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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.