To compute the fixed points of a sine map, I need to solve
$$x = r\sin\pi x$$.
The question asks me to find the value of r for which the non-trivial fixed point (a second solution of the above equation) appears. The trivial point is x = 0, but I am trying to find the non-trivial one.
The question I am attempting also specifies $0 \leq x \leq 1$ and $0 \leq r \leq 1$.
I know that for intersection to occur, $y = x$ must first become a tangent to $r\sin\pi x$. This means $1 = r\sin\pi x$. But then I am not sure where to go from here. Please could anyone explain the next steps?
Let $f(x)=r\sin(\pi x)$. Then since $f(x)$ is concave, if $f'(0)<1$ then the equation will not have a second solution in the closed interval $[0,1]$. Hence $f'(0)=\pi r>1$, and thus $r>\frac{1}{\pi}$. Conversely, if $r>\frac{1}{\pi}$ then it will have a nontrivial solution.