Unsigned Stirling Number of the First Kind of 0
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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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(48)$