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Question Number 132302 by Lordose last updated on 13/Feb/21

Simplify ((𝚪((p/2))𝚪((1/2)))/(𝚪((p/2)+(1/2))))

$$\boldsymbol{\mathrm{Simplify}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$

Answered by Ar Brandon last updated on 13/Feb/21

Γ(m)Γ(m+(1/2))=((√π)/2^(2m−1) )Γ(2m) Γ((p/2)+(1/2))=((√π)/2^(p−1) )∙((Γ(p))/(Γ((p/2)))) ((𝚪((p/2))𝚪((1/2)))/(𝚪((p/2)+(1/2))))=(((√π) Γ((p/2)))/( (√π)Γ(p)))∙Γ((p/2))2^(2p−1) =((Γ^2 ((p/2)))/(Γ(p)))2^(p−1) =Γ^2 ((p/2))(2^(p−1) /((p−1)!))

$$\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\Gamma\left(\mathrm{2m}\right) \\ $$$$\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{p}−\mathrm{1}} }\centerdot\frac{\Gamma\left(\mathrm{p}\right)}{\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)} \\ $$$$\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}\:\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)}{\:\sqrt{\pi}\Gamma\left(\mathrm{p}\right)}\centerdot\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)\mathrm{2}^{\mathrm{2p}−\mathrm{1}} \\ $$$$=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{p}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{p}\right)}\mathrm{2}^{\mathrm{p}−\mathrm{1}} =\Gamma^{\mathrm{2}} \left(\frac{\mathrm{p}}{\mathrm{2}}\right)\frac{\mathrm{2}^{\mathrm{p}−\mathrm{1}} }{\left(\mathrm{p}−\mathrm{1}\right)!} \\ $$

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