Signed Stirling Number of the First Kind of Number with Greater
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Theorem
Let $n, k \in \Z_{\ge 0}$
Let $k > n$.
Let $\map s {n, k}$ denote a signed Stirling number of the first kind.
Then:
- $\map s {n, k} = 0$
Proof
By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $\map s {n, k}$ which satisfy the equation:
- $\ds x^{\underline n} = \sum_k \map s {n, k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the left hand side and right hand side are polynomials in $x$ of degree $n$.
Hence the coefficient $\map s {n, k}$ of $x^k$ where $k > n$ is $0$.
$\blacksquare$