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F(x)=1/B(p,q)×X^(p-1) /(1+X)^(p+q) |
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Answer» F(x) = \(\frac{1}{B(p,q)}\) \(\frac{X^{p-1}}{(1+X)^{p+q}}\) Harmonic mean = \(\frac{\displaystyle\sum_{n=0}^{\infty} F(X_n)}{\displaystyle\sum_{n=0}^{\infty} \frac{F(X_n)}{X}}\) and we know sum change into integration in continuous series. Harmonic mean of F(x) = \(\frac{\frac{1}{B(p,q)}\int_0^\infty \frac{X^{p-1}}{(1+X)^{p+q}}}{\frac{1}{B(p,q)}\int_0^\infty \frac{X^{p-1}}{X(1+X)^{p+q}}}\) = \(\frac{\int_0^\infty\frac{X^{p-1}}{(1+X)^{p+q}}}{\int_0^\infty\frac{X^{(p-1)-1}}{(1+X)^{p+q+1}}}\) = \(\frac{B(p,q)}{B(p-1,q+1)}\) \(\Big(\because \int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} = B(m,n)\Big)\) = \(\frac{p!q!}{(p+q)!}\) x \(\frac{(p+q)!}{(p-1)!(q+1)!}\) = \(\frac{p(p-1)!q!}{(p-1)!(q+1)q!}\) = \(\frac{p}{q+1}\) |
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