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How interpret the combination

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

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

3 Answers 3

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The general definition of $\binom\alpha k$ for $k\in\Bbb N$ and $\alpha\in\Bbb C$ is $$\binom\alpha k=\frac1{k!}\prod_{h=0}^{k-1} (\alpha-h)$$

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

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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 $$ \binom{1/2}{n}=\frac{(-1)^{n-1}}{2^{2n}(2n-1)}\frac{(2n)!}{(n!)^2} $$

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