Converting a cubic to a perturbation problem I'm trying to learn about Perturbation, but feel like I'm confused before I've even started. 
Right now I'm focused on using them to find solutions to polynomial equations. 
The initial example I've been given has  $x^3 - 4.001x + 0.002 = 0$, the numbers clearly lend towards $\epsilon$ = 0.001, and you can then have $x^3 - (4 + \epsilon)x + 2\epsilon = 0$
Where I'm confused, is how to apply this to a cubic, when there isn't an obvious value for $\epsilon$. 
So, for example, IF we take away the .001 from the first equation & simplify, we have: $x^3 + 4x + 2 = 0$ as our starting equation,  how would we then decide a reasonable value for $\epsilon$?
Do we just pick anything reasonably small? (though how small is reasonably?)
Cheers, 
Belle

the idea, as i understand it, is to identify a small or large parameter in the problem. you have identified in yours as $\epsilon.$  the non perturbed problem, that is with $\epsilon = 0$ has easy solution. in your case this is $x^3 - 4x = 0.$ the solutions are $$x = 0 , 2, -2$$  we can three roots of the perturbed problem $$x^3 -(4+\epsilon) x + 2\epsilon=0$$ near any one of the solution $x_0$ by looking for $$x = x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots$$  subbing in the equation and expanding we find $$(x_0+\epsilon x_1+\cdots)^3 -(4+\epsilon)(x_0+\epsilon x_1 + \cdots)+2\epsilon= 0$$ at order $\epsilon$ you get $$3x_0^2x_1-x_0 - 4x_1 + 2=0 \to x_1 = \frac{2-x_0}{4-3x_0^2} $$ the first order corrections to the roots are 
$$x = 0+\frac 12 \epsilon , 2+0\epsilon, -2-\frac 12\epsilon$$
