Find the first three non-zero terms in the asymptotic expansion for each of the three roots of the cubic equation $$\epsilon x^3+x^2-1=0, |\epsilon|<<1.$$

I found the expansions $1-\frac{\epsilon}{2}+\frac{5}{8}\epsilon^2$ and $-1-\frac{\epsilon}{2}-\frac{5}{8}\epsilon^2$ by substituting the ansatz $x=x_0+x_1\epsilon+x_2\epsilon^2$ into the equation and setting like terms equal.

I have trouble finding the expansion of the third root. I had to use many terms in the ansatz as I get zero coefficients. Rescaling $x=\frac{X}{\epsilon^n}$ and dominant balance give $n=1$ and the equation $X^3+X^2-\epsilon^2=0$. Substituting $X=x_0+x_1\epsilon+x_2\epsilon^2$ and solving for $x_0,x_1,…$ gives $$\begin{cases}x_0^3+x_0^2=0\\2x_0x_1+3x_0^2x_1=0 \\ 2x_0x_2+3x_0^2x_2+x_1^2+3x_0x_1^2-1=0 \end{cases}$$ Solving this I get $x_0=-1$ ($x_0=0$ gives the previous solutions), $x_1=0$,$x_2=1$. Since I have a zero coefficient I used two more terms in the ansatz and then I also get (using Wolfram Alpha) an equation $2x_0x_3+3x_0^2x_3+2x_1x_2+6x_0x_1x_2+x_1^3=0$ so that also $x_3=0$ and $2x_0x_4+3x_0^2x_4+2x_1x_3+6x_0x_1x_3+3x_1^2x_2+x_2^2+3x_0x_2^2$ so that $x_4=2$.

Thus the expansion is $x=-\frac{1}{\epsilon}+\epsilon+2\epsilon^3$. This would be very tedious to do by hand, so I wonder if I am doing something wrong or if there is a quicker way?

  • 1
    $\begingroup$ Don't assume $X$ should be an expansion in integer powers of $\epsilon$. Instead, write $X=X_0+\epsilon^\alpha X_1$ and use dominant balance to find $\alpha=2$. $\endgroup$
    – David
    Sep 26 '18 at 15:50
  • $\begingroup$ Using Newton's identities the product of all three roots is $1/\epsilon$, so if you already have two roots, the third is easy to find, $\endgroup$
    – Somos
    Sep 26 '18 at 18:20

In my opinion it is probably better to make the expansion step by step and work at each step on the "remainder" terms rather than substituting the ansatz in one step.

The equation is $X^3+X^2-\epsilon^2=0$ as you noticed the first term is $x_0=-1$.

Let $x=x_0+y_0$ with $y_0=o(1)$. Then the equation rewrite as: $$(-1+y_0)^3+(-1+y_0)^2-\epsilon^2=0$$ i.e $$y_0^3-2y_0^2+y_0-\epsilon^2=0$$ so the first order (in $y_0$) is rather simple, $y_0=\epsilon^2+o(\epsilon^2)$.

Once again let $y_0=\epsilon^2(1+y_1)$ with $y_1=o(1)$.

Then the equation in $y_1$ is: $$\epsilon^6(1+y_1)^3-2\epsilon^4(1+y_1)^2+\epsilon^2(1+y_1)-\epsilon^2=0$$ i.e $$\epsilon^4 y_1^3+(3 \epsilon^4-2 \epsilon^2) y_1^2 +(3 \epsilon^4-4 \epsilon^2+1)y_1+(\epsilon^4-2 \epsilon^2)=0$$ so $y_1=2\epsilon^2(1+y_2)$ with $y_2=o(1)$ etc.


Maybe you want to experience the joy of tedious calculation, but I make my computer slave do this for me and reward him with extra time to bounce virtual polygons across the screen.

Simply calculate the cubic roots explicitly and expand them in a power series. In Mathematica the code looks like:

rootlis=Solve[e*x^3+x^2-1==0, x];


The answer is: {1-e/2+5*e^2/8-e^3, -1/e+e+2*e^3,-1-e/2-5*e^2/8-e^3}


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