# Monthly Archives: December 2016

## 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 … Continue reading

## Proof of beta function using convolution

Prove the following $$\beta(x, y)=\int^{1}_{0}t^{x-1}\, (1-t)^{y-1}\,dt= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma{(x+y)}}$$ $$\textit{proof}$$ Let us choose some functions $f$ and $g$ $$f(t) = t^{x} \,\, , \, g(t) = t^y$$ Hence we get $$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds$$ So by the convolution rule we have the … Continue reading

$$\Gamma(z)\,= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}$$ $\textit{where } \gamma \textit{ is the Euler constant}$ $$\textit{proof}$$ Take logarithm to the Euler representation $$\log z\Gamma(z) =\lim_{n\to \infty}z\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right)-\sum_{k=1}^n\log\left(1+\frac{z}{k}\right)$$ Note the alternating sum $$\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right) = \log(n+1)$$ Hence we have … Continue reading
$$\Gamma(z) = \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = \frac{1}{z}\prod^\infty_{k=1}\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}$$ $$\textit{proof}$$ Note that $$\Gamma(z+n+1) = \Gamma(z+1) \prod^{n}_{k=1}(k+z)$$ Which indicates that $$\prod^{n}_{k=1}(k+z) = \frac{\Gamma(z+n+1)}{z\Gamma(z)}$$ Also note that $$\prod^{n}_{k=1}k = n!$$ Hence we have \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = … Continue reading