Stirling Numbers of the First Kind/Examples/5th Falling Factorial

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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

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