#1:
4.1:	Many people thought that \beta = a was an assumption and not
something they needed to prove.

4.2:	Mostly well done.

#2:	A common mistake here was when computing x4 = 7.0603e-5 and x5 =
-2.3470e-13, many people had x4 = 0.0001 and x5 = -0.0000, leading people to
conclude that Newton's Method returns the exact minimum in 5 iterations.

#3:	Not everyone realized that Newton's Method was producing iterates
that were outside of the domain of the function.

#4:
(a)	Most people were able to show that if Q is positive semidefinite and
b is in the range of Q, then q(x) is bounded below.  Also, many students
were able to argue by contradiction that if q(x) is bounded below, then Q is
positive semidefinite.  However, no one was able to show that b must be in
the range of Q, which relied on the fact that the range of Q is the
orthogonal complement of the null space of Q.

(b)	Almost everyone did this question assuming that Q is invertible, but
did not consider the case when Q is not invertible.  Newton's Method still
works in this case, but we have to use the equation 0 = f'(x0) + f''(x0)(x1
- x0) instead.

#5:
4.15: This question was fairly well done, with only minor calculation errors
here and there.

4.16: Most people got this by using the fact from Assignment #1 that if h(t)
= f(x+td), then h'(t) = (grad f)(x+td)^Td.

4.17: Just like question 4b, but since A is assumed to be positive definite,
and is therefore invertible, students had no problems.

4.19: Many students were not able to calculate the step length in the
gradient method since it required them to solve a 3rd-degree polynomial.
However, except for some calculation mistakes, the Newton Method was well
done.