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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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    $\begingroup$ Does this arise from a modelling problem, or is it an exercise somewhere? $\endgroup$
    – Yemon Choi
    Commented Feb 7, 2012 at 5:48
  • $\begingroup$ 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. $\endgroup$
    – Matt Brenneman
    Commented Feb 7, 2012 at 6:00
  • $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Commented Feb 7, 2012 at 8:38
  • $\begingroup$ I would have thought that you want $x$ to be a small perturbation of $-1$... $\endgroup$
    – Yemon Choi
    Commented Feb 7, 2012 at 8:40
  • $\begingroup$ unknown: a good opportunity for you to use the Lagrange inversion formula $\endgroup$
    – Pietro Majer
    Commented Feb 7, 2012 at 9:12

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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.

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  • $\begingroup$ 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 $1/e$. $\endgroup$
    – Pietro Majer
    Commented Feb 7, 2012 at 9:21
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    $\begingroup$ That is such a beautiful theorem. I knew the solution worked but I didn't know why. I knew there had to be a "formal" justification for why this power series worked, and this is spot on. Thank you so much for introducing me to such a beautiful result!! $\endgroup$
    – Matt Brenneman
    Commented Feb 9, 2012 at 16:40
  • $\begingroup$ To Yemon, 1) Yes, x is close to -1, this is the point of the problem (which is that regular perturbation series will not work if one tries x=-1+e*x1+...) 2) I don't understand why my post is not appropriate as I: a) am not a student b) am not trying to start a discussion c) I did not ask for a definition of a term (i.e. my problem had an answer) d) Did not ask a question about MathOverLow. Please correct me if I am wrong or violated the site's policy in some other way, as I respect MathOverflow and do not want to abuse it in any way. Thanks, Matt $\endgroup$
    – Matt Brenneman
    Commented Feb 9, 2012 at 16:46

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