SOLUTION: Let f(x) = {x - sqrt{3}}/{xsqrt{3} + 1}. What is f^{2012}(x), where the function is being applied 2012 times? This notation indicates repeated composition of functions, not exp

Algebra ->  Rational-functions -> SOLUTION: Let f(x) = {x - sqrt{3}}/{xsqrt{3} + 1}. What is f^{2012}(x), where the function is being applied 2012 times? This notation indicates repeated composition of functions, not exp      Log On


   

Question 1178101: Let
f(x) = {x - sqrt{3}}/{xsqrt{3} + 1}. What is f^{2012}(x), where the function is being applied 2012 times?
This notation indicates repeated composition of functions, not exponentiation of functions. For example,f^2(x)=f(f(x))and not f(x) * f(x). Similarly,f^3(x)=f(f(f(x))).
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

https://www.quora.com/If-f-x-frac-x-sqrt-3-x-sqrt-3-+-1-What-is-f-2012-x

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let f(x) = {x - sqrt{3}}/{xsqrt{3} + 1}.
What is f^{2012}(x), where the function is being applied 2012 times?
This notation indicates repeated composition of functions, not exponentiation of functions.
For example,f^2(x)=f(f(x)) and not f(x) * f(x). Similarly,f^3(x)=f(f(f(x))).
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            This problem is very original --- I never saw such or similar problems before in my practice.

            It allows absolutely unexpected and elegant solution. 


Let  x = tan(a)  (which means "let a = arctan(x)").  Then


    f(x) = %28tan%28a%29+-+sqrt%283%29%29%2F%28tan%28a%29%2Asqrt%283%29%2B1%29 = %28tan%28a%29+-+tan%28pi%2F3%29%29%2F%281%2Btan%28a%29%2Atan%28pi%2F3%29%29 = tan%28a-pi%2F3%29.


It means that taking f(x) for x= tan(a) returns tan%28a-pi%2F3%29.


Obviously, that taking the composition (fof)(x) = f(f(x))      will return  tan%28a-2pi%2F3%29;

                taking the composition (fofof)(x) = f(f(f(x))) will return  tan%28a-3pi%2F3%29;


                . . . . and so on . . . 


                taking the composition f^{2012}(x) = f(f(f...f(x)))...)  will return  tan%28a-2012pi%2F3%29.


It easy to calculate:  2012pi%2F3 = 670%2Api + 2pi%2F3;   THEREFORE  


    tan%28a-2012pi%2F3%29 = tan%28a-670pi+-+2pi%2F3%29 = tan%28a-2pi%2F3%29 = %28tan%28a%29+-+tan%282pi%2F3%29%29%2F%281+%2B+tan%28a%29%2Atan%282pi%2F3%29%29 = %28tan%28a%29-%28-sqrt%283%29%29%29%2F%281%2Btan%28a%29%2A%28-sqrt%283%29%29%29 = %28x%2Bsqrt%283%29%29%2F%281-sqrt%283%29%2Ax%29.    ANSWER



ANSWER.  f^{2012}(x) = %28x%2Bsqrt%283%29%29%2F%281-sqrt%283%29%2Ax%29.


Solved.


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Without any doubts,  it is a  FULL  SCALE  Math  OLYMPIAD  level problem  (!)