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

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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:

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

$\blacksquare$


Sources

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