Function composition: $f^{653}(56)=?$

Question

Let $f(x) = \frac1{(1-x)}$.

Define the function $f^r$ to be $f^r(x) = f(f(f(...f(f(x)))))$.

Find $f^{653}(56)$.

What I've done:

I started with r=1,2,3 and noticed the following pattern: $$f^r(x)= \left\{ \begin{array}{c} \frac1{1-x}, when \ r\equiv 1\pmod 3 \\ \frac{x-1}x, when \ r\equiv 2\pmod 3 \\ x, \ when \ r\equiv 0\pmod 3 \end{array} \right. $$

As $653\equiv 2\pmod 3$, $\\$ $f^{653}(56) = \frac{55}{56}$

BUT how can I prove that I'm right? By induction? I don't know what to do then, when I go from $r$ to $r+1$.

Could you please share with me your reasoning by solving this problem?

PS: This problem is from the book "How to think like a mathematician" by Kevin Houston.

Answer 1

The conclusion $f^3(x)=x$ $(=f^0(x))$ applies whatever $x$ is.

Use induction on $r$ to show that $f^{3r+s}(x)=f^s(x)$

Answer 2

The ingredient that you’re missing is the precise meaning of congruence modulo 3. Since $653=3k+2$, where the particular value of $k$ doesn’t concern us, and since $f^3(x)=x$, you get $f^{653}(x)=f^{3k+2}(x)=(f^3)^k[f^2(x)]$; but since $f^3$ is the identity, its k-fold iterate is identity too, and we can just erase that part of the expression: $f^{653}(x)=f^2(x)$, and you’re done.