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\begin{document}


{\bf Question}

Show that the small $\varepsilon$ expansion of the roots of

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

are given by

$$x=\left.\begin{array}{l}2+\ds\frac{1}{9}\varepsilon+O(\varepsilon^2)\\-1\pm
\sqrt{\ds\frac{2}{3}\varepsilon}+O(\varepsilon)
\end{array}\right.$$

Sketch the behaviour of the roots as $\varepsilon\to 0^+$.




\vspace{.5in}

{\bf Answer}

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

Put $\varepsilon=0$

$x^3-3x-2=0 \rightarrow$ obvious root $x=-1$

Therefore

$\left.\begin{array}{rcl} (x+1)(x^2-x-2) & = & 0\\ (x+1)(x+1)(x-2)
& = & 0 \end{array} \right\}$ so $x=-1$ \un{twice} and 2

Therefore 2 degenerate roots at $\varepsilon=0$ so try

$x=x_0+\varepsilon^{\frac{1}{2}}x_1+O(\varepsilon)$ to capture
these.

Other root is regular so can use $x=x_0+\varepsilon
x_1+O(\varepsilon^2)$.

\un{Substitute}:

$[x_0+\varepsilon^{\frac{1}{2}}x_1+O(\varepsilon)]^3-
(3+\varepsilon)(x_0+\varepsilon^{\frac{1}{2}}x_1+O(\varepsilon))-2+\varepsilon=0$

$x_0^3+3x_0^2x_1\varepsilon^{\frac{1}{2}}-3x_0-3x_1\varepsilon^{\frac{1}{2}}-2+O(\varepsilon)=0$

Balance at

$O(\varepsilon^0): x_0^3-3x_0-2=0 \Rightarrow x_0=-1$ twice as
above (and 2 but this gives regular root)

$O(\varepsilon^{\frac{1}{2}}): 3x_0^2x_1-3x_1=0 \Rightarrow 0
\cdots x_1=0,\ \un{x_1=?}$

Can't find $x_1$ at this order. Need to go to $O(\varepsilon)$.

After more algebra:

$O(\varepsilon): 1-x_0+3x_0^2x_2-3x_2+3x_0x_1^2=0$

$\Rightarrow 2-3x_1^2=0 \Rightarrow x_1=\pm
\sqrt{\ds\frac{2}{3}}$. ($x_2$ got from
$O(\varepsilon^{\frac{3}{2}})$ etc.)

Therefore degenerate root splits as: $x=-1\pm
\sqrt{\ds\frac{2\varepsilon}{3}}+O(\varepsilon)$

Other root: $x=x_0+\varepsilon x_1+O(\varepsilon^2)$

$[x_0+\varepsilon
x_1+O(\varepsilon^2)]^3-(3+\varepsilon)(x_0+\varepsilon
x_1+O(\varepsilon^2))-2+\varepsilon=0$

$x_0^3+3x_0^2x_1\varepsilon-3x_0-3x_1\varepsilon-\varepsilon
x_0-2+\varepsilon+O(\varepsilon^2)=0$

Balance at:

$O(\varepsilon^0): x_0^3-3x_0-2=0 \rightarrow$ roots as above,
pick $x_0=2$

$O(\varepsilon^1): 3x_0^2x_1-3x_1-x_0+1=0 \rightarrow x_0=2
\Rightarrow 9x_1-1=0 \Rightarrow x_1=\ds\frac{1}{9}$

so root is $x=2+\ds\frac{1}{9}\varepsilon+O(\varepsilon^2)$

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