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{\bf Question}

Find the small $\varepsilon$ expansion of the roots of
$\varepsilon x^2-4x+1=0$.




\vspace{.5in}

{\bf Answer}

$\varepsilon x^2-4x+1=0$

This is a singular perturbation as $\varepsilon$ multiplies
highest power of $x$: when $\varepsilon=0$ obtain only one root
$x=\ds\frac{1}{4}$.

CHEAT!

Exact solution is

\begin{eqnarray*} x &amp; = &amp;
\ds\frac{4\pm\sqrt{16-4\varepsilon}}{2\varepsilon}\\ &amp; = &amp;
\ds\frac{2\pm\sqrt{4-\varepsilon}}{\varepsilon}\\ &amp; = &amp;
\ds\frac{2}{\varepsilon}\pm\ds\frac{2}{\varepsilon}\left
(1-\ds\frac{\varepsilon}{4}\right)^{\frac{1}{2}}\\ &amp; = &amp;
\ds\frac{2}{\varepsilon}\pm\ds\frac{2}{\varepsilon}\mp
\ds\frac{1}{4}\mp\ds\frac{\varepsilon}{64}+O(\varepsilon^2)\\ &amp; =
&amp; \left\{\begin{array}{l}
\ds\frac{4}{\varepsilon}-\ds\frac{1}{4}cdot\ds\frac{-\varepsilon}{64}+O(\varepsilon^2)\\
\ds\frac{1}{4}+\ds\frac{\varepsilon}{64}+O(\varepsilon^2)\end{array}\right.\end{eqnarray*}

so now try it the other (proper) way:

Use ansatz

\ $x=\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon^2) \rightarrow$
for singular root $\rightarrow x_0=0 \Rightarrow$ regular root
discord or $x_0=4$ actual one.

\ $x=x_0+\varepsilon x_1+O(\varepsilon^2) \rightarrow$ for regular
root

Substitute into equation and balance at each order to show
coefficients above.



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