Power of Complex Number as Summation of Stirling Numbers of Second Kind/Proof

Theorem

Let $z \in \C$ be a complex number whose real part is positive.

Then:

$z^r = \ds \sum_{k \mathop \in \Z} {r \brace r - k} z^{\underline {r - k} }$

where:

$\ds {r \brace r - k}$ denotes the extension of the Stirling numbers of the second kind to the complex plane

$z^{\underline {r - k} }$ denotes $z$ to the $r - k$ falling.

Proof

This theorem requires a proof. In particular: It is unclear exactly how the Stirling numbers are extended to the complex plane. Knuth's exposition is uncharacteristically non-explicit. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. 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 {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $65$