Perturbation problem This is a mathematica exercise that I have to do, where $y(x) = x - \epsilon \sin(2y)$ and it wants me to express the solution $y$ of the equation as a power series in $ \epsilon$. 
 A: We're looking for a perturbative expansion of the solution $y(x;\epsilon)$ to
$$ y(x) = x-\epsilon \sin(2y)$$
I don't know if you're asking for an expansion to all orders. If so, I have no closed form to offer. But, for illustration, the first four terms in the series may be found as follows by use of the addition theorem (this procedure may be continued to any desired order). Expand $$y(x;\epsilon)=y_o(x)+\epsilon y_1(x)+\epsilon^2 y_2(x)+\epsilon^3 y_3(x) + o(\epsilon^3).$$
Then clearly, $y_o(x)=x$ and $$\epsilon \,y_1(x) +o(\epsilon)=-\epsilon\sin(2x+\epsilon 2y_1+o(\epsilon))=-\epsilon\sin(2x)\underbrace{\cos(\epsilon\, 2y_1+o(\epsilon))}_{\sim 1}\\ 
\phantom{tttttttttttttttttttttttttttttttttttttttttttttttttt}+\epsilon\cos(2x)\underbrace{\sin(\epsilon\, 2y_1+o(\epsilon))}_{\sim 2\epsilon y_1 = O(\epsilon)}. \\$$
This implies $y_1(x)= -\sin(2x)$. Similarly, at the next two orders we find $y_2(x)=2\sin(2x)\cos(2x)=\sin(4x)$ and $y_3(x)=2\sin^3(x)-2\cos(2x)\sin(4x)$, if I haven't made a mistake in the algebra. Hence
$$y(x;\epsilon) = x - \epsilon \sin(x) + \epsilon^2 \sin(2x)\cos(x)+\epsilon^3 (2\sin^3(x)-2\cos(2x)\sin(4x)) +o(\epsilon^3).  $$
