Stirling Numbers of the First Kind/Examples/5th Falling Factorial
Jump to navigation
Jump to search
Example of Stirling Numbers of the First Kind
- $x^{\underline 5} = x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x$
and so:
- $\dbinom x 5 = \dfrac 1 {120} \paren {x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}$
Proof
Follows directly from Stirling's triangle of the first kind (unsigned):
\(\ds {5 \brack 5}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds {5 \brack 4}\) | \(=\) | \(\ds 10\) | ||||||||||||
\(\ds {5 \brack 3}\) | \(=\) | \(\ds 35\) | ||||||||||||
\(\ds {5 \brack 2}\) | \(=\) | \(\ds 50\) | ||||||||||||
\(\ds {5 \brack 1}\) | \(=\) | \(\ds 24\) |
Also from Stirling's triangle of the first kind (signed):
\(\ds \map s {5, 5}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \map s {5, 4}\) | \(=\) | \(\ds -10\) | ||||||||||||
\(\ds \map s {5, 3}\) | \(=\) | \(\ds 35\) | ||||||||||||
\(\ds \map s {5, 2}\) | \(=\) | \(\ds -50\) | ||||||||||||
\(\ds \map s {5, 1}\) | \(=\) | \(\ds 24\) |
By definition of binomial coefficient:
- $\dbinom x 5 = \dfrac {x^{\underline 5} } {5!} = \dfrac {x^{\underline 5} } {120}$
Hence:
- $\dbinom x 5 = \dfrac 1 {120} \paren {x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients