Eulerian polynomials and numbers

Here we look at another group of special polynomials and numbers first studied by Euler. These “Eulerian” polynomials must be distinguished from the similar-sounding “Euler” polynomials.

A generating function for the Eulerian polynomials is: \[\frac{t-1}{t-e^{(t-1)z}} = \sum\limits_{n=0}^\infty A_n(t) \frac{z^n}{n!}\]

Multiplying out the denominator and using the series expansion of the exponential allows the derivation of a recurrence relation: \[A_n(t)=\sum\limits_{m=0}^{n-1}\binom{n}{m}A_m(t)(t-1)^{n-m-1}\]

The initial term is \(A_0(t)=1\). Apart from this term, the degrees of the polynomials are \(n-1\). In fact the highest power is contributed by the \(m=0\) term of the sum, \(A_0(t)(t-1)^{n-1}=(t-1)^{n-1}\), since the remaining terms have degree \(n-2\). The leading degree term is obviously \(t^{n-1}\), which makes it “monic”.

Absorbing the \((t-1)\) in the exponential of the generating function: \[\frac{t-1}{t-e^{z}} = \sum\limits_{n=0}^\infty A_n(t) \frac{(z/(t-1))^n}{n!}\]

\[\implies \frac{e^z}{1-te^z}=\sum\limits_{n=0}^{\infty}\frac{A_n(t)}{(1-t)^{n+1}}\frac{z^n}{n!}\]

A new form for the Eulerian polynomials arises from: \[\frac{e^z}{1-te^z}=\sum\limits_{m=0}^{\infty}t^m e^{(m+1)z}=\sum\limits_{m=0}^{\infty}\sum\limits_{n=0}^{\infty}t^m \frac{[(m+1)z]^n}{n!}\]

Comparing with the previous equation: \[\frac{A_n(t)}{(1-t)^{n+1}}=\sum\limits_{m=0}^{\infty}t^m (m+1)^n\]

Multiplying by \(t\) and differentiating we obtain: \[\frac{A_{n+1}(t)}{(1-t)^{n+2}}=\frac{d}{dt}\left(\frac{t A_n(t)}{(1-t)^{n+1}}\right)\]

This leads to a new recurrence relation: \[A_{n+1}(t)=t(1-t)A_n'(t)+(1-n t)A_n(t)\]