Modular forms: Fourier coefficients of Eisenstein series

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  • Опубліковано 20 гру 2024

КОМЕНТАРІ • 31

  • @Zeitgeist9000
    @Zeitgeist9000 3 роки тому +28

    I'm not even a math student but I'm finding myself excited each day for a new video on this channel. It's this and the latest Marvel episode.

    • @evankim4096
      @evankim4096 3 роки тому +17

      I haven't watched the first episode of the Falcon and the Winter Soldier, but I've watched all three videos of modular forms so far

    • @justanotherman1114
      @justanotherman1114 3 роки тому +7

      I dont want to spoil for you but future episodes of modular form will be great.

    • @Zeitgeist9000
      @Zeitgeist9000 3 роки тому +1

      @@justanotherman1114 Haha, looking forward to them!

  • @FractalMannequin
    @FractalMannequin Рік тому +4

    I was on the bed with flu, kind of bored, so I picked up a pen, some paper and started watching your modular forms lectures. Everything is motivated and linked to the other concepts. This is how math should be taught.

    • @goldjoinery
      @goldjoinery Рік тому +2

      Prof Borcherds is a class act.

  • @gavinyu4351
    @gavinyu4351 Рік тому +1

    Best explanation for j-invariant I have seen so far! Weight 0 actually turns it into an invariant.

  • @stenzenneznets
    @stenzenneznets 3 роки тому +3

    Thank you Dr. Borcherds!
    It is possible to find more information about this "historical mistake"? 22:58

    • @positivearrow
      @positivearrow Рік тому +2

      Check the last video in the series, "Modular forms: Petersson inner product" at 8 min.

  • @rgicquaud
    @rgicquaud 3 роки тому

    I was a bit skeptical at the beginning of the video expecting it to be a rather unpleasant pile of calculations but the video became an incredible teaser for the rest of the lectures. I am looking forward the rest of the lectures!

  • @midwesteigensheaf
    @midwesteigensheaf Рік тому +1

    I believe the 1/m term in the series for pi/tan(pi*z) at 4:30 should be 1/(tau+m), and then one can either multiply by 1/2 or rewrite as 1/tau+ sum_{m=1}^infty [1/(tau+m)+1/(tau-m)].

  • @rosieshen8431
    @rosieshen8431 3 роки тому +2

    13:26 minor typo - (-1)^{k/2} + 1 and no 2 in the denominator

  • @aa-lr1jk
    @aa-lr1jk 3 роки тому +6

    Fantastic.

  • @anthonymurphy5689
    @anthonymurphy5689 3 роки тому +5

    9:50 minor typo - sin and cos should be reversed. Rest of expression ok.

    • @rosieshen8431
      @rosieshen8431 3 роки тому +2

      Also e^{i \tau} should be e^{i \pi \tau}, and 2 pi i should be pi i. Although it really doesn't matter...

  • @nonindividual
    @nonindividual 3 роки тому +6

    I _feel_ that the 1728 somehow is connected to Ramanujan's observation about 1729. But I'm not equipped with enough knowledge to check if this hunch is correct.

    • @theflaggeddragon9472
      @theflaggeddragon9472 3 роки тому +4

      1728=12^3 and Ramanujan observed that 1729=12^3+1^3=10^3+9^3 is the smallest number that is a sum of two cubes in two different ways. Haven't seen anything deep arise from the "taxicab" property, but it sure is neat!

    • @nonindividual
      @nonindividual 3 роки тому

      @@theflaggeddragon9472 Yeah, I think you're right, it's likely got nothing much to do with Ramanujan's observation, and it's just that 1728 = 12^3. 12^3 is certainly not a surprise.

    • @ecv257
      @ecv257 3 роки тому

      @@theflaggeddragon9472 1729 is the second Carmichael number en.wikipedia.org/wiki/Carmichael_number so who knows? there might be a relationship between sum of two cubes, Carmichael numbers and modular forms

  • @antoinebrgt
    @antoinebrgt 3 роки тому +3

    Jubilation when you don't have to look at your notes to write the coefficient of q^2 in the j function :)
    I know by heart 196884 but didn't manipulate those enough to remember the next coefficient !

  • @rosieshen8431
    @rosieshen8431 3 роки тому

    19:50 This is obtained by equating coefficient of q^2? Then it should be 240 sigma_3(2) * 2 + 240^2 = 480 sigma_7(2).

  • @yatima1158
    @yatima1158 3 роки тому +2

    I find it beautiful that you use the action of SL(2, R) on the upper half plane with the element \tau -> \tau + 1 to parametrise the summation. Is there other ‘naturally important’ series that come from other subgroups of SL(2,R)?

    • @Entropize1
      @Entropize1 3 роки тому +2

      There are! But I think you mean SL_2(Z). Not to plug myself, but I have a new series of lectures up that explain some of these things.

    • @mathmo
      @mathmo 3 місяці тому

      @@Entropize1 Woah! Cool channel. Subscribed!

    • @Entropize1
      @Entropize1 3 місяці тому

      @@mathmo I appreciate you!

  • @paesanng
    @paesanng 3 роки тому +1

    May I know what is your camera set up? This is a very efficient way of teaching.

    • @cogito2958
      @cogito2958 3 роки тому +3

      You can find it in the description of the channel. I think he made a community post about this as well.

  • @ecv257
    @ecv257 3 роки тому +1

    I wonder what other funcion should have been in the numerator to avoid the historical mistake? E_6^2? E_12?

  • @ShenghuiYang
    @ShenghuiYang 3 роки тому

    Must have for analytic number theory!

  • @migarsormrapophis2755
    @migarsormrapophis2755 3 роки тому +2

    ye

  • @abnishkumarverma2136
    @abnishkumarverma2136 3 роки тому

    Sir please make videos on real analysis please sir