different… a new characterization of
03:55:230Paolo Guiotto: the C differentiability. So to do this, we start noticing that if we have a function f of complex variable that defined on a domain D of the complex plane.
04:11:960Paolo Guiotto: So this means that, for every Z, F of Z is a complex number.
04:17:529Paolo Guiotto: F of Z is a complex number, and therefore, we can decompose this by its real part and its imaginary part.
04:26:770Paolo Guiotto: So there will be the real part of the number F of Z plus i times the imaginary part of F of Z. These two quantities are real numbers, no? The real part and the imaginary part are both real numbers.
04:43:500Paolo Guiotto: So, in other words, two new functions arise from F. So, the function that associates to Z, the real part of F of Z,
04:56:550Paolo Guiotto: and the function that associates to Z, the imaginary part of F of Z.
05:02:890Paolo Guiotto: Now, these functions are defined on C, and they are real value.
05:11:540Paolo Guiotto: since they are real-valued, we can also imagine that this function, as you will see in a moment, can be also seen as function from R2 to R. Let's see how, because if you take this first function.
05:28:50Paolo Guiotto: Z goes into the real part of Z. You can represent Z equal X plus IY,
05:35:450Paolo Guiotto: So, to each Z, there is an X… there is a pair, XY, so complex numbers can be identified in a complex plane. As you know, the number X plus IY
05:51:750Paolo Guiotto: is represented as the point with coordinates X and Y, where X is the real part and Y is the imaginary part, no?
06:01:190Paolo Guiotto: So we… we can think that the function that associates z to the real part of F of Z
06:12:500Paolo Guiotto: can be seen, actually, as a function that associates a point XY to a real number, which is the real part of F of Z, which is X plus IY.
06:24:700Paolo Guiotto: So, this function here is a function that goes from R2…
06:30:560Paolo Guiotto: or actually better, you imagine that there is a domain D, where the function is defined. The domain D can be identified with the set of points of R2, so I will still use the letter D, even if D is formally a set of complex numbers.
06:48:250Paolo Guiotto: And this goes into R, so I see this as a function from R2 to R, and I will call this U.
06:58:820Paolo Guiotto: And the same for the other part, so the function Z goes into imaginary part of F of Z can be equivalently represented as a function that associates to a point XY,
07:15:180Paolo Guiotto: A real number, the imaginary part of…
07:20:370Paolo Guiotto: F of X plus IY, and I will call this function V.
07:25:890Paolo Guiotto: So, in other words, this is, just, a way,
07:31:390Paolo Guiotto: a way to write the function. So my function f of z, evaluated at point X plus IY, can be written as U of XY plus iV of XY.
07:47:310Paolo Guiotto: takes this form.
07:49:610Paolo Guiotto: So when… whenever we use these two letters, U and V, in this context, we mean UXY is the real part.
07:59:310Paolo Guiotto: of F of X plus IY, and V, X, Y… Is the imaginary part.
08:09:90Paolo Guiotto: of FX plus AY.
08:13:170Paolo Guiotto: So I take my F, I decompose into real imaginary part, I get this. For example, let's see a few examples.
08:20:980Paolo Guiotto: Examples.
08:22:820Paolo Guiotto: For example, the first non-trivial function is F of Z equals Z, huh? Take this function. So what are u and v for this? Well, this means that F of the point, a complex number X plus IY is X plus IY, right?
08:43:159Paolo Guiotto: Well, here you see immediately that this is U of XY, and this is V of XY.
08:51:760Paolo Guiotto: So these are the two functions, okay? The real and the imaginary part of the function.
08:58:860Paolo Guiotto: Let's take something a little bit more complicated, like F of Z equals Z squared.
09:06:300Paolo Guiotto: In this case, if I represent the number Z in the standard algebraic form, X plus AY,
09:15:840Paolo Guiotto: I have to do X plus IY squared. Now, what is written here is not the algebraic representation of this number, because you don't distinguish
09:25:710Paolo Guiotto: The real and the imaginary part, they are under the square. To see the real imaginary part, I have to develop the square. So when I do that, I get X square, then I do the square of IY, which comes minus Y squared, then I have the double product I2XY. I write this way to emphasize that this is the real part, you see here.
09:50:700Paolo Guiotto: And this is the imaginary part. So these are, respectively, the UXY and VXY.
10:00:90Paolo Guiotto: You see?
10:02:350Paolo Guiotto: So, let's say that for these functions, it is simple to do. You have to do, for these algebraic functions of Z, for example, z cubed is what? It is F of X plus IY equal
10:19:390Paolo Guiotto: X plus IY cubed.
10:23:550Paolo Guiotto: Now.
10:24:440Paolo Guiotto: To determine the real and imaginary part of this quantity, I have to do the cube. So, I have the cube of the first term, so X cubed. Then we have plus 3 times the square of the first times the second, so X squared times IY.
10:44:260Paolo Guiotto: Plus 3 times the first times the square of the second.
10:49:830Paolo Guiotto: and then plus the cube of the second. This is the expansion of A plus B to power 3.
10:58:950Paolo Guiotto: Now, let's rearrange to separate the real part and the imaginary part. As you can see, for example, this one is real, this one is imaginary, you see there is that eye.
11:15:60Paolo Guiotto: This one, the third, is I square that will give minus 1, so this is real. Now we will rewrite, but just to identify who are the real and who are the imaginary numbers. And here we have i cubed, which is I square times i. I square is minus 1, so it is minus i, so it is, again, imaginary. And in fact, if we separate this, we get
11:40:400Paolo Guiotto: X cubed, then we take this term, which is, I square is minus, so minus 3XY squared. As you can see, it is real.
11:50:280Paolo Guiotto: So this is the real part. Plus, on the other terms, we factorize the i. So this is i times 3x square y.
12:00:750Paolo Guiotto: And from the last one, Y, IY cubed, i cubed, we said it is minus i. Y cubed is Y cubed, so the i is factorized minus Y cubed. And here you read, this is UXY, and this is VXY.
12:21:870Paolo Guiotto: Okay.
12:22:930Paolo Guiotto: Now, this thing is a little bit more complicated, for example, when we have functions like, let's say, let's see, take the exponential function, E to Z.
12:35:00Paolo Guiotto: What are the real and imaginary part of this? How can I say this? Well, let's do the same job. So this is E2X plus IY, right?
12:45:880Paolo Guiotto: Then, we know that, exponential,
12:50:290Paolo Guiotto: of a sum is the product of the exponentials, so this is E2X times e to iY. And by Euler formulas, this is cos y plus i sine y.
13:04:660Paolo Guiotto: So we have E2X times this. And now we can separate the real and imaginary part. We have E2X cos y.
13:13:430Paolo Guiotto: plus I e tox sine.
13:17:470Paolo Guiotto: Y. So we can now say that this is the UXY
13:24:80Paolo Guiotto: for the exponential, and this is DVXY.
13:28:540Paolo Guiotto: Still for the exponential. Okay?
13:32:740Paolo Guiotto: And you can continue like that for the other functions. So the idea is that, in general, whatever is the function f of variable Z,
13:46:60Paolo Guiotto: Now, since I can always write a complex number in algebraic form, the values of the function
13:55:180Paolo Guiotto: can be represented by, as in algebraic form, something plus I something else, where these two numbers are real. So this is the real part, this is the imaginary part of what? Of this number, which is F of Z. But if I look at that as, let's say, depending…
14:14:150Paolo Guiotto: in terms of X and Y over here, the imaginary part of the complex numbers, these can be seen as two functions.
14:21:490Paolo Guiotto: that are functions of XY, which are now reals, yeah, no? And they are real-valued. So they are functions of the first part of the course, let's say, no? Functions of two variables, real-valued.
14:36:900Paolo Guiotto: Now, A crucial point is…
14:40:710Paolo Guiotto: What does it mean that F is C differentiable for these two functions, U and V? What should we expect on these two functions?
14:52:290Paolo Guiotto: Yeah, that's the first reasonable guess.
14:55:210Paolo Guiotto: But what is surprising, and you cannot imagine by looking at this, is that not only the two functions must be differentiable, but their partial derivative must verify certain equations. And this is the theorem now we are going to see, which is the fundamental theorem of today.
15:17:320Paolo Guiotto: Okay, so, let's… F… function of complex variable Z, defined on the domain D.
15:29:30Paolo Guiotto: C valued…
15:31:420Paolo Guiotto: Let's say… let's write, also, let's introduce these two guys, U and V, so F of X plus AY,
15:43:590Paolo Guiotto: equal U.
15:45:920Paolo Guiotto: XY… plus IVXY.
15:52:610Paolo Guiotto: So, I… I… I repeat, whenever we write this thing, means that U is the real part and V is the imaginary part, because I could always write this with U and V complex numbers, okay? But we assume that U is the real part of F,
16:11:870Paolo Guiotto: And V is the imaginary part of F with this right.
16:16:750Paolo Guiotto: Then… D… following.
16:25:420Paolo Guiotto: Properties… are equivalent.
16:31:540Paolo Guiotto: So we are going to give a characterization of C differentiability.
16:35:780Paolo Guiotto: The number 1 is F is C differentiable, at point…
16:45:880Paolo Guiotto: Z equal X plus IY.
16:51:840Paolo Guiotto: Number 2 is the UNV.
16:55:990Paolo Guiotto: R.
16:57:70Paolo Guiotto: I will write here to make clear that here, U and V, since I shouldn't write, but since we have two definitions of differentiables.
17:07:79Paolo Guiotto: a definition for functions of complex variables, but also differentiability for functions of real variables. And U and V are functions of real variables. So, U and V are… are differentiable, so they are differentiable in the traditional sense we have seen
17:24:230Paolo Guiotto: A couple of months… no, months ago, one…
17:28:850Paolo Guiotto: So, let's say, a few weeks ago, some weeks ago, At point.
17:35:690Paolo Guiotto: XY.
17:38:910Paolo Guiotto: And, and this is the important, this is not sufficient.
17:44:80Paolo Guiotto: And this explains, let's say, why
17:47:570Paolo Guiotto: This phenomenon happens, because if you look at this function here.
17:52:840Paolo Guiotto: The real part of Z, F of X plus IY is X. If you want, we can now say plus I0.
18:00:320Paolo Guiotto: Where this is U, and this is V.
18:03:880Paolo Guiotto: You see? So the U is UXY equal X.
18:08:190Paolo Guiotto: It's differentiable, right? It's a polynomial. V is flat, is constantly equal to 0, it is differentiable, but F is not differentiable. So it means that the simple differentiability of U and V is not sufficient.
18:22:540Paolo Guiotto: And that's, that's his plane, D.
18:26:940Paolo Guiotto: The… the following…
18:35:580Paolo Guiotto: Relations…
18:40:480Paolo Guiotto: Hold.
18:42:980Paolo Guiotto: So… There are two conditions that says Maybe the first timer.
18:49:130Paolo Guiotto: You have to think a second how to remind of this, but you take the real part, and you do the derivative with respect to the first variable.
18:59:680Paolo Guiotto: So the derivative of U with respect to X at point XY must coincide with the derivative with respect to Y of V at point XY.
19:12:160Paolo Guiotto: And second, the derivative with respect to Y of U at point XY must coincide not with the derivative with respect to X of V, but with minus the derivative with respect to X of V at point XY.
19:26:260Paolo Guiotto: These two equations are called Cauchy… Riemann.
19:37:250Paolo Guiotto: equations.
19:39:990Paolo Guiotto: Or, it depends how you look at them, conditions.
19:48:470Paolo Guiotto: So they must be verified by any C-differentiable function. If they are not verified.
19:56:20Paolo Guiotto: the function cannot be differentiable. Before we enter into the proof, let's do this remark.
20:03:20Paolo Guiotto: So, just to…
20:04:810Paolo Guiotto: illustrate this on this example here. That explains why the real part of that is not C-differentiable. Well, we know that it… the explanation is here. We computed the limit, we tried to compute the limit, and we have seen that the limit does not exist. This already explains why it is not C-differentiable. But now, we have a different,
20:29:530Paolo Guiotto: let's say a different explanation of why this function is not differentiable by showing that these two conditions are never verified for that case. So, if the function f of Z is the real part of Z,
20:47:120Paolo Guiotto: So it means that… let's identify the U and V. We already done one second ago, but let's repeat. So F of X plus I y is just X. So it means that it is X plus I0. We're worried that this is UXY,
21:05:120Paolo Guiotto: And this is VXY.
21:07:730Paolo Guiotto: Okay? So UXY is X, VXY is 0.
21:13:970Paolo Guiotto: Clearly…
21:17:460Paolo Guiotto: If we look at this condition 2, that should be, according to this theorem, equivalent to the C differentiability, we must verify two conditions. One is that the two functions must be C differentiable, and the second is that the Cauchyriemann
21:34:910Paolo Guiotto: Equations, holes are verified.
21:38:450Paolo Guiotto: So, number one is true in this case, because clearly, these two functions, UXY equal X,
21:47:420Paolo Guiotto: and VXY equals 0, R… Always.
21:56:250Paolo Guiotto: differentiable.
21:58:490Paolo Guiotto: They are polynomials.
22:08:700Paolo Guiotto: Okay? So, they are differentiable.
22:12:680Paolo Guiotto: But, if we look now at the Kosher-Riemann equations, a differential. But… D.
22:22:880Paolo Guiotto: Koshi Riemann.
22:25:90Paolo Guiotto: Equations… R?
22:33:280Paolo Guiotto: So, let's start already rewriting so that you memorize without the need of staying hours on these equations and memorizing. So, say, DXU equals DYV,
22:49:550Paolo Guiotto: So, you see, it's a sort… does it remind anything to you?
22:59:320Paolo Guiotto: Yeah, it has to do with green formula that has to do with…
23:07:00Paolo Guiotto: Yeah, that's something… in fact, there is a deep connection that should come.
23:12:50Paolo Guiotto: Later. So, and this is DYU equal minus DXV.
23:19:130Paolo Guiotto: So this is a cross-derivative test, somehow. It smells like that. And in fact, it is like that with a suitable interpretation. Well, for this case, what is DXU?
23:33:660Paolo Guiotto: One. One. What is the YV?
23:36:910Paolo Guiotto: 0. So you see, this is false. And this is false whatever is the point XY is never verified. Let's check the second one. DYU
23:47:700Paolo Guiotto: 0. Minus DXV.
23:51:90Paolo Guiotto: This one is verified, but the two must be verified together, so this is false.
24:00:20Paolo Guiotto: So, this means that… the Kosher-Riemann equations… are. Not.
24:10:430Paolo Guiotto: verified.
24:15:230Paolo Guiotto: So this means that even if this U and V are differentiable.
24:20:930Paolo Guiotto: This is not sufficient to have that the function they compose, U plus IV, which is the real part of Z, is differentiable, because these two conditions are not verified.
24:34:270Paolo Guiotto: are never verified, so there is no point XY where this is true, where the Kosheriemann system is verified, and this means that there is no point XY where the function is c-differentiable. So the function is never C-differentiable.
24:50:120Paolo Guiotto: are not verified at every point, XY,
24:54:980Paolo Guiotto: in R2, and this is equivalent to, say, at every Z equal X plus IY in C.
25:04:890Paolo Guiotto: So the Cartesian plane is the complex plane. So, this means that function f of z equal real part of Z
25:17:370Paolo Guiotto: is never… C. Differentiable.
25:23:220Paolo Guiotto: Never means, is not differentiable at any point, at every point of the complex plane.
25:30:870Paolo Guiotto: Okay?
25:33:410Paolo Guiotto: Okay, now let's go back to the statement and see how the proof works, okay?
25:39:600Paolo Guiotto: it's sort of not particularly complicated proof, it's just an application of definitions, and it will refresh how it works, the differentiability for functions of… from R2 to R, basically.
26:00:720Paolo Guiotto: So,
26:07:940Paolo Guiotto: profumes.
26:16:100Paolo Guiotto: Well, I start setting things like if we have to show that if F is C differentiable, then the condition 2 holds, but as you will see, it will be an if and only if, okay? So let's say that, let's start from, F…
26:34:80Paolo Guiotto: is, C, differentiable.
26:37:650Paolo Guiotto: at point…
26:39:500Paolo Guiotto: Z if and all if, and this is the definition, it's the unique thing we have at this point, the limit when H goes to 0 of F of Z plus H minus F of Z
26:54:810Paolo Guiotto: divided by H is this number, F prime of Z, complex number.
27:03:600Paolo Guiotto: Okay?
27:04:970Paolo Guiotto: Now, we can rewrite this in this way. So, carry this on the left-hand side. Put inside the limit, and you have, equivalently, it's an if and all if, the limit when H goes to 0 of, well, let's do slowly, F of Z
27:24:330Paolo Guiotto: plus H minus F of Z divided by H minus F prime of Z, This limit is zero.
27:34:560Paolo Guiotto: Now, let's do the common denominator. This is equivalent to the limit H goes to 0 of fraction. Downstairs we have an H, and in the numerator, we have Fz plus H minus F of Z minus F prime z times H.
27:54:720Paolo Guiotto: This limit is 0. Now, you have a limit of a fraction where both numerator and denominator go to zero.
28:03:830Paolo Guiotto: So, when this happens, the fraction goes to zero, it means that numerator is a little o of the denominator. That's the definition of what is a little o. So, this is equivalent to say that F of Z plus H minus F of Z
28:22:230Paolo Guiotto: minus F prime z times h is a little o of h.
28:29:520Paolo Guiotto: So these are the equivalences.
28:32:840Paolo Guiotto: Okay, now let's introduce U and V. So, let's rewrite this thing in terms of U and V. Let's see what does it mean, okay? So, we have to do a little bit of,
28:46:50Paolo Guiotto: restyling of this formula, it will become more complicated, because I will introduce… I will split each value of F into U plus IV, and I have to introduce also coordinates XY, so we have to do a little work. So let's say that Z is equal to X plus IY.
29:05:920Paolo Guiotto: Let's write the H, the increment, is a complex number, so let's call it, how can we call…
29:18:420Paolo Guiotto: Yeah, but,
29:28:730Paolo Guiotto: Okay, because I wanted to use, there is a reason for the… You cannot use… okay, let's follow your advice. A plus AB. Okay.
29:39:240Paolo Guiotto: So, this means that F of Z, let's start from this one, it is F of X plus IY, this is U of XY plus I of V of XY, okay?
29:54:720Paolo Guiotto: Then, what is F of Z plus H?
29:58:950Paolo Guiotto: Well, inside F, I have Z, which is this one, and H, which is this one. I do the sum, I get the number. X plus A,
30:12:490Paolo Guiotto: plus IY plus B, right?
30:17:80Paolo Guiotto: So this will be, in terms of U and V, it will be U of X plus A…
30:23:380Paolo Guiotto: comma Y plus B.
30:27:50Paolo Guiotto: plus I, V, the same argument, X plus A, Y plus B.
30:34:840Paolo Guiotto: Okay?
30:36:390Paolo Guiotto: Now, we have to do this product. This is the crucial point. So let's say that that number, F prime of Z, is a complex number.
30:47:20Paolo Guiotto: And, we call it, so since we already used A and B, I wanted to use this alpha and beta, but…
30:55:600Paolo Guiotto: That would be… What letters could I use?
31:05:500Paolo Guiotto: Well, let's say, now gamma remains of a… of a curve,
31:12:100Paolo Guiotto: Yeah, but those are normally used for integers, so they… let's use alpha and beta. So this is the complex number, F prime of Z, alpha plus A beta.
31:24:820Paolo Guiotto: So when I do F prime of Z times H, I'm doing alpha plus i beta.
31:30:930Paolo Guiotto: times H, which is A plus IB.
31:36:440Paolo Guiotto: So when we do the product of this, we get alpha A minus beta B plus I,
31:46:900Paolo Guiotto: Then we have beta. A… plus alpha B.
31:54:620Paolo Guiotto: So this is the product.
31:57:40Paolo Guiotto: Okay, and also, finally, there is this little O, this is a little o, which is a little O, it's a complex number, because here everything is a complex number, so that little O of H will be a real part that will be still a little o of h, so let's call it O1H plus I02
32:20:560Paolo Guiotto: age.
32:23:180Paolo Guiotto: Okay, it's clear that if O of H is a little o, it will have a real and an imaginary part that will still be little O.
32:31:850Paolo Guiotto: Okay, so now we can say that this one, let's give a name, star, becomes equivalent. I plug all these things into that formula, okay? So, star is equivalent.
32:46:860Paolo Guiotto: And remind that this is a chain of equivalencies that started from this F is C differentiable at point Z.
32:53:670Paolo Guiotto: So, it is equivalent to… So, number 1, f of z plus h, it is this.
32:59:930Paolo Guiotto: Okay?
33:01:150Paolo Guiotto: Now, I will do this graphic trick. I will write in a color everything which is real, and in another color, everything which is imaginary, so you can visualize a little bit better. So let's say that we start writing in light blue the real parts. So, F of Z plus H is this one, so I take this.
33:24:440Paolo Guiotto: Youth.
33:27:300Paolo Guiotto: X plus A, Y plus B,
33:31:400Paolo Guiotto: Then I have, blast.
33:34:190Paolo Guiotto: I, let's write it in red, V, X plus A.
33:39:580Paolo Guiotto: Y plus B. And this is… this guy here is F of Z plus H. Then I have to do minus F of Z.
33:51:610Paolo Guiotto: So, minus…
33:53:690Paolo Guiotto: F of Z, well, since it is a complex number, F of Z is U plus AV at point XY, so we have U, X, Y,
34:06:360Paolo Guiotto: plus… I, V, X, Y.
34:12:739Paolo Guiotto: It's a mess, but we have to do only one time. Okay, so now I have written F of X z plus h minus F of Z. Then, next, there is this minus F prime z times h.
34:27:40Paolo Guiotto: So I will continue down here. So, minus… sorry, minus F prime Z times H.
34:35:449Paolo Guiotto: We already separated, it is written here, so we already separated the real and the imaginary part, so we still keep blue for the real. Alpha A minus beta B.
34:49:570Paolo Guiotto: plus… I, then we have this, beta A plus alpha B.
34:58:440Paolo Guiotto: Okay, now we have written the left-hand side of this star identity. This equal to the little o.
35:07:520Paolo Guiotto: equal to the little o of h that we split it into a real and an imaginary little O. So we will have O1H
35:19:260Paolo Guiotto: plus I, or 2, age.
35:24:740Paolo Guiotto: Okay? Now, this is an identity between two complex numbers written in algebraic form.
35:33:290Paolo Guiotto: So they are the same numbers, if and only if the real part are the same and the imaginary part are the same. You see? So here, here…
35:45:360Paolo Guiotto: We have… an identity, of type… Say, a number… And I'm, C.
36:02:160Paolo Guiotto: plus I eta.
36:05:380Paolo Guiotto: is equal to a number, I don't know, delta… plus… Aye.
36:14:210Paolo Guiotto: Omega.
36:16:230Paolo Guiotto: Now, this is possible if and only if the two blue numbers coincide and the two red numbers coincide, where, of course, all these numbers are real, no? Where this is real, this is real, this is real, in such a way that they are really the real and imaginary parts.
36:35:510Paolo Guiotto: when these numbers are the same. Now, these are identified by points when the points are the same, so if and only if the real parts coincide
36:45:630Paolo Guiotto: And the imaginary part coincide.
36:49:610Paolo Guiotto: So now what we are doing, what we are going to do here.
36:53:870Paolo Guiotto: the star, which is equivalent to all this, is also equivalent to a couple of conditions. One is the real part, so the blue quantities at left is equal to the blue quantity at right, and the red quantity set left is the red quantity at Y.
37:11:680Paolo Guiotto: So, star… It's equivalent, and we gain a system now.
37:19:580Paolo Guiotto: So, let's copy the blue part. We have u of x plus AY plus B.
37:26:590Paolo Guiotto: U of X plus a… Y plus B.
37:31:650Paolo Guiotto: minus… U of XY.
37:39:20Paolo Guiotto: Then there is…
37:41:460Paolo Guiotto: this is… there is a minus in front of this number, okay? Because this is minus F prime z times H, so minus the blue parenthesis.
37:54:580Paolo Guiotto: So, minus… What is, alpha A minus beta B?
38:05:250Paolo Guiotto: This is equal to little o of h… well, still H is going to disappear. Little O of H
38:13:930Paolo Guiotto: little O of H here, the O1H, is… can be… it's basically something that goes to 0 faster than H, or faster than the absolute value of H, because the size is the same.
38:29:860Paolo Guiotto: So we prefer to use this, because if we write the modulus of H, and H is,
38:36:680Paolo Guiotto: A plus IB, the modulus is the square root of A squared plus B squared.
38:43:600Paolo Guiotto: This is modulus of H.
38:47:590Paolo Guiotto: And the similar condition is, obtained by equating the red parts. So we have, for the red parts, V of X plus, so this is horrible V.
39:01:950Paolo Guiotto: V of X plus A, Y plus B, minus V of XY.
39:08:950Paolo Guiotto: minus, let's copy this quantity, beta alpha plus alpha B.
39:14:750Paolo Guiotto: beta A plus alpha B.
39:17:620Paolo Guiotto: beta A plus alpha P.
39:23:60Paolo Guiotto: Alpha B. Again, this is a little of O2 of modulus of H, so we can still write O2 of root of A squared plus B squared.
39:36:330Paolo Guiotto: Now, I claim that these two conditions are at once equivalent to the conclusion.
39:43:150Paolo Guiotto: That is, U and V are differentiable, and their partial derivative verify the Cauchier-Riemann equations. Let's see why. Let's start from the first one, the blue condition.
39:58:110Paolo Guiotto: Now, you see that we have u of x plus ay plus B,
40:02:520Paolo Guiotto: So, this can be also seen as U at point XY plus an increment AB, right?
40:12:520Paolo Guiotto: minus U of XY, Now, what do we need to have there?
40:20:590Paolo Guiotto: for the differentiability.
40:23:190Paolo Guiotto: I know it's something we have done a lot of time ago, but…
40:27:620Paolo Guiotto: remind that a function U, let's focus on… let's give it a definition for you. U as function of XY
40:36:880Paolo Guiotto: Ease.
40:37:930Paolo Guiotto: Differentiable, is real, art, differentiable.
40:42:830Paolo Guiotto: at point XY,
40:46:340Paolo Guiotto: If and only if the definition is basically the McLaurin formula, the Taylor formula. It says U of… you take your point, X, Y, plus an increment, let's take the increment AB,
41:04:700Paolo Guiotto: minus U of XY, Minus what is here.
41:10:280Paolo Guiotto: The…
41:13:940Paolo Guiotto: What should I write here?
41:16:320Paolo Guiotto: The increment of the function minus… Not…
41:23:200Paolo Guiotto: No, the point is XY. This is the rate increment. This is the point, this is the increment.
41:30:260Paolo Guiotto: Let's go back, let's hope to find this definition.
41:36:620Paolo Guiotto: Well, fortunately, that's the defini… no, this is the differentiability test. That's not the definition. Probably the definition was given…
41:52:780Paolo Guiotto: Yeah.
41:55:720Paolo Guiotto: That's the definition. What you see, you have your function, forget the arrays, no? Function at point plus increment, minus function at point minus derivative of the function times the increment, where this is a product matrix times vector.
42:13:150Paolo Guiotto: For the case of numerical functions, you remind that That formula, where is it?
42:21:480Paolo Guiotto: This is the general formula for… okay, here we are. For the case of a numerical function, which is this one, the product, the Jacobian matrix, is just the gradient vector.
42:34:400Paolo Guiotto: Okay? So, that product becomes just the product, one line times one column.
42:42:670Paolo Guiotto: So what is… the vector here should be the gradient… let me write in… gradient of U at point XY,
42:52:650Paolo Guiotto: times the vector, which is the increment, AB.
42:58:390Paolo Guiotto: this equal to the little o of norm of the vector. What is norm of AB of the increment? It is exactly the root of A square. So that relation in blue over there says that the function u
43:16:190Paolo Guiotto: Is differentiable with gradient.
43:19:700Paolo Guiotto: the vector that now we have to identify in such a way that we get exactly this expression when we do the product between the gradient and vector AB.
43:30:490Paolo Guiotto: So what is the vector, the gradient factor?
43:34:160Paolo Guiotto: where gradient U… gradient U is the vector what? You have to imagine, to take a vector.
43:45:40Paolo Guiotto: that multiply the line by columns with AB gives that thing rounded in yellow. So what should be the vector that I get here with the two entries, such a way that when I do the first times A plus the second times B, I get that positive. What is the vector?
44:03:570Paolo Guiotto: It is the vector alpha minus beta. Exactly.
44:08:490Paolo Guiotto: Where gradient 2U is the vector.
44:12:220Paolo Guiotto: Alpha.
44:13:390Paolo Guiotto: minus beta, because, you see, when you do gradient 2 applied to vector AB, line by column, you are doing alpha minus beta.
44:23:790Paolo Guiotto: times AB, and when you do this product, it comes exactly alpha, A minus beta B.
44:31:350Paolo Guiotto: So this says that the number one, the first of these two conditions, says that U is differentiable.
44:39:630Paolo Guiotto: And the gradient is that vector.
44:43:880Paolo Guiotto: So, let's, let's call, S this system here.
44:51:710Paolo Guiotto: So, D.
44:56:630Paolo Guiotto: First, equation of S.
45:04:60Paolo Guiotto: Says… That.
45:07:680Paolo Guiotto: First, U is differentiable at point XY, and the gradient of U
45:18:830Paolo Guiotto: The gradient of U is the vector, we said, alpha minus beta.
45:26:750Paolo Guiotto: And for the same reason, the second, The second… equation of the system S, says…
45:38:870Paolo Guiotto: If you look, it's similar. There is the increment, the red equation now. The increment of V minus something that we can always see as product line by column of what metrics now?
45:56:20Paolo Guiotto: Minus bit alpha, are you sure?
45:58:870Paolo Guiotto: No, the minus is outside, the min…
46:03:920Paolo Guiotto: beta alpha, exactly. So this says that the increment of V minus that vector times AB is a little o of the norm of AB. So this says V is differentiable.
46:18:860Paolo Guiotto: V is differentiable.
46:22:870Paolo Guiotto: at point XY, and the gradient of V
46:29:850Paolo Guiotto: Is, we said, beta alpha, right?
46:37:160Paolo Guiotto: The gradient of V is beta alpha.
46:39:850Paolo Guiotto: So now we can summarize, because star is equivalent to the initial point, the C differentiability.
46:48:280Paolo Guiotto: So, C differentiability is equivalent to this condition star, which is just the McLaurin formula, the Taylor formula for the function F. This one, after introducing the real imaginary parts of everything, is equivalent to a system of these two equations, the blue and red.
47:10:620Paolo Guiotto: The first one, the blue, says that U is differentiable at point XY in real sense, in the traditional sense, and the gradient is that vector alpha minus beta. The second one says that V is differentiable at XY, and gradient V is beta alpha. So we can now trace the conclusion, conclusion.
47:33:320Paolo Guiotto: So we can say that F is C differentiable.
47:39:320Paolo Guiotto: at point Z equal X plus IY, If, and only if.
47:47:990Paolo Guiotto: No? Because it is an equivalence, we have that both U and V are differentiable, so U and V are…
47:57:580Paolo Guiotto: Both.
47:58:970Paolo Guiotto: Differentiable. So, real differentiable, at point.
48:04:720Paolo Guiotto: XY.
48:07:100Paolo Guiotto: And we have a condition here, because you see that when we say that the gradient of U must be alpha minus beta, and gradient of V must be beta alpha, it means that there is some relation between the derivatives of U and V, because gradient of U is the vector made of dx u, d, y, u, while gradient of V
48:31:720Paolo Guiotto: is the XV, DYV.
48:34:820Paolo Guiotto: So what do you deduce from this? You see that the XU is alpha, which is this alpha, which is DYV, and that's the first of the Kosheriemann condition, and the XU at point XY,
48:51:920Paolo Guiotto: must coincide with DYV at the same point, XY.
48:56:970Paolo Guiotto: But not only, because when you look at DYU, well, let's use a different color, DYU is minus beta, but beta is DXV, so we can say that DYU
49:13:830Paolo Guiotto: DY.
49:15:470Paolo Guiotto: U, at point XY, is minus beta minus beta is DXV.
49:25:900Paolo Guiotto: And these are the costume and conditions. Now, the proof is completed.
49:32:330Paolo Guiotto: Okay? So, let's say that the proof is not extremely complicated, because the idea is simple, we write the definition, let's see what happens. At a certain point, we have to split the definition into real imaginary part, and draw some conclusion on this to real and imaginary part.
49:52:970Paolo Guiotto: Okay, now we have done this, we can take a short break, 5 minutes, and then we see some exercises on the use of these conditions.
50:10:50Paolo Guiotto: Bye, Diablo.
50:16:380Paolo Guiotto: Okay, so, example…
51:19:300Paolo Guiotto: the mic?
51:22:210Paolo Guiotto: Recording, yes, so everything is okay, but why?
51:29:560Paolo Guiotto: Okay, so let's do some, examples, exercises on this. For example, let's check… check…
51:40:620Paolo Guiotto: If the function f of z equals Z conjugate, Is, C differentiable?
51:51:850Paolo Guiotto: Well, we can do by using the definition, so we have to compute the limit to see if the limit exists, and if yes, to compute the limit, we have the derivative, but here we will use the Cauchy-Riemann conditions, so…
52:08:750Paolo Guiotto: Sweet.
52:09:790Paolo Guiotto: use… the Kosheriemann… conditions.
52:16:270Paolo Guiotto: So first, we have to determine the real imaginary part. Here, the function f
52:24:140Paolo Guiotto: of Z, written Z as X plus IY, it is Z conjugate, so it is X minus IY. So you can see that from this, U of XY is
52:38:440Paolo Guiotto: X, and V of XY is… No, minus Y.
52:45:80Paolo Guiotto: It's important, because if you put Y in, you get the wrong answer, yeah. So, clearly.
52:51:640Paolo Guiotto: never forget that there is also this condition, because often it happens that you reduce this equivalence to saying that F is C differentiable if and if the Kosheriemann equations are verified. But that's false.
53:12:300Paolo Guiotto: Why? Because it can be possible, there is an example in the exercises, the 7116,
53:20:280Paolo Guiotto: It is… you know that it is possible to have partial derivatives and not being differentiable, no? That's the problem, basically. So you can have U and V that they have partial derivatives, they fulfill the Kosheriman condition, but they are not differentiable, and so your F is not differentiable. That's the example of exercise 7-116.
53:41:530Paolo Guiotto: So… Always check the differentiability.
53:45:590Paolo Guiotto: So clearly, Clearly.
53:51:730Paolo Guiotto: U-N-V-R… are differentiable.
53:57:550Paolo Guiotto: They are polynomials, etc.
54:00:820Paolo Guiotto: They… Polynomials.
54:09:860Paolo Guiotto: So, and… the gradient of U
54:15:570Paolo Guiotto: is the vector, derivative with respect to X is 1, derivative with respect to Y is 0. The gradient of V is the vector, derivative with respect to X is 0, derivative with respect to Y is minus 1.
54:29:190Paolo Guiotto: So… F equal U plus IV.
54:34:400Paolo Guiotto: Is C differentiable?
54:37:900Paolo Guiotto: Now, if and only if, since I already checked the differentiability.
54:43:180Paolo Guiotto: If, not if, Kosher-Riemann equations.
54:48:630Paolo Guiotto: are verified.
54:52:890Paolo Guiotto: And this yields what?
54:55:150Paolo Guiotto: Now, the first condition is DX, let's rewrite… do this five times, and then you will see that you don't need to memorize anything. DXU equals DYV, and DYU
55:08:960Paolo Guiotto: Equal minus DXV.
55:11:460Paolo Guiotto: And so I wrote the DYU, but that's DYV.
55:15:720Paolo Guiotto: That means the XU is 1, that must be equal to DYV, which is minus 1, so I can forget of the second equation, this is impossible.
55:28:530Paolo Guiotto: And this is never verified, so there is not some XY for which this is true. So this means that this function f
55:37:550Paolo Guiotto: is, not… C.
55:41:580Paolo Guiotto: differentiable.
55:44:140Paolo Guiotto: at each.
55:46:410Paolo Guiotto: Point.
55:48:40Paolo Guiotto: Z of C.
55:50:300Paolo Guiotto: And this, again, it's quite surprising, because you may say, it looks like F of Z equals Z, it's Z conjugate, but that innocent conjugate makes this function not differentiable.
56:02:270Paolo Guiotto: Okay?
56:03:570Paolo Guiotto: Now, you have some exercises, of the… of this type, the exercise DO7,
56:12:370Paolo Guiotto: 11.4, same check of this one.
56:16:880Paolo Guiotto: And the 7116er.
56:21:470Paolo Guiotto: Example… of F… Such that… U and V fulfill
56:35:680Paolo Guiotto: the Koshy-Riemann equations But… F is not.
56:44:820Paolo Guiotto: C, differentiable.
56:47:320Paolo Guiotto: Once again, let me repeat. You may think, why that? You say that if they verify the Kushriemann equation, the function, this is equivalent to C differentiability. That's not what I said.
56:59:660Paolo Guiotto: The theorem… not what I said, what the theorem says, the theorem says that the C differentiability is equivalent to the real differentiability of the real imaginary part, so the standard differentiability, plus the Cauchier-Riemann equation.
57:16:620Paolo Guiotto: So you need the two conditions together, not just one of them.
57:21:60Paolo Guiotto: So in particular, in particular, as we said before this statement, the simple differentiability of U and V is not at all sufficient to have the differentiability of F. You must have also the Cauchyriemann conditions.
57:37:150Paolo Guiotto: Now, the exercise 7-11-5… is of this type. For each… for each… of the following U.
57:54:60Paolo Guiotto: XY… So it is given U.
57:58:740Paolo Guiotto: determine… V, function of XY, Such that, F equal U plus IV.
58:10:700Paolo Guiotto: B.
58:13:340Paolo Guiotto: Well, it is written holomorphic, we have not yet introduced this term. However, let's say it's a little bit more than C-differentiable. So, C-differentiable…
58:25:610Paolo Guiotto: determining… also… the function F.
58:31:70Paolo Guiotto: So let's take, for example, the number… well, let's do the number 1, which is particularly easy, so we understand the idea, then we…
58:41:70Paolo Guiotto: We can do some more complicated exercises. The number one says UXY is equal to X.
58:49:410Paolo Guiotto: So, what is this problem? It says, in this problem, you know what is the real part of a function, and you want to say if there is an imaginary part that I can combine with you in such a way that I get a C-differentiable function.
59:06:810Paolo Guiotto: Okay? And in that case, I want also to write what is the function f as function of Z here.
59:14:760Paolo Guiotto: as a function… of Z.
59:21:480Paolo Guiotto: So let's see, let's think about this, no?
59:26:340Paolo Guiotto: So, what is, what is the… it's just to review the theorem we have seen in a slightly different way. Now, we know that
59:42:520Paolo Guiotto: a function F equal U plus IV, where I suppose that U and V are the real and the imaginary part of F, so U is the real part of F, V is the imaginary part of F.
59:57:850Paolo Guiotto: is C differentiable?
00:01:560Paolo Guiotto: If, null if, we just proved, number 1, U and V are real differentiable.
00:13:100Paolo Guiotto: And number two, UNV fulfill… the Cauchier-Riemann equations.
00:23:700Paolo Guiotto: Okay.
00:25:720Paolo Guiotto: So, we have you.
00:29:100Paolo Guiotto: because U is given, it is this one. We do not have V. We have to determine V, it's a sort of equation, no? Where we have to determine unknown. It's a problem with an unknown with that V, which is unknown.
00:42:370Paolo Guiotto: So, about U, we of course know that U alone is R differentiable. That's evident, no?
00:50:560Paolo Guiotto: Because UXY is X.
00:53:910Paolo Guiotto: So, clearly, U is R differentiable.
01:05:660Paolo Guiotto: And we also… it will be useful, we notice that the gradient of U is what? It's the vector, derivative with respect to X is 1, derivative with respect to Y is 0. That's the gradient.
01:17:590Paolo Guiotto: So this is the DXU.
01:21:40Paolo Guiotto: DYU.
01:22:700Paolo Guiotto: So… We are looking…
01:31:510Paolo Guiotto: 4.
01:32:960Paolo Guiotto: a V, Such that… What does… what this V should be? Should be R differentiable, So, V is…
01:43:560Paolo Guiotto: are differentiable.
01:46:750Paolo Guiotto: And second, V, together with this U, verify the Kusheriemann equation. So, V.
01:55:640Paolo Guiotto: Together.
01:59:780Paolo Guiotto: with… U.
02:02:760Paolo Guiotto: verifies…
02:06:150Paolo Guiotto: Kaushi-Riemann equations, that now, since we don't know what is V, they become a system of equations for V.
02:15:240Paolo Guiotto: Look.
02:16:170Paolo Guiotto: The Cauchy-Riemann equation are usually DX written in this form, dxu equals DYV.
02:24:140Paolo Guiotto: Nice. DYV…
02:27:930Paolo Guiotto: and DYU equal minus DXV, right? But now we can rewrite this as an equation where DYV is DXU, and DXV is minus DYU.
02:45:280Paolo Guiotto: Well, we know these are the right-hand sides, because the DXU is 1, and the DYU is 0, so minus this is minus 0 equals 0.
02:56:70Paolo Guiotto: So we have to look at a V, which must be R- differentiable, such that DXV is 0, and DYV is 1.
03:08:870Paolo Guiotto: What kind of problem reminds this?
03:17:150Paolo Guiotto: We are looking… for a function V whose derivatives are given.
03:25:440Paolo Guiotto: What is this?
03:29:230Paolo Guiotto: It's a problem we have already discussed in this course. Let's think about…
03:37:510Paolo Guiotto: It's the same, the same, let's say, the same, part where you said before, you said, irritational norm, referring to the pair UV.
03:49:840Paolo Guiotto: V is what?
03:52:200Paolo Guiotto: The function V that verifies this condition is…
03:58:260Paolo Guiotto: Now, you're still foxed on the area where you have been touched by this formula, but it's a bit before. The function V is the potential of that, of the field made by the two components 0, 1, no? Because we are looking for V such that the derivative with respect to X of V is known.
04:17:840Paolo Guiotto: and the derivative of V with respect to Y. So the gradient of V is given. So the V is the potential of something. So how do we solve? In the same manner. So take the first equation. This says that
04:31:480Paolo Guiotto: V of XY is… so we integrate this in the variable X, so it's a primitive of 0 in the variable X plus a constant in X, so potentially a function of Y, so C of Y. Now, the primitive of 0 is 0, so function of Y.
04:52:00Paolo Guiotto: Then we plug this into the second equation, and we get C prime of y equal 1. This means C of Y equals Y plus a true constant C.
05:04:370Paolo Guiotto: This is constant.
05:07:130Paolo Guiotto: So, at the end, we get VXY is equal to Y plus a constant.
05:15:930Paolo Guiotto: Okay? So, now, this says that in order V, together with U, verifies these Cauchier Riemann equations, necessarily V is this thing. So we got that V must be this one.
05:32:970Paolo Guiotto: Now, since this V is clearly differentiable, Since… this… V is, clearly, real differentiable.
05:47:640Paolo Guiotto: It's, again, a polynomial, no?
05:51:320Paolo Guiotto: So you see that we were looking for a V such that V is real differentiable, and V together with U verifies the Koshiniman conditions. But the Koshiniman conditions yield this type of V, which is real differentiable, so we finished, because we found that all possible V
06:10:920Paolo Guiotto: for which they are differentiable, and together with you, they verify that Kosheriemann equations are of this type. So this is the unique possible shape of it. The unique, it's not unique, because there is the constant, which is an arbitrary real number.
06:26:530Paolo Guiotto: Okay, so we found that V is this one, and therefore F of Z
06:32:860Paolo Guiotto: now we, respond to… we answer to this, second part, now determine also F as a function of Z.
06:41:400Paolo Guiotto: So F of Z is F of X plus IY.
06:45:790Paolo Guiotto: It is U plus IV.
06:49:340Paolo Guiotto: But we found U was given, it was equal to X plus IV. V is Y plus constant, so we have that this is X plus IY plus I constant. As you can see, this X plus IY is exactly Z.
07:07:60Paolo Guiotto: So it means that the function f that verifies this condition is this. F of Z is equal to Z plus an imaginary number where C is real, because IC is imaginary if C is real. So this is the final answer of this problem, okay?
07:28:300Paolo Guiotto: So let's repeat the same type of problem.
07:32:810Paolo Guiotto: On a more complicated example, so like the… Number, 5.
07:42:940Paolo Guiotto: So here we have U of X, Y,
07:47:110Paolo Guiotto: equal X squared plus Y squared.
07:52:190Paolo Guiotto: Same… problem.
07:56:980Paolo Guiotto: So the problem consists in determining whether or not there is a function V such that U plus IV, where U is known, V is not known, is c-differentiable. So we can… this part is the same. So, F equals U plus IV.
08:16:319Paolo Guiotto: is… C, differentiable.
08:21:270Paolo Guiotto: If and only if… We have that U.
08:26:300Paolo Guiotto: NVR… are differentiable.
08:33:120Paolo Guiotto: And of course, for U, if this is U, it is true, X squared plus Y squared, so this is true.
08:42:930Paolo Guiotto: for… You?
08:45:439Paolo Guiotto: It's evident, we do not have to justify, it's a polynomial, so we don't justify this. And second, U and V fulfill
08:56:590Paolo Guiotto: Because you're given equations.
09:00:300Paolo Guiotto: So this means that, the derivative with respect to X of U is the derivative with respect to Y of V.
09:08:430Paolo Guiotto: And the derivative with respect to Y of U is minus the derivative with respect to X of V. That we switch into a system for V, so we have DXV is minus DYU, but DYU is 2Y, so this will be minus 2Y.
09:26:930Paolo Guiotto: and DYV will be DXU, which is 2X. So at the end, we have to determine a V such that DXV is minus 2Y,
09:41:529Paolo Guiotto: And DYV is 2X.
09:45:970Paolo Guiotto: And that's the problem of determining a potential for a certain vector field. The field is minus 2Y2X. So the first one says that V of XY
09:59:200Paolo Guiotto: must be, since the derivative with respect to X is minus 2Y, it must be minus 2Y times X, plus a function which is constant in X, so it can be a function of Y.
10:13:520Paolo Guiotto: And if we plug this into the second condition, we get this second becomes… derivative with respect to Y is minus 2X plus C prime of Y equal to 2X.
10:27:680Paolo Guiotto: So this means that C prime of Y is 4X.
10:35:570Paolo Guiotto: And this says that…
10:47:340Paolo Guiotto: No, because… This function is independent of X. This function C was a function of Y.
10:57:360Paolo Guiotto: Okay? So it cannot be the derivative that is a function of X. So this means that there cannot be such a C,
11:06:290Paolo Guiotto: So this is impossible.
11:09:540Paolo Guiotto: So what does it mean? It means that there is not a V such that the two conditions are verified, so there is not a V for which the Cauchyriemann equations are verified. And if this is the case, it means that there is not a V for which U plus IV is C-differentiable.
11:28:970Paolo Guiotto: So this means that… that cannot… Et vie, are differentiable.
11:38:500Paolo Guiotto: Such that, U and V.
11:42:800Paolo Guiotto: verify… the Kosher-Riemann equations.
11:49:170Paolo Guiotto: And since this is equivalent to have U plus AV differentiable, so it means that there is not a V,
11:58:30Paolo Guiotto: Such that F equals U plus IV is C differentiable.
12:06:750Paolo Guiotto: So, is there a V such that U plus IV is C differentiable? No. The answer is negative here.
12:15:920Paolo Guiotto: I would like to… Illustrate maybe one, because these are… classical exam exercises.
12:28:900Paolo Guiotto: But I have to… would give me a second that I… Go… Indeed.
12:39:970Paolo Guiotto: a model to… to see the… Exam exercises.
12:46:740Paolo Guiotto: We showed one of them.
12:53:890Paolo Guiotto: So… Be patient, one… still one second… Where is it?
13:02:310Paolo Guiotto: Solution finally sounds.
13:06:10Paolo Guiotto: Okay… So, let's take,
13:22:840Paolo Guiotto: Let's take this one. This is the Exercise 19.
13:32:380Paolo Guiotto: from… Exam.
13:37:160Paolo Guiotto: Set.
13:39:280Paolo Guiotto: So it says… We have given now V to let.
13:45:780Paolo Guiotto: V, X, Y.
13:48:300Paolo Guiotto: be equal to Y cubed Minus 3… X square Y.
13:57:910Paolo Guiotto: plus 4.
14:00:340Paolo Guiotto: XY… minus X.
14:05:180Paolo Guiotto: for XY in R2.
14:13:360Paolo Guiotto: So, the exercise ask… well, it's split in two questions. Number one, determine Say, if any.
14:22:280Paolo Guiotto: all possible.
14:26:950Paolo Guiotto: U.
14:28:80Paolo Guiotto: function of XY.
14:30:790Paolo Guiotto: Such that, F equal U plus IV.
14:37:620Paolo Guiotto: B, C, differentiable. It's written holomorphic, but it's a little bit more than C differentiable, so it's C differentiable on…
14:49:210Paolo Guiotto: on C.
14:52:560Paolo Guiotto: And, say, question two… In… in this case…
15:01:730Paolo Guiotto: determine F as a function of Z.
15:10:330Paolo Guiotto: Okay.
15:13:830Paolo Guiotto: It's not a particularly difficult exercise, it's the same music, no? So, F equal, question 1.
15:21:390Paolo Guiotto: F equals U plus IV.
15:25:170Paolo Guiotto: Is C differentiable?
15:29:640Paolo Guiotto: on… C.
15:32:410Paolo Guiotto: If and all if… The two UNV, R.
15:41:600Paolo Guiotto: are differentiable.
15:44:700Paolo Guiotto: Well, C in terms of X, Y, which are the variables, for U and V means R2.
15:54:00Paolo Guiotto: So this would be for every Z in C, this would mean for every Z corresponds to point XY in R2.
16:04:920Paolo Guiotto: And… the Kosher-Riemann equations.
16:12:530Paolo Guiotto: are verified.
16:17:860Paolo Guiotto: still for every XY natural.
16:20:320Paolo Guiotto: Now, clearly, That, we have V here, we have not U, but…
16:25:900Paolo Guiotto: That's not change too much. We have… that is V, it's supposed to be V, so we clearly see that that V is a polynomial, so it's clearly an R differentiable function. So, since…
16:41:10Paolo Guiotto: V is… a polynomial.
16:49:260Paolo Guiotto: in XY.
16:53:420Paolo Guiotto: V is… R… differentiable.
16:59:250Paolo Guiotto: on R2.
17:02:410Paolo Guiotto: So… we look… for a V, U, are differentiable.
17:16:180Paolo Guiotto: And such that… The Kosheriemann condition holds. So, here we have DXU equal DYU equal.
17:29:220Paolo Guiotto: Now, we copied the derivative, so this is DYV. DYV is what?
17:34:480Paolo Guiotto: V is this, so we have 3Y squared.
17:43:860Paolo Guiotto: Then we have, minus 3x squared.
17:51:740Paolo Guiotto: then plus 4X, and the minus X is 0, so plus 4X.
17:59:430Paolo Guiotto: While the derivative with respect to Y of U should be minus the derivative with respect to X of V,
18:07:90Paolo Guiotto: So this means minus what? Derivative with respect to X of U cubed is 0. The second term, it's minus 6xy.
18:21:00Paolo Guiotto: Then, the next term is still deriving respect to X for Y.
18:27:150Paolo Guiotto: So, plus 4Y.
18:29:610Paolo Guiotto: And the last one yields minus 1. So we have this.
18:33:720Paolo Guiotto: So we have to look for this U that verifies these two conditions. The XU is 3Y square minus 3X squared plus 4X.
18:45:330Paolo Guiotto: And for the DYU, we get 6XY minus 4Y plus 1.
18:55:350Paolo Guiotto: Okay?
18:56:920Paolo Guiotto: Now, we start with the first equation. This says that derivative with respect to X is this thing.
19:03:930Paolo Guiotto: So, it means that from this one, we get UXY,
19:10:620Paolo Guiotto: is a primitive of this thing, 3Y square, minus 3X squared plus 4X.
19:20:360Paolo Guiotto: Still, since we are taking the derivative with respect to X, this will be a primitive in the X variable, plus, potentially, a constant in X, so a function of the other variables. Here we have another variable only.
19:34:300Paolo Guiotto: So this is… now, 3Y squared is a constant, so we get 3Y square times the primitive of 1, which is X minus 3X squared
19:45:70Paolo Guiotto: The primitive of 3x squared in X variable is X cubed.
19:50:560Paolo Guiotto: Now, because the derivative with respect to X is exactly 3X squared. Plus, 4x is the derivative of… well, 2X is the derivative of X squared, so I can say 2X squared.
20:03:470Paolo Guiotto: plus C of Y.
20:05:850Paolo Guiotto: So this is the possible U, the unique possible U. Now we plug this into the second condition to impose also this one, and we get, so…
20:20:860Paolo Guiotto: DYU will be for… with this U here, We have.
20:27:570Paolo Guiotto: 6yx.
20:31:120Paolo Guiotto: then the dy of minus X cubed is 0, the dy of 2x squared is 0, and then we have plus C prime of Y. This must be equal to the right-hand side of this equation, 6.
20:44:600Paolo Guiotto: XY minus 4Y plus 1.
20:48:650Paolo Guiotto: As you can see, this 6XY simplifies, so we get that this is if and only if C prime of Y is minus 4Y plus 1,
21:01:480Paolo Guiotto: And this is good, because it's just a function of Y, so we can determine C of Y.
21:07:560Paolo Guiotto: is a primitive with respect to Y of this function, minus 4Y plus 1 in the variable y, plus a constant, which is now a true constant, okay? So this is a constant.
21:21:400Paolo Guiotto: is not a function of X, okay? Because DC of Y wasn't depending of x, so the constant cannot be depending on X, cannot reintroduce X. So we have a minus 2Y square plus Y plus C.
21:38:560Paolo Guiotto: So, now we have determined the C of Y, and we plug into U, and finally we get this. So, the argument is.
21:49:320Paolo Guiotto: We look for U, which is R differentiable, and such that the U verifies the Koshiriemann equations together with V. This imposes these two equations for U. We solved… we found that U of XY
22:06:580Paolo Guiotto: is equal to… So… 3Y square X.
22:15:610Paolo Guiotto: minus… X cubed… plus 2X squared.
22:25:190Paolo Guiotto: Then we have, plus C of Y, which is this thing, minus 2Y squared.
22:31:370Paolo Guiotto: plus Y plus a constant.
22:34:210Paolo Guiotto: And this is the U, okay?
22:37:00Paolo Guiotto: And since this U is clearly differentiable, And since… This you is clearly.
22:49:80Paolo Guiotto: are differentiable.
22:52:900Paolo Guiotto: Well, we closed the argument, because, you see,
22:57:670Paolo Guiotto: So, F equals U plus IV is C differentiable on C, if and only if there are U and V are differentiable on F2, for which the Cauchyriemann conditions are verified. V is already… are differentiable. We found the unique possible U…
23:15:340Paolo Guiotto: such that together with V, the Kosheriemann equations are verified, and the U is this one. And since this is differentiable, the conclusion is that, is our differentiable on R2,
23:28:190Paolo Guiotto: that the function F equals U plus AV is C.
23:35:360Paolo Guiotto: Differentiable.
23:38:00Paolo Guiotto: on… C.
23:42:150Paolo Guiotto: Now, to respond to the question, too…
23:46:70Paolo Guiotto: So determine the function f as a function of Z. So we have that f of x plus IY is U of XY
23:56:950Paolo Guiotto: plus IV of XY. Now, we have to recognize,
24:03:30Paolo Guiotto: how it depends on Z, okay? Because here, it's, if we write the 2U and V, we get for U, we have these three.
24:11:580Paolo Guiotto: Y squared X minus X cubed plus 2X squared minus 2Y squared plus Y plus C, this is U.
24:23:200Paolo Guiotto: plus IV,
24:25:840Paolo Guiotto: V is, what is… what was given in the assignment. Y cubed minus 3X square Y.
24:34:480Paolo Guiotto: So, Y cubed minus 3X squared Y.
24:39:350Paolo Guiotto: plus 4XY.
24:43:330Paolo Guiotto: for XY minus X.
24:49:600Paolo Guiotto: Now, this is, yes, it is correct, there is nothing wrong here, but if I want to see this, how is this function in terms of Z,
25:00:720Paolo Guiotto: So I have to recognize Z in that expression, so can I do that?
25:06:990Paolo Guiotto: You see?
25:11:860Paolo Guiotto: So maybe there are terms here.
25:16:320Paolo Guiotto: that could, could stay together. So, in fact, if you take the… these are, they, they should come from some cubic, expansion, these 4 times.
25:27:750Paolo Guiotto: So if we put together… well, let's start to put the cube minus X cubed…
25:34:990Paolo Guiotto: Then we have, the term with X squared is minus 3, X squared Y.
25:40:590Paolo Guiotto: Then we have a… well, with an I here, this, this is with the I. Then we have a plus 3XY squared.
25:50:220Paolo Guiotto: And then we have plus IY cubed.
25:54:450Paolo Guiotto: Well, and then we have, this is quadratic, and also this one.
26:00:200Paolo Guiotto: So, we should say… if we factor the 2, it is X squared minus Y squared.
26:08:00Paolo Guiotto: plus I2XY.
26:11:210Paolo Guiotto: This is easier than we have a Y.
26:15:110Paolo Guiotto: plus Y minus IX.
26:20:60Paolo Guiotto: And plus, finally, a constant. Now, we have to recognize all this, what is in terms of Z. Let's start from this one.
26:27:760Paolo Guiotto: Remind that Z is X plus IY.
26:33:580Paolo Guiotto: If we factorize the minus i, this becomes X.
26:39:320Paolo Guiotto: So, how do you get plus Y?
26:44:490Paolo Guiotto: If, because if I do like this, I have minus i. I want to have Y without the I, and you see that if I put I here.
26:55:310Paolo Guiotto: This product yields exactly minus iX plus minus i square y. I squared is minus 1 with a minus plus, so it is plus I. And this is Z.
27:07:840Paolo Guiotto: So we can have a first term here, down here, minus iZ, while the constant remains a constant.
27:15:570Paolo Guiotto: About this, this comes from the square of Z, because if you do Z square, you get X squared minus Y squared plus the double product I2XY.
27:28:730Paolo Guiotto: So that's exactly Z squared. So here we have plus 2Z squared.
27:35:50Paolo Guiotto: And this should come, maybe, from this… the cube of Z.
27:39:940Paolo Guiotto: Well, if we do the cube of Z,
27:44:610Paolo Guiotto: the cube of X plus IY, we start with X cubed, and here we have a minus X cubed, so it's not the cube of X plus IY, maybe minus X here.
28:00:150Paolo Guiotto: Now, when we do the cube, it comes with the minus, then we have,
28:04:620Paolo Guiotto: the triple product of the square of this minus X squared, so it gives X squared, times this second IY. But you see there is a minus here, so I need also a minus down here to make this exact.
28:22:520Paolo Guiotto: So now I get minus IY, which is exactly this term here.
28:27:940Paolo Guiotto: And continuing with this, then I have a 3-point product, the first, which is minus X, times the square of the second. Now, the minus square is 1. I square is minus 1, so minus Y squared, so we get from this 3XY squared, which is this one.
28:47:80Paolo Guiotto: So it's perfect, this. And finally, the cube of minus i
28:52:270Paolo Guiotto: Y is minus, because the cube remains…
28:56:280Paolo Guiotto: the minus remains. I cubed is minus i, so minus i, y cubed. And this comes Y… IY cubed, so this is exactly minus…
29:08:720Paolo Guiotto: X minus IY cubed, which is, if you carry outside the minus, this becomes plus.
29:16:930Paolo Guiotto: And that's Z. So we have minus Z cubed plus 2Z squared plus, minus… IZ.
29:28:830Paolo Guiotto: Plus constant, and that's the solution.
29:33:810Paolo Guiotto: Okay, guys, time is over, so, do this exercise, the… the…
29:42:120Paolo Guiotto: There are several exercises of this type, maybe I will write an email
29:48:70Paolo Guiotto: Because I want also, to… to… to tell you to start, do these exercises from this, exam set.
29:57:570Paolo Guiotto: So, I will proceed in this way, maybe… I will tell you, let's say, not day by day, but…
30:06:700Paolo Guiotto: Now, we have also the Christmas days, but maybe with a couple of days of interruption, I will give you a first set of exercises to do, then I will publish solution, then I will redo the assignment.
30:21:90Paolo Guiotto: Still based on the exam exercises and published solution.
30:25:340Paolo Guiotto: Maybe we will try to do two or three times before the restart of classes, so in such a way that you can practice even on all the exercises, so you can start practicing for the exam, and checking solutions, okay?
30:43:00Paolo Guiotto: Okay, have a nice holidays, and see you on January the 7th, right? Okay.
30:55:40Paolo Guiotto: Bye-bye.