Saved Bookmarks
| 1. |
Prove that \( \beta(m, 1 / 2)=2^{2 m-1} \beta(m, m) \) |
|
Answer» β(m, 1/2) = \(\frac{Γ(m)Γ(1/2)}{Γ(m+1/2)}\) \(\left(\because\beta=\frac{\Gamma(m)(n)}{\Gamma(m+n)}\right)\) \(=\cfrac{\Gamma(m)\Gamma(m)\Gamma(\frac12)}{\Gamma(m)\Gamma(m+\frac12)}\) (Multiplying numerator & denominator by \(\Gamma(m)\)) \(=\cfrac{\Gamma(m)\Gamma(m)\Gamma(\frac12)}{\frac{\sqrt{\pi}}{2^{2m-1}}\Gamma(2m)}\) (\(\because\) \(\Gamma(m)\Gamma(m+\frac12)=\frac{\pi}{2^{2m-1}}\Gamma(2m)\)) \(=2^{2m-1}\frac{\Gamma(m)\Gamma(m)}{\Gamma(2m)}\frac{\sqrt\pi}{\sqrt\pi}\) (\(\because\Gamma(\frac12)=\sqrt\pi\)) = 22m-1β (m, m) \((\because \beta(m, m) = \frac{\Gamma(m)\Gamma(m)}{\Gamma(2m)})\) Hence Proved |
|