Cauchy Riemann equations

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  • Опубликовано: 27 окт 2024

Комментарии • 81

  • @blackpenredpen
    @blackpenredpen 7 лет назад +51

    It's the Oreo-Cauchy-Riemann-Peyam Equations!!

  • @Kingradek2
    @Kingradek2 7 лет назад +37

    "X not"
    "Why not?"
    :DDD
    I always think that he is having a weird conversation with himself :'DDD

    • @drpeyam
      @drpeyam  7 лет назад +7

      Kingradek2 Who says I'm not? Hahaha

    • @koenth2359
      @koenth2359 6 лет назад

      9:00 Why... ¿Why not?

  • @ibonitog
    @ibonitog 7 лет назад +9

    Awesome video! Love your style of teaching, I'd be so grateful if I had a Prof like you!

  • @tomatrix7525
    @tomatrix7525 4 года назад +1

    I’m on a marathon watching your videos, Really enjoying them.

  • @abhishekchakraborty2316
    @abhishekchakraborty2316 4 года назад +1

    Your enthusiasm is contagious. I love it.

  • @Fematika
    @Fematika 7 лет назад +20

    Yes!! Now prove it in the other direction!

    • @gregorio8827
      @gregorio8827 6 лет назад

      What does the other direction says?

    • @xdandrewbxify
      @xdandrewbxify 5 лет назад +3

      @@gregorio8827 If u and v have continuous derivative in neighborhood of z and they satisfy Cauchy- Riemann equations, then f has derivative in z.

  • @Bbb78651
    @Bbb78651 9 месяцев назад

    Brilliant, and hillarious. I haven't ever laughed more than in your lecture, Dr. Peyam! As another comment said, your enthusiasm is amazing. Glad we made it through the marathon! SUCH A good video. Thank you Dr. Peyam.

  • @dalitas
    @dalitas 6 лет назад

    This kind of maths is exactly what I enjoy, I always though numberphile did a too broad a perspective, but your channel has all the complex analysis I could ever need

  • @iilugs
    @iilugs 7 лет назад +2

    Hey doctor Peyam! great video as always, I love them! Well, I have a question for you. It´s a little bit paradoxical, and I wolud love if you would tell me where's the mathematical flaw.
    So, the taylor Serie for e^x = 1/0! + x/1! + x^2/2! + x^3/3! ... and so on.Based on that, substituting x by 1, we can say that e is equal to the sum of the reciprocals of the factorials of the natural numbers, including 0.
    Thus (i feel very eloquent when I use this expression), we can say that:
    e = 1 + 1 + 1/2! + 1/3! + 1/4!.... (adding -2 in both sides and, on the right side, multiplying and dividing by 1/2)
    e - 2 = 1/2(1+1/3+ 1/(4*3) + 1/(5*4*3) . . . ) (multiplying and dividing by 1/3)
    e - 2 = 1/2(1+1/3(1+1/4 + 1/(5*4) + 1/(6*5*4) . . .)
    if we kept dividing and multiplying by each reciprocal of natural number, as n factorial includes every natural number up to n, we can assure that said reciprocal will be one of the factors.
    With this in mind, we can say that :
    e-2 = 1/2/1+1/3(1+1/4(1+1/5(1+1/6(......
    Multiplying a number times a fraction is the same as dividing the number by the reciprocal of the fraction, so, e can say that
    e-2 = ...)+1)/4+1)/3+1)/2 + 2 (sorry, I couldn't manage to upload a pic of the fraction to the commet.
    Ok, now that we've established that e = 2 + infinite fraction lets save that knowledge for later.
    Now, let me talk about a particularly adorable piece of math, that is ramanujans funny square roots.
    Srinivasa Ramanjan (an indian mathematician from the 20th century, as you probably know) once wrote that:
    sqrt(1+2*sqrt(1+3*sqrt(1+4*sqrt(1+.... = 3
    It is a mathematical fact, if you want to see the proof, follow the link ruclips.net/video/leFep9yt3JY/видео.html .To prove this apparently ridiculous claim, ramanujan kind of decomposed 3 in an infinity of square roots (view the link for more information). He started from finitest of numbers, and turned it into the result of an infinty of processes. I propose we do this but, instead of square roots, we use fractions
    Well, 3 = 2 +1 = 2+ 2/2 = 2 + (1+1)/2 = 2 + (1+3/3)/2 = 2 + (1+(1+2)/3)/2 = 2 + (1+(1+8/4)/3)/2 = 2 + (1+(1+(1+7)/4)/3)/2 ... and so on.
    If we repeated this process to infinity , we would get the same fraction that equals e !!! That means that 3 = e, and we could do this "fraction expansion" with any other natural number bigger that 3 and get the exact same fraction! that means e = 4, and e=5, and that is impossible, and I can't find the error in my reasoning.
    I would love if you could point out the error.
    Anyways, thanks a lot for reading, and please continue doing your amazing work, you are a super likeable person, and I love your demonstrations. Also, the video about the half derivative absolutely blew my mind! Great Job!
    Ps: I would lovve if you could prove L'Hôpital's Rule, I cant get the demonstration that's on wikipedia
    PPs: I would also love to see you proof Euler's solution for the Basil Problem ((pi^2)/6 = sum of the reciptocals of the perfect squares)

    • @drpeyam
      @drpeyam  7 лет назад +7

      Hi Luis!
      Wow, this is an absolutely amazing idea, and believe it or not, but I actually read through everything :) You are right, strictly speaking there are no algebra mistakes in your demonstration, but although I'm not 100 % sure, I think the problem is that if you continue like that, your terms are actually getting bigger and bigger, because notice that the numerators in your fractions are blowing up: First you have 8/4, and then in 1+7 you get 35/5, and then you get (1 + 34)/5 = (1 + 204/6))/5, so it seems at first sight that you're expanding out e, but I think your process will just blow up to infinity, and at each term you'll have a positive remainder, which is why at the end you don't get 3 = e, but 3 = e + a positive number (which should be hopefully 3-e). Hope that helps!

  • @cedricp.4941
    @cedricp.4941 7 лет назад +7

    Hi ! Nice video :)
    I have two questions:
    - Is the converse of this proposition true ?
    - You can only split the limit if you know that the two limits you get exist. Here, how do you prove that Ux and Vx exist ?
    Thank you !

    • @leonardromano1491
      @leonardromano1491 7 лет назад +7

      Actually the converse is generally not true. It's only true if the function is also a harmonic function meaning it solves d²/dx² u +d²/dy² v=0.
      Also the limit can be split because f is differentiable and therefore it's partially differentiable (It needs to be to be differentiable), so the limits exist (because they are just the partial derivatives).

    • @desmondc2706
      @desmondc2706 7 лет назад

      Leonard Romano the Laplace Equation should be Uxx + Uyy = 0 right?

    • @leonardromano1491
      @leonardromano1491 7 лет назад

      Yes in this notation :)

    • @fanism90
      @fanism90 5 лет назад +1

      On your second question, that's how the metric works in Complex numbers. A fuction f while z->z0 has limit a complex number x +iy if and only if the real part of f limits x and the imaginary part of f limits y. That's what he did on the video, he split the limit on the real and imaginary part of the quotient function which both have limit because the original has limit.
      More generally, a sequence of complex numbers z_n, n=1,2,... limits z_0 if and only if the real part of z_n limits the real of z_0 and the imaginary of z_n limits the imaginary of z_0.

  • @dashbaljinbishuubazar1466
    @dashbaljinbishuubazar1466 5 лет назад +1

    Thank you very much for the lecture. love the efforts you are putting in. Appreciate it.

  • @Channel-zb1fi
    @Channel-zb1fi 9 месяцев назад

    Why is it sufficient to only check the vertical and horisontal direction in order to guarantee the differentiability of a function. Should it not be necessary to check all possible paths to z0?

  • @deeptochatterjee532
    @deeptochatterjee532 6 лет назад +1

    Wow, I just finished Calc III and I'm amazed that I understood this

  • @BuddyVQ
    @BuddyVQ 3 года назад

    Love your enthusiasm! Keep it up!

  • @camilomuianga7865
    @camilomuianga7865 2 года назад

    Hey Dr peyam, in first hand i'm congratulation for this interesting vidie, however, i Just came out with a doubt and i'd be glad with you clarify it!
    Well, cauchy-Riemann equations we can related it with analytic functions right?
    With we succed to see that the Cauchy-Riemann equations was sutisfied is that enough to we say is analytic functions?

    • @camilomuianga7865
      @camilomuianga7865 2 года назад

      If* sorry by mistaking here i swiched if by with.

  • @yousifsalam
    @yousifsalam 3 года назад

    is it sufficient to choose two paths and if equal we say the limit exists and therefore the derivative exists? Or do we need more to prove that the limit is the same across all paths?

  • @francribaj3501
    @francribaj3501 6 лет назад +2

    Thank you!Lesson very much appreciated subbed.:)

  • @BarkanUgurlu
    @BarkanUgurlu 5 лет назад +1

    Hi Dr. Peyam, great explanation. Could you please make a video on Lipschitz condition? Greetings from Istanbul.

    • @drpeyam
      @drpeyam  5 лет назад +1

      The ODE Existence Uniqueness Theorem Video has some Lipschitz in it

    • @BarkanUgurlu
      @BarkanUgurlu 5 лет назад

      @@drpeyam Thank you for the prompt reply. I will check it out.

  • @weinihao3632
    @weinihao3632 5 лет назад +1

    Dear Dr. Peyam! Your videos are wonderful and I would love to watch them all. Leider sind einige davon, darunter auch die gesamte Liste zum Fehlerintegral, nicht abrufbar, da sie als privat markiert wurden. Cela est extremement regrettable! Do you anticipate to publish these videos in the future?
    Sigan con el buen trabajo- Vous etes un merveilleux professeur! PS: Wie heisst ihr Kaninchen?

    • @drpeyam
      @drpeyam  5 лет назад +1

      Ja genau, diese werden später veröffentlicht! Und mein Hase heißt Oreo 🐰

    • @drpeyam
      @drpeyam  5 лет назад

      Is there one you’re interested in particular? I could send you the link!

    • @weinihao3632
      @weinihao3632 5 лет назад

      @@drpeyam I'm particularly interested in topology & functional analysis, but all your videos are well worth watching and I'm very happy to hear that you did not de-publish the currently private ones but are rather going to publish them later :) Greetings to Oreo (now I finally understand the related jokes in the comments), I'm sorry that I mistook him for a rabbit.

  • @leonine5216
    @leonine5216 3 года назад

    Thanks you so much, the explanation is very helpful 🙏

  • @desmondc2706
    @desmondc2706 7 лет назад +4

    wow!Thats so clear. Thank you! Can we have something like Cauchy integral formula next time?

    • @drpeyam
      @drpeyam  7 лет назад +1

      Desmond chen Great idea! Thank you :)

  • @michaelpurtell4741
    @michaelpurtell4741 2 года назад

    Is that Black pen filming and if so do you both teach at same institution. By the way excellent presentation (must be cause I understood it and I am not so good at math)

    • @drpeyam
      @drpeyam  2 года назад

      Yep he’s filming it! And we’re not quite at the same institution, he’s in LA and I used to be in Irvine

  • @OndrejPopp
    @OndrejPopp 4 года назад

    So, if the limits approaching Z0 are equal for both dimensions, does this also mean that the limit approaching Z0 for any linear combination of the unity vectors of these dimensions remains the same same? So, I really needed to think about that, and after I did, I say, this only holds if the unity vectors are independent of each other, which is the case here, but otherwise, if you have a dependent coordinate system it may not be enough to treat these dimensions separately... That's what I think. Right?

    • @drpeyam
      @drpeyam  4 года назад

      Mmmmh, in general, the limit existing in the x and y directions does not imply the limit exists in all directions, such as lim (x,y) goes to (0,0) of xy / x^2 + y^2

    • @OndrejPopp
      @OndrejPopp 4 года назад

      @@drpeyam Ok, tx!

  • @plaustrarius
    @plaustrarius 5 лет назад

    Little late to the party, I'm trying to learn a bit about holomorphic functions. I'm trying to convince myself of the polar form of the CRE, would love to see a video like this that proves it from more basic principals! (likely you have one already haha!)
    Blown away that z^i is holomorphic and so is a^z! (only did the example for a being positive and real so far.)

  • @isaaconyach9401
    @isaaconyach9401 6 лет назад +2

    I like your enthusiasm, now i feel like scoring some coke

  • @koenth2359
    @koenth2359 4 года назад

    Around 11:00 you split one limit into two limits. Is this always a valid step? What if the limit of the sum exists, but the separated limits do not?

  • @jayjayf9699
    @jayjayf9699 3 года назад

    I was thinking if the criteria of the partial derivatives have to equate from two paths, why isnt this applied also in vector calculus, as both situations are similar

  • @coffiberengerhoundefo1259
    @coffiberengerhoundefo1259 4 года назад

    i like this video thanks, please you give some explanation on stereographique projection?

  • @y2536524
    @y2536524 7 лет назад

    Is there any chance that you can you give an overview of either fourier analysis or functional analysis if possible

  • @anon8109
    @anon8109 6 лет назад

    What is the definition of limits in complex numbers? Does the usual epsiolon-delta definition apply, taking absolute values of complex rather than real numbers?

    • @drpeyam
      @drpeyam  6 лет назад +1

      Correct! It’s the same def as for real numbers, except | | means absolute value for complex numbers

    • @anon8109
      @anon8109 6 лет назад

      So is the statement in the video @3:45, that no matter how you approach the limit must be the same, a named theorem, or is it so trivial that it doesn't deserve a name?

  • @wyvernmonarch7
    @wyvernmonarch7 7 лет назад

    Can you prove, without complex numbers, that the Taylor series of sin and cos centered at 0 equal the cos and sin function as the number of terms approaches infinity

  • @Myrslokstok
    @Myrslokstok 6 лет назад

    Can you prov the other part, the backward thing.

  • @seoexperimentations6933
    @seoexperimentations6933 7 лет назад +3

    Riemann bigger beast than oreo

  • @helloitsme7553
    @helloitsme7553 6 лет назад

    Challenge: find a formula for sin(a+bi) and cos(a+bi) . (I used Euler's identity and the angle sum formula)

  • @IceCreams62
    @IceCreams62 7 лет назад

    Why do you check on the paper what are you writing on table ?

    • @drpeyam
      @drpeyam  7 лет назад

      IceCreams62 It's just to make sure I didn't make a mistake when writing it down on the blackboard and also to remind myself what the next step is!

    • @IceCreams62
      @IceCreams62 7 лет назад +1

      Thanks a lot for clarification :-) !!!

    • @DendrocnideMoroides
      @DendrocnideMoroides 2 года назад

      It is called blackboard not "table"

  • @maximilianmueller4707
    @maximilianmueller4707 5 лет назад

    I have Gänsehaut i am so excited because it make so much Sense i totally Love it thanks 3,14-am you are the best Teacher i ever had
    In Addition it is so funny y0 Sounds like why Not like why Not we i can do it this way

  • @xdandrewbxify
    @xdandrewbxify 5 лет назад

    Why are so many mathematicians lefthanded?

  • @piyalikarmakar5099
    @piyalikarmakar5099 3 года назад

    cauchy came up with the 1st eqn and riemann with the 2nd one.....

  • @damiandassen7763
    @damiandassen7763 5 лет назад

    9:44 why not?

  • @TheMauror22
    @TheMauror22 6 лет назад +1

    Next semester I'll take complex analysis wish me luck!!!

    • @drpeyam
      @drpeyam  6 лет назад +2

      May your luck be holomorphic :)

  • @RoTShift
    @RoTShift 6 лет назад

    The line through your Q at the beginning is on the wrong side lol

  • @sedmochill4707
    @sedmochill4707 5 лет назад +1

    Kya hai..kya hai ye🤣🤣
    Intro toh baut khtarnak hai

  • @harshitbamotra2013
    @harshitbamotra2013 4 года назад

    What is "ZOT ZOT"

    • @drpeyam
      @drpeyam  4 года назад

      It’s what we say at UC Irvine

    • @harshitbamotra2013
      @harshitbamotra2013 4 года назад

      @@drpeyam could you please confirm something I have sent you.

    • @harshitbamotra2013
      @harshitbamotra2013 4 года назад

      please check your twitter dm

  • @swift3564
    @swift3564 4 года назад

    15:02

  • @Amantheparadise
    @Amantheparadise 3 месяца назад

    Dr.pie yyum

  • @HERŞEYKÖTÜ
    @HERŞEYKÖTÜ 6 лет назад

    nenen öle peyam, hele şu ağza bak hiç ayran içmiş

  • @hamzalaraib9468
    @hamzalaraib9468 4 года назад

    plz whatsapp its notes to me

    • @drpeyam
      @drpeyam  4 года назад +1

      Brown and Churchill Chapter 1

    • @hamzalaraib9468
      @hamzalaraib9468 4 года назад

      thanks but if you send to me it is your goodness.

    • @hamzalaraib9468
      @hamzalaraib9468 4 года назад

      plz tell me book name and its edition