# Thread: Sterling Numbers of the Second Kind

1. ## Sterling Numbers of the Second Kind

I recently learned about the Stirling Numbers, and I'm trying to get my head around them. I understand that the 2nd Kind numbers are the coefficients that will convert a polynomial in the standard (monomial power) basis to the falling power basis. I found a piece on Planetmath that presents a proof of this, but I can't follow it.

There must be somebody out there who knows more about the Stirling Numbers than I do. I hope you can explain this proof to me, or tell me where the author went wrong, if that's the case.

Thanks in advance. I've attached two pdfs. One is the Planetmath piece, the other is by me an explains my difficulties (I hope).

2. ## Re: Sterling Numbers of the Second Kind

Originally Posted by zhandele
I recently learned about the Stirling Numbers, and I'm trying to get my head around them. I understand that the 2nd Kind numbers are the coefficients that will convert a polynomial in the standard (monomial power) basis to the falling power basis. I found a piece on Planetmath that presents a proof of this, but I can't follow it.
There must be somebody out there who knows more about the Stirling Numbers than I do. I hope you can explain this proof to me, or tell me where the author went wrong, if that's the case.

Frankly, I have no idea (although I have written on this topic) what you are asking about.
Have a look at this webpage.

This idea is also closely related to Bell Numbers.

Now here is my rant.
YOU can post the questions that you have.
Learning to post in LaTeX is relatively easy.
If you are serious about this study then you are going to be required to learn it.

3. ## Re: Sterling Numbers of the Second Kind

Dear Sir,

If you had opened my second pdf, which I wrote and generated myself using SWP, and which spells out in detail just what my question is and where my difficulty lies, you would know that everything you say and imply is simply not true.

Even without that, you should know in general what I'm "asking about," based on what I wrote in the posting window.

Let me try again, with an example.

$x^4 = x + 7x(x-1) + 6x(x-1)(x-2) +x(x-1)(x-2)(x-3)$

This statement 1) is true, 2) expresses an element of the standard power basis for polynomials as a linear combination of falling factorials, and 3) the coefficients 1, 7, 6, 1 are the Stirling Numbers of the 2nd Kind, fourth degree.

Any element of the power basis can be expressed that way, and the coefficients are always the Stirling Numbers, 2nd kind. Indeed, these numbers can be defined as just those coefficients. Surely this is not news to you.

I could have simply memorized this fact, but I declined to take the easy way. I wanted to know how one might prove that this definition of the Stirling Numbers is logically equivalent to the more common definitions, recursive or in terms of combinatorics. This is all "above and beyond" for me; I have no reason to believe I will ever be required to produce such a proof, but I'm looking for a better understanding.

I found a proof, but it seems to me incorrect. Most likely this proof is correct after all, and I'm missing something, but after four days of worrying the problem, I still don't know what.

I hope this is more clear.