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If $f_1(x)=||x|-2|$ and $f_n(x)=|f_{n-1}(x)-2|$ for all $n\geq 2$ then find the no. of solutions for $f_{2015}=2$.

I tried solving these equations till 2013 but couldn't find a pattern in the number of solutions. I don't know what approach I should adopt to solve such questions. It would be great if I could get a hint for finding the answer.

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We have $f_{2015}=2$ if and only if $f_{2014}=0$ or $f_{2014}=4$.

You can prove that $f_n$ is a function of period $4$ on $[-2n,2n]$, $f_n(-2n)=f_n(2n)=0$, and $f_n$ is strictly monotone on $(-\infty,-2n]$ and $[2n,\infty)$, so $f_n(x)=4$ has only two solutions, and $f_n(x)=0$ has $\frac{2n.(-2n)}{4}+1=n+1$ solutions. Then $f_n(x)=0$ or $f_n(x)=4$ for $2+n+1=n+3$ values of $x$. By the first paragraph, $f_{2015}=2$ has $2014+3$ solutions.

To prove induction, note that the graph of $f_n$ is obtained from the graph of $f_{n-1}$, moving it two units down, and later you apply the absolute value to the resulting graph.

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    That's when I know that there are $2^{3n}$ values...how did you get to that? – 2017-02-24
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    I had made a mistake – 2017-02-24
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    But how did you get to that? – 2017-02-24
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    @OsheenSachdev Mathematical induction. – 2017-02-24
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    The correct answer is supposed to be 2017 – 2017-02-24
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    I apologize for the successive errors. Review the new answer. – 2017-02-24