For the functions

f1(x)=(2x-1)/(x+1)

f(n+1)(x)=f1(fn(x)) for n>=1

it can be shown that f35=f5.Find the simple expression for f28.

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- Mar 16th 2011, 05:22 AMchris86Find simple expression for f28
For the functions

f1(x)=(2x-1)/(x+1)

f(n+1)(x)=f1(fn(x)) for n>=1

it can be shown that f35=f5.Find the simple expression for f28. - Mar 16th 2011, 09:26 AMOpalg
If $\displaystyle f_{35} = f_5$ then $\displaystyle f_1$ composed with itself 35 times is the same as $\displaystyle f_1$ composed with itself 5 times. So $\displaystyle f_1$ composed with itself 30 times must be the identity function. Therefore $\displaystyle f_{28} = f_{-2}$. So find the inverse function of $\displaystyle f_1$ and take its composition with itself. That will give you $\displaystyle f_{-2}.$