Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m
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Theorem
Let $m, n \in \Z_{\ge 0}$.
- $\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
where:
- $\sqbrk {n \ge m}$ is Iverson's convention
- $\ds {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
- $\ds {k \brace m}$ denotes a Stirling number of the second kind.
Proof
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \) | \(\) | \(\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \delta_{1 \paren {k + 1} } {k \brace m} \paren {-1}^{k - m}\) | Unsigned Stirling Number of the First Kind of 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \delta_{0 k} {k \brace m} \paren {-1}^{k - m}\) | Definition of Kronecker Delta | |||||||||||
\(\ds \) | \(=\) | \(\ds {0 \brace m} \paren {-1}^{- m}\) | All terms but where $k = 0$ vanish | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{0 m} \paren {-1}^{- m}\) | Stirling Number of the Second Kind of 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{0 m}\) | multiplier irrelevant when $m \ne 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {0 \ge m} \dfrac {0!} {m!}\) | Factorial of Zero |
So $\map P 0$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_k \sqbrk {r + 1 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r \ge m} \dfrac {r!} {m!}$
from which it is to be shown that:
- $\ds \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {\paren {r + 1} {r + 1 \brack k + 1} + {r + 1 \brack k} } {k \brace m} \paren {-1}^{k - m}\) | Definition of Unsigned Stirling Numbers of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {r + 1} \sum_k {r + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {r \ge m} \dfrac {\paren {r + 1} r!} {m!} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \delta_{m \paren {r + 1} }\) | First Inversion Formula for Stirling Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {cases} \dfrac {\paren {r + 1}!} {m!} & : r + 1 > m \\ 1 & : r + 1 = m \\ 0 & : r + 1 < m \end {cases}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}\) |
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
$\blacksquare$
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: Exercise $61$