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

Example of Falling Factorial as Sum of Stirling Numbers of First Kind by Powers

 * $x^{\underline 5} = x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x$

and so:
 * $\dbinom x 5 = \dfrac 1 {120} \left({x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}\right)$

Proof
Follows directly from Stirling's triangle of the first kind (unsigned):


 * $\displaystyle \left[{5 \atop 5}\right] = 1$
 * $\displaystyle \left[{5 \atop 4}\right] = 10$
 * $\displaystyle \left[{5 \atop 3}\right] = 35$
 * $\displaystyle \left[{5 \atop 4}\right] = 50$
 * $\displaystyle \left[{5 \atop 4}\right] = 24$

Also from Stirling's triangle of the first kind (signed):


 * $s \left({5, 5}\right) = 1$
 * $s \left({5, 4}\right) = -10$
 * $s \left({5, 3}\right) = 35$
 * $s \left({5, 2}\right) = -50$
 * $s \left({5, 1}\right) = 24$

By definition of binomial coefficient:
 * $\dbinom x 5 = \dfrac {x^{\underline 5} } {5!} = \dfrac {x^{\underline 5} } {120}$

Hence:
 * $\dbinom x 5 = \dfrac 1 {120} \left({x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}\right)$