Question Number 195357 by SirMUTUKU last updated on 31/Jul/23
Answered by Spillover last updated on 31/Jul/23
![(E/(1−π))=(√((h−0.5)/(1−h))) ((E/(1−π)))^2 =((h−0.5)/(1−h)) (E^2 /((1−π)^2 ))=((h−0.5)/(1−h)) E^2 (1−h)=(h−0.5)(1−π)^2 E^2 −E^2 h=h(1−π)^2 −0.5(1−π)^2 E^2 +0.5(1−π)^2 =E^2 h+h(1−π)^2 E^2 +0.5(1−π)^2 =h[E^2 +(1−π)^2 ] h=((E^2 +0.5(1−π)^2 )/(E^2 +(1−π)^2 ))](https://www.tinkutara.com/question/Q195384.png)
$$\frac{{E}}{\mathrm{1}−\pi}=\sqrt{\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}}}\: \\ $$$$\left(\frac{{E}}{\mathrm{1}−\pi}\right)^{\mathrm{2}} =\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}} \\ $$$$\frac{{E}^{\mathrm{2}} }{\left(\mathrm{1}−\pi\right)^{\mathrm{2}} }=\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}} \\ $$$${E}^{\mathrm{2}} \left(\mathrm{1}−{h}\right)=\left({h}−\mathrm{0}.\mathrm{5}\right)\left(\mathrm{1}−\pi\right)^{\mathrm{2}} \\ $$$${E}^{\mathrm{2}} −{E}^{\mathrm{2}} {h}={h}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} −\mathrm{0}.\mathrm{5}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} \\ $$$${E}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} ={E}^{\mathrm{2}} {h}+{h}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} \\ $$$${E}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} ={h}\left[{E}^{\mathrm{2}} +\left(\mathrm{1}−\pi\right)^{\mathrm{2}} \right] \\ $$$${h}=\frac{{E}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}\left(\mathrm{1}−\pi\right)^{\mathrm{2}} }{{E}^{\mathrm{2}} +\left(\mathrm{1}−\pi\right)^{\mathrm{2}} } \\ $$
Answered by Frix last updated on 31/Jul/23
$${h}=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{2}\left(\left({E}−\mathrm{1}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} \right)} \\ $$