\subsection{\bf More Mathematics Deriving Gaussian}
\subsection{\bf Integration Formulae}

\begin{slide}

\begin{itemize}
\item To convince yourself, choose a particular complete set of
polynomials of degree $2n+1$.
\begin{description}
\begin{description}
\item{$\bullet$} a) Firstly, $P_{n+1}(x)$ multiplied into 1, $x$,
$x^2$, $\ldots x^n$

\item{$\bullet$} For these, $I$ is zero because $x_i = x_i^{(n+1)}$
are at zeros of $P_{n+1}(x)$.
\vspace{10pt}

\item{$\bullet$} b) Secondly, 
\begin{eqnarray*}
& K_{n+1}\,\left( x_i^{(n+1)}, x\right)\, \left(
x-x_0^{(n+1)}\right)\, \left( x-x_1^{(n+1)}\right) \\
& \ldots \left( x-x_{i-1}^{(n+1)}\right)\, \left( x-x_{i+1}^{(n+1)}\right)\,
\ldots \left( x-x_n^{(n+1)}\right) \\
\end{eqnarray*}

{\bf Note:}\ Omit $\left( x-x_i^{(n+1)}\right)$ factor

\item{$\bullet$} By construction, the only term that contributes to
$I$ for this polynomial is that with $r=i$ and this gives
formula for $w_i$.
\end{description}
\end{description}

\item This is quick derivation of Gaussian Integration.  Key point
is that $x_i$ are zeroes of the orthogonal polynomial $P_{n+1}(x)$.
\end{itemize}
\end{slide}

