Let \(f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}.\) What is \(f^{2012}(x)\), where the function is being applied 2012 times?
Note: This notation indicates repeated composition of functions, not exponentiation of functions. For example, f^2(x)=f(f(x)) and not \(f^{2012}(x)\). Similarly,f^3(x)=f(f(f(x))).
Please help! I'm really confused and would like to learn how to do this!
+118608
\(f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f(f(x)) = \frac{\frac{x - \sqrt{3}}{x\sqrt{3} + 1} - \sqrt{3}}{\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\sqrt{3} + 1}\\ \text{This simplifies to}\\ f(f(x)) =\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f(f(f(x)) )=\frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3\frac{x+\sqrt3}{1-\sqrt3x}}\\~\\ \text{This simplifies to}\\ f(f(f(x)) )=x\\~\\ \text{now there is a loop happening.}\\~\\ f^0(x)=x\\ f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f^2(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f^3(x)=x\\~\\ \)
2012=2(mod3)
so it seems that
\(f^{2012}(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\\)
Coding
f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\
f(f(x)) = \frac{\frac{x - \sqrt{3}}{x\sqrt{3} + 1} - \sqrt{3}}{\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\sqrt{3} + 1}\\
\text{This simplifies to}\\
f(f(x)) =\frac{x+\sqrt3}{1-\sqrt3x}\\~\\
f(f(f(x)) )=\frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3\frac{x+\sqrt3}{1-\sqrt3x}}\\~\\
\text{This simplifies to}\\
f(f(f(x)) )=x\\~\\
\text{now there is a loop happening.}\\~\\
f^0(x)=x\\
f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\
f^2(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\
f^3(x)=x\\~\\
