# Regular Perturbation Series soln to eqn

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

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Does this arise from a modelling problem, or is it an exercise somewhere? – Yemon Choi Feb 7 '12 at 5:48
It is an exercise from Logan's applied math book. I am trying to learn perturbation methods and he seems to have a pretty decent intro to the subject. To be specific, it is problem 7, on page 101. Incidentally, if you happen to have the text, there is what appears to be a very fun problem directly preceding it, which I believe is a generalization of this problem, but which I have not yet been able to crack either. I am not sure the soln admits of a regular perturbation series, but if you play around with it a bit, it seems like it must be. – Matt Brenneman Feb 7 '12 at 6:00
This is not my area, and I don't know the book, but I'm not sure the question is really on topic for the site, see mathoverflow.net/faq#whatnot – Yemon Choi Feb 7 '12 at 8:38
I would have thought that you want $x$ to be a small perturbation of $-1$... – Yemon Choi Feb 7 '12 at 8:40
unknown: a good opportunity for you to use the Lagrange inversion formula – Pietro Majer Feb 7 '12 at 9:12

Since $e$ is small, the solution $x$ is close to $-1$. So write $x=-1+u$ and write your equation as $u(1-u)^{-1/3}=-e^{1/3}$. Then use the Lagrange inversion formula.
rmk: This gives a series expansion of $x=-1+u$ in powers of $e^{1/3}$. For large values of $e$ write instead the equation as $x(1+x)^{-3}=1/e$, and get an expansion of the solution in powers of $1/e$. – Pietro Majer Feb 7 '12 at 9:21