Unsigned Stirling Number of the First Kind of Number with Self
Jump to navigation
Jump to search
Theorem
- $\ds {n \brack n} = 1$
where $\ds {n \brack n}$ denotes an unsigned Stirling number of the first kind.
Proof
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\ds {n \brack n} = 1$
Basis for the Induction
$\map P 0$ is the case:
\(\ds {0 \brack 0}\) | \(=\) | \(\ds \delta_{0 0}\) | Unsigned Stirling Number of the First Kind of 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Definition of Kronecker Delta |
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds {k \brack k} = 1$
from which it is to be shown that:
- $\ds {k + 1 \brack k + 1} = 1$
Induction Step
This is the induction step:
\(\ds {k + 1 \brack k + 1}\) | \(=\) | \(\ds k {k \brack k + 1} + {k \brack k}\) | Definition of Unsigned Stirling Numbers of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds k \times 0 + {k \brack k}\) | Stirling Number of Number with Greater | |||||||||||
\(\ds \) | \(=\) | \(\ds {k \brack k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Induction Hypothesis |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: {n \brack n} = 1$
$\blacksquare$
Also see
- Signed Stirling Number of the First Kind of Number with Self
- Stirling Number of the Second Kind of Number with Self
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(48)$