Evaluating gamma functions.

Question

I am a bit lost on Gamma functions and would like to have the following evaluation from a textbook explained to me:

$\frac{\Gamma(k+1/2)}{\sqrt{\pi}} = \frac{\Gamma(k+\frac{1}{2})}{\Gamma(\frac{1}{2})} = (k-\frac{1}{2})(k-\frac{3}{2})...\frac{1}{2} = \frac{1}{2^k}(2k-1)!! = \frac{(2k)!}{4^k*k!}$

The first two equalities I understand, but when it comes to evaluating the factorials I'm not sure, since I assume that $\Gamma(1/2)$ is the irrational constant $sqrt(\pi)$ and can't really be neatly factored out.

Answer 1

We have that: $$\frac{\Gamma(x+½)}{\Gamma(x-½)}=x-½$$ for any $x>0$.

So you can write: \begin{align} \frac{\Gamma(k+½)}{\Gamma(½)}&=(k-½)\frac{\Gamma(k-½)}{\Gamma(½)}\\ &=\frac{2k-1}2\frac{\Gamma(k-½)}{\Gamma(½)} \end{align} which reorganizes as the form you have in your question.

Answer 2

Recall the recursive property of the Gamma function: $$\Gamma(z+1)=z\Gamma(z)$$

This means,

$$\Gamma(1/2+1)=\frac{1}{2}\Gamma(1/2) \\ \Gamma(1/2+2)=\Gamma(1/2+1+1)=\left(\frac{1}{2}+1\right)\Gamma(1/2+1)=\frac{3}{2}\frac{1}{2}\Gamma(1/2)$$ I'll do one more to make sure the pattern is clear. $$\Gamma(1/2+3)=\Gamma(1/2+2+1)=\left(\frac{1}{2}+2\right)\Gamma(1/2+2)=\frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(1/2)$$ Clearly, $$\Gamma(1/2+n)=\Gamma(1/2)\prod_{k=1}^n \frac{2k-1}{2}$$

Therefore, and using the special value $\Gamma(1/2)=\sqrt \pi$,

$$\frac{\Gamma(n+1/2)}{\sqrt \pi}=\color{red}{\frac{1}{\Gamma(1/2)}\Gamma(1/2)}\prod_{k=1}^n \frac{2k-1}{2} \\ =\left(\prod_{k=1}^n\frac{1}{2}\right)\cdot\left(\prod_{k=1}^n(2k-1)\right) \tag{1}$$ $$=\frac{1}{2^n}\prod_{k=1}^n(2k-1)\tag 2$$

The object $\prod_{k=1}^n(2k-1)$ is often written as $(2n-1)!!$ though personally I detest this notation and prefer $(2n-1)!_2$. Anyway, now comes the tricky part. We have to show that $$\prod_{k=1}^n(2k-1)=\frac{(2n)!}{2^nn!}$$

The easiest way to do this is to use a specific example that highlights how this works in general. So, I'll pick $n=4$ to demonstrate. See, $$\prod_{k=1}^4(2k-1)=(2\cdot 1-1)\cdot(2\cdot 2-1)\cdot(2\cdot 3-1)\cdot (2\cdot 4-1)=1\cdot 3\cdot 5\cdot 7$$ What we do now is add in the rest of the numbers on the top and bottom: $$1\cdot 3\cdot 5\cdot 7=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \cdot 7\cdot 8}{2\cdot 4\cdot 6\cdot 8}=\frac{8!}{2\cdot 4\cdot 6\cdot 8}$$ Now take out a two from each factor in the bottom: $$\frac{8!}{2\cdot 4\cdot 6\cdot 8}=\frac{8!}{(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)}$$ Now take all the twos to the outside: $$\frac{8!}{(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)}=\frac{8!}{2^4\cdot (1\cdot 2\cdot 3\cdot 4)}=\frac{8!}{2^44!}=\frac{(2\cdot 4)!}{2^4\cdot 4!}$$ So we have shown $$\prod_{k=1}^{\color{green}{4}}(2k-1)=\frac{(2\cdot \color{green}{4})!}{2^{\color{green}{4}}\color{green}{4}!}$$ And so in general $$\prod_{k=1}^n(2k-1)=\frac{(2n)!}{2^n n!}\tag 3$$ Putting this back into $(2)$ we get $$\frac{1}{2^n}\prod_{k=1}^n(2k-1)=\frac{(2n)!}{4^n n!}$$ And plugging this back into $(1)$ we get $$\boxed{\frac{\Gamma(1/2+n)}{\Gamma(1/2)}=\frac{(2n)!}{4^n n!}}\tag 4$$ As desired.

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By the way, this is just a reshuffled version of Legendre's duplication formula. Substitute $m!=\Gamma(m+1)$: $$\frac{\Gamma(1/2+n)}{\Gamma(1/2)}=\frac{\Gamma(2n+1)}{4^n\Gamma(n+1)}$$ Again use the recursive property of the Gamma function: $$\frac{\Gamma(2n+1)}{4^n\Gamma(n+1)}=\frac{2n\cdot\Gamma(2n)}{2^{2n}n\cdot\Gamma(n)}=\frac{\Gamma(2n)}{2^{2n-1}\Gamma(n)}$$ Substituting $n\to z$ and combining the previous two equations we get $$\frac{\Gamma(1/2+z)}{\Gamma(1/2)}=\frac{\Gamma(2z)}{2^{2z-1}\Gamma(z)}$$ Which is usually written as $$\Gamma(2z)=\frac{2^{2z-1}\Gamma(z)\Gamma(1/2+z)}{\sqrt \pi}$$ As seen on the linked page.