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Question Number 655 by 123456 last updated on 19/Feb/15
f(z)=((1−((1−z^2 )/(2+z^2 )))/(2+((1−z^2 )/(2+z^2 ))))  ((f(0)+f(1))/2)−f((1/2))=?
$${f}\left({z}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{2}+{z}^{\mathrm{2}} }}{\mathrm{2}+\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{2}+{z}^{\mathrm{2}} }} \\ $$$$\frac{{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)}{\mathrm{2}}−{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$
Answered by prakash jain last updated on 19/Feb/15
f(0)=((1−(1/2))/(2+(1/2)))=(1/5)  f(1)=(1/2)  f(0)=((1−(1/3))/(2+(1/3)))=(2/7)  ((f(0)+f(1))/2) − f((1/2))=(((1/5)+(1/2))/2)−(2/7)=(7/(20))−(2/7)=(9/(140))
$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\frac{{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)}{\mathrm{2}}\:−\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{7}}=\frac{\mathrm{7}}{\mathrm{20}}−\frac{\mathrm{2}}{\mathrm{7}}=\frac{\mathrm{9}}{\mathrm{140}} \\ $$

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