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Proposition: Inversion Formulas For Stirling Numbers
Let n,m\ge 0 be integers. The Stirling numbers of the first and second kind obey the following inversion formulas:
\begin{align}
\sum_{k=0}^\infty \left[\begin{array}{c}n\\k\end{array}\right]\left\{\begin{array}{c}k\\m\end{array}\right\}(-1)^{n-k}&=[m=n]\nonumber,\\
\sum_{k=0}^\infty \left\{\begin{array}{c}n\\k\end{array}\right\}\left[\begin{array}{c}k\\m\end{array}\right](-1)^{n-k}&=[m=n]\nonumber.
\end{align}
Above, the
Iverson notation [m=n] is used and means that the sums equal
1 if and only if
m=n, otherwise they equal
0. Please note that the above sums are, in fact, finite, since all summands are
0 if the index
k is above
\min(m,n).
Table of Contents
Proofs: 1
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References
Bibliography
- Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition
