I have a function as the following:
$f(n) = \begin{cases} \frac{1}3 + 2x, & 0 ≤ x ≤ \frac{1}3 \\ 2 - 3x, & 1/3 ≤ x ≤ \frac{2}3 \\ 2x - \frac{4}{3}, & \frac{2}3 ≤ x ≤ 1 \end{cases}$
I would like to work out $F o F$, the composite of itself. Can someone help?
Thanks,
Think about what happens to each value of $x$ as it passes through $f \circ f$, and which questions you have to ask about it. For example, we first have to ask "is $x$ less than $1/3$, between $1/3$ and $2/3$, or greater than $2/3$?"
If $0 \leq x \leq 1/3$, then we use the first line and $f(x) = \frac{1}{3} + 2x$. Remember, $x$ is between $0$ and $1/3$, so $\frac{1}{3} + 2x$ is between $1/3$ and $1$ - which means that on the next pass, we use either the second or third lines. We use the second line if $f(x)$ is between $1/3$ and $2/3$, and the third line otherwise. So the first two pieces of our final composited function are:
$2 - 3(\frac{1}{3} + 2x)$ if $0 \leq x \leq 1/3$ and $1/3 \leq \frac{1}{3} + 2x \leq 2/3$
$2(\frac{1}{3} + 2x) - \frac{4}{3}$ if $0 \leq x \leq 1/3$ and $2/3 \leq \frac{1}{3} + 2x \leq 1$
Now, you should be able to find a better way of saying this - for which values of $x$ is $1/3 \leq \frac{1}{3} + 2x \leq 2/3$? And you've still got a bunch of other cases to handle. But hopefully this is enough to get you started.