Complex analysis: Zeta function functional equation

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  • Опубликовано: 28 дек 2024

Комментарии • 18

  • @gunhasirac
    @gunhasirac 3 года назад +12

    One of the best thing about this channel is knowing one of the greatest mathematicians also get confused with signs and some definition often not even trying to hide it. Best respect.

  • @indiscernables
    @indiscernables 3 года назад +16

    This really is the best math channel out there now. We are so lucky.
    (And Borcherds cannot but be pretty hilarious at times. Idkw, but accidental comic relief is so much more satisifying than all the people out there writing scripts.)

  • @brian8507
    @brian8507 3 года назад +22

    You have quickly risen to my favorite math channel. Top notch content bruh!!!

  • @abhinavnatarajan4180
    @abhinavnatarajan4180 3 года назад +34

    "If I don't tell you which way I'm going around the contour you can't catch my sign errors" lol

  • @theflaggeddragon9472
    @theflaggeddragon9472 3 года назад +8

    Thank you for your explanations of the contours. I got totally lost reading Edward's "Riemann's zeta function" for this exact reason.

  • @ethanjahan780
    @ethanjahan780 3 года назад +1

    I love how it's not at all explained rigorously, but still exact

  • @ricardoa70
    @ricardoa70 2 года назад +5

    How do you reconcile the fact that for the integral over the small circle to vanish Re(s)>0 and later for the integral over the big arc to vanish Re(s)

  • @ricardoa70
    @ricardoa70 2 года назад +3

    Profwssor, if I am not mistaking, this is not a Bromwich contour (which is used in inverse Laplace) but a Hankel contour.

  • @uy-ge3dm
    @uy-ge3dm 3 года назад +1

    At 6:55 should the fraction be equal to the geometric series 1 + e^(-z) + e^(-2z) + ... ? Did you forget the 1 or does it not matter

    • @diribigal
      @diribigal 3 года назад +2

      Since there's no 1 term, we have e^(-z)/(1-e^(-z)) which becomes the written 1/(e^z-1) when you multiply by 1 in the form e^z/e^z.

    • @uy-ge3dm
      @uy-ge3dm 3 года назад

      @@diribigal Ah, I see, thank you. I didn't see that negative sign.

  • @kharnakcrux2650
    @kharnakcrux2650 Год назад

    This fascinated me years ago, when i was able to relate the Sine function in terms of reflections of Gamma and Zeta. Creepy almost... that in a way, prime numbers are a subtle hidden element in the Trig functions.

  • @stigenbeat568
    @stigenbeat568 Год назад

    WE LOVE YOU RICHARDS

  • @Salmanul_
    @Salmanul_ 3 года назад +2

    I feel confused, I keep getting sign errors or are there sign errors in the video?

    • @brian8507
      @brian8507 3 года назад +8

      Probably both

    • @valeriobertoncello1809
      @valeriobertoncello1809 5 месяцев назад

      He did say preemptively that he gets the sign wrong 50% of the times 😂

  • @Sharpgamingvideos
    @Sharpgamingvideos 3 года назад +1

    Awesome

  • @migarsormrapophis2755
    @migarsormrapophis2755 3 года назад +1

    yeee