# Stuck on perturbation theory for finding a root of polynomial, with rescaling

I have been given the polynomial $$\epsilon x^3+x-2=0,$$ where epsilon is very small and I need to find the roots using perturbation theory. So far I have found the first root, 2, using the direct method and then have rescaled the x value with a new variable $x= \bar x \delta$ in order to find the other two roots. When this is put into the original equation, I get $$\epsilon\delta^3\bar x^3+\delta\bar x-2=0$$ I then looked at the coefficients of these terms and decided it was best to use $\delta=\epsilon^{-0.5}$ which gives the new equation $$\epsilon^{-0.5}\bar x^3+\epsilon^{-0.5}\bar x-2=0$$ but when I use the perturbation method with this, and equate the coefficients of different orders of epsilon, I get (for $O(\epsilon^0))$: $$\bar x_0^3+\bar x_0=0$$ which has roots 0, i and -i which surely can't be right? I don't have a clue where I'm going wrong. Any help would be greatly appreciated.

The first root should be $2-8\epsilon$ as it should be up to order 1 not order 0, and what you have for $\bar {x_0}$ is correct except we reject the 0 value as that stops our approximation being of moderate size. You then need to also find $\bar {x_1}$

Yeah that's fine, note that your new direct perturbation should be in powers of $\epsilon^{\frac{1}{2}}$ for $\bar {x}$

You choose both $\bar{x_0}=i,-i$, in each case you get a different root, so by this method we get two more roots. i.e. we have a total of 3, which is good because the equation is cubic.

• so it's okay that imaginary numbers have come into play? xD i think that's what threw me off
– Lucy
Commented Mar 24, 2014 at 19:38
• Yeah, check my edit Commented Mar 24, 2014 at 19:51
• do i choose $\bar x_0$ to be i or -i?
– Lucy
Commented Mar 24, 2014 at 19:58
• Both, see my edit Commented Mar 24, 2014 at 20:01
• Is that all good? Commented Mar 24, 2014 at 20:21