How is this huge function useful??

[@MichaelPennMath]

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[@QuantumHistorian]

That final formula does not look like it would give natural number values for all n. And yet, given the definition of Bell numbers, it must!

[@alexpotts6520]

I'm guessing this is just a thing for combinatorial sequences, given that the formula of Fibonacci numbers similarly has a bunch of irrational stuff in it that magically always spits out integers.

[@lih3391]

Why doesn't it give natural numbers in desmos? I think that thing is too irrational to be integers. The function 1/e*e^e^x doesnt, but the sum does, oh nevermind, B(x) is not the bell bumber series itself

[@lorenzosaudito]

@@lih3391 Yep exactly, you have to be very careful to not mix up the original function and its generating function

[@Alex_Deam]

I managed to convince myself it makes sense via series manipulation for small values of n. E.g. for n=2, you can do: sum[(k^2)/k!] = sum[k/(k-1)!] = sum[{(k-1)+1}/(k-1)!] = sum[1/(k-2)!] + sum[1/(k-1)!] = 2e You can then show you'll always get an integer multiple of e for all n using induction, but I had to use strong induction by assuming it was true for all values below. I wonder if there's a way to do it with a weaker form of induction?

[@chickencyanide9964]

@@lorenzosaudito may I then ask how we find the original function?

[@landsgevaer]

Explaining Bell numbers, finding a recursive form, and then an explicit (albeit infinite) sum. This is one nice video! Hoping you could manage something similar for the partition numbers... 🙂

[@seneca983]

1:45 "How can you partition a zero element set?" Easy! It's {} and that's the only possible partition so 1 is indeed the right answer.

[@driksarkar6675]

Technically, the empty set isn’t non-empty, so I’m not sure that’s a valid partition.

[@seneca983]

@@driksarkar6675 It's its members that are not supposed to be empty (and the empty set has no empty members).

[@Hyakurin_]

Now it would be nice if you did a video about Stirling numbers and the number of surjection between two finite sets, and maybe its relationship with Bell numberz

[@5alpha23]

I never expected combinatorics to be so aesthetically pleasing - thanks for showing me that, it certainly widened my horizon!

[@tolberthobson2610]

I find it pretty cool how the closed-form explicitly-defined generating function for bell numbers involved a series and a transcendental number, but is able to 'spit out' natural numbers. Pretty mind-bending.

[@STbender]

Last formula was unexpected Also, as bell number is sum of S(n,k) I read somewhere that for large n S(n,k)~k^n/k! Now ik why : ) So please make video s(n,k) like max of it

[@Fightclub1995]

That generating function is the same generating function for a Poisson distribution with lambda = 1. Any relation?

[@lexyeevee]

Phew! Instead of having to do a large finite amount of recursive work, now we can simply do an infinite amount of iterative work

[@seaassasin1855]

The thumbnail is so good

[@TheMemesofDestruction]

I was gonna save this then saw it was a Professor Penn Video! ^.^

[@Queen-be6md]

Great content! and very good edition by the video editor

[@oida10000]

Shouldn't the inner sum only run up to the current index of the outer sum? So k 0 to n and m 0 to l?

[@Francisco-vl5ub]

14:00 the sum from k = 0 to n changed to k = 0 to infinity

[@edskev7696]

Yup, and that change appears to have a big impact on the rest of the calculation.

[@Francisco-vl5ub]

@@edskev7696 it works out correctly, but definitely obscures how Cauchy’s formula is used here

[@landsgevaer]

Isn't that because Bn for negative n and/or n-choose-k outside Pascal's triangle are zero, so do not contribute?

[@Francisco-vl5ub]

@@landsgevaer ​ I don’t believe that to be the case. Note 14:38 RHS of Cauchy formula, the sum defining c_k is limited above

[@landsgevaer]

@@Francisco-vl5ub Yeah, but then nCk has k>n so you are outside Pascal's triangle, and those terms are zero. So the extra terms in the sum are simply zero. I'll admit that I was too lazy to wind back, but I did write it had to be either the Bn or the nCk.

[@aweebthatlovesmath4220]

I can't believe that last sum is always an integer!!

[@ethanbartiromo2888]

I noticed that you put in B(0) = 1, but shouldn’t it be B(0) = 0? Since we are plugging 0 in for x not n

[@Hiltok]

B(x=0) = sum 0 to inf of Bn*x^n/n! All terms in the sum go to zero except for the zero-th term, since 0^0=1, so B(x=0) = B0*0^0/0! = B0*1/1 = B0 = 1

[@stoneman172]

The last chain of equalities gives an assertion about equality of limits. How can you conclude term-wise equality of the two series from this?

[@ddystopia8091]

14:07 Why second sum started to go to infinity??

[@ukaszpawlak6953]

That's also moment generating function of Poisson distribution, I think there were some deeper connection between it and Bell numbers

[@josephmathmusic]

I am a probabilist so I immediately think of Poisson distribution when I see a double exponential...

[@jamiepianist]

THE FAST FORWARD PFFFFT

[@peon17]

Another fun set of numbers are Catalan numbers. They're not as exotic as Bell numbers, but they are notorious for how many different things they count.

[@crazycat1503]

And first 4 Bell numbers are actually the same as 4 first Catalan numbers

[@zachbills8112]

It's interesting how much simpler the closed form of the nth bell number is than that of the nth number of integer partitions.

[@emanuellandeholm5657]

This was ... beautiful. Made my morning really :)

[@Maths_3.1415]

Hey Michael penn You are my favourite teacher :)

[@M.athematech]

B0 = 1 because there is indeed precisely one partition of the empty set, namely the empty partition.

[@Milan_Openfeint]

But an empty subset is forbidden by definition...? Feels like defining 0/0=1 or something.

[@MisterGhosh]

What is the empty partition?

[@YO-in2uw]

@@Milan_Openfeint Yes, so {{}} is not a solution. The only valid partition is {}, which contains no empty subsets.

[@schweinmachtbree1013]

@@MisterGhosh The partition consisting of zero parts

[@MisterGhosh]

@@schweinmachtbree1013 okay, think I got... You can take all zero parts and exhaust the empty set, it makes sense.

[@MooImABunny]

it's pretty unbelievable that these sums all converge to e times an integer. it's like finding out the zeta function at every integer converged to a constant, irrational as it may be, times a rational. that would be crazy unlikely.

[@johannesmoerland5438]

They don't. The n-th derivative at x=0 however gives you the n-th bell number

[@the-avid-engineer]

The second sum from k = 0 to n suddenly changes to k = 0 to infinity.. is that a mistake or did you just not explain it?

[@Happy_Abe]

@14:32 why is the second sum from 0 to infinity when it was from 0 to n before?

[@Happy_Abe]

And how would dividing by (n-k)! work when k>n So doesn’t look like the sun should go to infinity, just to n

[@vaevfunc]

And that’s the perfect video to watch

[@david-hogarty]

It's great to see some combinatorics!

[@AJ-et3vf]

great video. thank you

[@TronSAHeroXYZ]

Thanks Michael.

[@gael8828]

Hum, how can the last result give the rational numbers seen at the beginning of the video ?

[@davidgillies620]

Isn't the partition here also called an exact cover?

[@YTSPoster]

This is so cool what the heck

[@pharaohgarmar5611]

I had a wry laugh at this. You go overboard with your signposting that you are about to embark on an example but at 14:18, you give a “simplification” in a formula which is clearly a massive leap and looks wrong (how can k range from 1 to infinity in (n-k)! ?). While I have appreciated many of your videos there is a tendency to gloss over critical explanations like this.

[@thomaspeck4537]

I think the infinity is actually a mistake. Looking at Cauchy's formula, it is indexed to n. It seems he was looking ahead, and accidentally wrote two infinite sums a step too early.

[@forheuristiclifeksh7836]

4:00

[@PotatoImaginator]

He always know the good place to stop :)

[@jonathantorres913]

God

[@okoyoso]

The rotating transition animation is a little annoying. It feels like my phone screen just rotated and makes me want to rotate back.

[@albertozuanon3874]

I totally agree!

[@memesThatDank]

here before good place to stop

[@yahav897]

I'll be hijacking this comment section: I'm a first year, and I barely got a pass in my linear algebra I course, and failed my real analysis one. I'll be taking the re-exam for both of them in the next two weeks. Any tips you might come up with? I hope I pass this time.

[@user-le1oc9js4h]

In my first semester of uni I almost failed both linear algebra and real analysis and pass by some miracle. The problem was that I tried to memorize the formulas and theorems instead of trying to understand them and recreate them during exam based on my understanding. Although it is a very hard task, you should try to remain calm and read through the material that you got and explain to yourself every little detail until you got it. You have not that much time, so you won’t be able to do it with all the material, but it’s good to get the hang of it just so the professor would see that you at least understood something. And of course, get all the help you can - microheadphones, cheatsheets, try to put some pressure on feelings of the examiner. Good luck!

[@yahav897]

@@user-le1oc9js4h that is good advice. While I think I understand the material, I probably don't understand it well enough - so I should go over that again, and solve more problems.

[@kono152]

Second comment!

[@kozokosa9289]

First comment!