# Relation between Signed and Unsigned Stirling Numbers of the First Kind

## 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 $\paren {-1}^{2 n} = 1$

$\blacksquare$