Regular Perturbation Series Problem I want to find the a 3 term perturbation solution of $$(1+x)^3 = \epsilon x\tag{i}$$
where $\epsilon \ll1$.
Direct substitution of the regular perturbation series $x = x_0 + \epsilon x_1 + \epsilon^2x_2$ into (i) does not work.
I think the solution has the form: $x = x_0 + \epsilon^{1/3}\cdot x_1 + \epsilon^{2/3}\cdot x_2$. This seems to work, but I'm not sure it is correct.
TIA, Matt.

We want to solve 
$$(1+x)^3=\epsilon x \tag{1}, $$
where $\epsilon \ll 1$. We solve order by order. Instead of doing the naive expansion 
$$x=x_o+\epsilon x_1+\epsilon^2 x_2 + o(\epsilon^2),$$
we first posit $x=x_o+o(1)$, where $o(1)$ is some term which is asymptotically small compared to $1$ whose exact order remains to be determined. Evaluating (1) at order $1$ gives $$(1+x_o)^3=0,$$ i.e. $x_o=-1$. Next we need to find out what order the subdominant term is. Let $$x=-1 + \epsilon^\alpha x_1+o(\epsilon^\alpha),$$ where $\alpha> 0$. Then evaluating (1) at order $\epsilon$ gives $\alpha=1/3$ and $x_1 = -1$ (evaluating (1) at any order between 1 and $\epsilon$ will not yield any nontrivial term). Therefore $$x=-1-\epsilon^{1/3}+\epsilon^\beta x_2+o(\epsilon^\beta).$$ Evaluating (1) at order $\epsilon^{4/3}$ gives $3x_2=-1$, i.e.
$$x = -1 -\epsilon^{1/3} - \frac{1}{3}\epsilon^{2/3}+o(\epsilon^{2/3}) $$
I hope that makes sense for you.
