Question 30
Given the linear fractional transformation $$
f_1(x)=\frac{2x-1}{x+1},
$$ define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,$ $2,$ $3,$ $\ldots.$ Assuming $f_{35}(x)=f_5(x),$ it follows that $f_{28}(x)$ is equal to:
f_1(x)=\frac{2x-1}{x+1},
$$ define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,$ $2,$ $3,$ $\ldots.$ Assuming $f_{35}(x)=f_5(x),$ it follows that $f_{28}(x)$ is equal to: