Deriving a DUPLICATION FORMULA by beta functions

Question

May you please prove that $B(n,n+1)=2^{-2n}B(n,1/2)$ and hence derive the duplication formula. I'm totally clueless on how to prove that because I lack enough resources to master the concept. Please help!

Answer 1

Recalling the relationship between the Beta and Gamma functions, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, we can write

$$\begin{align} B(n,n+1)&=\frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}\\\\ &=\frac{n!\,(n-1)!}{(2n)!}\tag1 \end{align}$$

Next, using the functional relationship $\Gamma(x+1)=x\Gamma(x)$ repeatedly reveals

$$\begin{align} \Gamma(n+1/2)&=(n-1/2)(n-3/2)\cdots (1/2)\Gamma(1/2)\\\\ &=\left(\frac{(2n-1)(2n-3)\cdots (1)}{2^n}\right)\,\Gamma(1/2)\\\\ &=\left(\frac{(2n)(2n-1)(2n-2)(2n-3)\cdots (2)(1)}{(2^n)(2^n\,n!)}\right)\,\Gamma(1/2)\\\\ &=\left(\frac{(2n)!}{2^{2n}n!}\right)\Gamma(1/2)\tag 2 \end{align}$$

Using $(2)$ in $(1)$ we obtain

$$B(n,n+1)=2^{-2n}\frac{\Gamma(n)\Gamma(1/2)}{\Gamma(n+1/2)}=2^{-2n}B(n,1/2)\tag 3$$

as was to be shown!

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From $(3)$, we can derive the duplication formula $\Gamma(n)\Gamma(n+1/2)=2^{1-2n}\Gamma(1/2)\Gamma(2n)$ in a straightforward fashion. From $(3)$ we have

$$\frac{\Gamma(n+1)}{\Gamma(2n+1)}=2^{-2n}\frac{\Gamma(1/2)}{\Gamma(n+1/2)} \tag 4$$

Rearranging $(4)$, we find that

$$\Gamma(n+1)\Gamma(n+1/2)=2^{-2n}\Gamma(1/2)\Gamma(2n+1) \tag 5$$

Using the functional relationship $\Gamma(x+1)=x\Gamma(x)$ in $(5)$ yields

$$n\Gamma(n)\Gamma(n+1/2)=2^{-2n}\Gamma(1/2)(2n)\Gamma(2n)\tag 6$$

Simplifying $(6)$, we obtain

$$\Gamma(n)\Gamma(n+1/2)=2^{1-2n}\underbrace{\Gamma(1/2)}_{=\sqrt{\pi}}\Gamma(2n)$$

which can be rearranged to give

$$\Gamma(2n)=2^{2n-1}\frac{\Gamma(n)\Gamma(n+1/2)}{\sqrt \pi}$$

And we are done!

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{B}\pars{n,n + 1} & = {\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} = {\Gamma\pars{n}\Gamma\pars{n + 1} \over \pars{2\pi}^{-1/2}\,2^{2n + 1/2}\,\Gamma\pars{n + 1/2}\Gamma\pars{n + 1}} \label{1}\tag{1} \\[5mm] & = 2^{-2n}\root{\pi}{\Gamma\pars{n} \over \Gamma\pars{n + 1/2}} = 2^{-2n}\,{\Gamma\pars{n}\Gamma\pars{1/2} \over \Gamma\pars{n + 1/2}} = 2^{-2n}\,\mrm{B}\pars{n,{1 \over 2}} \label{2}\tag{2} \end{align}

In \eqref{1}, I used the Gamma Duplication Formula and in \eqref{2} I used the well known result $\ds{\Gamma\pars{1 \over 2}} = \root{\pi}$. See the A & S Table.