Find an explicit formula (as explicit as possible) for $a_k$

Question

Let $a_k$ be defined as follows: $$(1+x)^n = \sum_{k=0}^na_k(x)_k$$ where $(x)_k = x(x-1)(x-2)\dots(x-k+1)$. Find a formula that describes $a_k$ as explicitly as possible, in terms of a famous number family (Stirling numbers in this case). I approached it the following way:

$$(1+x)^n=\sum_{k=0}^n {n\choose k}x^k$$ and $$\sum_{k=0}^na_k(x)_k=\sum_{k=0}^na_k\sum_{i=0}^ks(k,i)x^i$$

Here, $s(k,i)$ denotes the Stirling numbers of the first kind, as $(x)_k=\sum_{i=0}^ks(k,i)x^i$. Writing out some terms by hand, and rearranging, I found the following:

$$\sum_{k=0}^na_k\sum_{i=0}^ks(k,i)x^i = \sum_{k=0}^n x^k\bigg[\sum_{i=k}^na_is(i,k)\bigg]$$ which, with the first (original) equality, gives us: $${n\choose k} =\sum_{i=k}^ka_is(i,k)$$ as these are the respective coefficients of $x^k$ in the defined sum.

This is as far as I got, and I'm not even sure if it's correct. Does anyone know how to go from here, or is this regarded as 'explicit enough'?

Answer 1

The problem does not specify which kind of Stirling numbers is to use. Here is a variant based upon ${n\brace k}$, the Stirling numbers of the Second kind. We use $x^{\underline{k}}=x(x-1)\cdots(x-k+1)$ to denote the falling factorials.

The Stirling numbers of the second kind ${n\brace k}$ are given for non-negative integers $n$ as \begin{align*} \sum_{k=0}^n {n\brace k}x^{\underline{k}}=x^n\tag{1} \end{align*}

We obtain from (1) for integral $n\geq 0$ \begin{align*} \sum_{k=0}^{n+1}{n+1\brace k}x^{\underline{k}}&=x^{n+1}\tag{2}\\ \sum_{k=1}^{n+1}{n+1\brace k}(x-1)^{\underline{k-1}}&=x^{n}\tag{3}\\ \sum_{k=0}^{n}{n+1\brace k+1}(x-1)^{\underline{k}}&=x^{n}\tag{4}\\ \sum_{k=0}^n\color{blue}{{n+1\brace k+1}}x^{\underline{k}}&=(1+x)^n\tag{5} \end{align*} and we conclude $\color{blue}{a_k={n+1\brace k+1}}$.

Comment:

• In (2) we use (1) and set $n\to n+1$.

• In (3) we use $x^{\underline{k}}=x\cdot (x-1)^{\underline{k-1}}$ with $k>0$ and divide the identity by $x$. We set the lower limit of the sum to $1$, since ${n+1\brace 0}=0$ for $n>0$.

• In (4) we shift the index $k$ by one to start from $0$.

• In (5) we shift $x$ by one $x\to x+1$.