\documentclass[12pt,letterpaper]{beamer}
\usepackage[utf8]{inputenc}
\usepackage{framed}
\author{hknaust}
\title{Computation of the Daubechies-4 Coefficients}
\usetheme{JuanLesPins}
\begin{document}
\begin{frame}
Here is our system of four equations:
\begin{equation}h_0^2+h_1^2+h_2^2+h_3^2=1\end{equation}
\begin{equation}h_0 h_2+h_1 h_3=0\end{equation}
\begin{equation}h_0-h_1+h_2-h_3=0\end{equation}
\begin{equation}h_1-2h_2+3h_3=0\end{equation}
\end{frame}
\begin{frame}
We first solve (3) and (4) for $h_2$ and $h_3$.
\begin{equation}h_2=-h_0+h_1+h_3\end{equation}
Substituting (5) in the last equation yields
\begin{equation}h_1-2(-h_0+h_1+h_3)+3h_3=0\end{equation}
and thus
\begin{framed}
\begin{equation}h_3=h_1-2h_0\end{equation}
\end{framed}
Going back to (5) we obtain:
\begin{framed}
\begin{equation}
h_2=2h_1-3h_0
\end{equation}
\end{framed}
\end{frame}
\begin{frame}
Next we substitute (7) and (8) in (2) and solve for $h_1$:
\begin{equation}h_0(2h_1-3h_0)+h_1(h_1-2h_0)=0\end{equation}
This yields
\begin{equation}
h_1^2=3h_0^2
\end{equation}
Discarding one of the two solutions we obtain
\begin{framed}
\begin{equation}
h_1=\sqrt{3}h_0
\end{equation}
\end{framed}
\end{frame}
\begin{frame}
Finally we substitute (7), (8) in (1):
\begin{equation}
h_0^2+h_1^2+(2h_1-3h_0)^2+(h_1-2h_0)^2=1
\end{equation}
and obtain
\begin{equation}
14h_0^2+6h_1^2-16h_0h_1=1
\end{equation}
\end{frame}
\begin{frame}
Using (11) we get:
\begin{equation}
14h_0^2+18h_0^2-16\sqrt{3}h_0^2=1
\end{equation}
Solving for $h_0$ (again discarding one of the two solutions) we obtain:
\begin{framed}
\begin{equation}
h_0=\frac{1}{4 \sqrt{2-\sqrt{3}}}
\end{equation}
\end{framed}
\end{frame}
\begin{frame}
Backsolving we can compute all 4 coefficients:
\begin{framed}
\begin{eqnarray*}
h_0&=&\frac{1}{4 \sqrt{2-\sqrt{3}}}\\
h_1&=&\frac{\sqrt{3}}{4 \sqrt{2-\sqrt{3}}}\\
h_2&=&\frac{2\sqrt{3}-3}{4 \sqrt{2-\sqrt{3}}}\\
h_3&=&\frac{\sqrt{3}-2}{4 \sqrt{2-\sqrt{3}}}\\
\end{eqnarray*}
\end{framed}
\end{frame}
\begin{frame}
Finally note that
\begin{eqnarray*}
&&h_0+h_1+h_2+h_3\\
&=&\frac{1}{4 \sqrt{2-\sqrt{3}}}(1+\sqrt{3}+2\sqrt{3}-3+\sqrt{3}-2)\\
&=&\frac{1}{4 \sqrt{2-\sqrt{3}}}(-4+4\sqrt{3})=\frac{\sqrt{3}-1}{\sqrt{2-\sqrt{3}}}\\
&=&\frac{(\sqrt{3}-1)(\sqrt{2+\sqrt{3}})}{(\sqrt{2-\sqrt{3}})(\sqrt{2+\sqrt{3}})}\\
&=&\sqrt{4-2\sqrt{3}}\sqrt{2+\sqrt{3}}\\
&=&\sqrt{2}\sqrt{(2-\sqrt{3})(2+\sqrt{3})}=\sqrt{2}
\end{eqnarray*}
\end{frame}
\end{document}