# How to deal with repeated roots when finding asymptotic expansion of roots

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?

• 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. Aug 12 '19 at 17:10

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 $$X = \pm \sqrt {\epsilon/3}$$.