Talk:Eulerian number

Notation
There doesn't seem to be a standard notation, I've seen E(n,k), A(n,k) and angle brackets used in different sources. So for the sake of consistency I'm picking E(n,k) to be used in the article with a mention of the other notations at the beginning of the article.--RDBury (talk) 18:08, 15 May 2009 (UTC)
 * After review of some additional sources I'm changing to A(n,k). It's been used in more places than the others and it's more consistent with the An(x) notation for the polynomials.--RDBury (talk) 18:56, 15 May 2009 (UTC)

source of the Euler-snippet
Hi I wanted to find the context of the Euler-snippet; wirkstoff, could you please provide the page-number?

Thanks - Gotti 08:04, 31 August 2009 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs) Found it myself Gotti 20:23, 1 September 2009 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs)


 * page 485f. --84.130.170.107 (talk) 22:21, 6 September 2011 (UTC)

Identities
The first of the listed equations in the "Identities"-section was wrong. To be more precise, in the case where n = 0 the left-hand side is a geometric series, yielding $$\frac{x}{1-x}$$ as its limit. But the right-hand side of the equation is zero (we have an empty sum there). I adapted the sum on the right-hand side (the elements are now summed to n not n-1). This corrects the mistake for n = 0 and doesn't change anything for n > 0 because for the index m = n the summand involves a factor A(n,n), which is zero and is this zero itself, contributing nothing to the sum. Now, I'm not an expert in combinatorics and never really worked with Euler numbers. I hope someone, who has experience with these can confirm my correction or look it up in the literature. --TheLaeg (talk) 14:21, 29 March 2010 (UTC)


 * I adjusted indices of summation in the power-sum identity so it will be valid for n &ge; 0. Zaslav (talk) 20:11, 7 December 2013 (UTC)

Eulerian numbers vs Eulerian polynomials
Premise: the definition and the notation for the Eulerian numbers and polynomials is by no means standard. That said, I see no advantage in the choice of the given definition : "[Eulerian numbers of the first kind] are the coefficients of the Eulerian polynomials:
 * $$A_{n}(x) = \sum_{m=0}^{n} A(n,m)\ x^{n-m}.$$"

An obvious instance of standardization wants that if certain numbers C(n,m), for 0≤m≤n, are called the X-numbers, the corresponding X-polynomial of order n is
 * $$\sum_{m=0}^{n} C(n,m)\ x^m.$$

Allowing inconsistencies between the notation for the coefficients and the notation for polynomials, and adopting free variations such as $$\sum_{m=0}^{n} C(n,m)\ x^{n-m}$$ or $$\sum_{m=0}^{n} C(n+1,m)\ x^m$$ or $$\sum_{m=0}^{n}(-1)^m C(n,m)\ x^m$$ may be of some little local advantage, but turns out into a complete mess as soon as one treats polynomials of several type at one time (suppose we have to write sum down the sum of three special polynomials An(x)+Bn(x)+Cn(x), each with its own notation for denoting the coefficients, giving for granted that we are able to keep them all by hart). On the same lines, I don't think that it is an optimal choice having the same letter for the coefficients and the polynomials. In this case, they are too similar, with some risk of confusion. I would go for Knuth solution $$\left \langle {n\atop m} \right \rangle, $$ which is also of more practical use. --pm a 17:12, 19 November 2010 (UTC)

More on Identities
I've added a third identity on the page, along with how to get it. It relates to the picture posted at the introduction of this page (Euler's Eulerian polynomials). I discovered it a couple of weeks ago and was wondering if anyone has seen it before? I have not, but I am not expert on the subject.

1:26 Dec 07 2010 (PST). OK but you may as well start from


 * $$\sum_{k=1}^{\infty}k^n x^k = \frac{\sum_{m=0}^{n}A(n,m)x^{m+1}}{(1-x)^{n+1}},$$

multiply both sides by the coefficient $$c_n$$ of any entire power series $$f(x)$$, and sum over $$n$$; of course for $$|x|<r$$ you can exchange order of summation in the LHS say, provided $$f(k)\le Mr^{-k}$$), and obtain


 * $$\sum_{k=1}^{\infty}f(k) x^k = \sum_{n=0}^\infty c_n \frac{\sum_{m=0}^{n}A(n,m)x^{m+1}}{(1-x)^{n+1}},$$

but what is the point? You can write down a lot of these identities... --pm a 00:44, 8 December 2010 (UTC)

Oh I see! I guess this one is kind of special because it's the sum of the actual Eulerian polynomials themselves and it relates with the constant e.

5:32am Dec 08 2010. —Preceding unsigned comment added by 98.164.252.55 (talk) 13:33, 8 December 2010 (UTC)

This article is still lacking of something more relevant: e.g. the recurrence relations for the E. polynomials of first and second kind; it's also lacking of the g.F. of both. So, when this article will be rectified, there is something to add and something to remove! --pm a 01:03, 9 December 2010 (UTC)

That's very strange! I remember earlier this year they did in fact have the recursive definition for them... check in the history, it should be there... [Edit: Hm or maybe not!]

Sorry if I wasted your time, please remove the identity if you feel it is useless.


 * I think this identity is nice but the proof should be removed. Either it is original research, or it is known and can be cited in the literature.  I will place the text below here so it is still available if needed.  It might be desirable to remove the identity altogether as I don't see a reason to prefer it to other identities, but I won't make that decision.  Zaslav (talk) 21:27, 7 December 2013 (UTC)

Another interesting identity is given by the following manipulation:
 * $$e^k=\sum_{n=0}^\infty \frac{k^n}{n!}$$
 * $$(x^k)(e^k)=\sum_{n=0}^\infty \frac{(x^k)(k^n)}{n!}$$
 * $$\sum_{k=1}^\infty(x^k)(e^k)=\sum_{k=1}^\infty\sum_{n=0}^\infty \frac{(x^k)(k^n)}{n!}$$

For $$0\le x<\frac{1}{e}$$, the terms on the right side are positive, so we may switch the sum. The terms on the left make a geometric series, and we know that converges. After all of that, we use the above identity to finish the manipulation:
 * $$\sum_{k=1}^\infty(xe)^k=\sum_{k=1}^\infty\sum_{n=0}^\infty \frac{(x^k)(k^n)}{n!}$$
 * $$\frac{ex}{1-ex}=\sum_{n=0}^\infty\sum_{k=1}^\infty \frac{(x^k)(k^n)}{n!}=\sum_{n=0}^\infty\frac{\sum_{m=0}^{n}A(n,m)x^{m+1}}{n!(1-x)^{n+1}}$$

Finally, for $$0\le x<\frac{1}{e}$$ we get
 * $$\frac{e}{1-ex}=\sum_{n=0}^\infty\frac{\sum_{m=0}^{n}A(n,m)x^{m}}{n!(1-x)^{n+1}}.$$

Notice that the numerator on the right-hand side is the Eulerian polynomial shown at the top of this page.

Analogy to normal distribution & Pascal's triangle?
As n increases, do the values of A(n,k)/n! turn out to approximate a well known function arbitrarily well? Thanks, Rich peterson198.189.194.129 (talk) 02:50, 16 March 2012 (UTC)


 * Good question, to which I don't know the answer. One observation: you don't have the scaling quite right; we really want to scale vertically by something like $$\frac{\sqrt{n}}{n!}$$ (just based on numerics), which is actually very similar to the Pascal case.  (If you just divide by n!, everything goes to 0.)  --Joel B. Lewis (talk) 02:49, 17 March 2012 (UTC)
 * Update: it's normal. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2636886#p2636886 --Joel B. Lewis (talk) 19:33, 22 March 2012 (UTC)
 * thanks!198.189.194.129 (talk) 18:16, 23 March 2012 (UTC)

Discrepancy of the tables of Eulerian numbers regarding $A_{0,0}$
When I use the generating-function for the introduction of the article to create the table of Eulerian numbers, I get that table (indexes (r,c)=(0,0)...(n,m))

[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] [1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0] [1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0] [1 11 11 1 0 0 0 0 0 0 0 0 0 0 0 0]

where the element A_{0,0}=1.

When I use the "explicite formula" (from that so-named section) I get

[0 0 0 0 0 0] [1 0 0 0 0 0] [1 1 0 0 0 0] [1 4 1 0 0 0] [1 11 11 1 0 0]

where A_{0,0}=0. The second version seems to be in use by combinatorists, and even with an prefixed column [1,0,0,0,...], (and called "shifted version", moreover giving it a unit-diagonal), while the first version is also occuring if we use the "polylog"-(or "Li") function rescaled by x and powers of (1-x). Could someone please look at this with the goal to   a) mention it,      b) make a comment of          b1) the history of and          b2) the relevance/rationale in it?

Note also that the generalization to negative (and even fractional) row-indexes is only possible when the first version is used. — Preceding unsigned comment added by Druseltal2005 (talk • contribs) 07:45, 19 February 2021 (UTC)

Removal of python code
I don't see that having python code in the article adds anything particularly useful; it seems to be original research. I will remove it shortly; but below is a copy for historical reference. 67.198.37.16 (talk) 23:55, 5 January 2024 (UTC)


 * The following is a Python implementation.