Signed Stirling Number of the First Kind of 0
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Theorem
- $\map s {0, n} = \delta_{0 n}$
where:
- $\map s {0, n}$ denotes a signed Stirling number of the first kind
- $\delta_{0 n}$ denotes the Kronecker delta.
Proof
By definition of signed Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
Thus we have:
\(\ds x^{\underline 0}\) | \(=\) | \(\ds 1\) | Number to Power of Zero Falling is One | |||||||||||
\(\ds \) | \(=\) | \(\ds x^0\) | Definition of Integer Power |
Thus, in the expression:
- $\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
we have:
- $\map s {0, 0} = 1$
and for all $k \in \Z_{>0}$:
- $\map s {0, k} = 0$
That is:
- $\map s {0, k} = \delta_{0 k}$
$\blacksquare$