Stirling Number of the Second Kind of n with n-1

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Theorem

Let $n \in \Z_{> 0}$ be an integer greater than $0$.

Then:

$\ds {n \brace n - 1} = \binom n 2$

where:

$\ds {n \brace n - 1}$ denotes a Stirling number of the second kind
$\dbinom n 2$ denotes a binomial coefficient.


Proof

The proof proceeds by induction.


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

$\ds {n \brace n - 1} = \binom n 2$


Basis for the Induction

$\map P 1$ is the case:

\(\ds {1 \brace 0}\) \(=\) \(\ds \delta_{1 0}\) Stirling Number of the Second Kind of 1
\(\ds \) \(=\) \(\ds 0\) Definition of Kronecker Delta
\(\ds \) \(=\) \(\ds \binom 1 2\) Definition of Binomial Coefficient


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds {k \brace k - 1} = \binom k 2$


from which it is to be shown that:

$\ds {k + 1 \brace k} = \binom {k + 1} 2$


Induction Step

This is the induction step:


\(\ds {k + 1 \brace k}\) \(=\) \(\ds k {k \brace k} + {k \brace k - 1}\) Definition of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds k \times 1 + {k \brace k - 1}\) Stirling Number of the Second Kind of Number with Self
\(\ds \) \(=\) \(\ds k + \binom k 2\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k + \frac {k \paren {k - 1} } 2\) Binomial Coefficient with Two
\(\ds \) \(=\) \(\ds \frac {k \paren {k - 1} + 2 k} 2\)
\(\ds \) \(=\) \(\ds \frac {k^2 - k + 2 k} 2\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1} k} 2\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 2\) Binomial Coefficient with Two


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


Therefore:

$\ds \forall n \in \Z_{\ge 1}: {n \brace n - 1} = \binom n 2$

$\blacksquare$


Also see


Sources

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