A Bell-Derangement Identity (

Поделиться
HTML-код
  • Опубликовано: 4 ноя 2024

Комментарии • 11

  • @enstucky
    @enstucky Год назад +5

    Hey! Found your video on the SoME voting page. My process didn't allow me to write official feedback, but I wanted to give you my thoughts anyway. As a combinatorialist, I always think there's more room in this world for lovingly rendered animations to complement combinatorial proofs. The "story" interpretation you used for Bell numbers really fit with this goal.
    If I had to criticize... without going back to watch the previous video you mentioned, the motivation might be a bit weak? But personally the probability was compelling enough for me. And I don't actually know that I've seen the distribution that you alluded to in the conclusion, so you got my attention there ^_^

    • @enumerable345
      @enumerable345  Год назад +1

      The motivation is sort of jokingly weak, I admit! The *real* motivation was when I decided to see what would happen when I computed the inner product of the character of the V= representation of S_n with itself. Too much for a comment--maybe I'll make a video about *that* some time.

  • @YTomS
    @YTomS Год назад +3

    Great video and an interesting topic. One small note - music is quite distracting when it abruptly increases in volume when you're not talking, consider keeping it low.

    • @enumerable345
      @enumerable345  Год назад +1

      Yeah, sorry about that. Rushed the sound editing to make the #SoME3 deadline just under the wire.

  • @proofsgoofs5057
    @proofsgoofs5057 Год назад +1

    wooooo bijections!!!!

  • @OscarCunningham
    @OscarCunningham Год назад +3

    A probability distribution with Bell numbers as its moments? Would that be a Bell curve? :-)

  • @alexsere3061
    @alexsere3061 2 месяца назад

    But wait, if n=1, don't we always just get one fixed point? Meaning that the variance is 0?

    • @enumerable345
      @enumerable345  2 месяца назад

      Yes! The formula we derived here only works for m ≤ n, and for a variance computation you'll need the 2nd moment, i.e. m=2. So yes, the variance will be 1 for n at least 2. And in that one case when n=1, the variance is indeed 0.

  • @MichaelDarrow-tr1mn
    @MichaelDarrow-tr1mn 2 месяца назад

    Hang on. Did Did use cards for the bell numbers because one of the bell numbers is 52?

    • @enumerable345
      @enumerable345  2 месяца назад

      Huh...I didn't even notice that. Just a coincidence--like how C(14, 6)=C(15, 5).

    • @MichaelDarrow-tr1mn
      @MichaelDarrow-tr1mn 2 месяца назад

      @@enumerable345 C(14,6) equals C(15,5) because 10 times 9 equals 15 times 6.