🔔 Bell Numbers and Its Recurrence Relation (Proof)

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  • Published on Sep 23, 2018
  • We define Bell Numbers and derive its recurrence relation.
    For exponential generating function: usclip.net/video/1I4VpemnWmg/video.html
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Comments • 11

  • Sumit Kumar
    Sumit Kumar 7 months ago

    Excellent.

  • Jayanth Kumar Duvvada
    Jayanth Kumar Duvvada 11 months ago +1

    Tq very much

  • James Wilson
    James Wilson Year ago +1

    Sweet! I love this kind of stuff!

  • Vighnesh Raut
    Vighnesh Raut Year ago +1

    When you said partitions, I was thinking about the partitions that Ramanujan generalised. But these are different.

  • Konrad Kamiński
    Konrad Kamiński Year ago +3

    According to given definition B(0) =0 there is no partitioning of empty set without using empty set, but theorem works only when we set B(0)=1.

    • Konrad Kamiński
      Konrad Kamiński Year ago

      Or you can say that partition is maximal set of disjoint subsets. It doesn't change B for n greater than 0 but it defines B(0)=1 correctly. But then we have to live with every partition containing empty set. ;) Ehh, life is hard!

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +3

      One may argue that we can consider {empty set} as a "partition" of empty set, but you are right--from the definition I gave, B(0) = 0. Perhaps I should have mentioned that our definition applies only for positive integer n, then defined B(0) as 1 at the beginning of the video.

  • Logan Hargrove
    Logan Hargrove Year ago +3

    Thank you! I have been working on a certain infinite series and I just now realized that the result can be expressed in terms of Bell numbers.

  • adandap
    adandap Year ago +6

    Nice, though essentially an induction proof.

  • Xander Gouws
    Xander Gouws Year ago +11

    Your vids are really warming me up to combinatorics

  • GENIUS
    GENIUS Year ago

    Awesome!!!!!!I absolutely love your videos!!!Keep them coming.Oh,and by the way,could you solve this integral that's been bugging me:int from 0 to pi/4 of(1-(x)^2+(x)^4-(x)^6...)/((cosx)^2+(cosx)^4+(cosx)^6+....).You can find it here around the 1:34:00 mark in the integration bee here: usclip.net/video/UNpa3-EGxGY/video.html