Stirling Numbers of the Second Kind/Examples/5th Power

Example of Stirling Numbers of the Second Kind

 * $x^5 = x^{\underline 5} + 10 x^{\underline 4} + 25 x^{\underline 3} + 15 x^{\underline 2} + x^{\underline 1}$

and so:
 * $x^5 = 120 \dbinom x 5 + 240 \dbinom x 4 + 150 \dbinom x 3 + 30 \dbinom x 2 + \dbinom x 1$

Proof
From the definition of Stirling numbers of the second kind:


 * $\displaystyle x^n = \sum_k \left\{ {n \atop k}\right\} x^{\underline k}$

Reading the values directly from Stirling's triangle of the second kind:


 * $\displaystyle \left\{ {5 \atop 5}\right\} = 1$
 * $\displaystyle \left\{ {5 \atop 4}\right\} = 10$
 * $\displaystyle \left\{ {5 \atop 3}\right\} = 25$
 * $\displaystyle \left\{ {5 \atop 4}\right\} = 15$
 * $\displaystyle \left\{ {5 \atop 4}\right\} = 1$

By definition of binomial coefficient:


 * $x^{\underline 5} = 5! \dbinom x 5 = 120 \dbinom x 5$
 * $x^{\underline 4} = 4! \dbinom x 4 = 24 \dbinom x 4$
 * $x^{\underline 3} = 3! \dbinom x 3 = 6 \dbinom x 3$
 * $x^{\underline 2} = 2! \dbinom x 2 = 2 \dbinom x 2$
 * $x^{\underline 1} = 1! \dbinom x 1 = \dbinom x 1$

Hence:
 * $x^5 = 120 \dbinom x 5 + 240 \dbinom x 4 + 150 \dbinom x 3 + 30 \dbinom x 2 + \dbinom x 1$