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:


 * $\ds x^n = \sum_k {n \brace k} x^{\underline k}$

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

By definition of binomial coefficient:

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