$\displaystyle f_{o}(x)=\frac{1}{1-x}$
Let $\displaystyle f_{r}(x)=f(f(f(...(f(x))...))$
Where there are r iterations of the function
So $\displaystyle f_{2}(x)=f(f(f(x)))$
Evaluate $\displaystyle f_{2013}(\pi )$

Hey smokesalot.

Have you looked at finding a continued fraction form of your function?

No, not really. The problem is that I don't completely understand the question
Make that count 205 and help me here!

Originally Posted by smokesalot
$\displaystyle f_{o}(x)=\frac{1}{1-x}$
Let $\displaystyle f_{r}(x)=f(f(f(...(f(x))...))$
Where there are r iterations of the function
So $\displaystyle f_{2}(x)=f(f(f(x)))$
Evaluate $\displaystyle f_{2013}(\pi )$
This has a really nice simple answer.

First calculate the explicit forms of $\displaystyle f_1,~f_2,~\&~f_3$.

For example: $\displaystyle f_1(x)=f_0(f_0(x))=\frac{x-1}{x}$.

Originally Posted by Plato
This has a really nice simple answer.

First calculate the explicit forms of $\displaystyle f_1,~f_2,~\&~f_3$.

For example: $\displaystyle f_1(x)=f_0(f_0(x))=\frac{x-1}{x}$.
$\displaystyle f_{1}(x)=\frac{x-1}{x}$

$\displaystyle f_{2}(x)=x$

$\displaystyle f_{3}(x)=\frac{1}{1-x}$

$\displaystyle f_{4}(x)=\frac{x-1}{x}$

I am working on it now. I think I can do it!!!

does that mean f_{0} = f_{2013}? then the answer is $\displaystyle \frac{1}{1-\pi }$
Am I correct?

does that mean f_{0} = f_{2013}? then the answer is $\displaystyle \frac{1}{1-\pi }$
$\displaystyle f_N(x)=f_{mod(N,3)}(x)$