Rising Factorial in terms of Falling Factorial

From ProofWiki
Jump to navigation Jump to search

Theorem

$x^{\overline n} = \paren {x + n - 1}^{\underline n}$

where:

$x^{\underline n}$ is the $n$th falling factorial power of $x$
$x^{\overline n}$ is the $n$th rising factorial power of $x$.


Proof

\(\ds x^{\overline n}\) \(=\) \(\ds \prod_{j \mathop = 0}^{n - 1} \paren {x + j}\) Definition of Rising Factorial
\(\ds \) \(=\) \(\ds x \paren {x + 1} \cdots \paren {x + n - 1}\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \paren {x + n - 1} \paren {x + n - 2} \cdots \paren {x + 1} x\) reversing the order
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^{n - 1} \paren {\paren {x + n - 1} - j}\) Definition of Falling Factorial
\(\ds \) \(=\) \(\ds \paren {x + n - 1}^{\underline n}\) Definition of Continued Product

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Rising_Factorial_in_terms_of_Falling_Factorial&oldid=662644"