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Let $$f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}.$$ What is $f^{2012}(x)$, where $f^{2012}$ is the function we get when we compose $f$ with itself 2012 times?

I can't just do direct calculations, I need to manipulate the function. But how? I'm really confused right now.

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  • $\begingroup$ Did you try $f^2$ and see if you can find any similarities with $f$? $\endgroup$
    – Riccardo Orlando
    Sep 1, 2016 at 23:26
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    $\begingroup$ It's worth looking up "Möbius transformations" to understand what this has to do with linear algebra. $\endgroup$
    – Thomas Andrews
    Sep 2, 2016 at 0:08

2 Answers 2

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Notice that $f^3(x) = x$. To see that you just have to plug in the previous $f$ and simplify the expressions.

$$f^1(x) = \frac{x+\sqrt{3}}{x\sqrt{3}+1} \\ f^2(x) = \frac{x-\sqrt{3}}{1-x\sqrt{3}} \\ f^3(x) = x $$ Next, notice that $2012 \equiv 2 \pmod{3}$. So $$ f^{2012}(x) = f^2(x) = \frac{x-\sqrt{3}}{1-x\sqrt{3}} $$

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  • $\begingroup$ Love it! Clear and precise! $\endgroup$
    – Dreamer
    Sep 2, 2016 at 18:19
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We certainly don't want to plug the function into itself $2012$ times, so we test the first few to see if there is a pattern:

$f^1(x) = f(x) = \frac{x-\sqrt3}{x\sqrt3+1}$ $f^2(x) = \frac{\left(\frac{x-\sqrt3}{x\sqrt3+1}\right)-\sqrt3}{\left(\frac{x-\sqrt3}{x\sqrt3+1}\right)\sqrt3+1} = \frac{x+\sqrt3}{1-x\sqrt3}$ $f^3(x) = \frac{\left(\frac{x-\sqrt3}{x\sqrt3+1}\right)+\sqrt3}{1-\left(\frac{x-\sqrt3}{x\sqrt3+1}\right)\sqrt3} = x $

This means that $f^4(x) = f(x)$, so the terms repeat every $3$.

$2012/3=670 \text{ remainder } 2$, so $f^{2012}=f^2 = \boxed{\frac{x-\sqrt3}{1-x\sqrt3}}$.

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