- This assignment was pretty well done by everyone.  Everyone seems to
understand Newton's method and steepest descent very well.  

- Difficulties were encountered in #12.  Please see the solutions.  The
two most common mistakes were:

	-  Part a: a lot of people said: " If f is convex, then f has a
global minimizer" This isn't true.  As a counter example, think of the
function f(x) = e^x.  This function is strictly convex as f"(x) = e^x >0
for all x, but it has no global minimizer.  What is true is " f is convex
implies that a global minimimum, IF IT EXISTS, can be found among the
critical points of f.  In other words, if f has critical points (recall f
convex) then
the global min can be found among those (of course I assume f continuous
and differentiable).
	 Another key factor in part A was to notice A positive definite
implies that its inverse exists.  So the unique critical point (and
therefore global min by above) is

			x* = -A^(-1)*b

	- For part b, most people got the general idea, but some proofs
had errors, e.g.

- inverting vectors (makes no sense)
- squaring vectors (probably the norm square is what they meant)
- dividing expressions by vector (makes no sense)
- Saying that the steplenght was A^(-1).  The steplength is a scalar, a
real number, not a matrix.