Definition of $\binom{\frac{1}{2}}{1}$? How interpret the combination
$\binom{\frac{1}{2}}{n}$ when $n$ is a positive integer ?
 A: The general definition of $\binom\alpha k$ for $k\in\Bbb N$ and $\alpha\in\Bbb C$ is $$\binom\alpha k=\frac1{k!}\prod_{h=0}^{k-1} (\alpha-h)$$
A: Same way as usual binomial.  $\binom{\frac{1}{2}}{n}=\frac{\frac{1}{2}\times\frac{-1}{2}...\times(\frac{1}{2}-n+1)}{n!}$
A: 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}-1\right)\left(\frac{1}{2}-2\right)\dotsm\left(\frac{1}{2}-n+1\right)
&=\frac{(-1)^{n-1}}{2^n}\bigl(1\cdot3\cdot\dots\cdot(2n-3)\bigr) \\
&=\frac{(-1)^{n-1}}{2^n}\bigl(1\cdot3\cdot\dots\cdot(2n-3)\bigr)
  \frac{2\cdot4\cdot\dots\cdot(2n-2)}{2^{n-1}(1\cdot 2\cdot\dots\cdot(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
$$
\binom{1/2}{n}=\frac{(-1)^{n-1}}{2^{2n}(2n-1)}\frac{(2n)!}{(n!)^2}
$$
