# Definition of $\binom{\frac{1}{2}}{1}$?

How interpret the combination

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

• In what context does this come up? Nov 16, 2018 at 21:57
• @MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed. Nov 16, 2018 at 22:01

The general definition of $$\binom\alpha k$$ for $$k\in\Bbb N$$ and $$\alpha\in\Bbb C$$ is $$\binom\alpha k=\frac1{k!}\prod_{h=0}^{k-1} (\alpha-h)$$
Same way as usual binomial. $$\binom{\frac{1}{2}}{n}=\frac{\frac{1}{2}\times\frac{-1}{2}...\times(\frac{1}{2}-n+1)}{n!}$$
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 $$\binom{1/2}{n}=\frac{(-1)^{n-1}}{2^{2n}(2n-1)}\frac{(2n)!}{(n!)^2}$$