Let $$f(x) = 1 -|1- 2x |.$$ Find the number of solutions of the equation $$f ( f ( f ( f ( f ( f ( f ( f ( f ( f (x))))))))))=x,$$
i.e., $f^{(10)}(x)=x$.
And what about if there is an arbitrary number of iterations?
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Find the analytical expression of $f(f(x))$, then $f(f(f(x)))$ and generalize. – Jun 01 '14 at 09:01
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one solution is when $f(x) = x$ – Petite Etincelle Jun 01 '14 at 09:03
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You should seriously consider writing $f^{(2)} = f(f(x))$ and so on for the sake of clarity. – user88595 Jun 01 '14 at 09:04
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$0$ is a solution, but since $f$ is not a contraction, we cannot guarantee uniqueness. – Ellya Jun 01 '14 at 09:05
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Consider solutions in the intervals $x>1/2$ and $x\leq 1/2$ separately. – poolpt Jun 01 '14 at 09:08
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Draw the graph of $f$. The solutions of $f(x)=x$ are the interseccions with the diagonal of the first quadrant.
Next draw the graph of $f(f(x))$. It looks like two mountains (or tents). Hoy many times does it cross the diagonal?
Repeat for $f(f(f(x)))$. Can you see a pattern?
Julián Aguirre
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The main question here is what solutions exist for $f(x)=x$, as I commented, an obvious solution is $x=0$, but this may not be unique, so let us see if there are more.
So we consider $1-|1-2x|=x$, if $-\infty\lt x\le \frac{1}{2}$, then we have:
$1-(1-2x)=x\Rightarrow 2x=x$ so $x=0$.
If $1/2\lt x\lt\infty$, then we have:
$1-(2x-1)=x\Rightarrow 2-2x=x\Rightarrow x=2/3$
So we have $x=0,2/3$ as solutions.
You may be interested to know, that on $[0,1]$, this map exhibits behaviour a lot like the tent map, which is a chaotic map.
Ellya
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Well $f^{10}(x)=x$ implies that $f$ has to be one-on-one for some $x$ since for a composition to be injective at least one function has to be injective since all functions are $f$ it implies $f$ is injective.From this we know that $f(f(x))=x$ for $$x\leq\frac{1}{2}\\f(x)=2x\\4x=x\implies x=0\\x>\frac{1}{2}\\f(x)=2-2x\\2-2(2-2x)=x\\2-4+4x=x\\4x-2=x\\3x=2\\x=\frac{2}{3}$$
kingW3
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