Gauss' Mean Value Property and the Maximum Modulus -- Complex Analysis 11

Поделиться
HTML-код
  • Опубликовано: 13 дек 2024

Комментарии • 24

  • @vladimir10
    @vladimir10 2 года назад +16

    13:40
    Shouldn't that limit be taking r to 1 from below rather than to 0 from above?
    Obviously, it doesn't change the calculations, but I don't get the logic behind it.

    • @StratosFair
      @StratosFair Год назад

      Yeah I'm confused at this part too. I think you're probably right

    • @GroundThing
      @GroundThing 4 месяца назад +2

      He didn't explain it well, but essentially it's that Gauss' MVP is true for any arbitrary radius within the domain. So ln(1-1*e^(-i*2*theta)) = ln(1-r*e^(-i*2*theta)) for an arbitrary r in the domain, so can be taken by a limit to 0+

    • @catfromlothal8506
      @catfromlothal8506 4 месяца назад

      I mean if we even try to take limit r->0+ without changing the negative sign
      can't we just take r = 1

  • @lucasm.b.4390
    @lucasm.b.4390 2 года назад +7

    It is usually not true that Log(z1z2)=Log(z1)+log(z2), because the sum of arguments could be outside the main branch (-π,π]. I am not sure I understood why your example of Log(sinθ) holds.

  • @eamon_concannon
    @eamon_concannon 2 года назад +1

    18:33 u takes its maximum value at z_0 not z_1 since the value of λ chosen in defining the function u was to make u(z_0) = M.
    20:13 The first set is contained in the second set. They are not necessarily equal. If u(z)

    • @catfromlothal8506
      @catfromlothal8506 3 месяца назад

      I sorta see λ as λ_0 for z_0, λ_1 for z_1...
      so u(z) is like a function which finds any λ that works automatically for it's input and hence a constant in essence.

  • @PunmasterSTP
    @PunmasterSTP 2 года назад +1

    Maximum modulus? More like "More great videos for us." Thanks for putting together everything on this channel!

  • @StratosFair
    @StratosFair Год назад +1

    13:40 can anyone explain how this limit makes sense ? If we made r go to 1 I would agree but to zero it doesn't make sense to me...

  • @half_pixel
    @half_pixel 2 года назад +1

    Very confused about 24:53. What does "achieved this setup" mean?

    • @nicholasvazquez9765
      @nicholasvazquez9765 2 года назад +1

      By "achieved this setup", he is referring to the fact that z being in the boundary of D satisfies the desired condition to prove the Maximum principle immediately. We wanted to show that our function achieves its maximum modulus on the boundary, and this case considers only times when z is on the boundary. Hopefully that helps!

    • @half_pixel
      @half_pixel 2 года назад +1

      @@nicholasvazquez9765 ah, I see - I think the thing that was confusing me is that the M in the proof is defined as the maximum of h on the closure of D, where the M in the theorem as written is the maximum of h on the boundary. Thanks for the explanation!

  • @David-bh7hs
    @David-bh7hs 2 года назад +4

    I really like how you explain which theorems and definitions you use for your proofs! The only critique I can think of is your reliance on terms like "but then" or "but that means", etc. It just prefer saying "so that" instead because it makes me more certain in my wording, but either way, great videos!

  • @pharaohgarmar5611
    @pharaohgarmar5611 2 года назад +2

    I think it would have been helpful in the exposition of the strict maximum value principle if you could have reminded us that the real part function is harmonic. Also some signposting in that exposition would have been helpful.
    I think also there is an error at around minute 30 because the modulus of 3+2isin(theta) equals sqrt(9-4sin2(theta)) not sqrt(9+4sin2(theta)).

  • @robshaw2639
    @robshaw2639 Год назад +1

    There's something I don't get about this Mean Value Property.... It's a circle around z0. Doesn't the Cauchy Integral formula for a closed loop say the value should be 0 and not f(z0)? What am I missing?

    • @fartsniffa8043
      @fartsniffa8043 Год назад +2

      There are two theorems involving integrals and Cauchy's name, namely (no pun intended):
      1. The Cauchy Integral Theorem, which states as you said, that the integral of an analytic function in a domain D over a closed loop is equal to 0. Provided that that loop is entirely in the domain D (this theorem is used to prove the next theorem).
      2. The Cauchy Integral Formula (which is used to derive this Mean Value Property), which says that the value of an analytic function f(z), is equal to the closed integral over a loop (again provided that the loop is in the domain in which the function is analytic) of (1/2*pi*i * f(w)/w-z) where w is the dummy variable being integrated over; the key thing here being that f(w)/w-z is clearly not analytic at z, so there is no contradiction with (1).
      So in summary, (1)C.I.T is used to prove (2)C.I.F which is then used to prove G.M.V.P . Now your statement is perfectly reasonable, and it leads me to believe that the integrand used in the Gauss's Mean Value Property is not analytic, which makes sense since the exponential in the function is not 1-1 in the domain.

    • @robshaw2639
      @robshaw2639 Год назад

      @@fartsniffa8043 Thanx for the detailed explanation - with CIF, we are picking a function that intentionally has a problem inside the loop, and then don't get 0, but instead a different specfic value...

  • @cah220
    @cah220 2 года назад +2

    Is your "Strict Maximum Principle" simply a restatement of Liouville's theorem, namely that a bounded entire function is constant?

    • @ConManAU
      @ConManAU 2 года назад +4

      It’s a generalisation which allows the domain to be something other than the whole complex plane.

  • @JoeBrown-w1g
    @JoeBrown-w1g Месяц назад

    The notation of u(z_0) is really misleading (and confusing) because z_0 is a complex, while u is supposed to be a real function (i.e., a function allowing real numbers (or elements of R^n) as input). It makes it seem like u sometimes accept complex numbers as input.

  • @romajimamulo
    @romajimamulo 2 года назад +1

    11:54 oops, you doubled the range by accident
    EDIT: fortunately fixed

  • @romajimamulo
    @romajimamulo 2 года назад +2

    20:17 but can't U be negative? We only took the real part, no absolute values

  • @raiza.chakufora5271
    @raiza.chakufora5271 8 месяцев назад

    I have attempted to find the maximum value of |z^4+z^2+1| and getting √5 as the maximum value. Can it be correct?

  • @eamon_concannon
    @eamon_concannon 2 года назад

    6:15 Shouldn't we have dx = -rsinθ dθ + cosθdr and dy = rcosθdθ + sinθdr? I guess dr = 0 here as we keep r fixed as we integrate w.r.t. θ.