# ðŸ”” 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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Sumit Kumar4 months agoExcellent.

Duvvada Jayanth Kumar8 months ago^{+1}Tq very much

James Wilson9 months ago^{+1}Sweet! I love this kind of stuff!

Vighnesh Raut9 months ago^{+1}When you said partitions, I was thinking about the partitions that Ramanujan generalised. But these are different.

Konrad KamiÅ„ski9 months 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 KamiÅ„ski9 months agoOr 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!

LetsSolveMathProblems9 months 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 Hargrove9 months 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.

Paco Gomez-Paz9 months agoI am trying to prove that the cardinality of a bell set is greater than that of the original set for both infinite and non infinite sets non infinite is quite easy to prove but I am having troubles for infinite sets . Anyone have any ideas

Paco Gomez-Paz8 months agoI meant greater than or equal to also I already proved it but thank you for taking the time to respond

Konrad KamiÅ„ski9 months agoWhat is Bell set? Is it set of all partitions? Then it's not true for n=1? For infinite just notice that Bell set's power is greater or equal than Power set's cardinality (using partitions of 2 sets) and Power set's cardinality is strictly greater than cardinality of original set.

If Bell's set is something different, plz tell your definition.

PS. Actually now I see that B(Kappa)=2^(Kappa) for Kappa greater or equal Aleph_0.

adandap9 months ago^{+6}Nice, though essentially an induction proof.

Xander Gouws9 months ago^{+11}Your vids are really warming me up to combinatorics

GENIUS9 months agoAwesome!!!!!!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