I didn't write it down again but of course, in the last statement, you still need the total differentiability of f_R to get the equivalence. This is something you should not forget :)
Really liked how you briefly summarized the concepts from the previous videos and slowly built up the theorem in such an understandable way, sets a new standard for teaching mathematics.
This is so much more elegant than the way I learned in in a mathematics-for-physicists course (where you just demand that f'(z0) has to be the same whether you approach z0 parallel to the x-axis or the y-axis). Very nice series so far!
I love the approach relying strongly on normed vector spaces. It is so clean and modern. Hopefully there will be space to talk about C^n -> C^n in the future and how C -> C generalizes or not in that case. Thanks!
I'm currently taking a class on complex variables, but it is not a proof based one which results in formulations like this being thrown around seemingly out of nowhere. As a student who took real analysis I was struggling a lot because I just wanted to understand the subject. This series is proving incredibly helpful, and I am finally able to get it.
Hey there ! I really wanted to start learning Complex analysis during the summer and really just couldn't settle down to actually read the books and watch the videos. But now , I am able to ! Despite being busy with college and those pesky finals , the fact that those videos are in bite-sized pieces really helps a lot in getting introduced to concepts in complex analysis. Thank you ! :)
@@PunmasterSTP I am still preparing for them! Tommorow I have my Circuit Analysis course , Hopefully it will go fine! I got the LKV and LKC by my side! Hopefully Matrices will help solving the rest!
The connection between the field of R^2 and that of C blew my mind away ! I had to check where id seen that before and it was in a nice algebra book at the chapter of matrices. Connecting my beloved linear algebra with complex analysis is something unexpected and beautiful ! Thanks ! ❤
This is really interesting. Because for real function if the function has no gaps or sharp edges it is always differentiable. But for complex functions as it turns out this is not the case. A simple counterexample is u(x,y) = x with v(x,y) = -y.
The content is on top as always, but haven't you though about adding timestamps to your videos? When you deal with those longer than 10 minutes you may easily forget something important and havin a timestamp to quickly revive the idea could save a lot of time and make all the courses easy to navigate.
@@brightsideofmaths well, since I'm going to watch all of your Complex Analysis playlist (at least I plan to), I can post timestamps in the comments, but in order to let everyone see them you'll have to add them to the description. If that's ok with you, I'll be happy to help
Hello. Thanks for another great video! I have a question; would you mind helping me understand? From what I gather, if f is complex differentiable at a+ib, then it implies that g is totally differentiable at (a,b) where g((x,y)) := [Re{f(x+iy)}, Im{f(x+iy)}], but it's not the case the other way around right (ie g totally differentiable at (a,b) implies f complex differentiable at a+ib)? Also, is there an easier way to tell if a multivariable function is totally differentiable at a point? For example, will a function be totally differentiable at a point if all of its partial derivatives exist at that point? Thanks again for all of your hard work and clear explanations.
Thank you very much. Regarding the first question, you are absolutely right: the other way around you need more. For your second question: totally differentiable is more than just the existence of the partial derivatives.
There is something that is not clear to me. I understood that if f is differentiable then the Cauchy-Riemann equations must be satisfied, but I don’t understand why the converse is true.
If CR equations are satisfied and f_R is totally differentiable, then the Jacobian matrix of f_R is of the symmetric form mentioned in the video. However, this means that we can see the differential as a multiplication with complex numbers. And this is exactly needed for complex differentiability.
Maybe you confuse this with a sufficient criterion: if all partial derivatives exist and are continuous, then the function is totally differentiable. However, this is the actual implication and not the equivalence.
I don't understand the point about complex multiplication. You said that "if we want c-differentiability, we need the jacobian to have this form" implied the jacobian should represent complex multiplication of two complex numbers. But why is it enough at all ? We can imagine complicated complex functions that cannot be represented as multiplications, yet still be differentiable (or maybe not ?) What i understand is that we proved this fact: If the jacobian represent complex multiplication (i omit all technical details), then f is c differentiable. But why should the converse be true at all i don't understand ? Great content anyways, I'll proceed and admit this result for now x)
I apologize if it's not the place, but have you already seen this video on complex integration? ua-cam.com/video/EyBDtUtyshk/v-deo.html I assume it'd be cool if you'd reference it when you get to that subject. Coupled with your lucid explanation of all the lemmas and proofs, some visual intuition is exactly what's needed to create a killer package for an all time best complex analysis course that UA-cam'd ever seen)
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
I didn't write it down again but of course, in the last statement, you still need the total differentiability of f_R to get the equivalence. This is something you should not forget :)
Thank you!
That's because the Jacobian can still be defined (because all partial derivatives are defined) but the function might not be totally differentiable.
Really liked how you briefly summarized the concepts from the previous videos and slowly built up the theorem in such an understandable way, sets a new standard for teaching mathematics.
This is so much more elegant than the way I learned in in a mathematics-for-physicists course (where you just demand that f'(z0) has to be the same whether you approach z0 parallel to the x-axis or the y-axis). Very nice series so far!
00:00 Intro
00:16 Two notions of differentiability
6:39 When we can use vector-matrix multiplication?
9:04 Deriving the Cauchy-Riemann equations
I love the approach relying strongly on normed vector spaces. It is so clean and modern. Hopefully there will be space to talk about C^n -> C^n in the future and how C -> C generalizes or not in that case. Thanks!
I'm currently taking a class on complex variables, but it is not a proof based one which results in formulations like this being thrown around seemingly out of nowhere. As a student who took real analysis I was struggling a lot because I just wanted to understand the subject. This series is proving incredibly helpful, and I am finally able to get it.
Thank you very much! I am glad that these videos can help you :)
This is pretty elegant and I like the connection with the total derivative in R2.
Thank you!
Hey there ! I really wanted to start learning Complex analysis during the summer and really just couldn't settle down to actually read the books and watch the videos. But now , I am able to ! Despite being busy with college and those pesky finals , the fact that those videos are in bite-sized pieces really helps a lot in getting introduced to concepts in complex analysis. Thank you ! :)
How have your finals been going/how did they go?
@@PunmasterSTP I am still preparing for them! Tommorow I have my Circuit Analysis course , Hopefully it will go fine!
I got the LKV and LKC by my side! Hopefully Matrices will help solving the rest!
@@fouadio4108 I wish you the best of luck! Feel free to let me/everyone know how you did in the comments. Or no pressure if you’d rather not share…
Well! It went better than expected! Hopefully I will pass this class!
@@fouadio4108 I am glad to hear that!
The connection between the field of R^2 and that of C blew my mind away ! I had to check where id seen that before and it was in a nice algebra book at the chapter of matrices. Connecting my beloved linear algebra with complex analysis is something unexpected and beautiful ! Thanks ! ❤
Yes, C is just R^2 with an additional multiplication :)
Good video! I really wanted to understand the Cauchy-Riemann equations but struggled with it. However, it felt almost trivial for me in the video!
Glad it was helpful!
Your work is great sir it very helpful for me💯💯 (engineering students)
Your channel is a wonderland, I'll use my brain...💖💖💖
This is really interesting. Because for real function if the function has no gaps or sharp edges it is always differentiable. But for complex functions as it turns out this is not the case. A simple counterexample is u(x,y) = x with v(x,y) = -y.
It seems that this course was published before the multivariable calculus course! nice review!
Thanks! The multivariable calculus course explains a lot more :)
The content is on top as always, but haven't you though about adding timestamps to your videos? When you deal with those longer than 10 minutes you may easily forget something important and havin a timestamp to quickly revive the idea could save a lot of time and make all the courses easy to navigate.
Very good idea! I am always happy when viewers contribute suitable timestamps. Otherwise, I will add them over time.
@@brightsideofmaths well, since I'm going to watch all of your Complex Analysis playlist (at least I plan to), I can post timestamps in the comments, but in order to let everyone see them you'll have to add them to the description. If that's ok with you, I'll be happy to help
@@NewDeal1917 Of course, I happy to copy them :) Thanks!
Could you make a lecture series about the Riemann-Zetafunction?
finally understood where the riemann equations come from
Very nice explanation. which software is used for these presentation?
Xournal :)
@@brightsideofmaths iPad or digital tablet?
@@aarian4908 Xournal is the software and I use it in Linux, in combination with a graphic tablet.
it would be awesome if we could have a video every day :))
That would be awesome for sure. However, making a video takes a lot of time. At the moment I produce as many videos as I can.
How is the total differentiability supposed to add up when J((x,y,) - (x0, y0)) is a 2x2 matrix and the other entries are 2 element vectors?
J ((x,y,) - (x0, y0)) is a 2d-vector. Only J is a 2x2 matrix.
Are you following a textbook when you do these videos? i.e. how do you decide what to cover?
No, I don't use a particular book. I have a lot of my own lecture notes and a bunch of books about different topics.
@@brightsideofmaths thank you 😊
Would you mind creating a video explaining the terms we are using here, most of it is new even after the real analysis series
My multivariable calculus series should help here: tbsom.de/s/mc
@@brightsideofmaths thank you
Cauchy-Riemann? More like "Gee, that was fun!" Thanks for another very well put-together and intriguing video.
Hello. Thanks for another great video! I have a question; would you mind helping me understand? From what I gather, if f is complex differentiable at a+ib, then it implies that g is totally differentiable at (a,b) where g((x,y)) := [Re{f(x+iy)}, Im{f(x+iy)}], but it's not the case the other way around right (ie g totally differentiable at (a,b) implies f complex differentiable at a+ib)? Also, is there an easier way to tell if a multivariable function is totally differentiable at a point? For example, will a function be totally differentiable at a point if all of its partial derivatives exist at that point? Thanks again for all of your hard work and clear explanations.
Thank you very much. Regarding the first question, you are absolutely right: the other way around you need more.
For your second question: totally differentiable is more than just the existence of the partial derivatives.
It was awesome, thank you very much!
Thank you too!
Great video! Thank you!
Glad you liked it!
Thank you! Brilliant as always :)
Why is complex analysis so fun?!
You will see :)
@@brightsideofmaths Cauchy's integral formula is my favorite result in elementary complex analysis
I don't know, I don't know
You stick around, now he may show
@@jaimelima2420 I had to look that reference up!
I'm not sure there is just a simple answer. I think the reason may be more...complex 😎🤦♂
There is something that is not clear to me. I understood that if f is differentiable then the Cauchy-Riemann equations must be satisfied, but I don’t understand why the converse is true.
If CR equations are satisfied and f_R is totally differentiable, then the Jacobian matrix of f_R is of the symmetric form mentioned in the video. However, this means that we can see the differential as a multiplication with complex numbers. And this is exactly needed for complex differentiability.
u and v must have their partial derivatives continuous at (x0, y0) too for it to be an equivalence, otherwise it is just an implication, I think.
Why do you think that?
@@ilyboc No, for the total differentiability, we don't need continuous partial derivatives.
Maybe you confuse this with a sufficient criterion: if all partial derivatives exist and are continuous, then the function is totally differentiable. However, this is the actual implication and not the equivalence.
@@brightsideofmaths Oh wait f_R is declared as a totally differentiable map so it's implied that the partial derivatives of u and v are continuous.
@@ilyboc As I said: your implication is not correct :)
im 15 rn learning complex analysis and i gotta say its the most fun type of math i have ever done lol
in the proof of CR equation we only consider equality of derivative from two independent directionly(x and y) what about other direction?
I have a video about directional derivatives in my Multivariable Calculus playlist. This might clear things up!
I don't understand the point about complex multiplication.
You said that "if we want c-differentiability, we need the jacobian to have this form" implied the jacobian should represent complex multiplication of two complex numbers.
But why is it enough at all ? We can imagine complicated complex functions that cannot be represented as multiplications, yet still be differentiable (or maybe not ?)
What i understand is that we proved this fact:
If the jacobian represent complex multiplication (i omit all technical details), then f is c differentiable.
But why should the converse be true at all i don't understand ?
Great content anyways, I'll proceed and admit this result for now x)
Thank you very much :)
Differentiability means that there is a good linear approximation. This is described with using a Jacobian matrix.
I apologize if it's not the place, but have you already seen this video on complex integration? ua-cam.com/video/EyBDtUtyshk/v-deo.html
I assume it'd be cool if you'd reference it when you get to that subject. Coupled with your lucid explanation of all the lemmas and proofs, some visual intuition is exactly what's needed to create a killer package for an all time best complex analysis course that UA-cam'd ever seen)