Second Inversion Formula for Stirling Numbers
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Theorem
For all $m, n \in \Z_{\ge 0}$:
- $\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
where:
- $\ds {n \brace k}$ denotes a Stirling number of the second kind
- $\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
- $\delta_{m n}$ denotes the Kronecker delta.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \sum_k {0 \brace k} {k \brack m} \paren {-1}^{0 - k}\) | \(=\) | \(\ds \sum_k \delta_{k 0} {k \brack m} \paren {-1}^{-k}\) | Definition of Stirling Numbers of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds {0 \brack m}\) | as all other terms vanish by $\delta_{k 0}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{m 0}\) | Definition of Unsigned Stirling Numbers of the First Kind |
Thus $\map P 0$ has been shown to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true for all $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
- $\ds \forall m \in \Z_{\ge 0}: \sum_k {r \brace k} {k \brack m} \paren {-1}^{r - k} = \delta_{m r}$
from which it is to be shown that:
- $\ds \forall m \in \Z_{\ge 0}: \sum_k {r + 1 \brace k} {k \brack m} \paren {-1}^{r + 1 - k} = \delta_{m \paren {r + 1} }$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_k {r + 1 \brace k} {k \brack m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {k {r \brace k} + {r \brace k - 1} } {k \brack m} \paren {-1}^{r + 1 - k}\) | Definition of Stirling Numbers of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k k {r \brace k} {k \brack m} \paren {-1}^{r + 1 - k} + \sum_k {r \brace k - 1} {k \brack m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k k {r \brace k} \frac 1 k \paren { {k + 1 \brack m} - {k \brack m - 1} } \paren {-1}^{r + 1 - k}\) | Definition of Unsigned Stirling Numbers of the First Kind | |||||||||||
\(\ds \) | \(\) | \(\ds + \sum_k {r \brace k - 1} {k \brack m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {r \brace k} {k + 1 \brack m} \paren {-1}^{r + 1 - k} - \sum_k {r \brace k} {k \brack m - 1} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(\) | \(\ds + \sum_k {r \brace k - 1} {k \brack m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {r \brace k} {k + 1 \brack m} \paren {-1}^{r + 1 - k} - \sum_k {r \brace k} {k \brack m - 1} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(\) | \(\ds + \sum_k {r \brace k} {k + 1 \brack m} \paren {-1}^{r + 1 - k + 1}\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {r \brace k} {k + 1 \brack m} \paren {-1}^{r + 1 - k} - \sum_k {r \brace k} {k \brack m - 1} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(\) | \(\ds -\sum_k {r \brace k} {k + 1 \brack m} \paren {-1}^{r + 1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {r \brace k} {k \brack m - 1} \paren {-1}^{r - k}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{\paren {m - 1} r}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{m \paren {r + 1} }\) | Definition of Kronecker Delta |
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
$\blacksquare$
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: $(47)$