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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!

 Sep 15, 2021
 #2
avatar+114826 
+2

 

 

\(f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f(f(x)) = \frac{\frac{x - \sqrt{3}}{x\sqrt{3} + 1} - \sqrt{3}}{\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\sqrt{3} + 1}\\ \text{This simplifies to}\\ f(f(x)) =\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f(f(f(x)) )=\frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3\frac{x+\sqrt3}{1-\sqrt3x}}\\~\\ \text{This simplifies to}\\ f(f(f(x)) )=x\\~\\ \text{now there is a loop happening.}\\~\\ f^0(x)=x\\ f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f^2(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f^3(x)=x\\~\\ \)

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\\~\\

 Sep 16, 2021
 #3
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+1

Thank you so much Melody! Your solution was really clear and easy to understand! :D 

Guest Sep 19, 2021
 #4
avatar+114826 
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Thank you,  I am pleased that it helped.   laugh

Melody  Sep 19, 2021
 #5
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I'm a bit confused as to how you got that $f(f(f(x)))=x\dots$ I put it into a calculator and didn't get that

Guest Sep 25, 2021
 #6
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Nvm, I see now sorry

Guest Sep 25, 2021

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