# First Inversion Formula for Stirling Numbers

## Theorem

For all $m, n \in \Z_{\ge 0}$:

$\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$

where:

$\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind
$\displaystyle \left\{ {k \atop m}\right\}$ denotes a Stirling number of the second kind
$\delta_{m n}$ denotes the Kronecker delta.

## Proof

The proof proceeds by strong induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \sum_k \left[{0 \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{0 - k}$ $=$ $\displaystyle \sum_k \delta_{k 0} \left\{ {k \atop m}\right\} \left({-1}\right)^{-k}$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle \left\{ {0 \atop m}\right\}$ as all other terms vanish by $\delta_{k 0}$ $\displaystyle$ $=$ $\displaystyle \delta_{m 0}$ Definition of Stirling Numbers of the Second Kind

Thus $P \left({0}\right)$ has been shown to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({j}\right)$ is true for all $j$ such $0 \le j \le r$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \forall j: 0 \le j \le r: \forall m \in \Z_{\ge 0}: \sum_k \left[{j \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{j - k} = \delta_{m j}$

from which it is to be shown that:

$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{ {r + 1} \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} = \delta_{m \left({r + 1}\right)}$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_k \left[{ {r + 1} \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k}$ $\displaystyle$ $=$ $\displaystyle \sum_k \left({\left[{r \atop k - 1}\right] + r \left[{r \atop k}\right]}\right) \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k}$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle \sum_k \left[{r \atop k - 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} - r \sum_k \left[{r \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r - k}$ $\displaystyle$ $=$ $\displaystyle \sum_k \left[{r \atop k - 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} - r \delta_{m r}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_k \left[{r \atop k - 1}\right] \left({\left\{ {k - 1 \atop m - 1}\right\} + m \left\{ {k - 1 \atop m}\right\} }\right) \left({-1}\right)^{r + 1 - k} - r \delta_{r m}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle \sum_k \left[{r \atop k - 1}\right] \left\{ {k - 1 \atop m - 1}\right\} \left({-1}\right)^{r - \left({k - 1}\right)} + m \sum_k \left[{r \atop k - 1}\right] \left\{ {k - 1 \atop m}\right\} \left({-1}\right)^{r - \left({k - 1}\right)} - r \delta_{r m}$ $\displaystyle$ $=$ $\displaystyle \delta_{r \left({m - 1}\right)} + m \delta_{r m} - r \delta_{r m}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \delta_{m \left({r + 1}\right)} + \left({m - r}\right) \delta_{r m}$ $\displaystyle$ $=$ $\displaystyle \delta_{m \left({r + 1}\right)}$ as $\left({m - r}\right) \delta_{r m}$ vanishes for all $r$ and $m$

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$