<s>[INST] Coefficient power series problem #49 What is the coefficient of $x^n$ in the power series form of $(1-2x)^{1/3}$? This problem is taken from bona chapter 4, third edition.
 [/INST] If $\sum_{n=0}^\infty a_n x^n$ is the Taylor expansion of $f$ about zero, then $a_n = \frac{f^{(n)} (0)}{n!}$.
Here $f(x) = (1-2x)^{\frac{1}{3}}$. A little work shows that $f^{(n)}(x) = (-2)^n (1-2x)^{( \frac{1}{3}-n )} \left( (\frac{1}{3}) (\frac{1}{3}-1) \cdots (\frac{1}{3}-(n-1)) \right)$, and so
$a_n = \frac{f^{(n)} (0)}{n!} = (-2)^n  \frac{ (\frac{1}{3}) (\frac{1}{3}-1) \cdots (\frac{1}{3}-(n-1)) }{n!}$.
Alternatively (although basically the same thing), we have $(1+x)^{\frac{1}{3}} = \sum_{n=0}^\infty \binom{\frac{1}{3}}{n}x^n$, and so $(1-2x)^{\frac{1}{3}} = \sum_{n=0}^\infty \binom{\frac{1}{3}}{n}(-2)^nx^n$, hence $a_n = (-2)^n \binom{\frac{1}{3}}{n}$. The term $\binom{\alpha}{n}$ is defined in http://en.wikipedia.org/wiki/Binomial_series, for example.
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