TAOCP 1.2.6 Exercise 44

Exercise 44. [HM20] Using the generalized binomial coefficient suggested in exercise 42, show that

$$ \binom{r}{1/2} = \frac{2^{2r+1}}{\binom{2r}{r}\pi}. $$

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Solution

Exercise 42 suggests extending the binomial coefficient to arbitrary arguments by means of the beta function. Since exercise 41 established

$$ B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, $$

and equation (3) gives

$$ \binom{r}{k} = \frac{r(r-1)\cdots(r-k+1)}{k!}, $$

the corresponding gamma-function form is

$$ \binom{r}{k} = \frac{\Gamma(r+1)} {\Gamma(k+1)\Gamma(r-k+1)}. $$

Substituting $k=\tfrac12$ yields

$$ \binom{r}{1/2} = \frac{\Gamma(r+1)} {\Gamma(3/2)\Gamma(r+1/2)}. $$

Since

$$ \Gamma(3/2) = \frac12\Gamma(1/2) = \frac{\sqrt{\pi}}{2}, $$

this becomes

$$ \binom{r}{1/2} = \frac{2,\Gamma(r+1)} {\sqrt{\pi},\Gamma(r+1/2)}. \tag{1} $$

Now apply the gamma-function representation to $\binom{2r}{r}$:

$$ \binom{2r}{r} = \frac{\Gamma(2r+1)} {\Gamma(r+1)^2}. \tag{2} $$

The duplication formula for the gamma function gives

$$ \Gamma(r+1)\Gamma!\left(r+\frac12\right) = 2^{-2r}\sqrt{\pi},\Gamma(2r+1). \tag{3} $$

Solving (3) for $\Gamma(r+1/2)$,

$$ \Gamma!\left(r+\frac12\right) = \frac{2^{-2r}\sqrt{\pi},\Gamma(2r+1)} {\Gamma(r+1)}. $$

Substitute this into (1):

$$ \binom{r}{1/2} = \frac{2,\Gamma(r+1)} {\sqrt{\pi}} \cdot \frac{\Gamma(r+1)} {2^{-2r}\sqrt{\pi},\Gamma(2r+1)}. $$

After simplification,

$$ \binom{r}{1/2} = \frac{2^{2r+1}\Gamma(r+1)^2} {\pi,\Gamma(2r+1)}. $$

By equation (2),

$$ \frac{\Gamma(r+1)^2}{\Gamma(2r+1)} = \frac{1}{\binom{2r}{r}}. $$

Therefore

$$ \binom{r}{1/2} = \frac{2^{2r+1}} {\binom{2r}{r}\pi}. $$

Hence

$$ \boxed{ \binom{r}{1/2} = \frac{2^{2r+1}} {\binom{2r}{r}\pi} }. $$

This completes the proof.

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