
\subsection{\bf Full Conjugate Gradient Algorithm}

\begin{slide}

\begin{itemize}
\item Initialize in true conjugate gradient
\begin{eqnarray*}
\underline{r}^{(0)} & = & \underline{b} - A\underline{x}^{(0)} \\
k=1 \hspace{5em} p^{(1)} & = & \underline{r}^{(0)} \\
k>1 \hspace{5em} p^{(k)} & = & \underline{r}^{(k-1)} + \beta_k
\underline{p}^{(k-1)}
\end{eqnarray*}

\item This subtracts off $\underline{p}^{(k-1)}$ component in
$\underline{r}^{(k-1)}$

\hspace{1em} $\bullet$ as long as: 
\[ \beta_k = - \frac{\underline{p}^{(k-1)T} A \underline{r}^{(k-1)}}
{\underline{p}^{(k-1)T} A \underline{p}^{(k-1)}}
\]

\item Then as before, solve for $\alpha_k$ so that
$\underline{x}^{(k-1)} + \alpha_k \underline{p}^{(k)}$ minimizes
the quadratic form $- 1/2 \underline{x}^T A\underline{x} -
\underline{b}^T \cdot \underline{x}$

\item This gives
\begin{eqnarray*}
\alpha_k & = & \underline{p}^{(k)T} \underline{p}^{(k)} /
\underline{p}^{(k)T} A\underline{p}^{(k)} \\
\underline{x}^{(k)} & = & \underline{x}^{(k-1)} + \alpha_k
\underline{p}^k \\
\underline{r}^{(k)} & = & b - A x^{(\underline{k})}
\end{eqnarray*}
\end{itemize}

\end{slide}
