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A function $f(x)$ is defined by $f(x) = f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}.$

Find $f^{2021}(x)$

 Jul 17, 2021
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Solution:

$f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$

\(\begin{align*} f^2(x) &= \frac{\frac{x-\sqrt{3}}{x\sqrt{3}+1}-\sqrt{3}}{\sqrt{3}\left(\frac{x-\sqrt{3}}{x\sqrt{3}+1}\right)+1} \\ &= \frac{\frac{x-\sqrt{3}}{\sqrt{3}x+1}-\sqrt{3}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}+1} \\ &= \frac{\frac{-2x-2\sqrt{3}}{\sqrt{3}x+1}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}+1} \\ &= \frac{-2x-2\sqrt{3}}{\left(x\sqrt{3}+1\right)\left(\frac{\left(x-\sqrt{3}\right)\sqrt{3}}{x\sqrt{3}+1}+1\right)} \\ &= \frac{-2x-2\sqrt{3}}{\frac{2\sqrt{3}x-2}{\sqrt{3}x+1}\left(\sqrt{3}x+1\right)} \\ &= \frac{-2x-2\sqrt{3}}{2\sqrt{3}x-2} \\ &= -\frac{2\left(x+\sqrt{3}\right)}{2\left(\sqrt{3}x-1\right)} \\ &= -\frac{x+\sqrt{3}}{\sqrt{3}x-1} \end{align*}\)

\(\begin{align*} f^3(x) &= f \left(-\frac{x+\sqrt{3}}{\sqrt{3}x-1} \right) \\ &= -\frac{\frac{x-\sqrt{3}}{x\sqrt{3}+1}+\sqrt{3}}{\sqrt{3}\left(\frac{x-\sqrt{3}}{x\sqrt{3}+1}\right)-1} \\ &= \frac{\frac{x-\sqrt{3}}{\sqrt{3}x+1}+\sqrt{3}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1} \\ &= -\frac{\frac{4x}{\sqrt{3}x+1}}{\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1} \\ &= -\frac{4x}{\left(\sqrt{3}x+1\right)\left(\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\sqrt{3}x+1}-1\right)} \\ &= -\frac{4x}{\left(-\frac{4}{\sqrt{3}x+1}\right)\left(\sqrt{3}x+1\right)} \\ &= \frac{4x}{\left(x\sqrt{3}+1\right)\frac{4}{x\sqrt{3}+1}} \\ &= \frac{4x}{4} \\ &= x \end{align*}\)

\(\begin{align*} f^4(x) &= f(f^3(x)) \\ &= f(x) \\ &= \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \end{align*}\)

$f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$
$f^2(x) = -\frac{x+\sqrt{3}}{\sqrt{3}x-1}$
$f^3(x) = x$
$f^4(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$

$2021 \pmod 3 \equiv 2 \pmod 3$

$f^{2021}(x) = f^2(x) = \boxed{-\frac{x+\sqrt{3}}{\sqrt{3}x-1}}$

 Jul 17, 2021

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