1
$\begingroup$

I want to find the a 3 term perturbation soln of (i) $(1+x)^3 = ex$ where $e\ll1$

Direct substitution of the regular perturbation series $x = x_0 + ex_1 + e^2x_2$ into (i) does not work

I think soln has the form: $x = x_0 + e^{1/3}*x_1 + e^{2/3}*x_2$ Seems to work, but not sure it is correct

TIA, Matt

  • 0
    You probably meant Lagrange inversion formula? I hope you understand that you can also solve this problem, just as easily, by purely perturbation techniques.2012-03-15

1 Answers 1

1

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.