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:

$\displaystyle {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$

where:

$\displaystyle {n \brack m}$ denotes an unsigned Stirling number of the first kind
$\map s {n, m}$ denotes a signed Stirling number of the first kind.


Proof

\(\displaystyle \sum_k \map s {n, k} x^k\) \(=\) \(\displaystyle x^{\underline n}\) Definition of Signed Stirling Numbers of the First Kind
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \paren {-1}^{n - k} {n \brack k} x^k\) Definition of Unsigned Stirling Numbers of the First Kind
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map s {n, m}\) \(=\) \(\displaystyle \paren {-1}^{n - m} {n \brack m}\) Comparing coefficients of $x^m$
\(\displaystyle \leadsto \ \ \) \(\displaystyle {n \brack m}\) \(=\) \(\displaystyle \paren {-1}^{m - n} \map s {n, m}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {-1}^{n + m} \map s {n, m}\) as $\paren {-1}^{2 n} = 1$

$\blacksquare$