# 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:

 $\ds {5 \brace 5}$ $=$ $\ds 1$ $\ds {5 \brace 4}$ $=$ $\ds 10$ $\ds {5 \brace 3}$ $=$ $\ds 25$ $\ds {5 \brace 2}$ $=$ $\ds 16$ $\ds {5 \brace 1}$ $=$ $\ds 1$

By definition of binomial coefficient:

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

Hence:

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

$\blacksquare$