Vrftjkry

How interpret the combination

$binom{frac{1}{2}}{n}$ when $n$ is a positive integer ?

The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.

In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$

The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$

Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$