What is the fixed point for the following function? $f(x) = 2sin(\pi x) + x = 0$ between $[1,2]$
I expressed $f(x)$ as $x=g(x)$ such that $g(x)$ can be $- 2sin(\pi x)$ or $\frac{1}{\pi}sin^{-1}(\frac{-x}{2})$. However, in both cases, the method is not coverging to a root lying between in the interval $[1, 2]$. What to do? Please help me.
This is the visual interpretation of @Robert Israel's answer.
Captured from my old exercise book ($\epsilon$ used is $0.0005$).
Basically, the problem lies within how we truncate the sine-shaped graph when we draw arcsine graph $y = \frac{1}{\pi} sin^{-1}(-\pi/2)$, while in fact we are solving an equation involving sine, that is, $2sin(\pi x)+x=0 $. The $y_1$ and $y_2$ in the picture above is the "extended version" of the arcsine graph.
Note that if $1 \le x \le 2$, you don't want the principal branch of arcsin. You could try $x = 2 + \dfrac{\arcsin(-x/2)}{\pi}$ (for the solution in $[1.5,2]$, or $x = 1 + \dfrac{\arcsin(x/2)}{\pi}$ (for the solution in $[1,1.5]$).
Rewrite, as you did, as $\sin(\pi x)=-\frac{x}{2}$.
Then $\pi x=\pi +\arcsin(x/2)$ leads to one of the solutions.
