Unsigned Stirling Number of the First Kind of Number with Greater
Theorem
Let $n, k \in \Z_{\ge 0}$.
Let $k > n$.
Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
- $\ds {n \brack k} = 0$
Proof 1
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
- $\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
where $x^{\underline n}$ denotes the $n$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 $\ds {n \brack k}$ of $x^k$ where $k > n$ is $0$.
$\blacksquare$
Proof 2
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
- $\ds k > n \implies {n \brack k} = 0$
Basis for the Induction
$\map P 0$ is the case:
- $\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
- $\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brack k} = 0$
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:
- $\ds k > r \implies {r \brack k} = 0$
from which it is to be shown that:
- $\ds k > r + 1 \implies {r + 1 \brack k} = 0$
Induction Step
This is the induction step:
| \(\ds {r + 1 \brack k}\) | \(=\) | \(\ds r {r \brack k} + {r \brack k - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds r \times 0 + {r \brack k - 1}\) | Induction Hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds r \times 0 + 0\) | Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brack k} = 0$
$\blacksquare$