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:

$\map s {0, k} = \delta_{0 k}$

$\blacksquare$


Also see