# Stirling number of the first kind expansion?

I'm trying to expand this expression (for $1 \leq j \leq n+1$)

$$\prod_{i=1 ; i\neq j}^{n+1} (x-i)$$

in terms of Stirling numbers. The falling factorial $(x)_{n}$ is defined as

$$(x)_{n}=\prod_{i=0}^{n-1} (x-i)=\sum_{i=0}^{n}(-1)^{n-i} \begin{bmatrix} n\\ i \end{bmatrix}x^i =\sum_{i=0}^{n}s(n,i)x^i$$ where $s(n,i)$ is the Stirling number of the first kind. This gives me the expression.. $$\prod_{i=1 ; i\neq j}^{n+1} (x-i)=\frac{1}{x-j}\sum_{i=1}^{n+2}s(n+2,i)x^{i-1}$$ Formula fixed.

Edit:

Formula is fixed, this solution works.

Lets get the definition of the Stirling numbers (of the first kind) in terms of falling factorials right \begin{eqnarray*} (x)_{n}=\prod_{i=0}^{\color{red}{n-1}} (x-i)=\sum_{i=0}^{n}(-1)^{n-i} \begin{bmatrix} n\\ i \end{bmatrix}x^{\color{red}{i}} =\sum_{\color{red}{i=1}}^{n}(-1)^{n-i} \begin{bmatrix} n\\ i \end{bmatrix}x^{i}. \end{eqnarray*} The zeroth term is zero $\begin{bmatrix} n\\ 0 \end{bmatrix}=0$ and can be included or ommited (for convenience.) Have a quick look at the numbers in Wiki https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Table_of_values_for_small_n_and_k
In your formula there is no need for $i \neq j$ and it should look like this \begin{eqnarray*} \prod_{i=1}^{n} (x-i)=\sum_{i=1}^{n+1}(-1)^{n+1-i} \begin{bmatrix} n+1 \\ i \end{bmatrix}x^{i-1}. \end{eqnarray*}
• I am trying to get the expression for $i \neq j$ though, i have updated the question with your formulas.. – Baklava Gain Oct 30 '17 at 1:27