Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k
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Theorem
Let $m, n \in \Z_{\ge 0}$.
- $\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
where:
- $\ds {k \brace m}$ denotes a Stirling number of the second kind
- $\dbinom n k$ denotes a binomial coefficient.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \sum_k {k \brace m} \binom 0 k\) | \(=\) | \(\ds \sum_k {k \brace m} \delta_{0 k}\) | Zero Choose n | |||||||||||
\(\ds \) | \(=\) | \(\ds {0 \brace m}\) | all terms vanish except for $k = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{0 m}\) | Stirling Number of the Second Kind of 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds \delta_{1 \paren {m + 1} }\) | Definition of Kronecker Delta | |||||||||||
\(\ds \) | \(=\) | \(\ds {1 \brace m + 1}\) | Stirling Number of the Second Kind of 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds {0 + 1 \brace m + 1}\) |
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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_k {k \brace m} \binom r k = {r + 1 \brace m + 1}$
from which it is to be shown that:
- $\ds \sum_k {k \brace m} \binom {r + 1} k = {r + 2 \brace m + 1}$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_k {k \brace m} \binom {r + 1} k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {k \brace m} \paren {\binom r k + \binom r {k - 1} }\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k {k \brace m} \binom r k + \sum_k {k \brace m} \binom r {k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m + 1} + \sum_k {k \brace m} \binom r {k - 1}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m + 1} + \sum_k \paren {m {k - 1 \brace m} + {k - 1 \brace m - 1} } \binom r {k - 1}\) | Definition of Stirling Numbers of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m + 1} + m \sum_k {k - 1 \brace m} \binom r {k - 1} + \sum_k {k - 1 \brace m - 1} \binom r {k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m + 1} + m \sum_k {k \brace m} \binom r k + \sum_k {k \brace m - 1} \binom r k\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds {r + 1 \brace m + 1} + m {r + 1 \brace m + 1} + {r + 1 \brace m}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 1} {r + 1 \brace m + 1} + {r + 1 \brace m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {r + 2 \brace m + 1}\) | Definition of Stirling Numbers of the Second Kind |
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall m, n \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(52)$