So I need to use Mathematica to find the solution of $y=x- \epsilon \sin(2y)$ as a power series in terms of $\epsilon$. I'd assume I'd need to create an equation $f=x-y- \epsilon \sin(2y)$, then express $y(\epsilon)=\sum_n a_n \epsilon^n$, then input into series, but I can't seem to get it to work. Some help would be appreciated.
Let define the equation to solve $f=x-y\epsilon\sin(2 x)\equiv 0$ and series expansion of $y$ in powers of $\varepsilon$.
Then expand the equation and solve for any value of $\varepsilon$ parameter
If we are interested in the expansion coefficients $a_n$ in terms of $x$ then following code will refine the solution
One should also do check
First, one can express $\varepsilon$ as a function of $y$:
$$-(x-y) \csc (2 y).$$
Then, expand into series around $y=x$, since it is a solution for $\varepsilon=0$. Applying
$$x+\varepsilon \sin (2 x)+\varepsilon ^2 \sin (4 x)+\varepsilon ^3 \sin (2 x) (3 \cos (4 x)+1)+O\left(\varepsilon ^4\right).$$