If $f_{0} (x) = \frac{1}{1 - x}$ $f_0(x)=1/(1-x)$ and $f_{k} (x) = f_{0} (f_{k - 1} (x))$ $f_k(x)=f_0(f_(k-1)(x))$ what is the value of $f_{2016} (2016)$ $f_(2016)(2016)$ ?

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If $f_{0} (x) = \frac{1}{1 - x}$ $f_0(x)=1/(1-x)$ and $f_{k} (x) = f_{0} (f_{k - 1} (x))$ $f_k(x)=f_0(f_(k-1)(x))$ what is the value of $f_{2016} (2016)$ $f_(2016)(2016)$ ?

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Answer 1 · 2016-10-22T19:27:04

$f_{2016} (2016) = - \frac{1}{2015}$ $f_2016(2016)=-1/2015$

Explanation:

$f_{2} (x) = \frac{1}{1 - f_{1} (x)}$ $f_2(x) = 1/(1-f_1(x))$

$= \frac{1}{1 - \frac{1}{1 - f_{0} (x)}}$ $=1/(1-1/(1-f_0(x))$

$= \frac{1 - f_{0} (x)}{1 - f_{0} (x) - 1}$ $=(1-f_0(x))/(1-f_0(x)-1)$

$= \frac{1 - f_{0} (x)}{f_{0} (x)}$ $=(1-f_0(x))/f_0(x)$

$= 1 - \frac{1}{f_{0} (x)}$ $=1-1/f_0(x)$

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

$= 1 - (1 - x)$ $=1-(1-x)$

$= x$ $=x$

Note, then, that
$f_{3} (x) = f_{0} (x)$ $f_3(x) = f_0(x)$
$f_{4} (x) = f_{1} (x)$ $f_4(x) = f_1(x)$
$f_{5} (x) = f_{2} (x) = x$ $f_5(x) = f_2(x) = x$
$...$

In general:
$f_k(x) = {(f_0(x) if k -= 0" (mod 3)"), (f_1(x) if k -= 1" (mod 3)"), (x if k -= 2" (mod 3)"):}$

As $2016$ is divisible by $3$ , we have

$f_2016(2016) = f_0(2016)$

$=1/(1-2016)$

$=-1/2015$

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If $f_{0} (x) = \frac{1}{1 - x}$ $f_0(x)=1/(1-x)$ and $f_{k} (x) = f_{0} (f_{k - 1} (x))$ $f_k(x)=f_0(f_(k-1)(x))$ what is the value of $f_{2016} (2016)$ $f_(2016)(2016)$ ?

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If $f_{0} (x) = \frac{1}{1 - x}$ $f_0(x)=1/(1-x)$ and $f_{k} (x) = f_{0} (f_{k - 1} (x))$ $f_k(x)=f_0(f_(k-1)(x))$ what is the value of $f_{2016} (2016)$ $f_(2016)(2016)$ ?