\documentstyle[psbox,ps,11pt,fullpage]{article}

\input epsf

\begin{document}
\header{9.520}{1}
\problem{1}
\part{a}

Assuming we set the RBF's to interpolate around the points where we
set the conditions (ie, $x_1 = 0, x_2 = 3, x_3 = 4$), we get:

\begin{eqnarray*}
3c_2 + 4c_3 = 1\\
3c_1 + c_3 = 4\\
4c_1 + c_2 = 3
\end{eqnarray*}

Therefore, $c_1 = 1, c_2 = -1,$ and $c_3 = 1$, and the resulting
interpolation can be seen in figure \ref{fig1}.

\begin{figure}[hbpt]
\begin{center}
\mbox{\epsfysize=3in \epsfbox{fig1.eps}}
\end{center}
\caption{One-dimensional RBF iterpolation of 3 points with 3 RBF's}
\label{fig1}
\end{figure}

\part{b}

A modified set of radial basis functions of the form $f(x) =
\sigma_{i=1}^N y_i\phi_i(x)$ can be found.  One such set of basis
functions $\phi_i(x)$ are:

\begin{eqnarray*}
\phi_1(x) = |x|\\
\phi_2(x) = |x-3|\\
\phi_3(x) = \frac{|x-4| - 5|x-3|}{3}
\end{eqnarray*}

\problem{2}

As $\sigma \rightarrow 0$, the Gaussians $G_\sigma({\bf x} - {\bf
x_i})$ become ``tall and skinny''.  This causes $f_\sigma({\bf x})$ to
become equivalent to nearest neighbor.  This is because for any point
closest to $x_j$ and next closest to $x_k$, as $\sigma \rightarrow 0$,
$\frac{e^\frac{|{\bf x}-{\bf x_k}|^2}{\sigma^2}}{e^\frac{|{\bf x}-{\bf
x_k}|^2}{\sigma_2}} \rightarrow \infty$, which causes $f_\sigma({\bf
x}) \rightarrow x_j$.
\par

As $\sigma \rightarrow \infty$, the Gaussians become completely flat,
yielding a constant $f_\sigma({\bf x})$ equal to the sample mean of
the $y_i$.

\problem{3}



\problem{4}

The approximation error caused by projecting a dataset $\{{\bf
x}_\alpha\}^N_{\alpha=1}$, onto the first $k$ principal components in
the $d$ dimensional space is

\begin{equation}
E(k) = \frac{1}{N} \sum_{\alpha=1}^{N}({\bf x}_\alpha - \sum_{i=1}^{k}({\bf x}_\alpha \cdot {\bf d}_i){\bf d}_i)^2
\end{equation}

This has the effect of removing the components of the ($k+1$)th
through $d$th eigenvectors.  This means $E(k)$ can be rewritten as:

\begin{equation}
E(k) = \frac{1}{N} \sum_{i=k+1}^{d} \sum_{\alpha=1}^{N} (({\bf x}_\alpha \cdot {\bf d}_i){\bf d}_i)^2
\end{equation}

and the inner most term can be combined with the $\frac{1}{N}$ to get:

\begin{equation}
E(k) = \sum_{i=k+1}^{d} \lambda_i
\end{equation}

\end{document}
