Singularities of Analytic Functions -- Complex Analysis 20

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  • เผยแพร่เมื่อ 16 ต.ค. 2024

ความคิดเห็น • 26

  • @ingobojak5666
    @ingobojak5666 2 ปีที่แล้ว +32

    The limit at 21:21 should be r->0, not r->infinity. Then M times a positive power of r (for negative n) goes to zero as required.

    • @thomashoffmann8857
      @thomashoffmann8857 2 ปีที่แล้ว +2

      You were few minutes quicker than my post 😉

    • @JM-us3fr
      @JM-us3fr 2 ปีที่แล้ว +2

      Agreed, r was chosen to be less than R, so it doesn't make sense if it surpasses R towards infinity.

    • @ediltonbrandao8407
      @ediltonbrandao8407 ปีที่แล้ว +1

      I was stuck in this part of the video, I went to the comments to see if anyone else had found the error

    • @StratosFair
      @StratosFair 11 หลายเดือนก่อน +1

      Thanks, got me confused for a second

  • @Jeff_Saunders
    @Jeff_Saunders 9 หลายเดือนก่อน +2

    Nitpick about 22:30 (the second of the three equivalent statements): g isn't defined at z_0. Shouldn't we instead say that g has a removable singularity at z_0?

  • @TheEpicProOfMinecraf
    @TheEpicProOfMinecraf 2 ปีที่แล้ว +2

    I just wanted to say that these course videos have been awesome. I haven't been able to keep up super well with the material during my term, but now that the summer is rolling around...
    :)

  • @JM-us3fr
    @JM-us3fr 2 ปีที่แล้ว +2

    I hope we have a video on common meromorphic function examples. Not like exponential or trig functions, but weird functions like the Gamma function

  • @thomashoffmann8857
    @thomashoffmann8857 2 ปีที่แล้ว +2

    21:40 r is less than R, so it shouldn't go to infinity. For n

  • @raoufallani5142
    @raoufallani5142 2 ปีที่แล้ว +2

    Thank you professor great work

  • @jimallysonnevado3973
    @jimallysonnevado3973 2 ปีที่แล้ว +3

    I'm a bit confused on the definition of isolated singularity. Wouldn't that make any point in an analytic function be an isolated singularity? Because, if a function is analytic and we just consider the point z_0 ( even if f is analytic on z_0), we can definitely find a punctured disk where f is also analytic. Or am I missing something? Does the fact that we are considering a punctured disk mean that we don't really care whether f is analytic on z_0 or not?

    • @jimallysonnevado3973
      @jimallysonnevado3973 2 ปีที่แล้ว +1

      To clarify my question. Take for example f(z)=z^2 and take the point z=0. Is 0 an isolated singularity? Since f(z) is analytic everywhere, it must also be analytic on the punctured disk of unit 1 from origin because we just removed the point 0 from the disk and it is obviously analytic in the other points from there? Does this also mean that if a function is analytic everywhere then every point is isolated singularity?

    • @strikeemblem2886
      @strikeemblem2886 2 ปีที่แล้ว +1

      @@jimallysonnevado3973 That is more or less correct. The correct phrasing is: z0=0 is a removable singularity of the function f:C\{0} -> C, f(z) = z^2... So if you were tasked to extend f to an analytic function from C->C, you will set f(0) = 0. But this is slightly cheating, because we started with C->C and deleted the point z0=0.
      .
      In the business of isolated singularities, you should really keep in mind a different setup: You are first given f:D\{z0}->C, i.e. f is only defined on a punctured disc D\{z0}. We call z0 a singularity because f is undefined at z0. f(z) = complicated. You are then asked to identify what kind of isolated sing is z0. If z0 turns out to be a removable sing, you are further tasked to extend f to D->C.

    • @jimallysonnevado3973
      @jimallysonnevado3973 2 ปีที่แล้ว +1

      Ohh, I got it already. The way he defined f, implicitly stated that the domain of f is the punctured disk. My confusion arosed because what I thought was that f has some domain where it is analytic (This domain is potentially bigger than the punctured disk.) and the punctured disk is only inteded to be a subset of this domain. My understanding is that if you can find a subset of the domain which is a punctured disk around some z_0 where it is analytic then z_0 is an isolated singularity.

    • @anshumanagrawal346
      @anshumanagrawal346 ปีที่แล้ว +1

      I guess the assumption is that f doesn't exist at z0

    • @manstuckinabox3679
      @manstuckinabox3679 ปีที่แล้ว +1

      @@jimallysonnevado3973 it has to be a singularity for it to be an isolated one. I had the same thought.

  • @sinecurve9999
    @sinecurve9999 2 ปีที่แล้ว

    Riemann Zeta function here we come!

  • @thomashoffmann8857
    @thomashoffmann8857 2 ปีที่แล้ว

    Around 28:00
    The preconditions of the theorem is maybe not enough. It has to be an isolated singularity which is not removable. Otherwise the limit doesn't need to be infinity (?)

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 ปีที่แล้ว

      The theorem says that if the isolated singularity z_0 is a pole then the limit is infinity. Here 'pole' means pole of positive order so a 'pole' is by definition not a removable singularity.
      (removable singularities can be thought of as "poles of order 0", but that is not being considered here)

  • @broccoloodle
    @broccoloodle ปีที่แล้ว

    30:32 It looks like Michael assumed that 1/|f(z)| has a isolated zero at z_0?

    • @DendrocnideMoroides
      @DendrocnideMoroides ปีที่แล้ว

      yes if f(z) has an isolated singularity at z_0 then 1/f(z) has an isolated zero at z_0, pretty obvious

  • @Invalid571
    @Invalid571 2 ปีที่แล้ว +1

    Thank you for the awesome lessons/content!
    Can we also expect to see a course on differential geometry?
    I know I'd love to.
    :)

  • @noumanegaou3227
    @noumanegaou3227 2 ปีที่แล้ว

    Please abstract linear algebra

  • @hyperduality2838
    @hyperduality2838 2 ปีที่แล้ว

    Convergence is dual to divergence.
    Poles (eigenvalues) are dual to zeroes -- optimized control theory.
    Analytic (mathematics) is dual to synthetic (physics) -- Immanuel Kant.
    "Always two there are" -- Yoda.

    • @focusmaestro4013
      @focusmaestro4013 ปีที่แล้ว

      ok sir, noted

    • @hyperduality2838
      @hyperduality2838 ปีที่แล้ว

      @@focusmaestro4013 Symmetric matrices (real eigenvalues) are dual to anti-symmetric matrices (complex eigenvalues) -- linear algebra.
      Real numbers are dual to complex numbers -- complex numbers are dual.
      Bosons or symmetric wave functions are dual to Fermions or anti-symmetric wave functions -- spin statistics theorem.
      Bosons are dual to Fermions -- atomic duality or wave/particle duality.
      Homology (convergence, syntropy) is dual to co-homology (divergence, entropy).
      Syntropy (prediction) is dual to increasing entropy -- the 4th law of thermodynamics!
      The 4th law of thermodynamics is hardwired into mathematics and mathematical thinking.
      Duality creates reality!
      Teleological physics (syntropy) is dual to non-teleological physics (entropy).