Difference between revisions of "MIE 2015/Day 2/Problem 2"

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Latest revision as of 16:28, 11 January 2018

Problem 2

Let the functions $f_n$, for $n\in\{0,1,2,3,...\}$, such that $f_0(x)=\frac{1}{1-x}$ and $f_n(x)=f_0(f_{n-1}(x))$, for every $n\geq1$.

Compute $f_{2016}(2016)$.


Solution

First, let see the case $n=1$


$f_1(x)=f_0(f_{1-1}(x))$


$f_1(x)=\frac{1}{1-\frac{1}{1-x}}$


$f_1(x)=\frac{x-1}{x}$


Now, when $n=2$


$f_2(x)=f_0(f_1(x))$


$f_2(x)=\frac{1}{1-\frac{x-1}{x}}$


$f_2(x)=x$


Now, when $n=3$


$f_3(x)=\frac{1}{1-x}$


At this point it's easy to see the pattern. So, we just find the remainder of 2016 by 3.


$f_{2016}(x)=f_0(x)$


$f_{2016}(2016)=\frac{1}{1-2016}$


$f_{2016}(2016)=-\frac{1}{2015}$

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