In the limit as $\epsilon \to 0$, obtain a three-term asymptotic solution to the roots of the following equation.
$$\epsilon x^3+x^2-2x+1=0$$
I'm not too sure where to start
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1Welcome to Math.SE! Can you tell us what you do know about this question? – Hrodelbert Oct 07 '15 at 16:00
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Hi, thank you, i have tried setting $\epsilon = 0 \Rightarrow x=1$ hence letting $x = x_0 +a_1\epsilon +a_2\epsilon^2+ a_3\epsilon^3 + O(\epsilon^4)$ and substituting this in – jonny Oct 07 '15 at 16:05
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1Hint because your pertubation problem is singular, you need two types of series. – tired Oct 07 '15 at 16:07
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My original close vote on this question was mistaken. The linked question is asking about $\epsilon \to \infty$ while this one is asking about $\epsilon \to 0$. – Antonio Vargas Oct 19 '15 at 09:37
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This is not a complete answer, but rather meant to give you an idea of what to think about.
The first thing to recognize is that the problem is singular: for $\epsilon=0$ there are only two roots (counting multiplicities) but for $\epsilon>0$ there are three roots (counting multiplicities). As a result, you should expect two roots which are close to the roots of the quadratic, and one root which is very large. The quadratic problem is easy to solve: there is a double root at $1$. So you should try to write a series expansion in powers (maybe not integer powers!) of $\epsilon$, whose constant term is $1$. Note that the perturbation is likely to split the double root into two single roots (so you may need two series). It may even split the double root into two complex roots (so you may need complex coefficients).
To find the other root, you need to rescale the variables so that you are looking at ranges where the cubic term is significant. It will turn out that this happens with $x=\epsilon^{-1}y$, where $y$ is bounded as $\epsilon \to 0$. This works because you get
$$\epsilon^{-2}y^3+\epsilon^{-2}y^2-2\epsilon^{-1}y+1=0.$$
Now the first two coefficients are of the same order, so that you are really looking at
$$y^3+y^2-2\epsilon y + \epsilon^2=0.$$
This is now a regular perturbation problem, since the small parameters are multiplying the lower order terms. Thus your third root is $\epsilon^{-1}$ times a perturbation of one of the roots of $y^3+y^2=0$.
To work this out more systematically, you can consider $x=\epsilon^p y$, then
$$\epsilon^{3p+1}y^3+\epsilon^{2p}y^2-2\epsilon^py+1=0$$
Clear the leading coefficient:
$$y^3+\epsilon^{-1-p}y^2-2\epsilon^{-1-2p}y+\epsilon^{-1-3p}=0.$$
So to have all exponents nonnegative, you need $p \leq -1$. On the other hand, if $p<-1$ then all but the leading coefficient are a multiple of a positive power of $\epsilon$, then your unperturbed problem has a root at zero, which is undesirable (why?) So you take $p=-1$ and it works out.
Ian
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Hi, when you say "So you should try to write a series expansion in powers (maybe not integer powers!) of $\epsilon$, whose constant term is $1$." ... what do you mean? could you expand? – jonny Oct 07 '15 at 16:16
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@jonny The roots which are perturbed from the roots of the quadratic should be of the form $1+a_1\epsilon^{p_1}+a_2\epsilon^{p_2}+\dots$. The powers need not always be integer powers (in particular, in this case $p_1$ is not an integer). You need to figure out what powers work by yourself. – Ian Oct 07 '15 at 16:19
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For $\epsilon >0$ there is only one real root and only for $\epsilon < 0$ there are two roots around 1 and one negative root. – A.S. Oct 07 '15 at 16:19
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@A.S. As it happens you are correct; but I actually hinted at that in the main paragraph, and I don't think it really matters (it sounds like jonny needs to consider all three roots of the cubic). Though I did assume that this problem was only about $\epsilon \geq 0$, even though strictly speaking that was not given. – Ian Oct 07 '15 at 16:20
