Relation between Signed and Unsigned Stirling Numbers of the First Kind
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Theorem
Let $m, n \in \Z_{\ge 0}$ be positive integers.
Then:
- $\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
where:
- $\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind
- $\map s {n, m}$ denotes a signed Stirling number of the first kind.
Proof
\(\ds \sum_k \map s {n, k} x^k\) | \(=\) | \(\ds x^{\underline n}\) | Definition of Signed Stirling Numbers of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {-1}^{n - k} {n \brack k} x^k\) | Definition of Unsigned Stirling Numbers of the First Kind | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map s {n, m}\) | \(=\) | \(\ds \paren {-1}^{n - m} {n \brack m}\) | Comparing coefficients of $x^m$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {n \brack m}\) | \(=\) | \(\ds \paren {-1}^{m - n} \map s {n, m}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n + m} \map s {n, m}\) | as $\paren {-1}^{2 n} = 1$ |
$\blacksquare$