Sum of Indices of Rising Factorial
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Theorem
- $x^{\overline {m + n} } = x^{\overline m} \paren {x + m}^{\overline n}$
where $x^{\overline m}$ denotes $x$ to the $m$ rising.
Proof
\(\ds x^{\overline {m + n} }\) | \(=\) | \(\ds \prod_{j \mathop = 0}^{m + n - 1} \paren {x + j}\) | Definition of Rising Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = m}^{m + n - 1} \paren {x + j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = 0}^{n - 1} \paren {x + \paren {m + j} }\) | Translation of Index Variable of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds x^{\overline m} \paren {x + m}^{\overline n}\) | Definition of Rising Factorial |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $25$