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I'm working on this question and I have been told that whenever a repeated root is involved with the first term in the asymptotic expansion of a root that you should try a different method to find x1.

The question is

Let $0<\epsilon<<1.$Show that the cubic equation $x^3 - \epsilon x^2 -3x + 2 =0$ has one root of the form $x = -2 + \frac{4}{9} \epsilon + O(\epsilon^2).$ Find the asymptotic expansions for the two remaining roots.

So I used the method of trying $$x = \epsilon^\alpha X $$ and then balancing the terms and I found alpha to be one which is just the usual assumption to try. In the lecturers solution just goes straight to powers of a half which is standard for repeated roots but I should have got alpha equal to a half using my method. What have I done wrong?

Question My workings Lecturers solution

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  • $\begingroup$ Please include the question and other references into your question. Make it easy on would-be answerers to do so! Besides, the linked images could go away any time. $\endgroup$
    – vonbrand
    Aug 12 '19 at 17:10
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You just need to shift by $x_0$ first. If you take $x = 1 + X \epsilon^\alpha$, the two lowest powers of $\epsilon$ will be $1$ and $2 \alpha$. Equating the lowest powers gives $\alpha$, and equating the coefficients gives $X$.

Alternatively, if $x = 1 + X$, the equation becomes $$X^3 - \epsilon X^2 + 3 X^2 - 2 \epsilon X - \epsilon = 0.$$ The side of the Newton polygon which is closest to the origin goes through the vertices corresponding to $X^2$ and $\epsilon$, which gives $X = \pm \sqrt {\epsilon/3}$.

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