binomial theoremcombinatoricsfactorialstirling-numbers
Consider the falling factorial
$$
x^{\underline n} = x (x-1) \dots (x-n+1)
$$
This is clearly a polynomial of degree $n$. Is there an expressions for the coefficients of that polynomial? I tried taking derivatives and substituting $x=0$ to get the coefficients but the expressions I am getting are very complicated.
The present discussion is polluted by the confusion of unfortunate symbols for falling and rising factorials as well as the Pochhammer symbol. This has been well underlined in the comments. In order to make it clear to everybody, an excerpt from WolframMathWorld is attached below. So, without using the Pochhammer symbol, the formula is : $$ x(x+1)(x+2)…(x+n-1) = s(n,n)x^n-s(n,n-1)x^{n-1}+…+(-1)^{n-1}s(n,1)x $$ $$ x(x-1)(x-2)…(x-n+1) = s(n,n)x^n+s(n,n-1)x^{n-1}+…+s(n,1)x $$ where $s(n,k)$ is the Stirling number of the first kind. Using the Stirling’s symbol $s(n,k)$ is the simplest way to represent analytically the coefficients of the polynomial. Of course, if the goal is to simplify the numerical computation, using the Stirling numbers is not an advantage in the practical viewpoint.
We use D.E. Knuth's preferred notation $k^{\underline r}=k(k-1)\cdots(k-r+1)$ to denote the falling factorials and consider \begin{align*} k^r=\sum_{j=0}^r a_{r,j}(m-k)^{\underline{j}}\tag{1} \end{align*}
Substituting $k$ with $m-k$ gives \begin{align*} \sum_{j=0}^r \color{blue}{a_{r,j}}k^{\underline{j}}&=(m-k)^r\tag{2}\\ &=\sum_{p=0}^r\binom{r}{p}(-1)^pk^pm^{r-p}\tag{3}\\ &=\sum_{p=0}^{r}\binom{r}{p}(-1)^p\sum_{q=0}^p{p \brace q}k^{\underline{q}}m^{r-p}\tag{4}\\ &=\sum_{0\leq q\leq p\leq r}(-1)^p\binom{r}{p}{p \brace q}k^{\underline{q}}m^{r-p}\tag{5}\\ &=\sum_{q=0}^{r}\left(\color{blue}{\sum_{p=q}^{r}\binom{r}{p}{p \brace q}(-1)^pm^{r-p}}\right)k^{\underline{q}}\tag{6} \end{align*}
Comment:
In (3) we apply the binomial theorem.
In (4) we use the representation of powers of $k$ as Stirling numbers of the second kind.
In (5) we use a somewhat more convenient representation of the index range as preparation for the next step.
In (6) we exchange the sums from (4).
Coefficient comparison of (2) and (6) gives \begin{align*} \color{blue}{a_{r,q}=\sum_{p=q}^{r}\binom{r}{p}{p \brace q}(-1)^pm^{r-p}\qquad\qquad 0\leq q\leq r}\\ \end{align*}
Best Answer
Yes indeed, Notice that you can either get an $x$ or a number when you unfold it, hence the coefficient of $x^k$ in that expression is $$(-1)^{n-k}\sum _{\substack{S\subseteq [n-1]\\|S|=n-k}}\prod _{s\in S}s,$$ this is not very simple to compute, but is a closed formula. For example, when $k=1$ you get $(-1)^{n-1}(n-1)!.$ This numbers have a name, they are called Stirling numbers of the first kind. They satisfy a recurrence that can be explicit from the definition of the falling factorial.