# Stirling Number of the Second Kind of Number with Greater

## Theorem

Let $n, k \in \Z_{\ge 0}$.

Let $k > n$.

Let $\displaystyle {n \brace k}$ denote a Stirling number of the second kind.

Then:

$\displaystyle {n \brace k} = 0$

## Proof 1

By definition, the Stirling numbers of the second kind are defined as the coefficients $\displaystyle {n \brace k}$ which satisfy the equation:

$\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$

where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.

Both of the expressions on the left hand side and right hand side are polynomials in $x$ of degree $n$.

Hence the coefficient $\displaystyle {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.

$\blacksquare$

## Proof 2

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\displaystyle k > n \implies {n \brace k} = 0$

### Basis for the Induction

$\map P 0$ is the case:

$\displaystyle {0 \brace k} = \delta_{0 k}$

So by definition of Kronecker delta:

$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle {0 \brace k} = 0$

and 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 $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:

$\displaystyle k > r \implies {r \brace k} = 0$

from which it is to be shown that:

$\displaystyle k > r + 1 \implies {r + 1 \brace k} = 0$

### Induction Step

This is the induction step:

 $\displaystyle {r + 1 \brace k}$ $=$ $\displaystyle k {r \brace k} + {r \brace k - 1}$ $\displaystyle$ $=$ $\displaystyle r \times 0 + {r \brace k - 1}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle r \times 0 + 0$ Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ $\displaystyle$ $=$ $\displaystyle 0$

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

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$

$\blacksquare$