# Gamma duplication formula proof

$$\Gamma \left(\frac{1}{2}+n\right) = {(2n)! \over 4^n n!} \sqrt{\pi}$$

$$\textit{proof}$$

For the proof we use induction by assuming $n\geq 0$. If $n=0$ we have our basic identity

$$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$

Now we need to prove that

$$\Gamma\left(\frac{1}{2}+n\right) = {(2n)! \over 4^n n!} \sqrt{\pi} \, \, \implies \Gamma\left(\frac{1}{2}+n+1\right) = {(2n+2)! \over 4^{n+1} (n+1)!} \sqrt{\pi}$$

Now we use the reduction formula

$$\Gamma\left(\frac{1}{2}+n+1\right) = \frac{1+2n}{2}\Gamma\left(\frac{1}{2}+n\right)$$

By the inductive step we have

$$\frac{1+2n}{2}\Gamma\left(\frac{1}{2}+n\right) =\frac{1+2n}{2}\cdot {(2n)! \over 4^n n!} \sqrt{\pi}$$

We can multiply and divide by $2n+2$

$$\frac{1+2n}{2}\cdot {(2n)! \over 4^n n!} \sqrt{\pi} \cdot \frac{2n+2}{2n+2} = {(2n+2)! \over 4^{n+1} (n+1)!} \sqrt{\pi}\,$$

This entry was posted in Gamma function and tagged , , . Bookmark the permalink.