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{\bf Question}

Determine two terms in the small $\varepsilon$ expansion of the
roots of

$$x^3-2x^2-x+2=\varepsilon(3x^2+2x+2)$$



\vspace{.5in}

{\bf Answer}

$x^3-2x^2-x+2=\varepsilon(3x^2+2x+2)$

Not a singular perturbation as $\varepsilon \times O(x^2)$ and
polynomial is 3rd order.

$\varepsilon=0 \Rightarrow x^3-2x^2-x+2=0$ with roots $x=1$
(obvious) by substitution, or $x^2-x-2=0 \Rightarrow x=2$ or $-1$.

Therefore no degenerate roots at $\varepsilon=0$

$\Rightarrow$ try ansatz $x=x_0+\varepsilon x_1+O(\varepsilon^2)$

where $x_0=1,\ -1$ pr 2. Find $x_1$ by substitution:

$\left.\begin{array}{rclcl}x^3 & = & (x_0+\varepsilon
x_1)+O(\varepsilon^2))^3 & = & x_0^3+3\varepsilon
x_1x_0^2+O(\varepsilon^2)\\ -2x^2 & = & -2(x-0+\varepsilon
x_1+O(\varepsilon^2))^2 & = & -2x_0^2-4\varepsilon
x_0x_1+O(\varepsilon^2)\\ -x & = & -(x_0+\varepsilon
x_1+O(\varepsilon^2)) & = & -x_0-\varepsilon
x_1+O(\varepsilon^2)\\ +2 & = & +2 & = & +2
\end{array} \right\}$

LHS=$x_0^3-2x_0^2-x_0+2+\varepsilon(3x_1x_0^2-4x_0x_1-x_1)+O(\varepsilon^2)$

$\left.\begin{array} {rclcl} 3\varepsilon x^2 & = &
3\varepsilon(x_0+\varepsilon x_1+O(\varepsilon^2))^2 & = &
3\varepsilon x_0^2+O(\varepsilon^2)\\ 2\varepsilon x & = &
2\varepsilon(x_0+\varepsilon x_1+O(\varepsilon^2)) & = &
2\varepsilon x_0+O(\varepsilon^2)\\ 2\varepsilon & = &
2\varepsilon & = & 2\varepsilon \end{array} \right\}$

RHS=$\varepsilon(2+3x_0^2+2x_0)+O(\varepsilon^2)$

LHS=RHS $\Rightarrow
x_0^3-2x_0^2-x_0+2+\varepsilon(3x_1x_0^2-4x_0x_1-x_1-2-3x_0^2-2x_0)+O(\varepsilon^2)=0$

Balancing:

$O(\varepsilon^0)$: Gives roots $x_0=\pm$ and 2 as above

$O(\varepsilon^1): 3x_1x_0^2-4x_0x_1-x_1-2-3x_0^2-2x_0=0$

\ \ \ so $x_0=\left\{\begin{array}{rcl} +1 & \Rightarrow &
x_1=-\ds\frac{7}{2}\\ -1 & \Rightarrow & x_1=+\ds\frac{1}{2}\\ 2 &
\Rightarrow & x_1=+6 \end{array} \right.$

Therefore
$x=\left\{\begin{array}{l}1-\ds\frac{7}{2}\varepsilon+O(\varepsilon^2)\\
-1+\ds\frac{1}{2}\varepsilon+O(\varepsilon^2)\\
2+\ds\frac{1}{6}\varepsilon+O(\varepsilon^2) \end{array} \right.$


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