\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\parindent=0pt
\begin{document}


{\bf Question}

Solve $x^2-2.01x+1=0$

\begin{description}
\item[(i)]
Exactly

\item[(ii)]
Using a perturbation series to an accuracy of 0.0001.

\end{description}



\vspace{.5in}

{\bf Answer}

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

${}$

$\begin{array}{rcl} x^2-2.01x+1=0 \Rightarrow x & = &
\ds\frac{2.01\pm\sqrt{4.0401-4}}{2}\\ & = &
\ds\frac{(2.10\pm0.200249844)}{2}\\ & = & 1.105124922\cdots\\ & &
\un{or} 0.904875078\cdots \end{array}$

\item[(ii)]
Roots not distinct at $\varepsilon=0$, so try perturbation on
$\undb{x^2-2x+1}=0$

\hspace{3.45in} $(x-1)^2=0$

$$x^2-(2+\varepsilon)x+1=0\ \rm{where}\ \varepsilon=0.01$$

Ansatz:
$x=x_0+x_1\varepsilon^{\frac{1}{2}}+x_2\varepsilon+O(\epsilon^{\frac{3}{2}})$

since 2 coincident roots at $\varepsilon=0$. Check notes and see.

Substitute:

$$[x_0+x_1\varepsilon^{\frac{1}{2}}+x_2\varepsilon
+O(\epsilon^{\frac{3}{2}})]^2-(2+\varepsilon)[x_0+x_1\varepsilon^{\frac{1}{2}}
+x_2\varepsilon+O(\varepsilon^{\frac{3}{2}})]+1=0$$

Expand:

$\begin{array}{cl} &
x_0^2+x_1^2\varepsilon+x_2^2\varepsilon^2+O(\varepsilon^3)\\ & \
\rm{probably\ ok\ as}\ \varepsilon^2=0.\un{0001}\ \rm{so\
truncate\ at}\ O(\varepsilon^2)\\ & \ \rm{and\ hope\ that\
implied\ constant\ is}\ <1\\ + &
2x_0x_1\varepsilon^{\frac{1}{2}}+2x_0x_2\varepsilon
+2x_0x_3\varepsilon^{\frac{3}{2}}+O(\varepsilon^{\frac{5}{2}}\\+ &
2x_1x_2\varepsilon^{\frac{3}{2}}+2x_1x_3\varepsilon^{\frac{5}{2}}
+O(\varepsilon^{\frac{5}{2}}\\+ &
2x_2x_3\varepsilon^{\frac{5}{2}}+O(\varepsilon^{\frac{7}{2}}\\- &
2x_0-2x_1\varepsilon^{\frac{1}{2}}-2x_2\varepsilon-2x_3\varepsilon^{\frac{3}{2}}
- 2x_4\varepsilon^2+O(\varepsilon^5)\\- & \varepsilon
x_0-x_1\varepsilon^{\frac{3}{2}}-x_2\varepsilon^2+O(\varepsilon^\frac{5}{2})\\+
& 1=0 \end{array}$

Balance at $O(\varepsilon^n)$:

$O(\varepsilon^0): x_0^2-2x_0+1=0 \Rightarrow x_0\un{=1}$\un{
twice}

$O(\varepsilon^{\frac{1}{2}}: 2x_1x+0-2x_1=0 \Rightarrow
x_1(\undb{x_0-1})=0 \Rightarrow x_1=$ anything! What do we do?
Carry on and see.

$O(\varepsilon): -2x_2+2x_0x_2+x_1^2-x_0=0 \Rightarrow
-2x_2+2x_2+x_1^2-1=0 \Rightarrow \un{x_1=\pm 1}$

$O(\varepsilon^{\frac{3}{2}}: -x_1+x_1x_2-2x_3+2x_0x_3=0
\Rightarrow \mp1\pm 2x_2=0 \Rightarrow \un{x_2=+\ds\frac{1}{2}}$

Actually need to go to $O(\varepsilon^2)$

$O(\varepsilon^2): x_2^2-2x_4-x_2+2x_0x_4+2x_1x_3=0 \Rightarrow
\ds\frac{1}{4}-\ds\frac{1}{2}\pm2x_3=0 \Rightarrow
x_3=\pm\ds\frac{1}{8}$

Thus we have a perturbation expansion:

$x=1\pm\varepsilon^{\frac{1}{2}}+\ds\frac{1}{2}\varepsilon\pm\ds\frac{1}{8}\varepsilon^{\frac{3}{2}}+O(\varepsilon^2)$

This turns out to be sufficient as $O(\varepsilon^2)=\un{0.0001}$
when $\varepsilon=0.01$

Therefore

$\begin{array}{rcl} x & \approx &
1\pm(0.01)^{\frac{1}{2}}+\ds\frac{0.01}{2}\pm\ds\frac{1}{8}(0.001)^{\frac{3}{2}}\\
& = & 1.10512_\uparrow 5\ \rm{or}\ 0.904875_\uparrow \end{array}$

which is certainly accurate to 0.0001 by comparison with exact
result.
\end{description}

\un{Note}: Here we had to calculate the $x_i$ at the
$O(\varepsilon^{\frac{i+1}{2}})$ level, due to the strange
behaviour at $O(\varepsilon^{\frac{1}{2}})$. Not a problem, just
an example of how you may have to go to a higher order (and hence
do more work) to find coefficients of lower orders.

Nasty question to start with but good practice!


\end{document}
