Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$.
I was able to prove that $$ \Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$ using the Legendre's duplication formula. But I can't do the same to $$\Gamma\left ( -k+\frac{1}{2} \right )=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}.\tag{$k\geq 1$}$$ If possible, I'd like you to give a hint. If it is not possible to use Legendre's duplication formula, then I tried this way, for $n\geq 1$, \begin{align*} \Gamma\left ( -n+\frac{1}{2} \right )&=\left ( -n-\frac{1}{2} \right )\Gamma\left ( -n-\frac{1}{2} \right )\\ &=\dots\\ &=\left ( \frac{1}{2} \right )\left ( \frac{3}{2} \right )\cdots\left ( \frac{-2n-3}{2} \right )\left ( \frac{-2n-1}{2} \right )\Gamma\left ( \frac{1}{2} \right ) \end{align*} and not anymore. How do I count how many factors there are to right side beside $\Gamma(1/2)$?
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$\begingroup$I think that the reflection formula
$$ \Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin{(\pi z)}} \qquad z \not \in \mathbb{Z}, $$ should work better here, given that you already have computed $\displaystyle{\Gamma\left(k + \frac{1}{2} \right)}$.
$\endgroup$ 1 $\begingroup$If you are willing to accept that for $z \notin \Bbb Z$
$$\Gamma (z) \Gamma (1-z) = \frac \pi {\sin \pi z}$$
then letting $z = \frac 1 2 - n$ leads to
$$\Gamma \left( \frac 1 2 - n \right) \Gamma \left( \frac 1 2 + n \right) = \frac \pi {\sin \left( \frac \pi 2 - n \pi \right)} = \frac \pi {\cos n \pi} = (-1)^n \pi$$
and since you already know
$$\Gamma \left( n + \frac 1 2 \right) = \frac {1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} {2^n} \sqrt{\pi}$$
then
$$\Gamma \left( \frac 1 2 - n \right) = \frac {(-1)^n 2^n \sqrt \pi} {1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} .$$
Alternatively, your own attempt based upon $\Gamma (z) = (z-1) \Gamma (z-1)$ is fine, too, because
$$\Gamma \left( \frac 1 2 \right) = \left( \frac 1 2 - 1 \right) \Gamma \left( \frac 1 2 - 1 \right) = \\ \left( \frac 1 2 -1 \right) \left( \frac 1 2 - 2 \right) \Gamma \left( \frac 1 2 - 2 \right) = \dots \\ \underbrace { \left( \frac 1 2 -1 \right) \left( \frac 1 2 - 2 \right) \dots \left( \frac 1 2 - n \right) } _{n \text{ factors}} \Gamma \left( \frac 1 2 - n \right) = \\ \underbrace { \left( - \frac 1 2 \right) \left( - \frac 3 2 \right) \dots \left( - \frac {2n-1} 2 \right) } _{n \text{ factors}} \Gamma \left( \frac 1 2 - n \right) = (-1)^n \frac {1 \cdot 3 \cdot \dots \cdot (2n-1)} {2^n} \Gamma \left( \frac 1 2 - n \right)$$
whence the desired result follows.
$\endgroup$ 2 $\begingroup$Is this Possible? Since, $\Gamma(a)=\frac{\Gamma(a+n)}{(a)_n}$ You can deduce that \begin{eqnarray*} \Gamma(-m+\frac{1}{2})&=&\frac{\Gamma(\frac{1}{2})}{(-m+\frac{1}{2})_m}\\ &=&\frac{\sqrt{\pi}}{\left(\frac{-2m+1}{2}\right)\cdot\left(\frac{-2m+3}{2}\right)\cdot\left(\frac{-2m+5}{2}\right)\cdots\left(\frac{1}{2}\right)}\\ &=&\frac{(-1)^m\cdot 2^m\cdot\sqrt{\pi}}{(2m-1)\cdot(2m-3)\cdots 3\cdot 2\cdot 1} \end{eqnarray*}
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