
\be
\item Suppose that $f:\R^2 \to \R$ is defined by
$\ds f(\vecx)=\frac{1}{2}\vecx^t A \vecx +\vecb^t\vecx$, where
$\ds A=\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}$ and
$\ds \vecb=\begin{bmatrix} -3 \\ 2 \end{bmatrix}$, and let
$\ds \vecx_0=\begin{bmatrix} 5 \\ 5 \end{bmatrix}$.
    \be
    \item Calculate the global minimum of $f$ on $\R^2$.
    \item Determine the optimal descent direction, $\vecd$, of $f$ at $\vecx_0$.
    \item Calculate the smallest positive local minimum of
    the function $\ds \phi(t)=f\left(\vecx_0+t\vecd\right)$.
    \item Use your previous result to calculate the first Steepest Descent Method
    iterate, $\vecx_1$.
    \item Calculate the next four Steepest Descent Method iterates, and verify
    that formulas (\ref{E:SteepestDescent_ConvergenceRate_Inequality1}) and
    (\ref{E:SteepestDescentConvergenceRateInequality2}) are valid.
    \item How many iterations are required before $\ds \left\| \nabla f(\vecx_k) \right\| < .01$?
    \ee
\item Repeat the previous question using  $\ds A=\begin{bmatrix} 10 & 0 \\ 0 & .1 \end{bmatrix}$,
$\ds \vecb=\begin{bmatrix} -3 \\ 2 \end{bmatrix}$, and
$\ds \vecx_0=\begin{bmatrix} 5 \\ 5 \end{bmatrix}$. How does the rate of convergence to the global
minimum compare with that in the previous question? What accounts for any major differences?
\item Suppose that $f:\R^2 \to \R$ is defined by
\newline
$\ds f(x_1,x_2)=x_1^4+2x_2^4+5x_1^2+4x_2^2-10x_1^2x_2$. By choosing different initial values, demonstrate
how the Steepest Descent Method leads to sequences converging to a local minimum, a global minimum, and
a saddle point of $f$.

\item Assume that $f$ is the \CM objective function from (\ref{E:SteepestDescent_CMC})
and that
\newline
$\ds \vecx_0=\begin{bmatrix} 10 \\ 10 \end{bmatrix}$. Verify that the first
five Steepest Descent Method iterates satisfy inequality
(\ref{E:SteepestDescent_ConvergenceRate_Inequality1}).

\item Suppose that $f:\R^2 \to \R$ is defined by
$\ds f(x_1,x_2)=\ln(1+x_1^2)+x_2^2$ and that $\ds \vecx_0=\begin{bmatrix} 2 \\ 2 \end{bmatrix}$. Observe
that $f$ is bounded below by zero.
\be
    \item Plot the lower set,
    \begin{equation*}
    S_0=\left\{\vecx \mid \vecx \in S \mbox{ and } f(\vecx) \leq f(\vecx_0) \right\}.
    \end{equation*}
    Note: Maple's \mf{implicitplot} command can assist in this process. If the function, $f$, has been defined and the \mf{plots} package has been loaded, enter the following command:
    \newline
   \mf{implicitplot(f(x1,x2)<=evalf(f(2,2)),x1=-20..20,x2=-5..5,
   \newline
    coloring=[grey,white],filled=true);}
    \item Verify that there is an open set $\mathcal{O}$ containing $S_0$, in which
    $\ds \rho \left(H_f(\vecx)\right)$ is bounded.
    \item Using $\vecx_0$ as an initial value, verify that the sequence of Steepest Descent Method
    iterates converges to the origin, which is in fact a global minimum.
    \ee
\item Repeat the previous question using $f:\R^2 \to \R$ defined by
\newline
$\ds f(x_1,x_2)=x_1^4-8x_1^3+24x_1^2-32x_1+4x_2^2-4x_2$ and
$\ds \vecx_0=\begin{bmatrix} 2 \\ 2 \end{bmatrix}$.

\ee









%\item Establish the validity of all five statements marked \checkmark in the proof of
%Theorem \ref{Th:SteepestDescent_ConvergenceRate}. (Hint: For the second statement,
%start by using the identity
%\begin{multline*}
%f(\vecx+\vecy)=\frac{1}{2}\left(\vecx +\vecy\right)^t A \vecx +\vecb^t\left(\vecx +\vecy\right) +c \\
%= \frac{1}{2}\vecx^t A \vecx + \frac{1}{2}\vecy^t A \vecy + \vecx^t A \vecy
%+ \vecb^t\vecx + \vecb^t\vecy +c \\
%= f(\vecx)+ \frac{1}{2}\vecy^t A \vecy + \vecx^t A \vecy + \vecb^t\vecy +c.
%\end{multline*}
%Set $\vecx=\vecx_k$ and $\vecy=t_k \vecd_k$, in which case
%$\vecx+\vecy= \vecx_k+ t_k \vecd_k$. Simplify $f(\vecx_{k+1})-f(\vecx_k)$
%as much as possible to achieve the desired result.)

%\item Prove Kantorovich's Inequality. That is, prove that for a $n$ by $n$ positive definite matrix $A$ having condition
%number $\ds \xi=\frac{\lambda_n}{\lambda_1}$,
%\begin{equation*}
%\frac{\|\vecx\|^4}{\left(\vecx^t A \vecx\right)\left(\vecx^t A^{-1} \vecx\right)}
% \geq 1- \left(\frac{\xi-1}{\xi+1}\right)^2.
%\end{equation*}
%Do so by carrying out the following steps:
%\be
%	\item Establishing the desired inequality is equivalent to proving
%	\begin{equation*}
%	\frac{\left(\vecx^t A \vecx\right)\left(\vecx^t A^{-1} \vecx\right)}{\left(\vecx^t\vecx\right)^2}  \leq \frac{(\lambda_1+\lambda_n)^2}{4\lambda_1\lambda_n}.
%	\end{equation*}
%	\item Since $A$ is symmetric, it can be diagonalized using the Spectral Theorem and thus written as
%	$A=PDP^t$, where the columns of $P$ are eigenvectors of $A$ forming an orthonormal basis of $\R^n$ and the matrix $D$ is diagonal, with
%	entries given by the distinct, positive eigenvalues of $A$. We may order these so that $\lambda_1 < \lambda_2 \ldots <\lambda_n$.
%	Substitute this diagonalization into
%	$\ds \frac{\left(\vecx^t A \vecx\right)\left(\vecx^t A^{-1} \vecx\right)}{\left(\vecx^t\vecx\right)^2}$, and simplify as much as possible.
%	\item Suppose we define
%	\begin{equation*}
%	t_i = \frac{\left[P^t\vecx \right]^2_{ii}}{\left[P^t\vecx \right]^t\left[P^t\vecx \right]}
%	\end{equation*}.
%	Explain why $t_i \geq 0$ and $\ds \sum_{i=1}^{n}t_i=1$. Then use this notation to show that the end result from (2) may be used to conclude
%	\begin{equation*}
%	\frac{\left(\vecx^t A \vecx\right)\left(\vecx^t A^{-1} \vecx\right)}{\left(\vecx^t\vecx\right)^2}
%	= \frac{ \sum_{i=1}^{n}t_i \lambda_i^{-1}}{ \left(\sum_{i=1}^{n} t_i \lambda_i  \right)^{-1}}.
%	\end{equation*}
%	\item The function $\phi(x)=\frac{1}{x}$ is convex on $(0,\infty)$ and, hence on $\left[\lambda_1,\lambda_n \right]$. The denominator of
%	the quotient on the right side of the preceding sum represents an output of $\phi$ using an input on this latter interval. The numerator
%	is on or below the chord connecting the points $\left(\lambda_1,\phi(\lambda_1)\right)$ and $\left(\lambda_n,\phi(\lambda_n)\right)$. Show that
%	the ratio of two such quantities is maximized at the midpoint of $\left[\lambda_1,\lambda_n \right]$.
%	\ee





