Unsigned Stirling Number of the First Kind of 0

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds {0 \brack n} = \delta_{0 n}$

where:

$\ds {0 \brack n}$ denotes an unsigned Stirling number of the first kind
$\delta_{0 n}$ denotes the Kronecker delta.


Proof

By definition of unsigned Stirling number of the first kind:

$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$

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:

$\ds x^{\underline 0} = \sum_k \paren {-1}^{-k} {0 \brack k} x^k$

we have:

$\ds {0 \brack 0} = 1$

and for all $k \in \Z_{>0}$:

$\ds {0 \brack k} = 0$

That is:

$\ds {0 \brack k} = \delta_{0 k}$

$\blacksquare$


Also see


Sources