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- Thread starter Poirot
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- Jan 26, 2012

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There are two roots near 0 and one root near 1, you have already dealt with than near 1. Then for the roots near 0 guess a trial solution: \(x=aw^k\) (which is a two term logarithmic expansion: \(\log(x)=A+B\log(w)\) ).Question: obtain 2-term expansions for the roots of x^3+x^2-w=0 , 0<w<<1.

I assumed an expansion of the form x=a+bw+.... and from this obtained a=-1, b=1 as one solution. How do I work out the form of the other 2 expansions?

Thanks.

CB

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- Jan 26, 2012

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Sorry Poirot, are you familiar with singular perturbation expansions?sorry Captain Black, are you familiar with perturbation expansions? That is what I'm doing, I should have said the above solution as

x= -1+w+....... as it is an infinite (perturbation) series. I want to know how to arrive at the correct asymptotic sequence.

When you find the correct exponent for the trial solution you can use a perturbation expansion for \(a=a_0+a_1w^{k}+...\), where \(a_0\) is one or other of the two zeroth-order coefficients found using the trial solution.

CB

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- Jan 26, 2012

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It is not a guess. There can be no constant term since the root is going to zero as \(w\) goes to zero, so the simplest candidate is a multiple of some power of \(w\). Try the candidate , find the coefficient and exponent for an initial approximate solution and take it from there.I take it you thought I was trying to be funny. Anyway, is there any way to arrive at the correct answer without just guessing?

CB

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