Sum over k of m choose k by -1^m-k by k to the n

From ProofWiki
Jump to navigation Jump to search

Theorem

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

$\ds \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$

where:

$\dbinom m k$ denotes a binomial coefficient
$\ds {n \brace m}$ etc. denotes a Stirling number of the second kind
$m!$ denotes a factorial.


Proof

The proof proceeds by induction on $m$.


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

$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \) \(\) \(\ds \sum_k \binom 0 k \paren {-1}^{0 - k} k^n\)
\(\ds \) \(=\) \(\ds \sum_k \delta_{0 k} \paren {-1}^{- k} k^n\) Zero Choose n
\(\ds \) \(=\) \(\ds \paren {-1}^{- 0} 0^n\) all terms vanish except for $k = 0$
\(\ds \) \(=\) \(\ds \delta_{0 n}\) $0^n = 0$ except when $n = 0$, as $0^0 = 1$
\(\ds \) \(=\) \(\ds 0! {n \brace 0}\) Definition of Stirling Numbers of the Second Kind

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:

$\ds \sum_k \binom r k \paren {-1}^{r - k} k^n = r! {n \brace r}$


from which it is to be shown that:

$\ds \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n = \paren {r + 1}! {n \brace r + 1}$


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n\)
\(\ds \) \(=\) \(\ds \sum_k \paren {\binom r k + \binom r {k - 1} } \paren {-1}^{r + 1 - k} k^n\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sum_k \binom r k \paren {-1}^{r + 1 - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r + 1 - k} k^n\)
\(\ds \) \(=\) \(\ds -\sum_k \binom r k \paren {-1}^{r - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n\)
\(\ds \) \(=\) \(\ds -r! {n \brace r} + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n\) Induction Hypothesis
\(\ds \) \(=\) \(\ds -r! {n \brace r} + \sum_k \binom r k \paren {-1}^{r - k} \paren {k + 1}^n\) Translation of Index Variable of Summation


This needs considerable tedious hard slog to complete it.
In particular: The next step to take is not clear.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


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


Therefore:

$\ds \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_over_k_of_m_choose_k_by_-1%5Em-k_by_k_to_the_n&oldid=511178"