# Exponential Generating Function for Bell Numbers (Proof)

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• Published on Sep 24, 2018
• We encapsulate bell numbers in a gorgeous formula: e^(e^x-1).
The previous video on Bell Numbers and Its recurrence formula: usclip.net/video/fkdTwxGwA5U/video.html.
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• B(x) denotes the exponential generating function for bell numbers (that is, it enumerates bell numbers within its coefficients), NOT the function that outputs bell numbers. That is, B(x) and B(n) used in the video denote two separate entities: the former refers to the generating function and the latter refers to the sequence (actual values) of bell numbers. I apologize if the notation led to confusion. Perhaps I should have used G(x), not B(x), to denote the exponential generating function.

• Will Bishop 11 months ago +2

This is confusing at the end, when you say B(0) = 1 by convention, which refers to the B(n) formula from the previous video, but then you plug in 0 to the B(x) or G(x) formula in order to calculate C. Of course, the two are the same because G(0) = B(0) + B(1)*0 + B(2)*0/2! + ..., but this is an extra step it would be good to mention.

• JustinsRealmMC Year ago +1

You should try solving IMO problems. Hehe it’d be a show

• James Wilson Year ago +1

You showed us how easy that was :)

• loopingdope Year ago +3

Can you make a video about the new RH paper/proof? It's not usual from you but would be interesting to watch. But ur cool anyhow, love u bb

• loopingdope Year ago +1

@LetsSolveMathProblems ty for reply, u cool man

• LetsSolveMathProblems  Year ago +4

I would love to, but I am not experienced enough with advanced topics in complex analysis and Atiya's proof to provide a fruitful, thorough discussion. I apologize. Thank you for your suggestion, though. =)

• It's getting complicated. Over my head now.

• LetsSolveMathProblems  Year ago +3

• δτ Year ago

If B(x) is the generating function for Bell Numbers, then it should give out integer values for integer inputs, right?
Or have you forgotten to apply the floor function to it?

• δτ Year ago

@佐藤裕也
@Robert Davis
Oh, so I have just misunderstood how to generate the Bell Numbers from the given function.
Thanks for the explanation.

• 佐藤裕也 Year ago

B(x) and B(n) are different things.B(x) is the function whose coefficient of x^n in the Taylor series is B(n)/n! and B(n) is nth Bell number.

• Robert Davis Year ago

No, the coefficient of x^n in the generating function B(x) is B(n)/n! Since the coefficient is divided by n!, it is not necessarily an integer. Since the generating function is a formal power series, you wouldn't normally plug in a value for x.

• Jaleb Year ago +3

Are B(x) and B(n) the same function? Started getting lost on if you're comparing two different functions one over x=R->R while the other is limited to n=Z->Z

• Paul Sampson Year ago +1

G(x) would indeed have been customary :)

• LetsSolveMathProblems  Year ago

Absolutely! Taking nth derivative gets us B'(x) = B(n) + x*(some formal power series), so B'(0) = B(n). However, it may be more intuitive to think of B(n) as the values the generating function B(x) encapsulates in its coefficients.

• Jaleb Year ago

Could we consider your B(n) as the nth derivative of B(x) evaluated at 0?

• LetsSolveMathProblems  Year ago +5

I apologize for the possible confusion. As Robert Davis suggested, perhaps I should have used subscripts for bell numbers or some other letter (like G(x)) for the generating function.

• Robert Davis Year ago +3

B(x) is the generating function. B(n) is the n'th Bell number. Might have been better to use subscripts for the latter to avoid confusion.

• No no no ! You've to atleast listen me please, it's my last wish that you make video on standard question 58 , only specifically 58 I'm saying! , and yea I viewed your every vid , it's awesome as expected

• LetsSolveMathProblems  Year ago +3

To maximize the quality of my videos, I strive to explain the contents that I either know well or feel passionate about. I apologize again, but I cannot just reiterate your transcript to make a video. For the final time, please try consulting Art of Problem Solving website.

• Siddhartha Shree kaushik Year ago

@LetsSolveMathProblems I'm giving you script of whole video , now you've to do it please ,

• LetsSolveMathProblems  Year ago +2

I appreciate that you watched every one of my recent videos, but as I specifically told you before, I do not have enough leisure time and, for some very challenging requests, sufficient mathematical maturity in that particular topic to make every video my viewers suggest to me (I truly wish I could). Again, please try posting the problem in Art of Problem Solving Forum, which is arguably the best forum for discussing math contests and problems. You will probably be able to receive some thorough explanations or insights there.

• madhav 's talks Year ago

Please make a video on partitions..