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\begin{document}


{\bf Question}

For small $|\varepsilon|$ show that the roots of the equation

$$x^3-(2+\varepsilon)x^2+(1-\varepsilon)x+2+3\varepsilon=0$$

are given by

$$x=\left\{\begin{array}{l}
1+\ds\frac{3}{2}\varepsilon+O(\varepsilon^2)\\
-1-\ds\frac{1}{6}\varepsilon+O(\varepsilon^2)\\
2-\ds\frac{1}{3}\varepsilon+o(\varepsilon^2) \end{array}
\right\}$$

If $\varepsilon$ is switched on from 0 such that
$\varepsilon=\alpha i,\ 0<\alpha<\ll 1$, sketch how the solutions
move in the complex plane.



\vspace{.5in}

{\bf Answer}

Use hint that $x=x_0+x_1\varepsilon+O(\varepsilon^2)$ is a good
ansatz.

Substitute:

\begin{description}
\item[(A)]

${}$

$\begin{array}{rcl} x_3 & = &
[x_0+x_1\varepsilon+O(\varepsilon^2)]^3\\ & = &
x_0^3+x_1^3\varepsilon^3+O(\varepsilon^9)+3x_0^2x_1\varepsilon
+O(\varepsilon^2)+3x_0x_1^2\varepsilon^2+O(\varepsilon^4)\\ & = &
x_0^3+3x_0^2x_1\varepsilon+O(\varepsilon^2) \end{array}$

We only want answers to $O(\varepsilon^2)$.

\item[(B)]

${}$

$\begin{array}{rcl}-(2+\varepsilon)x^2 & = &
-(2+\varepsilon)(x_0+\varepsilon x_1+O(\varepsilon^2))\\ & = &
-(2+\varepsilon)(x_0^2+2x_0x_1\varepsilon+O(\varepsilon^2))\\ & =
& -2x_0^2-4x_0x_1\varepsilon-\varepsilon
x_0^2+O(\varepsilon^2)\end{array}$

\item[(C)]

${}$

$\begin{array}{rcl}-(1-\varepsilon)x & = &
-(1-\varepsilon)(x_0+\varepsilon x_1+O(\varepsilon^2))\\ & = &
-(x_0+\varepsilon x_1-\varepsilon x_0+O(\varepsilon^2))
\end{array}$

\item[(D)]

${}$

$2+3\varepsilon=2+3\varepsilon$

$\rm{(A)}+\rm{(B)}+\rm{(C)}+\rm{(D)} = (x_0^3-2x_0^2-x_0 +2)$

$+\varepsilon(3x_0^2x_1-4x_0x_1+x_0^2-x_1+x_0 +3)
+O(\varepsilon^2)=0$ (original equation!)

Therefore at $O(\varepsilon^0): x_0^3-2x_0^2-x_0 +2=0$
(unperturbed equation)

Use hints in question to observe solutions are $x_0=\pm 1,\ 2$.

Therefore

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

$$x_1=\ds\frac{(3+x_0-x_0^2)}{(1+4x_0-3x_0^2)}$$

$\left.\begin{array}{lcl} x_0=1 & \Rightarrow &
x_1=\ds\frac{3}{2}\\ x_0=-1 & \Rightarrow & x_1=-\ds\frac{1}{6}\\
x_0=2 & \Rightarrow & x_1=-\ds\frac{1}{3}\end{array}\right\}\
x=\left\{\begin{array}{l}
+1+\ds\frac{3}{2}\varepsilon+O(\varepsilon^2)\\
-1-\ds\frac{1}{6}\varepsilon+O(\varepsilon^2)\\
2-\ds\frac{1}{3}\varepsilon+O(\varepsilon^2) \end{array}\right.$

${}$

${}$

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\end{picture} $\varepsilon=\alpha i$

\end{description}



\end{document}
