- #1

Joppy

MHB

- 284

- 22

I suspect this is trivial, but I couldn't find any info onlin.

Consider the folowing map: $\phi_{n+1} = f(\phi_n ; \Theta, a) = (\phi_n + \Theta + a \sin \phi_n) \mod 2\pi$.

I need to check if is invertible: $\phi_n = f^{-1} (\phi_{n+1}; \Theta, a)$ when a = 1/2 or 3/2.

[DESMOS=-0.626372053161002,7.429191250095138,-0.30385709978601394,7.751706203470127]y=\operatorname{mod}\left(x\ +\ b+\frac{3}{2}\sin\left(x\right),2\pi\right);b=0;y=\operatorname{mod}\left(x\ +\ b+\frac{1}{2}\sin\left(x\right),2\pi\right);[/DESMOS]

I figure show that the map is injective, and then find it's inverse as we would with any other function (from the graph we see that for $a = 1/2$, the map is injective).

Which would just showing: $\phi_n + a \sin (\phi_n) = \phi_{n+1} + a \sin (\phi_{n+1}) \implies \phi_n = \phi_{n+1}$? Is there some neat substitution which will help out this step?